Summary of the course

Nonlinear partial differential equations IStefan Muller

Bonn UniversityFall term 2018–2019

This is only a summary of the main results and arguments discussed in classand not a complete set of lecture notes. These notes can thus not replacethe careful study of the literature. As discussed in class, among others thefollowing two books are recommended:

[Ev] L.C. Evans, Partial differential equations, Amer. Math. Soc., 1998(2nd edition 2010).

[GT] D.Gilbarg and N.S. Trudinger, Elliptic partial differential equations ofsecond order, Springer, 1998 (reprinted as Classics in Mathematics).

These notes are based on the books mentioned above and further sourceswhich are not always mentioned specifically. These notes are only for theuse of the students in the class ’Nonlinear partial differential equations I’ atBonn University, Fall term 2018–2019.

Please send typos and corrections to sm@hcm.uni-bonn.de.

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Contents

1 Introduction 41.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 The Navier-Stokes equation . . . . . . . . . . . . . . . 41.1.2 Mean curvature flow . . . . . . . . . . . . . . . . . . . 51.1.3 Ricci flow . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.4 Minimal surfaces . . . . . . . . . . . . . . . . . . . . . 71.1.5 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . 71.1.6 Nonlinear elasticity . . . . . . . . . . . . . . . . . . . . 8

1.2 Some general strategies . . . . . . . . . . . . . . . . . . . . . 91.2.1 Soft functional analytic and topological methods . . . 91.2.2 Weak convergence methods and compactness . . . . . 151.2.3 Variational methods . . . . . . . . . . . . . . . . . . . 161.2.4 Convex integration . . . . . . . . . . . . . . . . . . . . 18

1.3 Overview of a priori estimates . . . . . . . . . . . . . . . . . . 201.4 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Elliptic equations and systems 222.1 Energy methods: Lp and Cα estimates for elliptic equations

and systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.2 Existence of weak solutions . . . . . . . . . . . . . . . 262.1.3 Reverse Poincare inequality and pointwise estimates . 322.1.4 Morrey and Campanato spaces . . . . . . . . . . . . . 342.1.5 Regularity theory in Campanato spaces, interior esti-

mates . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.1.6 Estimates up to the boundary . . . . . . . . . . . . . . 492.1.7 Flattening the boundary . . . . . . . . . . . . . . . . . 582.1.8 Global estimates . . . . . . . . . . . . . . . . . . . . . 612.1.9 Ck,α theory, Schauder estimates up to the boundary . 632.1.10 An alternative route to Schauder estimates by rescal-

ing and blow-up . . . . . . . . . . . . . . . . . . . . . 662.1.11 The space BMO . . . . . . . . . . . . . . . . . . . . . 672.1.12 Lp theory . . . . . . . . . . . . . . . . . . . . . . . . . 732.1.13 A short look back on elliptic regularity . . . . . . . . . 91

2.2 Harnack inequality and the DeGiorgi-Moser-Nash theorem . . 932.2.1 Weak Harnack inequalities and Holder estimates . . . 932.2.2 Exponential integrability in general Lipschitz domains 105

2.3 Meyers’ estimate for systems with L∞ coefficients . . . . . . . 1072.4 A substitute for L1: maximal functions and the Hardy space

H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1132.5 Regularity theory for nonlinear pde . . . . . . . . . . . . . . . 121

2.5.1 Harmonic maps and Hardy space estimates . . . . . . 121

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2.5.2 Harmonic maps with general targets . . . . . . . . . . 1402.5.3 Regularity for minimizers of variational problems . . . 1482.5.4 Monotone operators and the Minty-Browder trick . . . 164

2.6 Maximum principles and existence for quasilinear elliptic equa-tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1682.6.1 Classcial maximum principles for linear equations . . . 1682.6.2 Quasilinear elliptic equations: overview . . . . . . . . 1752.6.3 Maximum and comparison principles for quasilinear

elliptic equations . . . . . . . . . . . . . . . . . . . . . 1762.6.4 Existence of classical solutions . . . . . . . . . . . . . 1862.6.5 C1,α estimates . . . . . . . . . . . . . . . . . . . . . . 1932.6.6 Gradient bounds at the boundary . . . . . . . . . . . 1982.6.7 Global gradient bounds . . . . . . . . . . . . . . . . . 204

3 A brief look back 2073.1 Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2073.2 Regularity for nonlinear PDE . . . . . . . . . . . . . . . . . . 2083.3 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2083.4 Some topics that were not covered . . . . . . . . . . . . . . . 209

A Interpolation between BMO and L2: proofs 213

B Regularity for two-dimensional systems with W 1,1 coefficients216

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[8.10. 2018, Lecture 1]

1 Introduction

1.1 Examples

1.1.1 The Navier-Stokes equation

Let n ≥ 2, ν > 0 and let

v : [0,∞)× Rn → Rn ’velocity’, (1.1)

p : [0,∞)× Rn → Rn ’pressure’. (1.2)

The Navier-Stokes equations are given by

∂

∂tvj +

n∑i=1

vi∂

∂xivj − ν∆vj +

∂

∂xjp = 0, ∀j = 1, . . . n, (1.3)

div v = 0, (1.4)

where div v :=∑n

i=1∂∂xivi and ∆ =

∑ni=1

∂2

∂2xi, with initial condition

v(0, x) = v0(x). (1.5)

The sixth Clay Millennium problem states1: prove or disprove for n = 3that

If v0 ∈ C∞(Rn) with rapid decay, i.e., |∂αu(x)| ≤ CαK(1 + |x|)−K for all αand K, then there exists (v, p) ∈ C∞([0,∞)× Rn;R3 × R) which satisfies

(1.3)–(1.5).

What is known is that there exist solutions for a small time interval[0, T ) for general smooth initial data v0 with sufficiently rapid decay (whereT depends on v0) and solutions on an infinite time interval for small initialdata. These results are proved by a perturbation argument based on goodestimates for the linear heat equation ∂tv = ∆v and on the Helmholtzdecomposition of a vector field into a gradient and a divergence free part(this is used to eliminate the pressure)2.

An important breakthrough was J. Leray’s paper in 1934 where heshowed the global existence of weak solutions. This reduces the question

1See http://www.claymath.org/millennium/Navier-Stokes Equations/navierstokes.pdffor the precise statement

2Over time such results have been proved in various function spaces. A nearly optimalresult is H. Koch, D. Tataru, Well-posedness for the Navier-Stokes equations, Adv. Math.157 (2001), 22–35.

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of existence of smooth solutions to the question to prove that weak solu-tions are already regular. Leray himself already showed that his solutionsare smooth except on closed set in time of one dimensional Lebesgue measure0. The estimate on the size of the set of possible singularities in space-timewhere much later improved by Scheffer and Cafarelli-Kohn-Nirenberg.

1.1.2 Mean curvature flow

Let Mn be a compact n-dimensional manifold and consider a family of mapsF : [0,∞)×M → Rn+1. We say that F evolves by mean curvature if

∂tF = −Hν (1.6)

where ν denotes the normal of the hypersurface F (t, ·) and H its meancurvature (more precisely the sum of the principal curvatures). If F0 is asphere then one easily sees that the mean curvature flow yields a family ofshrinking spheres whose radii satisfy the ODE

R(t) = − n

R(t)(1.7)

so that R2(t) = R(0)2 − 2nt and the flow has a point singularity at T =R2(0)/2n.

If the surfaces are (locally) graphs of a function u : [0,∞)× Ω→ R theequation is equivalent (up to reparametrisation due to tangential motion)to

∂tu =√

1 + |∇u|2 div∇u√

1 + |∇u|2=

(δij −

∂iu∂ju

1 + |∇u|2

)︸ ︷︷ ︸

=aij(x)

∂i∂ju. (1.8)

If |∇u| ≤M , i.e., if the slope of the graph is bounded then

∞∑i,j=1

aij(x)ξiξj ≥1

1 +M2|ξ|2, (1.9)

i.e., the equation (1.8) is parabolic. Moreover in the limit of vanishing slopethe equation reduces to the linear heat equation. This suggests that meancurvature flow at least for small times behaves similar to the heat equation.

The following global results are known for the mean curvature flow.Embedded curves shrink to a round point, see M. A. Grayson, J. Diff. Geom.26 (1987), 285–314. Convex hypersurfaces shrink to a round point, seeG. Huisken, Invent. Math. 84 (1986), 463–480. Dumbbells can developsingularities, see M. A. Grayson, Duke Math. J. 58 (1989), 555–558. Thefull classification of singularities is open despite important partial results,

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see, e.g., Huisken and Sinestrari, Acta Math. 183 (1999), 45–79 and Invent.Math. 175 (2009), 137–221.

In the analysis of the mean curvature flow maximum principles play acrucial role. The simplest one is a geometric maximum principle. If F ′0 iscontained in the open bounded set enclosed by F0 then the relation holdsat all later times (as long as both flows exist and are smooth). There aremore subtle maximum principles for the curvatures of the surfaces. Roughlyspeaking these guarantee that the surfaces become rounder along the flow(i.e., the ratio of the largest and smallest principal curvature approaches 1).

1.1.3 Ricci flow

The mean curvature flow describes the evolution of a family of surfaces in anambient space. The Ricci flow describes the evolution of a family of metricson a fixed manifold. Under suitable conditions both make the geometrynicer as the flow evolves.

Consider a family of Riemannian metrics gt on an n-dimensional mani-fold M . Then the Ricci flow is given by the equation

∂tgt = −2 Ricg (1.10)

where Ricg is the Ricci curvature of the metric g which depends affinely onthe second derivatives of g. The Ricci flow should be viewed as a kind of heatequation for the metric g. Nonetheless the equation (1.10) is in general nota parabolic equation because the right hand side is not a positive definiteexpression in the second derivatives of g (this is related to the fact thatthe flow is invariant under a change of coordinates). If one makes a goodchoice of coordinates (a ’gauge’) then the equation becomes parabolic. Inthe simplest case n = 2 one can always choose conformal coordinates, i.e.,one can choose coordinates for which the metric has the form

gαβ = e2pδαβ. (1.11)

Then (1.10) becomes∂tp = e−2p∆p = ∆gp, (1.12)

where ∆g denotes the Laplace-Beltrami operator for the metric g. TheGauss curvature K and the scalar curvature R are given by

R = 2K = −2e−2p∆p (1.13)

and thus∂tR = R2 + e−2p∆R = R2 + ∆gR. (1.14)

In this form one sees the competition between the ODE ∂tR = R2 whichwould lead to blow-up in finite time and which enhances differences betweendifferent spatial values of R and the heat equation ∂tR = ∆R which makes R

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smoother in space as time evolves and reduces the difference between valuesat R at different spatial points. Fundamental results about the Ricci flowcentered around the idea that after suitable rescaling the flow converges toa metric of constant (scalar) curvature and that possible singularities canbe controlled.

The most spectacular application of Ricci flow is certainly Perelman’sproof of the Poincare conjecture (see Tao’s paper arXiv:math/0610903 fora high-level overview of Perelman’s proof from a PDE perspective and thebook ’Ricci flow and the Poincare conjecture’ by J. Morgan and G. Tian fora full exposition).

Another striking recent application of the Ricci flow is the proof of thedifferentiable version of the Klingenberg sphere theorem (every simply con-nected compact manifold with section curvatures strictly between 1/4 and1 is diffeomorphic to the sphere3).

1.1.4 Minimal surfaces

These are by definition stationary points of the mean curvature flow, i.e.,surfaces which satisfy

H = 0. (1.15)

For graphs this reduces to the PDE

div∇u√

1 + |∇u|2= 0, (1.16)

which is elliptic as long as |∇u| ≤M .The equation (1.16) arises as the Euler-Lagrange equation of the area

functional

A(u) =

∫Ω

√1 + |∇u|2 dx. (1.17)

1.1.5 Harmonic maps

Let Ω ⊂ Rn be open, let M ⊂ Rm be a submanifold and consider mapsu : Ω → M . A harmonic map is a stationary point (i.e., a solution of theEuler-Lagrange equation) of the Dirichlet integral

E(u) =1

2

∫Ω|∇u|2 dx. (1.18)

If M = Rn the Euler-Lagrange equation reduces to −∆u = 0 (hence thename). In general a (sufficiently regular) map is harmonic if and only if

−∆u = A(u)(∇u,∇u), (1.19)

3see S. Brendle, R. Schoen, Bull. Amer. Math. Soc. 48 (2011), 1–32 and J. Amer. Math.Soc. 22 (2009), 287–307, C. Bohm and B. Wilking, Ann. Math. 167 (2008), 1079–1097.

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where A(p) is a certain quadratic form with values in the normal space(TpM)⊥. If M = Sm−1 then u is a harmonic map if and only if

−∆u = |∇u|2u. (1.20)

It has long been open whether weak solutions of this equation are smooth.Finally in the 1990’s F. Helein showed that this is indeed the case if n = 2(C.R. Acad. Sci. Paris Math. 311 (1990), 519–524, while Riviere con-structed a counterexample with a dense set of singularities for n = 3 (ActaMath. 175 (1995), 197–226). Under a mild additional assumption for n ≥ 3one can show that the set of singularities is at most a closed set whose n−2dimensional Hausdorff measure is zero. We will discuss these results later inthe course. An important feature which makes the analysis of the regularityof solutions very delicate is that (1.20) is invariant under rescaling. Moreprecisely if we define

ur(x) := u(rx) (1.21)

then ur satisfies the same equation as u. If instead we consider the equation

−∆v = |∇v|pv with 1 ≤ p < 2 (1.22)

then the rescaled functions vr satisfy the equation

−∆vr = r2−p|∇vr|pv, (1.23)

i.e., by rescaling small scales back to 1 the equation becomes more linear andhence easier. For geometric problems, however, often scaling invariance is anatural feature of the problem and one often needs to develop new analyticaltools since one is at the borderline of classical compactness and regularitymethods.

1.1.6 Nonlinear elasticity

Let Ω ⊂ Rn. To an elastic deformation u : Ω→ Rn we assign an energy

E(u) =

∫ΩW (∇u) dx. (1.24)

The corresponding Euler Lagrange equation is

−divDW (∇u) = 0 (1.25)

or, in components,

−n∑

α=1

∂

∂xα∂W

∂Fiα(∇u(x)) = 0 for all i = 1, . . . , n. (1.26)

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This is a system of PDE and we will discuss in the next section when thissystem is elliptic (the harmonic map equations are also a system of PDEbut in this case the term in the second order derivatives is diagonal).

The corresponding time dependent equation is given by

ρ0(x)∂2t u(t, x) = divDW (∇u). (1.27)

If W (∇u) = 12 |∇u|

2 and ρ0 ≡ 1 this becomes a system of n decoupledwave equations ∂2

t uj = ∆uj for j = 1, . . . , n. In general the system (1.27)

is a nonlinear system of wave equations. We will consider such hyperbolicsystems in NPDE II in the summer term.

As a side remark note that the physical interpretation of (1.27) is justNewton’s third law. On the right hand side DW (∇u) is the Piola-Kirchhoffstress tensor and its divergence is a force density (force per unit volume).On the left hand side ρ0 is a mass density (mass per unit volume) and ∂2

t u isthe acceleration. Thus the equation is a continuum version of Newton’s law:force equals mass times acceleration. A systematic derivation of (1.27), theNavier-Stokes equation and related equations was discussed in the summerterm 2018 in the course ’PDE and modeling’. A good reference is the book’Introduction to continuum mechanics’ by M.E. Gurtin, Academic Press.

1.2 Some general strategies

1.2.1 Soft functional analytic and topological methods

The idea is to view the left hand side of a PDE as a map between Banachspaces and to use the inverse function theorem or topological fixed pointtheorems. The basic idea from the point of PDE is that good estimatesimply existence of solutions.

We begin with a simply result for a family for linear operators whichillustrates this idea.

Lemma 1.1. Let X be a Banach space and let Y be normed space and letL0 and L1 be bounded linear operators from X to Y . For each t ∈ [0, 1] set

Lt = (1− t)L0 + tL1

and suppose that there exists a constant C such that

‖x‖X ≤ C‖Ltx‖Y ∀x ∈ X, t ∈ [0, 1].

Then L1 maps X onto Y if and only if L0 maps X onto Y .

Proof. See [GT], Theorem 5.2. Idea: Assume that Ls is onto for somes ∈ [0, 1]. Now rewrite the equation Ltx = y as

x = L−1s y + (t− s)L−1

s (L0 − L1)x

and use the Banach fixed point theorem to show that this equation is solvablefor |s−t| < δ. Finally subdivide [0, 1] into intervals of length less than δ.

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New we recall the inverse function theorem which shows that a nonlinearproblem can be solved (for small data) if we have a good estimate for thelinearized problem.

[8.10. 2018, Lecture 1][12.10. 2018, Lecture 2]

Theorem 1.2. Let X and Y be Banach spaces. Let U ⊂ X be open andassume that A ∈ Cr(U, Y ) with r ≥ 1, i.e., A is r times Frechet differentiablewith continuous derivatives. Let u0 ∈ U , f0 := A(u0) and assume that thedifferential

DA(u0) : X → Y is an invertible map. (1.28)

Then there exist ε, δ > 0 such that

For each f ∈ Bε(f0) ⊂ Y there exists a unique u ∈ Bδ(u0) ⊂ X with

Au = f (1.29)

Moreover the map f 7→ u is a Cr map.

Application to PDEs. Let Ω ⊂ Rn be bounded with smooth boundary.Let g ∈ C1(R) with bounded derivative and consider the boundary valueproblem

−∆u+ g(u) = f in Ω, (1.30)

u = 0 on ∂Ω. (1.31)

We first make the choice

X = u ∈ C2(Ω) : u|∂Ω = 0, (1.32)

Y = C(Ω) (1.33)

and we setA(u) = −∆u+ g u. (1.34)

It is easy to check that A ∈ C1(X,Y ) and that the differential at u0 = 0 isgiven by

Lv := DA(0)v = −∆v + g′(0)v. (1.35)

The invertibility of L is equivalent to the following assertion for a linearboundary value problem. For every f ∈ Y there exists a unique v ∈ X suchthat

−∆v + g′(0)v = f in Ω, (1.36)

v = 0 on ∂Ω. (1.37)

Moreover there exists a constant C such that for all f ∈ Y

‖v‖X ≤ C‖f‖Y . (1.38)

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Assume first thatg′(0) = 0. (1.39)

It turns out that with the above choice of X and Y the linear map L is notinvertible (see, e.g., the lecture notes [FAPDE], Proposition 7.9).

As we shall see in the next section the operator L is invertible if we workinstead in the Holder spaces

X = u ∈ C2,α(Ω) : u|∂Ω = 0, (1.40)

Y = C0,α(Ω) (1.41)

with 0 < α < 1 (note that a Holder continuous function in Ω has a uniqueextension to Ω so that the condition u|∂Ω = 0 makes sense). Thus thereexist ε, δ > 0 such that

If ‖f‖C0,α < ε then the problem (1.30)– (1.31) (with g′(0) = 0)has a unique solution u with

‖u‖C2,α < δ. (1.42)

If g′(0) 6= 0 the same conclusion holds by the Fredholm alternative forelliptic second order PDE (see [GT], Thm. 8.6 or [Ev], Chapter 6, Thm. 4and 5 or the lecture notes [FAPDE]), as long as g′(0) is not an eigenvalueof the operator ∆. Strictly speaking the Fredholm alternative first providesa solution in the Sobolev space W 1,2

0 . The fact that the solution is in C2,α

then follows from the regularity theory we will develop in the next section.In a similar way one can use the inverse function theorem to show that

there exist solutions for the equations of nonlinear elasticity (1.25) which inC2,α are close to the identity. In fact until J.M. Ball’s paper on a variationalapproach in 1977 this was the only way to show the existence of solutionsunder physically realistic conditions on W .

Remark. 1. There is a natural version of the inverse function theoremfor Banach manifolds. This allows one to handle problems with constraintssuch as the harmonic map problem.2. It suffices that DA(u0) is bijective. Then the boundedness of the inversefollows from the open mapping theorem.3. In some cases the operator L := DA(u0) is not invertible as a map fromX to Y , but only has an inverse which maps into a weaker space than X (’loss of regularity’). Under additional assumptions it is still possible to solvethe nonlinear problem through the so called ’hard inverse function theorem’of Nash and Moser4.

4see R.S. Hamilton, Bull. Amer. Math. Soc. 7 (1982), 65–222 or the book S. Alinhac,P. Gerard, Operateurs pseudo-differentiels et theoreme de Nash-Moser.

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Warning The assumption A ∈ C1(U, Y ) is not always trivial. Note forexample that the map

u 7→ sin u (1.43)

is differentiable (and in fact C∞) as map from L∞(0, 1) to itself, but thatthis map is not differentiable as a map from L2(0, 1) to itself. In fact wehave

Proposition 1.3. Let E ⊂ Rn be measurable with 0 < Ln(E) < ∞, let1 ≤ p <∞, let f : R→ R and consider the Banach space X = Lp(E). Thenthe map

u 7→ f u (1.44)

is Frechet differentiable as a map from X to X if and only if f is affine.

Remark. Similarly one sees that if Ln(E) =∞ then f must be linear.

Proof. The ’if’ part is clear. For the ’only if’ part first note that by consid-ering constant functions u one sees that f must be a differentiable function.By subtracting an affine function we may assume that f(0) = f ′(0) = 0. Itremains to show that f ≡ 0. This is left as an exercise.

There is a useful criterion for Frechet differentiability which is based ondirectional derivatives. A map A : U ⊂ X → Y is called Gateaux differen-tiable at a point x0 ∈ U if there exists a bounded linear map Lx0 : X → Ysuch that for all v ∈ X the one dimensional function t 7→ f(x0 + tv) isdifferentiable at t = 0 and d

dt |t=0f(x0 + tv) = Lv. Using the one-dimensional

Taylor formula with remainder term one can easily derive the following cri-terion for continuous Frechet differentiability: if f is Gateaux differentiablein each point in an open set V and the map x 7→ Lx is continuous as a mapfrom X to the space L(X,Y ) of bounded linear maps from X to Y (equippedwith the operator norm) then f is Frechet differentiable in V (and in factin C1(V ;Y )).

Roughly speaking, the inverse function theorem can only be used inspaces which are already sufficiently strong with respect to the nonlinearitiesinvolved.

The inverse function theorem gives existence and uniqueness of solutions(in the open set where it applies). This is good, but may be more than wecan expect in particular if we also want to look for ’large’ solutions. Inthis case topological fixed point theorems which only give existence are veryuseful. We start with a fixed point theorem in finite dimensional spaces.

Theorem 1.4 (Brouwer’s fixed point theorem). Let B ⊂ Rn be the closedunit ball. Assume that f : B → B is continuous. Then f has a fixed point,i.e., there exists x ∈ B such that

f(x) = x. (1.45)

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Remark. Clearly B can be replaced by a set which is homeomorphic toB.

Proof. For a short PDE proof see, e.g., [Ev] Chapter 8.1, Thm. 3. Seealso Problems 4. and 5. in Chapter 8 for the definition of the topologicaldegree.

Now we look for fixed point theorems in an (infinite-dimensional) Banachspace X. Here in addition to continuity we need some compactness.

Theorem 1.5 (Schauder). Let S ⊂ X be compact, convex, and not empty.Let T : S → S be continuous. Then T has a fixed point.

Proof. See [GT], Thm. 11.1

Remark. It suffices that S is not empty and homeomorphic to a compactand convex set.

Corollary 1.6. Let S ⊂ X be closed, convex and not empty. Let T : S → Sbe continuous and assume that T (S) is precompact. Then T has a fixedpoint.

Proof. See [GT], Cor. 11.2

Theorem 1.7 (Leray-Schauder, special case). Suppose that T : X → X iscompact, i.e., the for every bounded set in X the image under T is precom-pact. Suppose that there exists R > 0 with the following property.

∀σ ∈ (0, 1) if x = σTx then ‖x‖ < R. (1.46)

Then T has a fixed point.

Proof. Define

T (x) :=

Tx if ‖Tx‖ ≤ RR‖Tx‖Tx if ‖Tx‖ > R.

(1.47)

Then T is continuous and compact and maps the closed ball B(0, R) toitself. Hence T has a fixed point x. If ‖Tx‖ ≤ R then Tx = T x = x and weare done. If ‖Tx‖ > R then ‖T x‖ = R and

x = T x =R

‖Tx‖Tx. (1.48)

Hence by assumption (with σ := R/‖Tx‖ < 1)

R = ‖T x‖ = ‖x‖ < R. (1.49)

This contradiction finishes the proof.

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Application to PDEs. Let Ω ⊂ Rn be open and bounded with smoothboundary and let ϕ ∈ C2,α(Ω) with 0 < α < 1. Let g ∈ C1(R) with boundedderivative. Consider the boundary value problem

−∆u+ g(u) = 0 in Ω, (1.50)

u = ϕ on ∂Ω. (1.51)

For v ∈ C0,α(Ω) consider the linear PDE for u

−∆u+ g(v) = 0 in Ω, (1.52)

u = ϕ on ∂Ω. (1.53)

Assume for a moment that this linear PDE has a unique solution u. Wenow define a map T by Tv := u.

Then u is a solution of the nonlinear problem if and only if it is a fixedpoint of T , i.e., if

T u = u. (1.54)

It is easy to see that g v in C0,α(Ω) and that the map v 7→ g v isa continuous map from C0,α(Ω) to itself. It follows from the linear theorywhich we will develop in the next section that T is a continuous map fromC0,α(Ω) to C2,α(Ω). The embedding C2,α(Ω) → C0,α(Ω) is compact (Arzela-Ascoli !) and thus

T : C0,α(Ω)→ C0,α(Ω) is compact. (1.55)

Thus it suffices to show that

∃R > 0 ∀σ ∈ (0, 1) Tu =u

σ=⇒ ‖u‖C0,α < R. (1.56)

By the definition of T we thus need to show that if

−∆u

σ+ g(u) = 0 in Ω, (1.57)

u

σ= ϕ on ∂Ω (1.58)

then ‖u‖C0,α < R. Or, equivalently: there exists R > 0 such that if σ ∈ (0, 1)and if u solves

−∆u+ σg(u) = 0 in Ω,u = σϕ on ∂Ω

(1.59)

then ‖u‖C0,α < R. Thus the problem of existence is reduced to the problemof showing a uniform estimate if a solution exists. Such estimates are knownas a priori estimates. Proving a priori estimates will be a key element ofthis course.

14 [January 28, 2019]

1.2.2 Weak convergence methods and compactness

Here is a general strategy to solve nonlinear PDE written symbolically asA(u) = f

(i) Consider approximative problems (e.g. finite dimensional approxima-tion, regularization by higher order terms, . . . )

Ak(uk) = fk with fk → f, Ak → A in a suitable sense. (1.60)

(ii) Derive a priori estimates

‖uk‖X ≤ C. (1.61)

(iii) Deduce (weak) convergence of a subsequence in a suitable topology τ

ukτ→ u. (1.62)

(iv) Show A(u) = f (compactness/ stability argument).

The crucial difficulty in the last step is that usually weak convergencedoes not commute with the action of nonlinear functions. This problem canbe overcome in several ways.

We can try to prove a priori estimates in a strong enough space to ensurestrong converge of the quantities which enter in the nonlinearities, e.g., ifthe nonlinearities only involve terms with lower order derivatives.

We can also use the fact that uk is an approximate solution to get strongconvergence. Think of the result that in a Hilbert space weak convergenceuk u plus convergence of the norm ‖uk‖ → ‖u‖ implies strong conver-gence. This follows simply by expanding the ‖uk − u‖2. The deeper reasonbehind this result is that Hilbert spaces are uniformly convex and the as-sertion holds more generally in uniformly convex Banach spaces like Lp (for1 < p < ∞). Convexity arguments or closely related monotonicity argu-ments5, are very powerful tools to show that A(u) = f , but convexity ormonotonicity are often not available.

We can also try to use additional structures to show that not all, butcertain nonlinear quantities pass to the limit. In the course PDE and mod-eling we showed, e.g., that the weak convergence uk u in W 1,p(Rn,Rn)implies that det∇k det∇u in Lp/n if p > n. This is the idea of themethod of compensated compactness developed by F. Murat and L. Tartar(the original French name ’compacite par compensation’ describes the ideaof the theory better).

The proof of existence via subsolutions and supersolutions (Perron’smethod) can be seen in that framework, too. The approximation consists

5see Section 2.5.4

15 [January 28, 2019]

in replacing the equation by an inequality (understood in a rather weaksense). The necessary bounds on the class of all subsolutions come from amaximum principle (or more generally from a comparison principle). Weakconvergence is replaced by taking the supremum of all subsolutions. Thenthe key point is to show that this yields actually a solution.

In this course we will see a number of strategies to implement weakconvergence methods. A beautiful short survey of a number of importantmethods can be found in the book L.C. Evans, Weak convergence methodsfor nonlinear partial differential equations, Amer. Math. Soc., 1990. Theclassic book J.L. Lions, Quelques methodes de resolution des problemes auxlimites non lineaires, Dunod; Gauthier-Villars, 1969, which also emphasizesthe aspect of compactness is still very enjoyable to read.

1.2.3 Variational methods

Let X be a Banach space and let I : X → R be a functional on X. We saythat u ∈ X is a stationary point of I if I is (Frechet) differentiable at u andthe derivative satisfies

DI(u) = 0. (1.63)

If u0 is a minimizer of I (in X or at least in some open neighbourhood of u0)and if I is differentiable at u0 then u0 is a stationary point. Many interestingPDE arise as stationary points of suitable functionals (these PDE are thencalled variational).

Example. Let Ω ⊂ Rn be bounded and open, let X = W 1,20 (Ω) and let

G ∈ C1(Rn). For simplicity of notation let n ≥ 3 and assume that

|G(s)| ≤ C(1 + |s|2∗), where 2∗ =2n

n− 2(1.64)

is the critical exponent in the Sobolev embedding theorem W 1,20 (Ω) →

L2∗(Ω). Consider on X the functional

I(u) =

∫Ω

1

2|∇u|2 +G(u) dx. (1.65)

If we assume in addition that

|G′(s)| ≤ C(1 + |s|2∗−1) (1.66)

then one can show (exercise !) that I : X → R is Frechet differentiable andthe derivative is given by

DI(u)ϕ =

∫Ω∇u · ∇ϕ+G′(u)ϕdx ∀ϕ ∈W 1,2

0 . (1.67)

16 [January 28, 2019]

Thus if u is a minimum of I then u is a weak solution of the PDE

−∆u+G′(u) = 0. (1.68)

This equation is called the Euler-Lagrange equation of I.Here is a fine technical point. One can make a connection between

minima of I and the PDE even if we replace (1.66) by the weaker condition

|G′(s)| ≤ C(1 + |s|2∗). (1.69)

Under this condition one can still show that I has directional derivatives ingood directions. More precisely for all ϕ ∈ C1

c (Ω) the functions t 7→ I(u+tϕ)are differentiable (exercise !) and

d

dt |t=0I(u+ tϕ) =

∫Ω∇u · ∇ϕ+G′(u)ϕdx ∀ϕ ∈ C1

c (Ω). (1.70)

Thus under assumptions (1.69) and (1.64) a minimizer u of I is still a dis-tributional solution of (1.68).

What are the advantages of the variational formulation? One point isthat the variational formulation is much more robust. It only involves firstderivatives and not second derivatives and the existence of minimizers (evenfor rather ’rough’ integrands) can often be established by applying the di-rect method of the calculus of variations in the Sobolev space W 1,p(Ω) (ora subspace thereof). Another point is that solutions of the Euler-Lagrangeequation which come from minimizers (or local minimizers) often enjoy im-portant additional properties. We will see this later, e.g., in the discussionof the regularity of harmonic maps.

[12.10. 2018, Lecture 2][15.10. 2018, Lecture 3]

There are other important stationary points beyond minimizer. If X isone dimensional then between two minima of a function there is always alocal maximum. In higher dimensions one can argue that between two localminima there is a saddle point or mountain-pass. We say that x is a strictlocal minimum of a function E : X → R if there exists an open ball B(x, r)such that E(x) ≤ E(y) for all y ∈ B(x, r) and inf∂B(x,r)E > E(x).

Theorem 1.8 (Mountain pass theorem). Suppose that E ∈ C1(Rn) is coer-cive, i.e., limx→∞E(x) =∞. Suppose further that E has two distinct strictlocal minima x1 and x2. Then E possesses a third critical point x3 which isnot a local minimum and characterized by the minmax principle

E(x3) = infp∈P

maxx∈p

E(x) (1.71)

where

P := p ⊂ Rn : x1, x2 ∈ p, p is compact and connected. (1.72)

17 [January 28, 2019]

There is a version of the theorem for C1,1 functionals E (i.e. maps whosefirst Frechet derivative is Lipschitz continuous) on Banach spaces providedthat E satisfies a so-called Palais-Smale condition. We say that sequence aumm∈N ⊂ X is a Palais-Smale sequence if

|E(um)| ≤ C and limm→∞

‖DE(um)‖X∗ = 0 (1.73)

The Palais-Smale condition is:

every Palais-Smale sequence contains a convergent subsequence. (1.74)

Note that in finite dimensional spaces this follows from coercivity since thenthe closure of bounded sets is compact.

An excellent reference on minmax method and on variational methods ingeneral is the book M. Struwe, Variational methods, Springer, 4th edition,2008. This book also contains many beautiful applications to problems ingeometry.

1.2.4 Convex integration

Ideas related to what is now called convex integration first appear in J.Nash’s seminal work on ’wild’ isometric immersion with minimal regular-ity and were later developed into a systematic theory by Gromov who alsocoined the name6. The method was first applied mostly to geometric equa-tions but recently also led to deep new insights in PDEs coming from physics,including in particular the Euler equations of hydrodynamics7. L. Szekely-hidi’s lecture notes8 provide an excellent short survey. Recently this ap-proach has also lead to surprising nonuniqueness results for weak solutionsof the Navier-Stokes equations9.

The idea is initially somewhat counterintuitive since it relies to someextend on a certain lack of compactness while we usually argue that com-pactness is crucial for existence in nonlinear PDE. One replaces the originalproblem first by a ’relaxed’ problem, usually some sort of convexificationof the original problem which is much easier to solve. Then one iterativelyadds the right (usually one-dimensional) oscillations to the solution of the

6see the book M. Gromov, Partial differential relations, Springer, 1986, which is toughreading but contains a wealth of deep ideas.

7C. DeLellis, L. Szekelyhidi, The Euler equations as a differential inclusion, Ann. Math.170 (2009), 1417–1436; C. DeLellis, L. Szekelyhidi, Dissipative Euler flows, Invention.Math. 193 (2013), 377–407; T. Buckmaster, C. DeLellis, L. Szekelyhidi, V. Vicol, On-sager’s conjecture for admissible weak solutions, arXiv 1701.08678v1; P. Isett, A proof ofOnsager’s conjecture, Ann. Math. 188 (2018), 1–93.

8From isometric embeddings to turbulence, Preprint MPI MIS Leipzig,http://www.mis.mpg.de/preprints/ln/lecturenote-4112.pdf

9T. Buckmaster and V. Vicol, Nonuniqueness of weak solutions to the Navier-Stokesequation arXiv:1709.10033v3, arXiv.org.

18 [January 28, 2019]

relaxed problem to obtain a solution of the original problem. The initiallysurprising fact is that this sequence of iterations actually converges stronglyto a solution of the original problem (one might expect that due to the os-cillations one only gets weak convergence and everything is lost again in thelimit). For those who know probability this argument is similar to the proofthat martingales converge.

To explain the idea let us just consider a toy example. Let Ω ⊂ Rn beopen and bounded. We look for a (Lipschitz) function u : Ω→ R such that

|∇u| = 1 a.e. in Ω, (1.75)

u = 0 on ∂Ω. (1.76)

The first condition can be written as a partial differential relation

∇u(x) ∈ Sn−1 ⊂ Rn. (1.77)

If we replace Sn−1 by its convex hull, the closed unit ball, we obtain therelaxed problem

|∇u| ≤ 1 a.e. in Ω, (1.78)

u = 0 on ∂Ω. (1.79)

This has the trivial solution u0 = 0. Now we want to add oscillations to u0

to obtain a solution of the original problem. There are many ways to dothat for this concrete problem. Here is one possibility. Consider for ease ofimagination the case n = 2. Let Ωk := x ∈ Ω : dist (x,Rn \ Ω) ≥ 2−k. Inthe first step choose finitely many disjoint triangles (contained in Ω) whichcover the compact set Ω1. On each triangle T choose a piecewise linear hatfunction vT which satisfies |∇vT | = 1 and agrees with u0 on the boundaryof T . Replacing u0 by vT on all triangles we obtain the next iterationu1. For the next step add further triangles so that Ω2 is covered and dothe replacement on this triangles. In the limit this yields a function whichsolves the original problem |∇u∞| = 1 a.e. and u∞ = 0 on ∂Ω.

This toy problem is actually too simple. We could have done the replace-ment in one step, e.g. by taking u as the distance function u = dist (x, ∂Ω).What makes this problem easy is that we can do directly a multidimen-sional replacement and in one step replace an affine function on a triangleby a piecewise affine one with the same boundary conditions such that thenew function already solves the original problem. In typical applications ofconvex integration one can only add oscillation in one spatial direction attime. This makes it impossible to satisfy affine boundary conditions on thetriangle exactly. Thus some additional errors arise in a thin region near theboundary of the triangle. To make sure that these errors are at least com-patible with the relaxed problem (and thus can be eventually removed inlater steps) we can usually not choose the one-dimensional piecewise affine

19 [January 28, 2019]

maps so that the gradients are exactly on set where we want them to be(in our example Sn−1) but we have to stay a little inside the interior of therelaxed set (in our example the unit ball Bn). Thus in each step on eachtriangle the gradient is only pushed closer to our desired set and the trian-gles have to be refined for the next step. Then the final strong convergenceof the gradients (a.e.) is less obvious, but also not difficult to show if thenew oscillations are always chosen much faster than the previous ones.

There is also a variant of this construction where one adds smooth oscil-lations rather than piecewise affine ones (this is indeed what Gromov does)but the issues are very similar.

[15.10. 2018, Lecture 3][19.10. 2018, Lecture 4]

1.3 Overview of a priori estimates

Let A denote a partial differential operator. As we have seen a priori esti-mates of the form

Au = f, ‖f‖... ≤ R1 =⇒ ‖u‖... ≤ R2. (1.80)

play a crucial role in the theory.Below is a rough overview of some important spaces and techniques.

This list is be no means exhaustive. To prove estimates in the Holder spacesCα one can, e.g., also so the representation of the solution in terms of aGreen’s function (see, e.g., [GT]).

L2 use test functions, most robustL∞ max. principle or Sobolev embedding

Lp, 1 < p <∞ harmonic analysis orCampanato spaces + interpolation

Cα Campanato spaces,or blow-up & compactness

BMO substitute for L∞, Campanato spacesor harmonic analysis

H1 substitute for L1, harmonic analysisor duality with BMO

parabolic space-time elliptic + semigroup ideas,harmonic analysis,damping of high frequencies

hyperbolic space-time L2 + Strichartz est., wave packetsdisperse due tp speed differences

While a lot can be achieved for nonlinear PDE using mostly good esti-mates for linear PDE for the most difficult and interesting nonlinear PDE

20 [January 28, 2019]

one often has to find and use additional estimates which depend on thenonlinear structure.

For the Navier-Stokes equation, e.g., multiplication of the equation byv, integration in x and integration by parts yields the crucial global energyestimate

1

2

∫Rn|v|2(T, x) dx+

∫ T

0

∫Rn|∇v|2 dxdt ≤ 1

2

∫Rn|v0|2(x) dx ∀T > 0.

(1.81)This estimate depends upon the fact that for the most dangerous term weget (after a clever integration by parts and using that div v = 0)∫

Rn

n∑j=1

n∑i=1

[vi

∂

∂xivj]vj dx = 0. (1.82)

There is no similar estimate for the derivatives of v. One only gets the weakerestimates d

dt12‖v‖

2Wk,2 ≤ C‖v‖3Wk,2 − ν‖∇v‖2Wk,2 . These estimates are useful

to construct local-in-time solutions but they do not rule out blow-up of thehigher norms in finite time.

Important nonlinear estimates are often motivated by insights from phy-sics about the conservation or increase or decrease of certain quantities suchas energy or entropy. One should note, however, that the derivation ofestimates like (1.81) assumes a certain degree of regularity and that theydo not automatically hold for weak solutions. Indeed for the Euler equationfor which the same calculation shows that ‖v(t, ·)‖L2 should be constant int there exist weak solutions which very strongly violate the energy estimate.

For the Ricci flow one key insight of Perelman was the constructionof certain nonlinear expression (in the spirit of a physical entropy) whichdecrease along every solution. Using these he could in particular rule outcertain potential singularities by showing that to form these this quantitywould have to go to ∞.

1.4 Prerequisites

Some basic knowledge in Functional analysis and the theory of Sobolevspaces and weak solutions is required. For participants with no previousexperience in these areas it is recommend to take the course ’FunctionalAnalysis and PDE’ in parallel. The material needed is more or less coveredin

• [GT] Chapter 5, Chapter 7.1-7.7 and 7.10 and Chapter 8.1 and 8.2, or

• [Ev] Chapter 5.1-5.8, Chapter 6.1 and 6.2 and Appendices C, D andE.

Other useful books include

21 [January 28, 2019]

• H.W. Alt, Linear Funktionalanalysis, Springer, 5th edition, 2006.

• H. Brezis, Functional analysis, Sobolev spaces and partial differentialequations, Springer, 2011.

2 Elliptic equations and systems

2.1 Energy methods: Lp and Cα estimates for elliptic equa-tions and systems

This subsection follows very closely Chapter III of the book

• M. Giaquinta, Multiple integrals in the calculus of variations and non-linear elliptic systems, Princeton Univ. Press, 1983.10

In the following we always assume that

Ω ⊂ Rn is open and bounded. (2.1)

Sometimes additional assumptions on Ω are needed. These will be statedexplicitly.

2.1.1 Overview

In the following we disucss the approach of Campanato to the regularitytheory for elliptic systems. The whole approach is based on three simpleinequalities: the L2 energy estimate, the reverse Poincare inequality andthe Poincare inequality. Their combination turns out to be surprisinglypowerful. We first illustrate the idea by looking at the Poisson equation andharmonic functions.

Energy estimate for the boundary value problem Suppose that f ∈L2(Ω) and that u is a weak solution of the boundary value problem

−∆u = −div f in Ω,

u = 0 on ∂Ω.

The weak form of this boundary value problem is u ∈W 1,20 (Ω) and∫

Ω∇u · ∇ϕdx =

∫Ωf · ∇ϕdx ∀ϕ ∈W 1,2

0 (Ω) (2.2)

10A more detailed exposition can be found in the original paper Sergio Campanato,Equazioni ellittiche del II0 ordine e spazi L2,λ, Annali di Matematica Pura Appl (4) 69(1965), 321–381.

22 [January 28, 2019]

and the choice ϕ = u yields, in combination with the Cauchy-Schwarz in-equality, ∫

Ω|∇u|2 dx =

∫Ωf · ∇u dx ≤ ‖f‖L2 ‖∇u‖L2 (2.3)

and thus the energy estimate∫Ω|∇u|2 dx ≤

∫Ω|f |2 dx. (2.4)

Reverse Poincare estimate Now let Br = B(x0, r) ⊂ Rn be an openball in Rn. We assume that v is weakly harmonic in Br , i.e., a weak solutionof

−∆v = 0 in Br, (2.5)

but we do not impose any boundary conditions on v. The weak form of theequation is ∫

Br

∇v · ∇ϕdx = 0 ∀ϕ ∈W 1,20 (Ω). (2.6)

Now we cannot choose ϕ = v since we do not now that v has zero boundaryconditions. Thus we consider s ∈ (0, r) and a cut-off function η ∈ C∞c (Br)with

0 ≤ η ≤ 1, η = 1 in Bs, |∇η| ≤ 2

r − s. (2.7)

Let a ∈ R and let ϕ = η2(v − a). Since ∇v = ∇(v − a) we get

∇(η2v) = η2∇v + 2η(v − a)∇η. (2.8)

and the weak form of the equation yields∫Br

η2|∇v|2 dx = −∫Br

2∇v · η(v − a)∇η dx

≤ 2

(∫Br

η2|∇v|2 dx)1/2(∫

Br

|∇η|2|v − a|2 dx)1/2

.(2.9)

Using that η = 1 in Bs and thus ∇η = 0 in Bs it follows that∫Bs

|∇v|2 dx ≤ 16

(r − s)2

∫Br\Bs

|v − a|2 dx ∀a ∈ R, s ∈ (0, r). (2.10)

This inequality is called reverse Poincare inequality because it provides anestimate of ∇v in terms of v whereas the usual Poincare inequality (seebelow) provides an estimate for v in terms of ∇v. Some authors also callthis inequality a Cacciopoli inequality.

23 [January 28, 2019]

Poincare inequality

Theorem 2.1 (Poincare inequality for zero mean). Let Ω ⊂ Rn be open,bounded and connected, with Lipschitz boundary. Let 1 ≤ p ≤ ∞. Thenthere exists a constant C(Ω, p) such that

‖u− uΩ‖Lp(Ω) ≤ C(Ω, p)‖∇u‖Lp(Ω) ∀u ∈W 1,p(Ω), (2.11)

where

uΩ :=1

Ω

∫Ωu dx . (2.12)

In particular for 1 ≤ p <∞ there exist constants C1(p) such that∫Br\Br/2

|u− uBr\Br/2 |p dx ≤ C1(p)rp

∫Br\Br/2

|∇u|p dx (2.13)

∫Br

|u− uBr |p dx ≤ C1(p)rp∫Br

|∇u|p dx. (2.14)

Remark. This can be combined with the Sobolev embedding theorem.For p ≤ n, 1

q ≤1p −

1n and q 6= ∞ we thus obtain the Sobolev-Poincare

inequality

‖u− uΩ‖Lq(Ω) ≤ C(Ω, p)‖∇u‖Lp(Ω) ∀u ∈W 1,p(Ω), (2.15)

Notation: we write

ux0,r as a shorthand for uB(x0,r). (2.16)

Proof. We may assume that uΩ = 0 since otherwise we can consider thefunction v = u− uΩ.

SetK = u ∈W 1,2(Ω) : uΩ = 0.

Then K is a closed cone in W 1,2(Ω). Moreover the conditions u ∈ K and∇u = 0 imply u = 0. Thus the result follows from the homework problemwhich establishes the Poincare inequality in this setting.

Now (2.13) and (2.14) hold for r = 1. To obtain the result for generalr it suffices to apply the result for r = 1 the the rescaled function ur(z) =u(x0 + rz).

Consequences of the reverse Poincare inequality

24 [January 28, 2019]

Widman’s hole filler If we combine the inverse Poincare inequalitywith s = r/2, a = uBr\Br/2 with the Poincare inequality with zero mean weget ∫

Br/2

|∇u|2 dx ≤ C

r2

∫Br\Br/2

|u− uBr\Br/2 |2 dx ≤ C

∫Br\Br/2

|∇u|2 dx

(2.17)Now we ’fill the hole’, i.e., we add C

∫Br/2|∇u|2 dx to both sides. This yields

(C + 1)

∫Br/2

|∇u|2 dx ≤ C∫Br

|∇u|2 dx (2.18)

and thus∫Br/2

|∇u|2 dx ≤ θ∫Br

|∇u|2 dx, where θ =C

C + 1< 1. (2.19)

In dimension n = 2 this inequality has very powerful consequences. Ifthis inequality (and the reverse Poincare inequality) holds for all balls thenwe have (see Homework sheet 1)

• Regularity: u is in C0,α for some α > 0 (which only depends on θ),

• Liouville theorem: If u is bounded then u is constant.

Remark. There is often a close relation between Liouville theorems andregularity.We shall see later that the assertion on regularity hold for solutions of scalarelliptic equations with L∞ coefficients also in dimensions n ≥ 3 (DeGiorgi,Nash, Moser theorem), but for systems the conclusion in general do not holdfor n ≥ 3

Reverse Holder inequality Let n ≥ 2 and 1p = 1

2 + 1n . Then 1 ≤ p <

2. Then the Sobolev Poincare inequality yields

‖u− uB1‖L2(B1) ≤ C‖∇u‖Lp(B1). (2.20)

If we combine this with the reverse Poincare inequality we get∫B1/2

|∇u|2 dx ≤ C∫B)

|u− uB1 |2 dx ≤ C(∫

B1

|∇u|p dx)2/p

, (2.21)

i.e., we can bound the L2 norm of the gradient on a smaller ball, by aweaker Lp norm on a larger ball. Such an inequality is called a reverseHolder inequality.

25 [January 28, 2019]

By scaling by obtain the corresponding inequality for arbitrary radii:(1

Ln(Br/2)

∫Br/2

|∇u|2 dx

)1/2

≤ C(

1

Ln(Br)

∫Br

|∇u|p dx)1/p

, (2.22)

where 1p = 1

2 + 1n .

A nontrivial results states that the increase in integrability from p < 2to 2 can be lifted into an increase from 2 to some q > 2.

Theorem 2.2 (Higher integrability). Assume that u ∈ W 1,2loc (Rn) and that

there exists a C > 0 and p < 2 such that (2.22) holds for all balls inB(x0, r) ⊂ Rn. Then there exists a q > 2 (which only depends on p, Cand n such that u ∈W 1,q

loc (Rn) and for all balls(1

Ln(Br/2)

∫Br/2

|∇u|q dx

)1/q

≤ C ′(

1

Ln(Br)

∫Br

|∇u|2 dx)1/2

(2.23)

Proof. See Giaquinta, [Gi], Chapter V, Theorem 1.1 and Theorem 1.2.

Such higher integrability estimates play an important role in the so calleddirect approach to regularity of nonlinear PDE, see Giaquinta [Gi], ChapterVI.

[19.10. 2018, Lecture 4][22.10. 2018, Lecture 5]

2.1.2 Existence of weak solutions

We now begin the study of general elliptic equations and system. To warmup we recall the results for scalar equations proved in the course FA andPDE. Let Ω ⊂ Rn be open and bounded. Let f : Ω→ Rn and g : Ω→ R begiven. We seek solutions u : Ω→ R of the boundary value problem

Lu := −n∑

α,β=1

∂αaαβ∂βu = g −

n∑α=1

∂αfα in Ω, (2.24)

u = 0 on ∂Ω, (2.25)

where the coefficients aαβ are bounded and elliptic, i.e.,

aαβ ∈ L∞(Ω), (2.26)

∃ν > 0 ∀ξ ∈ Rn∑

aαβξαξβ ≥ ν|ξ|2. (2.27)

To simplify the notation we will use from now on the so called summa-tion convention, i.e.,

26 [January 28, 2019]

we always sum over indices which appear twice.

Definition 2.3. A function u ∈W 1,2 is a weak solution of (2.24) if∫Ωaαβ∂βu∂αϕdx =

∫Ωfα∂αϕ+ gϕ ∀ ϕ ∈W 1,2

0 (Ω;Rn). (2.28)

Moreover u is a weak solution of the boundary value problem (2.24)–(2.25)if in addition u ∈W 1,2

0 (Ω).

Theorem 2.4. Assume that f ∈ L2(Ω,Rn), g ∈ L2(Ω) and assume thatthe coefficients satisfy (2.26) and (2.27) Then the boundary value problem(2.24)–(2.25) has a unique weak solution. Moreover this solution satisfies∫

Ω|∇u|2 dx ≤ 1

ν2

(‖f‖L2(Ω;Rn) + C(Ω)‖g‖L2(Ω)

)2(2.29)

where C(Ω) is the constant in the Poincare inequality (2.32)

The proof is based on the Lax-Milgram theorem and the Poincare in-equality in W 1,2

0 (Ω).

Theorem 2.5 (Lax-Milgram). Let X be a real Hilbert space. Let a : X ×X → K be a bilinear form. Suppose that there exist constants c0, C0 > 0such that

(i) (continuity) |a(u, v)| ≤ C0‖u‖ ‖v‖ ∀u, v ∈ X;

(ii) (coercivity) Re a(u, u) ≥ c0‖u‖2.

Let T be a continuous linear functional on X. Then there exists one andonly one u ∈ X such that

a(u, v) = T (v) ∀v ∈ X. (2.30)

Moreover

‖u‖X ≤1

c0‖T‖X∗ (2.31)

Theorem 2.6 (Poincare inequality for W 1,p0 (Ω)). For each bounded open

set Ω ⊂ Rn there exists a constant C(Ω) > 0 such that

‖u‖Lp(Ω) ≤ C(Ω)‖∇u‖Lp(Ω) ∀u ∈W 1,p0 (Ω). (2.32)

If Ω ⊂∏ni1

(ai, ai + li) we may take

C(Ω) = mini=1,...,n

li. (2.33)

27 [January 28, 2019]

Sketch of proof. Assume for simplicity p <∞. For n = 1 and Ω = (0, l) oneuses the fundamental theorem of calculus and the Holder inequality to get

|u(t)| ≤∫ t

0|u′(s)| ds ≤ t

p−1p ‖u′‖Lp . (2.34)

The assertion follows by taking the p-th power and integrating over t. Forn > 1 we extend u be zero to a function in W 1,p(Rn) and we use the onedimensional result and Fubini’s theorem. The proof for p = ∞ is similar.Details were discussed in the course FA and PDG.

Proof of Theorem 2.4. Consider the quadratic form

a(ϕ, u) :=

∫Ωaαβ∂βu∂αϕdx (2.35)

and the linear functional

T (ϕ) :=

∫Ωfα∂αϕ+ gϕ (2.36)

Then a is bounded on W 1,20 (Ω) and T is a bounded linear functional on this

space. Indeed by the Poincare inequality and the Cauchy-Schwarz inequality

T (ϕ) ≤ ‖f‖L2‖∇ϕ‖L2 + C(Ω)‖g‖L2‖∇ϕ‖L2 (2.37)

It thus remains to show that a is coercive. By the ellipticity we have

a(u, u) ≥ ν∫

Ω|∇u|2 dx (2.38)

and the Poincare inequality yields

‖u‖2W 1,2 = ‖u‖2L2 + ‖∇u‖2L2 ≤ (1 + C(Ω)2)‖∇u‖2L2 . (2.39)

Thus a is coercive. The estimate for ∇u follows directly from (2.37) withϕ = u and (2.38).

We now turn to elliptic systems. For i, j = 1, . . . , N and α, β = 1, . . . , nconsider functions fαi : Ω→ R and gi : Ω→ R and coefficients

Aαβij ∈ L∞(Ω). (2.40)

We look for solutions u : Ω→ RN of the elliptic system

(Lu)i := −∂α(Aαβij ∂βuj) = gi − ∂αfαi ∀i = 1, . . . , N. (2.41)

Here we use again the summation convention. The Greek indices α and βare summed from 1 to n and Latin index j is summed from 1 to N . Wewrite this system in concise notation as

Lu := −divA∇u = −div f + g (2.42)

28 [January 28, 2019]

where (AG)αi = Aαβij Gjβ, (∇u)jβ = ∂βu

j and where the divergence is taken

columnwise, i.e. (divG)i = ∂αGαi . On the space RN×n of N × n matrices

we use the Euclidean scalar product an norm, i.e.,

(F,G) = F ·G = F iαFiα, |F |2 = F iαF

iα. (2.43)

We also define

‖A‖ := sup(F,AG) : |F | ≤ 1, |G| ≤ 1 (2.44)

We say that u ∈W 1,2(Ω;RN ) is a weak solution of (2.41) if∫ΩAαβij ∂βu

j∂αϕi dx =

∫Ωgiϕ

i+fαi ∂αϕi dx = 0 ∀ϕ ∈W 1,2

0 (Ω;RN ). (2.45)

We say that the coefficients Aαβij are strongly elliptic (with ellipticityconstant ν) if

∃ν > 0 Aαβij(x)ξαξβζiζj ≥ ν|ξ|2 |ζ|2 ∀ξ ∈ Rn, ζ ∈ RN and a.e. x ∈ Ω.

(2.46)This condition is also called the Legendre-Hadamard condition.

Sometimes we require a stronger condition, namely

∃ν > 0 Aαβij (x)F iαFjβ ≥ ν|F |

2 ∀F ∈ RN×n and a.e. x ∈ Ω. (2.47)

This is condition is sometimes called super-strong ellipticity.To outline the difference between these two conditions assume for sim-

plicity that the coefficients are independent of x consider the bilinear formB : RN×n × RN×n → R given by

B(F,G) = Aαβij FiαG

jβ. (2.48)

Super-strong ellipticity is equivalent to the condition that the quadratic formB is positive definite. Strong-ellipticity is equivalent to the weaker conditionthat B is positive on matrices of the form F = ζ⊗ξ, i.e., on rank-1 matrices.

Theorem 2.7. Assume that f iα ∈ L2(Ω), gi ∈ L2(Ω) and that either

the coefficients Aαβ are strongly elliptic and constant (2.49)

orthe coefficients Aαβij satisfy (2.47) and (2.40) (2.50)

Then there exists a unique weak solution u ∈ W 1,20 (Ω;RN ) of the elliptic

system−divA∇u = g − div f (2.51)

Moreover the solution u satisfies∫Ω|∇u|2 dx ≤ 1

ν2

(‖f‖L2(Ω;Rn) + C(Ω)‖g‖L2(Ω)

)2. (2.52)

29 [January 28, 2019]

Proof. Consider the quadratic form

a(ϕ, u) =

∫ΩAαβij ∂βu

j∂αϕi dx. (2.53)

We only need to show that

a(u, u) ≥ ν∫

Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.54)

Then the rest of the proof proceeds as in the scalar case. Under assump-tion (2.50) this estimate follows from the pointwise estimate (2.47). Underassumption we state this estimate is contained in the following lemma.

