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.


$l_1$ 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:


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:

  1. How do I show that $\Sigma_{n=1}^\infty|z_n| <\infty$?
  2. How do I show that for a given $\epsilon$, there is a $K'$, so that if $K \ge K'$:


  • $\begingroup$ see this math.stackexchange.com/questions/328479/… $\endgroup$ – rlartiga Mar 26 '14 at 17:20
  • $\begingroup$ It is much easier to show that $l_1$ is complete directly. $\endgroup$ – copper.hat Mar 26 '14 at 17:26
  • $\begingroup$ @copper.hat I tried that, but I get the same problem. I get that every component is Cauchy in our Cauchy sequence. So they must converge pointwise. But It is difficult to prove that the limit is in $l_1$, and that it converges in the $l_1$-norm. I did not understand so much of the link I was provided. $\endgroup$ – user119615 Mar 26 '14 at 17:29

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 $\|x_n-x\|_1 \to 0$.

  • $\begingroup$ It was very smart what you did that when you should show that it holds for an infinite sum, you show that it holds for all finite sums instead, and hence it must hold for all infinite sums. I hope it is ok if I ask you one thing about this, if you were not allowed to use trick of only summing up to L, would you have been able to complete the proof? $\endgroup$ – user119615 Mar 26 '14 at 19:30
  • 1
    $\begingroup$ It is a fairly standard trick, and no, the above proof would not work (it works because I only need a finite number of $x_m(i)$ to converge). $\endgroup$ – copper.hat Mar 26 '14 at 19:32
  • $\begingroup$ I just have one more question application of the L-$\infty$ trick. Lets say you wanted to use the trick on the m's instead of the n's,That is, if you want to use the trick on $\Sigma_{i=1}^\infty|x_m(i)-x(i)|$, instead(I get this when i start with $\infty$ on the left side). Now both L and m are variables, but would it be ok to say that for any L, we could get $\Sigma_{i=1}^L|x_m(i)-x(i)|$ as small as we wanted by increasing m, and hence for a large enough m $\Sigma_{i=1}^\infty|x_m(i)-x(i)|<\epsilon?$, or will the trick only work when we only have variable in L, and not in the sequence(m)? $\endgroup$ – user119615 Mar 26 '14 at 20:23
  • 1
    $\begingroup$ No. You only know $x_m(i) \to x(i)$, (and hence for a finite number of $i$ as well), so you can't leap to some form of uniform convergence. Also, note that there are two 'tricks' above: (1) Showing that any finite sum is bounded by a uniform bound and (2) the fact that the $m$ on the right hand side is in some way 'independent' of the $n,L$ on the right hand side (so we have some freedom to choose $m$ to make it as small as we need, for example). $\endgroup$ – copper.hat Mar 26 '14 at 20:31
  • $\begingroup$ I see, thanks again for your time, it really helped me! $\endgroup$ – user119615 Mar 26 '14 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.