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Problem: Let $Y= \{f\in L^{2}[0,1] \mid f(x)\geq x \text{ a.e.}\}$. Show that $Y$ is weakly closed in $L^{2}$.

My thought about solving this problem is that consider a sequence $\{f_{n}\}$ which converges to some $f$ weakly and to show that $f$ is also in the set $Y$. According to the definition of weakly convergence, $f_n \stackrel w\longrightarrow f$ if and only if for any $T\in (L^2)^*, T(f_{n})\longrightarrow T(f)$, and by Riesz Representation Theorem, we can express $T$ in terms of some $g$ in $L^2$. But I have no idea about how to proceed. I am new in functional analysis and I will be grateful to any inspiring replies. Thanks!

kahen
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Brain Zhang
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    Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. Lastly, some would consider your post rude because it is a command ("Show..."), not a request for help, so please consider rewriting it. – Zev Chonoles Jun 01 '13 at 06:02

2 Answers2

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Let $f_0(x):=x$, and $S=\{f\in L_2[0,1]:f\geq 0\quad\text{ a.e.}\}$.

For each $g\in L_2[0,1]$ consider functional $$ \phi_g:L_2[0,1]\to\mathbb{R}:f\mapsto\langle f,g\rangle $$ One can show that $$ f\geq f_0 \quad\text{ a.e.}\Longleftrightarrow \int_0^1 f(x)g(x)dx\geq \int_0^1 f_0(x)g(x)dx\quad\text{ for all }\quad g\geq 0\quad\text{ a.e.} $$ Hence $$ f\in Y\Longleftrightarrow \phi_g(f)\geq \phi_g(f_0)\quad\text{ for all }\quad g\in S $$ This means that $$ Y=\bigcap\limits_{g\in S}(\phi_g)^{-1}\left([\phi_g(f_0),+\infty)\right) $$ Note that $(\phi_g)^{-1}\left([\phi_g(f_0),+\infty)\right)$ is weakly closed as preimage of closed set $[\phi_g(f_0),+\infty)$ under action of continuous linear functional $\phi_g$. Thus $Y$ is weakly closed as intersection weakly closed sets.

Norbert
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The set $Y$ is convex and closed, hence weakly closed.

gerw
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  • okay~but I have a question, why "convex and closed" leads to "weakly closed"? Can you give me a hint of how to prove it? That will be of much help! Thank you in advance! – Brain Zhang Jun 02 '13 at 01:28
  • This follows by the geometric version of Hahn-Banach (the second bullet in http://en.wikipedia.org/wiki/Hahn–Banach_theorem#Hahn.E2.80.93Banach_separation_theorem). Take a sequence ${x_k} \subset Y$ with weak limit $x$. Then, set $A = {x}$ and $B = Y$ in the above version of Hahn-Banach. – gerw Jun 02 '13 at 15:32
  • okay!Thanks a lot!! – Brain Zhang Jun 03 '13 at 10:54