Bakry meets Villani Fabrice Baudoin Purdue University Purdue - - PowerPoint PPT Presentation

Bakry meets Villani

Fabrice Baudoin

Purdue University

Purdue Probability Seminar

Dominique Bakry

Dominique Bakry is a professor at the University Paul Sabatier in Toulouse, France.

Dominique Bakry

Dominique Bakry is a professor at the University Paul Sabatier in Toulouse, France. His works and the methods he introduced in semigroup theory have revolutionized the theory of functional inequalities and of their applications to PDEs and geometry.

Cedric Villani

Cedric Villani was awarded the Fields medal in 2009 for his works

• n the Boltzmann equation and the kinetic theory of gases.

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function.

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function. Let Pt = etL be the semigroup generated by L. Our basic questions are:

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function. Let Pt = etL be the semigroup generated by L. Our basic questions are: Does Pt converge to an equilibrium ?

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function. Let Pt = etL be the semigroup generated by L. Our basic questions are: Does Pt converge to an equilibrium ? In which sense ?

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function. Let Pt = etL be the semigroup generated by L. Our basic questions are: Does Pt converge to an equilibrium ? In which sense ?At which speed ?

The Bakry-Emery criterion

Consider on Rn the Ornstein-Uhlenbeck type operator Lf = ∆f − ∇U, ∇f where U : Rn → R is a smooth function. Let Pt = etL be the semigroup generated by L. Our basic questions are: Does Pt converge to an equilibrium ? In which sense ?At which speed ? In other words, if (Xt)t≥0 is the Markov process with generator L, Can we find conditions on U so that, in distribution, Xt → X∞ ?

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable.

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable. For general U’s, the answer is intimately related to the convexity properties of U:

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable. For general U’s, the answer is intimately related to the convexity properties of U: This is the Bakry-Emery criterion.

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable. For general U’s, the answer is intimately related to the convexity properties of U: This is the Bakry-Emery criterion. The first observation is that L is symmetric with respect to the measure dµ(x) = e−U(x)dx, that is

• Rn fLgdµ = −
• Rn∇f , ∇gdµ =
• Rn gLfdµ.

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable. For general U’s, the answer is intimately related to the convexity properties of U: This is the Bakry-Emery criterion. The first observation is that L is symmetric with respect to the measure dµ(x) = e−U(x)dx, that is

• Rn fLgdµ = −
• Rn∇f , ∇gdµ =
• Rn gLfdµ.

The semigroup Pt is therefore self-adjoint in L2(µ):

• Rn(Ptf )gdµ =
• Rn f (Ptg)dµ

The Bakry-Emery criterion

For instance, if U(x) = x2, there is a convergence with exponential rate to a Gaussian random variable. For general U’s, the answer is intimately related to the convexity properties of U: This is the Bakry-Emery criterion. The first observation is that L is symmetric with respect to the measure dµ(x) = e−U(x)dx, that is

• Rn fLgdµ = −
• Rn∇f , ∇gdµ =
• Rn gLfdµ.

The semigroup Pt is therefore self-adjoint in L2(µ):

• Rn(Ptf )gdµ =
• Rn f (Ptg)dµ

and the spectral theorem implies that in L2(µ): Ptf → E0(f ).

The Bakry-Emery criterion

◮ If µ(Rn) = +∞, then E0(f ) = 0, and the rate of convergence

is typically polynomial; Think of the case U = 0, for which we have the ultracontractive estimate: Ptf 2 ≤ C tn/4 f 1.

The Bakry-Emery criterion

◮ If µ(Rn) = +∞, then E0(f ) = 0, and the rate of convergence

is typically polynomial; Think of the case U = 0, for which we have the ultracontractive estimate: Ptf 2 ≤ C tn/4 f 1.

