Syetrization of Lévy processes and applications 1 Pedro J. Méndez Hernández Universidad de Costa Rica July 2010, resden 1 Joint work with R. Bañuelos, Purdue University Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 1 / 25
Isoperietric inequalities Classical Inequality Aong all figures of equal perieter, the circle encloses the largest area. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 2 / 25
Isoperietric inequalities Classical Inequality Aong all figures of equal perieter, the circle encloses the largest area. Equivalently: Aong all regions of equal area, the disk has the sallest perieter. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 2 / 25
Isoperietric inequalities Classical Inequality Aong all figures of equal perieter, the circle encloses the largest area. Equivalently: Aong all regions of equal area, the disk has the sallest perieter. Generalized Isoperietric Inequalities Aong all regions of fixed easure, there is a large class of quantities that are axiized, or iniized, by the corresponding quantities for the ball: surface area, eigenvalues, capacities, exit ties, etc. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 2 / 25
Isoperietric proble for exit ties Let be a doain in R d with finite easure, and let τ the first exit tie of B t fro a doain. Brownian otion killed upon leaving! z B t Question Assuing sae easure, which of the two figures have a largest survival tie Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 3 / 25
Isoperietric proble for exit ties Let be a doain in R d with finite easure, and let τ the first exit tie of B t fro a doain. Brownian otion killed upon leaving! z B t Question Assuing sae easure, which of the two figures have a largest survival tie Answer: Obviously the ball Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 3 / 25
Isoperietric proble for exit ties efine = B( 0, R( the open ball centered at the origin 0 of sae volue as. Theore Let be a doain of finite area, then for all z P z (τ > t P 0 (τ > t Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 4 / 25
Isoperietric proble for exit ties efine = B( 0, R( the open ball centered at the origin 0 of sae volue as. Theore Let be a doain of finite area, then for all z P z (τ > t P 0 (τ > t The isoperietric theore, deeply rooted in our experience and intuition so easy to conjecture, but not so easy to prove, is an inexhaustible source of inspiration. G. Pólya: Matheatics and Plausible Thinking. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 4 / 25
Isoperietric proble for exit ties Let X t be a syetric α-stable processes in R d, and τ α the first exit tie of X t fro. P z ( τ α > t = Pz ( X s, 0 s < t = li Pz ( X jt, j = 1,,. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 5 / 25
Isoperietric proble for exit ties Let X t be a syetric α-stable processes in R d, and τ α the first exit tie of X t fro. Thus to prove P z ( τ α > t = Pz ( X s, 0 s < t It is enough to prove that = li Pz ( X jt, j = 1,, P z (τ α > t P0 (τ α > t, for all z Rd. P z ( X t P 0 ( X t, X 2t, X 2t,, X t,, X t.. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 5 / 25
Isoperietric proble for exit ties Let X t be a syetric α-stable processes in R d, and τ α the first exit tie of X t fro. Thus to prove P z ( τ α > t = Pz ( X s, 0 s < t It is enough to prove that = li Pz ( X jt, j = 1,, P z (τ α > t P0 (τ α > t, for all z Rd. P z ( X t P 0 ( X t, X 2t, X 2t,, X t,, X t.. Main idea: Reduce the proble to an inequality of finite diensional distributions of X t. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 5 / 25
