Convexity of Intersection Bodies

Convexity of Intersection Bodies Gautier Berck P&V January 11, 2014

Introduction Classic descriptions Let K be a convex body centered at the origin of the n-dimensional space V. Classic definition, V = R n The boundary of the intersection body of K, IK R n, is characterized by: Intrinsic definition x (IK) x = Area(K x ) The boundary of the intersection body of K, IK Λ n 1 V, is characterized by: a (IK) ω Λ n 1 V, ω(a) = ω K a (P&V) Intersection Bodies January 11, 2014 2 / 15

Introduction On the dual space Let Ω Λ n V be a fixed volume form. IK class V = R n x x IK V IK intr Λ n 1 V a Ω(a ) Definition Let ξ V and ω ξ Λ n 1 V be such that Ω = ξ ω ξ. Then, ( ξ IK := K ker ξ ω ξ ) 1. (P&V) Intersection Bodies January 11, 2014 3 / 15

Introduction On the dual space Let Ω Λ n V be a fixed volume form. IK class V = R n x x IK V IK intr Λ n 1 V a Ω(a ) Definition The intersection body of K, IK V, is characterized by: ξ IK ξ IK 1. (P&V) Intersection Bodies January 11, 2014 3 / 15

Introduction Convexity Busemann s theorem The intersection body of a centered convex body is convex. Equivalently, IK is a norm on V. (P&V) Intersection Bodies January 11, 2014 4 / 15

Introduction Two proofs Geometric proof The intersection body is the limit of duals of floating bodies; these are obviously convex. Analytical proof This proof extends directly to the whole family of L p -intersection bodies. (P&V) Intersection Bodies January 11, 2014 5 / 15

Floating bodies and floating hypersurfaces Basics K V, convex, centered. Floating body K α V : floating body with parameter α [0, 1/2]. K K α αvol(k) (P&V) Intersection Bodies January 11, 2014 6 / 15

Floating bodies and floating hypersurfaces Basics K V, convex, centered. Floating body K α V : floating body with parameter α [0, 1/2]. Easy properties K α is convex and centered. K 0 = K. K 1/2 = {0}. (P&V) Intersection Bodies January 11, 2014 6 / 15

Floating bodies and floating hypersurfaces Basics K V, convex, centered. Floating hypersurface The floating hypersurface S α V is defined by ξ S α Ω = α Vol(K). K {ξ 1} Property Kα S α K α S α (P&V) Intersection Bodies January 11, 2014 6 / 15

Floating bodies and floating hypersurfaces Basics K V, convex, centered. Theorem (Schutt-Werner) The floating hypersurface is convex. Hence Kα = S α K α K α S α S α (P&V) Intersection Bodies January 11, 2014 6 / 15

Floating bodies and floating hypersurfaces Distributions (V, Ω) ϕ C 0 (V ) a test function. Notation f D a distribution on V f, ϕ or V f ϕω (P&V) Intersection Bodies January 11, 2014 7 / 15

Floating bodies and floating hypersurfaces Distributions (V, Ω) ϕ C 0 (V ) a test function. Definition f is homogeneous of degree α if f, ϕ( /λ) = λ α+n f, ϕ Example δ is homogeneous of degree n. (P&V) Intersection Bodies January 11, 2014 7 / 15

Floating bodies and floating hypersurfaces Distributions Definition Let ξ V and ω ξ Ω n 1 (V ) be such that Ω = ξ ω ξ. The Radon transform of ϕ w.r.t ξ, R ξ (ϕ), is the test function on R defined by R ξ (ϕ)(s) = ϕω ξ. ξ x=s Definition The pull-back on V of a distribution f on R, πξ f, is defined by duality: π ξ f, ϕ := f, R ξ(ϕ). Proposition If f is homogeneous of degree α, so are its pull-back π ξ f and the map ξ π ξ f. (P&V) Intersection Bodies January 11, 2014 7 / 15

Floating bodies and floating hypersurfaces Examples Intersection body ξ IK = π ξ δ, 1 K 1 (P&V) Intersection Bodies January 11, 2014 8 / 15

Floating bodies and floating hypersurfaces Examples Intersection body ξ IK = π ξ δ, 1 K 1 Translated Heaviside On R, H t (s) := { 1 s t 0 s < t Proposition The floating hypersurface S α = (K α) is the level set at height α Vol(K) of the function ξ π ξ H 1, 1 K. (P&V) Intersection Bodies January 11, 2014 8 / 15

