19
$\begingroup$

Let $\phi : \mathbb{R}^d \to \mathbb{R}$ be a convex function, and assume that it grows at most linearly at infinity for simplicity. Denote by $\gamma$ the standard Gaussian measure on $\mathbb{R}^d$, and assume further that $\int e^\phi \mathrm{d} \gamma = 1$. Define the probability measure $$ d \mu := e^\phi \mathrm{d} \gamma. $$ The following inequality, which I will call (sls) for "strange log-Sobolev", is true $$ 2 \int \phi(x) \mathrm{d} \mu(x) \le \int (\phi(x) - \phi(y))^2 \mathrm{d}\mu(x) \mathrm{d} \mu(y). $$ Surprisingly, despite significant effort, the only proof I could find of (sls) uses arguments from stochastic control (which will be outlined below).

Q1: can one give a proof of (sls) that does not require, say, to know what a progressively measurable stochastic process is?

I will now explain the name and clarify why (sls) looks at least to some extent like the standard log-Sobolev inequality. Recall that the Gaussian log-Sobolev inequality states that, for every (say smooth) function $f : \mathbb{R}^d \to \mathbb{R}_+$ such that $\int f \mathrm{d}\gamma = 1$, we have $$ 2 \int f \log f \mathrm{d} \gamma \le \int \frac{|\nabla f|^2}{f} \mathrm{d} \gamma. $$ Applying this inequality with $f = e^\phi$, we obtain $$ 2 \int \phi \mathrm{d} \mu \le \int|\nabla \phi|^2 \mathrm{d} \mu, $$ which is surprisingly similar to (sls).

Q2: is this a fruitful analogy? Shouldn't inequality (sls), which does not depend explicitly on the dimension, be useful in some contexts?

The validity of (sls) is equivalent to the convexity of the following function: $$ \lambda \mapsto \frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) \qquad (\lambda > 0). $$

Since this is not the point, I will not explain it in details, but for those familiar with it, let me sketch briefly the stochastic-control proof that the mapping above is convex. Denote by $(B_t)$ a standard $d$-dimensional Brownian motion, and write $$ \frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) = \frac 1 \lambda \sup_{h} \mathbb{E}\left[ \lambda \phi \left( B_1 + \int_0^1 h_s \mathrm{d}s \right) - \frac 1 2 \int_0^1 h_s^2 \mathrm{d} s \right], $$ where the supremum is over suitable progressively measurable $(h_s)$. Replacing $h$ by $\lambda h$, we find that $$ \frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) = \sup_{h} \mathbb{E}\left[\phi \left( B_1 + \lambda \int_0^1 h_s \mathrm{d}s \right) - \frac \lambda 2 \int_0^1 h_s^2 \mathrm{d} s \right]. $$ This is a supremum of convex functions, so we are done.

(EDIT: I have completely rewritten the question on January 27 2020 to emphasize (sls), which initially did not appear at all.)

$\endgroup$
6
  • $\begingroup$ Do you have a reference to the stochastic-control proof ? $\endgroup$
    – jjcale
    Nov 1 '19 at 22:16
  • $\begingroup$ The representation can be taken from the paper of Boué and Dupuis "A variational representation for certain functionals of Brownian motion", Annals of Probability 1998. Intuitively, you can understand it as follows: you can always write log E[exp(f)] as the supremum over probability measures Q of Q[f] minus the relative entropy of Q with respect to E. For Brownian motion, we know an explicit description of the set of absolutely continuous probability measures, via the Girsanov theorem. $\endgroup$
    – Elwood
    Nov 3 '19 at 19:22
  • $\begingroup$ Fleming and Souganidis paper "On The Existence of Value Functions of Two-Player, Zero-Sum Stochastic Differential Games" (1988) contains Bellman--Isaac equations (Theorem 2.6). Let $g(x,t)$ be solving backward heat equation $g(x,1)=g(x)$, and $g_{t}+\frac{g_{xx}}{2}=0$, $t\in [0,1]$. Then $V(t,x)=\ln g(t,x)$ satisfies Hamilton-Jacobi-Bellman-Isaac PDE: $V_{t}+\sup_{b\in \mathbb{R}}\left\{\frac{V_{xx}}{2}+V_{x}\cdot b - \frac{b^{2}}{2} \right\} = 0$. So one could use representation (1.10) from Fleming--Souganidis paper to recover the value $V(0,0) = \ln \int g d\gamma$. $\endgroup$ Nov 3 '19 at 23:03
  • $\begingroup$ In the special case of $\phi(z) = z^2/2$, and restricting to $\lambda < 1$ for definiteness, the function we want to see is convex is proportional to $-\frac{\log(1-\lambda)}{\lambda}$. This function is indeed convex, but it does not strike me as something obvious! $\endgroup$
    – Elwood
    Dec 19 '19 at 15:58
  • $\begingroup$ 27 Jan 2021.......?? $\endgroup$ Jan 27 '20 at 18:12
1
$\begingroup$

