Let $X=L_2(0,1)$ and $S=\lbrace x\in X: \int_0^{1/2}x^2 dt\le1\rbrace$

Question

Let $X=L_2(0,1)$ and $S=\lbrace x\in X: \int_0^{1/2}x^2 dt\le1\rbrace$.

I'm trying to show wheter S can be written as $\cup_{n=1}^\infty S_n$ where $S_n\in X$ and $Int\bar{S_n}=\emptyset$

I tried a few things for $S_n$ like $S_n=\lbrace x\in X:x=(\sqrt{2}-{1/n}).\mathcal{X}_{[0,1/2]}+y.\mathcal{X}_{(1/2,1]}$ where $y\in X\rbrace$ and some ridicolus stuff. Have no idea to approach to problem.

Any help is appriciated

Answer 1

The set $S$ is closed in $X$ because the map $F\colon x\mapsto \int_{(0,1/2)}x(t)^2\mathrm dt$ is continuous. Indeed, if $\lVert x_n-x\rVert_2\to 0$, then $\int_{(0,1/2)}(x_n(t)-x(t))^2\mathrm dt\to 0$ and expanding the square and using the convergence $\int_{(0,1/2)}x(t)\cdot x_n(t)\mathrm dt\to \int_{(0,1/2)} x(t)^2\mathrm dt$ we obtain $F(x_n)\to F(x)$. Now, two scenarii:

• if $X$ has an empty interior, we are done;
• if not, we conclude using Baire's theorem.

Actually, the interior of $X$ is not empty. Indeed, if $\lVert x\rVert_2\leqslant 1/2$, then necessarily, $\int_{(0,1/2)}x(t)^2\leqslant 1/4$, hence $x$ is an element of $S$.