introduction to partial di erential equations thierry cazenave · introduction to partial di...

Introduction to partial differential equations

Thierry Cazenave

Sorbonne Universite & CNRS, Laboratoire Jacques-Louis Lions,B.C. 187, 4 place Jussieu, 75252 Paris Cedex 05, France

Email address: [email protected]

Version: October 30, 2019

mailto:[email protected]

Contents

Foreword 5

Notation 7

Chapter 1. Laplace’s equation 111.1. Existence and interior regularity 111.1.1. On RN 111.1.2. On a domain (connected open set), bounded or not 121.2. The maximum principle 161.2.1. The weak maximum principle 171.2.2. The strong maximum principle 171.3. Lp(Ω) et C0(Ω) regularity 191.3.1. Lp regularity 191.3.2. C0 regularity 251.4. Spectral decomposition of the Laplacian 27

Chapter 2. The heat equation 352.1. The heat equation on RN 352.1.1. Existence and regularity 352.1.2. The heat semigroup 392.1.3. The nonhomogeneous equation and Duhamel’s formula 412.2. The heat equation on a bounded domain 432.2.1. Existence and regularity 432.2.2. The heat semigroup 462.2.3. The nonhomogeneous equation and Duhamel’s formula 472.3. The maximum principle 492.3.1. The weak maximum principle 492.3.2. The strong maximum principle 502.3.3. Some applications of the maximum principle 53

Chapter 3. The one-dimensional wave equation 553.1. The wave equation on the line 553.2. The wave equation on an interval 58

Appendix A. Some useful results 61A.1. Functional analysis 61A.2. Inequalities 66

Appendix B. Sobolev spaces 69B.1. Definitions and basic properties 69B.2. The chain rule and applications 76B.3. Sobolev’s inequalities 80B.4. Compactness properties 96B.5. The case of complex-valued functions 101B.6. The Fourier transform and Sobolev spaces 103

3

4 CONTENTS

B.6.1. The Fourier transform on RN 103B.6.2. The Schwartz space S(RN ) 105B.6.3. The Fourier transform on L2(RN ) 108B.6.4. Tempered distributions: the space S ′(RN ) 108B.6.5. Fractional order Sobolev spaces. The L2 case 113B.6.6. Fractional order Sobolev spaces. The general case 115

Appendix C. Vector integration 121C.1. Measurable functions 121C.2. Integrable functions 123C.3. The space Lp(I,X) 126

Bibliography 129

Foreword

Partial differential equations (PDEs) are a fundamental tool for modeling nat-ural phenomena, and anyone working in the field of applied mathematics shouldknow its bases. The purpose of these lectures is to present three classes of PDEs(elliptic, parabolic, hyperbolic), to introduce their fundamental properties, and tointroduce the mathematical tools that are necessary for their study. Prerequisitesfor these lectures are the bases of functional analysis, of distribution theory, and ofSobolev spaces.

PDEs have been used since the mid-18th century for modeling various phenom-ena arising in particular in mechanics, physics, biology, chemistry, etc. It is anextremely vast domain of study, even if one consider only linear equations.

We consider here typical examples in the three main classes of PDEs: elliptic,parabolic, hyperbolic. More specifically, we consider Laplace’s equation

−∆u+ λu = f

the heat equation∂u

∂t−∆u = f

and the one-dimensional wave equation

∂2u

∂t2− ∂2u

∂x2= f

We note when these equations are set on a bounded domain, one needs bound-ary conditions, which are in general imposed by the physical phenomena that theseequations model. For instance, a vibrating string fixed at its ends. Classical bound-ary conditions are Dirichlet (one imposes u on the boundary of the domain), Neu-mann (one imposes the normal derivative of u on the boundary of the domain), andRobin (a combination of the preceding two conditions). The boundary conditionscan be homogeneous (for instance u = 0 in the case of Dirichlet) or nonhomoge-neous (for instance u = g, for a given function g, in the case of Dirichlet). In theselectures, we only use homogeneous Dirichlet boundary conditions, which are ingeneral technically simpler and do not require regularity conditions on the domain.

For each of these three examples, we first consider the case of the equation seton the whole space, in which one can use explicit formulas given by the Fouriertransform; then the case of a bounded domain with boundary conditions, in whichone can use Fourier series.

For Laplace’s equation, we study the following questions: existence, regularity,maximum principle, spectral decomposition of the Laplacian.

For the heat equation, we study the following questions: existence, regularity,smoothing effect, maximum principle, large time behavior.

For the one-dimensional wave equation, we study the following questions: ex-istence, regularity, finite speed of propagation, large time behavior.

Nonlinear PDEs is an extremely vast domain, and many different phenomenacan take place. If we have time, we will study a few fundamental phenomena, onlyfor semilinear equations. More precisely, we will study the existence (by variational

5

6 FOREWORD

methods) for cettain semilinear elliptic equations. For the semilinear heat equation,we will study local existence and the blowup alternative, global existence, finite-time blowup.

Notation

a.a. almost alla.e. almost everywhereu ? v the convolution in RN , i.e.

u ? v(x) =

∫RN

u(y)v(x− y) dy =

∫RN

u(x− y)v(y) dy

F the Fourier transform in RN , defined by

Fu(ξ) =

∫RN

e−2πix·ξu(x) dx

F = F−1, given by Fv(x) =

∫RN

e2πiξ·xv(ξ) dξ

u = FuE the closure of the subset E of the topological space XC(E,F ) the space of continuous functions from the topological space E to the

topological space FCk(E,F ) the space of k times continuously differentiable functions from the

topological space E to the topological space FCb(E,F ) the Banach space of continuous, bounded functions from the topo-

logical space E to the Banach space F , with the topology of uniformconvergence

Cc(E,F ) the space of continuous, compactly supported functions from thetopological space E to the topological space F

L(E,F ) the Banach space of linear, continuous operators from the Banachspace E to the Banach space F , equipped with the norm topology

L(E) the space L(E,E)X? the topological dual of the space XX → Y if X ⊂ Y with continuous injectionΩ an open subset of RNΩ the closure of Ω in RN∂Ω the boundary of Ω, i.e. ∂Ω = Ω \ Ωω ⊂⊂ Ω if ω ⊂ Ω and ω is compactx+ = maxx, 0 for x ∈ R. The positive part of x, i.e. x+ = x if x ≥ 0

and x+ = 0 if x ≤ 0x− = max−x, 0 for x ∈ R. The negative part of x, i.e. x− = 0 if x ≥ 0

and x− = −x if x ≤ 0|x| = x+ + x− for x ∈ R. The absolute value of x, i.e. |x| = x if x ≥ 0

and |x| = −x if x ≤ 0

xα =

N∏j=1

xαjj for α = (α1, . . . , αN ) ∈ NN and x ∈ RN

∂iu = uxi =∂u

∂xi

7

8 NOTATION

∂ru = ur =∂u

∂r=

1

rx · ∇u, where r = |x|

Dα =∂α1

∂xα11

· · · ∂αN

∂xαNNfor α = (α1, . . . , αN ) ∈ NN

∇u (∂1u, · · · , ∂Nu)

∆ =

N∑i=1

∂2

∂x2i

Cc(Ω) the space of continuous functions Ω→ R with compact supportCkc (Ω) the space of functions of Ck(Ω) with compact supportCb(Ω) the Banach space of continuous, bounded functions Ω→ R, equipped

with the topology of uniform convergenceCmb (Ω) the Banach space of u ∈ Cb(Ω) such that Dαu ∈ Cb(Ω) for all

α ∈ NN with |α| ≤ m, equipped with the norm ‖u‖Cmb (Ω) =∑|α|≤m ‖Dαu‖L∞

C(Ω) the space of continuous functions Ω → R. When Ω is bounded,C(Ω) is a Banach space when equipped with the topology of uniformconvergence

Cb,u(Ω) the Banach space of uniformly continuous and bounded functions

Ω→ R equipped with the topology of uniform convergenceCmb,u(Ω) the Banach space of functions u ∈ Cb,u(Ω) such that Dαu ∈ Cb,u(Ω),

for every multi-index α such that |α| ≤ m. Cmb,u(Ω) is equipped with

the norm of Wm,∞(Ω)Cm,α(Ω) for 0 ≤ α < 1, the Banach space of functions u ∈ Cmb,u(Ω) such that

‖u‖Cm,α = ‖u‖Wm,∞ + supx,y∈Ω|β|=m

|Dβu(x)−Dβu(y)||x− y|α

<∞.

D(Ω) = C∞c (Ω), the Frechet space of C∞ functions Ω → R (or Ω → C)compactly supported in Ω, equipped with the topology of uniformconvergence of all derivatives on compact subsets of Ω

C0(Ω) the closure of C∞c (Ω) in L∞(Ω). C0(Ω) is the set of u ∈ C(Ω) suchthat u(x) = 0 for all x ∈ ∂Ω and such that u(x)→ 0 as |x| → ∞ (ifΩ is unbounded)

C0(RN ) the closure of C∞c (RN ) in L∞(Ω). C0(RN ) is the set of u ∈ C(RN )such that u(x)→ 0 as |x| → ∞

Cm0 (Ω) the closure of C∞c (Ω) in Wm,∞(Ω)D′(Ω) the space of distributions on Ω, that is the topological dual of D(Ω)S(RN ) the Schwartz space, i.e. the space of u ∈ C∞(RN ,R) (or C∞(RN ,C))

such that for every nonnegative integer m,

pm(u) = sup|α|≤m

supx∈RN

(1 + |x|2)m/2|Dαu(x)| <∞.

S(RN ) is a Frechet space when equipped with the seminorms pmS ′(RN ) the space of tempered distributions on RN , that is the topological

dual of S(RN ). S ′(RN ) is a subspace of D′(RN )

p′ the conjugate of p given by1

p+

1

p′= 1

Lp(Ω) the Banach space of (classes of) measurable functions u : Ω→ R (or

Ω→ C) such that

∫Ω

|u(x)|p dx <∞ if 1 ≤ p <∞, or ess supx∈Ω

|u(x)| <

NOTATION 9

∞ if p =∞. Lp(Ω) is equipped with the norm

‖u‖Lp =

(∫

Ω|u(x)|p dx

) 1p

if p <∞ess supx∈Ω |u(x)| if p =∞.

Lploc(Ω) the set of measurable functions u : Ω → R (or Ω → C) such thatu|ω ∈ Lp(ω) for all ω ⊂⊂ Ω

Wm,p(Ω) the space of (classes of) measurable functions u : Ω→ R (or Ω→ C)such that Dαu ∈ Lp(Ω) in the sense of distributions, for every multi-index α ∈ NN with |α| ≤ m. Wm,p(Ω) is a Banach space whenequipped with the norm ‖u‖Wm,p =

∑|α|≤m ‖Dαu‖Lp

Wm,ploc (Ω) the set of measurable functions u : Ω → R (or Ω → C) such that

u|ω ∈Wm,p(ω) for all ω ⊂⊂ ΩWm,p

0 (Ω) the closure of C∞c (Ω) in Wm,p(Ω)

W−m,p′(Ω) the topological dual of Wm,p

0 (Ω)Hm(Ω) = Wm,2(Ω). Hm(Ω) is equipped with the equivalent norm

‖u‖Hm =( ∑|α|≤m

∫Ω

|Dαu(x)|2 dx) 1

2

,

and Hm(Ω) is a Hilbert space for the scalar product (u, v)Hm =∫Ω

Re (u(x)v(x)) dx

Hmloc(Ω) = Wm,2

loc (Ω)

Hm0 (Ω) = Wm,2

0 (Ω)H−m(Ω) = W−m,2(Ω)|u|m,p,Ω =

∑|α|=m ‖Dαu‖Lp(Ω)

Hs,p(RN ) for s ∈ R and 1 ≤ p ≤ ∞, the set of u ∈ S ′(RN ) such thatF−1[(1 + | · |2)

s2 u] ∈ Lp(RN ). Hs,p(RN ) is a Banach space for the

norm ‖u‖Hs,p = ‖F−1[(1 + | · |2)s2 u]‖Lp

`p(N) the Banach space of sequences (un)n∈N such that

∞∑n=1

|un|p < ∞ if

1 ≤ p < ∞, or supn≥1 |un| < ∞ if p = ∞. `p(N) is equipped withthe norm

‖u‖`p =

(∑∞

n=1 |un|p) 1p

if p <∞supn≥1 |un| if p =∞.

`p(Z) similar to `p(N), but for sequences (un)n∈Z

CHAPTER 1

Laplace’s equation

1.1. Existence and interior regularity

1.1.1. On RN . Given λ ∈ R, we consider the model Laplace equation

−∆u+ λu = f (1.1.1)

set on the whole space RN . Suppose that for some f ∈ S ′(RN ), there is a solutionu ∈ S ′(RN ) of (1.1.1) . Applying the Fourier transform, we deduce that

(λ+ 4π2|ξ|2)u = f (1.1.2)

in S ′(RN ). (See Section B.6.) Classically, we look for “localized” solutions, forinstance u ∈ L2(RN ). Consider a very nice f , for instance the Gaussian f(x) =

e−πa|x|2

. It follows that f(ξ) = a−N2 e−

πa |ξ|

2

. (See formula (B.6.9).) Therefore, ifthere exists a solution u ∈ S ′(RN ), then

u(ξ) = a−N2 (λ+ 4π2|ξ|2)−1e−

πa |ξ|

2

.

If λ < 0, then u is too singular at |ξ| =√−λ2π to be in L2(RN ). If λ = 0, one can

also find f ∈ L2(RN ) for which there is no solution in L2(RN ). Therefore, we onlyconsider the case λ > 0. We have the following existence result.

Theorem 1.1.1. Suppose λ > 0.

(i) Given f ∈ S ′(RN ), there exists a unique solution u ∈ S ′(RN ) of (1.1.1), givenby (1.1.2).

(ii) If f ∈ Lp(RN ) for some 1 < p < ∞, then u ∈ W 2,p(RN ). More generally,if f ∈ Hs,p(RN ) for some s ∈ R and 1 < p < ∞, then u ∈ Hs+2,p(RN ).Moreover, there exists a constant C independant of f such that ‖u‖Hs+2,p ≤C‖f‖Hs,p .

Proof. As observed above, if u ∈ S ′(RN ) is a solution of (1.1.1), the u is givenby (1.1.2). This proves uniqueness. Moreover, given f ∈ S ′(RN ), formula (1.1.2)determines an element u ∈ S ′(RN ). (See Remark B.6.10 (v).) This proves prop-erty (i).

Suppose now f ∈ Hs,p(RN ) for some s ∈ R and 1 < p < ∞. It follows

in particular that F−1[(λ + 4π2| · |2)s2 f ] ∈ Lp(RN ). (See Proposition B.6.28.)

Applying (1.1.2), we deduce that

F−1[(λ+ 4π2| · |2)s+22 u] = F−1[(λ+ 4π2| · |2)

s2 f ] ∈ Lp(RN )

so that u ∈ Hs+2,p(RN ) and ‖u‖Hs+2,p ≤ C‖f‖Hs,p . This completes the proof.

Remark 1.1.2. Formula (1.1.2) yields interesting informations on the solutionu of (1.1.1). Indeed, let λ > 0 and f ∈ S ′(RN ). If u is the solution u of (1.1.1),then u is given by formula (1.1.2), i.e.

u = (λ+ 4π2| · |2)−1f

It follows that (see Theorem B.6.14 (iv))

u = F−1[(λ+ 4π2| · |2)−1] ? f

11

12 1. LAPLACE’S EQUATION

With the notation in Remark B.6.16, we have F−1[(λ+ 4π2| · |2)−1] = F 1λ , which is

a radially symmetric and decreasing function of L1(RN ). Therefore, if f ∈ Lp(RN )for some 1 ≤ p <∞ and f is nonnegative, radially symmetric and decreasing, thenthe solution u of (1.1.1) is also nonnegative, radially symmetric and decreasing.See Proposition A.2.2.

The following result describes the local regularity of u in terms of the localregularity of f .

Theorem 1.1.3. Let λ ∈ R and f ∈ S ′(RN ), and suppose u ∈ S ′(RN ) sat-isfies (1.1.1) in S ′(RN ). If u ∈ Lploc(RN ) and f ∈ Wm,p

loc (RN ) for some m ≥ 0

and 1 < p < ∞, then u ∈ Wm+2,ploc (RN ). In particular, if f ∈ C∞(RN ), then

u ∈ C∞(RN ).

Proof. Let 1 < p < ∞ and suppose u ∈ W `,ploc (RN ) for some 0 ≤ ` ≤ m + 1.

Given θ ∈ C∞c (RN ), we have in S ′(RN )

−∆(θu) + θu = θf −∇θ · ∇u− u∆θ + (1− λ)θu

Since u ∈W `,ploc (RN ), we see that −∇θ · ∇u− u∆θ+ (1− λ)θu ∈W `−1,p(RN ); and

since f ∈ Wm,ploc (RN ), we also have θf ∈ Wm,p(RN ). Applying Theorem 1.1.1 (ii)

(with λ = 1), we deduce that θu ∈ W `+1,p(RN ). θ being arbitrary, it follows that

u ∈W `+1,ploc (RN ).

We now argue by induction. Starting with ` = 0, we obtain u ∈ W 1,ploc (RN ).

Iterating, it follows that u ∈W `+1,ploc (RN ) for all ` ∈ N such that ` ≤ m+ 1. Hence

the result follows.

Remark 1.1.4. Note that Theorem 1.1.3 does not impose any estimate of the

derivatives of f . For instance, let f(x) = (1 + |x|2)−N sin(e|x|2

) et λ = 1. Since f ∈L2(RN ), there exists a solution u ∈ H2(RN ) of (1.1.1). The derivatives of f becomevery large as |x| → ∞, however f ∈ L2 ∩ C∞, so that u ∈ H2(RN ) ∩ C∞(RN ).

Remark 1.1.5. Note that we can consider real-valued functions as well ascomplex-valued functions.

1.1.2. On a domain (connected open set), bounded or not. Let Ω bean open, connected subset of RN . We look for solutions of

−∆u+ λu = f in Ω

u = 0 on ∂Ω(1.1.3)

If Ω is sufficiently smooth (of class C2) and u ∈ C1(Ω,RN ), then we have thefollowing Green’s formula∫

Ω

∇ · u(x) dx =

∫∂Ω

u(σ) · n(σ) dσ

where n(σ) is the outwards normal at σ ∈ ∂Ω and dσ is the surface measure on∂Ω. Suppose u is a classical solution of (1.1.3), i.e. u ∈ C2(Ω), and let ϕ ∈ C1(Ω).We have

−ϕ∆u = −∇ · (ϕ∇u) +∇u · ∇ϕso that Green’s formula yields

−∫

Ω

ϕ∆u =

∫Ω

∇u · ∇ϕ−∫∂Ω

ϕ(σ)∇u(σ) · n(σ) dσ

Therefore, if ϕ ∈ C1c (Ω), then ϕ(σ) = 0 for all σ ∈ ∂Ω so that

−∫

Ω

∆uϕ =

∫Ω

∇u · ∇ϕ

1.1. EXISTENCE AND INTERIOR REGULARITY 13

Thus we see that if u is a classical solution of (1.1.3), then∫Ω

(∇u · ∇ϕ+ λuϕ− fϕ) = 0 (1.1.4)

for all ϕ ∈ C1c (Ω), hence by density for all ϕ ∈ H1

0 (Ω).Note that the term

∫Ωuf is well defined for all f ∈ L2(Ω), and that∫

Ω

uf = 〈u, f〉H10 ,H

−1

where 〈·, ·〉H10 ,H

−1 is the duality bracket H10 −H−1. (Here, we identify L2(Ω) with

its dual, and we use the fact that (H10 (Ω))? = H−1(Ω).)

Thus we see that if u is a classical solution of (1.1.3), then∫Ω

(∇u · ∇ϕ+ λuϕ) = 〈ϕ, f〉H10 ,H

−1 (1.1.5)

for all ϕ ∈ H10 (Ω). Note that identity (1.1.5) makes sense for all u, ϕ ∈ H1

0 (Ω) andf ∈ H−1(Ω), where Ω is any domain.

We use identity (1.1.5) as the definition of “weak solution”: Given f ∈ H−1(Ω),we say that u is a weak solution of (1.1.3) iff

u ∈ H10 (Ω)

(1.1.5) holds for all ϕ ∈ H10 (Ω)

(1.1.6)

The important fact at this point is that if Ω is smooth and u is a classical solutionof (1.1.3), then u is a weak solution. The boundary condition u|∂Ω = 0 is understood

in the weak sense u ∈ H10 (Ω).

The motivation of this approach is that we will see that, under very generalconditions, there always exists a unique weak solution, whereas there does notalways exist a classical solution.

Several simple methods are available for proving the existence of a weak solu-tion, such as Lax-Milgram’s theorem ([20, Theorem 2.1]), or minimization (varia-tional methods). In the present case, the operator ∆ is symmetric, and one can usean even simpler technique, based on the Riesz representation theorem.

We set

λ(Ω) = inf∫

Ω

|∇u|2; u ∈ H10 (Ω) et ‖u‖L2 = 1

(1.1.7)

so that λ(Ω) ≥ 0 and ∫Ω

|∇u|2 ≥ λ(Ω)

∫Ω

|u|2

for all u ∈ H10 (Ω). On the other hand, we know that λ(RN ) = 0 (Remark B.3.21),

and that λ(Ω) > 0 if Ω is bounded (by Poincare’s inequality (B.3.73)). In addition,we will see later (Section 1.4) that if Ω is bounded, then λ(Ω) is the first eigenvalueof −∆ in H1

0 (Ω).Given 0 ≤ θ ≤ 1, we have∫

(|∇u|2 + λ|u|2) = θ

∫|∇u|2 + (1− θ)

∫|∇u|2 + λ

∫|u|2

≥ θ∫|∇u|2 + [(1− θ)λ(Ω) + λ]

∫|u|2

= θ

∫|∇u|2 + [λ(Ω) + λ− θλ(Ω)]

∫|u|2

for all u ∈ H10 (Ω). Suppose

λ > −λ(Ω) (1.1.8)


and let

θ =

1 λ ≥ 1λ+λ(Ω)

1+λ(Ω)−λ(Ω) < λ < 1

It follows that θ > 0 and

λ(Ω) + λ− θλ(Ω) =

λ λ ≥ 1

θ −λ(Ω) < λ < 1

In particular, λ(Ω) + λ− θλ(Ω) ≥ θ, so that∫(|∇u|2 + λ|u|2) ≥ θ

∫(|∇u|2 + |u|2) = θ‖u‖2H1 (1.1.9)

Therefore, the scalar product

((u, v)) =

∫(∇u · ∇v + λuv)

defines on H10 (Ω) a norm which is equivalent to the canonical norm. Thus we may

equip H10 (Ω) with the scalar product ((·, ·)) provided (1.1.8) holds.

Theorem 1.1.6. Assume (1.1.8). Given f ∈ H−1(Ω), there exists a uniqueweak solution u of (1.1.3). In addition, there exist constants 0 < c < C < ∞independent of f such that

c‖f‖H−1 ≤ ‖u‖H1 ≤ C‖f‖H−1 (1.1.10)

Proof. Let f ∈ H−1(Ω), and note that u is a weak solution of (1.1.3) iff

((u, ϕ)) = 〈ϕ, f〉H10 ,H

−1 (1.1.11)

for all ϕ ∈ H10 (Ω). By the Riesz representation theorem, there exists a unique

u ∈ H10 (Ω) such that (1.1.11) holds for all ϕ ∈ H1

0 (Ω). This proves the existenceand uniqueness part. Moreover, it follows from Propositions B.6.29 and B.6.24 that

‖f‖H−1 = ‖ −∆u+ λu‖H−1 ≤ ‖∆u‖H−1 + |λ| ‖u‖H−1 ≤ C‖u‖H1

which proves the first inequality in (1.1.10). Letting ϕ = u in (1.1.11), we obtain

‖u‖2H1 ≤ C((u, u)) ≤ C‖u‖H1‖f‖H−1

which yields the second inequality in (1.1.10).

Remark 1.1.7. Here are some comments on Theorem 1.1.6.

(i) We do not impose any condition on the domain Ω, neither of smoothness, norof boundedness.

(ii) If Ω then λ(Ω) > 0 so we may choose λ = 0.(iii) We may apply the above very elementary technique because the Laplacian is

a symmetric operator, so that ((u, v)) defined a scalar product on H10 (Ω). For

non-symmetric operators, on can apply for instance Lax-Milgram’s theorem.(iv) A classical solution is a weak solution, and weak solutions are unique. It is

therefore natural to consider weak solutions. It may happen that there is noclassical solution, while there is a weak solution. For example, let U be theunit ball in R3, and Ω = U \ 0. Let u ∈ C∞c (U) such that u ≡ 1 in aneighborhood of 0, and let f = −∆u ∈ C∞c (Ω). We have u ∈ H1

0 (Ω), sinceH1

0 (Ω) = H10 (U). Therefore, u is a weak solution of

−∆u = f

u|∂Ω = 0(1.1.12)

1.1. EXISTENCE AND INTERIOR REGULARITY 15

Since u does not vanish on ∂Ω (because u does not vanish at 0), this meansthat there is no classical solution of (1.1.12). (Otherwise, it would equal u,by uniqueness.)

Remark 1.1.8. If we consider Laplace’s equation with Neumann boundaryconditions, i.e. −∆u+ λu = f in Ω

∂u

∂n= 0 on ∂Ω

(1.1.13)

then Green’s formula leads to the following definition of a weak solutionu ∈ H1(Ω)∫

Ω

(∇u · ∇ϕ+ λuϕ− fϕ) = 0 ∀ϕ ∈ H1(Ω)

Applying again the Riesz representation theorem, it is immediate that if λ > 0 andf ∈ L2(Ω), then there exists a unique weak solution u.

In analogy with Theorem 1.1.1, one might expect extensions of Theorem 1.1.6in two directions. On the one hand, if f is more regular, for instance f ∈ Hm(Ω)for some m ≥ 0, then one might expect accordingly u to be more regular, i.e.u ∈ Hm+2(Ω). On the other hand, one might consider for instance f ∈ Lp(Ω) withp 6= 2 and expect the existence of a weak solution, in some appropriate sense, inan Lp-based space such as W 1,p

0 (Ω). It turns out that both these extensions arepossible only under certain smoothness assumptions on Ω. Moreover, the proofsare considerably more difficult than in the case of the equation set on the wholespace RN . On these issues, see for instance [14] and the references therein.

We present below a result on the interior regularity, similar to Theorem 1.1.3.

Theorem 1.1.9. Let λ ∈ R, f ∈ H−1(Ω), and suppose u ∈ H10 (Ω) satis-

fies (1.1.5). If f ∈Wm,ploc (Ω) for some m ≥ 0 and 1 < p <∞, then f ∈Wm+2,p

loc (Ω).In particular, if f ∈ C∞(Ω), then u ∈ C∞(Ω).

Proof. Consider θ ∈ C∞c (Ω). Since u ∈ H10 (Ω), we see that θu ∈ H1

0 (Ω) and

∇(θu) = θ∇u+ u∇θ.Define w : RN → R by

w(x) =

θ(x)u(x) x ∈ Ω

0 x 6∈ Ω.

It follows that w ∈ H1(RN ) and

∇w =

∇(θu) Ω

0 RN \ Ω

Given ϕ ∈ S(RN ), we have∫RN∇w · ∇ϕ =

∫Ω

∇(θu) · ∇ϕ =

∫Ω

(θ∇u+ u∇θ) · ∇ϕ

=

∫Ω

∇u · (θ∇ϕ) +

∫Ω

u∇θ · ∇ϕ

=

∫Ω

∇u · ∇(θϕ)−∫

Ω

∇u · (ϕ∇θ) +

∫Ω

u∇θ · ∇ϕ

Applying (1.1.5) with ϕ replaced by (θϕ)|Ω ∈ H10 (Ω), we calculate the first term in

the right-hand side of the last inequality as follows∫Ω

∇u · (θ∇ϕ) = −λ∫RN

wϕ+

∫Ω

fθϕ


so that∫RN∇w · ∇ϕ = −λ

∫RN

wϕ+

∫Ω

fθϕ−∫

Ω

∇u · (ϕ∇θ) +

∫Ω

u∇θ · ∇ϕ

hence∫RN

(∇w · ∇ϕ+ wϕ) = (1− λ)

∫RN

wϕ+

∫Ω

fθϕ−∫

Ω

∇u · (ϕ∇θ) +

∫Ω

u∇θ · ∇ϕ

We integrate the last term by parts, which is possible because u∇θ ∈ (H10 (Ω))N .

Since ∇ · (u∇θ) = ∇u · ∇θ + u∆θ, we obtain∫RN

(∇w · ∇ϕ+ wϕ) = (1− λ)

∫RN

uθϕ+

∫Ω

fθϕ−∫

Ω

∇u · (ϕ∇θ)

−∫

Ω

(∇u · ∇θ + u∆θ)ϕ =

∫RN

Fϕ

where

F (x) =

fθ − 2∇u · ∇θ + [(1− λ)θ −∆θ]u Ω

0 RN \ Ω

We first show that u ∈ W 1,ploc (Ω). Since u ∈ H1

0 (Ω), there is nothing to prove ifp ≤ 2, so we assume 2 < p <∞. We consider the unique integer ` ≥ 1 such that

1

2− `

N<

1

p≤ 1

2− `− 1

N

and we define qj ≥ 2 for 0 ≤ j ≤ ` by

1

qj=

1

2− j

N

In particular, q0 = 2, so that u ∈ W 1,q0loc (Ω). We show by induction that u ∈

W 1,q`loc (Ω). Suppose u ∈ W 1,qj

loc (Ω) for some 0 ≤ j ≤ ` − 1. Note that all terms inF constructed above have a compact support in Ω. Therefore, since fθ ∈ Lp(RN ),

we have fθ ∈ Lqj (RN ). Moreover, u ∈ u ∈ W 1,qjloc (Ω), so that −2∇u · ∇θ + [(1 −

λ)θ − ∆θ]u ∈ Lqj (RN ). Therefore, F ∈ Lqj (RN ), so that by Theorem 1.1.1 (ii),w ∈W 2,qj+1(RN ). By Sobolev’s embedding, we deduce that w ∈W 1,qj (RN ). Since

θ and ϕ are arbitrary, this implies that u ∈ W 1,qj+1

loc (Ω). Thus u ∈ W 1,q`loc (Ω), and

applying once more this regularity argument, we obtain u ∈W 1,ploc (Ω).

We finally argue as in the proof of Theorem 1.1.3. Suppose u ∈ W `,ploc (Ω) for

some 1 ≤ ` ≤ m+ 1. It follows that F constructed above belongs to W `−1,p(RN ).

Therefore, w ∈ W `+1,p(RN ) by Theorem 1.1.1 (ii), so that u ∈ W `+1,ploc (Ω). Iterat-

ing, we deduce that u ∈Wm+2,ploc (Ω), which is the desired result.

Remark 1.1.10. All the results in this section are also true, with the sameproofs, for complex-valued functions. Note that the new scalar product on H1

0 (Ω)is defined in this case by

((u, v)) = <∫

Ω

(∇u · ∇v + λuv)

for uv ∈ H10 (Ω).

1.2. The maximum principle

The maximum principle is a fundamental property of the Laplace equation, andwe establish below several forms of it. Throughout this section, we assume that Ωis an open, connected subset of RN , bounded or not.

1.2. THE MAXIMUM PRINCIPLE 17

1.2.1. The weak maximum principle. We begin with the following weakmaximum principle, which is very simple but on which are based the more elabo-rated forms of the maximum principle.

We recall that if f ∈ H−1(Ω), then f ≥ 0 (in the sense of H−1) if

〈f, ϕ〉H−1,H10≥ 0

for all ϕ ∈ H10 (Ω) such that ϕ ≥ 0 a.e.

Theorem 1.2.1. Suppose (1.1.8). Let u ∈ H1(Ω) and assume

−∆u+ λu ≥ 0 (respectively, ≤ 0)

in the sense of H−1(Ω). If u− ∈ H10 (Ω)(respectively, u+ ∈ H1

0 (Ω)), then u ≥ 0(respectively, u ≤ 0) a.e. in Ω.

Proof. We prove the first part of the result, the second follows by changingu to −u. Since u− ≥ 0, we have

〈−∆u+ λu,−u−〉H−1,H10≤ 0

which we rewrite, using formuls (B.1.8) and (B.1.7), in the form∫Ω

[∇u · ∇(−u−) + λu(−u−)] ≤ 0.

Note that u = u+ − u−, so that u(−u−) = (u−)2; and that ∇u = ∇u+ −∇u−, sothat ∇u · ∇(−u−) = |∇u−|2 (by (B.2.4). It follows that∫

Ω

[|∇(u−)|2 + λ|u−|2] ≤ 0.

Since λ > −λ(Ω), we deduce by applying (1.1.9) that u− = 0, so that u ≥ 0 a.e.

Remark 1.2.2. Here are some comments on the preceding result.

(i) Since u ∈ H1(Ω), we know that u− ∈ H1(Ω), by Remark B.2.4. Therefore,the condition u− ∈ H1

0 (Ω) is a weak way of saying u− = 0 on ∂Ω, whichmeans u ≥ 0 on ∂Ω.

(ii) Let f ∈ H−1(Ω), and u ∈ H10 (Ω) the weak solution of (1.1.3) given by The-

orem 1.1.6. It follows that if f ≥ 0, then u ≥ 0 a.e. on Ω. Indeed, sinceu ∈ H1

0 (Ω), we have u− ∈ H10 (Ω), by Remark B.2.4.

1.2.2. The strong maximum principle. There are several forms of thestrong maximum principle. We begin with the simplest, which is valid in anydomain.

Theorem 1.2.3. Assume (1.1.8). Let u ∈ H1(Ω)∩C(Ω) satisfy −∆u+λu ≥ 0(respectively, ≤ 0) in the sense of H−1(Ω). If u− ∈ H1

0 (Ω) (respectively, u+ ∈H1

0 (Ω)) and if u 6≡ 0, then u > 0 (respectively, u < 0) in Ω.

The proof of Theorem 1.2.3 is based on the following simple lemma.

Lemma 1.2.4. Let 0 < ρ < R < ∞ and set ω = ρ < |x| < R. Let λ ∈ Rand suppose β > max0, N − 2 satisfies β(β −N + 2) ≥ |λ|R2. If v is defined byv(x) = |x|−β −R−β for ρ ≤ |x| ≤ R, then the following properties hold.

(i) v ∈ C∞(ω).(ii) v(x) = 0 if |x| = R.(iii) ρ−β > v(x) ≥ βR−(β+1)(R− |x|) if ρ ≤ |x| ≤ R.(iv) −∆v + λv ≤ 0 in ω.


Proof. Properties (i), (ii) and (iii) are immediate. Next,

−∆v + λv = −β(β −N + 2)|x|−(β+2) + λ|x|−β − λR−β

≤ −β(β −N + 2)R−2|x|−β + |λ| |x|−β ,

and (iv) easily follows.

Proof of Theorem 1.2.3. We prove the first part of the result, the secondfollows by changing u to −u. We first note that by Theorem 1.2.1 (and Re-mark 1.2.2 (ii)), u ≥ 0 a.e. in Ω. Since u ∈ C(Ω) and u 6≡ 0, the set

O = x ∈ Ω; u(x) > 0,is a nonempty open subset of Ω. Ω being connected, we need only show that O isa closed subset of Ω. Suppose (yn)n≥0 ⊂ O and yn → y ∈ Ω as n→∞. Let R > 0be such that B(y, 2R) ⊂ Ω, and fix n0 large enough so that |y − yn0

| < R. Sinceu(yn0

) > 0, there exist 0 < ρ < R and ε > 0 such that u(x) ≥ ε for |x − yn0| = ρ.

Set U = ρ < |x− yn0 | < R and let w(x) = u(x)− ερβv(x− yn0) for x ∈ U , whereβ and v are as in Lemma 1.2.4. It follows that w ∈ H1(U) ∩ C(U). Moreover,−∆w+λw ≥ 0 by property (iv) of Lemma 1.2.4. Also, since u ≥ 0 in Ω, we deducefrom property (ii) of Lemma 1.2.4 that w(x) ≥ 0 if |x − yn0

| = R. Furthermore,w(x) ≥ 0 if |x − yn0

| = ρ by property (iii) of Lemma 1.2.4 and because u(x) ≥ ε.Thus we may apply Theorem 1.2.1 and we deduce that w(x) ≥ 0 for x ∈ U . Inparticular, u(y) ≥ ερβv(y − yn0

) > 0 by property (iii) of Lemma 1.2.4, so thaty ∈ O. Therefore, O is closed, which completes the proof.

We now state a stronger version of the maximum principle, which requires acertain amount of regularity of the domain. More precisely, we assume the followinggeometric condition. (See Figure 1.)

Ω is a connected, open, bounded subset of RN , and

∃η, ν > 0 s.t. ∀x ∈ Ω with d(x, ∂Ω) ≤ η,∃y ∈ Ω

s.t. x ∈ B(y, η), B(y, η) ⊂ Ω and η − |x− y| ≥ νd(x, ∂Ω)

(1.2.1)

Assumption (1.2.1) means that one cannot have outwards corners. It is easy to seethat it is satisfied if Ω is of class C2. Indeed, let γ(z) denote the unit outwardsnormal to ∂Ω at z ∈ ∂Ω. Since Ω is bounded, ∂Ω is uniformly C2, so that thereexists η > 0 such that B(z − ηγ(z), η) ⊂ Ω for every z ∈ ∂Ω. If x ∈ Ω andd(x, ∂Ω) ≤ η, let z ∈ ∂Ω be such that |x − z| = d(x, ∂Ω). It follows that x − z isparallel to γ(y). Thus, if we set y = z−ηγ(z), we see that x ∈ B(y, η), B(y, η) ⊂ Ωand η − |x− y| = |z − x| = d(x, ∂Ω).

Theorem 1.2.5. Assume (1.2.1) and (1.1.8). Suppose u ∈ H1(Ω) ∩ C(Ω)satisfies −4u+λu ≥ 0 (respectively, ≤ 0) in H−1(Ω). If u− ∈ H1

0 (Ω) (respectively,u+ ∈ H1

0 (Ω)) and if u 6≡ 0, then there exists µ > 0 such that u(x) ≥ µd(x, ∂Ω)(respectively, u(x) ≤ −µd(x, ∂Ω)) in Ω, where d(x, ∂Ω) is the distance of x to ∂Ω.

Proof. We prove the first part of the result, the second follows by changingu to −u. Let 0 < ε ≤ η/2 and consider Ωε = x ∈ Ω; d(x, ∂Ω) ≥ ε. We fix ε > 0sufficiently small so that Ωε is a nonempty, compact subset of Ω. It follows fromTheorem 1.2.3 that there exists δ > 0 such that

u(x) ≥ δ for all x ∈ Ωε. (1.2.2)

We now consider x0 ∈ Ω such that d(x0, ∂Ω) < ε, and we let y0 ∈ Ω satisfy (1.2.1).Since B(y0, η) ⊂ Ω and η ≥ 2ε, we see that d(z, ∂Ω) ≥ ε for all z ∈ B(y0, η/2). Itthen follows from (1.2.2) that

u(z) ≥ δ for all z ∈ B(y0, η/2). (1.2.3)

1.3. Lp(Ω) ET C0(Ω) REGULARITY 19

Ω

x×

y×

η

η

B(y, η)

Figure 1. The geometric condition (1.2.1)

We let ρ = η/2, R = η and U = ρ < |x − y0| < R, so that x0 ∈ U . Letw(x) = u(x) − ερβv(x − y0) for x ∈ U , where β and v are as in Lemma 1.2.4. Itfollows that w ∈ H1(U) ∩ C(U). Moreover, −∆w + λw ≥ 0 by property (iv) ofLemma 1.2.4. Also, w(x) ≥ 0 if |x− y0| = R by property (ii) of Lemma 1.2.4 andbecause u ≥ 0 in Ω. Furthermore, w(x) ≥ 0 if |x − y0| = ρ by property (iii) ofLemma 1.2.4 and (1.2.3). Thus we may apply Theorem 1.2.1 and we deduce thatw(x) ≥ 0 for x ∈ U . In particular,

u(x0) ≥ βερβR−(β+1)(R− |x0 − y0|) ≥ νβερβR−(β+1)d(x0, ∂Ω),

where the first inequality above follows from of Lemma 1.2.4 (iii) and the secondfrom (1.2.1). Since x0 ∈ Ω \ Ωε is arbitrary, we see that there exists µ > 0 suchthat u(x) ≥ µd(x, ∂Ω) for all x ∈ Ω \ Ωε. On the other hand, (1.2.2) implies thatthere exists µ′ > 0 such that u(x) ≥ µ′d(x, ∂Ω) for all x ∈ Ωε. This completes theproof.

1.3. Lp(Ω) et C0(Ω) regularity

Throughout this section, Ω is an open, connected subset of RN . Unless other-wise specified, Ω may be bounded or not.

1.3.1. Lp regularity. We begin with a simple estimate.

Theorem 1.3.1. Soient λ > 0, f ∈ H−1(Ω) et soit u ∈ H10 (Ω) la solution

de (1.1.5). Si f ∈ Lp(Ω) pour un certain p ∈ [1,∞], alors u ∈ Lp(Ω) et λ‖u‖Lp ≤‖f‖Lp .

Proof. Let ϕ ∈ C1(R,R). Assume that ϕ(0) = 0, ϕ′ ≥ 0 and ϕ′ ∈ L∞(R). Itfollows from Proposition B.2.1 that ϕ(u) ∈ H1

0 (Ω) and that ∇ϕ(u) = ϕ′(u)∇u a.e.


in Ω. We let ϕ = ϕ(u) in (1.1.5), and we obtain∫Ω

ϕ′(u)|∇u|2 dx+ λ

∫Ω

uϕ(u) dx =

∫Ω

fϕ(u) dx

so that

λ

∫Ω

uϕ(u) dx ≤∫

Ω

fϕ(u) dx

Assuming further |ϕ(u)| ≤ |u|p−1, we see that |ϕ(u)| ≤ (uϕ(u))p−1p , so that

λ

∫Ω

uϕ(u) dx ≤ ‖f‖Lp‖ϕ(u)‖L

pp−1≤ ‖f‖Lp

(∫Ω

uϕ(u) dx) p−1

p

Therefore,

λ(∫

Ω

uϕ(u) dx) 1p ≤ ‖f‖Lp (1.3.1)

We now consider separately two cases.

Case 1: 1 ≤ p ≤ 2 Given ε > 0, let ϕ(u) = u(ε + u2)p−22 . It is easy to see

that ϕ satisfies the previous conditions, and we deduce from (1.3.1) that

λ(∫

Ω

u2(ε+ u2)p−22 dx

) 1p ≤ ‖f‖Lp

Letting ε ↓ 0 and applying Fatou, we obtain the desired estimate

Case 2: 2 < p ≤ ∞ We apply a duality argument. Let h ∈ C∞c (Ω) andv ∈ H1

0 (Ω) the solution of (1.1.5) with f replaced by h. It follows that∫Ω

uh = 〈u,−∆v + λv〉H10 ,H

−1 = 〈−∆u+ λu, v〉H−1,H10

= 〈f, v〉H−1,H10

=

∫Ω

fv

Therefore, ∣∣∣∫Ω

uh∣∣∣ ≤ ‖f‖Lp‖v‖Lp′ ≤ 1

λ‖f‖Lp‖h‖Lp′

by Case 1, since p′ < 2. Since h ∈ C∞c (Ω) is arbitrary, we deduce that ‖u‖Lp ≤λ−1‖f‖Lp .

One can improve the above estimates by using Sobolev’s inequalities.

Theorem 1.3.2. Let λ > 0, f ∈ H−1(Ω) and u ∈ H10 (Ω) the solution of (1.1.5).

Assume f ∈ Lp(Ω) for some 1 < p <∞.

(i) If N ≥ 3 and p = N/2, then u ∈ Lr(Ω) for all r ∈ [p,∞), and there exists aconstant C(r) independent of f such that ‖u‖Lr ≤ C(r)‖f‖Lp .

(ii) If N ≥ 3 and 1 < p < N/2, then u ∈ Lp(Ω) ∩ LNpN−2p (Ω), and there exists

constant C independent of f such that ‖u‖Lr ≤ C‖f‖Lp for all r ∈ [p, NpN−2p ].

Proof. Let 1 < q < ∞ such that (q − 1)p′ ≥ 1. We obtain an estimate byusing ϕ = |u|q−2u in (1.1.5). In fact, as in the preceding theorem, a regularizationis necessary, and one should really use the test function

ϕε =

u(ε+ |u|2)

q−22 q ≤ 2

u|u|q−2(1 + ε|u|2)2−q2 q > 2

then ε ↓ 0 in the estimate that one obtains. For simplicity, we only give the formalcalculations, letting ϕ = |u|q−2u in (1.1.5). Since ∇ϕ = (q− 1)|u|q−2∇u, we obtain

∇u · ∇ϕ = (q − 1)|u|q−2|∇u|2


On the other hand,

∇(|u|q−22 u) =

q

2|u|

q−22 ∇u

so that

∇u · ∇ϕ =4(q − 1)

q2|∇(|u|

q−22 u)|2

Identity (1.1.5) therefore gives

4(q − 1)

q2

∫Ω

|∇(|u|q−22 u)|2 + λ

∫Ω

|u|q =

∫Ω

f |u|q−2u ≤ ‖f‖Lp‖u‖q−1

L(q−1)p′

Applying Sobolev’s inequality to the function |u|q−22 u, we obtain∫

Ω

|∇(|u|q−22 u)|2 ≥ c‖ |u|

q−22 u‖2

L2NN−2

= c‖u‖q2NqN−2

so that

‖u‖qL

NqN−2

≤ C q2

q − 1‖f‖Lp‖u‖q−1

L(q−1)p′ , (1.3.2)

for all 1 < q < ∞ satisfying (q − 1)p′ ≥ 1. To prove (i), we choose p = N2 and we

apply (1.3.2) with q > N2 . We obtain

‖u‖qL

NqN−2

≤ C q2

q − 1‖f‖LN/2‖u‖

q−1

LN(q−1)N−2

. (1.3.3)

Applying Holder’s inequality and Theorem 1.3.1, we deduce

‖u‖q−1

LN(q−1)N−2

≤ ‖u‖(2q−N)q2q−N+2

LNqN−2

‖u‖N−2

2q−N+2

LN2

≤ ‖u‖(2q−N)q2q−N+2

LNqN−2

‖f‖N−2

2q−N+2

LN2

.

Substitution into (1.3.3) yields

‖u‖L

NqN−2≤ C(q)‖f‖

LN2.

Estimate (i) is a consequence of the above inequality, combined with Theorem 1.3.1,

since q can be arbitrarily large. Finally, to prove (ii), we lat q = (N−2)pN−2p , so that

1 < Nq(N−2) = (q − 1)p′ = Np

(N−2p) . Inequality (1.3.2) yields

‖u‖L

NpN−2p

≤ C ′‖f‖Lp

which, together with Theorem 1.3.1, proves (ii).

The preceding results do not apply when λ ≤ 0. One can, however, obtaincertain estimates by more elaborated techniques. We begin with an L∞ estimate.

Theorem 1.3.3. Assume (1.1.8). Let f ∈ H−1(Ω) and u ∈ H10 (Ω) the solution

of (1.1.5). If f ∈ Lp(Ω) for some max1, N2 < p ≤ ∞, then u ∈ L∞(Ω). Moreover,given any 1 ≤ r <∞, there exists a constant C independant of f such that

‖u‖L∞ ≤ C(‖f‖Lp + ‖u‖Lr ) (1.3.4)

In particular,

‖u‖L∞ ≤ C(‖f‖Lp + ‖f‖H−1) (1.3.5)

Proof. By homogeneity, we may assume that ‖u‖Lr + ‖f‖Lp ≤ 1. Since −usolves the same equation as u, with f replaced by −f (which satisfies the sameassumptions), it is sufficient to estimate ‖u+‖L∞ . Set T = ‖u+‖L∞ ∈ [0,∞] andassume that T > 0. For t ∈ (0, T ), set

v(t) = (u− t)+


We It follows from Corollary B.2.6 that v(t) ∈ H10 (Ω) and

∇v(t) =

∇u u > t

0 u ≤ t(1.3.6)

Set now

α(t) = |x ∈ Ω, u(x) > t|for t > 0. Note that α(t) is always finite. In particular, since v(t) ∈ L2(Ω) issupported in x ∈ Ω, u(x) > t, we have v(t) ∈ L1(Ω). We set

β(t) =

∫Ω

v(t) dx

Integrating the measurable function 1u>s(x) on (t,∞)× Ω and applying Fubini,we obtain

β(t) =

∫ ∞t

α(s) ds

so that β ∈W 1,1loc (0,∞) and

β′(t) = −α(t) (1.3.7)

for a.a. t > 0.The idea of the proof is to obtain a differential inequality on β(t) which implies

that β(t) must vanish for t large enough. Lettin ϕ = v(t) in (1.1.5), we obtain∫Ω

∇u · ∇v(t) + λ

∫Ω

uv(t) = 〈f, v(t)〉H−1,H10

Applying (1.3.6) and the property v(t) ∈ L1(Ω), we see that∫Ω

|∇v(t)|2 + λ|v(t)|2 dx =

∫Ω

(f − tλ)v(t) dx

By (1.1.9), this yields

‖v(t)‖2H1 ≤ C∫

Ω

(f − tλ)v(t) dx ≤ C∫

Ω

(|f |+ t|λ|)v(t) dx (1.3.8)

Note that ∫Ω

|f |v(t) ≤ ‖f‖Lp‖v(t)‖Lp′ ≤ ‖v(t)‖Lp′

so that, applying (1.3.8),

‖v(t)‖2H1 ≤ C(1 + t)(‖v(t)‖Lp′ + ‖v(t)‖L1) (1.3.9)

Since p > min1, N2 , we may fix

2p′ < ρ <2N

(N − 2)+

so that H10 (Ω) → Lρ(Ω). On the other hand, it follows from Holder’s inequality

that ‖v(t)‖L1 ≤ α(t)1− 1ρ ‖v(t)‖Lρ and ‖v(t)‖Lp′ ≤ α(t)

1p′−

1ρ ‖v(t)‖Lρ . Therefore, we

deduce from (1.3.9) that

‖v(t)‖2Lρ ≤ C(1 + t)(α(t)1p′−

1ρ + α(t)1− 1

ρ )‖v(t)‖Lρ

Since β(t) = ‖v(t)‖L1 ≤ α(t)1− 1ρ ‖v(t)‖Lρ , we obtain

β(t) ≤ C(1 + t)(α(t)1+ 1

p′−2ρ + α(t)2− 2

ρ )

which we rewrite in the form

β(t) ≤ C(1 + t)F (α(t))


where F (s) = s1+ 1

p′−2ρ + s2− 2

ρ . It follows that

− α(t) + F−1( β(t)

C(1 + t)

)≤ 0 (1.3.10)

Since −α(t) = β′(t) by (1.3.7), we deduce from (1.3.10) that

z′ +ψ(z(t))

C(1 + t)≤ 0

where z(t) = β(t)C(1+t) and ψ(s) = F−1(s) + Cs. This yields∫ t

s

dσ

C(1 + σ)≤∫ z(s)

z(t)

dσ

ψ(σ),

If 0 < s < t < T . Si T ≤ 1, then ‖u+‖L∞ ≤ 1. Otherwise, we obtain∫ t

1

dσ

C(1 + σ)≤∫ z(1)

z(t)

dσ

ψ(σ)

for 1 < t < T , which implies∫ T

1

dσ

C(1 + σ)≤∫ z(1)

0

dσ

ψ(σ)

Note that F (s) ≈ s1+ 1

p′−2ρ as s ↓ 0, and 1 + 1

p′ −2ρ > 1, so that 1

ψ is integrable

at 0. Since 11+σ is not integrable at infinity, this proves that T = ‖u+‖L∞ < ∞.