◮ If µ(Rn) < +∞, then E0(f ) = 1 µ(Rn)

• Rn fdµ and the rate of

convergence is typically exponential: Ptf − E0(f )2 ≤ e−λtf − E0(f )2

The Bakry-Emery criterion

Since Ptf − E0(f )2 ≤ e−λtf − E0(f )2 is equivalent to the fact that the spectrum of −L lies in {0} ∪ [λ, +∞), the convergence to equilibrium with an exponential rate λ is also equivalent to the Poincaré inequality:

• Rn f 2dµ −
• Rn fdµ

2 ≤ 1 λ

• Rn ∇f 2dµ.

The Bakry-Emery criterion

Problem: How do we prove the Poincaré inequality ?

The Bakry-Emery criterion

Problem: How do we prove the Poincaré inequality ? The main

• bservation by Bakry and Emery is the equivalence between the

following two assertions:

◮ ∇2U ≥ ρ in the sense of quadratic forms;

The Bakry-Emery criterion

Problem: How do we prove the Poincaré inequality ? The main

• bservation by Bakry and Emery is the equivalence between the

following two assertions:

◮ ∇2U ≥ ρ in the sense of quadratic forms; ◮ ∇Ptf ≤ e−ρtPt∇f

The Bakry-Emery criterion: Poincaré inequality

The gradient bound ∇Ptf ≤ e−ρtPt∇f easily implies the Poincaré inequality indeed

• Rn f 2dµ −
• Rn fdµ

2 = − +∞ d dt

• Rn(Ptf )2dµdt

= 2 +∞

• Rn ∇Ptf 2dµdt

≤ 2 +∞

• Rn e−2ρtPt∇f 2dµdt

≤ 1 ρ

• Rn ∇f 2dµ.

The Bakry-Emery criterion: The log-Sobolev inequality

For physical reasons, we may also be interested in the entropic convergence to equilibrium of (Pt)t≥0, which is much stronger than the L2 convergence.

The Bakry-Emery criterion: The log-Sobolev inequality

For physical reasons, we may also be interested in the entropic convergence to equilibrium of (Pt)t≥0, which is much stronger than the L2 convergence. The entropic convergence of the heat flow Pt to equilibrium: Entµ(Ptf ) ≤ e− 4t

C Entµ(f )

The Bakry-Emery criterion: The log-Sobolev inequality

For physical reasons, we may also be interested in the entropic convergence to equilibrium of (Pt)t≥0, which is much stronger than the L2 convergence. The entropic convergence of the heat flow Pt to equilibrium: Entµ(Ptf ) ≤ e− 4t

C Entµ(f )

is equivalent to the log-Sobolev inequality Entµ(f 2) ≤ C

• Rn ∇f 2dµ,

where Entµ(f ) =

• Rn f ln fdµ.

The Bakry-Emery criterion: The log-Sobolev inequality

For physical reasons, we may also be interested in the entropic convergence to equilibrium of (Pt)t≥0, which is much stronger than the L2 convergence. The entropic convergence of the heat flow Pt to equilibrium: Entµ(Ptf ) ≤ e− 4t

C Entµ(f )

is equivalent to the log-Sobolev inequality Entµ(f 2) ≤ C

• Rn ∇f 2dµ,

where Entµ(f ) =

• Rn f ln fdµ.

Log-Sobolev = ⇒ Poincaré inequality

Gross’ logarithm Sobolev inequality

Gross’ logarithm Sobolev inequality

The celebrated Gross’ log-Sobolev inequality (1975) asserts that if dγ(x) =

1 (2π)n/2 exp

• − x2

2

• dx denotes the standard Gaussian

measure on Rn, then, if

• Rn f 2dγ = 1,
• Rn f 2 ln f 2dγ ≤ 2
• Rn ∇f 2dγ

Federbush’ logarithm Sobolev inequality ?

On 03/21/2011, I received the following email from Pr. Federbush (University of Michigan):

Federbush’ logarithm Sobolev inequality ?

On 03/21/2011, I received the following email from Pr. Federbush (University of Michigan): " Dear Prof. Baudoin , I am a rather good mathematical physicist, and at age 77 am still doing research. I am writing to you because I am unfortunate enough to have many years ago invented log-Sobolev inequalities, and my name is not Gross [...]

Federbush’ logarithm Sobolev inequality ?

