Is $L^2(\mathbb R)$ isometrically isomorphic with $\ell^2(\mathbb Z)?$

Question

Is $L^2(\mathbb R)$ isometrically isomorphic with $\ell^2(\mathbb Z)?$

My thoughts:
We can define an operator $\mathcal L:L^2(\mathbb R)\rightarrow \ell^2(\mathbb Z)$ : $\mathcal Lf=\{\hat f(ξ)\}_{ξ\in \mathbb Z}$
(obviously $\mathcal L$ is linear & $1-1$ by uniqueness) and by the Parseval identity we have that $\lVert f\rVert_{L^2(\mathbb R)}^2=\lVert \hat f\rVert_{\ell^2(\mathbb Z)}^2$ Hence we have an isometry.
is that enough?
Also, can we claim that $:L^2(\mathbb R)≅ \ell^2(\mathbb Z)?$
Thanks you.
EDIT: how about $L^2([a,b])?$

Answer 1

Recall that $L^2(\Bbb R)$ is a separable Hilbert space and hence has a countable Orthonormal Basis say $\{e_n\}_{n \in \Bbb Z}$. This implies that $L^2(\Bbb R)$ is isometrically isomorphic to $\ell^2(\Bbb Z)$, by the linear map $f \mapsto \{\langle f,e_n\rangle\}_{n \in \Bbb Z}$ . For more clarity and details of the argument check here. As a matter of fact, there exists only one separable infinite-dimensional Hilbert space upto isometric isomorphism, namely $\ell^2(\Bbb Z)$

Answer 2

The normalized Hermite functions form an orthonormal basis of $L^2(\mathbb{R})$. These are $$ H_n(x)=(-1)^n(2^n n!\sqrt{\pi})^{-1/2}e^{x^2/2}\frac{d^n}{dx^n}e^{-x^2},\;\;\; n=0,1,2,3,\cdots. $$ $\{ H_n \}$ is a countable orthonormal basis of $L^2(\mathbb{R})$. It is not hard to use this in order to map $L^2(\mathbb{R})$ unitarily onto $\ell^2(\mathbb{Z})$.

Answer 3

Another explicit example of an orthonormal basis for $L^2(\mathbb R)$ is $(e^{2\pi i m x} I_{n \le x < n+1})_{(m,n) \in \mathbb Z\times\mathbb Z}$. And since $\mathbb Z \times \mathbb Z$ is in one to one correspondence with $\mathbb Z$, this gives another isometry from $L^2(\mathbb R)$ to $\ell^2(\mathbb Z)$.