Let a convex $ U \subset_{op} \mathbb{R^n} , n \geq 2$, with the usual inner product. A function $F: U \rightarrow \mathbb{R^n} $ is monotone if $ \langle F(x) - F(y), x-y \rangle \geq 0, \forall x,y \in \mathbb{R^n}.$
Let $f:U \rightarrow \mathbb{R}$ differentiable. Show that $f$ is convex $\iff \nabla f:U \rightarrow \mathbb{R^n}$ is monotone.
My attempt on the right implication: I already proved that if $f$ is convex and 2-differentiable then $f''(x) \geq 0$. But this exercise only says f is 1-differentiable.Then I tried the following:$f$ is convex $\iff \forall x,y \in U $ the function $\varphi:[0,1] \rightarrow \mathbb{R}$, defined by $ \varphi(t) = f((1-t)x+ty)$ is convex. Then $\varphi'$ is non-decreasing, then $\nabla \varphi(x) \geq 0$... but I'm stucked here.
My attempt on the left implication:
$ |\nabla \varphi (x) - \nabla \varphi (y)|| x-y| \geq | \langle \nabla \varphi (x) - \nabla \varphi (y), x-y \rangle | \geq 0$
And so $ |\nabla \varphi (x) - \nabla \varphi (y)| \geq 0 $ then $\nabla \varphi $ is non-increasing and then (By an already proved Theorem) it is convex.
Can someone please verify what I did and give me a hint?
Thanks.