Problem Setting. Let $\mu$ be a probability measure on some space $\Omega$. Consider the subset of $L^\infty(\mu)$ defined by $S=\{f\in L^\infty(\mu):\alpha\le f\le \beta\}$, where $\alpha\le\beta$ are positive constants. Identify $L^\infty(\mu)$ with the continuous dual of $L^1(\mu)$.
My question: is the functional $f\mapsto\int \frac{1}{f}d\mu$ weak-* continuous on $S$?
Additional Material. I'm trying to fill in the gaps of Talagrand's one-paragraph proof of the following inequality:
Assume $g$ is a measurable function on $\Omega$ satisfying $e^{-t}\le g(\omega)\le 1$ for some $t>0$. Then \begin{align}\int \frac1gd\mu\int gd\mu\le\frac12+\frac{e^{-t}+e^t}{4}. \end{align}
The proof goes as fixing $\int gd\mu$ and optimizing the convex functional $g\mapsto\int g^{-1}d\mu$, claiming that it attains its maximum at some extreme point of $S$. Of course, a standard result following from Krein-Milman says that continuous convex functional on a compact convex set attains its minimum at some extreme points. For our case, $S$ is weak-* compact by Banach-Alaoglu theorem, so it remains to show that the convex functional of interest is weak-* continuous, which resisted several of my attempts.
I appreciate any answer to my question or an alternative way to fill in the gaps in Talagrand's proof.
The functional is not weak-$*$ continuous.
Consider the example with $\Omega=(0,1)$, $\mu$ the Lebesgue measure, $\alpha=1,\beta=2$. We define the sequence $$ f_n = 1+\chi_{A_n} \quad \text{with} \quad A_n = \{\omega\in\Omega : sin(2\pi n \omega)>0\}. $$ Then it is possible to show that $f_n$ converges weakly-$*$ to $f_0:=3/2$. However, for the function values we have $$ \int \frac1{f_n} \mathrm d\mu = \tfrac12+\tfrac12\cdot\tfrac12=\tfrac34 $$ and $$ \int\frac1{f_0}\mathrm d\mu = \tfrac23\neq\tfrac34. $$