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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 $L^2([a,b])?$

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    $\begingroup$ @LeechLattice Actually, $L^2(\mathbb{R})$ is complete. For any measure space $(X,\Sigma,\mu)$ and for any $p\in [1,\infty]$ the space $L^p(X)$ is a Banach space. Read the comments and answers to the question in your link very carefully. $\endgroup$
    – Mark
    Commented Jul 30, 2020 at 17:40
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    $\begingroup$ Rather than trying to construct individual isometries by hand, you should try to prove that any two infinite dimensional separable Hilbert spaces are isometric. You can do this by showing that any infinite dimensional separable Hilbert space $H$ is isometric to $\mathbb \ell^2(\mathbb N)$. To do this, let $(h_i)_{i=1}^\infty$ be a complete orthonormal system for $H$ (prove that such a thing exists!), and let $(e_i)_{i=1}^\infty$ be one for $\ell^2(\mathbb N)$, and consider the continuous linear map that sends $h_i$ to $e_i$ for each $i$. $\endgroup$ Commented Jul 30, 2020 at 17:55
  • $\begingroup$ You are right, i've already proved it,but $L^2$ is a space with functions and $\ell ^2$ a space of sequences, can we claim that they are isomorphic? ,thank you $\endgroup$ Commented Jul 30, 2020 at 18:01
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    $\begingroup$ @John Mars Why should it matter how we call the elements? Obviously, all Hilbert spaces are different from each other, but they still might be isomorphic. $\endgroup$
    – Mark
    Commented Jul 30, 2020 at 19:29
  • $\begingroup$ Thank you, now its clear. $\endgroup$ Commented Jul 30, 2020 at 19:32

3 Answers 3

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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 $\ell^2(\Bbb Z)$

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    $\begingroup$ should be "only one separable INFINITEDIMENSIONAL Hilbert space ...", obviously $l^2(\{1,...,n\})$ is separable. $\endgroup$ Commented Jul 30, 2020 at 18:02
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    $\begingroup$ @DominikKutek Do we really bother about finite dimensional spaces while doing Analysis? $\endgroup$
    – Brozovic
    Commented Jul 30, 2020 at 18:03
  • $\begingroup$ @Brozovic I know that , but 𝐿2 is a space with functions and ℓ2 a space with sequences, can we claim that theorem in this case? Thank you. $\endgroup$ Commented Jul 30, 2020 at 18:46
  • $\begingroup$ @JohnMars I don't see why you can't. All I am concerned is with that both are separable Hilbert spaces hence both have countable Orthonormal bases. $\endgroup$
    – Brozovic
    Commented Jul 30, 2020 at 18:50
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    $\begingroup$ @Brozovic: Do we really bother about finite dimensional spaces while doing Analysis? Yes, we do. $\endgroup$ Commented Jul 31, 2020 at 2:21
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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 $\ell^2(\mathbb Z)$.

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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 $\ell^2(\mathbb{Z})$.

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