Let $\langle X, X' \rangle$ be a dual pair equipped with the weak and weak\* topologies.


Let $C$ be a weak\* compact subset of $X'$ with nonempty interior. For each $x \in X$, let $M(x)$ be the set of minimisers of $\langle x, \cdot \rangle$ on $C$. Each $M(x)$ is non-empty, convex, and closed.


My question is: What additional conditions on $C$ imply that $M$ admits a continuous selection, i.e. a continuous function $f: X \to X'$ such that $f(x) \in M(x)$?



How to choose minimisers in a continuous way
I think the answer is: only in the trivial case when each $M(x)$ is a singleton $\{f(x)\}$. Indeed, if $M(x)$ has more than one point, it has at least $2$ extremal points. An extremal point of a convex compact set $C$ of a LCTVS (here, $X’$ with its weak\* topology) is the unique minimiser on $C$ for some continuous linear form (thus, here, an evaluation). So for $i\in\{1,2\}$ there are $x\_i$ in $X$ and $y\_i$ in $M(x)$ such that $M(x\_i)=\{y\_i\}$. Moreover, $M(x\_i)=M(tx\_i+(1-t)x)$ for any proper convex combination of $x\_i$ and $x$. This forbids the existence of a continuous selection for $M$, for any selection has to jump at $x$ from $y\_1$ to $y\_2$.


In fact, for any $x\in X$ and for any selection $f$ of $M$ the whole set $M(x)$ can be recovered as closed convex hull of the discontinuity set of $f$ at $x$:
$$M(x)=\bigcap\_{U\text{ nbd of } x}\overline{\text{co}}f(U)= \overline{\text{co}}\Big( \bigcap \_{U\text{ nbd of } x} f(U)\Big).$$
In particular, there exists a continuous selection of the multi $M$ if and only if there is only one selection, that is $M(x)$ is a singleton for all $x$: in other words, every point of $\partial C$ is extremal, that is $C$ is strictly convex ($\partial C$ contains no non-empty open segment).


