For $x \in \ell^1$ we set $\Vert x\Vert = \sup\limits_{N \in \mathbb{N}}|\sum\limits_{n=1}^{N}x_n|$.
One can easily see that this is a norm on $\ell^1$. I was wondering if this space is now complete. I tried finding an absolutely convergent series that does not converge, but I did not find anything.


The space $\ell^1$ is not complete with this norm. To find a counterexample, you may consider the sequence of sequences $x_n \in \ell^1$ defined for $n \ge 1$ by

$$x_n = (x_{n,k})_{k \ge 1} := \left( \frac{(-1)^k}{k}\right)_{1 \le k \le n}$$


$$x = (x_k)_{k \ge 1} := \left( \frac{(-1)^k}{k}\right)_{1 \le k}$$

Note that we have $x_n \in \ell^1$, but $x \notin \ell^1$.

We may check the sequence $(x_n)_{n \ge 1}$ converges to $x$ with regards to the norm $\| \cdot \|$. Indeed, let $\epsilon > 0$. We have

$$\left|\sum_{k = 1}^N x_{n,k} - x_{k}\right| = \left|\sum_{k = n+1}^{N} \frac{(-1)^k}{k} \right|$$

and since the series $\sum_k \frac{(-1)^k}{k}$ is convergent, we may find $n_0$ such that all those partial sums are smaller than $\epsilon$ for every $n \ge n_0$. Which gives us $\|x_{n}-x\| \le \epsilon$ for $n \ge n_0$. From this, we can deduce that $(x_n)_{n \ge 1}$ is a Cauchy sequence in $\ell^1$, but it has no limit in $\ell^1$.

  • $\begingroup$ this has not much to do with the question, but do you know if this norm has a name ? $\endgroup$ – user9620780 May 2 at 3:51
  • $\begingroup$ This norm is essentially the $\ell^{\infty}$ norm for series. Given some sequence $x = (x_k)_{k \ge 1}$ let us denote $\sum x = \left(\sum_{i = 1}^k x_i\right)_{k \ge 1}$ for the associatied series. Then we have : $$\|x\| = \left\|\sum x\right\|_{\infty}$$ The space of bounded series (which contains $\ell^1$) is complete for this norm, but $\ell^1$ is not closed inside of it (so it was possible to take a sequence of $\ell^1$ whose limit was not in $\ell^1$). $\endgroup$ – Joel Cohen May 2 at 12:09

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.