Lemma 2.8. (i) Assume (2.49). Then∫ΩAαβij ∂βu

j∂αui dx ≥ ν

∫Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.55)

(ii) Assume the estimate in (2.46) holds for some x0 ∈ Ω, i.e.,

∃ν > 0 Aαβij (x0)ξαξβζiζj ≥ ν|ξ|2 |ζ|2 ∀ξ ∈ Rn, ζ ∈ RN (2.56)

and assume thatsupx∈Ω‖A(x)−A(x0)‖ ≤ ν

2(2.57)

Then∫ΩAαβij ∂βu

j∂αui dx ≥ ν

2

∫Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.58)

Proof. To prove the first assertion extend u by zero to Rn and let Fu denotethe Fourier transform of u defined by

Fu(ξ) :=

∫Rnu(x)e−ix·ξ dx. (2.59)

Then

(F∂αuj)(ξ) = i ξα(Fuj)(ξ) and thus (F∇u)(ξ) = i (Fu)(ξ)⊗ ξ. (2.60)

Thus using Plancherel’s theorem, strong ellipticity, the identity |a ⊗ b|2 =|a|2|b|2 and again Plancherel’s theorem we get

a(u, u) =1

(2π)n

∫RnAαβij ξβ(Fuj)(ξ) ξα(Fui)(ξ) dξ (2.61)

≥ 1

(2π)n

∫Rnν|ξ|2|Fu(ξ)|2 dξ (2.62)

=1

(2π)n

∫Rnν|F∇u|2 dξ (2.63)

=ν

∫Rn|∇u|2 dx. (2.64)

30 [January 28, 2019]

The second assertion follows from the first and the pointwise estimate

Aαβij (x)∂βuj∂αu

i ≥ Aαβij (x0)∂βuj∂αu

i − ν

2|∇u|2. (2.65)

For general systems with non-constant strongly elliptic coefficients theconclusion does not hold11. If the coefficients are continuous one can obtain,however, a slightly weaker estimate which is almost as useful.

Theorem 2.9 (Garding inequality). Suppose that

the coefficients Aαβij are continuous in Ω and strongly elliptic. (2.66)

Then there exists a constant K > 0 such that∫ΩAαβij ∂βu

j∂αuj dx ≥ ν

2

∫Ω|∇u|2 dx−K

∫Ω|u|2 dx ∀u ∈W 1,2

0 (Ω;RN ).

(2.67)

Proof. Homework sheet 2. Hint: use Lemma 2.8 on balls with sufficientlysmall radius r0, a partition of unity

∑k ϕ

2k = 1 in Ω, where

ϕk ∈ C∞c (B(xk, r0)), and estimate the difference

(ϕk∇u,Aϕk∇u)− (∇(ϕku), A∇(ϕku)). (2.68)

The proof shows that the constant K can be estimated as

K ≤ C

r20

‖A‖2

ν, (2.69)

where r0 is such that

|x− y| ≤ r0 =⇒ ‖A(x)−A(y)‖ ≤ ν

2, (2.70)

and where C only depends on the dimension n

It follows from the Garding inequality and the Lax-Milgram theoremthat the problem Lu + Ku = −div f + g has a unique weak solution inW 1,2

0 (Ω). Moreover the Fredholm alternative and its proof (see the courseFA and PDG or [GT], Thm. 8.6. or [Ev], Section 6.2.3, Thm. 4) imply thatthere exists an at most countable set Σ ⊂ (−K,∞) such that the problem

Lu− σu = −div f + g has a unique weak soln. in W 1,20 (Ω;Rn) if σ /∈ Σ.

(2.71)

11see, H. Le Dret, An example of H1-unboundedness of solutions to strongly elliptic sys-tems of partial differential equations in a laminated geometry, Proc. Roy. Soc. EdinburghSect. A, 105 (1987), 77–82.

31 [January 28, 2019]

2.1.3 Reverse Poincare inequality and pointwise estimates

Theorem 2.10. Suppose that the coefficients Aαβij are constant and strongly

elliptic, that fαi ∈ L2(Ω) and that u ∈W 1,2(Ω;RN ) is a weak solution of

−divA∇u = −div f in Ω. (2.72)

Assume that Br = B(x0, r) ⊂ Ω. Then for 0 < s < r∫Bs

|∇u|2 dx ≤ C

(r − s)2

∫Br\Bs

|u− a|2 dx+C

ν2

∫Br

|f |2 dx ∀a ∈ R.

(2.73)If moreover

f = 0 (2.74)

then

supBr/2

|∇u|2 ≤ C

rn+2

∫Br

|u− ur|2 dx ≤C

rn

∫Br

|∇u|2 dx, (2.75)

supBr/2

|∇2u|2 ≤ C

rn+2

∫Br

|∇u− (∇u)r|2 dx. (2.76)

Moreover the constants C depend only on the dimension n and the ratio‖A‖/ν.

Proof. We may assume thatν = 1. (2.77)

Indeed if the result has been proved for ν = 1 the general case follows byreplacing A by A/ν and f by f/ν. We first assume that f = 0.

Let η ∈ C∞c (Br) with 0 ≤ η ≤ 1, η = 1 in Bs and |∇η| ≤ 2/(r − s).Then

∇(ηu) = η∇u+ u⊗∇η. (2.78)

Using Lemma 2.8 and applying the definition of weak solution with ϕ = η2uwe thus get

∫Br

|∇(ηu)|2 dx ≤∫Br

(∇(ηu), A∇(ηu)) dx

=

∫Br

(∇(ηu), Aη∇u) + (∇(ηu), A(u⊗∇η)) dx

=

∫Br

(∇(η2u), A∇u)) dx︸ ︷︷ ︸=0

−∫Br

(ηu⊗∇η,A∇u) dx+

∫Br

(∇(ηu), A(u⊗∇η)) dx

≤2‖A‖ ‖∇(ηu)‖L2 ‖u⊗∇η‖L2 + ‖A‖‖u⊗∇η‖2L2

≤1

2‖∇(ηu)‖2L2 + 2‖A‖2 ‖u⊗∇η‖2L2 + ‖A‖ ‖u⊗∇η‖2L2 , (2.79)

32 [January 28, 2019]

where for the last but one inequality we used the estimate |η∇u| ≤ |∇(ηu)|+|u⊗∇η| and for the last inequality we used Young’s inequality ab ≤ 1

2a2+ 1

2b2.

Now the first term of the right hand side can be absorbed into the left handside and we get∫

Br

|∇(ηu)|2 dx ≤ C(‖A‖) 1

(r − s)2‖u‖2L2(Br\Bs) (2.80)

Since η = 1 in Bs the assertion (2.73) follows for a = 0. Now if u is aweak solution so is u− a and hence the general form of (2.73) follows.

[22.10. 2018, Lecture 5][26.10. 2018, Lecture 6]

Now consider the case f 6= 0. The only difference is that the term withvanished for f = 0 is now estimated as∫

Br

(∇(η2u), A∇u)) dx =

∫Br

(∇(ηηu), f) dx

≤‖∇(ηu)‖L2‖f‖L2 + ‖u⊗∇η‖L2‖f‖L2

≤1

4‖∇(ηu)‖2L2 + 2‖f‖2L2 +

1

2‖u⊗∇η‖2L2 +

1

2‖f‖2L2 , (2.81)

where we used that 2ab ≤ 12a

2 + 2b2. Now the additional term 14‖∇(ηu)‖2L2

can still be absorbed into the left hand side of (2.79) and the argument isfinished as for f = 0.

To prove the other two assertions we use the fact that the coefficientsare constant and hence the derivatives ∂γu solve the same elliptic equation.

The rigorous justification was discussed in the course FA and PDE: one

first considers the difference quotients ∂hγu(x) :=u(x+heγ)−u(x)

h . It is easy tosee that these satisfy the weak form of the equation (in the set Ωh := x ∈Ω,dist (x, ∂Ω) > h. Then (2.73) provides estimates for ∇∂hγu which areuniformly in h. Now by the relation between difference quotients and weakderivatives (see, e.g., [GT], Lemma 7.24) we see that for each i the functions∂hi ∂ju are bounded in L2(Bs) uniformly for h < h0 and thus u ∈W 2,2(Bs).Moreover passage to the limit h→ 0 yields∫

Bs

|∇∂γu|2 dx ≤C

(r − s)2

∫Br\Bs

|∂γu|2 dx ∀s ∈ (0, r). (2.82)

Iterative application of this argument with radii ri = 1− i2k shows that∫

B1/2

|∇ku|2 dx ≤ Ck∫B1

|u− a|2 dx ∀a ∈ R. (2.83)

The Sobolev embedding theorem implies that W k,2(B1/2) → C1(B1/2) fork > n/2+1. Thus we get the first inequality in (2.75). The second inequality

33 [January 28, 2019]

in (2.75) follows by the Poincare inequality. The statement for general rfollows by considering the rescaled function ur(z) = u(x0 + rz). Finally toget (2.76) we use that the derivatives of u are also weak solutions and hence(2.75) can be applied with u replaced by ∇u.

2.1.4 Morrey and Campanato spaces

The main idea is that scaled L2 estimates can be used to derive Cα andeven Lp estimates.

Definition 2.11. Let λ > 0. The Morrey space L2,λ(Ω) consists of allfunctions f ∈ L2(Ω) for which the seminorm [f ]L2,λ defined by

[f ]2L2,λ = supx∈Ω

supr<diamΩ

1

rλ

∫B(x,r)∩Ω

|f |2 dy (2.84)

is finite.The Campanato space L2,λ(Ω) consists of all functions f ∈ L2(Ω) for

which the seminorm [f ]L2,λ defined by

[f ]2L2,λ = supx∈Ω

supr<diamΩ

1

rλ

∫B(x,r)∩Ω

|f − fx,r|2 dy (2.85)

is finite. Here we set

fx,r :=1

Ln(B(x, r) ∩ Ω)

∫B(x,r)∩Ω

f dy (2.86)

One can easily check these spaces are Banach spaces with the norms

‖f‖2L2,λ = ‖f‖2L2 + [f ]2L2,λ , ‖f‖2L2,λ = ‖f‖2L2 + [f ]2L2,λ , (2.87)

respectively.

Examples.

f ∈ L∞ =⇒ f ∈ L2,n,

f ∈W 1,∞ =⇒ f ∈ L2,n+2

f ∈ C0,α =⇒ f ∈ L2,n+2α

If Ω has Lipschitz boundary the Poincare inequality implies that

∇f ∈ L2,λ =⇒ f ∈ L2,λ+2.

Note also that∫B(x,r)∩Ω

|f − fx,r|2 dx ≤∫B(x,r)∩Ω

|f − a|2 dx ∀a ∈ R. (2.88)

34 [January 28, 2019]

Taking a = 0 we see in particular that L2,λ ⊂ L2,λ. Another consequence of(2.88) is the following monotonicity property

Ω ⊂ Ω′ =⇒∫B(x,r)∩Ω

|f − fΩ∩B(x,r)|2 dx ≤∫B(x,r)∩Ω

|f − fΩ′∩B(x,r)|2 dx

≤∫B(x,r)∩Ω′

|f − fΩ′∩B(x,r)|2 dx (2.89)

Here

fΩ∩B(x,r) :=1

Ln(B(x, r) ∩ Ω)

∫B(x,r)∩Ω

f dy,

fΩ′∩B(x,r) :=1

Ln(B(x, r) ∩ Ω′)

∫B(x,r)∩Ω′

f dy.

An immediate consequence of (2.89) is

Ω ⊂ Ω′ =⇒ [f ]L2,λ(Ω) ≤ [f ]L2,λ(Ω′). (2.90)

Theorem 2.12 (Characterization of Campanato spaces). Suppose that thereexists an A > 0 such that

Ln(B(x, r) ∩ Ω) ≥ Arn ∀x ∈ Ω, ∀r < diamΩ. (2.91)

Then

(i) L2,λ(Ω) = L2,λ(Ω) if 0 < λ < n,

(ii) L2,n+2α(Ω) = C0,α(Ω)

with equivalence of the norms involved.

As usual the second assertion has to be understood in the sense thatf has an Holder continuous representative. We will always identify f withthat representative.

More precisely, one has the trivial estimates

[f ]L2,λ ≤ [f ]L2,λ for all λ > 0, (2.92)

[f ]L2,n+2α ≤ 2(Ln(B1))1/2[f ]C0,α for all α ∈ (0, 1) (2.93)

and the proof below shows that one also has the non-trivial estimates

[f ]L2,λ ≤ C(A)[f ]L2,n+2α + C(A)(diamΩ)−n/2‖f‖L2(Ω) (2.94)

(with the convention (diamΩ)−n/2 = 0 if Ω is unbounded),

[f ]C0,α ≤ C(A)[f ]L2,n+2α for α ∈ (0, 1). (2.95)

Moreover writing f(x) = f(x)− fx,r + fx,r one easily gets the estimate

supΩ|f | ≤ inf

0<r≤diamΩ[f ]C0,αrα +A−1/2r−n/2‖f‖L2 . (2.96)

35 [January 28, 2019]

Proof. The main idea is to compare the averages of f ∈ L2,λ over balls ofradius 2−kr. By scaling Ω we may assume without loss of generality thatdiamΩ = 1. We have for x ∈ Ω and for s < r∫

B(x,s)∩Ω|fx,r − fx,s|2 dx

≤2

∫B(x,s)∩Ω

|f − fx,r|2 dx+

∫B(x,s)∩Ω

|f − fx,s|2 dx

≤2(rλ + sλ) [f ]2L2,λ . (2.97)

If we take s = r/2 and use (2.91) we get

|fx,r − fx,r/2|2 ≤ C(A)rλ−n[f ]2L2,λ (2.98)

If λ < n this implies that

|fx,r − fx,r0 | ≤ C(A)rλ−n

2 [f ]L2,λ . (2.99)

for all 0 < r < r0 (indeed let k be such that 12r0 < 2kr ≤ r0 and apply (2.98)

with 2l−1r and 2lr for l ≤ k, finally apply (2.97) with s = 2kr and r0). Thuswe get∫

B(x,r)∩Ω|f |2 dy =

∫B(x,r)∩Ω

|f − fx,r|2 dy + Ln(B(x, r) ∩ Ω)|fx,r|2

≤C(A)rλ[f ]2L2,λ + C(A)rn|fx,r0 |2 (2.100)

and since by (2.91) we have |fx,1| ≤ C‖f‖L2 . Now by (2.91) we have |fx,r0 | ≤C(A)r

−n/20 ‖f‖L2 . If diamΩ = ∞ we can let r0 → ∞ and get [f ]L2,λ ≤

C[f ]L2,λ . If diamΩ <∞ we conclude by taking r0 = diamΩ.Now assume that α ∈ (0, 1) and λ = n + 2α. By the Lebesgue point

theorem we havelimr→0

fx,r = f(x) for a.e. x ∈ Ω. (2.101)

On the other hand it follows from (2.98) that the continuous functions x 7→fx,2−k converge uniformly to some limit g. Hence g is continuous and g = fa.e. In the following we do not distinguish between f and g. By (2.98) wehave

|fx,r − f(x)| ≤ C(A)rα[f ]L2,n+2α . (2.102)

Now consider two points x, y ∈ Ω and take r = |x− y|. To prove Holdercontinuity we write

|f(x)− f(y)| ≤ |f(x)− fx,2r|+ |fx,2r − fy,2r|+ |fy,2r − f(y)| (2.103)

36 [January 28, 2019]

The first and the last term on the right hand side are controlled by (2.102).To control the middle term note that

Ln(B(x, 2r) ∩B(y, 2r) ∩ Ω)|fx,2r − fy,2r|2

=

∫B(x,2r)∩B(y,2r)∩Ω

|fx,2r − fy,2r|2 dz

≤2

∫B(x,2r)∩B(y,2r)∩Ω

|f(z)− fx,2r|2 dz + 2

∫B(x,2r)∩B(y,2r)∩Ω

|f(z)− fy,2r|2 dz

≤C(A)rn+2α[f ]L2,n+2α (2.104)

Now by the choice of r we have B(x, 2r) ∩ B(y, 2r) ⊃ B(x, r) and by as-sumption Ln(B(x, r) ∩ Ω) ≥ Arn. Thus

|fx,2r − fy,2r|2 ≤ C(A)r2α[f ]L2,n+2α (2.105)

and the proof is finished.

Lemma 2.13. Let 0 < λ < n+ 2. Let Ω ⊂ Rn be open. Then

(i) C∞(Ω) is not dense in L2,λ(Ω).

(ii) Assume that f ∈ L2loc(Rn) and [f ]L2,λ(Rn) < ∞. Then there exist

fk ∈ C∞(Rn) with

fk → f in L2loc(Rn), lim sup

k→∞[fk]L2,λ(Rn) ≤ [f ]L2,λ(Rn). (2.106)

(iii) Assume in addition that Ω is bounded with Lipschitz boundary. Letf ∈ L2,λ(Ω). Then there exist fk ∈ C∞(Ω) with

fk → f in L2(Ω), lim supk→∞

‖fk‖L2,λ(Ω) ≤ C(Ω)‖f‖L2,λ(Ω).

(2.107)

(iv)

fk f in L2loc(Ω) =⇒ [f ]L2,λ(Ω) ≤ lim inf

k→∞[fk]L2,λ(Ω), (2.108)

where both sides are allowed to take the value +∞.

Notation. We say that f ∈ L2loc(E) if f ∈ L2(K) for all compact sets

K ⊂ E. We say that fk and converges weakly or strongly to f in L2loc(E) if

the convergence holds in all L2(K), where K ⊂ E is compact.

Proof. (i), (ii), (iv) see Homework. Hints:(i): Denote by Xλ the closure of C∞(Ω) in L2,λ(Ω). Show first that

f ∈ Xλ =⇒ limr→0

1

rλ

∫B(x,r)

|f − fx,r|2 dy = 0 ∀ x ∈ Ω (2.109)

37 [January 28, 2019]

(consider first f ∈ C∞(Ω)). Then assume without loss of generality that0 ∈ Ω and show that g ∈ L2,λ(Ω) \Xλ where

g(x) =

|x|

λ−n2 if λ 6= n,

− ln |x| if λ = n.(2.110)

(ii): Let η ∈ C∞c (B(0, 1)) with η ≥ 0 and∫Rn η dx = 1 and let ηk(x) =

knη(kx). Set fk = ηk ∗ f . Apply Jensens’s inequality (for the probabilitymeasure ηk(x) dx), i.e.,∣∣∣∣∫

Rng(z)ηk(z) dz

∣∣∣∣2 ≤ ∫Rn|g(z)|2ηk(z) dz (2.111)

to

g(z) = f(y − z)− 1

Ln(B(x, r))

∫B(x,r)

f(y − z) dy (2.112)

and integrate the left hand side of (2.111) to deduce that [fk]2L2,λ ≤ [f ]2L2,λ .

(iv): Argue by contradiction. First, passing to a subsequence if necessarywe may assume that L := limk→∞[fk]L2,λ(Ω) exists and L < ∞ (since forL = ∞ the assertion is trivially true). If the assertion does not hold thenthere exist x ∈ Ω, r > 0 and δ > 0 such that

1

rλ

∫B(x,r)∩Ω

|f(y)− fx,r|2 dy > L2 + δ. (2.113)

Now derive a contradiction to the assumption fk f in L2loc(Ω) (for sim-

plicity assume first that the convergence holds in L2(Ω)).Assertion (iii) can be deduced from (ii) and the following extension result.

There exists a constant C(Ω) such that for each f ∈ L2,λ(Ω) there exists anF ∈ L2,λ(Rn) with F|Ω = f and ‖F‖L2,λ(Rn) ≤ C(Ω)|f‖L2,λ(Ω). In additionone uses the following trivial estimates for x ∈ Ω and fk = Fk |Ω∫

B(x,r)∩Ω|fk − (fk)x,r|2 dy ≤

∫B(x,r)∩Ω

|fk − (Fk)B(x,r)|2 dy

≤∫B(x,r)

|Fk − (Fk)B(x,r)|2 dy.

[26.10. 2018, Lecture 6][29.10. 2018, Lecture 7]

38 [January 28, 2019]

2.1.5 Regularity theory in Campanato spaces, interior estimates

Theorem 2.14 (Interior estimates in L2,λ for constant coefficients). Let

B = B(x0, R), B′ = B(x0, θR), where 0 < θ < 1. (2.114)

Let 0 < λ < n + 2 and suppose that fαi ∈ L2,λ(B). Assume that the coeffi-

cients Aαβij are constant and strongly elliptic and let u ∈ W 1,2(B;RN ) be aweak solution of the system

−divA∇u = −div f in B. (2.115)

Then∇u ∈ L2,λ(B′;RN×n). (2.116)

and

[∇u]2L2,λ(B′) ≤C

ν2[f ]2L2,λ(B) + C

1

Rλ

∫B|∇u− (∇u)B|2, (2.117)

where C depends on θ, λ, ‖A‖/ν and the dimension n.

Remark. Taking λ = n+ 2µ we see in particular that f ∈ C0,µ(B(x0, R))implies that ∇u ∈ C0,µ(B(x0, θR) with a quantitative estimate of the Holderseminorms.

Proof. Note first it suffices to prove the result in the case

ν = 1, B(x0, R) = B(0, 1), (∇u)B = 0. (2.118)

Indeed to recover the result for general ν > 0 (it suffices replace A byA/ν and f by f/ν). To recover the result for general R > 0 and x0 defineuR(z) = R−1u(x0 + Rz) and fR(z) = f(x0 + Rz), observe that uR is aweak solution of −divA∇uR = fR in B(0, 1) and that [∇u]2

L2,λ(B(x0,R))=

Rn−λ[∇uR]2L2,λ(B(0,1))

and right hand side of (2.117) scales in the same way.

To see that we can assume that (∇u)B = 0 note if u is a weak solutionof (2.116) then so is the function v defined by v(x) = u(x)− (∇u)Bx (sinceA is constant). Now ∇vB = 0 and once we have the estimate for v weimmediately deduce the estimate for u since [∇u]L2,λ = [∇v]L2,λ and ∇u−(∇u)B = ∇v.

Let x1 ∈ B(0, θ). Set Br := B(x1, r), gr = gx1,r. We establish a decayestimate for

φ(r) :=

∫Br

|∇u− (∇u)r|2 dx (2.119)

39 [January 28, 2019]

for r ≤ 1− θ. This is sufficient because for r ≥ 1− θ we can use the trivialestimates

r−λ∫Br∩B(0,1)

|∇u− (∇u)r|2 dx ≤ r−λ∫Br∩B(0,1)

|∇u|2 dx

≤(1− θ)−λ∫B(0,1)

|∇u|2 dx. (2.120)

We first derive a functional inequality for φ. Together with a simpleiteration lemma this will give the decay of ϕ. In order to derive the functionalinequality we write u = v + w, where v solves an elliptic system with zeroright hand side (and hence has good decay) and where w has zero boundaryconditions. More precisely we define w as the unique weak solution of thesystem

−divA∇w = −div (f − fr) in Br, w ∈W 1,20 (Br;RN ) (2.121)

and we set v = u − w. Note that fr is a constant and hence div fr = 0. Itfollows that v is a weak solution of

−divA∇v = 0 in Br. (2.122)

By Theorem 2.10 we have

supBr/2

|∇2v|2 ≤ C

rn+2

∫Br

|∇v − (∇v)r|2 dx (2.123)

This implies the following excess decay estimate∫Bρ

|∇v − (∇v)ρ|2 dx ≤ Cρn+2

rn+2

∫Br

|∇v − (∇v)r|2 dx (2.124)

for all 0 < ρ ≤ r/2, where C depends on ‖A‖/ν and the dimension n. Ifwe take C ≥ 2n+2 the estimate holds trivially also for ρ ∈ (r/2, r]. Here weused ρ 7→

∫Bρ|∇v − (∇v)ρ|2 dx is monotone. Indeed by (2.88) we have for

ρ < r∫Bρ

|∇v−(∇v)ρ|2 dx ≤∫Bρ

|∇v−(∇v)r|2 dx ≤∫Br

|∇v−(∇v)r|2 dx (2.125)

Regarding w, the energy estimate in Theorem 2.7 yields∫Br

|∇w|2 dx ≤∫Br

|f − fr|2 ≤ rλ[f ]2L2,λ . (2.126)

40 [January 28, 2019]

Since v = u−w this provides an estimate for the right hand side of (2.124)∫Br

|∇v − (∇v)r|2 dx ≤ 2

∫Br

|∇u− (∇u)r|2 dx+ 2

∫Br

|∇w − (∇w)r|2 dx

≤2

∫Br

|∇u− (∇u)r|2 dx+ 2rλ[f ]2L2,λ (2.127)

The energy estimate (2.126) implies trivially that for all 0 < ρ < r∫Bρ

|∇w − (∇w)ρ|2 dx ≤∫Bρ

|∇w|2 dx ≤∫Br

|∇w|2 dx ≤ rλ[f ]2L2,λ . (2.128)

Combining this estimate with (2.124) and (2.127) we get

φ(ρ) ≤ 2

∫Bρ

|∇v − (∇v)ρ|2 dx+ 2

∫Bρ

|∇w − (∇w)ρ|2 dx

≤2Cρn+2

rn+2

(2φ(r) + 2rλ[f ]2L2,λ

)+ 2rλ[f ]2L2,λ). (2.129)

This gives the desired functional inequality for φ

φ(ρ) ≤ C1ρn+2

rn+2φ(r) + C2r

λ[f ]2L2,λ ∀ 0 < ρ ≤ r ≤ 1− θ. (2.130)

where C1 and C2 depend only on ‖A‖ and n. It now follows from Lemma2.15 below that

φ(ρ) ≤ C(φ(1− θ)(1− θ)λ

+ [f ]2L2,λ

)ρλ (2.131)

where C depends on λ, ‖A‖ and n. Since (1− θ)−λφ(1− θ) is controlled by(2.120) with r = 1− θ this finishes the proof (note that argued at the verybeginning of the proof that it suffices to consider the case (∇u)B = 0).

Remark. (Not discussed in class) Theorem 2.14 can be extended toequations of the form

−divA∇u = −div f + g (2.132)

withg ∈ L2,λ−2(B;RN ) (2.133)

provided that we add a term C[g]2L2,λ−2(B)

to the right hand side of (2.117)

Here we use the following conventions for λ ≤ 2

L2,λ−2(Ω) = L2(Ω), [g]2L2,λ(Ω) = (diamΩ)2−λ∫

Ω|g|2 dx if 0 < λ ≤ 2.

(2.134)

41 [January 28, 2019]

The only difference in the proof is that w is now defined by solving−divA∇u = −div f + g and the energy estimate becomes∫Br

|∇w|2 ≤ (‖f −fr‖L2 +C(Br)‖g‖L2)2 ≤ 2rλ[f ]2L2,λ +2C(Br)2rλ−2‖g‖2L2,λ

(2.135)where C(Br) = rC(B1) is the constant in the Poincare inequality with zeroboundary values.

Lemma 2.15 (Iteration lemma). Let 0 < β < α. Let φ : (0, r0) → [0,∞)be an increasing function and suppose that

φ(ρ) ≤ A[ρα

rα+ ε

]φ(r) +Brβ, whenever 0 < ρ ≤ r ≤ r0 (2.136)

Then there exists a constant ε0 > 0 (which depends on A, α and β) suchthat if ε < ε0 then

φ(ρ) ≤ C(ρβ

rβφ(r) +Bρβ

)whenever 0 < ρ ≤ r ≤ r0, (2.137)

where C depends on A, α and β.

Proof. For 0 < τ < 1 and r ≤ r0 we have

φ(τr) ≤ Aτα[1 + ετ−α]φ(r) +Brβ. (2.138)

Now let γ ∈ (α, β) and choose τ and ε0 such that

Aτα ≤ 1

2τγ , ε0 = τα. (2.139)

Thenφ(τr) ≤ τγφ(r) +Brβ. (2.140)

By induction we get for all integers k ≥ 0:

φ(τk+1r) ≤ φ(τ τkr)

≤ τγφ(τkr) +Bτkβrβ

≤ τγ(τγφ(τ (k−1)r) +Bτ (k−1)βrβ

)+Bτkβrβ

≤ τ (k+1)γφ(R) +Bτkβrβk∑j=0

τ j(γ−β)

≤ C ′τ (k+1)β[φ(r) +Brβ].

This gives the assertion for the discrete values ρ = τkr. Now the followingstandard argument shows that we get the estimate for all ρ ≤ r, at theexpense of a slightly larger constant.

42 [January 28, 2019]

Given ρ ≤ τr choose k such that τk+2r < ρ ≤ τk+1r. Using thatτk+1 ≤ ρ/(τr) and the fact φ is non-decreasing

φ(ρ) ≤ φ(τk+1r) ≤ C ′( ρτr

)β[φ(r) +Brβ] =

C ′

τβ

(ρβ

rβφ(r) + ρβ

). (2.141)

For τr < ρ ≤ r one uses the trivial estimate

φ(ρ) ≤ φ(r) ≤ τ−β ρβ

rβφ(r). (2.142)

Corollary 2.16 (Global estimates in Rn). Assume that the coefficients Aαβijare constant and strongly elliptic.

(i) Let f ∈ L2(Rn;Rn×N ). Then there exists a weak solutionu ∈W 1,2

loc (Rn;RN ) of

−divA∇u = −div f, with ∇u ∈ L2(Rn;RN×n). (2.143)

Moreover u is unique up to the addition of constants and∫Rn|∇u|2 dx ≤ 1

ν2

∫Rn|f |2 dx. (2.144)

(ii) Let 0 < λ < n+ 2. Assume that f ∈ L2loc(Rn;Rn×N ) with [f ]L2,λ(Rn) <

∞. Then there exists a weak solution u ∈W 1,2loc (Rn;RN ) of

−divA∇u = −div f, with [∇u]L2,λ(Rn) <∞. (2.145)

Moreover u is unique up to the addition of affine functions and

[∇u]L2,λ(Rn) ≤ C[f ]L2,λ(Rn), (2.146)

where C only depends on n and ‖A‖/ν.

Proof. Homework. Hint: As before we may assume ν = 1. To show unique-ness up to constant and affine maps use (2.75) and (2.76), respectively, andtake the limit r →∞.(i): Apply the Lax-Milgram theorem to show that the problem

−divA∇uk +1

ku = −div f (2.147)

has a unique solution in W 1,2(Rn;RN ) and ‖∇uk‖L2 ≤ ‖f‖L2 . Deducethat (a subsequence of) ∇uk converges weakly in L2 and hence uk − (uk)0,1

converges weakly in W 1,2loc to the desired solution u.

43 [January 28, 2019]

Alternatively observe that ‖∇u‖ is a norm on C∞c (Rn;RN ) (and not justa seminorm) and define

D1,20 := closure of C∞c (Rn;RN ) in ‖∇u‖. (2.148)

Then apply the Lax-Milgram theorem in D1,20 .

(ii): Step 1. Assume in addition that f = 0 in Rn \B(0, R0).Then f ∈ L2. Let u be the solution in (i). Deduce the estimate for[∇u]L2,λ(Rn) from the interior estimate and (2.144) by exploiting that theinterior estimate holds for every R0 > 0.

Step 2. General f .Let η ∈ C∞c (B(0, 1) with η = 1 on B(0, 1/2) and let ηk(x) = η(2−kx). Letfk = ηk(f − f0,2k) and let uk be the solution of −divA∇u = fk which existsby Step 1. Show that

[fk]L2,λ(Rn) ≤ C[f ]L2,λ(Rn). (2.149)

Then Step 1 yields[∇uk]L2,λ(Rn) ≤ C[f ]L2,λ(Rn). (2.150)

Deduce that (for ρ ≥ 1)∫B(0,ρ)

|∇uk − (∇uk)0,ρ|2 dx ≤ Cρλ[f ]2L2,λ(Rn),

|(∇uk)0,ρ − (∇uk)0,1| ≤ C(ρ)[f ]L2,λ(Rn). (2.151)

Deduce that there exists a subsequence of ∇uk − (∇uk)0,1 which convergesweakly in L2

loc(Rn) to some v. Then show that uk = uk− (uk)0,1− (∇uk)0,1x

converges weakly in W 1,2loc (Rn) to some u. It is then easy to see that u is

a weak solution of the PDE in Rn and the estimate for [u]L2,λ(Rn) followsfrom (2.108).

Remark: in fact it is easy to see that the whole sequence ∇uk− (∇uk)0,1

converges strongly in L2loc by using that

−divA∇(uk − uk+1) = 0 in B(0, 2k−1)

and the decay estimate for solutions of the homogeneous equation.

[29.10. 2018, Lecture 7][2.11. 2018, Lecture 8]

Theorem 2.17 (Interior estimates with continuous coefficients). Let

B = B(x0, R), B′ = B(x0, θR), where 0 < θ < 1. (2.152)

44 [January 28, 2019]

Let 0 < λ < n and suppose that fαi ∈ L2,λ(B). Assume that the co-

efficients Aαβij are uniformly continuous in B and strongly elliptic. Let

u ∈W 1,2(B;RN ) be a weak solution of the system

−divA∇u = −div f in B. (2.153)

Then∇u ∈ L2,λ(B′;RN×n), (2.154)

and

[∇u]2L2,λ(B′) ≤C

ν2[f ]2L2,λ(B) +

C

Rλ

∫B|∇u|2, (2.155)

where C depends on θ, λ, sup ‖A‖/ν, the dimension n and the modulus ofcontinuity of the rescaled function z 7→ A(x0 +Rz)/ν.

Remark. For 0 < λ < n we have L2,λ = L2,λ with equivalent norms. Wehave written [u]L2,λ and [f ]L2,λ because these are the seminorms we actuallyuse in the proof.

Notation The modulus of continuity of a function g is the function ωggiven by

ωg(r) := sup|g(x)− g(y)| : x, y ∈ Ω, |x− y| ≤ r. (2.156)

A function g is uniformly continuous if and only if limr↓0 ωg(r) = 0.

Proof. As in the proof of Theorem 2.14 we may assume that

ν = 1, B = B(x0, R) = B(0, 1). (2.157)

We set

ω(r) = sup‖A(x)−A(y)‖ : x, y ∈ Ω, |x− y| ≤ r. (2.158)

Let x1 ∈ B′. We set Br = B(x1, r). We have to prove decay of thequantity

φ(r) =

∫Br

|∇u|2 dx. (2.159)

It suffices to control φ for r ≤ r0 ≤ 1− θ since for r > r0 we have the trivialestimate

r−λ0

∫Br∩B

|∇u|2 ≤ r−λ0

∫B|∇u|2 dx. (2.160)

The main idea is to ’freeze the coefficients’, i.e., we observe that u is alsoa weak solution of the constant coefficient PDE

−divA(x1)∇u = −divF, where F = (f − fr) + [A(x1)−A]∇u.

45 [January 28, 2019]

We write again u = v + w where w ∈ W 1,20 (Br) is the weak solution of the

constant coefficient system

−divA(x1)∇w = −divF, (2.161)

and where v solves −divA(x1)∇v = 0. Now using (2.75) we get∫Bρ

|∇v|2 dx ≤ Cρn

rn

∫Br

|∇v|2 dx (2.162)

for ρ ≤ r/2. By taking C ≥ 2n we can assume that this estimate holds forall ρ ≤ r ≤ 1− θ. The energy estimate for w gives∫Br

|∇w|2 dx ≤ 2

∫Br

|f |2 dx+ 2ω2(r)

∫Br

|∇u|2 ≤ 2[f ]2L2,λrλ + 2ω2(r)φ(r).

(2.163)

This yields

φ(ρ) ≤ C(ρn

rn+ ω(r)2

)φ(r) + C[f ]2L2,λr

λ ∀ ρ ≤ r ≤ 1− θ, (2.164)

where C depends on ‖A(x1)‖ and n.Now let ε0 > 0 be as in the iteration lemma. Since A is uniformly

continuous there exists r0 > 0 such that ω2(r) ≤ ε0 for all r ≤ r0. Thusthe iteration lemma implies that there exists a constant (depending on λ,‖A(x1)‖ and n) such that

φ(ρ) ≤ Cρλ

rλφ(r) + C[f ]2L2,λr

λ ∀ ρ ≤ r ≤ r0. (2.165)

Taking r = r0 and using the trivial estimate (2.160) we obtain the as-sertion.

Theorem 2.18 (Interior estimates with Holder continuous coefficients). Let0 < µ < 1, 0 < λ ≤ n+ 2µ and let

B = B(x0, R), B′ = B(x0, θR), 0 < θ < 1. (2.166)

Assume that the coefficients Aαβij are in C0,µ(B) and strongly elliptic. Let

fαi ∈ L2,λ(B) and let u ∈W 1,2(B;RN ) be a weak solution of

−divA∇u = −div f in B. (2.167)

Then∇u ∈ L2,λ(B′,RN×n) (2.168)

and

[∇u]2L2,λ(B′) ≤C

ν2[f ]2L2,λ(B)+

C

Rλ

∫B|∇u−(∇u)B|2 dx+

C[A]2C0,µR2µ

ν2Rλ

∫B|∇u|2 dx,

(2.169)where C depends on supB ‖A‖/ν, λ, n and on Rµ[A]C0,µ/ν.

46 [January 28, 2019]

Remark. Since L2,n+2µ(B) = C0,µ(B) this shows in particular that

A, f ∈ C0,µ(B) =⇒ ∇u ∈ C0,µ(B′). (2.170)

The corresponding estimate is called a Schauder estimate.

Proof. Step 1. Simplifications.We claim that again it suffices to show the result for

ν = 1, B = B(0, 1), (∇u)B = 0, λ ≥ n. (2.171)

The restriction to ν = 1 is clear. As regards the restriction to R = 1 notethat the function uR(z) = R−1u(x0 +Rz) solves

−divAR∇uR = −div fR, (2.172)

whereAR(z) = A(x0 +Rz), fR(z) = f(x0 +Rz). (2.173)

Now [AR]C0,µ(B(0,1)) = Rµ[A]C0,µ(B(x0,R)) and

∇uR(z) = ∇u(x0 +Rz)

so that all the terms in (2.169) scale in the right way.Finally assume that we have shown the estimate for function which

satisfy in addition (∇u)B = 0. Take B = B(0, 1) and note that v(x) =uB(x)− (∇u)Bx satisfies (∇v)B = 0 and solves

−divA∇v = −div f , f := f −A(∇u)B. (2.174)

Since diamB(0, 1) = 2 we have

[A(∇u)B]L2,λ ≤ [A]L2,λ |(∇uB)| ≤ 2n+2µ−λ[A]L2,n+2µ |(∇uB)|

≤C[A]C0,µ

(∫B|∇u|2 dx

)1/2

(2.175)

Thus [f ]2L2,λ is controlled by twice the sum of the first and the last term onthe right hand side of (2.169). This gives the desired bound for [∇v]L2,λ =[∇u]L2,λ .

We finally note that the assertion for 0 < λ < n follows from Theorem2.17 (in this case the last term on the right hand side of (2.169) is notneeded).

To obtain the result for λ ≥ n will use a so called bootstrap argument,i.e., we will successively improve the regularity. More precisely we will provethe estimates first for λ < n, then for n ≤ λ < n + 2µ and finally forλ = n+ 2µ on a sequence of decreasing balls

B(0, 1− τ) ⊃ B(0, 1− 2τ) ⊃ B(0, 1− 3τ) = B(0, θ), (2.176)

where τ = (1− θ)/3.

47 [January 28, 2019]

Step 2. Estimate of [∇u]L2,λ(B(0,1−2τ)) for n ≤ λ < n+ 2µ.By Theorem 2.17 applied with λ− 2µ < n instead of λ we have

[∇u]2L2,λ−2µ(B(0,1−τ)) ≤ C[f ]2L2,λ−2µ(B) + C

∫B|∇u|2 dx. (2.177)

We want to prove decay of

φ(r) =

∫Br

|∇u− (∇u)r|2 dx

where Br = B(x1, r), x1 ∈ B(0, 1− 2τ) and r ≤ τ (for r ≥ τ we can use thetrivial estimate φ(τ) ≤

∫B |∇u|

2).We freeze the coefficients, i.e., we observe that u is also a weak solution

of the constant coefficient PDE

−divA(x1)∇u = −divF, where F = (f − fr) + [A(x1)−A]∇u.

We argue as in the proof of Theorem 2.17 but now use (2.76) for v insteadof (2.75) and get

φ(ρ) ≤ Cρn+2

rn+2φ(r) + C

∫Br

|F |2 dx (2.178)

and thus

φ(ρ) ≤ Cρn+2

rn+2φ(r) + Cr2µ

∫Br

|∇u|2 dx+ C[f ]2L2,λrλ (2.179)

≤Cρn+2

rn+2φ(r) + C[∇u|2L2,λ−2µ(Br)

rλ + C[f ]2L2,λrλ ∀ ρ ≤ r ≤ τ (2.180)

Then the desired estimate follows from the iteration lemma (with ε = 0)and the bound (2.177) on [∇u|2

L2,λ−2µ(B(0,1−τ)).

Step 3. λ = n+ 2µ.From the previous step we know that ∇u ∈ L2,n+2µ−δ(B(0, 1− 2τ)), for allδ > 0 and that the corresponding norm is bounded by the right hand side of(2.169). Let B′′ = B(0, 1− 2τ). Then by definition of the Holder seminorm

|u(x)− u(y)| ≤ |x− y|µ/2[u]C0,µ/2(B′′) ≤ 2µ/2[u]C0,µ/2(B′′).

Integrating over y and using Jensen’s inequality we get

|u(x)− uB′′ | ≤ 2µ/2[u]C0,µ/2(B′′).

48 [January 28, 2019]

Now we take δ = µ and use that L2,n+µ = C0,µ/2 (with equivalentseminorms). This yields

supB′′|∇u| ≤ [∇u]C0,µ/2(B′′) +

1

Ln(B′′)

∫B′′∇u dx

≤ [∇u]C0,µ/2(B′′) + C‖∇u‖L2(B′′), (2.181)

where B′′ = B(0, 1 − 2τ). It follows that supB′′ |∇u|2 is controlled by theright hand side of (2.169).

Returning to (2.180) (now for x1 ∈ B′ = B(0, 1−3τ) and r ≤ τ) we thusget

φ(ρ) ≤ Cρn+2

rn+2φ(r) + C sup

Br

|∇u|2rλ + C[f ]2L2,λrλ (2.182)

and the iteration lemma shows that ∇u ∈ L2,λ and that (2.169) holds.

[2.11. 2018, Lecture 8][5.11. 2018, Lecture 9]

2.1.6 Estimates up to the boundary

We now prove estimates near the boundary. We first consider functionsu which vanish on the boundary of the half-space Rn+ and we show thatthe estimates for balls which were derived in the previous section can beextended to half-balls. Later we will map a neighbourhood of a boundarypoint of a general open set with C1,α boundary to a half ball (’flattening theboundary’). We consider the half space

Rn+ := x ∈ Rn : xn > 0 = Rn−1 × (0,∞) with ∂Rn+ = Rn−1 × 0(2.183)

and the half balls

B+(x0, r) = B(x0, r) ∩ Rn+, where x0 ∈ Rn−1 × 0. (2.184)

We want to define the space of Sobolev functions which vanish on the lowerboundary of B+, i.e., on the set x ∈ B(x0, r) : xn = 0. On possibility isto first set

V+ := u ∈W 1,2(B+(x0, r)) : ∃δ > 0 u = 0 for xn < δ (2.185)

and then to define

W 1,2+ (B+(x0, r)) = closure of V+ in W 1,2(B+(x0, r)). (2.186)

Note in particular that (assuming still x0 ∈ ∂Rn+)

u ∈W 1,2+ (B+(x0, r)), η ∈ C∞c (B(x0, r)) =⇒ ηu ∈W 1,2

0 (B(x0, r))(2.187)

49 [January 28, 2019]

Functions u ∈W 1,2+ (B+(x0, r)) satisfy the Poincare inequality∫B+(x0,r)

|u|2 dx ≤ Cr2

∫B+(x0,r)

|∂nu|2 dx. (2.188)

This can be easily deduced from the one-dimensional Poincare inequality(applied in direction xn) and Fubini’s theorem or by a compactness argumentas in the proof of the Poincare inequality with zero mean, see homework fora general abstract version of the Poincare inequality. In this case on firstconsiders r = 1 and then argues by scaling.

Theorems 2.10, 2.17 and 2.18 can easily be adapted to half-balls.

Theorem 2.19. Let B+r = B+(0, r). Suppose that the coefficients Aαβij are

constant and strongly elliptic and that u ∈W 1,2+ (B+

r ) is a weak solution of

−divA∇u = f in B+r . (2.189)

Then for 0 < s < r∫B+s

|∇u|2 dx ≤ C

(r − s)2

∫B+r \B+

s

|u|2 dx+C

ν2

∫B+r

|f |2 dx (2.190)

If moreoverf = 0 (2.191)

then

supB+r/2

|∇u|2 ≤ C

rn+2

∫B+r

|u|2 dx ≤ C

rn

∫B+r

|∂nu|2 dx, (2.192)

supB+r/2

|∇2u|2 ≤ C

rn+2

∫Br

|∂nu− λ|2 dx ∀ λ ∈ RN , (2.193)

Moreover the constants C depend only on the dimension n and the ratio‖A‖/ν.

Remark. One can slightly improve (2.190) by replacing∫Br\Bs |u|

2 dx with

infa∈RN∫Br\Bs |u−a|

2 dx, but this is less important than in the interior esti-

mates. Here we have a Poincare inequality even for a = 0, see (2.188) sinceu has zero boundary conditions on xn = 0 while for the interior estimatesone only has a Poincare inequality for a suitable choice of a.

Proof. In view of (2.187) the first estimate is proved in the same way as forballs.

For the second an third estimate we need control of higher derivatives.On the half-ball we can still use that all the tangential derivatives ∂iu,i = 1, . . . , n− 1 also solve the PDE. This gives the estimate∫

B+s

|∇∂iu|2 dx ≤C

(r − s)2

∫Br

|∇iu|2 dx (2.194)

50 [January 28, 2019]

The left hand side gives control of all second derivatives ∂j∂iu and ∂n∂iufor i, j = 1 . . . , n − 1. The only missing second derivative is ∂n∂nu. Thisderivative can be controlled by the equation. Consider first scalar equations∑

α,β aαβ∂α∂βu = 0. With the choice ξ = en the ellipticity conditions gives

ann ≥ ν. Thus we get from the equation

∂n∂nu = − 1

ann

∑(α,β)6=(n,n)

aαβ∂α∂βu. (2.195)

For systems define the matrices Bαβ = Aαβij . Then

Bnn∂n∂nu = −∑

(α,β) 6=(n,n)

Bαβ∂α∂βu (2.196)

The choice ξ = en in the strong ellipticity condition implies that

Bnnij ζ

iζj ≥ ν|ζ|2 ∀ ζ ∈ RN . (2.197)

Hence the matrix Bnn is invertible (this is true, even if Bnn is not symmetric;use, e.g., the Lax-Milgram theorem on RN ). Thus we can obtain the estimatefor ∂n∂nu from (2.196)

For the higher derivative one argues by induction. First one gets∫B+

1/2

|∇∂i1 . . . ∂ik−1u|2 dx ≤ Ck

∫B+

1

|u|2 dx (2.198)

for ij = 1, . . . , n − 1. Thus gives control of all k-th order derivatives whichcontain at most one derivative in direction xn. By taking k − 2 tangentialderivatives of (2.196) we get those partial derivatives which contain at most2 derivatives in direction xn. Now we inductively take l derivatives of or(2.196) in direction xn and k − 2− l derivatives in tangential directions forl = 1, . . . , k−2 to obtain and L2 estimates for all k-th order derivatives. Nowthe Poincare inequality for functions in W 1,2

+ and the Sobolev embeddingtheorem (2.192) and (2.193) for r = 1 and λ = 0. As usual the assertion forgeneral r follows by scaling.

Assume finally that λ 6= 0 and let v(x) = u(x)−λxn. Then v ∈W 1,2+ (B+

r )and v solves the same PDE as u. Thus we can apply (2.193) to v. Since∇2v = ∇2u and ∂nv = ∂nu− λ the estimate (2.193) holds for all λ ∈ R.

The regularity results in L2,λ(B+) follow from the estimates in Theorem2.19 and two simply ingredients:

• a decay estimate for half-balls and

• a simple geometric observation which allows one to combine the esti-mates and half-balls and the interior estimates.

51 [January 28, 2019]

We directly state the results for variable coefficients.

Lemma 2.20 (Decay on half balls with continuous coefficients). Let B+R =

B+(0, R), let 0 < λ < n, f ∈ L2,λ(B+R), Aαβij ∈ C0(B+

R) and let u ∈W 1,2

+ (B+R) be a weak solution of

−divA∇u = f in B+R . (2.199)

Then for 0 < r ≤ R∫B+r

|∇u|2 dx ≤ C

ν2[f ]L2,λ(B+

R)rλ + C

rλ

Rλ

∫B+R

|∇u|2 dx (2.200)

Proof. This is very similar to the proof of Theorem 2.17. As before we mayassume that ν = 1 and R = 1. We have to prove decay of

ϕ(r) :=

∫B+r

|∇u|2 dx. (2.201)

It suffices to control ϕ(r) for r ≤ r0. As before set

ω(r) := sup‖A(x)−A(y)‖ : |x− y| ≤ r (2.202)

We again freeze the coefficients, i.e., we note that u is a weak solution of

−divA(0)∇ = −divF, where F = (f − fr) + [A(0)−A]∇u. (2.203)

As before write u = v + w, where w ∈ W 1,20 (B+

r ) is the weak solution of−divA(0)∇w = −divF and where −divA(0)∇v = 0. Then∫

B+r

|∇w|2 dx ≤∫B+r

|F |2 dx ≤ 2

∫B+r

|f |2 dx+ 2ω(r)2ϕ(r). (2.204)

and by (2.192) ∫B+ρ

|∇v|2 dx ≤ Cρn

rn

∫B+r

|∇v|2 dx (2.205)

for ρ ≤ r/2. As before, taking C ≥ 2n we may assume that this estimateholds for all ρ ≤ r. As before this yields

ϕ(ρ) ≤ C[ρn

rn+ ω2(r)

]ϕ(r) + C[f ]L2,λrλ. (2.206)

Let ε0 be the constant in the iteration lemma and choose r0 such thatω2(r0) ≤ ε0. Then the iteration lemma yields

ϕ(ρ) ≤ Cρλ

rn0ϕ(r0) + C[f ]L2,λρλ ∀ ρ ≤ r0. (2.207)

This implies the assertion since for r0 ≤ ρ ≤ 1 we can use the trivial estimateρ−λφ(ρ) ≤ r−λ0 φ(1).

52 [January 28, 2019]

[5.11. 2018, Lecture 9][9.11. 2018, Lecture 10]

Theorem 2.21 (Boundary estimates with continuous coefficents). Let

B+ = B+(0, R), B′,+ = B+(0, θR), where 0 < θ < 1. (2.208)

Let 0 < λ < n and suppose that fαi ∈ L2,λ(B+). Assume that the co-

efficients Aαβij are uniformly continuous in B+ and strongly elliptic. Let

u ∈W 1,2+ (B+;RN ) be a weak solution of the system

−divA∇u = −div f in B+. (2.209)

Then∇u ∈ L2,λ(B′,+;RN×n). (2.210)

and

[∇u]2L2,λ(B′,+) ≤C

ν2[f ]2L2,λ(B+) +

C

Rλ

∫B+

|∇u|2, (2.211)

where C depends on θ, λ, sup ‖A‖/ν,the dimension n and the modulus ofcontinuity of z 7→ A(Rz)/ν.

Proof. As before we may assume ν = 1, R = 1. Let 2δ = 1− θ,

a ∈ B+(0, θ) = B+(0, 1− 2δ). (2.212)

If suffices to estimate1

ρλ

∫B(a,ρ)

|∇u|2 (2.213)

forρ ≤ δ/2.

We distinguish three geometric situations

Case 1: an ≥ δ (i.e., a is far away from ∂Rn+)Then B(a, δ) ⊂ B+ and we can use the interior regularity result (Theorem2.17) with x0 = a, R = δ and θ = 1

2 and get

1

ρλ

∫B(a,ρ)

|∇u|2 ≤ [u]L2,λ(B(a,δ/2))

≤C[f ]L2,λ(B(a,δ)) + C1

δλ

∫B(a,δ)

|∇u|2 dx. (2.214)

Case 2: an ≤ ρ (i.e., a is very close to ∂Rn+)Then B(a, ρ) ⊂ B+(a′, 2ρ), where a′ = (a1, . . . , an−1, 0) and we can use the

53 [January 28, 2019]

Table 1: Case 1

decay estimate on half balls. Since |a′| ≤ |a| < 1− 2δ we have B+(a′, 2δ) ⊂B+ and thus by Lemma 2.20∫

B(a,ρ)|∇u|2 ≤

∫B+(a′,2ρ)

|∇u|2 dx

≤C[f ]2L2,λ(B+)(2ρ)λ + C(2ρ)λ

(2δ)λ

∫B+(a′,2δ)

|∇u|2 dx. (2.215)

Case 3: ρ < an < δ (i.e., a has an intermediate distance from theboundary).Set s = an. In this case we use first the interior estimate for the ballsB(a, s/2) ⊂ B(a, s) ⊂ B+. and the fact that the Campanato norm andthe Morrey norm are equivalent on the unit ball for λ < n. This gives forρ ≤ s/2

1

ρλ

∫B(a,ρ)

|∇u|2 dx ≤ C[f ]2L2,λ(B+) +C

sλ

∫B(a,s)

|∇u|2 dx. (2.216)

For s/2 ≤ ρ ≤ s the estimate holds trivially as long as C ≥ 2λ. Then wenote that B(a, s) ⊂ B+(a′, 2s) ⊂ B+(a′, 2δ) ⊂ B+ and we use the decayestimate on half balls∫

B(a,s)|∇u|2 ≤

∫B+(a′,2s)

|∇u|2 dx

≤C[f ]L2,λ(B+)(2s)λ +

(2s)λ

(2δ)λ

∫B+(a′,2δ)

|∇u|2 dx. (2.217)

Together with (2.216) this concludes the proof.

54 [January 28, 2019]

Table 2: Case 2

Theorem 2.22 (Boundary estimates with Holder continuous coefficients).Let 0 < µ < 1, 0 < λ ≤ n+ 2µ and let

B+ = B+(x0, R), B′,+ = B+(x0, θR), 0 < θ < 1. (2.218)

Assume that the coefficients Aαβij are in C0,µ(B+) and strongly elliptic. Let

fαi ∈ L2,λ(B+) and let u ∈W 1,2+ (B+;RN ) be a weak solution of

−divA∇u = −div f in B+. (2.219)

Then∇u ∈ L2,λ(B′,+,RN×n) (2.220)

and

[u]2L2,λ(B′,+) ≤C

ν2[f ]2L2,λ(B+) +

C

Rλ

∫B+

|∇u|2 dx (2.221)

where C depends on supB ‖A‖/ν, λ, n and on Rµ[A]C0,µ/ν.

Remark. From this local estimate on can again derive a global estimatefor systems in Rn+ with constant coefficients analogous to the estimate inCorollary 2.16 for system in Rn. More precisely f ∈ L2(Rn+). Then one canagain pass to the limit in the approximations −divA∇uk + 1

kuk = −div fthat there exists a unique distributional solution of

−divA∇u = −div f in Rn+ with u = 0 on ∂Rn+ (2.222)

and ∇u = 0. Moreover

‖∇u‖L2(Rn+) ≤ C‖f‖L2(Rn+) (2.223)

55 [January 28, 2019]

Table 3: Case 3

If in addition f ∈ L2,λ then we can apply (2.221) with x0 = 0, and letR→∞. This yields the global estimate

[∇u]L2,λ(Rn+) ≤ C[f ]L2,λ(Rn+) (2.224)

Proof. This proof was not discussed in detail in classAgain we may assume that ν = 1, R = 1. If λ < n the assertion follows

from Theorem 2.21. Thus assume that λ ≥ n.As in the proof of the corresponding interior estimate we set 3τ = 1− θ

and we prove increasingly better estimates on the decreasing half balls

B+(0, 1− τ) ⊃ B+(0, 1− 2τ) ⊃ B+(0, 1− 3τ) = B(0, θ). (2.225)

Step 1. n ≤ λ < n+ 2µ, estimate in L2,λ(B+(0, 1− 2δ).By Theorem 2.21 applied with λ− 2µ < n instead of λ we have

[∇u]2L2,λ−2µ(B+(0,1−δ)) ≤ C[f ]2L2,λ−2µ(B+) +C

Rλ−2µ

∫B+

|∇u|2 dx, (2.226)

The main point is to establish an improved version of the decay estimatefor half-balls, namely∫

B+(a′,s)|∇u− (∇u)a′,s|2 dx

≤ C[f ]2L2,λ(B+(0,1))sλ + C[A]2C0,µ [u]2L2,λ−2µ(B+(0,1−δ))s

λ

∀ a′ ∈ ∂Rn+, |a′| ≤ 1− 2τ, 0 < s < τ. (2.227)

To prove this let

φ(s) =

∫B+(a′,s)

|∇u− (∇u)a′,s|2 dx (2.228)

56 [January 28, 2019]

We have

−divA(a′)∇u = −divF, F = (f − fa′,r) + (A(a′)−A)∇u. (2.229)

Using Theorem 2.19 we can write u = v + w where w ∈ W 1,20 (B+(a′, r)),

v ∈W 1,2+ (B+(a′, r)), −divA(a′)∇v = 0 and∫

B+(a′,r)|∇w|2 dx ≤

∫B+(a′,r)

|F |2 dx

≤2

∫B+(a′,r)

|f − fa′,r|2 dx+ 2[A]2C0,µr2µ

∫B+(a′,r)

|∇u|2 dx

≤2[f ]L2,λrλ + 2[A]2C0,µrλ[u]2L2,λ−2µ(B+(0,1−δ)) (2.230)

as well as∫B+(a′,ρ)

|∇v− (∇v)a′,ρ|2 dx ≤ Cρn+2

rn+2

∫B+(a′,r)

|∇v− (∇v)a′,r|2 dx. (2.231)

This yields in the usual way

φ(ρ) ≤ Cρn+2

rn+2φ(r) + C[f ]L2,λrλ + C[A]2C0,µ [u]2L2,n−δ(B+(0,1−δ))r

λ (2.232)

and (2.227) follows from the iteration lemma.Now the estimate for [∇u]2L2,λ(B(0,1−2δ)

follows by combining the half-ball

decay estimate (2.227) with the interior estimate in the same way as in theproof for Theorem 2.21.

Step 2. λ = n+ 2.Let 0 < β < µ. From the estimates in Step 1 we get in particular the bound

[∇u]2C0,β(B+(0,1−2τ)) ≤ C[f ]L2,λ(B+) + C

∫B+(0,1)

|∇u|2 dx (2.233)

where the constant depends also on [A]C0,µ . By definition of the Holderseminorm there exists a matrix M such that

supB+(0,1−2τ)

|∇u−M | ≤ [∇u]C0,β(B+(0,1−2τ)).

Since u = 0 on ∂Rn+∩B(0, 1) and thus ∂iu = 0 on ∂Rn+∩B(0, 1) for i ≤ n−1we may choose M of the form M = b⊗ en with b ∈ RN . Thus

supB(0,1−2τ)

|∇u−b⊗en|2 ≤ C[∇u]2C0,β(B+(0,1−2τ)) ≤ C[f ]2L2,λ(B+)+C

∫B+(0,1)

|∇u|2 dx

(2.234)Combining this with the estimate (2.226) (with λ = n+ µ) we deduce this

|b|2 ≤ C[f ]2L2,λ(B+) + C

∫B+(0,1)

|∇u|2 dx.