On 03/21/2011, I received the following email from Pr. Federbush (University of Michigan): " Dear Prof. Baudoin , I am a rather good mathematical physicist, and at age 77 am still doing research. I am writing to you because I am unfortunate enough to have many years ago invented log-Sobolev inequalities, and my name is not Gross [...] [...] I guess it is part of human nature for a researcher to want credit for their work, though I believe it would be much more ideal to only need the joy of making discoveries, giving contributions to the advancement of knowledge. Good luck with your research career,"

Bakry-Emery and log-Sobolev

• f ln fdµ −
• fdµ ln
• fdµ

Bakry-Emery and log-Sobolev

• f ln fdµ −
• fdµ ln
• fdµ

= − +∞ d dt Ptf ln Ptfdtdµ

Bakry-Emery and log-Sobolev

• f ln fdµ −
• fdµ ln
• fdµ

= − +∞ d dt Ptf ln Ptfdtdµ = − +∞ LPtf ln Ptfdtdµ

Bakry-Emery and log-Sobolev

• f ln fdµ −
• fdµ ln
• fdµ

= − +∞ d dt Ptf ln Ptfdtdµ = − +∞ LPtf ln Ptfdtdµ = +∞ ∇Ptf 2 Ptf dtdµ

Bakry-Emery and log-Sobolev

• f ln fdµ −
• fdµ ln
• fdµ

= − +∞ d dt Ptf ln Ptfdtdµ = − +∞ LPtf ln Ptfdtdµ = +∞ ∇Ptf 2 Ptf dtdµ ≤ +∞ e−2ρt (Pt∇f )2 Ptf dtdµ

Bakry-Emery and log-Sobolev

Now, from Cauchy-Schwarz inequality, (Pt∇f )2 ≤ (Ptf )Pt ∇f 2 f

Bakry-Emery and log-Sobolev

Now, from Cauchy-Schwarz inequality, (Pt∇f )2 ≤ (Ptf )Pt ∇f 2 f

• This yields
• f ln fdµ −
• fdµ ln
• fdµ

Bakry-Emery and log-Sobolev

Now, from Cauchy-Schwarz inequality, (Pt∇f )2 ≤ (Ptf )Pt ∇f 2 f

• This yields
• f ln fdµ −
• fdµ ln
• fdµ

≤ +∞ e−2ρtPt ∇f 2 f

• dtdµ

Bakry-Emery and log-Sobolev

Now, from Cauchy-Schwarz inequality, (Pt∇f )2 ≤ (Ptf )Pt ∇f 2 f

• This yields
• f ln fdµ −
• fdµ ln
• fdµ

≤ +∞ e−2ρtPt ∇f 2 f

• dtdµ

≤ 1 2ρ ∇f 2 f dµ.

The kinetic Fokker-Planck

The Bakry-Émery criterion and the associated semigroup commutation ∇Ptf ≤ e−ρtPt∇f require some form of ellipticity of the operator L.

The kinetic Fokker-Planck

The Bakry-Émery criterion and the associated semigroup commutation ∇Ptf ≤ e−ρtPt∇f require some form of ellipticity of the operator L. Studying the convergence to equilibrium for semigroups generated by hypoelliptic operators is a difficult problem which has been extensively studied in the literature. A primary example is given by the kinetic Fokker-Planck equation.

The kinetic Fokker-Planck equation

Let V : Rn → R be a smooth function. The kinetic Fokker-Planck equation with confinement potential V is the parabolic partial differential equation: ∂h ∂t = ∆vh − v · ∇vh + ∇V · ∇vh − v · ∇xh, (x, v) ∈ R2n.

The kinetic Fokker-Planck equation

Let V : Rn → R be a smooth function. The kinetic Fokker-Planck equation with confinement potential V is the parabolic partial differential equation: ∂h ∂t = ∆vh − v · ∇vh + ∇V · ∇vh − v · ∇xh, (x, v) ∈ R2n. It is the Kolmogorov-Fokker-Planck equation associated to the stochastic differential system

• dxt = vtdt

dvt = −vtdt − ∇V (xt)dt + dBt, where (Bt)t≥0 is a Brownian motion in Rn.