Here is a simple proof that $$\frac1\lambda \mapsto \frac1\lambda \, \log \int \exp(\lambda \, \phi(x)) \, \mathrm d \gamma(x)$$ is convex. This does not need any assumptions on $\phi$ or $\gamma$. Maybe this can be used to prove convexity of your function (see below).

Let $M$ denote the set of measurable functions from $\mathbb R$ to $\mathbb R$. An application of Hölder's inequality shows that the LogIntExp-functional \begin{equation*} M \ni u \mapsto \log \int \exp(u(x)) \, \mathrm d \gamma(x) \end{equation*} is convex. Hence, \begin{equation*} \mathbb R \ni s \mapsto \log \int \exp(s \, \phi(x)) \, \mathrm d \gamma(x) \end{equation*} is convex. Consequently, its perspective \begin{equation*} \mathbb R^+ \times \mathbb R \ni (t,s) \mapsto t \, \log \int \exp(t^{-1} \, s \, \phi(x)) \, \mathrm d \gamma(x) \end{equation*} is convex. By fixing $s = 1$, \begin{equation*} \mathbb R^+ \ni t \mapsto t \, \log \int \exp(t^{-1} \, \phi(x)) \, \mathrm d \gamma(x) \end{equation*} is convex. If this function would be increasing in $t$ (here your assumptions may come into play), this would imply convexity of your function.

$\endgroup$
4
  • $\begingroup$ This function would be increasing if and only if the original function (whose convexity I am asking about) would be decreasing. This is not so. The original function goes to a constant as $\lambda \to 0$, but, assuming, say, that $\phi(x) \sim cx$ as $x \to \infty$, one finds that the function grows like $c^2 \lambda/2$ as $\lambda \to \infty$. $\endgroup$
    – Elwood
    Oct 21 '19 at 12:25
  • $\begingroup$ Unless I am missing something, the final function is nonincreasing in $t$ by basic properties of the logarithm, exponential and Jensen's inequality. $\endgroup$
    – Steve
    Oct 21 '19 at 14:08
  • $\begingroup$ You're right Steve. In any case, there is no simple way to go from gerw's comment to the statement I want. $\endgroup$
    – Elwood
    Oct 21 '19 at 18:02
  • 3
    $\begingroup$ The fact that $d\gamma$ is Gaussian is crucial here. For example, if you replace $d\gamma$ by $d\mu = \frac{1}{2}\delta_{0} + \frac{1}{2} \delta_{1}$, and take $\varphi(x)=x$ then the map $\lambda \mapsto \frac{1}{\lambda} \ln \int \exp (\lambda \varphi(x)) d\mu$ is strictly concave. $\endgroup$ Oct 21 '19 at 21:46
1
$\begingroup$

For $p>0$ we may define the $p$-norm of a complex function as $$\Vert f\Vert_p=\Bigl(\int |f(x)|^p\,d\mu(x)\Bigr)^{1/p}.$$ your function is then $\log\Vert g\Vert_{\lambda}$ for an adequate function and measure.

The convexity is known in this context everywhere the function is defined. In your case for all $\lambda>0$.

I have now not a good reference, but in many books you find it. For example, this is problem 1.1.16 in page 15 of the book

L. Grafakos, Classical Fourier Analysis, 2nd edition, Springer, 2008.

$\endgroup$
3
  • $\begingroup$ The assertion in your reference claims convexity in $1/\lambda$, not in $\lambda$. $\endgroup$
    – jjcale
    Nov 30 '19 at 16:48
  • $\begingroup$ I confirm what jjcale said. Recall that the statement is false if your replace the Gaussian measure by an arbitrary probability measure. $\endgroup$
    – Elwood
    Nov 30 '19 at 17:51
  • $\begingroup$ @jjcale Yes you have reason, I was wrong. $\endgroup$
    – juan
    Nov 30 '19 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.