Contradiction to Baire's theorem with $\{f\in L^2([0,1]):\int_0^1|f_j|^2\leq j\}$?

Question

For $j\in \mathbb{N}$ let $$M_j=\{f\in L^2([0,1]):\int_0^1 |f|^2 dx \leq j\}$$ (a) Establish that $L^2([0,1])=\cup_{j\in \mathbb{N}}M_j$.

(b) Show that each $M_j$ is a closed subset in $L^1([0,1])$.

(c) Show that the interior of each $M_j$ in the norm topology of $L^1([0,1])$ is empty.

(d) From (a)-(c) it appears that $L^2([0,1])$ is the countable union of sets with empty interior. Explain why this does not contradict Baire's theorem.

I believe (a) is obvious.

For (b) I need to show if $\int |f_n|^2 \leq j$ for $j\in \mathbb{N}$ and $\int |f_n-f| \to 0$ then also $\int |f|^2\leq j$. To relate $L^2$ and $L^1$ I was thinking of using Cauchy-Schwarz and saying $\int |f_n-f|\leq (\int |f_n-f|^2)^{\frac12}$ but I need the inequality to go the other way.

For (c) I assume there exists an $M_j$ such that $O\in M_j$ where $O$ is open. Then there exists $f \in M_j$ and a sequence $g_j \in M_j$ such that $\int |f-g_j|<\epsilon$. I somehow want to obtain a contradiction.

I found this helpful post Set with empty interior in $L^1([0,1])$ but there $f\in L^1([0,1])$ in the definition of $M_j$ so I am not sure if I can use it. There they argue if $M_j \ni f_k\to f$ in $L^1$ then for some subsequence $f_{k_n}\to f$ a.e. Then by Fatou $\int |f|^2\leq \lim \inf \int |f_{k_n}|^2\leq j$ so $f\in M_j$. Wouldn't this argument show $M_j$ is closed in any $L^p$-space then as we can always extract an almost everywhere converging subsequence?

For (d) is the problem that $M_j$ is closed and of empty interior in $L^1([0,1])$ but it is defined as a subset of $L^2([0,1])$?

Answer 1

In b, say you have a sequence in $M_j$ converging in $L^1$, pass to a subsequence to get an a.e. convergent subsequence, then Fatou's Lemma does what you want.

In c, find a family $f_{\epsilon,A}$ with $\| f_{\epsilon,A}\|_{L^1}=\epsilon$ but $\| f_{\epsilon,A} \|_{L^2}=A$, then for each $\epsilon$ and each $f\in M_j$, perturb $f$ using an appropriate function of this family to conclude that $M_j$ does not contain the ball centered at $f$ of radius $\epsilon$ in the $L^1$ norm.

In d the point is that $L^2([0,1])$ with the $L^1$ norm is not a complete metric space. Indeed one can use my suggestion in part c to see that.

Answer 2

You can prove (b) without touching any integral: $L^2[0,1]$ is reflexive (as a Hilbert space) so that the closed balls $M_j$ are weakly compact. The inclusion $i:L^2[0,1]\hookrightarrow L^1[0,1]$ is continuous and hence weakly continuous (just by abstract nonsense) so that $M_j=i(M_j)$ is weakly compact in $L^1[0,1]$, hence weakly closed and thus also norm closed.