Moreover, ‖u+‖L∞ is estimated in terms of z(1). Since

z(1) =1

C

∫Ω

(u− 1)+ ≤ 1

C

∫u>1

u ≤ 1

C

∫u>1

ur ≤ 1

C

this proves the desired estimate.

Corollary 1.3.4. Let λ > 0, f ∈ H−1(Ω) and u ∈ H10 (Ω) the solution

of (1.1.5). If f ∈ Lp(Ω) for some 1 < p < ∞, p > N2 , then u ∈ L∞(RN ) ∩ C(Ω)

and there exists a constant C independent of f such that ‖u‖L∞ ≤ C‖f‖Lp .

Proof. It follows from Theorem 1.3.1 that λ‖u‖Lp ≤ ‖f‖Lp , so that by The-orem 1.3.3 u ∈ L∞(Ω) and

‖u‖L∞ ≤ C(‖f‖Lp + ‖u‖Lp) ≤ C(1 + λ−1)‖f‖Lp (1.3.11)

To prove that u ∈ C(Ω), consider (fn)n≥1 ⊂ C∞c (Ω) such that fn → f in Lp(Ω), andlet (un)n≥1 be the corresponding solutions of (1.1.5). It follows from Theorem 1.1.9that un ∈ C(Ω). Since un → u in L∞(Ω) by (1.3.11), we obtain u ∈ C(Ω).

In the case where Ω has finite measure, we deduce from Theorem 1.3.3 thatTheorem 1.3.2 holds under assumption (1.1.8), and not just for λ > 0.

Corollary 1.3.5. Assume N ≥ 3, (1.1.8) and |Ω| <∞. Let f ∈ H−1(Ω) andu ∈ H1

0 (Ω) the solution of (1.1.5). If f ∈ Lp(Ω) for some 1 < p ≤ ∞, then thefollowing properties hold.

(i) If p = N2 , then u ∈ Lr(Ω) for all r ∈ [p,∞), and there exists a constant C(r)

independent of f such that ‖u‖Lr ≤ C(r)(‖f‖H−1 + ‖f‖Lp).

(ii) If 1 < p < N/2, then u ∈ LNpN−2p (Ω), and there exists a constant C independent

of f such that ‖u‖L

NpN−2p

≤ C(‖f‖H−1 + ‖f‖Lp).


Proof. It follows from Theorem 1.1.6 that ‖u‖H1 ≤ C‖f‖H−1 . Therefore,‖u‖

L2NN−2≤ C‖f‖H−1 by Sobolev’s embedding. We fix

1 < q0 <N

2, q0 ≤

2N

N − 2, q0 ≤ p (1.3.12)

and, since |Ω| <∞, we deduce that

‖u‖Lq0 ≤ C‖f‖H−1 (1.3.13)

We now write equation (1.1.3) in the form −∆u + u = g with g = f + (1 −λ)u. Given any 1 < q < N

2 , q ≤ p, we have ‖f‖Lq ≤ C‖f‖Lp (since |Ω| < ∞).Therefore, if u ∈ Lq(Ω), then g ∈ Lq(Ω) and ‖g‖Lq ≤ C(‖f‖Lp + ‖u‖Lq ), so that

by Theorem 1.3.2 (ii), u ∈ LNqN−2q (Ω) and

‖u‖L

NqN−2q

≤ C(‖f‖Lp + ‖u‖Lq ) (1.3.14)

Since NqN−2q ≥

NN−2q and |Ω| <∞, we deduce that if 1 < q < N

2 , then

‖u‖L

NN−2

q ≤ C(‖f‖Lp + ‖u‖Lq ) (1.3.15)

Let k ∈ N be defined by( N

N − 2

)kq0 ≤ p <

( N

N − 2

)k+1

q0

where q0 is given by (1.3.12), and set

qj =( N

N − 2

)jq0

for 0 ≤ j ≤ k + 1. It follows from (1.3.15) that if u ∈ Lqj (Ω) for some 0 ≤ j ≤ k,then u ∈ Lqj+1(Ω) and ‖u‖Lqj+1 ≤ C(‖f‖Lp + ‖u‖Lqj ). Since u ∈ Lq0(Ω) and‖u‖Lq0 ≤ C‖f‖H−1 by (1.3.13), we deduce from an obvious iteration argumentthat u ∈ Lqk+1(Ω) and ‖u‖Lqk+1 ≤ C(‖f‖Lp + ‖f‖H−1). Since qk+1 ≥ p and|Ω| <∞, we deduce in particular that u ∈ Lp(Ω) and ‖u‖Lp ≤ C(‖f‖Lp +‖f‖H−1).Properties (i) abd-(ii) now follows easily from (1.3.14).

Remark 1.3.6. The previous results do not apply to the case p = 1. If N = 1,then ‖u‖L∞ ≤ C‖f‖H−1 , simply by Sobolev’s embedding H1

0 (Ω) → L∞(Ω). IfN ≥ 2 and λ > 0, one can deduce certin estimates by a duality argument. Fixq > N

2 . Let h ∈ C∞c (Ω) and ϕ ∈ H10 (Ω) the weak solution of −∆ϕ + λϕ = h. It

follows from Corollary 1.3.4 that ‖ϕ‖L∞ ≤ C‖h‖Lq . On the other hand

〈f, ϕ〉H−1,H10

= 〈−∆u+ λu, ϕ〉H−1,H10

= 〈−∆ϕ+ λϕ, u〉H−1,H10

= 〈h, u〉H−1,H10

so that ∣∣∣∫Ω

hu∣∣∣ ≤ ‖f‖L1‖ϕ‖L∞ ≤ C‖f‖L1‖h‖Lq

Since h ∈ C∞c (Ω) is arbitrary, we deduce that ‖u‖Lq′ ≤ C‖f‖L1 . Moreover, q > N2

is also arbitrary, so that ‖u‖Lr ≤ C(r)‖f‖L1 for 1 ≤ r < NN−2 .

Remark 1.3.7. Theorems 1.3.2 and 1.3.3 can be established by proving W 2,p

regularity and the using Sobolev embedding W 2,p(Ω) → Lq(Ω). See [14], forinstance. This last approach is, however, considerably more difficult, and requiresregularity assumptions on Ω.


Ω

y(x0)×

x0×

ρ

B(y(x0), ρ)

Figure 2. The geometric condition (1.3.16)

1.3.2. C0 regularity. We show, assuming a mild geometric assumption on Ω(which means that there is no invards corner), that the solutions of (1.1.5) withλ > 0 and f ∈ L∞(Ω), are estimated by the distance to ∂Ω.

Theorem 1.3.8. Assume either N = 1, or else N ≥ 2 and∃ρ > 0 s.t. ∀x0 ∈ ∂Ω,∃y(x0) ∈ RN

with |x0 − y(x0)| = ρ and B(y0, ρ) ∩ Ω = ∅(1.3.16)

(See Figure 2.) Let λ > 0, f ∈ H−1(Ω) and u ∈ H10 (Ω) the solution of (1.1.5). If

f ∈ L∞(Ω), then

|u(x)| ≤ C‖f‖L∞d(x, ∂Ω), (1.3.17)

for all x ∈ Ω, where the constant C is independent of f .

Proof. By homogeneity, we may assume |f | ≤ 1, so that |u| ≤ λ−1 by Theo-rem 1.3.1. Suppose further N ≥ 2, the case N = 1 being immediate. We constructa local barrier at each point of ∂Ω. Given c > 0, set

w(x) = h(|x|)where

h(r) =

14 (ρ2 − r2) + c log(r/ρ) N = 21

2N (ρ2 − r2) + c(ρ2−N − r2−N ) N ≥ 3

(Here ρ > 0 is given by (1.3.16).) It is easy to see that

−∆w = 1 in RN \ 0and that for if c is sufficiently large, then there exist ρ < ρ0 < ρ1 and a constantK such that

w(x) > 0 for ρ < |x| ≤ ρ1, w(x) ≥ λ−1 for ρ0 ≤ |x| ≤ ρ1 (1.3.18)


K(r − ρ)

λ−1

h(r)

r× ×

ρ ρ0 ρ1×

Figure 3. The function h(r)

and

w(x) ≤ K(|x| − ρ) for ρ ≤ |x| ≤ ρ1. (1.3.19)

(See Figure 3.) Let now x ∈ Ω such that 2d(x, ∂Ω) < ρ1 − ρ, and let x0 ∈ ∂Ω

be such that |x − x0| ≤ 2d(x, ∂Ω). Set Ω = x ∈ Ω; ρ < |x − y(x0)| < ρ1 and

v(x) = w(x − y(x0)) for x ∈ Ω. We note that |x − y(x0)| > ρ by the geometriccondition (1.3.16). Moreover,

|x− y(x0)| ≤ |x− x0|+ |x0 − y(x0)| < ρ1 − ρ+ ρ = ρ1,

so that x ∈ Ω. Next, it follows from (1.3.18)-(1.3.19) that v > 0 on Ω and that

0 ≤ v(x) ≤ K(|x− y(x0)| − ρ) ≤ K(|x− x0|+ |x0 − y(x0)| − ρ)

= K|x− x0| ≤ 2Kd(x, ∂Ω)

On the other hand

−∆(u− v) + λ(u− v) = f − (1 + λv) ≤ f − 1 ≤ 0,

in Ω. We claim that

(u− v)+ ∈ H10 (Ω). (1.3.20)

It then follows from the maximum principle that u ≤ v in Ω. In particular, u(x) ≤v(x) ≤ 2Kd(x, ∂Ω). Changing u to −u, one obtains as well that −u ≤ v, so that

|u(x)| ≤ 2Kd(x, ∂Ω) for a.a. x ∈ Ω. Therefore, |u(x)| ≤ 2Kd(x, ∂Ω). Since x isarbitrary, we deduce that if x ∈ Ω such that 2d(x, ∂Ω) < ρ1 − ρ, then |u(x)| ≤2Kd(x, ∂Ω). For x ∈ Ω such that 2d(x, ∂Ω) ≥ ρ1 − ρ, we have u(x) ≤ λ−1 ≤2λ−1(ρ1 − ρ)−1d(x, ∂Ω), and the result follows.

It now remains to establish the claim (1.3.20). Let ϕ ∈ C∞c (RN ) satisfy 0 ≤ ϕ ≤1, ϕ ≡ 1 on the set |x− y(x0)| ≤ ρ0 and ϕ ≡ 0 on the set |x− y(x0)| ≥ ρ1. We

note that by (1.3.18), u ≤ λ−1 ≤ v, thus ϕu− v ≤ u− v ≤ 0 on Ω ∩ |x− y(x0)| ≥ρ0. Therefore, (ϕu − v)+ = (u − v)+ = 0 on Ω ∩ |x − y(x0)| ≥ ρ0. On

Ω ∩ |x − y(x0)| ≤ ρ0, ϕu − v = u − v, so that (u − v)+ = (ϕu − v)+ in Ω.Let now (un)n≥0 ⊂ C∞c (Ω) satisfy un → u in H1(Ω) as n → ∞. It follows (see

Proposition B.2.3) that (ϕun − v)+ → (ϕu − v)+ = (u − v)+ in H1(Ω). Thus, we

need only verify that (ϕun − v)+ ∈ H10 (Ω). This follows from Remark B.1.10 (i),

because ϕun = 0 and v ≥ 0 on ∂Ω.

1.4. SPECTRAL DECOMPOSITION OF THE LAPLACIAN 27

1.4. Spectral decomposition of the Laplacian

Throughout this section Ω is an open, connected, bounded subset of RN , and weequip H1

0 (Ω) with the norm ‖u‖H10

= ‖∇u‖L2 (see Corollary B.3.20), and H−1(Ω)with the dual norm.

We recall that if H is an infinite-dimensional Hilbert space, and if L ∈ L(H) isa self-adjoint, compact, nonnegative (in the sense that (Au, u) ≥ 0 for all u ∈ H)operator, then the eigenvalues of L are a nonincreasing sequence (µn)n≥1 ⊂ (0,∞)such that µn → 0 as n → ∞, and there exists a Hilbert basis (ϕn)n≥1 of Hcomposed of eigenvectors of L, i.e. Lϕn = µnϕn. (See for instance [5, Theorems 6.8and 6.11].) Note that eigenvalues may be repeated in the sequence (µn)n≥1, sincesome eigenvalues may be multiple.

Theorem 1.4.1. Let Ω be a bounded domain of RN . There exists a nonde-creasing sequence (λn)n≥1 ⊂ (0,∞) such that λn → ∞ as n → ∞, and a Hilbertbasis (ϕn)n≥1 of L2(Ω) such that (ϕn)n≥1 ⊂ H1

0 (Ω) and

−∆ϕn = λnϕn (1.4.1)

in H−1(Ω). In particular,

‖∇ϕn‖2L2 = λn (1.4.2)

In addition, ϕn ∈ L∞(Ω)∩C∞(Ω), for all n ≥ 1, and there exist a constant C andan integer ` such that

‖ϕn‖L∞ ≤ C(1 + λj)` (1.4.3)

for all n ≥ 1.

Proof. Let f ∈ H−1(Ω), and u ∈ H10 (Ω) the solution of −∆u = f in H−1(Ω).

We set u = Kf . Theorem 1.1.6 shows that K is bounded H−1(Ω)→ H10 (Ω), hence

L2(Ω) → H10 (Ω). Since Ω is bounded, the embedding H1

0 (Ω) → L2(Ω) is compact(see Theorem B.4.4), therefore K is a compact operator L2(Ω)→ L2(Ω). We claimthat K is self-adjoint. Indeed, let f, g ∈ L2(Ω) and set u = Kf , v = Kg. It followsthat

(Kf, g)L2 − (f,Kg)L2 = (u, g)L2 − (f, v)L2

= (−∆v, u)H−1,H10− (−∆u, v)H−1,H1

0= 0

by formula (B.1.8). Next, if f ∈ L2(Ω) and u = Kf , then

(Kf, f)L2 = (u,−∆u)H10 ,H

−1 =

∫Ω

|∇u|2 ≥ c‖u‖2H1 ≥ c‖f‖2H−1

so that K is a positive operator. Therefore, the eigenvalues of K are a nonincreasingsequence (µn)n≥1 ⊂ (0,∞) such that µn → 0 as n→∞, and there exists a Hilbertbasis (ϕn)n≥1 of H composed of eigenvectors of K, i.e. Kϕn = µnϕn. In particular,µnϕn = Kϕn ∈ H1

0 (Ω), so that (ϕn)n≥1 ⊂ H10 (Ω). In addition, Kϕn = µnϕn

means that (1.4.1) holds, where

λn =1

µn−→n→∞

∞

Formula (1.4.2) follows immediately from (1.4.1) and (B.1.8).To prove that ϕn ∈ L∞(Ω) with the estimate (1.4.3), we write (1.4.1) in the

form

−∆ϕn + ϕn = (λn + 1)ϕn (1.4.4)

We claim that there exist p > max1, N2 , a constant C and an integer ` such that

‖ϕn‖Lp ≤ C(1 + λj)` (1.4.5)


for all n ≥ 0. The conclusion then follows from Corollary 1.3.4. We now prove theclaim (1.4.5). Since ‖ϕn‖L2 = 1, this is verified with p = 2 if N ≤ 3, so we nowassume N ≥ 4. Let the integer k ≥ 1 be defined by

2( N

N − 2

)k−1

≤ N

2< 2( N

N − 2

)kfix 1 < q ≤ 2 such that

q( N

N − 2

)k−1

<N

2< q( N

N − 2

)k(1.4.6)

and set

qj = q( N

N − 2

)j−1

, j = 1, · · · k + 2 (1.4.7)

Suppose ϕn ∈ Lqj (Ω) for some 1 ≤ j ≤ k + 1. Since 1 < qj <N2 , It follows

from Theorem 1.3.2 that ϕn ∈ LNqj

N−2q−J (Ω) and ‖ϕn‖L

NqjN−2qj

≤ C(1 + λn)‖ϕn‖Lqj .

Moreover, qj ≥ 1, so that

NqjN − 2qj

≥ N

N − 2qj = qj+1

Thus we see that ϕn ∈ Lqj+1(Ω) and ‖ϕn‖Lqj+1 ≤ C(1 + λn)‖ϕn‖Lqj . Since‖ϕn‖Lq1 ≤ C‖ϕn‖L2 ≤ C, an obvious induction shows that ϕn ∈ Lqk+1(Ω) and‖ϕn‖Lqk+1 ≤ C(1 + λn)k. Since qk+1 >

N2 , the claim (1.4.5) follows.

Finally, since ϕn ∈ H1loc(Ω), it follows from (1.4.4) and Theorem 1.1.9 that

ϕn ∈ H3loc(Ω). Applying again Theorem 1.1.9, we obtain ϕn ∈ H5

loc(Ω). An ob-vious iteration shows that ϕn ∈ Hm

loc(Ω) for all m ≥ 0, and so ϕn ∈ C∞(Ω) byCorollary B.3.17.

Corollary 1.4.2. Let Ω be a bounded domain of RN , and let (λn)n≥1 ⊂(0,∞) and (ϕn)n≥1 be given by Theorem 1.4.1. If Ω satisfies the geometric condi-tion (1.3.16), then there exist a constant C and an integer ` such that

|ϕn| ≤ C(1 + λn)`d(·, ∂Ω) (1.4.8)

for all n ≥ 1, and in particular ϕn ∈ C0(Ω).

Proof. Estimate (1.4.8) follows from Theorem 1.3.8 applied to the equa-tion (1.4.4), together with estimate (1.4.3). Moreover, ϕn ∈ C(Ω) by Theorem 1.4.1,and continuity at the boundary follows from (1.4.8).

Proposition 1.4.3. Let Ω be a bounded domain of RN , and let (λn)n≥1 ⊂(0,∞) and (ϕn)n≥1 be given by Theorem 1.4.1. Let u ∈ H−1(Ω), and set

αj = 〈u, ϕj〉H−1,H10

(1.4.9)

for j ≥ 1.

(i) u ∈ L2(Ω)⇔ the series u =∑αjϕj is convergent in L2(Ω)⇔

∑∞j=1 α

2j <∞.

In this case, ‖u‖2L2 =∑∞j=1 α

2j , i.e. ‖u‖L2 = ‖(αj)j≥1‖`2 .

(ii) u ∈ H10 (Ω) if and only if

∑∞j=1 λjα

2j < ∞. In this case, the series u =∑

αjϕj is convergent in H10 (Ω), and

∑∞j=1 λjα

2j = ‖∇u‖2L2 , i.e. ‖u‖H1

0=

‖(λ12j αj)j≥1‖`2 .

(iii) The series u =∑αjϕj is convergent in H−1(Ω),

∑∞j=1 λ

−1j α2

j < ∞, and

‖u‖2H−1 =∑∞j=1 λ

−11 a2

j , i.e. ‖u‖H−1 = ‖(λ−12

j αj)j≥1‖`2 .


Proof. Since (ϕj)j≥1 is a Hilbert basis of L2(Ω), Property (i) is immediate.Consider now the isometric isomorphism T : `2(N)→ L2(Ω) defined by

TA =

∞∑j=1

αjϕj ,

where A = (αj)j≥1. Consider the space

V = A ∈ `2(N);∑

λjα2j <∞,

equipped with the norm ‖A‖V = (∑∞j=1 λjα

2j )

12 . It is clear that V is a Banach (in

fact, a Hilbert) space. We first observe that T (V) ⊂ H10 (Ω) and that

‖A‖V = ‖∇TA‖L2 , (1.4.10)

for all A ∈ V. Indeed, given A ∈ V, let (An)n≥0 ⊂ V be defined by αn,j = αjif j ≤ n and αn,j = 0 if j > n. We see that An → A in V (hence in `2(N)). Inaddition,

‖∇TAn‖2L2 = (−∆TAn, TAn)H−1,H10

=( n∑j=1

λjαjϕj ,

n∑j=1

αjϕj

)H−1,H1

0

=( n∑j=1

λjαjϕj ,

n∑j=1

αjϕj

)L2

=

n∑j=1

λjα2j = ‖An‖2V .

(1.4.11)

Similarly, if m > n ≥ 1, then

‖∇(TAm − TAn)‖2L2 =

m∑j=n+1

λjα2j −→n→∞

0. (1.4.12)

Since TAn ∈ H10 (Ω) (finite sum of elements of H1

0 (Ω)), (TAn)n≥1 is a Cauchysequence in H1

0 (Ω); and since TAn → TA in L2(Ω), we deduce that TA ∈ H10 (Ω)

and An → A in V. Identity (1.4.10) is obtained by letting n → ∞ in (1.4.11).We now prove that TV = H1

0 (Ω). Since V is a Banach space and T : V → H10 (Ω)

is isometric, TV is a closed subspace of H10 (Ω). If V 6= H1

0 (Ω), then there existsw ∈ H1

0 (Ω), w 6= 0 such that (TA,w)H10

= 0 for all A ∈ V. Given j ≥ 1, we may

choose TA = ϕj , so that (ϕj , w)H10

= 0. Since

(ϕj , w)H10

= (∇ϕj ,∇w)L2 = (−∆ϕj , w)H−1,H10

= λj(ϕj , w)H−1,H10

= λj(ϕj , w)L2 ,

and j ≥ 1 is arbitrary, we deduce that w = 0. This contradiction shows thatTV = H1

0 (Ω), from which Property (ii) easily follows.Finally, we observe that if we identify `2(N) with its dual, then the dual of V

is W = A ∈ `2(N);∑λ−1j α2

j < ∞, equipped with the natural scalar product.One deduces easily that T can be extended to an isometric isomorphism W →(H1

0 (Ω))? = H−1(Ω). Hence Property (iii).

Remark 1.4.4. Here are some comments on Proposition 1.4.3.

(i) Given j ≥ 1, the map u 7→ αj , where αj is defined by (1.4.9), is con-tinuous H−1(Ω) → R. It follows easily that if I ⊂ R is an interval andu ∈ C(I,H−1(Ω)), then the map t 7→ αj(t) defined by (1.4.9) is continu-ous. Moreover, if u is differentiable at some t ∈ I, then every j ≥ 1, αj isdifferentiable at t, and

dαjdt

=⟨dudt, ϕj

⟩H−1,H1

0

(1.4.13)


Therefore, if u ∈ C1(I,H−1(Ω)), then the map t 7→ αj(t) is C1 and (1.4.13)holds for all t ∈ I.

(ii) Let u1, u2 ∈ H−1(Ω), and let a`j = 〈u`, ϕj〉H−1,H10. It follows that

(u1, u2)H−1 =

∞∑j=1

λ−1j a1

ja2j

(u1, u2)L2 =

∞∑j=1

a1ja

2j if u1, u2 ∈ L2(Ω)

(u1, u2)H10

=

∞∑j=1

λja1ja

2j if u1, u2 ∈ H1

0 (Ω)

〈u1, u2〉H−1,H10

=

∞∑j=1

a1ja

2j if u2 ∈ H1

0 (Ω)

Proposition 1.4.5. Let Ω be a bounded domain of RN and λ(Ω) be definedby (1.1.7). It follows that there exists ϕ ∈ H1

0 (Ω) ∩ L∞(Ω) ∩ C(Ω), ϕ > 0 in Ω,‖ϕ‖L2 = 1, such that

−∆ϕ = λ(Ω)ϕ, (1.4.14)

in H−1(Ω). In addition, the following properties hold.

(i) ϕ is the unique nonnegative solution of

u ∈ S, J(u) = infv∈S

J(v), (1.4.15)

where

J(v) =1

2

∫Ω

|∇v|2, (1.4.16)

and S = v ∈ H10 (Ω); ‖v‖L2 = 1.

(ii) If ψ ∈ H10 (Ω) is a solution of (1.4.14), then there exists a constant c ∈ R such

that ψ = cϕ.

Proof. We proceed in five steps.

Step 1. Problem (1.4.15) has a solution u ∈ H10 (Ω), u ≥ 0. We recall that

J ∈ C1(H10 (Ω),R) and J ′(u) = −∆u for all u ∈ H1

0 (Ω). (See Corollary B.1.21.)Moreover, one verifies easily that if F (u) = ‖u‖2L2 , then F ∈ C1(H1

0 (Ω),R) andF ′(u) = 2u for all u ∈ H1

0 (Ω). On the other hand, it follows from (1.1.7) that

infv∈S

J(v) =λ(Ω)

2. (1.4.17)

Let (vn)n≥0 be a minimizing sequence of (1.4.15). Setting un = |vn|, it followsthat un ≥ 0 and that (un)n≥0 is also a minimizing sequence. (Apply (B.2.4).)In addition, ‖un‖L2 = 1, so that (un)n≥0 is bounded in H1

0 (Ω). Therefore, thereexist a subsequence (unk)k≥1 and u ∈ H1

0 (Ω) such that un → u in L2(Ω) and‖∇u‖L2 ≤ lim inf ‖∇un‖L2 . In particular, u ≥ 0 and u is a solution of (1.4.15).

Step 2. If u ∈ H10 (Ω) and ‖u‖L2 = 1, then u is a solution of (1.4.15) iff u is a

solution of equation (1.4.14). Suppose u is a solution of (1.4.15). Then there existsa Lagrange multiplier λ such that −∆u = λu. One can see this easily as follows.Let w ∈ H1

0 (Ω) such that (w, u)L2 = 0, and set v = (1 + t2‖w‖2L2)−12 (u+ tw), with

t ∈ R. It follows that ‖v‖L2 = 1, so that J(u) ≤ J(v). This means that

t(J(u)− J(v)) ≤∫

Ω

∇u · ∇w


Letting t ↓ 0, we obtain ∫Ω

∇u · ∇w ≥ 0

then replacing w by −w ∫Ω

∇u · ∇w = 0

Consider now ϕ ∈ H10 (Ω). We may write ϕ = ηu + w with (w, u)L2 = 0 (choose

η = (ϕ, u)L2). Setting λ = ‖∇u‖2L2 , we see that

〈−∆u− λu, ϕ〉H−1,H10

=

∫Ω

(∇u · ∇ϕ− λuϕ) = η

∫Ω

|∇u|2 − ηλ = 0

Since ϕ ∈ H10 (Ω) is arbitrary, this shows that −∆u = λu in H−1(Ω). Taking

the H−1 − H10 duality product with u, we obtain 2J(u) = λ, so that λ = λ(Ω)

by (1.4.17). Therefore, u is a solution of (1.4.14). Conversely, suppose u is asolution of (1.4.14). Taking the H−1 − H1

0 duality product with u, we obtain2J(u) = Λ, so that by (1.4.17), u is a solution of (1.4.15).

Step 3. If u ∈ H10 (Ω), u ≥ 0, u 6≡ 0, is a solution of (1.4.14), then u ∈

L∞(Ω) ∩C(Ω) and u > 0 on Ω. Regularity is proved like for Theorem 1.4.1. Thefact that u > 0 is a consequence of the strong maximum principle.

Step 4. If u, v ≥ 0 are two solutions of (1.4.15), then u = v. Let w = u−v andassume by contradiction the w 6≡ 0. Since u and v are solutions of (1.4.14), so is w.It follows from Step 2 that w/‖w‖L2 is a solution of (1.4.15), so that z = |w|/‖w‖L2

is also a solution. Steps 2 et 3 show that z > 0 in Ω. In particular, w does notvanish in Ω. Therefore, either w > 0, or else w < 0. Thus either 0 ≤ v < u, or else0 ≤ u < v, which is absurd since ‖u‖L2 = ‖v‖L2 .

Step 5. Conclusion. Let ϕ = u with u given by Step 1. Steps 2 et 3 showthat ϕ is a solution of (1.4.14), that ϕ ∈ H1

0 (Ω) ∩ L∞(Ω) ∩ C(Ω), ϕ > 0 in Ωand ‖ϕ‖L2 = 1. Property (i) is then a consequence of Step 4. Finally, supposeψ ∈ H1

0 (Ω), ψ 6≡ 0 is a solution of (1.4.14). Setting z = |ψ|/‖ψ‖L2 , we deduce fromStep 2 that z is a positive solution of (1.4.15). Hence z = ϕ, so that |ψ| = ‖ψ‖L2ϕ.In particular, ψ has constant sign. Therefore, ψ = ±‖ψ‖L2ϕ, which proves (ii).

Corollary 1.4.6. Let Ω be a bounded domain of RN , and let (λn)n≥1 ⊂ (0,∞)and (ϕn)n≥1 be given by Theorem 1.4.1.

(i) λ1 = λ(Ω), where λ(Ω) is defined by (1.1.7).(ii) λ1 is a simple eigenvalue, and either ϕ1 > 0 or else ϕ1 < 0 in Ω.(iii) Suppose Ω satisfies the geometric conditions (1.3.16) and (1.2.1). If ϕ1 is

chosen such that ϕ1 > 0, then

cd(·, ∂Ω) ≤ ϕ1(·) ≤ Cd(·, ∂Ω) (1.4.18)

for some constants 0 < c ≤ C <∞.(iv) If ψ ∈ H1

0 (Ω) satisfies −∆ψ = λψ for some λ ∈ R, and if ψ ≥ 0, then thereexists a constant c ∈ R such that ψ = cϕ1.

Proof. It follows from Proposition 1.4.5 that λ(Ω) is an eigenvalue of −∆ onH1

0 (Ω), so λ(Ω) = λ` for some ` ≥ 1. In particular, λ(Ω) ≥ λ1. On the other hand,‖ϕ1‖L2 = 1 and ‖∇ϕ1‖2L2 = λ1 by (1.4.2), so that λ(Ω) ≤ λ1 by (1.1.7). Thus

λ(Ω) = λ1, which proves (i).By Proposition 1.4.5 (ii), all nontrivial solutions of −∆u = λ1u are multiples of

a given one, so that λ1 is a simple eigenvalue. Moreover, again by Proposition 1.4.5,these solutions are either positive or else negative on Ω, which proves (ii).


The upper estimate in (1.4.18) follows from (1.4.8), and the lower estimatefollows from Theorem 1.2.5 (applied to the equation −∆ϕ1 + ϕ1 = (1 + λ1)ϕ1),since ϕ1 > 0.

Finally, suppose ψ ∈ H10 (Ω) satisfies −∆ψ = λψ for some λ ∈ R, and ψ ≥ 0,

ψ 6≡ 0. It follows that ψ is an eigenvector of −∆. If λ 6= λ1, then∫

Ωψϕ1 = 0, which

is absurd because ϕ1 > 0 and ψ ≥ 0, ψ 6≡ 0. Therefore, λ = λ1 and Property (iv)follows from Property (1.4.18).

Remark 1.4.7. In dimension 1, let Ω = (0, `) with ` > 0. Let

ϕj(x) =

√2

`sin(jπ`x)

for j ≥ 1. It follows that ϕ′′j = λjϕj with λj = ( jπ` )2. Moreover, every solution

u ∈ H10 (Ω) of the equation −u′′ = λu is given by the above formula for some j ≥ 1.

This shows that the spectral decomposition of the Laplacian on H10 (0, `) is (ϕj , λj)

where ϕj and λj are as above. In particular, if Ω = (0, π), then ϕj(x) = sin(jx)and λj = j2, and the above analysis corresponds to the Fourier series.

Remark 1.4.8. Suppose N = 3, and let Ω = BR, the ball of center 0 andradius R in R3. Let

ϕ(r) =1

r(2πR)

−12 sin

( πRr)

so that ϕ > 0 and ‖ϕ‖L2(Ω) = 1. It is not difficult to check that ϕ ∈ H10 (Ω) and

that −∆ϕ = ( πR )2ϕ. Since ϕ > 0, it follows that ϕ = ϕ1 and ( πR )2 = λ1. Note

that for every integer k ≥ 1, ψ(r) = 1r (2πR)

−12 sin(kπR r) is an eigenvector of −∆

corresponding to the eigenvalue λ = (kπR )2. However, this does not give all theeigenvectors and eigenvalues of −∆, since there are eigenvalues corresponding tonon-radial eigenvectors.

It follows from Remarks 1.4.7 and 1.4.8 that the first eigenvector of −∆ inH1

0 (BR) is radially symmetric and decreasing in dimensions N = 1 and N = 3. Infact, this property is true in any dimension, as shows the following result.

Proposition 1.4.9. Let Ω = BR, the ball of center 0 and radius R in RN ,and let λ1 be the first eigenvalue of −∆ in H1

0 (Ω) and ϕ1 > 0 a corresponding firsteigenvector.

(i) ϕ1 ∈ C∞(Ω) is radially symmetric and decreasing.(ii) There exist two constants c = c(N) and C = C(N) such that

cR−1(R− |x|) ≤ ϕ1(x) ≤ CR−1(R− |x|) (1.4.19)

for all x ∈ Ω if ϕ1 is normalized by the condition ϕ1(0) = 1, and

cR−N+2

2 (R− |x|) ≤ ϕ1(x) ≤ CR−N+2

2 (R− |x|) (1.4.20)

for all x ∈ Ω if ϕ1 is normalized by the condition ‖ϕ1‖L2 = 1(iii) There exists a constant λ = λ(N) such that λ1 = R−2λ.

Proof. Consider the solution of the ODEz′′ + N−1

r z′ + u = 0

z(0) = 1, z′(0) = 0(1.4.21)

We claim that

∃ρ > 0 s.t. z is positive, decreasing on (0, ρ), z(ρ) = 0 and z′(ρ) < 0 (1.4.22)

We complete the proof, assuming the claim (1.4.22). It follows that there exist twoconstants 0 < c < C <∞ (depending only on N) such that

c(ρ− r) ≤ z(r) ≤ C(ρ− r) 0 ≤ r ≤ ρ


Setting

ψ(x) = z( ρR|x|)

we deduce thatcρ

R(R− |x|) ≤ ψ(x) ≤ Cρ

R(R− |x|) 0 ≤ r ≤ R (1.4.23)

Moreover, ψ is radially symmetric and decreasing, and one verifies easily that ψ ∈C∞(Ω) (so that ψ ∈ H1

0 (Ω), since ψ|∂Ω = 0), and that −∆ψ = ( ρR )2ψ. Therefore,

ψ is a multiple of the ϕ1 and λ1 = ( ρR )2, by Corollary 1.4.2 (iv). Since ψ(0) = 1,estimate (1.4.19) follows from (1.4.23). Estimate (1.4.20) also follows from (1.4.23)

(replacing ψ by ψ‖ψ‖L2

), since ‖ψ‖L2 ≈ RN2 . This completes the proof.

We now prove the claim (1.4.22). It follows easily from the equation (1.4.21)that z′(r) < 0 for r > 0 sufficiently small, and that z′ cannot vanish while z remains> 0. If z vanishes, then z′ 6= 0 (otherwise z ≡ 0), and (1.4.22) is proved. Otherwise,z′ < 0 and z > 0 for all r > 0. It follows that z decreases to a nonnegative limit asr →∞, and it one deduces easily from the equation (1.4.21) that this limit must be0. Next, we write (1.4.21) in the form (rN−1z′)′ = −rN−1z. Integrating on (0, r)and since z is decreasing, we obtain

−rN−1z′(r) =

∫ r

0

sN−1z(s) ≥ z(r)∫ r

0

sN−1 =1

NrNz(r)

This means that z′ + rz ≤ 0, so that (er2

2 z)′ ≤ 0. Thus 0 ≤ z(r) ≤ e−r2

2 , and inparticular

erz(r) −→r→∞

0

On the other hand,

[erz(r)]′ = er[z′′ + 2z′ + z] = er(

2− N − 1

r

)z′ ≤ 0

for r ≥ N−12 , since z′ < 0. We deduce that erz(r) is a positive, concave function

for r large, that converges to 0 at infinity. This is absurd, so the claim (1.4.22) isproved.

Remark 1.4.10. Suppose Ω is a bounded domain of RN and let λ satisfy (1.1.8)(i.e. λ > −λ1). Given f ∈ H−1(Ω), it follows from Theorem 1.1.6 that there existsa unique weak solution u ∈ H1

0 (Ω) of (1.1.8). The solution u can be expressedin terms of the spectral decomposition of the Laplacian. More precisely, let βj =〈f, ϕj〉H−1,H1

0and αj = 〈u, ϕj〉H−1,H1

0for j ≥ 1. Letting ϕ = ϕj in (1.1.5) and

applying (B.1.8), we deduce that (λ+ λj)αj = βj . In other words,

u =

∞∑j=1

〈f, ϕj〉H−1

λ+ λjϕj

where the series is convergent in H10 (Ω) (see Proposition 1.4.3).

CHAPTER 2

The heat equation

The heat equation is the prototype parabolic equation. Among its fundamentalproperties are the maximum principle, and the smoothing effect. We first studythe equation set on the whole space RN , by using the Fourier transform. Then westudy the equation set on a bounded domain, with Dirichlet boundary conditions,by using Fourier series associated with the spectral decomposition of the Laplacian.

2.1. The heat equation on RN

We consider the following (Cauchy) problemut = ∆u t > 0, x ∈ RN

u(0, x) = u0(x) x ∈ RN(2.1.1)

2.1.1. Existence and regularity. We begin with an existence and unique-ness result in the space S ′(RN ).

Theorem 2.1.1. Given any u0 ∈ S ′(RN ), there exists a unique solution u ∈C1([0,∞),S ′(RN )) of equation (2.1.1). This solution is given, for all t > 0, by

u(t) = e−4π2t|ξ|2 u0 (2.1.2)

where u = Fu is the Fourier transform of u or, equivalently, by

u(t, x) = Gt ? u0 (2.1.3)

where G ∈ C∞((0,∞),S(RN )) is defined by

Gt(x) = (4πt)−N2 e−

|x|24t (2.1.4)

(Gt is called the Gauss kernel.) Moreover, the following properties hold.

(i) u ∈ C∞([0,∞),S ′(RN ))(ii) For all t > 0, u(t) is a function of C∞(RN ).(iii) if u0 ∈ S(RN ), then u ∈ C∞([0,∞),S(RN ))(iv) ∂mt u = ∆mu for all t ≥ 0 and all m ∈ N.

Remark 2.1.2. Note that the convolution (2.1.3) is well defined for all t >0 (convolution of a tempered distribution with a fonction of S(RN ), see Theo-rem B.6.14).

Proof of Theorem 2.1.1. We first prove uniqueness. This is the more deli-cate part of the proof, as we claim uniqueness in the large class C1([0,∞),S ′(RN )).We consider u, v ∈ C1([0,∞),S ′(RN )) two solutions of (2.1.1). Setting w = u − vand w = Fw, it follows that w ∈ C1([0,∞),S ′(RN )) (see Remark B.6.12) and thatw solves the equation

wt + 4π2|ξ|2w = 0

(See (B.6.42).) Moreover w(0) = 0. If the above equation were an ODE, it would beimmediate that w = 0. However, the equation makes sense in C1([0,∞,S ′(RN )),

35

36 2. THE HEAT EQUATION

so some care must be taken. We fix θ ∈ S(RN ) and τ > 0, and we define ϕ ∈C1([0, τ ],S(RN )) by

ϕ(t, ξ) = e−4π2|ξ|2(τ−t)θ(ξ)

for 0 ≤ t ≤ τ and ξ ∈ RN , so that ϕ′(t) = 4π2| · |2ϕ(t). It follows from (B.6.50)that

d

dt〈w(t), ϕ(t)〉 = 〈w′(t), ϕ(t)〉+ 〈w(t), ϕ′(t)〉

= −〈4π2| · |2w(t), ϕ(t)〉+ 〈w(t), 4π2| · |2ϕ(t)〉 = 0

by (B.6.33). Therefore, 〈w(τ), ϕ(τ)〉 = 〈w(0), ϕ(0)〉 = 0, since w(0) = 0. Thismeans that 〈w(τ), θ〉 = 0. Since θ ∈ S(RN ) and τ > 0 are arbitrary, we concludethat w = 0. Therefore, w = 0, hence u = v.

We next prove existence. Let u0 ∈ S ′(RN ), so that u0 ∈ S ′(RN ). It follows

from Remark B.6.10 (v) that e−4π2t|ξ|2 u0 ∈ S ′(RN ) for all t ≥ 0. Therefore,formula (2.1.2) defines u(t) ∈ S ′(RN ) for all t ≥ 0. In addition, it is not difficult tocheck that, given any ϕ ∈ S(RN )

the map t 7→ e−4π2t|ξ|2ϕ is C∞ : [0,∞)→ S(RN ) (2.1.5)

It follows (see e.g. Proposition B.6.15) that formula (2.1.2) defines a functionu ∈ C∞([0,∞),S ′(RN )), hence u ∈ C∞([0,∞),S ′(RN )). Moreover, u satisfies theequation ut + 4π2|ξ|2u = 0, which means that ut = ∆u. Since u(0) = u0, we haveu(0) = u0, so that u is a solution of (2.1.1).

So far, we have established existence, uniqueness, formula (2.1.2), and theregularity property (i). Property (iii) is an immediate consequence of (2.1.5) (withϕ = u0).

Next, we prove formula (2.1.3) and the regularity property (ii). We apply theinverse Fourier transform to formula (2.1.2), and we note that for all t > 0

F−1[e−4π2t|·|2 ] = Gt

where Gt is defined by (2.1.4). (Apply (B.6.9) with a = 4πt.) Since

u = F−1u = F−1[e−4π2t|·|2 ] ? F−1u0 = F−1[e−4π2t|·|2 ] ? u0

(see Theorem B.6.14 (iv)) we see that for every t > 0, (2.1.2) is equivalent to (2.1.3).It follows in particular (see Theorem B.6.14 (i)) that u(t) ∈ C∞(RN ) for all t > 0.

Finally, we prove Property (iv). Let t ≥ 0 and h 6= 0 such that t+h ≥ 0. Givenϕ ∈ S(RN ),

〈ut(t+ h)− ut(t), ϕ〉 = 〈∆u(t+ h)−∆u(t), ϕ〉 = 〈u(t+ h)− u(t),∆ϕ〉

Letting h→ 0, we obtain

〈∂2t u, ϕ〉 = 〈ut,∆ϕ〉 = 〈∆u,∆ϕ〉 = 〈∆2u, ϕ〉

Thus we see that ∂2t u = ∆2u, and Property (iv) follows by an obvious iteration

argument.

Remark 2.1.3. Here are some comments on Theorem 2.1.1

(i) Let u0 ∈ S ′(RN ) and u the corresponding solution of (2.1.1). Given τ > 0,it follows that v(t) = u(τ + t) satisfies vt = ∆v in C([0,∞),S ′(RN )) andv(0) = u(τ). By uniqueness, it follows that the solution v of (2.1.1) with theinitial condition v(0) = u(τ) is given by v(t) = u(t+ τ).

(ii) The proof of uniqueness shows in fact the following stronger uniqueness prop-erty. If T > 0 and u ∈ C1([0, T ],S ′(RN )) satisfies ut = ∆u for all 0 ≤ t ≤ Tand if u(0) = 0, then u(t) = 0 for all 0 ≤ t ≤ T .

2.1. THE HEAT EQUATION ON RN 37

We now describe some regularity properties that follow from formulas (2.1.2)and (2.1.3). The results below show that the regularity in Sobolev spaces is pre-served by the heat equation. Moreover, they show that the heat equation hasa remarkable smoothing effect, in the sense that for t > 0, u(t) is considerablysmoother than the initial value u(0).

Corollary 2.1.4. Let u0 ∈ S ′(RN ) and u ∈ C1([0,∞),S ′(RN )) the cor-responding solution of (2.1.1). If u0 ∈ Lp(RN ) for some 1 ≤ p ≤ ∞, thenu(t) ∈ Lq(RN ) all t > 0 and all p ≤ q ≤ ∞. Moreover,

‖u(t)‖Lq ≤ (4πt)−N2 ( 1

p−1q )‖u0‖Lp (2.1.6)

for all t > 0 and all p ≤ q ≤ ∞.

Proof. Note that

‖Gt‖Lr = r−N2r (4πt)−

N2 (1− 1

r ) ≤ (4πt)−N2 (1− 1

r ) (2.1.7)

for all 1 ≤ r ≤ ∞. Applying Young’s inequality to formula (2.1.3), we obtain

‖u(t)‖Lq ≤ ‖Gt‖Lr‖u0‖Lq

where r is given by 1− 1r = 1

p −1q . Hence the result.

Remark 2.1.5. Here are some comments on Corollary 2.1.4.

(i) If u0 ∈ Lp(RN ) with 1 ≤ p ≤ ∞, u0 ≥ 0, u0 6≡ 0, then formula (2.1.3) showsthat u(t, x) > 0 for all t > 0 and x ∈ RN .

(ii) Note that u(t) is the convolution of u0 with the radially symmetric and de-creasing kernel Gt. It follows that if u0 is radially symmetric, then so is u(t)for all t ≥ 0; and if u0 is radially symmetric and decreasing, then so is u(t)for all t ≥ 0. See Proposition A.2.2.

(iii) It follows from (2.1.6) that if, for instance, u0 ∈ L1(RN ), then ‖u(t)‖L∞ ≤Ct−

N2 . This decay cannot, in general, be improved. Indeed, if u0(x) = e−|x|

2

,

then u(t, x) = (1+4t)−N2 e−

|x|21+at (see Remark 2.1.10 (iv) below). In particular,

‖u(t)‖L∞ = (1 + 4t)−N2 , which is precisely the decay given by (2.1.6).

(iv) Let u0 ∈ L1(RN ) and u the corresponding solution of (2.1.1). It followsfrom (2.1.6) that ‖u(t)‖Lp → 0 as t → ∞ for every p > 1. In general,‖u(t)‖L1 6→ 0 as t→∞. Indeed,∫RN

u(t, x) dx =

∫RN

∫RN

Gt(y)u0(x− y) =(∫

RNGt

)(∫RN

u0

)=

∫RN

u0

If u0 ≥ 0 and u0 6≡ 0 then u(t) > 0 by Property (i), so that ‖u(t)‖L1 =∫u(t) =

∫u0 > 0.

Corollary 2.1.6. Let u0 ∈ S ′(RN ) and u ∈ C1([0,∞),S ′(RN )) the corre-sponding solution of (2.1.1). If u0 ∈ Hs,p(RN ) for some s ∈ R and 1 < p < ∞,then the following properties hold.

(i) u ∈ C([0,∞), Hs,p(RN ))) ∩ C((0,∞), Hs,q(RN ))) for all p ≤ q ≤ ∞, and

‖u(t)‖Hs,q ≤ (4πt)−N( 1p−

1q )‖u0‖Hs,p (2.1.8)

for all t > 0.(ii) u ∈ Ck([0,∞), Hs−2k,p(RN )) ∩ Ck((0,∞), Hs−2k,q(RN )) for all k ∈ N and

p ≤ q ≤ ∞, and

‖∂kt u(t)‖Hs−2k,q ≤ Ct−N( 1p−

1q )‖u0‖Hs,p (2.1.9)

for all t > 0, where the constant C is independent of t > 0 and u0.


(iii) u ∈ C∞((0,∞), Hσ,q(RN )) for all σ ∈ R and q ≥ p. In addition

‖Dβ∂kt u(t)‖Hs,q ≤ Ct−N2 ( 1

p−1q )− |β|2 −k‖u0‖Hs,p (2.1.10)

for all multi-indices β, all k ∈ N and all t > 0. In particular, u ∈ C∞((0,∞)×RN ). Moreover,

‖∂kt u(t)‖Hs+m,q ≤ Ct−N2 ( 1

p−1q )−k(1 + t−

m2 )‖u0‖Hs,p (2.1.11)

for all m, k ∈ N and all t > 0.

Proof. Fix s ∈ R and 1 ≤ p < ∞. Suppose first u0 ∈ S(RN ), so thatu ∈ C∞([0,∞),S(RN )). Let β be a multi-index. It follows from (2.1.3) that

Dβu(t) = (DβGt) ? u0 (2.1.12)

Let σ ∈ R and p ≤ q ≤ ∞, and let 1 ≤ r ≤ ∞ be given by

1

r=

1

q− 1

p+ 1 (2.1.13)

Since ∂µt u = (∆)µu by Theorem 2.1.1 (iv), it follows from (B.6.98) that

‖Dβ∂µt u(t)‖Hσ−2µ,q = ‖(∆)µDβu(t)‖Hσ−2µ,q ≤ ‖Dβu(t)‖Hσ,qApplying (2.1.12) and (B.6.91) we deduce that

‖Dβ∂µt u(t)‖Hσ−2µ,q ≤ ‖DβGt‖Hσ−s,r‖u0‖Hs,p (2.1.14)

for all t > 0. (Note that DβGt ∈ S(RN ) ⊂ Hσ−s,r(RN ), so that ‖DβGt‖Hσ−s,r <∞.) Since H0,r(RN ) = Lr(RN ), we see by letting σ = s in (2.1.14) that

‖Dβ∂µt u(t)‖Hs−2µ,q ≤ ‖DβGt‖Lr‖u0‖Hs,p (2.1.15)

In particular, if β = 0 and µ = 0, we obtain by applying (2.1.7) and (2.1.13) that

‖u(t)‖Hs,q ≤ (4πt)−N2 ( 1

p−1q )‖u0‖Hs,p (2.1.16)

for all t > 0. Moreover, writing Gt(x) = t−N2 Φ( x√

t), where Φ(y) = (4π)−

N2 e−

|y|24 ,

we see that DβGt(x) = t−N2 t−

|β|2 DβΦ( x√

t). Hence

‖DβGt‖Lr = t−N2 (1− 1

r )t−|β|2 ‖DβΦ‖Lr = t−

N2 ( 1

p−1q )t−

|β|2 ‖DβΦ‖Lr

and it follows from (2.1.15) that

‖Dβ∂µt u(t)‖Hs−2µ,q ≤ Ct−N2 ( 1

p−1q )t−

|β|2 ‖u0‖Hs,p (2.1.17)

for all t > 0.We now consider u0 ∈ Hs,p(RN ) with s ∈ R and 1 ≤ p < ∞. By density of

S(RN ) in Hs,p(RN ) (see Remark B.6.23 (vii)), there exists a sequence (un0 )n≥1 ⊂S(RN ) such that

‖un0 − u0‖Hs,p −→n→∞

0 (2.1.18)

We denote by un the corresponding solutions of (2.1.1). It follows from (2.1.2) that

un(t) −→n→∞

u(t) in S ′(RN ) (2.1.19)

for all t ≥ 0. We first apply (2.1.16) with q = p, and u0 replaced by un0 , then byum0 − un0 , and we obtain

‖un(t)‖Hs,p ≤ ‖un0‖Hs,p (2.1.20)

‖um(t)− un(t)‖Hs,p ≤ ‖um0 − un0‖Hs,p (2.1.21)

It follows from (2.1.21) that, given any T > 0, (un)n≥1 is a Cauchy sequencein the Banach space C([0, T ], Hs,p(RN )). Applying (2.1.19), we deduce that u ∈C([0, T ], Hs,p(RN )) and un → u in C([0, T ], Hs,p(RN )). Letting n→∞ in (2.1.20)


yields (2.1.8) for q = p (since T > 0 is arbitrary). This proves the case q = p ofProperty (i). For q > p, we apply (2.1.16) with u0 replaced by un0 and um0 − un0 ,respectively, and we obtain

‖un(t)‖Hs,q ≤ (4πt)−N2 ( 1

p−1q )‖un0‖Hs,p (2.1.22)

‖um(t)− un(t)‖Hs,q ≤ (4πt)−N2 ( 1

p−1q )‖um0 − un0‖Hs,p (2.1.23)

It follows that, given any 0 < τ < T < ∞, (un)n≥1 is a Cauchy sequence in theBanach space C([τ, T ], Hs,q(RN )), and we conclude as above that Property (i) holdsas well in the case q > p.

