Are $L^\infty$ bounded functions compact in $L^2$?

Question

Is the set $\{ m \in L^2(0,1) : |m|_{L^\infty}\leq A \}$, (i.e. the set of $L^2$ functions with bounded $L^\infty$ norm) a compact subset of $L^2$?

(Compact in the topology induced by the $L^2$-norm)

Maybe useful: By Question we know already that this set is closed...

Answer 1

No, because it contains the sequence $\{ A \sin(n \pi x) \}_{n=1}^\infty$. The Riemann-Lebesgue lemma implies this converges to $0$ weakly in $L^2$, so the strong limit can only be $0$, yet the norms are constant.

Answer 2

Let $S$ be the set in question and $x_n(t) = A e^{i2 \pi n t}$. If $n\neq m$ we have $\|x_n-x_m\| = \sqrt{2}A$. Hence $U= \{e_n\}_m$ is a closed subset of $S$.

Then the collection of sets $B(x_n, {A \over \sqrt{2}})$ form an open cover of $U$ which has no finite subcover. Hence $U$ is not compact and so it follows that $S$ is not compact.

Aside: Here is a straightforward way to show that $S$ is closed: Suppose $ x \notin S$. Then $\|x\|_\infty > A$. Choose $\alpha,\beta$ such that $\|x\|_\infty > \alpha > \beta > A$. Let $C = \{ t | |x(t)| > \alpha \}$ and note that $\mu C >0$.

Choose $\epsilon>0$ such that $\mu C - ({\epsilon \over \alpha-\beta})^2 >0$, and suppose $\|x-y\|_2 < \epsilon$.

Let $D = \{t | |y(t) | > \beta\}$. Note that $\epsilon^2> \|x-y\|_2^2 \ge \int_{C \setminus D} |x-y|^2 d \mu \ge \int_{C \setminus D} ||x|-|y||^2 d \mu \ge (\alpha-\beta)^2 \mu ({C \setminus D})$.

Then we have $\mu D \ge \mu (C \cap D) = \mu C - \mu ({C \setminus D}) > \mu C - ({\epsilon \over \alpha-\beta})^2 >0$, from which it follows that $\|y\|_\infty > A$, and so $S^c$ is open.