# Weak and pointwise convergence in a $L^2$ space

Let $I$ be a measured space (typically an interval of $\Bbb R$ with the Lebesgue measure), and let $(f_n)_n$ a sequence of function of $L^2(I)$.

Assume that the sequence $(f_n)$ converge pointwise and weakly. How to prove that the pointwise limit and the weak limit are the same ?

Here's a functional analytic approach:

A weakly convergent sequence in a Hilbert space $H$ is bounded, and by the Banach-Saks theorem has a subsequence whose Cesàro averages converge strongly in $H$ to the same limit. Almost sure convergence is preserved by taking subsequences and Cesàro averages. So, without loss of generality you may assume that your weakly convergent sequence is actually strongly convergent.

Both strong $L^2$ convergence and almost sure convergence imply convergence locally in measure, so you only need to show that such limits are unique, which is easy.

• I like that, thanks ! – Lierre Jun 19 '12 at 13:21
• I'm glad to help. – user940 Jun 19 '12 at 13:22

Read the proof on page 266 of this book

• Thanks ! Is the requirement that $E$ has finite measure necessary ? – Lierre Jun 19 '12 at 12:34
• Good question. I don't think so, but Egorov's theorem does not work without this assumption. – Siminore Jun 19 '12 at 12:36
• After some thoughts, no it's not necessary, at least if $E$ is $\sigma$-finite, because then the result in the book can be applied to restriction to every subset of a covering of $E$ by set with finite measure. – Lierre Jun 19 '12 at 12:39
• I suspect that everything works provided continuous functions with compact support are dense in $L^2$. – Siminore Jun 19 '12 at 12:59

Following Byron Schmuland we can say:

1) By the Banach-Saks theorem, a weakly convergent sequence, in a Banach space, has a subsequence whose Cesàro averages converge strongly to the same limit.

2) In a $L^p$ space, strong convergence implyes pointwise convergence a.e. for a subsequence;

3) Pointwise convergence a.e. is preserved by taking subsequences and Cesàro averages.

4) For Pointwise convergence a.e. we have unicity of limits.

So we can conclude that, in $L^p$, weak convergence to $f$ and pointwise convergence to $g$, imply $f=g$.

Assume that $f_n \rightharpoonup g$ and $f_n \rightarrow f$ a.e. Then $$| \int_{I} (f-g)| \le \int_{I} |f_n - f| + \int_{I} |f_n - g|$$ As $f_n \rightharpoonup g$, e $1 \chi \{I\} \in L^{2}(I)$ we have $$\int_{I} |f_n - g| \rightarrow 0.$$ Now, notice that $$|f_n -f| \le |f_n| + |f| < 2|f| + \varepsilon.$$ for $n>>1$. Then by Dominated convergence Theorem $$\int_{I} |f_n - f| \rightarrow 0.$$ Hence $f=g$ a.e.

• As far as I understand, there is something wrong. How can you state that $f_n \to g$ strongly on $I$? – Siminore Jun 19 '12 at 12:56
• You seem right. $\lim_{n} \int_{I} f_n \chi \{I\} \rightarrow \int_{I} f \chi \{I\}.$ Then, $\lim_{n} \int_{I} (f_n -g) \le 0$. But not necessarily $\lim_{n} \int_{I} (g - f_n ) \le 0$. – user29999 Jun 19 '12 at 13:08
• I think that the flaw in your argument is that $f_n \rightharpoonup g$, and $1 \chi \{I\} \in L^{2}(I)$ implies $\int_I(f_n-g)\to0$ but not $\int_I|f_n-g|\to0$, since the latter is not the pairing between $\chi$ and $f_n-g$... – bartgol Jun 20 '12 at 15:11