The kinetic Fokker-Planck equation

Let V : Rn → R be a smooth function. The kinetic Fokker-Planck equation with confinement potential V is the parabolic partial differential equation: ∂h ∂t = ∆vh − v · ∇vh + ∇V · ∇vh − v · ∇xh, (x, v) ∈ R2n. It is the Kolmogorov-Fokker-Planck equation associated to the stochastic differential system

• dxt = vtdt

dvt = −vtdt − ∇V (xt)dt + dBt, where (Bt)t≥0 is a Brownian motion in Rn. Heuristically, d2xt dt = −dxt dt − ∇V (xt) + dBt dt , describes the random motion of a particle in a force field with a white noise perturbation.

The kinetic Fokker-Planck equation

The operator L = ∆v − v · ∇v + ∇V · ∇v − v · ∇x is not elliptic but only hypoelliptic.

The kinetic Fokker-Planck equation

The operator L = ∆v − v · ∇v + ∇V · ∇v − v · ∇x is not elliptic but only hypoelliptic. The operator L admits for invariant measure the measure dµ = e−V (x)− v2

2 dxdv.

The kinetic Fokker-Planck equation

The operator L = ∆v − v · ∇v + ∇V · ∇v − v · ∇x is not elliptic but only hypoelliptic. The operator L admits for invariant measure the measure dµ = e−V (x)− v2

2 dxdv.

L is not symmetric with respect to µ.

Villani and the kinetic Fokker-Planck equation

Due to its importance in mathematical physics, there has been a considerable amount of research concerning the convergence to equilibrium for the kinetic Fokker-Planck semigroup Pt = etL.

Villani and the kinetic Fokker-Planck equation

Due to its importance in mathematical physics, there has been a considerable amount of research concerning the convergence to equilibrium for the kinetic Fokker-Planck semigroup Pt = etL. One

• f the most general results is due to Villani.

Villani and the kinetic Fokker-Planck equation

Due to its importance in mathematical physics, there has been a considerable amount of research concerning the convergence to equilibrium for the kinetic Fokker-Planck semigroup Pt = etL. One

• f the most general results is due to Villani.

Assume that ∇2V ≤ c(1 + ∇V ) and that µ satisfies the classical Poincaré inequality

• R2n ∇f 2dµ ≥ κ
• R2n f 2dµ −
• R2n fdµ

2 .

Villani and the kinetic Fokker-Planck equation

Due to its importance in mathematical physics, there has been a considerable amount of research concerning the convergence to equilibrium for the kinetic Fokker-Planck semigroup Pt = etL. One

• f the most general results is due to Villani.

Assume that ∇2V ≤ c(1 + ∇V ) and that µ satisfies the classical Poincaré inequality

• R2n ∇f 2dµ ≥ κ
• R2n f 2dµ −
• R2n fdµ

2 . Then, there exist constants C > 0 and λ > 0 such that for every f , with

• R2n fdµ = 0,
• R2n(Ptf )2dµ+
• R2n ∇Ptf 2dµ ≤ Ce−λt
• R2n f 2dµ +
• R2n ∇f 2dµ

Villani and the kinetic Fokker-Planck equation

The proof by Villani is intricate and difficult.

Villani and the kinetic Fokker-Planck equation

The proof by Villani is intricate and difficult. It relies on the deep concept of hypocoercivity that he specifically introduced to study this equation.

Villani and the kinetic Fokker-Planck equation

The proof by Villani is intricate and difficult. It relies on the deep concept of hypocoercivity that he specifically introduced to study this equation. Recently, I proposed an alternative approach which is based on a suitable extension of the Bakry-Emery criterion.

Villani and the kinetic Fokker-Planck equation

The proof by Villani is intricate and difficult. It relies on the deep concept of hypocoercivity that he specifically introduced to study this equation. Recently, I proposed an alternative approach which is based on a suitable extension of the Bakry-Emery criterion.We associate to L the carré du champ operator, Γ(f , g) = 1 2 (L(fg) − fLg − gLf ) = ∇vf · ∇vg and its iteration Γ2(f , g) = 1 2 (LΓ(f , g) − Γ(f , Lg) − Γ(g, Lf )) .

