Does there exist a strongly convex function that is strongly convex with respect to norm $\|\cdot\|_p$ for any $p > 2$?

Question

A function $f$ is said to be strongly convex with respect to a norm $\|\cdot\|_p$ if for all $x,y$, $$f(x) \geq f(y) + \nabla f(y)^T(x-y) + \frac{1}{2}\|x-y\|^2_p.$$

There are a bunch of functions used in machine learning, statistics, etc. that are extremely well known to be strongly convex with respect to the $2$ or $1$ norm

Examples:

$\sum_{j = 1}^m x_j^2$ is 2-strongly convex with respect to $\|\cdot\|_2$

$\sum_{j = 1}^m x_j \log(x_j) $ is 1-strongly convex with respect to $\|\cdot\|_1$

Does there exist any strongly convex function with respect to a norm $\|\cdot\|_p, p>2$?

Answer 1

It depends on what you mean by that. In finite dimensional spaces, as other people noted, all norms are equivalent. However, the constants relating them can scale with the dimension.

If you are interested in functions that would have a strong convexity constant that is independent of the dimension (while having dimension independent divergence on sets of dimension-independent diameters), then no such function exists for $p > 2.$ For proof, see Example 4.1 in https://arxiv.org/pdf/1301.0465.pdf.