# Is $\ell^1$ complete with this norm?

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

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

and

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

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 $$\ell^1$$.

• this has not much to do with the question, but do you know if this norm has a name ? – user9620780 May 2 at 3:51
• 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 $\ell^1$). – Joel Cohen May 2 at 12:09