Upper bound for the dual norm of a subgradient of a convex function?

Question

Assume $f: \mathbb{R}^n\rightarrow \mathbb{R}$ is a nondifferentiable convex function on $X$. Let represent a subgradient of the function at point $x \in X$ by $x^* \in \partial f(x)$, that is $$ f(y) \geq f(x) + \langle x^*, y-x \rangle $$ Is there any $\lambda$ such that we have the following? $$ \| x^* -y^* \|_* \leq \lambda \| x-y \| $$ where $\| \cdot \|_*$ is the dual norm of $\| \cdot \|$.

Answer 1

No, I don't think this is true. I'm going to show it first with a differentiable function. Take $$f : \mathbb{R} \to \mathbb{R} : x \mapsto e^{x^2}.$$ The function is convex, as $f''(x) = 2(2x^2 + 1)e^{x^2} > 0$. Moreover, we have $f'(x) = 2xe^{x^2}$: the unique element of $\partial f(x)$. Then $$\frac{\|f'(n + 1) - f'(n)\|}{\|(n + 1) - n\|} = 2(n+1)e^{(n+1)^2} - 2ne^{n^2} > e^{(n+1)^2} \to \infty$$ as $n \to \infty$. So, no such bound holds for this (differentiable) $f$.

If you insist on a non-differentiable function, take $g(x) = \max \lbrace f(x), 2 \rbrace$. The argument above still holds since $g(n) = f(n)$ for all integers $n$, and $g$ is definitely still convex as it is the supremum of two convex functions.