# Convergence in the L^2 norm

1. Apr 8, 2012

### AxiomOfChoice

Suppose there exists a sequence $f_n$ of square-integrable functions on $\mathbb R$ such that $f_n(x) \to f(x)$ in the L^2-norm with $x \ f_n(x) \to g(x)$, also in the L^2-norm. We know from basic measure theory that there's a subsequence $f_{n_k}$ with $f_{n_k}(x) \to f(x)$ for a.e. x. But my professor seems to be claiming that this somehow implies $x \ f_{n_k}(x) \to g(x)$ for a.e. x. I don't see why this is. Obviously, we know that $x \ f_{m_k} \to g$ a.e. for SOME subsequence of $f_n$...but how do we know it works for the SAME subsequence? Can someone offer some guidance? Thanks!

2. Apr 8, 2012

### Fredrik

Staff Emeritus
If I understand you correctly, there's a set E such that $\mu(E)=0$ and $xf_n(x)\to g(x)$ for all x in $E^c$. This means that for all $x\in E^c$, $\langle xf_n(x)\rangle_{n=1}^\infty$ is a convergent sequence in $\mathbb R$, and as you know (or can easily prove), every subsequence of a convergent sequence in $\mathbb R$ converges to the limit of the sequence.

3. Apr 8, 2012

### AxiomOfChoice

Well, if we have $f_{n_k}(x) \to f(x)$ off of a set $E\subset \mathbb R$ with $\mu(E) = 0$, then $|f_{n_k}(x) - f(x)| \to 0$ as $k\to \infty$ whenever $x\notin E$. But I want to somehow show that this implies the existence of a $E'\subset \mathbb R$ (which may or may not be the same as $E$) with $\mu(E') = 0$ such that $|x \ f_{n_k}(x) - g(x)| \to 0$ as $k\to \infty$ whenever $x\notin E'$. How in the world is one supposed to get from the first statement to the one we want to prove, if the only other thing you know is that $\| x \ f_n(x) - g(x) \|_{L^2(\mathbb R)} \to 0$ as $n\to \infty$?

I hope that clarifies my question a bit!

4. Apr 8, 2012

### Fredrik

Staff Emeritus
OK, I wasn't paying enough attention to when you were using the L^2-norm and when you were just talking about convergence almost everywhere. I will think about it.

5. Apr 8, 2012

### micromass

Staff Emeritus
Isn't it obvious that if $f_n$ converges both to f and g in $L^2$ that then f=g a.e.?? That would imply it.

6. Apr 8, 2012

### Fredrik

Staff Emeritus
Hehe, after my last post yesterday, I realized that I was much too tired to do any math. I decided to give it another shot "tomorrow", i.e. today, in the unlikely event that micromass wouldn't already have posted the solution. I should have realized that there was no chance that he wouldn't already have done that.

7. Apr 8, 2012

### micromass

Staff Emeritus
Haha. Next time I'll let you finish it up!!

8. Apr 9, 2012

### morphism

But it's xf_n(x) that's converging to g, and not f_n.

Something does seem fishy about the argument in the OP. Maybe we need more context?

9. Apr 9, 2012

### micromass

Staff Emeritus
Oh, I see. I misread there.

Anyway. If $xf_n(x)\rightarrow g$ in $L^2$, then the $xf_{n_k}\rightarrow g$ in $L^2$. Thus there is a subsequence $xf_{n_{k_l}}$ that converges to g a.e.

Evidently, the sequence $xf_{n_k}(x)$ converges a.e. (to xf(x)). And since a subsequence converges to g, it means that the sequence $xf_{n_k}(x)$ converges to g a.e.

Did I do something stupid?

10. Apr 9, 2012

### morphism

That works!

11. Apr 10, 2012

### Fredrik

Staff Emeritus
Oh, don't worry about that. I don't mind at all. I would probably need 7 hours to do what you can do in 7 minutes anyway.