57 [January 28, 2019]

Thus

supB(0,1−2τ)

|∇u|2 ≤ C[∇u]2C0,β(B+(0,1−2τ)) ≤ C[f ]L2,λ(B+) + C

∫B+(0,1)

|∇u|2 dx

(2.235)Now revisiting the derivation of (2.227) we easily obtain the half-ball

decay estimate∫B+(a′,s)

|∇u− (∇u)a′,s|2 dx ≤ C[f ]2L2,λ(B+(0,1))sλ + C[A]2C0,µ sup

B(0,1−2τ)|u|2 sλ

∀ a′ ∈ ∂Rn+, |a′| ≤ 1− 3τ, 0 < s < τ (2.236)

for λ = n+ 2µ.Combining this estimate with the interior estimate as before we finally

get (2.221)

[9.11. 2018, Lecture 10][12.11. 2018, Lecture 11]

2.1.7 Flattening the boundary

Definition 2.23. Let Ω ⊂ Rn be open and bounded. We say that Ω hasa Ck,µ boundary if for each point x0 ∈ ∂Ω there exists a r > 0 and aCk,µ function γ : Rn−1 → R such that - upon relabeling and reorienting thecoordinates if necessary - we have

Ω ∩B(x0, r) = x ∈ B(x0, r) : xn > γ(x1, . . . , xn−1). (2.237)

Likewise we define open bounded sets with Lipschitz, Ck, C∞ or analyticboundary.

Remark. For x ∈ Rn let x′ = (x1, . . . , xn − 1). It suffices to assume thatγ is defined on the n − 1 dimensional ball x′ : |x′ − x0| < r. Then γ canbe extended to Rn−1 as a C1,α function.

To ’flatten out’ the boundary we define a map Φ : Rn → Rn by

Φi(x) = xi if i = 1, . . . , n− 1, (2.238)

Φn(x) = xn − γ(x1, . . . , xn−1). (2.239)

and writey = Φ(x). (2.240)

Note that

Φ(B(x0, r) ∩ Ω) ⊂ Rn+, Φ(B(x0, r) ∩ ∂Ω) ⊂ ∂Rn+ (2.241)

58 [January 28, 2019]

Similarly we define a map Ψ : Rn → Rn by

Ψi(x) = xi if i = 1, . . . , n− 1, (2.242)

Ψn(x) = xn + γ(x1, . . . , xn−1). (2.243)

and writex = Ψ(y). (2.244)

Note that Ψ and Φ are Lipschitz and invertible with Lipschitz inverse, indeedΨ Φ = Φ Ψ = id Rn . In particular

Ψ(Φ(B(x0, r) ∩ Ω)) = B(x0, r) ∩ Ω, Ψ(Φ(B(x0, r) ∩ ∂Ω) = B(x0, r) ∩ ∂Ω.(2.245)

Moreover we havedetDΦ = detDΨ = 1. (2.246)

The Campanato spaces are invariant under a composition with Bilips-chitz maps.

Lemma 2.24. Let U,Ω ⊂ Rn be open.

(i) Assume that Φ : U → Rn is a C1 diffeomorphism from U to Φ(U) and

1

L|x− y| ≤ |Φ(x)− Φ(y)| ≤ L|x− y| ∀x, y ∈ U. (2.247)

Let 0 < λ < n+2. Then f ∈ L2,λ(Φ(U)) if and only if f Φ ∈ L2,λ(U)and

[f Φ]L2,λ(U) ≤ L(n+λ)/2[f ]L2,λ(Φ(U)), ‖f Φ‖L2(U) ≤ Ln/2‖f‖L2(Φ(U)).(2.248)

If detDΦ = 1 then the constants in these estimates can be improvedto Lλ/2 and 1 respectively.

(ii) Assume that k ∈ N, 0 < β < 1 and f, g ∈ Ck,β(Ω). Then fg ∈ Ck,β(Ω)and

‖fg‖Ck,β(Ω) ≤ C‖f‖Ck,β(Ω) ‖g‖Ck,β(Ω), (2.249)

where ‖f‖Ck,β(Ω) := [∇kf ]C0,β(Ω) +∑k

l=0 supΩ |∇lg|.

(iii) Assume in addition that Ω is bounded and that

Ln(B(x, r) ∩ Ω) ≥ Arn ∀r < diam(Ω). (2.250)

Let 0 < µ < 1 and 0 < λ ≤ n + 2µ. Suppose that f ∈ L2,λ(Ω) andg ∈ C0,µ(Ω). Then fg ∈ L2,λ(Ω) and

‖fg‖L2,λ(Ω) ≤ C‖f‖L2,λ(Ω) ‖g‖C0,µ(Ω), (2.251)

where ‖g‖C0,µ(Ω) := [g]C0,µ(Ω) + supΩ |g|.

59 [January 28, 2019]

Remark. For part (i) the assumption that Φ ∈ C1 is not needed. Itsuffices that Φ is Bilipschitz, i.e., (2.247). To see this one uses that Lipschitzmaps are differentiable almost everywhere and that for Bilipschitz maps thechange of variables formula still holds, if one uses the a.e. derivative in thisformula.Part (iii) is obvious for λ < n (indeed in this case it suffices to assume thatg ∈ L∞(Ω). For λ = n+ 2β with 0 < β ≤ µ Part (iii) follows from Part (ii)with k = 0 since Ln+2β(Ω) = C0,β(Ω).

Proof. (i): It suffices to show that f ∈ L2,λ(Φ(U)) =⇒ f Φ ∈ L2,λ(U)since Φ−1 has the same properties as Φ. Details: Homework. Hint: use thechange of variables formula.

(ii): This is a short calculation for k = 0 and the general case follows byinduction and the product rule.

(iii): By the remark after the lemma it suffices to consider the caseλ = n. See Campanato’s original paper. The key calculation is the same asin Homework 5, Problem 1

Lemma 2.25. Let 0 < µ < 1 and 0 < λ ≤ n + 2µ and assume that Ω isan open and bounded set with C1,µ boundary. Then for each x0 ∈ ∂Ω thereexists an r1 > 0 with the following property. If the coefficients Aαβij belong

to C0,µ(Ω), if f ∈ C0,µ and if u ∈W 1,20 (Ω) is weak solution of

−divA∇u = −div f in Ω (2.252)

Then ∇u ∈ C0,µ(B(x0, r1)) and

[∇u]C0,µ(B(x0,r1)∩Ω)∩Ω ≤ C(‖f‖C0,µ(Ω) + ‖∇u‖L2(Ω)), (2.253)

where C depends on Ω (and in particular on r1), µ, λ, n, sup ‖A‖/ν and[A]C0,µ(Ω)).

Proof. Let Φ be the flattening map defined above. Then

Ω := Φ(B(x0, r) ∩ Ω) ⊂ Rn+. (2.254)

Defineu : Ω→ RN by u(y) = u Ψ. (2.255)

Step 1. Derivation of the PDE for u. We have∫ΩAαβij (x)∂βu

j(x)∂αϕi(x) dx =

∫Ωfαi (x)∂αϕ

i(x) dx (2.256)

for all ϕ ∈W 1,20 (Ω;RN ). Let ϕ ∈ C∞c (Ω;RN ) and set

ϕ = ϕ Φ. (2.257)

60 [January 28, 2019]

Then ϕ ∈ C1c (Ω;RN ) ⊂W 1,2

0 (Ω;RN ). Since u = uΦ we get from the chainrule and the transformation formula (with x = Ψ(y), y = Φ(x))∫

ΩAαβij (x) (∂βu)(y)∂βΦβ(x) (∂αϕ)(y)∂αΦα(x) dx

=

∫Ωfαi (x) (∂αϕ)(y)∂αΦα(x) dx (2.258)

or ∫Φ(Ω)

Aαβij (y) ∂βu(y) ∂αϕ(y) dy =

∫Φ(Ω)

fαi (y) ∂αϕ(y) dy (2.259)

for all ϕ ∈ C∞c (Ω;RN ), where

Aαβij (y) =[Aαβij ∂β Φβ ∂αΦα

](Ψ(y)) detDΨ(y),

f αi =[fαi ∂α Φα

](Ψ(y)) detDΨ(y). (2.260)

By density we see that (2.259) also holds for all ϕ ∈W 1,20 (Ω;RN ).

Step 2. Regularity of u.Let a0 = Φ(x0) = (x′0, 0). Since Ψ is continuous there exist a radius r0 suchthat Ψ(B(a0, 2r0) ⊂ B(x0, r). Then Ψ(B+(a0, 2r0) ⊂ B(x0, r) ∩ Ω. UsingLemma 2.24 we see that f ∈ L2,λ(Ω;Rn×N ) and A ∈ C0,µ(Ω). Moreover u ∈W 1,2(Ω;RN ) implies that u ∈W 1,2

+ (B+(a0, 2r0);RN ). We finally verify thatthat the transformed coefficients are still elliptic. Indeed, since detDΨ = 1,

Aαβij (y)ξαξβζiζj = [Aαβij ∂αΦα∂βΦβ](Ψ(y)) detDΨ(y)ξαξβζ

iζj

= Aαβij (Ψ(y))(∂αΦα(Ψ(y))ξα

) (∂βΦβ(Ψ(y))ξβ

)ζiζj

≥ ν∣∣((∇Φ)T (Ψ(y)))ξ

∣∣2 |ζ|2≥ ν

∣∣∣((∇Φ)T (Ψ(y)))−1∣∣∣−2|ξ|2 |ζ|2 ,

where in the last step we used that |ξ| ≤∣∣∣((∇Φ)T (Ψ(y))

)−1∣∣∣ |∇Φ(Ψ(y))ξ|.

Thus by Lemma 2.22 we see that ∇u ∈ L2,λ(B+(a0, r0)).

Step 3. Regularity of u.Since Φ is continuous there exists r1 ∈ (0, r) such that Φ(B(x0, r1)) ⊂B(a0, r0). Then Φ(B(x0, r1)∩Ω) ⊂ B+(a0, r0). Using again Lemma 2.24 wesee that ∇u ∈ L2,λ(B+(a0, r0)) implies that ∇u ∈ L2,λ(B(x0, r1) ∩ Ω).

2.1.8 Global estimates

Theorem 2.26. Let 0 < µ < 1 and 0 < λ ≤ n + 2µ. Assume thatΩ is bounded and open with C1,µ boundary. Assume that the coefficients

61 [January 28, 2019]

Aαβij are in C0,µ(Ω;RN ) and strongly elliptic, that f ∈ L2,λ(Ω;Rn×N ), g ∈L2,λ−2(Ω;RN ) and the u ∈W 1,2

0 (Ω) is a weak solution of

−divA∇u = −div f + g in Ω. (2.261)

Then ∇u ∈ L2,λ(Ω;RN ) and

[∇u]L2,λ(Ω) ≤ C[f ]L2,λ(Ω) + C[g]L2,λ−2(Ω) + C‖∇u‖L2(Ω). (2.262)

Remark. If k ≥ 2 and if in addition ∂Ω ∈ Ck,µ, A ∈ Ck−1,µ, ∇lf ∈ L2,λ

for all l ≤ k and ∇lg ∈ L2,λ for l ≤ k− 1 one can similarly obtain estimatesfor [∇ku]L2,λ by locally differentiating the equation (see the proof of Theorem2.30 below).In particular for k = 2, A ∈ C1,µ and f = 0 and g ∈ L2,λ one obtains

[∇2u]L2,λ(Ω) ≤ C[g]L2,λ(Ω) + C‖g‖2L2 + C‖∇u‖L2(Ω). (2.263)

Proof. This follows by combining the interior estimates and the boundaryestimates through a covering argument. By Lemma 2.27 below we mayassume g = 0.

For each x ∈ ∂Ω there exists r1(x) > 0 such that the conclusion ofLemma 2.25 holds. Then the collection of balls B(x, r1(x)) for x ∈ ∂Ω covers∂Ω. Since ∂Ω is compact there exists finitely many sets Vi = B(ai, ri) withri = r1(ai) such that

∂Ω ⊂m⋃i=1

Vi. (2.264)

Set

Ω0 = Ω \m⋃i=1

Vi. (2.265)

Then Ω0 is a compact subset of Ω and thus

dist (Ω0, ∂Ω) = infx∈Ω0

dist (x, ∂Ω) = minx∈Ω0

dist (x, ∂Ω) > 0. (2.266)

Set

r :=1

2min(dist (Ω0, ∂Ω), min

i=1,...Mri). (2.267)

If x ∈ Ω0 we can use the interior estimate with R0 = 2r, θ = 12 to

estimate

φ(x, r) :=

∫B(x,r)∩Ω

|∇u− (∇u)x,r|2 dz (2.268)

for r ≤ r.If x ∈ Vi we use the boundary estimate, Lemma 2.25, to estimate φ(x, r)

for r ≤ r ≤ 12ri.

Finally if r > r we use the trivial estimate φ(x, r) ≤ r−λ‖∇u‖2L2(Ω).

62 [January 28, 2019]

Lemma 2.27. Let 0 < λ < n+2. Let Ω ⊂ Rn be bounded and open Assumethat g ∈ L2,λ−2(Ω) (where we use the convention (2.134) if λ < 2). Thenthere exists f ∈ L2,λ(Ω;Rn) such that

−div f = g (2.269)

[f ]L2,λ(Ω) ≤ C(A)[g]L2,λ−2(Ω). (2.270)

Proof. Homework. Hint: LetR = 2 diam(Ω), x0 ∈ Ω. Then Ω ⊂ B(x0, R/2).Extend g by zero to B(x0, R), let v ∈ W 1,2

0 (B(x0, R)) be the weak solutionof

−∆v = g in B(x0, R) (2.271)

and set f = ∇v.The assertion now follows from the energy estimate for ∇v, the defini-

tion of the Morrey seminorm and the interior estimate for equations withconstant coefficients. see the remark after Theorem 2.14).

[12.11. 2018, Lecture 11][16.11. 2018, Lecture 12]

2.1.9 Ck,α theory, Schauder estimates up to the boundary

Theorem 2.28. Let 0 < µ < 1. Assume that Ω is bounded and open withC1,µ boundary. Let ϕ ∈ C1,µ(Ω). Assume that the coefficients Aαβij are in

C0,µ(Ω;RN ) and strongly elliptic and that u ∈ W 1,2(Ω) is a weak solutionof

−divA∇u− divBu+ C∇u+Du = −div f + g in Ω (2.272)

withu = ϕ on ∂Ω, (2.273)

where

(Bu)αi (x) = Bαij(x)uj(x), (C∇u)i(x) = Cαij(x)∂αu(x)j , (Du)i = Dij(x)uj(x)

(2.274)and

Bαij ∈ C0,µ(Ω), Cαij ∈ L∞(Ω), Dij ∈ L∞(Ω), (2.275)

f ∈ C0,µ(Ω), g ∈ L∞(Ω). (2.276)

Then u ∈ C1,µ(Ω) and

‖u‖C1,µ(Ω) ≤ C(‖f‖C0,µ + ‖g‖L∞ + ‖ϕ‖C1,µ(Ω) + ‖u‖L2). (2.277)

63 [January 28, 2019]

Remark. (i) The boundary condition is understood in the sense that u−ϕ ∈W 1,2

0 (Ω;RN ).(ii) It suffices to assume that C,D, g ∈ L2,n+2µ−2.(iii) If we have an a priori estimate in W 1,2 then the term with u on theright hand side can be dropped. Example: if B = C = D = 0, ϕ = 0and N = 1 the energy estimate and the Poincare inequality imply that‖u‖L2 ≤ C‖∇u‖L2 ≤ C(‖f‖L2 + ‖g‖L2).(iv) In general the term ‖u‖L2(Ω) cannot be dropped because there can benontrivial solutions for f = g = 0 (i.e. eigenfunctions of the operator on theleft with eigenvalue zero) . Consider e.g. the case n = 1 and the equation−u′′ − π2u = 0 on (0, 1) which has the solution u(x) = sinπx.

Proof. We may assume that ϕ = 0. Indeed if ϕ 6= 0 we use that v := u− ϕsatisfies a similar equation (with f replaced by f−A∇ϕ−Bϕ and g replacedby g − C∇ϕ−Dϕ), use the estimate for v and deduce the estimate for u.

We use a bootstrap argument to show that a weak solution u actuallybelongs to C1,µ(Ω;RN ). Let

G = g − C∇u−Du (2.278)

F = f −Bu. (2.279)

Then−divA∇u = −divF +G. (2.280)

Note that for an open bounded set Ω with C1,µ boundary and λ ≤ n + 2µand u ∈ L2(Ω) we have the implication

∇u ∈ L2,λ−2(Ω) =⇒ u ∈ L2,λ.

We can improve the regularity of a a weak solution u ∈W 1,2 by succes-sive application of of Theorem 2.26. Indeed u ∈ W 1,2 implies that G ∈ L2

and F ∈ L2,2. Thus Theorem 2.26 yields ∇u ∈ L2,2. We can repeat thisargument to get ∇u ∈ L2,2k where k is the largest integer ≤ (n+ 2µ)/2. Inparticular 2(k + 1) > n+ 2µ. Thus ∇u ∈ L2,2k implies that F ∈ Ln+2µ andG ∈ Ln+2µ−2. Using once more Theorem 2.26 we get ∇u ∈ Ln+2µ = C0,µ.

We now show the slightly weaker estimate

[∇u]C1,µ(Ω) ≤ C(‖f‖C0,µ + ‖g‖L∞ + C supΩ|u|+ sup

Ω|∇u|). (2.281)

Then the full estimate follows from a simple interpolation estimate, seeLemma 2.29 below. Indeed given (2.281) it suffices to take ε = 1

2C in theLemma 2.29 . Then the assertion follows.

To prove (2.281) we apply Theorem 2.26 with λ = n + 2µ. SinceL2,λ+2µ = C0,µ this yields

[∇u]C0,µ ≤ C([F ]C0,µ + [G]L2,n+2µ−2 + ‖∇u‖L2). (2.282)

64 [January 28, 2019]

Since [G]L2,n+2µ−2 ≤ Cdiam(Ω)2−2µ‖G‖L∞ and [u]C0,µ ≤ C[u]C0,1 ≤≤ C supΩ |∇u| (for the last estimate see Lemma 2.29(i)) this implies (2.281).

To absorb the lower order terms we have used the following result.

Lemma 2.29. Let Ω be a bounded domain with Lipschitz boundary.

(i) [u]C0,1(Ω) ≤ C(Ω) supΩ |∇u|.

(ii) Let 0 < µ < 1, let k ∈ N, k ≥ 1. Then for each ε > 0 there exists aCε > 0 (which also depends on µ and k) such that

k∑l=0

supΩ|∇lu| ≤ ε[∇ku]0,µ + Cε‖u‖L2 . (2.283)

Proof. Homework sheet 6. Hints:(i) If the line segment from [x, y] is contained in Ω this follows by calculatingddtu((1− t)x+ ty). If x and y are close to the boundary one can first map theregion containing x and y to a half-ball and use the same argument there.

(ii) This can be proved in the usual way by contradiction and compact-ness (which in this case comes from the Arzela-Ascoli theorem).

Theorem 2.30. Let k ≥ 2. Let µ, Ω, A, B, C, D, f , g and u be as inTheorem 2.28 and assume in addition that

ϕ ∈ Ck,µ(Ω), A,B, f ∈ Ck−1,µ(Ω), C,D, g ∈ Ck−2,µ(Ω), ∂Ω ∈ Ck,µ.(2.284)

Then ∇u ∈ Ck−1,µ(Ω) and

‖u‖Ck,µ(Ω) ≤ C(‖f‖Ck−1,µ(Ω) + ‖g‖Ck−2,µ(Ω) + ‖ϕ‖Ck,µ(Ω) + ‖u‖L2). (2.285)

Proof. Again it suffices to consider the case ϕ = 0. Let V1, . . . , Vm and Ω0

and r > 0 be as in the proof of Theorem 2.26.Assume first that B = C = D = 0. We will show that estimate

[∇ku]L2,n+2µ(Ω) ≤ C([f ]Ck−1,µ(Ω) + ‖g‖Ck−2,µ(Ω) + ‖ϕ‖Ck,µ(Ω) + ‖u‖Ck(Ω)).(2.286)

If x ∈ Ω0 then B(x0, 2r) ⊂ Ω. Taking difference quotients and passingto the limit we see by induction thatv := ∂γ1 . . . ∂γk−1

u is a weak solution in W 1,2 of

−divA∇v = −div ∂γ1 . . . ∂γk−1f + ∂γ1

(∂γ2 . . . ∂γk−1

g)− div f (2.287)

where f is a sum of products of l derivatives of A and k− l derivatives of uwith 1 ≤ l ≤ k − 1. Thus

[f ]C0,µ ≤ C‖A‖Ck−1,µ‖u‖Ck−1,µ ≤ C‖A‖Ck−1,µ‖u‖Ck . (2.288)

65 [January 28, 2019]

Thus the interior estimate yields the desired estimate for r−n−2µ∫B(x,r) |∇v−

(∇v)x,r|2 dx and hence for ∇ku. More precisely, we argue by induction overk using Theorem 2.28 for the initial estimate for k = 2.

If x ∈ Vi we argue similarly with the following modifications. We firstshow the estimate for the transformed function u = u Ψ. We differen-tiate the equation for u (k − 1) times in tangential direction. This yieldsthe estimates for all k-th partial derivatives of u which contain at most onederivative in normal direction. To recover the higher order normal deriva-tives we use the transformed PDE. This yields (with the same notation asbefore)

Bnn∂n∂nu = ∂n( Bnn ∂nu)− (∂nBnn)∂nu,

∂n(Bnn∂nu) = −∑

(α,β) 6=(n,n)

∂αBαβ∂βu (2.289)

Taking inductively tangential and normal derivatives of these identities weobtain control of all partial derivatives of u of order k in L2,n+2µ. Finallythe chain rule yields control of all partial derivatives of order k of u = u Φ.

In the generally case we define F and G as in the proof Theorem 2.28.It follows from Theorem 2.28 that G ∈ C0,µ and F ∈ C1,µ. The resultfor B = C = D = 0 thus implies that u ∈ C2,µ and hence F ∈ C2,µ andG ∈ C1,µ. Iterating the see that ∇u ∈ Ck,µ. Finally the estimate for [u]k,µfollows from interpolation as in the proof of Theorem 2.28.

2.1.10 An alternative route to Schauder estimates by rescalingand blow-up

There is a different argument, only based on scaling, which can be used toprove an estimate which only depends on the data of the problem (i.e., f ,A and Ω) under the assumption (!) that we already have a C1,µ solution.

By slight additional arguments one can also show the existence of C1,µ

solutions and that every weak solutions is already in C1,µ. This approach isbeautifully developed in the paper

• L. Simon, Schauder estimates by scaling, Calculus of Variations andPartial Diff. Equations 5 (1997), 391–407.

The notation is at first glance a bit overwhelming, because Simon treatssimultaneously elliptic equations like −∆u = −div f or ∆2u = div div f andparabolic equations like ∂t −∆u = −div f . For a first reading it completelysuffices to take κ = (1, . . . , 1) and I = J = 1 so that each term Dγ and Dδ

just contains one partial derivative.

66 [January 28, 2019]

2.1.11 The space BMO

Definition 2.31. Let Ω ⊂ Rn be open. We say that f belongs to the spaceBMO (Ω) (the space of function of bounded mean oscillation) if f is inte-grable over each bounded subset of Ω and

[f ]BMO (Ω) := sup 1

Ln(Q)

∫Q|f − fQ| dx : Q ⊂ Ω axi-paralell cube <∞.

(2.290)Here

fQ :=1

Ln(Q)

∫Qf dx. (2.291)

By writing f − fQ = (f − a)− (f − a)Q we see that∫Q|f − fQ| dx ≤ 2 inf

a

∫Q|f − a| dx.

The space BMO (Ω) ∩ L1(Ω) is a Banach space with the norm

‖f‖BMO (Ω) := ‖f‖L1(Ω) + [f ]BMO (Ω). (2.292)

The expression [f ]BMO (Ω) is only a seminorm since it vanishes on func-tions which are constant on each connected component of Ω. In particularBMO (Rn) strictly speaking consists of equivalence classes of functions whichonly differ by a constant.

The Cauchy-Schwarz inequality yields

1

Ln(Q)

∫Q|f − fQ| dx ≤

(1

Ln(Q)

∫Q|f − fQ| dx

)1/2

(2.293)

and it follows easily that

L2,n(Ω) ⊂ BMO (Ω), [f ]BMO (Ω) ≤ C[f ]L2,n(Ω). (2.294)

We shall see below that in fact L2,n(Ω) = BMO (Ω) with equivalent normsfor nice domains, in particular for cubes, Rn, Rn+. More generally one canshow this identity bounded open sets with Lipschitz boundary.

The following examples show that BMO is closely related to L∞, butmore flexible. In particular, the critical Sobolev embedding holds for BMOand in two dimensions the Greens function for −∆ belongs to BMO .

Examples:

(i) L∞(Ω) ⊂ BMO (Ω) and [f ]BMO ≤ 2‖f‖L∞ ;

(ii) (BMO scales like L∞) if f ∈ BMO (Ω) and fλ(x) = f(x/λ) then[fλ]BMO (λΩ) = [f ]Ω;

67 [January 28, 2019]

(iii) (Critical Sobolev embedding) if Ω ⊂ Rn then W 1,n(Ω) ⊂ BMO (Ω)and [f ]BMO ≤ Cn‖∇f‖Ln ;

(iv) Let f(x) = ln |x|. Then f ∈ BMO (Rn).

Properties (i) and (ii) follow immediately from the definition. For (iii)use the Poincare inequality∫

Q(a,r)|f − fQ| dx ≤ r

∫Q(a,r)

|∇f | dx

and Holder’s inequality. For (iv) write φ(a, r) = r−n∫Q(a,r) |f − fQ(a,r)| and

show first that φ(λa, λr) = φ(a, r) for all λ > 0. Then estimate φ(a, 1) bydistinguishinging the cases Q(a, 1) ∩ Q(0, 1) = ∅ and Q(a, 1) ∩ Q(0, 1) 6= ∅.In the first case use that f is Lipschitz outside Q(0, 1). In the second caseuse that Q(a, 1) ⊂ Q(0, 2).

The following fundamental result shows that the example in (iv) is op-timal.

Lemma 2.32. Let Q0 ⊂ Rn be a cube and assume that u ∈ BMO (Q0). Fort > 0 let

St := x ∈ Q0 : |f(x)− fQ0 | > t. (2.295)

ThenLn(St) ≤ Ae−σt/[f ]BMO Ln(Q0), (2.296)

where A and σ depend only on n. Indeed we may take A = e and σ =1/(2ne).

Lemma 2.33. (Calderon-Zygmund decomposition) Let Q0 ⊂ Rn be a cubeand let h ∈ L1(Q0). Let s > 0 be such that

1

Ln(Q)

∫Q0

|h| dx < s. (2.297)

Then there exist at most countable many disjoint open cubes Ik such that

(i) |h| ≤ s a.e. in Q0 \⋃k Ik,

(ii) s ≤ 1Ln(Ik)

∫Ik|h| dx < 2ns

(iii)∑

k Ln(Ik) ≤ 1s

∫Q0|h| dx.

Proof. This is taken literally from [JN], pp. 418–419 (with a few extra wordsfor the proof of (i)).DivideQ0 (by halving each edge) into 2n cubes of equal size and let I11, I12, . . .be those open cubes over which the average value of of |h| is greater than orequal to s. Then in view of the assumption (2.297) we have on those cubes

sLn(I1k) ≤∫I1k

|h| dx < 2nsLn(I1k).

68 [January 28, 2019]

Next subdivide each remaining cube, over which the average of |h| is strictlyless than s, into 2n cubes of equal size, and denote by I21, I22, . . . those cubesover which the average of |h| is greater than or equal to s. Again subdividethe remaining cubes, etc. In this way we obtain a sequence of cubes Iik,which we rename Ik, such that

sLn(Iik) ≤∫Iik

|h| dx < 2nsLn(Iik)

Clearly property (ii) is satisfied. Furthermore, summing the left inequalityover k we obtain (iii).

Finally let x ∈ Q0 \⋃Ik. Then there exist closed cubes Ji such that

x ∈ Ji, the length li of Ji satisfies li = 2−il0 where l0 is the length of Q0 and

1

Ln(Ji)

∫Ji

|h| dy < s. (2.298)

Now by the Lebesgue point theorem we have, for a.e. x ∈ Q0 \⋃Ik,

limr→0

r−n∫B(x,r)

|h(x)− h| dy = 0. (2.299)

Since Ji ⊂ B(x,√nli) we get

|h(x)| = 1

Ln(Ji)

∫Ji

|h(x)| dy ≤ 1

Ln(Ji)

∫Ji

|h|+ |h(x)− h| dy

≤s+1

Ln(Ji)

∫B(x,

√nli)|h(x)− h| dy

where we used (2.298). Thus taking the limit i → ∞ and using (2.299) weget property (i).

[16.11. 2018, Lecture 12][19.11. 2018, Lecture 13]

Proof of Lemma 2.32. It suffices to prove the result if [f ]BMO = 1. Indeed,for a general function g ∈ BMO we can first consider f = g/[g]BMO . Thenthe assertion for g follows from the assertion for f since |g| > t if and onlyif |f | > t/[g]BMO . Moreover we may assume that fQ0 = 0.

For t > 0 let F (t) be the smallest number such that

Ln(St) = Ln(x ∈ Q0 : |f(x)| > t) ≤ F (t)Ln(Q0) (2.300)

for all cubes Q0 and all f ∈ L1(Q0) with

[f ]BMO (Q0) ≤ 1, fQ0 = 0. (2.301)

69 [January 28, 2019]

Note that trivially F (t) ≤ 1 for all t > 0.The key estimate is: if t ≥ 2n then

F (t) ≤ 1

sF (t− 2ns) ∀s ∈ [1, 2−nt]. (2.302)

To prove this estimate first note that by (2.301)∫Q0

|f | dx =

∫Q0

|f − fQ0 | dx ≤ Ln(Q0). (2.303)

Now apply the Calderon-Zygmund decomposition with h = f . Since s < twe have |f | ≤ t almost everywhere in Q0 \

⋃k Ik. Hence

Ln(St) ≤∑k

Ln(x ∈ Ik : |f(x)| > t). (2.304)

Now by assertion (ii) in Lemma 2.33 we have |fIk | < 2ns. Thus |f(x)| > timplies that |f − fIk | > t− 2ns and therefore

Ln(x ∈ Ik : |f(x)| > t) ≤ Ln(x ∈ Ik : |f − fIk | > t− 2ns)≤F (t− 2ns)Ln(Ik) (2.305)

where we used the definition of F in the second inequality.Using first (2.304), (2.305) and then property (iii) in Lemma 2.33 and

(2.303) we get

Ln(St) ≤ F (t− 2ns)∑k

Ln(Ik) ≤1

sF (t− 2ns)Ln(Q0).

Since Q0 and u were arbitrary (subject to (2.301)) we get (2.302).Now set s = e in (2.302) and let σ = 1/(2ne). Then

F (t) ≤ Ae−σt =⇒ F (t+ 2ne) ≤ 1

eAe−σt = Ae−σ(t+2ne).

It follows that if the inequality F (t) ≤ Ae−σt holds for t in some interval oflength 2ne it holds for all larger t. Using the trivial estimate

F (t) ≤ 1 ≤ e e−σt for 0 ≤ t ≤ 2ne (2.306)

we see that (2.296) holds with A = e and σ = 1/(2ne).

Theorem 2.34 (John-Nirenberg estimate for functions in BMO ). Let Q0 ⊂Rn be a cube.

(i) If f ∈ BMO (Q0) then∫Q0

exp

(σ1|f − fQ0 |

[f ]BMO

)≤ A1Ln(Q0) (2.307)

for some σ1 > 0 and A1 > 0 which depend only on the space dimensionn. We may take σ1 = 1/(2n+1e) and A1 = e+ 1.

70 [January 28, 2019]

(ii) If f ∈ BMO (Q0) then f ∈ Lp(Q0) for all 1 < p <∞ and

1

Ln(Q)

∫Q|f − fQ|p ≤ C(p, n)[f ]pBMO (Q) for all cubes Q ⊂ Q0.

(2.308)Moreover the Lp norms grow at most linearly in p, i.e.,

lim supp→∞

C(p, n)1/p

p≤ C ′(n). (2.309)

(iii) We have L2,n(Q0) = BMO (Q0) with equivalent norms. More preciselythere exists a constant B which only depends only on n such that

B−1[u]L2,n ≤ [u]BMO ≤ B[u]L2,n , (2.310)

Ln(Q0)−1/2‖u‖L2 ≤ B(

[u]BMO + Ln(Q0)−1‖u‖L1

), (2.311)

Ln(Q0)−1‖u‖L1 ≤ Ln(Q0)−1/2‖u‖L2 . (2.312)

Remark. The example x 7→ ln |x| shows that linear growth of the Lp normin p is optimal.

To deduce (2.307) from Lemma 2.32 we use the following general formulawhich was proved in Analysis 3.

Lemma 2.35. Let E ⊂ Rn be measurable and let f : E → [0,∞) be mea-surable. Let g ∈ C0[0,∞) ∩C1((0,∞) and assume that g ≥ 0 and g(0) = 0.Let

µf (t) := Ln(x ∈ E : f(x) > t. (2.313)

Then ∫Eg(f(x)) dx =

∫ ∞0

g′(t)µf (t) dt (2.314)

in the sense that if one of the integrals exists then the other exists andequality holds.

Proof. By the fundamental theorem of calculus

g(f(x)) =

∫ f(x)

0g′(t) dt (2.315)

LetF := (x, t) ∈ Rn+1 : f(x) > t

Then F is a measurable subset of Rn+1 since the function f(x) − t is mea-surable in Rn+1. Thus by Fubini’s theorem∫

g(f(x)) dx =

∫Rn

∫RχF (x, t)g′(t) dt dx =

∫R

∫RnχF (x, t) dx g′(t) dt,

=

∫Rµf (t)g′(t) dt.

71 [January 28, 2019]

Proof of Theorem 2.34. (i): We may assume without loss of generality that[f ]BMO ≤ 1 and fQ0 = 0. Then by Lemma 2.32 we have

µf (t) ≤ ALn(Q0)e−σt.

If we apply Lemma 2.35 with g(t) = eσ1t − 1 and 0 < σ1 < σ we get∫Q0

(eσ1f − 1

)dx =

∫ ∞0

σ1eσ1tALn(Q0)e−σt dt ≤ ALn(Q0)

σ1

σ − σ1.

Hence we may take σ1 = 12σ = 1/(2n+1e) and A′ = A+ 1 = e+ 1.

(ii): It suffices to consider the case Q = Q0 since otherwise we can applythe result to f = f|Q and use that [f ]BMO (Q) ≤ [f ]BMO (Q0). By scaling wecan also assume that Ln(Q0) = 1. Then the assertion follows again fromLemma 2.35. To get the asymptotics for C(p, n) we can use the change ofvariables τ = σt and the identities (for k ∈ N, k ≥ 1)∫ ∞

0ptp−1e−σt dt = σ−p

∫ ∞0

pτp−1e−τ dτ = σ−ppΓ(p) = σ−pΓ(p+ 1)

(2.316)where Γ is the Γ function. For integers k ≥ 1 we have Γ(k + 1) = k! ≤ kk.Since Γ is an increasing function we have

lim sup1

pΓ(p+ 1)1/p ≤ lim sup

k→∞,k∈N

1

k − 1Γ(k + 1)1/k−1 ≤ 1.

(iii): All the estimates are invariant under the scaling u(x) 7→ u(λx). Hencewe may assume that Q0 = (0, 1)n. The inclusion L2,n(Q0) ⊂ BMO (Q0)and the second inequality in (2.310) were already established in (2.294).Estimate (2.312) is just the Cauchy-Schwarz inequality.

For the reverse inclusion L2,n(Q0) ⊃ BMO (Q0) and the remaining es-timates we use assertion (ii) and the following observation. If x ∈ Q0 =(0, 1)n and r ≤ 1 then there exist an open cube Q of size ≤ 2r such thatB(x, r) ∩ Q0 ⊂ Q ⊂ Q0. Indeed by reflection on the symmetry planes ofQ0 we may assume that 0 < xi ≤ 1

2 . If r ≤ 12 take Q = a + (−r, r)n where

ai = max(xi, r). If r > 12 take Q = Q0. Thus

r−n∫B(x,r)∩Q0

|f − fB(x,r)|2 dy ≤ r−n∫B(x,r)∩Q0

|f − fQ|2 dy

≤r−n∫Q|f − fQ|2 dy ≤ 2nC(n, 2)[f ]2BMO

where we used (2.88) in the first inequality and assertion (ii) and the estimateLn(Q) ≤ (2r)n in the last. This proves the first inequality in (2.310).

Finally we have∫Q0

|f |2 dx = |fQ0 |2 +

∫Q0

|f − fQ0 |2 dx ≤ ‖f‖2L1 + C(n, 2)[f ]2BMO

and this implies (2.310). Thus BMO (Q0) ⊂ L2,n(Q0).

72 [January 28, 2019]

2.1.12 Lp theory

The heart of the matter is the following interpolation result.

Theorem 2.36 (Lp estimates from L2 and BMO estimates). (i) Let Ω bean open bounded set in Rn and suppose that Q ⊂ Rn is a cube. Supposethat T : L2(Ω;Rd1)→ L2(Q;Rd2) is a linear operator such that

‖Tf‖L2(Q) ≤ M2‖f‖L2(Ω) ∀f ∈ L2, (2.317)

[Tf ]L2,n(Q) ≤ M∞‖f‖L∞(Ω) ∀f ∈ L∞. (2.318)

Then for all 2 < q <∞

‖Tf‖Lq(Ω) ≤ C(q, n,M2,M∞,Ln(Ω)/Ln(Q)) ‖f‖Lq(Ω) ∀ f ∈ Lq.(2.319)

(ii) Let Ω = U = Rn or Ω = U = Rn+ and suppose that T : L2(Ω;Rd1) →L2(U ;Rd2) is a linear operator such that

‖Tf‖L2 ≤ M2‖f‖L2 ∀f ∈ L2, (2.320)

[Tf ]L2,n ≤ M∞‖f‖L∞ ∀f ∈ (L2 ∩ L∞). (2.321)

Then for all 2 < q <∞

‖Tf‖Lq ≤ C(q, n,M2,M∞) ‖f‖Lq ∀ f ∈ (L2 ∩ Lq) (2.322)

and T has a unique extension to a bounded operator from Lq(Ω;Rd1)to Lq(U ;Rd2).

(iii) The assertion in (i) holds if the cube Q is replaced by a bounded openset with Lipschitz boundary.

Proof. A short proof of (i) (with L2,n replaced by BMO ) can be found in[Ca66]. It uses Lemma 3 of [JN]. In the form stated assertion (i) then followsfrom (2.294).

If Ω = U = Rn then assertion (ii) follows directly from the proof of (i)in [Ca66]. This proof shows that ‖Tf‖Lq(Rn) ≤ C‖u‖Lq(Rn) if u ∈ Lq(Rn) ∩L2(Rn). Since this space is dense in Lq the estimate T has a unique extensionto a bounded operator from Lq(Rn) to Lq(Rn).

To treat the case Ω = U = Rn+ we define an extension operator E :Lp(Rn+)→ Lp(Rn) by Ef(x′, xn) = f(x′, |xn|), i.e., by reflection. Then it iseasy to show that [Ef ]L2,n(Rn) ≤ 2(n+1)/2[f ]L2,n(Rn+) (see the proof of (2.325)

below).Now let R : Lp(Rn) → Lp(Rn+) be the restriction operator, i.e., Rf =

f|Rn+ and let T = ETR. Then T satisfies the assumptions for Ω = U = Rn

and hence is a bounded operator from Lq(Rn) to Lq(Rn). Now T = RTEand hence T is also a bounded operator from Lq(Rn+) to Lq(Rn+)

73 [January 28, 2019]

For part (iii) one uses that there exists a linear extension operator Ewhich maps L2,n(U) to L2,n(Q) where Q is a cube which contains U . In factone can even extend to BMO (Rn), see Lemma 2.37. Then one can apply (i)to the operator T = ET and we conclude that

‖T f‖Lp(Q) ≤ Cp‖f‖Lp(Ω). (2.323)

Now T f = ETf = Tf in U and thus we have the trivial estimate

‖Tf‖Lp(U) ≤ ‖T f‖Lp(Q). (2.324)

Lemma 2.37. Let U ⊂ Rn be a bounded domain with Lipschitz boundary.Then there exists a linear bounded operator E : L2,n(Ω) → L2,n(Rn) with(Ef)|U = f . Moreover E is also bounded as an operator from L2(Ω) toL2(Rn).

Proof. Only the proof of f (2.325).was discussed in detail in class. The restof the argument was only sketched.To construct this operator one first considers the situation in the half spaceand then uses a local flattening of the boundary and a partition of unity.First, if v ∈ L2,n(Rn+) and if one extends v by reflection to Rn, i.e., if onedefines (ERn+v)(x′, xn) = v(x′, |xn|) then

‖ERn+v‖L2,n(Rn) ≤ 2(n+1)/2‖v‖L2,n(Rn+). (2.325)

Indeed we need to estimate

ω(x0, r) := r−n mina∈R

∫B(x0,r)

|Ev − a|2 dx

If B(x0, r) ∩ ∂Rn+ = ∅ and B(x0, r) is contained in the upper and lowerhalf space and the definition of the Campanato seminorm gives ω(x0, r) ≤[v]2L2,n(Rn+). If B(x0, r)∩ ∂Rn+ 6= ∅ define x′0 = ((x0)1, . . . , (x0)n−1, 0) ∈ ∂Rn+.

ThenB(x0, r) ⊂ B(x′0, 2r).

Set B± = B(x′0, 2r) ∩ Rn±. Then the averages vB+ and (Ev)B− agree and∫B−|Ev − (Ev)B− |2 dx =

∫B+

|v − vB+ |2 dx ≤ (2r)n[v]2L2,n(Rn+).

Thus with a = vB+ = vB−

ω(x0, r) ≤ r−n∫B+∪B−

|Ev − a|2 dx ≤ 2n+1[v]2L2,n(Rn+).

74 [January 28, 2019]

This concludes the proof of (2.325).Now assume that x ∈ ∂Ω and ∂Ω∩B(x, r) is a Lipschitz graph and Ω∩

B(x, r) lies above this graph (see Definition 2.23). Assume that u ∈ L2,n(Ω)and suppu ⊂ B(x, r). Then by Lemma 2.24 we have uΨ ∈ L2,n(Rn+) (notethat by the remark after that lemma it suffices that Ψ is Bilipschitz). Hencewe define an extension operator by

EB(x,r)u := [ERn+(u Ψ)] Φ. (2.326)

Then another application of Lemma 2.24 shows that

‖EB(x,r)u‖L2,n(Rn) ≤ C‖u‖L2,n(Ω) if suppu ⊂ B(x, r). (2.327)

Here C may depend on x and r.Now by compactness of ∂Ω there exist finitely many balls B(xi, ri) as

above such that⋃mi=1B(xi, ri/2) ⊃ Ω. Set

Ω0 = Ω \m⋃i=1

B(xi, ri/2). (2.328)

Now we consider a partition of unity subordinate to these sets. Let ηi ∈C∞c (B(xi, ri) with ηi = 1 in B(xi, ri/2) and η0 ∈ C∞c (Ω) with η0 = 1 in Ω0.Then set

ηi =ηi∑mj=0 ηj

. (2.329)

Then∑m

i=0 ηi = 1 in Ω and the ηi are smooth since∑m

j=0 ηj ≥ 1 in Ω. Nowdefine

Eu :=m∑i=0

Ei(ηiu), (2.330)

where Ei = EB(xi,ri) for i ≥ 1 and where E0 denotes the extension by zerooutside Ω, i.e. (E0v)(x) = v(x) for v ∈ Ω and (E0v)(x) = 0 if x ∈ Rn \ Ω.Let

r :=1

2min(dist (Ω0, ∂Ω), min

i=1,...mri) (2.331)

Then one can easily check that ‖ηiu‖L2,n(Ω) ≤ C(r)‖u‖L2,n(Ω) for i ≥ 0and ‖E0(η0u)‖L2,n(Rn) ≤ ‖η0u‖L2,n(Ω). The estimates in L2 are similar buteasier.

[19.11. 2018, Lecture 13][23.11. 2018, Lecture 14]

Theorem 2.38. Suppose that the coefficients Aαβij are constant and stronglyelliptic. Let 1 < p <∞.

75 [January 28, 2019]

(i) Assume that f ∈ Lp(Rn;Rn×N ). Then the problem

−divA∇u = −div f (2.332)

has a distributional solution u ∈ W 1,ploc (Rn) with ∇u ∈ Lp(Rn;RN×n)

and u is unique up to the addition of constants. Moreover

‖∇u‖Lp ≤ C(p, n,A)‖f‖Lp . (2.333)

(ii) Let g ∈ Lp(Rn;RN ). Then the problem

−divA∇u = g (2.334)

has a distributional solution u ∈W 2,ploc (Rn;RN ) with ∇2u ∈ Lp(Rn;RN×n×n)

and u is unique up to the addition of affine functions. Moreover

‖∇2u‖Lp ≤ C(p, n,A)‖g‖Lp . (2.335)

Remark. The constant C(p, n,A) depends on A only through ‖A‖ andthe ellipticity constant ν.We sometimes write T1 for the solution operator f 7→ ∇u and T2 for thesolution operator g 7→ ∇2u.

Proof. Step 1. Proof of (i) for 2 < p <∞.This is a direct consequence of Corollary 2.16 and the interpolation estimateTheorem 2.36 (ii).

In more detail one argues as follows. Let f ∈ L2. Then Corollary 2.16states there exists a unique (up to constants) weak solution u (with∇u ∈ L2)of −divA∇u = −div f . Define

Tf = ∇u. (2.336)

Then by Corollary 2.16 (and the trivial inclusion L2,n ⊂ L2,n = L∞)

‖Tf‖L2 ≤ C‖f‖L2 ∀f ∈ L2, (2.337)

[Tf ]L2,n ≤ C‖f‖L∞ ∀f ∈ L2 ∩ L∞ (2.338)

Hence‖Tf‖Lp ≤ Cp‖f‖Lp ∀f ∈ L2 ∩ Lp. (2.339)

If f ∈ Lp there exist fk ∈ Lp ∩L2 such that fk → f in Lp. Thus ∇uk = Tfkis a Cauchy sequence in Lp. After subtracting a suitable constant from ukwe may assume that uk → u in W 1,p

loc . It follows that u is a distributionalsolution of −divA∇u = −div f and

‖∇u‖Lp ≤ Cp‖f‖Lp (2.340)

76 [January 28, 2019]

It remains to show uniqueness. Assume that u1 and u2 are distributionalsolutions with ∇ui ∈ Lp. Then u = u1 − u2 is a distributional solution of−divA∇u = 0. Since p > 2 we have u ∈ W 1,2(B(0, r)) for every r > 0 andu is a weak solution in B(0, r). The decay estimate (2.75) in combinationwith the Holder inequality implies that

supB(0,r/2)

|∇u|2 ≤ Cr−n∫B(0,r)

|∇u|2 dx ≤ Cr−nrn(1− 2p

)

(∫B(0,r)

|∇u|p dx

) 2p

.

(2.341)Taking the limit r →∞ we see that ∇u ≡ 0.

Step 2 Proof of (i) for 1 < p < 2.This follows by duality. Let f ∈ Lp and assume that u with ∇u ∈ Lp is adistributional solution of −divA∇u = f , i.e.,∫

RnA∇u · ∇ϕdx =

∫Rnf · ∇ϕdx ∀ϕ ∈ C∞c (Rn;RN ). (2.342)

Let (AT )αβij = Aβαji , let g ∈ Lp′. By Step 1 there exists a distributional

solution v of −divAT∇v = −div g with ∇v ∈ Lp′ , i.e.,∫RnAT∇v · ∇ψ dx =

∫Rng · ∇ψ dx ∀ψ ∈ C∞c (Rn;RN ). (2.343)

Now by density (see Lemma 2.40) the identity in (2.342) also holds withϕ = v and in (2.343) we may take ψ = u. Since

AT∇v · ∇u = A∇u · ∇v (2.344)

this yields the duality relation∫Rng · ∇u dx =

∫Rnf · ∇v dx. (2.345)

From the estimate in Step 1 we deduce that∫g · ∇u dx ≤ ‖f‖Lp ‖v‖Lp′ ≤ C(p′, n,A)‖f‖Lp ‖g‖Lp′ . (2.346)

Taking g = ∇u|∇u|p−2 and noting that p′ = pp−1 we deduce that

‖∇u‖Lp ≤ C(p′, n,A)‖f‖Lp . (2.347)

This implies in particular uniqueness (up to constants) since f = 0 impliesthat ∇u = 0.

To show existence we first consider f ∈ Lp ∩ L2. Then by Corollary2.16 there exists a solution u of (2.342) with ∇u ∈ L2 and u is unique up

77 [January 28, 2019]

to the addition of constants. As before we write ∇u = Tf . Similarly forg ∈ Lp′∩L2 by Step 1 there exists a solution v of (2.343) with ∇v ∈ Lp′∩L2.Thus using Lemma 2.40 with q = 2 we get again the duality relation (2.345)and the estimate in Step 1 yields∫

g · Tf dx ≤ C(p′, n,A)‖f‖Lp ‖g‖Lp′ ∀ g ∈ Lp′ ∩ L2. (2.348)

Now Lemma 2.39 implies that

‖Tf‖Lp ≤ C(p′, A)‖f‖Lp ∀ f ∈ Lp ∩ L2. (2.349)

To show that for each f ∈ Lp there exists a solution u of (2.342) with∇u ∈ Lp such that (2.349) holds we use an approximation argument asabove. Let f ∈ Lp. Then there exist fk ∈ Lp ∩ L2 such that fk → f in Lp.For each fk there exists a unique solution uk of (2.342) with ∇uk ∈ Lp ∩L2

and by (2.349) the maps ∇uk are a Cauchy sequence in Lp. Hence (afterpossible subtraction of constants) uk → u in W 1,p

loc , ∇uk → ∇u in Lp, themap u is a solution of (2.342) and estimate in (2.349) holds.

Step 3. Proof of (ii). The estimate follows directly from (i). Indeed, if uis a distributional solution of −divA∇u = g and ∇2u ∈ Lp then ∂γu is adistributional solution of −divA∇∂γu = ∂γg and thus by (i) (applied withf δi = gi and fαi = 0 if α 6= δ)

‖∇∂γu‖Lp ≤ C(p, n,A)‖g‖Lp (2.350)

This yields the estimate for ∇2u. This estimate shows in particular thatdifference of two solutions must satisfy ∇2u = 0 and hence be an affinefunction.

To show existence of u we note that by (i) for each γ = 1, . . . , n theequation

−divA∇vγ = ∂γg (2.351)

has a unique distributional solution with ∇vγ ∈ Lp. Moreover

‖∇vγ‖Lp ≤ C(p,A)‖g‖Lp (2.352)

We want to show that there exists a u such that ∂γu = vγ and u is a solutionof the original problem.

The function w := ∂γvδ − ∂δvγ belongs to Lp(Rn;RN ) and satisfies−divA∇w = 0 (in the sense of distributions). We want to show thatw = 0. Let ρ ∈ C∞c (Rn) with

∫ρ dx = 1 be the usual mollifier. Let

ρk(x) = k−nρ(x/k). Then wk := ρk ∗ w ∈ W 1,p(Rn;RN ). Since the coef-ficients A are constant we easily see that −divA∇wk = 0. The unique-ness result in (i) implies that ∇wk = 0 and thus wk = const. Sincewk ∈ Lp(Rn;RN ) we get wk = 0 and hence w = 0.

78 [January 28, 2019]

Since ∂γvδ − ∂δvγ = 0 for all γ, δ it follows that there exists a u ∈ W 2,ploc

such that vγ = ∂γu. Hence ∇2u ∈ Lp. Moreover

∂γ(−divA∇u− g) = 0 (2.353)

in the sense of distributions and hence

−divA∇u− g = const (2.354)

in the sense of distributions. Since ∇2u ∈ Lp and g ∈ Lp we must haveconst = 0.

Alternative proof of (ii). The estimate for ‖∇u‖Lp is proved as before.The main point is to show that for g ∈ L2 there exists a solution in W 2,2

loc .To show the existence of such a solution consider the problems

−divA∇uk +1

kuk = g. (2.355)

By Fourier transform we have∫Rn

(A∇u,∇u) dx ≥ ν∫Rn|∇u|2 dx, (2.356)

where ν > 0 is the ellipticity constant. Thus by the Lax-Milgram theoremthe equation (2.355) has a unique solution in W 1,2(Rn) and by using 1

kuk asa test function we see that

1

kν‖∇uk‖2L2 +

1

k2‖uk‖2L2 ≤ ‖g‖L2

1

k‖uk‖L2 ≤

1

2‖g‖2L2 +

1

2k2‖uk‖2L2 . (2.357)

By taking difference quotients and passing to the limit we see that ∂γuk isa weak solution of

−divA∇∂γuk +1

kuk = ∂γg. (2.358)

Using ∂γuk as a test function we get the estimate

ν‖∇∂γuk‖2L2 +1

k‖∂γuk‖2L2 ≤ ‖g‖L2‖∂2

γuk‖L2 ≤1

2ν‖g‖2L2 +

ν

2‖∇∂γuk‖2L2 .

(2.359)Hence

‖∇2uk‖2L2 +1

k‖∇uk‖2L2 +

1

k2‖uk‖L2 ≤ C‖g‖2L2 . (2.360)

If we letuk(x) = uk(x)− (uk)0,1 − (∇uk)0,1x (2.361)

Then ∇2uk = ∇2uk is uniformly bounded in L2 and by the Poincare inequal-ity there exists a subsequence such that uk u in W 2,2

loc and ∇2uk ∇2uin L2. Moreover taking a further subsequence we may assume that

1

kuk w in L2, ∇1

kuk → 0 in L2, (2.362)

79 [January 28, 2019]

where we used that k−1/2∇uk is bounded in L2. It follows that ∇w = 0, i.e.,w = const. Passing to the limit in (2.355) we see that u is a distributionalsolution of

−divA∇u+ const = g (2.363)

Since ∇2u ∈ L2 and g ∈ L2 it follows that the constant must be zero, sinceno non-trivial constant function is in L2(Rn).

Now let g ∈ L2 ∩ Lp and let u ∈ W 2,2loc with ∇2u ∈ L2 be the solution

just constructed. Then vγ := ∂γu is the unique distributional solution of

−divA∇vγ = ∂γg (2.364)

with ∇vγ ∈ L2. Thus ∇vγ = Tf (γ) where (f (γ))γi = gi and (f (γ))αi = 0 ifα 6= γ. Thus by (2.339) and (2.349)

‖∇vγ‖Lp ≤ C(p, n,A)‖g‖Lp . (2.365)

And this shows that ‖∇2u‖Lp ≤ C(p, n,A)‖g‖Lp . Finally for g ∈ Lp asolution u is constructed as before by approximating g by gk ∈ L2 ∩ Lp.

Lemma 2.39. Let 1 ≤ p < 2. Let f ∈ Lploc(Rn;Rd) and assume f · g is

integrable for each g ∈ (L2 ∩ Lp′)(Rn;Rd) that there exists a constant Msuch that∫

Rnf · g dx ≤M‖g‖Lp′ ∀ g ∈ (L2 ∩ Lp′)(Rn;Rd). (2.366)

Then f ∈ Lp(Rn;Rd) and‖f‖Lp ≤M. (2.367)

Proof. This proof was only sketched briefly in class.If f ∈ Lp and if (2.366) holds forall g ∈ Lp

′then one can simply take

g = f |f |p−2. To prove the result under the assumptions stated consideran increasing sequence of compact sets Kk ⊂ Rn with

⋃kKk = Rn. Let

Ek := x ∈ Kk : |f(x)| ≤ k. Let fk = χEkf . Then fk ∈ L1 ∩ L∞ and itfollows from (2.366) that ∫

Rnfkg ≤M‖g‖Lp′ (2.368)

for all g ∈ L2 ∩ Lp′ . Taking g = fk|fk|p−2 we see that ‖fk‖pLp ≤ Mp. Nowthe assertion follows by taking k →∞ and using the monotone convergencetheorem.

Lemma 2.40. Let 1 ≤ q < ∞. Assume that ψ ∈ W 1,qloc (Rn) with ∇ψ ∈

Lq(Rn;Rn). Then there exist ψk ∈ C∞c (Rn) such that

∇ψk → ∇ψ in Lq(Rn;Rn). (2.369)

80 [January 28, 2019]

Proof. Homework. Idea: since C∞c (Rn) is dense in W 1,q(Rn) it suffices toshow that there exist ψk ∈W 1,q(Rn) such that (2.369) holds.

Consider annuli Ak := B(0, 2k) \B(0, 2k−1) and scaled cut-off functionsηk(x) = η(x/2k) and define ψk = η(ψ− ak) with a suitable choice of ak anduse the Poincare inequality.

Theorem 2.41. Suppose that the coefficients Aαβij are constant and stronglyelliptic. Let 1 < p <∞.

(i) Assume that f ∈ Lp(Rn+;Rn×N ). Then the problem

−divA∇u = −div f in Rn+ with u = 0 on ∂Rn+ (2.370)

has a unique distributional solution u ∈W 1,ploc (Rn+) with ∇u ∈ Lp(Rn+;RN×n).

Moreover‖∇u‖Lp ≤ C(p, n,A)‖f‖Lp . (2.371)

(ii) Let g ∈ Lp(Rn+;RN ). Then the problem

−divA∇u = g in Rn+ with u = 0 on ∂Rn+ (2.372)

has a distributional solution u ∈W 2,ploc (Rn+;RN ) with ∇2u ∈ Lp(Rn+;RN×n×n)

and u is unique up to the addition of linear functions of the form cxn.Moreover

‖∇2u‖Lp ≤ C(p, n,A)‖g‖Lp . (2.373)

Remark. Note that the uniqueness statement is slightly stronger thanfor solutions in Rn since the boundary condition rules out the addition ofconstants or of linear functions of x1, . . . , xn−1.

Proof. This is proved like Theorem 2.38 using (2.223) and (2.224) instead ofCorollary 2.16. In this case the alternative proof of part (ii) applies directly.The other proof can be modified by first only looking that the equation fortangential derivatives vγ = ∂γu. This defines u uniquely up to the additionof a function h(xn). Then the PDE for u becomes an ODE for h and onecan easily see that this ODE has solution (which is unique up to addingterms of the form cxn).

