infimum of a convex function over an open domain

Question

Let $f: \cal D_0 \to [0, \infty]$ be a convex function on a compact set $\cal{D}_0$ and let $\cal D \subseteq \cal D_0$. I think the following holds: $$ \inf_{x \in \cal{D}}\ f(x) = \inf_{x \in \overline{\cal{D}}}\ f(x) \quad \textrm{if} \quad \inf_{x \in \cal{D}}\ f(x) < \infty.$$ Is this correct and can anybody give me a hint on how to show this?

Answer 1

We must have $\inf_{x \in \overline{\cal D}} f(x) \le \inf_{x \in {\cal D}} f(x)$.

I am assuming that the topological structure is such that if $\bar{x} \in \overline{\cal D}$ and $x \in {\cal D}$, then $tx+(1-t) \bar{x} \in {\cal D}$ for all $t \in [0,1)$.

Let $\bar{x} \in \overline{\cal D}$ be such that $f(\bar{x}) < \infty$. Since $\inf_{x \in {\cal D}} f(x) < \infty$, there is some point $x_0 \in {\cal D}$ such that $f(x_0) < \infty$. Consider $\phi:[0,1] \to [0,\infty)$ defined by $\phi(t) = f(t x_0 + (1-t) \bar{x})$ (note that the range of $\phi$ is $[0,\infty)$, not $[0, \infty]$ since $f$ is convex). Since $\phi$ is convex, it is upper semi-continuous. Since $\phi(t) \ge \inf_{x \in {\cal D}} f(x)$ for all $t \in [0,1)$, we have $\phi(1) \ge \inf_{x \in {\cal D}} f(x)$.

Hence we have $f(\bar{x}) \ge \inf_{x \in {\cal D}} f(x)$ for all $\bar{x} \in \overline{\cal D}$.