Around the Brunn-Minkowski inequality

Around the Brunn-Minkowski inequality Andrea Colesanti Technische Universität Berlin - Institut für Mathematik January 28, 2015

Summary

Summary The Brunn-Minkowski inequality

Summary The Brunn-Minkowski inequality The isoperimetric inequality

Summary The Brunn-Minkowski inequality The isoperimetric inequality Infinitesimal form of Brunn-Minkowski inequality

Summary The Brunn-Minkowski inequality The isoperimetric inequality Infinitesimal form of Brunn-Minkowski inequality Inequalities of Brunn-Minkowski type

The Brunn-Minkowski inequality

The Brunn-Minkowski inequality Thm. A, B R n, compact; λ [0, 1]; then V n ((1 λ)a + λb) 1/n (1 λ)v n (A) 1/n + λv n (B) 1/n. (BM) V n = volume (Lebesgue measure); (1 λ)a + λb = {(1 λ)a + λb : a A, b B}. Equivalently: The functional Vn 1/n is concave in the class of compact sets of R n, equipped with the vector addition. An excellent survey (much better than this talk): R. Gardner, The Brunn-Minkowski inequality, Bull. A.M.S., 2002.

(BM) privileges convex sets

(BM) privileges convex sets In the inequality V n ((1 λ)a + λb) 1/n (1 λ)v n (A) 1/n + λv n (B) 1/n, equality holds iff A is convex and B is a homothetic copy of A (up to subsets of volume zero).

(BM) privileges convex sets In the inequality V n ((1 λ)a + λb) 1/n (1 λ)v n (A) 1/n + λv n (B) 1/n, equality holds iff A is convex and B is a homothetic copy of A (up to subsets of volume zero). Why? If you plug A = B in (BM) in general you don t get an equality, because

Many equivalent forms

Many equivalent forms Classic V n ((1 λ)a+λb) 1/n (1 λ)v n (A) 1/n +λv n (B) 1/n. (BM)

Many equivalent forms Classic V n ((1 λ)a+λb) 1/n (1 λ)v n (A) 1/n +λv n (B) 1/n. (BM) Elegant V n (A + B) 1/n V n (A) 1/n + V n (B) 1/n. (BM )

Many equivalent forms Classic V n ((1 λ)a+λb) 1/n (1 λ)v n (A) 1/n +λv n (B) 1/n. (BM) Elegant V n (A + B) 1/n V n (A) 1/n + V n (B) 1/n. (BM ) Multiplicative V n ((1 λ)a + λb) V n (A) 1 λ V n (B) λ. (BM 0 )

A general fact about homogeneous functional Let F be a real-valued functional defined on a convex cone C; α-homogeneous: F(λx) = λ α F(x), x C, λ > 0 (α > 0); non-negative.

A general fact about homogeneous functional Let F be a real-valued functional defined on a convex cone C; α-homogeneous: F(λx) = λ α F(x), x C, λ > 0 (α > 0); non-negative. F 1/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].

An elementary proof of (BM) - I

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality).

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality). Let f, g, h : R n R + be measurable functions,

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality). Let f, g, h : R n R + be measurable functions, and let λ [0, 1].

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality). Let f, g, h : R n 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 R n.

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality). Let f, g, h : R n 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 R n. Then ( ) 1 λ ( ) λ fdz gdx hdy. R n R n R n

An elementary proof of (BM) - I Lemma (Prékopa-Leindler inequality). Let f, g, h : R n R + be measurable functions, and let λ [0, 1]. Assume that Then Proof. f ((1 λ)x + λy)) g(x) (1 λ) h(y) λ x, y R n. ( ) 1 λ ( ) λ fdz gdx hdy. R n R n R n Prove the 1-dimensional case (using just the so-called layer cake, or Cavalieri s, principle);

A proof of (BM) - II

A proof of (BM) - II Given A, B R n and λ [0, 1], let

A proof of (BM) - II Given A, B R n and λ [0, 1], let f = characteristic function of (1 λ)a + λb,

A proof of (BM) - II Given A, B R n and λ [0, 1], let f = characteristic function of (1 λ)a + λb, g = charact. function of A, h = charact. function of B.

A proof of (BM) - II Given A, B R n and λ [0, 1], let f = characteristic function of (1 λ)a + λb, g = charact. function of A, h = charact. function of B. Then: f ((1 λ)x + λy)) g(x) (1 λ) h(y) λ x, y R n. By Prékopa-Leindler inequality V n ((1 λ)a + λb) = R n fdz

The isoperimetric inequality

The isoperimetric inequality Thm. Among all subsets of R n with given perimeter, the ball having such perimeter maximizes the volume.