More precisely one can argue as follows. For γ ≤ n − 1 let vγ be thesolution of

−divA∇vγ = ∂γg

whose existence is guaranteed by (i). Then ‖∇vγ‖Lp ≤ ‖g‖Lp . Using con-volution in Rn−1 one shows as before ∂δvγ − ∂γvδ = 0 in Lp and hence a.e.for all γ, δ ∈ 1, . . . , n − 1. By Fubini’s theorem for a.e. xn ∈ (0,∞) themaps x′ 7→ vγ(x′, xn) are in W 1,p

loc (Rn−1) with ∇′v ∈ Lp(Rn−1). Thus for a.e.

xn ∈ (0,∞) there exists maps x′ 7→ U(x′, xn) in W 2,ploc (Rn−1;RN ) such that

81 [January 28, 2019]

∂γU(·, xn) = vγ(·, xn). For a.e. xn the map U(·, xn) is determined uniquelyup to a constant. We can select a unique U , for example by imposing thecondition ∫

B′(0,1)U(x′, xn) dx′ = 0 where B′(0, 1) ⊂ Rn−1.

.We next use the difference quotient method to show that for each bounded

interval I ⊂ (0,∞) we have U ∈W 1,p(B′(0, 1)× I). It follows from Fubini’stheorem that the weak derivatives in the tangential directions exist and aregiven by the vγ

Regarding the difference quotients in normal direction the Poincare in-equality implies that∥∥∥∥U(·, xn + h)− U(·, xn)

h

∥∥∥∥Lp(B′(0,1))

≤ Cn−1∑γ=1

∥∥∥∥vγ(·, xn + h)− vγ(·, xn)

h

∥∥∥∥Lp(B′(0,1)

.

Hence∥∥∥∥U(·, xn + h)− U(·, xn)

h

∥∥∥∥pLp(B′(0,1)×(0,∞)

≤ Cn−1∑γ=1

∥∥∥∥vγ(·, xn + h)− vγ(·, xn)

h

∥∥∥∥Lp(B′(0,1)×(0,∞)

.

The right hand side is bounded independent of h and I since ∂nvγ ∈ Lp(Rn).Thus U ∈W 1,p(B′(0, 1)× I) and ‖∂nU‖Lp(B′(0,1)×(0,∞)) <∞.

It follows from the definition of vγ that

∂γ(−divA∇U − g) = 0 for all γ ≤ n− 1

in the sense of distributions. Thus

−divA∇U − g = H(xn)

in the sense of distributions. Testing this against functions of the formϕ(x′)ψ(xn) where ϕ ∈ C∞c (B′(0, 1)) with

∫Rn−1 ϕdx

′ = 1 and ψ ∈ C∞c ((0,∞))we see that∫ ∞

0H(xn)ψ(xn) dxn =

∫a(xn)ψ(xn)− b(xn)ψ′(xn) dxn

with

a(xn) = −∫Rn−1

g(x′, xn)φ(x′)−n−1∑γ,δ=1

Aγδ∂γvδ(x′, xn)φ(x′) dx′

− 2

∫Rn−1

n−1∑γ=1

Anγ∂nvγ φ(x′) dx′

82 [January 28, 2019]

and

b(xn) = −∫B′(0,1)

Ann∂nU(x′, xn)φ(x′) dx′.

Thus a, b ∈ Lp((0,∞)). We now want to show that there is a solution h ∈W 1,p((0,∞)∩C([0,∞) with h(0) = 0 of the distributional ODE Annh′′ = H.Indeed it suffices to take

h(xn) =

∫ xn

0

∫ t

0a(s) ds dt+

∫ xn

0b(t) dt.

Finally setu(x′, xn) = U(x′, xn) + h(xn).

Then−divA∇u = g in Rn+

and u = 0 on ∂Rn+. By construction ∂i∂γu = ∂i∂γU = ∂ivγ ∈ Lp(Rn+;RN )for all γ ≤ n − 1 and all i ≤ n. Hence Ann∂2

nu ∈ Lp and thus ∂2nu ∈ Lp.

Hence u is the desired solution.

Corollary 2.42. Let 1 < p < ∞. Then there exists a constants Cp,n andCp,q,n with the following properties.

(i) If g ∈ Lp(B(0, R)) then there exists f ∈W 1,p(B(0, R);Rn) such that

−div f = g in B(0, R) (2.374)

1

R‖f‖Lp + ‖∇f‖Lp ≤ Cp,n‖g‖Lp . (2.375)

(ii) Assume in addition that ∫B(0,R)

g dx = 0. (2.376)

Let G = g in B(0, R) and G = 0 in Rn \B(0, R). Assume that q <∞and 1

q ≥1p −

1n Then there exist f ∈ Lq(Rn) such that

−div f = G in Rn, (2.377)

1

R1−n/p+n/q ‖f‖Lq ≤ Cp,q,n‖g‖Lp . (2.378)

Proof. By scaling we may assume that R = 1. Define G as in (ii).(i): There exist u ∈W 1,p

loc (Rn) such that

−∆u = G, ‖∇2u‖Lp ≤ Cn,p‖G‖Lp . (2.379)

Set f = (∇u−(∇u)B(0,1) then div f = G and the estimate for f inW 1,p(B(0, 1))follows from the Poincare inequality. The Lq estimate for f then follows fromthe Sobolev embedding theorem.

83 [January 28, 2019]

(ii): The main point is to show that the solution u decays sufficientlyrapidly at ∞. One way to show this is to express u as the convolution of Gwith the fundamental solution.

For variety we give an argument that just uses the L2 estimates andinterior decay estimates we have already derived. A duality argument allowsone to convert the interior decay estimates to exterior decay estimates.

Indeed we have under the additional assumption g ∈ L2∫Rn\B(0,s)

|∇u|2 dx ≤ C

sn‖g‖L1 for s ≥ 4 (2.380)

and

|∇u(x)| ≤ C

|x|n‖g‖L1(B(0,1)) ≤

C

|x|n‖g‖Lp(B(0,1)) ∀x ∈ Rn \B(0, 8),

(2.381)see Homework 7. Hint: the estimate (2.381) is an easy consequence of(2.381) and the estimate in Theorem 2.10 for supB(x,r/2) ‖∇u‖ with a suitablechoice of r. To prove (2.381) one can use a duality argument. Set f =∇uχRn\B(0,s) and let v ∈ W 1,2

loc (Rn) be the weak solution of −∆v = −div fwith ∇v ∈ L2(Rn). Argue that we have (f, f) = (f,∇u) = (∇v,∇u) =(v, g) = (v − (v)B(0,1), g) ≤ C supB(0,1) |∇v| ‖g‖L1 and ‖∇v‖L2 = ‖f‖L2 .Then use Theorem 2.10 again.

Since L2(B(0, 1)) is dense in Lp(B(0, 1)) it follows by approximationthat the estimate (2.381) holds also without the additional assumption g ∈L2(B(0, 1)). We now can use a Poincare-Sobolev inequality of the form

∫B(0,9)

|h|q dx ≤ C∫B(0,9)\B(0,8)

|h|q dx+ C

(∫B(0,9)

|∇h|p dx.

)q/p(2.382)

This inequality can be proved in the usual way by contradiction usingthe compact Sobolev embedding W 1,p(B(0, 9)) → Lp(B(0, 9)) as well theSobolev embedding W 1,p(B(0, 9)) → Lq(B(0, 9)). Taking h = ∇u we getfrom (2.379) ∫

B(0,9)|∇h|p dx ≤ C‖G‖pLp . (2.383)

and (2.381) implies that∫Rn\B(0,8)

|h|q dx ≤ C‖G‖qLp . (2.384)

This implies the desired estimate.

[23.11. 2018, Lecture 14][26.11. 2018, Lecture 15]

84 [January 28, 2019]

We will make the following general assumptions for the rest of the sub-subsection

The coefficients Aαβij are strongly elliptic (2.385)

Ω ⊂ Rn is bounded and open. (2.386)

We consider the differential operator

Lu = −divA∇u+ div (Bu) + C∇u+Du (2.387)

Lemma 2.43 (Improvement of regularity). Let 1 < s ≤ q < ∞. Assumethat ∂Ω ∈ C1, A ∈ C0(Ω). Assume that u ∈W 1,s

0 (Ω;RN ) is a distributionalsolution of

Lu = −div f + g (2.388)

Iff ∈ Lq(Ω), g ∈ Lt (2.389)

with t > 1 and 1t ≤

1q + 1

n then u ∈W 1,q0 (Ω) and

‖u‖W 1,q ≤ C(‖f‖Lq + ‖g‖Lt + ‖u‖Lq). (2.390)

Proof. We first note that it suffices to consider the case g = 0. Indeedlet B(0, R) be a ball which contains Ω and extend g by zero to that ball.Then by Corollary 2.42 and the Sobolev embedding theorem there exists anf ∈ Lq(B(0, R) such that g = div f and ‖f‖Lq ≤ C‖g‖Lt .

Step 1. A priori estimate for q = s and B = C = D = 0.We first derive an interior estimate for functions u ∈W 1,s

0 (Ω) which satisfy

−divA∇u = −div f. (2.391)

Let B(x0, R) ⊂ Ω. Then we claim that

‖∇u‖Ls(B(x0,R/2)) ≤C‖f‖Ls(B(x0,R)) +C

R‖u‖Ls(B(x0,R)

+ C supB(x0,R)

|A−A0| ‖∇u‖Ls(B(x0,R)) (2.392)

where A0 = A(x0). To see this first note that

−divA0∇u = −div f , where f = f + (A0 −A)∇u. (2.393)

Now let 12 < θ < 1 and let η ∈ C∞c (B(x0, R) with η = 1 on B(x0, θR) and

set v = ηu. Then

−divA0(ηu) = −∂α[Aαβ0 (η∂βu+ u∂βη)]

= −η∂α[Aαβ0 ∂βu]−Aαβ0 ∂αη∂βu− ∂α[Aαβ0 u∂βη]

= −ηdiv f − ∂β[Aαβ0 u∂αη] +Aαβ0 u∂α∂βη − ∂α[Aαβ0 u∂βη].

85 [January 28, 2019]

Since −ηdiv f = −div (ηf) + fα∂αη we get (after exchanging α and β in thesecond term on the right hand side)

−divA0(ηu) = −div f + g (2.394)

where

fα = ηfα +Aβα0 u∂βη +Aαβ0 u∂βη,

g = fα∂αη +Aαβ0 u∂α∂βη. (2.395)

Note that f and v have compact support in B(x0, R). Hence if we extendv, f and g by zero we see that v is a distributional solution of

−divA0∇v = −div f + g in Rn. (2.396)

Note that∫Rn g = 0. Indeed for every ϕ ∈ C∞c (Rn) with ϕ = 1 on supp η∫

Rng dx =

∫Rnϕg dx =

∫Rn∇ϕ · (A0(∇v)− f) dx = 0. (2.397)

By Corollary 2.42(ii) there exist f such that g = −div f and

‖f‖Ls ≤ CR‖g‖Ls . (2.398)

By Theorem 2.38 we get

‖∇(ηu)‖Ls = ‖∇v‖Ls ≤ ‖f + f‖Ls (2.399)

and together with (2.395) this implies the estimate (2.392).Now for x0 ∈ ∂Ω one can obtain a similar estimate in B(x0, R/2)∩Ω (in

this case one get supB(x0,R)∩Ω |A− A0| instead of supB(x0,R) |A−A0| where

A are the coefficients in the flattened domain).Finally since A is uniformly continuous and ∂Ω is C1 for each ε > 0

there exists an R0 such that for all R ≤ R0 we have supB(x0,R) |A−A0| < ε

if B(x0, R) ⊂ Ω and supB(x0,R)∩Ω |A− A0| < ε if x0 ∈ ∂Ω and R ≤ R0.We first cover ∂Ω by finitely many balls B(yi, R0/2) with yi ∈ ∂Ω. Let

Ω0 = Ω \⋃iB(yi, R0/2) and R1 := min(R0, dist (Ω0, ∂Ω)) and cover Ω0 by

finitely many balls B(yj , R1/2) (then B(yj , R1) ⊂ Ω). This can be done insuch a way that each point is only contained in a finite number M of theballs. Then taking the local estimates to the power s and summing over allballs we get∫

Ω|∇u|s dx ≤ 3sCM

(∫Ω|f |s dx+R−s

∫Ω|u|s dx+ εs

∫Ω|∇u|s dx

).

(2.400)By choosing ε small enough we get the estimate for q = s and B = C =D = 0, i.e., we get the implication

−divA∇u = −div f =⇒ ‖∇u‖Ls ≤ C(‖f‖Ls + ‖u‖Ls), (2.401)

provided that u ∈W 1,s0 (Ω).

86 [January 28, 2019]

Step 2. A priori estimate for q = s.We write the equation Lu = −div f as

−divA∇u = −div f + div (Bu)− C∇u−Du. (2.402)

For g ∈ Lp(Ω) we define ∇−1g = ∇u − (∇u)Ω, where u is the solution of−∆u = g in Rn (with g extended by zero). Note that u is in general onlydefined up to the addition of an affine function but that ∇u − (∇u)Ω doesnot change if we add an affine function to u. Let

Tu = −∇−1(C∇u+Du). (2.403)

Then the equation Lu = −div f can be written as

−divA∇u = −div f , where f = f −Bu+ Tu. (2.404)

By Corollary 2.42(i) the operator T is bounded from W 1,s(Ω) to itself.Since 1 < s <∞ the space W 1,s

0 (Ω) is reflexive and therefore we have12

uk 0 in W 1,s0 =⇒ Tuk 0 in W 1,s

0 . (2.405)

We claim that this implies that for each ε > 0 there exists a Cε such that

‖Tu‖Ls ≤ ε‖∇u‖Ls + Cε‖u‖Ls . (2.406)

Indeed if this was not true there exists an ε > 0 and a sequence such that‖Tuk‖Ls = 1, ‖∇uk‖ ≤ ε−1 and ‖uk‖Ls → 0. Thus uk u in W 1,s

0 and

hence Tuk 0 in W 1,s0 . By the compact Sobolev embedding this implies

Tuk → 0 in Ls and this is a contradiction. Now we can apply (2.401) to(2.404). This yields

‖∇u‖Ls ≤ C(‖f‖Ls +C‖u‖Ls + ‖Tu‖Ls) ≤ C(‖f‖Ls +Cε‖u‖Ls + ε‖∇u‖Ls).(2.407)

By choosing ε > 0 sufficiently small we can absorb the last term on the righthand side and this finishes the proof of the estimate for q = s.

Step 3. Higher regularity.Again we first prove interior regularity. Let B(x0, R) ⊂ Ω, letη ∈ C∞c (B(x0, R) and let

v = ηu. (2.408)

12See Functional Analysis and PDE. Sketch: let X = W 1,s0 and define the dual operator

by T ∗ : X ′ → X ′ by T ∗ϕ(u) = ϕ(Tu). Then T ∗ is bounded. Since X is reflexivethere exists a subsequence such that Tukj v in X. Thus ϕ(v) = limj→∞ ϕ(Tukj ) =limj→∞ T

∗ϕ(uk) = 0. It follows that v = 0 and by the usual argument one shows thatthe whole sequence Tuk converges weakly to zero.

87 [January 28, 2019]

We may extend v by zero to Rn. Using the abbreviations

(u⊗∇η)jβ = uj∂βη, [A(∇η ⊗∇u)]i = Aαβij ∂α∇η∂βuj (2.409)

we get−divA0∇v = −div (A0 −A)∇v − divA(∇v) (2.410)

and

−divA∇v = −divAη∇u− divA(u⊗∇η)

= −ηdivA∇u−A(∇η ⊗∇u)− divA(u⊗∇η). (2.411)

Now

−divA∇u = −div (f −Bu)− C∇u−Du,−ηdivA∇u = −div (ηf − ηBu) + (fα − (Bu)α) ∂αη − ηC∇u− ηDu,

and thus−divA∇v = −div f + g, (2.412)

where

f = η(f −Bu) +A(u⊗∇η),

gi = fα∂αη − (Bu)α∂αη − ηC∇u− ηDu−A(∇η ⊗∇u) (2.413)

Let1

r= max(

1

q,1

s− 1

n). (2.414)

Then by the Sobolev embedding theorem f ∈ Lr(B(x0, R)). Moreover g ∈Ls(B(x0, R)). Note also that f and v have compact support in B(x0, R).Thus

∫B(x0,r)

g dx = 0 (see Step 1). Hence by Corollary 2.42(ii) there exists

f ∈ Lr(Rn) ∩ Ls(Rn) with div f = g in Rn where g is extended by zero.Extending v and f by zero to Rn we thus get

−A0∇v = −div (A0 −A)∇v − div h (2.415)

with h = f + f ∈ Lr(Rn) ∩ Ls(Rn).Let T = T1 be the solution operator in Theorem 2.38(i) and set Sg =

T (A−A0)g. Then (2.415) implies that

∇v = S∇v + Th. (2.416)

We now show that S is a contraction on Lr and on Ls. By Theorem 2.38

‖Sg‖Lr ≤ C‖(A−A0)g‖Lr ≤ C supB(x0,R)

|A−A0| ‖g‖Lr . (2.417)

88 [January 28, 2019]

and the same estimate holds in Ls. Since A is uniformly continuous thereexists R0 such that for R ≤ R0

‖Sg‖Lr ≤1

2‖g‖Lr and ‖Sg‖Ls ≤

1

2‖g‖Ls . (2.418)

Again by Theorem 2.38 we have ‖Th‖Lr ≤ C‖h‖Lr and ‖Th‖Ls ≤ C‖h‖Ls .Thus one can use (2.416) to express ∇v by the Neumann series of (Id−S)−1,i.e.,

∇v =∞∑k=0

Sk Th (2.419)

where the series converges in Ls. Since ‖Sg‖Lr ≤ 12‖g‖Lr and Th ∈ Lr the

series on the right hand side also converges in Lr and thus ∇v ∈ Lr and

‖∇v‖Lr ≤ 2‖∇Th‖Lr ≤ C‖h‖Lr ≤ C(‖f‖Lq + ‖u‖W 1,s) (2.420)

and be Step 1 we get

‖∇v‖Lr ≤ C(‖f‖Lq + ‖u‖Ls) ≤ C(‖f‖Lq + ‖u‖Lq) (2.421)

Now if η = 1 on B(x0, θR) we get u ∈W 1,r(B(x0, θr)).If r = q the interior estimate is proved. Otherwise define iteratively

1ri+1

= max(1q ,

1ri− 1

n). Then we get u ∈ W 1,ri(B(x0, θiR) and the corre-

sponding estimate. If i is the smallest integer that 1s −

in <

1q . Then ri = q

and the interior estimate is established.The boundary estimate is proved in the same way and the estimates

can be combined to a global estimate by covering Ω with sufficiently smallballs.

Theorem 2.44. Assume that Ω is a bounded open set with C1 boundary.and asssume that the coefficients A are uniformly continuous and stronglyelliptic and that B,C,D ∈ L∞. Assume that 0 is not an eigenvalue of L,i.e.,

for all for f ∈ L2(Ω) there exists

a unique weak solution u ∈W 1,20 (Ω) of Lu = −div f .

Then for all p ∈ (1,∞) and all f ∈ Lp(Ω) the problem

Lu = −div f (2.422)

has a unique solution and there exists a constant C (depending on p, n, Ω,ν, sup ‖A‖, ‖B‖L∞, ‖C‖L∞, ‖D‖L∞ and the modulus of continuity of A)such that

‖u‖W 1,p ≤ C‖f‖Lp . (2.423)

89 [January 28, 2019]

Proof. For p > 2 the existence and the estimate follow directly from Lemma2.43. The uniqueness in W 1,p

0 follows from the uniqueness in W 1,20 since

W 1,p0 ⊂W 1,2

0 .For 1 < p < 2 we first consider uniqueness. If u1 and u2 are solutions

then u = u1 − u2 ∈ W 1,p0 and Lu = 0. By Lemma 2.43 (applied with

s = p, q = 2) we get u ∈W 1,20 and hence u = 0 by assumption.

To prove existence we use uniqueness to show that solutions of Lu =−div f satisfy the improved estimate

‖u‖W 1,p ≤ C‖f‖Lp . (2.424)

By Lemma 2.43 (with q = p) is suffices to show that

‖u‖Lp ≤ C‖f‖Lp . (2.425)

If this was not true there would exist sequences fk → 0 in Lp and uk with‖uk‖Lp = 1 and Luk = −div fk. By Lemma 2.43 it follows that ‖uk‖W 1,p ≤C. Hence by the compact Sobolev embedding there exists a subsequence(not relabeled) such that uk → u in Lp and uk u in W 1,p

0 . Then Lu = 0and ‖u‖Lp = 1. This contradicts uniqueness.

Now let f ∈ Lp and let fk ∈ Lp ∩ L2 such that fk → f in Lp. Byassumption there exists uk ∈ W 1,2

0 with Luk = −div fk. It follows from

(2.424) that uk is a Cauchy sequence in W 1,p and hence uk → u in W 1,p0 and

u is a (distributional) solution of Lu = −div f .

[26.11. 2018, Lecture 15][30.11. 2018, Lecture 16]

Similarly one can obtain estimates in W 2,p and W k,p. The followingresults were not discussed in detail in class.

Lemma 2.45. Let 1 < p ≤ q < ∞. Assume that Ω is a bounded open setwith C2 boundary, that the coefficients A ∈ C1(Ω) are strongly elliptic thatB ∈ C0,1, C,D ∈ L∞ and that u ∈ (W 2,p ∩W 1,p

0 )(Ω;RN ) is a distributionalsolution of Lu = g. If

g ∈ Lq(Ω) (2.426)

then u ∈ (W 2,q ∩W 1,q0 )(Ω) and

‖u‖W 2,q ≤ C(‖g‖Lq + ‖u‖Lq + ‖∇u‖Lq). (2.427)

Remark. The term ‖∇u‖Lq can be removed by using an interpolationinequality ‖∇u‖Lp ≤ ε‖∇2u‖Lp + Cε‖u‖Lp .

Proof. In the interior one considers the equation for the derivatives ∂γu andargues as in the proof of Lemma 2.43. At the boundary one first flattens the

90 [January 28, 2019]

boundary, then considers the equation for the tangential derivatives ∂γ u forthe transformed function and applies the reasoning in Lemma 2.43. Finallythe double normal derivative is recovered in the usual way by using theequation.

Theorem 2.46. Assume that Ω is a bounded open set with C2 boundary.and asssume that the coefficients A are strongly elliptic and satisfy A ∈C1(Ω) and that B ∈ C0,1, C,D ∈ L∞. Assume that 0 is not an eigenvalueof L, i.e.,

for all for f ∈ L2(Ω) there exists

a unique weak solution u ∈W 1,20 (Ω) of Lu = −div f .

Then for all p ∈ (1,∞) and all g ∈ Lp(Ω) the problem

Lu = g (2.428)

has a unique solution in u ∈W 2,p ∩W 1,p0 and there exists a constant C (de-

pending on p, n, Ω, ν, ‖A‖C1, ‖B‖, ‖C‖, ‖D‖ and the modulus of continuityof A) such that

‖u‖W 2,p ≤ C‖g‖Lp . (2.429)

Proof. By the global W 2,2 estimates for g ∈ L2 there exists a solution inW 2,2 ∩W 1,2

0 . Thus for p > 2 one can use Lemma 2.45. For p < 2 one arguesas in the proof of Theorem 2.44.

Remark. Similarly one can obtain W k,p estimates if g ∈ W k−2,p and thecoefficients lie in the appropriate spaces.

Remark. The following remark was not discussed in classOne can obtain slightly better results by looking at operators in non-divergenceform, i.e.,

(Lu)i := Aαβij (x)∂α∂βuj +Bα

ij(x)∂αuj + Cij(x)uj . (2.430)

Then for the ‖∇2u‖Lp estimate it suffices to assume that A ∈ C0(Ω), B,C ∈L∞(Ω) and ∂Ω ∈ C1,1. This can be proved in essentially the same way, see[GT], Chapter 9.2 to 9.6 for the details.

2.1.13 A short look back on elliptic regularity

This review section was not discussed in classWe have studied solutions of the second order systems in an open set

Ω ⊂ Rn

Lu := −div (A∇u)− div (Bu) + C∇u+Du = −div f + g (2.431)

91 [January 28, 2019]

under the assumption that A are bounded and strongly elliptic, i.e., thereexists a ν > 0 such that

(A(x)(ζ ⊗ ξ), ζ ⊗ ξ) = A(x)αβij ζiζjξαξβ ≥ ν|ζ|2|ξ|2 ∀ ζ ∈ RN , ξ ∈ Rn, x ∈ Ω

(2.432)The basic results are

• In the Holder spaces Ck,α, with 0 < α < 1 and the Sobolev spaces W k,p

with 1 < p < ∞ the solution u is as good as the data f, g, A,B,C,D(and the regularity of ∂Ω for global estimates) allow.

• There is no such theory for α = 0, α = 1, p = 1 or p =∞.

• All the estimates can be derived from L2 type estimates in the Cam-panato spaces L2,λ (and suitable perturbation/ compactness argu-ments)

One subtlety is that one usually needs A ∈ C0 in cases where one mightthink that A ∈ L∞ is sufficient. The reason is that one usually has toapproximate at least the leading order term −divA∇ by an operator withconstant coefficients and then uses a fixed point argument (or an iterationargument for the Campanato estimates) to treat the case of variable coeffi-cients. This requires that the oscillation of the coefficient is small in smallballs. The only exception are the W 1,2 estimates for scalar equations (orfor systems which satisfy the super strong ellipticity condition). For scalarequations there are some deep results which still hold for L∞ coefficients.We will discuss these in the next section.

We have derived different though closely related results. For illustrationwe consider the W 1,p estimates.

(i) Local or global a priori estimates. If u is a solution and if u ∈W 1,s

0 (Ω) then

‖u‖W 1,s ≤ C(‖f‖Ls + ‖g‖Lt) + C‖u‖Ls (2.433)

In these estimates usually a weaker norm of u still appears on the righthand side. In local estimates the norm on the right hand is usually ona slightly large set then the norm of the left hand side. The norm of uon the right hand side can in general not be avoided since the problemLu = 0 can have nontrivial solutions.

(ii) Improvement of regularity estimates. If u ∈W 1,s0 (Ω) (or in an even

weaker space) and if f ∈ Lq with q > s then u ∈ W 1,q0 . Again there

are local and global versions.

(iii) Existence and uniqueness of solutions (for problems with boundaryconditions)

92 [January 28, 2019]

(iv) Global estimates for problems with boundary conditions. Thesecan often be deduced from the apriori estimate using a compactnessargument to eliminate the weaker norm on the right hand side.

2.2 Harnack inequality and the DeGiorgi-Moser-Nash theo-rem

This subsection follow closely [GT], Sections 8.5–8.9

2.2.1 Weak Harnack inequalities and Holder estimates

So far we have developed a theory which tells us that if the coefficients inthe leading order term are sufficiently regular then in appropriate spaces thesolution of the PDE is as good as the coefficients and the right hand sideallows.

In this subsection we consider elliptic problems where the coefficients inthe leading order term are only in L∞ but not continuous. In this case thereis a big difference between scalar equations and systems. For scalar equationthe DeGiorgi-Nash-Moser theorem states that we have C0,α estimates forsome small α > 0. The corresponding results for systems fails, even if weassume super strong ellipticity, see [Gi] for the history of the subject andvarious counterexamples. More recent counterexamples can be found the inpapers [SY00] and [SY02].

Apart from its intrinsic interest C0,α estimates for solutions of linearequations with L∞ coefficients are of crucial importance for the regularitytheory for nonlinear problems. Consider for example the problem of mini-mizer the functional

I(u) =

∫Ωf(∇u) dx (2.434)

where f is C∞ and uniformly convex, i.e., there exists ν > 0 such that

∂2f

∂ξi∂ξjξiξj ≥ ν|ξ|2 ∀ξ ∈ Rn, (2.435)

and that the second derivatives of f are uniformly bounded, i.e.,∣∣∣∣ ∂2f

∂ξi∂ξj

∣∣∣∣ ≤ C. (2.436)

By the direct method of the calculus of variations one sees that for givenboundary data u0 ∈ W 1,2(Ω) the functional I has a (unique) minimizer uwith u−u0 ∈W 1,2

0 (Ω). Moreover u is a weak solution of the Euler-Lagrangeequation

−∂i∂f

∂ξi(∇u) = 0. (2.437)

93 [January 28, 2019]

If we formally take a derivative in direction xk we obtain that the functionv = ∂ku is a solution of the linear PDE

−∂i∂2f

∂ξi∂ξj(∇u(x))︸ ︷︷ ︸

aij(x)

∂jv = 0. (2.438)

The assumptions that f is uniformly convex and that the second derivativesare bounded imply that the coefficients aij are elliptic and in L∞. The fact

that v ∈ W 1,2loc (Ω) and that v is a weak solution of (2.437) can be justified

by the difference quotient method (Homework).Now if we can show that solutions of (2.437) are in C0,α

loc (Ω) then we canderive full regularity of v from the theory developed so far by a bootstrapargument. Indeed v ∈ C0,α

loc implies that u ∈ C1,αloc . Hence the coefficients aij

are in C0,αloc and the Schauder theory implies that v ∈ C1,α

loc . Hence aij ∈ C1,αloc

and thus v ∈ C2,αloc , etc.

To focus on the essential points we consider only the leading order termsand we look at the operator

Lu := −∂iaij(x)∂ju. (2.439)

in a bounded and open set Ω ⊂ Rn. We write Ω′ ⊂⊂ Ω if Ω′ is open, and ifΩ′ is compact with Ω

′ ⊂ Ω. We assume that there exist Λ > 0, and ν > 0such that ∑

i,j

a2ij(x) ≤ Λ2 for a.e. x, (2.440)

∑i,j

aij(x)ξiξj ≥ ν|ξ|2 for all ξ ∈ Rn and a.e. x. (2.441)

Theorem 2.47 (DeGiorgi-Nash-Moser). Assume that (2.440) and (2.441)hold and that u ∈W 1,2(Ω) is a weak solution of

Lu = 0. (2.442)

Then for any open set Ω′ ⊂⊂ Ω we have

‖u‖C0,α(Ω′) ≤ C‖u‖L2(Ω) (2.443)

where

α = α(n,Λ/ν) > 0, C = C(n,Λ, ν, d′) with d′ = dist (Ω′,Ω). (2.444)

94 [January 28, 2019]

Remark. (i) As mentioned above such an estimate does not hold for sys-tems.(ii) For each α ∈ (0, 1) there exists a solution of an elliptic equation whichis C0,α but not in C0,β for any β > α. Indeed for n = 2 and z = x1 + ix2 thefunction u(z) = Re z

|z|1−α is the solution of an elliptic equation (see Home-

work 1).(iii) For a version with lower order terms and solutions of Lu = −div f + gwith f ∈ Lq(Ω), g ∈ Lq/2(Ω) for q > n see Gilbarg-Trudinger [GT].

We follow Moser’s proof and will derive the Cα estimate from a Harnackinequality for weak solutions. In fact it is convenient to state separate Har-nack type inequalities for subsolutions and supersolutions. We say that u isa weak subsolution in Ω and write

Lu ≤ 0 (2.445)

if u ∈W 1,2(Ω) and∫Ω

∑i,j

aij∂ju ∂iϕ ≤ 0 ∀ϕ ∈W 1,20 (Ω) with ϕ ≥ 0. (2.446)

Similar we define a weak supersolution, Lu ≥ 0. An easy way to rememberthe signs is the implication

u convex =⇒ u subsolution.

Theorem 2.48 (Weak Harnack inequality). Assume that (2.440) and (2.441)hold, let u ∈W 1,2(Ω) and assume that B(y, 4R) ⊂ Ω.

(i) IfLu ≤ 0 and u ≥ 0 (2.447)

then for all p > 1

supB(y,R)

u ≤ CR−n/p‖u‖Lp(B(y,2R)) (2.448)

(ii) IfLu ≥ 0 and u ≥ 0 (2.449)

then for all p ∈ [1, nn−2)

R−n/p‖u‖Lp(B(y,2R)) ≤ C infB(y,R)

u. (2.450)

Here C = C(n,Λ/ν, p).

Proof. For ease of notation we assume that Lu = 0. By scaling and trans-lation we may assume that R = 1. We write Br := B(y, r).

95 [January 28, 2019]

Key idea. To prove (i) show an estimate for the form

‖u‖Lp(Bs) ≤ C(s, r, p, p)‖u‖Lp(Br)

for p > p and s < r. Then iterate this estimate and finally take the limitp → ∞. The procedure is know as Moser iteration. For (ii) one uses that

1inf u = sup 1

u and estimates negative powers of u.We also note that Lu = 0 implies L(u + δ) = 0. Hence we may assume

u ≥ δ > 0 on B(y, 4R) and in the end let δ → 0. Thus we can use negativepowers of u without further discussion.

Step1: the basic estimate. Let β 6= 0 and use the test function

v = η2uβ, where η ∈ C∞c (B4) with 0 ≤ η ≤ 1.

Assume first that u is a weak subsolution and β > 0. Then

0 ≥∫B4

∑i,j

aij∂ju ∂iv dx

=

∫B4

∑i,j

aij∂ju η2βuβ−1∂iu dx+

∑i,j

aij∂ju 2η∂iη βuβ dx

Ellipticity and the bound on the coefficients imply that

ν

∫B4

η2uβ−1|∇u|2 dx ≤ 2Λ

|β|

∫B4

η|∇η|uβ|∇u| dx. (2.451)

The same estimate holds if u is a weak supersolution and β < 0.Now we want to write the term uβ−1|∇u|2 as |∇w|2 (up to a constant

factor). Assuming for the moment

β 6= −1

we setw = u

β+12

and obtain

∇w =β + 1

2uβ−1

2 ∇u,

uβ−1|∇u|2 =

(2

β + 1

)2

|∇w|2,

uβ|∇u| = 2

|β + 1|u|β+1|

2 |∇w| = 2

β + 1w|∇w|

96 [January 28, 2019]

Thus with C(β) = C |β+1||β|∫

B4

η2|∇w|2 ≤ C(β)

∫B4

η|∇η|w|∇w|

≤ C(β)‖∇η w‖L2 ‖η∇w‖L2

and therefore

‖η∇w‖L2 ≤ C(β) ‖∇η w‖L2

‖∇(ηw)‖L2 ≤ (C(β) + 1) ‖∇η w‖L2 .

Roughly speaking we have transferred the gain in derivatives for solutions ofelliptic PDE from u to w = u(β+1)/2. Now the Sobolev embedding theoremgives

‖ηw‖L2κ ≤ (C(β) + 1)‖∇η w‖L2 , where κ =n

n− 2(2.452)

(for n = 2 we can take any κ <∞).Now let 0 < s < r < 4 and choose η such that

η ∈ C∞c (Br), η = 1 in Bs, sup |∇η| ≤ C

r − s. (2.453)

Then

‖w‖L2κ(Bs) ≤C

r − s(C(β) + 1)‖w‖L2(Br) (2.454)

Finally set γ = β + 1. Then we obtain the basic improvement estimate(∫Bs

|u|κγ dx) 1

2κ

≤ C

r − s(|γ|+ 1)

(∫Br

|u|γ) 1

2

if |β| = |γ − 1| ≥ d > 0.

(2.455)This estimate holds for

subsolutions if γ − 1 = β > 0,

supersolutions if γ − 1 = β < 0.

Step 2: iteration. Set

Φ(p, r) :=

(∫Br

|u|p dx) 1p

. (2.456)

Then

supBr

u = limp→∞

Φ(p, r) =: Φ(∞, r),

infBru = lim

p→−∞Φ(p, r) =: Φ(−∞, r).

97 [January 28, 2019]

With this notation the basic estimate (2.455) becomes

Φ(κγ, s) ≤(

C

r − s(|γ|+ 1)

) 2γ

Φ(γ, r) if γ > 0 (2.457)

and

Φ(κγ, s) ≥(

C

r − s(|γ|+ 1)

) 2γ

Φ(γ, r) if γ < 0, (2.458)

as long as 0 < s < r and |γ−1| ≥ d > 0. Here (2.458) holds for subsolutions,while (2.457) holds for subsolutions if 0 < γ < 1 and for supersolutions ifγ > 1.

To prove assertion (i) of Theorem 2.48 take

1 < p ≤ 2, γm = κmp, rm = 1 + 2−m

and apply (2.457) with γ = γm−1, r = rm−1, s = rm. Not that |γ|+1| ≤ 2|γ|and the factor 2 can be included in the constant C in (2.457). Thus (2.457)yields

ln Φ(γm, rm) ≤ 2

γm(C +m+ ln γm) + ln Φ(γm−1, rm−1). (2.459)

Now∑m

2

γm(C +m+ ln γm) =

∑m

2

κmp(C +m+m lnκ+ ln p) ≤ C. (2.460)

HencesupB1

u = Φ(∞, 1) ≤ CΦ(p, 2). (2.461)

This proves assertion (i).[30.11. 2018, Lecture 16][3.12. 2018, Lecture 17]

Proof of assertion (ii) Let 1 < p < κ = nn−2 . We will show that for each

p0 = κ−Mp ∈ (0, 1) we have

Φ(p, 2) ≤ CΦ(p0, 3), (2.462)

Φ(−∞, 1) ≥ Φ(−p0, 3). (2.463)

Here the constant may depend on p and M . Since infB1 u = Φ(−∞, 1)assertion (ii) then follows if we can show that for sufficiently small p0 > 0(which depends only on n and and Λ/ν) we have

Φ(p0, 3) ≤ CΦ(−p0, 3) (2.464)

98 [January 28, 2019]

To prove (2.462) we apply the estimate (2.457) with

γm = κmp0, rm = 2 + 2−m

and we deduce (2.462) after M iterations. Note that γm < κ−1 if m ≤M−2and hence |γ − 1| ≥ 1− κ−1 so at each step in the iteration (2.457) can beapplied since u is a supersolution. For m = M − 1 we have γM−1 = p

κ < 1but γM−1 may be very close to one if p is close to κ. Hence the condition|γ − 1| ≥ d may be violated. Nonetheless by going back to the original

estimate (2.452) (with C(β) = |β+1||β| and β = γM−1 − 1) we still get the

desired bound Φ(p, rM ) ≤ C(p)Φ(γM−1, rM−1).To prove (2.463) we apply (2.458) with

γm = −κmp0, rm = 1 + 2−m+1

and the convergence follows again from (2.460), now with γm replaced by|γm| and p replaced by p0.

Step 3: closing the gap: Φ(p0, 3) ≤ CΦ(−p0, 3). We first show that thefunction

w := lnu

satisfies ∇w ∈ L2,n−2(B3) where L2,n−2 denotes the Morrey space.We use (2.451) with β = −1. Since ∇w = u−1∇u this gives∫

B4

η2|∇w|2 dx ≤ C

∫B4

η|∇η| |∇w| dx

≤ C

(∫B4

|∇η|2 dx) 1

2(∫

B4

η2|∇w|2 dx) 1

2

.(2.465)

Now let z ∈ B3, r ≤ 12 and η ∈ C∞c (B(z, 2r) with η = 1 in B(z, r) and

‖∇η| ≤ Cr . Then (2.465) implies that∫

B(z,r)|∇w|2 dx ≤

∫B(z,2r)

|∇η|2 dx ≤ Crn−2. (2.466)

Similarly for η ∈ C∞c (B4) with η = 1 in B3 we get∫B3

|∇w|2 dx ≤ C. (2.467)

Thus ∇w ∈ L2,n−2(B3) and ‖∇w‖L2,n−2 ≤ C(n,Λ/ν). The Poincare in-equality implies that

[w]L2,n(B3) ≤ C‖∇w‖L2,n−2 ≤ C(n,Λ/ν). (2.468)

99 [January 28, 2019]

Now it follows from the John-Nirenberg theorem (see Theorem 2.50 be-low) that ∫

B3

exp

(σ|w − a|[w]L2,n

)≤ C (2.469)

where a = wB(0,3) . Let p0 = σ/[w]L2,n . Then using that ±(w−a) ≤ |w−a|we deduce from (2.469) that∫

B3

up0e−p0a dx ≤ C,∫B3

u−p0ep0a dx ≤ C

Hence ∫B3

up0 dx

∫B3

u−p0 dx ≤ C

and taking this inequality to the power 1/p0 we obtain the assertionΦ(p0, 3) ≤ CΦ(−p0, 3).

Step 4: Justification of the choice of test function This step wasonly sketched in class.The main point of Step 1 is the implication

uγ ∈ L1(Br), supp η ⊂ Br =⇒ (ηuγ/2)2κ ∈ L1 and (2.455) holds.

To explain the main line of the argument we used the test function η2uβ

with β = γ − 1 without further discussion. This is no problem for β ≤ 1,since by our assumption u ≥ δ we then have v = η2uβ ∈ W 1,2

0 (B4) so wecan use v in the definition of sub- oder supersolution.

If, however, β > 1 it is initially not clear whether v ∈ W 1,2. We nowgive a rigorous argument which shows that

uγ ∈ L1(Br), supp η ⊂ Br

=⇒∫B4

η2|u|γ−2|∇u|2 dx <∞, ηuγ/2 ∈W 1,20 (Br) and (2.451) holds.

(2.470)

From this it follows that all the remaining calculation in Step 1 are justified.To prove (2.470) we consider the test function v = η2hM (u) where, for

γ > 2 the function hM (t) is a truncation of tγ−1, namely,

hM (t) =

tγ−1 if 0 ≤ t < M ,

(2− γ)Mγ−1 + (γ − 1)Mγ−2t if t ≥ nM .

Then hM is a C1 function on (0,∞) and h′M is bounded. Thus hM (u) ∈W 1,2. We have

h′M (t) =

(γ − 1)tγ−2 if 0 ≤ t < M ,

(γ − 1)Mγ−2 if t ≥ nM .

100 [January 28, 2019]

Note that h′M is monotonically increasing in M . Using η2hM (u) as a testfunction and using ellipticity and boundedness of the aij we get∫

B4

η2h′M (u)|∇u|2 dx ≤ C∫B4

η|∇η|hM (u)|∇u| dx (2.471)

Now one can check directly from the formulae for hM and h′M that

hM (t) ≤ th′M (t).

Using this estimate and the Cauchy-Schwarz inequality we get∫B4

η|∇η|hM (u)|∇u| dx ≤∫B4

η|∇η|uh′M (u)|∇u| dx

≤(∫

B4

|∇η|2h′M (u)u2 dx

) 12(∫

B4

η2h′M (u)|∇u|2 dx) 1

2

Thus (2.471) implies that∫B4

η2h′M (u)|∇u|2 dx ≤ C∫B4

|∇η|2h′M (u)u2 dx

Now h′M (u)|u|2 ≤ (γ−1)|u|γ . Thus the right hand side is bounded indepentlyof M . By the monotone convergence theorem we can take the limit M →∞on the left hand side and we get∫

B4

η2(γ − 1)uγ−2|∇u|2 dx <∞ (2.472)

From this we deduce that

ηuγ−1∇u = ηuγ2 u

γ2−1∇u ∈ L1(B4)

since uγ2 ∈ L2(Br) by assumption. Thus the right hand side of (2.471) is

bounded independently of M . Using monotone convergence we can pass tothe limit M →∞ and we obtain (2.451).

Finally to see that ηuγ/2 ∈W 1,2 we approximate again uγ/2 by functionsgM (u) where gM has linear growth at infinity and we use the estimate 2.472and monotone convergence to pass to the limit M →∞.

We will first show that the weak Harnack inequality implies C0,α reg-ularity. Then we will note that the weak Harnack inequality implies theclassical Harnack inequality. Finally we prove the John-Nirenberg resulton exponential integrability of BMO functions which was crucial to ’closethe gap’ between Φ(−p0, 3 and Φ(p0, 3) in the proof of the weak Harnackinequality.

Proof that Theorem 2.48 =⇒ Theorem 2.47.

101 [January 28, 2019]

Step 1. L∞ estimate.Let u ∈W 1,2(Ω) be a weak solution of Lu = 0. Assume that B(y, 4R) ⊂ Ω.We first show that u is bounded in B(y,R) by showing that convex functionsof u are subsolutions.13

Assume thatf ∈ C2(R), sup

R|f ′|+ |f ′′| <∞. (2.473)

We claim thatf u ∈W 1,2(Ω), L(f u) ≤ 0. (2.474)

Choosing f(t) =√t2 + δ2 and using Theorem 2.48 (i) for f u and p = 2

we then get

supB(y,R)

|u| ≤ supB(y,R)

f u ≤ CR−n/2‖f u‖L2(B(y,2R).

Taking the limit δ → 0 we conclude that

supB(y,R)

|u| ≤ CR−n/2‖u‖L2(B(y,2R)). (2.475)

and thus u ∈ L∞loc(Ω).The first assertion in (2.474) follows from the chain rule. To prove that

f u is a subsolution we first consider ϕ ∈ (W 1,20 ∩ L∞)(Ω) with ϕ ≥ 0.

Then ∫Ω

∑i,j

aij∂jf(u) ∂iϕdx =

∫Ω

∑i,j

aijf′(u)∂ju ∂iϕ

=

∫Ω

∑i,j

aij∂ju ∂i(f′(u)ϕ) dx−

∫Ω

∑i,j

aij∂ju f′′(u)∂iu︸ ︷︷ ︸

≥0

ϕdx ≤ 0.

Here we used that the first term in the last line is zero since u is a weaksolution of Lu = 0 and since v = f ′(u)ϕ is in W 1,2

0 (Ω). Since ϕ ∈ W 1,20 ∩

L∞(Ω) : ϕ ≥ 0 is dense in ϕ ∈ W 1,20 (Ω) : ϕ ≥ 0 it follows that f u is a

subsolution.14

Step 2. C0,α estimate.Set

ω(r) = oscB(y,r)

u := supB(y,r)

u− infB(y,r)

u. (2.476)

13Alternatively the estimate (2.475) below can be proved by using Moser iteration withthe test functions |u|β−1u or more precisely a suitable truncation of these functions as inStep 4 of the proof of the weak Harnack inequality.

14To prove density note that there exist ϕk ∈ C∞c (Ω) such that ϕk → ϕ in W 1,2 andlet ϕ+

k = max(ϕk, 0). Then ϕ+k → ϕ+ = ϕ in W 1,2.

102 [January 28, 2019]

We claim that there exist a 0 < θ < 1 such that

B(y, 4r) ⊂ Ω =⇒ ω(r) ≤ θω(4r). (2.477)

From this one easily deduces that there exist α > 0 and C > 0 such forB(y, r) ⊂ B(y,R) ⊂ Ω we have

oscB(y,r)

u ≤ C( rR

)αosc

B(y,R)u. (2.478)

and together with the local estimates for supu and − inf u this implies theconclusion of Theorem 2.47 To prove the claim set Bρ = B(y, ρ) and define

M4 = supB4ru, M1 = supBr u,

m4 = infB4r u, m1 = infBr u.

ThenM4 − u ≥ 0, L(M4 − u) = 0 in B4r

u−m4 ≥ 0, L(u−m4) = 0 in B4r.

Thus Theorem 2.48 (ii) with p = 1 yields

r−n∫B2r

(M4 − u) ≤ C infBr

(M4 − u) ≤ C(M4 −M1),

r−n∫B2r

(u−m4) ≤ C infBr

(u−m4) ≤ C(m1 −m4).

Adding these two inequalities we get

2nLn(B1)(M4 −m4) ≤ C(M4 −M1) + C(m1 −m4)

and thus

(M1 −m1) ≤ (1− 1

C2nLn(B1))(M4 −m4).

Taking θ = 1− 1C 2nLn(B1) we get (2.477).

Theorem 2.49 (Harnack inequality). Assume that (2.440) and (2.441) holdand that u ∈W 1,2(Ω) is a weak solution of

Lu = 0. (2.479)

Assume further thatu ≥ 0 in B(y, 4R) ⊂ Ω. (2.480)

ThensupB(y,R)

u ≤ C infB(y,R)

u (2.481)

where C = C(n,Λ/ν).

Proof. This follows directly from the weak Harnack inequality, Theorem 2.48.

103 [January 28, 2019]

Remark. The Harnack inequality can be seen as a quantitative version ofthe strong maximum principle. The strong maximum principle implies thatif u(x) = 0 for some x ∈ B(y,R) then u ≡ 0 in B(y,R). In other words thestrong maximum principle implies that

infB(y,R)

u = 0 =⇒ supB(y,R)

u = 0. (2.482)

The Harnack inequality provides a quantitative estimate between the supre-mum and the infimum.

Remark. Weak Harnack inequalities at the boundaryAdded Jan 28, 2019. This was not discussed in class.There is a version of the weak Harnack inequalities which holds also nearthe boundary. Recall that u+ = max(u, 0) and for m,M ∈ R define

u+M (x) =

max(u(x),M) x ∈ Ω,

M x /∈ Ω(2.483)

and

u−m(x) =

min(u(x),m) x ∈ Ω,

m x /∈ Ω.(2.484)

Assume that u ∈W 1,2(Ω). Then for any ball B(y,R) ⊂ Rn we have

Lu ≤ 0 in Ω ∩B(y, 2R) =⇒ supB(y,R)

u+M ≤ CR

−n/p‖u+M‖Lp(B(y,2R).

(2.485)where

M = sup∂Ω∩B(y,2R)

u+. (2.486)

Moreover if u > 0 in B(y, 4R)

Lu ≥ 0 in Ω ∩B(y, 4R) =⇒ infB(y,R)

u+M ≥ cR

−n/p‖u+M‖Lp(B(y,4R).

(2.487)where

m = inf∂Ω∩B(y,2R)

. (2.488)

To prove (2.485) one uses the test function

v = η2((u+M )β −Mβ) with η ∈ C∞c (B(y, 4R)).

Note that v ∈ W 1,20 (Ω) since u ≤ M on ∂Ω. Note also ∇v = 0 on the set

u ≤ M . Moreover on the set u > M we have ∇u = ∇u+M . Note also that

∇u+M = 0 in Rn \ 0. Hence we get

0 ≥∫

Ω

∑i,j

aij∂ju∂iv dx =

∫B(y,4R)

∑i,j

aij∂ju+M∂iv dx. (2.489)

104 [January 28, 2019]

From this inequality we can derive as before the key estimate (2.451) with

u replaced by u+M . If we set w = (u+

M )β+1

2 we obtain as before

‖∇(ηw)‖L2 = (C(β) + 1)‖∇ηw‖L2

where, as before, C(β) = C β+1β . One minor difference is that we no longer

have ηw ∈ W 1,20 (Ω) (since η need not vanish on ∂Ω). Nonetheless we can

use the Sobolev inequality in B(y, 4R) in the form

‖W‖L2κ(B(y,4R) ≤ C(R−1‖W‖L2(B(y,4R) + ‖∇W‖L2(B(y,4R)).

Taking W = ηw and R = 1 we arrive again at (2.454) and finally at the basicimprovement estimate (2.455). Now (2.485) follows by iteration exactly asbefore.

To prove (2.487) one uses similarly the test function

v = η2((u−m)β −mβ) with η ∈ C∞c (B(y, 4R)).

2.2.2 Exponential integrability in general Lipschitz domains

By using the extension operator in Lemma 2.37 the John-Nirenberg theo-rem (Thm. 2.34) can be extended to general open and bounded sets withLipschitz boundary.

The details were not discussed in class

Theorem 2.50. Let Ω ⊂ Rn be an open and bounded set with Lipschitzboundary. Then exists constants σ2 > 0 and A2 > 0 which only depend onthe dimension n and on Ω with the following property. If f ∈ L2,n(Ω) then∫

Ωexp

(σ2|f − fΩ|

[f ]L2,n

)≤ A2(diam Ω)n (2.490)

Remark. (i) By scaling one sees that σ2 and A2 do not change if Ω isreplaced by a dilation λΩ. In particular if Ω is a ball then the constants σ2

and A2 can be chosen indepent of the radius of the ball.(ii) The result holds even under the weaker condition that Ω is bounded andopen and that there exists an a > 0 such that

Ln(B(x, r) ∩ Ω) ≥ arn ∀x ∈ Ω, r < diamΩ. (2.491)

The constants σ2 and A2 depend only on n and a. In this case one use theextension

Ef(x) :=1

Ln(B(x, 2d(x)) ∩ Ω)

∫B(x,2d(x))∩Ω

f dy

for x ∈ Rn \ Ω where d(x) = dist (x,Ω).

105 [January 28, 2019]

Note that the condition (2.491) implies that Ln(∂Ω) = 0. Indeed otherwiseby the Lebesgue point theorem there exists x ∈ Ω such that

limr→0

Ln(∂Ω ∩B(x, r))

Ln(B(x, r))= 1 and thus lim

r→0r−nLn(Ω ∩B(x, r)) = 0

and this leads to a contradiction with (2.491).

Proof of Theorem 2.50. It suffices to prove the Theorem for function withfΩ = 0 (if the result is know for such function it can be applied to thefunction g = f − fΩ). Moreover by scaling we may assume that diamΩ = 1

2and by translation we may assume that

Ω ⊂ Q0 := (0, 1)n.

Finally we may assume[f ]L2,n(Ω) = 1. (2.492)

By the definition of the L2,n seminorm we have for any point y ∈ Ω∫Ω|f |2 dx =

∫Ω|f − fΩ|2 dx =

∫Ω∩B(x, 1

2

|f − fΩ|2 dx ≤ 2−n. (2.493)

since [f ]L2,n(Ω) = 1. Thus ‖f‖L2,n ≤ 2.Now let E be extension operator in Lemma 2.37 and let

g = Ef|Q0(2.494)

Then‖g‖L2,n(Q0) ≤ CE‖f‖2L2,n ≤ 2CE , (2.495)

where CE = CE(Ω) is the norm of the extension operator. In particular wehave

|gQ0 | ≤∫Q0

|g| dx ≤(∫

Q0

|g|2 dx)1/2

≤ 2CE . (2.496)

Finally there exist B = B(n) ≥ 1 such that

[g]BMO (Q0) ≤ B‖g‖L2,n(Q0) ≤ 2BCE . (2.497)

Thus

|g| ≤ |g − gQ|+ |gQ| ≤ 2BCE|g − gQ|[g]BMO

+ 2CE . (2.498)

Now setσ2 =

σ1

2BCE. (2.499)

Then by Theorem 2.34∫Q0

exp(σ2|g|) dx ≤∫Q0

exp

(σ1|g − gQ|[g]BMO

)eσ22CE dx

≤ A1eσ1 .

106 [January 28, 2019]

Since σ22CE = σ1/B ≤ σ1. Since Ω ⊂ Q0 and g = f on Ω this implies theassertion with σ2 = σ1/(2BCE) and A2 = A1e

σ1 .

[3.12. 2018, Lecture 17][7.12. 2018, Lecture 18]

2.3 Meyers’ estimate for systems with L∞ coefficients

The DeGiorgi-Nash-Moser theorem shows that for scalar elliptic equationswith L∞ coefficients the solution u is much better than L2, namely C0,α, butgives no information about better properties of ∇u. Meyers’ Theorem showsthat ∇u ∈ Lp where p is slightly large than 2. This results even holds forsystems with super-strongly elliptic L∞ coefficients. The text below followsclosely the original paper [Me63]. The discussion of Riesz-Thorin theoremis taken from [?].

Theorem 2.51 (Meyers’ estimate). Suppose that Ω ⊂ Rn is a bounded

open set with C1 boundary. Suppose that the Aαβij ∈ L∞(Ω) and satisfy thesuper-strong ellipticity condition, i.e., there exists a ν > 0 such that

(AF,F ) ≥ ν|F |2. (2.500)

Then there exists a p0 > 2 such that for p ∈ (p′0, p0) and f ∈ Lp the problem

−divA∇u = −div f, u ∈W 1,p0 (Ω) (2.501)

has a unique solution and

‖∇u‖Lp ≤ C‖f‖Lp . (2.502)

Here p0 and C depend only on n, ‖A‖L∞/ν and Ω.

Remark. (i) One easily obtains the following local version. If −divA∇u =−div f in Ω, if u ∈ W 1,2(Ω) and f ∈ Lp(Ω) with 2 < p < min(p∗, 2

∗) thenu ∈W 1,p

loc (Ω). To see this note that by the Sobolev embedding theorem andby Corollary 2.42 the function v = ηu solves −divA∇v = −div f − div hwhere h ∈ L2∗(Ω).(ii) The result is optimal, even for n = 2, scalar equations and f = 0. Indeedfor each α > 0 the function u = x1

|x|1−α is the solution of an elliptic equation

(see Homework 1) and u /∈ Lp(B(0, 1)) if p ≥ 2/(1− α).(iii) This result allows one to extend the regularity results for minimizers ofthe integral functionals (2.434) to systems if n = 2. Indeed if the integrandf is uniformly convex and has bounded second derivatives we get by thedifference quotient method and Meyers estimate ∇2u ∈ Lp for some p > 2.Now for n = 2 the Sobolev embedding theorem implies that ∇u ∈ C0,α for

107 [January 28, 2019]

some α > 0 and then the previous bootstrap argument in Schauder spacesworks.(iv) The paper by Meyers’ also covers the case of complex coefficients (thisrequires only obvious changes in the argument in Step 3 of the proof).

Proof. This is done by a clever perturbation argument. We only show theestimate for 2 < p < p′0. The other case can be handled by duality andapproximation (see [Me63] for the details).

Step 1. Linear algebra.Recall that ‖A(x)‖ := sup‖AF‖ : |F | ≤ 1 denotes the operator norm. Bydividing the equation by ‖A‖L∞ we may assume that

‖A(x)‖ ≤ 1 a.e. (2.503)

To focus on the main idea we assume for simplicity in addition that

A(x) is symmetric,

i.e., (A(x)F,G) = (F,A(x)G) for all F,G ∈ RN×n and a.e. x. Now the PDEcan be written in the equivalent form

−∆u = −div (Id −A)∇u− div f (2.504)

Using ellipticity and the bound ‖A(x)‖ ≤ 1 we get

((Id −A(x))F, F ) ≤ (1− ν)|F |2 and ((Id −A(x))F, F ) ≥ 0.

Thus the eigenvalues of the symmetric operator Id −A(x) lie in the interval[0, (1− ν)]. Therefore we obtain the crucial estimate

‖Id −A(x)‖ ≤ 1− ν for a.e. x. (2.505)

Step 2. Fixed point argument in W 1,p0 (Ω).

