Let $P\subseteq \mathbb{R}^n$ be a convex polytope (cut out by finitely many linear inequalities) and $O\subseteq \mathbb R^n$ be an open set such that $O\cap P$ contains the (relative) interior of $P$. Suppose $f: O\cup P \to \mathbb R$ is a function that is
- continuously differentiable on $O$,
- convex on $P$,
- for all $y \in P \setminus O$ and sequences $x_n \in P\cap O$ approaching $y$, we have $\lim_{n} \|\nabla f(x_n)\| = \infty$.
Is it guaranteed that if $x^* \in \mathrm{argmin}_{x \in P} f(x)$, then $x^* \in O \cap P$?
I only have the result on an interval. I am not sure if you can exploit it for higher dimensions. The condition $\|\nabla f(x_n)\|\to\infty$ might be not enough if $P$ is not full dimensional.
For $n=1$:
The convex polytope $P$ is just an interval (may include none, one, or both endpoints) and $O\cap P$ is just the interior of $P$. If $P$ itself is open, then there is nothing to show. Assume without loss of generality $\max P = y < \infty$.
Now, let $x_n\in O\cap P$ with $x_n\uparrow y$. Notice that $f$ is convex and thus $f'$ is (non-strictly) increasing. Hence, $|f'(x_n)|\to\infty$ implies that $f'(x_n)$ is positive for $n$ sufficiently large. By convexity of $f$, we have $$ f(y) \ge f(x_n) + \underbrace{f'(x_n)(y - x_n)}_{>0} > f(x_n). $$ That is, $y$ is not an minimizer of $f$.