separable Hilbert spaces are bijective with ell-2

Theorem

If $H$ is an infinite-dimensional separable Hilbert space, then $H$ is isometrically isomorphic to $ℓ^{2}$ .

ie, there exists a bijective bounded linear operator $T : H \to ℓ^{2}$ such that for all $u, v \in H$ we have

| | T u | |_{ℓ^{2}} = | | u | |_{H} and ⟨ T u, T v ⟩_{ℓ^{2}} = ⟨ u, v ⟩_{H}

The finite case is easier to deal with- we can just show that they are isometrically isomorphic to

C^{n}

for some

n

Proof (sketch)

\begin{aligned} u & = \sum_{n = 1}^{\infty} ⟨ u, e_{n} ⟩ e_{n} \\ ⟹ | | u | | & = {(\sum_{n = 1}^{\infty} | ⟨ u, e_{n} ⟩ |^{2})}^{1 / 2} \end{aligned}

By Parseval's identity. So we can define our map $T$ as

T u := {⟨ u, e_{n} ⟩}_{n}

ie, $T u$ is the sequence of coefficients in the expansion. And this sequence is in $ℓ^{2}$ (ell-2).

$T$ is linear by construction,
It is surjective because every sum $\sum_{n = 1}^{\infty} c_{n} e_{n}$ is Cauchy in $H$
It is one-to-one because every $u$ is expanded in this way, so any two expansions that are the same will be have the same infinite sum.

References