Recall that X t has transition densities p α (t, x, y such that with f X decreasing. We want = P z 0 = ( X t j=1 p α (t, x, y = f α ( x y, ( t p α, z j z j 1 dz 1 dz, X 2t ( P 0 X t, X 2t p α,, X t,, X t ( t 1, z ( t p α, z j z j 1 dz 1 dz. j=2 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 6 / 25
Using rearrangeent inequalities one can prove that, ( t p α, z j z j 1 dz 1 dz j=1 p α ( t 1, z ( t p α, z j z j 1 dz 1 dz. j=2 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 7 / 25
Using rearrangeent inequalities one can prove that, ( t p α, z j z j 1 dz 1 dz j=1 p α ( t 1, z ( t p α, z j z j 1 dz 1 dz. j=2 Theore (Luttinger 73 Let f j, 1 j r be nonnegative functions in R d and fj be the syetric decreasing rearrangeent of f j. Then for any z 0 R d we have f j (z j z j 1 dz 1 dz j=1 f 1 (z 1 j=2 f j (z j z j 1 dz 1 dz. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 7 / 25
Syetric decreasing rearrangeents Given f 0, its syetric decreasing rearrangeent f is the function satisfying f (x = f (y, if x = y, and for all t 0. In particular If {f > t} <, then f (x f (y, if x y, {f > t} = {f > t}, sae distribution f (x = (χ = χ. 0 χ { f >t} (x dt. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 8 / 25
Theore (Brascap-Lieb-Luttinger Let p j, 1 j r be nonnegative functions in R d and pj be the syetric decreasing rearrangeent of p j. Let a jk be a r real atrix. R n R n R n j=1 r ( p j a j,k z k dz 1 dz r pj R n j=1 k=1 ( a j,k z k dz 1 dz. k=1 Thus, if X t is a syetric α-stable processes in R d, for all z ( P z X t, X 2t,, X t P 0 ( X t, X 2t,, X t. Proved by any, first appearance for Brownian Motion Aizenan and Sion 80. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 9 / 25
Question Under what conditions, if X t is a Lévy processes in R d, for all z. P z ( X t P 0 ( X t, X 2t, X 2t,, X t,, X t. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 10 / 25
Question Under what conditions, if X t is a Lévy processes in R d, for all z. P z ( X t P 0 ( X t, X 2t, X 2t,, X t,, X t Sae proof: If X t has transition densities such that p X (t, x, y = f X ( x y, with f X decreasing (Isotropic Uniodal. Then P z ( X t P 0 ( X t, X 2t, X 2t,, X t,, X t.. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 10 / 25
How far can you extend this inequality Not true in general. Using the sae arguent if X t has transition densities p X (t, x, y. We need ( = P z 0 X t j=1 p X ( t, z j z j 1 dz 1 dz, X 2t [ p X ( t, z 1,, X t ] j=2 [ p X ( t, z j z j 1 ] dz1 dz. Proble: We cannot even ensure that [ p(t,, y], is the transition density of a Lévy processes. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 11 / 25
Syetrization of Lévy processes Let X t be a Lévy process in R d such that where Ψ(ξ = i b, ξ + 1 2 A ξ, ξ + E x [ e iξ Xt ] = e tψ(ξ iξ x, R d [ 1 + i ξ, y χ B e i ξ y ] φ(y dy. Consider X t the Lévy process in R d given by [ ] E x e iξ X t = e tψ (ξ iξ x, where and A = (det A 1/d I d. Ψ (ξ = 1 [ ] 2 A ξ, ξ + 1 e i ξ y φ (y dy. R d Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 12 / 25
Theore (Bañuelos-Méndez 10 Suppose X t is a Lévy process with Lévy easure absolutely continuous with respect to the Lebesgue easure and let Xt be the syetrization of X t constructed as above. Let f 1,, f be nonnegative lower seicontinuous functions. Then for all z R d, [ ] [ ] E z f i (X ti χ i (X ti E 0 fi (Xt i χ i (Xt i, i=1 for all 0 t 1 t. i=1 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 13 / 25