On W 1,p0 (Ω) we can consider the norm

‖u‖p := ‖∇u‖Lp . (2.506)

It follows from the Poincare inequality for W 1,p0 that this is indeed a norm

and that this norm is equivalent to the usual W 1,p norm. Theorem 2.44implies that for every p ∈ (1,∞) and every h ∈ Lp(Ω) the problem

−div∇v = −div h, v ∈W 1,p0 (Ω)

has a unique solution. We write define the operator R by Rh := v. Then,again by Theorem 2.44, we have

‖∇Rh‖Lp ≤Mp‖h‖Lp . (2.507)

108 [January 28, 2019]

Now (2.504) is equivalent to

u = Su+Rf where Su = R[(Id −A)∇u].

Thus‖Su‖p ≤Mp‖(A− Id )∇u‖Lp ≤Mp(1− ν)‖u‖p. (2.508)

We are done if we can show that S is a contraction. For this it sufffices that

‖Mp(1− ν)‖ < 1. (2.509)

We know that M2 = 1 (test the equation for v with v). Let p ∈ (2,∞), letθ ∈ (0, 1) and let

1

p= (1− θ)1

2+ θ

1

p.

By the Riesz-Thorin Theorem (stated below)

Mp ≤M1−θ2 M θ

p = M θp . (2.510)

If Mp ≤ 1/(1 − ν) then (2.509) holds for all p ∈ (2, p). If Mp > 1/(1 − ν)define θ∗ by the relation M θ∗

p = 1/(1 − ν) and set 1p∗

= (1 − θ∗)12 + θ∗

1p .

Then (2.509) holds for all p ∈ (2, p∗).

Step 3. Nonsymmetric A. This part was not discussed in classLet 0 < ν ≤ 1 and assume that

min|F |=1

(AF,F ) ≥ ν and ‖A‖ ≤ 1.

We show that there exist θ > 0, λ > 0 and M > 0, a symmetric operator Sand an operator R such that

A = S +R, inf|F |=1

(SF, F ) ≥ λ, ‖S‖ ≤M, ‖R‖ ≤ (1− θ)λ. (2.511)

Here θ, λ and M depend only on ν. For fixed ν the operators S and R de-pends linearly on A. To prove this define the symmetric and skew symmetricpart of A as

symA :=A+AT

2, skwA :=

A−AT

2.

Then A = symA+ skwA. We claim that

(symAF, skwAF ) = 0 and thus ‖AF‖2 = ‖ symAF‖2 + ‖ skwAF‖2.

To prove the first identity note that

(symAF, skwAF ) =(AF,AF ) + (ATF,AF )− (AF,ATF )− (ATF,ATF )

=([ATA−AAT ]F, F ) + ([ATAT −AA]F, F )

109 [January 28, 2019]

Now the operators ATA − AAT and ATAT − AA are skew symmetric andhence the right hand side vanishes. In particular we have

‖ symA‖ ≤ ‖A‖, ‖ skwA‖ ≤ ‖A‖ (2.512)

For c ≥ 0 write

A = (symA+ cId ) + (skwA− cId )

By (2.512)‖ symA+ cId ‖ ≤ ‖A‖+ c ≤ 1 + c

Since (skwAF,F ) = 0 we have

‖(skwA− cId )F‖2 =(skwAF, skwAF )− 2c (skwAF,F )︸ ︷︷ ︸=0

+c2|F |2

≤‖A‖2 + c2 ≤ 1 + c2

where we used (2.512) in the last inequality. Now define θ by

1− θ = infc≥0

g(c), where g(c) :=

√1 + c2

ν + c(2.513)

For c ≥ 1/ν we have (c+ ν)2 > c2 + 1. Hence infc≥0 g(c) < 1 and one easilysees that the infimum is a minimum since the limits of g(c) at 0 and ∞ are≥ 1. Let c be such that g(c) = min g and set

S = A+ c Id , R = skwA− c Id . (2.514)

Then

min|F |=1

(SF, F ) ≥ (ν + c), ‖S‖ ≤ 1 + c, ‖R‖ ≤√

1 + c2 = (1− θ)(ν + c)

and the assertion follows with λ = ν + c, M = 1 + c.

To finish the proof we rewrite the PDE as

−M∆u = −div (M Id − S −R)∇u− div f

or

−∆u = −div

(Id − 1

MS − 1

MR

)∇u− 1

Mdiv f.

Let λ′ = λ/M . Then the estimate (SF, F ) ≥ λ|F |2 implies that∥∥∥∥Id − 1

MS

∥∥∥∥+

∥∥∥∥ 1

MR

∥∥∥∥ ≤ 1− λ′ + (1− θ)λ′ ≤ 1− θλ′.

Now we can conclude as in Step 1.

110 [January 28, 2019]

Theorem 2.52 (Riesz-Thorin interpolation theorem). Suppose that T is alinear map which maps measurable functions on E to measurable functionson E′. Assume that T is of type (pi, qi) for i = 0, 1, i.e. T is bounded mapfrom Lpi(E) to Lqi(E′), and

‖Tf‖Lq0 ≤ k0‖f‖Lp0‖Tf‖Lq1 ≤ k1‖f‖Lp1

Then T is of type (pt, qt) and

‖Tf‖Lqt ≤ k1−t0 k1t‖f‖Lpt

provided

1

pt= (1− t) 1

p0+ t

1

p1and

1

qt= (1− t) 1

q0+ t

1

q1

with the usual convention 1/∞ = 0.

Remark. This theorem is also known as M. Riesz’s convexity theorembecause it shows that:(i) The set of points (1

p ,1q ) for which T is of type (p, q) is a convex subset of

[0, 1]2

(ii) If ϕ(t) denotes the logarithm of the operator norm of T as an operatorfrom Lpt to Lqt then ϕ is a convex function on [0, 1].

Lemma 2.53 (Three line lemma). Suppose that F : [0, 1] × R → C isbounded and continuous and that F is holomorphic in (0, 1)× R. If

|F (iy)| ≤ m0 and |F (1 + iy)| ≤ m1 ∀y ∈ R

then|F (x+ iy)| ≤ m1−x

0 mx1 ∀x ∈ [0, 1], y ∈ R.

Thus if function k(x) := supy∈R |F (x + iy)| then x 7→ ln k(x) is a convexfunction on [0, 1].

Proof. The proof was only discussed very briefly in class.It suffices to prove the result for m0 > 0 and m1 > 0. Indeed if theresult holds in this case the assertion that mθ = 0 if m0 = 0 followsby letting m0 ↓ 0. Similarly one argues if m1 = 0. Replacing F (z)by F (z) = F (z)/(m1−z

0 mz1) we see that it suffices to consider the case

m0 = m1 = 1.Note that the real and imaginary part of F are harmonic functions.

Thus |F |2 = (ReF )2 + (ImF )2 is subharmonic. If we assume in additionthat lim|y|→∞ F (x+iy) = 0 uniformly for x ∈ [0, 1] then the assertion followsfrom the maximum principle for |F |2.

111 [January 28, 2019]

For general F we consider the functions

Fk(z) := F (z)e(z2−1)/k.

with z = x + iy. Since z2 = x2 − y2 + 2ixy it follows that |Fk(z)| ≤e(1−y2)/k sup |F | and hence we can apply the previous reasoning to Fk andget |Fk(x + iy)| ≤ 1 for all (x, y) ∈ [0, 1] × R. Let k → ∞ we obtain theassertion.

Proof of Theorem 2.52. This proof was only sketched in class.We have to show∫

E′Tf g dx ≤ k1−t

0 kt1 if ‖f‖Lpt ≤ 1 and ‖g‖Lq′t≤ 1 (2.515)

where 1q′t

= 1 − 1qt

. To do so we embed f and g in an holomorphic family

and use the three line theorem lemma. We set

fz = |f |α(z) f

|f |, gz = |g|β(z) g

|g|

where

α(z) = pt[(1− z)1

p0+ z

1

p1], β(z) = q′t[(1− z)

1

q′0+ z

1

q′1].

Then α(t) = 1 = β(t) and thus f = ft and g = gt. Moreover α(0) = pt/p0

and α(1) = pt/p1. For a > 0 we have |aiy| = | exp(iy ln a)| = 1 and we thusget

|fiy|p0 = |f |pt , |f1+iy|p1 = |f |pt

and similarly|giy|q

′0 = |g|q′t , |g1+iy|q

′1 = |g|q′t .

Set

F (z) =

∫E′Tfz gz dx

Then the conditions ‖f‖Lpt ≤ 1 and ‖g‖Lq′t≤ 1 and the assumptions on T

imply that

|F (iy)| ≤ k0 and |F (1 + iy)| ≤ k1 ∀y ∈ R.

Since ft = f and gt = g the expression on the left hand side of (2.515) isjust F (t) and the desired estimate follows from the three line lemma.

To be sure that F is really bounded on the strip [0, 1] × R on can firstconsider functions f and g which a finite linear combination of characteristicfunctions of sets with finite measure (and hence in all Lp spaces), prove(2.515) for these function and then use that these functions are dense in Lpt

and Lq′t .

112 [January 28, 2019]

2.4 A substitute for L1: maximal functions and the Hardyspace H1

The part about the Hardy-Littlewood maximal function is standard and canbe found, e.g., in [St70]. For an introduction to Hardy spaces, see Semmes’primer on Hardy spaces (Section 1 in [Se]), Chapters 3 and 4 in the bookby Stein [St93] or the original paper [FS].

Another important application of the Marcinkiewicz interpolation theo-rem are the Lp estimates for the maximal function. For f ∈ L1

loc(Rn) theHardy-Littlewood maximal function Mf is defined as

Mf(x) := supr>0

1

Ln(B(x, r))

∫B(x,r)

|f | dy. (2.516)

Note that‖Mf‖L∞ ≤ ‖f‖L∞ . (2.517)

Theorem 2.54. There exists a constant A1 such that

Ln(x ∈ Rn : |Mf(x)| > t ≤ A1

t‖f‖L1 . (2.518)

Moreover for p ∈ (1,∞) there exist Ap such that

‖Mf‖Lp ≤ Ap‖f‖Lp . (2.519)

Proof. The second estimate follows from the first estimate, (2.517) and theMarcinkiewicz interpolation theorem (note that the maps f 7→Mf is sublin-ear). The first estimate follows from Vitali covering theorem (see Analysis3 or [St70], Chapter 1, Paragraph 1).

Sketch of proof: Let At := x : Mf(x) > t. For each x ∈ At thereexists a ball B(x, rx) such that∫

B(x,rx)|f | ≥ t

2Ln(B(x, rx)). (2.520)

By the Vitali covering theorem there exist a subfamily of disjoint ballsB(xi, ri) such that

⋃iB(xi, 5ri) ⊃ At. Thus

Ln(At) ≤ 5n∑i

Ln(B(xi, ri) ≤ 5n2

t

∫⋃iB(xi,ri)

|f | dy ≤ 5n2

t‖f‖L1 .

The map f 7→ Mf is not of type (1, 1). Take for example B = B(0, 1),f = χB. Then for |x| > 1 we have B(0, 1) ⊂ B(|x|, |x| + 1) and thusMf(x) ≥ (|x|+1)−n. ThusMf /∈ L1(Rn). If we now take fk(x) = knfk(x) =knχB(0,1/k) then ‖fk‖ = 1 but ‖Mfk‖L1(B) ∼ ln k →∞. One can show that

Mf ∈ L1(B) ⇐⇒ |f | ln(2 + |f |) ∈ L1(B) if f = 0 on Rn \B, (2.521)

113 [January 28, 2019]

see [St70], Chapter 1, Paragraph 5.2.[7.12. 2018, Lecture 18]

[10.12. 2018, Lecture 19]Let

ϕ ∈ C∞c (B(0, 1)) with ϕ ≥ 0,

∫B(0,1)

ϕdx = 1, ϕ =1

ω nin B(0, 1/2)

(2.522)where

ωn := Ln(B(0, 1))

and set

ϕr(z) =1

rnϕ(zr

)(2.523)

Definition 2.55. Let f ∈ L1loc(Rn). We define the regularized maximal

function by15

f∗(x) := supr>0|(ϕt ∗ f)(x)| = sup

r>0

∫Rn

1

rnϕ

(x− yr

)f(y) dy. (2.524)

We say that f belongs to the Hardy space H1(Rn) if

f∗ ∈ L1(Rn) (2.525)

and we set‖f‖H1 := ‖f∗‖L1 (2.526)

Remark. By the Lebesgue point theorem we have limr→0(ϕr ∗ f)(x) =f(x) for a.e. x. Thus |f | ≤ f∗ a.e. and ‖f‖L1 ≤ ‖f‖H1 .

Lemma 2.56 (Elementary properties of H1).

(i) Scaling. Let ft(x) = t−nf(xt ). Then

(ft)∗ = (f∗)t and thus ‖ft‖H1 = ‖f‖H1 . (2.527)

(ii) Cancellation.

f ∈ H1(Rn) =⇒∫Rnf dx = 0. (2.528)

15Alternatively one can define a ’grand maximal function’ by additionally taking asupremum over a class of test function ϕ, e.g., ϕ ∈ T := ψ ∈ C∞c (B(0, 1)) : |∇ψ| ≤ 1.This is the original approach of Fefferman. Semmes [Se] explains why this is more natural.In the end, however, this yields the same space (with equivalent) norms as the use of asingle function ϕ.

114 [January 28, 2019]

(iii) Local cancellation. Let a ∈ Rn, a 6= 0 and suppose that 0 < ρ <|a|/20. Let

f(x) = ρ−nχB(0,ρ) − ρ−nχB(a,ρ).

Then

‖f‖H1 ∼ ln|a|ρ.

(iv) ’Atoms’. Suppose that p > 1. Then

f ∈ Lp(Rn), supp f ⊂ B(0, R),

∫f dx = 0

=⇒ f ∈ H1(Rn) and R−n‖f‖H1 ≤ C(p)R−n/p‖f‖Lp . (2.529)

Proof. (i) This is a straightforward calculation using the linear change ofvariables ξ = tx, η = ty, ρ = r/t. We have∫

Rn

1

rnϕ

(x− yr

)ft(y) dy =

∫Rn

1

rnϕ

(x− yr

)t−nf

(yt

)dy

=

∫Rn

1

tntn

rnϕ

(t(ξ − η)

r

)f(η) dη =

1

tn

∫Rn

1

ρnϕ

(ξ − ηρ

)f(η) dη

≤ 1

tnf∗(ξ) =

1

tnf∗(xt

).

Taking the supremum over r we get f∗t (x) ≤ t−nf∗(xt

). Since f = (ft)1/t

we also obtain the reverse inequality.(ii): Assume that α :=

∫Rn f dx 6= 0. We claim that then there exist an

R0 such that

f∗(x) ≥ |α|2 4nωn

1

|x|nif |x| ≥ R0. (2.530)

This immediately implies that f∗ /∈ L1(Rn).Let

ω(R) :=

∫Rn\B(0,R)

|f | dx.

Then limR→∞ ω(R) = 0 since f ∈ L1(Rn). To prove (2.530) we use r = 4|x|in the definition of the regularized maximal function. If |y| ≤ |x| thenϕ((x− y)/4|x|) = ωn and thus∣∣∣∣∫

Rnϕ

(x− y4|x|

)f(y) dy

∣∣∣∣≥

∣∣∣∣∣ 1

ωn

∫B(0,|x|)

f(y) dy

∣∣∣∣∣−∫Rn\B(0,|x|)

supϕ |f(y)| dy

≥ 1

ωn(|α| − ω(|x|))− supϕ ω(|x|).

115 [January 28, 2019]

The right hand side is ≥ |α|/(2ωn) if |x| ≥ R0. This implies (2.530).(iii): Homework. Hint: By (i) one may assume that ρ = 1. For a lower

bound on the H1 norm show that f∗(x) ≥ c|x|−n for 2 < |x| < |a|/8 by usingr = 4|x| in the definition of f∗. For the upper bound use the argument inthe proof of (iv) for large r.

(iv): By (i) and the usual scaling of the Lp norm we may assume thatR = 1. We have f∗(x) ≤ CMf(x) (with C ≤ supϕ/ωn). Thus

‖f∗‖L1(B(0,2)) ≤ C‖f∗‖Lp(B(0,2) ≤ C‖Mf‖Lp(Rn) ≤ C(p)‖f‖Lp(Rn).

Hence it suffice to estimate f∗(x) for |x| ≥ 2. In this case (ϕr ∗ f)(x) = 0 ifr ≤ |x|/2. For r > |x|/2 we have∣∣∣∣∫

Rnϕ

(x− yr

)f(y) dy

∣∣∣∣ =

∣∣∣∣∫Rn

[ϕ

(x− yr

)− ϕ

(xr

)]f(y) dy

∣∣∣∣≤∫Rn

sup |∇ϕ| |y|r|f(y)| dy ≤ C

rn+1‖f‖L1(B(0,1) ≤

2n+1C

|x|n+1‖f‖Lp(B(0,1).

Here in the first identity we used that∫Rn f(y) dy = 0 and in the second

inequality we used that |y| ≤ 1 and the set where f is not zero. Thus itfollows that f∗ ∈ L1(Rn \ (B(0, 2)) and the proof is finished.

Lemma 2.57 (Approximation in H1).

(i) For η ∈ L1(Rn) and f ∈ H1(Rn) we have

(η ∗ f)∗(x) ≤ (|η| ∗ f∗)(x). (2.531)

(ii) (Lp ∩H1)(Rn) is dense in H1(Rn) for all p ∈ (1,∞).

Remark. One can also show that (C∞c ∩H1)(Rn) is dense in H1(Rn). Inthis case the proof is easier if one defines f∗ as the ’grand maximal function’,see [Se] Proposition 1.42 and in particular Lemma 1.51. Alternatively onecan apply the standard convolution argument to finitely many terms in theatomic decompostion (introduced below).

Proof. (i) We have

|(ϕr ∗ η ∗ f)(x)| = |(η ∗ ϕr ∗ f)(x)| ≤ (|η| ∗ |ϕr ∗ f |)(x) ≤ |η| ∗ f∗(x).

Taking the supremum over r > 0 we get the assertion.(ii) Homework. Hint:

Let ηε(x) = ε−nη(x/ε) denote the usual family of mollifiers. Show first that

h(x) := lim supε→0

(ηε ∗ f − f)∗(x) = 0 a.e.

116 [January 28, 2019]

To do so note that h(x) ≤ aδ(x) + bδ(x) where

aδ(x) = lim supε→0

sup0<t≤δ

|(ηε ∗ ϕt ∗ f)(x)− (ϕt ∗ f)(x)|,

bδ(x) = lim supε→0

supt>δ|(ηε ∗ ϕt ∗ f)(x)− (ϕt ∗ f)(x)|.

Then show that bδ(x) = 0 for every x and every δ > 0 and that limδ→0 aδ(x) =0 for every Lebesgue point x, i.e., for a.e. x. Finally use (i) and a suitableversion of the Lebesgue dominated convergence theorem to conclude thatlimε→0 ‖(ηε ∗ f − f)∗‖L1 = 0.

A fundamental result of Fefferman states that the dual of H1(Rn) isBMO (Rn) and thatH1(Rn) is the dual of VMO (Rn) (’functions of vanishingmean oscillations’), the closure of Cc(Rn) in BMO (Rn). Here BMO (Rn) isunderstood as a space of equivalence classes with f ∼ g if f − g = const andwe use the BMO seminorm on this space of equivalence classes. Since theH1 functions have integral zero the expression

∫fg dx depends only on the

equivalence class of g if f ∈ H1.

Theorem 2.58 (Fefferman; H1 − BMO duality). For f ∈ H1(Rn) andg ∈ L∞(Rn) we have ∣∣∣∣∫

Rnfg dx

∣∣∣∣ ≤ C‖f‖H1 [g]BMO . (2.532)

Moreover if f ∈ L1loc then

‖f‖H1 ≤ C sup∫Rnfg dx : g ∈ C∞c (Rn), [g]BMO ≤ 1. (2.533)

Here C depends only on n (and the choice of the function ϕ).

Warning. If f ∈ H1 and g ∈ BMO then the product fg is in general notin L1.

One way to prove (2.532) is to show that each f ∈ H1 has an atomicdecomposition, i.e., f(x) =

∑k∈N αkak(x) with

∑k |αk| <∞ and each ak is

an atom, i.e.,

ak is supported in a ball Bk, |ak| ≤1

Ln(Bk)a.e.,

∫Bk

ak = 0.

The construction of the atomic decomposition is similar, but more subtle,than the construction of the Calderon-Zygmund decomposition, see [St93],

117 [January 28, 2019]

Chapter III.2, pp. 101–112. Note that if a is an atom supported on a ballB then that∫

Bag dx =

∫Ba(g − gB) dx ≤ C‖a‖L∞Ln(B)[g]BMO ≤ C[g]BMO .

Theorem 2.59 (Elliptic estimates in H1). Assume that the coefficients Aαβijare constant and strongly elliptic.

(i) Let f ∈ H1(Rn;Rn×N ). Then there exists a distributional solutionu ∈W 1,1

loc (Rn;RN ) of

−divA∇u = −div f (2.534)

with‖∇u‖H1 ≤ C‖f‖H1 . (2.535)

Moreover u is unique up to the addition of constants.

(ii) Similarly if g ∈ H1(Rn) then there exists a distributional solution u ∈W 2,1loc (Rn;RN ) of

−divA∇u = g, (2.536)

with‖∇2u‖H1 ≤ C‖g‖H1 . (2.537)

Moreover u is unique up to the addition of affine functions.

Here C only depends on n,N, ‖A‖ and the ellipticity constant ν.

Proof. (i): This follows directly from the corresponding BMO estimate inCorollary 2.16 by duality (recall that BMO = L2,n with equivalent semi-norms).

Let f ∈ (H1 ∩ L2)(Rn). Then there exist u ∈ W 1,2loc such that ∇u ∈ L2

and u is a distributional solution of −divA∇u = f , i.e.,∫RnA∇u · ∇ϕdx =

∫Rnf · ∇ϕdx ∀ϕ ∈ C∞c (Rn;RN ). (2.538)

Let (AT )αβij = Aβαji , let g ∈ C∞c (Rn). Then in particular g ∈ (BMO ∩L2)(Rn) and by Corollary 2.16 there exists a distributional solution v of−divAT∇v = −div g with ∇v ∈ (BMO ∩ L2)(Rn). Thus∫

RnAT∇v · ∇ψ dx =

∫Rng · ∇ψ dx ∀ψ ∈ C∞c (Rn;RN ). (2.539)

118 [January 28, 2019]

For future reference we note that in addition ∇v ∈ L∞(Rn). Indeed, byusing the equations for the derivatives of v we see that ∇v ∈ W k,2(Rn) forall k ∈ N. Hence by the Sobolev embedding

‖∇v‖2L∞(B(x,1)) ≤ C(k, n)‖∇v‖2Wk,2(B(x,1)) ≤ C(k, n)‖∇v‖2Wk,2(Rn)

for k > n2 and all x ∈ Rn.

Now by density (use Lemma 2.40 with q = 2) the identity in (2.342) alsoholds with ϕ = v and in (2.539) we may take ψ = u. Since

AT∇v · ∇u = A∇u · ∇v (2.540)

this yields the duality relation∫Rng · ∇u dx =

∫Rnf · ∇v dx. (2.541)

Now ∇v ∈ L∞(Rn) and we deduce from (2.532) (applied to each productfi∂vi), that ∫

Rng · ∇u dx ≤ C[f ]H1 [∇v]BMO ≤ C[f ]H1 [g]BMO

Now it follows from (2.533) that

‖∇u‖H1 ≤ C[f ]H1 ∀f ∈ (H1 ∩ L2)(Rn) (2.542)

Since (H1∩L2)(Rn) is dense in H1(Rn) (see Lemma 2.57) we easily concludeby approximation that for each f ∈ H1(Rn) there exists a distributionalsolution u ∈ W 1,1

loc (Rn) of −divA∇u = −div f such that ∇u ∈ H1(Rn) andthe bound in (2.542) holds.

To prove uniqueness it suffices to consider the case f = 0. Assumetherefore that u ∈W 1,1

loc with ∇u ∈ H1(Rn) and∫RnA∇u · ∇ϕdx = 0 ∀ϕ ∈ C∞c (Rn;RN ). (2.543)

We have in particular ∇u ∈ L1. Thus one easily sees by approximation thatthis identity holds for all ϕ ∈ C∞(Rn;RN ) with ∇ϕ ∈ L∞. Thus we maytake ϕ = v where v is the solution of the dual problem constructed above.This yields ∫

Rng · ∇u dx = 0 ∀g ∈ C∞c (Rn).

Since ∇u ∈ L1(Rn) it follows that ∇u = 0.(ii): If g ∈ (H1∩L2) then there exists a solution in u ∈W 2,2

loc with ∇2u ∈L2 and the derivative vk := ∂ku are distributional solutions of −divA∇vk =−∂kg. Thus for g ∈ (H1∩L2) the assertion follows from (i). Now the generalresult follows by approximation.

119 [January 28, 2019]

Another breakthrough came in 1990 when Coifman, Lions, Meyer andSemmes showed that certain quantities which arise naturally in PDEs arenot only in L1 but in H1.

Theorem 2.60 (Special quantities are in H1). Assume that u ∈ W 1,2(Rn)and v ∈ L2(Rn;Rn) with

div v = 0 (2.544)

in the sense of distributions. Then ∇u · v ∈ H1(Rn), with the bound

‖∇u · v‖H1 ≤ C‖∇u‖L2 ‖v‖L2 . (2.545)

Proof. This proof follows Evans [Ev]. For additional proofs and more generalstatements see the original paper [CLMS].Fix x ∈ Rn. Set ψr(y) = r−nϕ

(x−yr

). We have∫

Rn∇u · v ψr dy = −

∫B(x,r)

(u− (u)x,r) v · ∇ψr dy

because v is divergence free. Thus∣∣∣∣∫Rn∇u · v ψr dy

∣∣∣∣ ≤ C

rn+1

∫B(x,r)

|u− (ux,r)| |v| dy.

Choose any p ∈ (2, 2∗) where 2∗ = 2nn−2 and let q = p

p−1 ∈ (1, 2) be the dualexponent. Then∣∣∣∣ 1

rn

∫Rn∇u · v ψr dy

∣∣∣∣ ≤ C

rn+1

(∫B(x,r)

|u− (ux,r)|p dy

) 1p(∫

B(x,r)|v|q dy

) 1q

≤Cr

(1

Ln(B(x, r))

∫B(x,r)

|u− (ux,r)|p dy

) 1p(

1

Ln(B(x, r))

∫B(x,r)

|v|q dy

) 1q

≤ C

(1

L(B(x, r))

∫B(x,r)

|∇u|s dy

) 1s(

1

Ln(B(x, r))

∫B(x,r)

|v|q dy

) 1q

where s∗ = p, i.e., 1s = 1

p + 1n > 2. Thus s ∈ (1, 2). Thus∣∣∣∣∫

Rn∇u · v ψr dy

∣∣∣∣ ≤CM(|∇u|s)1s (x) M(|v|q)

1q (x).

This implies that the regularized maximal function satisfies

(∇u · v)∗(x) ≤ C[M(|∇u|s)2s (x) +M(|v|q)

2q (x)]

Now |∇u|s ∈ L2s and |v|q ∈ L

2q . Thus by the Lp estimate by the Hardy-

Littlewood maximal function

‖M(|∇u|s)2s ‖L

2s≤ C‖|∇u|s‖

L2s

120 [January 28, 2019]

and therefore ∫Rn

(M(|∇u|s)2s dx ≤ C

∫Rn|∇u|2 dx.

Similarly ∫Rn

(M(|v|q)2q dx ≤ C

∫Rn|v|2 dx.

It follows that ∇u · v ∈ H1(Rn) and

‖∇u · v‖H1 ≤ C(‖∇u‖2L2 + ‖v‖2L2

)Thus the proof is finished if ‖∇u‖L2 = ‖v‖L2 = 1. To obtain the full resultapply this assertion to u = u/‖∇u‖L2 and v = v/‖v‖L2 . Note that (2.545)holds trivially if ∇u = 0 or v = 0.

Historical comment: Both the invention of BMO and a precursor toTheorem 2.60 have their origins in the theory of elastiticity. The spaceBMO appears in the paper [Jo] by Fritz John. In linear elasticity onedeals with displacements u : Ω ⊂ Rn → Rn and due to linearized frameindifference the energy only depends on the symmetric part of the gra-dient sym∇u = [∇u + (∇u)T ]/2. Korn’s second inequality states thatfor a Lipschitz domain Ω one has ‖∇u − (∇u)Ω‖L2 ≤ C(Ω)‖ sym∇u‖L2 .If sym∇u ∈ L∞ this (and the natural scaling by dilation) implies that∇u ∈ BMO .Motivated by question in nonlinear elasticy in [Mu] it was shown that ifΩ ⊂ Rn, if u ∈W 1,n(Ω;Rn) and if det∇u ≥ 0 then det∇u ln(2 + det∇u) ∈L1loc(Ω) which is by logarithm better than the obvious estimate coming from| detF | ≤ C|F |n. To prove this one shows that maximal function of det∇uis in L1

loc and uses (2.521). In [CLMS] the deep connections with harmonicanalysis were revealed. In particular the result holds without any sign re-striction if one uses the regularized maximal functions and the authors deriveHardy space estimates for a very general class of ’compensated compactnessquantities’.

[10.12. 2018, Lecture 19][14.12. 2018, Lecture 20]

2.5 Regularity theory for nonlinear pde

2.5.1 Harmonic maps and Hardy space estimates

Let M be a submanifold of RN . A harmonic map is a critical point of theDirichlet integral

I(u) =1

2

∫Ω|∇u|2 dx (2.546)

121 [January 28, 2019]

among all maps u : Ω ⊂ Rn → RN with u(x) ∈M for a.e. x.To keep the notation as simple as possible we focus on the case that

M is the N − 1 dimensional unit sphere SN−1. To motivate the precisedefinition of weakly harmonic maps we derive the usual necessary first ordercondition (or Euler-Lagrange equations) for a map u which minimizes theDirichlet integral (subject to suitable boundary conditions). Thus let u bea minimizer of the Dirichlet integral among maps in W 1,2(Ω;RN ) whichsatisfy the constraint

|∇u|2 = 1 a.e. in Ω. (2.547)

To derive the Euler-Lagrange equations we usually consider variationsu+ tϕ. Here such variations are in general not admissible since they do notsatisfy the constraint |u+ tϕ| = 1 a.e. Thus we start from the relation

I(u) ≤ I(ut) where ut =u+ tϕ

|u+ tϕ|with ϕ ∈ (W 1,2

0 ∩ L∞)(Ω). (2.548)

The condition ϕ ∈ L∞ guarantees that |u+ tϕ| ≥ 12 for |t| ≤ 1

2‖ϕ‖L∞so that

ut is well defined for small t and ut ∈W 1,2(Ω;RN ). We have

∂iut =∂iu+ t∂iϕ

|u+ tϕ|− u+ tϕ

|u+ tϕ|3(u+ tϕ) · (∂iu+ t∂iϕ). (2.549)

Using that |u| = 1 and

u · ∂iu =1

2∂i|u|2 = 0 (2.550)

we deduce that

d

dt |t=0∂iut = ∂iϕ− ∂iu (u · ϕ)− u (ϕ · ∂iu)− u (u · ∂iϕ).

Thus using again (2.550) we see that t 7→ I(ut) is differentiable and

0 =d

dt |t=0I(ut) =

∫Ω

n∑i=1

∂iu · ∂iϕ− |∂iu|2u · ϕdx. (2.551)

With the Euclidean scalar product on matrices we have ∇v ·∇ϕ =∑n

i=1 ∂iv ·∂iϕ and we say that u is a weakly harmonic map to SN−1 if u ∈W 1,2(Ω;RN )with |u| = 1 a.e and∫

Ω∇u · ∇ϕ− |∇u|2u · ϕdx = 0 ∀ϕ ∈ (W 1,2

0 ∩ L∞)(Ω;RN ). (2.552)

In particular a weakly harmonic map is a distributional solution of the sys-tem of following system of PDEs

−∆u = u|∇u|2 (2.553)

122 [January 28, 2019]

Even though a lot of deep results have been obtained for geometric equa-tion of this type in the 1960’s and 70’s the question whether weakly harmonicmaps are smooth has been open until 1990. We will see the answer shortly.Let us first see why this question is delicate. A key point is that the equation(2.553) is critical in at least two ways.

(i) The equation is invariant under the natural rescaling. If u solves(2.553) and ur(z) = u(rz) then ur is a solution of the same equation.Thus the nonlinearity does not get weaker under blow-up.

(ii) The usual elliptic estimates do not give an improvement of the initialregularity u ∈ W 1,2 ∩ L∞. Indeed take n = 2. The initial regularityimplies that the right hand side is in L1. Hence the best we couldhope for is that u is in W 2,1. By the Sobolev embedding theorem thiswould give u ∈W 1,2 which is no improvement.In fact the situation is slightly worse because if v is a distributionalsolution −∆v = f with f ∈ L1 then this does not imply that∇v ∈ L2

loc

(see Homework 8). One only gets that ∇v is (locally) in weak-L2, i.e.,|∇v| > t ≤ Ct−2. For n ≥ 3 the situation is even worse.

If there is a regularity result for weakly harmonic maps it must use somespecial structure of the equation. It is not enough to use that the righthand side is a quadratic term in ∇u. Indeed, the seemingly simpler scalarequation

−∆v = |∇v|2 in B(0, 1/2) ⊂ R2

has the distributional solution

v = ln ln1

|x|

and v ∈W 1,2(B(0, 1/2)) (see Homework 9).

A general principle for critical elliptic equations is that one gets regu-larity if a suitable scale invariant quantity is small. In this spirit we firstshow the classical result that weakly harmonic maps which are continuousare already smooth.

Theorem 2.61. Let u be a weakly harmonic map from Ω to SN−1. Ifu ∈ C0(Ω;RN ) then u ∈ C∞(Ω;RN ).

Proof.

Step 1. ∇u ∈ L2,λloc (Ω;RN ) for 0 < λ < n.

Let B(x, r) ⊂ Ω. As before decompose u as u = v + w where ∆w = ∆u

123 [January 28, 2019]

and w has zero boundary conditions and ∆v = 0. More precisely we definew ∈W 1,2

0 (B(x, r)) as the unique weak solution of∫B(x,r)

∇w · ∇ϕdx =

∫B(x,r)

∇u · ∇ϕdx ∀ϕ ∈W 1,20 (B(x, r);RN ) (2.554)

(existence and uniqueness follow from the Lax-Milgram theorem) and we set

v = u− w.

Then ∫B(x,r)

∇v · ∇ϕdx = 0 ∀ϕ ∈W 1,20 (B(x, r)), (2.555)

i.e., v is weakly harmonic.If we can show that in addition w is in L∞(B(x, r);RN ) then by the

definition of weakly harmonic maps we can use ϕ = w in (2.554) and we get∫B(x,r)

|∇w|2 dy =

∫B(x,r)

u|∇u|2 · w dy ≤ supB(x,r)

|w|∫B(x,r)

|∇u|2 dy (2.556)

For this to be useful we need that supB(x,r) |w| is small. To show thiswe use the weak maximum principle (see Lemma 2.62) for each componentvi. Let Mi = maxB(x,r) u

i. Then ui −Mi ≤ 0 and thus vi −Mi = ui −Mi − wi belongs to the space W 1,2

− (B(x, r)) of functions which are ≤ 0 onthe boundary in a weak sense (see (2.563) and (2.564)). Thus vi ≤ Mi =maxB(x,r) u

i a.e. Similarly vi ≥ minB(x,r) ui. Since w = u− v it follows that

|wi| ≤ maxB(x,r)

ui − minB(x,r)

ui a.e. (2.557)

Since u is continuous it follows that for each ε > 0 there exists an r0 > 0such that ∫

B(x,r)|∇w|2 dy ≤ ε

∫B(x,r)

|∇u|2 dy (2.558)

Moreover we have by (2.75)∫B(x,ρ)

|∇v|2 dy ≤ C(ρr

)n ∫B(x,r)

|∇v|2 dy.

for all ρ ≤ r (note that for ρ ≥ r/2 the estimate holds trivially with C = 2n).Now we can combine the w and the v estimate in the usual way and use

the iteration lemma to conclude. Here are the details: assume that ε ≤ 1and set

φ(r) :=

∫B(x,r)

|∇u|2 dy.

124 [January 28, 2019]

Then ∫B(x,r)

|∇v|2 dy ≤ 2

∫B(x,r)

|∇u|2 dy + 2

∫B(x,r)

|∇w|2 dy ≤ 4φ(r)

and thus

φ(ρ) ≤2

∫B(x,ρ)

|∇v|2 dy + 2

∫B(x,ρ)

|∇w|2 dy

≤C(ρr

)n ∫B(x,r)

|∇v|2 dy + 2εφ(r)

≤4C[(ρr

)n+ ε]φ(r)

(here we assume without loss of generality that C ≥ 1). Now by Lemma2.15 for each λ < n there exists a constant C ′ which only depends on C, λ,n such that

φ(ρ) ≤ C ′(ρ

r0

)λφ(r0),

for all ρ ≤ r0. Thus ∇u ∈ L2,λloc (Ω).

Step 2. ∇u ∈ L2,µloc (Ω;RN ) for all µ < n+ 1.

By Step 1 and the Poincare inequality we have u ∈ Lλ+2loc (Ω;RN ) for all

λ < n. Hence u ∈ C0,αloc (Ω;RN ) for all α < 1. Let n < µ < n + 1 and let

α ∈ (0, 1) be such that λ := µ− α < n.Define v and w as in Step 1 and set

φ(r) :=

∫B(x,r)

|∇u− (∇u)x,r|2 dy.

The Holder continuity of u and (2.557) imply that

sup |wi| ≤ Crα. (2.559)

Using that u ∈ L2,λ(B(x, r);RN ) and (2.556) we get∫B(x,r)

|∇w|2 dy ≤ Crα rλ = Crµ.

By (2.76) we have∫B(x,ρ)

|∇v − (∇v)x,ρ|2 dy ≤ C(ρr

)n+2∫B(x,r)

|∇v − (∇v)x,r|2 dy.

Combining the v and the w estimates as above and using that∫B(x,ρ)

|∇w − (∇w)x,ρ|2 dy ≤∫B(x,ρ)

|∇w|2 dy

125 [January 28, 2019]

we get

φ(ρ) ≤ A(ρr

)n+2φ(r) +Brµ.

Since µ < n+ 2 it follows from Lemma 2.15 that there exists a constant Cwhich only depends on A, B, µ and n such that

φ(ρ) ≤ C(ρr

)µφ(r)

for all ρ ≤ r. Thus ∇u ∈ L2,µloc (Ω).

Step 3. u ∈ C∞(Ω;RN ).

By Step 2 and Theorem 2.12 we have ∇u ∈ C0,βloc (Ω) for all β < 1

2 . We show

by induction that ∇u ∈ Ck,βloc (Ω;RN ) for all k ∈ N. Indeed the Schaudertheory gives

∇u ∈ Ck,βloc =⇒ u|∇u|2 ∈ Ck,βloc =⇒ u ∈ Ck+2,βloc =⇒ ∇u ∈ Ck+1,β

loc .

In the argument above we have used the following form of the maximumprinciple for weak subsolutions. Recall that for a ∈ R we use the notationa+ = max(a, 0).

Lemma 2.62 (Weak maximum principle). Assume that Ω ⊂ Rn is boundedand open that the coefficients aij ∈ L∞(Ω) are elliptic. Define

W 1,2− (Ω) := v ∈W 1,2(Ω) : v+ ∈W 1,2

0 (Ω). (2.560)

If v ∈W 1,2− (Ω) is weakly subharmonic, i.e., if∫Ω

∑i,j

aij(x)∂jv ∂iϕdx ≤ 0 ∀ϕ ∈W 1,20 (Ω) with ϕ ≥ 0 a.e.

thenv ≤ 0 a.e.

Proof. See Homework 7. Hint: Take ϕ = v+ and use that ∇v+ = χv>0∇va.e. (see Functional Analysis and PDE or [GT]). Then

ν

∫Ω|∇v+|2 dx ≤

∫Ωaij∂jv

+ ∂iv+ =

∫Ωaij∂jv ∂iv

+ ≤ 0.

Hence ∇v+ = 0. Since v+ ∈ W 1,20 (Ω) by assumption it follows that v+ = 0

a.e.

126 [January 28, 2019]

Remark. The space W 1,2− (Ω) should be viewed as the space of function

which satisfy the condition v ≤ 0 on ∂Ω in a weak sense. One has thefollowing characterization (see Homework 7).

v ∈W 1,2− (Ω) ⇐⇒

∃ Kk ⊂ Ω compact, vk → v in W 1,2(Ω), vk ≤ 0 in Ω \Kk (2.561)

and from this one easily derives that (see Homework 7)

v ∈W 1,2(Ω) ∩ C0(Ω), v ≤ 0 on ∂Ω =⇒ v ∈W 1,2− (Ω), (2.562)

and

W 1,20 (Ω) ⊂ W 1,2

− (Ω), (2.563)

v, w ∈W 1,2− (Ω)− =⇒ v + w ∈W 1,2

− (Ω). (2.564)

We now come to the heart of the matter. The key observation is thatsince 2u · ∂ku = ∂k|u|2 = 0 the nonlinearity can be written in the form

ui|∇u|2 =∑j

(ui∇uj − uj∇ui) · ∇uj (2.565)

and that

div (ui∇uj − uj∇ui) = 0 in the sense of distributions.

Thus the nonlinearity is not only in L1 but in (a suitable local version of)the Hardy space H1. From this we will easily conclude that u is continuousand hence smooth.

Lemma 2.63. Assume that u is a weakly harmonic map to SN−1. Thenfor each i, j ∈ 1, . . . , N the vectorfield ui∇uj −uj∇ui is divergence free inthe sense of distributions, i.e.,∫

Ω(ui∇uj − uj∇ui) · ∇η dx = 0 ∀ η ∈ C∞c (Ω).

Proof. Let ei be the i-th unit vector and use the test function ϕ = ujη eiin the definition of a weakly harmonic map. Since ∇(ujη) = ∇uj η + uj∇ηthis yields∫

Ω∇ui · ∇uj η dx+

∫Ωuj∇ui · ∇η dx =

∫Ωui|∇u|2 ujη dx.

Exchanging i and j we also get∫Ω∇uj · ∇ui η dx+

∫Ωui∇uj · ∇η dx =

∫Ωuj |∇u|2 uiη dx.

The assertion follows by substracting the two equations.

127 [January 28, 2019]

Lemma 2.64 (Localization of H1 − BMO estimates). Let Br = B(x, r) ⊂Rn.

(i) Assume thatb ∈ L2(B2r;Rn), div b = 0

in the sense of distributions in B2r. Then there exists an extension

b ∈ L2(Rn;Rn), b = b in Br, b = 0 in Rn \B2r

‖b‖L2 ≤ C‖b‖L2 , div b = 0

in the sense of distributions in Rn. Here C depends only on n.

(ii) Suppose that a ∈ W 1,2(B2r), b ∈ L2(B2r;Rn) and div b = 0 in thesense of distributions. Then there exists f ∈ H1(Rn) such that

f = b · ∇a in Br, ‖f‖H1 ≤ C‖∇a‖L2(B2r) ‖b‖L2(B2r),

where C only depends on Rn.

(iii) Suppose that a ∈W 1,20 (Br), b ∈ L2(B2r;Rn), g ∈ L2,n(B2r)∩L∞(B2r)

with div b = 0 in the sense of distributions. Then∫Br

b · ∇a g dx ≤ C‖∇a‖L2(Br) ‖b‖L2(B2r) [g]L2,n(B2r) (2.566)

Remark. In part (i) it is important that the ball is simply connected.The function b(x) = x

|x|2 is smooth and divergence free in the annulus A =

B(0, 2) \ B(0, 1) ⊂ R2 and it is not difficult to show that there exists nodivergence free extension b ∈ L2(R2;R2) (Exercise. Hint: first assume thatb is smooth and show that

∫∂B(0,R) b·ν dH

1 = 2π and use Jensen’s inequality.

Then approximate).

Proof. By scaling and we easily see that it suffices to consider the case r = 1.(i): Let ϕ ∈ C∞c (B2) with ϕ = 1 on B1 and |∇ϕ| ≤ 2. Assume first in

addition that b is C1 on suppϕ. Then div (ϕb) = b · ∇ϕ in B2 \ B1. Let Ube the weak solution of the Neumann problem

∆U = b · ∇ϕ in B2 \B1,

∂U

∂ν= 0 on ∂(B2 \B1),

i.e., ∫B2\B1

∇U · ∇ψ dx = −∫B2\B1

b · ϕψ dx ∀ψ ∈W 1,2(B2 \B1).

128 [January 28, 2019]

Note that since div b = 0 we have∫B2\B1

b · ∇ϕdx =

∫∂(B2\B1)

b · ν ϕ dH1 = −∫∂B1

b · ν dH1 =

∫B1

div b = 0.

Here we used that on ∂B1 the outer normal of B2 \ B1 is the inner normalof B1. Thus the Neumann problem has a solution, the solution is unique upto the addition of constants, and using the test function ψ = U we get∫B2\B1

|∇U |2 dx = −∫B2\B1

b · ϕ(U − (U)B2\B1

)dx ≤ 2CP ‖b‖L2‖∇U‖L2 ,

where CP is the Poincare constant of B2 \ B1. Thus ‖∇U‖L2 ≤ C‖b‖L2 .Finally extend h = ∇U by zero from B2 \B1 to Rn, extend ϕb by zero fromB2 to Rn and set

b = ϕb− h.

Then ‖b‖L2 ≤ C‖b‖L2 and b is divergence free in the sense of distributions.Indeed for every η ∈ C∞c (Rn) we have∫

Rnϕb · ∇ηdx = −

∫B2

div (ϕb) dx = −∫B2\B1

b · ϕdx,∫Rnh∇η dx =

∫B2\B1

∇U · ∇η = −∫B2\B1

b · ϕdx

and the assertion follows by substracting the two equations.If b is only in L2 we can consider approximations bk = ρk ∗ b where

ρk(x) = knρ(kx) are the usual mollifying kernels. For large enough k themaps bk are smooth on suppϕ. Define bk as before. Then ‖bk − bl‖L2 ≤C‖bk − bl‖L2 . Thus bk is a Cauchy sequence in L2 and the limit b has thedesired properties.

[14.12. 2018, Lecture 20][17.12. 2018, Lecture 21]

(ii): Let b and ϕ as in (i), let a = ϕa and extend a by zero from B2 toRn. Then

b · ∇a = b · ∇a, in B1

We may asssume that∫B2a dx = 0 since the quantity we are interested

in and the right hand side of the estimate depend only on ∇a. Thus by thePoincare inequality

‖∇a‖L2(Rn) ≤ C‖a‖W 1,2(B2) ≤ C‖∇a‖L2(B2)

Now by Theorem 2.60 the function f = ∇a · b belongs to H1 and

‖∇a · b‖H1 ≤ C‖∇a‖L2 ‖b‖L2 ≤ C‖∇a‖L2(B2) ‖b‖L2(B2).

129 [January 28, 2019]

(iii): Let b be as in (i), let a denote the extension of a by zero to fromB1 to Rn and let g = ϕg where ϕ is as in (i) and extend g by zero from B2

to Rn. Then ∫B1

b · ∇a g dx =

∫Rnb · ∇a g dx

We may assume that g has average zero on B2 since∫B1

b · ∇a dx =

∫B1

b · ∇a dx = 0.

Then it is easy to show that there exists a constant C (which only dependson n) such that

[ϕg]L2,n(Rn) ≤ C[g]L2,n(B2)

Now the assertion follows from Theorem 2.60 and the H1 − BMO dualityapplied to a, b and g and the equivalence of the BMO seminorm and theL2,n seminorm on Rn.

Theorem 2.65. (Helein) Let u : Ω ⊂ R2 → RN be a weakly harmonic mapto SN−1. Then u ∈ C∞(Ω).

Proof. In view of Theorem 2.61 it suffices to show that u is continuous. Forthis it suffices to show the following: if B(x, 2r) ⊂ Ω then u is continuousin B(x, r).

By (2.565) the component ui is a weak solution of

−∆ui =∑j

(ui∇uj − uj∇ui) · ∇uj

and by Lemma 2.63div (ui∇uj − uj∇ui) = 0

in the sense of distributions in Ω. Assume that B(x, 2r) ⊂ Ω. Then byLemma 2.64 (ii) there exist functions fj ∈ H1(R2) such that

−∆ui =∑j

fij in B(x, r)

By Theorem 2.59 there exists U i ∈W 2,1loc (R2) such that

−∆U i =∑j

fij in Rn

By Lemma 2.66 function in W 2,1loc (R2) are continuous. Finally ui − U i is

weakly harmonic in B(x, r) and hence smooth. Thus ui is continuous inB(x, r).

130 [January 28, 2019]

Lemma 2.66. Suppose that u ∈ W 2,1loc (R2) and ∇2u ∈ L1(R2). Then u is

continuous and there exist constants a ∈ R and b ∈ R2 such that

supx∈Rn

|u(x)− a− b · x| ≤∫R2

|∂1∂2u| dy (2.567)

Proof. See Homework 8. Hint: First show that the assertion holds for v ∈W 2,1(R2). To see this note that for v ∈ C∞c (R2)

v(x) =

∫ x1

−∞

∫ x2

−∞∂1∂2v(y1, y2) dy2 dy1.

and argue by density. To show that u is continuous at x consider a cut-off function ψ ∈ Cc∞(Rn) which is 1 in a neighbourhood of x. Then ψu ∈W 2,1(Rn) and hence ψu is continuous. Thus u is continuous at x. To provethe full result argue in a similar way as in Lemma 2.40 to approximate u byfunctions uk ∈W 2,1(R2) such that ∇2uk → ∇2u.

We now consider the regularity of weakly harmonic maps Ω ⊂ Rn →SN−1 for n ≥ 3. In this case singularities can arise. Indeed

u(x) :=x

|x|is a weakly harmonic map

from B(0, 1) ⊂ Rn to Sn−1 for all n ≥ 3.

This can be seen by a straightforward calculation (see Homework 9). Amuch more subtle argument shows that this map u is actually a minimizerof the Dirichlet integral among all W 1,2 maps v with v = u = id on ∂B(0, 1)and |v(x)| = 1 a.e. (see [?]). In general the situation is much worse.

Theorem 2.67 (Riviere). There exists a weakly harmonic map from B(0, 1) ⊂R3 to S2 for which the set of points of discontinuity is dense.

Proof. See [Ri95].

At this point we could give up. Alternatively we can argue that thenotion of a weakly harmonic map is simply too weak to select ’good’ har-monic maps. In fact one important feature of the harmonic map equationis invariance an rescaling of the independent variable: if u is a solution andut(x) = u(x/(1 + t)) is also a solution.

This suggests to consider inner variations which are obtained by varia-tions of the independent variable. Let

ut = u Φt, i.e., ut(x) = u(Φt(x)).

131 [January 28, 2019]

Here Φ0 = Id and Φt is a smooth one-parameter family of diffeomorphismsfrom Ω to itself, i.e.,

(x, t) 7→ Φt(x) is a C∞ map and

Φt : Ω→ Ω is a diffeomorphism for each t ∈ (−δ, δ).

We set

Ψt = Φ−1t and η(x) =

d

dt |t=0Ψt(x).

In the following we assume that the first and second derivatives of Φt andΨt in x and t are uniformly bounded.

The additional condition we want to impose on u is that

d

dt |t=0

∫Ω|∇ut|2 dx = 0.

To express this in a simpler form we compute

∇ut(x) = ∇ut(Φt(x))∇Φt(x)

and we use the change of variable y = Φt(x) and thus x = Ψt(y) and thechain rule ∇Ψ(y)∇Φ(x) = Id to get∫

Ω|∇ut(x)|2 dx =

∫Ω∇u(y)(∇Ψt(y))−1 · ∇u(y)(∇Ψt(y))−1 det∇Ψt dy.

Now if t 7→ A(t) is a C1 map into the space of invertible n× n matrices wehave

d

dtA−1(t) = −A−1(t)

d

dtA(t)A−1(t),

d

dtdetA(t) = detA(t) trA−1(t)

d

dtA(t)

Since ∇Ψ0 = Id we get

d

dt |t=0

∫Ω|∇ut(x)|2 dx =

∫Ω−(∇u∇η) ·∇u − ∇u · (∇u∇η)+ |∇u|2Id ·∇η dy

Now we use the following relation16 for n× n matrices F and G

FG · F = F TF ·G.

This yields

d

dt |t=0

∫Ω|∇ut(x)|2 dx =

∫Ω

(−2(∇u)T∇u+ |∇u|2Id

)· ∇η dy

16Proof:∑i,β,α F

iβG

βα F

iα =

∑i,β,α(FT )βi F

iα G

βα

132 [January 28, 2019]

Finally we note that if η ∈ C∞c (Ω;Rn) and if we define Ψt(x) as thesolution of the ODE

d

dtΨt(x) = η(Ψt(x)), Ψ0(x) = x

then Ψt is a smooth one parameter family of diffeomorphisms and Ψt(x) = xon Ω\ supp η and thus Ψ maps Ω to itself and is the identity in a neighbour-hood of ∂Ω.

Definition 2.68 (Stationary harmonic map). A weakly harmonic map iscalled a stationary harmonic map if∫

Ω

(2(∇u)T∇u− |∇u|2Id

)· ∇η dx = 0 ∀η ∈ C∞c (Ω;Rn). (2.568)

Here [(∇u)T∇u

]αβ

=N∑i=1

∂αui ∂βu

i = ∂αu · ∂βu

and Id denotes the identity map on Rn.

Remark. If u is sufficiently regular (e.g. W 2,2 ∩ W 1,∞) then (2.568)follows from the fact that u is weakly harmonic. It suffices to use the testfunction ϕ = η · ∇u =

∑α η

α∂αu. If u is only in W 1,2 ∩ L∞, however, ingeneral ϕ is not an admissible test function because the assumptions do notsuffice to ensure that ϕ ∈W 1,2.

Lemma 2.69 (Monotonicity formula). Assume that B(x,R) ⊂ Ω and thatu ∈W 1,2(Ω;RN ) satisfies (2.568). Then

1

ρn−2

∫B(x,ρ)

|∇u|2 dy ≤ 1

rn−2

∫B(x,r)

|∇u|2 dy ∀ 0 < ρ < r ≤ R.

Proof. For simplicity of notation assume that x = 0 and write Br = B(0, r).First note that by Fubini’s theorem (and transformation to polar coor-

dinates) the maps r 7→∫B(x,r) |∇u|

2 dx is absolutely continuous (i.e., the

primitive of an L1 function) and

d

dr

∫Br

|∇u|2 dx =

∫∂Br

|∇u|2 dx for a.e. r < R

Hence it suffices to show that

d

dr

(r2−n

∫Br

|∇u|2 dx)

=r1−n∫Br

(2− n)|∇u|2 dx+ r2−n∫∂Br

|∇u|2 dHn−1 ≥ 0 for a.e. r < R.

(2.569)

133 [January 28, 2019]

Now observe that by density (2.568) holds for all η ∈W 1,∞0 (Ω;RN ).17

Let 0 < s < r and apply (2.568) with

η(x) = ψ(|x|)x

with

ψ(t) =

1 if t < s,r−tr−s if s ≤ t ≤ r,0 if t > r.

Then ∇η(x) = Id for |x| < s and ∇η = 0 for |x| > r. Moreover

∇η(x) = ψ(|x|) Id − 1

r − sx⊗ x

|x|if s ≤ |x| ≤ r.

Thus (2.568) and the identities F TF · Id = |F |2 and F TF · z ⊗ z = |Fz|2imply that18

0 =

∫Br

ψ(|x|)(2− n)|∇u|2 − 1

r − s

∫Br\Bs

|x|

(2

∣∣∣∣∇u x

|x|

∣∣∣∣2 − |∇u|2)dx.

Taking the limit s ↑ r we get, for a.e. r < R,

0 =

∫Br

(2− n)|∇u|2 +

∫∂Br

r

(−2

∣∣∣∣∇u x|x|∣∣∣∣2 + |∇u|2

)dx

≤∫Br

(2− n)|∇u|2 + r

∫∂Br

|∇u|2 dx

Multiplying by r1−n we see that (2.569) holds.

[17.12. 2018, Lecture 21][21.12. 2018, Lecture 22]

Lemma 2.70 (ε-regularity lemma). Let n ≥ 3. There exists ε0 > 0 (whichonly depends on n) with the following property. If u is a stationary harmonicmap from Ω to SN−1, if B(a, 4R) ⊂ Ω and if

1

(4R)n−2

∫B(a,4R)

|∇u|2 dy ≤ ε20 (2.570)

thenu ∈ C∞(B(a,R);RN ).

Proof.

17Note that C∞c (Ω;RN ) is not dense in W 1,∞0 (Ω;RN ), but for each η ∈ W 1,∞

0 (Ω;RN )there exist ηk ∈ C∞c (Ω;RN ) such that ‖ηk‖W1,∞ is uniformly bounded and ∇ηk → ∇ηa.e. By the Lebesgue dominated convergence theorem this is enough to pass to the limitin (2.568).

18Proof of the second identity: FTF ·z⊗z =∑i,α,β F

iαF

iβzαzβ =

∑i(Fz)

i(Fz)i = |Fz|2

134 [January 28, 2019]

Step 1. Let B′ = B(a, 3R). Then u ∈ L2,n(B′,RN ) and [u]L2,n(B′) ≤ Cε0

where C depends only on n.Let x ∈ B′, 0 < r ≤ R. Then B(x, r) ⊂ B(x,R) ⊂ B(a, 4R). By (2.89), thePoincare inequality, the monotonicity formula and the inclusion B(x,R) ⊂B(a, 4R) we get

1

rn

∫B(x,r)∩B′

|u− uB(x,r)∩B′ |2 dy

≤ 1

rn

∫B(x,r)

|u− uB(x,r)|2 dy

≤ CPrn−2

∫B(x,r)

|∇u|2 dy

≤ CPRn−2

∫B(x,R)

|∇u|2 dy ≤ 4n−2CP ε20

where CP denotes the constant in the Poincare estimate for the unit ball.For the r ≥ R use (2.89) and the Poincare inequality to obtain

1

rn

∫B(x,r)∩B′

|u− uB(x,r)∩B′ |2 dy

1

rn

∫B(a,4R)

|u− uB(a,4R)|2 dy

1

Rn(4R)2CP

∫B(a,4R)

|∇u|2 dy ≤ 4nCP ε20.

Step 2. ∇u ∈ L2,λ(B(a,R);RN×n) for all λ < n.This implies that u ∈ L2,λ+2(B(a,R);RN ) for all λ < n. Hence u is Holdercontinuous in B(a,R) and thus C∞ by Theorem 2.61.

The proof is very similar to Step 1 in the proof of Theorem 2.61. Themain point is that we use the local H1 − BMO estimate (2.566) to replacesmallness of w in L∞ by smallness in L2,n.