Property (ii) is proved like Property (i), but instead of applying (2.1.16), oneapplies (2.1.17) with β = 0.

To prove Property (iii), we first fix ε > 0, σ ∈ R and p ≤ q ≤ ∞. Apply-ing (2.1.14) with β = 0, µ = 0 and t = ε, we obtain

‖u(ε)‖Hσ,q ≤ ‖Gε‖Hσ−s,r‖u0‖Hs,p (2.1.24)

Arguing by density as above, we deduce that u(ε) ∈ Hσ,q(RN ). Note that ε > 0,σ ∈ R and q ≥ p are arbitrary, so that the first statement of Property (iii) followsfrom Property (ii), since the solution v(t) of (2.1.1) with the initial value v(0) = u(ε)is v(t) = u(t+ ε). (See Remark 2.1.3.) To prove the estimate (2.1.10), we note thatDβ∂kt u(t) = Dβ∆ku(t), so that by (2.1.17) (applied with µ = 0 and β replaced by

β with |β| = |β| + 2k) yields (2.1.10). Finally, (2.1.11) follows from (2.1.10) andthe fact that ‖u‖Hs+m,p ≈

∑|β|≤m ‖Dβu‖Hs,p by Proposition B.6.30.

Corollary 2.1.7. Let u0 ∈ C0(RN ) and u ∈ C1([0,∞),S ′(RN )) the corre-sponding solution of (2.1.1). It follows that u ∈ C([0,∞), C0(RN )).

Proof. We first consider the case u0 ∈ C∞c (RN ). In particular, u0 ∈ HN (RN ),so that u ∈ C([0,∞), HN (RN )) by Corollary 2.1.6. Since HN (RN ) → C0(RN ), wesee that u ∈ C([0,∞), C0(RN )).

The general case follows by density and estimate (2.1.6). Indeed, let u0 ∈C0(RN ). Let (un0 )n≥1 ⊂ C∞c (RN ) satisfy un0 → u0 in C0(RN ) and let un be thecorresponding solutions of (2.1.1). We have seen that un ∈ C([0,∞), C0(RN )).On the other hand, it follows from (2.1.6) (with p = q = ∞) that un → u inL∞((0,∞)×RN ). In particular, un → u in L∞((0, T )×RN ) for every 0 < T <∞.Since C([0, T ], C0(RN )) is a Banach space with the norm of L∞((0, T ) × RN ), weconclude that u ∈ C([0, T ], C0(RN )) for all 0 < T <∞.

2.1.2. The heat semigroup. The heat semigroup, often denoted (et∆)t≥0, isthe family of operators S ′(RN )→ S ′(RN ) defined as follows for every t ≥ 0: givenu0 ∈ S ′(RN ), T (t)u0 is the solution at time t of equation (2.1.1).

Proposition 2.1.8. If (T (t))t≥0 is as defined above, then the following prop-erties hold.

(i) T (t) ∈ L(S ′(RN )) and T (t) ∈ L(S(RN )) for all t ≥ 0.(ii) T (0) = I.(iii) T (t+ s) = T (t)T (s), for all s, t ≥ 0.(iv) ∀u0 ∈ S ′(RN ), the map t 7→ T (t)u0 is C∞([0,∞),S ′(RN )) and d

dtT (t)u0 =∆T (t)u0 for all t ≥ 0.

(v) ∀u0 ∈ S(RN ), the map t 7→ T (t)u0 is C∞([0,∞),S(RN )).(vi) 〈T (t)ϕ,ψ〉S′,S = 〈ϕ,T (t)ψ〉S′,S for all ϕ ∈ S ′(RN ) and ψ ∈ S(RN ).

Proof. Properties (ii), (iv) and (v) follow from Theorem 2.1.1, and Prop-erty (iii) from Remark 2.1.3 (i). Next, we note that given t > 0, T (t)u = Gt ? uby formula (2.1.3). Since convolution with a function of S(RN ) is a continuous


map S(RN ) → S(RN ) and S ′(RN ) → S ′(RN ) (see Remark B.6.4 (v) and Theo-rem B.6.14), Property (ii) follows. Finally, given ϕ ∈ S ′(RN ), ψ ∈ S(RN ) andt > 0, it follows from Theorem B.6.14 that

〈T (t)ϕ,ψ〉S′,S = 〈Gt ? ϕ, ψ〉S′,S = 〈ϕ,Gt ? ψ〉S′,S = 〈ϕ,T (t)ψ〉S′,Swhich proves Property (vi).

We now describe the regularity and smoothing properties of the heat equationin terms of the heat semigroup.

Proposition 2.1.9. If (T (t))t≥0 is as defined above, and s ∈ R, 1 < p < ∞,then the following properties hold.

(i) (T (t))t≥0 ⊂ L(Hs,p(RN )) and ‖T (t)‖L(Hs,p) ≤ 1.

(ii) ∀u0 ∈ Hs,p(RN ), the map t 7→ T (t)u0 is Ck : [0,∞) → Hs−2k,p(RN ) for all

k ∈ N, and dk

dtkT (t)u0 = ∆kT (t)u0.

(iii) For all t > 0, σ ≥ s and q ≥ p, T (t) ∈ L(Hs,p(RN ), Hσ,q(RN )). In addition,if p ≤ q <∞, then there exists a constant C such that

‖T (t)‖L(Hs,p,Hσ,q) ≤ Ct−N2 ( 1

p−1q )(1 + t−

σ−s2 ) (2.1.25)

for all t > 0.(iv) Given any u0 ∈ Hs,p(RN ),

T (h)− Ih

u0−→h↓0

∆u0 (2.1.26)

in Hs−2,p(RN ).

Proof. This follows immediately from Corollary 2.1.6, except for the esti-mate (2.1.25) which we prove now. The case σ = s follows from (2.1.9), so wesuppose now σ > s. Let n ≥ σ− s be an integer. Applying (2.1.11) with k = s = 0,first with m = 0, then with m = n, we obtain

‖T (t)‖L(Lp,Lq) ≤ Ct−N2 ( 1

p−1q ), ‖T (t)‖L(Lp,Hn,q) ≤ Ct−

N2 ( 1

p−1q )(1 + t−

n2 )

Moreover, we deduce from (B.6.101) (with s replaced by σ − s, ` = 0, m = n, andp replaced by q) that

‖u‖Hσ−s,q ≤ C‖u‖n−σ+sn

Lq ‖u‖σ−sn

Hn,q

Therefore, we obtain

‖T (t)‖L(Lp,Hσ−s,q) ≤ Ct−N2 ( 1

p−1q )[(1 + t−

n2 )]

σ−sn ≤ Ct−

N2 ( 1

p−1q )(1 + t−

σ−s2 )

This means that

‖T (t)v0‖Hσ−s,q ≤ Ct−N2 ( 1

p−1q )(1 + t−

σ−s2 )‖v0‖Lq

for all v0 ∈ Lq. Let now u0 ∈ Hs,p(RN ) and v0 = Jsu0 ∈ Lp(RN ), with thenotation (B.6.80). Applying (B.6.83), we obtain

‖T (t)Jsu0‖Hσ−s,q ≤ Ct−N2 ( 1

p−1q )(1 + t−

σ−s2 )‖Jsu0‖Lq

= Ct−N2 ( 1

p−1q )(1 + t−

σ−s2 )‖u0‖Hs,q

Next, it follows easily from (B.6.80) and (2.1.2) that T (t)Jsu0 = JsT (t)u0. Ap-plying now (B.6.85), we finally obtain

‖T (t)u0‖Hσ,q = ‖JsT (t)u0‖Hσ−s,q = ‖T (t)Jsu0‖Hσ−s,q

≤ Ct−N2 ( 1

p−1q )(1 + t−

σ−s2 )‖u0‖Hs,q

which proves (2.1.25).

Remark 2.1.10. Here are some particular solutions of (2.1.1) .


(i) Given any c ∈ R, u(t, x) ≡ c is the solution of (2.1.1) with u0(x) ≡ c. In otherwords, T (t)c = c.

(ii) Since ∆(|x|2) = 2N , it follows that u(t, x) ≡ 2Nt+|x|2 is the solution of (2.1.1)with u0(x) ≡ |x|2. In other words, T (t)|x|2 = 2Nt+ |x|2.

(iii) Let δ0 be the Dirac mass at 0. In particular, δ0 ∈ S ′(RN ) and 〈δ0, ϕ〉S′,S =ϕ(0). As is well known, f ? δ0 = f for all f ∈ S(RN ) and in particular,Gt ? δ0 = Gt. It follows that the solution of (2.1.1) with u0 = δ0 is u(t) = Gt.In other words, T (t)δ0 = Gt.

(iv) Let u0 be a Gaussian, i.e. u0(x) = e−a|x|2

for some a > 0. Setting τ = 14a ,

we have u0 = ( aπ )−N2 Gτ . On the other hand, T (t)Gτ = Gτ+t by Propo-

sition 2.1.8 (iii) and Property (iii) above, so that T (t)u0 = ( aπ )−N2 Gτ+t.

Therefore, T (t)e−a|x|2

= (1 + 4at)−N2 e−

a|x|21+4at .

2.1.3. The nonhomogeneous equation and Duhamel’s formula. Weconsider the nonhomogeneous heat equation

ut = ∆u+ f 0 ≤ t ≤ T, x ∈ RN

u(0, x) = u0(x) x ∈ RN(2.1.27)

where T > 0 and f is a given function [0, T ]× RN → R.

Theorem 2.1.11. Let T > 0, 1 < p <∞ and σ ∈ R. Given u0 ∈ Hs,p(RN ) andf ∈ C([0, T ], Hs,p(RN )), there exists a unique solution u ∈ C([0, T ], Hs,p(RN )) ∩C1([0, T ], Hs−2,p(RN )) of equation (2.1.27), and u given by Duhamel’s formula

u(t) = T (t)u0 +

∫ t

0

T (t− s)f(s) ds (2.1.28)

for all 0 ≤ t ≤ T . Moreover,

‖u(t)‖Hs,p ≤ ‖u0‖Hs,p +

∫ t

0

‖f(s)‖Hs,pds (2.1.29)

for all 0 ≤ t ≤ T .

For the proof of Theorem 2.1.11, we will use the following lemma.

Lemma 2.1.12. Let T > 0, 1 < p <∞, σ ∈ R and f ∈ C([0, T ], Hs,p(RN )). If

w(t) =

∫ t

0

T (t− s)f(s) ds (2.1.30)

for all 0 ≤ t ≤ T , then w ∈ C([0, T ], Hs,p(RN ))∩C1([0, T ], Hs−2,p(RN )), w(0) = 0and wt = ∆w + f for all 0 ≤ t ≤ T . Moreover,

‖u(t)‖Hs,p ≤∫ t

0

‖f(s)‖Hs,pds (2.1.31)

for all 0 ≤ t ≤ T . In addition, if p ≤ q <∞ and σ ≥ s satisfy

N

2

(1

p− 1

q

)+σ − s

2< 1 (2.1.32)

then w ∈ C([0, T ], Hσ,q(RN )) and

‖u(t)‖Hσ,q ≤ C∫ t

0

(t− s)−N2 ( 1

p−1q )(1 + (t− s)−

σ−s2 )‖f(s)‖Hs,pds (2.1.33)



Proof. Given 0 < t ≤ T , it follows from Proposition 2.1.9 that the map s 7→T (t− s)f(s) is continuous on [0, t]→ Hs,p(RN ), so that the integral in (2.1.30) de-fines an element w(t) ∈ Hs,p(RN ). Moreover, we deduce from Proposition 2.1.9 (i)that estimate (2.1.31) holds. We now show that w ∈ C([0, T ), Hs,p(RN )). Given0 ≤ t < t+ h ≤ T , it follows from (2.1.30) and the semigroup property T (h)T (t−s) = T (t+ h− s) (see Proposition 2.1.8 (iii)) that

w(t+ h)− w(t) = [T (h)− I]w(t) +

∫ t+h

t

T (t+ h− s)f(s) ds (2.1.34)

It follows easily (applying once again Proposition 2.1.9) that, given any 0 ≤ t < T ,w(t + h) → w(t) in Hs,p(RN ) ad h → 0. Hence w ∈ C([0, T ), Hs,p(RN )), andw(0) = 0. Continuity at T requires a slightly different argument. Given 0 ≤ h ≤ T ,we write

w(T )− w(T − h) =

∫ T

T−hT (T − s)f(s) ds+

∫ T

0

Ψh(s) ds

where

Ψh(s) =

T (T − s)f(s)− T (T − h− s)f(s) 0 ≤ s ≤ T − h0 T − h ≤ s ≤ T

Therefore,

‖w(T )− w(T − h)‖Hs,p ≤ h sup0≤s≤T

‖f(s)‖Hs,p +

∫ T

0

‖Ψh(s)‖Hs,pds

Since ‖Ψh(s)‖Hs,p ≤ 2 sup0≤s≤T ‖f(s)‖Hs,p and ‖Ψh(s)‖Hs,p → 0 as h → 0, conti-nuity at T follows by dominated convergence. Next, given 0 ≤ t < t + h ≤ T , wededuce from (2.1.34) that

u(t+ h)− u(t)

h=

T (h)− Ih

u(t) +1

h

∫ t+h

t

T (t− s)f(s) ds

Letting h ↓ 0 and applying Proposition 2.1.9, we obtain

limh↓0

u(t+ h)− u(t)

h= ∆u(t) + f(t)

where the limit is in Hs−2,p(RN ). This shows that u : [0, T ) → Hs−2,p(RN ) is

right-differentiable at every 0 ≤ t < T and d+udt = ∆u + f . Since ∆u + f ∈

C([0, T ], Hs−2,p(RN )), we deduce (see Theorem A.1.3 and Remark A.1.4) that u ∈C1([0, T ], Hs−2,p(RN )) and du

dt = ∆u+ f . This proves the first part of the lemma.Let now p ≤ q <∞ and σ ≥ s satisfy (2.1.32). Given 0 < t ≤ T , it follows from

Proposition 2.1.9 that the map s 7→ T (t−s)f(s) is continuous on [0, t)→ Hσ,q(RN )and (see (2.1.25))

‖T (t− s)f(s)‖Hσ,q ≤ C(−s)t−N2 ( 1

p−1q )(1 + (t− s)−

σ−s2 )

Thus we see that condition (2.1.32) ensures that the function s 7→ T (t − s)f(s)is integrable (0, t) → Hσ,q(RN ). Thus w(t) ∈ Hσ,q(RN ) for all t ∈ [0, T ] and wsatisfies estimate (2.1.33). Continuity is proved as above.

Proof of Theorem 2.1.11. We first prove uniqueness. Given two solutionsu, v ∈ C1([0, T ],S ′(RN )), we set w = u− v. It follows that w ∈ C1([0, T ],S ′(RN ))satisfies wt = ∆w on [0, T ] and w(0) = 0. Therefore, w = 0 by Remark 2.1.3 (ii),hence u = v.

To prove existence and Duhamel’s formula (2.1.28), it suffices to consider thecases f(t) ≡ 0 and u0 = 0 independently. (The general solution is then the sum of

2.2. THE HEAT EQUATION ON A BOUNDED DOMAIN 43

these two solutions.) The result in the case f(t) ≡ 0 follows from Proposition 2.1.9,while in the case u0 = 0 it follows from Lemma 2.1.12.

We now state a smoothing effect for the solutions of (2.1.27).

Theorem 2.1.13. Let T > 0, 1 < p < ∞ and s ∈ R. Let u0 ∈ Hs,p(RN )and f ∈ C([0, T ], Hs,p(RN )), and let u ∈ C([0, T ], Hs,p(RN )) be the correspondingsolution of (2.1.27). If q ≤ p < ∞ and s ≤ σ < s + 2 satisfy (2.1.32), then u ∈C((0, T ], Hσ,q(RN )). In particular, u ∈ C((0, T ], Hσ,p(RN )) for all s ≤ σ < s + 2

and u ∈ C((0, T ], Hs,q(RN )) for all p ≤ q < Np(N−2p)+ .

Proof. This follows from Corollary 2.1.6 and Lemma 2.1.12.

2.2. The heat equation on a bounded domain

We consider the heat equation with Dirichlet boundary conditionsut = ∆u t > 0, x ∈ Ω

u(t, x) = 0 t > 0, x ∈ ∂Ω

u(0, x) = u0(x) x ∈ Ω

(2.2.1)

on a bounded domain Ω ⊂ RN . As for the Laplace equation, the boundary condi-tion is interpreted in a weak sense. We will use the spectral decomposition of theLaplacian (see Section 1.4) to construct solutions of (2.2.1). In particular, we con-sider the eigenvalues (λj)j≥1 and a corresponding sequence of eigenvectors (ϕj)j≥1

as given by Theorem 1.4.1. Moreover, we equip H10 (Ω) with the (equivalent) norm

‖u‖H10

= ‖∇u‖L2 , and H−1(Ω) with the corresponding dual norm.

2.2.1. Existence and regularity. Our first result of this section is as follows.

Theorem 2.2.1. Given any u0 ∈ H−1(Ω), there exists a unique solution

u ∈ C([0,∞), H−1(Ω)) ∩ C((0,∞), H10 (Ω)) ∩ C1((0,∞), H−1(Ω)) (2.2.2)

of (2.2.1). In addition, u is given by the formula

u(t) =

∞∑j=1

e−tλja0jϕj (2.2.3)

where the sequence (a0j )j≥1 is defined by

a0j = 〈u0, ϕj〉H−1,H1

0j ≥ 1 (2.2.4)

Moreover, u ∈ C∞((0,∞), H10 (Ω)), with the estimate∥∥∥dkudtk

∥∥∥H1

0

≤ (k + 1)!

tk+1‖u0‖H−1 (2.2.5)

Furthermore, the following properties hold.

(i) ‖u(t)‖H−1 ≤ e−tλ1‖u0‖H−1 for all t > 0.(ii) If u0 ∈ L2(Ω), then u ∈ C([0,∞), L2(Ω)) and ‖u(t)‖L2 ≤ e−tλ1‖u0‖L2 for all

t > 0.(iii) If u0 ∈ H1

0 (Ω), then u ∈ C([0,∞), H10 (Ω)) and ‖u(t)‖H1

0≤ e−tλ1‖u0‖H1

0for

all t > 0.

Proof. We first prove uniqueness, so we consider u0 ∈ H−1(Ω) and a solutionu of (2.2.1) in the class (2.2.2). Let

aj(t) = 〈u(t), ϕj〉H−1,H10

j ≥ 1


so that aj ∈ C([0,∞))∩C1((0,∞)) by Remark 1.4.4. Moreover, ut = ∆u in H−1(Ω)for all t > 0 so that, applying (1.4.13),

a′j = 〈∆u, ϕj〉H−1,H10

= 〈u,∆ϕj〉H−1,H10

= −λj〈u, ϕj〉H−1,H10

= −λjajfor all t > 0. Integrating the above equation yields

aj(t) = e−tλja0j (2.2.6)

Therefore, u is given by (2.2.3), which proves uniqueness.We now prove the existence and regularity properties. Let u0 ∈ H−1(Ω),

(a0j )j≥1 defined by (2.2.4), and (aj(t))j≥1 defined by (2.2.6). We see that aj ∈

C∞([0,∞)) and

dkajdtk

= (−1)kλkj e−tλja0

j (2.2.7)

In particular,

|aj(t)| ≤ e−tλj |a0j | ≤ e−tλ1 |a0

j | (2.2.8)

since λj ≥ λ1 for all j ≥ 1. In the rest of the proof, we apply repeatedly Proposi-tion 1.4.3 without further mention. Since∑

λ−1j |a

0j |2 = ‖u0‖2H−1 <∞ (2.2.9)

we deduce from (2.2.8) that e2tλ1∑λ−1j |aj(t)|2 ≤ ‖u0‖2H−1 . Thus we see that

formula (2.2.3) defines a function u(t) ∈ H−1(Ω) for all t ≥ 0, and ‖u(t)‖H−1 ≤e−tλ1‖u0‖H−1 . Moreover,

‖u(t)− u(s)‖2H−1 =

∞∑j=1

λ−1j |aj(t)− aj(s)|

2

Applying (2.2.8) and dominated convergence (for sequences), we deduce that u ∈C([0,∞), H−1(Ω)). Next, m!ex ≥ xm for all x ≥ 0 and m ∈ N, so that e−x ≤m!x−m. Therefore, it follows from (2.2.7) that∣∣∣dkaj

dtk

∣∣∣ ≤ λkj e−tλj |a0j | ≤ λkj (k + 1)!(tλj)

−k−1|a0j | = (k + 1)!t−k−1λ−1

j |a0j |

Therefore,∞∑j=1

λj

∣∣∣dkaj(t)dtk

∣∣∣2 ≤ [(k + 1)!t−k−1]2∞∑j=1

λ−1j |a

0j |2

= [(k + 1)!t−k−1]2‖u0‖2H−1

(2.2.10)

For k = 0, this yields∞∑j=1

λj |aj |2 ≤ t−2‖u0‖2H−1

This proves that u(t) ∈ H10 (Ω) for all t > 0, as well as estimate (2.2.5) for k = 0,

then by dominated convergence for sequences, that u ∈ C((0,∞), H10 (Ω)). Next,

given s, t > 0,

‖u(t)− u(s)‖2H1 =

∞∑j=1

λj |aj(t)− aj(s)|2 =

∞∑j=1

λj

∣∣∣∫ t

s

dajdt

∣∣∣2Applying (2.2.10), we conclude easily that u : (0,∞) → H1

0 (Ω) is differentiable atevery t > 0. Moreover (see Remark 1.4.4 (i)),

du

dt=

∞∑j=1

dajdtϕj


One concludes as above that u ∈ C1((0,∞), H10 (Ω)) and that estimate (2.2.5) holds

for k = 1. An obvious iteration argument shows that u ∈ Ck((0,∞), H10 (Ω)) for all

k ≥ 2 and that estimate (2.2.5) holds.So far, we have proved the fists part of the theorem, as well as Property (i).

We finally prove Properties (ii) and (iii). We deduce from (2.2.3) and(2.2.8) that

‖u(t)‖2L2 =

∞∑j=1

|aj(t)|2 ≤ e−2λ1

∞∑j=1

|a0j |2 = ‖u0‖2L2

Moreover, it follows easily from formula (2.2.3) and dominated convergence forsequences that u ∈ C([0,∞), L2(Ω)). This proves Property (ii), and Property (iii)is proved likewise.

Remark 2.2.2. Here are some comments on Theorem 2.2.1.

(i) Property (i) (as well as (ii) and (iii)) show that all solutions of (2.2.1) decaylike e−tλ1 as t → ∞. This is in contrast with the case of the heat equationset on RN , where the decay is in general power-like. (See Remark 2.1.5 (iii).)

(ii) Estimate (2.2.5) quantifies the smoothing effect of the heat equation. Forinstance, for k = 0, ‖u(t)‖H1

0≤ t−1‖u0‖H−1 . One can deduce from for-

mula (2.2.3) other estimates of the same type. For instance,√ye−y ≤ 1 for

y ≥ 0, so that√λje−tλj ≤ t−

12 . With the notation (2.2.6), this implies that√

λj |aj(t)| ≤ t−12 |a0

j |. It follows easily that ‖u(t)‖L2 ≤ t−12 ‖u0‖H−1 ; and if

u0 ∈ L2(Ω), then ‖u(t)‖H10≤ t− 1

2 ‖u0‖L2 .

(iii) The proof of uniqueness shows in fact a stronger uniqueness property: If T > 0and u ∈ C([0, T ), H−1(Ω)) ∩ C((0, T ), H1

0 (Ω)) ∩ C1((0, T ), H−1(Ω)) satisfiesut = ∆u in H−1(Ω) for all 0 ≤ t ≤ T and if u(0) = 0, then u(t) = 0 for all0 ≤ t ≤ T .

We next prove an interior regularity property.

Proposition 2.2.3. Let Ω be any open subset of RN . Let T > 0, and letu ∈ C((0, T ), H1(Ω)) ∩ C1((0, T ), H−1(Ω)) satisfy ut = ∆u in H−1(Ω) for all0 < t < T . It follows that u ∈ C∞((0, T )× Ω).

Proof. Let ϕ ∈ C∞c ((0, T )× Ω) and define the function v on [0, T )× RN by

v(t, x) =

ϕ(t, x)u(t, x) x ∈ Ω

0 x 6∈ Ω(2.2.11)

It follows easily that v ∈ C([0, T ), H1(RN )) ∩ C1([0, T ), H−1(RN )), v(0) = 0, and

vt −∆v = f (2.2.12)

where

f(t, x) =

u(ϕt −∆ϕ)− 2∇u · ∇ϕ 0 < t < T, x ∈ Ω

0 0 < t < T, x 6∈ Ω(2.2.13)

In particular, f ∈ C([0, T ], L2(RN )). Applying Theorem 2.1.13, we deduce that

v ∈ C([0, T ), H32 (RN )).

We now prove by induction on m that for all m ∈ N and ϕ ∈ C∞c ((0, T )× Ω),the function v defined by (2.2.11) belongs to C([0, T ), H

m2 (RN )). Indeed, we have

seen above that the property is true for m = 3. Assume it holds up to somem ≥ 3, and let ϕ ∈ C∞c ((0, T ) × Ω). Consider ψ ∈ C∞c ((0, T ) × Ω) such thatψ = 1 on the support of ϕ. By the induction assumption, ψu (extended by 0outside Ω) belongs to C([0, T ), H

m2 (RN )). Moreover, since ψ = 1 on the support

of ϕ, we may replace u by ψu in (2.2.13). Since ∇(ψu) ∈ C([0, T ), Hm2 −1(RN )), it


follows easily that f ∈ C([0, T ), Hm2 −1(RN )). Applying Theorem 2.1.13, we obtain

v ∈ C([0, T ), Hm2 + 1

2 (RN )), which closes the induction argument.It follows from what precedes that, given any ϕ ∈ C∞c ((0, T ) × Ω), we have

∆v + f ∈ C([0, T ), H`(RN )) for all ` ∈ N, and equation (2.2.12) implies thatv ∈ C1([0, T ), H`(RN )) for all ` ∈ N. Then one deduces easily that, given anyϕ ∈ C∞c ((0, T ) × Ω), we have ∆v + f ∈ C1([0, T ), H`(RN )) for all ` ∈ N; and sov ∈ C2([0, T ), H`(RN )) for all ` ∈ N. An obvious iteration argument shows that,given any ϕ ∈ C∞c ((0, T )× Ω), we have v ∈ C∞([0, T ), H`(RN )) for all ` ∈ N. BySobolev’s embedding, we conclude that v ∈ C∞((0, T )×RN ), which completes theproof since ϕ ∈ C∞c ((0, T )× Ω) is arbitrary.

The following property gives some information on the behavior at t = 0.

Proposition 2.2.4. Let Ω be any open subset of RN . Let u0 ∈ C∞c (Ω), T > 0,and let u ∈ C([0, T ), H1(Ω)) ∩ C1((0, T ), H−1(Ω)) satisfy ut = ∆u in H−1(Ω) forall 0 < t < T and u(0) = u0. It follows that u(t) → u0 as t ↓ 0 in L∞(ω) for anyopen set ω ⊂⊂ Ω.

Proof. The proof is very similar to the proof of Proposition 2.2.3 above. Letϕ ∈ C∞c (Ω) and define the function v on [0, T )× RN by

v(t, x) =

ϕ(x)u(t, x) x ∈ Ω

0 x 6∈ Ω(2.2.14)

It follows easily that v ∈ C([0, T ), H1(RN )) ∩ C1((0, T ), H−1(RN )), v(0) = 0, and

vt −∆v = f (2.2.15)

where

f(t, x) =

−u∆ϕ− 2∇u · ∇ϕ 0 < t < T, x ∈ Ω

0 0 < t < T, x 6∈ Ω(2.2.16)

In particular, f ∈ C([0, T ), L2(RN )). Applying Theorem 2.1.13, we deduce that

v ∈ C([0, T ), H32 (RN )).

We now prove by induction on m that for all m ∈ N and ϕ ∈ C∞c (Ω), thefunction v defined by (2.2.14) belongs to C([0, T ), H

m2 (RN )). Indeed, we have seen

above that the property is true for m = 3. Assume it holds up to some m ≥ 3, andlet ϕ ∈ C∞c (Ω). Consider ψ ∈ C∞c (Ω) such that ψ = 1 on the support of ϕ. By theinduction assumption, ψu (extended by 0 outside Ω) belongs to C([0, T ), H

m2 (RN )).

Moreover, since ψ = 1 on the support of ϕ, we may replace u by ψu in (2.2.16). Since∇(ψu) ∈ C([0, T ), H

m2 −1(RN )), it follows easily that f ∈ C([0, T ), H

m2 −1(RN )).

Applying Theorem 2.1.13, we obtain v ∈ C([0, T ), Hm2 + 1

2 (RN )), which closes theinduction argument.

Finally, form > N2 , we haveHm(RN ) → L∞(RN ), and we deduce that ϕu(t)→

ϕu0 in L∞(Ω) as t ↓ 0, for every ϕ ∈ C∞c (Ω), which proves the desired result.

2.2.2. The heat semigroup. The heat semigroup on Ω with Dirichlet bound-ary conditions, often denoted (et∆)t≥0, is the family of operators H−1(Ω) →H−1(Ω) defined as follows for every t ≥ 0: given u0 ∈ H−1(Ω), T (t)u0 is thesolution at time t of equation (2.2.1), given by Theorem 2.2.1. In particular,

T (t)u0 =

∞∑j=1

e−tλj 〈u0, ϕj〉H−1,H10ϕj (2.2.17)

by formula (2.2.3).

Proposition 2.2.5. Let (T (t))t≥0 be as above, and let X be either of the spacesH−1(Ω), L2(Ω), H1

0 (Ω).


(i) (T (t))t≥0 ⊂ L(X ) and ‖T (t)‖L(X ) ≤ e−tλ1 .(ii) T (0) = I and T (t+ s) = T (t)T (s) in L(X ), for all s, t ≥ 0.(iii) ∀u0 ∈ X , the map t 7→ T (t)u0 is continuous [0,∞)→ X .(iv) (T (t))t≥0 ⊂ L(H−1(Ω), H1

0 (Ω)). Moreover, ‖T (t)‖L(H−1,H10 ) ≤ t−1 and

‖T (t)‖L(H−1,L2) ≤ t−12 .

(v) (T (t))t≥0 ⊂ L(L2(Ω), H10 (Ω)) and ‖T (t)‖L(L2,H1

0 ) ≤ t−12 .

(vi) (T (t)u0, v0)X = (u0,T (t)v0)X for all t ≥ 0 and u0, v0 ∈ X . In other words,T (t) is self-adjoint on X .

Proof. Properties (i) and (iii), (v) and (v) follow from Theorem 2.2.1 and Re-mark 2.2.2 (ii); and Property (ii) is an immediate consequence of uniqueness (cf. Re-mark 2.1.3 (i)). To prove Property (vi), we note that by formula (2.2.17), T (t)u0 =∑∞j=1 e

−tλja0jϕj , where a0

j = 〈u0, ϕj〉H−1,H10. Therefore, if b0j = 〈v0, ϕj〉H−1,H1

0, so

that v0 =∑∞j=1 j

0jϕj , then (see Remark 1.4.4 (ii))

(T (t)u0, v0)L2 =

∞∑j=1

e−tλja0jb

0j = (u0,T (t)v0)L2

Self-adjointness in H−1(Ω) and in H10 (Ω) are proved similarly.

2.2.3. The nonhomogeneous equation and Duhamel’s formula. Weconsider the nonhomogeneous heat equation

ut = ∆u+ f 0 < t < T, x ∈ Ω

u(t, x) = 0 t > 0, x ∈ ∂Ω

u(0, x) = u0(x) x ∈ Ω

(2.2.18)

where T > 0 and f is a given function [0, T ]× Ω→ R.

Theorem 2.2.6. Given T > 0, u0 ∈ L2(Ω) and f ∈ C([0, T ], L2(Ω)), thereexists a unique solution

u ∈ C([0, T ], L2(Ω)) ∩ C((0, T ), H10 (Ω)) ∩ C1((0, T ), H−1(Ω)) (2.2.19)

of equation (2.2.18), and u given by Duhamel’s formula

u(t) = T (t)u0 +

∫ t

0

T (t− s)f(s) ds (2.2.20)

for all 0 ≤ t ≤ T . Moreover,

‖u(t)‖L2 ≤ e−tλ1‖u0‖L2 +

∫ t

0

e−(t−s)λ1‖f(s)‖L2ds (2.2.21)


Proof. We first prove uniqueness. Let u, v in the class (2.2.19) be two solu-tions of (2.2.18). It follows that w = u − v satisfies wt = ∆w in H−1(Ω) for all0 < t < T , and w(0) = 0, so that w = 0 by Remark 2.2.2 (iii).

We now prove existence and formula (2.2.20). As a matter of fact, we needonly prove that formula (2.2.20) defines u in the class (2.2.19) and that u is asolution of (2.2.18). Moreover, T (t)u0 belongs to the class (2.2.19), satisfies thehomogeneous heat equation, and the estimate ‖T (t)u0‖L2 ≤ e−tλ1‖u0‖L2 by The-orem 2.2.1, so we may assume u0 = 0 and consider

u(t) =

∫ t

0

T (t− s)f(s) ds (2.2.22)


Note that, given any 0 < t ≤ T , the map s 7→ T (t − s)f(s) is continuous [0, t] →L2(Ω), so that the integral (2.2.22) defines an element u(t) ∈ L2(Ω). In addition,since 〈T (t− s)f(s), ϕj〉H−1,H1

0= e−(t−s)λj 〈f(s), ϕj〉H−1,H1

0, it follows that

〈u(t), ϕj〉H−1,H10

=

∫ t

0

e−(t−s)λj 〈f(s), ϕj〉H−1,H10

In other words, setting

aj(t) = 〈u(t), ϕj〉H−1,H10

bj(t) = 〈f(t), ϕj〉H−1,H10

we have

aj(t) =

∫ t

0

e−(t−s)λj bj(s) ds (2.2.23)

for all 0 ≤ t ≤ T and j ≥ 1. In particular,

‖u(t)‖L2 = ‖(aj(t))j≥1‖`2 ≤∫ t

0

‖(e−(t−s)λj bj(s))j≥1‖`2

≤∫ t

0

e−(t−s)λ1‖(bj(s))j≥1‖`2 =

∫ t

0

e−(t−s)λ1‖f(s)‖L2

which proves (2.2.21). Next, it follows from (2.2.23) that

|aj(t)| ≤ Ajwhere

Aj =

∫ T

0

|bj(s)| ds

We have

‖(Aj)j≥1‖`2 ≤∫ T

0

‖(bj)j≥1‖`2 <∞

On the other hand, each bj is a continuous function of t, hence so is aj , and wededuce by dominated convergence (for sequences) that (aj)j≥1 ∈ C([0, T ], `2(N)),

i.e. u ∈ C([0, T ], L2(Ω)). Similarly, using the property λ12j e−(t−s)λj ≤ C(t − s)− 1

2 ,we see that

λ12j |aj(t)| ≤ Bj

where

Bj =

∫ T

0

(t− s)− 12 |bj(s)| ds

We have

‖(Bj)j≥1‖`2 ≤∫ T

0

(t− s)− 12 ‖(bj)j≥1‖`2 <∞

and we deduce by dominated convergence that (λ12j aj)j≥1 ∈ C([0, T ], `2(N)), i.e.

u ∈ C([0, T ], H10 (Ω)). Moreover,

a′j = −λjaj + bj

and it follows that (λ− 1

2j a′j)j≥1 ∈ C([0, T ], `2(N)), i.e. u ∈ C1([0, T ], H−1(Ω)). Thus

we see that u belongs to the class (2.2.19). Finally,

〈ut −∆u− f, ϕj〉H−1,H10

= a′j + λjaj + bj = 0

for all j ≥ 1, so that ut −∆u− f = 0 in H−1(Ω) for all 0 ≤ t ≤ T . Since u(0) = 0(because aj(0) = 0), we see that u is a solution of (2.2.18).


2.3. The maximum principle

Throughout this section, we assume Ω is an open, connected subset of RN ,bounded or not.

2.3.1. The weak maximum principle. We begin with a weak form of themaximum principle. As for Laplace’s equation, this first result, together with theconstruction of explicit solutions, yields stronger forms of the maximum principle.

Theorem 2.3.1. Let T > 0, and let u ∈ C((0, T ), H1(Ω))∩C1((0, T ), H−1(Ω))and f ∈ C((0, T ), H−1(Ω)) satisfy

ut −∆u = f (2.3.1)

in H−1(Ω) for all t ∈ (0, T ). Assume further that:

(i) u ∈ C([0, T ), L2(Ω))(ii) There exists v ∈ C((0, T ), H1

0 (Ω)) such that u(t) ≤ v(t) a.e. on Ω, for all0 < t < T

(iii) f = g + h, with g ∈ C((0, T ), H−1(Ω)), g(t) ≤ 0 for a.a. t ∈ (0, T ), andh ∈ C((0, T ), L2(Ω)), h(t) ≤ C|u(t)| a.e. on (0, T )× Ω

(iv) u(0) ≤ 0 a.e. on Ω

It follows that u(t) ≤ 0 a.e. on Ω, for all t ∈ (0, T ).

The proof of Theorem 2.3.1 uses the following result, in the spirit of Corol-lary A.1.6.

Lemma 2.3.2. Let T > 0 and u ∈ C((0, T ), H1(Ω))∩C1((0, T ), H−1(Ω)). Sup-pose there exists v ∈ C((0, T ), H1

0 (Ω)) such that u(t) ≤ v(t) a.e. on Ω for allt ∈ (0, T ). It follows that u+ ∈ C((0, T ), H1(Ω)), that the map t 7→ ‖u+(t)‖2L2 isC1 on (0, T ), and

d

dt‖u+(t)‖2L2 = 2〈ut(t), u+(t)〉H−1,H1

0(2.3.2)

for all t ∈ (0, T ).

Proof. Note that u+(t) ∈ H1(Ω) for all 0 < t < T , by Proposition B.2.3.Moreover, since |u+| = u+ ≤ v, it follows from Corollary B.2.5 (replacing u byu+) that u+(t) ∈ H1

0 (Ω) for all t ∈ (0, T ). Applying again Proposition B.2.3, weconclude that u+ ∈ C((0, T ), H1(Ω)).

Next, it follows from an easy calculation (considering separately the possiblesigns of x, y) that ∣∣(y+)2 − (x+)2 − 2(y − x)x+

∣∣ ≤ (y − x)2

for all x, y ∈ R. We deduce that, given t ∈ (0, T ) and h 6= 0 such that t+h ∈ (0, T ),∣∣∣‖u+(t+ h)‖2L2 −‖u+(t)‖2L2 − 2

∫Ω

(u(t+ h)− u(t))u+(t) dx∣∣∣ ≤ ∫

Ω

(u(t+ h)− u(t))2

which we rewrite in the form∣∣‖u+(t+ h)‖2L2 − ‖u+(t)‖2L2−2〈u(t+ h)− u(t)), u+(t)〉H−1,H10

∣∣≤ 〈u(t+ h)− u(t), u(t+ h)− u(t)〉H−1,H1

0

The conclusion easily follows by dividing through by h and letting t− s→ 0.

Proof of Theorem 2.3.1. Since u satisfies the assumptions of Lemma 2.3.2,it follows in particular that u+ ∈ C((0, T ), H1

0 (Ω)). Therefore, we may take theH−1 −H1

0 duality product of (2.3.1) with u+ and we obtain

〈ut(t), u+(t)〉H−1,H10− 〈∆u(t), u+(t)〉H−1,H1

0= 〈f(t), u+(t)〉H−1,H1

0


We deduce from (B.1.8) and (B.2.2) that

〈∆u(t), u+(t)〉H−1,H10

= −∫

Ω

∇u+ · ∇u = −∫

Ω

|∇u+|2 ≤ 0

Moreover, by assumption (iii),

〈f(t), u+(t)〉H−1,H10≤ 〈h(t), u+(t)〉H−1,H1

0

≤ C∫

Ω

|u(t)|u+(t) dx = C

∫Ω

u+(t)2 dx.

Applying Lemma 2.3.2, we find

d

dt

∫Ω

u+(t)2 dx ≤ C∫

Ω

u+(t)2 dx,

for all t ∈ (0, T ). We fix t ∈ (0, T ) and integrate on (s, t) with 0 < s < t to obtain

‖u+(t)‖2L2 ≤ ‖u+(s)‖2L2 +

∫ t

s

‖u+(σ)‖2L2dσ

It follows from assumption (iv) that ‖u+(s)‖2L2 → 0 as s ↓ 0, so that

‖u+(t)‖2L2 ≤∫ t

0

‖u+(σ)‖2L2dσ

Applying Gronwall’ lemma (Lemma A.2.1) we deduce that u+(t) = 0 all t ∈ (0, T ),hence u ≤ 0.

2.3.2. The strong maximum principle. We apply Theorem 2.3.1 to obtainstronger versions of the maximum principle.

Theorem 2.3.3. Let T > 0, and let u ∈ C((0, T ), H1(Ω))∩C1((0, T ), H−1(Ω))satisfy ut −∆u ≥ 0 in H−1(Ω) for all t ∈ (0, T ). Suppose further

(i) u(t) ∈ C(Ω) for all 0 < t < T(ii) u ≥ 0 on (0, T )× Ω(iii) u ∈ C([0, T ), L2(Ω))(iv) u(0) ≥ 0 a.e. on Ω and u(0) 6≡ 0.

It follows that u(t, x) > 0 on (0, T )× Ω.

We will use the following lemma.

Lemma 2.3.4. Given 0 < ρ < R, there exists a function v : [0,∞) × BR → Rwith the following properties.

(i) v ∈ C([0,∞), L2(BR)) ∩ C1((0,∞), H10 (BR)) ∩ C∞((0,∞)×BR)

(ii) v(0) ≤ 1Bρ(iii) vt −∆v ≤ 0 on (0,∞)×BR(iv) v(t) is radially symmetric and decreasing, for all t > 0(v) There exist a constant c > 0 depending only on N such that

v(t, x) ≥ cρNR−1e−CR2 tt−

N2 e−

R2

t (R− |x|) (2.3.3)

on (0,∞)×BR

Proof. We let z0 = 1Bρ and consider the solution z ofzt = ∆z t > 0, x ∈ RN

z(0) = z0

In particular, z ∈ C∞((0,∞)×RN by Corollary 2.1.6, and z(t) is radially symmetricand decreasing for all t ≥ 0, by Remark 2.1.5 (ii). Moreover, since z(t) = Gt ? z0


with the notation (2.1.4), it follows from elementary calculations that there existsa constant c > 0 depending only on N such that

z(t, x) ≥ ct−N2 e−R2

t ρN t > 0, x ∈ BR (2.3.4)

Next, we let λ be the first eigenvalue of −∆ in H10 (BR), and ϕ > 0 a corresponding

eigenvector, normalized by ϕ(0) = 1. In particular, ϕ ∈ C∞(BR) is radially sym-metric and decreasing, and there exist two constants c = c(N) and C = C(N) suchthat

ϕ(x) ≥ cR−N+2

2 (R− |x|) (2.3.5)

for all x ∈ BR and

λ ≤ C

R2(2.3.6)

by Proposition 1.4.9. We now let

v(t, x) = e−tλϕ(x)z(t, x) > 0 t > 0, x ∈ BRIt follows from (2.3.4), (2.3.5) and (2.3.6) that v satisfies (2.3.3). Moreover, oneverifies easily that v is in the regularity class (i), v(0) = ϕ1Bρ ≤ 1Bρ , and

vt −∆v = e−tλ[(−∆ϕ− λϕ)z + ϕ(zt −∆z)− 2∇ϕ · ∇z]

= −2e−tλ∇ϕ · ∇zBoth ϕ and z being radially symmetric and decreasing, we see that ∇ϕ · ∇z =∂rϕ∂rz ≥ 0; and so, vt −∆v ≤ 0. This completes the proof.

Proof of Theorem 2.3.3. We proceed in three steps.

Step 1. Let 0 < t0 < T , x0 ∈ Ω, and R > 0 such that B(x0, R) ⊂ Ω. Ifu(t0, x0) > 0, then u(t, x) > 0 for all t > t0 and x ∈ B(x0, R). To prove this, wemay assume without loss of generality that x0 = 0 (by space-translation invarianceof the equation). We first observe that, since u(t0) ∈ C(Ω) by assumption (i), andu(t0, 0) > 0, there exist 0 < ρ ≤ R and ε > 0 such that

u(t0, x) ≥ ε, x ∈ Bρ (2.3.7)

We consider the function v given by Lemma 2.3.4 and we set

w(t) = −u(t)|BR + εv(t− t0)

It easily follows that w ∈ C1((t0, T ], H1(BR)) ∩ C([t0, T ], L2(BR)), wt − ∆w ≤ 0in H−1(BR), w(t0) ≤ 0. Moreover, w(t) ≤ εv(t − t0) and v ≥ 0, so that w+(t) ≤εv(t − t0). Since v ∈ C((0, T ), H1

0 (BR)), we deduce (see Corollary B.2.5) thatw+ ∈ C((t0, T ), H1

0 (BR)). Thus we may apply Theorem 2.3.1 and we deduce thatw ≤ 0. Therefore, u(t) ≥ εv(t − t0) on (0, T ) × BR, which proves the desiredconclusion.

Step 2. We show that for every 0 < t < T , there exists x0 ∈ Ω such thatu(t, x0) > 0. Indeed, let 0 < t < T . By assumptions (ii), (iii) and (iv), thereexists 0 < τ ≤ t such that u(τ) ≥ 0 and u(τ) 6≡ 0. Therefore, there exists x0 suchthat u(τ, x0) > 0. By Step 1, u(t, x0) > 0.

Step 3. Conclusion. Fix 0 < t0 < T . We need to show that u(t0) > 0 onΩ. We have u(t0) ∈ C∞(Ω) and, by Step 2, there exists x ∈ Ω such that u(t0, x).Therefore,

O = x ∈ Ω; u(t0, x) > 0is a nonempty, open subset of Ω. Since Ω is connected, the result follows if weshow that O is closed. Let (xn)n≥1 ⊂ O such that xn → x0 ∈ Ω, and let R > 0be sufficiently small so that B(x0, R) ⊂ Ω. Fix n0 sufficiently large so that |x0 −xn0 | ≤ R

2 . Since u(t0, xn0) > 0 (because xn0

∈ O) and u ∈ C∞((0, T ) × Ω), there


exists 0 < τ < t0 such that u(τ, xn0) > 0. Note that B(xn0,

R2

) ⊂ B(x0, R) ⊂ Ω.

Therefore, u(t0, x) > 0 for all x ∈ B(xn0 ,R2 ) by Step 1. In particular, u(t0, x0) > 0,

hence x0 ∈ O. Thus O is closed, which completes the proof.

We now give a stronger version of the maximum principle, under a geometricassumption on the domain Ω.

Theorem 2.3.5. Suppose Ω satisfies the geometric condition (1.2.1). Under theassumptions of Theorem 2.3.3, it follows that there exists a function c : (0, T ) →(0,∞) such that

u(t, x) ≥ c(t)d(x, ∂Ω), x ∈ Ω, t ∈ (0, T ) (2.3.8)

where d(x, ∂Ω) is the distance of x to ∂Ω.

Proof. Let η, ν be given by assumption (1.2.1). Let 0 < ε ≤ η/2 and considerΩε = x ∈ Ω; d(x, ∂Ω) ≥ ε. We fix ε > 0 sufficiently small so that Ωε is anonempty, compact subset of Ω. Fix 0 < t0 < T . It follows from Theorem 2.3.3and u(t0) ∈ C(Ω) that there exists δ > 0 such that

u(t0, x) ≥ δ, x ∈ Ωε (2.3.9)

and

u(t0/2, x) ≥ δ, x ∈ Ωε (2.3.10)

We now consider x0 ∈ Ω such that d(x0, ∂Ω) < ε, and we let y0 ∈ Ω satisfy (1.2.1).Since B(y0, η) ⊂ Ω and η ≥ 2ε, we see that d(z, ∂Ω) ≥ ε for all z ∈ B(y0, η/2). Inparticular, z ∈ Ωε, and it follows from (2.3.10) that

u(t0/2, z) ≥ δ for all z ∈ B(y0, η/2). (2.3.11)

We let ρ = η/2, R = η, we consider the function v given by Lemma 2.3.4 and weset

w(t, x) = −u(t+ t0/2, y) + δv(t, y − y0), y ∈ B(y0, R), 0 ≤ t < T − t0/2

It easily follows (see Step 1 of the proof of Theorem 2.3.3) that w satisfies theassumptions of Theorem 2.3.1, and we deduce that w ≤ 0. In particular, fort = t0/2,

u(t0, y) = u(t0/2 + t0/2, y) ≥ δv(t0/2, y − y0)

for |y| ≤ R so that, applying (2.3.3),

u(t0, y) ≥ δcρNR−1e−Ct02R2 (t0/2)−

N2 e−

2R2

t0 (R− |y − y0|)

Since |x0 − y0| < η = R by (1.2.1), we may let x = x0 in the above inequality.Moreover, it follows from (1.2.1) that

R− |x0 − y0| = η − |x0 − y0| ≥ νd(x0, ∂Ω)

and we obtain

u(t0, x0) ≥ νδcρNR−1e−Ct02R2 (t0/2)−

N2 e−

2R2

t0 d(x0, ∂Ω) (2.3.12)

Let now x0 ∈ Ω. If d(x0, ∂Ω) ≥ ε, it follows from (2.3.9) that u(t0, x0) ≥ δ ≥µd(x0, ∂Ω), where

µ =δ

supx∈Ω d(x, ∂Ω)

If d(x0, ∂Ω) < ε, then we deduce from (2.3.12) that u(t0, x0) ≥ µ′d(x0, ∂Ω), where

µ′ = νδcρNR−1e−Ct02R2 (t0/2)−

N2 e−

2R2

t0 . This completes the proof.


2.3.3. Some applications of the maximum principle. We begin with asimple, but very useful, application of the weak maximum principle.

Proposition 2.3.6. Let Ω be a bounded domain and (T (t))t≥0 the correspond-ing heat semigroup, as defined in Proposition 2.2.5. If u0 ∈ L2(Ω)∩Lp(Ω) for some1 ≤ p ≤ ∞, then T (t)u0 ∈ Lq(Ω) for all p ≤ q ≤ ∞, and

‖T (t)u0‖Lq ≤ (4πt)−N2 ( 1

p−1q )‖u0‖Lp (2.3.13)

for all t > 0.

