Prove $\{f\in L^\infty : \|f\|_\infty \leq 1- \epsilon\}$ for $\epsilon\in (0, 1)$ is $w^*$- closed

Question

I want to prove the following statement

Prove $\{f\in L^\infty : \|f\|_\infty \leq 1- \epsilon\}$ for $\epsilon\in (0, 1)$ is $w^*$- closed

I've started:

suppose $f_n \xrightarrow{w^*}f$ so tht we have $f_n(g) \to f(g)$ for all $f\in L_1$ Icould'nt able to continuo from here

Any help is much appreciated! Thank you!

Answer 1

Let $r=1-\varepsilon$. Let $(f_j)_{j\in J}$ be a net (*) in $L^{\infty}(\Omega)$ with $\|f_j\|_{\infty}\leq r$ and $f_j\rightharpoonup^* \bar{f}$. We then have $$\int_{\Omega} f_j g\to \int_{\Omega} \bar{f}g\qquad \forall g\in L^1(\Omega) \qquad (1)$$ We need to prove that $\|\bar{f}\|_{\infty}\leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|\bar{f}|\geq r+\varepsilon$ on $E$ for some $\varepsilon>0$. We may also assume that $E$ has finite measure (as long as $\Omega$ is $\sigma$-finite, which is a rather standard assumption).

Now apply $(1)$ with $g=\chi_E$, which is in $L^1(\Omega)$ because $|E|<\infty$. Since $|f_j|\leq r$ a.e., we have $$r|E|\geq\int_E f_j\to \int_E \bar{f}\geq (r+\varepsilon)|E| $$ which is a contradiction.

Anyway, it is worth to notice that in general

Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin $$ B_r:=\left\{f\in X^*:\|f\|\leq r\right\}$$ is weakly-star closed.

($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{\infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.