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Around the Brunn-Minkowski inequality

Andrea Colesanti

Technische Universitat Berlin - Institut fur Mathematik

January 28, 2015

Summary

I The Brunn-Minkowski inequality

I The isoperimetric inequality

I Infinitesimal form of Brunn-Minkowski inequality

I Inequalities of Brunn-Minkowski type

Summary

The Brunn-Minkowski inequality

Thm. A,B ⊂ Rn, compact; λ ∈ [0, 1]; then

Vn((1− λ)A + λB)1/n ≥ (1− λ)Vn(A)1/n + λVn(B)1/n. (BM)

I Vn = volume (Lebesgue measure);

(1− λ)A + λB = {(1− λ)a + λb : a ∈ A, b ∈ B}.

Equivalently: The functional V1/nn is concave in the class of

compact sets of Rn, equipped with the vector addition.

An excellent survey (much better than this talk):R. Gardner, The Brunn-Minkowski inequality, Bull. A.M.S., 2002.

(1− λ)A + λB = {(1− λ)a + λb : a ∈ A, b ∈ B}.

(BM) privileges convex sets

In the inequality

Vn((1− λ)A + λB)1/n ≥ (1− λ)Vn(A)1/n + λVn(B)1/n,

equality holds iff A is convex and B is a homothetic copy of A (upto subsets of volume zero).

Why? If you plug A = B in (BM) in general you don’t get anequality, because

(1− λ)A + λA 6= A ((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

In the inequality

(1− λ)A + λA 6= A ((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

In the inequality

(1− λ)A + λA 6= A ((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

In the inequality

(1− λ)A + λA 6= A

((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

In the inequality

(1− λ)A + λA 6= A ((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

In the inequality

(1− λ)A + λA 6= A ((1− λ)A + λA ⊃ A).

But if A is convex

(1− λ)A + λA = A ∀λ ∈ [0, 1].

Many equivalent forms

I Classic

Vn((1−λ)A+λB)1/n ≥ (1−λ)Vn(A)1/n +λVn(B)1/n. (BM)

I Elegant

Vn(A + B)1/n ≥ Vn(A)1/n + Vn(B)1/n. (BM∗)

I Multiplicative

Vn((1− λ)A + λB) ≥ Vn(A)1−λVn(B)λ. (BM0)

I Minimal

Vn((1− λ)A + λB) ≥ min{Vn(A),Vn(B)}. (BM−∞)

I Classic

I Elegant

I Multiplicative

I Minimal

I Classic

I Elegant

I Multiplicative

I Minimal

I Classic

I Elegant

I Multiplicative

I Minimal

I Classic

I Elegant

I Multiplicative

I Minimal

A general fact about homogeneous functional

Let F be a real-valued functional

I defined on a convex cone C;

I α-homogeneous: F(λx) = λαF(x), ∀x ∈ C, ∀λ > 0 (α > 0);

I non-negative.

F1/α concave ⇔ {F ≥ t} is convex ∀ t.

The last condition (quasi-concavity) is equivalent to

F((1− λ)A + λB) ≥ min{F(A),F(B)} ∀A,B ∈ C, ∀λ ∈ [0, 1].

In our case: C = {compact sets}, F = Vn, α = n.

I non-negative.

An “elementary” proof of (BM) - I

Lemma (Prekopa-Leindler inequality). Let

f , g , h : Rn → R+

be measurable functions, and let λ ∈ [0, 1]. Assume that

f ((1− λ)x + λy)) ≥ g(x)(1−λ) h(y)λ ∀ x , y ∈ Rn.

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

I Prove the 1-dimensional case (using just the so-called layercake, or Cavalieri’s, principle);

I the n-dimensional case follows by induction and Fubini’stheorem.

Lemma (Prekopa-Leindler inequality).

f , g , h : Rn → R+

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

be measurable functions,

and let λ ∈ [0, 1]. Assume that

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

be measurable functions, and let λ ∈ [0, 1].