Γ2 and the kinetic Fokker-Planck equation

A straightforward computation shows that:

Lemma

For f ∈ C ∞(R2n), Γ2(f ) = ∇2

vf 2 + Γ(f ) + ∇xf · ∇vf ,

=

n

• i,j=1

∂2f ∂vi∂vj 2 +

n

• i=1

∂f ∂vi 2 +

n

• i=1

∂f ∂xi ∂f ∂vi

Γ2 and the kinetic Fokker-Planck equation

A straightforward computation shows that:

Lemma

For f ∈ C ∞(R2n), Γ2(f ) = ∇2

vf 2 + Γ(f ) + ∇xf · ∇vf ,

=

n

• i,j=1

∂2f ∂vi∂vj 2 +

n

• i=1

∂f ∂vi 2 +

n

• i=1

∂f ∂xi ∂f ∂vi The term ∇xf · ∇vf makes impossible to bound from below Γ2 by Γ alone. As a consequence the Bakry-Émery curvature of L is −∞.

Γ2 and the kinetic Fokker-Planck equation

The idea is then to introduce a carefully chosen vertical gradient and to compute the corresponding curvature of L in this vertical direction.

Γ2 and the kinetic Fokker-Planck equation

The idea is then to introduce a carefully chosen vertical gradient and to compute the corresponding curvature of L in this vertical

• direction. For i = 1, · · · , n, we denote

Zi = 2 ∂ ∂xi + ∂ ∂vi .

Γ2 and the kinetic Fokker-Planck equation

The idea is then to introduce a carefully chosen vertical gradient and to compute the corresponding curvature of L in this vertical

• direction. For i = 1, · · · , n, we denote

Zi = 2 ∂ ∂xi + ∂ ∂vi . We define then ΓZ(f , g) =

n

• i=1

ZifZig

Γ2 and the kinetic Fokker-Planck equation

The idea is then to introduce a carefully chosen vertical gradient and to compute the corresponding curvature of L in this vertical

• direction. For i = 1, · · · , n, we denote

Zi = 2 ∂ ∂xi + ∂ ∂vi . We define then ΓZ(f , g) =

n

• i=1

ZifZig and ΓZ

2 (f , g) = 1

2

• LΓZ(f , g) − ΓZ(f , Lg) − ΓZ(g, Lf )
• .

Γ2 and the kinetic Fokker-Planck equation

Lemma

For f ∈ C ∞(R2n), ΓZ

2 (f ) = ∇vZf 2 + 1

2ΓZ(f ) + 1 2∇vf · Zf − 2∇2V (∇vf , Zf ) =

n

• i,j=1

∂ ∂vi Zjf 2 + 1 2ΓZ(f ) + 1 2

n

• i=1

∂f ∂vi Zif − 2

n

• i,j=1

∂2V ∂xi∂xj ∂f ∂vi Zjf .

Γ2 and the kinetic Fokker-Planck equation

As a consequence of the previous computations, we deduce

Theorem

For every 0 < η < 1

2, there exists K(η) ≥ − 1 2 such that for every

f ∈ C ∞(R2n), Γ2(f ) + ΓZ

2 (f ) ≥ −K(η)Γ(f ) + ηΓZ(f ).

Γ2 and the kinetic Fokker-Planck equation

Theorem

Assume that the normalized invariant measure dµ = 1

Z e−V (x)− v2

2 dxdv is a probability measure that satisfies the

Poincaré inequality

• R2n(Γ(f ) + ΓZ(f ))dµ ≥ κ
• Rn f 2dµ −
• Rn fdµ

2 .

