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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})?

JustWannaKnow
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  • Use the Japanese bracket instead of the Euclidean norm. Also all you need is a bijection that grows polynomially (as well as its inverse). It is a reflection of the fact volumes of spheres being $\sim r^{d-1}$ grow polynomially in the radius. Here this holds in a discrete fashion. You can split $\mathbb{Z}^d$ in terms of spheres $S(r)$ for $\ell^{\infty}$ norm, i.e., boundaries of cubes. Then you build $T$ by enumerating the elements of $S(0)$, and then $S(1)$, $S(2)$, etc. – Abdelmalek Abdesselam Nov 16 '21 at 14:24
  • @AbdelmalekAbdesselam what is the Japanese bracket? Your next suggestion seems like a very smart trick, I will try to fill in the details and post an answer. – Just dropped in Nov 16 '21 at 15:11
  • The Japanese bracket is $\langle x\rangle=\sqrt{1+|x|^2}$. I don't know where the name comes from. The trick is not smart at all. If you unpack the definitions and see what conditions you need on the relabeling bijection you immediately see the polynomial growth requirement. Then the construction with pealing off spheres is the first thing that comes to mind. – Abdelmalek Abdesselam Nov 16 '21 at 16:15
  • @AbdelmalekAbdesselam does the 1 play any role here? I tried with $\langle x \rangle = \sqrt{x_{1}^{2}+\cdots + x_{d}^{2}}$ (just changed the notation), and I've sketched some ideas but I don't know how to complete the proof. For instance, I took $T$ as you mentioned, defined, for each $n \in \mathbb{N}$ the spheres $S(n) := {x \in \mathbb{Z}^{d}: ||x||_{\infty} = n}$ and defined $T$ as an enumeration of $\mathbb{Z}^{d}$ by following your rule $S(0),S(1),...$. I haven't got any explicit map from this (but I don't think this is the idea, right?) but I noticed that (continues) – JustWannaKnow Nov 17 '21 at 09:10
  • if $x \in \mathbb{Z}^{d}$ is such that $Tn_{0} = x$, then $||x||{\infty} \le n{0}$. This, for instace, can get you the following. $ K \ge \sum_{n\in \mathbb{N}}n^{2m}|\varphi(n)|^{2} \ge n_{0}^{2m}|\varphi(n_{0})|^{2} \ge ||x||_{\infty} ^{2m}|\varphi(x)|^{2} \ge C \langle x \rangle^{2m}|\varphi(x)|^{2}$, which is condition (\ref{1}). However, this holds only if $m \ge 1$. But maybe I'm getting the wrong/most difficult path. If you have the time, could you give me some more hints, please? – JustWannaKnow Nov 17 '21 at 09:15
  • The norms with $m<0$ are useless and can throw them away completely. It would take too long to explain why it is better to use $\langle\cdot\rangle$ instead of the Euclidean norm. All I can do is to urge you to study https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-mathscr-d-omega/3511753#3511753 on the example of sequence spaces. For practice, show that you can use $\ell^p$ norms, $p\in[1,\infty]$, to give equivalent systems of seminorms, i.e., which define the same locally convex topology. – Abdelmalek Abdesselam Nov 17 '21 at 12:48
  • @AbdelmalekAbdesselam you mean consider condition (\ref{2}) only for $m \ge 0$? Tks for the link btw. – JustWannaKnow Nov 17 '21 at 13:24
  • I misread you question. I thought you were trying to show isomorphism of the spaces $s$ of rapidly decaying sequences or multisequences, whereas you are considering their duals $s'$, i.e., (multi)sequences of at most polynomial growth. In that case these are wrong norms to use. You would need, for say your favorite $p\in[1,\infty]$, $\ell^p$ norms weighted by nonnegative sequences in $s$. This is an essentially uncountable collection of seminorms which reflects the fact $s'$-type spaces are nonmetrizable. – Abdelmalek Abdesselam Nov 17 '21 at 17:42
  • The question is not entirely clear to me. Is the bijection given or can you choose it as you wish? Are you asking only about (1)$\implies$(2) or also about (2)$\implies$(1)? – supinf Nov 19 '21 at 13:01
  • I'm trying to prove $(1) \iff (2)$ – JustWannaKnow Nov 19 '21 at 13:38
  • Can you choose $T$ as you wish? Does it have to be the same for both directions? – supinf Nov 24 '21 at 11:12
  • In the notation of my post $T$ is just an enumeration of $\mathbb{Z}^{d}$ because I thought it was just a matter of rewritting the indices. But it seems more natural to construct a mapping $T$ from one space of sequences to the other, in which case $T$ must hold for both directions indeed. – JustWannaKnow Nov 24 '21 at 13:11
  • I think the correctness of the result probably depends on how this $T$ looks. For some enumerations $T$, the claim is probably false. – supinf Nov 24 '21 at 14:26

1 Answers1

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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 $c=d$.

supinf
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  • The key question now is whether there is a $T$ satisfying those growth conditions. I tried to show those conditions for a few well-known enumerations and failed. – Ewan Delanoy Nov 24 '21 at 16:13
  • @EwanDelanoy I have added a paragraph to explain existence of such a $T$. Which $T$ have you tried? – supinf Nov 25 '21 at 11:27
  • @supinf I am sorry I wasn't able to read and study your answer carefully yet because I have been really busy these days but I appreciate the answer and details. One little thing that caught my attention was about the existence of $T$. When you say to sort all points $x \in \mathbb{Z}^{d}$ according to its norm and get an increasing sequence out of it, do you mean a non-decreasing sequence? Because we can have distinct points with the same norm which, in this case, would be consecutive elements of the sequence right? – JustWannaKnow Nov 28 '21 at 20:29
  • @IamWill yes, you are right, it is not strictly increasing. And elements with the same norm are consecutive elements. – supinf Nov 28 '21 at 22:54
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    @supinf now I have worked out the details of your answer and it seems pretty nice to me! Thanks you again. Just a quick comment: when you say that $\varphi(0) = 1 $ and $\varphi(x) = 0$ for $x \neq 0$ satisfies (2) but not (1), this is true indeed; this is probably the reason why Abdelmalek told me to use $\langle x \rangle := \sqrt{1 + ||x||^{2}}$ instead of $||x||$ in (1). This $\langle x\rangle$ norm allows us to consider the case $x = 0$ and the arguments in your answer remain true with basically no modifications. – JustWannaKnow Nov 30 '21 at 11:41