Let $f: [a,b] \to [0,\infty]$ be an extended-real-valued function defined on a closed (non-degenerate) interval in $\mathbb R$. Say that $f$ is strictly convex if for all distinct $x,y \in [a,b]$ and all $\lambda \in (0,1)$,
$$f(\lambda x + (1 – \lambda)y) < \lambda f(x) + (1 – \lambda)f(y).$$
Observation. If $f(x) = \infty$, then $x \in \{a,b\}$. For suppose $x \in (a,b)$, then $x = \lambda a + (1 – \lambda) b$ for some $\lambda \in (0,1)$, so we can't have $f(x) = \infty$ without violating strict convexity.
Question. What is the most general theorem of which the observation above is an instance? For example, is it true that if $X$ is a locally convex topological vector space, and $C$ is a compact and convex subset of $X$, then a strictly convex function $f: C \to [0,\infty]$ can take the value $\infty$ only on the boundary of $C$?
I currently don't see how to generalize beyond the finite dimensional case. In general, if $x \in int C$, then local convexity implies that $x$ is contained within a closed convex subset $N$ of $int C$. But if $N$ is not finite dimensional, then it need not be the case that $x$ is equal to a convex combination of finitely many elements of $N$.
Best Answer
Your observation is right in most cases.
However, this will not holds for any topological vector space $X$. For example, let $X$ be normed vector space but non-Hausdorff at $0$ (over $\mathbb{R}$) and $f(0)=\infty$. You can find some $y\in \text{cl}\{0\}$ such that $\{\lambda y:\lambda\in[0,1]\}$ is compact and convex, however $0\notin \partial \{\lambda y:\lambda\in[0,1]\}$.
Narici, Lawrence; Beckenstein, Edward, Topological vector spaces, Pure and Applied Mathematics (Boca Raton) 296. Boca Raton, FL: CRC Press (ISBN 978-1-58488-866-6/hbk). xvii, 610 p. (2011). ZBL1219.46001.