separable Hilbert spaces are bijective with ell-2

[[concept]]

Topics

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Theorem

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

ie, there exists a bijective bounded linear operator $T:H\to {\ell }^2$ such that for all $u,v\in H$ we have $||Tu|{|}_{{\ell }^2}=||u|{|}_H\text{ and }⟨Tu,Tv{⟩}_{{\ell }^2}=⟨u,v{⟩}_H$

The finite case is easier to deal with- we can just show that they are isometrically isomorphic to ${\mathbb{C}}^n$ for some $n$

Proof (sketch)

Since $H$ is separable, it has an orthonormal basis (Hilbert spaces are separable if and only if they have an orthonormal basis). And we can write the Fourier-Bessel series for each element $u\in H$:

u &= \sum_{n=1}^\infty \langle u, e_{n} \rangle e_{n} \\ \implies \lvert \lvert u \rvert \rvert &= \left( \sum_{n=1}^\infty \lvert \langle u, e_{n} \rangle \rvert^2 \right)^{1/2} \end{align}$$ By [[Concept Wiki/Parseval's identity]]. So we can define our map $T$ as $$T\,u := \{ \langle u, e_{n} \rangle \}_{n}$$ ie, $T\,u$ is the sequence of coefficients in the expansion. And this sequence is in $\ell^2$ ([[Concept Wiki/l-p vector space\|ell-2]]). - $T$ is [[Concept Wiki/linear function\|linear]] by construction, - It is [surjective](https://en.wikipedia.org/wiki/Surjective_function) because every sum $\sum_{n=1}^\infty c_{n }e_{n}$ is Cauchy in $H$ - It is [one-to-one](https://en.wikipedia.org/wiki/Injective_function) because every $u$ is expanded in this way, so any two expansions that are the same will be have the same infinite sum.

References

References

See Also

Mentions

Mentions

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