0

Let $$l^2=\left\{(x_n):\sum_{n=1}^{\infty}x_n^2<\infty\right\}$$

equipped with the norm $$\|(x_n)\|=\left(\sum_{n=1}^{\infty}x_n^2\right)^{1/2}.$$

Prove that $l^2$ is complete with respect to the norm $\|\cdot\|$.

Can somebody please check my working?

Here is my working and some of my doubts:

We need to show that every Cauchy sequence in $l^2$ has a limit in $l^2$.

I first define the metric as (am I correct to define it in this way?): $$d((x_n),(x_m))=\|(x_n)-(x_m)\|=\left(\sum_{k=1}^{\infty}((x_n)_k-(x_m)_k)^2\right)^{1/2}$$

Pick any Cauchy sequence and we write it as $d((x_n),(x_m))\to0$.

So $d((x_n),(x_m))=\sqrt{\sum_{k=1}^{\infty}((x_n)_k-(x_m)_k)^2}$
$=\sqrt{\sum_{k=1}^{\infty}(x_n)_k^2-2\sum_{k=1}^{\infty}(x_n)_k(x_m)_k+\sum_{k=1}^{\infty}(x_m)_k^2}$

Since $\sum_{k=1}^{\infty}(x_n)_k^2$ and $\sum_{k=1}^{\infty}(x_m)_k^2$ are both finite, so the sum is finite.

But here are the parts that got me stuck:

Is $\sum_{k=1}^{\infty}(x_n)_k(x_m)_k$ finite? If it is finite we are done isn't it? We have shown that there is a limit.

And my last doubt is how can we show that the limit lies in $l^2$? Do we have to show that it can be written as $\sum_{j=1}^{\infty}x_j^2$ for some $(x_j)$?

EDITTED: I think it is not a duplicated question, I asked some different points.

I really need some help. Many thanks!

Martin Sleziak
  • 56,060
user71346
  • 4,301
  • When do you come up to the point in the title? You can define the metric as you did using the norm. What do you mean by "Pick any Cauchy sequence and we write it as $d((x_n),(x_m))\to0$."? Your proof lacks a possible limit point, a sequence the Cauchy sequence would converge.The problem in the link above, by Crostul, proposes such a possible sequence in one of the answers. Showing $\sum_{k=1}^{\infty}(x_n)_k(x_m)_k$ is finite would not be enough to solve the problem but it would only show that your metric is well defined, it gives finite distances between any to points in $l^2$. @user71346 – Deniz Sargun Feb 06 '16 at 12:29

1 Answers1

0

The statement in the title is true, since by the second condition the only sequence that applies is the zero sequence. However, the title seems unrelated to the question.

$$ \left|\sum^∞_{k=1}(x_n)_k(x_m)_k\right|\le\|x_n\|·\|x_m\| $$ per the Cauchy-Schwarz inequality.

For any scalar product and its norm, the CSI follows from $$ 0\le\Bigl\|\;\|v\|·u ± \|u\|·v\;\Bigr\|^2=2·\|u\|·\|v\|·\Bigl(\|u\|·\|v\|±\langle u,\,v\rangle\Bigr) $$

To prove finiteness the first step is to recognize that the Cauchy convergence also applies term-wise, so that the term-wise limit exists.

Lutz Lehmann
  • 131,652