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

Thank you.

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Welcome to Math.stackexchange. People can help you better if you tell us what you have tried to do, what your thoughts are, your guesswork, or any relevant information, really! – Ravi May 2 '12 at 20:59
Also answered here: – Jonas Meyer May 2 '12 at 21:07
generic solution to all such problems: use the open mapping theorem. 2nd category implies onto. – mike May 2 '12 at 21:09

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 dominated convergence \[ \int_0^1 |g|^2\, dx = \lim_k \int_0^1 |g_k|^2 \, dx \le n \] 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.

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If $B_n$ had interior points in $L^1$ the inclusion $L^2 \to L^1$ would have to be onto, which it isn't. – t.b. May 2 '12 at 21:09
Oh yes, I see. Nice use of the open mapping theorem. But we don't need it in full strength here, do we, as we can easily use $L^1\setminus L^2\ne \emptyset$ (i. e. the inclusion is not onto) to prove that $\mathrm{int}\, B_n =\emptyset$ without using it. – martini May 2 '12 at 21:12

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