# Prove ${1\over x}\int_0^x f(t)\rm{d}t$ is convex.

Let $f$ be a convex function in $[0,+\infty)$, prove $F(x)={1\over x}\int_0^x f(t)\mathrm{d}t$ is convex in $(0,+\infty)$.

I just use the definition ${f(x_2)-f(x_1)\over x_2-x_1}\leq {f(x_3)-f(x_1)\over x_3-x_1}$ for $x_1<x_2<x_3$. But when I change $f$ to $F$ in this inequality, things become too complex to handle it. Or is there any other equivalent definition easy for this question?

• Is $f$ differentiable or something? – Jimmy R. Oct 2 '17 at 7:01
• Do you know how to verify convexity of a function? – StubbornAtom Oct 2 '17 at 7:01
• @JimmyR. Since a convex function is piecewise motonous, it must be almost everywhere differentiable. But there is no reason to assume more than that, I think. For instance, $|x-1|$ should be an entirely valid $f$. – Arthur Oct 2 '17 at 7:02
• @Arthur I've thought a lot but none of my thought seem to be usefull. I totally have no idea. – yahoo Oct 2 '17 at 7:03
• @JimmyR No requirement for $f$ . – yahoo Oct 2 '17 at 7:11

Obviously, substituting $t=ux$, we get $$F(x)=\frac1x\int_0^x f(t)\,dt=\int^1_0f(ux)\,du,$$ and since $f(ux)$ is a convex function of $x$ for every $u\in[0,1]$, $F(x)$ is convex.