Theore (Bañuelos-Méndez 10 Suppose X t is a Lévy process with Lévy easure absolutely continuous with respect to the Lebesgue easure and let Xt be the syetrization of X t constructed as above. Let f 1,, f be nonnegative lower seicontinuous functions. Then for all z R d, [ ] [ ] E z f i (X ti χ i (X ti E 0 fi (Xt i χ i (Xt i, i=1 for all 0 t 1 t. i=1 The proof is based on the fact that X t is the week liit of processes of the for X n t = C n t + G n t. where C n t is a Copound Poisson Processes and G n t is a non-singular Gaussian processes. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 13 / 25
Consequences: Let τ X be the first exit tie of X t for, then If V 0 and f are continuous, E z{ ( f (X t exp E z{ ( f (X t exp where V = ( V. t 0 t 0 V (X s ds V (X s ds ; τ X > t } If ψ is a nonnegative increasing function, then [ ( ] [ ( ] E z ψ τ X E 0 ψ τ X, ; τ X > t } In particular, for all 0 < p <, [ ( p ] E z τ X E 0 [ ( τ X p ]. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 14 / 25
If p X (t, x, y is the transition density of X t killed upon leaving, then f (y p X (t, x, y dy f (y p X (t, 0, y dy. (1 In particular for all x, y p X (t, x, y px (t, 0, 0, If X t and Xt are transient then f (z G X (x, zdz f (zg X (0, zdz. (2 For all increasing convex function φ : R + R + ( ( φ p X (t, x, y dy φ p X (t, 0, y dy ( ( φ G X (x, z dz φ G X (0, z dz. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 15 / 25
Isoperietric inequality for the trace i=1 e tλk,x = p X (t, z, zdz p X (t, z, zdz = i=1 e tλk,x. Rayleigh-Faber-Krahn Inequality The Gaa function, for all s > d α λ 1,X λ 1,X. ( i=1 1 λ k,x s ( i=1 1 λ k,x s Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 16 / 25
Consider a Lévy process of the for where Then X t = C t + G t, G t is a Gaussian process with covariance atrix A, and ean. C t is an independent copound Poisson process with characteristic function ( E e iξ Ct { = exp c R d X t = C t + G t, } [ 1 exp(iξ y ] φ(y dy, where Gt is a Gaussian process with covariance atrix A = (det A 1/d I d. Ct is an independent copound Poisson process with characteristic function ( { } E e iξ C t = exp c [ 1 exp(iξ y ] φ (y dy. R d Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 17 / 25
Let B be a Borel subsets of R d, then E x [ C t + G t B] = f A, (t, u x x 0 dµ t (x 0 du, where µ t is the distribution of C t, and f A, is the density of G t. B Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 18 / 25
Let B be a Borel subsets of R d, then E x [ C t + G t B] = f A, (t, u x x 0 dµ t (x 0 du, B where µ t is the distribution of C t, and f A, is the density of G t. One can prove that f A, (t, u x x 0 dµ t (x 0 du B = P [N t = 0] f A, (t, u du (3 + B P [N t = k] k=1 B f A, (t, u x x 0 q k (t, x 0 dx 0 du where N t is a Poisson process and f A, (t, u x x 0 q k (t, x 0 dx 0 du = B B f A, (t, u x x 0 k φ(x i x i 1 dx 1 dx k dx 0 du. i=1 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 18 / 25
Then using B-L-L B = B B f A, (t, x x 0 f A, (t, x 0 f A,0(t, x 0 k φ(x i x i 1 dx 1 dx k dx 0 i=1 k φ (x i x i 1 dx 1 dx k dx 0 i=1 k φ (x i x i 1 dx 1 dx k dx 0. i=1 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 19 / 25
Then using B-L-L B = We conclude B B f A, (t, x x 0 f A, (t, x 0 f A,0(t, x 0 k φ(x i x i 1 dx 1 dx k dx 0 i=1 k φ (x i x i 1 dx 1 dx k dx 0 i=1 k φ (x i x i 1 dx 1 dx k dx 0. i=1 E x {G t1 + C t1,, G t + C t } E 0 { G t 1 + C t 1,, G t + C t }. Then take a sequence X n t = C n t + G n t that converges weakly to X t. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 19 / 25
Extreal for convex doains of fixed inradius If X t has transition densities such that p X (t, x, y = f X ( x y, with f X decreasing. There are siilar results for exit ties of Lévy processes X t fro convex doains with fixed inner inradius r. Assue the biggest ball inside is centered at 0. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 20 / 25
