How can I convert "norms" using the bijection between $\mathbb{N}$ and $\mathbb{Z}^{d}$? Suppose $\varphi$ is a sequence $\varphi = \{\varphi(x)\}_{x\in \mathbb{Z}^{d}}$ satisfying the following condition. There exists $k \in \mathbb{N}$ and $C \ge 0$ such that:
$$|\varphi(x)| \le C ||x||^{k} \tag{1}\label{1}$$
for every $x \in \mathbb{Z}^{d}$. Here, $||x|| = \sqrt{x_{1}^{2}+\cdots x_{d}^{2}}$.
Since $\mathbb{Z}^{d}$ is countable, I can actuallt treat $\varphi$ to be a sequence indexed by $\mathbb{N}$ instead of $\mathbb{Z}$, say, $\varphi = \{\varphi_{n}\}_{n\in \mathbb{N}}$. This is what I'm trying to prove.
Claim 1: There exists $m \in \mathbb{Z}$ such that:
$$\sum_{n\in \mathbb{N}}n^{2m}|\varphi_{n}|^{2} < \infty \tag{2}\label{2}.$$
Claim 2: If (\ref{2}) holds, then (\ref{1}) also hold.
Of course, I have to use condition (\ref{1}) to get (\ref{2}) and vice-versa, but I don't know how exactly does the $||x||$ is afected by the bijection $T: \mathbb{N} \to \mathbb{Z}^{d}$. In other words, when $\{\varphi(x)\}_{x\in \mathbb{Z}^{d}}$ becomes $\{\varphi_{n}\}_{n\in \mathbb{N}}$, how is condition (\ref{1}) changed, so I can use it to prove (\ref{2})?
 A: As an assumption, we will need that $T$ and its inverse grow at most polynomial,
that is, there exist a constants $c,\hat C>0$ with
$$
\| T(n) \| \leq n^c,
\forall n\in\Bbb N
\qquad
T^{-1}(x) \leq 1+\hat C\|x\|^c,
\forall x\in \Bbb Z^d.
$$
Such a $T$ exists, see below.
(1) $\implies$ (2):
We have
$$
\sum_{n\in\Bbb N} n^{2m} |\varphi_n|^2
=
\sum_{n\in\Bbb N} n^{2m} |\varphi(T(n))|^2
\leq
\sum_{n\in\Bbb N} n^{2m} C^2\|(T(n))\|^{2k}
\leq
\sum_{n\in\Bbb N} C^2 n^{2m} n^{2kc}
=C^2\sum_{n\in\Bbb N} n^{2m+2kc}.
$$
If we choose $m$ such that $2m+2kc<-1$, this sum will be finite.
(2) $\implies$ (1):
This is not true:
If we choose $\varphi$ such that $\varphi(0)=1$ and $\varphi(x)=0$ for all $x\neq0$,
then (1) is never satisfied, but (2) is satisfied.
However, if we ignore the $x=0$ case, then this direction can be shown:
From (2) we obtain $n^{2m}|\varphi_n|^2\leq1$ for large $n$,
or $|\varphi_n|\leq n^{-m}$.
Then, one has
$$
|\varphi(x)| = |\varphi_{T^{-1}(x)}|
\leq T^{-1}(x)^{-m}
\leq (1+\hat C\|x\|^c)^{-m}
\\\qquad
\leq (1+\hat C\|x\|^c)^{\max(-m,0)}
\leq \tilde C\| x\|^{\max(-cm,0)}
\leq \tilde C\| x\|^{k},
$$
for all but finitely many $x\in\Bbb Z^d$, where $\tilde C$ is a suitable constant
and $k>\max(-cm,0)$.
Existence of $T$ with growth conditions:
There are many ways to construct such a bijection.
One such possibility was already suggested in the comments.
A similar possibility is to sort all points $x\in\Bbb Z^d$ according to $\|x\|$,
which gives a sequence $x_n$ such that $\|x_n\|$ is non-decreasing.
Then we set $T(n)=x_n$.
For the first estimate we then have
$\|T(n)\| \leq \|(n,0,\dots,0)\| = n \leq n^d$.
For the second estimate, let $x\in\Bbb Z^d$ be given.
There exists less than $1+C \|x\|^d$ points
$y\in\Bbb Z^d$ with $\|y\|\leq \|x\|$,
where $C>0$ is a suitable constant.
Due to the sorting by the norm, this implies
$T^{-1}(x)\leq 1+C\|x\|^d$.
In summary, the growth conditions are satisfied with $c=d$.