Floating bodies and floating hypersurfaces Convexity: proof 1 Theorem Up to normalization, the dual of floating bodies converge to the intersection body: lim t S 1/2 t = (IK). t 0 Vol(K) Hence the intersection body is convex. (P&V) Intersection Bodies January 11, 2014 9 / 15

Floating bodies and floating hypersurfaces Convexity: proof 1 Proof The hypersurface t S 1/2 t is characterized by: which is equivalent to π ξ H t, 1 K = (1/2 t) Vol(K) π ξ H 0 H t, 1 K = Vol(K). t Taking the limit as t goes to 0, it comes π ξ δ, 1 K = Vol(K) which characterizes (IK)/ Vol(K). (P&V) Intersection Bodies January 11, 2014 9 / 15

Definitions L p -intersection bodies For 1 < p 0 and K V, the L p -intersection-body of K, I p K V, is the body whose Minkowski functional is ξ IpK = ( x K + ξ (ξ x) p ) 1/p. K K + ξ ξ = 0 ξ > 0 (P&V) Intersection Bodies January 11, 2014 10 / 15

Definitions Alternative definition For 1 < p 0 and K V, the L p -intersection-body of K, I p K V, is the body whose Minkowski functional is ξ IpK = π ξ sp +, 1 K 1/p with s p +, ϕ := s p ϕ(s)ds. 0 (P&V) Intersection Bodies January 11, 2014 10 / 15

Easy properties Property For p 1, I p K is convex even if K is a non-symmetric star-body. Property since lim p 1 I p K = IK 1/p (1 + p) lim (1 + p 1 p)sp + = δ. (P&V) Intersection Bodies January 11, 2014 11 / 15

Brunn-Minkowski for moments Definition Let K V and η V s.t. η 0 on K. For r 0, the r-th moment of K w.r.t η is M r (K) := (η x) r Ω. x K (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Definition Let K V and η V s.t. η 0 on K. For r 0, the r-th moment of K w.r.t η is M r (K) := (η x) r Ω. Theorem (Hadwiger) x K Let η be non-negative on two convex bodies K 0 and K 1. Then M r (K 0 + K 1 ) 1 n+r M r (K 0 ) 1 n+r + M(K 1 ) 1 n+r. (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Definition Let K V be symmetric and η V s.t. η 0 on K. For r 0, the symmetric r-th moment of K w.r.t η is M s r (K) := η x r Ω. x K (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Definition Let K V be symmetric and η V s.t. η 0 on K. For r 0, the symmetric r-th moment of K w.r.t η is M s r (K) := η x r Ω. x K Brunn s theorem for moments Let K be a centered convex body in V, p 0 and H t a family of parallel hyperplanes defined by the equations ξ x = t. Then any linear form η 0 on H 0 may be extended to a linear form η on V s.t. the symmetric p-th moment M s p(k H t ) is maximal at t = 0. (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments K η = 0 ξ = 0 Idea of the proof Choose η arbitrarily and consider K {η 0}. (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Idea of the proof Choose η arbitrarily and consider K {η 0}. For this half-body, the moments of hyperplane sections to the right power is a concave function of t. (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Idea of the proof Choose η arbitrarily and consider K {η 0}. For this half-body, the moments of hyperplane sections to the right power is a concave function of t. One can choose η so that the moment to the right power is maximal at t = 0, hence the moment itself is maximal at t = 0 (but not necessarily a concave function). (P&V) Intersection Bodies January 11, 2014 12 / 15

Brunn-Minkowski for moments Idea of the proof Choose η arbitrarily and consider K {η 0}. For this half-body, the moments of hyperplane sections to the right power is a concave function of t. One can choose η so that the moment to the right power is maximal at t = 0, hence the moment itself is maximal at t = 0 (but not necessarily a concave function). The result follows by symmetry. (P&V) Intersection Bodies January 11, 2014 12 / 15

Distributions s r + For r > 1 s r +, ϕ = 0 s r ϕ(s)ds (P&V) Intersection Bodies January 11, 2014 13 / 15

Distributions s r + For k 1 < r < k with k N s r +, ϕ = 0 ( ) s r ϕ(s) T0 k 1 ϕ(s) ds (P&V) Intersection Bodies January 11, 2014 13 / 15

