Consider a twice differentiable 1-strongly convex function $f:\mathbb{R}^n \to \mathbb{R}$.

Is it true that there exists $\alpha>0$ **independent of $n$** such that, for all $x \in \mathbb{R}^n$:
\begin{equation}
\label{prop}
\tag{P}\qquad \alpha \lVert x-x^*\rVert_{\infty} \leq \lVert\nabla f(x)\rVert_{\infty},
\end{equation}
where $x^*$ is the unique global minimizer of $f$.

If the answer is no, what would be a sufficient condition to verify property \eqref{prop}?

The reason why I am asking the question is that, using the equivalence of norms, it is simple to show that (P) holds with $\alpha = \frac{1}{\sqrt{n}}$. But I think it is possible to do better, but cannot prove it. For example, it holds with $\alpha = 1$ for $n=1$, and for $f=x\mapsto \frac{1}{2}\lVert x\rVert_{2}^2$ for every $n$.

Edit. * Reminder*:

- As $f$ is twice differentiable, it is 1-strongly convex iff $\nabla^2 f \succcurlyeq I_{n \times n}$.
- $\lVert x\rVert_{\infty} \triangleq \max_{1 \leq i \leq n}\lvert x_i\rvert$.