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I have read an article about convex function and I have problems in this article. In this article said that

"It is well known that if $F(x, y)$ is increasing relative to $y$ and $y = h(x)$ is convex on $[a, b]$, then $F(x, h(x))$ is convex on $[a, b]$."

For my try, I define $F : [-1 , 0] \times [-1 , 0] \to \mathbb{R}$ by $$F(x,y) = -y^{2}.$$ We see that $F(x, y)$ is increasing relative to $y$. And if I let $y = h(x) =x$, we have also $h(x)$ is a convex function $[-1 , 0]$. So now, we have $F(x, h(x)) = -x^{2}$ is not a convex function on $[-1, 0]$.

Do I'm right ?

An another problem, if $F(x, h(x))$ is a convex function on $[a, b]$, then $F(x, h(x))$ is differentiable ?

I think it is not. For my example is, let $F(x, h(x)) = h(x)$ and $h(x) = |x|$ on $[-1, 1]$.

Thank you for your answers and comments.

share|improve this question
I hasard the guess that there was a condition that $F$ itself be convex that wasn't placed as prominently as it should have been. –  Daniel Fischer Oct 10 '13 at 10:37
I also think that it is not enough condition. –  nameless Oct 10 '13 at 10:49
That's for sure. The question is, which condition(s) should be added, and whether they are totally absent from the paper or only too well hidden, so that you overlooked them. –  Daniel Fischer Oct 10 '13 at 11:05
Are my counter example okay ? –  nameless Oct 10 '13 at 13:00
Yes, the counterexample is okay. You could use $h(x) = x^2$ to get a counterexample with a strictly convex $h$, in case anybody might think that could be the missing condition. –  Daniel Fischer Oct 10 '13 at 13:05

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