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Special Properties of Hilbert Spaces?

  1. Nov 6, 2014 #1

    WWGD

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    Hi All,
    AFAIK, the key property that separates/differentiates a Hilbert Space H from your generic normed inner-product vector space is that , in H, the norm is generated by an inner-product, i.e., for every vector ##v##, we have## ||v||_H= <v,v>_H^{1/2} ##, and a generalized version of the Pythagorean theorem is satisfied (Polarization Identity, Parallelogram Law). Why is this so special, and, what is it this property gives rise to that is not the case with more general inner-product normed vector spaces?
     
    Last edited: Nov 6, 2014
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  3. Nov 6, 2014 #2

    mathman

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    The most important distinction is that a Hilbert space is complete in the norm, i.e. every Cauchy sequence has a limit in the space. Note that the nom is defined by the square root of the inner product.
     
  4. Nov 6, 2014 #3

    WWGD

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    But isn't any metric/normed space as good as a complete space anyway, since one can always do a metric completion of the space? And , yes, I did define the norm as the square root of the inner-product, I don't understand your comment in that respect.
     
  5. Nov 7, 2014 #4

    dextercioby

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    With regards to the first post, you've got it a little wrong, or definitly not clear, see the exact comment in the post below.

    Mathematics tells us that a pre-Hilbert space is a pre-Banach space and this is just a consequence of algebra (nothing topological here), once you define the norm in the pre-Hilbert space in terms of its scalar product. But any pre-Banach is space is automatically a metric space, again this follows from algebra, once you define the metric with respect to the norm, thus with respect of the scalar product which originated the norm, hence this chain of spaces:

    [tex]\{\mbox{pre-Hilbert spaces}\} \subset \{\mbox{pre-Banach spaces}\} \subset \{\mbox{metric spaces}\} [/tex]

    The completion theorem already brings you into the realms of topology. A theorem of completion is valid for all types of spaces and you can only issue one proof, not 3, the other follow automatically.
     
    Last edited: Nov 7, 2014
  6. Nov 7, 2014 #5

    dextercioby

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    More specifically, it should have read: the key property that separates/differentiates a Hilbert Space H from your generic normed inner-product vector space is that H is complete with respect to the metric induced by the norm which in its turn was induced by the scalar product.
     
  7. Nov 7, 2014 #6

    WWGD

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    I am not sure I get your point; as far as I know, a pre-Hilbert space is a normed inner-product vector space , whose norm originates from an inner-product. Once we have a norm, this determines the topology ; we have a metric topology originating from the open sets of the metric defined by d(x,y):=||x-y|| . So I don't see how to define a pre-Hilbert space without appeal to the underlying metric, and the metric implies a topology. How do we define a Hilbert space without referring to the metric/topological properties?
     
  8. Nov 7, 2014 #7

    WWGD

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    But isn't any metric space as good as a complete space? Actually, in, say, L^p spaces for ## p \neq 2 ## , none
    of the norms are induced by an inner-product , i.e., there are no inner-products < , >_p so that || f ||_p =
    <f,f>_p for the standard norm ||f||_p := \Int (f^p)^{1/p}. This property is unique to Hilbert spaces. The completeness
    of a metric is not a special property, I believe, because any metric can always be made complete (while, of
    course, preserving the properties of the space).
     
  9. Nov 7, 2014 #8

    dextercioby

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    I believe we can't see eye to eye here. The properties from the definition of a pre-Hilbert space don't include the topology. The metric topology comes as a consequence, once you define the metric directly from the scalar product or indirectly through the norm. Actually, you don't need to give a pre-Hilbert space the metric topology. You can give it the so-called weak topology which only includes scalar products without appealing to the induced norm/metric. A comment in post #7 is interesting. It addresses the existence of (pre-)Banach spaces which are not (pre-)Hilbert.
     
