Let $X'$ be the topological dual of a normed TVS $X$, and let $f$ be a real, strictly convex function on $X'$
If $f$ attains a minimum on the closed unit ball of $X'$, does it attain a minimum on every weak*-compact subset of $X'$?
Attempt. My guess is that it's not true.
Let $X = X' = \mathbb R$, so that the closed unit ball in $X'$ is $[1,-1]$. The idea now is to define $f$ so that it's continuous on $[-1,1]$, and therefore achieves a minimum, but introduce a discontinuity elsewhere so that it doesn't achieve a minimum on some other closed interval.
For example, I was thinking something like $f(x) = (x-2)^2$ if $x \in (-\infty,2)$ and $f(x)$ is a linear on $[2,\infty)$ with, say, $f(2)=1$. Then $f$ doesn't attain a minimum on $[1,2]$.
I'm not sure I can define $f$ in a way that makes it strictly convex, however. (Turns out I can't. See diracdeltafunk's comment below.)