A Banach space of sequences where convergence to zero depend only on their norm

Question

Does there exist a Banach space $(C,\|.\|)$ with $C\subset\mathbb{R}^{\mathbb{Z}}$ satisfying:

for each $\varepsilon>0$, there exists $k\in\mathbb{N}$ such that if $v=(v_n)_{n\in\mathbb{Z}}\in C$ has $\|v\|\leq1$, then $$|v_n|\leq\varepsilon \,\,\, \text{whenever} \,\,\, |n|\geq k?$$

Roughly speaking this means that the rate of convergence of sequences in $C$ depend basically on their norm.

Answer 1

Yes.

For example, for every $w>0$, $$ \ell^2_w(\mathbb Z)=\Big\{(a_k)_{k\in\mathbb Z}: \sum_{k\in\mathbb Z}w^{2|k|}|a_k|^2<\infty\Big\} $$ is a Hilbert space (and hence Banach space), with inner product $$ \langle(a_k),(b_k)\rangle=\sum_{k\in\mathbb Z}w^{2|k|}a_kb_k. $$ If $w>1$ and $\|(a_k)\|\le 1$, then $|w^{|k|}|a_k|\le 1$, and hence $$ |a_k|<\varepsilon, $$ if $w^{|k|}>1/\varepsilon$ or $|k|>\frac{\log(1/\varepsilon)}{\log w}$.