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Consider the space of eventually zero sequences: $$c_{00} = \left\{ x = (x^{(1)},x^{(2)},\dots,x^{(k)},\dots)\in\ell^\infty \,\middle|\, \exists k_0 \text{ such that $x^{(k)}=0$ for $k>k_0$}\right\}$$

What would be an example of a linear function from $c_{00}$ to $\mathbb R$ that isn't continuous? I'm thinking of something of the form $\frac 1 x$, because it would not be defined if $x$ equaled zero, but I'm not sure where to go from there.

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  • $\begingroup$ When $x$ is a sequence $(x^{(1)}, x^{(2)}, \dots)$ of real numbers, what do you even mean by $\frac 1 x$? $\endgroup$
    – Christoph
    Oct 14 '15 at 19:27
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Hint: (if you'll allow me to put the indices back in their usual place as subscripts): is $(x_1, x_2, x_3, \ldots) \mapsto \sum_n nx_n$ continuous? (It's certainly well-defined and linear if only finitely many $x_n$ are non-zero.)

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  • $\begingroup$ All you need is $\sum_n x_n$. $\endgroup$ Oct 14 '15 at 21:11
  • $\begingroup$ @DavidC.Ullrich: agreed, my example factors through a linear but discontinuous map on the domain, so I thought that it made it a bit more interesting. $\endgroup$
    – Rob Arthan
    Oct 14 '15 at 21:34
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Hint: use the fact that continuous is equivalent to bounded in normal spaces. I think you're onto this with $1/x $, but it needs to be more clear. Pick a sequence in c00 that has bounded norm and make its image in $\mathbb{R}$ blow up.

Edit: to give a concrete answer, note that $x_n = (1,\frac 1 2, \frac 1 3, \dots, \frac 1 n, 0, 0, \dots)$ is a bounded sequence (in the sup norm), but the sequence $A x_n = \sum_{i=1}^n \frac 1 i$ is unbounded.

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