# Convex and Lipschitz implies "bounded" subgradient in Banach spaces

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?

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 $$\langle x^*, h\rangle = \langle x^*, (x+h) - x\rangle \le f(x+h) - f(x) \le L \| (x+h) - h)\|_X = L \|h\|_X.$$ Thus, $$\|x^*\|_{X^*} = \sup_{\|h\|_X\le1}\langle x^*,h\rangle \le L.$$