How to prove the convexity of this function?

Question

If $f(x)$ is a convex function defined in $(0,+\infty)$. I want to prove the convexity of $F(x)=\dfrac{1}{x}\int_0^xf(t)\mathrm d t$. Note that $F(x)=\int_0^1 f(xt) \mathrm d t$. The convexity of $F(x)$ follows obviously. My question is, if we further assume that $f(x)\in C^2$, I want to prove the convexity of $F$ by proving $F''\geq 0$, which seems harder to prove. How to prove it by this way? Any ideas will be greatly appreciated.

Answer 1

If $f$ has a continuous second derivative then one can differentiate under the integral sign and conclude that $$ F''(x)=\int_0^1 t^2 f''(xt) \, dt \, . $$ In particular, $f'' \ge 0$ implies that $F'' \ge 0$, and $f'' > 0$ implies that $F'' > 0$, i.e. (strict) convexity of $f$ implies (strict) convexity of $F$.