Let x ∈ B(a,R). Then B(x, 2R) ⊂ B′ = B(a, 3R). Let Br := B(x, r)and

φ(r) :=

∫Br

|∇u|2 dy. (2.571)

As before write u = v + w where w ∈W 1,20 (B(x, r),RN ) is the solution of∫

B(x,r)∇w · ∇ϕdy =

∫B(x,r)

∇u · ∇ϕdy ∀ϕ ∈W 1,20 (B(x, r),RN ).

Then v is weakly harmonic. Moreover |u| = 1 and hence the weak maximumprinciple shows that |vi| ≤ 1 in B(x, r). Thus w is in L∞ and taking ϕ = w

135 [January 28, 2019]

and using the definition of weakly harmonic maps we get

‖∇w‖2L2(Br)=

∫Br

u|∇u|2 · w dy =∑i,j

∫Br

(ui∇uj − uj∇ui

)· ∇ujwi dy

=−∑i,j

∫Br

(ui∇uj − uj∇ui

)· ∇wiuj dy

where in the last step we used that ∇ujwi = ∇(ujwi) − ∇wiuj and thediv ui∇uj − uj∇ui = 0. Now we apply (2.566) to each term in the last sumand using that |uk| ≤ 1 we get

‖∇w‖2L2(Br)≤ CN2‖∇u‖L2(B2r)‖∇w‖L2(Br)[u]L2,n(B2r).

By Step 1 (and (2.90)) it follows that∫Br

|∇w|2 dy ≤ Cε20

∫B2r

|∇u|2 dy = Cε20φ(2r) (2.572)

where C depends on n and N . Together with the usual estimate∫Bρ

|∇v|2 dy ≤ C(ρr

)n ∫Br

|∇v|2 dy

we deduce that

φ(ρ) ≤ C(( ρ

2r

)n+ ε2

0

)φ(2r) ∀0 < ρ ≤ r ≤ R.

Replacing C by 2nC we see that this estimate also holds for ρ ≤ 2r. Thus itfollows from the iteration lemma (Lemma 2.15) (with r replaced by 2r)that

φ(ρ) ≤ C ′( ρ

2R

)λφ(2R)

provided that Cε20 is sufficiently small. Here C ′ depends on λ, n and C.

Now

φ(2R) =

∫B(x,2R)

|∇u|2 dy ≤∫B(a,4R)

|∇u|2 dy

Since the right hand side is independent of the point x ∈ B(a,R) it followsthat ∇ ∈ L2,λ(B(a,R)).

Theorem 2.71 (Partial regularity of stationary harmonic maps for n ≥ 3).Let n ≥ 3 and let u : Ω ⊂ Rn → RN be a stationary harmonic map to SN−1.Then there exists a closed set S such

u ∈ C∞(Ω \ S;RN ) and Hn−2(S) = 0 (2.573)

where Hs denotes the s dimensional Hausdorff measure.

136 [January 28, 2019]

Proof. Let ε0 be as in Lemma 2.70 and define the set U by

U := x ∈ Ω : lim supr→0

r2−n∫B(x,r)

|∇u|2 < ε20.

We claim that U is open. Let a ∈ U . Then there exists an r0 > 0 such thatB(a, r0) ⊂ Ω and

r2−n0

∫B(a,r0)

|∇u|2 dy < ε20.

Thus by the ε-regularity Lemma u ∈ C∞(B(a, r0/4);RN ). In particular wehave for every x ∈ B(a, r0/4)

lim supr→0

r2−n∫B(x,r)

|∇u|2 dy = 0.

Thus B(a, r0/4) ⊂ U . Hence U is open and u ∈ C∞(U ;RN ). Finally letS = Ω \U . Then S is closed and by Lemma 2.72 we have Hn−2(S) = 0.

Lemma 2.72. Let Ω ⊂ Rn be open and let f ∈ L1(Ω).

(i)

Ln(x ∈ Ω : lim supr→0

r−n∫B(x,r)

|f | dy =∞) = 0

(ii) Let 0 < α < n and let ε > 0. Then

Hα(x ∈ Ω : lim supr→0

r−α∫B(x,r)

|f | ≥ ε) = 0

where Hα denotes the α-dimensional Hausdorff measure.

Proof. (i) This follows from the Lebesgue point theorem.(ii) Let

S = x ∈ Ω : lim supr→0

r−α∫B(x,r)

|f | ≥ ε).

By (i) we have Ln(S) = 0. Let σ > 0. We claim that there exists U ⊂ Ωsuch that

S ⊂ U, U open,

∫U|f | dy < σ.

Indeed there exists η > 0 such that for every set E ⊂ Ω with Ln(E) < η wehave

∫E |f | dy < σ. Since Ln(S) = 0 there exists an open set U ⊃ S with

Ln(U) < η.For δ > 0 consider the following family of balls

Fδ = B(x, r) : x ∈ S, 0 < r < δ, B(x, r) ⊂ U,∫B(x,r)

|f | dy > 1

2εrα.

137 [January 28, 2019]

Then Fδ is a fine cover of S, i.e., for every x ∈ S and every r0 there existsB(x, r) ∈ Fδ with r < r0. By the Vitali covering theorem there existcountable many disjoint balls Bi = B(xi, ri) ∈ Fδ such that

S ⊂∞⋃i=1

B(xi, 5ri).

Note that diamB(xi, 5ri) ≤ 10ri ≤ 10δ. Thus the Hausdorff premeasure ofS satisfies

Hα10δ(S) ≤∑

C(α)(5ri)α ≤ C(α)5α

ε

∞∑i=1

∫Bi

|f | dy

≤ C(α)5α

ε

∫U|f | dy ≤ C(α)5α

εσ.

Taking δ → 0 we get

Hα(S) ≤ C(α)5α

εσ

Since σ > 0 was arbitrary this yields the assertion.

The role of symmetries A key ingredient in the proof of (partial) reg-ulartiy of stationary harmonic maps is the identity

div (ui∇uj − uj∇ui) = 0.

in Lemma 2.63. The proof was short but gave little insight why such anestimate should be true.

In fact this identity follows from Noether’s theorem that every symmetryof a variational problem leads to a conservation law (i.e., the statement thatthe divergence of a suitable quantity vanishes). Here the symmetry is therotational symmetry of SN−1 or more precisely the invariance of SN−1 andthe Dirichlet integral under the action of SO(N). If u : Ω → SN−1 and ifQ ∈ SO(N) then Qu : Ω→ SN−1 and∫

Ω|∇(Qu)|2 dx =

∫Ω|∇u|2 dx.

We now look for the infinitesimal version of this symmetry. Considervariations of the form

ut = Q(t)u, where Q(t) ∈ SO(N) and Q(0) = Id .

Set W := Q′(0). Then W is skew-symmetric (i.e, Wji = −Wij) and

d

dt t=0ut = Wu

138 [January 28, 2019]

This motivates to consider more generally variations of the form

ut(x) = u(x) + tW (x)u(x) where W ∈ C∞c (Ω;RN×N ) and Wji = −Wij .

Since u · (Wu) = 0 we get

|ut|2 = |u(x)|2 + t2|Wu(x)|2 = 1 + t2|Wu(x)|2.

Thus we should really divide ut by 1 + t2|Wu(x)|2 but for calculating firstorder necessary conditions the contributions from this term will drop out.Thus we get the following necessary first order condition for a minimizer u

0 =

∫Ω

∑k

(∂ku · ∂k(Wu) ) dx

=

∫Ω

∑k

(∂ku ·W∂ku) dx+

∫Ω

∑k

(∂ku · (∂kW )u) dx

We have (∂ku ·W∂ku) = 0 since W is skew symmetric. Using the skewsymmetry of W the other term can be rewritten as

∂ku · (∂kW )u =∑i,j

∂kuj (∂kWji)u

i =∑i<j

(ui∂kuj − uj∂ui)∂kWji.

Taking the sum over k we thus get

0 =

∫Ω

∑i<j

(ui∇uj − uj∇ui) · ∇Wji.

Since the n(n−1)2 functions Wji ∈ C∞c (Ω) which correspond to indices i < j

can be chosen independently we thus get div (ui∇uj − uj∇ui) = 0 for alli < j (and hence by symmetry for all i, j).

Similarly the monotonicity formula is related to the symmetry underrescaling. Set

ut(x) = ut(x

1 + t).

Thend

dtut(x) = −x · ∇u(x) = −

∑α

xα∂αu.

If u is sufficiently regular one obtains the monotonicity formula multiplyingthe strong form of the equation by x · ∇u(x) or equivalently by using thetest function ϕ(x) = ψ(|x|) x · ∇u(x), where ψ is an approximation of thecharacteristic function of (0, r).

139 [January 28, 2019]

2.5.2 Harmonic maps with general targets

This subsection was not discussed in classOverview: ff the target is SN−1 we can exploit the symmetry which

comes from the key symmetry is the isometric action of SO(N) on SN−1.For a general N − 1 dimensional submanifold of RN we cannot expect tohave such a global action. Instead we will expression the equation in thesuitable orthonormal basis of the tangent space. The choice of this basisis not canonical. Each orthogonal matrix induced a change of basis. This(gauge) action of SO(N−1) is the fundamental symmetry action in Helein’soptimal frame method. The key idea is that a good choice of orthogonalbasis (or ’frame’) leads to a particularly nice form of the equation. Later,for n ≥ 3 Riviere and Struwe found an even more natural way to exploit theSO(N − 1) action on the tangent space, see Theorem 2.75 below and thediscussion preceeding it.

We now describe the optimal frame method in more detail. Let M ⊂ RNbe a smooth N − 1 dimensional submanifold of RN . Let Ω ⊂ Rn be open.We look for critical points of the Dirichlet integral∫

Ω|∇u|2 dx

among maps u : Ω→ RN which satisfy u(x) ∈M a.e.To define admissible variations we consider the closest point projection

π : U →M which is defined on a sufficiently small neighbourhood U of M .We consider the variations

ut := π (u+ tϕ).

Thend

dt |t=0ut = Dπ(u)ϕ = ϕ− (ϕ · ν u) ν u,

where ν(p) denotes for normal at a point p ∈ M . Thus u is a weaklyharmonic map to M if∫

Ω∇u ·∇[ϕ− (ϕ ·ν u) ν u] dx = 0 ∀ϕ ∈ (W 1,2

0 ∩L∞)(Ω;Rm). (2.574)

One easily sees that equivalently u is a weakly harmonic map if∫Ω∇u·∇ψ dx = 0 ∀ψ ∈ (W 1,2

0 ∩L∞)(Ω;Rm) with ψ(x) ∈ Tu(x)M a.e.,

(2.575)where TpM denotes the tangent space of M at p. Indeed (2.574) implies(2.575) since for the ψ in (2.575) we have ψ · ν = 0 and hence we cantake ϕ = ψ in (2.574). For the converse implication it suffices to note that

140 [January 28, 2019]

ν u ∈ W 1,2 ∩ L∞, that ϕ − (ϕ · ν u) ν u ∈ Tu(x)M a.e. and thatf, g ∈W 1,2 ∩ L∞ implies that fg ∈W 1,2 ∩ L∞.

Now in contrast to the sphere which is invariant under rotations a generalsubmanifold M has in general no global symmetries. Hence we can not hopeto obtain a conservation law like div (ui∇uj − uj∇ui) = 0 in a obvious wayfrom global symmetries.

Helein realized that one can instead use local (or infinitesimal) symme-tries of the tangent space of M . To explain his idea we assume for simplicitythat the manifoldM is parallelizable, i.e., that there existN−1 smooth fieldsεi : M → RN such that for each p ∈M the vectors ε1(p), . . . , εN−1(p) are anorthonormal basis of the tangent space TpM . Define the maps ei : Ω→ RNby

ei(x) = εi(u(x)) (2.576)

Then ei ∈W 1,2∩L∞ and for a.e. x ∈ Ω the vectors e1(x), . . . , eN−1 form anorthonormal basis of Tu(x). For brevity we call e1, . . . , em−1 an orthonormalframe of the M (more precisely these vectors form an orthonormal frame ofthe pull-back bundle u∗(TM)). Using test functions of the form ψ = ηieiwe easily see that (2.575) is the weak form of the equation

∆u · ei = 0 for i = 1, . . . , N − 1. (2.577)

Since we are interested in local properties we may assume that

and Ω is a ball B

To express this equation in a more convenient form we assume from now on

n = 2 (2.578)

and expand all tangent vector fields in the basis e1, . . . eN−1. Since ∂αu(x) ∈Tu(x)M we have

∂αu =N−1∑j=1

(∂αu · ej)ej . (2.579)

We want to derive a system of PDE for the coefficients ∂αu · ei. First, bythe product rule,

∂β(∂αu · ei) = (∂β∂αu · ei) + (∂αu · ∂βei) = (∂β∂αu · ei) + (∂αu · ej)(ej · ∂βei).

We now introduce the abbreviations

∇u · ei :=

(∂1u · ei∂2u · ei

), ωij := ∇ei · ej :=

(∂1ei · ej∂2ei · ej

)With the notation we get

div (∇u · ei) = (∆u · ei)︸ ︷︷ ︸=0

+∑j

ωij · (∇u · ej) (2.580)

141 [January 28, 2019]

where the · after ωij denotes the scalar product in R2. Recall that forv : Ω ⊂ R2 → R2 we define

curl v = ∂2v1 − ∂1v2, v⊥ =

(−v2

v1

).

Thus

curl (∇u · ei) = (∂1∂2u− ∂2∂1u · ei)︸ ︷︷ ︸=0

+∑j

ω⊥ij · (∇u · ej). (2.581)

Now (2.580) and (2.581) is a first order system for the 2(N − 1) coefficients∂αu · ei. It is convenient to rewrite this as a second order system using theHodge (or Helmholtz) decomposition.

Before we do so we note that for ease of notation we have done thecalculation using the strong form of the equation. The same assertion canbe directly derived in the weak form by using the test functions ψ = ηei withη ∈ (W 1,2

0 ∩ L∞)(B) in (2.575). With the product rule and the expansion(2.579) this yields

0 =

∫B∇u · (ei ⊗∇η) +∇u · ∇ei η dx

=

∫B

(∇u · ei) · ∇η +∑j

(∇u · ej)(ej · ∇ei), η dx,

i.e., the weak form of (2.580). Similarly to get the weak form of (2.581) onestarts from the identity

0 =

∫B∂1u · ∂2ψ − ∂2u · ∂1ψ dx.

This identity original clearly holds for ψ ∈ C2c (B;RN ) since ∂1∂2ψ = ∂2∂1ψ =

0. Since ∇u ∈ L2 the identity also holds for ψ ∈W 1,20 (B;RN ). Now we take

again ψ = ηei and we get

0 =

∫B

(∂1u · ei)∂2η − (∂2u · ei)∂1η dx

+

∫B

∑j

(∂1u · ej)(ej · ∂2ei)η − (∂2u · ej)(ej · ∂1ei)η dx.

This is the weak form of (2.581).

Lemma 2.73 (Hodge decomposition in two dimensions). Let Ω ⊂ R2 beconvex and bounded. Let v ∈ L2(Ω;R2). Then

(i) If curl v = 0 in the sense of distributions the there exists a C ∈W 1,2(Ω)such that

v = ∇V.

142 [January 28, 2019]

(ii) If div v = 0 in the sense of distributions the there exists a D ∈W 1,2(Ω)such that

v = ∇⊥D.

(iii) There exist C and D such that

v = ∇C +∇⊥D

and∆C = div v, ∆D = curl v

in the sense of distributions.

The assumption that Ω is convex can be replaced by the assumption thatΩ is simply connected.

Proof. (i): If v ∈ C1(Ω) we have shown this in Analysis 2. The general casefollows by approximation.(ii): If div v = 0 and curl v⊥ = 0 and the result follow from (i).(iii): Let C ∈ W 1,2

0 (Ω) be the weak solution of ∆C = div v. Then div (v −∇C) = 0 and the assertion follows from (ii).

By Lemma 2.73 there exist Ci, Di for i = 1, . . . , N − 1 such that

∇u · ei = ∇Ci +∇⊥Di (2.582)

and

−∆Ci = −div (∇u·ei) = −∑j

ωij ·(∇Cj+∇⊥Dj) =∑j

(−ωij ·∇Cj+ω⊥ij ·∇Dj)

Similarly

−∆Di = −curl (∇u·ei) = −∑j

ω⊥ij ·(∇Cj+∇⊥Dj) =∑j

(−ω⊥ij ·∇Cj−ωij ·∇Dj)

Finally we set W = (C1, D1, . . . , CN−1, DN−1). Then we can write thesecond order system in the compact form

−∆Wi =

2N−2∑j=1

Aij · ∇Wj

or even more concisely as

−∆W = A · ∇W. (2.583)

where the Aij are linear functions of the ωij .

143 [January 28, 2019]

This is an interesting reformulation of the harmonic map equation be-cause the only geometric information which enters are the connection co-efficients ωij , but from the analytical point of view we so far have gainednothing. We only know that the coefficients ωij (and hence the Aij) are inL2 and that ∇W ∈ L2. Thus the equation (2.583) is still critical. Indeed,if ∇W is in L2 then the right hand side is in L1 and the best we can hopefor is that the solution W is in W 2,1. By the Sobolev embedding this wouldgive again ∇W ∈ L2. In fact without further structure we only get ∇2W inweak L1 and ∇W in weak L2 (see below for the second assertion).

Infinitesimal symmetries and Helein’s moving frame method Thekey idea of Helein was that there are many possible frames e1, . . . , eN−1 andthat one improve the properties of the connection coefficients by choosingan optimal frame. To focus on the main idea we assume now in additionthat the manifold M is two dimensional, i.e.

N − 1 = 2.

Since ∇(ei · ej) = 0 we have ωji = −ωij . Thus for N − 1 = 0 there is wehave

ω11 = ω22 = 0, ω12 = −ω21 and we set ω = ω12.

Now if e1, e2 is an orthonormal frame and R : Ω→ SO(2) is a W 1,2 map(when considered as a map with values in R2×2) then the pair e1, e2 definedby

ei =∑j

Rijej

is also an orthonormal frame.A key observation is that ω = ∇e1 · e2 can easily be computed from ω.

Every element of SO(2) is of the form

R =

(cosψ − sinψsinψ cosψ

)and thus

e1 = cosψ e1 − sinψ e2, e2 = sinψ e1 + cosψ e2.

A short calculation shows that

(ω)α = (∂αe1) · e2 = ωα − ∂αψ

or, written more compactly,

ω = ω −∇ψ. (2.584)

144 [January 28, 2019]

Now Helein’s idea is to look for an optimal frame, i.e., one which mini-mizes the Dirichlet energy∫

Ω

∑i,j

(∇ei · ej)2 dx = 2

∫Ω|ω|2 dx = 2

∫Ω|ω −∇ψ|2 dx.

Since φ is an arbitrary W 1,2 function an optimal frame e1, e2 must satisfy∫Ωω · ∇η dx =

∫Ω

(ω −∇φ) · ∇η dx = 0 ∀η ∈W 1,20 (Ω).

Thusdiv ω = 0 (2.585)

in the sense of distributions.We now show that this implies that the optimal frame has slightly better

regularity than obvious property ω ∈ L2. By Hodge decomposition thereexists φ ∈W 1,2(Ω) such that

∇e1 · e2 = ω = ∇⊥φ (2.586)

and

−∆φ =− curl ω = −curlω = −∂1(∂2e1 · e2) + ∂2(∂1e1 · e2)

=− ∂2e1 · ∂1e2 + ∂1e1 · ∂2e2 =N∑k=1

−∂2ek1 ∂1e

k2 + ∂1e

k1 ∂2e

k2. (2.587)

Now for each k we have

div

(∂2e

k1

−∂1ek1

)= 0.

Thus each term in the sum has the div-grad structure in Lemma 2.64 Itfollows that there exists a function f ∈ H1(R2) such that −∆φ = f in B.Hence φ ∈ W 2,1(B′) for any ball B′ strictly contained in B. In particularwe have

ω ∈W 1,1(B′).

This is an improvement over the trivial estimate ω ∈ L2. Note that forn = 2 we have the Sobolev embedding W 1,1 → L2 so on the level of theSobolev embedding there seems to be no improvement. We will see belowhowever, that W 1,1 functions are in a space which is slightly better than L2.

The regularity now follows from the formulation (2.583) and the followingresult.

Theorem 2.74. Let B ⊂ R2 be a ball, let Aij ∈ W 1,1(B;Rm) for i, j ∈1, . . . ,m and let W ∈W 1,2(B;Rm) be a weak solution of

−∆W = A ·W in B. (2.588)

Let B′ ⊂ B be a concentric subball. Then W ∈ C0,α(B′,Rm) and ∇W ∈L2,2α(B′;Rm) for any α ∈ (0, 1).

145 [January 28, 2019]

Note that this result implies that∇u =∑N−1

i=1 (∇u·ei)ei ∈ L2,2α and thusu ∈ L2,2+2α = C0,α (which in turns yields u ∈ C∞ by the same reasoning asin Theorem 2.61).

A sketch of the proof can be found in the appendix.

The case N − 1 > 2. In this case we can still define a new frame by

ei =N−1∑k=1

Rikek

if R(x) ∈ SO(N − 1) for a.e. x. In this case two matrices in SO(N − 1) nolonger commute and the new connection coefficients are given by

ω = (∇R)R−1 +RωR−1. (2.589)

The frame which minimizes∫

Ω

∑i,j |ωi,j |2 dx still satisfies

div ωij = 0. (2.590)

Using this we still get ωij ∈W 1,1 and obtain regularity using Theorem 2.74.

Partial regularity for stationary harmonic maps to M for n ≥ 3.If M is sufficiently regular this can also be proved by the moving framemethod, see [?]. One uses a higher dimensional counterpart of the Hodgedecomposition 19 and shows a decay estimate for the BMO norm of u. Theargument is similar to Step 2 in the proof of the corresponding result forM = SN−1. In this setting one does not control the L2 norm of all relevantgradients but one can use that in view of the Poincare inequality the BMOnorm of f in a ball B is also controlled by[

supB(x,r)⊂B

1

rn−p

∫B(x,r

|∇f |p dx

]1/p

.

Another symmetry For n ≥ 3 Rivere and Struwe recently found a veryelegant proof of partial regularity of stationary harmonic maps which onlyrequires minimal regularity of the target manifold M (namely M ∈ C2).This proof uses another symmetry (see [RiSt] which follows the earlier work[Ri07]). Returning to the form (2.574) of the harmonic map equation andsetting w = ν u we have∫

Ω∇u · ∇[ϕ− (ϕ · w)w] dx = 0 ∀ϕ ∈ (W 1,2

0 ∩ L∞)(Ω;Rm). (2.591)

19A natural way to express this higher dimensional Hodge decomposition in higherdimensions is the language of differential forms. One can, however, also write it in astandard way.

146 [January 28, 2019]

Now using ∇u · w = 0 since ∇u is tangential and w is normal to M , thiscondition for u be rewritten as∫

Ω∇u · ∇ϕ−∇u · ∇w (ϕ · w) dx = 0.

The strong form of this equation is

−∇ui =∑j

wi∇wj · ∇uj .

Using again that∑

j wj∂αu

j = 0 (because ∂αu is a tangent vector) we canrewrite the harmonic map equation as

−∆ui =∑j

Ωij · ∇uj (2.592)

whereΩij = wi∇wj − wj∇wi.

Now the only symmetry which is used is that

Ωji = −Ωij , (2.593)

i.e., the skew symmetry of Ω. In fact under this assumption alone can showthe following ε regularity result.

Theorem 2.75. Let B be a ball. Then there exists an ε0 > 0 with thefollowing properties. If u ∈W 1,2(B;RN ) is a weak solutions of (2.592) andif

‖∇u‖2L2,n−2(B) < ε20 and ‖Ω‖L2,n−2(B) < ε2

0

then u is locally Holder continuous in B.

For stationary harmonic maps u : Ω → M one can then use the mono-tonicity formula as before to obtain regularity in Ω \ S where S is a closedset with Hn−2(S) = 0.

To exploit the skew symmetry of Ω one uses again a change of coordinatesor a ’gauge transformation’ (similar to the choice of the optimal frame inHelein’s approach). First one shows if u solves the equation −∆u = Ω · ∇u(with a skew-symmetric Ω) in a ball B and if P : B → SO(N) then

−div (P−1∇u) = [P−1∇P + P−1ΩP ] · P−1∇u (2.594)

Here · denotes the scalar product in Rn and the other operations a matrixtimes vector or matrix times matrix multiplication20 in RN . Note that

20Without the · notation the equations (2.592) reads

−n∑α=1

∂2αu

i =

n∑α=1

N∑j=1

Ωijα∂αuj

147 [January 28, 2019]

this formula is exactly analogous to (2.589) if one makes the substitutionR = P−1.

Then one proves (this similar to but more subtle argument than thechoice of the optimal frame) that there exists a map P : B → SO(n) forwhich

div [P−1∇P + P−1ΩP ] = 0. (2.595)

The choice of this P is called the Coloumb gauge. Now the right hand side of(2.594) almost has a div-grad structure (except for the extra P−1 in front of∇u). One uses a Hodge decomposition of P−1∇u and then one can handlethe extra term can by use Morrey space and BMO estimates, duality andthe usual iteration lemma.

[21.12. 2018, Lecture 22][7.1. 2019, Lecture 23]

2.5.3 Regularity for minimizers of variational problems

We now study the regularity of minimizers of variational integrals

I(v) =

∫Ωf(∇v) dx.

Here Ω ⊂ Rn is open and bounded and v : Ω→ RN . We say that

u ∈W 1,2(Ω;RN ) is a minimizer of I if

I(u) ≤ I(v) for all v with v − u ∈W 1,20 (Ω;RN ). (2.596)

We assume throughout this subsection that

f ∈ C2(RN×n). (2.597)

Now we assume first that f is uniformly convex and D2f is bounded,i.e., we assume that there exists ν > 0 and M > 0 such that

D2f ≥ ν Id , (2.598)

|D2f | ≤ M. (2.599)

or

−n∑α=1

∂2αu =

n∑α=1

Ωα∂αuj

where Ωα is an N ×N matrix with entries Ωijα. Then (2.594) and (2.595) become

−n∑α=1

∂α(P−1∂αu) =n∑α=1

[P−1∂αP + P−1ΩαP ]P−1∂αu,

n∑α=1

∂α[P−1∂αP + P−1ΩαP ] = 0.

148 [January 28, 2019]

More explicitly these conditions read

D2f(A)(F, F ) ≥ ν|F |2 ∀A,F ∈ RN×n,D2f(A)(F,G) ≤ M |F | |G| ∀A,F,G ∈ RN×n.

Differentiating I(u + tϕ) at t = 0 we see that minimizers of I are weaksolutions of the Euler-Lagrange equations

−divDf(∇u) = 0. (2.600)

Using the uniform convexity of f and the difference quotient method wesee that u ∈ W 2,2

loc (Ω;RN ) and that each derivative w = ∂γu satisfies thelinearized equation21

−divD2f(∇u)∇w = 0.

For N = 1, i.e., scalar problems, the DeGiorgi-Nash-Moser theoremimplies that w ∈ C0,α

loc (Ω) and thus ∇u ∈ C1,αloc (Ω;Rn) for some α > 0. Then

a bootstrap argument implies that u ∈ C∞(Ω) if f ∈ C∞(Rn). (see thebeginning of Section 2.2).

If N > 1 then the DeGiorgi-Nash-Moser theorem cannot be applied. Forn = 2, there is a different argument which yields regularity. Indeed, byMeyers’ theorem, Theorem 2.51, we get ∇w ∈ Lploc(Ω;RN ) for some p > 2

and then the Sobolev embedding theorem gives again w ∈ C0,αloc (Ω;RN ).

For n ≥ 3 in general singularities arise may arise for systems. The firstcounterexample (for n andN large) is due Necas [Ne], see also Chapter II.3 in[Gi]. Later Sverak and Yan constructed new counterexamples, including onefor n = 3 and N = 5 and one for n = 4 and N = 3 [SY00, SY02]. RecentlyMooney and Savin found a counterexample in the optimal dimensions n = 3and N = 2 [MS16].

Thus for n ≥ 3 and the case of systems (N ≥ 2) it is natural to look forpartial regularity, i.e., regularity outside a ’small’ closed set. In a numberof interesting applications of RN valued map (e.g., in nonlinear elasticity)convexity is a too restrictive condition and quasiconvexity is a more nat-ural condition (see, e.g., the lecture notes of PDE and modelling, [Mu99]or [Da08] for a discussion of the various convexity notions). For (partial)

21In index notation we have for each i = 1, . . . , N

−∑α

∂α∂f

∂F iα(∇u) = 0.

and differentiating with respect to xγ yields

−∑α

∂α∑β,j

∂2f

∂F iα ∂Fjβ

(∇u) ∂γ(∇u)jβ = 0.

Now ∂γ(∇u)jβ = ∂γ∂βuj = ∂β∂γu

j = ∂βwj = (∇w)jβ .

149 [January 28, 2019]

regularity it is natural to assume that f is uniformly quasiconvex, i.e., thereexists a ν > 0 such that∫

B(x,r)f(A+∇ϕ)− f(A) dy ≥ ν

2

∫B(x,r)

|∇ϕ|2 dy

∀A ∈ RN×n, ϕ ∈ C1c (B(x, r);RN ), (2.601)

and all x ∈ Rn, r > 0 (indeed if the condition holds for one open andbounded set U with Ln(∂U) = 0 it holds for all such sets). It is easy tosee that (2.598) implies (2.601). Under the assumption (2.599) it follows bydensity that ∫

B(x,r)f(A+∇ϕ)− f(A) dy ≥ ν

2

∫B(x,r)

|∇ϕ|2 dy

∀A ∈ RN×n, ϕ ∈W 1,20 ((B(x, r);RN ). (2.602)

Lemma 2.76 (Uniform quasiconvexity implies strong ellipticity). Assumethat f ∈ C2(RN×n) satisfies (2.601) then

D2f(A)(a⊗ b, a⊗ b) ≥ ν|a|2 |b|2 ∀A ∈ RN×n, a ∈ RN , b ∈ Rn. (2.603)

In particular for N = 1 and f ∈ C2 the uniform quasiconvexity condition(2.601) is equivalent to the uniform convexity condition (2.598).

Proof. The main point is to use highly oscillatory test functions. Let η ∈C∞c (B(x, r)) and let

ϕk(x) = εka

ksin(kb · x)η(x), with εk → 0.

Then∇ϕk = εk(a⊗ b) cos(kb · x)η +

εkk

(a⊗∇η) sin(kb · x)

and a short calculation, which uses that∫B(x,r)∇ϕdy = 0, shows that

limk→∞

1

ε2k

∫B(x,r)

f(A+∇ϕk)− f(A) dy

= limk→∞

∫B(x,r)

1

2D2f(A) [(a⊗ b) cos(kb · y)η, (a⊗ b) cos(kb · y)η] dy

= limk→∞

1

2D2f(A)(a⊗ b, a⊗ b)

∫B(x,r)

cos2(kb · y)η2 dy

=1

2D2f(A)(a⊗ b, a⊗ b)

∫B(x,r)

1

2η2 dy.

Similarly we get

limk→∞

1

ε2k

∫B(x,r)

|∇ϕk|2 dy = |a⊗ b|2∫B(x,r)

1

2η2 dy.

Now |a⊗ b|2 = |a|2 |b|2 and the assertion follows from (2.601).

150 [January 28, 2019]

We use the following notation for the average.

–

∫E

f dx :=1

Ln(E)f dx.

Theorem 2.77 (Evans 1986). Assume that f ∈ C2(RN×n) satisfies (2.601)and (2.599) and that D2f is uniformly continuous. Let u be a minimizer ofI. Then there exists a closed set S such that

u ∈ C1,αloc (Ω \ S;RN ), Ln(S) = 0. (2.604)

If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S;RN ). We may take

S := x ∈ Ω : lim supr→0

–

∫B(x,r)

|∇u− (∇u)x,r|2 dy > 0. (2.605)

Remark. (i) The assumption that D2f is uniformly continuous is notneeded for partial regulartiy (see Corollary 2.82 below). Without this as-sumption one may take

S := x ∈ Ω : lim supr→0

–

∫B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→0

–

∫B(x,r)

|∇u| dy =∞.

(ii) It is important that u is a minimizer. There exist Lipschitz continuousweak solutions of the Euler-Lagrange equations −divDf(∇u) = 0 which arenowhere C1 (see [MS03]).(iii) If f is uniformly convex the estimate for the singular set can be improvedto Hn−p(S) = 0 for some p > 2 (see below).(iv) It is not known whether the estimate for the singular set can be improvedfor uniformly quasiconvex integrands. The best known result states that ifwe assume in addition that u ∈ W 1,∞

loc then there exists an ε > 0 such thatHn−ε(S) = 0, see [MiKr].

We first show that every minimizer satisfies a reverse Poincare inequalityor Cacciopoli inequality.

Lemma 2.78 (Cacciopoli inequality/ reverse Poincare inequality). Assumethat f ∈ C2(Rn×N ) satisfies (2.601) and (2.599). Then there exists a con-stant C such for every minimizer u of I, for every ball B(x, r) ⊂ Ω andevery A ∈ RN×n and every a ∈ RN

–

∫B(x,r/2)

|∇u−A|2 dy ≤ C

r2–

∫B(r)

|u− a−Ay|2 dy. (2.606)

Here C depends only on n, µ and M .

151 [January 28, 2019]

Proof. We may assume x = 0 and we write B(r) = B(0, r). To illustrate howthe combination of uniform quasiconvexity and the minimization propertyleads to an estimate for ∇u − A let us first consider a toy problem andassume for a moment that u is affine on the boundary ∂B(r), i.e., thereexists A and a such that u(y) = Ay + a on ∂B(r).

Then we can apply the uniform quasiconvexity condition with ϕ = u −a−Ay and we get

ν

2

∫B(r)|∇u−A|2 dy ≤

∫B(r)

f(∇u)− f(A) dy

On the other hand we can apply the minimizing property of u with v =a+Ay = u− ϕ. This yields∫

B(r)f(∇u) dy ≤

∫B(r)

f(A) dy.

Combining the two estimates we get

ν

2

∫B(r)|∇u−A|2 dy = 0,

i.e., u is affine in B(r).In general u does not equal an affine function on the boundary so we

need to truncate u − a − Ay near the boundary. Let 0 < t < s < r and letζ ∈ C∞c (Ω) be a cut-off function with

ζ = 1 on B(t), ζ = 0 on Ω \B(s),

0 ≤ ζ ≤ 1, |∇ζ| ≤ C

s− t.

Setϕ = ζ(u− a−Ay), ψ = (1− ζ)(u− a−Ay).

Then∇ϕ+∇ψ = ∇(ϕ+ ψ) = ∇u−A.

The key idea is to bound ∇ϕ (which controls ∇u−A in the inner ball)by ∇ψ (which depends ∇u − A and u − Ay − a in an annulus near theboundary). Then a ’hole-filling’ argument leads to a decay inequality fors 7→

∫B(s) |∇u−A|

2 and an iteration lemma finishes the proof. Note that inhe toy problem we have ψ = 0.

We have ϕ ∈ W 1,20 (B(s);RN ) and thus uniform quasiconvexity implies

thatν

2

∫B(s)|∇ϕ|2 dy ≤

∫B(s)

f(A+∇ϕ)− f(A) dy.

152 [January 28, 2019]

Since u is a minimizer and u− ϕ = u on Ω \B(s) we get∫B(s)

f(A+∇ψ +∇ϕ︸ ︷︷ ︸=∇u

) dy ≤∫B(s)

f(A+∇ψ︸ ︷︷ ︸=∇(u−ϕ)

) dy.

Adding −f(A + ∇ψ + ∇ϕ) + f(A + ∇ϕ) − f(A) to the integrand in thesecond inequality and using the first inequality we get

ν

2

∫B(s)|∇ϕ|2 dy ≤

∫B(s)

f(A+∇ψ)−f(A+∇ψ+∇ϕ)+f(A+∇ϕ)−f(A) dy.

Now by Taylor expansion and the assumption |D2f | ≤M we have

| − f(A+∇ψ +∇ϕ) + f(A+∇ϕ) +Df(A+∇ϕ)∇ψ| ≤ 1

2M |∇ψ|2,

|f(A+∇ψ)− f(A)−Df(A)∇ϕ| ≤ 1

2M |∇ψ|2,

|Df(A)−Df(A+∇ψ)| ≤ M |∇ψ|.

Adding the three terms inside the absolute values on the left hand side andusing the triangle inequality we get

|f(A+∇ψ)− f(A+∇ψ +∇ϕ) + f(A+∇ϕ)− f(A)|

≤M |∇ψ|2 +M |∇ψ| |∇ϕ| ≤(M +

M2

ν

)|∇ψ|2 +

ν

4|∇ϕ|2.

where we used Young’s inequality ab ≤ 12εa

2 + ε2b

2 with ε = ν/2. Thus weget the desired estimate of ∇ϕ in terms of ∇ψ:∫

B(s)|∇ϕ|2 dy ≤ C

∫B(s)|∇ψ|2 dy, (2.607)

where C = 4Mν + 4M2

ν2 .Now ∇ψ = 0 in B(t) and

|∇ψ| ≤ |∇u−A|+ C

s− t|u− a−Ay| in B(s) \B(t).

Thus (2.607) implies that∫B(t)|∇u−A|2 dy ≤ C

∫B(s)\B(t)

|∇u−A|2 dy+C

(s− t)2

∫B(s)\B(t)

|u−a−Ay|2 dy.

Now ’fill the hole’, i.e., add C∫B(t) |∇u− A|

2 dy to both sides an divide byC + 1 to conclude∫B(t)|∇u−A|2 dy ≤ θ

∫B(s)\B(t)

|∇u−A|2 dy+C

(s− t)2

∫B(s)\B(t)

|u−a−Ay|2 dy

153 [January 28, 2019]

for all r/2 ≤ s < t ≤ r where

θ =C

C + 1< 1.

Finally the assertion follows from Lemma 2.79 below withA′ = C∫B(r) |u−

a−Ay|2 dy, B = 0, α = 2 and ρ = r0 = r/2, r1 = r.

Remark. One can simplify the algebra leading to (2.607) by observingthat it suffices to consider A = 0, a = 0 and that we may assume thatW (0) = 0 and W (0) = 0. Indeed set

W (F ) = W (A+ F )−W (A)−DW (A)F, u = u− a−Ay,

I(v) =

∫ΩW (∇v) dx.

Then u is a minimizer of I if and only if u is a minimizer of I, and W satisfies(2.599) and (2.5.3). Moreover W (0) = 0, DW (0) = 0. Then the estimatesreduce to

| − W (∇ψ +∇ϕ) + W (∇ϕ)| ≤ |DW (∇ψ)∇ϕ|+ 1

2M |∇ψ|2

≤M |∇ϕ| |∇ψ|+ 1

2M |∇ψ|2,

|W (∇ψ)− W (0)| = |W (∇ψ)| ≤ 1

2M |∇ψ|2.

[7.1. 2019, Lecture 23][11.1. 2019, Lecture 24]

Lemma 2.79. Let 0 < r0 < r1 and let f : [r0, r1] → [0,∞) be a boundedfunction. Assume that there exists θ ∈ [0, 1), α > 0 and A′, B ≥ 0 such thatfor all r0 ≤ s < t ≤ r1

f(t) ≤ θf(s) +A′(s− t)−α +B.

Then there exists a constant c which depends only on α and θ such that

f(ρ) ≤ c[A′(R− ρ)−α +B] forall r0 < ρ < R ≤ r1.

Proof. Let τ ∈ (0, 1), let r0 < ρ < R ≤ r1 and consider the sequence tidefined by

t0 = ρ, ti+1 − ti = (1− τ)τ i(R− ρ).

Then ti ∈ (ρ,R) and ti → R as i→∞. By iteration we get

f(t0) ≤ θkf(tk) +

[A′

(1− τ)α(R− ρ)α +B

] k∑i=0

θiτ−iα.

Now choose τ ∈ (0, 1) such that τ−αθ < 1. Taking the limit k →∞ we getthe assertion with c(α, τ) = (1− τ)−α(1− τ−αθ)−1.

154 [January 28, 2019]

Now we come to the heart of the matter. Define

U(x, r) := –

∫B(x,r)

|∇u− (∇u)x,r|2 dy, where (∇u)x,r := –

∫B(x,r)

∇u dy

The quantity U(x, r) measures how close u is to an affine function and isoften called the excess.

Lemma 2.80 (Excess decay). Assume that f ∈ C2(RN×n satisfies |D2f | ≤M and (2.601) and that D2f is uniformly continuous. Let u be a minimizerof I. Then there exists a constant C2 (which only depends on n, M and theconstant ν in (2.601) ) with the property that for each τ ∈ (0, 1

4) there existsan ε(τ) (ε may also depend on n, M and ν) such that for every B(x, r) ⊂ Ωthe relation

U(x, r) ≤ ε(τ) (2.608)

implies thatU(x, τr) ≤ C2τ

2U(x, r). (2.609)

Proof. Let C2 be a constant, which will be determined later, and fix τ ∈(0, 1

4). To show that (2.608) =⇒ (2.609) we argue by contradiction.

Step 1. Blow-up.If the implication (2.608) =⇒ (2.609) was false, there would exist for m =1, 2, . . . balls B(xm, rm) ⊂ Ω such that

U(xm, rm) = λ2m → 0 (2.610)

butU(xm, τrm) ≥ C2τ

2λ2m. (2.611)

Define

am := (u)xm,rm , bm := (u)xm,2τrm ,Am := (∇u)xm,rm , Bm := (∇u)xm,2τrm .

(2.612)

and set

vm(z) =u(xm + rmz)− am − rmAmz

λmrmfor z ∈ B = B(0, 1). (2.613)

Then

∇vm(z) =∇u(xm + rmz)−Am

λm(2.614)

and(vm)1 = 0, (∇vm)1 = 0. (2.615)

155 [January 28, 2019]

Here we write (vm)r for the average over the ball B(r) = B(0, r). Define

dm := (vm)2τ , em := (vm)τ ,Dm := (∇vm)2τ , Em := (∇vm)τ ,

(2.616)

Now (2.610) implies that

–

∫B

|∇v|2 dz = 1. (2.617)

Since (vm)1 the Poincare inequality yields

–

∫B

|v|2 dz ≤ C. (2.618)

On the other hand (2.611) gives

–

∫B(τ)

|∇v − Em|2 ≥ C2τ2. (2.619)

By assumption we have|D2f(Am)| ≤M. (2.620)

Together with (2.617) and (2.618) and the compact Sobolev embeddingit follows that there exists a subsequence, which upon relabeling we alsoindex by m, such that

vm → v in L2(B;RN ),∇vm ∇v in L2(B;RN×n),∂2f

∂F iα∂Fjβ

(Am) → Lαβij .

for some v ∈W 1,2(B;RN ) and Lαβij ∈ R. For future reference we set

d := (v)2τ , e := (v)τ ,D := (∇v)2τ , E := (∇v)τ ,

(2.621)

We now claim that v is a weak solution to the linear PDE22

−divL∇v = 0 (2.622)

and

|L| := supLF ·G : |F | ≤ 1, |G| ≤ 1 ≤M, (2.623)

L(a⊗ b) · (a⊗ b) ≥ ν|a|2|b|2 ∀a ∈ RN , b ∈ Rn. (2.624)

22In index notation this reads −∑α ∂α

∑β,j L

αβij ∂βv

j = 0.

156 [January 28, 2019]

These estimates follow directly from the corresponding estimatesD2f(Am)(F,G) ≤M |F ||G| and D2f(Am)(a⊗ b, a⊗ b) ≥ ν|a|2|b|2. To prove(2.622) we note that the weak form of the equation for u implies that∫

B(xm,rm)Df(∇u) · ∇ψ dx = 0 ∀ψ ∈ C1

c (B(xm, rm);RN ).

By (2.613) it follows that

1

λm

∫B

[Df(Am + λm∇vm)−Df(Am)] · ∇ϕdz = 0 ∀ϕ ∈ C1c (B;RN ).

Hence ∫B

∫ 1

0D2f(Am + sλm∇vm)∇vm · ∇ϕds dz = 0. (2.625)

We have λm∇vm → 0 in L2. Thus for a further subsequence we may assumethat λm∇vm → 0 a.e. in B. Since D2f is uniformly continuous we deducethat

sups∈[0,1]

D2f(Am + sλm∇vm)−D2f(Am)→ 0 a.e. in B. (2.626)

Boundedness of D2f and the dominated convergence theorem imply that

D2f(Am + sλm∇vm)− L =

=(D2f(Am + sλm∇vm)−D2f(Am)

)+ (D2f(Am)− L)→ 0

in L2(B;L(RN×n;RN×n)), uniformly in s. Thus after exchanging the inte-gration in s and z in (2.625) we get∫

BL∇v · ∇ϕ = 0 ∀ϕ ∈ C1

c (B;RN ).

By density this also holds for all ϕ ∈ W 1,20 (B;RN ) and hence v is a weak

solution of (2.622).

Step 2. Decay estimate for v.By the regularity estimate (2.76) for linear equations with constant, stronglyelliptic coefficients we have

supB( 1

2)

|D2v| ≤ C∫B|∇v|2 dz ≤ C

and thus

–

∫B(ρ)

|∇v − (∇v)ρ|2 dz ≤ Cρ2 ∀ρ ∈ (0,1

2]. (2.627)

Here we use the fact that the constant C depends only on n,M and ν.

157 [January 28, 2019]

Step 3. Transfer of the decay from v to vm.Since vm → v strongly in L2(B;RN ) we have, using the Poincare inequalityand (2.627) with ρ = 2τ ,

limm→∞

–

∫B(2τ)

|vm − dm −Dmz|2 dz = –

∫B(2τ)

|v − d−Dz|2

≤Cτ2 –

∫B(2τ)

|∇v −D|2 dz ≤ Cτ4. (2.628)

The Cacciopoli inequality, Lemma 2.78, provides the estimate

–

∫B(xm,τrm)

|∇u−Bm|2 dy ≤C

(2τrm)2–

∫B(xm,2τrm)

|u− bm −Bm(y − xm)|2 dy.

Dividing by λ2m and rescaling yields

–

∫B(τ)

|∇vm −Dm|2 dz ≤C1

τ2–

∫B(2τ)

|vm − dm −Dmz|2 dz.

Combining this with (2.628) we get

lim supm→∞

–

∫B(τ)

|∇vm − Em|2 dz ≤ lim supm→∞

–

∫B(τ)

|∇vm −Dm|2 dz

≤C1

τ2lim supm→∞

–

∫B(2τ)

|vm − dm −Dmz|2 dz ≤ C3τ2.

This contradicts (2.619), provided we chose C2 > C3.

Remark on the structure of the proof. There are three key ingredi-ents:

• Blow-up and weak convergence to a function v which satisfies a linearequation with constant coefficient

• Decay estimate for solutions of such equations

• Transfer of the decay from v to vm

The first step, which may look complicated initially, is actually usually easy.In particular the first two steps still work if instead of uniform quasiconvex-ity we just assume the strong elliptictiy condition D2f(A)(a ⊗ b, a ⊗ b) ≥ν|a|2 |b|2.

158 [January 28, 2019]

For the last step we used the Cacciopoli inequality. More abstractly forthe last step one can use a compactness argument: if vm are minimizers(of the rescaled problem) and vm v in W 1,2(B;RN ) then vm → v inW 1,2(B(τ);RN ) (see [EvG] for a proof of this compactness result). Thusexcess decay and hence (partial) regularity is very closely related to com-pactness of minimizers or of solutions to the PDE.

Proof of Theorem 2.77. Let

Ω0 = Ω \ S = x ∈ Ω : limr→0

–

∫B(x,r)

|∇u− (∇u)x,r|2 dy = 0.

Let α ∈ (0, 1). We will show that

x0 ∈ Ω =⇒ ∇u ∈ C0,α(B(x0, ρ);RN ) for some ρ > 0. (2.629)

This implies that

limr→0

–

∫B(x,r)

|∇u− (∇u)x,r|2 dy = 0 ∀x ∈ B(x0, ρ).

Hence Ω0 is open and ∇u ∈ C0,αloc (Ω0). Since α ∈ (0, 1) was arbitrary this

proves (2.604).To prove (2.629) let C2 be the constant in Lemma 2.80 and choose τ ∈

(0, 14) such that

C2τ2 ≤ τ2α.

By definition of Ω0 there exists r > 0 (with r < dist (x0, ∂Ω)) such that

U(x0, r) = –

∫B(x0,r)

|∇u− (∇u)x0,r|2 dy < ε2(τ).

Since x 7→ U(x, r) is continuous there exist ρ > 0 such that

U(x, r) < ε2(τ) ∀x ∈ B(x0, ρ).

Let x ∈ B(x0, ρ). Then by Lemma 2.80 we have U(x, τr) ≤ τ2αU(x, r) <ε(τ)2. Hence by induction

U(x, τkr) ≤ τ2kαε(τ)2 ∀k ∈ N.

Since s →∫B(x,s) |∇u − (∇u)x,s|2 dy is nondecreasing (see (2.89)) it follows

that

U(x, s) ≤ 1

τn

(sr

)2αε(τ)2 = s2α ε(τ)2

τnr2α.

Thus ∇u belongs to the Campanato space L2,n+2α(B(x0, ρ);RN×n) whichagrees with C0,α(B(x0, ρ);RN×n).

The higher regularity follows from Lemma 2.81 below and Schauder the-ory (see the beginning of Section 2.2).

159 [January 28, 2019]

Lemma 2.81. Assume that f ∈ C3(RN×n) and

|D2f | ≤M,

D2f(a⊗ b, a⊗ b) ≥ ν|a|2 |b|2 ∀A ∈ RN×n, a ∈ RN , b ∈ Rn

with ν > 0. Let Ω ⊂ RN be open and assume that u ∈ C1,αloc (Ω;RN ) is a

distributional solution of

−divDf(∇u) = 0.

Then u ∈ W 2,2loc (Ω;RN ) and each derivative w = ∂γu is a distributional

solution of−divD2f(∇u)∇w = 0. (2.630)

Remark. The assumptions f ∈ C3 and u ∈ C1,αloc can be weakened to

f ∈ C2 and u ∈ C1, respectively.

Proof. Homework.Hint: Let B(x, r) ⊂ Ω. It suffices to show that u ∈W 2,2(B(x, r/4);RN )

and that w is a weak solution of (2.630) in B(x, r/4).To prove this you may use that the identity

Df(G)−Df(F ) =

∫ 1

0D2f((1− s)F + sG)(G− F ) ds

implies that for h < r/2 the difference quotients

wh(x) :=w(x+ heγ)− w(x)

h

are weak solutions of

−div

∫ 1

0D2f((1− s)∇u(x) + s∇u(x+ heγ)) ds︸ ︷︷ ︸

=Ah(x)

∇wh = 0 (2.631)

in B(x, r/2) (see Homework 6, Problem 1). Now show that [Ah]C0,α(B(x,r/2)

is bounded independent of h, use the test function η2wh whereη ∈ C∞c (B(x, r/2)) with η = 1 on B(x, r/4) and apply Garding’s inequality,Theorem 2.9 to obtain a uniform bound for ‖wh‖W 1,2(B(x,r/4);RN ). Finallypass to the limit h→ 0 in (2.631).

160 [January 28, 2019]

Removal of the condition that D2f is uniformly continuous

Corollary 2.82. Assume that f ∈ C2(RN×n) satisfies |D2f | ≤ M and(2.601). Let u be a minimizer of I. Then there exists a closed set S suchthat

u ∈ C1,αloc (Ω \ S;RN ), Ln(S) = 0. (2.632)

Here we may take

S := x ∈ Ω : lim supr→0

–

∫B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→∞

–

∫B(x,r)

|∇u| dy =∞.

(2.633)If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S;RN ).

To show this we use the following variant of the excess decay lemma.

Lemma 2.83. Assume that f satisfies |D2f | ≤M and (2.601). Then thereexists a constant C2 (which only depends on n, M and the constant ν in(2.601) ) with the property that for each τ ∈ (0, 1

4) and each L > 0 thereexists an ε(τ, L) (ε may also depend on n, M and ν such that for everyB(x, r) ⊂ Ω and every minimizer u of I the relations

U(x, r) ≤ ε(τ) and |(∇u)x,r| ≤ L (2.634)

imply thatU(x, τr) ≤ C2τ

2U(x, r). (2.635)

Proof. The proof is the same as for Lemma 2.80. Uniform continuity wasonly used to show (2.626). The assumptions now guarantee that |Am| ≤ L.SinceD2f is uniformly continuous on compact subsets we obtain (2.626).

Iteration of the excess decay lemma gives the following result.

Lemma 2.84. Let f and C2 be as in Lemma 2.83. Let α ∈ (0, 1) and letτ ∈ (0, 1

4) be such thatτ2α = C2τ

2.

For L > 0 letε(L) = min(ε(τ, 2L), [(1− τα)τnL]2)

Then for any minimizer u of I and any ball B(x, r) ⊂ Ω the relation

U(x, r) ≤ ε(L)2 and |(∇u)x,r| ≤ L (2.636)

imply that

U(x, τkr) ≤ τ2kαU(x, r), |(∇u)x,τkr| ≤ (2− τ−kα)L. (2.637)

161 [January 28, 2019]

Proof. We argue by induction. For k = 0 there is nothing to show. Thus as-sume that (2.637) holds for k. Since ε(L) ≤ ε(2L) it follows from Lemma 2.83and the induction hypothesis that

U(x, τk+1r) ≤ τ2(k+1)αU(x, r).

To show the second inequality note that for any ball B(x, ρ) ⊂ Ω

|(∇u)x,τρ − (∇u)x,ρ| ≤ –

∫B(x,τρ)

|∇u− (∇u)x,ρ| dy

≤ 1

τn–

∫B(x,ρ)

|∇u− (∇u)x,ρ| dy ≤1

τnU1/2(x, ρ).

In particular the condition

U(x, τkr) ≤ τ2kαε(L)

implies that

|(∇u)x,τk+1r − (∇u)x,τkr| ≤ τkα(1− τα)L = [τkα − τ (k+1)α]L.

Thus|(∇u)x,τk+1r| ≤ (2− τ−(k+1)α)L

as desired.

Proof of Corollary 2.82. Let x0 ∈ Ω \ S. Then

L := 1 + lim supr→0

–

∫B(x0,r)

|∇u| dy <∞.

Thus there exists r0 > 0 such that

|(∇u)x0,r| < L ∀r ∈ (0, r0).

Moreover there exists r1 ∈ (0, r0) such that

U(x0, r1) < ε(L).

By continuity there exists a ρ > 0 such that

|(∇u)x,r1 | < L, U(x, r1) < ε(L) ∀x ∈ B(x0, ρ).

Now Lemma 2.84 implies that ∇u ∈ C0,α(B(x0, ρ);RN ). Hence Ω \ S isopen and ∇u ∈ C0,α

loc (Ω \ S;RN ).

The rest of this subsection was not discussed in class

162 [January 28, 2019]

Better estimates on the singular set of uniformly convex f

Corollary 2.85. Assume that f ∈ C2(RN×n) satisfies (2.598) and (2.599).Let u be a minimizer of I. Then there exists a closed set S and a q > 2 suchthat

u ∈ C1,αloc (Ω \ S;RN ), Hn−q(S) = 0. (2.638)

If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S;RN ).

Proof. By Meyers’ theorem there exists a p > 2 with the following property.If U ⊂⊂ Ω is open and compactly contained in Ω then∇u ∈W 1,p(U ;RN×n).Now consider Uj ⊂⊂ Ω with

⋃j Uj = Ω. Corollary 2.82 implies that ∇u ∈

C1,αloc (Uj \ Sj) where

Sj := x ∈ Uj : lim supr→0

–

∫B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→∞

–

∫B(x,r)

|∇u| dy =∞.

Now by Lemma 2.86 we have Hq(Sj) = 0 for all q < p. Hence Hq(⋃j Sj) =

0.

Lemma 2.86. Let U ⊂ Rn be open. Let g ∈ L2(U) and let ∇g ∈ Lp(U ;Rn)with 2 ≤ p ≤ n. Let

S1 := x ∈ U : lim supr→0

–

∫B(x,r)

|g − gx,r|2 dy > 0,

S2 := x ∈ U : lim supr→0

–

∫B(x,r)

|g| dy =∞.

ThenHn−p(S1) = 0, Hn−q(S2) = 0 ∀q < p.

Proof. Homework.Hint: To estimate S1 for p < n use the Poincare inequality, Holder’s in-equality and Lemma 2.72 with ε = 1

k . For p = n the argument is slightlysimpler.To estimate S2 first show an estimate for gx,r/2 − gx,r in terms of ∇g andargue like in the proof that L2,n+2α = C0,α.

Generalizations As we have seen in the course PDE and modelling forsystems (e.g., in nonlinear elasticity) one often wants to consider integrandsf which satisfy

f(F ) ≥ |F |q − Cwith q > n. This is incompatible with the assumption |D2f | ≤M . One canshow that partial regularity also holds under the weaker condition

|D2f(A)| ≤M(1 + |A|q−2)

163 [January 28, 2019]

with q > 2 if one strengthens the quasiconvexity to condition to∫B(x,r)

f(A+∇ϕ)− f(A) dy ≥ ν

2

∫B(x,r)

(1 + |∇ϕ|q−2)|∇ϕ|2 dy

for all ϕ ∈ C1c (B(x, r);RN ), and all A, x, r > 0, see [Ev86]

The results can also be generalized to minimizer of functionals of theform ∫

Ωf(x, v(x),∇v(x)) dx,

see, e.g., [FH86, GM86] and there is a large literature on the regularity ofglobal and local minimizers.

2.5.4 Monotone operators and the Minty-Browder trick

This section was not discussed in class.Let Ω ⊂ Rn be bounded and open, let g ∈ L2(Ω;RN ) and assume that

f is convex

and that there exist c > 0 and C > 0 such that23

f(F ) ≥ c|F |2 − C ∀F ∈ RN×n. (2.639)

Consider the functional

I(v) =

∫Ωf(∇v)− g · v dx

It follows from the Poincare inequality that infW 1,2

0 (Ω;RN )I > −∞ and by

the direct method of the calculus of variations I has minimizer u among allmaps in W 1,2

0 (Ω;RN ). If, in addition, f ∈ C1(RN×n) and the exist M > 0and C > 0 such that

|Df(F )| ≤M |F |+ C ∀F ∈ RN×n (2.640)

then the minimizer u is a weak solution of the PDE24

−divDf(∇u) = g, u ∈W 1,20 (Ω;RN ).

