Are All Hilbert Spaces Isomorphic to a Subspace of Another?

[benorin]

The questions reads:

If H₁ and H₂ are two Hilbert spaces, prove that one of them is isomorphic to a subspace of the other. (Note that every closed subspace of a Hilbert space is a Hilbert Space.)

What I'm thinking is that every separable Hilbert space is isomorphic to L². If I recall, a Hilbert space is separable iff it has a countable basis, so what of uncountable bases? Do I then consider a subspace of that space spanned by a countable subset of the uncountable basis?

 

[matt grime]

What has separability got to do with it?

Given two vector spaces V and W one is isomorphic to a subspace of the other.

 

[benorin]

Duh! Thanks Matt.

 

[Hurkyl]

(Don't forget you have slightly more to prove here -- a Hilbert space has a little more structure than a vector space!)

 

[HallsofIvy]

What has separability got to do with it?

Given two vector spaces V and W one is isomorphic to a subspace of the other.

Is that really true? Certainly if V and W are finite dimensional then the one with smaller dimension is isomorphic to a subspace of the other. But I thought "If two spaces have the same dimension then they are isomorphic" was only true for finite dimensional spaces.

 

[matt grime]

A vector space is determined up to isomorphism by the cardinality of its basis, and since the cardinals are well ordered (modulo the axiom of choice) I think we're okay.

I am leaving a lot unanswered (this is just my thought as to how to start the proof), since there is more structure to verify than is 'a vector space' but as long as the image is closed, then we're ok.

 

[benorin]

The definition of a Hilbert space isomorphism is:

Two Hilbert spaces, say H₁ and H₂, are isomorphic iff there exists a one-to-one linear mapping $$\Lambda$$ of H₁ onto H₂ such that $$\left( \Lambda x, \Lambda y\right) = (x,y), \\ \forall x,y\in H_1$$.

 

[LinnsburghF]

A vector space is determined up to isomorphism by the cardinality of its basis, and since the cardinals are well ordered (modulo the axiom of choice) I think we're okay.

I am leaving a lot unanswered (this is just my thought as to how to start the proof), since there is more structure to verify than is 'a vector space' but as long as the image is closed, then we're ok.

I agree with HallsOfIvy. The polynomials with domain [0,1] form a vector space with countable basis, as does the l² space of the natural numbers. These vector spaces are not isomorphic, since the first is not complete. A Hilbert space is determined up to isomorphism by the cardinality of its basis.

My suggestion for the proof is as follows: If H is an arbitrary Hilbert space over the field F with orthonormal basis B, show that H is isomorphic to the l² space of B. The isomorphism is exactly what you'd expect. To define a linear map, you need only define it on the basis B. If v is a basis element, let T take v to the function f_v, where
f_v:B -> F is just the characteristic function of {v}, so f_v(u)=1 if u=v, and f_v=0 otherwise. This is a linear map. Surjectivity is pretty easy from the definition of an l² space, injectivity comes from the fact that B is a basis, and the isometry <Tu,Tv>=<u,v> is immediate once you write as a combination of basis vectors.

Once you've done this, you've shown that a Hilbert space is determined up to isomorphism by the card of its basis. Then for two Hilbert spaces H,K, assume H has the lesser dimension. Take an orthonormal set S in K with cardinality equal to the cardinality of an ONB for H. Once you've established that span S is closed, it's a Hilbert space, and you're done.

 

[Hurkyl]

Welcome to PF, LinnsburghF!

While we do appreciate having more people to give homework help, we strongly discourage giving complete (or near-complete) solutions here -- the student will learn much more if they did the exercise themselves (with our help) rather than watch someone else do the exercise (which is why it was given as an exercise in the first place!)... and giving answers amounts to cheating anyways.

But the question is nearly four years old, so no harm done!