Is every monotone map the gradient of a convex function?

Question

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.

Answer 1

Not all fields $F$ are gradients. If $F=(F_1,\dots,F_n)$ is $C^1$, a necessary condition for $F$ to be a gradient is that $$ \frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i},\quad 1\le i<j\le n. $$

Answer 2

Consider a $C^1$ monotone vector field $f=(f_1,\ldots,f_n)$ on $\mathbb R^n$. If there exists a function $G$ on $\mathbb R^n$ such that $G'=f$ than $G$ is automatically convex. So the existance of $G$ is the only thing one should verify.

Prop. The function $G$ exists if and only if $\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i}$ for all $i$ and $j$.

The space $\mathbb R^n$ in contraclible. So $H^1(\mathbb R^n)=0$. Consequently $f=dG$ when $df=0$. It is equivalent to condition $\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i}$.