Does strictly convex imply invertible gradient?

Question

If $f:\mathbb R^n \to \mathbb R$ is strictly convex and continuously differentiable, does this imply that $\nabla f$ is a one-to-one mapping? To be precise, can we say that $x, y \in \mathbb R^n$ and $\nabla f(x) = \nabla f(y)$ implies $x = y$?

Answer 1

No. Try $x \mapsto \mathrm{e}^x$.

Answer 2

Suppose that there exist $x, y \in \mathbb R^n$ such that $\nabla f(x) = \nabla f(y)$ and $x \neq y$.

Then, by the strict convexity of $f$, we can write \begin{equation} \nabla f(x) \cdot (y-x) < f(y) - f(x) \end{equation} and similarly \begin{equation} \nabla f(y) \cdot (x-y) < f(x) - f(y). \end{equation} Multiplying both sides of the latter inequality by $-1$ and substituting $\nabla f(x)$ in place of $\nabla f(y)$, we obtain \begin{equation} \nabla f(x) \cdot (y - x) > f(y) - f(x), \end{equation} which contradicts the first inequality.

Thus, if $x, y \in \mathbb R^n$ satisfy $\nabla f(x) = \nabla f(y)$, then $x =y$.