# This simple function is convex. What is the simple proof?

Let $$\phi : \mathbb{R} \to \mathbb{R}$$ be a convex function, and say that it grows at most linearly at infinity for simplicity. Denote by $$\gamma$$ the standard Gaussian measure on $$\mathbb{R}$$. The following function is convex: $$\lambda \mapsto \frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) \qquad (\lambda > 0).$$ There is a proof that uses a stochastic control representation of the function. Is there an alternative proof? I wrote in the title that I was interested in a "simple" proof, but in fact I am interested in any alternative proof. For instance, I tried to represent the function as the value of a PDE and see if I could prove it in this way, and did not succeed. The statement is false if you replace $$\gamma$$ by an arbitrary probability measure. It is equivalent to the statement that the third derivative of the mapping $$\lambda \mapsto \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x)$$ is nonnegative. In particular, it implies that the third cumulant of $$\phi(z)$$, where $$z$$ is a Gaussian, is nonnegative.

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. Denote by $$(B_t)$$ a standard Brownian motion, and write $$\frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) = \frac 1 \lambda \sup_{h} \mathbb{E}\left[ \lambda \phi \left( B_1 + \int_0^1 h_s \mathrm{d}s \right) - \frac 1 2 \int_0^1 \dot h_s^2 \mathrm{d} s \right],$$ where the supremum is over suitable progressively measurable $$(h_s)$$. Replacing $$h$$ by $$\lambda h$$, we find that $$\frac 1 \lambda \log \int e^{\lambda \phi(x)} \mathrm{d} \gamma(x) = \sup_{h} \mathbb{E}\left[\phi \left( B_1 + \lambda \int_0^1 h_s \mathrm{d}s \right) - \frac \lambda 2 \int_0^1 \dot h_s^2 \mathrm{d} s \right].$$ This is a supremum of convex functions, so we are done.

• Do you have a reference to the stochastic-control proof ? – jjcale Nov 1 at 22:16
• 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. – Elwood Nov 3 at 19:22
• 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 $V(0,0) = \ln \int g d\gamma$. – Paata Ivanishvili Nov 3 at 23:03

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.

• 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 $\lambda \to \infty$. – Elwood Oct 21 at 12:25
• Unless I am missing something, the final function is nonincreasing in $t$ by basic properties of the logarithm, exponential and Jensen's inequality. – Steve Oct 21 at 14:08
• You're right Steve. In any case, there is no simple way to go from gerw's comment to the statement I want. – Elwood Oct 21 at 18:02
• 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. – Paata Ivanishvili Oct 21 at 21:46

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.

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