Prove that if g,h convex functions and g is positive ,then (goh) is a convex function ,too. So, I am stuck and I can't think of an answer to the question above. Any help?
Note : g is convex and positive and h is just convex.** We want to prove that (goh) is convex**
Note2 : We do not know if g and h have derivatives.
 A: Your statement is false. Indeed, let define the following functions:
$$g:x\in\mathbb{R}\mapsto|x|\in\mathbb{R},$$
$$h:x\in\mathbb{R}\mapsto x^2-1\in\mathbb{R}.$$
$g$ and $h$ are convex and $g$ is positive. Nevertheless, $g\circ h$ isn't convex.
Let assume $g$ and $h$ are $\mathcal{C}^2$, then, one has: $$(g\circ h)''=h''g'\circ h'+(h')^2g''\circ h.$$
Since, $g$ and $h$ are convex, $g''\geqslant 0$ and $h''\geqslant 0$, but one has no information on the sign of $g'\circ h'$, even if $g$ is positive. However, if $g$ is an increasing function, $g'\geqslant 0$ and $(g\circ h)''\geqslant 0$. Therefore, $g\circ h$ is convex.
Henceforth, let's take away any hypothesis on the smoothness of $g$ and $h$ and assume $g$ is an increasing function, then $g\circ h$ is still convex. Indeed, one has: $$\forall x,\forall y,\forall t\in[0,1],h(tx+(1-t)y)\leqslant th(x)+(1-t)h(y).$$
Since, $g$ is an increasing function, it follows: $$\forall x,\forall y,\forall t\in[0,1],g(h(tx+(1-t)y))\leqslant g(th(x)+(1-t)h(y)).$$
Besides, since $g$ is convex, one has: $$\forall x,\forall y,\forall t\in[0,1],g(th(x)+(1-t)h(y))\leqslant tg(h(x))+(1-t)g(h(y)).$$
From there, one has: $$\forall x,\forall y,\forall t\in[0,1],g(h(tx+(1-t)y))\leqslant tg(h(x))+(1-t)g(h(y)).$$
$g\circ h$ is convex.
