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.
Best Answer
1) If $f$ is convex, then $$ f(y)\geq f(x) + \nabla f(x)\cdot (y-x) $$
and $$ f(x)\geq f(y) + \nabla f(y)\cdot (x-y) $$
so that by adding $$ (y-x)\cdot( \nabla f(x) - \nabla f(y)) \leq 0 $$
2) Assume that $\nabla f$ is monotone : Define $A =\{ x| f(x)\leq a\}$. If $A$ is not convex, then there are $x,\ y\in \partial A$ s.t. $$ \nabla f(x)\cdot (y-x),\ \nabla f(y)\cdot (x-y) >0 $$ Hence $$ (\nabla f(x) -\nabla f(y))\cdot (y-x) >0 $$
It is a contradiction.