$\ell^2(\mathbb{Z}) \simeq \ell^2(\{ \sqrt{n} : n \in \mathbb{N}\})$

Here is a result of Viazovska from 2017:

There exists a collection of even Schwartz functions $ a_n: \mathbb{R} \to \mathbb{R}$ such that for any even function $ f: \mathbb{R} \to \mathbb{R}$ and any $ x \in \mathbb{R}$ we have: $ $ f(x) = \sum_{n = 0}^\infty a_n(x) \, f(\sqrt{n}) + \sum_{n=0}^\infty \hat{a}_n(x)\, \hat{f}(\sqrt{n}) $ $ where the right hand side converges absolutely.

Then she defines a space $ \mathfrak{s}$ of rapidly decaying sequences of numbers. There is an isomorphism between the space of even Schwartz functions and the kernel of an operator $ L$ :

  • $ \Psi: f \mapsto (f(\sqrt{n}))_{n > 0} \oplus (\hat{f}(\sqrt{n}))_{n > 0} $
  • $ L: (x,y) \mapsto \sum_{n \in \mathbb{Z}} x_{n^2} – \sum_{n \in \mathbb{Z}} y_{n^2}$

In Fourier Analysis class, we learn that $ L^2(\mathbb{R}/\mathbb{Z}) \simeq \ell^2(\mathbb{Z})$ is that correct? I think we also have that all Hilbert spaces are Unitary equivalent. Here’s the statement from Stein’s textbook on Real Analysis:

Cor 2.5 Any two infinite dimensional Hilbert spaces are unitarily equivalent.

The example is that $ \ell^2(\mathbb{N})\simeq \ell^2(\mathbb{Z})$ for all countable sets, the proof is to re-order the basis.

Then we sholud also have $ \ell^2(\mathbb{Z})\simeq $ for the square root $ M = \{ \sqrt{n} : n > 0, \; n \in \mathbb{Z}\}$ or even $ M = \{ \sqrt{m} + \sqrt{n} : m,n > 0 ,\; m,n \in \mathbb{Z} \}$ .

The above theorem discusses the behavior of elements of $ f \in L^2(\mathbb{R})$ on the subset of the square-roots of integers: $ \{ \sqrt{n} : n \in \mathbb{N}\} \subset \mathbb{R}$ and discusses the behaviors of $ f$ and $ \hat{f}$ combined.


What is it called when we handle both $ f \oplus \hat{f}$ combined? In physics the same wavefunction $ f$ could be evaluated at a point $ \langle f|x\rangle$ or a momentum $ <f|p\rangle = \hat{f}(x)$ and that $ (x,p) \simeq \mathbb{R}^2$ could be combined in to a single plane. Something like symplectic geometry.

The Hamiltonian of the Harmonic oscillator could be written $ H = x^2 + p^2$ and is symmetric under a rotations of the $ (x,p)$ plane. This “physics” approach is likely to come up short compared to Viazovska’s careful analysis. Could this theorem deduced using microlocal analysis?

Limit of sequence of vectors in $\ell^2$ with coefficients approaching $0$

Let $ \{v_m\}_{m \in \mathbb{N}} \subset \ell^2$ be a sequence in $ \ell^2$ over the complex plane $ \mathbb{C}$ such that: $ \{v_m\}_{m \in \mathbb{N}}$ is linearly independend and $ v_m \to v$

Let $ V= \operatorname{span} \{v_m\}_{m \in \mathbb{N}}$

Let $ \{u_p\}_{p \in \mathbb{N}} \subset V$ be a sequence in $ V$ such that $ u_p \to u$ so we have $ $ \forall p \in \mathbb{N}: u_p = \sum_{m=1}^\infty \left( a_{p,m} \cdot v_m \right) $ $ with $ a_{m,p} \in \mathbb{C}$ and for each fixed $ p \in \mathbb{N}$ there are only finitely many m with $ a_{p,m} \neq 0$

Further, we have for each fixed $ m \in \mathbb{N}$ $ $ \lim_{p \to \infty} a_{p,m} =0 $ $ My question is if it is true that: $ $ \lim_{p \to \infty} u_p = a \cdot v $ $ with $ a \in \mathbb{C}$

Thanks.