# Functional analysis and limits

1. Oct 6, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data

2. Relevant equations
$$\lim_n a_n := \lim_{n \rightarrow \infty} a_n$$

3. The attempt at a solution
I'm stuck at exercise (c). Since if n heads to infinity the m doesn't play the role the limit must be one. So the mistake is somewhere on the left and I think it is at the part where both limits are taken at the same time.

Or is the limit 1/2? Can someone help me?

2. Oct 6, 2008

### Dick

Does f_n really converge to 1? What's ||f_n-1|| for any n?

3. Oct 6, 2008

### morphism

Hint: we have to be careful when we say things like "$f_n \to \textbf{1}[/tex]" nonchalantly. What does "[itex]\to$" mean in this setting?

4. Oct 7, 2008

### dirk_mec1

$$||f_n-1|| = \frac{m}{m+n}$$ for any n. But if I let n tend to infinity this would go to zero, right? So the limit is one? I guess I miss the point here because the n stands for the nth sequence and the m is just the element in that sequence, right?

If I look in one sequence (so for fixed n) and let m tend to infinity I would always get zero do you mean that?

It means keeping m fixed while letting $$n \rightarrow \infty$$ right?

Last edited: Oct 7, 2008
5. Oct 7, 2008

### Dick

Don't take n to infinity. Just focus on a single n for a minute. I think the limit m->infinity of m/(m+n) is one, not zero. So I would say ||f_n-1||=1. For ALL n. So I would say while it converges pointwise, the sequence f_n does not converge in l_infinity.

6. Oct 7, 2008

### dirk_mec1

So there holds: $$\lim_{m \rightarrow \infty} f_n =1$$ imlpying pointwise convergence.

A sequence converges in $$l^{\infty}$$ if $$\lim_{j \rightarrow} ||f_j-f||_{\infty} \rightarrow 0$$ but by noticing the pointwise convergence one can note that this will not head to zero but 1, right?

Last edited: Oct 7, 2008
7. Oct 7, 2008

### Dick

I think you've got the right idea. I think your first statement should say lim n->infinity f_nm=1. Each term m of the sequences tends to 1 as n->infinity. But that's not enough to make the f_n converge.