Around the Brunn-Minkowski inequality

Transcription

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

2 Summary

3 Summary The Brunn-Minkowski inequality

4 Summary The Brunn-Minkowski inequality The isoperimetric inequality

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

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

7 The Brunn-Minkowski inequality

8 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}.

9 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.

10 (BM) privileges convex sets

11 (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).

12 (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

13 (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 (1 λ)a + λa A

14 (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 (1 λ)a + λa A ((1 λ)a + λa A).

15 (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 But if A is convex (1 λ)a + λa A ((1 λ)a + λa A). (1 λ)a + λa = A λ [0, 1].

16 Many equivalent forms

17 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)

18 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 )

19 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 )

20 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 ) Minimal V n ((1 λ)a + λb) min{v n (A), V n (B)}. (BM )

21 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.

22 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].

23 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]. In our case: C = {compact sets}, F = V n, α = n.

24 An elementary proof of (BM) - I

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

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

27 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].

28 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.

29 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

30 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);

31 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); the n-dimensional case follows by induction and Fubini s theorem.

32 A proof of (BM) - II

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

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

35 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.

36 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.

37 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

38 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) = fdz R ( n ) 1 λ ( gdx hdy R n R n ) λ

39 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) = fdz R ( n ) 1 λ ( ) λ gdx R n hdy R n = V n (A) 1 λ V n (B) λ,

40 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) = fdz R ( n ) 1 λ ( ) λ gdx R n hdy R n = V n (A) 1 λ V n (B) λ, i.e. the multiplicative form of (BM).

41 The isoperimetric inequality

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

43 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.

44 (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}

45 (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.

46 (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

47 Proof of the isoperimetric inequality

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

49 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

50 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.

51 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

52 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).

53 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). V n (A) n 1 n 1 nv n (B) 1/n Hn 1 ( A)

54 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). V n (A) n 1 n 1 nv n (B) 1/n Hn 1 ( A) = c(n)h n 1 ( A).

55 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). V n (A) n 1 n 1 nv n (B) 1/n Hn 1 ( A) = c(n)h n 1 ( A). When A is a ball this becomes an equality.

56 The infinitesimal form of (BM)

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

58 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.

59 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;

60 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; a prototype is φ 2 dh n 1 c(n) φ 2 dh n 1, S n 1 S n 1

61 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; a prototype is φ 2 dh n 1 c(n) φ 2 dh n 1, S n 1 S n 1 φ C 1 (S n 1 ), verifying some zero-mean condition.

62 Convex bodies

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

64 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 }.

65 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.

66 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. The Brunn-Minkowski inequality holds in particular in K n.

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

68 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.

69 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.

70 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.

71 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:

72 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 ),

73 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).

74 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). C := {h C 2 (S n 1 ) : (h ij + hδ ij ) > 0 on S n 1 }

75 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). C := {h C 2 (S n 1 ) : (h ij + hδ ij ) > 0 on S n 1 } = {support functions of C 2 + convex bodies}.

76 A representation formula for the volume

77 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

78 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

79 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 By the Brunn-Minkowski inequality, F is concave in C.

80 The second variation of F.

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

82 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 ):

83 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)

84 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.

85 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

86 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

87 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 (c ij ) > 0, (c ij ) depends on h.

88 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 (c ij ) > 0, (c ij ) depends on h. for every φ verifying a zero-mean condition..

89 A special case

90 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.

91 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.

92 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. (C. 2008; Saorín-Gomez, C. 2010).

93 Inequalities of Brunn-Minkowski type

94 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.

95 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.

96 Examples

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

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

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

100 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...).

101 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).

102 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).

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

104 Other examples

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

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

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

108 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.

109 Hints

110 Hints Is there some general phenomenon behind these examples?

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

112 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:

113 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;

114 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;

115 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;

116 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; additivity (or valuation property): G(K L) = G(K) + G(L) G(K L), for every K, L K n such that K L K n.

117 A result in this direction

118 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.

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