I proving l^2 is a complete metric space

[michael.wes]

Homework Statement

Prove that the sequence space l^2 (the set of all square-summable sequences) is complete in the usual l^2 distance.

Homework Equations

No equations.. just the definition of completeness and l^2.

The Attempt at a Solution

I have a sample proof from class to show that the space of bounded sequences l^infinity is complete in the sup-norm, but I'm having trouble adapting it. I asked some friends, and they linked me some difficult looking proofs... this is one of the early questions on my assignment so I think the modification of the proof should be straightforward. I have some intuition about cauchy sequences in l^2, but I can't seem to finish the proof.

I don't expect anyone to post a complete proof obviously, but I want to move on to the other questions soon. I would appreciate someone giving me an idea to complete this question.

Thanks,
M

 

[michael.wes]

There's apparently a proof here:
http://www.math.umn.edu/~jodeit/course/ell2.pdf

I am really hoping there is an easier way.

 

[hgfalling]

Here is a thin sketch of a proof. Then you can fill in the details and ask for help on steps you don't know how to do.

1) Take a Cauchy sequence c_n in l2. Construct a new sequence X by treating each coordinate as a Cauchy sequence in R (or C).
2) Show that X is in l2.
3) Show that c_n converges to X in norm.
4) Win.

 

[michael.wes]

Here is a thin sketch of a proof. Then you can fill in the details and ask for help on steps you don't know how to do.

1) Take a Cauchy sequence c_n in l2. Construct a new sequence X by treating each coordinate as a Cauchy sequence in R (or C).
2) Show that X is in l2.
3) Show that c_n converges to X in norm.
4) Win.

OK, it's 2 and 3 that I am having trouble with... maybe after I see some of the machinery for one I can do the other. How do we get on with step 3? (since we could do 2 or 3 at this point..) My friend suggested lifting some inequalities from the pointwise limits, but I don't think that works. That is, let epsilon > 0, and "temporarily fix n". Then:
\exists N\in\mathbb{N} s.t. n\geq N \Rightarrow |x_n^{(k)}-x_n|< \frac{\epsilon}{n}

The trouble with this approach, I think, is that the "N" we get depends on "n" that we temporarily fixed, so the whole argument is bogus even before you try to work with the inequalities and try to show that x^(k) -> x in the l^2 norm.

 

[michael.wes]

Nevermind, I got it.

If anyone is wondering about this in the future: you CAN adapt the proof for "l infinity". Fix k, and choose N large s.t. |x_k^n-x_n|< epsilon / (2^k/2). Then using this inequality in the l^2 norm will work, and the result falls into your lap.