$M=\{f\in L^2(0,1): \int_0^1 \frac{f^2(x)}{x^2}dx<+\infty\}$ show that $M$ is closed in $H$

Asked Jan 17 '25 at 07:20

Active Jan 17 '25 at 14:20

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Let $H=L^2(0,1)$ and $M=\{f\in L^2(0,1): \int_0^1 \frac{f^2(x)}{x^2}dx<+\infty\}$ show that $M$ is closed in $H$ and find $M^{\perp}$.

If $f_n\in M \rightarrow f \in L^2(0,1)$, I have to show that $f \in M$.

I have tried to use Fatou lemma in this way $\exists f_{n_k}\rightarrow f$ pintwise.

So $\int_0^1\frac{f^2}{x^2}\leq\liminf_{n \rightarrow \infty}\int_0^1\frac{f_n^2}{x^2}$ but the limit on the right could depend on n .

I have tried using continuity of norm because if $f_n\rightarrow_{L^2} f$ $\lVert f_n \rVert_{L^2}\rightarrow \lVert f \rVert_{L^2}$ but here I have $\lVert \frac{f_n}{ x} \rVert_{L^2} $.

edited Jan 17 '25 at 14:20

Mittens

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asked Jan 17 '25 at 07:20

wela el

This is not at all true. Take $f_n(x)=\sqrt x$ for $x >\frac 1 n$ and $0$ for $x \le \frac 1 n$, $f(x)=\sqrt x$. – Kavi Rama Murthy Jan 17 '25 at 07:49
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In fact $M$ is dense https://math.stackexchange.com/questions/3166217/multiplication-operator-on-l2-is-densely-defined – Evangelopoulos Foivos Jan 17 '25 at 08:01
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There are typos, please edit – Sine of the Time Jan 17 '25 at 09:31
@geetha290krm why your example works? $f_n\rightarrow f$ in $L^2(0,1)$ but $\int_0^1 \frac{f^2}{x}=1 $ maybe something like $\frac{1}{x^\frac{1}{3}}$? – wela el Jan 17 '25 at 10:55
You have to edit the question. In the title (as well as 5th line)you had $\int_0^1 \frac{f^2(x)}{x^2}dx$ and my example is based on that. But you can easily modify it for $\int_0^1 \frac{f^2(x)}{x}dx$ – Kavi Rama Murthy Jan 17 '25 at 11:21
@geetha290krm aaah I am so sorry thanks – wela el Jan 17 '25 at 11:44
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@welael: incidentally, this problem gives you an example of a subspace $M$ in $H$ that is dense but not closed in $H$: $\overline{M}=(M^\perp)^\perp\neq M$. – Mittens Jan 17 '25 at 14:38

$M=\{f\in L^2(0,1): \int_0^1 \frac{f^2(x)}{x^2}dx<+\infty\}$ show that $M$ is closed in $H$

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