Proof. Let u0 6≡ 0 and set u(t) = T (t)u0. Suppose first p < ∞. By density,we need only consider the case u0 ∈ H1

0 (Ω)∩Lp(Ω). Let ψ ∈ H1(RN )∩Lp(RN ) bedefined by

ψ(x) =

|u0(x)| x ∈ Ω

0 x ∈ RN \ Ω

and let v(t) = Gt ?ψ where Gt is defined by (2.1.4). It follows from Corollary 2.1.6that v ∈ C([0,∞), H1(RN )) ∩ C1([0,∞), H−1(RN )) ∩ C∞((0,∞) × RN ) and vt −∆v = 0. In addition, v(t) > 0 on (0,∞)× RN (since Gt > 0). Therefore, if we setw(t) = v(t)|Ω, then w ∈ C([0,∞), H1(Ω)) ∩ C1([0,∞), H−1(Ω)), wt − ∆v = 0 in

H−1(Ω) and w(t) > 0 on (0,∞)×Ω. It follows easily that both z1(t) = u(t)−w(t)and z2(t) = −u(t)−w(t) satisfy the assumptions of Theorem 2.3.1 with f = 0 andv = |u|. Consequently, z1(t), z2(t) ≤ 0, so that |u(t)| ≤ w(t) a.e. on Ω for all t ≥ 0.In particular, ‖u(t)‖Lq(Ω) ≤ ‖w(t)‖Lq(Ω) ≤ ‖v(t)‖Lq(RN ), and the result is then aconsequence of Corollary 2.1.4. In the case p = ∞ (hence q = ∞), we apply theinequality with p <∞ and q =∞, then we let p ↑ ∞.

Corollary 2.3.7. Let Ω be a bounded domain and (T (t))t≥0 the correspondingheat semigroup, as defined in Proposition 2.2.5. Given any 1 ≤ p < ∞, it followsthat for any t ≥ 0, T (t) can be uniquely extended to an operator of L(Lp(Ω))such that ‖T (t)‖L(Lp) ≤ 1, which we also denote by T (t). Moreover, the mapt 7→ T (t)u0 is continuous [0,∞)→ Lp(Ω), for every u0 ∈ Lp(Ω).

Proof. That T (t) can be uniquely extended to an operator of L(Lp(Ω)) suchthat ‖T (t)‖L(Lp) ≤ 1 is an immediate consequence of estimate (2.3.13) with q = p,

and the fact that L2(Ω) ∩ Lp(Ω) is dense in Lp(Ω). To prove the continuity of themap t 7→ T (t)u0, consider u0 ∈ Lp(Ω), t ≥ 0, and (tn)n≥1 ⊂ [0,∞) such thattn → t as n → ∞. We need to show that ‖T (tn)u0 − T (t)u0‖Lp → 0. Let ε > 0.Since Cc(Ω) is dense in Lp(Ω), there exists v0 ∈ Cc(Ω) such that ‖u0 − v0‖Lp ≤ ε

4 ;and so

‖T (tn)(u0 − v0)− T (t)(u0 − v0)‖Lp ≤ε

2(2.3.14)

by (2.3.13). Moreover, again by (2.3.13), ‖T (tn)v0−T (t)v0‖L∞ ≤ 2‖v0‖L∞ . Since‖T (tn)v0 − T (t)v0‖L2 → 0 as n → ∞ by Proposition 2.2.5 (iii), we conclude that‖T (tn)v0 − T (t)v0‖Lp → 0. Thus ‖T (tn)v0 − T (t)v0‖Lp ≤ ε

2 for all sufficientlylarge n, and the result follows by combining with (2.3.14).

We now apply the weak maximum principle to prove that the heat semigroupoperates on C0(Ω).

Proposition 2.3.8. Let Ω be a bounded domain and (T (t))t≥0 the correspond-ing heat semigroup, as defined in Proposition 2.2.5. Suppose further Ω satisfies thegeometric condition (1.3.16). It follows that for any t ≥ 0, T (t) can be uniquelyextended to an operator of L(C0(Ω)) such that ‖T (t)‖L(C0(Ω)) ≤ 1, which we alsodenote by T (t). Moreover, the map t 7→ T (t)u0 is continuous [0,∞)→ C0(Ω), forevery u0 ∈ C0(Ω).


Proof. We first consider the case u0 ∈ C∞c (Ω), and we set u(t) = T (t)u0. Letλ1 be the first eigenvalue of −∆ in H1

0 (Ω), and ϕ1 > 0 a corresponding eigenvector.(See Section 1.4.) Recall that (see (1.4.18))

ϕ1(·) ≤ Cd(·, ∂Ω) (2.3.15)

for some constant C < ∞. Moreover, ϕ1 > 0 in Ω and ϕ1 ∈ C∞Ω). Since u0 hascompact support, it follows that there exists a constant A such that |u0| ≤ Aϕ1.Since T (t)ϕ1 = e−tλ1ϕ1, we deduce from the weak maximum principle (Theo-rem 2.3.1) that

|u(t)| ≤ Ce−tλ1ϕ1 (2.3.16)

Estimates (2.3.16) and (2.3.15) yield

|u(t, x)| ≤ ACd(·, ∂Ω) (2.3.17)

On the other hand, it follows from Propositions 2.2.3 and 2.2.4 that, given any openset ω ⊂⊂ Ω, u ∈ C([0,∞), C(ω)). This, together with the estimate (2.3.17) (whichcontrols what happens near ∂Ω), imply that u ∈ C([0,∞), C0(Ω)).

Since C∞c (Ω) is dense in C0(Ω), the result easily follows from the above prop-erty, together with estimate (2.3.13) (with p = q = ∞). (See the proof of Corol-lary 2.1.7.)

Remark 2.3.9. One can apply Proposition 2.3.8 (under the assumption thatΩ satisfies the geometric condition (1.3.16)) to extend the heat semigroup to thespace M(Ω) of bounded measures. We recall that M(Ω) is the topological dual ofC0(Ω), equipped with the weak? topology. Given u0 ∈M(Ω) and t ≥ 0, one definesT (t)u0 ∈M(Ω) by

〈T (t)u0, ϕ〉M,C0= 〈u0,T (t)ϕ〉M,C0

for all ϕ ∈ C0(Ω). It is not difficult to verify that T (t)u0 defined as above isindeed an element of M(Ω), and that this definition is consistent with the previousdefinition. (That is, if u0 ∈M(Ω)∩H−1(Ω), then T (t)u0 defined above is the sameas T (t)u0 defined in Section 2.2.2.) Moreover, the map t 7→ T (t)u0 is continuous[0,∞)→M(Ω). In addition, T (t) ∈ C0(RN ) for all t > 0 and

‖T (t)u0‖L∞ ≤ (4πt)−N2 ‖u0‖M(Ω) (2.3.18)

To prove this last statement, we use the property that C∞c (Ω) is dense in M(Ω) forthe strong topology. If u0 ∈ C∞c (Ω), then T (t)u0 ∈ C0(Ω) by Proposition 2.3.8,and (2.3.18) is (2.3.13) with p = 1 and q = ∞, since ‖u0‖M(Ω) = ‖u0‖L1 . Bydensity, we conclude that T (t)u0 ∈ C0(Ω) for all u0 ∈ M(Ω) and that (2.3.18)holds.

CHAPTER 3

The one-dimensional wave equation

3.1. The wave equation on the line

We consider the wave equation on the line, for which we can use explicit for-mulas due to D’Alembert. Since the wave equation is of second order in time, wemust prescribe both the initial value and the initial velocity. Thus we consider thefollowing Cauchy problem

utt = uxx t ∈ R, x ∈ Ru(0, x) = u0(x) x ∈ Rut(0, x) = v0(x) x ∈ R

(3.1.1)

In order to make the presentation simpler, we consider “localized solutions”, i.e. inL2(R), although this is not necessary due to the finite speed of propagation. Wewill use the following elementary result

Lemma 3.1.1. Let s ≥ 0 and ϕ ∈ Hs(R), and let u(t) ∈ Hs(R) for t ∈ Rbe defined by u(t, ·) = ϕ(· + t) (respectively, u(t, ·) = ϕ(· − t)). It follows thatu ∈ C(R, Hs(R))∩C1(R, Hs−1(R)), and ut = ϕ′(·+ t) (respectively, ut = ϕ′(·− t)).

Proof. This is easily seen by using Fourier. We consider the case u(t) =ϕ(x+ t), the other case being similar. Since

u(t, ξ) =

∫Re−2πixξϕ(x+ t)dx

we see by setting y = x+ t that

u(t, ξ) =

∫Re−2πi(y−t)ξϕ(y)dy = e2πitξϕ(ξ)

Therefore

‖u(t+ h)− u(t)‖2Hs =

∫R|e2πi(t+h)ξ − e2πitξ|(1 + ξ2)s|ϕ(ξ)|2

=

∫R|e2πihξ − 1|(1 + ξ2)s|ϕ(ξ)|2

Since ϕ ∈ Hs(R), we have (1 + ξ2)s2 ϕ(ξ) ∈ L2(R), so the right-hand side converges

to 0 as h → 0, by dominated convergence. Hence u ∈ C(R, Hs(R)). Moreover,ut = ϕ′(x + t), and ϕ′ ∈ Hs−1(R). By the preceding result (replacing s by s − 1)we deduce that ut ∈ C(R, Hs−1(R)).

Remark 3.1.2. Applying twice Lemma 3.1.1, we see that if ϕ ∈ Hs(R), s ≥ 1,then u ∈ C(R, Hs(R)) ∩ C1(R, Hs−1(R)) ∩ C2(R, Hs−2(R)), and utt = ϕ′′(· ± t).Since uxx = ϕ′′(· ± t), we see that u is a solution of the wave equation.

We have the following existence and uniqueness result.

Theorem 3.1.3. Given u0 ∈ H1(R) and v0 ∈ L2(R), there exists a uniquesolution

u ∈ C(R, H1(R)) ∩ C1(R, L2(R)) ∩ C2(R, H−1(R)) (3.1.2)

55

56 3. THE ONE-DIMENSIONAL WAVE EQUATION

of (3.1.1). Moreover, the following properties hold.

(i) u is given by the formula

u(t, x) =1

2(u0(x+ t) + u0(x− t)) +

1

2

∫ x+t

x−tv0(s) ds (3.1.3)

for all t, x ∈ R.(ii) u satisfies the following estimates

‖u(t)‖2L2 ≤ ‖u0‖2L2 + t2‖v0‖2L2 (3.1.4)

‖ux(t)‖2L2 ≤ ‖∂xu0‖2L2 + ‖v0‖2L2 (3.1.5)

‖ut(t)‖2L2 ≤ ‖∂xu0‖2L2 + ‖v0‖2L2 (3.1.6)

for all t ∈ R(iii) If, in addition, u0 ∈ H2(R) and v0 ∈ H1(R), then u ∈ C(R, H2(R)) ∩

C1(R, H1(R)) ∩ C2(R, L2(R)).(iv) We have conservation of energy, that is∫

R(|ux|2 + |ut|2) =

∫R

(|(u0)x|2 + |v0|2) (3.1.7)

for all t ∈ R

Proof. We first prove uniqueness, and we consider u, v in the class (3.1.2) twosolutions of (3.1.1). Setting w = u−v, it follows that w belongs to the class (3.1.2)and that

wtt = ∆w

w(0) = wt(0) = 0

Setting z = (1 + 4π2|ξ|2)−12 w, we deduce that z ∈ C2([0,∞), L2(R)) and that z

satisfiesztt + 4π2|ξ|2z = 0

in L2(R) for all t ∈ R (see (B.6.42)), with the initial conditions z(0) = zt(0) = 0.We integrate the equation twice in time to obtain

z(t, ξ) = −4π2|ξ|2∫ t

0

∫ s

0

z(τ, ξ) dτds

The above equation is, for every t ∈ R, an identity in L2(R). In particular, we mayintegrate it on BR with 0 < R <∞, and obtain for t ≥ 0∫

BR

|z(t, ξ)| dξ ≤ 4π2R2∣∣∣∫BR

∫ t

0

∫ s

0

z(σ, τ) dτdsdξ∣∣∣

≤ 4π2R2

∫ t

0

∫ s

0

∫BR

|z(τ, ξ)| dξdτds

≤ 4π2R2

∫ t

0

∫ t

0

∫BR

|z(τ, ξ)| dξdτds

≤ 4π2R2t

∫ t

0

∫BR

|z(τ, ξ)| dξdτ

We deduce by applying Gronwall’s lemma that z(t, ξ) = 0 a.e on (0,∞)×BR. SinceR > 0 is arbitrary, this implies that z(t, ξ) = 0 a.e on (0,∞) × R. One shows bysimilar calculations that z(t, ξ) = 0 a.e on (−∞, 0)×R. This implies that w(t) = 0for all t ∈ R, hence w(t) ≡ 0. This completes the proof of uniqueness.

We now prove the existence part. Set

U(t, x) =1

2(u0(x+ t) + u0(x− t))

3.1. THE WAVE EQUATION ON THE LINE 57

It follows from Lemma 3.1.1 that U ∈ C(R, H1(R))∩C1(R, L2(R))∩C2(R, H−1(R))and U is a solution of the wave equation. Moreover, U(0) = u0 and Ut(0) = 0.Next, set

V (t, x) =1

2

∫ x+t

x−tv0(s) ds

It follows that V (0) = 0 and, for t > 0∫R|V (t, x)|2 ds =

1

4

∫R

(∫ x+t

x−tv0(s) ds

)2

dx ≤ t

2

∫Rdx

∫ x+t

x−t|v0(s)|2ds

=t

2

∫Rds

∫ −s+t−s−t

|v0(s)|2dx = t2∫R|v0(s)|2

(3.1.8)

where we used Fubini. A similar calculation for t < 0 shows that V (t) ∈ L2(R) forall t ∈ R. One shows a well that V ∈ C(R, L2(R)). Moreover,

Vx(t, x) =1

2(v0(x+ t) + v0(x− t))

and it follows from Lemma 3.1.1 that Vx ∈ C(R, L2(R)). Thus we conclude thatV ∈ C(R, H1(R)). In addition

Vt(t, x) =1

2(v0(x+ t)− v0(x− t))

and so Vt(0) = v0; and Lemma 3.1.1 shows that Vt ∈ C(R, L2(R))∩C1(R, H−1(R)).It is immediate that V is a solution of the wave equation, thus we see that u = U+Vhas the desired properties. This proves the existence part, as well as formula (3.1.3).

Estimates (3.1.5) and (3.1.6) are immediate consequences of formula (3.1.3).Estimate (3.1.4) follows from formula (3.1.3), together with (3.1.8) for the estimateof ‖V (t)‖L2 . This proves Property (ii).

Property (iii) is a consequence of formula (3.1.3) and Lemma 3.1.1.It remains to prove the conservation of energy. We first suppose u0 ∈ H2(R)

and v0 ∈ H1(R). The equation is satisfied in L2(R), so we may multiply it byut ∈ H1(R). It follows that ∫

Ruttut +

∫Ruxuxt = 0

i.e.1

2

d

dt

∫R

(|ux|2 + |ut|2) = 0

which proves (3.1.7). In the general case, we approximate u0 in H1 by a sequence(un0 )n≥1 ⊂ H2(R), and v0 in L2 by a sequence (vn0 )n≥1 ⊂ H1(R). We thereforeobtain a sequence of solutions (un)n≥1. It follows from (3.1.4)–(3.1.6) (applied withu0 and v0 replaced by u0 − un0 and v0 − vn0 ) that un → u in C([−T, T ], H1(R)) ∩C1([−T, T ], L2(R)) for all T < ∞. Therefore, we can apply the conservation ofenergy to un, and then let n→∞ to obtain the conservation of energy for u.

Remark 3.1.4. Here are some observations.

(i) Formula (3.1.3) shows that, given t, x ∈ R, the solution u(t, x) depends on theinitial data u0 and v0 in the interval [x− |t|, x+ |t|].

(ii) It follows from the preceding observation that if u0 and v0 have compactsupport, say in [−R,R], then u(t) also have compact support, more preciselyin [−R− |t|, R+ |t|].

(iii) Given 0 < R < ∞, it follows easily from formula (3.1.3) that, as t → ±∞,‖ux(t)‖L2(−R,R) + ‖ut(t)‖L2(−R,R) → 0. This means that, on every boundedinterval, the solution becomes more and more flat as t→ ±∞.

58 3. THE ONE-DIMENSIONAL WAVE EQUATION

Remark 3.1.5. Some special solutions.

(i) Given ϕ ∈ H1(R), u(t, x) = ϕ(x± t) is a solution of the wave equation. It isa given profile, that travels at constant velocity, either to the right or to theleft.

(ii) Given ϕ ∈ H1(R), u(t, x) = 12 (ϕ(x − t) + ϕ(x + t)) is a solution of the wave

equation. It is the sum of two half profiles, one of which travels to the left,the other to the right, with constant velocity.

3.2. The wave equation on an interval

We now consider the wave equation on the interval Ω = (0, `), with Dirichletboundary conditions

utt = uxx t ∈ R, x ∈ Ω

u(t, 0) = u(t, `) = 0 t ∈ Ru(0, x) = u0(x) x ∈ Ω

ut(0, x) = v0(x) x ∈ Ω

(3.2.1)

We recall that in this case, the eigenvalues of −∆ in H10 (Ω) are λj = ( jπ` )2 and the

corresponding eigenfunctions

ϕj(x) =

√2

`sin(jπ`x)

See Remark 1.4.7.

Theorem 3.2.1. Let u0 =∑a0jϕj ∈ H1

0 (Ω) and v0 =∑b0jϕj ∈ L2(Ω).

It follows that (3.2.1) has a unique solution u ∈ C(R, H10 (Ω)) ∩ C1(R, L2(Ω)) ∩

C2(R, H−1(Ω)). Moreover, u is given by the formula

u(t) =

∞∑j=1

aj(t)ϕj (3.2.2)

with

aj(t) = a0j cos

(jπ`t)

+`b0jπj

sin(jπ`t)

(3.2.3)

In addition, there is conservation of energy, i.e.∫ `

0

(|ux|2 + |ut|2) =

∫ `

0

(|(u0)x|2 + |v0|2) (3.2.4)

for all t ∈ R

Proof. We first prove uniqueness. Let u be a solution. In particular, thefunction t 7→ 〈u(t), ϕj〉H−1,H1

0is C2 and

d2

dt2〈u, ϕj〉H−1,H1

0= 〈uxx, ϕj〉H−1,H1

0= 〈u, ∂xxϕj〉H−1,H1

0= −λj〈u, ϕj〉H−1,H1

0

Therefore, if we write u in the form (3.2.2), then aj = 〈u, ϕj〉H−1,H10, so that

aj ∈ C2(R) and a′′j + λjaj = 0. It follows that aj is given by formula (3.2.3).Therefore, aj is uniquely determined, hence u is uniquely determined.

Let now u be defined by (3.2.2)-(3.2.3). It follows easily (see the proof ofTheorem 2.2.1) that u has the required regularity and that u is a solution of (3.2.1).

We finally prove conservation of energy. Since a′′j + λjaj = 0, we have λja2j +

(a′j)2 = λj(a

0J)2 + (b0J)2; and so

∞∑j=1

λja2j + (a′j)

2 =

∞∑j=1

λj(a0J)2 + (b0J)2

3.2. THE WAVE EQUATION ON AN INTERVAL 59

`/2

3`/8

`/4

`/8

t = 0

t = `/8

t = `/4

t = 3`/8

t = `/2 `

Figure 1. A solution of the wave equation

which yields (3.2.4).

Corollary 3.2.2. For every u0 ∈ H10 (Ω) and v0 ∈ L2(Ω), the corresponding

solution u ∈ C(R, H10 (Ω)) ∩ C1(R, L2(Ω)) ∩ C2(R, H−1(Ω)) of (3.2.1) is periodic

(in time) with period 2`.

Proof. u is given by formula (3.2.2)-(3.2.3). Note that cos( jπ` t) and sin( jπ` t)

are both periodic with period 2`j , hence also periodic with period 2`. Therefore, all

the coefficients aj in (3.2.3) are 2`-periodic, hence so is u.

Remark 3.2.3. Here are some particular solutions of the wave equation on(0, `).

(i) u(t, x) = cos( jπ` t) sin( jπ` x) is a solution, corresponding to u0 = sin( jπ` x) andv0 = 0.

(ii) Let

θ(x) =

x 0 ≤ x ≤ `

2

`− x 0 ≤ `2 ≤ x ≤ `

The solution of the wave equation with u0 = θ and v0 = 0 is given by (seeFigure 1)

u(t, x) =

x 0 ≤ x ≤ `

2 − t`2 − t

`2 − t ≤ x ≤

`2 + t

`− x 0 ≤ `2 + t ≤ x ≤ `

for 0 ≤ t ≤ `2 . For `

2 ≤ t ≤ `, it is given by

u(t) = −u(`− t)for ` ≤ t ≤ 2`, it is given by

u(t) = u(2`− t)and then it is reproduced by 2`-periodicity.

APPENDIX A

Some useful results

We collect in this appendix some useful results.

A.1. Functional analysis

Theorem A.1.1 (The Banach fixed point theorem). Let (E,d) be a nonemptycomplete metric space and F : E → E. Suppose F is a strict contraction, i.e. thereexists a constant 0 ≤ k < 1 such that d(F (x), F (y)) ≤ kd(x, y) for all x, y ∈ E. Itfollows that there exists a unique x ∈ E such that F (x) = x.

Proof. Fix x0 ∈ E and define xn recursively by xn+1 = F (xn). Given n ≥1, we have d(xn+1, xn) = d(F (xn), F (xn−1)) ≤ kd(xn, xn−1), and an immediateiteration argument shows that

d(xn+1, xn) ≤ knd(x1, x0), n ≥ 0

Given 0 ≤ n < m, we deduce that

d(xm, xn) ≤m−n−1∑j=0

d(xn+j+1, xn+j) ≤ d(x1, x0)

m−n−1∑j=0

kn+j ≤ kn

1− kd(x1, x0)

It follows that (xn)n≥0 is a Cauchy sequence, so it has a limit x ∈ E. Passing to thelimit in the equality xn+1 = F (xn), we deduce that x = F (x). Uniqueness followsfrom the contraction property.

Theorem A.1.2 (Lax-Milgram [20, Theorem 2.1]). Let H be a Hilbert spaceand consider a bilinear functional b : H ×H → R. If there exist C <∞ and α > 0such that

|b(u, v)| ≤ C‖u‖ ‖v‖, (u, v) ∈ H ×H (continuity),

|b(u, u)| ≥ α‖u‖2, u ∈ H (coerciveness),

then, for every f ∈ H? (the dual space of H), the equation

b(u, v) = 〈f, v〉H?,H for all v ∈ H, (A.1.1)

has a unique solution u ∈ H.

Proof. By the Riesz representation theorem, there exists ϕ ∈ H such that

〈f, v〉H?,H = (ϕ, v)H ,

for all v ∈ H. Furthermore, given u ∈ H, the map v 7→ b(u, v) defines an elementof H?; and so, by the Riesz representation theorem, there exists an element of H,which we denote by Au, such that

b(u, v) = (Au, v)H ,

for all v ∈ H. It is clear that A : H → H is a linear operator such that‖Au‖H ≤ C‖u‖H ,(Au, u)H ≥ α‖u‖2H ,

61

62 A. SOME USEFUL RESULTS

for all u ∈ H. We see that (A.1.1) is equivalent to Au = ϕ. Given ρ > 0, this lastequation is equivalent to

u = Tu, (A.1.2)

where Tu = u + ρϕ − ρAu. It is clear that T : H → H is continuous. Moreover,Tu− Tv = (u− v)− ρA(u− v); and so,

‖Tu− Tv‖2H = ‖u− v‖2H + ρ2‖A(u− v)‖2H − 2ρ(A(u− v), u− v)H

≤ (1 + ρ2C2 − 2ρα)‖u− v‖2H .

Choosing ρ > 0 small enough so that 1 + ρ2C2− 2ρα < 1, T is a strict contraction.By Banach’s fixed point theorem, we deduce that T has a unique fixed point u ∈ H,which is the unique solution of (A.1.2).

Theorem A.1.3. Let X be a Banach space, T > 0 and f ∈ C([0, T ), X). It f is

right-differentiable for all t ∈ [0, T ) and d+fdt ∈ C([0, T ), X), then f ∈ C1([0, T ), X)

and dfdt = d+f

dt .

Proof. Set

g(t) = f(t)− f(0)−∫ t

0

d+f

dtds,

for all t ∈ [0, T ). It follows that g ∈ C([0, T ), X), g(0) = 0, g is right-differentiable

for all t ∈ [0, T ) and d+gdt = 0. Let now ξ ∈ X?, and set h(t) = 〈ξ, g(t)〉X?,X .

We have h ∈ C([0, T )), h(0) = 0, h is right-differentiable for all t ∈ [0, T ) andd+hdt = 0. We show that h ≡ 0. To see this, let ε > 0, set hε(t) = h(t)− εt, and let

us show that hε ≤ 0. Otherwise, there exists t ∈ [0, T ) such that hε(t) > 0. Letτ = inft ∈ [0, T ); hε(t) > 0. We have hε(τ) = 0, and there exists tn ↓ τ such thathε(tn) > 0. It follows that

lim supt↓τ

hε(t)− hε(τ)

t− τ≥ 0.

On the other hand, we have d+hεdt = −ε, which is a contradiction. Therefore,

hε ≤ 0. Since ε > 0 is arbitrary, we have h ≤ 0. Applying the same argument to−h, we obtain as well h ≥ 0, hence h ≡ 0. Therefore, given t ∈ [0, T ), we have〈ξ, g(t)〉X?,X = 0 for all ξ ∈ X?; and so, g(t) ≡ 0. The result follows easily.

Remark A.1.4. In Theorem A.1.3 above, if d+fdt ∈ C([0, T ], X), then f ∈

C1([0, T ], X). This follows easily from the formula

f(t) = f(0) +

∫ t

0

f ′(s) ds

for all 0 ≤ t < T .

Theorem A.1.5. If X and Y are two Banach spaces such that X → Y withdense embedding, then, the following properties hold.

(i) Y ? → X?, where the embedding e is defined by 〈ef, x〉X?,X = 〈f, x〉Y ?,Y , forall x ∈ X and f ∈ Y ?.

(ii) If X is reflexive, then the embedding Y ? → X? is dense.(iii) If the embedding X → Y is compact and X is separable, then the embedding

Y ? → X? is compact. More precisely, if (y′n)n≥0 ⊂ Y ? and ‖y′n‖Y ? ≤ M ,then there exist a subsequence (nk)k≥0 and y′ ∈ Y ? with ‖y′‖Y ? ≤ M suchthat y′nk → y′ in X? as k →∞.

A.1. FUNCTIONAL ANALYSIS 63

Proof. (i) Given y′ ∈ Y ?, set ey′(x) = 〈y′, x〉Y ?,Y for all x ∈ X → Y . Since

|ey′(x)| ≤ ‖y′‖Y ?‖x‖Y ≤ C‖y′‖Y ?‖x‖X ,we see that e ∈ L(Y ?, X?). Suppose that ey′ = ez′, for some y′, z′ ∈ Y ?. Itfollows that 〈y′ − z′, x〉Y ?,Y = 0, for every x ∈ X. By density, we deduce that〈y′ − z′, y〉Y ?,Y = 0, for every y ∈ Y ; and so y′ = z′. Thus e is injective and (i)follows.

(ii) Assume to the contrary that Y ? 6= X?. Then there exists x0 ∈ X?? = Xsuch that 〈y′, x0〉X?,X = 0, for every y′ ∈ Y ? (see e.g. [5, Corollary 1.8, p. 8]). LetE = Rx0 ⊂ Y , and let f ∈ E? be defined by f(λx0) = λ, for λ ∈ R. We have‖f‖E? = 1, and by the Hahn-Banach theorem (see e.g. [5, Corollary 1.2, p. 3]) thereexists y′ ∈ Y ? such that ‖y′‖Y ? = 1 and 〈y′, x0〉Y ?,Y = 1, which is a contradiction,since 〈y′, x0〉Y ?,Y = 〈y′, x0〉X?,X = 0.

(iii) Let BX? (respectively, BX , BY ? , BY ) be the unit ball of X? (respectively,X, Y ?, Y ). Consider a sequence (y′n)n≥0 ⊂ BY ? . Since Y ? is the dual of a separableBanach space, it follows (see e.g. [5, Corollary 3.30, p. 76]) that there exist asubsequence, which we still denote by (y′n)n≥0, and an element y′ ∈ BY ? such thaty′n → y′ in Y ? weak?. We show that ‖y′n − y′‖X? → 0, which proves the desiredresult. We note that

‖y′n − y′‖X? = supx∈BX

|〈y′n − y′, x〉X?,X | = supx∈BX

|〈y′n − y′, x〉Y ?,Y |, (A.1.3)

by (i). Let ε > 0. Since BX is a relatively compact subset of Y , we see that thereexists a (finite) sequence (xj)1≤j≤` ⊂ BX such that for every x ∈ BX , there exists1 ≤ j ≤ ` such that ‖x − xj‖Y ≤ ε. Given x ∈ BX and 1 ≤ j ≤ ` as above, wededuce that

|〈y′n − y′, x〉Y ?,Y | ≤ |〈y′n − y′, x− xj〉Y ?,Y |+ |〈y′n − y′, xj〉Y ?,Y |≤ ε‖y′n − y′‖Y ? + |〈y′n − y′, xj〉Y ?,Y |≤ 2ε+ |〈y′n − y′, xj〉Y ?,Y |.

Aplying now (A.1.3), we deduce that

‖y′n − y′‖X? ≤ 2ε+ sup1≤j≤`

|〈y′n − y′, xj〉Y ?,Y |.

Since y′n → y′ in Y ? weak?, we conclude that

lim supn→∞

‖y′n − y′‖X? ≤ 2ε.

Since ε > 0 is arbitrary, the result follows.

Corollary A.1.6. Let H be a Hilbert space, and identify H? = H by the Rieszrepresentation theorem. Let X be a Banach space such that X → H with denseembedding, so that

X → H = H? → X?

by Theorem A.1.5. If T > 0 and u ∈ C((0, T ), X) ∩ C1((0, T ), X?), then the mapt 7→ ‖u(t)‖2H is C1 on (0, T ), and

d

dt‖u(t)‖2H = 2〈ut(t), u(t)〉X?,X (A.1.4)

for all t ∈ (0, T ).

Proof. Set F (t) = ‖u(t)‖2H . Since u ∈ C((0, T ), X) and X → H, we seethat F ∈ C((0, T )). In addition, given 0 < s, t < T , s 6= t, it follows fromTheorem A.1.5 (i) that

F (t)− F (s)

t− s=(u(t)− u(s)

t− s, u(t) + u(s)

)H

=⟨u(t)− u(s)

t− s, u(t) + u(s)

⟩X?,X


which proves (A.1.4).

If X is a separable Banach space, then its dual X? needs not be separable. (Forexample X = L1(Ω) is separable, but it dual L∞(Ω) is not). However, X? is weak?

separable, as shows the following result.

Lemma A.1.7. Let X be a separable Banach space and let X? be its dual.There exists a sequence (x′n)n∈N ⊂ X? such that for every x′ ∈ X?, there exists asubsequence (x′nk)k∈N with the following properties:

(i) x′nk → x′ weak? as k →∞.(ii) ‖x′nk‖X? ≤ ‖x

′‖X? .(iii) ‖x′nk‖X? → ‖x

′‖X? as k →∞.

Proof. B′ = x′ ∈ X?; ‖x‖X? ≤ 1 equipped with the weak? topology of X?,is a compact metric space. In particular, B′ is separable and we denote by (y′n)n∈Na dense sequence in B′. Given x′ ∈ X?, there exists a sequence (nk)k∈N such thaty′nk → x′/‖x′‖X? weak? as k → ∞. Consider now a sequence (λk)k∈N such thatλk → ‖x′‖X? as k →∞ and 0 < λk ≤ ‖x′‖X? . It follows that λky

′nk→ x′ weak? as

k → ∞. Furthermore, ‖λky′nk‖X? ≤ |λk|‖y′nk‖X? ≤ ‖x′‖X? . Since also ‖x′‖X? ≤

lim inf ‖λky′nk‖X? as k →∞, the result follows with (x′n)n∈N = ∪λ∈Qn∈Nλy′n.

Lemma A.1.8. Let X → Y be two Banach spaces and let (xn)n∈N be a sequenceof X. If xn x in X, as n→∞, then xn x in Y , as n→∞.

Proof. The embedding is continuous X → Y ; and so, it is also continuousX → Y for the weak topologies. The result follows.

Lemma A.1.9. Let X → Y be two Banach spaces and let (xn)n∈N be a boundedsequence of X such that xn y in Y , as n→∞, for some y ∈ Y . If X is reflexive,then y ∈ X and xn y in X, as n→∞.

Proof. Let us first prove that y ∈ X. There exists x ∈ X and a subsequencenk such that xnk x in X, as k → ∞. Therefore, by Lemma A.1.8, xnk x inY , as k →∞. It follows that y = x ∈ X.

Let us prove that xn y in X by contradiction. If not, there exists x′ ∈ X?,ε > 0 and a subsequence nk such that |〈x′, xnk − y〉| ≥ ε, for every k ∈ N. Onthe other hand, there exists x ∈ X and a subsequence nkj such that xnkj x in

X as j → ∞. In particular, x = y; and so xnkj y in X as j → ∞, which is a

contradiction.

Corollary A.1.10. Let X → Y be two Banach spaces. If Y is separable andX is reflexive, then X is separable.

Proof. Let B be the closed unit ball of X. Since B ⊂ Y and Y is separable,it follows that B is separable for the Y norm. Therefore, there exists a sequence(xn)n∈N ⊂ B such that for every x ∈ X, there exists a subsequence (xnk)k∈N whichconverges to x strongly in Y , hence weakly in X by Lemma A.1.9. Therefore, B iscontained in, hence equal to the weak X closure of the set (xn)n∈N. In particular,B is also the weak X closure of the convex hull C of the set (xn)n∈N. Since theweak and strong closures of convex sets coincide, it follows that C is strongly Xdense in B. Since the convex hull of a countable set is clearly separable, it followsthat B is separable, which completes the proof.

Corollary A.1.11. Let X → Y be two Banach spaces, let I be a bounded,open interval of R, and let u : I → Y be a weakly continuous function. Assumethat there exists a dense subset E of I such that

(i) u(t) ∈ X, for all t ∈ E,

A.1. FUNCTIONAL ANALYSIS 65

(ii) sup‖u(t)‖X , t ∈ E = K <∞.

If X is reflexive, then u(t) ∈ X for all t ∈ I and u : I → X is weakly continuous.

Proof. Let t ∈ I and let (tn)n∈N ⊂ E converge to t, as n → ∞. Sinceu(tn) u(t) in Y , it follows from Lemma A.1.9 that u(t) ∈ X and that

‖u(t)‖X ≤ lim infn→∞

‖u(tn)‖X ≤ K.

Let now t ∈ I and let (tn)n∈N ⊂ I converge to t, as n→∞. Since u(tn) u(t) inY and u(tn) is bounded in X, it follows from Lemma A.1.9 that u(tn) u(t) inX. Hence the result.

Theorem A.1.12. Let Y be a Banach space, and X be a topological vectorspace If X → Y with dense embedding and I is a bounded, closed interval of R thenC(I,X) is a dense subspace of C(I, Y ).

Proof. Let f ∈ C(I, Y ). One can approximate f in C(I, Y ) by a sequence ofpiecewise linear functions I → Y . Thus we may assume f ∈ C(I, Y ) is piecewiselinear. It is now easy to approximate f in C(I, Y ) by a sequence of piecewise linearfunctions I → X.

It is sometimes convenient to consider the intersection and sum of Banachspaces. Consider two Banach spaces X1 and X2 that are subsets of a Hausdorfftopological vector space X . Let

X1 ∩X2 = x ∈ X ; x ∈ X1, x ∈ X2,

and

X1 +X2 = x ∈ X ; ∃x1 ∈ X1, ∃x2 ∈ X2, x = x1 + x2.Define

‖x‖X1∩X2= ‖x‖X1

+ ‖x‖X2, for x ∈ X1 ∩X2,

and

‖x‖X1+X2 = Inf‖x1‖X1 + ‖x2‖X2 ; x = x1 + x2, for x ∈ X1 +X2.

We have the following result.

Proposition A.1.13. (X1∩X2, ‖ ‖X1∩X2) and (X1+X2, ‖ ‖X1+X2) are Banachspaces. If furthermore X1 ∩ X2 is a dense subset of both X1 and X2, then thefollowing properties hold:

(i) (X1 ∩X2)? = X?1 +X?

2 and (X1 +X2)? = X?1 ∩X?

2 ;(ii) 〈f, x1 + x2〉X?1∩X?2 ,X1+X2

= 〈f, x1〉X?1 ,X1+ 〈f, x2〉X?2 ,X2

, for all f ∈ X?1 ∩X?

2

and (x1, x2) ∈ X1 ×X2;(iii) 〈f1 + f2, x〉X?1 +X?2 ,X1∩X2 = 〈f1, x〉X?1 ,X1 + 〈f2, x〉X?2 ,X2 , for all (f1, f2) ∈ X?

1 ×X?

2 and x ∈ X1 ∩X2;(iv) if X1 and X2 are reflexive, then X1 ∩X2 and X1 +X2 are reflexive.

Proof. Property (i), (ii) and (iii) follow from [4, Lemma 2.3.1, Theorem 2.7.1,proof of Theorem 2.7.1]. To prove Property (iv) we note that (by (i)) it is sufficientto show that X1∩X2 is reflexive. By applying Eberlein-Smulian’s theorem, we needto show that every bounded sequence (xn)n∈N ⊂ X1 ∩X2 has a weakly convergentsubsequence. Since xn is bounded in both X1 and X2, there exists x ∈ X1 ∩ X2

and a subsequence, which we still denote by (xn)n∈N, such that xn x, in X1 andin X2. Given (f1, f2) ∈ X?

1 ×X?2 , we have

〈f1, xn〉X?1 ,X1+ 〈f2, xn〉X?2 ,X2

−→n→∞

〈f1, x〉X?1 ,X1+ 〈f2, x〉X?2 ,X2

.

Applying Property (iii), we deduce that xn x in X1 ∩X2.


Remark A.1.14. It is clear that the definition of the spaces X1 ∩ X2 andX1 +X2 as well as their properties described in Proposition A.1.13 are independentof the Hausdorff space X . It follows that an element of X1 + X2 is equal to zeroif and only if it is equal to zero in some Hausdorff space containing X1 ∪ X2. Inparticular, if X1 and X2 are spaces of distributions on some open set Ω ⊂ RN , thenan element of X1 +X2 is equal to zero if and only if it is equal to zero in D′(Ω).

We recall below a quite useful result concerning Lp spaces.

Lemma A.1.15. Let Ω be an open subset of RN , N ≥ 1 and let 1 < p ≤ ∞.Consider u : Ω → R and let (un)n∈N be a bounded sequence of Lp(Ω) such thatun → u a.e. as n → ∞. It follows that u ∈ Lp(Ω) and un → u as n → ∞ inLq(Ω′), for every Ω′ ⊂ Ω of finite measure and every q ∈ [1, p). In particular,un → u as n→∞, in Lp(Ω) weak if p <∞, and in L∞(Ω) weak? if p =∞.

Proof. By extending the functions by 0 outside Ω, we may assume that Ω =RN . Observe that by Fatou’s lemma, we have u ∈ Lp(RN ). Let Ω′ ⊂ RN have afinite measure and let q ∈ [1, p). Consider ε > 0. By Egorov’s theorem, there existsa measurable subset E of Ω′ such that un → u uniformly on Ω′ \ E and

|E|p−qp sup

n≥0

(∫RN|un − u|p

) qp ≤ ε/2.

Let n0 be large enough so that |un− u|q ≤ ε/2|Ω′| on Ω′ \E, for n ≥ n0. It followsthat ∫

Ω′|un − u|q =

∫E

|un − u|q +

∫Ω′\E

|un − u|q

≤ |E|p−qp

(∫E

|un − u|p) qp

+ |Ω′ \ E| supΩ′\E

|un − u|q ≤ ε.

Hence the result, since ε is arbitrary.

A.2. Inequalities

Below is the standard form of Gronwall’s lemma (see [15, Lemma, p. 293], [3,Lemma 1]).

Theorem A.2.1 (Gronwall’s lemma). Let T > 0, A ≥ 0 and let f ∈ C([0, T ])be a nonnegative function. If the nonnegative function ϕ ∈ C([0, T ]) satisfies

ϕ(t) ≤ A+

∫ t

0

f(s)ϕ(s) ds

for every t ∈ [0, T ], then

ϕ(t) ≤ A exp(∫ t

0

f(s) ds),

for every t ∈ [0, T ]. In particular, if

ϕ(t) ≤∫ t

0

f(s)ϕ(s) ds

for every t ∈ [0, T ], then ϕ(t) ≡ 0.

Proof. Set h(t) = ψ(t) exp(−∫ t

0f(s) ds), where ψ(t) = A+

∫ t0f(s)ϕ(s) ds. It

follows that ψ, h ∈ C1([0, T ]), ψ, h ≥ 0, and

h′(t) = (ψ′(t)− f(t)ψ(t)) exp(−∫ t

0

f(s) ds)

= (f(t)ϕ(t)− f(t)ψ(t)) exp(−∫ t

0

f(s) ds)≤ 0.

A.2. INEQUALITIES 67

It follows that h(t) ≤ h(0), from which the result follows.

We complete this section with a useful property of the convolution. We recallthat the convolution f ? g of two functions f, g defined on RN is given by

f ? g(x) =

∫RN

f(x− y)g(y) dy

The following proposition provides useful information.

Proposition A.2.2. Let 1 ≤ p, q < ∞ such that 1p + 1

q ≥ 1. Let f ∈ Lp(RN )

and g ∈ Lq(RN ), so that f ? g ∈ Lr(RN ) with 1r = 1

p + 1q − 1 by Young’s inequality.

(i) If f, g ≥ 0, then f ? g ≥ 0(ii) If f and g are radially symmetric, then so is f ? g(iii) If f and g are radially symmetric and radially decreasing, then so is f ? g

Proof. Parts (i) and (ii) are quite standard. Part (iii) is not, and is takenfrom [29].

By density and Young’s inequality, we need only consider the case f, g ∈Cc(RN ). Property (i) is immediate. To prove (ii), suppose f, g are radially sym-metric, and consider a rotation R. Since |detR| = 1, we see that

f ? g(Rx) =

∫RN

f(Rx− y)g(y) dy =

∫RN

f(R(x−R−1y))g(RR−1y) dy

=

∫RN

f(x−R−1y)g(R−1y) dy =

∫RN

f(x− z)g(z) dz = f ? g(x)

The rotation R being arbitrary, this proves Property (ii).We now prove (iii), so we assume that f and g are radially symmetric and

nonincreasing. In particular, f ? g is radially symmetric by Property (ii).We first consider the one-dimensional case N = 1. We need to prove that

h = f ? g is nonincreasing on (0,∞). Given x, s ≥ 0, we have

h(x+ 2s)− h(x) =

∫ ∞−∞

[f(x+ 2s− y)− f(x− y)]g(y) dy

=

∫ ∞−∞

[f(y + s)− f(y − s)]g(x+ s− y) dy

= I− + I+

where

I− =

∫ 0

−∞[f(y + s)− f(y − s)]g(x+ s− y) dy

I+ =

∫ ∞0

[f(y + s)− f(y − s)]g(x+ s− y) dy

Changing y to −y in I−, then using the property that f is even, we obtain

I− =

∫ ∞0

[f(−y + s)− f(−y − s)]g(x+ s+ y) dy

=

∫ ∞0

[f(y − s)− f(y + s)]g(x+ s+ y) dy

so that

h(x+ 2s)−h(x) =

∫ ∞0

[f(y+ s)− f(y− s)][g(x+ s− y)− g(x+ s+ y)] dy (A.2.1)


We claim that

f(y + s)− f(y − s) ≤ 0 y, s ≥ 0 (A.2.2)

g(x+ s− y)− g(x+ s+ y) ≤ 0 x, y, s ≥ 0 (A.2.3)

Formulas (A.2.1), (A.2.2) and (A.2.3) imply that h(x+ 2s) ≤ h(x) for all x, s ≥ 0,which is the desired conclusion. To prove (A.2.2), we observe that if y ≥ s, then0 ≤ y − s ≤ y + s, so that f(y − s) ≤ f(y + s) since f is radially decreasing. If0 ≤ y ≤ s, then 0 ≤ s − y ≤ y + s, so that f(y + s) ≤ f(s − y) = f(y − s), againbecause f is radially decreasing. Similarly, to prove (A.2.3), we observe that ify ≥ x+ s, then 0 ≤ x+ s− y ≤ x+ s+ y, so that g(x+ s− y) ≤ g(x+ s+ y) sinceg is radially decreasing. If 0 ≤ y ≤ x+ s, then 0 ≤ −x− s+ y ≤ x+ s+ y, so thatg(x+ s+ y) ≤ g(−x− s+ y) = g(x+ s− y), again because g is radially decreasing.

We finally consider the case N ≥ 2, and we set h = f ? g. Given x1 ∈ R, we letx = (x1, 0) ∈ RN . Since h is radially symmetric, we need only show that h(x) is anonincreasing function of x1 > 0. To see this, we write

h(x) =

∫RN−1

∫Rf(x1 − y1,−y′)g(y1, y

′) dy1dy′

=

∫RN−1

(∫Rf(x1 − y1,−y′)g(y1, y

′)dy1

)dy′

=

∫RN−1

fy′? gy

′(x1) dy′

where, for every y′ ∈ RN−1 fy′, gy

′R→ R are defined by

fy′(x1) = f(x1,−y′) gy

′(x1) = g(x1, y

′)

Since f and g are radially decreasing on RN , it follows that fy′(x1) and gy

′(x1)

are radially decreasing. Therefore, fy′? gy

′(x1) is radially decreasing, hence a

nonincreasing function of x1 ≥ 0. This completes the proof.

APPENDIX B

Sobolev spaces

Throughout this section, Ω is an open subset of RN . We study the basicproperties of the Sobolev spaces Wm,p(Ω) and Wm,p

0 (Ω), and particularly the spacesH1(Ω) and H1

0 (Ω) (which correspond to m = 1 and p = 2). For a more detailedstudy, see for example Adams and Fournier [1]. We consider the case of real-valuedfunctions, see Section B.5 for the necessary modification in the case of complex-valued function.

B.1. Definitions and basic properties

We begin with the definition of “weak” derivatives. Let u ∈ Cm(Ω), m ≥ 1. Ifα ∈ NN is a multi-index such that |α| ≤ m, it follows from Green’s formula that∫

Ω

Dαuϕ = (−1)|α|∫

Ω

uDαϕ, (B.1.1)

for all ϕ ∈ Cmc (Ω). We note that both integrals in (B.1.1) make sense since Dαuϕ ∈Cc(Ω) and uDαϕ ∈ Cc(Ω). As a matter of fact, the right-hand side makes sense assoon as u ∈ L1

loc(Ω) and the left-hand side makes sense as soon as Dαu ∈ L1loc(Ω).

This motivates the following definition.

Definition B.1.1. Let u ∈ L1loc(Ω) and let α ∈ NN . We say that Dαu ∈

L1loc(Ω) if there exists uα ∈ L1

loc(Ω) such that∫Ω

uαϕ = (−1)|α|∫

Ω

uDαϕ, (B.1.2)

for all ϕ ∈ C|α|c (Ω). Such a function uα is then unique and we set Dαu = uα.

If uα ∈ Lploc(Ω) (respectively, u ∈ Lp(Ω)) for some 1 ≤ p ≤ ∞, we say thatDαu ∈ Lploc(Ω) (respectively, Dαu ∈ Lp(Ω)).

The Sobolev spaces Wm,p(Ω) are defined as follows.

Definition B.1.2. Let m ∈ N and p ∈ [1,∞]. We set

Wm,p(Ω) = u ∈ Lp(Ω); Dαu ∈ Lp(Ω) for |α| ≤ m.

For u ∈Wm,p(Ω), we set

‖u‖Wm,p =∑|α|≤m

‖Dαu‖Lp ,

which defines a norm on Wm,p(Ω). We set

Hm(Ω) = Wm,2(Ω),

and we equip Hm(Ω) with the scalar product

(u, v)Hm =∑|α|≤m

∫Ω

DαuDαv dx,

69

70 B. SOBOLEV SPACES

which defines on Hm(Ω) the norm

‖u‖Hm =( ∑|α|≤m

‖Dαu‖2L2

) 12

,

which is equivalent to the norm ‖ · ‖Wm,2 .

Proposition B.1.3. Wm,p(Ω) is a Banach space and Hm(Ω) is a Hilbert space.If p <∞, then Wm,p(Ω) is separable, and if 1 < p <∞, then Wm,p(Ω) is reflexive.

Proof. Let k = 1 + N + · · · + Nm = (Nm+1 − 1)/(N − 1) (k = m + 1 ifN = 1), and consider the operator T : Wm,p(Ω)→ Lp(Ω)k defined by

Tu = (Dαu)|α|≤m.

It is clear that T is isometric and that T is injective. Therefore, Wm,p(Ω) can beidentified with the subspace T (Wm,p(Ω)) of Lp(Ω)k.

We claim that T (Wm,p(Ω)) is closed. Indeed, suppose (un)n≥0 is such thatun → u in Lp(Ω) and Dαun → uα in Lp(Ω) for 1 ≤ |α| ≤ m. Applying (B.1.1)to un and letting n → ∞, we deduce that Dαu ∈ Lp(Ω) and that Dαu = uα; andso, u ∈ Wm,p(Ω). Therefore, T (Wm,p(Ω)) is a Banach space, and so is Wm,p(Ω).If p < ∞, then Lp(Ω)k is separable. Thus so is T (Wm,p(Ω)), hence Wm,p(Ω).Finally, if 1 < p < ∞, then Lp(Ω)k is reflexive. Thus so is T (Wm,p(Ω)), henceWm,p(Ω).

Remark B.1.4. Here are some simple consequences of Definition B.1.2.

(i) It is not difficult to prove that if u ∈Wm,p(Ω) and if v ∈ Cm(Ω)∩Wm,∞(Ω),then uv ∈Wm,p(Ω) and Leibniz’s formula holds. More precisely,

Dα(uv) =∑

α1+α2=α

Cα1,α2Dα1uDα2v (B.1.3)

with constants Cα1,α2 independent of u, v. In particular, there exists a con-stant C such that

‖uv‖Wm,p ≤ C‖u‖Wm,p‖v‖Wm,∞ (B.1.4)

for all u ∈ Wm,p(Ω) and v ∈ Cm(Ω) ∩ Wm,∞(Ω). (The terms Dα1uDα2vin (B.1.3) are well defined, as the product of the function Dα1u ∈ Lp(Ω) withthe function Dα2v ∈ L∞(Ω).)