Distributions Easy properties ( s r + ) = rs r 1 + lim r 1 (r + 1)s r + = δ lim r 1 ( s r + ) = δ (P&V) Intersection Bodies January 11, 2014 13 / 15

Distributions Property Let ξ, η V, ξ 0, and f D (R), d q dt q ( π ξ+tη f ) t=0 = ηq π ξ f (q). (P&V) Intersection Bodies January 11, 2014 13 / 15

Convexity: proof 2 Busemann s theorem The intersection body of a centered convex body is convex. Proof: (P&V) Intersection Bodies January 11, 2014 14 / 15

Convexity: proof 2 Proof: One needs, for ν T ξ IK, d 2 dt 2 ( ξ + tν IK ) t=0 0. (P&V) Intersection Bodies January 11, 2014 14 / 15

Convexity: proof 2 Proof: One needs, for ν T ξ IK, Easier to deal with d 2 dt 2 ( ξ + tν IK ) t=0 0. 1 IK = π δ, 1 K. (P&V) Intersection Bodies January 11, 2014 14 / 15

Convexity: proof 2 Proof: One needs, for ν T ξ IK, d 2 dt 2 ( ξ + tν IK ) t=0 0. For any given λ R and η = ν + λξ, d 2 ( ξ + tη 1 dt 2 IK ) t=0 = 2λ2 d 2 dt 2 ( ξ + tν IK ) t=0. (P&V) Intersection Bodies January 11, 2014 14 / 15

Convexity: proof 2 Proof: One needs, for ν T ξ IK, d 2 dt 2 ( ξ + tν IK ) t=0 0. d 2 ( ) ξ + tη 1 dt 2 IK t=0 = π ξ δ, η 2 1 K (P&V) Intersection Bodies January 11, 2014 14 / 15

Convexity of L p -intersection bodies Theorem For 1 < p 0, the L p -intersection body of a centered convex body is convex. Proof: (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, d 2 ( ) ξ + tν IpK dt 2 t=0 0. (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, d 2 ( ) ξ + tν IpK dt 2 t=0 0. For any given λ R and η = ν + λξ, d 2 ( ξ + tη p dt 2 IK ) t=0 = p(p 1)λ2 + p d 2 dt 2 ( ξ + tν IK ) t=0. (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, d 2 ( ) ξ + tν IpK dt 2 t=0 0. d 2 ( ξ + tη p ) dt 2 IK t=0 = ( π ξ s p +), η 2 1 K (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, d 2 ( ) ξ + tν IpK dt 2 t=0 0. If p > 1, πξ sp +, η 2 1 K = p(p 1) x K + ξ (η x) 2 (ξ x) p 2 Ω (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, If 0 < p < 1, d 2 ( ) ξ + tν IpK dt 2 t=0 0. for ϕ = R ξ ( η 2 1 K ). πξ sp +, η 2 1 K = s p ( +, Rξ η 2 ) 1 K = p(p 1) 0 s p 2 (ϕ(s) ϕ(0)) ds (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, If 0 < p < 1, d 2 ( ) ξ + tν IpK dt 2 t=0 0. πξ sp +, η 2 1 K = s p ( +, Rξ η 2 ) 1 K for ϕ = R ξ ( η 2 1 K ). Choose η so that ϕ(0) ϕ(s). = p(p 1) 0 s p 2 (ϕ(s) ϕ(0)) ds (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, If 1 < p < 0, d 2 ( ) ξ + tν IpK dt 2 t=0 0. πξ sp +, η 2 1 K = p(p 1) for ϕ = R ξ ( η 2 1 K ). 0 s p 2 ( ϕ(s) ϕ(0) sϕ (0) ) ds (P&V) Intersection Bodies January 11, 2014 15 / 15

Convexity of L p -intersection bodies Proof: One needs, for ν T ξ IK, If 1 < p < 0, d 2 ( ) ξ + tν IpK dt 2 t=0 0. πξ sp +, η 2 1 K = p(p 1) 0 s p 2 ( ϕ(s) ϕ(0) sϕ (0) ) ds for ϕ = R ξ ( η 2 1 K ). Choose η so that ϕ(0) ϕ(s). By symmetry, ϕ (0) = 0. (P&V) Intersection Bodies January 11, 2014 15 / 15