Strictly Convex Implies Invertible Gradient? If $f:\mathbb R^n \rightarrow \mathbb R$ 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 $x = y$?
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) &lt; f(y) - f(x)
\end{equation}
and similarly
\begin{equation}
\nabla f(y) \cdot (x-y) &lt; 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) &gt; 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 $x =y$.
No. Try $x \mapsto \mathrm{e}^x$.