# All Hilbert spaces are isometric to $l^2(E)$ - how?

I was reading about Hilbert spaces and came across this line on Wikipedia:

By choosing a Hilbert basis (i.e., a maximal orthonormal subset of $$L^2$$ or any Hilbert space), one sees that all Hilbert spaces are isometric to $$ℓ^2(E)$$, where $$E$$ is a set with an appropriate cardinality.

My questions are:

1. What does the $$E$$ stands for? Is it the basis of $$l^2$$?
2. What is meant by "appropriate cardinality"?
3. Why is $$l^2$$ isometric to any other Hilbert space? Yes, the norm on $$l^2$$ is square root of sum of squares. But how do we know the isometry applies, even when some Hilbert spaces have elements with n coordinates, while $$l^2$$ has infinite coordinates? (Because elements are sequences and each sequence is infinite - has infinite "coordinates").

Thank you very much for your insights.

• $E$ is a set and $\ell^2(E)$ denotes the set of functions $f:E\rightarrow \mathbb{C}$ where $\sum_{n\in E}|f(n)|^2<\infty$. The isometry is related to the inner product of the Hilbert space and how linear functionals of Hilbert spaces are induced by the inner product. I don't have details off the top of my head, but you can see Chapter 5 of Folland's Real Analysis text.
– user443408
Aug 1, 2021 at 16:54
• $E$ can be chosen as the maximal orthonormal set. Aug 1, 2021 at 17:17
• Could you please give the link to the article you are referring to? Aug 1, 2021 at 17:57
• @Filippo en.wikipedia.org/wiki/Lp_space Aug 1, 2021 at 18:35
• @TerezaTizkova Thank you! I have added the link to your question. Aug 1, 2021 at 20:00

This is explained in Rudin's Real and Complex Analysis (3rd edition) Chapter 4, in the part of Orthonormal Sets. See pages 86 and 87 for more details. Recall that $$\ell^{2}(A)=L^{2}(A,\#),$$ where $$\#$$ is the counting measure in $$A$$. In particular, $$\ell^{2}(\mathbb{N})=\{y=(y_{n})_{n \in \mathbb{N}};\|y\|_{2}<\infty\}$$.

1. $$E$$ is the set of indexes used to index a orthonormal set which is maximal (with respect to inclusion).
2. The cardinality of $$E$$ (?). For example: for $$\mathbb{R}^{n}$$ you have $$|E|=n$$, and for $$\ell^{2}(\mathbb{N})$$ you have $$|E|=\aleph_{0}$$.
3. Let $$\{u_{\alpha}\}_{\alpha \in A}$$ be a maximal orthonormal set in the Hilbert space $$H$$. The isomorphism is given by \begin{align*} \varphi:H &\to \ell^{2}(A), \\ x & \mapsto \hat{x}, \end{align*} where $$\hat{x}(\alpha)=\langle x,u_{\alpha} \rangle.$$
• That is Parseval's theorem. Many books, even Rudin, give the proper attribution. Aug 1, 2021 at 17:19
• @Sebathon Thank you for your answer! May I ask, in #3., how is the isomorphism $\varphi$ connected to isometry? I searched for Parseval´s theorem, but only see it refering to Fourier transform. Does isomorphism imply isometry here? How do we check the metrics are preserved under $\varphi$? Aug 12, 2021 at 21:02
• @TerezaTizkova the version of Parseval's theorem that you are looking for is the following (Rudin's book Thm 4.18, page 85. See also en.wikipedia.org/wiki/Parseval%27s_identity and encyclopediaofmath.org/index.php?title=Parseval_equality): let $\{u_{\alpha}\}_{\alpha \in A}$ be an orthonormal set in $H$. Then for every $x,y \in H$ we have $$\sum_{\alpha \in A} \hat{x}(\alpha) \overline{\hat{y}(\alpha)}=\langle x,y \rangle_{H}.$$ Notice that the LHS is equal to $\langle \hat{x},\hat{y} \rangle_{\ell^{2}}$. Thus, $\varphi$ is an isometry. Aug 12, 2021 at 22:09

Every Hilbert space $$H$$ admits an orthonormal basis $$\{v_i : i \in E \}$$. Then $$H$$ is isometric to $$\ell^2(E)$$.

EDIT: The $$E$$ is dependent upon the Hilbert space. Moreover, if $$E, F$$ are two abstract sets, then $$\ell^2(E)$$ and $$\ell^2(F)$$ are isometric if and only if $$E, F$$ are of the same cardinality. The point of this question is that every Hilbert space can be represented as $$\ell^2$$ over an (essentially) unique set $$E$$.

