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On $\ell^{\infty}$, define a norm by: $$\| x \|=\sup_{N\in \mathbb{N}}\left (\frac{1}{N} \left |\sum_{k=1}^Nx_k \right |\right )$$ (I have proved that this is actually a norm.) I need to show that $\ell^{\infty}$ with this norm is a complete space.

Here's what I've got so far (TL;DR: I could only show convergence in each coordinate):

  • Let $\left \{ x^{(n)}\right \}\subset \ell^{\infty}$ be a Cauchy sequence. We know that for every $\varepsilon>0$ there exists $R\in \mathbb{N}$ such that for every $n,m>R$, $$\sup_{N\in \mathbb{N}}\left (\frac{1}{N} \left |\sum_{k=1}^N(x^{(n)}_k-x^{(m)}_k) \right |\right )<\varepsilon$$ which is like saying that for every $N\in \mathbb{N}$, $$\left |\sum_{k=1}^N(x^{(n)}_k-x^{(m)}_k) \right |<N.\varepsilon$$

  • Temporarily denoting $S_N=\sum_{k=1}^N(x^{(n)}_k-x^{(m)}_k) $, we
    get: $$|x^{(n)}_N-x^{(m)}_N|=|S_{N+1}-S_N|\leq |S_{N+1}|+|S_N|<(N+1)\varepsilon + N\varepsilon = (2N+1)\varepsilon.$$

  • So, to sum up, given any natural $N$, for every $\varepsilon>0$ there exists $R\in \mathbb{N}$ such that for every $n,m>R$, $|x^{(n)}_N-x^{(m)}_N|<(2N+1)\varepsilon$ (and it's actually the same $R$ for all of them, which I haven't used). That proves that
    $\left \{ x^{(n)}_N\right \}_{n=1}^{\infty}$ is a real Cauchy sequence, and therefore we can call it's limit $y_N$ and get a new
    sequence $\left \{ y_N\right \}$.

Now we need to show that the new sequence is bounded, and that $\left \|x^{(n)}-y \right \|\rightarrow 0$. That's where I got stuck.

I'd appriciate your suggestions about how to proceed from here.

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  • $\begingroup$ Why do you believe it's complete? $\endgroup$ – David C. Ullrich Nov 18 '17 at 22:01
  • $\begingroup$ That's what the question asks $\endgroup$ – 35T41 Nov 18 '17 at 22:04
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If you actually need to show that $\ell^\infty$ with this norm is complete you're in trouble, because it's not.

Hint: Define a sequence $x$ by $x_{2^j}=j$, $x_n=0$ for $n\ne 2^j$. Then $x\notin\ell^\infty$, although $||x||<\infty$. In fact $$\lim_{N\to\infty}\frac1N\left|\sum_1^Nx_n\right|=0.$$You can use this $x$ to construct a divergent Cauchy sequence in $\ell^\infty$.

Edit: Oh, a cooler way. Recall a consequence of the Open Mapping Theorem:

Suppose $X$ and $Y$ are Banach spaces. If $T:X\to Y$ is linear, invertible and bounded then $T^{-1}$ is bounded.

Define $T:(\ell^\infty, ||.||_\infty)\to(\ell^\infty,||.||)$ by $Tx=x$. Then $T$ is bounded but if $e_n$ is the sequence with all zeroes except for one one, in the $n$-th place, $||e_n||_\infty=1$ while $||e_n||=1/n$. So $T^{-1}$ is not bounded. Hence $(\ell^\infty,||.||)$ is not a Banach space.

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  • $\begingroup$ Oh man, that's quite irritating. Now I can't really trust the text's questions anymore... But on the other hand it's cool to see that I couldn't solve it for a reason. Thanks! $\endgroup$ – 35T41 Nov 19 '17 at 5:55
  • $\begingroup$ You should tell us what book it is... $\endgroup$ – David C. Ullrich Nov 19 '17 at 14:30
  • $\begingroup$ That's not a book, these are notes written by the lecturer. If it was a book I would of course share it... $\endgroup$ – 35T41 Nov 20 '17 at 5:33

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