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For $x=(x_j)_{j\in\mathbb N}\in \ell^1$ let

$$\|x\|=\sup_{n\in \mathbb N}\left \Vert \sum_{j=1}^{n}x_j\right\Vert$$

Show that $(\ell^1,\|\cdot\|)$ is a normed space, but it is not complete.

The first part was easy.

Now I try to find a sequence in $\ell^1$ such that it is a cauchy sequence, but not convergent.

Let me choose(try) $x_n=\frac{n}{j^2}$. For a fixed $n$ it is in $\ell^1$ because $\sum_{j=1}^{\infty} \frac{1}{j^2}$ converges.

WLOG $n>m$:

$$\|x_n-x_m\|=\sup_{k\in \mathbb N} \left \Vert \sum_{j=1}^{k}x_n-x_m \right\Vert| = \sup_{k\in \mathbb N} \sum_{j=1}^{k}\frac{n-m}{j^2}$$

Okay for me it seems that this is not even a cauchy-sequence..

Can someone help me? What kinds of sequences should I consider when I am facing problems like this?

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Observe that your norm is identical to the usual $\ell^1$ norm on sequences with all entries positive. So you'll have to think about alternating signs. The sequence with infinitely many terms of each sign which is closest to $\ell^1$ without being in it (this is an imprecise statement-just indicating why it's my guess) is $(1,-1/2,1/3,-1/4,...)$. The sequence of sequences $(1,0,0,...),(1,-1/2,0,0,...,),(1,-1/2,1/3,0,0,...),...$ is Cauchy in the new norm because the alternating harmonic sum converges. But it doesn't converge in $\ell^1$.

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  • $\begingroup$ Hello. Thanks for the answer. So we choose $\frac{(-1)^n}{n}$ because as a limit of a sequence (which are indeed in $\ell^1$, with the "new" norm") it is not in the "usual" $\ell^1$, hence this sequence is not convergent in $\ell^1$ with the new metric. Is this correct? $\endgroup$ – Duke Feb 3 '15 at 16:39
  • $\begingroup$ Yep, that's right. $\endgroup$ – Kevin Carlson Feb 3 '15 at 16:48
  • $\begingroup$ Thanks a lot! That helped me. $\endgroup$ – Duke Feb 3 '15 at 16:54
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A normed space is Banach if and only if whenever a series converges absolutely, it converges. Try to find a sequence $(x_n)$ with $$\sum \lVert x_n\rVert<\infty$$ but that cannot possibly converge.

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Hint: take some sequence which is just outside $\ell^1$ and play around with signs a little.

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This is a slightly overkill answer.

Consider the following theorem: Every one-to-one bounded linear operator from a Banach space onto a Banach space is an isomorphism.

Let $\ell^*$ denote the collection of $\ell^1$ sequences and endow it with the norm $\|\cdot\|$, for which it is a normed vector space. Suppose it is indeed complete, ie a Banach space (and $\ell^1$ is a Banach space already). Consider the canonical map $I : \ell^1 \to \ell^*$, which is one-to-one and onto. It is linear and bounded because obviously $\|\cdot\| \leq \|\cdot\|_1$. By the theorem, it is an isomorphism, ie its inverse $I^{-1}$ is also bounded (and linear anyway). This means $\|\cdot\|_1 \leq C \|\cdot\|$ for some constant $C \geq 0$. But this is nonsense: take the sequence $x^{(n)} = (1, -1, 1, \ldots, -1, 0, 0, \ldots)$ with $2 n$ nonzeros, half of them $+1$ and half of them $-1$, $n \in \mathbb{N}$. Then $\|x^{(n)}\|_1 = 2n$ but $\|x^{(n)}\| = 1$. Thus the assumption of completeness of $\ell^*$ was inappropriate.

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