A sequence pointwise convergent and equibounded in $L^2(\mathbb R)$ norm is weakly $L^2(\mathbb R)$ convergent

Question

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.

Perfect... that was what i was looking for.. But if i approximate every g in $L^2$ with functions with compact support? — uforoboa, Oct 04 '11 at 18:03
But step functions are dense. How are you going to apply LDCT? — JT_NL, Oct 04 '11 at 19:43

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

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

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

edited Apr 13 '17 at 12:21

Community

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answered Oct 04 '11 at 18:57

JT_NL

14,514

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 04 '11 at 19:09
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and moreover.. can you re explain please what is $D$? – uforoboa Oct 04 '11 at 19:18
Here is the link http://math.stackexchange.com/questions/11028/convergence-of-integrals-in-lp/11169#11169 – AD - Stop Putin - Oct 04 '11 at 19:28
@AD. Thanks! I'll add it. – JT_NL Oct 04 '11 at 19:29
@uforoboa: How are you going to apply LDCT? – JT_NL Oct 04 '11 at 19:30
Oh, right. I thnik I have fixed it now. – JT_NL Oct 04 '11 at 19:34
ok thanks to everybody for the reference. – uforoboa Oct 04 '11 at 19:35
@uforoboa: A little digest is good sometimes - it is a useful method. – AD - Stop Putin - Oct 04 '11 at 19:40
jonas.. can you fix your estimate $C^2/D$? – uforoboa Oct 04 '11 at 19:40
I don't see what there is wrong with it? $C$ comes from the $|f_n|_2$ and the $\frac1{\sqrt{D}}$ from the estimate on $\complement C_n$. – JT_NL Oct 04 '11 at 19:42
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 goodnight – uforoboa Oct 04 '11 at 19:44

A sequence pointwise convergent and equibounded in $L^2(\mathbb R)$ norm is weakly $L^2(\mathbb R)$ convergent

1 Answers1