Are $L^\infty$ bounded functions closed 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 closed subset of $L^2$?

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

Answer 1

Yes.

Call your set $M$. If $\{ f_n \}_{n=1}^\infty \subset M$ and $f_n \to f$ in the sense of $L^2$, then there is a subsequence $f_{n_k}$ which converges almost everywhere to a function $f$. (This is a nontrivial fact, and follows by the "rapidly Cauchy subsequence" argument. The Lebesgue case comes up on page 146 of Royden and Fitzpatrick; the general case comes up on page 397. I presume one could find it in any proof that $L^p$ is complete.) Take $N=\{ x : \exists k \: |f_{n_k}(x)| > A \text{ or } f_{n_k}(x) \not \to f(x) \}$. Then $N$ has measure zero. Off $N$ we get $|f(x)| \leq A$ by preservation of inequalities for real numbers, so $f \in M$.

This is not specific to $L^2$ with $L^\infty$. Provided the space has finite measure, exactly the same argument goes through to show that $L^\infty$-closed balls are closed in $L^p$ for $1 \leq p<\infty$. A similar argument goes through for $L^p$ and $L^r$ whenever $r>p$ and $X$ has finite measure. In the Lebesgue case and a number of related cases, $L^r$ balls are hollow in $L^p$, which means that $L^r$ is a first category subset of $L^p$ in these cases.