Assume that

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

f , g , h : Rn → R+

Then ∫Rn

fdz ≥(∫

)1−λ (∫Rn

Proof.

A proof of (BM) - II

Given A,B ⊂ Rn and λ ∈ [0, 1], let

f = characteristic function of (1− λ)A + λB,

g = charact. function of A, h = charact. function of B.

By Prekopa-Leindler inequality

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

)λ= Vn(A)1−λVn(B)λ,

i.e. the multiplicative form of (BM).

A proof of (BM) - IIGiven A,B ⊂ Rn and λ ∈ [0, 1], let

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

= Vn(A)1−λVn(B)λ,

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

Vn((1− λ)A + λB) =

≥(∫

)1−λ (∫Rn

The isoperimetric inequality

Thm. Among all subsets of Rn with given perimeter, the ballhaving such perimeter maximizes the volume.

Equivalently,

Vn(A)n−1n ≤ c(n)Hn−1(∂A)

for every set A (with sufficiently smooth boundary), where c(n) isa constant and Hn−1 is the (n − 1)-dimensional Hausdorffmeasure. Equality is attained when A is a ball.

Equivalently,

(BM) ⇒ isoperimetric inequality

Let A ⊂ Rn be a bounded domain with C 1 boundary. Then

Hn−1(∂A) = limε→0+

Vn(Aε)− Vn(A)

Aε = {x ∈ Rn : dist(x ,A) ≤ ε}

= A + εB,

andB = {x ∈ Rn : ‖x‖ ≤ 1} = unit ball.

Hn−1(∂A) = limε→0+

Vn(A + εB)− Vn(A)

Hn−1(∂A) = limε→0+

Vn(Aε)− Vn(A)

Aε = {x ∈ Rn : dist(x ,A) ≤ ε}= A + εB,

Hn−1(∂A) = limε→0+

Vn(Aε)− Vn(A)

Aε = {x ∈ Rn : dist(x ,A) ≤ ε}= A + εB,

Hn−1(∂A) = limε→0+

Proof of the isoperimetric inequality

Hn−1(∂A) = limε→0+

By (BM), for every ε > 0

Vn(A + εB)1/n ≥ Vn(A)1/n + Vn(εB)1/n = Vn(A)1/n + εVn(B)1/n.

Vn(B)1/n ≤ limε→0+

Vn(A + εB)1/n − Vn(A)1/n

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

When A is a ball this becomes an equality.

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

Vn(A + εB)1/n ≥ Vn(A)1/n + Vn(εB)1/n

= Vn(A)1/n + εVn(B)1/n.

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A)

= c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

Hn−1(∂A) = limε→0+

nVn(A)

1−nn Hn−1(∂A).

Vn(A)n−1n ≤ 1

nVn(B)1/nHn−1(∂A) = c(n)Hn−1(∂A).

The infinitesimal form of (BM)

I By the Brunn-Minkowski inequality V1/nn is a concave

functional.

I Hence the second variation (or second differential) of V1/nn

(whatever that means) must be negative semidefinite:

D2(V1/nn ) ≤ 0.

I If we restrict our attention to convex sets, this fact amountsto a class of functional inequalities of Poincare type on theunit sphere; a prototype is∫

Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

∀ φ ∈ C 1(Sn−1), verifying some zero-mean condition.

functional.

D2(V1/nn ) ≤ 0.

Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

functional.

D2(V1/nn ) ≤ 0.

Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

functional.

D2(V1/nn ) ≤ 0.

I If we restrict our attention to convex sets, this fact amountsto a class of functional inequalities of Poincare type on theunit sphere;

a prototype is∫Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

functional.

D2(V1/nn ) ≤ 0.

Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

functional.

D2(V1/nn ) ≤ 0.

Sn−1

φ2dHn−1 ≤ c(n)

∫Sn−1

|∇φ|2dHn−1,

Convex bodies

From now on we will only consider a special type of compact sets:convex bodies.A convex body is a compact convex subset of Rn. We set

Kn = {convex bodies in Rn}.