We now look for weak solutions of the equation

−div a(∇u) = g, u ∈W 1,20 (Ω;RN ).

23If f is C2 and Df2 ≥ νId with ν > 0 then (2.639) holds with c = ν/4. To seethis one can use that the function f(F ) = f(F ) − ν

2|F |2 is convex, which implies that

f(F ) ≥ f(0) + L · F , and use Young’s inequality |F | ≤ ε2|F |2 + 1

2ε.

24If f ∈ C2 and |D2f(F )| ≤ M for all F then (2.640) follows with C = |Df(0)|. Ifwe only assume that f is convex then (2.640) can be deduced from the bound f(F ) ≤M ′|F |2 + C′ for all F

164 [January 28, 2019]

and we want to identify sufficient condition on a which guarantee the exis-tence of a solutions.

One hint is the following observation. A function f ∈ C2(RN×n) isconvex if and only if Df is monotone

(Df(F )−Df(G)) · (F −G) ≥ 0.

We now show that monotonicity and combination with suitable upper andlower bounds on a is sufficient for the existence of a solutions.

Theorem 2.87 (Existence for monotone operators). Assume thata : RN×n → RN×n is continuous and there exist c > 0, M > 0 and C > 0such that

(a(F )− a(G)) · (F −G) ≥ 0 ∀F,G ∈ RN×n, (2.641)

a(F ) · F ≥ c|F |2 − C ∀F ∈ RN×n, (2.642)

|a(F )| ≤M |F |+ C ∀F ∈ RN×n, (2.643)

Let Ω ⊂ Rn be open and bounded, let g ∈ L2(Ω).For k ∈ N let Xk be a finite dimensional subspaces of W 1,2

0 (Ω;RN ) such

that Xk ⊂ Xk+1 for all k and⋃kXk is dense in W 1,2

0 (Ω;RN ). Equip W 1,20

(and the spaces Xk) with the scalar product (u, v) :=∫

Ω∇u · ∇v dx. Notethat by the Poincare inequality the corresponding norm is equivalent to theusual W 1,2 norm. Then

(i) there exists uk ∈ Xk such that∫Ωa(∇uk) · ∇ϕdx =

∫Ωg · ϕdx ∀ϕ ∈ Xk. (2.644)

Moreover ν‖uk‖ ≤ C‖f‖L2.

(ii) There exists u ∈W 1,20 (Ω;RN ) such that∫

Ωa(∇u) · ∇ϕdx =

∫Ωg · ϕdx ∀ϕ ∈W 1,2

0 (Ω). (2.645)

To prove (i) we use the following finite dimensional result.

Lemma 2.88. Let X be a finite dimensional Hilbert space and assume thatF : X → X is continuous and satisfies the following coercivity conditon:and that there exist c > 0 and C > 0 such that

F (x) · x ≥ c|x|2 − C. (2.646)

where a · b denotes the scalar product in H. Then for every y ∈ X the

equation F (x) = y has a solution which satisfies ‖x‖2 ≤ ‖y‖2

c2+ 2C

c .

165 [January 28, 2019]

Proof. Exercise.Hint: Rewrite the equation as x = x− F (x) + y and use Young’s inequalityx · y ≤ c

2 |x|2 + 1

2c |y|2 to show that T (x) := x − F (x) + y satisfies the

assumptions for Theorem 1.6 if R2 = ‖y‖2c2

+ 2Cc .

Proof of Theorem 2.87. Exercise.Hint for (i): Let u ∈ Xk. Use the Riesz representation theorem to showthat there exist gk ∈ Xk and Fk(u) ∈ Xk such that (gk, ϕ) =

∫Ω g ·ϕdx and

(F (uk), ϕ) =∫

Ω a(∇u) · ∇ϕdx for all ϕ ∈ Xk. Then apply Lemma 2.88Hint for (ii): To show existence note that a subsequence of the uk in (i)

converges weakly to a limit u. Let v ∈W 1,20 (Ω). First show that∫

Ωa(∇v) · (∇v −∇u) ≥

∫Ωg · (v − u) (2.647)

in the following way. Let vj ∈ Xj with vj → v in W 1,20 (Ω). Monotonicity

of a implies that∫

Ω (a(∇vj)− a(∇(uk))) · (∇vj − ∇uk) ≥ 0. If j ≤ k thatyields ∫

Ωa(∇vj) · (∇vj −∇uk) ≥

∫Ωg · (vj − uk) (2.648)

Now first take the limit k →∞ and then the limit j →∞. Finally let t > 0,consider v = u± tϕ in (2.647), divide by t and let t ↓ 0.

To show that a(∇vj) → a(∇v) in L2 as j → ∞, one can use that for asubsequence ∇vj → ∇v a.e. and then use (2.643) and a suitable version ofthe dominated convergence theorem.

Alternatively note that to pass to the limit in (2.647) one can argueas follows. By (2.643) the sequence a(∇vj) is bounded in L2 and hence a

subsequence has a weak limit a. Now by monotonicity for every w ∈ W 1,20

we have ∫Ω

(a(∇w)− a(∇vj)) · (∇w −∇vj) dx ≥ 0.

Since ∇vj converges strongly we can pass to the limit and get∫Ω

(a(∇w)− a) · (∇w −∇v) dx ≥ 0.

Now take again w = v ± tϕ divide the inequality by t and take the limitt ↓ 0 to conclude that ∫

Ω(a(∇v)− a) · ∇ϕdx = 0

for all ϕ ∈W 1,20 (Ω;RN ).

166 [January 28, 2019]

Remark. (i) The Minty Browder trick consists in replacing the nonlinearequation (2.645) for u by the linear inequalities (2.647) for u. Thus one canpass to the limit in (2.647) using only weak convergence of u.(ii) If we assume the uniform monotonicity condition (a(F ) − a(G)) · (F −G)ν|F −G|2 with ν > 0 then the solution is unique.

The existence result can be easily extended to more general boundaryconditions. Let u0 ∈W 1,2(Ω;RN ). Then there exists a weak solution of

−div a(∇u) = g, u− u0 ∈W 1,20 (Ω;RN ).

Regularity Solutions of monotone PDE have the same regularity as min-imizers of convex integrands. Let u be a weak solution of

−div a(∇u) = 0

and assume a ∈ C1 and that

|a(F )− a(G)| ≤ L|F −G|, (a(F )− a(G)) · (F −G) ≥ ν|F −G|2 (2.649)

for all F,G ∈ RN×n where ν > 0.Then u ∈ C1,α

loc (Ω\S) where S is closed and Hn−q(S) = 0 for some q > 2.For the Cacciopoli inequality one starts from the identity∫

B(x,r)[a(∇u)− a(A)] · ϕ = 0

and uses the choice ϕ = ζ2(u − a − Ay) and monotonicity to obtain theestimate

ν

∫B(x,r)

ζ2|∇u−A|2 ≤∫B(x,r)

[a(∇u)− a(A)] · ζ2(∇u−A) dx

=−∫B(x,r)

[a(∇u)− a(A)] · 2ζ(u− a−Ay)⊗∇ζ dx

and estimates the integrand on the right hand side by

2Lζ|∇u−A| |u− a−Ay||∇ζ| ≤ ν

2ζ2|∇u−A|2 +

2

νL2|u− a−Ay|2|∇ζ|2.

The proof of excess decay by blow-up and the estimate for the size of thesingular set can then be carried out in the same way as for convex f .

167 [January 28, 2019]

Beyond Hilbert spaces The results so far for convex variational inte-grals and monotone elliptic PDE are modelled on functionals with quadraticgrowth ∫

Ω|∇u|2 dx

and linear equations−∆u = 0

and the natural setting is the Hilbert space W 1,20 (Ω).

In order to treat functionals like∫Ω|∇u|p dx

for 1 < p < ∞ one can extend the setting to reflexive Banach spaces andagain monotonicity easily leads to existence results, see [Li69], Chapitre 2.2and [KS], Chapter III.1.

[11.1. 2019, Lecture 24][14.1. 2019, Lecture 25]

2.6 Maximum principles and existence for quasilinear ellip-tic equations

2.6.1 Classcial maximum principles for linear equations

This subsection follows [GT], Chapter 3.1 and 3.2, but note that in contrastto [GT] we put a minus sign in front of the leading order

∑aij∂i∂ju.

So far we mostly considered estimates which can be derived by choosingsuitable test functions, scaling and interpolation. These estimates are ratherrobust and hold for higher (even) order equations and for systems (with theexception of the results in Section 2.2). The maximum principle only appliesto scalar equations of second order, but for those equations it is a verypowerful tool and in particular immediately gives L∞ estimates. Before westudy the maximal principle for general linear elliptic second order operators,let us recall two versions of the maximum principle for −∆ which werealready proved in the course Einfuhrung in PDG.

Let Ω be open and bounded and u ∈ C2(Ω) ∩ C(Ω). The maximumprinciple states that

−∆u ≤ 0 in Ω =⇒ u ≤ max∂Ω

in Ω. (2.650)

The strong maximum principle provides a strict inequality if u is nontrivial

−∆u ≤ 0 in Ω, Ω connected =⇒ u < max∂Ω

in Ω or u = const.

(2.651)

168 [January 28, 2019]

The proof of the maximum principle was easy. If the conclusion is falsethen M := maxΩ > max∂Ω. Thus u must have a maximum at x0 ∈ Ω. Sinceu in C2 this implies that

D2u(x0) ≤ 0,

i.e, the symmetric matrix D2u(x0) is negative semidefinite. Hence

∆u(x0) = trD2u(x0) ≤ 0.

This is not yet a contradiction to the hypothesis −∆u ≤ 0. To get a con-tradiction we let ε > 0 and consider

u(x) = u(x) + ε|x|2.

If ε is small enough then we still have maxΩ u > max∂Ω. Thus u has amaximum at a point xε ∈ Ω and

D2u(xε) ≤ 0. (2.652)

Now we have−∆u = −∆u− 2ε < 0

which leads to a contradiction with (2.652). This argument is very robust.It works without change if

∑ij aij(x)∂i∂ju(x) ≤ 0 in Ω as long as aij = aji

and the matrix aij is positive definite for each x ∈ Ω.The proof of the strong maximum principle was more delicate. We de-

fined the set

V = x ∈ Ω : u(x) : M, where M = maxΩ

u.

The set V is (relatively) closed in Ω since u is continuous. We then used themean value inequality for subharmonic functions

−∆u ≤ 0 in Ω =⇒ u(x) ≤ –

∫B(x,r)

u(y) dy for all balls B(x, r) ⊂ Ω.

(2.653)Since u ≤ M it follows that x ∈ V implies B(x, r) ⊂ V . Hence V is alsoopen. Now Ω is connected and therefore either V = ∅ (then u < M in Ω)or V = Ω (then u = M in Ω). This argument looks much more fragile since(2.653) does not hold for functions which satisfy −

∑ij aij(x)∂i∂ju(x) ≤ 0.

A more subtle replacement which serves the same purpose (for equations indivergence form) is the weak Harnack inequality in Theorem 2.48 (ii) appliedto v = M − u. Below we follow instead the classical argument via the Hopfboundary lemma which historically preceeds Theorem 2.48.

We now consider general second order operators of the form

Lu = −∑i,j

aij∂i∂ju+∑i

bi∂iu+ cu.

169 [January 28, 2019]

Since ∂i∂ju = ∂j∂iu we may and will assume that the coefficients of theleading part are symmetric

aij(x) = aji(x).

Remark. In [GT] the operator is written as Lu =∑

i,j aij∂i∂ju+

∑bi∂iu+

cu. I preferred to keep the minus sign in front of the highest order termbecause then subsolutions are given by Lu ≤ 0 and because of consistencywith earlier sections in the notes. To compare the results below with thosein [GT] one has to

replace L by −L, c by −c and b by −b.

We use the following notation.

• λ(x) is the smallest eigenvalue of the symmetric matrix (aij(x))

• Λ(x) is the largest eigenvalue of the symmetric matrix (aij(x)).

Thenλ(x)|ξ|2 ≤

∑i,j

aij(x)ξiξj ≤ Λ(x)|ξ|2 ∀ξ ∈ Rn

and λ(x) and Λ(x) are the largest and smallest numbers, respectively, forwhich this inequality holds. We say that

• L is elliptic at x ∈ Ω if λ(x) > 0;

• L is elliptic in Ω if λ > 0 in Ω;

• L is strictly elliptic if λ ≥ λ0 > 0 in Ω;

• L is uniformly elliptic if λ > 0 and Λ/λ is bounded.

Note that this terminology slighty differs from our earlier terminolgy. Whatwe called ’elliptic’ before now corresponds to strictly and uniformly ellip-tic. The finer distinctions of ellipticity will also be useful when we considernonlinear equations.

We usually need some mild control of the coefficients of the lower orderterms in terms of the aij . In particular we will always assume that thereexists a constant C such that

|b(x)|λ(x)

≤ C <∞ ∀x ∈ Ω. (2.654)

When L is strictly elliptic this conditions is in particular satisfied if b =(b1, . . . , bn) is bounded.

170 [January 28, 2019]

Theorem 2.89 (Maximum principle). Let L be elliptic in the bounded andopen set Ω and assume (2.654). Assume

c = 0.

Let u ∈ C2(Ω) ∩ C(Ω). Then

Lu ≤ 0 in Ω =⇒ maxΩ

u ≤ max∂Ω

u.

Remark. Instead of (2.654) it suffices to assume that b/λ is bounded onevery compact set K ⊂ Ω. Indeed we need to show that L > max∂Ω u impliesu ≤ L. Let ΩL := x ∈ Ω : u(x) > L. If ΩL is not empty the closure of ΩL

is compact and contained in Ω. Hence we can apply the maximum principlein ΩL. Thus maxΩL u ≤ max∂ΩL

u = L. This gives the desired contradictionΩL = ∅.

Proof. Assume first Lu < 0 in Ω. If maxΩ u > max∂Ω u and u attains amaximum at x0 ∈ Ω, then

D2u(x0) < 0, ∇u(x0) = 0.

We now use the following fact about symmetric matrices. If A and B aresymmetric n× n matrices then25

A ≥ 0, B ≥ 0 =⇒∑i,j

AijBij ≥ 0.

Since L is elliptic and c = 0 this implies that Lu(x0) ≥ 0. This contradictsthe assumption.

Now we would like to reduce that case Lu ≤ 0 to the case Lu < 0. ForL = −∆ we could simply consider u = u + ε|x|2 because −∆|x|2 = −n.Now we no longer know that L|x|2 < 0 because we have no control on theterm |

∑i bi∂i|x|2| for large x. We thus seek a function ϕ such that Lϕ < 0.

Consider first the one dimensional problem. Then the inequality reduces to

−aϕ′′ + bϕ′ < 0.

Setting ψ = ϕ′ this becomes −aψ′ + bψ < 0. Assuming ψ > 0 and theellipticity condition a > 0 this is equivalent to

ψ′

ψ>b

aor (lnψ)′ >

b

a. (2.655)

25Proof: If C is a symmetric matrix with C ≥ 0 then Cii ≥ 0 for each diagonal element.Thus the assertion is clear if A is diagonal. In general there exists Q ∈ SO(n) such thatD = QAQT is diagonal. Then A = QTDQ as QT = Q−1. Moreover∑

i,j

AijBij = trAB = trQTDQB = trDQBQT ≥ 0

since B′ := QBQT is symmetric and B′ ≥ 0.

171 [January 28, 2019]

Now let us return to the multidimensional case. By the assumption (2.654)there exists a constant C such that |b1| ≤ Cλ and we have λ ≤ a11. Moti-vated by (2.655) we take ϕ = eγx1 . Then

Lϕ = (−γ2a11 + γb1) eγx1 < 0 if γ ≥ 2C.

Finally let u = u+εϕ with ε > 0. Then Lu < 0 and (since Ω is bounded)maxΩ u > max∂Ω u. Then we obtain a contradiction as shown above.

Corollary 2.90. [Upper bound for c ≥ 0] Let L be elliptic in the boundedand open set Ω and assume (2.654). Assume

c ≥ 0.

Let u ∈ C2(Ω) ∩ C(Ω). Then

Lu ≤ 0 in Ω =⇒ maxΩ

u ≤ max(max∂Ω

u, 0).

Proof. This will be reduced to Theorem 2.89 by using that cu ≥ 0 if u ≥ 0.Let M := maxΩ u and assume that M > 0. Let m = max(max∂Ω u, 0) andlet U = x ∈ Ω : u(x) > 0. Then u ≤ m on ∂U . Let

L′u = −∑ij

aij∂i∂ju+∑i

bi∂iu.

ThenL′u = Lu− cu ≤ Lu ≤ 0 in U.

By the maximum in principle applied to L′ in the set U we get M =maxU u ≤ max∂U u ≤ m.

The maximum principle guarantees that u attains its maximum at aboundary point x0. To work towards the strong maximum principle we nowshow that (under suitable conditions) the normal derivative at x0 does notvanish.

Definition 2.91. We say that an open set Ω satisfies an interior spherecondition at x0 ∈ ∂Ω if there exists an open ball B ⊂ Ω with x0 ∈ ∂B.

Lemma 2.92 (Hopf boundary lemma). Assume L is uniformly elliptic with(2.654) and c = 0. Let u ∈ C2(Ω) and assume that Lu ≤ 0 in Ω. Letx0 ∈ ∂Ω be such that

(i) u is continuous at x0;

(ii) u(x0) > u(x) for all x ∈ Ω;

(iii) Ω satisfies an interior sphere condition at x0.

Then the outer normal derivative of u at x0, if it exists, satisfies the strictinequality

∂u

∂ν(x0) > 0.

172 [January 28, 2019]

Remark. (i) If ∂Ω does not have a tangent space at x0, the positivity ofthe outer normal derivative should be understood as follows. If B(y, r) ⊂ Ωis a ball with x0 ∈ ∂B(y, r) then

limt↓0

u(x0)− u(x0 − tv)

t> 0, where v =

x0 − y|x0 − y|

.

If ∂Ω does not have a tangent space there may be more than one ’normal’vector v to which this reasoning can be applied but the importance is thatthere exists at least one.(ii) The interior sphere condition is important. Consider the first quadrantin R2, i.e., Ω = (x1, x2) : x1 > 0, x2 > 0 then u(x) = −x1x2 is harmonicin Ω, u < 0 in Ω, u(0) = 0 but ∇u(0) = 0. More generally let α > 1, letu = −Re zα. Then u is harmonic and u < 0 in Ωα = (r cosβ, r sinβ) : 0 <r <∞, 0 < β < π/α and ∇u(0) = 0.

Proof. We may assume without loss of generality that

u(x0) = 0. (2.656)

We try to ’squeeze in’ a function w for which we already know that ∂w∂ν > 0.

More precisely it suffices to find a function w with the following properties

(i) w(x0) = 0,

(ii) ∂w∂ν > 0,

(iii) u ≤ w on the segment (x0 − εν, x0) := x0 − tν : 0 < t < ε.

Then u− w ≤ 0 on (x0 − εν, x0) and hence

∂(u− w)

∂ν(x0) ≥ 0.

Thus ∂u∂ν (x0) ≥ ∂w

∂ν (x0) > 0.To satisfy (iii) we use the interior sphere condition. There exists a ball

B(y,R) ⊂ Ω with x0 ∈ ∂B(y,R). Instead of (iii) it will actually be easier toshow the stronger condition

u ≤ w in A := B(y,R) \B(y, ρ).

We first try to satisfy this relation on

∂A = ∂B(y,R) ∪ ∂B(y, ρ).

On ∂B(y, ρ) it is easy to satisfy the condition since u < 0 on ∂B(y, ρ) andthus

max∂B(y,ρ)

u < 0.

173 [January 28, 2019]

We also haveu ≤ 0 on ∂B(y,R).

Assume we have found w with u ≤ w on ∂A and w(x0) = 0. How do weget u ≤ w in A ?

The idea is to use the maximum principle. It suffices to show thatLw ≥ 0. Then L(u− w) = Lu− Lw ≤ 0 and hence u− w ≤ 0 in A.

Thus it suffices to find a function W ∈ C2(A) ∩ C(A) such that

W = 0 on ∂B(y,R), LW ≥ 0,∂W

∂ν> 0. (2.657)

Then min∂B(y,ρ)W > −∞ and we can take w = εW where

ε = max∂B(y,ρ)

u/ min∂B(y,ρ)

W.

It is natural to look for radially symmetric W . Take y = 0 and W (x) =ψ(|x|2) with

−ψ′′ ≥ 0, ψ′(R2) > 0, ψ(R2) = 0. (2.658)

Then with r = |x|

∂iW = 2ψ′(r2)xi, ∂j∂iW = 4ψ′′(r2)xixj + 2ψ′(r2)δij

LW =−∑i,j

4aijψ′′(r2)xixj − 2∑i

aiiψ′(r2) + 2ψ′(r2)∑i

bixi

≥− 4λr2ψ′′(r2)− 2nΛ|ψ′(r2)| − 2ψ′(r2)r|b|.

Try ψ(s) = −e−αs + e−αR2. If α is large enough then LW ≥ 0. Moreover

the conditions (2.658) are satisfied. Explicitly W is given by26

W (x) = −e−α|x|2 + e−αR2.

Theorem 2.93 (Strong maximum principle). Let Ω be open and connected(not necessarily bounded), let L be uniformly elliptic in Ω with (2.654) and

c = 0.

Assume that u ∈ C2(Ω) satisfies Lu ≤ 0. If u has an interior maximum inΩ then u is constant.

26[GT] takes v = −W and shows that u − u(x0) + εv ≤ 0 in A which yields again theassertion.

174 [January 28, 2019]

Proof. Assume that u is not constant and attains its maximum M at aninterior point of Ω. Let

Ω− := x ∈ Ω : u(x) < M.

Then Ω− is open and non-empty. Moreover ∂Ω− ∩ Ω 6= ∅ (otherwise Ω−

would be also relatively closed in Ω and thus Ω− = Ω contradicting theassumption that u(x) = M for some x in Ω).

Let y ∈ Ω− be a point with r = dist (y, ∂Ω−) < dist (y, ∂Ω). Then

B(y, r) ⊂ Ω−, B(y, r) ⊂ Ω and ∂B(y, r) contains a point x0 ∈ ∂Ω−.

Then u(x0) = M (otherwise x0 ∈ int Ω−).Now we can apply the Hopf boundary lemma to the open set Ω−. Thus

∂u∂ν (x0) > 0. But u attains a maximum at x0. Hence ∇u(x0) = 0. Thiscontradiction finishes the proof.

[14.1. 2019, Lecture 25][18.1. 2019, Lecture 26]

2.6.2 Quasilinear elliptic equations: overview

We want to show the existence of classical solutions of the Dirichlet problem

Qu = 0 in Ω,

u = ϕ on ∂Ω.

Here the quasilinear operator Q is given by

Qu = −∑i,j

aij(x, u,∇)∂i∂ju+ b(x, u,∇u) with aji = aij .

By a topological fixed point argument (see Thm. 1.7 and Thm. 2.99 inSection 2.6.4 below) this can be broken into two steps

• Existence results for linear equations with Cα coefficients (see Thm. 2.98below)

• Uniform C1,α estimates for solutions of the nonlinear problem

The main difficulty is to obtain the C1,α estimates. The derivation ofthese estimates is usually divided into four steps

(i) Estimate for supΩ |u|: this is obtained by a maximum principle/ com-parison principle for the quasilinear operator Q. If Qu− ≤ Qu ≤ Qu+

in Ω and u− ≤ u ≤ u+ on ∂Ω then u− ≤ u ≤ u+ in Ω, see Section 2.6.3.

175 [January 28, 2019]

(ii) Estimate for sup∂Ω |∇u|. This is obtained by a comparison principleusing suitable ’barrier’ functions near the boundary, see Section 2.6.6.

(iii) Estimate for supΩ |∇|. This can be obtained, for example, by a maxi-mum principle for |∇u|2, see Section 2.6.7.

(iv) Estimate for [∇u]C0,α . For equations in divergence form De Giorgi–Nash–Moser estimate, see Thm. 2.47 for the interior estimate and Sec-tion 2.6.5 for global estimates

The different steps can be carried out under slightly different assumptionon the quasilinear operator Q. We thus discuss them now in separate sub-sections.

The presentation is based on Chapters 10, 11, and 13–16 in [GT].

We remark in passing that there exists also a theory for fully nonlinearsecond order equations of the form

−A(x, u(x),∇u(x),∇2u(x)) = 0, (2.659)

see Chapter 17 in [GT]. Here the main hypothesis is that the map G 7→A(x, z, p,G) is strictly monotone on the set of symmetric n×n matrices (ora suitable subset), i.e.,

G > G′ =⇒ A(x, z, p,G) > A(x, z, p,G′).

Here G > G′ means that G − G′ is positive definite. One can show thatsolutions of (2.659) satisfy a comparison principle if A is independent of z(see Homework sheet 13). An example of map which is strictly monotone onthe set of positive definite matrices is given by f(G) = detG (see Homeworksheet 13).

2.6.3 Maximum and comparison principles for quasilinear ellipticequations

This subsubsection follows [GT], Chapter 10. Again we take a minus signin front of the leading order part. Thus to compare the statements in thisand the following subsubsections with [GT] one needs to

replace L by −L and b by −b.

Notation and definitions We consider second order quasilinear opera-tors Qu of the form

Qu = −∑i,j

aij(x, u,∇)∂i∂ju+ b(x, u,∇u) with aji = aij (2.660)

176 [January 28, 2019]

where x is contained in an open set Ω ⊂ Rn and, unless otherwise stated, thefunction u belongs to C2(Ω). The coefficients of Q, namely the functions aij

and b are assumed to be defined for all values (x, z, p) in the set Ω×R×Rn.Two operators Q1 and Q2 of the form (2.660) will be called equivalent ifthere is a fixed positive function h on Ω× R× Rn such that aij2 = haij1 andb2 = hb1. Equations Qu = 0 for equivalent operators will also be calledequivalent.

For (x, z, p) ∈ Ω× R× Rn we define:

• λ(x, z, p) is the smallest eigenvalue of the matrix (aij)(x, z, p),

• Λ(x, z, p) is the largest eigenvalue of the matrix (aij)(x, z, p).

Then

λ(x, z, p)|ξ|2 ≤∑i,j

aij(x, z, p)ξiξj ≤ Λ(x, z, p)|ξ|2 ∀ξ ∈ Rn.

Let U ⊂ Ω× R× Rn. We say that

• Q is elliptic in U if λ > 0 in U ;

• Q is uniformly elliptic in U if, in addition, Λ/λ is bounded in U .

If Q is elliptic in Ω × R × Rn we simply say that Q is elliptic in Ω andwe use the same convention for uniform ellipticity. If u ∈ C1(Ω) andλ(x, u(x),∇u(x)) > 0 for all x ∈ Ω we say that Q is elliptic with respect tou.

We define a scalar function E by

E(x, z, p) =∑i,j

aij(x, z, p)pipj . (2.661)

If Q is elliptic in U then

0 < λ(x, z, p)|p|2 ≤ E(x, z, p) ≤ Λ(x, z, p)|p|2 ∀(x, z, p) ∈ U with p 6= 0.

The operator Q is of divergence form if there exists a differentiable func-tionB(x, z, p) and a differentiable mapA(x, z, p) = (A1(x, z, p), . . . , An(x, z, p)such that

Qu = −divA(x, u,∇u) +B(x, u,∇u) for u ∈ C2(Ω), (2.662)

i.e., in (2.660),

aij(x, z, p) =1

2

(∂Aj

∂pi(x, z, p) +

∂Ai

∂pj(x, z, p)

).

177 [January 28, 2019]

Unlike the case of linear operators, a quasilinear operator with smooth co-efficients is not necessarily expressibly in divergence form.

The operator Q is variational if the equation Qu = 0 is the Euler-Lagrange equation of a variational integral∫

ΩF (x, u,∇u) dx

where F : Ω× R× Rn → R is a differentiable function. In this case Q is ofdivergence form and

Ai(x, z, p) =∂F

∂pi(x, z, p), B(x, z, p) =

∂F

∂z(x, z, p).

The ellipticity of of Q is equivalent to strict convexity of F as a function ofp (more precisely to the conditions that D2

pf is positive definite).

Examples (i) The minimal surface equation.This is the Euler-Lagrange operator for the area integral∫

Ω

√1 + |∇u|2 dx.

Thus

Qu =− div∇u√

1 + |∇u|2= −

∑i

∂i∂iu√

1 + |∇u|2

=1√

1 + |∇u|2

−∆u+∑ij

∂iu∂ju

1 + |∇u|2∂i∂ju

Thus Q is equivalent to the operator

Q = −∆u+∑ij

∂iu∂ju

1 + |∇u|2∂i∂ju

Here

aij(x, z, p) = δij − pipj1 + |p|2

=1

1 + |p|2[(1 + |p|2)δij − pipj

].

Thus the minimal and maximal eigenvalue of aij are given by

λ(x) =1

1 + |p|2, Λ(x) = 1.

Hence Q (and thus Q) is elliptic in Ω, but not uniformly elliptic. MoreoverQ is uniformly elliptic on every set Ω× R×B(0, R). We have

E(x, z, p) =|p|2

1 + |p|2

178 [January 28, 2019]

Geometrically div ∇u√1+|∇u|2

is the sum of the principle curvatures κi of

the graph of u, i.e., of the n dimensional surface (x, u(x) : x ∈ Ω. Wedefine the mean curvature27 of an n dimensional hypersurface as

H =1

n

n∑i=1

κi

Surfaces with H = 0 are called minimal surfaces and thus the equationQu = 0 is called the minimal surface equation.

Here the sign convention is as follows: we choose the sign of the normalsuch that νn+1 > 0. For a codimension 1 surface Γ with unit normal ν theprincipal curvatures are defined as follows. For a point x ∈ Γ and η in thetangent space TxΓ define the operator Bx by Bxη = −η · ∇ν. Then oneeasily sees that Bx maps TxΓ to itself and is symmetric (with respect to thescalar product on Tx which is obtained by restricting the Euclidean scalarproduct in Rn+1 to Tx). The principal curvatures κi(x) are the eigenvaluesof Bx. The sign convention is such that if Γ is the boundary of a boundedand open convex set U then the principle curvatures of Γ are non-negativeif we n choose ν as the inner normal. For further details see, e.g., [GT],Chapter 14.6.

Example Let Γ = (x, u(x)) : x ∈ B(0, 1) be the graph of a scalarfunction u. Let

ν =1√

1 + |∇u|2(−∇u, 1)

be the unit normal of Γ. If∇u(0) = 0 then TxΓ = Rn×0 andBx(0) = D2u.In particular the principle curvatures are the eigenvalue of D2u and the meancurvature at 0 is H(0) = 1

ntrD2u = 1n∆u.

(ii) Prescribed mean curvature equation.Let H : Ω → R be given. The equation of prescribed mean curvature (ormean curvature equation for short) is

−div∇u√

1 + |∇u|2= −nH (2.663)

This is equivalent to

Mu := −(1 + |∇u|2)∆u+∑ij

∂iu∂ju ∂i∂ju = −nH(1 + |∇u|2)3/2

27Some authors also use the definition H =∑ni=1 κi even though the word mean cur-

vature suggests to divide by n

179 [January 28, 2019]

For the operator M we have

λ(x, z, p) = 1, Λ(x, z, p) = 1 + |p|2, E(x, z, p) = |p|2

The mean curvature equation is the Euler-Lagrange equation of the func-tional

I(v) =

∫Ω

√1 + |∇u|2 dx+ n

∫ΩHudx.

The first integral is the area of the graphu and for constant H the secondterm is H times the (oriented) volume of the set bounded by graphu andΩ × 0. If H is positive and u = 0 on ∂Ω then the maximum principleimplies that u < 0 and∫

Ω(−u) dx = Ln+1(x, s) : x ∈ Ω, u(x) < s < 0.

Comparison principle

Theorem 2.94 (Comparison theorem for quasilinear elliptic operators). LetΩ be open and bounded. Assume that u, v ∈ C2(Ω) ∩ C(Ω) satisfy

Qu ≤ Qv in Ω, u ≤ v on ∂Ω, (2.664)

where

(i) the operator Q is elliptic with respect to u or v and λ−1(x, v(x),∇v(x))or λ−1(x, u(x),∇u(x)) is bounded on compact subset K ⊂ Ω;

(ii) the coefficients aij are independent of z;

(iii) the coefficient b is non-decreasing in z;

(iv) the coefficients aij and b are differentiable with respect to the p variableand the derivatives ∂pka

ij and ∂pkb are continuous in Ω× R× Rn.

Thenu ≤ v in Ω. (2.665)

If Qu < Qv in Ω,if Q is elliptic with respect to u or v and if (ii) and (iii)hold, (but not necessarily (iv)), we have strict inequality u < v in Ω.

Proof. Let us assume that Q is elliptic with respect to u. The idea is toshow that L(u − v) ≤ 0 whenever u − v ≥ 0 where L is a suitable linearelliptic operator. We have

0 ≥ Qu−Qv =∑i,j

−aij(x,∇u)∂i∂j(u− v)− [aij(x,∇u)− aij(x,∇v)]∂i∂jv

+ b(x, u,∇u)− b(x, u,∇v) + b(x, u,∇v)− b(x, v,∇v)︸ ︷︷ ︸≥0 if u−v≥0

.

(2.666)

180 [January 28, 2019]

For a continuously differentiable function g : Rn → R we have the followingversion of the mean value theorem

g(p)− g(q) = l · (p− q), with l =

∫ 1

0∇g((1− t)q + tp) dt. (2.667)

Thus writing

w = u− v,aij(x) = aij(x,∇u(x)),

−∑i,j

[aij(x,∇u)− aij(x,∇v)]∂i∂jv + [b(x, u,∇u)− b(x, u,∇v)] =∑i

βi(x)∂iw

we see that

Lw := −aij(x)∂i∂jw + βi∂iw ≤ 0 on Ω+ = x ∈ Ω : w(x) > 0

andw ≤ 0 on ∂Ω.

Thusw ≤ 0 on ∂Ω+.

Note that the functions βi(x) are bounded on every compact set K ⊂ Ω.This follows from (2.667) and assumption (iv) because u, v, ∇u, ∇v arebounded on compact subsets of Ω. By (i) λ−1(x, u(x),∇u(x) is bounded.Therefore we can apply the linear maximum principle, Theorem 2.89, andthe remark following that theorem, to L and Ω+ and we get w ≤ 0 in Ω+

and hence w ≤ 0 in Ω.If Qu < Qv and u − v has a maximum at x0 with u(x0) ≥ v(x0) then

∇u(x0) = ∇v(x0), D2(u− v)(x0) ≤ 0 and thus by conditions (i) and (ii)

−∑i.j

aij(x0,∇u(x0)) (∂i∂ju)(x0) ≥ −∑i.j

aij(x0,∇u(x0)) (∂i∂jv)(x0)

=−∑i.j

aij(x0,∇v(x0))(∂i∂jv)(x0).

Condition (iii) implies that

b(x0, u(x0),∇u(x0)) ≥ b(x, v(x0),∇u(x0)) = b(x0, v(x0),∇v(x0)).

Thus (Qu)(x0) ≥ (Qv)(x0) which contradicts the assumption Qu < Qv inΩ.

If Q is elliptic at v then we exchange the roles of u and v and similarlyfind a linear elliptic operator L such that L(v − u) ≥ 0 if v − u ≤ 0. Thuswe get again L(u− v) ≤ 0 if u− v ≥ 0 and we conclude as before.

181 [January 28, 2019]

Theorem 2.95. Let u, v ∈ C2(Ω) ∩ C(Ω) satisfy

Qu = Qv in Ω, u = v on ∂Ω

and suppose that conditions (i) to (iv) in Theorem 2.94 hold. Then u = v.

[18.1. 2019, Lecture 26][21.1. 2019, Lecture 27]

Maximum principles Assume that Q is elliptic, u ∈ C2(Ω) ∩ C(Ω) and

Qu ≤ 0.

We look for a bound for an upper bound for u of the form

maxΩ

u ≤ max∂Ω

u+ + C.

A natural idea is to prove an upper bound for u is to use the comparisontheorem, Theorem 2.94, and to construct a function v with Qv ≥ 0 in Ω andv ≥ u on ∂Ω. Then u ≤ v if the assumptions of the comparison theoremholds.

One drawback of the comparison is that the coefficients aij are not al-lowed to depend on z and b has to be non-decreasing in z. To free ourselvesfrom these restrictions we consider a new operator where the z dependenceis ’frozen’. We define an quasilinear operator

Qv := −∑ij

aij(x, u,∇v)∂i∂jv + b(x, u,∇v). (2.668)

More explicitly the coefficients aij and b of Q are defined by

aij(x, p) = aij(x, u(x), p), b(x, z, p) = b(x, u(x), p).

Then Q satisfies the assumptions (i)–(iii) of the comparison theorem. Notethat

Qu = Qu ≤ 0.

Thus we look for functions v with

0 < Qv = −∑ij

aij(x, u(x),∇v(x))∂i∂jv + b(x, u(x),∇v).

In the proof of the maximum principle for linear operators we have seenthat the function v(x) = −eξ·x is a good candidate (there we looked for

182 [January 28, 2019]

subsolutions and took the + sign rather than the minus sign). With thenotation q = −eξ·xξ this gives

Qv =∑ij

aij(x, u(x),−eξ·xξ)eξ·xξiξj + b(x, u(x),−eξ·xξ)

=∑ij

aij(x, u(x), q)e−ξ·xqiqj + b(x, u(x), q)

= E(x, u(x), q)e−ξ·x + b(x, u(x), q)

= e−ξ·x(E(x, u(x), q) + b(x, u(x), q)

|q||ξ|

)This suggests to impose the following structural condition

E(x, z, p) ≥ −µb(x, z, p)|p| if z ≥ 0

for some µ > 0. Then taking |ξ| > µ we get Qv > 0. The following theoremprovides an upper bound for u under a slightly more general condition.

Theorem 2.96 (Maximum principle). Let Q be elliptic in Ω (and assumethat the aij are continuous, or, more generally that λ−1 is bounded on com-pact subsets) and suppose that there exist µ1 ≥ 0 and µ2 ≥ 0 such that

b(x, z, p)

E(x, z, p)≥ −µ1|p|+ µ2

|p|2∀(x, z, p) ∈ Ω× [0,∞)× Rn. (2.669)

Then if u ∈ C2(Ω) ∩ C(Ω) satisfies

Qu ≤ 0 in Ω,

we havemax

Ωu ≤ max

∂Ωu+ + Cµ2

where u+ = max(u, 0) and C = C(µ1,diam Ω).

Remark. By applying this argument also to −u we obtain the followingresult. If

b(x, z, p)sgn z

E(x, z, p)≥ −µ1|p|+ µ2

|p|2∀(x, z, p) ∈ Ω× R× Rn

thenQu = 0 =⇒ max

Ω|u| ≤ max

∂Ω|u|+ Cµ2

183 [January 28, 2019]

Remark. The following example suggests28 that some restriction on thegrowth of b is needed. For the simple choice aij(x, z, p) = δij and b(x, z, p) =−|p|2 we get the equation

−∆u− u|∇u|2 = 0

and we have seen that this equation unbounded solution ln ln 1|x| inB(0, 1/2) ⊂

R2.

Proof. As above set

Qv = aij(x, u,∇v)∂i∂jv + b(x, u,∇v).

Assume thatΩ ⊂ x ∈ Rn : 0 < x1 < d

and µ2 > 0 and choose the comparison function

v(x) = max∂Ω

u+ + µ2(eαd − eαx1)

with α ≥ µ1 + 1. Using (2.669) we get in Ω+ = x ∈ Ω : u(x) > 0

|∇v| = αµ2eαx1

and

Qv = µ2α2a11(x, u,∇v)eαx1 + b(x, u,∇v)

≥ e−αx1

µ2E(x, u,∇v)− (µ1|∇v|+ µ2)

|∇v|2E(x, u,∇v)

≥ e−αx1

µ2E(x, u,∇v)

(1− µ1

α− e−αx1

α2

)> 0 ≥ Qu.

Hence, by Theorem 2.94, we have u ≤ v in Ω+ and hence in Ω. The resultfor µ2 = 0 follows by letting µ2 tend to zero.

The mean curvature equation Mu = −nH(x)(1 + |∇u|2)3/2 does notsatisfy the structure condition (2.669). Indeed in this case

E(x, z, p)p = |p|2, b(x, z, p) = nH(x)(1 + |p|2)3/2 ∼ |p|3 as p→∞.

Indeed, one can obtain an bound on supu only under a natural smallnesscondition on H (which arises by comparison with half-spheres).

28I say ’suggests’ and not ’proves’ because the singular solution below is not C2 so doesnot strictly fall under the theory considered here.

184 [January 28, 2019]

Theorem 2.97. Let Ω be an open and bounded set and assume that Ω ⊂B(y,R) for some y ∈ Rn and R > 0. Let H ∈ C0(Ω) and assume that

sup |H| ≤ 1

R.

Assume that u ∈ C2(Ω) ∩ C(Ω) satisfies

−div∇u√

1 + |∇u|2= −nH. (2.670)

Thenmax

Ω|u| ≤ max

Ω|u|+R.

Remark. (i) The result is optimal in the following sense. If Ω = B(y,R)and if H ≤ −A or H ≥ A with A > 1

R then there exists no C2 solutionof (2.670) which is bounded in Ω, see Homework 12, Problem 3 (in fact byconsidering a slight smaller ball B(y,R′) with 1

A < R′ < R we see that thereexist no C2 solution of (2.670) in Ω).(ii) For domains which are not balls L∞ bounds for u can be obtained underthe weaker condition

supΩ|H| <

(Ln(B(0, 1)

Ln(Ω)

)1/n

,

see [GT], Theorem 10.10 and (10.32). It even suffices to assume that∫Ω|H|n dx < Ln(B(0, 1)),

see [GT], (10.35) or Corollary 10.6.

Proof. It suffices to show the upper bound u ≤ max∂Ω u+R. Then the fullassertion follows by applying this estimate also to −u. For the proof of theupper bound see Homework 12, Problem 2. Hint: Let

v(x) =√R2 − |x|2 + sup

∂Ωu.

A short calculation (or a geometric argument) shows that

−div∇v√

1 + |∇v|2=n

R

Now use the comparison principle. The details are left as homework.

185 [January 28, 2019]

2.6.4 Existence of classical solutions

Let α ∈ (0, 1) Assume that Ω is a bounded open set and

∂Ω is of class C2,α, ϕ ∈ C2,α(Ω). (2.671)

We seek a solution u ∈ C2,α(Ω) of the boundary value problem

Qu = 0 in Ω, u = ϕ on ∂Ω.

We assume that the coefficients satisfy

aij ∈ C0,α(Ω×(−R,R)×B(0, R)), b ∈ C0,α(Ω×(−R,R)×B(0, R)) ∀R > 0.(2.672)

In particular the coefficients can be extended to continuous functions onΩ× R× Rn. We assume that

Q is elliptic in Ω, (2.673)

and that i.e. the lowest eigenvalue of the matrix of the extended coefficients(aij) satisfies λ > 0 in Ω× R× Rn.

To find such a solution we rewrite the boundary value problem for uas a fixed point problem involving the solution of a linear boundary valueproblem. Fix the boundary datum ϕ ∈ C2,α(Ω). Given u ∈ C1,β(Ω) forsome β > 0 define the linear partial differential operator

Lv = −∑i,j

aij(x, u,∇u)∂i∂jv + b(x, u,∇u). (2.674)

Then the coefficients aij(x) := aij(x, u(x),∇u(x)) and b(x) := b(x, u(x),∇u(x))are in C0,αβ(Ω). Moreover the lowest eigenvalue λ(x) of (aij) satisfies λ > 0in Ω. Since λ is continuous this implies that there exists νu > 0 such thatλ ≥ νu in Ω. In fact it follows from (2.672) that there exists νR such that

λ ≥ νR > 0 in Ω ∀u ∈ C1(Ω) with ‖u‖C1(Ω) ≤ R.

We now use the following existence and uniqueness results for linearelliptic equations in non divergence form.

Theorem 2.98. Let Ω ⊂ Rn be open and bounded with C2,α boundary, letf ∈ C0,α(Ω) and ϕ ∈ C2,α(Ω). Assume that aij ∈ C0,α(Ω) and that thereexists ν > 0 and M > 0 such that

‖aij‖C0,α(Ω) ≤M,∑ij

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

Then there exists one and only one u ∈ C2,α(Ω) such that

−∑ij

aij(x)∂i∂ju = f in Ω, u = ϕ on ∂Ω

186 [January 28, 2019]

Moreover there exists a constant C(Ω, ν,M, n) such that

‖u‖C2,α(Ω) ≤ C(Ω, ν,M, n)(‖f‖C0,α(Ω) + ‖ϕ‖C2,α(Ω)

).

A proof is given below. The corresponding result for divergence formoperators is Theorem 2.30.

If u in C1,β(Ω) and L is given by (2.674) then this result (with α replacedby γ = αβ) yields the existence and uniqueness of a solution v of the problem

Lv = 0 in Ω, v = ϕ on ∂Ω (2.675)

Define the operator T by Tu := v. (the solution also depends on ϕ but wesuppress this dependence in the notation because ϕ has been fixed).

Then u is a solution of the quasilinear boundary value problem if andonly if

u = Tu

Moreover the equationu = tTu

with 0 < t < 1 is equivalent to u/t = Tu or, more explicitly to

Lu

t= 0 in Ω,

u

t= ϕ on ∂Ω.

This can be rewritten as∑i.j

aij(x, u,∇u)∂i∂ju+ tb(x, u,∇u) = 0 in Ω, u = tϕ on ∂Ω. (2.676)

Now we will use the Leray-Schauder fixed point theorem, Theorem 1.7, toshow the following result.

Theorem 2.99. Let α ∈ (0, 1), let Ω ⊂ Rn be an open and bounded set.Assume that (2.671), (2.672) and (2.673) hold. Assume further that thereexist β ∈ (0, 1) and C > 0 such that

‖ut‖C1,β(Ω) ≤ C

whenever ut is a solution of the boundary value problem (2.676). Then thereexists a solution of the boundary value problem

Qu = 0 in Ω, u = ϕ on ∂Ω.

Proof. We first show that the operator T is a continuous and compact mapfrom C1,β(Ω) to itself. Indeed if u is in a bounded subset B of C1,β(Ω)the coefficients aij and b are in a bounded set of C0,αβ(Ω). By assumptionλ > 0 in Ω× R× Rn. Since λ is a continuous function it is strictly positiveon compact subsets of Ω × R × Rn and thus there exists a ν > 0 such

187 [January 28, 2019]

that λ(x, u(x),∇u(x) ≥ ν > 0 for all u ∈ B, x ∈ Ω. Thus the set TB isbounded in C2,αβ(Ω) and hence precompact in C2(Ω) and C1,β(Ω) by theArzela-Ascoli theorem. Thus T is compact.

To see that T is continuous assume that uk → u in C1,β and set vk = Tuk.Then

−∑i,j

aij(x, uk,∇uk)∂i∂jvk + b(x, uk,∇uk) = 0

and vk = 0 on ∂Ω By compactness there exists a subsequence such thatvk → v in C2(Ω). Thus

−∑i,j

aij(x, u,∇u)∂i∂jv + b(x, u,∇u) = 0.

and v = 0 on ∂Ω. Thus v = Tu. Since v is uniquely determined it followsthat the whole sequence vk converges to Tu. Hence T is continuous.

Thus the Leray-Schauder fixed point theorem, Theorem 1.7, implies thatthere exists a u ∈ C1,β(Ω) such that

u = Tu.

Above we have shown that Tu ∈ C2,αβ(Ω). Thus u ∈ C2,αβ(Ω). It followsthat aij , b ∈ C0,α(Ω). Thus, using Theorem 2.98 for the linear boundaryvalue problem (2.675) we get Tu ∈ C2,α(Ω).

Classical solutions for linear non divergence form equationsThis was not discussed in class. Here we prove Theorem 2.98.

Proof. Replacing u by u− ϕ and f by f +∑

ij aij(x)∂i∂jϕ we may assume

that ϕ = 0. Now the estimate for u follows from Lemma 2.100 below. Thisimplies in particular uniqueness. To show existence we consider the affinefamily for linear differential operators Lt

Ltv = −∑ij

aijt (x)∂i∂ju, with at(x) = (1− t)δij + taij(x).

Then L0 = −∆ is an invertible map from X = u ∈ C2,α(Ω) : u = 0 on ∂Ωto Y = C0,α(Ω). Indeed existence of a weak solution follows from the Lax-Milgram theorem and the C2,α regularity was obtained in Theorem 2.30.By Lemma 2.100 below we have

‖v‖X ≤ C(Ω, ν ′,M ′, n)‖Ltv‖Y ∀t ∈ [0, 1].

where ν ′ = min(ν, 1), M ′ = max(M, 1). Then by Lemma 1.1 the operatorsLt are invertible for all t ∈ [0, 1]. Invertibility of L1 implies the existence ofa solution u ∈ X for every f ∈ Y .

188 [January 28, 2019]

Lemma 2.100. Let Ω ⊂ Rn be a open and bounded with C2,α boundary, letν > 0, M > 0. Then there exists a constant C(Ω, ν,M, n) with the followingproperty. If f ∈ C0,α(Ω) and if u ∈ C2,α(Ω) satisfies

−∑ij

aij(x)∂i∂ju = f in Ω, u = 0 on ∂Ω

where

‖aij‖C0,α(Ω) ≤M,∑ij

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

Then‖u‖C2,α(Ω) ≤ C(Ω, ν,M, n)‖f‖C0,α(Ω).

Proof. The key estimate is

[D2u]0,α ≤ C ′(Ω, ν, ‖aij‖0,α, n)(‖u‖C2(Ω) + [f ]0,α

). (2.677)

We sketch a proof based on the blow-method of Simon [Si]. For analternative argument see the proof of Thm. 6.6 in [GT]. If the estimate doesnot hold there exist uk, fk, a

ijk such that

[D2uk]α = 1, ‖fk‖α → 0, ‖uk‖C2 → 0,

‖aijk ‖α ≤M,∑i,j

aijk ξiξj ≥ ν|ξ|2.

In particular there exist xk, yk with yk − xk → 0 such that

|D2uk(xk)−D2uk(yk)| ≥1

2|xk − yk|α.

Case 1. Assume that lim supk→∞ r−1k dist (xk, ∂Ω) =∞.

Upon passage to a subsequence we may assume that r−1k dist (xk, ∂Ω)→∞.

We have

−∑i,j

aijk (xk)∂i∂juk =∑i,j

[aijk (x)− aijk (xk)

]∂i∂juk + fk.

Since −∑

i,j aijk (xk)∂i∂juk(xk) = fk(xk) this can be rewritten as

−∑i,j

aijk (xk) [∂i∂juk − ∂i∂juk(xk)] =∑i,j

[aijk (x)− aijk (xk)

]∂i∂juk+[fk−fk(xk)].

(2.678)Define rk = |yk − xk| and

Uk(z) =1

r2+αk

(uk(xk + rkz − uk(xk)− rk(∇uk)(xk) · z − r2

k

1

2D2uk(xk)(z, z)

).

189 [January 28, 2019]

Aijk (z) = aijk (xk + rkz), Fk(z) =1

rαk(f(xk + rkzk)− f(xk)).

Then

D2Uk(z) =1

rαk

(D2uk(xk + rkz)−D2u(xk)

),

[D2Uk]α ≤ 1, Uk = 0, ∇Uk(0) = 0, D2Uk(0) = 0, (2.679)

[Aij ]C0,α ≤Mrαk , Fk(z) ≤ [fk]C0,α |z|α.

Dividing (2.678) by rαk we get

−∑i,j

aijk (xk)∂i∂jUk =∑i,j

Aijk (z)−Aijk (0)

rαk[(∂i∂ju)(xk) + rαk ∂i∂jUk] + Fk

(2.680)Now by assumption D2uk → 0 uniformly and it follows from the Arzela-Ascoli theorem and (2.679) that (for a subsequence) Uk → U in C2

loc(Rn).

Moreover |Aijk (z) − Aijk (0)| ≤ Mrαk |z|α and aijk (xk) → aij . Finally Fk → 0locally uniformly an thus

−∑i,j

aij∂i∂jU = 0 in Rn. (2.681)

Moreover it follows from (2.679) that |D2U(z)| ≤ C|z|α. Now every secondderivative ∂k∂lU satisfies the same equation and thus application of theestimate (2.75) with u = ∂k∂lU and passage to the limit r →∞ imply thatD3U = 0 and thus D2U = 0 since D2U(0) = 0. On the other hand wehave (again for a subsequence) zk := r−αk (yk − xk)→ z ∈ Sn−1 and the C2

loc

convergence of Uk implies that

|D2U(z)| = limk→∞

|D2Uk(zk)| = limk→∞

r−αk |D2u(yk)−D2u(xk)| ≥

1

2.

This contradiction finishes the proof in Case 1.

Case 2. Assume that lim supk→∞ r−1k dist (xk, ∂Ω) <∞.

The xk be the point on ∂Ω which is closest to xk (for large enough k this pointis even uniquely determined since ∂Ω is of class C2). Then by assumption|xk − xk| ≤ Crk. Define as before rk = |yk − xk| and set

Uk(z) =1

r2+αk

(uk(xk + rkz − uk(xk)− rk(∇uk)(xk) · z − r2

k

1

2D2uk(xk)(z, z)

),

Aijk (z) = aijk (xk + rkz), Fk(z) =1

rαk(f(xk + rkzk)− f(xk)).

190 [January 28, 2019]

By a translation and rotation we may assume that xk = 0 and that theouter normal to Ω at xk is −en. Since ∂Ω is of class C2,α there exists aψ ∈ C2,α(Rn−1) such that

Ω ∩B(0, ε) = (x′, xn) ∈ B(0, ε) : xn > ψ(x′).

and ∇ψ(0) = 0. Let

Ψk(z′) =

1

rkψ(rkz

′), then [D2Ψ]C0,α ≤ Cr1+αk .

Now we can again use the Arzela-Ascoli theorem to obtain again a subse-quence such that

Uk → U in C2(Ωk ∩B(0, R)) for every R > 0

whereΩk = (z′, zn) : zn > Ψk(z

′).In particular C2

loc(Rn+) since Ψk → 0 locally uniformly. Moreover as beforeFk → 0 in Ωk ∩B(0, R)) and we get

−∑i,j

aij∂i∂jU = 0 in Rn+.

We now show thatU = 0 on ∂Rn+. (2.682)

Indeed we have

Uk(z′,Ψk(z

′)) =1

r2+αk

uk(x′, ψ(x′))︸ ︷︷ ︸=0

−Tk(z′,Ψ(z′))

,

where Tk is the second order Taylor polynomial of uk at xk = 0. To estimatethe Taylor polynomial we differentiate the relation uk(x

′, ψ(x′)) twice anduse that ∇ψ(0) = 0. This gives the relations

∇uk(0) · (z′, 0) = 0, D2uk(0)(z′, z′) +∂uk∂xn

(0)D2ψ(0)(z′, z′) = 0. (2.683)

We also have ∇Ψ(0) = 0, |D2Ψ| ≤ Crk and [D2Ψ]C0,α ≤ Cr1+αk . This

implies

|Ψ(z′)| ≤ Crk|z′|2, |Ψ(z′)− 1

2D2Ψ(0)(z′, z′)| ≤ Cr1+α

k |z′|2. (2.684)

Thus

Tk(z′,Ψ(z′)) = rk

∂uk∂xn

(0)Ψ(z′) +1

2r2kD

2uk(0)(z′, z′)

+ r2kD

2uk(0)(z′,Ψ(z′)en) +1

2r2kD

2uk(0)(Ψ(z′)en,Ψ(z′)en)︸ ︷︷ ︸≤C(R)r3

k if |z′|≤R

191 [January 28, 2019]

Using the second identity in (2.683) and the second estimate in (2.684) weget for |z′| ≤ R

Tk(z′,Ψ(z′)) ≤ C

∣∣∣∣∂uk∂xn(0)

∣∣∣∣ r2+αk + Cr3

k.

Now by assumption sup |∇uk| → 0 and it follows that

Uk(z′,Ψk(z

′))→ 0 uniformly for z′ ∈ B(0, R)

Since Uk converges uniformly in B(0, R)∩Ωk and Ψk → 0 in B(0, R) we get(2.682).

Now we know that |DU(z)| ≤ C|z|1+α (since DUk(0) = D2Uk(0) = 0and [D2Uk]C0,α ≤ 1). Let k ∈ 1, . . . , n − 1. Then we can apply the halfspace estimate (2.193) to any tangential derivative u = ∂kU with λ = 0.This yields shows that D2∂kU = 0 and thus ∂j∂kU is constant for all j ≤ nand all k ≤ n− 1. By the PDE for U we get that ∂n∂nU is constant, too.Since D2Uk converges uniformly to D2U in B(0, R)∩Ωk this again leads toa contradiction with the assumption |D2uk(yk) −D2uk(xk)| ≥ 1

2 |xk − yk|2.

This finishes the proof of (2.677)

Given (2.677) it thus suffices to show that

‖u‖C2(Ω) ≤ C ′(Ω, ν, ‖aij‖0,α, n)‖f‖0,α. (2.685)

Assume that this estimate does not hold. Since the equation for u is linearthen there exist uk, fk, a

ijk such that

‖uk‖C2(Ω) = 1, ‖fk‖0,α → 0,

and‖aijk ‖0,α ≤M,

∑ij

aijk (x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

By (2.685) and the Arzela-Ascoli theorem it follows that for a subsequenceuk → u in C2(Ω). Moreover by assumption and the Arzela-Ascoli theoremfk → 0 uniformly and aijk → aij uniformly. In particular

∑ij a

ij(x)ξiξj ≥ν|ξ|2. Hence ∑

i,j

aij∂i∂ju = 0 in Ω, u = 0 on ∂Ω

Now by the maximum principle applied to ±u we get u = 0 in Ω. Hence‖uk‖C2(Ω) → 0. This contradicts the assumption ‖u‖C2(Ω) = 1.

[21.1. 2019, Lecture 27][25.1. 2019, Lecture 28]

192 [January 28, 2019]

2.6.5 C1,α estimates

We aim at proving global Holder estimates for the gradient of solutions ofquasilinear equations in divergence form. We follow [GT, Chapter 12] andassume that Ω ⊂ Rn is open, ∂Ω ∈ C2,α, and that Q is (equivalent to) anelliptic operator in divergence form. Precisely, there are A ∈ C1(Ω × R ×Rn;Rn) and B ∈ C0(Ω× R× Rn) such that for u ∈ C2(Ω),

Qu := −divA(x, u,∇u) +B(x, u,∇u). (2.686)

Assume that Q is elliptic in Ω, i.e., λ > 0, where λ(x, z, p) is the smallesteigenvalue of aij(x, z, p),

aij = aji =1

2

(∂Ai

∂pj+∂Aj

∂pi

). (2.687)

Plan: Given a solution u ∈ C1 ∩W 2,2, use that its derivatives satisfylinear elliptic equations, and apply DeGiorgi-Nash-Moser’s theorem.

