Zhangs inequality for log-concave functions B. Gonz alez Merino* - - PowerPoint PPT Presentation

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Zhangs inequality for log-concave functions B. Gonz alez Merino* - - PowerPoint PPT Presentation

Zhangs inequality for log-concave functions B. Gonz alez Merino* (joint with D. Alonso-Guti errez and J. Bernu es) Jena *Author partially funded by Fundaci on S eneca, proyect 19901/GERM/15, and by MICINN, project


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SLIDE 1

Zhang’s inequality for log-concave functions

  • B. Gonz´

alez Merino* (joint with D. Alonso-Guti´ errez and J. Bernu´ es)

Jena *Author partially funded by Fundaci´

  • n S´

eneca, proyect 19901/GERM/15, and by MICINN, project PGC2018-094215-B-I00, Spain. Departamento de An´ alisis Matem´ atico, Universidad de Sevilla

Convex, Discrete and Integral Geometry, 19th September 2019.

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First definitions

  • If M ⊂ Rn, let |M| be the volume (or n-Lebesgue measure) of M
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First definitions

  • If M ⊂ Rn, let |M| be the volume (or n-Lebesgue measure) of M
  • Let Bn

2 be the Euclidean unit ball

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SLIDE 4

First definitions

  • If M ⊂ Rn, let |M| be the volume (or n-Lebesgue measure) of M
  • Let Bn

2 be the Euclidean unit ball ,i.e.,

Bn

2 =

  • x ∈ Rn : |x| =
  • x2

1 + · · · + x2 n ≤ 1

  • .
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SLIDE 5

First definitions

  • If M ⊂ Rn, let |M| be the volume (or n-Lebesgue measure) of M
  • Let Bn

2 be the Euclidean unit ball ,i.e.,

Bn

2 =

  • x ∈ Rn : |x| =
  • x2

1 + · · · + x2 n ≤ 1

  • .
  • For any M ⊂ Rn, then χM is the characteristic of M

χM : Rn → {0, 1}

  • s. t.

χM(x) = 1 if x ∈ M

  • therwise.
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SLIDE 6

Geometric & Analytic ineqs.

Theorem 1 (Isoperimetric ineq. 1915) Let M ∈ C1 and M compact. Then n|Bn

2|

1 n |M| n−1 n

≤ S(M). ”=” iff M = Bn

2.

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SLIDE 7

Geometric & Analytic ineqs.

Theorem 1 (Isoperimetric ineq. 1915) Let M ∈ C1 and M compact. Then n|Bn

2|

1 n |M| n−1 n

≤ S(M). ”=” iff M = Bn

2.

Theorem 2 (Sobolev ineq. ’38) Let f : M → R with M ∈ C1 compact and f ∈ C1. Then n|Bn

2|

1 n

  • Rn |f |

n n−1

n−1

n

  • Rn |∇f |

”=” iff f = χBn

2.

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SLIDE 8

Definitions and Properties

  • K is a convex body, i.e., a convex compact set of Rn.
Powered by TCPDF (www.tcpdf.org)
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Definitions and Properties

  • K is a convex body, i.e., a convex compact set of Rn.
Powered by TCPDF (www.tcpdf.org)
  • Kn set of n-dimensional convex bodies.
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Definitions and Properties

  • If x ∈ Rn, then x⊥ is the hyperplane orthogonal to x.
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SLIDE 11

Definitions and Properties

  • If x ∈ Rn, then x⊥ is the hyperplane orthogonal to x. Moreover,

if M ⊂ Rn, then Px⊥M is the orthogonal projection of M onto x⊥.

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SLIDE 12

Definitions and Properties

  • If x ∈ Rn, then x⊥ is the hyperplane orthogonal to x. Moreover,

if M ⊂ Rn, then Px⊥M is the orthogonal projection of M onto x⊥.

  • The polar projection body Π∗(K) of K ∈ Kn is the unit ball of

the norm xΠ∗(K) := |x||Px⊥K|.

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Definitions and Properties

  • If x ∈ Rn, then x⊥ is the hyperplane orthogonal to x. Moreover,

if M ⊂ Rn, then Px⊥M is the orthogonal projection of M onto x⊥.

  • The polar projection body Π∗(K) of K ∈ Kn is the unit ball of

the norm xΠ∗(K) := |x||Px⊥K|.

