Some applications of optimal transport in functional inequalities · 2014. 4. 5. · Natha el Gozlan Applications of optimal transport in functional inequalities. Poincar e inequality

Poincaré inequality and dimension free concentrationBounds on the deficit in the log-Sobolev inequality

Variance estimates for log-concave probabilities

Some applications of optimal transport infunctional inequalities

Nathaël Gozlan∗

∗LAMA Université Paris Est – Marne-la-Vallée

Some applications of optimal transport in functional inequalitiesToulouse, April 2014

Nathaël Gozlan Applications of optimal transport in functional inequalities

Outline of Lecture 3

Part III -

Recent developments

III.1 Poincaré inequality and dimension free concentration.

III.2 Bounds on the deficit in the log-Sobolev inequality for the standardGaussian.

III.3 Transport proofs of variance estimates for log-concave probabilities.

Characterization of PI in terms of dimension freeconcentration

Theorem. [G.-Roberto-Samson ’12]

Suppose that µ satisfies the dimension free concentration property with aconcentration profile α such that α(t) < 1/2 for at least one value of t > 0,then µ satisfies PI(C) with the constant

C =

(inf

{r

Φ−1

(α(r)): r s.t. α(r) < 1/2

})2,

where

Φ(t) =1√2π

∫ +∞t

e−u2/2 du.

Proof of the theorem

Three ingredients :

An equivalent form of concentration of measure in terms of deviationinequalities for the functions Qt f .

The Hamilton Jacobi equation ∂Qt f (x)∂t

= − 14‖∇Qt f ‖2(x).

Central limit theorem.

Concentration and deviations of Lipschitz functions

Lemma

The probability µ satisfies the concentration property with the profile α if andonly if for all 1-Lipschitz function f the following deviation inequality holds

µ(f > mµ(f ) + r) ≤ α(r), ∀r ≥ 0,

where mµ(f ) is a median of f under µ.

(cf Lecture 1)

Concentration and inf-convolution.

For all t > 0, and all f : Rd → R bounded from below, define

Qt f (x) = infy∈Rd

{f (y) +

1

t‖x − y‖2

}, x ∈ Rd .

Lemma

The probability µ satisfies the concentration property with the profile α if andonly if for all function f : Rd → R bounded from below the following deviationinequality holds

µ(Qt f > mµ(f ) + r) ≤ α(√rt), t > 0, r ≥ 0,

For all t > 0, and all f : Rd → R bounded from below, define

Qt f (x) = infy∈Rd

{f (y) +

1

t‖x − y‖2

}, x ∈ Rd .

Lemma

The probability µ satisfies the concentration property with the profile α if andonly if for all function f : Rd → R bounded from below the following deviationinequality holds

µ(Qt f > mµ(f ) + r) ≤ α(√rt), t > 0, r ≥ 0,

For all t > 0, and all f : (Rd)n → R bounded from below, define

Qt f (x) = infy∈(Rd )n

{f (y) +

1

t‖x − y‖2

}, x ∈ (Rd)n.

Lemma

The probability µ satisfies the dimension free concentration property with theprofile α if and only if for all function f : (Rd)n → R bounded from below thefollowing deviation inequality holds

µn(Qt f > mµn (f ) + r) ≤ α(√rt), t > 0, r ≥ 0,

where mµn (f ) is a median of f under µ.

Proof of the lemma.

• [Dev. inq. ⇒ Conc.]Take A such that µ(A) ≥ 1/2, and define

fA = 0 on A and +∞ on Ac .

Then it holds Qt fA(x) =1td2(x ,A) and mµ(fA) = 0, and so

µ(d2( · ,A) > rt) ≤ α(√rt), ∀r ≥ 0.

• [Conc. ⇒ Dev. Inq.]Take f and consider A = {x ∈ Rd ; f ≤ mµ(f )} (µ(A) ≥ 1/2).If y ∈ Ar , then there is some x ∈ A such that ‖x − y‖ ≤ r ; so it holds

Qt f (y) ≤ f (x) +1

tr 2 ≤ mµ(f ) +

1

tr 2.