Kn is closed under addition and dilations: given K , L ∈ Kn andα, β ≥ 0,

αK + βL ∈ Kn.

The Brunn-Minkowski inequality holds in particular in Kn.

Convex bodies

From now on we will only consider a special type of compact sets:convex bodies.

A convex body is a compact convex subset of Rn. We set

αK + βL ∈ Kn.

Convex bodies

αK + βL ∈ Kn.

Convex bodies

αK + βL ∈ Kn.

Convex bodies

αK + βL ∈ Kn.

From sets to functions:the support function of a convex body

I The support function hK of a convex body K is defined by:

hK : Sn−1 → R , hK (u) = sup{(u, v) |v ∈ K} .

hK (u) is the distance from the origin of the hyperplanesupporting K , with outer unit normal u.

I The passage to support functions preserves the linearstructure on Kn:

hαK+βL = αhK + βhL .

for every K , L ∈ Kn α, β ≥ 0.

Convex bodies of class C 2+

A convex body is said to be of class C 2+ if:

I ∂K ∈ C 2,

I the Gauss curvature is strictly positive on ∂K .

In terms of the support function h of K :

h ∈ C 2(Sn−1) , (hij + hδij) > 0 on Sn−1

(hij = second covariant derivatives of h on Sn−1, δij =Kronecker’s symbols).

C := {h ∈ C 2(Sn−1) : (hij + hδij) > 0 on Sn−1}= {support functions of C 2

+ convex bodies}.

I ∂K ∈ C 2,

+ convex bodies}.

I ∂K ∈ C 2,

h ∈ C 2(Sn−1) ,

(hij + hδij) > 0 on Sn−1

+ convex bodies}.

I ∂K ∈ C 2,

+ convex bodies}.

I ∂K ∈ C 2,

C := {h ∈ C 2(Sn−1) : (hij + hδij) > 0 on Sn−1}

= {support functions of C 2+ convex bodies}.

I ∂K ∈ C 2,

+ convex bodies}.

A representation formula for the volume

If K is of class C 2+ and h is its support function, then

Vn(K ) =1

∫Sn−1

h det(hij + hδij) dHn−1.

Now we define a functional F : C → R+ as

F(h) =

∫Sn−1

h det(hij + hδij) dHn−1]1/n

= Vn(K )1/n.

By the Brunn-Minkowski inequality,

F is concave in C.

Vn(K ) =1

∫Sn−1

F(h) =

∫Sn−1

= Vn(K )1/n.

F is concave in C.

Vn(K ) =1

∫Sn−1

F(h) =

∫Sn−1

= Vn(K )1/n.

F is concave in C.

Vn(K ) =1

∫Sn−1

F(h) =

∫Sn−1

= Vn(K )1/n.

F is concave in C.

The second variation of F.

F(h) =

∫Sn−1

For every fixed h, D2F(h) is a bilinear symmetric form acting ontest functions φ ∈ C∞(Sn−1):

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

The condition(D2F(h)φ, φ) ≤ 0

turns out to be equivalent to a weighted Poincare inequality:∫Sn−1

trace(cij)φ2dHn−1 ≤

∫Sn−1

∑i ,j

cijφiφjdHn−1,

(cij) > 0, (cij) depends on h.

for every φ verifying a zero-mean condition. .

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ)

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

F(h) =

∫Sn−1

(D2F(h)φ, φ) =d2

ds2F(h + sφ)|s=0.

∫Sn−1

∑i ,j

cijφiφjdHn−1,

A special case

If we choose h ≡ 1 (the support function of the unit ball of Rn),we obtain (cij) =identity matrix, and we recover∫

Sn−1

φ2dHn−1 ≤ 1

n − 1

∫Sn−1

|∇φ|2dHn−1,

for every φ ∈ C 1(Sn−1) s.t.∫Sn−1

φdHn−1 = 0.

This is the standard Poincare inequality (with best constant) onSn−1.(C. 2008; Saorın-Gomez, C. 2010).

