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 on the Boltzmann equation and the kinetic theory of gases.

The Bakry-Emery criterion Consider on R n the Ornstein-Uhlenbeck type operator Lf = ∆ f − �∇ U , ∇ f � where U : R n → R is a smooth function.

The Bakry-Emery criterion Consider on R n the Ornstein-Uhlenbeck type operator Lf = ∆ f − �∇ U , ∇ f � where U : R n → R is a smooth function. Let P t = e tL be the semigroup generated by L . Our basic questions are:

The Bakry-Emery criterion Consider on R n the Ornstein-Uhlenbeck type operator Lf = ∆ f − �∇ U , ∇ f � where U : R n → R is a smooth function. Let P t = e tL be the semigroup generated by L . Our basic questions are: Does P t converge to an equilibrium ?

The Bakry-Emery criterion Consider on R n the Ornstein-Uhlenbeck type operator Lf = ∆ f − �∇ U , ∇ f � where U : R n → R is a smooth function. Let P t = e tL be the semigroup generated by L . Our basic questions are: Does P t converge to an equilibrium ? In which sense ?

The Bakry-Emery criterion Consider on R n the Ornstein-Uhlenbeck type operator Lf = ∆ f − �∇ U , ∇ f � where U : R n → R is a smooth function. Let P t = e tL be the semigroup generated by L . Our basic questions are: Does P t converge to an equilibrium ? In which sense ?At which speed ?

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

The Bakry-Emery criterion For instance, if U ( x ) = � x � 2 , there is a convergence with exponential rate to a Gaussian random variable.

The Bakry-Emery criterion For instance, if U ( x ) = � x � 2 , 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 ) = � x � 2 , 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 ) = � x � 2 , 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 d µ ( x ) = e − U ( x ) dx , that is � � � R n fLgd µ = − R n �∇ f , ∇ g � d µ = R n gLfd µ.

The Bakry-Emery criterion For instance, if U ( x ) = � x � 2 , 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 d µ ( x ) = e − U ( x ) dx , that is � � � R n fLgd µ = − R n �∇ f , ∇ g � d µ = R n gLfd µ. The semigroup P t is therefore self-adjoint in L 2 ( µ ) : � � R n ( P t f ) gd µ = R n f ( P t g ) d µ

The Bakry-Emery criterion For instance, if U ( x ) = � x � 2 , 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 d µ ( x ) = e − U ( x ) dx , that is � � � R n fLgd µ = − R n �∇ f , ∇ g � d µ = R n gLfd µ. The semigroup P t is therefore self-adjoint in L 2 ( µ ) : � � R n ( P t f ) gd µ = R n f ( P t g ) d µ and the spectral theorem implies that in L 2 ( µ ) : P t f → E 0 ( f ) .

The Bakry-Emery criterion ◮ If µ ( R n ) = + ∞ , then E 0 ( f ) = 0, and the rate of convergence is typically polynomial; Think of the case U = 0, for which we have the ultracontractive estimate: C � P t f � 2 ≤ t n / 4 � f � 1 .

The Bakry-Emery criterion ◮ If µ ( R n ) = + ∞ , then E 0 ( f ) = 0, and the rate of convergence is typically polynomial; Think of the case U = 0, for which we have the ultracontractive estimate: C � P t f � 2 ≤ t n / 4 � f � 1 . ◮ If µ ( R n ) < + ∞ , then E 0 ( f ) = 1 � R n fd µ and the rate of µ ( R n ) convergence is typically exponential: � P t f − E 0 ( f ) � 2 ≤ e − λ t � f − E 0 ( f ) � 2

The Bakry-Emery criterion Since � P t f − E 0 ( f ) � 2 ≤ e − λ t � f − E 0 ( 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: � 2 �� ≤ 1 � � R n f 2 d µ − R n �∇ f � 2 d µ. R n fd µ λ

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 observation by Bakry and Emery is the equivalence between the following two assertions: ◮ ∇ 2 U ≥ ρ in the sense of quadratic forms;

The Bakry-Emery criterion Problem: How do we prove the Poincaré inequality ? The main observation by Bakry and Emery is the equivalence between the following two assertions: ◮ ∇ 2 U ≥ ρ in the sense of quadratic forms; ◮ �∇ P t f � ≤ e − ρ t P t �∇ f �

The Bakry-Emery criterion: Poincaré inequality The gradient bound �∇ P t f � ≤ e − ρ t P t �∇ f � easily implies the Poincaré inequality indeed � + ∞ � 2 �� � d � R n f 2 d µ − R n ( P t f ) 2 d µ dt R n fd µ = − dt 0 � + ∞ � R n �∇ P t f � 2 d µ dt = 2 0 � + ∞ � R n e − 2 ρ t P t �∇ f � 2 d µ dt ≤ 2 0 ≤ 1 � R n �∇ f � 2 d µ. ρ

The Bakry-Emery criterion: The log-Sobolev inequality For physical reasons, we may also be interested in the entropic convergence to equilibrium of ( P t ) t ≥ 0 , which is much stronger than the L 2 convergence.

The Bakry-Emery criterion: The log-Sobolev inequality For physical reasons, we may also be interested in the entropic convergence to equilibrium of ( P t ) t ≥ 0 , which is much stronger than the L 2 convergence. The entropic convergence of the heat flow P t to equilibrium: Ent µ ( P t f ) ≤ e − 4 t C Ent µ ( f )

The Bakry-Emery criterion: The log-Sobolev inequality For physical reasons, we may also be interested in the entropic convergence to equilibrium of ( P t ) t ≥ 0 , which is much stronger than the L 2 convergence. The entropic convergence of the heat flow P t to equilibrium: Ent µ ( P t f ) ≤ e − 4 t C Ent µ ( f ) is equivalent to the log-Sobolev inequality � Ent µ ( f 2 ) ≤ C R n �∇ f � 2 d µ, where Ent µ ( f ) = � R n f ln fd µ .

The Bakry-Emery criterion: The log-Sobolev inequality For physical reasons, we may also be interested in the entropic convergence to equilibrium of ( P t ) t ≥ 0 , which is much stronger than the L 2 convergence. The entropic convergence of the heat flow P t to equilibrium: Ent µ ( P t f ) ≤ e − 4 t C Ent µ ( f ) is equivalent to the log-Sobolev inequality � Ent µ ( f 2 ) ≤ C R n �∇ f � 2 d µ, where Ent µ ( f ) = � R n f ln fd µ . Log-Sobolev = ⇒ Poincaré inequality

Gross’ logarithm Sobolev inequality

Gross’ logarithm Sobolev inequality The celebrated Gross’ log-Sobolev inequality (1975) asserts that if � � − � x � 2 1 ( 2 π ) n / 2 exp dx denotes the standard Gaussian d γ ( x ) = 2 measure on R n , then, if R n f 2 d γ = 1, � � � R n f 2 ln f 2 d γ ≤ 2 R n �∇ f � 2 d γ

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 [...]