(ii) If |Ω| < ∞ and p ≥ q, then Lp(Ω) → Lq(Ω). It follows that Wm,p(Ω) →Wm,q(Ω).

Remark B.1.5. One can show that if p <∞, then Wm,p(Ω)∩C∞(Ω) is densein Wm,p(Ω) (see [1, Theorem 3.17 p. 67]).

We now define the subspaces Wm,p0 (Ω) for p < ∞. Formally, Wm,p

0 (Ω) is thesubspace of functions of Wm,p(Ω) that vanish, as well of their derivatives up toorder m− 1, on ∂Ω.

Definition B.1.6. Let 1 ≤ p <∞ and let m ∈ N. We denote by Wm,p0 (Ω) the

closure of C∞c (Ω) in Wm,p(Ω), and we set Hm0 (Ω) = Wm,2

0 (Ω).

Remark B.1.7. It follows from Proposition B.1.3 that Wm,p0 (Ω) is a separable

Banach space and that Wm,p0 (Ω) is reflexive if p > 1. In addition, Hm

0 (Ω) is aseparable Hilbert space.

In general Wm,p0 (Ω) 6= Wm,p(Ω), however both spaces coincide when ∂Ω is

“small” (see [1, Sections 3.24–3.39]). In particular, we have the following result.

Theorem B.1.8. If 1 ≤ p <∞ and m ∈ N, then Wm,p0 (RN ) = Wm,p(RN ).

B.1. DEFINITIONS AND BASIC PROPERTIES 71

The proof of Theorem B.1.8 makes use of the following lemma.

Lemma B.1.9. Let ρ ∈ C∞c (RN ), ρ ≥ 0, with supp ρ ⊂ x ∈ RN ; |x| ≤ 1 and‖ρ‖L1 = 1. For n ∈ N, n ≥ 1, set ρn(x) = nNρ(nx). ((ρn)n≥1 is called a smoothingsequence.) Then the following properties hold.

(i) For every u ∈ L1loc(RN ), ρn ? u ∈ C∞(RN ).

(ii) If u ∈ Lp(RN ) for some p ∈ [1,∞], then ρn ? u ∈ Lp(RN ) and ‖ρn ? u‖Lp ≤‖u‖Lp . If p <∞ or if p =∞ and u ∈ Cb,u(RN ), then ρn ? u→ u in Lp(RN )as n→∞.

(iii) If u ∈ Wm,p(RN ) for some p ∈ [1,∞] and m ∈ N, then ρn ? u ∈ Wm,p(RN )and Dα(ρn ? u) = ρn ? D

αu for |α| ≤ m. In particular, if p <∞ or if p =∞and Dαu ∈ Cb,u(RN ) for all |α| ≤ m, then ρn ? u → u in Wm,p(RN ) asn→∞.

Proof. (i) Since

ρn ? u(x) =

∫RN

ρn(x− y)u(y) dy,

it is clear that ρn ? u ∈ C(RN ). One deduces easily from the above formula thatDα(ρn ? u) = (Dαρn) ? u, and the result follows.

(ii) The first part of property (ii) follows from Young’s inequality, since

‖ρn‖L1 =

∫RN

ρn(x) dx = nN∫RN

ρ(nx) dx =

∫RN

ρ(y) dy = 1.

Consider now u ∈ Cb,u(RN ) and set un = ρn ? u. We have

un(x) =

∫RN

ρn(y)u(x− y) dy, u(x) =

∫RN

ρn(y)u(x) dy;

and so,

un(x)− u(x) =

∫RN

ρn(y)(u(x− y)− u(x)) dy.

Therefore,

|un(x)− u(x)| ≤∫RN

ρn(y)|u(x− y)− u(x)| dy ≤ sup|y|≤1/n

|u(x− y)− u(x)|,

since supp ρn ⊂ y; |y| ≤ 1/n. Since u is uniformly continuous, we have

supx∈RN

sup|y|≤1/n

|u(x− y)− u(x)| −→n→∞

0;

and so, un → u uniformly. Consider next u ∈ Lp(RN ), with p <∞, and let ε > 0.There exists v ∈ Cc(RN ) such that ‖u− v‖Lp ≤ ε/3. Furthermore, it follows fromwhat precedes that for n large enough, we have ‖v − ρn ? v‖Lp ≤ ε/3. (Sinceρn ? v → v uniformly and ρn ? v is supported in a fixed compact subset of RN .)Finally, it follows from the inequality of (ii) that ‖ρn?v−ρn?u‖Lp ≤ ‖u−v‖Lp ≤ ε/3.Writing

u− ρn ? u = u− v + v − ρn ? v + ρn ? v − ρn ? u,we deduce that ‖u − ρn ? u‖Lp ≤ ε. Since ε > 0 is arbitrary, this completes theproof of property (ii).

(iii) For any v : RN → R, we set v(x) = v(−x). Given u ∈ Wm,p(RN ) andϕ ∈ Cmc (RN ), we have∫

RN(ρn ? u)Dαϕ =

∫RN

u(ρn ? Dαϕ) =

∫RN

uDα(ρn ? ϕ).


By definition of Dαu, we obtain∫RN

(ρn ? u)Dαϕ = (−1)|α|∫RN

Dαu(ρn ? ϕ) = (−1)|α|∫RN

(ρn ? Dαu)ϕ.

This means that Dα(ρn ? u) ∈ Lp(RN ) and that Dα(ρn ? u) = ρn ? Dαu; and so,

ρn ? u ∈Wm,p(RN ). The convergence property follows from property (ii).

Proof of Theorem B.1.8. Let u ∈ Wm,p(RN ) and ε > 0. It follows fromproperties (iii) and (i) of Lemma B.1.9 that there exists v ∈Wm,p(RN )∩C∞(RN )such that ‖u− v‖Wm,p ≤ ε/2. Fix now η ∈ C∞c (RN ) such that 0 ≤ η ≤ 1, η(x) = 1for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2. Set ηn(x) = η(x/n) and let vn = ηnv. It isclear that vn ∈ C∞c (RN ), and we claim that ηnv → v in Wm,p(RN ) as n → ∞.Indeed, it follows from Leibniz’ formula (see Remark B.1.4 (i)) that

Dα(ηnv) =∑

β+γ=α

DβηnDγu.

Since ‖Dγηn‖L∞ ≤ Cγn−|γ|, it follows that all the terms with |β| > 0 converge

to 0 as n → ∞. The remaining term in the sum is ηnDαv which, by dominated

convergence, converges to Dαv in Lp(RN ). We deduce that Dα(ηnv) → Dαv inLp(RN ), which proves the claim. Therefore, there exists w ∈ C∞c (RN ) such that‖v−w‖Wm,p ≤ ε/2. This implies that ‖u−w‖Wm,p ≤ ε, and the result follows.

Remark B.1.10. We describe below some useful properties of the Sobolev spaceWm,p

0 (Ω).

(i) If u ∈ Wm,p(Ω) and if suppu is included in a compact subset of Ω, thenu ∈Wm,p

0 (Ω). This is easily shown by using the regularization and truncationargument described above.

(ii) It follows easily from Remark B.1.4 (i) and Property (i) above that that ifu ∈ Wm,p

0 (Ω) and if v ∈ Cm(Ω) ∩Wm,∞(Ω), then uv ∈ Wm,p0 (Ω). Moreover,

formulas (B.1.3) and (B.1.4) hold.

(iii) If u ∈ W 1,p(Ω) ∩ C(Ω) and if u|∂Ω = 0, then u ∈ W 1,p0 (Ω). Indeed, if u has

a bounded support, let F ∈ C1(R) satisfy |F (t)| ≤ |t|, F (t) = 0 for |t| ≤ 1and F (t) = t for |t| ≥ 2. Setting un(x) = n−1F (nu(x)), it follows fromProposition B.2.1 below that un ∈ W 1,p(Ω). In addition, one verifies easily(see (B.2.1) below) that un → u in W 1,p(Ω) as n→∞. Since suppun ⊂ x ∈Ω; |u(x)| ≥ n−1, suppun is a compact subset of Ω, thus un ∈W 1,p

0 (Ω) by (i)

above; and so u ∈W 1,p0 (Ω). If suppu is unbounded, we approximate u by ξnu

where ξn ∈ C∞c (RN ) is such that ξn(x) = 1 for |x| ≤ n.

(iv) If u ∈ W 1,p0 (Ω) ∩ C(Ω) and if Ω is of class C1, then u|∂Ω = 0 (see [5, The-

orem 9.17, p. 288],). Note that this property is false if Ω is not smoothenough. For example, one can show that if Ω = RN \ 0 and N ≥ 2, thenH1

0 (Ω) = H1(Ω). In particular, if u ∈ C∞c (RN ) and u(0) 6= 0, then u ∈ H10 (Ω)

but u 6= 0 on ∂Ω.(v) Let u ∈ L1

loc(Ω) and define u ∈ L1loc(RN ) by

u(x) =

u(x) if x ∈ Ω,

0 if x 6∈ Ω.

If u ∈ W 1,p0 (Ω), then u ∈ W 1,p(RN ). This is immediate by the definition of

W 1,p0 (Ω). More generally, if u ∈ Wm,p

0 (Ω), then u ∈ Wm,p(RN ). Conversely,if u ∈W 1,p(RN ) and if Ω is of class C1 (as in part (iv) above, the smoothness

assumption on Ω is essential), then u ∈ W 1,p0 (Ω) (see [5, Proposition 9.18,

p. 289]).


Proposition B.1.11. Let 1 ≤ p ≤ ∞ and let u ∈ W 1,p(Ω). Let ω ⊂ Ω be aconnected, open set. If ∇u = 0 a.e. on ω, then there exists a constant c such thatu = c a.e. on ω.

Proof. Let x ∈ ω and let ρ > 0 be such that B(x, ρ) ⊂ ω. We claim thatthere exists c such that u = c a.e. on B(x, ρ). The result follows by Connectedness.To prove the claim, we argue as follows. Let 0 < ε < ρ and let η ∈ C∞c (RN )satisfy η ≡ 1 on B(x, ρ − ε), supp η ⊂ B(x, ρ), and 0 ≤ η ≤ 1. Setting v = ηu,

we deduce that v ∈ W 1,10 (B(x, ρ)) and that ∇v = 0 a.e. on B(x, ρ − ε). We

now extend v by 0 outside B(x, ρ) and we call v the extension. Let (ρn)n≥0 bea smoothing sequence and fix n > 1/(ρ− ε). We have wn = ρn ? v ∈ C∞c (RN ).Furthermore, since ∇wn = ρn ?∇v, and since supp ρn ⊂ B(0, 1/n), it follows that∇wn = 0 on B(x, ρ − ε − 1/n). In particular, there exists cn such that wn ≡ cnon B(x, ρ − ε − 1/n). Since wn → v in L1(RN ), we deduce in particular that forany µ < ρ− ε, there exists c(µ) such that v ≡ c(µ) on B(x, ρ− ε− µ). Therefore,c(µ) is independent of µ and we have v ≡ c on B(x, ρ− ε). for some constant c. Itfollows that c is independent of ε, and the claim follows by letting ε ↓ 0.

Proposition B.1.12. Let u ∈ Wm,∞(RN ) for some m ≥ 0. If Dαu ∈Cb,u(RN ) for all |α| ≤ m, then u ∈ Cmb,u(RN ). In other words, the distributionalderivatives of u are the classical derivatives.

Proof. Let (ρn)n≥0 be a smoothing sequence and set un = ρn ? u. It followsfrom Lemma B.1.9 that un ∈ C∞(RN ) ∩Wm,∞(RN ). Moreover, it is clear thatDαun = ρn ? (Dαu) is uniformly continuous on RN for all |α| ≤ m and n ≥ 0. Thus(un)n≥0 ⊂ Cmb,u(RN ). Moreover, it follows from Lemma B.1.9 that Dαun → Dαu

in L∞(RN ), i.e. un → u in Wm,∞(RN ). Since Cmb,u(RN ) is a Banach space, it is a

closed subset of Wm,∞(RN ), and we deduce that u ∈ Cmb,u(RN ).

We next introduce the local Sobolev spaces.

Definition B.1.13. Given m ∈ N and 1 ≤ p ≤ ∞, we set Wm,ploc (Ω) = u ∈

L1loc(Ω); Dαu ∈ Lploc(Ω) for all |α| ≤ m.

Proposition B.1.14. Let m ∈ N and 1 ≤ p ≤ ∞, let. If u ∈ L1loc(Ω), then the

following properties are equivalent.

(i) u ∈Wm,ploc (Ω).

(ii) u|ω ∈Wm,p(ω) for all ω ⊂⊂ Ω.(iii) φu ∈Wm,p

0 (Ω) (ϕ ∈Wm,∞(Ω) if p =∞) for all φ ∈ C∞c (Ω).

Proof. (i)⇒(ii). This is immediate.(ii)⇒(iii). Suppose u ∈ Wm,p

loc (Ω) and let φ ∈ C∞c (Ω). If ω ⊂⊂ Ω containssuppφ, then φu has compact support in ω. Since u ∈ Wm,p(ω), we know (seeRemark B.1.4 (i)) that φu ∈ Wm,p(ω). If p < ∞, then φu ∈ Wm,p

0 (Ω) by Re-mark B.1.10 (i).

(iii)⇒(i). Suppose φu ∈ Wm,p(Ω) for all φ ∈ C∞c (Ω). Given |α| ≤ m, wedefine uα ∈ Lploc(Ω) as follows. Let ω ⊂⊂ Ω and let φ ∈ C∞c (Ω) satisfy φ(x) = 1for all x ∈ ω. We set (uα)|ω = Dα(φu)|ω and we claim that (uα)|ω is independentof the choice of φ, so that uα is well-defined. Indeed, if ψ ∈ C∞c (Ω) is such thatψ(x) = 1 on ω, then for all ϕ ∈ C∞c (ω),∫

ω

Dα(ψu− φu)ϕ = (−1)|α|∫ω

(ψ − φ)uDαϕ = 0,


so that Dαψu = Dαφu a.e. in ω. It remains to show that uα = Dαu. Indeed, let

|α| ≤ m and ϕ ∈ C |α|c (Ω). Let φ ∈ C∞c (Ω) satisfy φ(x) = 1 on suppϕ. We have

(−1)|α|∫

Ω

uαϕ =

∫Ω

φuDαϕ =

∫Ω

uDαϕ,

and the result follows.

We now introduce the Sobolev spaces of negative index.

Definition B.1.15. Given m ∈ N and 1 ≤ p < ∞, we define W−m,p′(Ω) =

(Wm,p0 (Ω))?, and we denote by 〈·, ·〉W−m,p′ ,Wm,p

0the corresponding duality pair-

ing. For p = 2, we set H−m(Ω) = W−m,2(Ω) = (Hm0 (Ω))? and we denote by

〈·, ·〉H−m,Hm,0the corresponding duality pairing. .

Remark B.1.16. Here are some comments on Definition B.1.15.

(i) It follows from Remark B.1.7 that W−m,p′(Ω) is a Banach space. If p > 1,

then W−m,p′(Ω) is reflexive and separable. H−m(Ω) is a separable Hilbert

space.(ii) Let v ∈ Cm(Ω) ∩ Wm,∞(Ω) and u ∈ W−m,p

′(Ω). Given ϕ ∈ Wm,p

0 (Ω),it follows from Remark B.1.10 (ii) that vϕ ∈ Wm,p

0 (Ω) and ‖vϕ‖Wm,p ≤C‖v‖Wm,∞‖ϕ‖Wm,p . Therefore, one can define vu ∈W−m,p′(Ω) by

〈vu, ϕ〉W−m,p′ ,Wm,p0

= 〈u, vϕ〉W−m,p′ ,Wm,p0

for all ϕ ∈ Wm,p0 (Ω). Moreover, it follows from (B.1.4) that ‖vu‖W−m,p′ ≤

C‖v‖Wm,∞‖u‖W−m,p′(iii) It follows from the dense embedding C∞c (Ω) → Wm,p

0 (Ω) that W−m,p′(Ω) is

a space of distributions on Ω. In particular, we see that 〈u, ϕ〉W−m,p′ ,Wm,p0

=

〈u, ϕ〉D′,D for every u ∈W−m,p′(Ω) and ϕ ∈ C∞c (Ω). Like any distribution, an

element of W−m,p′(Ω) can be localized. Indeed, if u ∈ D′(Ω) and Ω′ is an open

subset of Ω, then one defines u|Ω′ as follows. Given any ϕ ∈ C∞c (Ω′), let ϕ ∈C∞c (Ω) be equal to ϕ on Ω′ and to 0 on Ω \Ω′. Then Ψ(ϕ) = 〈u, ϕ〉D′(Ω),D(Ω)

defines a distribution Ψ ∈ D′(Ω′), and one sets u|Ω′ = Ψ. Note that thisis consistent with the usual restriction of functions. Since ‖ϕ‖Wm,p

0 (Ω′) ≤‖ϕ‖Wm,p

0 (Ω), we see that if u ∈ W−m,p′(Ω), then u|Ω′ ∈ W−m,p

′(Ω′) and

‖u|Ω′‖W−m,p′ (Ω′) ≤ ‖u‖W−m,p′ (Ω).

(iv) An element f ∈ W−m,p′(Ω) being a distribution, it makes sense to say that

f ∈ W−m,p′(Ω) ∩ Lq(Ω) for some 1 ≤ q ≤ ∞. This means that there exists afunction g ∈ Lq(Ω) such that

〈f, ϕ〉W−m,p′ ,Wm,p0

=

∫Ω

gϕ

for all ϕ ∈Wm,p0 (Ω).

Definition B.1.17. Given m ∈ N and 1 ≤ p < ∞, we define W−m,p′

loc (Ω) =

u ∈ D′(Ω); u|ω ∈ W−m,p′(ω) for all ω ⊂⊂ Ω. (See Remark B.1.16 (iii) for the

definition of u|ω.) For p = 2, we set H−mloc (Ω) = W−m,2loc (Ω).

Proposition B.1.18. Let m ∈ N and 1 < p <∞. It follows that Wm,p0 (Ω) →

Lp(Ω) → W−m,p(Ω), with dense embeddings, where the embedding e : Lp(Ω) →W−m,p(Ω) is defined by

eu(ϕ) =

∫Ω

u(x)ϕ(x) dx, (B.1.5)

for all ϕ ∈Wm,p′

0 (Ω) and all u ∈ Lp(Ω).


Proposition B.1.18 is an immediate application of a general abstract result, seeTheorem A.1.5.

Remark B.1.19. Proposition B.1.18 calls for the following comments.

(i) Formula (B.1.5) means that if u ∈ Lp(Ω) and v ∈Wm,p′

0 (Ω), then

〈u, v〉W−m,p,Wm,p′

0

=

∫Ω

uv (B.1.6)

In particular,

〈u, v〉H−m,Hm0 =

∫Ω

uv (B.1.7)

for all u ∈ L2(Ω) and v ∈ Hm0 (Ω).

(ii) Note that any Hilbert space can be identified, via the Riesz representationtheorem, with its dual. By defining the embedding e : L2(Ω) → H−1(Ω)by (B.1.5), we implicitly identify L2(Ω) with its dual. If we identify H1

0 (Ω)with its dual, so that H−1(Ω) = H1

0 (Ω), then Proposition B.1.18 becomesabsurd. This means that we cannot, at the same time, identify L2(Ω) withits dual and H1

0 (Ω) with its dual, and use the canonical embedding H10 (Ω) →

L2(Ω).

Proposition B.1.20. If 1 < p <∞ and −∆ is defined by

〈−∆u, ϕ〉W−1,p,W 1,p′

0

=

∫Ω

∇u · ∇ϕ, (B.1.8)

for all ϕ ∈ W 1,p′

0 (Ω), then −∆ ∈ L(W 1,p(Ω),W−1,p(Ω)). In particular, −∆ ∈L(H1(Ω), H−1(Ω)).

Proof. We note that∣∣∣∫Ω

∇u · ∇ϕ∣∣∣ ≤ ‖∇u‖Lp‖∇ϕ‖Lp′ ≤ ‖u‖W 1,p‖ϕ‖

W 1,p′0

,

for all u ∈ W 1,p0 (Ω), ϕ ∈ W 1,p′

0 (Ω). It follows that (B.1.8) defines an element ofW−1,p(Ω) (note also that this definition is consistent with the classical definition)and that ‖ −∆u‖W−1,p ≤ ‖u‖W 1,p , i.e. −∆ ∈ L(W 1,p(Ω),W−1,p(Ω)).

Corollary B.1.21. Let

J(u) =1

2

∫Ω

|∇u|2, (B.1.9)

for u ∈ H10 (Ω). Then J ∈ C1(H1

0 (Ω),R) and

J ′(u) = −∆u, (B.1.10)

for all u ∈ H10 (Ω).

Proof. We have

J(u+ v)− J(u)− 〈−∆u, v〉H−1,H10

=1

2

∫Ω

|∇v|2,

from which the result follows.

We end this section with a useful density result.

Proposition B.1.22. Let m, j be nonnegative integers and let 1 ≤ p, q < ∞.The following properties hold:

(i) C∞c (Ω) is dense in Wm,p0 (Ω) ∩W j,q

0 (Ω);

(ii) if q > 1, then C∞c (Ω) is dense in Wm,p0 (Ω) ∩W−j,q′(Ω);

(iii) if p, q > 1, then C∞c (Ω) is dense in W−m,p′(Ω) ∩W−j,q′(Ω);

(iv) C∞c (Ω) is dense in Wm,p0 (Ω) ∩ C0(Ω);


(v) if p > 1, then C∞c (Ω) is dense in W−m,p′(Ω) ∩ C0(Ω);

Proof. Let X = Wm,p0 (Ω)∩W j,q

0 (Ω). It follows from Proposition A.1.13 that

X? = W−m,p′(Ω) +W−j,q

′(Ω).

Suppose that f ∈ X? is such that 〈f, ϕ〉X?,X = 0 for all ϕ ∈ C∞c (Ω) and write f =

f1+f2 with f1 ∈W−m,p′(Ω) and f2 ∈W−j,q

′(Ω). We have (see Proposition A.1.13)

〈f, ϕ〉X?,X = 〈f1, ϕ〉W−m,p′ ,Wm,p0

+ 〈f2, ϕ〉W−j,q′ ,W j,q0

= 〈f1, ϕ〉D′,D + 〈f2, ϕ〉D′,D= 〈f1 + f2, ϕ〉D′,D = 〈f, ϕ〉D′,D.

It follows that f = 0 in D′(Ω), hence in X? (see Remark A.1.14). Therefore,C∞c (Ω) is dense in X. This proves property (i), and properties (ii) and (iii) areproved by the same argument. (Note that if p > 1, then Wm,p

0 (Ω) is reflexive; and

so, (W−m,p′(Ω))? = Wm,p

0 (Ω).) Properties (iv) and (v) are also proved by the sameargument, since the dual of C0(Ω) is also a space of distributions (since C∞c (Ω) isdense in C0(Ω)).

B.2. The chain rule and applications

We now study the chain rule, and we begin with a simple result.

Proposition B.2.1. Let F ∈ C1(R,R) satisfy F (0) = 0 and ‖F ′‖L∞ = L <∞,and consider 1 ≤ p ≤ ∞. If u ∈W 1,p(Ω), then F (u) ∈W 1,p(Ω) and

∇F (u) = F ′(u)∇u, (B.2.1)

a.e. in Ω. Moreover, if p < ∞, then the mapping u 7→ F (u) is continuous

W 1,p(Ω) → W 1,p(Ω). Furthermore, if p < ∞ and u ∈ W 1,p0 (Ω), then F (u) ∈

W 1,p0 (Ω).

Proof. We proceed in four steps.

Step 1. The case u ∈ C1c (Ω). It is immediate that F (u) ∈ C1

c (Ω) andthat (B.2.1) holds.

Step 2. The case u ∈ W 1,p0 (Ω). Suppose p < ∞, let u ∈ W 1,p

0 (Ω) and let

(un)n≥0 ⊂ C∞c (Ω) satisfy un → u in W 1,p0 (Ω) as n→∞. By possibly extracting a

subsequence, we may assume that

|un|+ |∇un| ≤ f ∈ Lp(Ω),

and thatun → u, ∇un → ∇u,

a.e. in Ω. It follows from Step 1 that F (un) ∈ C1c (Ω) ⊂ W 1,p

0 (Ω) and that∇F (un) = F ′(un)∇un. In particular,

|∇F (un)| ≤ L|∇un| ≤ Lf.Since F ′(un)∇un → F ′(u)∇u a.e., we obtain ∇F (un)→ F ′(u)∇u in Lp(Ω). More-over, since |F (un) − F (u)| ≤ L|un − u|, we have F (un) → F (u) in Lp(Ω). This

implies that F (un)→ F (u) in W 1,p0 (Ω) and that (B.2.1) holds.

Step 3. The case u ∈ W 1,p(Ω). We have F (u) ∈ Lp(Ω). Furthermore, givenϕ ∈ C1

c (Ω), let ξ ∈ C1c (Ω) satisfy ξ = 1 on suppϕ. By Remark B.1.4 (i) and

Remark B.1.10 (i), we have ξu ∈ W 1,q0 (Ω) for all 1 ≤ q < ∞ such that q ≤ p. It

follows from Step 2 that∫Ω

F (u)∇ϕ =

∫Ω

F (ξu)∇ϕ = −∫

Ω

ϕF ′(ξu)∇(ξu) = −∫

Ω

ϕF ′(u)∇u.

B.2. THE CHAIN RULE AND APPLICATIONS 77

Since clearly F ′(u)∇u ∈ Lp(Ω), we deduce that F (u) ∈ W 1,p(Ω) and that (B.2.1)holds.

Step 4. Continuity. Suppose p < ∞ and un → u in W 1,p(Ω). We showthat F (un) → F (u) in W 1,p(Ω) by contradiction. Thus we assume that ‖F (un) −F (u)‖W 1,p ≥ ε > 0. We have F (un) → F (u) in Lp(Ω). By possibly extractinga subsequence, we may assume that un → u and ∇un → ∇u a.e. It follows bydominated convergence that F ′(un)∇un → F ′(u)∇u in Lp(Ω). Thus F (un)→ F (u)in W 1,p(Ω), which is absurd.

Remark B.2.2. One can prove the following stronger result. If F : R → Ris (globally) Lipschitz continuous and if F (0) = 0, then for every u ∈ W 1,p(Ω),we have F (u) ∈ W 1,p(Ω). Moreover, ∇F (u) = F ′(u)∇u a.e. This formula makessense, since F ′ exists a.e. and ∇u = 0 a.e. on the set x ∈ Ω; u(x) ∈ A whereA ⊂ R is any set of measure 0. Furthermore, the mapping u→ F (u) is continuous

W 1,p(Ω) → W 1,p(Ω) if p < ∞. Finally, if p < ∞ and u ∈ W 1,p0 (Ω), then F (u) ∈

W 1,p0 (Ω). The proof is rather delicate and makes use in particular of Lebesgue’s

points theory. (See [23].) We will establish below a particular case of that result.

Proposition B.2.3. Set u+ = maxu, 0 for all u ∈ R and let 1 ≤ p ≤ ∞. Ifu ∈W 1,p(Ω), then u+ ∈W 1,p(Ω). Moreover,

∇u+ =

∇u if u > 0,

0 if u ≤ 0,(B.2.2)

a.e. If p < ∞, then the mapping u 7→ u+ is continuous W 1,p(Ω) → W 1,p(Ω).

Furthermore, if u ∈W 1,p0 (Ω), then u+ ∈W 1,p

0 (Ω)

Proof. We proceed in four steps.

Step 1. If p < ∞ and u ∈ W 1,p0 (Ω), then u+ ∈ W 1,p

0 (Ω) and (B.2.2) holds.Given ε > 0, let

ϕε(u) =

√ε2 + u2 − ε if u ≥ 0,

0 if u ≤ 0.

It follows from Proposition B.2.1 that ϕε(u) ∈W 1,p0 (Ω) and∇ϕε(u) = ϕ′ε(u)∇u a.e.

We deduce easily that ϕε(u) → u+ and that ∇ϕε(u) converges to the right-hand

side of (B.2.2) in Lp(Ω) as ε ↓ 0. Thus u+ ∈W 1,p0 (Ω) and (B.2.2) holds.

Step 2. If u ∈ W 1,p(Ω), then u+ ∈ W 1,p(Ω) and (B.2.2) holds. Using Step 1,this is proved by the argument in Step 3 of the proof of Proposition B.2.1.

Step 3. If a ∈ R and u ∈W 1,p(Ω), then ∇u = 0 a.e. on the set x ∈ Ω; u(x) =a. Consider a function η ∈ C∞c (R) such that η(x) = 1 for |x| ≤ 1, η(x) = 0 for|x| ≥ 2 and 0 ≤ η ≤ 1. For n ∈ N, n ≥ 1, set

gn(x) = η(n(x− a)),

and

hn(x) =

∫ x

0

gn(s) ds.

It follows from Proposition B.2.1 that hn(u) ∈ W 1,p(Ω) and that ∇hn(u) =gn(u)∇u a.e. Therefore,

−∫Rhn(u)∇ϕ =

∫Rgn(u)ϕ∇u,


for all ϕ ∈ C1c (Ω). Since |hn| ≤ n−1‖η‖L1 , the left-hand side of the above inequality

tends to 0 as n→∞. Therefore,∫Rgn(u)ϕ∇u −→

n→∞0.

Note that gn(u)→ 1x∈Ω;u(x)=a. Since 0 ≤ gn ≤ 1, we deduce that∫R

1x∈Ω;u(x)=aϕ∇u = 0,

for all ϕ ∈ C1c (Ω); and so, 1x∈Ω;u(x)=a∇u = 0 a.e. The result follows.

Step 4. Continuity. Suppose p < ∞ and let un → u in W 1,p(Ω) as n → ∞.We have |u+ − u+

n | ≤ |u − un|, so that u+n → u+ in Lp(Ω). Therefore, we need

only show that for any subsequence, which we still denote by (un)n≥0, there existsa subsequence (unk)k≥0 such that ∇u+

nk→ ∇u+ in Lp(Ω) as k → ∞. We may

extract a subsequence (unk)k≥0 such that unk → u and ∇unk → ∇u a.e., and suchthat

|unk |+ |∇unk | ≤ f ∈ Lp(Ω).

Set

A0 = x ∈ Ω; u(x) = 0,A+ = x ∈ Ω; u(x) > 0, A+

k = x ∈ Ω; unk(x) > 0,A− = x ∈ Ω; u(x) < 0, A−k = x ∈ Ω; unk(x) < 0.

For a.a. x ∈ A+, we have x ∈ A+k for k large, thus ∇u+

nk(x) = ∇unk(x)→ ∇u(x) =

∇u+(x). For a.a. x ∈ A−, we have x ∈ A−k for k large, hence ∇u+nk

(x) = 0 =∇u+(x). For x ∈ A0, we have u(x) = 0, so that by Step 3, ∇u(x) = 0 a.e. Since∇unk → ∇u = 0 a.e. on A0, we deduce in particular that |∇u+

nk| ≤ |∇unk | → 0

a.e. in A0. Thus ∇u+nk→ 0 = ∇u+ a.e. on A0. It follows that

∇u+nk→

∇u if u > 0,

0 if u ≤ 0,

a.e., and the result follows by dominated convergence. This completes the proof.

Remark B.2.4. Let u− = max0,−u. Since u− = (−u)+, we may drawsimilar conclusions for u−. In particular, if u ∈ W 1,p(Ω), then u− ∈ W 1,p(Ω).Moreover,

∇u− =

−∇u if u < 0,

0 if u ≥ 0,(B.2.3)

a.e. If p < ∞, then the mapping u → u− is continuous W 1,p(Ω) → W 1,p(Ω).

Furthermore, if u ∈ W 1,p0 (Ω), then u− ∈ W 1,p

0 (Ω). Note in particular that (B.2.2)and (B.2.3) imply that

∇u+ · ∇u− = 0 (B.2.4)

a.e. Since |u| = u+ + u−, we deduce the following properties. If u ∈W 1,p(Ω), then|u| ∈W 1,p(Ω). Moreover,

∇|u| =

∇u if u > 0,

−∇u if u < 0,

0 if u = 0,

(B.2.5)

a.e. Note in particular that|∇|u| | = |∇u|, (B.2.6)

a.e. If p < ∞, then the mapping u → |u| is continuous W 1,p(Ω) → W 1,p(Ω).

Furthermore, if u ∈W 1,p0 (Ω), then |u| ∈W 1,p

0 (Ω).

B.2. THE CHAIN RULE AND APPLICATIONS 79

Corollary B.2.5. Let 1 ≤ p < ∞, let u ∈ W 1,p(Ω) and v ∈ W 1,p0 (Ω). If

|u| ≤ |v| a.e., then u ∈W 1,p0 (Ω).

Proof. It follows from Remark B.2.4 that |v| ∈ W 1,p0 (Ω). Let (wn)n≥0 ⊂

C∞c (Ω) satisfy wn → |v| in W 1,p(Ω) as n→∞. It follows that wn−u+ → |v| −u+

in W 1,p(Ω), so that (wn − u+)+ → (|v| − u+)+ in W 1,p(Ω) by Proposition B.2.3.Since (wn − u+)+ ≤ w+

n , we see that (wn − u+)+ has compact support; and so

(wn−u+)+ ∈W 1,p0 (Ω). We deduce that (|v|−u+)+ ∈W 1,p

0 (Ω). Since (|v|−u+)+ =

|v| − u+, we see that u+ ∈ W 1,p0 (Ω). One shows as well that u− ∈ W 1,p

0 (Ω), andthe result follows.

Corollary B.2.6. Let 1 ≤ p ≤ ∞ and let M ≥ 0. If u ∈ W 1,p(Ω), then(u−M)+ ∈ u ∈W 1,p(Ω) and

∇(u−M)+ =

∇u if u(x) > M,

0 if u(x) ≤M,(B.2.7)

a.e. in Ω. If p < ∞, then the mapping u 7→ (u −M)+ is continuous W 1,p(Ω) →W 1,p(Ω). Moreover, if u ∈W 1,p

0 (Ω), then (u−M)+ ∈W 1,p0 (Ω).

Proof. The last property is a consequence of Corollary B.2.5, since (u−M)+ ≤u+ ∈ W 1,p

0 (Ω). Next, observe that if Ω is bounded, then the conclusions are aconsequence of Proposition B.2.3, because u−M ∈W 1,p(Ω) whenever u ∈W 1,p(Ω).In particular, we see that for an arbitrary Ω, if u ∈ W 1,p(Ω), then (u −M)+ ∈W 1,p

loc (Ω) and (B.2.7) holds. In particular, |∇(u −M)+| ≤ |∇u| ∈ Lp(Ω). Since(u−M)+ ≤ u+ ∈ Lp(Ω), we see that (u−M)+ ∈W 1,p(Ω) and that (B.2.7) holds.

It now remains to show the continuity of the mapping u 7→ (u −M)+ whenp < ∞. By the above observation, we may assume that Ω is unbounded. GivenR > 0, let ΩR = x ∈ Ω; |x| < R and UR = Ω\ΩR. We argue by contradiction, andwe consider a sequence (un)n≥0 ⊂ W 1,p(Ω) and u ∈ W 1,p(Ω) such that un −→

n→∞u

in W 1,p(Ω) and ‖(un −M)+ − (u−M)+‖W 1,p ≥ ε > 0. Note that

|(un −M)+ − (u−M)+| ≤ |un − u| −→n→∞

0,

in Lp(Ω), so that we may assume ‖∇(un −M)+ − ∇(u −M)+‖Lp ≥ ε > 0. Bypossibly extracting a subsequence, we may also assume that there exists f ∈ Lp(Ω)such that |∇un|+ |∇u| ≤ f a.e. In particular, it follows from (B.2.7) that |∇(un −M)+−∇(u−M)+| ≤ |∇un|+ |∇u| ≤ f a.e. Therefore, by dominated convergence,we may choose R large enough so that

‖∇(un −M)+ −∇(u−M)+‖Lp(UR) ≤ε

4.

Finally, since ΩR is bounded, it follows that ‖∇(un−M)+−∇(u−M)+‖Lp(ΩR) → 0as n→∞. Therefore, for n large enough,

‖∇(un −M)+ −∇(u−M)+‖Lp(ΩR) ≤ε

4.

We deduce that ‖∇(un −M)+ − ∇(u −M)+‖Lp(Ω) ≤ ε/2, which yields a contra-diction. This completes the proof.

Corollary B.2.7. Let 1 ≤ p ≤ ∞, (un)n≥0 ⊂ W 1,p(Ω) and u ∈ W 1,p(Ω).If un → u in W 1,p(Ω) as n → ∞, then there exist a subsequence (unk)k≥0 andv ∈ W 1,p(Ω) such that |unk | ≤ v a.e. in Ω for all k ≥ 0. If, in addition, p < ∞and (un)n≥0 ⊂W 1,p

0 (Ω), then one can choose v ∈W 1,p0 (Ω).


Proof. Let the subsequence (unk)k≥0 satisfy ‖unk − u‖W 1,p ≤ 2−k−1, so that‖unk+1

− unk‖W 1,p ≤ 2−k. It follows from Remark B.2.4 that |unk+1− unk | ∈

W 1,p(Ω) and that ‖ |unk+1− unk | ‖W 1,p ≤ 2−k. Thus, the series

v = |un0|+∑j≥0

|unj+1− unj |,

is normally convergent in W 1,p(Ω). Since

unk+1= un0

+

k∑j=0

(unj+1− unj ),

we see that |unk+1| ≤ v. The result follows, using again Remark B.2.4 in the case

p <∞ and (un)n≥0 ⊂W 1,p0 (Ω).

Corollary B.2.8. Let 1 ≤ p <∞, 0 ≤ A,B ≤ ∞ and set

E = u ∈W 1,p0 (Ω); −A ≤ u ≤ B a.e.,

F = u ∈ C∞c (Ω); −A ≤ u ≤ B.

It follows that E = F , where the closure is in W 1,p0 (Ω). In particular, u ∈

W 1,p0 (Ω); u ≥ 0 a.e. is the closure in W 1,p

0 (Ω) of u ∈ C∞c (Ω); u ≥ 0.

Proof. We have F ⊂ E. Since E is clearly closed in W 1,p0 (Ω), we deduce that

F ⊂ E. We now show the converse inclusion. Let u ∈ E and let (un)n≥0 ⊂ C∞c (Ω)

be such that un → u in W 1,p0 (Ω). Set

vn = max−A,minun, B = un + (un +A)− − (un −B)+.

It follows from Corollary B.2.6 that vn ∈W 1,p0 (Ω) and that

vn −→n→∞

u+ (u+A)− − (u−B)+ = u,

in W 1,p0 (Ω). Thus if (vn)n≥0 ⊂ F , then the conclusion follows. Since clearly

vn ∈ Cc(Ω), we need only show the following property: if w ∈ E ∩ Cc(Ω), thenw ∈ F . To see this, let (ρn)n≥0 be a smoothing sequence and set wn = ρn ? w,where w is the extension of w by 0 outside Ω. Since w has compact support in Ω, wesee that if n is sufficiently large, then wn also has compact support in Ω. Moreover,wn ∈ C∞c (Ω), so that if wn = (wn)|Ω, then wn ∈ C∞c (Ω). In addition, wn → w in

W 1,p(RN ), so that wn → w in W 1,p0 (Ω). It remains to show that −A ≤ wn ≤ B,

which is immediate since −A ≤ w ≤ B. This completes the proof.

B.3. Sobolev’s inequalities

In this section, we establish some Sobolev-type inequalities and embeddings. Itis convenient to make the following definition.

Definition B.3.1. Given an integer m ≥ 0, 1 ≤ p ≤ ∞ and Ω and open subsetof RN , we set

|u|m,p,Ω =∑|α|=m

‖Dαu‖Lp(Ω).

When there is no risk of confusion, we set

|u|m,p = |u|m,p,Ω,

i.e. we omit the dependence on Ω.

We begin with inequalities for smooth functions on RN . The following resultis the main inequality of this section.

B.3. SOBOLEV’S INEQUALITIES 81

Theorem B.3.2 (Gagliardo-Nirenberg’s inequality). Consider 1 ≤ p, q, r ≤ ∞and let j,m be two integers, 0 ≤ j < m. If

1

p− j

N= a

(1

r− m

N

)+

1− aq

, (B.3.1)

for some a ∈ [j/m, 1] (a < 1 if r = N/(m − j) > 1), then there exists a constantC = C(N,m, j, a, q, r) such that

|u|j,p ≤ C|u|am,r‖u‖1−aLq , (B.3.2)

for all u ∈ Cmc (RN ).

The proof of Theorem B.3.2 uses various important inequalities. The funda-mental ingredients are Sobolev’s inequality (Theorem B.3.5), Morrey’s inequality(Theorem B.3.8), and an inequality for intermediate derivatives (Theorem B.3.10).We begin with the following first-order Sobolev inequality.

Theorem B.3.3. Let N ≥ 1. For every u ∈ C1c (RN ), we have

‖u‖L

NN−1≤ 1

2

N∏j=1

∥∥∥ ∂u∂xj

∥∥∥ 1N

L1. (B.3.3)

In particular,

‖u‖L

NN−1≤ 1

2N|u|1,1, (B.3.4)

for all u ∈ C1c (RN ).

Proof. We proceed in three steps.

Step 1. The case N = 1. Given x ∈ R, we have

u(x) =

∫ x

−∞u′(s) ds;

and so,

|u(x)| ≤∫ x

−∞|u′(s)| ds.

As well,

|u(x)| ≤∫ +∞

x

|u′(s)| ds,

so that by summing up the two above inequalities,

|u(x)| ≤ 1

2

∫ +∞

−∞|u′(s)| ds,

which yields (B.3.3) (and (B.3.4)) in the case N = 1.

Step 2. Proof of (B.3.3). We assume N ≥ 2. For any 1 ≤ j ≤ N , it followsfrom Step 1 that

|u(x)| ≤ 1

2

∫R|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| ds;

and so,

|u(x)|N ≤ 2−NN∏j=1

∫R|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| ds.


Taking the (N − 1)th root and integrating on RN , we obtain∫RN|u(x)|

NN−1 dx ≤

2−NN−1

∫RN

N∏j=1

(∫R|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| ds

) 1N−1

.

We observe that the right-hand side is the product of N functions, each of whichdepends only on N−1 of the variables x1, . . . , xN (with a permutation). Therefore,integrating in each of the variables x1, . . . , xN , we may apply Holder’s inequality∫

Ra

1N−1

1 · · · a1

N−1

N−1 ≤N−1∏`=1

(∫Ra`

) 1N−1

.

For example, if we first integrate in x1, we obtain∫Rdx1

N∏j=1

(∫R|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| ds

) 1N−1

=(∫

R|∂1u(s, x2, . . . , xN )| ds

) 1N−1

×∫R

N∏j=2

(∫R|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| ds

) 1N−1

≤(∫

R|∂1u(s, x2, . . . , xN )| ds

) 1N−1

×N∏j=2

(∫R2

|∂ju(x1, . . . , xj−1, s, xj+1, . . . , xN )| dsdx1

) 1N−1

.

Integrating successively in each of the variables x1, . . . , xN , we obtain finally theestimate (B.3.3).

Step 3. Proof of (B.3.4). We claim that if (aj)1≤j≤N ∈ RN with aj ≥ 0, then( N∏j=1

aj

) 1N ≤ 1

N

N∑j=1

aj . (B.3.5)

The estimate (B.3.4) is a consequence of (B.3.3) and (B.3.5). The claim (B.3.5)follows if show that

max|x|2=1

N∏j=1

x2j = N−N . (B.3.6)

To prove (B.3.6), we observe that if the maximum is achieved at x, then there existsa Lagrange multiplier λ ∈ R such that F ′(x) = λx, where F (x) = x2

1 . . . x2N . This

implies that

2xi∏j 6=i

x2j = λxi,

for all 1 ≤ i ≤ N . Since none of the xi vanishes (for the maximum is clearlypositive), this implies that x2

1 = · · · = x2N , from which (B.3.6) follows.

Corollary B.3.4. Let 1 ≤ r ≤ N (r < N if N ≥ 2). If r∗ > r is defined by

1

r∗=

1

r− 1

N,


then

‖u‖Lr∗ ≤ cN,r|u|1,r, (B.3.7)

for every u ∈ C1c (RN ), with cN,r = (N − 1)r/2N(N − r). (We use the convention

that (N − 1)/(N − 1) = 1 if N = 1.)

Proof. The case N = 1 follows from Theorem B.3.3, so we assume N ≥ 2.Let

t =N − 1

Nr∗ =

(N − 1)r

N − r.

Since r ≥ 1, we have t ≥ 1. We observe that

Nt

N − 1= (t− 1)r′ = r∗,

and we apply (B.3.4) with u replaced by |u|t−1u, and we obtain

‖u‖tLr∗ ≤ (2N)−1| |u|t−1u|1,1. (B.3.8)

It follows from (B.2.1) that ∂j(|u|t−1u) = t|u|t−1∂ju for all 1 ≤ j ≤ N . Therefore,by Holder’s inequality,

‖∂j(|u|t−1u)‖L1 ≤ t‖u‖t−1L(t−1)r′‖∂ju‖Lr = t‖u‖t−1

Lr∗‖∂ju‖Lr .

Thus | |u|t−1u|1,1 ≤ t‖u‖t−1Lr∗|u|1,r, and we deduce from (B.3.8) that

‖u‖tLr∗ ≤ (2N)−1t‖u‖t−1Lr∗|u|1,r, (B.3.9)

and (B.3.7) follows.

The following Sobolev’s inequality is now a consequence of Corollary B.3.4.

Theorem B.3.5 (Sobolev’s inequality). Let m ≤ N be an integer, let 1 ≤ r ≤N/m (r < N/m if N ≥ 2), and let r∗ > r be defined by

1

r∗=

1

r− m

N.

If

cN,m,r =[(N − 1)r]m

(2N)m∏

1≤`≤m

(N − `r), (B.3.10)

then

‖u‖Lr∗ ≤ cN,m,r|u|m,r, (B.3.11)

for all u ∈ Cmc (RN ). (We use the convention that (N − 1)/(N − 1) = 1 if N = 1.)

Proof. We argue by induction on m. By Corollary B.3.4, (B.3.11) holds form = 1. Suppose it holds up to some m ≥ 1. We suppose that m + 1 < N and weshow (B.3.11) at the level m+ 1. Let 1 ≤ r < N/(m+ 1) and let r∗ be defined by

1

r∗=

1

r− m+ 1

N.

Define p by1

p=

1

r∗+

1

N=

1

r− m

N, (B.3.12)

so that r < p < r∗. It follows from Corollary B.3.4 and the first identity in (B.3.12)that

‖u‖Lr∗ ≤ cN,p|u|1,p.Next, it follows from the second identity in (B.3.12) and (B.3.11) applied to ∂juthat

‖∂ju‖Lp ≤ cN,m,r|∂ju|m,r,


for all 1 ≤ j ≤ N . We deduce that

|u|1,p ≤ cN,m,r|u|m+1,r,

and (B.3.11) at the level m+1 follows with cN,m+1,r = cN,pcN,m,r, i.e. (B.3.10).

Remark B.3.6. Note that when N ≥ 2, the inequality ‖u‖L∞ ≤ C|u|1,N doesnot hold, for any constant C. Indeed, given 0 < θ < 1 − 1/N , let f ∈ C∞(0,∞)satisfy f(r) = | log r|θ for r ≤ 1/2 and f(r) = 0 for r ≥ 1. Let (fn)n≥1 ⊂C∞([0,∞)) be such that fn(r) = f(r) for r ≥ 1/n and 0 ≤ fn(r) ≤ f(r) and|f ′n(r)| ≤ |f ′(r)| for all r > 0. Setting un(x) = fn(|x|), one verifies easily that‖un‖L∞ → ∞ and lim sup ‖∇un‖LN < ∞ as n → ∞. More generally, a similarexample with 0 < θ < 1 − m/N shows that the inequality ‖u‖L∞ ≤ C|u|m,N/mdoes not hold, for any constant C if 1 ≤ m < N .

The following result, in the same spirit as Theorem B.3.3 (case N = 1) showsthat the inequality ‖u‖L∞ ≤ C|u|N,1 holds in any dimension.

Theorem B.3.7. Given any N ≥ 1,

‖u‖L∞ ≤ 2−N |u|N,1, (B.3.13)

for all u ∈ CNc (RN ).

Proof. Let y ∈ RN Integrating ∂1 · · · ∂Nu in x1 on (−∞, y1) yields

∂2 · · · ∂Nu(y1, x2, . . . , xN ) =

∫ y1

−∞∂1 · · · ∂Nu dx1.

Integrating successively in the variables x2, . . . , xN , we obtain

u(y) =

∫ y1

−∞· · ·∫ yN

−∞∂1 · · · ∂Nu dx1 · · · dxN .

Therefore,

|u(y)| ≤∫ y1

−∞

∫ y2

−∞· · ·∫ yN

−∞|∂1 · · · ∂Nu| dx1 · · · dxN . (B.3.14)

We observe that instead on integrating in x1 on (−∞, y1), we might have integratedon (y1,∞), thus obtaining

|u(y)| ≤∫ ∞y1

∫ y2

−∞· · ·∫ yN

−∞|∂1 · · · ∂Nu| dx1 · · · dxN . (B.3.15)

Summing up (B.3.14) and (B.3.15), we obtain

|u(y)| ≤ (1/2)

∫ ∞−∞

∫ y2

−∞· · ·∫ yN

−∞|∂1 · · · ∂Nu| dx1 · · · dxN .

Rereating this argument for each of the variables, we deduce that

|u(y)| ≤ 2−N∫RN|∂1 · · · ∂Nu| ≤ |u|N,1,

and the result follows since y is arbitrary.

In the case p > N , we have the following result.

Theorem B.3.8 (Morrey’s inequality). If r > N ≥ 1, then there exists aconstant c(N) such that

|u(x)− u(y)| ≤ c(N)r

r −N|x− y|1−Nr |u|1,r, (B.3.16)

for all u ∈ C1c (RN ). Moreover, if 1 ≤ q ≤ ∞ and a ∈ [0, 1) is defined by

0 = a(1

r− 1

N

)+

1− aq

,


then‖u‖L∞ ≤ c(N)

r

r −N|u|a1,r‖u‖1−aLq , (B.3.17)

for all u ∈ C1c (RN ).

Proof. In the following calculations, we denote by c(N) various constantsthat may change from line to line but depend only on N . Let z ∈ RN and ρ > 0,and set B = B(z, ρ). Consider x ∈ B. We have

u(y)− u(y) =

∫ 1

0

d

dtu(x+ t(y − x)) dt =

∫ 1

0

(y − x) · ∇u(x+ t(y − x)) dt,

for all y ∈ B. Integrating on B and dividing by |B|, we obtain

1

|B|

∫B

u(y) dy − u(x) =1

|B|

∫ 1

0

∫B

(y − x) · ∇u(x+ t(y − x)) dy dt.

Since ∣∣∣∫B

(y − x) · ∇u(x+t(y − x)) dy∣∣∣

≤(∫

B

|x− y|r′dy) 1r′(∫

B

|∇u(x+ t(y − x))|r dy) 1r

≤ c(N)ρ1+Nr′ t−

Nr

(∫t(B−x)

|∇u(x+ z)|r dz) 1r

≤ c(N)ρ1+Nr′ t−

Nr ‖∇u‖Lr ,

we deduce

1

|B|

∣∣∣∫ 1

0

∫B

(y − x) · ∇u(x+ t(y − x)) dy dt∣∣∣ ≤ c(N)

r

r −Nρ1−Nr ‖∇u‖Lr .

