# Understanding convergence in norm, uniform convergence

1. Oct 20, 2015

### Incand

1. The problem statement, all variables and given/known data
Find an example of a sequence $\{ f_n \}$ in $L^2(0,\infty)$ such that $f_n\to 0$ uniformly but $f_n \nrightarrow 0$ in norm.

2. Relevant equations
As I understand it we have norm convergence if
$||f_n-f|| \to 0$ as $n\to \infty$
and uniform convergence if there $\exists N$ so that
$|f_n-f| < \epsilon$ for every $n \ge N$ and $\epsilon \to 0$ as $n \to \infty$ .

3. The attempt at a solution
I'm having a hard time understanding these definitions not having seen them before.
As I understand it I have uniform convergence If I can choose $N$ independent of $x$ or equivalently that $f_n$ is bounded (by a function of $n$ that tends to zero) for each $n$.
So I'm thinking we could choose $f_n = \frac{1}{(1+x)n}$ which would be bounded by
$f_n \le \frac{1}{n}$ for all $x \ge 0$ so uniform convergent in $[0,\infty)$
However I don't ever see how to find a function that isn't also norm convergent satisfying this since if $f_n$ is in $L^2$ the norm can't be divergent and then I still have an $n$ in there somewhere forcing everything to go too zero. Am I misunderstanding the definitions?

2. Oct 20, 2015

### Krylov

This is not quite correct. The correct definition is: For every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \ge N$ implies that $|f_n(x) - f(x)| \le \epsilon$ for all $x \in (0,\infty)$. (Note that $N$ depends only on $\epsilon$ and not on $x$. In other words: the same $N$ works for all $x \in (0,\infty)$.)
Correct, but write $f_n(x)$ instead of just $f_n$. However, this sequence indeed does not serve as an example, for the reasons you indicate.

HINT: If we would look on any bounded domain of the form $(0,b)$ for some $b > 0$, then any uniformly convergent sequence also converges in $L^2$-norm. (You could try to prove that, it's not hard but besides the point here.) So, the example that you are looking for must somehow exploit the fact that the support of functions can grow beyond any bound...

3. Oct 20, 2015

### Incand

What's the difference too how I wrote it? I mean it's obviously different but say I have $N, \epsilon$ and $f_n$ that satisfy the definition I wrote would that mean something different to the real definition? I managed to find a book on real analysis (Baby Rudin) and they indeed state the definition almost identical word for word to your definition.

How about if I define
$f_n(x) = \begin{cases} 0\; \; \; \; \; 0 < x \le 1 \text{ or if n } \le 1\\ \frac{1}{\sqrt{n}} \; \; \; \; 1 < x < n\\ 0 \; \; \; \; x > n \end{cases}$
which is uniform convergent since $f_n \le \frac{1}{n}$. And we also have
$\int_0^\infty f_n^2(x) dx = \int_1^n \frac{1}{n} = ln(n) \to \infty$ when $n \to \infty$ however for each finite $n$ it's in $L^2[0,\infty)$. So does this work?

Btw. I'm aware of the theorem you spoke of, our book states and prove it although sadly without defining what uniform convergence is (I guess most readers already seen it in earlier courses).

4. Oct 21, 2015

### Incand

Oh I messed up so much, must have been sleeping when I was thinking yesterday. Got the wrong limit and thought i integrated on $n$ and everything. Here I think I got a correct example
$f_n(x) = \frac{1}{\sqrt{n}} \text{ for } 2^n < x < 2^{n+1}$ and $0$ elsewhere so
$\int_{2^n}^{2^{n+1}} \frac{1}{n}dx = (2\cdot 2^n-2^n)\frac{1}{n} = \frac{2^n}{n} \to \infty$. But don't trust me my mind seems to tired for this.

5. Oct 21, 2015

### Krylov

One of the problems I have with your definition is that it does not mention $x \in (0,1)$ anywhere. Also, the notation $|f_n - f|$ is meaningless. Either you write $\|f_n - f\|$ to denote the $L^2$-norm of the difference of the functions $f_n$ and $f$, or you write $|f_n(x) - f(x)|$ to denote the absolute value (also a norm) of the difference of the numbers $f_n(x)$ and $f(x)$.
Yes, almost. I would simply define $f_n := \frac{1}{\sqrt{n}}1_{(0,n)}$ for each $n \in \mathbb{N}$, where $1_{(0,n)}$ is the indicator on $(0,n)$, which is slightly easier than your definition but the principle is the same. Then you have $|f_n(x)| \le \frac{1}{\sqrt{n}}$ for all $x \in (0,\infty)$ and all $n \in \mathbb{N}$, proving uniform convergence. Next, due to fatigue you make a small mistake with the integral, but you know of course that $\int_0^{\infty}{|f_n(x)|^2\,dx} = 1$ for each $n \in \mathbb{N}$, so the sequence does not converge to zero in norm.
It was not that bad, your example was correct, you just make a mistake with the integral.
Yes, this is correct, too. It misbehaves even more than the first example: here the norms even blow up.

6. Oct 21, 2015

### Incand

Alright thanks!

Yea I later realised the second one works as well just me thinking I was integrating on $n$ for some reason. That's also why I excluded the $0 \to 1$ part. Thanks for explaining it too me!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted