Convolution of convex function and Gaussian is convex

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a convex function, i. e., for all $x_1, x_2 \in \mathbb{R}$ and $t \in [0, 1]$, $$ \qquad f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2).$$

I want to prove that the convolution $$g(x)=\int_{-\infty}^\infty e^{-(x-s)^2}f(s)\;ds $$ is also a convex function.

Answer 1

Hint:

Note $g(x)=\int_{-\infty}^\infty e^{-s^2}f(x-s)\;ds$ by a change of variable and $f((1-\lambda)x+\lambda y-s)=f((1-\lambda)(x-s)+\lambda (y-s))$ for fixed $x,y,s$.

Can you proceed from here using $f$ is convex?