The isoperimetric inequality Thm. Among all subsets of R n with given perimeter, the ball having such perimeter maximizes the volume. Equivalently, V n (A) n 1 n c(n)h n 1 ( A) for every set A (with sufficiently smooth boundary), where c(n) is a constant and H n 1 is the (n 1)-dimensional Hausdorff measure. Equality is attained when A is a ball.

(BM) isoperimetric inequality Let A R n be a bounded domain with C 1 boundary. Then where H n 1 V n (A ɛ ) V n (A) ( A) = lim, ɛ 0 + ɛ A ɛ = {x R n : dist(x, A) ɛ}

(BM) isoperimetric inequality Let A R n be a bounded domain with C 1 boundary. Then H n 1 V n (A ɛ ) V n (A) ( A) = lim, ɛ 0 + ɛ where A ɛ = {x R n : dist(x, A) ɛ} = A + ɛb, and B = {x R n : x 1} = unit ball.

(BM) isoperimetric inequality Let A R n be a bounded domain with C 1 boundary. Then H n 1 V n (A ɛ ) V n (A) ( A) = lim, ɛ 0 + ɛ where A ɛ = {x R n : dist(x, A) ɛ} = A + ɛb, and Hence B = {x R n : x 1} = unit ball. H n 1 V n (A + ɛb) V n (A) ( A) = lim. ɛ 0 + ɛ

Proof of the isoperimetric inequality

Proof of the isoperimetric inequality H n 1 V n (A + ɛb) V n (A) ( A) = lim. ɛ 0 + ɛ

Proof of the isoperimetric inequality H n 1 V n (A + ɛb) V n (A) ( A) = lim. ɛ 0 + ɛ By (BM), for every ɛ > 0 V n (A + ɛb) 1/n V n (A) 1/n + V n (ɛb) 1/n

Proof of the isoperimetric inequality H n 1 V n (A + ɛb) V n (A) ( A) = lim. ɛ 0 + ɛ By (BM), for every ɛ > 0 V n (A + ɛb) 1/n V n (A) 1/n + V n (ɛb) 1/n = V n (A) 1/n + ɛv n (B) 1/n.

Proof of the isoperimetric inequality H n 1 V n (A + ɛb) V n (A) ( A) = lim. ɛ 0 + ɛ By (BM), for every ɛ > 0 V n (A + ɛb) 1/n V n (A) 1/n + V n (ɛb) 1/n = V n (A) 1/n + ɛv n (B) 1/n. V n (B) 1/n lim ɛ 0 + V n (A + ɛb) 1/n V n (A) 1/n ɛ = 1 n V n(a) 1 n n H n 1 ( A).

The infinitesimal form of (BM)

The infinitesimal form of (BM) By the Brunn-Minkowski inequality Vn 1/n functional. is a concave

The infinitesimal form of (BM) By the Brunn-Minkowski inequality Vn 1/n is a concave functional. Hence the second variation (or second differential) of Vn 1/n (whatever that means) must be negative semidefinite: D 2 (V 1/n n ) 0. If we restrict our attention to convex sets, this fact amounts to a class of functional inequalities of Poincaré type on the unit sphere;

Convex bodies

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

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 R n. We set K n = {convex bodies in R n }.

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 R n. We set K n = {convex bodies in R n }. K n is closed under addition and dilations: given K, L K n and α, β 0, αk + βl K n.

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

From sets to functions: the support function of a convex body The support function h K of a convex body K is defined by: h K : S n 1 R, h K (u) = sup{(u, v) v K}. h K (u) is the distance from the origin of the hyperplane supporting K, with outer unit normal u. The passage to support functions preserves the linear structure on K n : for every K, L K n α, β 0. h αk+βl = αh K + βh L.

Convex bodies of class C 2 + A convex body is said to be of class C 2 + if: K C 2, the Gauss curvature is strictly positive on K.

Convex bodies of class C 2 + A convex body is said to be of class C+ 2 if: K C 2, the Gauss curvature is strictly positive on K. In terms of the support function h of K:

Convex bodies of class C 2 + A convex body is said to be of class C+ 2 if: K C 2, the Gauss curvature is strictly positive on K. In terms of the support function h of K: h C 2 (S n 1 ),

Convex bodies of class C 2 + A convex body is said to be of class C+ 2 if: K C 2, the Gauss curvature is strictly positive on K. In terms of the support function h of K: h C 2 (S n 1 ), (h ij + hδ ij ) > 0 on S n 1 (h ij = second covariant derivatives of h on S n 1, δ ij = Kronecker s symbols).

