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By Holder's Inequality we have $L^2[a,b] \subset L^1[a,b]$. Show that the set $L^2[a,b]$, considered as a subset of the complete metric space $L^1[a,b] $is of the first category.

I know that of the first category means that it is the union of a countable collection of nowhere dense sets in X. My idea is to take a collection $A_k = \{f \in L^1[a,b]: |\int| \leq 1\}$ and then consider a collection $B_n$ that is some intersection of $A_k$'s. Then each $B_n$ would be closed in $L^1[a,b]$ and I would just need to show that $L^2[a,b] \subset \cup B_n$ and that each $B_n$ has empty interior. Is this correct? I could use help filling in the missing details

Vinny Chase
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  • For $n\in \Bbb N$ let $S(n)$ be the closure in $L^1$ of ${f\in L^1: n<\int_a^b |f(x)|^2dx}.$... What is the interior of $S(n)$ in $L^1$? – DanielWainfleet Mar 23 '18 at 01:18
  • Pick $g \in L^{1}-L^{2}$. If $\int |f|^{2} \leq n$ then $f+\frac 1 n g \to f$ in $L^{1}$ and $\int |f+ \frac 1 n g|^{2} =\infty >n$. What does this tell you about interrior of ${f \in L^{1}: \int |f|^{2} \leq n}$? – Kavi Rama Murthy Mar 23 '18 at 07:47
  • @KaviRamaMurthy why you changed the set described in the first comment? –  Mar 18 '20 at 12:26

1 Answers1

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We know that $L^2[a,b]$ is dense in $L^1[a,b]$; take for $f \in L^1[a,b]$ the approximate sequence $f_n = f 1_{\{|f| \leq n\}}$. On the other hand, if $f \in L^1[a,b] \setminus L^2[a,b]$ and $(f_n)_{n \in \mathbb{N}} \subset L^2[a,b]$ with $f_n \rightarrow f$ in $L^1$, then we already have $\lim_{n \rightarrow \infty} \|f_n\|_{L^2[a,b]} = \infty$: For any subsequence of the natural numbers we can choose a subsequence which converges also $\lambda$-almost sure towards $f$. Using Fatou's Lemma, we see for this subsequence that $$\infty = \int_a^b |f|^2 \, \rm{d} x \leq \lim_{k \rightarrow \infty} \int_a^b |f_{n_k}|^2 \, \rm{d} x.$$ This is a helpful observation for the proof that $U := \{ f \in L^2[a,b]: \|f\|_{L^2[a,b]} \leq 1 \}$ is nowhere dense.

  1. $U$ is already closed in $L^1$: Let $(f_n)_{n \in \mathbb{N}} \subset U$ be a convergent sequence towards some $f \in L^1[a,b]$ in $L^1$-norm. Taking a subsequence we can assume that $f_n \rightarrow f$ $\lambda$-almost sure. Now use Fatou's-Lemma to show that $\int_a^b |f|^2 \, \rm{d} x \leq 1$.
  2. Show that $U$ contains no interior points: For $f \in U$ take $g \in L^1[a,b] \setminus L^2[a,b]$ and define $f_\varepsilon = f + \varepsilon g$.

As already indicated in the comments, we can write $L^2[a,b] = \bigcup_{n=1}^\infty n U$, i.e. $L^2[a,b]$ is of the first category as a subspace of $L^1[a,b]$.

Edit: Thanks for the comment. I have corrected the typo!

p4sch
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