A special case

Sn−1

φ2dHn−1 ≤ 1

n − 1

∫Sn−1

|∇φ|2dHn−1,

φdHn−1 = 0.

A special case

Sn−1

φ2dHn−1 ≤ 1

n − 1

∫Sn−1

|∇φ|2dHn−1,

φdHn−1 = 0.

This is the standard Poincare inequality (with best constant) onSn−1.

(C. 2008; Saorın-Gomez, C. 2010).

A special case

Sn−1

φ2dHn−1 ≤ 1

n − 1

∫Sn−1

|∇φ|2dHn−1,

φdHn−1 = 0.

Inequalities of Brunn-Minkowski type

Let G : Kn → R be s.t.:

I G(K ) ≥ 0 for every K ∈ Kn;

I G is α-homogeneous (α 6= 0):

G(tK ) = tα G(K ), ∀ t ≥ 0, K ∈ Kn.

—————–

We say that G verifies a Brunn-Minkowski type inequality if forevery K , L ∈ Kn, and for every λ ∈ [0, 1]:

G((1− λ)K + λL)1/α ≥ (1− λ)G(K )1/α + λG(L)1/α.

G(tK ) = tα G(K ), ∀ t ≥ 0, K ∈ Kn.

—————–

G((1− λ)K + λL)1/α ≥ (1− λ)G(K )1/α + λG(L)1/α.

G(tK ) = tα G(K ), ∀ t ≥ 0, K ∈ Kn.

—————–

G((1− λ)K + λL)1/α ≥ (1− λ)G(K )1/α + λG(L)1/α.

Examples

The following functionals verify a Brunn-Minkowski type inequality.

I Volume.

I Perimeter.

I Other functionals in convex geometry (intrinsic volumes,mixed volumes, 2-dim. affine surface area...).

I Principal frequency (= first Dirichlet eigenvalue of the Laplaceoperator) (Brascamp and Lieb; Borell).

I Electrostatic capacity (Borell; Caffarelli, Jerison and Lieb).

I Many other examples coming from the world of calculus ofvariations and elliptic PDE’s (torsional rigidity, p-capacity, ...).

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Examples

I Volume.

I Perimeter.

Other examples

Some well-known functional not obeying a Brunn-Minkowskiinequality.

I The diameter.

I The affine surface area in dimension n ≥ 3.

I The first Neumann eigenvalue of the Laplace operator.

Other examples

I The diameter.

Other examples

I The diameter.

Other examples

I The diameter.

Other examples

I The diameter.

Is there some general phenomenon behind these examples?Difficult (pointless?) to say.

Maybe simpler: understand the relation between theBrunn-Minkowski inequality and other basic features, such as:

I monotonicity;

I continuity;

I rigid motion invariance;

I additivity (or valuation property):

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

for every K , L ∈ Kn such that

K ∪ L ∈ Kn.

Is there some general phenomenon behind these examples?

Difficult (pointless?) to say.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

I monotonicity;

I continuity;

G(K ∪ L) = G(K ) + G(L)− G(K ∩ L),

K ∪ L ∈ Kn.

A result in this direction

Thm. (Hug, Saorın-Gomez, C., 2012). Let G : Kn → R be:additive, rigid motion invariant, continuous, (n − 1)-homogeneous,and assume that it verifies a Brunn-Minkowski type inequality.Then G is a mixed volume, and in particular is monotone.

Thm. (Hug, Saorın-Gomez, C., 2012). Let G : Kn → R be:additive, rigid motion invariant, continuous, (n − 1)-homogeneous,and assume that it verifies a Brunn-Minkowski type inequality.

Then G is a mixed volume, and in particular is monotone.

Thm. (Hug, Saorın-Gomez, C., 2012). Let G : Kn → R be:additive, rigid motion invariant, continuous, (n − 1)-homogeneous,and assume that it verifies a Brunn-Minkowski type inequality.Then G is a mixed volume, and in particular is monotone.

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