So

Ar ⊂{Qt f ≤ mµ(f ) +

r 2

t

}.

And soµ(Qt f > mµ(f ) + r

2/t) ≥ 1− α(r).

Proof of the lemma.

fA = 0 on A and +∞ on Ac .Then it holds Qt fA(x) =

1td2(x ,A) and mµ(fA) = 0, and so

µ(d2( · ,A) > rt) ≤ α(√rt), ∀r ≥ 0.

Qt f (y) ≤ f (x) +1

tr 2 ≤ mµ(f ) +

1

tr 2.

So

r 2

t

}.

2/t) ≥ 1− α(r).

Proof of the lemma.

µ(d2( · ,A) > rt) ≤ α(√rt), ∀r ≥ 0.

• [Conc. ⇒ Dev. Inq.]Take f and consider A = {x ∈ Rd ; f ≤ mµ(f )} (µ(A) ≥ 1/2).

If y ∈ Ar , then there is some x ∈ A such that ‖x − y‖ ≤ r ; so it holds

Qt f (y) ≤ f (x) +1

tr 2 ≤ mµ(f ) +

1

tr 2.

So

r 2

t

}.

2/t) ≥ 1− α(r).

Proof of the lemma.

µ(d2( · ,A) > rt) ≤ α(√rt), ∀r ≥ 0.

Qt f (y) ≤ f (x) +1

tr 2 ≤ mµ(f ) +

1

tr 2.

So

r 2

t

}.

2/t) ≥ 1− α(r).

Proof of the lemma.

µ(d2( · ,A) > rt) ≤ α(√rt), ∀r ≥ 0.

Qt f (y) ≤ f (x) +1

tr 2 ≤ mµ(f ) +

1

tr 2.

So

r 2

t

}.

2/t) ≥ 1− α(r).

Sketch of proof

By assumptions, it holds

µn (Qt f > mµn (f ) + r) ≤ α(√rt), ∀n, ∀f ,∀r , t > 0.

Apply this inequality with

fn(x) = h(x1) + h(x2) + · · ·+ h(xn),

where∫h dµ = 0. It holds

Qt fn(x) = Qth(x1) + Qth(x2) + · · ·+ Qth(xn).

Then chooser = u2

√n and t = 1/

√n

Sketch of proof

fn(x) = h(x1) + h(x2) + · · ·+ h(xn),

where∫h dµ = 0.

It holds

Then chooser = u2

√n and t = 1/

√n

Sketch of proof

fn(x) = h(x1) + h(x2) + · · ·+ h(xn),

Then chooser = u2

√n and t = 1/

√n

Sketch of proof

fn(x) = h(x1) + h(x2) + · · ·+ h(xn),

Then chooser = u2

√n and t = 1/

√n

Sketch of proof

P(∑n

i=1 Q1/√nh(Xi )√

n> mn + u

2

)≤ α(u), ∀n,∀u > 0,

where Xi are i.i.d of law µ and mn is the median of the random variable

1√n

n∑i=1

h(Xi ).

The idea is then to pass to the limit using the central limit theorem andHamilton-Jacobi equations :

1√n

n∑i=1

h(Xi )⇒ N (0,Varµ(h)), so mn → 0.

−√nE[Q1/√nh(X1)

]= E

[h(X1)− Q1/√nh(X1)

1/√n

]→ 1

4E[‖∇h‖2(X1)

]∑n

i=1 Q1/√nh(Xi )− E

[Q1/√nh(Xi )

]√n

⇒ N (0,Varµ(h))

Sketch of proof

P

(∑ni=1 Q1/

√nh(Xi )− E

[Q1/√nh(Xi )

]√n

> mn −√nE[Q1/√nh(X1)

]+ u2

)≤ α(u), ∀n, ∀u > 0,

1√n

n∑i=1

h(Xi ).

1√n

n∑i=1

h(Xi )⇒ N (0,Varµ(h)), so mn → 0.