It follows that if B = B(z, ρ) and x ∈ B, then∣∣∣ 1

|B|

∫B

u(y) dy − u(x)∣∣∣ ≤ c(N)

r

r −Nρ1−Nr ‖∇u‖Lr . (B.3.18)

Let now x1, x2 ∈ RN , x1 6= x2 and let z = (x1 + x2)/2 and ρ = |x1 − x2|. Ap-plying (B.3.18) successively with x = x1 and x = x2 and making the sum, weobtain

|u(x1)− u(x2)| ≤ c(N)r

r −N|x1 − x2|1−

Nr ‖∇u‖Lr ,

which proves (B.3.16).Consider now 1 ≤ q ≤ ∞. We have∣∣∣∫

B

u(y) dy∣∣∣ ≤ |B| 1q′ ‖u‖Lq ;

and so,

1

|B|

∣∣∣∫B

u(y) dy∣∣∣ ≤ |B|− 1

q ‖u‖Lq = N1q γ− 1q

N ρ−Nq ‖u‖Lq

≤ c(N)ρ−Nq ‖u‖Lq .

Therefore, we deduce from (B.3.18) that

|u(x)| ≤ c(N)ρ−Nq ‖u‖Lq + c(N)

r

r −Nρ1−Nr ‖∇u‖Lr .

We now choose ρ = ‖u‖αLq‖∇u‖−αLr , with 1 = α(1−N/r +N/q), and we obtain

|u(x)| ≤ c(N)r

r −N‖∇u‖aLr‖u‖1−aLq .

Since x ∈ RN is arbitrary, this proves (B.3.17).


For the proof of Theorem B.3.2, we will use the following (first-order) Gagliardo-Nirenberg’s inequality, which is a consequence of Sobolev and Morrey’s inequalities.

Theorem B.3.9. Let 1 ≤ p, q, r ≤ ∞ and assume

1

p= a

(1

r− 1

N

)+

1− aq

, (B.3.19)

for some a ∈ [0, 1] (a < 1 if r = N ≥ 2). It follows that there exists a constantC = C(N, p, q, r, a) such that

‖u‖Lp ≤ C|u|a1,r‖u‖1−aLq , (B.3.20)

for every u ∈ C1c (RN ).

Proof. We consider separately several cases.The case r > N . Note that in this case, p ≥ q, so that by Holder’s inequality,

‖u‖Lp ≤ ‖u‖p−qp

L∞ ‖u‖qp

Lq .

Estimating ‖u‖L∞ by (B.3.17), we deduce (B.3.20).The case r < N (thus N ≥ 2). Let r∗ = Nr/(N − r). It follows from

Holder’s inequality that‖u‖Lp ≤ ‖u‖aLr∗‖u‖

1−aLq ,

with a given by (B.3.19). (B.3.20) follows, estimating ‖u‖Lr∗ by (B.3.7).The case r = N . Suppose first N = 1. Then by Holder’s inequality,

‖u‖Lp ≤ ‖u‖aL∞‖u‖1−aLq ,

and the result follows from (B.3.4). In the case N ≥ 2 (thus a < 1) we cannot usethe same argument since ‖u‖L∞ is not estimated in terms of ‖∇u‖LN . Instead, weapply (B.3.4) with u replaced by |u|t−1u for some t ≥ 1. As in the proof of (B.3.9),we obtain

‖u‖tL

tNN−1≤ (2N)−1t‖u‖t−1

L(t−1)NN−1

|u|1,r. (B.3.21)

Suppose first that p ≥ q +N/(N − 1), and let t ≥ 1 be defined by

tN

N − 1= p.

It follows that (t− 1)N/(N − 1) ≥ q. By Holder’s inequality,

‖u‖L

(t−1)NN−1

≤ ‖u‖αLp‖u‖1−αLq , (B.3.22)

withN − 1

(t− 1)N=α(N − 1)

tN+

1− αq

.

It follows from (B.3.21)-(B.3.22) that

‖u‖tLp ≤ (2N)−1t‖u‖(t−1)αLp ‖u‖(t−1)(1−α)

Lq |u|1,r;and so,

‖u‖Lp ≤ (t/2N)1

t−(t−1)α ‖u‖(t−1)(1−α)t−(t−1)α

Lq |u|1

t−(t−1)α

1,r .

Since one verifies easily that

1

t− (t− 1)α=p− qp

= a,(t− 1)(1− α)

t− (t− 1)α=q

p= 1− a,

this yields (B.3.20), since t/2N ≤ p. For p < q + N/(N − 1), we apply Holder’sinequality

‖u‖Lp ≤ ‖u‖3(p−q)

2p

L3q ‖u‖3q−p2p

Lq .

(Note that 3q ≥ q + 2 ≥ p.) We estimate ‖u‖L3q by applying (B.3.20) with p = 3q,and the result follows.


We now study interpolation inequalities for intermediate derivatives.

Theorem B.3.10. Given an integer m ≥ 1, there exists a constant Cm withthe following property. If 0 ≤ j ≤ m and if 1 ≤ p, q, r ≤ ∞ satisfy

m

p=j

r+m− 1

q, (B.3.23)

then for every i ∈ 1, . . . , N,

‖∂ji u‖Lp ≤ Cm‖u‖m−jm

Lq ‖∂mi u‖

jm

Lr , (B.3.24)

for all u ∈ Cmc (RN ). Moreover,

|u|j,p ≤ Cm‖u‖m−jm

Lq |u|jmm,r, (B.3.25)

for all u ∈ Cmc (RN ).

The proof of Theorem B.3.10 is based on the following lemma.

Lemma B.3.11. If 1 ≤ p, q, r ≤ ∞ satisfy

2

p=

1

q+

1

r, (B.3.26)

then

‖u′‖Lp ≤ 8‖u‖12

Lq‖u′′‖

12

Lr , (B.3.27)

for all u ∈ C2c (R).

Proof. We first observe that we need only prove (B.3.27) for r > 1 and p <∞,since the general case can then be obtained by letting p ↑ ∞ or r ↓ 1. Thus we nowassume p <∞ and r > 1. Let 0 ≤ γ ≤ 2 be defined by

γ = 1 +1

p− 1

r. (B.3.28)

so that by (B.3.26),

− γ = −1− 1

q+

1

p. (B.3.29)

We observe that p ≤ 2r by (B.3.26), so that

γ ≥ 1/2. (B.3.30)

We now fix u ∈ C2c (R) and, given any interval I ⊂ R, we set

f(I) = |I|−γp‖u‖pLq(I), (B.3.31)

g(I) = |I|γp‖u′′‖pLr(I). (B.3.32)

We now proceed in five steps.

Step 1. The estimate

‖v′‖Lp(0,1) ≤ 4‖v‖Lq(0,1) + 2‖v′′‖Lr(0,1), (B.3.33)

holds for all v ∈ C2([0, 1]). Let ξ(x) = 1− 2x2 for 0 ≤ x ≤ 1/2, ξ(x) = 2(1− x)2

for 1/2 ≤ x ≤ 1. It follows that ξ ∈ C1([0, 1]) ∩ C2([0, 1] \ 1/2) and 0 ≤ ξ ≤ 1,ξ(0) = 1, ξ′(0) = ξ(1) = ξ′(1) = 0. Moreover, ξ′′(x) = −4 for 0 ≤ x ≤ 1/2,ξ′′(x) = 4 for 1/2 ≤ x ≤ 1. An integration by parts yields∫ 1

0

ξv′′ = −v′(0)− 4

∫ 1/2

0

v + 4

∫ 1

1/2

v;

and so,

|v′(0)| ≤ 4‖v‖L1(0,1) + ‖v′′‖L1(0,1).


Given 0 ≤ x ≤ 1, we deduce that

|v′(x)| ≤ |v′(0)|+∫ x

0

|v′′| ≤ 4‖v‖L1(0,1) + 2‖v′′‖L1(0,1).

x ∈ [0, 1] being arbitrary, we conclude that

‖v′‖L∞(0,1) ≤ 4‖v‖L1(0,1) + 2‖v′′‖L1(0,1).

The estimate (B.3.33) follows by Holder’s inequality.

Step 2. The estimate

‖u′‖Lp(a,b) ≤ 4(b− a)−γ‖u‖Lq(a,b) + 2(b− a)γ‖u′′‖Lr(a,b), (B.3.34)

holds for all −∞ < a < b <∞. Set v(x) = u(a+ (b− a)x), so that v ∈ C2([0, 1]).The estimate (B.3.34) follows by applying estimate (B.3.33) to v, then by us-ing (B.3.28)-(B.3.29).

Step 3. If f and g are defined by (B.3.31)-(B.3.32), then the estimate∫I

|u′|p ≤ 22p−1(2pf(I) + g(I)), (B.3.35)

holds for all finite interval I ⊂ R. This follows from (B.3.34) and the elementaryinequality (x+ y)p ≤ 2p−1(xp + yp).

Step 4. Given any δ > 0, there exist a positive integer ` and disjoints intervalsI1, . . . , I` such that ∪1≤j≤` Ij ⊃ suppu and with the following properties.

` ≤ 1 + |suppu|/δ, (B.3.36)either |Ij | = δ and f(Ij) ≤ g(Ij)

or else |Ij | > δ and f(Ij) = g(Ij),(B.3.37)

for all 1 ≤ j ≤ `. Indeed, set x0 = inf suppu and let I = (x0, x0 + δ). If f(I) ≤g(I), we let I1 = I. If f(I) > g(I), we observe that the functions ϕ(t) = f(x0, x0 +δ + t), ψ(t) = g(x0, x0 + δ + t) satisfy ϕ(0) > ψ(0) and ϕ(t) → 0, ψ(t) → ∞ ast → ∞ (we use (B.3.30)). Thus there exists t > 0 such that ϕ(t) = ψ(t) and welet I1 = (x0, x0 + δ + t). We then see that I1 satisfies (B.3.37). If suppu 6⊂ I1,we can repeat this construction. Since suppu is compact and |Ij | ≥ δ, we obtainin a finite number of steps, say `, a collection of disjoint open intervals Ij that allsatisfy (B.3.37) and such that

∪1≤j≤`−1

Ij ⊂ suppu ⊂ ∪1≤j≤`

Ij ,

which clearly imply (B.3.36).

Step 5. Conclusion. Fix δ > 0. It follows from Step 4 and (B.3.35) that∫R|u′|p ≤ 22p−1

∑j=1

[2pf(Ij) + g(Ij)], (B.3.38)

We let

A1 = j ∈ 1, · · · , `; |Ij | = δ,A2 = j ∈ 1, · · · , `; |Ij | > δ,

so that by (B.3.37)1, · · · , ` = A1 ∪A2. (B.3.39)

If j ∈ A1, then f(Ij) ≤ g(Ij) by (B.3.37), so that

2pf(Ij) + g(Ij) ≤ (2p + 1)g(Ij) ≤ (2p + 1)|Ij |γp‖u′′‖pLr(I)

≤ (2p + 1)δγp‖u′′‖pLr(R),(B.3.40)


where we applied (B.3.32). We deduce from (B.3.40) and (B.3.36) that∑j∈A1

[2pf(Ij) + g(Ij)] ≤ (2p + 1)(1 + |suppu|/δ)δγp‖u′′‖pLr(R). (B.3.41)

If j ∈ A2, then f(Ij) = g(Ij) by (B.3.37). Since

f(Ij)g(Ij) = ‖u‖pLq(Ij)‖u′′‖pLr(Ij)

,

we see that

f(Ij) = g(Ij) = ‖u‖p2

Lq(Ij)‖u′′‖

p2

Lr(Ij), (B.3.42)

for all j ∈ A2. It follows from (B.3.42) that∑j∈A2

[2pf(Ij) + g(Ij)] ≤ (2p + 1)∑j∈A2

‖u‖p2

Lq(Ij)‖u′′‖

p2

Lr(Ij). (B.3.43)

Using (B.3.26) and applying Holder’s inequality for the sum in the right-hand sideof (B.3.43), we deduce that∑

j∈A2

[2pf(Ij) + g(Ij)] ≤ (2p + 1)(∑j∈A2

‖u‖qLq(Ij)) p

2q(∑j∈A2

‖u′′‖rLr(Ij)

) p2r

,

which implies∑j∈A2

[2pf(Ij) + g(Ij)] ≤ (2p + 1)‖u‖p2

Lq(R)‖u′′‖

p2

Lr(R). (B.3.44)

We now deduce from (B.3.38), (B.3.39), (B.3.41) and (B.3.44) that∫R|u′|p ≤ 22p−1(2p + 1)×

[‖u‖p2

Lq(R)‖u′′‖

p2

Lr(R) + (1 + |suppu|/δ)δγp‖u′′‖pLr(R)]. (B.3.45)

Note that by (B.3.28)

γp = 1 + p− p

r> 1,

since r > 1. Letting δ ↓ 0 in (B.3.45) we obtain∫R|u′|p ≤ 22p−1(2p + 1)‖u‖

p2

Lq(R)‖u′′‖

p2

Lr(R).

Since 2p + 1 ≤ 2p+1, we see that 22p−1(2p + 1) ≤ 23p and the estimate (B.3.27)follows by taking the pth root of the above inequality.

Remark B.3.12. The proof of Lemma B.3.11 is fairly technical. Note, however,that some special cases of the inequality (B.3.27) can be established very easily. Forexample, if p = q = r, then setting f = −u′′ + u, we see that u = (1/2)e−|·| ? f ,so that u′ = φ ? f , with φ(x) = (x/2|x|)e−|x|. By Young’s inequality, ‖u′‖Lp ≤‖φ‖L1‖f‖Lp = ‖f‖Lp . Since ‖f‖Lp ≤ ‖u′′‖Lp + ‖u‖Lp , (B.3.27) follows. Anothereasy case is p = 2 (so that r = q′). Indeed, u′2 = (uu′)′ − uu′′, so that∫

u′2 = −∫uu′′ ≤ ‖u′′‖Lr‖u‖Lq ,

by Holder’s inequality, which shows (B.3.27). Note that in both these simple cases,one obtains (B.3.27) with the (better) constant 1.

Proof of Theorem B.3.10. The cases j = 0 and j = m being trivial, weassume 1 ≤ j ≤ m− 1 and we proceed in four steps.

Step 1. If 1 ≤ p, q, r ≤ ∞ satisfy (B.3.26) and i ∈ 1, . . . , N, then

‖∂iu‖Lp(RN ) ≤ 8‖u‖12

Lq(RN )‖∂2i u‖

12

Lr(RN ), (B.3.46)


for all u ∈ C2c (RN ). Indeed, assume first p < ∞ and let x = (x1, . . . , xN ) ∈ RN .

We apply (B.3.27) to the function

v(t) = u(x1, . . . , xi−1, t, xi+1, . . . , xN ),

and we deduce that∫R|v′(t)|p ≤ 8p

(∫R|v(t)|q

) p2q(∫

R|v′′(t)|r

) p2r

.

Integrating on RN−1 in the variables (x1, . . . , xi−1, xi+1, . . . , xN ) and applyingHolder’s inequality to the right-hand side (note that 2q/p + 2r/p = 1), we de-duce (B.3.46). The case p =∞ follows by letting p ↑ ∞ in (B.3.46).

Step 2. If m ≥ 2 and if 1 ≤ p, q, r ≤ ∞ satisfy

m

p=

1

r+m− 1

q, (B.3.47)

and i ∈ 1, . . . , N, then

‖∂iu‖Lp ≤ 82m−3‖u‖m−1m

Lq ‖∂mi u‖

1m

Lr , (B.3.48)

for all u ∈ Cmc (RN ). We argue by induction on m. By Step 1, (B.3.48) holds form = 2. Suppose it holds up to some m ≥ 2. Assume

m+ 1

p=

1

r+m

q, (B.3.49)

and let t be defined bym

t=

1

r+m− 1

p. (B.3.50)

In particular, minp, r ≤ t ≤ maxp, r, so that 1 ≤ t ≤ ∞. Applying (B.3.48) to∂iu, we obtain

‖∂2i u‖Lt ≤ 82m−3‖∂m+1

i u‖1m

Lr‖∂iu‖m−1m

Lp . (B.3.51)

Now, we observe that by (B.3.49) and (B.3.50), 2/p = 1/q + 1/t, so it followsfrom (B.3.46) that

‖∂iu‖Lp ≤ 8‖∂2i u‖

12

Lt‖u‖12

Lq . (B.3.52)

(B.3.51) and (B.3.52) now yield (B.3.48) at the level m+ 1.

Step 3. Proof of (B.3.24). We argue by induction on m ≥ 2. For m = 2, theresult follows from Step 1. Suppose now that up to some m ≥ 2, (B.3.24) holds forall 1 ≤ j ≤ m− 1. Assume 1 ≤ j ≤ m,

m+ 1

p=j

r+m+ 1− j

q, (B.3.53)

and let t be defined bym

p=j − 1

r+m+ 1− j

t. (B.3.54)

We first note that by (B.3.54) and (B.3.53),

m+ 1− jt

=m

m+ 1

m+ 1

p− j − 1

r

=m

m+ 1

m+ 1− jq

+m+ 1− j(m+ 1)r

≥ 0,

so that 0 ≤ t ≤ ∞. Also, by the above identity, and since q, r ≥ 1,

m+ 1− jt

=m

m+ 1

m+ 1− jq

+m+ 1− j(m+ 1)r

≤ m(m+ 1− j)m+ 1

+m+ 1− jm+ 1

= m+ 1− j,


so that t ≥ 1. Applying (B.3.24) (with j replaced by j − 1) to ∂iu, we obtain

‖∂ji u‖Lp ≤ Cm‖∂m+1i u‖

j−1m

Lr ‖∂iu‖m−j+1m

Lt . (B.3.55)

Now, we observe that by (B.3.53) and (B.3.54), (m+1)/t = 1/r+m/q, so it followsfrom (B.3.48) (applied with m replaced by m+ 1) that

‖∂iu‖Lt ≤ 82m−1‖∂m+1i u‖

1m+1

Lr ‖u‖mm+1

Lq . (B.3.56)

(B.3.55) and (B.3.56) now yield (B.3.24) at the level m+ 1.

Step 4. Proof of (B.3.25). We note that by (B.3.48),

|u|1,p ≤ 82m−3‖u‖m−1m

Lq |u|1mm,r,

whenever (B.3.47) holds. The proof is now parallel to the proof of estimate (B.3.24)in Step 3 above.

Proof of Theorem B.3.2. We proceed in three steps.

Step 1. The case (m− j)r < N . Let t be defined by

1

t=

1

r− m− j

N, (B.3.57)

so that r < t < ∞. It follows from Sobolev’s inequality (B.3.11) applied to jth

derivatives of u that|u|j,t ≤ C|u|m,r. (B.3.58)

Next, let s be defined bym

s=j

r+m− 1

q, (B.3.59)

so that minq, r ≤ s ≤ maxq, r. It follows from the interpolation inequal-ity (B.3.25) that

|u|j,s ≤ C‖u‖m−jm

Lq |u|jmm,r. (B.3.60)

It follows from (B.3.1), (B.3.57) and (B.3.59) that

1

p=θ

t+

1− θs

,

with

θ =ma− jm− j

.

Since j/m ≤ a ≤ 1, we see that 0 ≤ θ ≤ 1, and we deduce from Holder’s inequalitythat

|u|j,p ≤ |u|θj,t|u|1−θj,s ≤ C|u|θm,r(‖u‖

m−jm

Lq |u|jmm,r)

1−θ,

where we used (B.3.58) and (B.3.60). The estimate (B.3.2) follows.

Step 2. The case (m− j)r ≥ N and a = 1. Note that if a = 1, then by (B.3.1),

1

p=

1

r− m− j

N≤ 0.

The only possibility is (m− j)r = N and p =∞. This is allowed only if r = 1, andthe result is then a consequence of Theorem B.3.7.

Step 3. The case (m− j)r ≥ N and a < 1. Let t be defined by

m

t=j

r+m− 1

q, (B.3.61)

so that minq, r ≤ t ≤ maxq, r. It follows from the interpolation inequal-ity (B.3.25) that

|u|j,t ≤ C‖u‖m−jm

Lq |u|jmm,r. (B.3.62)


Next, let s be defined by

m− js

=1

r+m− j − 1

p, (B.3.63)

so that minp, r ≤ s ≤ maxp, r. It follows from the interpolation inequal-ity (B.3.25) applied to jth order derivatives of u that

|u|j+1,s ≤ C|u|1

m−jm,r |u|

m−j−1m−j

j,p . (B.3.64)

Next, let α ∈ [0, 1) be defined by

α =(m− j)(a− j/m)

1− a+ (m− j)(a− j/m), (B.3.65)

so that by (B.3.61), (B.3.63) and (B.3.1)

1

p= α

(1

s− 1

N

)+

1− αt

.

It follows from Theorem B.3.9 applied to jth order derivatives of u that

|u|j,p ≤ C|u|αj+1,s|u|1−αj,t . (B.3.66)

We deduce from (B.3.66), (B.3.64) and (B.3.62) that

|u|j,p ≤ C|u|α+(j/m)(1−α)(m−j)α+(1−α)(m−j)

m,r ‖u‖(1−j/m)(1−α)(m−j)α+(1−α)(m−j)

Lq .

Since by (B.3.65),

a =α+ (j/m)(1− α)(m− j)

α+ (1− α)(m− j),

this yields (B.3.2).

Corollary B.3.13 (Gagliardo-Nirenberg’s inequality). Let Ω ⊂ RN be anopen subset. Let 1 ≤ p, q, r ≤ ∞ and let j,m be two integers, 0 ≤ j < m. Assumethat (B.3.1) holds for some a ∈ [j/m, 1] (a < 1 if r = N/(m− j) > 1), and supposefurther that r < ∞. It follows that Dαu ∈ Lp(Ω) for all u ∈ Wm,r

0 (Ω) ∩ Lq(Ω) if|α| = j. Moreover, the inequality (B.3.2) holds for all u ∈Wm,r

0 (Ω) ∩ Lq(Ω).

Proof. We first consider the case Ω = RN . Let u ∈ Wm,r(RN ) ∩ Lq(RN )and let (un)n≥0 ⊂ C∞c (RN ) be the sequence constructed by regularization andtruncation in the proof of Theorem B.1.8, so that

un −→n→∞

u in Wm,r(RN ) and ‖un‖Lq ≤ ‖u‖Lq . (B.3.67)

Applying (B.3.2) to un − u`, we obtain

|un − u`|j,p ≤ C|un − u`|am,r‖un − u`‖1−aLq . (B.3.68)

Let α be a multi-index with |α| = j. It follows from (B.3.67)-(B.3.68) that Dαun isa Cauchy sequence in Lp(RN ). Thus Dαun has a limit v in Lp(RN ). In particular,∫

RNDαunϕ −→

n→∞

∫RN

vϕ,

for all ϕ ∈ Cjc (RN ). Since∫RN

Dαunϕ = (−1)j∫RN

unDαϕ −→

n→∞(−1)j

∫RN

uDαϕ,

by (B.3.67), we see that Dαu = v ∈ Lp(RN ) and

|un − u|j,p −→n→∞

0, (B.3.69)


which proves the first part of the result. Finally, we apply the inequality (B.3.2) toun. Letting n→∞ and using (B.3.67) and (B.3.69), we deduce that (B.3.2) holdsfor u.

In the general case Ω ⊂ RN , the result follows by extending u ∈Wm,p0 (Ω) by 0

outside Ω (see Remark B.1.10 (v)) and applying the result in RN .

We are now in a position to state and prove the Sobolev embedding theorems.We restrict ourselves to functions of Wm,p

0 (Ω). Similar statements hold for functionsof Wm,p(Ω), but they are obtained by using extension operators, so they require acertain amount of regularity of the domain. For functions of Wm,p

0 (Ω), instead, noregularity assumption on Ω is necessary. Furthermore, these results are sufficientfor our purpose. Our first result in this direction is the following.

Theorem B.3.14. Let Ω ⊂ RN be an open subset, let 1 ≤ r < ∞ and letm ∈ N, m ≥ 1.

(i) If mr < N , then Wm,r0 (Ω) → Lp(Ω) for all p such that r ≤ p ≤ Nr/(N −mr).

(ii) If m = N and r = 1, then Wm,r0 (Ω) → Lp(Ω) for all p such that r ≤ p ≤ ∞.

Moreover, Wm,r0 (Ω) → C0(Ω).

(iii) If mr = N and r > 1, then Wm,r0 (Ω) → Lp(Ω) for all p such that r ≤ p <∞.

(iv) If mr > N , then Wm,r0 (Ω) → Lp(Ω) for all p such that r ≤ p ≤ ∞. Moreover,

Wm,r0 (Ω) → C0(Ω).

Proof. The result follows from Corollary B.3.13 by taking j = 0, q = r anda = N(p− r)/mpr, except for the embeddings Wm,r

0 (Ω) → C0(Ω) in (ii) and (iv).These last embeddings follow from the density of C∞c (Ω) in Wm,r

0 (Ω) and theembedding Wm,r

0 (Ω) → L∞(Ω).

The next result is the general case of Sobolev’s embedding for functions ofWm,p

0 (Ω).

Theorem B.3.15. Let Ω ⊂ RN be an open subset, let 1 ≤ r < ∞ and letm, j ∈ N, m ≥ 1.

(i) If mr < N , then Wm+j,r0 (Ω) → W j,p

0 (Ω) for all p such that r ≤ p ≤Nr/(N −mr).

(ii) If m = N and r = 1, then Wm+j,r0 (Ω) → W j,p

0 (Ω) ∩W j,∞(Ω) for all p such

that r ≤ p <∞. Moreover, Wm+j,r0 (Ω) → Cj0(Ω).

(iii) If mr = N and r > 1, then Wm+j,r0 (Ω) → W j,p

0 (Ω) for all p such thatr ≤ p <∞.

(iv) If mr > N , then Wm+j,r0 (Ω) → W j,p

0 (Ω) ∩ W j,∞(Ω) for all p such that

r ≤ p <∞. Moreover, Wm+j,r0 (Ω) → Cj0(Ω).

Proof. We first prove (iv). Applying Theorem B.3.14 (iv) to Dαu with

|α| ≤ j, we deduce that Wm+j,r0 (Ω) → W j,p(Ω) for all r ≤ p ≤ ∞. The em-

bedding Wm+j,r0 (Ω) → W j,p

0 (Ω) if r ≤ p < ∞ follows from the density of C∞c (Ω)

in Wm+j,r0 (Ω) and the embedding Wm+j,r

0 (Ω) → W j,p(Ω). Next, the embedding

Wm+j,r0 (Ω) → Cj0(Ω) follows from the density of C∞c (Ω) in Wm+j,r

0 (Ω) and the

embedding Wm+j,r0 (Ω) → W j,∞(Ω). The proofs of (i), (ii) and (iii) are similar,

using properties (i), (ii) and (iii) of Theorem B.3.14, respectively.

We now apply Morrey’s inequality to obtain embedings in spaces of the typeCj,α(Ω).

Theorem B.3.16. Let Ω ⊂ RN be an open subset, let 1 ≤ r < ∞. Let m ≥ 1be the smallest integer such that mr > N . It follows that for all integers j ≥ 0,Wm+j,r

0 (Ω) → Cj0(Ω)∩Cj,α(Ω) with α = m−(N/r) if (m−1)r < N , α any numberin (0, 1) if (m− 1)r = N .


Proof. Let u ∈ C∞c (Ω). It follows from Theorem B.3.15 (iv) that

‖u‖W j,∞ ≤ C‖u‖Wm+j,r . (B.3.70)

Let α be a multi-index with |α| = j. Setting v = Dαu, we see that

‖v‖Wm,r ≤ ‖u‖Wm+j,r . (B.3.71)

Let p = Nr

N−(m−1)r if (m− 1)r < N,

maxr,N < p <∞ if (m− 1)r = N,

so that p > N . It follows from Theorem B.3.15 (i) and (B.3.71) that

‖v‖W 1,p ≤ C‖v‖Wm,r ≤ C‖u‖Wm+j,r . (B.3.72)

Finally, we deduce from (B.3.72) and Morrey’s inequality (B.3.16) that

|v(x)− v(y)| ≤ C|x− y|1−Np ‖u‖Wm+j,r ,

for all x, y ∈ RN . Applying (B.3.70), we conclude that ‖u‖Cj,α ≤ C‖u‖Wm+j,r ,with α = 1− (N/p), which is the desired estimate.

The following two results are applications of Sobolev’s embedding theorems.

Corollary B.3.17. Given any 1 ≤ r ≤ ∞, ∩m≥0

Wm,rloc (Ω) = C∞(Ω).

Proof. It is clear that C∞(Ω) ⊂Wm,rloc (Ω) for all m ≥ 0. Conversely, suppose

u ∈ Wm,rloc (Ω) for all m ≥ 0. Let ω ⊂⊂ Ω and let ϕ ∈ C∞c (Ω) satisfy ϕ(x) = 1 for

x ∈ ω. It follows from Proposition B.1.14 that v = ϕu ∈ Wm,10 (Ω) for all m ≥ 0,

so that v ∈ C∞(Ω) by Theorem B.3.15. Thus u ∈ C∞(ω) and the result follows,since ω is arbitrary.

Proposition B.3.18. Let 1 ≤ p ≤ ∞, m ∈ N, m ≥ 1 and u ∈ Wm,ploc (Ω). If

Dαu ∈ C(Ω) for all multi-index α with |α| = m, then u ∈ Cm(Ω).


Step 1. u ∈ C(Ω). Suppose u ∈ Lq0loc(Ω) for some q0 ≤ N and let q0 ≤ q1 <∞satisfy

1

q1≥ 1

q0− 1

N.

Let ϕ ∈ C∞c (Ω) and set v = ϕu, so that v ∈ Lq0(Ω). Since ∇v = ϕ∇u + u∇ϕ,we deduce that ∇v ∈ Lq0(Ω). v being compactly supported in Ω, it follows that

v ∈ W 1,q00 (Ω) (see Remark B.1.10 (i)). Applying Theorem B.3.14, we see that

v ∈ Lq1(Ω) and, since ϕ is arbitrary, we deduce that u ∈ Lq1loc(Ω). We now iterate theabove argument and, starting from q0 = 1, we construct q0 < · · · < qk−1 ≤ N < qksuch that u ∈ Lqjloc(Ω) for 0 ≤ j ≤ k. Finally, let ϕ ∈ C∞c (Ω) and set v = ϕu. Wesee as above that v ∈W 1,qk(Ω), and it follows from Theorem B.3.14 that v ∈ C(Ω).Since ϕ is arbitrary, we conclude that u ∈ C(Ω).

Step 2. The case m = 1. It follows from Step 1 that u ∈ C(Ω). Since ∇u ∈C(Ω) by assumption, we have in particular u ∈ W 1,∞

loc (Ω). Let ω ⊂⊂ Ω and letϕ ∈ C∞c (Ω) satisfy ϕ(x) = 1 for x ∈ ω. Set v = ϕu, so that v ∈ W 1,∞(Ω) byProposition B.1.14. Since v is supported in a compact subset of Ω, it follows thatif

w(x) =

v(x) if x ∈ Ω,

if x ∈ RN \ Ω,


then w ∈W 1,∞(RN ) ∩ C(RN ). Moreover, one verifies easily that

∇w =

u∇ϕ+ ϕ∇u in Ω,

0 in RN \ Ω,

so that ∇w ∈ C(RN ). Since w and ∇w have compact support, we see that w,∇w ∈Cb,u(RN ). Applying Proposition B.1.12, we deduce that w ∈ C1(RN ), and sincew = u in ω, it follows that u ∈ C1(ω). Hence the result, since ω is arbitrary.

Step 3. The case m ≥ 2. We proceed by induction on m. By Step 2, the resultholds for m = 1. Suppose it holds up to some m ≥ 1. Let u ∈ Wm+1,p

loc (Ω) satisfyDαu ∈ C(Ω) for all multi-index α with |α| = m + 1. Consider 1 ≤ j ≤ N andset v = ∂ju. It follows that v ∈ Wm,p

loc (Ω) and Dαu ∈ C(Ω) for all multi-indexα with |α| = m. Applying the result at the level m, we deduce that v ∈ Cm(Ω).Thus ∇u ∈ Cm(Ω). In particular, ∇u ∈ C(Ω) and we deduce from Step 2 thatu ∈ C1(Ω). Since ∇u ∈ Cm(Ω), we conclude that u ∈ Cm+1(Ω).

If |Ω| <∞, then ‖u‖Lr is dominated only in terms of ‖∇u‖Lr for functions of

W 1,r0 (Ω). This is the object of the following result.

Theorem B.3.19 (Poincare’s inequality). If |Ω| < ∞ and 1 ≤ r < ∞, thenthere exists a constant C = C(N, r) (independent of u and Ω) such that

‖u‖Lr ≤ C|Ω|1N ‖∇u‖Lr , (B.3.73)

for every u ∈W 1,r0 (Ω).

Proof. Let p = r(N+r)/N , so that by (B.3.20), ‖u‖Lp ≤ C‖∇u‖NN+r

Lr ‖u‖r

N+r

Lr .

Since ‖u‖Lr ≤ |Ω|1

N+r ‖u‖Lp by Holder’s inequality, the result follows.

Corollary B.3.20. Let 1 ≤ r < ∞, and suppose |Ω| < ∞. Then ‖u‖ =

‖∇u‖Lr defines an equivalent norm on W 1,r0 (Ω).

Remark B.3.21. Note that inequality (B.3.73) means that

inf‖∇u‖Lp ; u ∈W 1,p0 (Ω), ‖u‖Lp = 1 > 0.

Note that inequality (B.3.73) may fail if |Ω| = ∞. For instance, if Ω = RN , thenthe above infimum is 0, so that (B.3.73) fails. To see that, let u ∈ W 1,p(RN ),

‖u‖Lp = 1. Given λ > 0, let uλ ∈ W 1,p(RN ) be defined by uλ(x) = λNp u(λx).

Elementary calculations show that ‖uλ‖Lp = 1 and ‖∇uλ‖Lp = λ‖∇u‖Lp . Lettingλ ↓ 0, we see that the above infimum is 0.

We end this section with a result concerning the embedding of Lp spaces innegative order Sobolev spaces.

Corollary B.3.22. Suppose Ω ⊂ RN is an open set. Let 1 < r < ∞ and letm ≥ 1 be an integer. If

p =

∞ if mr ≥ N,Nr

N−mr if mr < N,

then Lp′(Ω) → W−m,r

′(Ω) with dense embedding for all r ≤ p ≤ p (and p < ∞ if

mr = N). If, in addition, |Ω| <∞, then the same property also holds for 1 ≤ p < r.

Proof. The last part of the result (the case |Ω| <∞) follows from the denseembedding Lp(Ω) → Lq(Ω) if 1 ≤ p ≤ q < ∞. The first part of the result followsfrom Theorem B.3.14 and Proposition A.1.5, except for the case p = ∞ (thus


mr > N), since L1(Ω) is not the dual of L∞(Ω). In this case, we argue directly asfollows. It follows from Theorem B.3.14 that Wm,r

0 (Ω) → L∞(Ω). Define

eu(ϕ) =

∫Ω

uϕ,

for all u ∈ L1(Ω) and ϕ ∈Wm,r0 (Ω). We have

|eu(ϕ)| ≤ ‖u‖L1‖ϕ‖L∞ ≤ ‖u‖L1‖ϕ‖Wm,r0

,

so that e defines a mapping L1(Ω) → W−m,r′(Ω). This mapping is injective,

because if (eu, ϕ)W−m,r′ ,Wm,r0

= 0 for all ϕ ∈Wm,r0 (Ω), then in particular

∫Ωuϕ = 0

for all ϕ ∈ C∞c (Ω), which implies u = 0. It remains to show that the embedding

e : L1(Ω) → W−m,r′(Ω) is dense. To prove this, we observe that by the density

of C∞c (Ω) in Lr′(Ω) and of Lr

′(Ω) in W−m,r

′(Ω) (see just above), it follows that

C∞c (Ω) is dense in in W−m,r′(Ω). The result follows, since C∞c (Ω) ⊂ L1(Ω).

B.4. Compactness properties

We now study the compact embeddings of W 1,r0 (Ω). We begin with a local

compactness result in RN .

Proposition B.4.1. Let 1 ≤ r < ∞, and suppose K be a bounded subset ofW 1,r(RN ). It follows that for every R < ∞, the set KR := u|BR ; u ∈ K isrelatively compact in Lr(BR), where BR = B(0, R).


Step 1. If (ρn)n≥1 is a smoothing sequence, then

‖u− ρn ? u‖Lr ≤C

n‖∇u‖Lr , (B.4.1)

for all u ∈ W 1,r(RN ), where C =(∫

RN |y|rρ(y) dy

) 1r . By density, we need only

show (B.4.1) for u ∈ C∞c (RN ). We claim that∫RN|u(x− y)− u(x)|r dx ≤ |y|r‖∇u‖rLr . (B.4.2)

Indeed,

u(x− y)− u(x) =

∫ 1

0

d

dtu(x− ty) dt =

∫ 1

0

y · ∇u(x− ty) dt;

and so,

|u(x− y)− u(x)| ≤ |y|∫ 1

0

|∇u(x− ty)| dt ≤ |y|(∫ 1

0

|∇u(x− ty)|r dt) 1r

.

(B.4.2) follows after integration in x. Next, since ‖ρn‖L1 = 1,

ρn ? u(x)− u(x) =

∫RN

ρn(y)(u(x− y)− u(x)) dy

=

∫RN

ρn(y)r−1r [ρn(y)

1r (u(x− y)− u(x))] dy.

By Holder’s inequality, we deduce

|ρn ? u(x)− u(x)|r ≤∫RN

ρn(y)|u(x− y)− u(x)|r dy.

Integrating the above inequality on RN and applying (B.4.2), we find

‖ρn ? u− u‖rLr ≤ ‖∇u‖rLr∫RN|y|rρn(y) dy.

B.4. COMPACTNESS PROPERTIES 97

Hence (B.4.1).

Step 2. If (ρn)n≥1 is as in Step 1, then

‖ρn ? u‖W 1,∞ ≤ nNr ‖ρ‖Lr′‖u‖W 1,r , (B.4.3)

for all u ∈ W 1,r(RN ). Since ∇(ρn ? u) = ρn ?∇u by Lemma B.1.9, it follows fromYoung’s inequality that

‖ρn ? u‖W 1,∞ ≤ ‖ρn‖Lr′‖u‖W 1,r ,

and the result follows.

Step 3. Conclusion. Let R > 0, let KR be as in the statement of theproposition, and let ε > 0. Given n ≥ 1, set Kn = ρn ? u; u ∈ K andKnR = u|BR ; u ∈ Kn. Fix n large enough so that

supu∈K‖u− ρn ? u‖Lr ≤

ε

2. (B.4.4)

Such a n exists by (B.4.1). It follows from (B.4.3) that Kn is a set of uniformly Lip-schitz continuous functions on RN . By Ascoli’s theorem, Kn

R is relatively compactin L∞(BR), thus in Lr(BR). Therefore, Kn

R can be covered by a finite number ofballs of radius ε/2 in Lr(BR). By (B.4.4), we see that KR can be covered by a finitenumber of balls of radius ε. Since ε > 0 is arbitrary, this shows compactness.

Corollary B.4.2. Let Ω ⊂ RN be an open subset, let 1 ≤ r < ∞ and let(un)n≥0 be a bounded sequence of W 1,r

0 (Ω). There exist u ∈ Lr(Ω) and a subse-quence (unk)k≥0 such that unk → u a.e. in Ω and in Lr(Ω ∩BR), for any R <∞,as k →∞.

Proof. We first consider the case Ω = RN . It follows from Proposition B.4.1applied with R = 1 that there exist n(1, k) → ∞ as k → ∞ and w1 ∈ Lr(B1)such that un(1,k) → w1 in Lr(B1) and a.e. in B1. We now apply Proposition B.4.1with R = 2 to the sequence (un(1,k))k≥0. It follows that there exist a subsequencen(2, k) → ∞ as k → ∞ and w2 ∈ Lr(B2) such that un(2,k) → w2 in Lr(B2) anda.e. in B2. By recurrence, we construct n(`, k) → ∞ as k → ∞ and (w`)`≥1

with w` ∈ Lr(B`) such that un(`,k) → w` in Lr(B`) and a.e. in B`. Moreover,(n(`, k))k≥0 is a subsequence of (n(m, k))k≥0 for ` > m, i.e. for every k ≥ 0, thereexists k′ ≥ k such that n(`, k) = n(m, k′). We set nk = n(k, k). Since n(k, k) is asubsequence of n(`, k) for any ` ≥ 1, we see that unk → w` in Lr(B`) and a.e. inB`. In particular, w` ≡ wm on Bm if ` ≥ m. We now set u ≡ w` on Bm, for ` ≥ m.We have unk → u in Lr(BR) and a.e. in BR, for any R <∞. In particular,

‖u‖Lr(BR) = limk→∞

‖unk‖Lr(BR) ≤ lim supn→∞

‖un‖Lr(RN ).

We deduce that u ∈ Lr(RN ). Since unk → u a.e. in BR for any R < ∞, weconclude that unk → u a.e. in RN .

We now consider the case of an arbitrary domain Ω ⊂ RN . Let (un)n≥0 be asabove and set

un(x) =

un(x) if x ∈ Ω,

0 if x ∈ RN \ Ω,

so that the sequence (un)n≥0 is bounded in W 1,r(RN ). (See Remark B.1.10 (v).)It follows from what precedes that there exist u ∈ Lr(RN ), supported in Ω, anda subsequence (unk)k≥0 such that unk → u as k → ∞ in Lr(BR) for any R < ∞and a.e. in RN . The result now follows by setting u = u|Ω. This completes theproof.


Lemma B.4.3. Let Ω be an open subset of RN . Let 1 ≤ r ≤ ∞, let (un)n≥0 bea bounded sequence of Lr(Ω) and let u ∈ L1

loc(Ω). Suppose that∫Ω

unϕ −→n→∞

∫Ω

uϕ, (B.4.5)

for all ϕ ∈ C∞c (Ω) (which is satisfied in particular if un → u in L1(ω) for everyω ⊂⊂ Ω). Then u ∈ Lr(Ω) and

‖u‖Lr ≤ lim infn→∞

‖un‖Lr . (B.4.6)

Moreover, if r > 1 then (B.4.5) holds for all ϕ ∈ Lr′(Ω). In addition, if 1 < r <∞and if ‖un‖Lr → ‖u‖Lr as n→∞, then un → u in Lr(Ω).

Suppose further that 1 < r < ∞ and that (un)n≥0 is a bounded sequence of

W 1,r0 (Ω). It follows that u ∈W 1,r

0 (Ω),∫Ω

∇unϕ −→n→∞

∫Ω

∇uϕ, (B.4.7)

for all ϕ ∈ Lr′(Ω) and

‖∇u‖Lr ≤ lim infn→∞

‖∇un‖Lr . (B.4.8)

If in addition ‖∇un‖Lr → ‖∇u‖Lr as n→∞, then ∇un → ∇u in Lr(Ω).

Proof. We claim that for all u ∈ L1loc(Ω),

‖u‖Lr = sup∣∣∣∫

Ω

uϕ∣∣∣; ϕ ∈ C∞c (Ω), ‖ϕ‖Lr′ = 1

. (B.4.9)

If r > 1, this is immediate because Lr′(Ω)? = Lr(Ω) and C∞c (Ω) is dense in

Lr′(Ω). Suppose now r = 1 and suppose u 6= 0 (the case u = 0 is immediate). Fix

0 < M < ‖u‖L1 ≤ ∞. There exists a compact set K ⊂ Ω such that∫K

|u| > M.

Let

h(x) =

u(x)|u(x)| if u(x) 6= 0 and x ∈ K,0 if u(x) = 0 or x 6∈ K.

We have h ∈ L∞(Ω), ‖h‖L∞ = 1. Moreover, h has compact support in Ω and∫Ω

uh =

∫K

|u| > M.

Let (ρn)n≥0 be a smoothing sequence and set hn = (ρn ? h)|Ω. For n large enough,we have hn ∈ C∞c (Ω). Moreover, up to a subsequence, hn → h a.e. In addition,‖hn‖L∞ ≤ ‖h‖L∞ = 1. By dominated convergence, we deduce∫

Ω

uhn −→n→∞

∫Ω

uh > M ;

and so,

sup∣∣∣∫

Ω

uϕ∣∣∣; ϕ ∈ C∞c (Ω), ‖ϕ‖Lp′ = 1

≥M.

Since M < ‖u‖L1 is arbitrary, we deduce

sup∣∣∣∫

Ω

uϕ∣∣∣; ϕ ∈ C∞c (Ω), ‖ϕ‖Lp′ = 1

≥ ‖u‖L1 .

The converse inequality being immediate, (B.4.9) follows. Now, since∣∣∣∫Ω

unϕ∣∣∣ ≤ ‖un‖Lr‖ϕ‖Lr′ ,

B.4. COMPACTNESS PROPERTIES 99

(B.4.6) follows from (B.4.5) and (B.4.9). The fact that if r > 1, then (B.4.5) holds

for all ϕ ∈ Lr′(Ω) follows by density of C∞c (Ω) in Lr′(Ω).

Suppose now that 1 < r < ∞, that ‖un‖Lr → ‖u‖Lr as n → ∞ and let usshow that un → u in Lr(Ω). If ‖u‖Lr = 0, then the result is immediate. Therefore,we may assume that ‖u‖Lr 6= 0, so that also ‖un‖Lr 6= 0 for n large. Let thenu = ‖u‖−1

Lr u and un = ‖un‖−1Lr un. It follows that

‖un‖Lr = ‖u‖Lr = 1.

Furthermore, (B.4.5) is satisfied with u and un replaced by u and un. Setting w =2u and wn = u+un, we deduce that (B.4.5) is satisfied with u and un replaced by wand wn. In particular, it follows from what precedes that ‖w‖Lr ≤ lim inf ‖wn‖Lr .Since ‖w‖Lr = 2 and ‖wn‖Lr ≤ ‖u‖Lr + ‖un‖Lr = 2, it follows that

‖wn‖Lr −→n→∞

2.

If r ≥ 2, we have Clarkson’s inequality (see e.g. [17])

‖un − u‖rLr ≤ 2r−1(‖u‖rLr + ‖un‖rLr )− ‖u+ un‖rLr .Therefore, ‖un − u‖Lr → 0, from which it follows that un → u in Lr(Ω). In thecase r ≤ 2, the conclusion is the same by using Clarkson’s inequality (see e.g. [17])

‖un − u‖rr−1

Lr ≤ 2(‖u‖rLr + ‖un‖rLr )1r−1 − ‖u+ un‖

rr−1

Lr .

Suppose finally that 1 < r < ∞ and that (un)n≥0 is a bounded sequence of

W 1,r0 (Ω). If ϕ ∈ C∞c (Ω), then for all j ∈ 1, . . . , N

−∫

Ω

∂un∂xj

ϕ =

∫Ω

un∂ϕ

∂xj−→n→∞

∫Ω

u∂ϕ

∂xj, (B.4.10)

by (B.4.5). Set now

fj(ϕ) = −∫

Ω

u∂ϕ

∂xj,

for ϕ ∈ C∞c (Ω). It follows from (B.4.10) and the boundedness of the sequence(un)n≥0 in W 1,r(Ω) that

|fj(ϕ)| ≤ C‖ϕ‖Lr′ .Therefore, f can be extended by continuity and density to a linear, continuousfunctional on Lr

′(Ω). Since Lr

′(Ω)? = Lr(Ω), there exists gj ∈ Lr(Ω) such that

fj(ϕ) =

∫Ω

gjϕ,

for all ϕ ∈ C∞c (Ω) and by density, for all ϕ ∈ Lr′(Ω). This implies that∫Ω

u∂ϕ

∂xj= −

∫Ω

gjϕ,

for all ϕ ∈ C∞c (Ω). Thus u ∈W 1,r(Ω). (B.4.7) follows from (B.4.10) and the aboveidentity. The last properties follow by using (B.4.7) and applying the first part ofthe result to ∇un instead of un.

We can now establish the compact sobolev embeddings.

Theorem B.4.4 (Rellich-Kondrachov). Let Ω ⊂ RN be a bounded, open set,

and let 1 ≤ r <∞. It follows that the embedding W 1,r0 (Ω) → Lr(Ω) is compact.

Proof. Let (un)n≥0 be a bounded sequence of W 1,r0 (Ω). It follows from Corol-

lary B.4.2 that there exist u ∈ Lr(Ω) and a subsequence (unk)k≥0 such that unk → uin Lr(Ω) as k →∞. This completes the proof.

In fact, we have the following stronger result.


Theorem B.4.5. Let Ω ⊂ RN be a bounded, open set, let 1 ≤ r <∞ and set

r =

∞ if r ≥ N,NrN−r if r < N.

If (un)n≥0 is a bounded sequence of W 1,r0 (Ω), then there is a subsequence (unk)k≥0

and u ∈ Lr(Ω) such that unk → u in Lr(Ω) as k → ∞. Moreover, the followingproperties hold.

(i) u ∈ Lr(Ω) (u ∈ Lρ(Ω) for all 1 ≤ ρ < r if r = N ≥ 2) and un → u in Lp(Ω)for all 1 ≤ p < r.

(ii) If r > 1, then u ∈W 1,r0 (Ω) and∫

Ω

∇unkϕ −→k→∞

∫Ω

∇uϕ,

for all ϕ ∈ Lr′(Ω). In particular,

‖∇u‖Lr ≤ lim infk→∞

‖∇unk‖Lr .

If, in addition, ‖∇u‖Lr = lim ‖∇unk‖Lr as k →∞, then unk → u in W 1,r0 (Ω).

Proof. The first part of the result follows from Theorem B.4.4. Next, exceptin the case r = N ≥ 2, it follows from Theorem B.3.14 that (un)n≥0 is boundedin Lr(Ω), from which we deduce u ∈ Lr(Ω). (See Lemma B.4.3.) In the caser = N ≥ 2, it follows from Theorem B.3.14 that (un)n≥0 is bounded in Lp(Ω) forany p < ∞, from which we deduce u ∈ Lp(Ω) for all p < ∞. Property (i) nowfollows from the Lr bound (or Lp bound for all p < ∞ if r = N ≥ 2) and theLr convergence by applying Holder’s inequality to unk − u. Finally, property (ii)follows from Lemma B.4.3.

Remark B.4.6. If Ω is not bounded, we still have a local compactness result.Given R > 0, set ΩR = x ∈ Ω; |x < R|. Given any bounded sequence (un)n≥0

of W 1,r0 (Ω), there exist a subsequence (unk)k≥0 and u ∈ Lr(Ω) such that unk → u

as k → ∞, a.e. in Ω and in Lr(ΩR) for every R < ∞. Moreover, the followingproperties hold.

(i) If r = N = 1, then u ∈ L∞(Ω) and unk → u in Lp(ΩR) for every p < ∞and every R < ∞. In addition, ‖u‖Lp ≤ lim inf ‖un‖Lp as n → ∞ for every1 ≤ p ≤ ∞.

