I want to show that a normed vector space is complete. I know that if you can show that every Cauchy sequence converges, then it is complete. But in a normed vector space, completeness is equivavlent to that every absolute convergent series, converges, and I want to use this approach. I have started, but I need some tricks to finish it, I think there will be some smart people here that can help me.
exercise:
$l_1$ is the set of all sequences $\textbf{x}=\{x_n\}_{n \in \mathbb{N}}$ of real numbers such that $\Sigma_{n=1}^\infty|x_n|$ converges. Use this norm:
$\|\textbf{x}\|=\Sigma_{n=1}^\infty|x_n|$.
Show that $l_1$ is complete.
my sollution:
As I said, I want to show that if $\Sigma \|\mathbb{Y}_k\|$, converges, then $\Sigma \mathbb{Y}_k$, converges.
So, since $\Sigma \|\mathbb{Y}_k\|$ converges, $\Sigma_k\Sigma_n|y_{kn}|$ converges, by interchanging the sums I get that $\Sigma_n\Sigma_k|y_{kn}|$, converges. This means that every component in the "vector" converges, and $\Sigma \mathbb{Y}_k \rightarrow \mathbb{Z}$. But now comes the problem. The convergence I have show is not the type of convergence in the $l_1$-norm, it is a type of "pointwise convergence". And I have also not actually shown that $\mathbb{Z}$ is in $l_1$. So my two questions are:
- How do I show that $\Sigma_{n=1}^\infty|z_n| <\infty$?
- How do I show that for a given $\epsilon$, there is a $K'$, so that if $K \ge K'$:
$\|\Sigma_{k=1}^K\mathbb{Y_k}-\mathbb{Z}\|<\epsilon$?
Best Answer
Given a Cauchy sequence $x_n$ it is straightforward to see that $x_n(i) $ converges to some $x(i)$ for all $i$.
Let $\epsilon>0$ and choose $N$ large enough so that if $n,m \ge N$, then $\|x_n-x_m\|_1 < {\epsilon \over 2}$. \begin{eqnarray} \sum_{i=1}^L |x_n(i)-x(i)| &\le & \sum_{i=1}^L |x_n(i)-x_m(i)| + \sum_{i=1}^L |x_m(i)-x(i)|\\ &\le & \|x_n-x_m\|_1 + \sum_{i=1}^L |x_m(i)-x(i)| \\ &\le & {\epsilon \over 2} + \sum_{i=1}^L |x_m(i)-x(i)| \end{eqnarray}
Since each component converges, for any fixed $L$ we can choose $m \ge N$ large enough so that $\sum_{i=1}^L |x_m(i)-x(i)| < {\epsilon \over 2}$, and hence we have $\sum_{i=1}^L |x_n(i)-x(i)| < \epsilon$ for any $L$, and so $\sum_{i=1}^\infty |x_n(i)-x(i)| \le \epsilon$ for all $n \ge N$.
Let $\epsilon = 1$, which gives some $N$. Then $\sum_{i=1}^L |x(i)| \le \sum_{i=1}^L |x_n(i)-x(i)| + \sum_{i=1}^L |x_n(i)| \le 1 + \|x_n\|$, and so $x \in l_1$.
Then the above shows that $\|x_n-x\|_1 \to 0$.