]= E

[h(X1)− Q1/√nh(X1)

1/√n

]→ 1

4E[‖∇h‖2(X1)

]∑n

i=1 Q1/√nh(Xi )− E

[Q1/√nh(Xi )

]√n

⇒ N (0,Varµ(h))

Sketch of proof

P

(∑ni=1 Q1/

√nh(Xi )− E

[Q1/√nh(Xi )

]√n

]+ u2

)≤ α(u), ∀n, ∀u > 0,

1√n

n∑i=1

h(Xi ).

1√n

n∑i=1

h(Xi )⇒ N (0,Varµ(h)), so mn → 0.

]= E

[h(X1)− Q1/√nh(X1)

1/√n

]→ 1

4E[‖∇h‖2(X1)

]∑n

i=1 Q1/√nh(Xi )− E

[Q1/√nh(Xi )

]√n

⇒ N (0,Varµ(h))

Sketch of proof

P

(∑ni=1 Q1/

√nh(Xi )− E

[Q1/√nh(Xi )

]√n

]+ u2

)≤ α(u), ∀n, ∀u > 0,

1√n

n∑i=1

h(Xi ).

1√n

n∑i=1

h(Xi )⇒ N (0,Varµ(h)), so mn → 0.

]= E

[h(X1)− Q1/√nh(X1)

1/√n

]→ 1

4E[‖∇h‖2(X1)

]

∑ni=1 Q1/

√nh(Xi )− E

[Q1/√nh(Xi )

]√n

⇒ N (0,Varµ(h))

Sketch of proof

P

(∑ni=1 Q1/

√nh(Xi )− E

[Q1/√nh(Xi )

]√n

]+ u2

)≤ α(u), ∀n, ∀u > 0,

1√n

n∑i=1

h(Xi ).

1√n

n∑i=1

h(Xi )⇒ N (0,Varµ(h)), so mn → 0.

]= E

[h(X1)− Q1/√nh(X1)

1/√n

]→ 1

4E[‖∇h‖2(X1)

]∑n

i=1 Q1/√nh(Xi )− E

[Q1/√nh(Xi )

]√n

⇒ N (0,Varµ(h))

Sketch of proof

Letting n→∞, we get

P(Y >

1

4

∫|∇h|2 dµ+ u2

)≤ α(u), ∀u ≥ 0,

where Y ∼ N (0,Varµ(h)).

Equivalently, with Φ(t) = 1√2π

∫ +∞t

e−u2/2 du, it holds

Φ

(14

∫|∇h|2 dµ+ u2√Varµ(h)

)≤ α(u)

And so

Φ−1

(α(u))√

Varµ(h) ≤1

4

∫|∇h|2 dµ+ u2

Replacing h by λh, λ > 0, and optimizing over λ yields

Φ−1

(α(u))√

Varµ(h) ≤

infλ>0

λ

4

∫|∇h|2 dµ+ u

2

λ

Sketch of proof

P(Y >

1

4

∫|∇h|2 dµ+ u2

)≤ α(u), ∀u ≥ 0,

∫ +∞t

Φ

(14

)≤ α(u)

And so

Φ−1

(α(u))√

Varµ(h) ≤1

4

∫|∇h|2 dµ+ u2

Φ−1

(α(u))√

Varµ(h) ≤ infλ>0

λ

4

∫|∇h|2 dµ+ u

2

λ

Sketch of proof

P(Y >

1

4

∫|∇h|2 dµ+ u2

)≤ α(u), ∀u ≥ 0,

∫ +∞t

Φ

(14

)≤ α(u)

And so

Φ−1

(α(u))√

Varµ(h) ≤1

4

∫|∇h|2 dµ+ u2

Φ−1

(α(u))√

Varµ(h) ≤

√u2∫|∇h|2 dµ

Sketch of proof

P(Y >

1

4

∫|∇h|2 dµ+ u2

)≤ α(u), ∀u ≥ 0,

∫ +∞t

Φ

(14

)≤ α(u)

And so

Φ−1

(α(u))√

Varµ(h) ≤1

4

∫|∇h|2 dµ+ u2

Φ−1

(α(u))√

Varµ(h) ≤

√u2∫|∇h|2 dµ

III.1 Bounds on the deficit in the log-Sobolevinequality for the standard Gaussian.