Extreal for convex doains of fixed inradius If X t has transition densities such that p X (t, x, y = f X ( x y, with f X decreasing. There are siilar results for exit ties of Lévy processes X t fro convex doains with fixed inner inradius r. Assue the biggest ball inside is centered at 0. Classical results: Using the axiu principle for the Laplacian it is proved that λ B ( r,r λb, (Hersh n = 2 60, Protter n 2 81 where B is Brownian otion. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 20 / 25
Extreal for convex doains of fixed inradius for B.M. Consider the infinite slab } S( = {(x 1,, x d R d : r < x 1 < r. Equivanlently λ B ( r,r = λb S(. Besides Bañuelos-Kröger 97 prove that P x (τ B > t P0 (τ B S( > t = P0 (τ B ( r,r > t. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 21 / 25
Extreal for convex doains of fixed inradius for B.M. Consider the infinite slab } S( = {(x 1,, x d R d : r < x 1 < r. Equivanlently λ B ( r,r = λb S(. Besides Bañuelos-Kröger 97 prove that P x (τ B > t P0 (τ B S( > t = P0 (τ B ( r,r > t. In a siilar way we can reduced this proble to finite diensional distributions. To obtained a siilar result for X t, it is enought to prove that P x ( X t P 0 ( X t, X 2t S(, X 2t,, X t S(,, X t S(. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 21 / 25
Theore Let be a convex doain in R d of finite inradius r and let S( be the infinite slab. Let p 1,, p be nonnegative nonincreasing radially syetric functions on R d. Then for any z 0 R d we have p j (z j z j 1 dz 1 dz S( S( j=1 p 1 (z 1 p j (z j z j 1 dz 1 dz. (4 j=2 ( Bañuelos-Lata la-méndez 00 d = 2, Méndez 02 d > 2 Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 22 / 25
Theore Let be a convex doain in R d of finite inradius r and let S( be the infinite slab. Let p 1,, p be nonnegative nonincreasing radially syetric functions on R d. Then for any z 0 R d we have p j (z j z j 1 dz 1 dz S( S( j=1 p 1 (z 1 p j (z j z j 1 dz 1 dz. (4 j=2 ( Bañuelos-Lata la-méndez 00 d = 2, Méndez 02 d > 2 Key in the proof: 1 can be assue to be a polyhedron 2 Any nonnegative radially syetric nonincreasing function f can be expressed in the for f (z = for soe nonnegative easure on (0, ] 0 I B(0,r (zdµ(r Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 22 / 25
Consequences For all z E z{ } τ X > t E 0{ τs( }. X > t Thus λ X S( λx. If ψ is a nonnegative increasing function, then [ ( ] [ E z ψ τ X E 0 ψ ( τ X S( In particular, for all 0 < p <, [ ( p ] [ ( E z τ X E 0 τs( X ], p ]. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 23 / 25
Inequalities for the heat content and the torsional rigidity p X (t, z, wdzdw ps( X (t, z, wdzdw, G X (z, wdzdw S( S( For all increasing function φ : R + R + ( φ p X (t, 0, y dy ( φ G X (0, z dz S( S( S( S( GS( X (z, wdzdw. ( φ ps( X (t, 0, y dy ( φ GS( X (0, z dz. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 24 / 25
Sharper version in R 2 Theore( Méndez 02 Let be a convex doain in R 2 of finite inradius r and of diaeter d (which ay be infinite. Let p 1,, p, q 1,, q be nonnegative nonincreasing radially syetric functions on R 2. Then for any z 0 H i,j we have that q j (z j p j (z j z j 1 dz 1 dz C( C( j=1 q j (z 1 p j (z 1 q j (z j p j (z j z j 1 dz 1 dz, j=2 where C( = ( r, r ( d + r, r + d. Question Are there higher diensional analogues? For instance can we replace S( by C( = ( r, r ( d + r, r d d 1. Pedro J. Méndez (UCR Syetrization & Lévy processes Levy 2010 25 / 25