  • The product |K|n−1|Π∗(K)| is an affine invariant.
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Petty projection inequality

Theorem 3 (Petty 1971) Let K ∈ Kn. Then |K|n−1|Π∗(K)| ≤ π

n 2 Γ

n+1

2

n Γ n+2

2

n . ”=” iff K is an ellipsoid.

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SLIDE 15

Petty projection inequality

Theorem 3 (Petty 1971) Let K ∈ Kn. Then |K|n−1|Π∗(K)| ≤ π

n 2 Γ

n+1

2

n Γ n+2

2

n . ”=” iff K is an ellipsoid. Petty projection ineq. ⇒ Isoperimetric ineq.

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SLIDE 16

Zhang’s inequality

Theorem 4 (Zhang 1991) Let K ∈ Kn. Then 2n

n

  • nn

≤ |K|n−1|Π∗(K)|. ”=” iff K is an n-simplex.

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SLIDE 17

Polar projection body of f

For every f ∈ W1,1 = {f ∈ L1(Rn) :

∂f ∂xi ∈ L1(Rn), i = 1, . . . , n}

with compact support, let Π∗(f ), the polar projection body of f , be the unit ball of the norm xΠ∗(f ) :=

  • Rn |∇f (y), x|dy.
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SLIDE 18

Functional affine inequalities

Theorem 5 (Zhang 1999) Let f ∈ C1 with compact support. Then f

n n−1 |Π∗(f )| 1 n ≤ π 1 2 Γ

n+1

2

n+2

2

. ”=” iff f = χE with E an ellipsoid.

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SLIDE 19

Functional affine inequalities

Theorem 5 (Zhang 1999) Let f ∈ C1 with compact support. Then f

n n−1 |Π∗(f )| 1 n ≤ π 1 2 Γ

n+1

2

n+2

2

. ”=” iff f = χE with E an ellipsoid. Notice that f

n n−1

  • Sn−1 ∇uf −n

1 du

1

n

≤ n

1 n π 1 2 Γ

n+1

2

n+2

2

  • .
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SLIDE 20

Log-concave functions

  • f : Rn → [0, ∞) is log-concave if f (x) = e−u(x) for some

u : Rn → (−∞, ∞] convex,

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SLIDE 21

Log-concave functions

  • f : Rn → [0, ∞) is log-concave if f (x) = e−u(x) for some

u : Rn → (−∞, ∞] convex, i.e., if f ((1 − λ)x + λy) ≥ f (x)1−λf (y)λ for every x, y ∈ Rn, λ ∈ [0, 1].

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Log-concave functions

  • f : Rn → [0, ∞) is log-concave if f (x) = e−u(x) for some

u : Rn → (−∞, ∞] convex, i.e., if f ((1 − λ)x + λy) ≥ f (x)1−λf (y)λ for every x, y ∈ Rn, λ ∈ [0, 1].

  • F(Rn) log-concave integrable functions in Rn.
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Log-concave functions

  • f : Rn → [0, ∞) is log-concave if f (x) = e−u(x) for some

u : Rn → (−∞, ∞] convex, i.e., if f ((1 − λ)x + λy) ≥ f (x)1−λf (y)λ for every x, y ∈ Rn, λ ∈ [0, 1].

  • F(Rn) log-concave integrable functions in Rn.
  • If f ∈ F(Rn) ∩ W1,1 then

xΠ∗(f ) = 2|x|

  • x⊥ Px⊥f (y)dy,

where Px⊥f (y) = maxs∈R f (y + sx).

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SLIDE 24

Functional affine inequalities

Theorem 6 (Alonso-Guti´ errez, Bernu´ es, G.M. +2018) Let f ∈ F(Rn). Then

  • Rn
  • Rn min{f (y), f (x)}dydx ≤ 2nn!f n+1

1

|Π∗(f )|. ”=” iff

f (x) f ∞ = e−x△ for a simplex △ ∋ 0.

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SLIDE 25

Functional affine inequalities

Theorem 6 (Alonso-Guti´ errez, Bernu´ es, G.M. +2018) Let f ∈ F(Rn). Then

  • Rn
  • Rn min{f (y), f (x)}dydx ≤ 2nn!f n+1

1

|Π∗(f )|. ”=” iff

f (x) f ∞ = e−x△ for a simplex △ ∋ 0.

Remark If f (x) = e−xK with K ∈ Kn then Thm. 6 becomes Thm. 4, i.e. 2n

n

  • nn

≤ |K|n−1|Π∗(K)|.