Joint work with S. Bobkov, C. Roberto, P-M Samson.

Independent work by E. Indrei and D. Marcon.

Log-Sobolev for the standard Gaussian measure

In all what follows γd denotes the standard Gaussian measure on Rd

γd(dx) =1

(2π)n/2e−

12‖x‖2 dx

Theorem.

For all f : Rd → R sufficiently smooth, it holds

Entγd (f2) ≤ 2

∫‖∇f ‖2(x) γd(dx)

Equivalently, for all probability measure ν on Rd , with a smooth density

H(ν|γd) ≤1

2I(ν|γd).

We recall that if ν = hγd ,

H(ν|γd) =∫

h(x) log(h)(x) γd(dx) and I(ν|γd) =∫‖∇h‖2

h(x) γd(dx).

Equality cases

It is easy to check that for all a ∈ Rd , the standard Gaussian measure withmean a, denoted by γad , is an equality case in LSI for γd

Theorem. [Carlen ’91]

The only equality cases in LSI for γd are the Gaussian measures γad , a ∈ Rd .

Equality cases

It is easy to check that for all a ∈ Rd , the standard Gaussian measure withmean a, denoted by γad , is an equality case in LSI for γd

The only equality cases in LSI for γd are the Gaussian measures γad , a ∈ Rd .

Deficit in LSI

We denote by δ(ν) the deficit in LSI for γd :

δ(ν) =1

2I(ν|γd)− H(ν|γd) ≥ 0

Carlen’s result can be restated as follows

(δ(ν) = 0)⇒ (∃a ∈ Rd , ν = γad)

We are interested in improving this statement in the following way :

“If δ(ν) is small then ν is close to γad for some a.”

Deficit in LSI

δ(ν) =1

2I(ν|γd)− H(ν|γd) ≥ 0

(δ(ν) = 0)⇒ (∃a ∈ Rd , ν = γad)

Deficit in LSI

δ(ν) =1

2I(ν|γd)− H(ν|γd) ≥ 0

(δ(ν) = 0)⇒ (∃a ∈ Rd , ν = γad)

Quantitative form of functional inequalities

N. Fusco, F. Maggi, and A. Pratelli ’08 quantitative isoperimetric inequality.

A. Figalli, F. Maggi, and A. Pratelli ’10 quantitative isoperimetric inequality via mass transport.

A. Cianchi, N. Fusco, F. Maggi, and A. Pratelli ’11 / Eldan ’13 / MosselNeeman ’13

quantitative isoperimetric inequality in Gauss space.

. . . many others . . .

Independent work by E. Indrei and D. Marcon on the deficit in LSI.

Carlen’s bound on the deficit

Define the Fourier transform F operator on L1(Rd , dx) by

F(g)(v) =∫

Rde2iπv·ug(u) du

Define also the isometric operator T : L2(Rd , γd)→ L2(Rd , dx)

T (f )(u) = 2d/4e−π‖u‖2

f (2√πu)

and then consider

W(f )(v) = T−1 ◦ F ◦ T (f )(v)

=1

2dπd/2e

14‖v‖2

∫e

i2v·ue−

14‖u‖2 f (u) du

For all f : Rd → R sufficiently smooth

Entγd (|f |2) + Entγd (|W(f )|

2) ≤ 2∫|∇f |2 dγd .

F(g)(v) =∫

Rde2iπv·ug(u) du

T (f )(u) = 2d/4e−π‖u‖2

f (2√πu)

and then consider

W(f )(v) = T−1 ◦ F ◦ T (f )(v)

=1

2dπd/2e

14‖v‖2

∫e

i2v·ue−

14‖u‖2 f (u) du

2) ≤ 2∫|∇f |2 dγd .

F(g)(v) =∫

Rde2iπv·ug(u) du

T (f )(u) = 2d/4e−π‖u‖2

f (2√πu)

and then consider

W(f )(v) = T−1 ◦ F ◦ T (f )(v)

=1

2dπd/2e

14‖v‖2

∫e

i2v·ue−

14‖u‖2 f (u) du

2) ≤ 2∫|∇f |2 dγd .

