Are there examples of sequences $(x_n)$ such that $(x_n) \in l^p(\mathbb{N})$ but $(x_n) \not\in l^p(\mathbb{Z})$?

Question

I'm curious about whether/how a bi-directional sequence can have stricter conditions for convergence than a sequence over $\mathbb{N}$. I assume that this is possible, but haven't been able to generate any examples on my own.

Answer 1

It's not possible. A bijection $\varphi:\mathbb{N} \to \mathbb{Z}$ induces an isomorphism $\ell^p(\mathbb{N}) \to \ell^p(\mathbb{Z})$ given by $(x_i)_{i \in \mathbb{N}} \mapsto (y_j = x_{\varphi^{-1}(j)})_{j \in \mathbb{Z}}$, sending each entry at position $i$ to become the entry at position $\varphi(i)$. You may check that this is an isometry by writing the definition of the measure in each side, and noting that we may always rearrange sums of positive numbers.

There is some ambiguity in the question, and maybe you meant a restriction to half a sequence instead, in the following sense: If you take the sequence $(x_n)_{n \in \mathbb{Z}}$ with $x_n = 0$ whenever $n \geq 0$ and $x_n = 1$ whenever $n < 0$, then the restricted sequence $(x_n)_{n \in \mathbb{N}}$ is in $\ell^p(\mathbb{N})$ while the full sequence is not in $\ell^p(\mathbb{Z})$.