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SLIDE 26

Proof of Theorem 6

Definition Let f ∈ F(Rn). Then Kt(f ) := {x ∈ Rn : f (x) ≥ e−tf ∞} ∀ t ≥ 0.

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Proof of Theorem 6

Definition Let f ∈ F(Rn). Then Kt(f ) := {x ∈ Rn : f (x) ≥ e−tf ∞} ∀ t ≥ 0. Lemma 1 Let f ∈ F(Rn). The covariogram g : Rn → R of f g(x) := ∞ e−t|Kt(f ) ∩ (x + Kt(f ))|dt =

  • Rn min

f (y) f ∞ , f (y − x) f ∞

  • dy

is even and g ∈ F(Rn).

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Proof of Theorem 6

Lemma 2 Let f ∈ F(Rn) and g its covariogram. For every 0 < λ0 < 1 then 2f 1Π∗(f ) =

  • 0<λ<λ0

K− log(1−λ)(g) λ .

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Proof of Theorem 6

  • Proof. Since K− log(1−λ)(g)/λ rewrites as
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Proof of Theorem 6

  • Proof. Since K− log(1−λ)(g)/λ rewrites as
  • x ∈ Rn :

∞ e−t|Kt ∩ (λx + Kt)|dt ≥ (1 − λ) ∞ e−t|Kt|dt

  • =
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Proof of Theorem 6

  • Proof. Since K− log(1−λ)(g)/λ rewrites as
  • x ∈ Rn :

∞ e−t|Kt ∩ (λx + Kt)|dt ≥ (1 − λ) ∞ e−t|Kt|dt

  • =
  • x ∈ Rn :

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ ∞ e−t|Kt|dt

  • .
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Proof of Theorem 6

  • Proof. Since K− log(1−λ)(g)/λ rewrites as
  • x ∈ Rn :

∞ e−t|Kt ∩ (λx + Kt)|dt ≥ (1 − λ) ∞ e−t|Kt|dt

  • =
  • x ∈ Rn :

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ ∞ e−t|Kt|dt

  • .

Since |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)| ≤ λ|x||Px⊥Kt|

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SLIDE 33

Proof of Theorem 6

  • Proof. Since K− log(1−λ)(g)/λ rewrites as
  • x ∈ Rn :

∞ e−t|Kt ∩ (λx + Kt)|dt ≥ (1 − λ) ∞ e−t|Kt|dt

  • =
  • x ∈ Rn :

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ ∞ e−t|Kt|dt

  • .

Since |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)| ≤ λ|x||Px⊥Kt|

Kt λx + Kt λ|x| |Kt| − |Kt ∩ (λx + Kt)| λ|x||Px⊥Kt|

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Proof of Theorem 6

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ |x| ∞ e−t|Px⊥Kt|dt

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SLIDE 35

Proof of Theorem 6

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ |x| ∞ e−t|Px⊥Kt|dt = xΠ∗(f ) 2f ∞ .

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SLIDE 36

Proof of Theorem 6

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ |x| ∞ e−t|Px⊥Kt|dt = xΠ∗(f ) 2f ∞ . Therefore if xΠ∗(f ) ≤ 2f ∞ ∞ e−t|Kt|dt = 2

  • Rn f (x)dx
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Proof of Theorem 6

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ |x| ∞ e−t|Px⊥Kt|dt = xΠ∗(f ) 2f ∞ . Therefore if xΠ∗(f ) ≤ 2f ∞ ∞ e−t|Kt|dt = 2

  • Rn f (x)dx

then x ∈ K− log(1−λ)(g)/λ

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Proof of Theorem 6

∞ e−t |Kt| − |Kt ∩ (λ|x| x

|x| + Kt)|

λ dt ≤ |x| ∞ e−t|Px⊥Kt|dt = xΠ∗(f ) 2f ∞ . Therefore if xΠ∗(f ) ≤ 2f ∞ ∞ e−t|Kt|dt = 2

  • Rn f (x)dx

then x ∈ K− log(1−λ)(g)/λ and thus 2f 1Π∗(f ) ⊂ K− log(1−λ)(g) λ for every 0 < λ < 1.

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Proof of Theorem 6

Definition (Ball 1988) Let f ∈ F(Rn) with f (0) > 0. Then ˜ K(f ) :=

  • x ∈ Rn : n

∞ f (rx)rn−1dr ≥ f (0)

  • fulfills ˜

K(f ) ∈ Kn and | ˜ K(f )| = f 1/f (0).

