# Does Lipschitz continuity of a convex imply boundedness of the domain of its Fenchel conjugate

Let $$g:\mathcal{H} \to \mathbb{R}$$ be a convex and $$L_{g}$$-Lipschitz continuous function on a Hilbert space $$\mathcal{H}$$. Is the domain of its Fenchel conjugate $$g^*$$, where $$g^*(y) := \sup_{x \in \mathcal{H}} \{ \langle y, x \rangle - g(x),$$ bounded?

Yes, the domain of its Fenchel conjugate is bounded. In particular $$$$\text{dom} \, g^* \subseteq B(0,L_{g}),$$$$ where $$B(0,L_{g})$$ denotes the ball with radius $$L_{g}$$ around the origin.
First note that for every Lipschitz continuous function $$$$\lVert g(x) \rVert \le \lVert g(x) - g(0) \rVert + \lVert g(0) \rVert \le C + L_g \lVert x \rVert, \quad \forall x \in \mathcal{H}$$$$ for some constant $$C > 0$$. Thus, for any $$y \in \mathcal{H}$$ with $$\lVert y \rVert > L_g$$ and any $$\lambda \in \mathbb{R}_+$$, \begin{aligned} g^*(y) = \sup_x \{ \langle y, x \rangle - g(y) \} \ge& \langle y, \lambda y \rangle - g(\lambda y) \\ \ge& \lambda \lVert y \rVert^2 - \lambda L_g \lVert y \rVert - C \\ =& \lambda \lVert y \rVert \left( \lVert y \rVert - L_g \right) - C, \end{aligned} proofing that $$g^*(y) = +\infty$$ by letting $$\lambda$$ go to $$+\infty$$.