How to show that the rotation map $f$ is not a gradient of a convex function?
Suppose (for a contradiction) that there exists a function $F:\mathbb R^2 \to \mathbb R$ such that $\nabla F(x) = f(x) = Rx$ for all $x \in \mathbb R^2$. Then the Hessian of $F$ is $R$. However, the Hessian must be symmetric, whereas $R$ is not symmetric. This is a contradiction.
Thus, there is no function $F:\mathbb R^2 \to \mathbb R$ such that the gradient of $F$ is $f$.