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Proof of Theorem 6

Definition (Ball 1988) Let f ∈ F(Rn) with f (0) > 0. Then ˜ K(f ) :=

  • x ∈ Rn : n

∞ f (rx)rn−1dr ≥ f (0)

  • fulfills ˜

K(f ) ∈ Kn and | ˜ K(f )| = f 1/f (0). Lemma 3 Let g ∈ F(Rn) with g(0) > 0. If 0 ≤ t ≤ n/e then t (n!)

1 n

˜ K(g) ⊂ Kt(g). ”=” iff g(x)

g(0) = e−xK for some K ∈ Kn with 0 ∈ K.

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Proof of Theorem 6

Proof of Theorem 6. By Lem. 1 we apply Lem. 3 to g and then for every 0 < λ0 < 1 − e− n

e we have

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Proof of Theorem 6

Proof of Theorem 6. By Lem. 1 we apply Lem. 3 to g and then for every 0 < λ0 < 1 − e− n

e we have

  • 0<λ<λ0

− log(1 − λ) (n!)

1 n λ

˜ K(g) ⊂

  • 0<λ<λ0

K− log(1−λ)(g) λ ,

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Proof of Theorem 6

Proof of Theorem 6. By Lem. 1 we apply Lem. 3 to g and then for every 0 < λ0 < 1 − e− n

e we have

  • 0<λ<λ0

− log(1 − λ) (n!)

1 n λ

˜ K(g) ⊂

  • 0<λ<λ0

K− log(1−λ)(g) λ , which by Lem. 2 is equivalent to

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Proof of Theorem 6

Proof of Theorem 6. By Lem. 1 we apply Lem. 3 to g and then for every 0 < λ0 < 1 − e− n

e we have

  • 0<λ<λ0

− log(1 − λ) (n!)

1 n λ

˜ K(g) ⊂

  • 0<λ<λ0

K− log(1−λ)(g) λ , which by Lem. 2 is equivalent to

  • 0<λ<λ0

− log(1 − λ) (n!)

1 n λ

˜ K(g) ⊂ 2f 1Π∗(f ).

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SLIDE 45

Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then

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Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then 1 (n!)

1 n

˜ K(g) ⊂ 2f 1Π∗(f ).

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Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then 1 (n!)

1 n

˜ K(g) ⊂ 2f 1Π∗(f ). Taking volumes we can then conclude

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Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then 1 (n!)

1 n

˜ K(g) ⊂ 2f 1Π∗(f ). Taking volumes we can then conclude 2nf n

1|Π∗(f )| ≥ 1

n!| ˜ K(g)|

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SLIDE 49

Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then 1 (n!)

1 n

˜ K(g) ⊂ 2f 1Π∗(f ). Taking volumes we can then conclude 2nf n

1|Π∗(f )| ≥ 1

n!| ˜ K(g)| = 1 n! 1 g(0)

  • Rn g(x)dx
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SLIDE 50

Proof of Theorem 6

Since h(λ) := − log(1 − λ)/λ is increasing in λ ∈ (0, 1) and limλ→0+ h(λ) = 1, then 1 (n!)

1 n

˜ K(g) ⊂ 2f 1Π∗(f ). Taking volumes we can then conclude 2nf n

1|Π∗(f )| ≥ 1

n!| ˜ K(g)| = 1 n! 1 g(0)

  • Rn g(x)dx

= 1 n! 1

  • f

f ∞

  • Rn
  • Rn min

f (y) f ∞ , f (x) f ∞

  • dydx.
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SLIDE 51

Bibliography

  • D. Alonso-Guti´

errez, J. Bernu´ es, B. Gonz´ alez Merino, Zhang’s inequality for log-concave functions, GAFA Seminar Notes 2019.

  • K. Ball, Logarithmically concave functions and sections of

convex sets in Rn, Studia Math. 88 (1) (1988), 69-84.

  • C. M. Petty, Isoperimetric problems, Proceedings of the

Conference on Convexity and Combinatorial Geometry, 26–41. University of Oklahoma, Norman (1971).

  • G. Zhang, Restricted chord projection and affine inequalities,
  • Geom. Dedicata 39 (2) (1991), 213–222.
  • G. Zhang, The affine Sobolev inequality, J. Differential Geom.

53 (1999), 183–202.

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Thank you for your attention!!

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Conference announcement: Geometry, Analysis & Convexity, Sevilla 22.6.20 to 26.6.20.