Concentration of Measure: Logarithmic Sobolev Inequalities

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1 Concentration o Measure: Logarithmic Sobolev Inequalities Aukosh Jagannath September 15, 13 Abstract In this note we'll describe the basic tools to understand Log Sobolev Inequalities and their relation to concentration. We will then briey discuss applications 1 Introduction In the ollowing the spaces we will be working with will be (M n, g, µ(dx)) where either M n = R n or is a compact riemannian maniold and dx is either the lebesuge measure or the normalized riemannian metric and µ(dx) = e Φ(x) Z dx where we assume Φ is smooth. For Φ we dene the operator L Φ = < Φ, > acting on C (M) D(L Φ ) where the terms are to be interpreted in the obvious way in the maniold setting. I'll drop out the Φ in the subscript rom now on. Denition 1. Let Γ n (, g) be dened as ollows, Γ (, g) = g, and Γ n (, g) = LΓ n 1 (, g) Γ n 1 (, Lg) Γ n 1 (g, L) In particular, when n = 1, this is called the carré du champs operator and when n = this is called the carré du champs itéré operator With this we then can dene LSI Denition. A space (M, g, µ) as above admits a Logarithmic Sobolev Inequality i there is a c > such that ( ) Ent µ ( ) := log dµ c Γ 1 (, )dµ dµ The main point o the LSI or us will be a consequence o the Herbst argument. Theorem 3. (Gross-Herbst) Let (M, g, µ) satisy and LSI with constant c. Then or all bounded 1-lipschitz unctions Ee λ( E) e cλ In particular, the space admits subgaussian concentration α (M,g,µ) (r) e λ 8c 1

2 Proo. Suppose that we have the log sobolev inequality. assume is dierentiable and bounded and mean zero. Let Λ(λ) = log Ee λ() then, letting F = e λ() d Λ(λ) dλ λ = 1 ( 1 λ = E( E)eλ() Eeλ() ) log E... λ 1 λ EF ( Eλ () F EF log EF ) = Ent(F ) λ EF LSI c 1Lip c λ F λ EF since is mean, the let hand side is at. integrating the inequality and exponentiating gives Ee λ() e cλ. subtracting the mean molliying extends the result to the ull class. Remark 4. Recall that by standard arguments this extends to all lipschitz unctions, which you can show must be in L 1 (µ) by subgaussian concentration. Logarithmic sobolev inequalities have several non trivial reasons why they are interesting. Perhaps the simplest is the dimension ree nature o the inequality under l products. That is,i we consider (M i, g i µ i ) and we take their measure product with the metric given by the product metric on riemannian maniolds, which is an l product, then we get that the new product gets a LSI with the optimal constant the smallest o the others. Theorem 5. (Eron-stein) Proo. Note that Ent() N i=1 Ent() = sup { Ent µi ()dµ N k=1 g, } e g dµ = 1 take g on the product and let g i e g(x 1 (x i,..., x n ) = log(,...,xn) dµ 1...dµ i 1 eg dµ 1...dµ i ) so that by jensens inequality, thus g i g i g i g i since e gi = 1, we see that maximizing and passing the sup through the sum gives the result. The Bakry-Emery condition Logarithmic sobolev inequalities hold in a variety o spaces. A simple condition that holds in a very large class o spaces is the Bakry-Emery condition

3 Denition 6. A space satises the Bakry-Emery condition i or some c > Γ (, ) 1 c Γ 1(, ) Remark 7. On M = R n, We will see below that this condition is equivalent to HessΦ cid We can relate the above operators to more classical objects. Recall the dirichlet orm Denition 8. The dirichlet orm or L is given by We can then show the ollowing ormulae. E(, g) = (, Lg) Proposition 9. The carré du champs operator satises and we have the Bochner-Bakry-Emery ormula Γ 1 (, ) = g(, ). Γ (, ) =< Hess(), Hess() > + (Ric + HessΦ) (, ). Furthermore, we have the integration by parts ormula E(, g) = Γ 1 (, g)dµ With this in hand we can now show the main theorem in this section. Theorem 1. I a space satises the BE condition with constant c then it has an LSI with constant c. Lemma 11. A space satises the BE condition with constant c i the carré du champ operator satises Γ 1 (P t, P t ) e c t P t Γ 1 (, ) Proo. Let ψ(s) = P s Γ 1 (P t s, P t s ) where D(L) in act we'll assume that is smooth and, i M n = R n then it can be taken to be at most poly growth at innity. Then ψ (s) = LP s Γ 1 (P t s, P t s ) P s Γ 1 (LP t s, P t s ) =: P s Γ (P t s, P t s ) 1 c P sγ 1 (P t s, P t s ) = 1 c ψ(s) solving the dierential inequality and plugging in s = t, gives the result. To go the other way, note that ( ) e c t P t Γ 1 (, ) Γ 1 (, ) /t (Γ 1 (P t, P t ) Γ 1 (, )) /t sending t c Γ 1(, ) + LΓ 1 (, ) (Γ 1 (L, )) 3