(ii) If N ≥ 2 and 1 ≤ r < N , then u ∈ LNrN−r (Ω) and unk → u in Lp(ΩR) for every

p < Nr/(N − r) and every R < ∞. In addition, ‖u‖Lp ≤ lim inf ‖un‖Lp asn→∞.

(iii) If N ≥ 2 and r = N , then u ∈ Lp(Ω) and unk → u in Lp(ΩR) for every p <∞and every R <∞. In addition, ‖u‖Lp ≤ lim inf ‖un‖Lp as n→∞.

(iv) If r > N , then u ∈ L∞(Ω) and unk → u in L∞(ΩR) for every R < ∞. Inaddition, ‖u‖Lp ≤ lim inf ‖un‖Lp as n→∞ for every r ≤ p ≤ ∞.

(v) If r > 1, then u ∈W 1,r0 Ω) and∫

Ω

∇unkϕ −→k→∞

∫Ω

∇uϕ,

for all ϕ ∈ Lr′(Ω). In particular,

‖∇u‖Lr ≤ lim infk→∞

‖∇unk‖Lr .

If moreover ‖∇u‖Lr = lim ‖∇unk‖Lr and ‖u‖Lr = lim ‖unk‖Lr as k → ∞,

then unk → u in W 1,r0 (Ω).

B.5. THE CASE OF COMPLEX-VALUED FUNCTIONS 101

Those properties are proved like Theorem B.4.5, except for the local conver-gence in (i)–(iv). This follows by applying Theorem B.4.5 to the sequence (ξun)n≥0,where ξ ∈ C∞c (RN ) is such that ξ(x) = 1 for |x| ≤ R.

Corollary B.4.7. Suppose Ω ⊂ RN is a bounded open set. Let 1 < r < ∞and let m ≥ 1 be an integer. If

p =

∞ if mr ≥ N,Nr

N−mr if mr < N,

then the embeddings Wm,r0 (Ω) → Lp(Ω) and Lp

′(Ω) →W−m,r

′(Ω) are compact for

all 1 ≤ p < p.

Proof. We first observe that by Theorem B.4.4, the embedding W 1,r0 (Ω) →

Lr(Ω) is compact, hence the embedding Wm,r0 (Ω) → L1(Ω) is also compact. Ap-

plying Theorem B.3.14 and Holder’s inequality, we deduce that if 1 ≤ p < p, thenthe embedding Wm,r

0 (Ω) → Lp(Ω) is compact. This proves the first part of theresult, and the second part follows from the abstract duality property of Proposi-tion A.1.5 (iii).

B.5. The case of complex-valued functions

So far in this section, we considered real-valued functions but the same theorycan be developped for complex-valued functions, with obvious modifications whichwe describe below.

One has to consider the spaces D(Ω,C) and Lp(Ω,C) instead of the spacesD(Ω,R) and Lp(Ω,R). In particular, a function f ∈ L1

loc(Ω,C) defines a distributionTf ∈ D′(Ω,C) by the formula

〈Tf , ϕ〉 =

∫Ω

Re (f(x)ϕ(x))dx, for all ϕ ∈ D(Ω,C).

In particular, Wm,p(Ω,C) ≈Wm,p(Ω,R)×Wm,p(Ω,R). In other words, a complex-valued function u belongs to Wm,p(Ω,C) if, and only if Re (u) ∈ Wm,p(Ω,R) andIm(u) ∈ Wm,p(Ω,R). As well, Wm,p

0 (Ω,C) ≈ Wm,p0 (Ω,R) ×Wm,p

0 (Ω,R), and it

follows in particular that W−m,p′(Ω,C) ≈ W−m,p

′(Ω,R) × W−m,p

′(Ω,R). The

scalar product on Hm(Ω,C) is defined by

(u, v)Hm =∑

0≤|α|≤m

∫Ω

Re (Dαu(x)Dαv(x)) dx. (B.5.1)

Formula (B.1.6) becomes

〈u, v〉W−m,p,Wm,p′

0

= Re

∫Ω

u(x)v(x) dx, (B.5.2)

and formula (B.1.8) becomes

〈∆u, v〉W−1,p,W 1,p′

0

= −Re

∫Ω

∇u(x) · ∇v(x) dx, for v ∈W 1,p′

0 (Ω,C). (B.5.3)

Therefore, most of the results that we established for real-valued functions stillhold for complex-valued functions, and are proved by considering separately thereal and imaginary parts. Notable exceptions are Proposition B.2.3, Remark B.2.4,and Corollaries B.2.6 and B.2.8 that do not make sense anymore. (See, however,Remark B.5.3 below.) Here is a chain rule that applies to complex-valued functions.

Theorem B.5.1. If F : C → C is a Lipschitz continuous function such thatF (0) = 0 and if 1 ≤ p ≤ ∞, then the following properties hold.

(i) F (u) ∈W 1,p(Ω,C), for every u ∈W 1,p(Ω,C).


(ii) If |F (z1) − F (z2)| ≤ L(z1, z2)|z1 − z2| for all z1, z2 ∈ C, where L : C × C →[0,∞) is some continous function, then |∇F (u)| ≤ L(u, u)|∇u| a.e. for everyu ∈W 1,p(Ω,C).

(iii) If F is C1 (considered as a function R2 → R2) except at a finite number ofpoints, then ∇F (u) = DF (u)∇u a.e. for every u ∈ W 1,p(Ω,C). If moreoverp <∞, then the mapping u 7→ F (u) is continous W 1,p(Ω,C)→W 1,p(Ω,C).

(iv) If p < ∞, then in properties (i) and (iii) above, one may replace W 1,p(Ω,C)

by W 1,p0 (Ω,C).

Proof. We proceed in five steps.

Step 1. Suppose F is C1 (considered as a function R2 → R2), then F (u) ∈W 1,p(Ω,C) and ∇F (u) = DF (u)∇u a.e. for every u ∈W 1,p(Ω,C). This is estab-lished as in [5, Proposition 9.5, p. 270]. The idea of the proof is to approximate uby a sequence (un)n∈N ⊂W 1,p(Ω,C) ∩ C∞(Ω,C).

Step 2. Proof of Property (i) . Consider a sequence of mollifiers (ρj)j∈N ⊂D(R2) and set Fj = ρj ? F . It follows that Fj → F uniformly on C as j → ∞.Moreover, |Fj(z1) − Fj(z2)| ≤ L|z1 − z2| where L is the Lipschitz constant of F .Given u ∈W 1,p(Ω,C), it follows from Step 1 that Fj(u) ∈W 1,p(Ω,C) and that

∇Fj(u) = DFj(u)∇u.In particular, |∇Fj(u)| ≤ L|∇u|. This implies that (up to a subsequence) ∇Fj(u)converges in Lp weak (weak? if p =∞) to some function ψ (apply Dunford-Pettis’theorem if p = 1). Since Fj(u) → F (u) in Lp(Ω,C), it follows that ψ = ∇F (u);and so, F (u) ∈W 1,p(Ω) .

Step 3. Proof of Property (ii). Let Fj be as in Step 2. We have |DFj(z)| ≤L(z, z); and so, |∇Fj(u)| ≤ L(u, u)|∇u|. Since ∇Fj(u) converges in Lp weak (weak?

if p = ∞) to ∇F (u) (see Step 2), we deduce that |∇F (u)| ≤ L(u, u)|∇u|. To seethis, we need only show that if a sequence (fn)n≥0 ⊂ Lp(Ω) satisfies fn → f as

n→∞ and |fn| ≤ g a.e., then |f | ≤ g a.e. Let ϕ ∈ Lp′(Ω), ϕ ≥ 0. We have∫Ω

gϕ ≥∫

Ω

fnϕ −→n→∞

∫Ω

fϕ;

and so, ∫Ω

(g − f)ϕ ≥ 0,

which implies that g ≥ f a.e.

Step 4. Proof of Property (iii). Let E = (xi)1≤i≤k be such that F ∈ C1(C \E,C), and let again Fj be as in Step 2. Note that DF ∈ L∞(C,C2), so thatF ′j = ρj ? F

′ (see [5, Lemme 9.1, p. 266]). It follows that F ′j → F ′ on C \ E.Since ∇u = 0 a.e. on ω = x ∈ Ω; u(x) ∈ E, we see that DFj(u)∇u convergesto DF (u)∇u a.e. It follows that ∇F (u) = DF (u)∇u. Suppose now un → u inW 1,p(Ω,C). We have

∇F (un)−∇F (u) = (DF (un)−DF (u))∇u+DF (u)(∇un −∇u).

Since DF (un(x)) → DF (u(x)) if x 6∈ ω and ∇u = 0 a.e. in ω, we see that∇F (un) → ∇F (u) a.e. If p < ∞, then it follows that ∇F (un) → ∇F (u) inLp(Ω,C) by dominated convergence.

Step 5. Proof of Property (iv). Let u ∈W 1,p0 (Ω,C) and let (un)n≥0 ⊂ D(Ω,C)

be such that un → u in W 1,p(Ω,C) as n→∞. Up to a subsequence, we may assumethat there exists ψ ∈ Lp(Ω) such that |∇un| ≤ ψ a.e. It follows Property (ii) that|∇F (un)| ≤ Lψ a.e., where L is the Lipschitz constant of F . We deduce as in Step 1

that F (un)→ F (u) in W 1,p(Ω,C) weak; and so F (u) ∈W 1,p0 (Ω,C).

B.6. THE FOURIER TRANSFORM AND SOBOLEV SPACES 103

Remark B.5.2. When F does not satisfy the assumption of Theorem B.5.1 (iii),we do not know if the mapping u 7→ F (u) is continuous W 1,p(Ω,C)→W 1,p(Ω,C).Note that the formula “∇F (u) = DF (u)∇u” does not hold in general, even forsmooth functions u. Indeed, take for example

F (u) =

u if |u| ≤ 1,u|u| if |u| ≥ 1.

F is C∞, except on the set |u| = 1 where DF is not defined. Taking for exampleu(x) = eia·x, with a ∈ RN , we see that F (u) = u, so that ∇F (u) = iaeia·x, butDF (u)∇u is not defined a.e. What happens is that (as opposed to the real-valuedcase) if E ⊂ C is a set of measure 0, then ∇u need not vanish a.e. on the setu ∈ E. Take for example u as above and E = |z| = 1.

Remark B.5.3. If u ∈W 1,p(Ω,C), it follows that Reu, Imu, |u| ∈W 1,p(Ω,R).In addition, one has a.e. ∇Re = Re∇u, ∇Im = Im∇u and

|∇|u| |2 =

0 if u = 0,

|∇u|2 −∣∣∣Im(u∇u|u| )∣∣∣2 if u 6= 0.

(In particular, one has |∇|u| | ≤ |∇u|, but in general |∇|u| | 6≡ |∇u|. Note that thisis in contrast with the real-valued case.) If p < ∞, then the mappings u 7→ Reu,u 7→ Imu and u 7→ |u| are continous W 1,p(Ω,C) → W 1,p(Ω,R). Moreover, if

u ∈W 1,p0 (Ω,C), then Reu, Imu, |u| ∈W 1,p

0 (Ω,R). This follows from properties (iii)and (iv) of Theorem B.5.1.

B.6. The Fourier transform and Sobolev spaces

See for instance [27, Chapter 7] and [2, Section 1.2] for details. Other usefulreferences on the subject include [18, Chapter VII], [21, Chapter 5], [22, Chap-ter 1], [31, Chapter I], [33, Chapter VI]. Throughout this section, we considercomplex-valued functions. Unless otherwise specified, all integrals are over RN . Itis convenient to introduce the following notation

u(x) = u(−x) (B.6.1)

[τyu](x) = u(x− y) (B.6.2)

B.6.1. The Fourier transform on RN . The Fourier transform on RN isdefined by

Fu(ξ) = u(ξ) =

∫RN

e−2πix·ξu(x) dx (B.6.3)

for u ∈ L1(RN ).

Proposition B.6.1. The following properties hold.

(i) Fu ∈ C0(RN ) and‖Fu‖L∞ ≤ ‖u‖L1 (B.6.4)

for all u ∈ L1(RN ).(ii) If u, v ∈ L1(RN ), then∫

u(x)v(x) dx =

∫u(x)v(x) dx. (B.6.5)

(iii) If u, v ∈ L1(RN ), then

F(u ? v) = FuFv. (B.6.6)

(iv) If u ∈ L1(RN ) and λ > 0, then v(x) = u(x/λ) satisfies v(ξ) = λN u(λξ) forall ξ ∈ RN .


(v) If u ∈ L1(RN ) and y ∈ RN , then

F [τyu](ξ) = e−2πiy·ξu(ξ) (B.6.7)

with the notation (B.6.2).

Proof. Property (i) follows easily by dominated convergence; and Proper-ties (ii) and (iii) are immediate applications of Fubini’s theorem. Finally, Proper-ties (iv) and (v) follow from elementary calculations.

Remark B.6.2. Here are some comments on the Fourier transform of radialfunctions.

(i) Suppose u ∈ L1(RN ) is radially symmetric, and write u(x) = u(|x|) = u(r)by abuse of notation. It follows that u is also radially symmetric (so we writeu(ξ) = u(|ξ|) = u(ρ) by abuse of notation). Indeed, let R is an orthogonaltransformation of RN . Note that R? = R−1 and that the Jacobian determi-nant of R is 1. Therefore,

F [u(R·)](ξ) =

∫RN

e−2πix·ξu(Rx) dx =

∫RN

e−2πiR−1y·ξu(y) dy

=

∫RN

e−2πiy·Rξu(y) dy = Fu(Rξ).

If u is radially symmetric, then u(R·) = u(·) for all orthogonal transformationR, so that Fu(R·) = Fu(·). Thus Fu is radially symmetric.

(ii) Suppose u ∈ L1(RN ) is radially symmetric, so that u is also radially symmetricby (i) above. It follows that

u(ρ) = 2πr−N−2

2

∫ ∞0

u(r)JN−22

(2πrρ)rN2 dr (B.6.8)

where the Jµ are the Bessel functions of the first kind. See [31, Chapter IV,Theorem 3.3, p. 155].

(iii) Since the Bessel functions Jµ are real valued, it follows from formula (B.6.8)that if u is radially symmetric and real valued, then u is also radially sym-metric and real valued.

Remark B.6.3. It is useful to calculate the Fourier transform of a Gaussian.It is given by

F [e−πa|·|2

](x) = a−N2 e−

π|x|2a (B.6.9)

for all a > 0 and x ∈ RN . Indeed, setting ρ = F [e−πa|ξ|2

], we see that

ρ(x) =

∫RN

e−2πix·ξe−πa|ξ|2

dξ

and so

x · ∇ρ =

∫RN

x · ∇x[e−2πix·ξe−πa|ξ|2

] dξ = −∫RN

2πix · ξe−2πix·ξe−πa|ξ|2

dξ

On the other hand

∇ξ · (e−2πix·ξe−πa|ξ|2

x) = ia(−2π

|x|2

a+ 2πix · ξ

)e−2πix·ξe−πa|ξ|

2

Integrating in ξ, we obtain

0 =

∫RN

(−2π

|x|2

a+ 2πix · ξ

)e2πix·ξe−πa|ξ|

2

dξ = −2π|x|2

aρ− x · ∇ρ

so that

x · ∇ρ(x) = −2π|x|2

aρ(x)


Therefore, given any x ∈ RN , if we set

f(s) = ρ(sx) s ≥ 0

then

f ′(s) = x · ∇ρ(sx) =1

ssx · ∇ρ(sx) = −2π

s

|sx|2

aρ(sx) = −2πs

|x|2

af(s)

which yields

f(s) = e−πs2|x|2

a f(0)

Letting s = 1, we obtain

ρ(x) = e−π|x|2a ρ(0) = e−

π|x|2a

∫RN

e−πa|ξ|2

dξ = e−π|x|2a a−

N2

from which (B.6.9) follows.

B.6.2. The Schwartz space S(RN ). The Schwartz space S(RN ) is the spaceof ϕ ∈ C∞(RN ,C) such that for every nonnegative integer m

pm(u) = sup|β|≤m

supx∈RN

(1 + |x|2)m2 |Dβϕ(x)| <∞.

S(RN ) is a Frechet space when equipped with the seminorms pm. In particular, if(ϕn)n≥1 ⊂ S(RN ), then ϕn → 0 in S(RN ) iff pm(ϕn)→ 0 for all m ∈ N, i.e.

sup|β|≤m

supx∈RN

(1 + |x|2)m2 |Dβϕn(x)| −→

n→∞0 (B.6.10)

for all m ∈ N.

Remark B.6.4. Here are some simple consequences of the definition of S(RN ).

(i) If ψ ∈ C∞(RN ) is such that for all integer m ≥ 0 there exist an integer km ≥ 0and a constant Cm such that

sup|β|≤m

|∂βψ(x)| ≤ Cm(1 + |x|2)km2 (B.6.11)

then ψϕ ∈ S(RN ) for all ϕ ∈ S(RN ) and the map ϕ 7→ ψϕ is continuousS(RN ) → S(RN ). Indeed, |∂β(ψϕ)| is estimated, by Leibniz’s formula, by asum of terms of the form ∂β1ψ∂β2ϕ where β1 + β2 = β. Therefore we deducefrom (B.6.11) that

pm(ψϕ) ≤ C supx∈RN

(1 + |x|2)m2 Cm(1 + |x|2)

km2 (1 + |x|2)−

m+km2 pm+km(ϕ)

≤ CCmpm+km(ϕ)(B.6.12)

and the result follows since m is arbitrary. Note that (B.6.11) is satisfied forinstance by ψ(x) = (1 + |x|2)

s2 with s ∈ R.

(ii) It follows from property (i) above that if ϕ,ψ ∈ S(RN ), then ϕψ ∈ S(RN ).Moreover, it follows from calculations similar to (B.6.12) that pm(ϕψ) ≤Cpm(ϕ)pm(ψ) with a constant C depending on m.

(iii) The space C∞c (RN ) is dense in S(RN ). This follows easily by applying thecalculations of property (i) above to appropriate cut-off functions ψ.

(iv) If ϕ ∈ S(RN ), then xαDβϕ ∈ S(RN ) for all multi-indices α, β. Moreover, themap ϕ 7→ xαDβϕ is continuous S(RN )→ S(RN ).

(v) Let ϕ,ψ ∈ S(RN ). Given m ∈ N and a multi-index β with |β| ≤ m,

(1 + |x|2)m2 Dβϕ ? ψ(x) =

∫RN

(1 + |x|2)m2 ϕ(y)Dβψ(x− y) dy


Since (1 + |x|2)m2 ≤ C(1 + |x− y|2)

m2 (1 + |y|2)

m2 , we deduce that

(1 + |x|2)m2 |Dβϕ ? ψ(x)| ≤ Cpm(ψ)

∫RN

(1 + |y|2)m2 ϕ(y) dy

≤ Cpm(ψ)pm+N+1(ϕ)

Thus we see that the map (ϕ,ψ) 7→ ϕ ? ψ is continuous S(RN ) × S(RN ) →S(RN ).

Proposition B.6.5. Given ϕ ∈ S(RN ), it follows that ϕ ∈ C∞(RN ). Moreover

ξγDβϕ = (−1)|β|(2πi)|β|−|γ|F [Dγ(xβϕ(x))] (B.6.13)

for all multi-indices β, γ. In addition,

Dβ(ξγϕ) = (−1)|β|(2πi)|β|−|γ|F(xβDγϕ(x)) (B.6.14)

for all multi-indices β, γ.

Proof. By dominated convergence, one can take derivatives with respect to ξin formula (B.6.3). One deduces the property ϕ ∈ C∞(RN ) and the formula

Dβϕ(ξ) = (−1)|β|(2πi)|β|∫RN

e−2πix·ξxβϕ(x) dx (B.6.15)

for every multi-index β. Let now γ be another multi-index, and note that

ξγe−2πix·ξ = (−1)|γ|(2πi)−|γ|Dγ(e−2πix·ξ) (B.6.16)

where the derivative Dγ is with respect to x. Multiplying equation (B.6.15) by ξγ

and integrating by parts, we obtain (B.6.13). Similarly, applying first (B.6.16) andintegrating by parts in (B.6.3) yields

ξγϕ(ξ) = (2πi)−|γ|∫RN

e−2πix·ξDγϕ(x) dx.

Taking derivatives with respect to ξ, we obtain (B.6.14).

Corollary B.6.6. Given ϕ ∈ S(RN ), it follows that ϕ ∈ S(RN ). Moreover,F : S(RN )→ S(RN ) is continuous.

Proof. This follows from formula (B.6.13).

Theorem B.6.7 (The inversion formula). Given ψ ∈ S(RN ), let Fψ ∈ S(RN )be defined by

Fψ(x) =

∫RN

e2πiξ·xψ(ξ) dξ. (B.6.17)

It follows that

ϕ = F [Fϕ] = F [Fϕ] (B.6.18)

for all ϕ ∈ S(RN ).

Proof. Let φ, θ ∈ S(RN ), λ > 0, and let f ∈ S(RN ) be defined by

f(x) = φ(xλ

). (B.6.19)

It follows from Proposition B.6.1 (iv) that

f(ξ) = λN φ(λξ). (B.6.20)

Applying (B.6.5) with u = θ and v = f , and using (B.6.20), we obtain∫θ(x)λN φ(λx) dx =

∫θ(x)φ

(xλ

)dx,


and, after a change of variables in the integral on the left∫θ(xλ

)φ(x) dx =

∫θ(x)φ

(xλ

)dx.

Letting λ→∞, we deduce that

θ(0)

∫φ(x) dx = φ(0)

∫θ(x) dx.

We now let φ(x) = e−π|x|2

, so that φ(x) = e−π|x|2

by (B.6.9), hence∫φ(x) dx = 1,

and we obtain

θ(0) =

∫θ(ξ) dξ (B.6.21)

for all θ ∈ S(RN ). Given now ϕ ∈ S(RN ) and x ∈ R, we let θ(·) = ϕ(· + x), and

we deduce from Proposition B.6.1 (v) that θ(ξ) = e2πix·ξϕ(ξ). Therefore, (B.6.21)yields

ϕ(x) =

∫e2πix·ξϕ(ξ) dξ.

Since x ∈ RN is arbitrary, this is the first identity in (B.6.17). The second identityfollows by an obvious change of variable.

Remark B.6.8. Here are some comments on Theorem B.6.7.

(i) Formula (B.6.17) can be written in the form

Fψ(x) = ψ(−x). (B.6.22)

Thus we see that F and F have similar properties. In particular, F : S(RN )→S(RN ) is continuous.

(ii) Formula (B.6.17) can be written in the form

Fψ = Fψ (B.6.23)

where the bars on the right-hand side refer to complex conjugation.(iii) It follows from Theorem B.6.7 and Property (i) above that F is an homeo-

morphism of S(RN ), with inverse

F−1 = F . (B.6.24)

Theorem B.6.9. The Parseval identity∫RN

u(x)v(x) dx =

∫RN

u(ξ)v(ξ) dξ (B.6.25)

holds for all u, v ∈ S(RN ), and the Plancherel formula

‖u‖L2 = ‖u‖L2 (B.6.26)

holds for all u ∈ S(RN ).

Proof. Let u, ψ ∈ S(RN ). Letting v = F−1ψ, we have v = ψ. Moreover,

by (B.6.24) and (B.6.23), v = F(ψ) = ψ. Therefore, applying (B.6.5),∫u(x)ψ(x) dx =

∫u(ξ)ψ(ξ) dξ

for all u, ψ ∈ S(RN ), which is (B.6.25). Finally, (B.6.26) follows by letting v = uin (B.6.25).


B.6.3. The Fourier transform on L2(RN ). So far, we have defined theFourier transform on L1(RN ). Using Plancherel’s formula (B.6.26), we may extendF to an isometry of L2(RN ), which we still denote by F . (By density of L2(RN )∩L1(RN ) in L2(RN ).) In particular, Plancherel’s formula (B.6.26) holds for all u ∈L2(RN ), and Parseval’s identity (B.6.25) as well as formula (B.6.5) hold for allu, v ∈ L2(RN ).

It follows in particular from (B.6.22) that F , which is in principle defined onS(RN ), can be extended (by density of S(RN ) in L2(RN )) to an isometry of L2(RN ).Therefore, the inversion formula (B.6.18) holds for all ϕ ∈ L2(RN ). Thus we seethat the Fourier transform is an homeomorphism of L2(RN ), and that (B.6.24)holds on L2(RN ), as well as formulas (B.6.22) and (B.6.23).

Since F is continuous L1(RN ) → L∞(RN ) with norm ≤ 1 by (B.6.4), andcontinuous L2(RN )→ L2(RN ) with norm 1 by (B.6.26), it follows from the Riesz-Thorin interpolation theorem (see e.g. [4, Theorem 1.1.1]) that F can be extended

to a continuous operator Lp(RN )→ Lp′(RN ) with norm ≤ 1, for all 1 ≤ p ≤ 2, i.e.

‖Fu‖Lp′ ≤ ‖u‖Lp (B.6.27)

for all u ∈ Lp(RN ).

B.6.4. Tempered distributions: the space S ′(RN ). The space S ′(RN ) oftempered distributions on RN is the topological dual of S(RN ), and we denote by〈·, ·〉 the corresponding duality bracket. Since C∞c (RN ) is a dense subset of S(RN )(see Remark B.6.4 (iii)), it follows that S ′(RN ) is a subspace of D′(RN ). Notethat a map u : S(RN ) → C defines an element u ∈ S ′(RN ) iff the following twoproperties hold

u :

S(RN )→ Cϕ 7→ u(ϕ)

is a C-linear map (B.6.28)

if (ϕn)n≥1 ⊂ S(RN ) and ϕn → 0 in S(RN ), then u(ϕn) −→n→∞

0 (B.6.29)

Moreover, given a sequence (un)n≥1 ⊂ S ′(RN ),

un −→n→∞

0 in S ′(RN )⇐⇒ 〈un, ϕ〉 −→n→∞

0 for all ϕ ∈ S(RN ) (B.6.30)

Remark B.6.10. Here are some comments on the definition of S ′(RN ).

(i) If un → u in S ′(RN ) and ϕn → ϕ in S(RN ) as n→∞, then 〈un, ϕn〉 → 〈u, ϕ〉.(See [27, Theorem 2.17, p. 51].)

(ii) If u ∈ L1loc(RN ) and (1 + | · |2)−

m2 u ∈ L1(RN ) for some m ∈ N, then u defines

a tempered distribution, via the formula

〈u, ϕ〉 =

∫RN

u(x)ϕ(x) dx (B.6.31)

for all ϕ ∈ S(RN ). Indeed,

|u(x)ϕ(x)| ≤ pm(ϕ)(1 + |x|2)−m2 u(x) ∈ L1(RN )

Therefore, the integral is well defined, and

|〈u, ϕ〉| ≤ pm(ϕ)‖(1 + | · |2)−m2 u‖L1

so that 〈u, ϕn〉 → 0 if ϕn → 0 in S(RN ).(iii) In connexion with (ii) above, we say that a tempered distribution u ∈ S ′(RN )

belongs to Lp(RN ), where 1 ≤ p ≤ ∞, if there exists a function f ∈ Lp(RN )such that (B.6.31) holds for all ϕ ∈ S(RN ).


(iv) Given u ∈ S ′(RN ) and a multi-index β, one defines Dβu ∈ S ′(RN ) by

〈Dβu, ϕ〉 = (−1)|β|〈u,Dβϕ〉 (B.6.32)

for all ϕ ∈ S(RN ). Note that this defines an element of S ′(RN ) by Re-mark B.6.4 (iv), and that this definition coincides with the standard definitionof derivatives if u ∈ S(RN ), by formula (B.6.31). Moreover, it follows fromRemark B.6.4 (iv) that the map u 7→ Dβu is continuous S ′(RN )→ S ′(RN ).

(v) We can multiply tempered distributions by C∞ functions whose derivativeshave at most a power growth at infinity. Indeed, if ψ ∈ C∞(RN ) satis-fies (B.6.11) for all integer m ≥ 0, then it follows from Remark B.6.4 (i) thatψϕ ∈ S(RN ) for all ϕ ∈ S(RN ). Therefore, we can define ψu ∈ S ′(RN ) by

〈ψu, ϕ〉 = 〈u, ψϕ〉 (B.6.33)

for all ϕ ∈ S(RN ). The fact that this defines an element of S ′(RN ) followsfrom (B.6.12). Note that this is consistent with standard multiplication offunctions, by formula (B.6.31). Moreover, it follows from Remark B.6.4 (i)that the map u 7→ ψu is continuous S ′(RN )→ S ′(RN ).

(vi) It follows in particular from (v) above that if γ is a multi-index, then for everyu ∈ S ′(RN ), xγu defined by

〈xγu, ϕ〉 = 〈u, xγϕ〉 (B.6.34)

is a tempered distribution and the map u 7→ xγu is continuous S ′(RN ) →S ′(RN ).

Definition B.6.11. In analogy with (B.6.5), the Fourier transform u = Fu ofa tempered distribution u ∈ S ′(RN ) is defined by

〈u, ϕ〉 = 〈u, ϕ〉 (B.6.35)

for all ϕ ∈ S(RN ).

Remark B.6.12. Here are some consequences of Definition B.6.11.

(i) It follows from Corollary B.6.6 that for every u ∈ S ′(RN ), u defined by (B.6.35)is a tempered distribution, i.e. u ∈ S ′(RN ); and that F : S ′(RN ) → S ′(RN )is continuous.

(ii) F is a bijection S ′(RN )→ S ′(RN ), and

F−1 = F (B.6.36)

where〈Fu, ϕ〉 = 〈u,Fϕ〉 (B.6.37)

with the notation (B.6.17). Indeed, it follows from identities (B.6.37), (B.6.35)and (B.6.24) that

〈FFu, ϕ〉 = 〈u,FFϕ〉 = 〈u, ϕ〉so that FF = I. Similarly, by (B.6.35), (B.6.37) and (B.6.24)

〈FFu, ϕ〉 = 〈Fu, ϕ〉 = 〈u,F ϕ〉 = 〈u, ϕ〉so that FF = I. Moreover, it follows from Remark B.6.8 that F−1 is contin-uous S ′(RN )→ S ′(RN ).

(iii) We define the complex conjugate of u ∈ S ′(RN ) by

〈u, ϕ〉 = 〈u, ϕ〉 (B.6.38)

for all ϕ ∈ S(RN ). (Note that this is consistent with the standard complexconjugation of functions, by (B.6.31).) It follows from (B.6.36) and (B.6.23)that

F−1u = Fu (B.6.39)


for all u ∈ S ′(RN ). Indeed, applying identities (B.6.36), (B.6.37), (B.6.23)and (B.6.38), we see that

〈F−1u, ϕ〉 = 〈Fu, ϕ〉 = 〈u,Fϕ〉 = 〈u,Fϕ〉 = 〈u,Fϕ〉 = 〈Fu, ϕ〉 = 〈Fu, ϕ〉for all ϕ ∈ S(RN ).

(iv) Given u ∈ S ′(RN ) and multi-indices β, γ, we have

ξγDβϕ = (−1)|β|(2πi)|β|−|γ|F [Dγ(xβϕ)] (B.6.40)

andDβ(ξγϕ) = (−1)|β|(2πi)|β|−|γ|F(xβDγϕ). (B.6.41)

In particular,F(Dβu) = (2πi)|β|ξβ u (B.6.42)

andDβu = (−2πi)|β|F(xβu). (B.6.43)

This follows easily by duality from formulas (B.6.13) and (B.6.14). As aparticular case, we see that

F(∆u) = −4π2|ξ|2u (B.6.44)

for all u ∈ S ′(RN ).

We note that if u ∈ L1(RN ) and ϕ ∈ S(RN ), then

u ? ϕ(x) =

∫u(y)ϕ(x− y) dy =

∫u(y)ϕ(y − x) dy

=

∫u(y)τxϕ(y) dy = 〈u, τxϕ〉

(B.6.45)

with the notation (B.6.1), (B.6.2) and (B.6.31).

Definition B.6.13. In analogy with identity (B.6.45), we define the convolu-tion of an element of S ′(RN ) with an element of S(RN ) by

u ? ϕ(x) = 〈u, τxϕ〉 (B.6.46)

for all u ∈ S ′(RN ) and ϕ ∈ S(RN ).

We have the following result. (See [27, Theorem 7.19, p. 195].)

Theorem B.6.14. Let u ∈ S ′(RN ) and ϕ ∈ S(RN ), and let u ? ϕ be definedby (B.6.46).

(i) u ? ϕ ∈ C∞(RN ) and

Dβ(u ? ϕ) = u ? (Dβϕ) = (Dβu) ? ϕ (B.6.47)

for every multi-index β.(ii) u?ϕ grows at most as a power of |x| as |x| → ∞. In particular, u?ϕ ∈ S ′(RN )

by Remark B.6.10 (ii), and

〈u ? ϕ, ψ〉 = 〈u, ϕ ? ψ〉 (B.6.48)

for all ψ ∈ S(RN ).(iii) (u ? ϕ) ? ψ = u ? (ϕ ? ψ) for all ψ ∈ S(RN ).(iv) F(u ? ϕ) = ϕu.(v) u ? ϕ = F(ϕu).

We will also use the following property.

Proposition B.6.15. Given an interval I ⊂ R and u : I → S ′(RN ), thefollowing properties hold.

(i) u ∈ C(I,S ′(RN )) if and only if the map t 7→ 〈u(t), ϕ〉 is continuous on I forall ϕ ∈ S(RN ).


(ii) u ∈ C1(I,S ′(RN )) if and only if the map t 7→ 〈u(t), ϕ〉 is C1 on I for allϕ ∈ S(RN ). In this case,

〈u′(t), ϕ〉 =d

dt〈u(t), ϕ〉 (B.6.49)

for all ϕ ∈ S(RN ) and t ∈ I.(iii) If u ∈ C1(I,S ′(RN )) and ϕ ∈ C1(I,S(RN )), then the map t 7→ 〈u(t), ϕ(t)〉 is

in C1(I), and

d

dt〈u(t), ϕ(t)〉 = 〈u′(t), ϕ(t)〉+ 〈u(t), ϕ′(t)〉 (B.6.50)

for all i ∈ I.

Proof. We consider the case where I = (0, T ) with T > 0, the other casesbeing similar. Property (i) follows from (B.6.30).

To prove Property (ii), suppose first u ∈ C1((0, T ),S ′(RN )) and let 0 < t < T .It follows that 1

h (u(t + h) − u(t)) − u′(t) → 0 in S ′(RN ) as h → 0. This means

that for every ϕ ∈ S(RN ), 1h (〈u(t + h), ϕ〉 − 〈u(t)), ϕ〉 − 〈u′(t), ϕ〉 → 0 as h → 0.

Thus we see that the map t 7→ 〈u(t), ϕ〉 is C1 on (0, T ) for all ϕ ∈ S(RN ), andthat (B.6.49) holds. Conversely, suppose the map t 7→ 〈u(t), ϕ〉 is C1 on (0, T ) forall ϕ ∈ S(RN ). Given ϕ ∈ S(RN ) and 0 < t < T , let Lt(ϕ) = d

dt 〈u(t), ϕ〉. We have

Lt(ϕ) = limh→0

Lt,h(ϕ)

where

Lt,h(ϕ) =1

h〈u(t+ h)− u(t), ϕ〉

Lt is clearly linear. Moreover, for all t ∈ 0, T ) and h 6= 0, u(t+h)−u(t) ∈ S ′(RN ),so that Lt,h : S(RN ) → C is continuous. Since S(RN ) is a Frechet space, itfollows from the Banach-Steinhaus theorem (see [27, Theorem 2.8]) that Lt iscontinuous S(RN ) → C. Therefore, Lt ∈ S ′(RN ), so that u ∈ C1((0, T ),S ′(RN ))and u′(t) = Lt.

Finally, we prove Property (iii), and we set f(t) = 〈u(t), ϕ(t)〉. Given t ∈ I andh 6= 0 such that t+ h ∈ I, we have

f(t+ h)− f(t) = 〈u(t+ h)− u(t), ϕ(t)〉+ 〈u(t+ h), ϕ(t+ h)− ϕ(t)〉Dividing by h, letting h → 0, and applying Remark B.6.10 (i), we obtain theidentity (B.6.50).

Remark B.6.16. Given ν > 0, set

fν(ξ) = (1 + 4π2|ξ|2)−ν (B.6.51)

so that fν ∈ S ′(RN ) by Remark B.6.10 (ii). (Note that fν ∈ L2(RN ) if ν > N2 ,

but fν 6∈ L2(RN ) if ν ≤ N2 .) The inverse Fourier transform F ν = F−1fν of fν is

in principle an element of S ′(RN ), but it turns out that F ν is in fact a function.More precisely,

F ν(x) =1

(4π)νΓ(ν)

∫ ∞0

e−π|x|2ρ e−

ρ4π ρ−1−N2 +νdρ (B.6.52)

where Γ is the Gamma function defined by

Γ(s) =

∫ ∞0

e−σσs−1dσ (B.6.53)

In particular, we see from formula (B.6.52) that F ν is positive, radially symmetricand radially decreasing. Moreover,

‖F ν‖L1 =

∫RN

F ν(x) dx = 1. (B.6.54)


(See [30, Chapter V, Proposition 2].) To see this, let s = ν in (B.6.53), and makethe change of variable σ = µ

4πρ, with µ > 0, to obtain

(4π)νΓ(ν)µ−ν =

∫ ∞0

e−µρ4π ρν−1dρ. (B.6.55)

For µ = 1 + 4π2|ξ|2, this yields

(4π)νΓ(ν)fν(ξ) =

∫ ∞0

e−ρ4π (1+4π2|ξ|2)ρν−1dρ (B.6.56)

On the other hand, given ϕ ∈ S(RN ) and ρ > 0, we deduce from (B.6.5) and (B.6.9)(with a = ρ) that ∫

RNe−ρπ|ξ|

2

ϕ(ξ) dξ = ρ−N2

∫RN

e−π|x|2ρ ϕ(x) dx (B.6.57)

so that, multiplying by e−ρ4π ρν−1∫

RNe−

ρ4π (1+4π2|ξ|2)ρν−1ϕ(ξ) dξ = e−

ρ4π ρ1−N2 +ν

∫RN

e−π|x|2ρ ϕ(x) dx (B.6.58)

Integrating in ρ and applying Fubini, we obtain∫RN

(∫ ∞0

e−ρ4π (1+4π2|ξ|2)ρν−1dρ

)ϕ(ξ) dξ

=

∫RN

(∫ ∞0

e−π|x|2ρ e−

ρ4π ρ−1−N2 +νdρ

)ϕ(x) dx

hence, applying (B.6.56),

(4π)νΓ(ν)

∫RN

fν(ξ)ϕ(ξ) dξ =

∫RN

(∫ ∞0

e−π|x|2ρ e−

ρ4π ρ−1−N2 +νdρ

)ϕ(x) dx

Since ϕ ∈ S(RN ) is arbitrary, the above identity, together with formula (B.6.5),yields (B.6.52). To prove (B.6.54), we integrate (B.6.52) over RN and apply Fubini,to obtain∫

RNF ν(x) dx =

1

(4π)νΓ(ν)

∫ ∞0

e−ρ4π ρ−1−N2 +ν

(∫RN

e−π|x|2ρ dx

)dρ

Since ∫RN

e−π|x|2ρ dx = ρ

N2

we obtain ∫RN

F ν(x) dx =1

(4π)νΓ(ν)

∫ ∞0

e−ρ4π ρ−1+νdρ = 1

where the last identity follows from (B.6.55) with µ = 1. Hence (B.6.54). Note alsothat, since fν is real-valued, then

fν = F ν (B.6.59)

by (B.6.39). More generally, given λ > 0, let

fνλ (ξ) = (λ+ 4π2|ξ|2)−ν (B.6.60)

and

F νλ = fνλ = F−1fνλ (B.6.61)

Since fνλ (ξ) = λ−ε2 fν(ξλ−

12 ), an obvious change of variables yields

F νλ (x) = λN−ε

2 F ν(xλ12 ) (B.6.62)


In particular, F νλ is positive, radially symmetric and radially decreasing and, ap-plying (B.6.62) and (B.6.54),

‖F νλ ‖L1 =

∫RN

F νλ (x) dx = λ−ν . (B.6.63)

B.6.5. Fractional order Sobolev spaces. The L2 case. It follows fromDefinitions B.1.2 and B.1.15 that, given m ∈ N, Hm(RN ) is the set of u ∈ L2(RN )such that Dαu ∈ L2(RN ) for all multi-indices β with |β| ≤ m, and H−m(RN ) =(Hm(RN ))?. Here is an equivalent definition, based on the Fourier transform andPlancherel’s formula.

Proposition B.6.17. Let m ∈ Z, and let the Sobolev space Hm(RN ) be givenby Definitions B.1.2 and B.1.15. It follows that

Hm(RN ) = u ∈ S ′(RN ); (1 + 4π2| · |2)m2 u ∈ L2(RN ) (B.6.64)

and

‖u‖Hm ∼ ‖(1 + 4π2| · |2)s2 u‖L2 . (B.6.65)

Proof. Set

Hm = u ∈ S ′(RN ); (1 + 4π2| · |2)m2 u ∈ L2(RN )

and

‖u‖Hm = ‖(1 + 4π2| · |2)s2 u‖L2 .

We first consider the case m ≥ 0. Let u ∈ Hm(RN ), j ∈ 1, · · · , N, and k ∈0, · · · ,m. It follows from formula (B.6.42) that

|ξj |k|u| = (2π)−k|F(∂kj u)|

so that by Plancherel’s formula (B.6.26)

‖|ξj |k|u|‖L2 ≤ (2π)−k‖∂kj u‖L2 ≤ ‖u‖Hm .

It follows easily that u ∈ Hm and ‖u‖Hm ≤ C‖u‖Hm with C independent of u.Conversely, suppose u ∈ Hm and let β be a multi-index with |β| ≤ m. Thepreceding calculations show that Dβu ∈ L2(RN ) and

‖Dβu‖L2 ≤ C‖u‖Hm

with a constant C independent of m. The result in the case m ≥ 0 easily follows.We now consider the case m < 0. Since Hm(RN ) = (H−m(RN ))?, we need

only show that Hm = (H−m)?. Any u ∈ Hm defines an element Fu ∈ (H−m)? by

〈Fu, v〉(H−m)?,H−m = ((1 + 4π2| · |2)m2 u, (1 + 4π2| · |2)−

m2 v)L2 (B.6.66)

for all v ∈ H−m, and it is easy to verify that ‖F (u)‖(H−m)? = ‖u‖Hm . This shows

that Hm → (H−m)?. Conversely, let F ∈ (H−m)?, and define

G(w) = 〈F,F−1[(1 + 4π2| · |2)m2 w]〉(H−m)?,H−m (B.6.67)

for w ∈ L2(RN ). Note that this makes sense. Indeed, if w ∈ L2(RN ), then F−1[(1+4π2| · |2)

m2 w] ∈ H−m and ‖w‖H−m = ‖F−1[(1 + 4π2| · |2)

m2 w]‖L2 . It follows that

G(w) ∈ (L2(RN ))?, so that by Riesz’s representation theorem, there exists z ∈L2(RN ) such that

G(w) = (z, w)L2 = (z, w)L2 (B.6.68)

for all w ∈ L2(RN ), where the last identity follows from Parseval’s identity (B.6.25).Set now u = F−1[(1 + 4π2| · |2)−

m2 z] ∈ Hm. It follows from (B.6.67), (B.6.68)


and (B.6.66) that, given any v ∈ H−m,

〈F, v〉(H−m)?,H−m = G(F−1[(1 + 4π2| · |2)−m2 v])

= (z, (1 + 4π2| · |2)−m2 v)L2

= ((1 + 4π2| · |2)m2 u, (1 + 4π2| · |2)−

m2 v)L2

= 〈Fu, v〉(H−m)?,H−m .

Therefore, F = Fu, which shows that (H−m)? → Hm and completes the proof.

We now extend the definition of the Sobolev space Hm(RN ) to non-integerindices. Given s ∈ R, we define

Hs(RN ) = u ∈ S ′(RN ); (1 + 4π2| · |2)s2 u ∈ L2(RN ) (B.6.69)

and‖u‖Hs = ‖(1 + 4π2| · |2)

s2 u‖L2 (B.6.70)

for u ∈ Hs(RN ). It follows from Proposition B.6.17 that this definition is consistentwith Definitions B.1.2 and B.1.15 for s ∈ Z. We note that (B.6.69) makes sense.Indeed, given u ∈ S ′(RN ), it follows that u ∈ S ′(RN ). Therefore, (1+4π2| · |2)

s2 u ∈

S ′(RN ) by Remark B.6.10 (v), so that it makes sense to say that (1 + 4π2|ξ|2)s2 u ∈

Lp(RN ) by Remark B.6.10 (iii).

Remark B.6.18. Here are some comments on the above definition.

(i) Given any s ∈ R, Hs(RN ) is a Hilbert space, with the scalar product

(u, v)Hs = ((1 + 4π2| · |2)s2 u, (1 + 4π2| · |2)

s2 v)L2 . (B.6.71)

for all u, v ∈ Hs(RN ).(ii) Let s ∈ R and 1 ≤ p ≤ ∞. The operators F and F−1, and the mul-

tiplication by (1 + |ξ|2)s2 are all continuous S(RN ) → S(RN ). (See Re-

mark B.6.4 (i) and Remark B.6.8.) Since S(RN ) → L2(RN ), it follows thatS(RN ) → Hs(RN ). Moreover, the embedding S(RN ) → Hs(RN ) is dense.Indeed, let u ∈ Hs(RN ), let v = F−1[(1 + 4π2| · |2)

s2 u] ∈ L2(RN ), and let

(vn)n≥1 ⊂ S(RN ) satisfy vn → v in L2. It follows that un → u in Hs,p, whereun = F−1[(1 + 4π2| · |2)−

s2 vn]. Moreover, un ∈ S(RN ), which completes the

proof.(iii) Given γ > 0, we see that 1 + 4π2|ξ|2 ∼ 1 + γ|ξ|2. It follows that

Hs(RN ) = u ∈ S ′(RN ); (1 + γ| · |2)s2 u ∈ L2(RN )

and‖u‖Hs ∼ ‖(1 + γ| · |2)

s2 u‖L2 .

(iv) Let s, s′ ∈ R, and suppose s ≥ s′. It follows immediately that Hs(RN ) →Hs′(RN ) and, more precisely,

‖u‖Hs′ ≤ ‖u‖Hs (B.6.72)

for all u ∈ Hs(RN ).

Proposition B.6.19. Let s ∈ R and β a multi-index. It follows that Dβu ∈Hs−|β|(RN ) for all u ∈ Hs,p(RN ). Moreover, there exists a constant C such that

‖Dβu‖Hs−|β| ≤ C‖u‖Hs (B.6.73)

for all u ∈ Hs(RN ). In particular, ‖∆u‖Hs−2 ≤ C‖u‖Hs for all u ∈ Hs(RN ).

Proof. We have F [Dβu] = (2πiξ)β u by (B.6.42), so that

|F [Dβu]| ≤ (2πξ)|β||u|. (B.6.74)

The result easily follows.


Proposition B.6.20. Let s ∈ R, m ∈ N, and u ∈ S ′(RN ). It follows thatu ∈ Hs+m(RN ) iff Dβu ∈ Hs(RN ) for all multi-indices β such that |β| ≤ m.Moreover, there exist constants 0 < c ≤ C <∞ such that

c‖u‖Hs+m ≤∑|β|≤m

‖Dβu‖Hs ≤ C‖u‖Hs+m (B.6.75)

for all u ∈ Hs+m(RN ).

Proof. The result easily follows from formula (B.6.74).

Remark B.6.21. The fractional order Sobolev spaces Hs(RN ) are complexinterpolation spaces of the standard Sobolev spaces Hm(RN ). More precisely, given`,m ∈ Z such that ` < m, and s ∈ (`,m),

Hs(RN ) = [H`(RN ), Hm(RN )][ s−`m−` ] (B.6.76)

See e.g. [4, Theorem 6.4.5 (7)]. Moreover, the interpolation inequality

‖u‖Hs ≤ ‖u‖m−sm−`H`‖u‖

s−`m−`Hm (B.6.77)

holds for all u ∈ Hm(RN ). Indeed,

(1 + 4π2| · |2)s2 |u| = [(1 + 4π2| · |2)

`2 |u|]

m−sm−` [(1 + 4π2| · |2)

m2 |u|]

s−`m−`

and 12 = m−s

2(m−`) + s−`2(m−`) , so it follows from Holder’s inequality that

‖u‖Hs = ‖(1 + 4π2| · |2)s2 u‖L2

≤ ‖(1 + 4π2| · |2)`2 u‖

m−sm−`L2 ‖(1 + 4π2| · |2)

m2 u‖

s−`m−`L2

= ‖u‖m−sm−`H`‖u‖

s−`m−`Hm

Remark B.6.22. Let s ∈ R and m ∈ N, m ≥ |s|. If θ ∈ Cm(RN )∩Wm,∞(RN ),then the map u 7→ θu is continuous Hs(RN )→ Hs(RN ), and

‖θu‖Hs ≤ C‖θ‖Wm,∞‖u‖Hs

This property is easy to prove if s = m ∈ Z, see Remark B.1.10 (ii) for the casem ≥ 0 and Remark B.1.16 (ii) for the case m < 0. See [2, Theorem 1.62] for aproof in the case s 6∈ Z.

B.6.6. Fractional order Sobolev spaces. The general case. Given s ∈ Rand 1 ≤ p ≤ ∞, we define

Hs,p(RN ) = u ∈ S ′(RN ); F−1[(1 + 4π2| · |2)s2 u] ∈ Lp(RN ) (B.6.78)

and

‖u‖Hs,p = ‖F−1[(1 + 4π2| · |2)s2 u]‖Lp (B.6.79)

for u ∈ Hs,p(RN ). We note that (B.6.78) makes sense. Indeed, given u ∈ S ′(RN ), itfollows that u ∈ S ′(RN ). Therefore, (1+4π2|·|2)

s2 u ∈ S ′(RN ) by Remark B.6.10 (v).

Hence F−1[(1 + 4π2| · |2)s2 u] ∈ S ′(RN ), so that it makes sense to say that F−1[(1 +

4π2| · |2)s2 u] ∈ Lp(RN ) by Remark B.6.10 (iii).

Remark B.6.23. Here are some comments on the above definition.

(i) It follows from (B.6.78) and (B.6.79) that H0,p(RN ) = Lp(RN ) with equalnorms for all 1 ≤ p ≤ ∞.


