# Completeness of Norm

Let us consider the sequence space $\ell_1$ endowed with the norm given by $\|x\|=\displaystyle\sup_n\left|\sum_{k=1}^nx(k)\right|$. Then how can I prove that $\ell_1$ is not complete with respect to this norm.

Thanks for any help.

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As defined, $\|\cdot\|$ is not a norm since $\|x\|$ might be negative, and $\|x\|\neq \|-x\|$. Do you mean $\|x\|=\sup_n\left|\sum_{k=1}^nx(k)\right|$? – Yury Feb 5 '13 at 6:20
@Yury Yes, there will be | | instead of { } – Ester Feb 5 '13 at 6:27

Consider the sequence $x_1 = (1, -1, 0, 0, \dots)$, $x_2 = (1, -1, 1/2, -1/2, 0,0, \dots)$, $x_3 = (1, -1, 1/2, -1/2, 1/3, -1/3, \dots)$ etc. That is, $$x_k = (1,-1,1/2,-1/2, \dots, 1/k,-1/k, 0,\dots).$$
Note that all $x_i$ lie in $\ell_1$ (since each $x_i$ has a finite support). Now $\|x_i - x_j\| \leq \max(1/i, 1/j)$. Therefore, $x_i$ is a Cauchy sequence w.r.t $\|\cdot\|$. It is easy to see that if $x_k \to x$ as $k\to\infty$ then $x$ must be equal to $$(1,-1,1/2,-1/2,1/3,-1/3, \dots, 1/n,-1/n, \dots)$$ but this sequence doesn't belong to $\ell_1$. We conclude that $x_k$ doesn't converge w.r.t. $\|\cdot\|$.