Non-trivial symmetric positive semidefinite matrix $M$ such that $(x_1 -x_2)^T(g_1-g_2) \ge \|x_1-x_2\|_M^2$ where $g_1,g_2$ are subgradients

Question

I read from a paper the following statement:

If $f$ is a closed convex function then there exists a symmetric positive semidefinite matrix $M$ such that $$(x_1 -x_2)^T(g_1-g_2) \ge (x_1-x_2)^TM(x_1-x_2)\quad \forall x_1,x_2\in\mathrm{dom}(f),g_1\in\partial f(x_1),g_2\in\partial f(x_2)$$ where $\partial f(x)$ denotes the subdifferential of $f$ at $x\in\mathrm{dom}(f)$.

It is straightforward to see that such a trivial matrix $M$ is $0$. My question: does there exist a non-trivial $M$ and how to determine it?

Thank you in advance for your discussions.

Answer 1

It depends on your function $f$. If $f(x)=\|x\|_2^2$ then $M=I$ is a valid choice. For $f(x)=\|x\|_1$, the only possible choice is $M=0$.