# A linear function on the space $c_{00}$ that is not continuous

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.

• When $x$ is a sequence $(x^{(1)}, x^{(2)}, \dots)$ of real numbers, what do you even mean by $\frac 1 x$? Oct 14 '15 at 19:27

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.)
• All you need is $\sum_n x_n$. Oct 14 '15 at 21:11
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.