If $f$ is strictly convex and $f(x) = \infty$, is $x$ a boundary point? 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 $N$.
 A: Your observation is right in most cases.

Statement.Let $X$ be a locally convex Hausdorff space, $C\subset X$ compact, and $f:C\rightarrow\mathbb{R}\cup\{-\infty,\infty\}$, with
usual order of extended real line. If $f(x)=\infty$ for some $x\in C$,
then $x\in\partial C$.
proof. Let $x\notin \partial C$. Then since $C$ is closed, $x\in C^\circ$ (by $X$ is Hausdorff), therefore there exists a open
neighborhood of $x$, $U_x\subset C^\circ$. Then because the topology is transition invariant, $$V_x:=U_x-x$$ is open, and $0\in V_x$, i.e. $V_x$ is open neighborhood of $0$, and thus it is an absorbing set (Theorem 4.3.6, Narici 2011). Therefore
there exists $y\in V_x$ such that $-y\in V_x$. Then
$x+y,x-y\in U_x$, and with the strict convexity we have a contradiction : $$\infty=f(x)=f(0.5(x+y)+0.5(x-y))< 0.5f(x+y)+0.5f(x-y)$$

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.
