# Is every monotone map the gradient of a convex function?

Recently in a seminar someone mentioned that monotone maps are equivalent to gradients of scalar convex functions, but it's not clear to me why this is true. One direction of the equivalence is straightforward but the other is not (as far as I can tell).

Definition. A map $F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is monotone on a convex set $C$ if $$(y-x)^T(F(y)-F(x))\ge0$$ for all $x,y \in C$.

One direction of the equivalence:

Prop. Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be convex and sufficiently differentiable. Then $\nabla f$ is monotone.

Pf. Convex differentiable functions satisfy $$f(y) \ge f(x) + \nabla f(x)(y-x).$$

By choosing the points in reverse, we also have, $$f(x) \ge f(y) + \nabla f(y)(x-y).$$ Add these inequalities and rearrange to get $(\nabla f(y)-\nabla f(x))(y-x) \ge 0$.∎

Now the other direction:

Prop. Let $F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be monotone and sufficiently differentiable. Then there exists a convex function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ such that $F=\nabla f$.

Pf. ???

It seems like this should be easy, but I'm stuck and google/wikipedia have been of little help. I'm actually starting to doubt whether it is true.

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