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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?

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  • $\begingroup$ Is $f$ differentiable or something? $\endgroup$ – Jimmy R. Oct 2 '17 at 7:01
  • $\begingroup$ Do you know how to verify convexity of a function? $\endgroup$ – StubbornAtom Oct 2 '17 at 7:01
  • $\begingroup$ @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 $f$. $\endgroup$ – Arthur Oct 2 '17 at 7:02
  • $\begingroup$ @Arthur I've thought a lot but none of my thought seem to be usefull. I totally have no idea. $\endgroup$ – yahoo Oct 2 '17 at 7:03
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    $\begingroup$ @JimmyR No requirement for $f$ . $\endgroup$ – yahoo Oct 2 '17 at 7:11
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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.

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