F(g)(v) =∫

Rde2iπv·ug(u) du

T (f )(u) = 2d/4e−π‖u‖2

f (2√πu)

and then consider

W(f )(v) = T−1 ◦ F ◦ T (f )(v)

=1

2dπd/2e

14‖v‖2

∫e

i2v·ue−

14‖u‖2 f (u) du

2) ≤ 2∫|∇f |2 dγd .

F(g)(v) =∫

Rde2iπv·ug(u) du

T (f )(u) = 2d/4e−π‖u‖2

f (2√πu)

and then consider

W(f )(v) = T−1 ◦ F ◦ T (f )(v)

=1

2dπd/2e

14‖v‖2

∫e

i2v·ue−

14‖u‖2 f (u) du

2) ≤ 2∫|∇f |2 dγd .

One word on the proof of Carlen’s result

Theorem. [Beckner]

If f ∈ Lp(Rd , dx), with p ∈ [1, 2], then

‖F(f )‖q ≤ Cp‖f ‖p,

where 1/p + 1/q = 1 and Cp = p1p / q

1q .

Differentiating at p = 2 yields to

Theorem. Beckner-Hirschman uncertainty principle

For all f ∈ L2(Rd , dx), with ‖f ‖2 = 1

h(|f |2) + h(|F(f )|2) ≥ d(1− log 2)

where h denotes the Shannon entropy :

h(ρ) = −∫ρ log ρ dx , ∀ρ : Rd → (0,∞),

∫ρ dx = 1.

Translating the Beckner-Hirschman uncertainty principle to the standardGaussian yields to Carlen’s result.

Theorem. [Beckner]

‖F(f )‖q ≤ Cp‖f ‖p,

1q .

h(|f |2) + h(|F(f )|2) ≥ d(1− log 2)

h(ρ) = −∫ρ log ρ dx , ∀ρ : Rd → (0,∞),

∫ρ dx = 1.

Translating the Beckner-Hirschman uncertainty principle to the standardGaussian yields to Carlen’s result.

Theorem. [Beckner]

‖F(f )‖q ≤ Cp‖f ‖p,

1q .

h(|f |2) + h(|F(f )|2) ≥ d(1− log 2)

h(ρ) = −∫ρ log ρ dx , ∀ρ : Rd → (0,∞),

∫ρ dx = 1.

Translating the Beckner-Hirschman uncertainty principle to the standardGaussian yields to Carlen’s result.Nathaël Gozlan Applications of optimal transport in functional inequalities

Bounding the deficit in terms of probability distances

We want to bound the deficit δ from below in terms of probability distanceslike for instance W2.

The following centering operation will play a role in the sequel :

Definition.

If ν is a probability measure on Rd , the probability ν̄ is defined as follows :

it is the law of the random vector X defined by

X = (X 1, . . . ,X d) with X i = Xi − E[Xi |X1, . . . ,Xi−1],

where X is a random vector of law ν.

Note in particular that X 1 = X1 − E[X1] and that (X i )i≤d is a sequence ofmartingale increments.

Moreover, if X is unconditional then X = X .Recall that X is unconditional if X has the same law as (ε1X1, . . . , εdXd) forany choice of εi = ±1.

Definition.

Theorem 1.

For all probability ν on Rd ,

δ(ν) ≥ c T2(ν̄, γd)

H(ν̄|γd),

where c is a universal constant and T is the optimal transport cost defined by

T (ν, µ) = infX∼ν Y∼µ

E

[d∑

i=1

∆(Xi − Yi )

], where ∆(t) = |t|−log(1+|t|).

Theorem 2.

If ν is a probability measure on Rd with a smooth density of the form e−V suchthat for some ε > 0, ∂2i V ≥ ε for all i ∈ {1, . . . , d}, then

δ(ν) ≥ c min(1; ε)W 22 (ν̄, γd).

Theorem 1.

