Let $f :\mathbb{R}^{n} \to \mathbb{R}$ and we are going to fix $u,v\in \mathbb{R}^{n}$ and we define a new function $\phi:\mathbb{R}^{*}_{+}\to\mathbb{R}$ by $\phi(x)=\frac{f(u+xv)-f(u)}{x}$. the question is to prove that if $F$ is convex so $\phi$ increasing.
I have a question this text is true or false because I choose the convex function can I get a contradiction.
when we take $F(x)=e^{x}$ and we fixe any $u,v$ we have function is not incrssing in $]0;1[$
If $0<x_1<x_2$ then $u+x_1v=(1-\frac {x_1}{x_2})u+\frac {x_1}{x_2}(u+x_2v)$. Hence,
$f(u+x_1v)\leq (1-\frac {x_1}{x_2})f(u)+\frac {x_1}{x_2}f(u+x_2v)$. Simple algebraic manipulation of this gives $\phi (x_1) \leq \phi (x_2)$.