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?

  • $\begingroup$ 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. $\endgroup$
    – Dirk
    Commented Sep 12, 2019 at 17:22
  • $\begingroup$ 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. $\endgroup$ Commented Sep 13, 2019 at 10:57
  • $\begingroup$ 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. $\endgroup$
    – Dirk
    Commented Sep 13, 2019 at 12:07
  • $\begingroup$ 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. $\endgroup$ Commented Feb 17, 2020 at 22:43
  • $\begingroup$ 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.) $\endgroup$ Commented Feb 17, 2020 at 22:50

1 Answer 1


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.

Theorem 10.2 in "Convex Analysis" by Rockafellar implies that any convex function defined on a finite-dimensional simplex is upper semicontinuous. This gives one direction.

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?


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