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

Question

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?

Answer 1

Yes, the domain of its Fenchel conjugate is bounded. In particular \begin{equation} \text{dom} \, g^* \subseteq B(0,L_{g}), \end{equation} where $B(0,L_{g})$ denotes the ball with radius $L_{g}$ around the origin.

First note that for every Lipschitz continuous function \begin{equation} \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} \end{equation} 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{equation} \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} \end{equation} proofing that $g^*(y) = +\infty$ by letting $\lambda$ go to $+\infty$.