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AS is said in the title, I'm given a sequence $\{f_n\}\in L^2(\mathbb R)$ and the following hypothesis:

$\{f_n\}\to 0$ pointwise and there exists a constant $C$ such that $\|f_n\|_{L^2(\mathbb R)} for every $n\in\mathbb N$. Now is it true that $f_n\to0\quad\text{weakly in}\quad L^2(\mathbb R)?$

My guess is that the answer is affirmative.

I've cancelled my further thoughts because they were wrong.

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    But step functions *are* dense. How are you going to apply LDCT?2011-10-04

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This is some crafty argument which I think I have seen somewhere on here (or something similar) Edit: AD. gave the link where I've got the argument from: Convergence of integrals in $L^p$.

First pick $D > 0$ and let choose the set $C_n := \{x \in \mathbf{R} : |f_n(x)g(x)| \leq D |g(x)|^2\}$. Then by Dominated Convergence we have that

$\int_{C_n} f_n g \, \text{d}\lambda \to 0.$

On the complement $\complement C_n$ we have that $|g(x)|^2 \leq D^{-1} |f_n(x) g(x)|$. So

$\int_{\complement C_n} |f_n g| \, \text{d}\lambda \leq \sqrt{\int_{\complement C_n} |f_n|^2 \, \text{d}\lambda}\sqrt{\int_{\complement C_n} |g|^2 \, \text{d}\lambda} \leq \frac{C}{\sqrt{D}} \sqrt{\int_{\complement C_n} |f_n g| \, \text{d}\lambda}.$

Hence,

$\int_{\complement C_n} |f_n| |g| \, \text{d}\lambda \leq \frac{C^2}{D}.$

So,

$\limsup_n \int_{\complement C_n} |f_n| |g| \, \text{d}\lambda \leq \frac{C^2}{D}.$

But $D$ was arbitrary, hence

$\lim_n \int |f_n g| \to 0.$

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    oh yes you are right.. sorry but i'm very tired :P however i understood why i couldn't apply LDCT also on $C_0(\mathbb R).$ Everything is clear... thanks and goodnight2011-10-04