Show that $c_0$ is not a Banach space in the $\| \cdot \|_2$-norm.

Question

Denote $c_0 = \{ \xi=(\xi)_{n \in \mathbb{N}} : \lim_{n \rightarrow \infty}\xi_n = 0 \}$ .

Question: Show that $c_0$ is not a Banach space in the $\| \cdot \|_2$-norm.

I wish to construct a Cauchy sequence $(\xi^{(i)})_{i \in \mathbb{N}}$ in $c_0$ such that it does not converge to am element in $c_0.$

First of all, I have trouble seeing $\xi^{(i)}.$ Should I have two indices for each $\xi^{(i)}?$ Because $\xi^{(i)} \in c_0$, meaning that $\lim_{n \rightarrow \infty}\xi^{(i)}_n = 0$ for each $i.$

I would be appreciated if someone can provide me a hint to solve the problem.

Answer 1

There is an even more fundamental reason why $c_0$ cannot be a Banach space with the $|| . ||_{2}$ norm, which is that some elements of $c_0$ do not have a finite $|| . ||_2$ norm!

For example, $$ \xi = (1, \tfrac 1 {\sqrt 2}, \tfrac 1 {\sqrt 3}, \tfrac 1 {\sqrt 4}, \dots ) \in c_0, $$ but $$ \sum_{n=1}^\infty \left( \tfrac 1 {\sqrt{n}} \right)^2= \infty.$$

Answer 2

Yes, we can't put l-2 norm in C0. As for any 1≤p< inf lp is a proper subspace of C0. Although C0 is a closed subspace of l-inf. So the question seems to be improper.