Γ2 and the kinetic Fokker-Planck equation

Theorem

Assume that the normalized invariant measure dµ = 1

Z e−V (x)− v2

2 dxdv is a probability measure that satisfies the

Poincaré inequality

• R2n(Γ(f ) + ΓZ(f ))dµ ≥ κ
• Rn f 2dµ −
• Rn fdµ

2 . For every f ∈ H1(µ), with

• R2n fdµ = 0,

(η + K(η))

• R2n(Ptf )2dµ +
• R2n(Γ(Ptf ) + ΓZ(Ptf ))dµ

≤e−λt

• (η + K(η))
• R2n f 2dµ +
• R2n(Γ(f ) + ΓZ(f ))dµ
• ,

where λ =

2ηκ κ+η+K(η).

Γ2 and the kinetic Fokker-Planck equation

Theorem

Assume that the normalized invariant measure dµ = 1

Z e−V (x)− v2

2 dxdv is a probability that satisfies the

log-Sobolev inequality

• R2n(f Γ(ln f ) + f ΓZ(ln f ))dµ ≥ κEntµ(f ).

Γ2 and the kinetic Fokker-Planck equation

Theorem

Assume that the normalized invariant measure dµ = 1

Z e−V (x)− v2

2 dxdv is a probability that satisfies the

log-Sobolev inequality

• R2n(f Γ(ln f ) + f ΓZ(ln f ))dµ ≥ κEntµ(f ).

For every positive and bounded f ∈ C ∞(R2n), such that √ f is Lipschitz and

• R2n fdµ = 1,

2(η + K(η))

• R2n Ptf ln Ptfdµ +
• R2n(Ptf Γ(ln Ptf ) + Ptf ΓZ(ln Ptf ))dµ

≤e−λt

• 2(η + K(η))
• R2n f ln fdµ +
• R2n(f Γ(ln f ) + f ΓZ(ln f ))dµ
• ,

where λ =

2ηκ κ+2(η+K(η)).

Γ2 and degenerate hypoelliptic operators

This approach may be generalized to more general situations.

Γ2 and degenerate hypoelliptic operators

This approach may be generalized to more general situations. Let X1, · · · , Xn be smooth vector fields on Rd. We consider on Rd a diffusion operator L that can be written as L = L0 + Y , where L0 = n

i=1 X 2 i and Y is a smooth vector field on Rd.

Γ2 and degenerate hypoelliptic operators

This approach may be generalized to more general situations. Let X1, · · · , Xn be smooth vector fields on Rd. We consider on Rd a diffusion operator L that can be written as L = L0 + Y , where L0 = n

i=1 X 2 i and Y is a smooth vector field on Rd.

We assume [Xi, Xj] =

• k

ωk

ijXk,

[Xi, [Y , Xj]] =

n

• k=1

αk

ijXk + n

• k=1

βk

ij[Y , Xk],

for some bounded and Lipschitz functions.

Γ2 and degenerate hypoelliptic operators

In that case, we can construct vector fields Z1, · · · , Zn such that:

Theorem

For every 0 < η < r, there exists K(η) ∈ R such that for every f ∈ C ∞(R2n), Γ2(f ) + ΓZ

2 (f ) ≥ −K(η)Γ(f ) + ηΓZ(f ).

Γ2 and degenerate hypoelliptic operators

As a consequence we obtain a nice criterium for convergence to equilibrium:

Theorem

Assume that L admits an invariant probability measure that satisfies the Poincaré inequality

• R2n(Γ(f ) + ΓZ(f ))dµ ≥ κ
• Rn f 2dµ −
• Rn fdµ

2 .

Γ2 and degenerate hypoelliptic operators

As a consequence we obtain a nice criterium for convergence to equilibrium:

Theorem

Assume that L admits an invariant probability measure that satisfies the Poincaré inequality

• R2n(Γ(f ) + ΓZ(f ))dµ ≥ κ
• Rn f 2dµ −
• Rn fdµ

2 . For every f ∈ H1(µ), with

• R2n fdµ = 0,

(η + K(η))

• R2n(Ptf )2dµ +
• R2n(Γ(Ptf ) + ΓZ(Ptf ))dµ

≤e−λt

• (η + K(η))
• R2n f 2dµ +
• R2n(Γ(f ) + ΓZ(f ))dµ
• ,

where λ =

2ηκ κ+η+K(η).