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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 $\|f_n\|_{L^2(\mathbb R)}<C$ 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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Can you write up in more detail what you have right now? – Jonas Teuwen Oct 4 '11 at 17:32
The indicator functions are not dense in $L^2$! – Jonas Teuwen Oct 4 '11 at 18:00
Perfect... that was what i was looking for.. But if i approximate every g in $L^2$ with functions with compact support? – uforoboa Oct 4 '11 at 18:03
But step functions are dense. How are you going to apply LDCT? – Jonas Teuwen Oct 4 '11 at 19:43
up vote 2 down vote accepted

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

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


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


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

But $D$ was arbitrary, hence

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

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did you think about my last comment? What if instead of working with an arbitrary $g\in L^2\mathbb(R)$ we verify the statement on the functions with compact support? this set is dense in our space $L^2$ or am I wrong again? – uforoboa Oct 4 '11 at 19:09
and moreover.. can you re explain please what is $D$? – uforoboa Oct 4 '11 at 19:18
Here is the link… – AD. Oct 4 '11 at 19:28
@AD. Thanks! I'll add it. – Jonas Teuwen Oct 4 '11 at 19:29
@uforoboa: How are you going to apply LDCT? – Jonas Teuwen Oct 4 '11 at 19:30

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