Example 2.101. Let u ∈ C1,α(Ω) be a solution to the minimal surfaceequation. For fixed 1 ≤ k ≤ n, we test the equation with ∂kψ for arbitraryψ ∈ C2

0 (Ω), and obtain that w := ∂ku is a solution of the linear equationLw := −

∑i,j ∂iaij∂jw = 0 with aij(x) := ∂pjA

i(∇u(x)). For arbitraryΩ′ ⊂⊂ Ω (cf. (2.674) and its discussion),

(i) the operator L is uniformly elliptic on Ω′, and

(ii) the coefficients aij are bounded.

We note that u ∈W 2,2loc (Ω) Hence the assumptions of Theorem 2.47 (DeGiorgi-

Nash-Moser) are satisfied, and we obtain local Holder regularity of ∂ku.

To derive global Holder estimates, we need the following generalizationof Theorem 2.47 (see also the remark following it). For the case f = g = 0the main ingredient is an extension of the weak Harnack inequalities to theboundary, see the remark at the end of subsection 2.2.1. For a detailedproof, see [GT, Theorem 8.29].

Theorem 2.102. Let Ω := B+R ⊂ Rn, R > 0, be a half ball, and let T

be contained in the flat part of its boundary, i.e., T ⊂ ∂Ω ∩ xn = 0.Suppose that the operator L satisfies (2.440) and (2.441), f ∈ Lq(Ω,Rn)and g ∈ Lq/2(Ω) for some q > n. Suppose that u ∈W 1,2(Ω) satisfies

Lu := −∑i,j

∂iaij(x)∂ju = g + div f in Ω,

193 [January 28, 2019]

and suppose that there exist κ and α0 > 0 such that for all x0 ∈ T and allR > 0,

osc∂Ω∩B(x0,R)

u ≤ κRα0 .

Then u ∈ Cαloc(Ω ∪ T ) for some α > 0, and for every Ω′ ⊂⊂ Ω ∪ T ,

‖u‖Cα(Ω′) ≤ C(‖u‖L2(Ω) + κ+ λ−1(‖f‖Lq(Ω;Rn) + ‖g‖Lq/2(Ω))),

where α and C depend on n, Λ/λ, dist (Ω′, ∂Ω \ T ) and α0.

We prove the following global Holder regularity result for the gradient,given that K = sup(|u|+ |∇u|) is bounded.

Theorem 2.103. Suppose Ω ⊂ Rn is open and bounded with ∂Ω ∈ C2.Suppose that u ∈ W 2,2(Ω) ∩ C1,α0(Ω) satisfies Qu = 0 in Ω and u = ϕ on∂Ω with ϕ ∈ C2(Ω), where Q is elliptic in Ω and of divergence form (2.686)with A ∈ C1(Ω × R × Rn;Rn) and B ∈ C0(Ω × R × Rn). Then there areα > 0 and C > 0 such that

[∇u]C0,α(Ω) ≤ C,

where C and α depend on n, K := supΩ(|u| + |∇u|), A, B, ‖ϕ‖C2 and Ω.(For explicit bounds see proof.)

Remark 2.104. To derive uniform Holder estimates, it suffices to assumeϕ ∈ W 2,q(Ω) for some q > n, ∂Ω ∈ C1,α0, α0 > 0, and u ∈ W 2,2(Ω) ∩C0,1(Ω).

Proof. By considering u− ϕ, we may assume ϕ = 0.Step 1. Consider Ω = B+

1 , and let r < 1, i.e., B+r ⊂ B+

1 , and set T := ∂B+r ∩

∂Rn+. Let k ∈ 1, . . . , n. Testing the equation with ∂kψ for ψ ∈ C20 (Ω), we

find that w := ∂ku satisfies

−∂i(aij∂jw) = −∂if ik in Ω, (2.688)

aij := ∂pjAi, f ik(x, z, p) := −pk∂zAi − ∂kAi + δikB.

In particular, in B+r , the coefficients aij are bounded and strongly elliptic

since u and A are continuously differentiable, and similarly f is bounded.Further, by the boundary condition u = 0 on ∂Ω, we have w = ∂ku = 0 on∂Ω ∩ Rn+ for k = 1, . . . , n − 1. In particular oscB(x0,R)∩∂Ωw = 0 for everyx0 ∈ T and R < 1− r. Hence, by Theorem 2.102,

[∂ku]C0,α(B+r+2

3

) ≤ C, k = 1, . . . , n− 1. (2.689)

It remains to bound ∂nu, for which we use the equation and Morrey’s lemma(cf. proof of Theorem 2.19) Let R ≤ 1−r

3 . Then for every y ∈ B+r , we have

194 [January 28, 2019]

B(y, 2R) ⊂ B(0, r+23 ) ⊂⊂ B(0, 1), and thus,

[∂ku]C0,α(B(y,2R)∩Rn+) ≤ C, k = 1, . . . , n− 1 (2.690)

uniformly for y ∈ B+(0, r) and R ≤ 1− r3

.

For y0 ∈ B+r and R ≤ (1 − r)/3, choose a standard cut-off function η ∈

C10 (B(y0, 2R)) with 0 ≤ η ≤ 1, η = 1 on B(y0, R) and |∇η| ≤ 2/R. Set

c :=

w(y0), if B(y0, 2R) ⊂ B+

r ,

0, if B(y0, 2R) ∩ ∂Rn+ 6= ∅.

Then (since w = 0 on ∂B+ ∩ ∂Rn+) we have ζ := η2(w− c) ∈W 1,20 (B+

1 ), i.e.,ζ is an admissible test function. We use that ∂iζ = 2η∂iη(w− c) + η2∂iw toderive∫B+

1

η2aijw∂jw∂iw ≤∫B+

1

|2η(w−c)aij∂iη∂jw|+|η2f ik∂iw|+|2η(w−c)f ik∂iη|,

which by ellipticity and Young’s inequality implies that∫B(y0,R)∩Rn+

|∇w|2 ≤ C

∫B+

1

(η2 + |∇η|2(w − c)2) dy

≤ CRn +Rn−2 supB(y0,2R)

(w − c)2.

By (2.690), we have sup(w − c)2 ≤ C|x− y|2α, and thus∫B(y0,R)∩Rn+

|∂i∂ju|2 ≤ CRn−2+2α j 6= n. (2.691)

Recall that aij and f ik are uniformly bounded and elliptic if K <∞. Then,solving the equation (2.688) for ∂n∂nu, we obtain (cf. (2.195) and its discus-sion)

∂n∂nu = bij∂i∂ju+ b, i = 1, . . . , n, j = 1, . . . , n− 1,

where bij and b are bounded in terms of K, ΛK/λK , and µK/λK with

0 < λK ≤ infΩλ(x, u,∇u), ΛK ≥ sup

Ω|aij |, µK ≥ sup

Ω|f ik|.

Thus by (2.691) ∫B(y0,R)

|∂n∂nu|2 ≤ CRn−2+2α.

195 [January 28, 2019]

Using the Poincare inequality for w = ∂nu and the equivalence of C0,α andL2,n+2α we see that

[∂nu]C0,α ≤ C.

Step 2. Assume now that Ω ⊂ Rn is a general bounded domain withC2-boundary. Then for every x0 ∈ ∂Ω, there exists a ball B := B(x0) anda bijective map ψ : B → D, with D ⊂ Rn open, and

ψ(B ∩ Ω) ⊂ Rn+, ψ(B ∩ ∂Ω) ⊂ ∂Rn+, ψ ∈ C2(B), ψ−1 ∈ C2(D).

We refer to the independent variables on the reference domain as y = ψ(x)(and x = ψ−1(y)), and to the dependent ones as v(y) := u(ψ−1(y)). Furtherset B+ = B ∩ Ω, and D+ = ψ(B+). We may assume that ψ(x0) = 0and B+

1 ⊂⊂ D+. Then, since u = 0 = const on ∂Ω, we have ∂ykv =0 on ∂D+ ∩ ∂Rn+ for k = 1, . . . , n − 1. By chain rule, since ψ is a C2-diffeomorphism, it follows from the equation Qu = 0 in B+ that

We will construct a quasilinear operator in divergence form Q such thatQu = 0 in B+ implies Qv = 0 in D+. Let

ϕ = ψ−1

and define

ai(x) := Ai(x, u(x),∇u(x)), b(x) = B(x, u(x),∇u(x)),

a(y) = det∇ϕ(y) (∇ϕ(y))−1a(ϕ(y)), b(y) = det∇ϕ(y) b(ϕ(y)).

We first claim that Qu = 0 in B+ implies that

−div a+ b = 0 in D+. (2.692)

To see this, let η ∈ C∞c (D+). Then the identity Qu = 0 implies that

0 =

∫B+

∑i

ai(x)∂i(η ψ)(x) + b(x)(η ψ)(x) dx

=

∫B+

∑i,j

ai(x)(∂jη)(ψ(x))∂iψj(x) + b(x)η(ψ(x)) dx

=

∫D+

∑j

(∇ψ(ϕ(y)) a(ϕ(y)) · ∇η(y) + b(ϕ(y))η(y)

det∇ϕ(y) dy.

Taking into account that (∇ψ(ϕ(y)) = (∇ϕ(y))−1 we get (2.692).Next we express a and b in terms of v and ∇v. We have v(y) = u(ϕ(y))

and thus u(x) = v(ψ(x)) and ∇u(x) = (∇ψ(x))T∇v(ψ(x)) and thus

∇u(ϕ(y)) = (∇ϕ(y))−T∇v(y).

196 [January 28, 2019]

Hence

a(y) = det∇ϕ(y) (∇ϕ(y))−1A(ϕ(y), v(y),∇ϕ(y)−T∇v(y))

andb(y) = det∇ϕ(y)B(ϕ(y), v(y),∇ϕ(y)−T∇v(y)).

Thus the desired operator Q is given by

(Qv)(y) = div A(y, v(y)∇v(y)) + B(y, v(y)∇v(y))

withA(y, z, p) = det∇ϕ(y) (∇ϕ(y))−1A(ϕ(y), z,∇ϕ(y)−T p),

andB(y, z, p) = det∇ϕ(y)B(ϕ(y), z,∇ϕ(y)−T p).

We finally verify that Q has the same ellipticity properties as Q. Set

aij =1

2

∂Ai

pj+

1

2

∂Aj

pi

and F = ∇ϕ(y) and note that

∇ppA(y, z, p) = detFF−1(∇pA)(y, z, F−T p)F−T .

Thus ∑ij

aij(y, z, p)ξiξj = detF∑ij

aij(y, z, F−T p)(F−T ξ)i(F

−T ξ)j .

Since F and F−1 are bounded in D+ the operator Q has the same ellipticityproperties as Q.

Now Step 1 applied to the operator Q gives C0,α estimates for ∇v andthese imply C0,α estimates for u (in ψ−1(B+(0, 1

2)). If x0 runs through ∂Ωthe preimages of the balls B+(0, 1

2) ⊂⊂ ψ(B(x0)) form an open cover of ∂Ω.Extracting a finite subcover (∂Ω compact since Ω bounded), and combiningthe result with the interior estimate, we conclude the proof.

Remark 2.105. (i) If n = 2, then the proof works for arbitrary ellipticequations (not necessarily in divergence form). Precisely, an arbitraryquasilinear elliptic equation is equivalent to

−a11

a22∂1∂1u−

2a12

a22∂1∂2u− ∂2∂2u+

b

a22= 0,

and thus the derivative w1 := ∂1u of a solution u ∈ C2 satisfies

∂1

(−a11

a22∂1w1 −

2a12

a22∂2w1

)− ∂2∂2w1 = −∂1

b

a22,

197 [January 28, 2019]

and similarly, for w2 := ∂2u,

∂2

(−a22

a11∂2w2 −

2a12

a11∂1w2

)− ∂1∂1w2 = −∂2

b

a11,

and we proceed as before.

(ii) To treat general equations in higher dimensions, one shows that certaincombinations of derivatives are subsolutions of linear elliptic equationsin divergence form, and uses weak Harnack inequalities. For detailssee [GT, Sections 12.3. and 12.4].

[25.1. 2019, Lecture 28][28.1. 2019, Lecture 29]

2.6.6 Gradient bounds at the boundary

As before we consider a quasilinear operator

Qu = −∑i,j

aij(x, u,∇u)∂i∂ju+ b(x, u,∇u), with aij = aji,

we denote by λ(x, z, p) and Λ(x, z, p) the smallest and largest eigenvalue ofthe symmetric matrix aij(x, z, p) and we assume that Q is elliptic, i.e.,

λ(x, z, p) > 0 ∀(x, z, p) ∈ Ω× R× Rn.

Theorem 2.106. Let Ω ⊂ Rn be open and bounded. Let ϕ ∈ C2(Ω) andassume that u ∈ C2(Ω) satisifies

Qu = 0 in Ωu = ϕ on ∂Ω

Assume in addition that

(i) Ω is convex and ∂Ω is of class C1;

(ii) aij and b depend on only on x and p;

(iii) There exists C1 > 0, p0 > 0 such that

Λ(x, p) ≤ C1λ(x, p)|p|2 if |p| ≥ p0,|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0,

(2.693)

Thenmax∂Ω|∇u| ≤ C(n,C1, p0, ‖ϕ‖C2(Ω), sup

Ω|u|)

198 [January 28, 2019]

Remark. (i) The result applies to the minimal surface operator M. Inthis case b = 0 and Λ(p)/λ(p) = 1 + |p|2. For the minimal surface operatorand n = 2 convexity of Ω is necessary to obtain an estimate for max∂Ω |∇u|for all boundary data ϕ. In higher dimensions the optimal condition is thatthe ∂Ω is mean convex, i.e., that the mean curvature of ∂Ω is ≥ 0.(ii) The result does not apply to the equation of prescribed mean curvatureMu = −nH(x)(1 + |∇u|2)3/2 since in this case λ(p) = 1, Λ(p) = 1 + |p|2,but b(x, p) ∼ |p|3 as p→∞. In this case on needs more stringent conditionson the (mean) curvature of ∂Ω, see [GT], Chapter 14.(iiI) One can obtain similar results for operators for aij and b depend alsoon z by applying the result above to the operator Q defined by

Qw = −∑i,j

aij(x, u,∇w)∂i∂jw + b(x, u,∇w),

compare the proof of the L∞ estimate, Theorem 2.96 and see [GT], Chapter14.

Proof. We first show the result for homogeneous boundary data

ϕ = 0.

In this case it suffices to assume that the second structure condition

|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0, (2.694)

holds. Note that |∇u| = |∂u/∂ν| on ∂Ω since u = 0 on ∂Ω and hence thetangential derivatives of u are zero. We will show that

max∂Ω|∇u| = max

∂Ω

∣∣∣∣∂u∂ν∣∣∣∣ ≤ p0e

2C1M where M := supΩ|u|. (2.695)

Step 1. Reduction to the construction of barriers.We argue similar as in the proof of the Hopf boundary lemma. If sufficesto show that for each point x0 ∈ ∂Ω there exists an open set N ⊂ Ω withB(x0, r)∩Ω ⊂ N for some small r and local barriers u± ∈ C1(N) such that

(i) u±(x0) = 0

(ii) u− ≤ u ≤ u+ in N ;

(iii) ∣∣∣∣∂u±∂ν (x0)

∣∣∣∣ ≤ S (2.696)

199 [January 28, 2019]

where S only depends on C1, p0, and M = supΩ |u|. The first two propertiesimply that the directional derivative in direction of the inner normal satisfies

−∂u−

∂ν≤ −∂u

∂ν≤ −∂u

+

∂ν

Hence the third property implies that∣∣∣∣∂u∂ν∣∣∣∣ ≤ S

Step 2. Ansatz for the barrier.In the following we fix a point x0 ∈ ∂Ω and we focus on u+. The argumentfor u− = −u+ is analogous. We write w = u+. If first condition (i) in Step1 is

w(x0) = 0. (2.697)

To obtain the inequality w ≥ u in N it suffices to show that

w ≥ u on ∂N (2.698)

Qw > 0 in N (2.699)

Then the comparison principle, Theorem 2.94, implies that w > u in N .Condition (2.698) is equivalent to the following two conditions

w ≥ 0 on ∂Ω ∩N , (2.700)

w ≥ u on ∂N ∩ Ω. (2.701)

To summarize we are looking for a function w which satisfies

(2.697), (2.700), (2.701), (2.699)

and ∣∣∣∣∂w∂ν (x0)

∣∣∣∣ ≤ S.Step 3. First attempt: distance from a touching hyperplane.Since Ω is convex the conditions (2.697) and (2.700) can be easily satisfiedby a linear function. Let P be a hyperplane which touches Ω at x0 (such aplane exists since Ω is convex and it is unique since ∂Ω is of class C1) andlet

v(x) = dist (x, P ).

Then v is an affine and nonnegative function in Ω. Indeed, v(x) = −(x −x0) ·ν where ν is the outer normal at x0. Thus if we take w = v then (2.697)and (2.700) hold. It is also easy to satisfy (2.701) if we set

N := x ∈ Ω : v(x) < a.

200 [January 28, 2019]

Then ∂N ∩ Ω = x ∈ Ω : v(x) = a and thus the choice w = αv(x)with α = supΩ |u|/a satisfies (2.700) and (2.701). If a is large then the set∂N ∩Ω can be empty, but in this case (2.701) holds trivially. Unfortunatelythis choice of w does in general not satisfy the remaining condition (2.699)since Qv = b(x,∇v).

Step 4. A monotone function of the distance function.Let ψ ∈ C2[0,∞) with ψ(0) = 0 and ψ′ > 0. If w = ψ(v) then

∂jw = ψ′(v)∂jv, ∂i∂jw = ψ′′(v)∂iv ∂jv + ψ′(v)∂i∂jv

∑i,j

−aij∂i∂jw = −∑i,j

ψ′aij ∂i∂jv︸ ︷︷ ︸=0

− ψ′′

(ψ′)2

∑i,j

aij∂iw∂jw

Assume now in addition that ψ′′ < 0. Then for v = dist (x, P ) the functionw = ψ(v) satisfies

Qw =− ψ′′

(ψ′)2

∑i,j

aij(x,∇w)∂iw∂jw + b(x,∇w)

≥− λ(x,∇w)ψ′′

(ψ′)2|∇w|2 + b(x,∇w).

Now the structure conditions (2.694) implies that Qw > 0 if

|∇w| ≥ p0 in N ,

− ψ′′

(ψ′)2= ν with ν > C1. (2.702)

Since |∇v| = |ν| = 1 we have |∇w| = ψ′(v) and since ψ′ is decreasing thefirst condition is equivalent to

ψ′(a) ≥ p0. (2.703)

Using that ∂N ∩ Ω = x ∈ Ω : v(x) = a we see that the condition (2.701)is satisfied if

ψ(a) ≥M := supΩ|u|. (2.704)

Thus w is a barrier if ψ(0) = 0, ψ′ > 0, and conditions

(2.702), (2.703) and (2.704)

hold.

201 [January 28, 2019]

Step 5. Solution of the ODE − ψ′′

(ψ′)2 = ν.

Let ν > 0 and let η = ψ′. Then

ν = − η′

(η)2=

(1

η

)′and thus 1

η (t) = νt + c and ψ(t) = ν−1 ln(νt + c) + c′. The conditions

w(x0) = 0 is equivalent to ψ(0) = 0 which yields ln c + νc′ = 0. Settingk := ν/c this finally yields

ψ(t) =1

νln(kt+ 1), (2.705)

where ν > 0 and k > 0. Then

ψ′(t) =k

ν

1

1 + kt, ψ′′(t) = − k2

(1 + kt)2.

Step 6. Verification of the conditions (2.702), (2.703) and (2.704).Let ψ be given by (2.705) and let ν = 2C1. Then (2.702) holds. The tworemaining conditions reduce to the inequalities

k

ν

1

1 + ka≥ p0 and

1

νln(1 + ka) ≥M.

To see that these two conditions can be satisfied simultaneously let

ka = eνM − 1

to ensure that the second condition holds. Note that the first condition canbe rewritten as

ap0 ≤1

ν

ka

1 + ka=

1

ν(1− e−νM )

Thus all conditions are satisfied if

a =1

νp0(1− e−νM ), k = νp0e

νM .

Then

|∇w(x0)| = ψ′(0) =k

ν= p0e

νM = p0e2C1M .

Hence we may take S = p0e2C1M in (2.696). This finishes the proof of

(2.695) under the assumption (2.694) for ϕ = 0.

202 [January 28, 2019]

Step 7. Extension to ϕ 6= 0.This is done by a change of variables. We set u = u− ϕ. Then

u = 0 on ∂Ω.

We will show that u is the solution of quasilinear PDE whose coefficientssatisfy the structure condition (2.694). We have

Qu = Q(u+ ϕ) = −∑i,j

aij(x,∇u+∇ϕ)(∂i∂j u+ ∂i∂jϕ) + b(x,∇u+∇ϕ)

Set

aij(x, p) = aij(x, p+∇ϕ(x)),

b(x, p) = −∑i,j

aij(x, p+∇ϕ(x))∂i∂jϕ(x) + b(x, p+∇ϕ(x)).

and define the quasilinear operator Q by

QU = −∑i,j

aij(x,∇U)∂i∂jU + b(x,∇U(x)).

ThenQu = Qu = 0 in Ω.

We will show that the coefficient functions aij and b satisfy the structurecondition

|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0 (2.706)

where λ(x, p) is the smallest eigenvalue of the symmetric matrix aij(x, p).Then by what we have already shown we get the estimate

max∂Ω|∇u| ≤ p0e

2C1M where M := supΩ|u| = sup

Ω|u− ϕ|

and the assertion follows since |∇u| ≤ |∇u|+ |∇ϕ|.To prove (2.706) set

p0 := max(2p0, 2 sup |∇ϕ|).

Then

|p| ≥ p0 =⇒ |p+∇ϕ(x)| ≥ |p| − supΩ|∇ϕ| ≥ 1

2|p| ≥ p0, (2.707)

|p| ≥ p0 =⇒ |p+∇ϕ(x)| ≤ 2|p|. (2.708)

203 [January 28, 2019]

For two symmetric matrices A and B we have |A ·B| ≤ Λ|B| where A ·B :=∑i,j AijBij , |A|2 := A · A and where Λ is the maximum of the absolute

values of the eigenvalues of A.29 Thus

|b(x, p)| ≤ Λ(x, p+∇ϕ(x))‖ϕ‖C2 + |b(x, p+∇ϕ(x))|≤

(2.693),(2.707)C1(‖ϕ‖C2 + 1)λ(x, p+∇ϕ(x)) |p+∇ϕ(x)|2

≤(2.708)

C1(‖ϕ‖C2 + 1)λ(x, p+∇ϕ(x)) 4|p|2

= 4C1(‖ϕ‖C2 + 1) λ(x, p) |p|2

where in the last step we used the relation λ(x, p) = λ(x, p+∇ϕ(x)) whichfollows from aij(x, p) = aij(x, p+∇ϕ(x)). Thus (2.706) holds with

C1 = 4C1(‖ϕ‖C2 + 1)

and the proof is finished

2.6.7 Global gradient bounds

Theorem 2.107. Let A : Rn → Rn be C1 and elliptic, i.e., the lowesteigenvalue λ(p) of the symmetric matrix with coefficients

1

2

(∂Ai

∂pj+∂Aj

∂pi

)satisfies λ(p) > 0 for all p ∈ Rn. Suppose that u ∈ C2(Ω) satisfies

−n∑i=1

∂iAi(∇u) = 0

in Ω. Thenmax

Ω|∇u|2 ≤ max

∂Ω|∇u|2

Remark. (i) Note that the assumption λ(p) > 0 for all p is equivalent to∑i,j

∂Ai

∂pj(p)ξiξj > 0 ∀p ∈ Rn, ξ ∈ Rn \ 0

(ii) The assumptions are satisfied for the minimal surface equation where

A(p) =p√

1 + |p|2

29This is clear if A is diagonal. For the general case observe that for any Q ∈ SO(n)we have A ·B = (QAQT ) ·QBQT and hence |QBQT | = |B|. Now use that there exists aQ ∈ SO(n) such that QAQT is diagonal.

204 [January 28, 2019]

Proof. First assume that u ∈ C3(Ω).General recipe: differentiate the PDE for u with respect to xk multiply

by ∂ku and sum over k to get a PDE for w = 12 |∇u|

2 to which the maximumprinciple can be applied.

Differentiation of the PDE with respect to xk gives by the chain rule

−∑i,j

∂i

[∂Ai

∂pj(∇u)∂j∂ku

]= 0

Multiplying this expression by ∂ku and using the product rule we get

−∑i,j

∂i

[∂Ai

∂pj(∇u)∂j∂ku ∂ku

]+∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂i∂ku (2.709)

Now apply the ellipticity condition with ξ = ∂ku. This shows that thesecond summand in (2.709) is non negative. Set

w =1

2|∇u|2.

Then ∂jw =∑

k ∂ku ∂j∂ku. Thus summing (2.709) over k we get

−∑i,j

∂i∂Ai

∂pj(∇u)∂jw ≤ 0.

Hence w is a subsolution of linear equation

−∑i,j

∂iaij∂jw ≤ 0

where

aij(x) =∂Ai

∂pj(∇u(x))

Since u ∈ C1(Ω) the gradient ∇u(x) takes values in a compact set andtherefore the coefficients aij are bounded and elliptic in the sense that∑

i,j

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

withν = minλ(p) : p ∈ ∇u(Ω) > 0.

Hence by the weak maximum principle

maxΩ

w ≤ max∂Ω

w.

This yields the assertion under the additional assumption u ∈ C3(Ω).

205 [January 28, 2019]

The following argument was not discussed in class.If we only assume u ∈ C2(Ω) then we can do the same calculation in theweak setting. Letting ψ ∈ C∞c (Ω) then using η = −∂kψ in the weak for ofthe equation we get, using integration by parts,

0 = −∫

Ω

∑i

Ai(∇u) ∂i∂kψ dx

=

∫Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂iψ dx

Now u ∈ C1(Ω) implies that ∂Ai

∂pj(∇u) ∈ C(Ω). Thus

[∂Ai

∂pj(∇u)∂j∂ku

]∈

C(Ω) and by density we have

0 =

∫Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂iψ dx ∀ψ ∈W 1,1

0 (Ω).

Now let ζ ∈ W 1,20 (Ω) and take ψ = ζ∂ku. Since ∂iψ = (∂iζ) ∂ku + ζ ∂i∂ku

we deduce that ∫Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku ∂ku

]∂iζ dx

=−∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂i∂ku ζ dx

As before ellipticity implies that the right hand side is≤ 0 if ζ ≥ 0. Summingover k and defining w as above we thus get∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂jw

]∂iζ dx ≤ 0 ∀ζ ∈W 1,2

0 (Ω) with ζ ≥ 0.

Thus w is a weak subsolution of a linear elliptic PDE and the assertionfollows again be the weak maximum principle.

[28.1. 2019, Lecture 29][1.2. 2019, Lecture 30]

206 [January 28, 2019]

3 A brief look back

3.1 Estimates

We essentially used three methods to derive estimates

(i) (L2 estimates) Multiply the equation by η2u and integrate by parts;

(ii) (Max. principle) If u has an interior maximum at x0 thenD2u(x0) ≤ 0;

(iii) (Moser iteration) Multiply the equation by η2|u|β−1u and let β → ±∞.

Strategy (i) also works for systems and higher order equations while (ii)and (iii) work for scalar second order equations; (i) and (iii) are most usefulfor equations in divergence form while (ii) also works for equations in non-divergence form and even fully nonlinear equations.

Consequences of (i):

(i) W 1,2 estimates and reverse Poincare inequality

(ii) Decay estimates for constant coefficients

(iii) Estimates in scaled L2 spaces (Morrey, Campanato, BMO ) by freezingthe coefficients, (i), (ii) and the iteration lemma

(iv) Schauder estimates by C0,α = L2,n+2α (possible alternative proof byblow-up)

(v) Lp estimates by interpolation between L2 and BMO and duality

(vi) Meyers’ Lp estimate for |p− 2| << 1 and L∞ coefficients

(vii) H1 estimates by duality with BMO

These estimates usually come in two very closely related variants. Localestimates (estimate of a stronger norm in a ball by the right hand sideand the solution in a weaker norm in a larger ball) and global estimates inRn. For the passage from local to global estimates see Corollary 2.16 forthe reverse passage see Lemma 2.43. For estimates up to the boundary oneusually uses a partition of unity and maps the neighbourhood of a boundarypoint to a half-ball (’flattening the boundary’). For estimates in a half-ballor halfspace one can still use tangential derivatives of the equation. Toestimate the second derivative in normal direction one uses the equation.Note:

L1 and L∞ are ’bad’ spaces for elliptic equations.The good replacements are the Hardy space H1 and BMO .

For right hand side in L1 one can obtain estimates in the weak-Lp spaces.

Consequences of (ii):

207 [January 28, 2019]

(i) Maximum principle: c = 0, Lu ≤ 0 =⇒ maxΩ u ≤ max∂Ω u

(ii) Hopf boundary lemma by squeezing in a supersolution on a touchingannulus

(iii) Strong maximum principle

(iv) Comparison principle for quasilinear or fully nonlinear equations

(v) Boundary gradient estimates by barrier functions

Consequences of (iii):

(i) Harnack inequality for weak sub- and supersolutions

(ii) C0,α estimates for equations in divergence form with L∞ coefficients

3.2 Regularity for nonlinear PDE

(i) From C0 or C1 to C∞ by linear theory

(ii) Initial (partial) regularity for harmonic maps byH1 estimates for div−curl quantities, H1 − BMO duality and the monotonicity formula

(iii) Role of symmetries and good gauge to get div − curl structure

(iv) Partial C1,α regularity for minimizers of quasiconvex integrals by ex-cess decay

(v) Excess decay by blow-up and Cacciopoli inequality/ compactness

3.3 Existence

(i) For elliptic equations without lower order terms by Lax-Milgram

(ii) For general elliptic equations by the Fredholm alternative

(iii) For elliptic systems with continuous coefficients by Garding’s inequal-ity and the Fredholm alternative

(iv) Existence in W 1,p for p > 2 by regularity, for 1 < p < 2 by duality.

(v) For Euler-Lagrange equations by of variational problems by the directmethod of the calculus of variations / weak lower semicontinuity (forconvex or quasiconvex integrands)

(vi) Existence for monotone equations by the Minty-Browder trick andweak convergence (not discussed in class)

(vii) Existence of classical solutions for quasilinear elliptic equations by theLeray-Schauder fixed point theorem and C1,α a priori estimates.

208 [January 28, 2019]

3.4 Some topics that were not covered

(i) Neumann boundary conditions for elliptic systems

(ii) The Stokes system −∆u+∇p = f, div u = 0 (not an ADN system)

(iii) The Alexandrov-Bakelmann-Pucci maximum principle (see [GT],Chapter 9.1)

(iv) C0,α estimates for equations in non divergence form (see [GT], Chapter9.8., 9.9)

(v) Classical solutions for fully nonlinear equations (see [GT], Chapter17).

(vi) Viscosity solutions for fully nonlinear equations (a notion of solutionfor C0 functions based on the maximum principle and Perron’s methodto obtain solutions as the supremum of subsolutions, see [CIL])

209 [January 28, 2019]

References

[Ca66] S. Campanato, Su un teorema di interpolazione di G. Stampacchia,Ann. Scuola Norm. Sup. Pisa 20 (1966), 649–652.

[CIL] M.G. Crandall, H. Ishii and P.-L. Lions, User’s guide to viscositysolutions of second order partial differential equations, Bull. Amer.Math. Soc. 27 (1992), 1–67.

[CLMS] R. Coifman, P.-L. Lions, Y. Meyer and S. Semmes, Compensatedcompactness and Hardy spaces, J. Math. Pures Appl. 72 (1993),247–286.

[Da08] B. Dacorogna, Direct methods in the calculus of variations,Springer, 2nd ed., 2008.

[Ev90] L.C. Evans, Weak convergence methods for nonlinear partial dif-ferential equations, Amer. Math. Soc., 1990

[Ev] L.C. Evans, Partial differential equations, Amer. Math. Soc., 1998.

[Ev86] L.C. Evans, Quasiconvexity and partial regularity in the calculusof variations, Arch. Rat. Mech. Anal. 95 (1986), 227–252.

[EvG] L.C. Evans and R.F. Gariepy, Blowup, compactness and partialregularity in the calculus of variations, Indiana Univ. Math. J. 36(1987), 361–371.

[FS] C. Feffermann and E.M. Stein, Hp spaces of several variables, ActaMath. 129 (1972), 137–193.

[FH86] N. Fusco, J.E. Hutchinson, C1,α partial regularity of functions min-imising quasi-convex integrals, Manuscr. Math. 54 (1986), 121–143.

[Gi] M. Giaquinta, Multiple integrals in the calculus of variations andnonlinear elliptic systems, Princeton Univ. Press, 1983.

[GM86] M. Giaquinta and G. Modica, Partial regularity of minimizers ofquasiconvex integrals. Ann. Inst. H. Poincare Anal. Non Lineaire3 (1986), 185–208.

[GT] D. Gilbarg and N.S. Trudinger, Elliptic partial differential equa-tions of second order, Springer, 1998.

[He] F. Helein, Harmonic maps, conservation laws and moving frames.Cambridge Tracts in Mathematics, vol. 150. Cambridge UniversityPress, Cambridge, 2002.

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[Jo] F. John, Rotation and strain, Comm. Pure Appl. Math. 14 (1961),391–413.

[JN] F. John and L. Nirenberg, On functions of bounded mean oscilla-tion, Comm. Pure Appl. Math. 14 (1961), 415–426.

[KS] D. Kinderlehrer and G. Stampacchia, An introduction to varia-tional inequalities and their applications, Academic Press, 1980(reprinted as Classics in Applied Mathematics, SIAM, 2000).

[K17] S. Klainerman, S, On Nash’s unique contribution to analysis injust three of his papers. American Mathematical Society. Bulletin.New Series, 54 (2016), 283–305.

[LD] H. Le Dret, An example of H1-unboundedness of solutions tostrongly elliptic systems of partial differential equations in a lam-inated geometry, Proc. Roy. Soc. Edinburgh Sect. A 105 (1987),77–82.

[Li69] J.L. Lions, Quelques methodes de resolutions des problemes auxlimites non lineaires, Dunod Gauthier-Villars, 1969.

[Me63] N.G. Meyers, An Lp estimate for the gradient of solutions of secondorder divergence equations, Ann. SNS Pisa 17 (1963), 189–206.

[MiKr] G. Mingione, J. Kristensen, The singular set of Lipschitzian min-ima of multiple integrals, Arch. Ration. Mech. Anal. 184 (2007),341–369.

[MS16] C. Mooney and O. Savin, Some singular minimizers in low dimen-sions in the calculus of variations. Archive for Rational Mechanicsand Analysis, 221 (2016), 1–22.

[Mu] S. Muller, A surprising higher integrabilit—appings with positivedeterminant, Bull. Amer. Math. Soc. 21 (1989), 245–248.

[Mu99] S. Muller, Variational models for microstructure and phase tran-sitions, in: Proc. Calculus of variations and geometric evolutionproblems (Cetraro 1996), Eds.: F. Bethuel, G. Huisken, S. Mullerand K. Steffen, Springer Lecture Notes in Mathematics 1713, 1999,pp. 85–210 (see also http://www.mis.mpg.de/publications/other-series/ln/lecturenote-0298.html)

[MS03] S. Muller and V. Sverak, Convex integration for Lipschitz map-pings and counterexamples to regularity, Ann. Math. 157 (2003),715–742.

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[Ne] J. Necas: Example of an irregular solution to a nonlinear ellip-tic system with an- alytic coefficients and conditions of regularity,Theory of Non Linear Operators, Abhandlungen Akad. der Wis-senschaften der DDR (1977), Proc. of a Summer School held inBerlin (1975).

[Ri95] T. Riviere, Everywhere discontinuous harmonic maps into spheres,Acta Math. 175 (1995), 197–226.

[Ri07] T. Riviere, Conservation laws for conformally invariant problems,Invent. Math. 168 (2007), 1–22.

[RiSt] T. Riviere and M. Struwe, Partial regularity for harmonic mapsand related problems, Comm. Pure Appl. Math. 61 (2008), 451–463

[Se] S. Semmes, A primer on Hardy spaces, and some remarks on atheorem of Evans and Muller, Comm. PDE 19 (1994), 277-319.

[Si] L. Simon, Schauder estimates by scaling, Calc. Var. 5 (1997), 391–407.

[St70] E.M. Stein, Singular integrals and differentiability properties offunctions, Princeton Univ. Press, 1970.

[St93] E.M. Stein, Harmonic analysis - Real-variable methods, orthogo-nality, and oscillatory integrals, Princeton Univ. Press, 1993.

[SY00] V. Sverak and X. Yan, Calc. Var. 10 (2000), 213–221.

[SY02] V. Sverak and X. Yan, Non Lipschitz minimizers of smooth uni-formly convex functionals, Proc. Natl. Acad. Sci. 99 (2002), 15269–15276.

212 [January 28, 2019]

A Interpolation between BMO and L2: proofs

This was not discussed in class

Lemma A.1 (Weak Lp estimate for BMO p). Let Q0 be a cube in Rn. Wesay that ∆ is a subdivision of Q0 if ∆ is a finite or countable collectionof subcubes of Q0 which have no common interior point and Q0 ⊂

⋃Q∈∆.

Assume that f ∈ L1(Q0) and that there exists a constant K > 0 such that∑Q∈∆

[1

Ln(Q)

∫Q|f − fQ| dx

]pLn(Q)

1p

≤ K (A.1)

for all subdivisions ∆. Then

Lnx ∈ Q0 : |f − fQ0 | > t ≤ AKp

tp(A.2)

where A only depends only on n and p.

Remark. Lemma A.1 and Lemma 2.32 also hold for functions with valuesin Rd if | · | denotes a norm in Rd. The constant A is independent of d.

Proof. This proved by induction, very similar to the proof of Lemma 2.32.See [JN], pp. 423–425.

We note that if f ∈ Lp(Q0) then condition (A.1) holds with K = 2‖f‖Lp .Indeed

∫Q |f − fQ| dx ≤ 2

∫Q |f | dx and thus by Jensens’s inequality[

1

Ln(Q)

∫Q|f − fQ| dx

]p≤ 2p

1

Ln(Q)

∫Q|f |p dx

and thus∑Q∈∆

[1

Ln(Q)

∫Q|f − fQ| dx

]pLn(Q) ≤ 2p

∑Q∈∆

∫Q|f |p dx = 2p

∫Q0

|f |p dx.

(A.3)

We now use Lemma A.1 to prove the interpolation result Theorem 2.36which was the basis of our approach to the Lp estimates. We first recall theMarcinkiewicz interpolation theorem.

Let E and E′ be measurable subsets of Rn. Assume that T maps mea-surable function on E to measurable functions on E′.We say that T is of type (p, q) if there exists a constant A such that

‖Tf‖Lq ≤ A‖f‖Lp ∀f ∈ Lp(E). (A.4)

213 [January 28, 2019]

Similarly, for q ∈ [1,∞), we say that T is of weak-type (p, q) if there existsa constant A such that

Ln(x ∈ E′ : |Tf(x)| > t) ≤(A‖f‖Lp

t

)q∀f ∈ Lp(E), t > 0. (A.5)

Note that if T is of type (p, q) it is in particular of weak-type (p, q) (forq <∞) since for any g ∈ Lq(E′) we have

Ln(x ∈ E′ : |g| > t) ≤(‖g‖Lpt

)q. (A.6)

We say that T is sub-additive if |T (f + g)(x)| ≤ |Tf(x)| + |Tg(x)|. ByL1(E) + Lr(E) we denote that space of functions which can be written inthe form f + g with f ∈ L1(E) and g ∈ Lr(E).

Theorem A.2 (Marcinkiewicz interpolation theorem). Let 1 ≤ p < r ≤ ∞.Suppose that T is a subadditive mapping from Lp(E)+Lr(E) to the space ofmeasurable functions on E′. If T is simultaneously of weak type (p, p) andweak type (r, r) then T is of type (q, q) for all q ∈ (p, r). More precisely if Tsatisfies

(i) |T (f + g)(x)| ≤ |Tf(x)|+ |Tg(x)| ∀f, g ∈ Lp(E) + Lr(E),

(ii) Ln(x ∈ E′ : |Tf(x)| > t) ≤(Ap‖f‖p

t

)p∀f ∈ Lp(E).

(iii) If r <∞ then

Ln(x ∈ E′ : |Tf(x)| > t) ≤(Ar‖f‖r

t

)r∀f ∈ Lr(E).

If r =∞ then ‖Tf‖L∞ ≤ A∞‖f‖L∞ ∀f ∈ L∞(E).

Then‖Tf‖Lq ≤ Aq‖f‖Lq ∀f ∈ Lq(E) (A.7)

for all q ∈ (p, r) where Aq depends only on Ap, Ar, p, q and r.

Remark. The result also holds for vector-valued functions. More preciselyif T maps Lp(E;Rd1) +Lr(E;Rd1) into measurable functions with values inRd2 then the result holds if we interpret | · | as the norm in Rd1 and Rd2 ,respectively. The constant Ap does not depend on d1 and d2.

Proof. See [St70], Theorem 4 in Section I.4, pp. 21-22 and Appendix B, pp.272–274. Note that it suffices to consider the case E = E′ = Rn since wecan extend all functions by zero to Rn.

214 [January 28, 2019]

Theorem A.3 (Lp estimate from L2 and BMO estimates). Let Ω be ameasurable set in Rn with finite measure and suppose that Q0 ⊂ Rn is acube. Assume that T : L2(Ω;Rd1) → L2(Q;Rd2) is a linear operator suchthat

‖Tu‖L2(Q) ≤ M2‖u‖L2(Ω) ∀u ∈ L2(Ω;Rd1) (A.8)

[Tu]BMO (Q) ≤ M∞‖u‖L∞(Ω) ∀u ∈ L∞(Ω;Rd1) (A.9)

Then for all 2 < q <∞

‖Tu‖Lq(Ω) ≤ C‖u‖Lq(Ω) ∀ f ∈ Lq(Ω;Rd1) (A.10)

where C depends on q, n,M2,M∞ and Ln(Ω)/Ln(Q).

Proof. This proof is taken from [Ca66].Let ∆ be a subdivision of the cube Q0. For f ∈ L2(Ω) define T u as thefollowing function which is constant on each Q ∈ ∆

(T u)(x) :=1

Ln(Q)

∫Q|Tu− (Tu)Q| ∀x ∈ Q. (A.11)

Then T is sublinear. Moreover

‖T (u)‖L∞ ≤ [Tu]BMO ≤M∞‖f‖L∞

and (use that∫Q |f − fQ|

2 ≤∫Q |f |

2 and compare also (A.3))

‖T u‖2L2 =∑Q∈∆

[1

Ln(Q)

∫Q|Tu− (Tu)Q| dx

]2

Ln(Q)

≤∑Q∈∆

∫Q|Tu|2 dx =

∫Q0

|Tu|2 dx ≤M2‖u‖2L2

Thus T is of type (2, 2) and of type (∞,∞). By the Marcinkiewicz interpo-lation theorem

‖T u‖Lr ≤Mr‖u‖Lr (A.12)

for all r ∈ (2,∞) were Mr depends only on M2, M∞ and r.30 In particularthe constant Mr is independent of the partition ∆.

Using the definition of T we get∑Q∈∆

[1

Ln(Q)

∫Q|Tu− (Tu)Q| dx

]rLn(Q) = ‖T u‖rLr ≤M r

r ‖u‖rLr (A.13)

30Actually the Riesz-Thorin interpolation theorem gives the more precise estimate Mr ≤M1−θ

2 Mθ∞ where 1/r = (1− θ)/2, but we do not need this.

215 [January 28, 2019]

Now Lemma A.1 implies that

Ln(x ∈ Q0 : |Tu− (Tu)Q0 | > t ≤ Ar(Mr‖u‖Lr

t

)r. (A.14)

Therefore the map u 7→ Tu− (Tu)Q0 is of weak type (r, r) for all r <∞. Byassumption this map is also of type (2, 2) since ‖Tu− (Tu)Q0‖L2 ≤ ‖Tu‖L2 .

Now let q ∈ (2,∞) and take r = 2q. It follows from the Marcinkiewiczinterpolation theorem that the map u 7→ Tu− (Tu)Q0 is of type (q, q) and

‖Tu− (Tu)Q0‖Lq ≤M ′q‖u‖Lq , (A.15)

whereM ′q = M ′q(q, n,M2,M∞). (A.16)

Finally we need to estimate (Tu)Q0 . For this we use the obvious esti-mates

1

Ln(Q0)

∫Q0

|(Tu)Q0 |q dx ≤ |(Tu)Q0 |q ≤(

1

Ln(Q0)

∫Q0

|Tu|2)q/2

≤M q2

(Ln(E)

Ln(Q0)

1

Ln(E)

∫E|u|2)q/2

≤M q2

(Ln(E)

Ln(Q0)

)q/2 1

Ln(E)

∫E|u|q dx

(A.17)

where we used Jensen’s inequality in the second and in the last step. Itfollows that

‖(Tu)Q0‖Lq ≤M2

(Ln(E)

Ln(Q0)

) 12− 1q

‖u‖Lq

and together with (A.15) this finishes the proof.

B Regularity for two-dimensional systems with W 1,1

coefficients

Here we sketch the proof of Theorem 2.74 which played a key role in theregularity of two-dimensional weakly harmonic map with general target. Wefirst restate the result.

Theorem B.1. Let B ⊂ R2 be a ball, let Aij ∈ W 1,1(B;Rm) for i, j ∈1, . . . ,m and let W ∈W 1,2(B;Rm) be a weak solution of

−∆W = A ·W in B. (B.1)

Let B′ ⊂ B be a concentric subball. Then W ∈ C0,α(B′,Rm) and ∇W ∈L2,2α(B′;Rm) for any α ∈ (0, 1).

216 [January 28, 2019]

Sketch of proof. Step 1. L2,∞ estimate.Let f ∈ L1(Ω) and 1 < p < 2. We claim that there exists one and only onedistributional solution w ∈W 1,p

0 of

−∆w = f in B

and that ∇w is in the weak L2 space L2,∞ and

‖∇w‖L2,∞(B) ≤ C‖f‖L1(B) (B.2)

where for a measurable function b we define

‖b‖L2,∞(B) := inf

M : L2(x ∈ B : |b(x)| > λ) ≤ M2

λ2

and C =

√8π/π. Note that ‖v‖L2,∞(B) is not really a norm (it satisfies the

triangle inequality only up to a factor), but it is equivalent to a norm.Proof: By scaling vr(z) = rv(rz) it suffices to show the result for the

unit ball. The main point is to show the estimate for sufficiently regular f .Let f ∈ C∞c (B). From Introduction to PDE we know that a classical (andhence distributional) solution is given by

w(x) =

∫BG(x, y)f(y) dy.

Here Greens function G is defined as

G(x, y) =1

2πln

1

|x− y|− 1

2πln

1

|y|2|x− y|, y =

y

|y|2

For |x| < 1 and |y| < 1 we have 31 |x− y| ≥ |x− y|. Thus

|∇xG(x, y)| ≤ 1

2π

1

|x− y|+

1

2π

1

|x− y|≤ 1

π

1

|x− y|.

Let

K(x, y) :=1

|x− y|.

Then the assertion (for smooth f) follows from the estimate

‖K ∗ f‖L2,∞(R2) ≤ C‖f‖L1(R2). (B.3)

To prove (B.3) let M = ‖f‖L1 and define

K1(z) :=

1|z| if |z| ≤ 2M

λ

0 else, K2 := K −K1.

31Proof: |y|2|x− y|2 = |y|2|x|2 − 2x · y + 1 = |x− y|2 + (1− |x|2)(1− |y|2) ≥ |x− y|2

217 [January 28, 2019]

Then |K2| ≤ λ/(2M) and hence |K2 ∗ f | ≤ λ/2. Now

x : |K ∗ f |(x) > λ ⊂ x : |K1 ∗ f |(x) > λ/2 ∪ x : |K2 ∗ f |(x) > λ/2

and thus

L2(x : |K ∗ f |(x) > λ) ≤ L2(x : |K1 ∗ f |(x) > λ/2)

≤ 2

λ‖K1 ∗ f‖L1 ≤

2

λ‖K1‖L1‖f‖L1 ≤

2

λ‖K1‖L1M.

Finally, using polar coordinates we get∫Rn|K1| dx = 2π

∫ 2M/λ

0

1

rr dr = 2π

2M

λ

and thus (B.3) holds with C =√

8π.The proof of the rest of the assertion is a standard approximation and

duality argument (this argument was not discussed in class).32

Step 2. A slight improvement of the embedding W 1,1(R2) → L2(R2).Assume that a ∈W 1,1(R2) and a = 0 outside a compact set33. Then

∞∑k=−∞

2k(L2(x ∈ R2 : |a(x)| > 2k)

)1/2≤ C‖∇a‖L1 (B.4)

The space of functions a for which the left hand side is bounded is theLorentz space L2,1. For the general definition of the Lorentz spaces Lp,q

32For f ∈ L1 we consider fk ∈ C∞c (B) with fk → f in L1. Since ‖g‖Lp(B) ≤Cp‖g‖L2,∞(B) for any p < 2 it follows that the corresponding solutions wk are a Cauchy

sequence in W 1,p0 (Ω) and hence converge to v in W 1,p

0 (Ω). One easily checks that v isa distributional solution of the equation and that ‖∇v‖L2,∞ ≤ lim infk→∞ ‖∇wk‖L2,∞ .Finally to show uniqueness it suffices to consider f = 0 and we argue by duality. Let p′

be the dual exponent of p. By assumption we have∫B

∇v · ∇ψ dx = 0

∫B

g · ∇η dx ∀ψ ∈ C∞c (B).

Since ∇w ∈ Lp this identity holds for all ψ ∈ W 1,p′

0 (B) Now let g ∈ Lp′(B;R2). Since

p′ ≥ 2 there exists a weak solution ϕ ∈W 1,20 (B) of −∆ϕ = −div g, i.e.,∫

B

∇ϕ · ∇η dx =

∫B

g · ∇η dx ∀η ∈W 1,20 (B).

For the Lq regularity theory we have ∇w ∈ W 1,p′(B) and hence this identity holds forη ∈W 1,p

0 (Ω). Taking η = w and ψ = ϕ we get∫B

g · ∇w dx =

∫B

ϕ · ∇v = 0.

Since this holds for all g ∈ Lp′(B,R2) we deduce that ∇w = 0 and hence w = 0.

33this condition is not really needed and can easily be removed by approximation

218 [January 28, 2019]

with p ∈ [1,∞) and q ∈ [1,∞] and their properties see, e.g., [?]. To comparethe left hand side of (B.4) with the usual Lp,q norm compare the sum to anintegral (using the monotinicity of λ 7→ L2(|a| > λ) and use the change ofvariables of variables t = 2k.

To show (B.4) we may assume that a ≥ 0 since |a| ∈ W 1,1(R2) and| ∇|a| | = |∇a| a.e. Define

ak =

a− 2k−1 if 2k−1 < a < 2k+1,

0 if a ≤ 2k−1,

2k+1 − 2k−1 if a ≥ 2k+1.

Note that ak can also be written as ak = min(max(a, 2k−1), 2k+1

)− 2k−1.

Thus

∇ak = ∇a χFk a.e. where Fk = x ∈ Rn : 2k−1 ≤ a(x) < 2k+1.

The sets Fk are not disjoint but each point x is contained in at most two ofthe sets Fk. Thus ∑

k

χFk ≤ 2. (B.5)

Moreover ak has compact support (since a has compact support anda = 0 implies that ak = 0). Hence by the Sobolev embedding theorem

‖ak‖L2 ≤ C∫R2

|∇ak| dx ≤ C∫R2

χFk |∇a| dx.

Now we have

L2(x ∈ R2 : a(x) > 2k) = L2(x ∈ R2 : ak(x) > 2k−1) ≤ 1

22(k−1)

∫R2

|ak|2 dx.

Taking the square root, multiplying by 2k−1, summing over k and using(B.5) we get

∞∑k=−∞

2k−1(L2(x ∈ R2 : a(x) > 2k)

)1/2≤∑‖ak‖L2 ≤

∫R2

∑k

χFk︸ ︷︷ ︸≤2

|∇a| dx.

This proves (B.4).Step 3. L2,1-L2,∞ duality.

Let Br be a ball of radius r. Let a ∈ W 1,1(Br) and assume that b is inL2,∞(Br), i.e.,

L2(x ∈ Br : |b(x)| ≥ λ) ≤ B2

λ2.

219 [January 28, 2019]

We claim that ab ∈ L1(Br) and∫Br

|ab| dx ≤ CB(‖∇a‖L1(Br) + ‖a‖L2(Br)). (B.6)

It suffices to prove the result for r = 1. Indeed one can define ar(z) = ra(rz),br(z) = rb(rz) and one easily sees that the assertion for ar and br impliesthe assertion for a and b.

Thus assume r = 1 and set B = Br. We can extend a to a functiona ∈W 1,1(R2) which vanishes outside a ball of radius 2 and which satisfies

‖a‖W 1,1(Rn) ≤ C‖a‖W 1,1(B) ≤ C(‖∇a‖L1(B) + ‖a‖L2(B)).

SetEk := x ∈ B : 2k ≤ |a(x)| < 2k+1.

Since a = a in B we get from Step 2 applied to a

∞∑k=0

2k(L2(Ek))1/2 ≤ C(‖∇a‖L1(B) + ‖a‖L2(B)). (B.7)

Now we claim that for any measurable set E ⊂ B∫E|b| dx ≤ 2‖b‖L2,∞(E)(L2(E))1/2 (B.8)

(in other words functions in weak L2 satisfy the same Holder inequality asfunctions in L2). To see this recall that the left hand side equals∫ ∞

0L2(x ∈ E : |b(x)| ≥ λ) dλ ≤

∫ ∞λ∗

‖b‖2L2,∞

λ2dλ+ λ∗L2(E)

≤‖b‖2L2,∞

λ∗+ λ∗L2(E)

for any λ∗ > 0. The choice λ∗ = ‖b‖L2,∞(L2(E))−1/2 gives the assertion.Thus∫

B|ab| dx ≤

∞∑k=−∞

∫Ek

2k+1|b| dx ≤∞∑

k=−∞2k+2‖b‖L2,∞(L2(Ek))

1/2

≤ C‖b‖L2,∞(‖∇a‖L1(B) + ‖a‖L2(B)).

Step 4. Decay of the L2,∞ norm.This is proved in the usual way by decomposing W into a harmonic functionand a function with zero boundary conditions and using the iteration lemma.

Let B′ ⊂ B be a ball which is compactly contained in B. Let x ∈ B′and let B(x,R) ⊂ B and let r ∈ (0, R). Write W = v + w where w is theunique solution of

−∆w = A · ∇W in B(x, r)

220 [January 28, 2019]

with zero boundary conditions in the sense of Step 1.By Step 3 we have

‖A · ∇W‖L1(B(x,r) ≤ Cµ(r)‖∇W‖L2,∞ ,

where

µ(r) := sup‖∇A‖L1(B(z,r) + ‖A‖L2(B(z,r) : B(z, r) ⊂ B.

Since A ∈W 1,1(B) ⊂ L2(B) we have

limr→0

µ(r) = 0.

Thus by using Step 1 for each component of w we get

‖∇w‖L2,∞(B(x,r)) ≤ Cµ(r)‖∇W‖L2,∞(B(x,r)). (B.9)

Now ∆v = 0 and thus for ρ ≤ r/2∫B(x,ρ)

|∇v|2 dx ≤ C(ρr

)2∫B(x,r/2)

|∇v|2 dx

≤ C(ρr

)2 1

r2

∫B(x,r)

|v − (v)x,r|2 dx.

Now we claim that for any function g with ∇g ∈ L2,∞(B(x, r)∫B(x,r)

|g − (g)x,r|2 dx ≤ Cr2‖∇g‖2L2,∞(B(x,r). (B.10)

Indeed by the Poincare-Sobolev estimate and (B.8) we have∫B(x,r)

|v−(v)x,r|2 dx ≤ C

(∫B(x,r

|∇v| dx

)2

≤ CL2(B(x, r)‖∇v‖2L2,∞(B(x,r).

Together with the trivial estimate ‖∇v‖L2,∞(B(x,ρ)) ≤ ‖∇v‖L2(B(x,ρ)) we con-clude that

‖∇v‖L2,∞(B(x,ρ)) ≤ C(ρr

)2‖∇v‖2L2,∞(B(x,r)). (B.11)

Combining this with (B.9) in the usual way we deduce that

‖∇W‖2L2,∞(B(x,ρ)) ≤ C[µ2(r) +

(ρr

)2]‖∇W‖2L2,∞(B(x,r)).

Since limr→0 µ(r) = 0 it follows from the iteration lemma (Lemma 2.15)that for each λ ∈ (0, 2) there exists a Cλ which depends on B′ and µ (butnot on x ∈ B′) such that

‖∇W‖2L2,∞(B(x,ρ)) ≤ Cλρλ‖∇W‖2L2,∞(B). (B.12)

221 [January 28, 2019]

Step 5. C0,α and L2,λ estimates.It follows from (B.10) and (B.12) that∫

B(x,r)|W − (W )x,r|2 dx ≤ Cr2‖∇W‖2L2,∞(B(x,r)) ≤ Cλr

2+λ‖∇W‖2L2,∞(B).

Thus W ∈ L2,2+λ(B′) = C0,λ/2(B′) for all λ < 2.Finally to get the L2,λ bound for∇W we argue as in the proof of Theorem

2.61. Using the weak maximum principle as in that proof and the C0,α

estimate for W we see that the function w defined in Step 4 satisfies

supB(x,r)

|w| ≤ Crα.

Thus using w as a test function we get∫B(x,r)

|∇W |2 dx ≤ Crα‖A‖L2‖∇W‖L2(B(x,r) ≤C

εr2α + ε‖∇W‖2L2(B(x,r)).

Together with the estimate for ‖∇v‖L2(B(x,ρ)) this yields

‖∇W‖2L2(B(x,ρ)) ≤[ε+ C

(ρr

)2]‖∇W‖2L2(B(x,r)) +

C

εr2α.

By the iteration lemma this implies that ∇W ∈ L2,2α(B′) for all α ∈ (0, 1).

222 [January 28, 2019]