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
$$
2 \int \phi \mathrm{d} \mu \le |\nabla \phi|^2 \mathrm{d} \mu,
$$
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
$$
 \frac 1 \lambda \log \int e^{\lambda \phi(x)}  \mathrm{d} \gamma(x)
 = \frac 1 \lambda \sup_{h} \mathbb{E}\left[ \lambda \phi \left( B_1 + \int_0^1 h_s \mathrm{d}s \right) - \frac 1 2 \int_0^1 \dot h_s^2 \mathrm{d} s \right],
$$
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.

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