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.