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Fall 2016 - Problem 4

Baire category theorem, Lp spaces

Let $L^1 = L^1\p{\br{0,1}}$ be the space of integrable and $L^2 = L^2\p{\br{0,1}}$ be the space of square-integrable functions on $\br{0,1}$ . Then $L^2 \subseteq L^1$ . Show that $L^2$ is a meager subset of $L^1$ , i.e., $L^2$ can be written as a countable union of sets in $L^1$ that are closed and have empty interior in $L^1$ .

Solution.

For $n \in \N$ , let

E_n = \set{f \in L^2 \mid \norm{f}_{L^2} \leq n}.

which is closed: suppose $\set{f_k}_k \subseteq E_n$ converges to $f$ in $L^1$ . Then there exists a subsequence $\set{f_{k_\ell}}_\ell$ which converges almost everywhere to $f$ . By Fatou's lemma,

\int \abs{f}^2 \leq \liminf_{\ell\to\infty} \int \abs{f_{k_\ell}}^2 \leq n^2,

so $f \in E_n$ . Since $L^2 \subseteq L^1$ , it also follows that

L^2 = \bigcup_{n=1}^\infty E_n,

so it remains to show that $E_n$ is nowhere dense.

First, observe that if $\phi\p{x} = x^{-1/2}$ , then $\norm{\phi}_{L^1} < \infty$ and $\norm{\phi}_{L^2} = \infty$ , i.e., $\phi \in L^1 \setminus L^2$ . Hence, given $f \in L^2$ and $\epsilon > 0$ , we see that $f - \epsilon\phi \in L^1 \setminus L^2$ , or else

\epsilon\phi = f - \p{f - \epsilon\phi} \in L^2.

Moreover,

\norm{f - \p{f - \epsilon\phi}}_{L^1} = \epsilon \norm{\phi}_{L^1},

i.e., any open neighborhood of $f \in L^2$ contains an element in $L^1 \setminus L^2$ . Thus, $E_n$ is nowhere dense for each $n$ , so $L^2$ is meager.