For all probability ν on Rd ,

δ(ν) ≥ c T2(ν̄, γd)

H(ν̄|γd),

where c is a universal constant and T is the optimal transport cost defined by

T (ν, µ) = infX∼ν Y∼µ

E

[d∑

i=1

∆(Xi − Yi )

], where ∆(t) = |t|−log(1+|t|).

Theorem 2.

If ν is a probability measure on Rd with a smooth density of the form e−V suchthat for some ε > 0, ∂2i V ≥ ε for all i ∈ {1, . . . , d}, then

δ(ν) ≥ c min(1; ε)W 22 (ν̄, γd).

Recovering equality cases

It is possible to recover the equality cases using Theorem 1.

Notation. If σ ∈ Sn (the symmetric group) and ν is the law of a random vectorX define νσ as the law of Xσ = (Xσ(1), . . . ,Xσ(n)).

Note that δ(ν) = δ(νσ) for all σ ∈ Sn.

Therefore the conclusion of Theorem 1 self improves into

δ(ν) ≥ supσ∈Sn

T 2(νσ, γd)H(νσ|γd)

If δ(ν) = 0, then necessarily νσ = γd for all σ.

. . . Identifying the densities yields to ν = γad , with a =∫x dν . . .

Step 1 : Relate the deficit in LSI to the deficit in T2.

Recall the HWI inequality (Lecture II):

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

Take ν1 = γd and ν0 = ν and write H = H(ν|γd), W = W2(ν, γd) andI = I(ν|γd) :

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

)2Because of T2 and LSI, it holds W ≤

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

)2 ≥ 14 δ2T2Hwith δT2 (ν) = deficit in T2 = 2 H(ν|γd)−W 22 (ν, γd).

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H

≥ 12I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

)2

Because of T2 and LSI, it holds W ≤√

2H ≤√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2

=1

2

(2H −W 2

)2(√2H + W

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

)2

≥ 14

δ2T2H

with δT2 (ν) = deficit in T2 = 2 H(ν|γd)−W 22 (ν, γd).

H(ν0|γd) ≤ H(ν1|γd) + W2(ν0, ν1)√

I(ν0|γd)−1

2W 22 (ν0, γd)

H ≤W√I − 1

2W 2.

Therefore,

δ =1

2I − H ≥ 1

2I −W

√I +

1

2W 2

=1

2

(√I −W

√2H ≤

√I and so

δ ≥ 12

(√2H −W

)2=

1

2

(2H −W 2

)2(√2H + W

Step 2 : Bound the deficit in T2 in dimension 1

Theorem. [Barthe-Kolesnikov ’08]

For all probability measure ν on R with mean 0, it holds

2 H(ν|γ)−W 22 (ν, γ) ≥ cT (ν, γ).

Proof. See later.

Step 3 : Bound δ in dimension 1

Inserting Barthe-Kolesnikov bound into the lower bound of δ yields to :

δ(ν) ≥ c T2(ν, γ)

H(ν|γ) ,

for all ν ∈ P(R) such that∫x ν(dx) = 0.

The deficit δ is translation invariant :if νa is the image of ν under the map x 7→ x − a, it holds

δ(ν) = δ(νa).

Therefore,

δ(ν) ≥ c T2(νa, γ)

H(νa|γ) , with a =∫

x ν(dx)

δ(ν) ≥ c T2(ν, γ)

H(ν|γ) ,

δ(ν) = δ(νa).

Therefore,

δ(ν) ≥ c T2(νa, γ)

x ν(dx)

δ(ν) ≥ c T2(ν, γ)

H(ν|γ) ,

δ(ν) = δ(νa).

Therefore,

δ(ν) ≥ c T2(νa, γ)

x ν(dx)

δ(ν) ≥ c T2(ν, γ)

H(ν|γ) ,

δ(ν) = δ(νa).