A representation formula for the volume

A representation formula for the volume If K is of class C+ 2 and h is its support function, then V n (K) = 1 h det(h ij + hδ ij ) dh n 1. n S n 1

A representation formula for the volume If K is of class C+ 2 and h is its support function, then V n (K) = 1 h det(h ij + hδ ij ) dh n 1. n S n 1 Now we define a functional F : C R + as F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1 = V n (K) 1/n. n S n 1

The second variation of F.

The second variation of F. F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1. n S n 1

The second variation of F. F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1. n S n 1 For every fixed h, D 2 F(h) is a bilinear symmetric form acting on test functions φ C (S n 1 ):

The second variation of F. F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1. n S n 1 For every fixed h, D 2 F(h) is a bilinear symmetric form acting on test functions φ C (S n 1 ): (D 2 F(h)φ, φ)

The second variation of F. F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1. n S n 1 For every fixed h, D 2 F(h) is a bilinear symmetric form acting on test functions φ C (S n 1 ): (D 2 F(h)φ, φ) = d 2 ds 2 F(h + sφ) s=0.

The second variation of F. F(h) = [ ] 1 1/n h det(h ij + hδ ij ) dh n 1. n S n 1 For every fixed h, D 2 F(h) is a bilinear symmetric form acting on test functions φ C (S n 1 ): The condition (D 2 F(h)φ, φ) = d 2 ds 2 F(h + sφ) s=0. (D 2 F(h)φ, φ) 0 turns out to be equivalent to a weighted Poincaré inequality: trace(c ij )φ 2 dh n 1 c ij φ i φ j dh n 1, S n 1 S n 1 i,j

A special case

A special case If we choose h 1 (the support function of the unit ball of R n ), we obtain (c ij ) =identity matrix, and we recover φ 2 dh n 1 1 φ 2 dh n 1, S n 1 n 1 S n 1 for every φ C 1 (S n 1 ) s.t. S n 1 φdh n 1 = 0. This is the standard Poincaré inequality (with best constant) on S n 1.

Inequalities of Brunn-Minkowski type

Inequalities of Brunn-Minkowski type Let G : K n R be s.t.: G(K) 0 for every K K n ; G is α-homogeneous (α 0): G(tK) = t α G(K), t 0, K K n.

Inequalities of Brunn-Minkowski type Let G : K n R be s.t.: G(K) 0 for every K K n ; G is α-homogeneous (α 0): G(tK) = t α G(K), t 0, K K n. We say that G verifies a Brunn-Minkowski type inequality if for every K, L K n, and for every λ [0, 1]: G((1 λ)k + λl) 1/α (1 λ)g(k) 1/α + λg(l) 1/α.

Examples

Examples The following functionals verify a Brunn-Minkowski type inequality.

Examples The following functionals verify a Brunn-Minkowski type inequality. Volume.

Examples The following functionals verify a Brunn-Minkowski type inequality. Volume. Perimeter.

Examples The following functionals verify a Brunn-Minkowski type inequality. Volume. Perimeter. Other functionals in convex geometry (intrinsic volumes, mixed volumes, 2-dim. affine surface area...).

Examples The following functionals verify a Brunn-Minkowski type inequality. Volume. Perimeter. Other functionals in convex geometry (intrinsic volumes, mixed volumes, 2-dim. affine surface area...). Principal frequency (= first Dirichlet eigenvalue of the Laplace operator) (Brascamp and Lieb; Borell). Electrostatic capacity (Borell; Caffarelli, Jerison and Lieb).

Other examples

Other examples Some well-known functional not obeying a Brunn-Minkowski inequality.

Other examples Some well-known functional not obeying a Brunn-Minkowski inequality. The diameter.

Other examples Some well-known functional not obeying a Brunn-Minkowski inequality. The diameter. The affine surface area in dimension n 3.

Other examples Some well-known functional not obeying a Brunn-Minkowski inequality. The diameter. The affine surface area in dimension n 3. The first Neumann eigenvalue of the Laplace operator.

Hints

Hints Is there some general phenomenon behind these examples?

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

Hints Is there some general phenomenon behind these examples? Difficult (pointless?) to say. Maybe simpler: understand the relation between the Brunn-Minkowski inequality and other basic features, such as: monotonicity; continuity; rigid motion invariance;

A result in this direction

A result in this direction Thm. (Hug, Saorín-Gomez, C., 2012). Let G : K n R be: additive, rigid motion invariant, continuous, (n 1)-homogeneous, and assume that it verifies a Brunn-Minkowski type inequality.