(ii) Given s ∈ R, define Jsu for u ∈ S ′(RN ) by

Jsu = F−1[(1 + 4π2| · |2)s2 u] (B.6.80)

(Js are the Bessel potentials.) It follows from Remark B.6.12 (i) and (ii), andRemark B.6.10 (v) that Js ∈ L(S ′(RN )) for all s ∈ R. Furthermore, givens, σ ∈ R,

Jσ[Jsu] = F−1[(1 + 4π2| · |2)σ2F(Jsu)]

= F−1[(1 + 4π2| · |2)σ2 (1 + 4π2| · |2)

s2 u] = Js+σu

for all u ∈ S ′(RN ), so that

JσJs = Js+σ (B.6.81)

Since J0 = I, it follows that Js : S ′(RN )→ S ′(RN ) is a bijection, and

J−1s = J−s (B.6.82)

for all s ∈ R. Note that, given u ∈ S ′(RN ) and 1 ≤ p ≤ ∞, we haveu ∈ Hs,p(RN ) iff Jsu ∈ Lp(RN ), and

‖u‖Hs,p = ‖Jsu‖Lp (B.6.83)

Therefore, Hs,p(RN ) = J−1s (Lp(RN )) = J−s(L

p(RN )).(iii) Let s ∈ R, β a multi-index, and u ∈ S ′(RN ). If Js is defined by (B.6.80), it

follows from (B.6.42) (where u is replaced by Jsu) that

DβJsu = (2πi)|β|F−1[ξβF(Jsu)] = (2πi)|β|F−1[(1 + 4π2|ξ|2)s2 ξβ u]

Applying once more (B.6.42), we obtain DβJsu = F−1[(1+4π2|ξ|2)s2F(Dβu)].

Thus we see that

DβJsu = JsDβu (B.6.84)

(iv) Let s, σ ∈ R, 1 ≤ p ≤ ∞ and u ∈ S ′(RN ). It follows from (ii) above thatu ∈ Hs+σ,p(RN ) iff Js ∈ uHσ,p(RN ), and in this case

‖u‖Hσ+s,p = ‖Jsu‖Hσ,p (B.6.85)

(v) It follows from (B.6.44) that, with the notation (B.6.80), J2u = −∆u+ u forall u ∈ S ′(RN ). In particular, we see that, given 1 ≤ p ≤ ∞, u ∈ H2,p(RN )iff −∆u+ u ∈ Lp(RN ). In this case,

‖u‖H2,p = ‖ −∆u+ u‖Lp (B.6.86)

(vi) It follows from (ii) above that Hs,p(RN ) is a Banach space. Indeed, if (un)n≥1

is a Cauchy sequence in Hs,p(RN ), then (Jsun)n≥0 is a Cauchy sequence inLp(RN ). Therefore, there exists v ∈ Lp(RN ) such that Jsun → v in Lp. Thismeans that un → u in Hs,p, where u = J−sv ∈ Hs,p(RN ).

(vii) Let s ∈ R and 1 ≤ p ≤ ∞. The operators F and F−1, and the multiplicationby (1 + |ξ|2)

s2 are all continuous S(RN )→ S(RN ). (See Remark B.6.4 (i) and

Remark B.6.8.) Since S(RN ) → Lp(RN ), it follows that S(RN ) → Hs,p(RN ).Moreover, if 1 ≤ p < ∞, then the embedding S(RN ) → Hs,p(RN ) is dense.Indeed, let u ∈ Hs,p(RN ) and let v = Jsu ∈ Lp(RN ). Since p < ∞, thereexists (vn)n≥1 ⊂ S(RN ) such that vn → v in Lp. Therefore, un → u inHs,p, where un = J−svn. Moreover, vn ∈ S(RN ), so that Fvn ∈ S(RN ); andso (1 + 4π2| · |2)−

s2Fvn ∈ S(RN ). Therefore, un = J−svn ∈ S(RN ), which

completes the proof.(viii) If 1 ≤ p < ∞ and s ∈ R, then (Hs,p(RN ))? = H−s,p

′(RN ) with equivalent

norms. Indeed, any f ∈ H−s,p′(RN ) defines an element F ∈ (Hs,p(RN ))? by

F (u) = 〈J−sf, Jsu〉Lp′ ,Lp (B.6.87)


as follows easily from (ii) above. Applying (B.6.83), we see that ‖F‖(Hs,p)? ≤‖f‖H−s,p′ . ThusH−s,p

′(RN ) → (Hs,p(RN ))?. Conversely, let F be an element

of (Hs,p(RN ))?. It follows from (ii) that the map w 7→ F (J−sw) is a linear,continuous form on Lp(RN ) with norm ‖F‖(Hs,p)? . Thus there exists g ∈Lp′(RN ) such that ‖g‖Lp′ = ‖F‖(Hs,p)? and

F (J−sw) = 〈g, w〉Lp′ ,Lp

for all w ∈ Lp(RN ). Given u ∈ Hs,p(RN ), applying the above formula withw = Jsu ∈ Lp(RN ) and using (B.6.82), we obtain

F (u) = 〈g, Jsu〉Lp′ ,Lp

Letting f = Jsg ∈ H−s,p′(RN ), this means that F has the form (B.6.87).

Since ‖f‖H−s,p′ = ‖g‖Lp′ = ‖F‖(Hs,p)? , we conclude that (Hs,p(RN ))? →H−s,p

′(RN ).

Proposition B.6.24. Suppose 1 ≤ p ≤ ∞ and s, s′ ∈ R. If s ≥ s′, thenHs,p(RN ) → Hs′,p(RN ). More precisely,

‖u‖Hs′,p ≤ ‖u‖Hs,p (B.6.88)

for all u ∈ Hs,p(RN ).

Proof. Let u ∈ Hs,p(RN ). We have

F−1[(1 + 4π2|ξ|2)s′2 u] = F−1[(1 + 4π2|ξ|2)−

s−s′2 (1 + 4π2|ξ|2)

s2 u]

= F−1[(1 + 4π2|ξ|2)−s−s′

2 ] ? F−1[(1 + 4π2|ξ|2)s2 u]

(B.6.89)

Moreover, it follows from (B.6.54) that

‖F−1[(1 + 4π2|ξ|2)−s−s′

2 ]‖L1 = 1. (B.6.90)

Estimate (B.6.88) then follows from (B.6.89), (B.6.90) and Young’s inequality.

Proposition B.6.25. Let θ ∈ S(RN ) and define L : S ′(RN ) → S ′(RN ) byLu = θ ? u for all u ∈ S ′(RN ). Let 1 ≤ p ≤ q ≤ ∞, s, σ ∈ R, and β a multi-index.If u ∈ Hs,p(RN ), then DβLu ∈ Hσ,q(RN ) and

‖DβLu‖Hσ,q ≤ ‖Dβθ‖Hσ−s,r‖u‖Hs,p (B.6.91)

where 1 ≤ r ≤ ∞ is given by 1r = 1

q −1p + 1. In particular (with s = σ)

‖DβLu‖Hs,q ≤ ‖Dβθ‖Lr‖u‖Hs,p (B.6.92)

and (with σ = s and q = p)

‖DβLu‖Hs,p ≤ ‖Dβθ‖L1‖u‖Hs,p (B.6.93)

Proof. By Theorem B.6.14 (i) and (iv), DβLu = (Dβθ) ? u = F−1(Dβθu), sothat

(1 + 4π2| · |2)σ2F(Lu) = (1 + 4π2| · |2)

σ2 θu = [(1 + 4π2| · |2)

σ−s2 θ][(1 + 4π2| · |2)

s2 u]

Therefore, by Theorem B.6.14 (iv),

F−1[(1 + 4π2| · |2)σ2F(Lu)] = F−1[θ(1 + 4π2| · |2)

σ−s2 ] ? F−1[(1 + 4π2| · |2)

s2 u]

Taking the Lr norm and applying Young’s inequality to the right-hand side yieldsestimate (B.6.91).

It turns out that for any 1 < p <∞, then the spaces Wm,p(RN ) and Hm,p(RN )coincide, with equivalent norms.


Theorem B.6.26. Given any m ∈ Z and 1 < p <∞, Wm,p(RN ) = Hm,p(RN ),and ‖u‖Wm,p ≈ ‖u‖Hm,p .

The proof of Theorem B.6.26 in the case p = 2 is rather elementary, see Propo-sition B.6.17. When p 6= 2, the proof is much more delicate. It is based on thefollowing result.

Theorem B.6.27 (The Mihlin multiplier theorem). Let ρ ∈ L∞(RN ) and let

` > N/2 be an integer. Suppose ρ ∈W `,∞loc (RN \ 0) and

sup|α|≤`

ess supξ∈RN

|ξ||α||∂αρ(ξ)| <∞.

It follows that for every 1 < p <∞, there exists a constant Cp such that

‖F−1(ρv)‖Lp ≤ Cp‖v‖Lp (B.6.94)

for all v ∈ S(RN ).

Theorem B.6.27 is a deep theorem in harmonic analysis. This is for instanceTheorem 6.1.6, p. 135 in [4]. The proof is delicate, and an essential ingredient isthe Marcinkiewicz interpolation theorem. In fact, only a simplified form of thistheorem is needed, namely the form stated in [30], §4.2, Theorem 5, p. 21. Asimple proof of this (simplified version of the) Marcinkiewicz interpolation theoremis given in [30], pp. 21–22.

Proof of Theorem B.6.26. We fix m ∈ Z and 1 < p < ∞, and we proceedin five steps.

Step 1. S(RN ) is dense inHm,p(RN ). Let u ∈ Hm,p(RN ) and set w = F−1[(1+4π2| · |2)

m2 u] ∈ Lp(RN ). S(RN ) being dense in Lp(RN ), there exists (wn)n≥0 ⊂

S(RN ) such that wn → w in Lp(RN ). Setting un = F−1[(1 + 4π2| · |2)−m2 wn] ∈

S(RN ), this means that un → u in Hm,p(RN ).

Step 2. If m ≥ 0, then Hm,p(RN ) →Wm,p(RN ). By Step 1, it suffices to showthat ‖u‖Wm,p ≤ C‖u‖Hm,p for all u ∈ S(RN ). Let α be a multi-index with |α| ≤ mand let

ρ(ξ) = ξα(1 + 4π2|ξ|2)−m2

One checks easily that ρ satisfies the assumptions of Theorem B.6.27. Applyingestimate (B.6.94) with v = F−1[(1+4π2| · |2)

m2 u], we deduce that ‖F−1(ξαu)‖Lp ≤

C‖u‖Hm,p . Since F−1(ξαu) = (2πi)−|α|Dαu, we obtain ‖Dαu‖Lp ≤ C‖u‖Hm,p .The result follows, since α with |α| ≤ m is arbitrary.

Step 3. If m ≥ 0, then Wm,p(RN ) → Hm,p(RN ). By density of S(RN ) inWm,p(RN ), it suffices to show that ‖u‖Hm,p ≤ C‖u‖Wm,p for all u ∈ S(RN ). Fix afunction θ ∈ C∞(R), θ ≥ 0, such that θ(t) = 0 for |t| ≤ 1 and θ(t) = 1 for |t| ≥ 2.Set

ρ(ξ) = (1 + 4π2|ξ|2)m2

(1 +

N∑j=1

θ(ξj)|ξj |m)−1

.

It is not difficult to show that ρ satisfies the assumptions of Theorem B.6.27. Ap-

plying (B.6.94) with v = F−1[(1 +∑Nj=1 θ(ξj)|ξj |m)u], we deduce that

‖u‖Hm,p ≤ C∥∥∥F−1

[(1 +

N∑j=1

θ(ξj)|ξj |m)u]∥∥∥Lp

≤ C(‖u‖Lp +

N∑j=1

‖F−1(θ(ξj)|ξj |mu)‖Lp).

(B.6.95)


Next, we observe that ρj(ξ) = θ(ξj)|ξj |mξ−mj satisfies the assumptions of Theo-

rem B.6.27. Applying (B.6.94) with ρ = ρj and v = u, successively for j = 1, · · · , N ,we deduce from (B.6.95) that

‖u‖Hm,p ≤ C(‖u‖Lp +

N∑j=1

‖F−1(ξmj u)‖Lp)

= C(‖u‖Lp + (2π)−m

N∑j=1

‖∂mj u‖Lp)≤ C‖u‖Wm,p ,

which is the desired property.

Step 4. The conclusion of the theorem holds for all m ≥ 0. This follows fromSteps 2 and 3.

Step 5. The conclusion of the theorem holds for all m < 0. Indeed, it followsfrom Step 4 that W−m,p

′(RN ) = H−m,p

′(RN ) with equivalent norms. Since, by

definition, (W−m,p′(RN ))? = Wm,p(RN ), and (H−m,p

′(RN ))? = Hm,p(RN ) by

Remark B.6.23 (viii), the result follows.

Below are some other useful applications of the Mihlin multiplier theorem.

Proposition B.6.28. Let 1 < p <∞, s ∈ R, and a, b > 0. It follows that

Hs,p(RN ) = u ∈ S ′(RN ); F−1[(a+ b| · |2)s2 u] ∈ Lp(RN ) (B.6.96)

and that there exists constants 0 < c ≤ C <∞ such that

c‖u‖Hs,p ≤ ‖F−1[(a+ b| · |2)s2 u]‖Lp ≤ C‖u‖Hs,p (B.6.97)

for u ∈ Hs,p(RN ).

Proof. By density of S(RN ) in Hs,p(RN ) (see Remark B.6.23 (vii)), it sufficesto prove (B.6.97) for all u ∈ S(RN ). We first consider

ρ(ξ) =(1 + 4π2|ξ|2)

s2

(a+ b|ξ|2)s2

It is not difficult to check that ρ satisfies the assumptions of Theorem B.6.27.Applying (B.6.94) with v = F−1[(a + b| · |2)

s2 u], we deduce the first inequality

in (B.6.97). The second inequality follows similarly by considering

ρ(ξ) =(a+ b|ξ|2)

s2

(1 + 4π2|ξ|2)s2

and applying (B.6.94) with v = F−1[(1 + 4π2| · |2)s2 u].

Proposition B.6.29. Let 1 < p < ∞, s ∈ R, and β a multi-index. It followsthat Dβu ∈ Hs−|β|,p(RN ) for all u ∈ Hs,p(RN ). Moreover, there exists a constantC such that

‖Dβu‖Hs−|β|,p ≤ C‖u‖Hs,p (B.6.98)

for all u ∈ Hs,p(RN ). In particular, ‖∆u‖Hs−2,p ≤ C‖u‖Hs,p for all u ∈ Hs,p(RN ).

Proof. By density of S(RN ) in Hs,p(RN ) (see Remark B.6.23 (vii)), it sufficesto prove (B.6.98) for all u ∈ S(RN ). This easily follows by applying (B.6.94) with

ρ(ξ) =(2πi)|β|ξβ

(1 + 4π2|ξ|2)|β|2

and v = F−1[(1 + 4π2| · |2)s2 u].


Proposition B.6.30. Let 1 < p < ∞, s ∈ R, m ∈ N, and u ∈ S ′(RN ). Itfollows that u ∈ Hs+m,p(RN ) iff Dβu ∈ Hs,p(RN ) for all multi-indices β such that|β| ≤ m. Moreover, there exist constants 0 < c ≤ C <∞ such that

c‖u‖Hs+m,p ≤∑|β|≤m

‖Dβu‖Hs,p ≤ C‖u‖Hs+m,p (B.6.99)

for all u ∈ Hs+m,p(RN ).

Proof. It follows from Remark B.6.23 (iv) that u ∈ Hs+m,p(RN ) iff Jsu ∈Hm,p(RN ). By Theorem B.6.26, this is equivalent to DβJsu ∈ Lp(RN ) for all|β| ≤ m, and

‖u‖Hs+m,p ≈∑|β|≤m

‖DβJsu‖Lp

Since DβJsu = JsDβu by (B.6.84), we deduce from (B.6.85) that ‖DβJsu‖Lp =

‖Dβu‖Hs,p , and the result follows.

Remark B.6.31. The fractional order Sobolev spaces Hs,p(RN ) are complexinterpolation spaces of the standard Sobolev spaces Wm,p(RN ). More precisely,given 1 < p <∞, `,m ∈ Z such that ` < m, and s ∈ (`,m),

Hs,p(RN ) = [W `,p(RN ),Wm,p(RN )][ s−`m−` ] (B.6.100)

See e.g. [4, Theorem 6.4.5 (7)]. In particular, the interpolation inequality

‖u‖Hs,p ≤ C‖u‖m−sm−`W `,p‖u‖

s−`m−`Wm,p (B.6.101)

holds for all u ∈ Wm,p(RN ). (Inequality (B.6.101) is immediate if p = 2, seeRemark B.6.21.)

Remark B.6.32. Let s ∈ R and m ∈ N, m ≥ |s|. If θ ∈ Cm(RN )∩Wm,∞(RN ),then the map u 7→ θu is continuous Hs,p(RN ) → Hs,p(RN ) for all 1 < p < ∞.Moreover,

‖θu‖Hs,p ≤ C‖θ‖Wm,∞‖u‖Hs,pThis property is easy to prove if s = m ∈ Z, see Remark B.1.10 (ii) for the casem ≥ 0 and Remark B.1.16 (ii) for the case m < 0. It seems that there is no reallysimple proof of the property if s 6∈ Z and p 6= 2. One of the several possible proofsconsists in considering two consecutive integers m1 < m2 such that m1 ≤ s ≤m2, and using interpolation (see e.g. [4, Theorems 4.1.2]) together with (B.6.100).See [2, Theorem 1.62] for a direct proof in the case p = 2.

APPENDIX C

Vector integration

Vector integration is an important tool in the study of evolution equation.Even though most existence and regularity results are stated in terms of continuousfunctions, weaker regularity classes often appear in intermediate steps.

Throughout this chapter, X is a Banach space with the norm ‖ · ‖ and Iis an open interval of R (bounded or unbounded) equipped with the Lebesguemeasure. We will use the basic theorems of real valued integration (Fatou’s lemma,the monotone convergence theorem, the dominated convergence theorem, Egorov’stheorem in particular).

C.1. Measurable functions

Definition C.1.1. A function f : I → X is measurable if there exists a setN ⊂ I of measure 0 and a sequence (fn)n∈N ⊂ Cc(I,X) such that lim

n→∞fn(t) = f(t),

for all t ∈ I \N .

Remark C.1.2. It follows easily from Definition C.1.1 that if f : I → X ismeasurable, then ‖f‖ : I → R is also measurable. Many properties of vector valuedmeasurable functions follow either immediately from the definition or else from theproperties of real valued measurable functions applied to ‖f − fn‖. In particular,one can show easily the following results.

(i) If f : I → X is measurable and if Y is a Banach space such that X → Y ,then f : I → Y is measurable.

(ii) If a sequence (fn)n∈N of measurable functions I → X converges a.a. (in theX topology) to a function f : I → X, then f is measurable.

(iii) If f : I → X and ϕ : I → R are measurable, then fϕ : I → X is measurable.In particular, if f : I → X is measurable and if J ⊂ I is an open interval,then f|J : J → X is measurable (take ϕ = 1J).

(iv) If (xn)n∈N is a family of elements of X and if (ωn)n∈N is a family of measurablesubsets of I such that ωi ∩ ωj = ∅ for i 6= j, then

∑∞n=0 xn1ωn : I → X is

measurable.

In Definition C.1.1 and Remark C.1.2, the strong topology of X is involved.However, in many applications, one needs to prove measurability of a function whichis only the limit in the weak topology of X of a sequence of measurable functions.For that purpose, a most useful tool is the following result.

Theorem C.1.3 (Pettis’ Theorem). Consider f : I → X. Then f is measurableif and only if it satisfies the following two conditions:

(i) f is weakly measurable (i.e. for every x′ ∈ X?, the function t 7→ 〈x′, f(t)〉 ismeasurable I → R);

(ii) there exists a set N ⊂ I of measure 0 such that f(I \N) is separable.

Proof. It is clear that measurability implies weak measurability; and so (i) isnecessary. If f is measurable and if (fn)n∈N ⊂ Cc(I,X) converges to f on I\N with|N | = 0, then fn(I \N) is separable; and so f(I \N) is also separable. Therefore (ii)is also necessary.

121

122 C. VECTOR INTEGRATION

Let now f satisfy (i) and (ii). By possibly replacing X by the smallest closedsubspace of X containing f(I \ N), we may assume that X is separable. We firstestablish that for every x ∈ X, the function t 7→ ‖f(t)− x‖ is measurable. Indeed,for a ≥ 0, we have

t ∈ I, ‖f(t)− x‖ ≤ a =⋂x′∈S′

t ∈ I, |〈x′, f(t)− x〉| ≤ a,

where S′ is the unit ball of X?. It follows from Lemma A.1.7 that there exists asequence (x′n)n∈N ⊂ S′such that

t ∈ I, ‖f(t)− x‖ ≤ a =⋂n∈Nt ∈ I, |〈x′n, f(t)− x〉| ≤ a.

The set on the right-hand side of the above identity is clearly measurable by as-sumption (i); and so the function t 7→ ‖f(t)− x‖ is measurable.

Consider now n ∈ N. The set f(I) being separable, it can be covered by acountable union of balls Bnj of center xnj and radius 1/n. Let fn : I → X defined

by fn =∑∞j=0 x

nj 1ωnj , where ωn0 = t ∈ I, f(t) ∈ Bn0 and ωnj = t ∈ I, f(t) ∈

Bnj \ ∩j−1

k=0Bnk , for j ≥ 1. It is immediate that ‖f(t) − fn(t)‖ ≤ 1/n, for all t ∈ I.

Furthermore, since the function t 7→ ‖f(t)−x‖ is measurable for all x ∈ X, it followsthat the sets ωnj are measurable; and so, by Remark C.1.2 (iv), fn is measurable.Therefore, by Remark C.1.2 (ii) f is measurable.

Corollary C.1.4. If f : I → X is weakly continuous (i.e. t 7→ 〈x′, f(t)〉X?,Xis continuous for every x′ ∈ X?), then f is measurable.

Proof. f is clearly weakly measurable; and so by Theorem C.1.3, it is sufficientto prove that f(I) is separable. It follows from the weak continuity of f thatf(I) ⊂ E, where E is the weak closure of the convex hull of f(I ∩Q). On the other

hand, E = f(I ∩Q); and so E is separable. Hence the result.

Corollary C.1.5. Let (fn)n∈N be a sequence of measurable functions I → Xand let f : I → X. If, for a.a. t ∈ I, fn(t) f(t) in X as n → ∞, then f ismeasurable.

Proof. Let x′ ∈ X?. Since 〈x′, fn(t)〉 → 〈x′, f(t)〉 a.a., it follows that thefunction t 7→ 〈x′, f(t)〉 is measurable; and so f is weakly measurable.

On the other hand, it follows from Theorem C.1.3 that for every n ∈ N, thereexists a set Nn of measure 0 such that fn(I \ Nn) is separable. Consider theset N = ∪∞n=0Nn, which is also of measure 0, and let C be the convex hull of

∪∞n=0fn(I\E). Clearly f(I\E) ⊂ C, where C is the weak closure of C. Furthermore,

C = C; and so C is separable. Hence the result, by Theorem C.1.3.

Corollary C.1.6. Let X → Y be two Banach spaces and let u : I → Y beweakly continuous. Assume that there exists a dense subset E of I such that

(i) u(t) ∈ X, for all t ∈ E,(ii) sup‖u(t)‖X , t ∈ E = K <∞.

If X is reflexive, then u(t) ∈ X, for all t ∈ I, and u : I → X is measurable.

Proof. The result follows from Corollaries A.1.11 and C.1.4.

Remark C.1.7. Consider two Banach spaces X → Y , and a measurable func-tion f : I → Y . Assume that f(t) ∈ X, for a.a. t ∈ I. It is natural to ask whetherf : I → X is measurable. In general, the answer is negative, as shows the followingexample. Let I = Ω = (0, 1) and consider the function u : I → L∞(Ω) given byu(t) = 1(0,t), for 0 < t < 1. One verifies easily that u ∈ C0,1/p(I, Lp(Ω)), for every

C.2. INTEGRABLE FUNCTIONS 123

p ∈ [1,∞). In particular, u : I → Lp(Ω) is measurable, for every p ∈ [1,∞). Fur-thermore, u(t) ∈ L∞(Ω) for all t ∈ I. However, u : I → L∞(Ω) is not measurable.To see this, observe that ‖u(t)−u(s)‖L∞ = 1, if t 6= s. Therefore, u(I) is a discretesubset of L∞(Ω); and so, given any non-countable subset A of I, u(A) ⊂ L∞(Ω) isdiscrete and non-countable, hence non-separable. In particular, given a subset Nof I of measure 0, u(I \N) is not a separable subset of L∞(Ω). Therefore, by Theo-rem C.1.3, u : I → L∞(Ω) is not measurable. Note that u is an elementary exampleof a non-measurable function. However, one can obtain measurability results underadditional assumptions. This is the object of the following result.

Proposition C.1.8. Let X → Y be two Banach spaces and let f : I → Ybe a measurable function. If f(t) ∈ X for a.a. t ∈ I and if X is reflexive, thenf : I → X is measurable.

Proof. By applying Theorem C.1.3 and by modifying f on a set of measure 0,we may assume that f(I) ⊂ X and that f(I) is a separable subset of Y . By replacingX by the smallest closed subspace of X containing f(I), then by replacing Y bythe closure of X in Y , we may assume that Y is separable and that the embeddingX → Y is dense. By applying Lemma A.1.10, it follows that X is separable.Therefore, by applying again Theorem C.1.3, we need only check that f is weaklymeasurable I → X. To see this, consider x′ ∈ X?. It follows from Theorem A.1.5that there exists (y′n)n∈N ⊂ Y ? such that y′n −→

n→∞x′ in X?. In particular,

〈y′n, f(t)〉X?,X −→n→∞

〈x′, f(t)〉X?,X , for all t ∈ I.

On the other hand, we have 〈y′n, f(t)〉X?,X = 〈y′n, f(t)〉Y ?,Y by Theorem A.1.5.Therefore, the mapping t 7→ 〈y′n, f(t)〉X?,X is measurable; and so, the mappingt 7→ 〈x′, f(t)〉X?,X is measurable. Hence the result.

C.2. Integrable functions

Definition C.2.1. A measurable function f : I → X is integrable if thereexists a sequence (fn)n∈N ⊂ Cc(I,X) such that

limn→∞

∫I

‖fn(t)− f(t)‖ dt = 0.

(Note that by Remark C.1.2, ‖fn−f‖ : I → R is a nonnegative measurable function,so that

∫I‖fn(t)− f(t)‖ dt makes sense.)

Lemma C.2.2. Let f : I → X be integrable. There exists i(f) ∈ X such thatfor any sequence (fn)n∈N ⊂ Cc(I,X) satisfying

limn→∞

∫I

‖fn(t)− f(t)‖ dt = 0,

one has

limn→∞

∫I

fn(t) dt = i(f),

the above limit being for the strong topology of X.

Proof. Let (fn)n∈N ⊂ Cc(I,X) satisfy the assumption of the lemma. We have∥∥∥∫I

fn(t) dt−∫I

fp(t) dt∥∥∥ ≤ ∫

I

‖fn(t)− fp(t)‖ dt

≤∫I

‖fn(t)− f(t)‖ dt+

∫I

‖fp(t)− f(t)‖ dt.


Therefore,∫Ifn(t) dt is a Cauchy sequence, that converges to some element x ∈ X.

Consider another sequence (gn)n∈N ⊂ Cc(I,X). We have∥∥∥∫I

gn(t) dt− x∥∥∥

≤∥∥∥∫

I

gn(t)− f(t) dt∥∥∥+

∥∥∥∫I

f(t)− fn(t) dt∥∥∥+

∥∥∥∫I

fn(t) dt− x∥∥∥

≤∫I

‖gn(t)− f(t)‖ dt+

∫I

‖fn(t)− f(t)‖ dt+∥∥∥∫

I

fn(t) dt− x∥∥∥

Therefore,

∫I

gn(t) dt converges also to x, as n → ∞. Hence the result, with

i(f) = x.

Definition C.2.3. The element i(f) defined by Lemma C.2.2 is called theintegral of f on I. We note

i(f) =

∫f =

∫I

f =

∫I

f(t) dt.

If I = (a, b), we also note

i(f) =

∫ b

a

f =

∫ b

a

f(t) dt.

As for real-valued functions, it is convenient to note∫ β

α

f(t) dt = −∫ α

β

f(t) dt,

if β < α.

Theorem C.2.4 (Bochner’s Theorem). Let f : I → X be measurable. Then fis integrable if and only if ‖f‖ : I → R is integrable. In addition,∥∥∥∫

I

f(t) dt∥∥∥ ≤ ∫

I

‖f(t)‖ dt,

for all integrable functions f : I → X.

Proof. Assume that f is integrable, and consider a sequence (fn)n∈N ⊂Cc(I,X) such that

limn→∞

∫I

‖fn(t)− f(t)‖ dt = 0.

We have

‖f‖ ≤ ‖fn‖+ ‖fn − f‖;and so ‖f‖ is integrable.

Conversely, suppose that f is measurable and that ‖f‖ is integrable. Let(gn)n∈N ⊂ Cc(I,R) be a sequence such that gn → ‖f‖ in L1(I) and a.a., andsuch that |gn| ≤ g a.a., for some g ∈ L1(I). Let (fn)n∈N ⊂ Cc(I,X) be a sequencesuch that fn → f a.e. Finally, let

hn =fn|gn|

‖fn‖+ 1/n.

It is clear that hn ∈ Cc(I,X), that ‖hn‖ ≤ g a.a. and that hn → f in X a.a., asn→∞. It follows from the dominated convergence theorem that

limn→∞

∫‖hn(t)− f(t)‖ dt = 0;

C.2. INTEGRABLE FUNCTIONS 125

and so f is integrable. Finally,∥∥∥∫ f(t) dt∥∥∥ = lim

n→∞

∥∥∥∫ hn(t) dt∥∥∥ ≤ lim

n→∞

∫‖hn(t)‖ dt ≤

∫‖f(t)‖ dt,

where the last inequality follows from the dominated convergence theorem. Thiscompletes the proof.

Remark C.2.5. Theorem C.2.4 allows one to deal with vector valued integrablefunctions like one deals with real valued integrable functions. It suffices in general toapply the usual convergence theorems to ‖f‖. For example, one can easily establishthe following results.

(i) If f : I → X is integrable and ϕ ∈ L∞(I), then fϕ : I → X is integrable. Inparticular, if f : I → X is integrable and if J ⊂ I is an open interval, thenf|J : J → X is integrable (take ϕ = 1J).

(ii) (The dominated convergence theorem) Let (fn)n∈N be a sequence of integrablefunctions I → X, let f : I → X and let g ∈ L1(I). If

‖fn(t)‖ ≤ g(t) for a.a. t ∈ I and all n ∈ N,limn→∞

fn(t) = f(t) for a.a. t ∈ I,

then f is integrable and

∫I

f(t) dt = limn→∞

∫I

fn(t) dt.

(iii) If Y is a Banach space, if A ∈ L(X,Y ), and if f : I → X is integrable, thenAf : I → Y is integrable and∫

I

Af(t) dt = A(∫

I

f(t) dt).

In particular, if X → Y and if f : I → X is integrable, then the integral of fin the sense of X coincides with the integral of f in the sense of Y .

Finally, we have the following important geometric property of integrable func-tions.

Proposition C.2.6. Suppose |I| < ∞, let K ⊂ X be a closed convex set, letf : I → X be integrable and let

y =1

|I|

∫I

f(t) dt.

If f(t) ∈ K for a.a. t ∈ I, then y ∈ K.

Proof. We argue by contradiction and we assume that y 6∈ K. It follows fromHahn-Banach’s theorem (see [5, Theorem 1.7 p.7]) that there exists x′ ∈ X? andε > 0 such that 〈x′, x〉X?,X ≤ 〈x′, y〉X?,X − ε, for all x ∈ K. In particular,

〈x′, f(t)〉X?,X ≤ 〈x′, y〉X?,X − ε,

for a.a. t ∈ I. Integrating that above inequality and applying Remark C.2.5 (iii),we obtain

〈x′, y〉X?,X =1

|I|〈x′,

∫I

f(t) dt〉X?,X =1

|I|

∫I

〈x′, f(t)〉X?,X dt

≤ 〈x′, y〉X?,X − ε,

which is a contradiction. Hence the result.


C.3. The space Lp(I,X)

Definition C.3.1. Let p ∈ [1,∞]. One denotes by Lp(I,X) the set of (classesof) measurable functions f : I → X such that the function t 7→ ‖f(t)‖ belongs toLp(I). For f ∈ Lp(I,X), one defines

‖f‖Lp(I,X) =

∫I‖f(t)‖p dt

1p

if p <∞,ess supt∈I ‖f(t)‖ if p =∞.

When there is no risk of confusion, we denote ‖ · ‖Lp(I,X) by ‖ · ‖Lp(I) or ‖ · ‖Lp or‖ · ‖p. One denotes by Lploc(I,X) the set of f : I → X such that f|J ∈ Lp(J,X), forevery bounded sub-interval J of I.

Remark C.3.2. The space Lp(I,X) enjoys most of the properties of the spaceLp(I) = Lp(I,R), by the same proofs or by applying the classical results to thefunction t 7→ ‖f(t)‖. In particular, one obtains easily the following results.

(i) ‖ · ‖Lp(I,X) is a norm on the space Lp(I,X). Lp(I,X) equipped with thatnorm is a Banach space. If p < ∞, then C∞0 (I,X) is dense in Lp(I,X).Indeed, it is not difficult to prove using the argument in second step of theproof of Bochner’s Theorem C.2.4 that Cc(I,X) is dense in Lp(I,X). Thenthe density of C∞0 (I,X) follows by the classical procedure of convolution witha smoothing sequence. In particular, if Y is a Banach space such that Y → Xwith dense embedding, then C∞0 (I, Y ) is dense in Lp(I,X) (since C∞0 (I, Y )is dense in Cc(I,X) for the norm of Cb(I,X)).

(ii) A measurable function f : I → X belongs to Lp(I,X) if and only if thereexists a function g ∈ Lp(I) such that ‖f‖ ≤ g a.a. on I.

(iii) If f ∈ Lp(I,X) and ϕ ∈ Lq(I) with 1p + 1

q = 1r ≤ 1, then ϕf ∈ Lr(I,X) and

‖ϕf‖Lr(I,X) ≤ ‖f‖Lp(I,X)‖ϕ‖Lq(I).

In particular, if f ∈ Lp(I,X) and if J is an open sub-interval of I, thenf|J ∈ Lp(J,X).

(iv) If f ∈ Lp(I,X) and g ∈ Lq(I,X?) with 1p + 1

q = 1r ≤ 1, then the func-

tion h defined by h(t) = 〈g(t), f(t)〉X?,X belongs to Lr(I) and ‖h‖Lr ≤‖f‖Lp(I,X) ‖g‖Lq(I,X?).

(v) If f ∈ Lp(I,X) ∩ Lq(I,X) with p < q, then f ∈ Lr(I,X), for every r ∈ [p, q],and

‖f‖Lr(I,X) ≤ ‖f‖θLp(I,X)‖f‖1−θLq(I,X),

where 1r = θ

p + 1−θq .

(vi) If I is bounded and p ≤ q, then Lq(I,X) → Lp(I,X) and

‖f‖Lp(I,X) ≤ |I|q−ppq ‖f‖Lq(I,X),

for all f ∈ Lq(I,X).(vii) Suppose f : I → X is measurable. If f ∈ Lp(J,X) for all J ⊂⊂ I and

if ‖f‖Lp(J,X) ≤ C for some C independent of J , then f ∈ Lp(I,X) and‖f‖Lp(I,X) ≤ C.

(viii) If Y is a Banach space and if A ∈ L(X,Y ), then for every f ∈ Lp(I,X) wehave Af ∈ Lp(I, Y ) and

‖Af‖Lp(I,Y ) ≤ ‖A‖L(X,Y ) ‖f‖Lp(I,X).

In particular, if X → Y and if f ∈ Lp(I,X), then f ∈ Lp(I, Y ).

C.3. THE SPACE Lp(I,X) 127

(ix) (The dominated convergence theorem) Let (fn)n∈N ⊂ Lp(I,X) and let g ∈Lp(I). If p <∞ and

‖fn(t)‖ ≤ g(t), for a.a. t ∈ I and all n ∈ N,limn→∞

fn(t) exists for a.a. t ∈ I,

then f := limn→∞

fn ∈ Lp(I,X) and limn→∞

fn = f in Lp(I,X).

(x) Let (fn)n∈N ⊂ Lp(I,X) and let f ∈ Lp(I,X). If fn → f in Lp(I,X) asn → ∞, then there exists g ∈ Lp(I) and a subsequence (nk)n∈N such that‖fnk(t)‖ ≤ g(t) for a.a. t ∈ I and for every k ∈ N.

Remark C.3.3. Duality theorems for the spaces Lp(I,X) are much more diffi-cult to obtain than for the spaces Lp(I). However, if X is reflexive and if 1 < p <∞,

then it known that Lp(I,X) is reflexive and that (Lp(I,X))? ≈ Lp′(I,X?) (see [9,

Chapter 13, Corollary 1 of Theorem 8, p. 252]. If 1 ≤ p <∞ and if X is reflexive or

if X? is separable, then (Lp(I,X))? ≈ Lp′(I,X?) (see [9]). Below are some specialcases, in which such results are easily obtained.

(i) If X is a Hilbert space with the scalar product (·, ·), then it follows thatL2(I,X) is a Hilbert space, for the scalar product

〈〈f, g〉〉 =

∫I

(f(t), g(t)) dt, for f, g ∈ L2(I,X).

Therefore, L2(I,X) is reflexive and by the Riesz representation theorem, wesee that (L2(I,X))? ≈ L2(I,X) (or (L2(I,X))? ≈ L2(I,X?) if one does notidentify X? with X).

(ii) Let Ω be an open subset of RN and let 1 ≤ p < ∞. It follows easilyfrom Fubini’s Theorem and a density argument that the operator T de-fined on Lp(I, Lp(Ω)) by Tu(t, x) = u(t)(x) is an isometry from Lp(I, Lp(Ω))onto Lp(I × Ω); and so, Lp(I, Lp(Ω)) is reflexive and (Lp(I, Lp(Ω)))? ≈Lp′(I, Lp

′(Ω)) for every 1 < p <∞.

(iii) The results of (ii) above are not anymore valid for p = ∞. For example, letI = Ω = (0, 1) and consider the function u : I → L∞(Ω) given by u(t) = 1(0,t),for 0 < t < 1. Evidently Tu ∈ L∞(I × Ω), but u 6∈ L∞(I, L∞(Ω)). Infact, u : I → L∞(Ω) is not even measurable, as follows from Remark C.1.7.(However, observe that u ∈ C0,1/p(I, Lp(Ω)), for every p ∈ [1,∞).) It followsin particular that (L1(I, L1(Ω)))? 6≈ L∞(I, L∞(Ω)) since the linear form f onL1(I, L1(Ω)) defined by

f(v) =

∫I

∫Ω

v(t)u(t) dx dt

is continuous but cannot be written as

f(v) =

∫I

∫Ω

v(t)z(t) dx dt

for some z ∈ L∞(I, L∞(Ω)). Indeed, the definition of f makes sense, since ifv ∈ L1(I, L1(Ω)), then vu ∈ L1(I, L1(Ω)); and on the other hand, if z wouldexist, we would obtain easily that Tz = Tu, hence z = u.

Theorem C.3.4. Let 1 ≤ p ≤ ∞ and let (fn)n∈N be a bounded sequence inLp(I,X). If there exists f : I → X such that fn(t) f(t) in X as n → ∞, fora.a. t ∈ I, then the following properties hold:

(i) f ∈ Lp(I,X) and ‖f‖Lp(I,X) ≤ lim infn→∞

‖fn‖Lp(I,X);

(ii) if p > 1, then∫Ifn(t)ϕ(t) dt

∫If(t)ϕ(t) dt as n→∞, for every ϕ ∈ Lp′(I).


Proof. By Corollary C.1.5, f is measurable. If p <∞, it follows from Fatou’slemma that ∫

I

lim infn→∞

‖fn(t)‖p dt ≤ lim infn→∞

∫I

‖fn(t)‖p dt.

By weak lower semicontinuity of the norm, we have∫I

‖f(t)‖p dt ≤∫I

lim infn→∞

‖fn(t)‖p dt;

and so, ∫I

‖f(t)‖p dt ≤ lim infn→∞

∫I

‖fn(t)‖p dt,

from which (i) follows. The case p =∞ follows from an obvious adaptation of thisargument. Hence property (i).

We now prove (ii). Consider first ϕ ∈ Cc(I). Let x′ ∈ X? and set

hn(t) = 〈x′, fn(t)− f(t)〉X?,Xϕ(t),

for a.a. t ∈ I. It follows that hn(t) → 0 as n → ∞ for a.a. t ∈ I and that hnis bounded in Lp(I), as n → ∞. Since hn is supported in a compact interval, wededuce (see Lemma A.1.15 below) that hn → 0 in L1(I). In particular,

〈x′,∫I

fn(t)ϕ(t) dt〉X?,X −→n→∞

〈x′,∫I

f(t)ϕ(t) dt〉X?,X ,

from which property (ii) follows, since x′ is arbitrary. In the general case ϕ ∈ Lp′(I),

let (ϕ`)`≥0 ⊂ Cc(I) be such that ϕ` → ϕ in Lp′(I) as ` → ∞. Given x′ ∈ X?, we

have∣∣∣〈x′,∫I

(fn(t)− f(t))ϕ(t) dt〉X?,X∣∣∣

≤∣∣∣〈x′,∫

I

(fn(t)− f(t))(ϕ(t)− ϕ`(t) dt〉X?,X∣∣∣

+∣∣∣〈x′,∫

I

(fn(t)− f(t))ϕ`(t) dt〉X?,X∣∣∣.

Given any ε > 0, we estimate the first term on the right-hand side of the aboveinequality by ‖x′‖X?(‖fn‖Lp(I,X)+‖f‖Lp(I,X))‖ϕ`−ϕ‖Lp′ ≤ ε/2 if ` is large enough.Given such a `, the second term on the right-hand side is smaller than ε/2 for nlarge enough by what precedes. Since ε > 0 and x′ ∈ X? are arbitrary, the resultfollows.

Lemma C.3.5. Let (fn)n∈N ⊂ Lp(I,X) and f ∈ Lp(I,X), where 1 ≤ p ≤ ∞.If fn f in Lp(I,X) as n→∞, then∫

I

fn(t)ϕ(t) dt

∫I

f(t)ϕ(t) dt,

as n→∞ for every ϕ ∈ Cc(I).

Proof. Without loss of generality, we may assume that f = 0. Considerϕ ∈ C∞0 (I) and x′ ∈ X? and define the linear functional F on Lp(I,X) by

F (g) = 〈x,∫I

g(t)ϕ(t) dt〉X?,X ,

for every g ∈ Lp(I,X). It follows from Remark C.3.2 (iii) that F is continuous.Therefore, F ∈ (Lp(I,X))?, which implies that F (fn)→ F (0) as n→∞. Since x′

is arbitrary, this implies the result.

Bibliography

[1] Adams R.A. and Fournier J.J.F. Sobolev spaces. Second edition. Pure and Applied Mathe-matics (Amsterdam) 140. Elsevier/Academic Press, Amsterdam, 2003. (MR2424078)

[2] Bahouri H., Chemin J.-Y. and Danchin R. Fourier analysis and nonlinear partial differen-

tial equations. Grundlehren der Mathematischen Wissenschaften, 343. Springer, Heidelberg,2011. (MR2768550) (doi: 10.1007/978-3-642-16830-7)

[3] Bellman R. The stability of solutions of linear differential equations, Duke Math. J.10 (1943),

no. 4, 643–647. (MR0009408) (doi: 10.1215/S0012-7094-43-01059-2)[4] Bergh J. and Lofstrom J. Interpolation spaces. An introduction. Grundlehren der Mathe-

matischen Wissenschaften 223. Springer-Verlag, Berlin-New York, 1976. (MR0482275) (doi:

10.1007/978-3-642-66451-9)[5] Brezis H. Functional analysis, Sobolev spaces and partial differential equations. Universitext.

Springer, New York, 2011. (MR2759829) (doi: 10.1007/978-0-387-70914-7)

[6] Cazenave T. An introduction to semilinear elliptic equations. Editora do IM-UFRJ, Riode Janeiro, 2006. ix+193 pp. ISBN: 85-87674-13-7. (link: https://www.ljll.math.upmc.fr/

~cazenave/77.pdf)[7] Cazenave T. and Haraux A. An introduction to semilinear evolution equations, Oxford Lec-

ture Series in Mathematics and its Applications 13, Oxford University Press, Oxford, 1998.

(MR1691574)

[8] Courant R. and Hilbert D. Methods of mathematical physics. Vol. II: Partial differential

equations. Interscience Publishers (a division of John Wiley & Sons), New York-London

1962. (MR0140802)[9] Dinculeanu N. Vector measures. International Series of Monographs in Pure and Applied

Mathematics, Vol. 95 Pergamon Press, Oxford-New York-Toronto, Ont.; VEB Deutscher

Verlag der Wissenschaften, Berlin 1967. (MR0206190)[10] Dunford N. and Schwartz J.T. Linear Operators. I. General Theory. Pure and Applied Math-

ematics, Vol. 7 Interscience Publishers, Inc., New York; Interscience Publishers, Ltd., London

1958. (MR0117523)[11] Edwards R.E. Functional analysis: theory and applications. Corrected reprint of the 1965

original. Dover Publications, Inc., New York, 1995. (MR1320261)[12] Evans L.C. Partial differential equations. Graduate Studies in Mathematics, 19. American

Mathematical Society, Providence, RI, 1998. (MR1625845)

[13] Friedman A. Partial differential equations, Holt, Rinehart and Winston, Inc., New York-Montreal, Que.-London, 1969. (MR0445088)

[14] Gilbarg D. and Trudinger N.S. Elliptic partial differential equations of the second or-

der. Reprint of the 1998 edition. Classics in Mathematics. Springer-Verlag, Berlin, 2001.(MR1814364) (doi: 10.1007/978-3-642-61798-0)

[15] Gronwall T.H. Note on the derivative with respect to a parameter of the solutions of a

system of differential equations, Ann. of Math. 20 (1919), no. 4, 292–296. (MR1502565) (doi:10.2307/1967124)

[16] Haraux A. Linear semigroups in Banach spaces, in Semigroups, theory and applications, Vol.

II (Trieste, 1984), Pitman Res. Notes in Math. Ser. 152, Longman Sci. Tech., Harlow, 1986,93–135. (MR0879306)

[17] Hewitt E. and Stromberg K. Real and abstract analysis. A modern treatment of the theoryof functions of a real variable. Springer-Verlag, New York 1965. (MR0188387) (doi: 10.1007/

978-3-642-88047-6)[18] Hormander L. The analysis of linear partial differential operators. I. Distribution theory and

Fourier analysis. Reprint of the second (1990) edition. Classics in Mathematics. Springer-Verlag, Berlin, 2003. (MR1996773) (doi: 10.1007/978-3-642-61497-2)

[19] Kavian O. Introduction a la theorie des points critiques et applications aux problemes ellip-tiques, Mathematiques & Applications 13, Springer-Verlag, Paris, 1993. (MR1276944)

129

http://www.ams.org/mathscinet-getitem?mr=MR2424078


http://dx.doi.org/10.1007/978-3-642-16830-7


http://dx.doi.org/10.1215/S0012-7094-43-01059-2


http://dx.doi.org/10.1007/978-3-642-66451-9

http://dx.doi.org/10.1007/978-3-642-66451-9


http://dx.doi.org/10.1007/978-0-387-70914-7

https://www.ljll.math.upmc.fr/~cazenave/77.pdf

https://www.ljll.math.upmc.fr/~cazenave/77.pdf









http://dx.doi.org/10.1007/978-3-642-61798-0


http://dx.doi.org/10.2307/1967124

http://dx.doi.org/10.2307/1967124



http://dx.doi.org/10.1007/978-3-642-88047-6

http://dx.doi.org/10.1007/978-3-642-88047-6


http://dx.doi.org/10.1007/978-3-642-61497-2


130 BIBLIOGRAPHY

[20] Lax P.D. and Milgram, A. N. Parabolic equations. In Contributions to the theory of par-

tial differential equations, pp. 167–190. Annals of Mathematics Studies, no. 33. Princeton

University Press, Princeton, N. J., 1954. (MR0067317)[21] Lieb E.H. and Loss M. Analysis. Second edition. Graduate Studies in Mathematics, 14.

American Mathematical Society, Providence, RI, 2001. (MR1817225) (doi: 10.1090/gsm/014)[22] Linares F. and Ponce G. Introduction to nonlinear dispersive equations. Second edition. Uni-

versitext. Springer, New York, 2015. (MR3308874) (doi: 10.1007/978-1-4939-2181-2)

[23] Marcus M. and Mizel V.J. Complete characterization of functions which act, via superposi-tion, on Sobolev spaces, Trans. Amer. Math. Soc. 251 (1979), 187–218. (MR0531975) (doi:

10.1090/S0002-9947-1979-0531975-1)

[24] Pazy A. Semi-groups of linear operators and applications to partial differential equa-tions, Applied Math. Sciences 44, Springer, New-York, 1983. (MR0710486) (doi: 10.1007/

978-1-4612-5561-1)

[25] Quittner P. and Souplet P. Superlinear parabolic problems. Blow-up, global existence andsteady states, Birkhauser Advanced Texts, Birkhauser Verlag, Basel, 2007. 584 p.+xi. ISBN:

978-3-7643-8441-8 (MR2346798) (doi: 10.1007/978-3-7643-8442-5)

[26] Rudin W. Real and complex analysis. Third edition. McGraw-Hill Book Co., New York, 1987.(MR0924157)

[27] Rudin W. Functional Analysis. Second edition. International Series in Pure and AppliedMathematics. McGraw-Hill, Inc., New York, 1991. (MR1157815)

[28] Shatah J. and Struwe M. Geometric wave equations. Courant Lecture Notes in Mathematics,

2 . New York University, Courant Institute of Mathematical Sciences, New York; AmericanMathematical Society, Providence, RI, 1998. (MR1674843)

[29] Smets D. Personal communication.

[30] Stein E.M. Singular integrals and differentiability properties of functions, Princeton Univer-sity Press, Princeton, 1970. (MR0290095)

[31] Stein E.M. and Weiss G. Introduction to Fourier analysis on Euclidean spaces. Princeton

Mathematical Series, No. 32. Princeton University Press, Princeton, N.J., 1971. (MR0304972)[32] Struwe M. Variational methods. Applications to nonlinear partial differential equations and

Hamiltonian systems, Fourth edition, Ergebnisse der Mathematik und ihrer Grenzgebiete.

3. Folge. A Series of Modern Surveys in Mathematics, 34. Springer-Verlag, Berlin, 2008.(MR2431434) (doi: 10.1007/978-3-540-74013-1)

[33] Yosida K. Functional Analysis. Reprint of the sixth (1980) edition. Classics in Mathematics.

Springer-Verlag, Berlin, 1995. (MR1336382) (doi: 10.1007/978-3-642-61859-8)



http://dx.doi.org/10.1090/gsm/014


http://dx.doi.org/10.1007/978-1-4939-2181-2


http://dx.doi.org/10.1090/S0002-9947-1979-0531975-1

http://dx.doi.org/10.1090/S0002-9947-1979-0531975-1


http://dx.doi.org/10.1007/978-1-4612-5561-1

http://dx.doi.org/10.1007/978-1-4612-5561-1


http://dx.doi.org/10.1007/978-3-7643-8442-5







http://dx.doi.org/10.1007/978-3-540-74013-1


http://dx.doi.org/10.1007/978-3-642-61859-8

introduction to partial di erential equations thierry cazenave · introduction to partial di...

Documents