# When do convexity and lower semicontinuity imply continuity?

Let $$X$$ be a nonempty compact convex subset of a locally convex space. Say $$X$$ has the convex function property if every convex, lower semicontinuous $$f:X\to\mathbb R$$ is also continuous.

Question: Which $$X$$ have the convex function property?

• We know some $$X$$ do. For instance, this is clearly true if $$X$$ is one-dimensional because every convex function on $$[0,1]$$ is continuous on $$(0,1)$$ and so (being convex) is upper semicontinuous. More generally, using continuity on the relative interior and inducting on dimension, any $$X$$ with finitely many extreme points has the convex function property.
• We know some $$X$$ don't. Indeed, an answer to this question https://math.stackexchange.com/questions/772841/convex-closed-and-unclosed-functions-and-lower-semicontinuity can be restricted to $$X=\{(a,b)\in[0,1]^2:\ b\geq a^2\}$$ for an example.

Is there a general characterization?

• There is a misunderstanding. The function in the example is not defined on a compact set. The domain of definition is a compact set with one point on the boundary removed. A convex function is continuous at some point, if it is finite in a neighborhood. So a convex function on a compact set is continuous everywhere.
– Dirk
Commented Sep 12, 2019 at 17:22
• I'm confused. Let $X:=\{(a,b)\in [0,1]^2:\ b\geq a^2\}$, a compact convex set. Define the function $f:X\to\mathbb R$ by letting $$f(a,b):= \begin{cases} \tfrac{a^2}b &:\ (a,b)\neq (0,0) \\ 0 &:\ (a,b)= (0,0). \end{cases}$$ This function is defined everywhere on $X$. It's clearly continuous on the convex set $X\setminus\{(0,0)\}$, and calculus shows it's convex there too. For any straight line in $X$ including the origin, $f$ is continuous along that straight line and therefore (being convex in the relative interior of said line), convex on the line. Commented Sep 13, 2019 at 10:57
• Ah, I see. So you deal with the case where you really want a compact set as domain of definition. I am so used to have convex function defined on the whole space (extended by $+\infty$ everywhere where it is not defined) that I got the question wrong.
– Dirk
Commented Sep 13, 2019 at 12:07
• The usual proof rests on the core-int lemma (for a closed convex set, algebraic interior equals topological interior), which in turn is usually derived from the Baire category theorem. So it is sufficient that $X$ is complete, but the core-int lemma might hold under weaker assumptions, too. Commented Feb 17, 2020 at 22:43
• Also, just for the record (you probably know this), the usual statement is that a convex lower semicontinuous function (on a Banach space) is continuous on the interior of its effective domain -- so if $X$ is not open, you also don't have the "convex function property". (This is of course the counterexample you gave in your comment.) Commented Feb 17, 2020 at 22:50

Let me resolve a special case of the question, by leveraging the examples mentioned above. Suppose $$(K,\rho)$$ is a compact metric space and $$X$$ is the set of Borel probability measures on $$K$$ endowed with its weak* topology. I claim that $$X$$ has the convex function property if and only if $$K$$ is finite.
Conversely, suppose $$K$$ is infinite. Letting $$k_\infty$$ be an accumulation point of $$K$$ (which exists because $$K$$ is an infinite compact metrizable space), define the affine continuous function $$\varphi:X\to\mathbb R^2$$ given by $$\varphi(x):=\int_K \left(\rho(k,k_\infty), \rho(k,k_\infty)^2\right)\text{ d}x(k)$$. Then, the convex function $$f:X\to\mathbb R$$ which takes $$f(\delta_{k_\infty}):=0$$ and $$f(x):=\tfrac{\varphi_1(x)^2}{\varphi_2(x)}$$ for every $$x\neq\delta_{k_\infty}$$ should work. A failure of continuity is witnessed along sequence $$(\delta_{k_n})_{n=1}^\infty$$ where, $$\{{k_n}\}_{n=1}^\infty\subseteq K\setminus\{k_\infty\}$$ is a sequence converging to $$k_\infty$$.
A natural conjecture is now that a general $$X$$ will have the convex function property if and only if $$X$$ has finitely many extreme points. Is this true?