Therefore,

δ(ν) ≥ c T2(ν̄, γ)

H(ν̄|γ)

Step 4 : Tensorizing the deficit bound

Recall the tensorizing formula for H, I and T :

H(ν|µ1 × µ2) = H(ν1|µ1) +∫

H(ν2( · |x1) |µ2) ν1(dx1)

I(ν|µ1 × µ2) ≥ I(ν1|µ1) +∫

I(ν2( · |x1) |µ2) ν1(dx1)

T (ν|µ1 × µ2) ≤ T (ν1|µ1) +∫T (ν2( · |x1) |µ2) ν1(dx1),

where

µ1 ∈ P(Rd1 ), µ2 ∈ P(Rd2 )ν ∈ P(Rd1+d2 ) and

ν(dx1dx2) = ν2(dx2|x1)ν1(dx1)

Consequence: Taking µ1 = γd1 , µ2 = γd2 , one gets

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |x1)) ν1(dx1).

Recall the tensorizing formula for H, I and T :

H(ν|µ1 × µ2) = H(ν1|µ1) +∫

H(ν2( · |x1) |µ2) ν1(dx1)

I(ν|µ1 × µ2) ≥ I(ν1|µ1) +∫

I(ν2( · |x1) |µ2) ν1(dx1)

T (ν|µ1 × µ2) ≤ T (ν1|µ1) +∫T (ν2( · |x1) |µ2) ν1(dx1),

where

µ1 ∈ P(Rd1 ), µ2 ∈ P(Rd2 )ν ∈ P(Rd1+d2 ) and

ν(dx1dx2) = ν2(dx2|x1)ν1(dx1)

Consequence: Taking µ1 = γd1 , µ2 = γd2 , one gets

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |x1)) ν1(dx1).

By induction : Assume that δ(ν) ≥ c T2(ν̄,γd−1)

H(ν̄|γd−1), for all ν ∈ P(Rd−1).

Take ν ∈ P(Rd) ; write z1 = (x1, . . . , xd−1) and z2 = xd .- let ν1 be the marginal on the first d − 1 coordinates,- let ν2( · |z1) be the conditional distribution of z2 knowing z1.

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)

δ(ν) ≥ δ(ν1) +∫δ(ν2( · |z1)) ν1(dz1)

≥ c T2(ν1, γd−1)

H(ν1 | γd−1)+ c

∫ T 2 (ν2( · |z1), γ)H(ν2( · |z1) | γ

) ν1(dz1)≥ c T

2(ν1, γd−1)

H(ν1 | γd−1)+ c

(∫T (ν2( · |z1), γ) ν1(dz1)

)2∫

H(ν2( · |z1) | γ) ν1(dz1)

≥ c

(T (ν1, γd−1) +

∫T (ν2( · |z1), γ) ν1(dz1)

)2H(ν1 | γd−1) +

∫H(ν2( · |z1) | γ) ν1(dz1)

≥ c T2(ν, γd)

H(ν|γd)Nathaël Gozlan Applications of optimal transport in functional inequalities

III.2 Transport proofs of variance estimates forlog-concave probabilities.

Joint work with D. Cordero-Erausquin.

Log-concave random vectors

Recall that a probability µ on Rd is log-concave if it has a density with respectto Lebesgue of the form e−V with a convex function V : Rd → R ∪ {+∞}.

Basic example : µ uniform probability on a (bounded) convex domain.

A random vector X is said log-concave if its law is log-concave.It is called isotropic if E[X ] = 0 and E[XiXj ] = δij .

The KLS and Variance conjectures

Recall the KLS conjecture.

KLS conjecture

There exists a universal constant c > 0 such that any isotropic log-concaverandom vector X satisfies

Var(f (X )) ≤ cE[‖∇f (X )‖2], for all f smooth enough.

Considering the special case f (x) = ‖x‖2, the KLS conjecture implies thevariance conjecture

Variance conjecture

There exists a universal constant c > 0 such that any d-dimensional isotropiclog-concave random vector X satisfies

Var(‖X‖2) ≤ cd .

KLS conjecture

Variance conjecture

Var(‖X‖2) ≤ cd .

KLS conjecture

Variance conjecture

Var(‖X‖2) ≤ cd .

Some applications of optimal transport in functional inequalities · 2014. 4. 5. · Natha el Gozlan Applications of optimal transport in functional inequalities. Poincar e inequality

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