4 Remark 1. Note that using the standard grownwall inequality argument we can already get subgaussian concentration without passing to the log-sobolev arguments. Theorem 13. The BE condition implies Ergodicity (Assuming that your maniold is geodesically complete). That is, P t L µ and almost surely Proo. Let be some suciently regular unction (see the previous proo), and let t (x) = P t, then i γ(s) is a geodesic starting at y and ending at x, t (x) t (y) = 1 1 ( 1 < t (γ(s)), γ(s) > ds t (γ(s)) gγ(s) γ(s) g(γ(s)) ds lem. e t c Γ 1 ( t, t ) d(x, y) ( 1 ) 1 Γ 1 (, ) d(x, y). assuming your maniold is cpt or that your unction is lipschitz gives a lipschits constant o e t/c Lip. sending t gives the almost sure convergence. since µ P t +ɛ µ +ɛ by jensens, (take C on R n ) we have by uniorm integrability and thus the result or convergence in L. We can extend this result urther by obvious means. Remark 14. What this result tells us is that it gives us a contiuum analogue o the glauber dynamics result, that is i we propose a suciently regular measure, then we have a dynamics to recover it as a stationary distribution. With these results we can now prove the main result o the section Proo. (Log-Sobolev) First note that by the above arguments and 19, i we assume ɛ Ent() = = = ( P log P dµ ( L= = = P log P dµ P s log P s dµ) ds ) LP s log P s + LP s dµ ds ( ) LP s log P s Γ 1 (P s, log P s )dµds 4

5 Now Γ 1 (P s, log P s )dµ symm. = Γ 1 (, P s (log(p s )))dµ ( c.s. Γ1 ( (, ) ( e s Γ1 ( (, ) c ( symm = e s Γ1 ( (, ) c ( Γ1 ( (, ) e s c Γ 1 (P s log P s, P s log P s ) P s Γ 1 (log P s, log P s ) P s Γ 1 (log P s, log P s ) Γ 1 (P s, log P s ) so that, replacing with Ent( ) = 4 e s c e c = c Γ1 (, ) dµ Γ 1 (, )dµ now again we can extend to the ull class regularizing + ɛ 3 Examples Γ 1 (, )dµds Proposition 15. The space (R n, µ Φ ) with HessΦ cid where c satises the Bakry-emery condition with constant c. In act, i we have Ric + HessΦ cg with then we have the BE conditons with constant c. Proo. Recall that Γ (, ) = (Hess(), Hess()) + (Ric + HessΦ) (, ). To get the rst, note that Ric = and so i we take = a i x i this condition is equivalent to Hess(Φ) 1 c Id. The second argument is clear. Example 16. Gauss space (R n, γ n ) admits a logarithmic sobolev inequality with optimal constant c = 1. To see this note that Φ(x) = x so we get the above conditions with c = 1. Example 17. Maniolds with lower ricci bounds and the normalized volume orm have no hessian term so, Ric cg. Thus it satises the BE conditions with constant c and thus an LSI with constant c. Note that or the classical homogenous spaces, the quotients o SO(n), we can show that Ric = ( ) N 4 g so that we have subgaussian concentration. 4 Application: Random Matrices LSI's lend themselves even to correlated settings Theorem 18. Consider dµ = e Ntr(V (X)) dvol Z where V N (x) is twice dierentiable and strictly convex with V c and dvol is the volume measure on the real symmetric N N matrices. Then the ensemble has LSI. In particular or all lipschitz : R N R, P( (λ 1,... λ N ) E r) e Nr c lip 5

6 Proo. Working in coordinates one can see that trv is also twice dierentiable and strictly convex with the same constant, giving us the LSI result. the map to the eigenvalues is lipschitz by the Homan-Weilandt lemma giving us the other result by the contraction principle. Appendix Proposition 19. A Markov Semigroup is symmetric (i.e. (P t, g) = (, P t g) ) i and only i the innitesimal operator is sel adjoint Proo. going rom symmetry o the operator to seladjointness is just the denition o the operator. To go the other way, recall that the key element o the proo o the Hille-Yosida theorem is the construction o the Yosida regularizers L λ = Lλ(λ L) 1 B(H) which will converge to L as λ point wise on its domain, rom there one goes about constructing the semigroup by using the exponential series or L λ. use this to show symmetry on the domain o D(L) and extend by density. 6