• You are answering by stating the problem as a fact. Of course this is true, and the result has a name or attribution: Parseval's theorem. Aug 1, 2021 at 17:15
• @OliverDiaz I believe OP’s confusion was with how -if you don’t see that $E$ depends on $H$- then this statement seems to imply that every Hilbert space is isometric to every other.
– AJY
Aug 1, 2021 at 17:18
• I believe I answered all three questions.
– AJY
Aug 1, 2021 at 17:19
• I don't thinks is being done correctly. At least construct the map from $H$ to $\ell^2(E)$. Parseval's implies a unique representation $x=\sum_{n\in E}\langle x,e_n\rangle e_n$, where $\{e_n:n\in E\}$ is a maximal orhonoprmal system (this exists by Zorn's lemma). Define $X:H\rightarrow\mathbb\ell^2(E)$ be letting $X(x)$ be the map from E to $\mathbb{C}$ given by $X(x)(n)=\langle x,e_n\rangle$, Show that this is indeed an isomorphic isometry. Aug 1, 2021 at 17:27
• @OliverDiaz I see you already have, and I see no point in emulating an answer that already exists on the same post. My hope is that my answer clears up any remaining confusions for OP, who is the actual arbiter of what is or isn’t helpful.
– AJY
Aug 1, 2021 at 17:39

A theorem by Parseval states that

Theorem: Any Hilbert $$H$$ space admits a maximal collection $$\mathscr{E}\subset H$$ of orthonormal vectors such that $$H=\overline{\operatorname{span}(\mathscr{E})}$$. Moreover, $$x\in H$$ can be expressed uniquely as $$x=\sum_{\boldsymbol{e}\in\mathscr{E}}\langle x,\boldsymbol{e}_\alpha\rangle \boldsymbol{e}_\alpha,\qquad \|x\|^2=\sum_{\boldsymbol{e}\in\mathscr{E}}|\langle x,\boldsymbol{e}\rangle|^2$$

The convergence of the sums above are in terms of nets over finite subsets of $$\mathscr{E}$$ (see note 1 below; also, this posting may be helpful)

Q1. For any nonempty set $$R$$, the space $$\ell^2(R)$$ is the collection of all functions $$X:R\rightarrow\mathbb{C}$$ such that $$\sum_{r\in R}|X(r)|^2<\infty$$ where as above, convergence is in the sense of nets. It can be seen that on $$\ell^2(R)$$, the map $$(\cdot,\cdot):\ell^2(R)\times\ell^2(R)\rightarrow\mathbb{C}$$ given by $$(X,Y)=\sum_{r\in R}X(r)\overline{Y(r)}$$ defines an inner product. The space of interest for the OP is $$\ell^2(\mathscr{E})$$, where $$\mathscr{E}$$ is a maximal orthonormal collection in the Hilbert space $$H$$.

Q2. For each $$\boldsymbol{e}\in\mathscr{E}$$, the map $$E_\boldsymbol{e}:\mathscr{E}\rightarrow\mathbb{C}$$ defined as $$E_\boldsymbol{e}(\boldsymbol{e}')=\delta_{\boldsymbol{e},\boldsymbol{e}'}$$ is clearly an element of $$\ell^2(\mathscr{E})$$; moreover, for any $$\boldsymbol{e},\boldsymbol{d}\in\mathscr{E}$$ $$(E_{\boldsymbol{e}},E_{\boldsymbol{d}})=\delta_{\boldsymbol{e},\boldsymbol{d}}$$ The dimension of $$\ell^2(\mathscr{E})$$ is defined as the cardinality of $$\mathscr{E}$$.

Q3. Appealing to Parseval's theorem and Bessel's inequality, one can se that the map $$\Lambda:H\rightarrow\ell^2(\mathcal{E})$$ defined as $$(\Lambda x)(\boldsymbol{e})=\langle x,\boldsymbol{e}\rangle$$ is a linear isometric isomorphism between $$H$$ and $$\ell^2(\mathcal{E})$$, that is $$\Lambda$$ is onto and $$(\Lambda x,\Lambda x)=\|x\|^2=\sum_{\boldsymbol{e}\in\mathscr{E}}|\langle x,\boldsymbol{e}\rangle|^2$$

Notes:

1. The potentially uncountable sums are understood in as limits of nets (instead of sequences). Order the collection $$\mathcal{F}(\mathcal{E})$$ of all finite subsets of $$\mathcal{E}$$ by inclusion. A function $$\xi:\mathcal{F}(\mathcal{E})\rightarrow H$$ (or net in $$\mathcal{C}(\mathcal{E})$$) converges to $$h\in H$$, iff for any $$\varepsilon>0$$ there is $$C_\varepsilon\in\mathcal{F}(\mathcal{E})$$ such that for all $$D\in\mathcal{F}(\mathcal{E})$$, $$C_0\subset D$$ implies $$|\xi(D)-h|<\varepsilon$$

2. When $$H$$ is separable, one can take $$\mathcal{E}=\mathbb{N}$$ and all the net abstract nonsense reduces to the familiar real of sequences.

3. It is possible to define $$\ell^2(\mathscr{E})$$ as the $$L^2$$ space of a measure space. Define $$\mathcal{P}(\mathscr{E}$$ as the collection of all subsets of $$\mathscr{E}$$, and $$\mu$$ is the counting measure on $$(\mathscr{E},\mathcal{P}(\mathscr{E}))$$, that is, $$\mu(A)=n\in\mathbb{Z}_+$$ if $$A$$ is finite and has $$n$$ elements, and $$\mu(A)=\infty$$ otherwise. Then, it can be shown that $$\ell^2(\mathscr{E})$$ is the same as $$L_2(\mathscr{E},\mathcal{P}(\mathscr{E}),\mu)$$.

4. It can be show that if $$A$$ and $$B$$ are sets of the same cardinality, then $$\ell^2(A)$$ and $$\ell^2(B)$$ are isometric isomorphic.