Question #114845
Prove that Hilbert space is seperable
1
Expert's answer
2020-05-12T18:28:55-0400

Theorem: A Hilbert space H is separable (that is, has a countable dense subset) if and only if it has one countable orthonormal basis if and only if every orthonormal basis for H is countable. 


Proof:     \implies: Suppose H is a separable and consider any orthonormal basis SHS\subset H. Then if ϕ1,ϕ2S\phi_1,\phi_2\in S and ϕ1=/ϕ2\phi_1{=}\mathllap{/\,}\phi_2 we conclude ϕ1ϕ2=(ϕ1ϕ2,ϕ1ϕ2)=2||\phi_1-\phi_2||=\sqrt{(\phi_1-\phi_2,\phi_1-\phi_2)}=\sqrt{2}. Thus SHS\subset H has the discrete topology.

Now HH a separable metric space     S\implies S is a separable metric space     S\implies S has a countable dense subset. But SS is the only subset of itself that is dense in SS (since SS has the discrete topology) and so SS must be countable.


    \impliedby: Suppose for the converse that there is one countable orthonormal basis S={ϕ1,ϕ2,...}S=\{\phi_1,\phi_2,...\} for HH. We look at the case K=R\Kappa=\R (HH is a real Hilbert space) first. It is quite easy to check that the sets

Dn={D_n=\{\sumnj=1n\atop j=1 qjϕj:qjQq_j \phi_j:q_j\in Q for 1jn}1\leq j\leq n\}

are each countable and that their union D=D=\bigcupn=1\infin\atop n=1DnD_n is countable and dense in HH. The closure of each DnD_n is easily seen to be the R\R-linear span of ϕ1,ϕ2,...,ϕn\phi_1,\phi_2,...,\phi_n and so the closure of DD includes all finite linear combinations \sumnj=1n\atop j=1xjϕjx_j\phi_j. But, each xHx\in H is a limit of such finite linear combinations. Hence the closure of DD is all of HH. As DD is countable, this shows that HH must be separable.

In the complex case (HH is a Hilbert space over K=CK=\Complex we must take qjQ+iQq_j\in Q+iQ instead, so that we can get all finite C\Complex-linear combinations of the ϕj\phi_j n the closure of DD (and there is no other difference in the proof). \blacksquare


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Comments

Assignment Expert
13.05.20, 16:15

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Sheela John
13.05.20, 07:07

Thank you so much for your help assignment expert

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