# Continuity of the Legendre transform of a Lipschitz function

Let $$p > 1$$, $$f$$ defined on $$L^p(\mathbb{R}^n,\mathbb{R})$$, locally Lipschitz and concave, with $$f(x) = 0$$ when $$x \geq 0$$ a.e. We define, with $$q$$ dual to $$p$$, for any $$y\in L^q$$ , $$g(y) := \sup_{x \in L^p} (f(x) - \int x\cdot y)$$, having the property that it is finite on $$L^{q,\ge 0}$$ (which is the set of a.e positive functions), but it is $$+\infty$$ on $$L^{q,\le 0}$$. We know that $$g$$ is convex and lower semi-continuous on $$L^q$$.

Is $$g$$ continuous on $$L^{q,\ge 0}$$ ? Which simple assumption one could add to make it continuous ?

• A convex function is locally Lipschitz on the interior of its domain. So if g is finite everywhere, it is continuous. – Dirk Dec 9 '19 at 20:43
• But we don't know the interior of its domain, and $g$ is infinite for negative functions – Alfred Dec 9 '19 at 21:47