$L^2()$ is a set of first category in $L^1()$?

How to show that $L^2([0,1])$ is a set of first category in $L^1([0,1])$?

Thank you.

Solutions Collecting From Web of "$L^2()$ is a set of first category in $L^1()$?"

For $n \in \mathbb N$ set $B_n = \{f \in L^2[0,1]\mid \|f\|_2 \le n \}$. We will show that $B_n$ is nowhere dense in $L^1$. Let $g \in L^1[0,1]\setminus L^2[0,1]$ and $f \in B_n$, then $f + \frac 1k g \to f$ in $L^1$ but $f+\frac 1k g \not\in B_n$ for all $k$. Hence $f \not\in \mathring{B_n}$ and $B_n$ doesn’t have inner points. On the other hand, $B_n$ is closed in $L^1$: Let $g \in L^1$ and $g_k \in B_n$, $g_k \to g$. Then $g_{k_\ell} \to g$ almost everywhere for some subsequence, it follows by Fatou’s Lemma
\[
\int_0^1 |g|^2\, dx \le \liminf_k \int_0^1 |g_{k_\ell}|^2 \, dx \le n^2
\]
so $g \in B_n$.

As $L^2[0,1] = \bigcup_n B_n$ is a countable union of nowhere dense sets, it is of first category.