Convex and Lipschitz implies "bounded" subgradient in Banach spaces

Question

In this question: Lipschitz implies bounded gradient it is shown that if $f: \mathbb{R}^n \to \mathbb{R}^n$ is convex and L-Lipschitz the gradient is bounded by L. I wonder if this holds in Banach spaces? If $f$ is a L-Lipschtiz, convex functional from a Banach space $X$ to $\mathbb{R}$ and $\lambda \in \partial f(x) = \{ x^*: f(x) - f(u) \leq \langle x*,x-u\rangle, \forall u \in X\} $, is it true that $\lVert \lambda \rVert_* \leq L$?

If I consider the inverse duality mapping $J^{-1}(\lambda) = \{ \in X: \langle \lambda,u\rangle = \lVert u \rVert^2= \lVert \lambda \rVert_*^2\}$ and choose $y$ such that $x-y \in J^{-1}(\lambda)$ then I think the proof can be modified:

$$ L \lVert \lambda_t \rVert_* = \lVert x-y \rVert L \geq | \ell_t(x) - \ell_t(y)| \geq | \left(\lambda_t, x-y \right) |= \lVert \lambda_t \rVert_*^2, $$ and then divide both sides by $\lVert \lambda_t \rVert_* $. Is this correct or am I missing something?

Answer 1

In the case that $X$ is not reflexive, $J^{-1}(\lambda)$ might be empty.

However, we can argue directly. For arbitrary $h \in X$ we have $$\langle x^*, h\rangle = \langle x^*, (x+h) - x\rangle \le f(x+h) - f(x) \le L \| (x+h) - x\|_X = L \|h\|_X. $$ Thus, $$ \|x^*\|_{X^*} = \sup_{\|h\|_X\le1}\langle x^*,h\rangle \le L.$$