What is $l^2(\mathbb{Z}^2)$?

Question

From a textbook in Harmonic Analysis:

What is meant by the term "$l^2(\mathbb{Z}^2)$", and how is it different from $l^2(\mathbb{Z})$?

Answer 1

By definition $l^2(\Bbb Z^2)$ is the set of square-summable families of complex numbers $\{u_{m,n}\}_{(m,n)\in\Bbb Z^2}$.

By square-summable we mean $$\sum |u_{m,n}|^2<\infty$$

It is endowed with an inner product the same way as $l^2(\Bbb Z)$ which makes it a separable Hilbert space (the proof is identical to the $\Bbb Z$ case).

Fact: there is only one separable infinite-dimensional Hilbert space, up to isomorphism, so $l^2(\Bbb Z^2)$ has to be isomorphic to $l^2(\Bbb Z)$. One way to think about this is the following:

Theorem:

Let $\{u_i\}_{i\in I}$ be a sequence of non-negative real numbers indexed by a countable set $I$. Then the non-decreasing sequence $$S_N=\sum_{n=1}^N u_{\varphi(n)}$$ has the same limit (hence denoted by $\sum_{i\in I}u_i $) for all bijections $\varphi:\Bbb N\to I$.

In other words, the order in which you add the elements does not affect the status of convergence of the series, or its sum.

Therefore, since $\Bbb Z$ and $\Bbb Z^2$ are both countable, any bijection between them will induce an isomorphism between $l^2(\Bbb Z)$ and $l^2(\Bbb Z^2)$.

In some cases, however, it may seem more natural to consider one notation rather than the other. But both are really simply spaces of countable sequences whose square is an absolutely convergent series.