  10. Nov 7, 2014 #9

    WWGD

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    We may be using different definitions; different authors have different layouts/assumptions. Still, re the original question: what do we get in a Hilbert space that we do not get in a regular inner-product vector space as a result of having a norm induced by an inner-product so that the norm is either complete or not ? i.e., in your layout, Dexter, what specially property do we get in Hilbert spaces that we do not get on other inner-product normed spaces?
     
    Last edited: Nov 7, 2014
  11. Nov 8, 2014 #10
    If you are going to say it's not important that it's complete because we can always make it complete, that seems to point to the conclusion that the important thing is that it is possible to complete it, thus reaping the benefits of completeness. However, you still have to keep in mind that all your theorems involving completeness are working in the bigger space, so that you are forced to refer to the bigger space to get things to work and not just work in the smaller space with the understanding that the completion takes care of everything. It's analogous to asking what's the big deal about the real numbers? We can just take the rationals and complete them. Well, that IS the real numbers.

    So, what are the benefits of completeness? Taking projections is one of the big things, and the whole theory, really (the spectral theorem involves integrating projection operators, whose existence is provided for by completeness, for example). I haven't thought through exactly where it's needed at every point in the theory, though.

    The key point, involving the construction of a minimizing vector going from a point to a subspace, involves the parallelogram identity to show that the sequence you construct is Cauchy and completeness to then get that it converges. This is how projection operators to a subspace are defined. Add the minimizing vector to the point and it takes it to its projection. Once you have your hands on that, a lot more things follow.
     
  12. Nov 8, 2014 #11

    dextercioby

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    @homeomorphic, if I'm not mistaking the uniqueness of the projection is needed to prove the Riesz-Fischer representation theorem for (anti-)linear bounded functionals on Hilbert spaces. So if completeness fails, then R-F would fail, right?
     
  13. Nov 8, 2014 #12

    Stephen Tashi

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    Does that phrase refer to a space with a norm and an inner product such that the norm is not "induced" by the inner product? If so, we need a clear definition of the relation "not induced by".
     
  14. Nov 8, 2014 #13
    I think so. The proof I have in mind definitely uses the existence of projections (onto the kernel and its orthogonal complement). There may be other proofs, but it should be possible to come up with a counter-example if you lack completeness.
     
  15. Nov 8, 2014 #14

    WWGD

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    I guess we need to have a predefined norm ||.|| we will work with. Then:

    This means that there is no inner-product < , > we can define ( meaning an operator defined on ## V \times V \rightarrow \mathbb R ## satisfying the axioms of an inner-product so that <v,v>^{1/2} =||v|| )

    For example, if in the L^p spaces we use the norm ## ||f||_p := (\int f^p)^{1/p} ## , we can show there is no inner-product <,> we can define so that ##<f,f>^{1/2} = \int f^p)^{1/p} ## . One way of showing this is using the result that there is an inner-product that gives rise to the norm iff the norm satisfies the polarization identity.
     
  16. Nov 8, 2014 #15

    WWGD

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    But I think this ends up being a semantic issue; the Hilbert space is the completion ( if necessary) of the inner-
    product space whose norm is induced by the inner-product. Yes, we would have to check that this property
    of the norm is preserved under the completion, but I think that is not too hard. So the bigger space here is the
    Hilbert space; the smaller one is the pre-Hilbert space. But then this depends on where/how you start defining terms.

    And, yes, I had forgotten the special property of the existence of a unique shortest distance between a point p and a subspace S , this being reached between p and its orthogonal projection into S.

    Dextercioby: Which of the versions of R-F are you referring-to ? Is it the completeness of L^p

    EDIT: It would seem, from the context of the discussion that you are referring to the Riesz
    representation theorem whereby every functional in H* can be represented/described by the inner-product against a fixed element x.
     
    Last edited: Nov 8, 2014
  17. Nov 8, 2014 #16

    mathman

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    A simple example may help clarify the issue.
    Let A=set of all finite series. An inner product and resultant norm can be defined on this space. A is not a Hilbert space and needs to include infinite series which have the property of finite sum of terms squared to be a Hilbert space.
     
  18. Nov 8, 2014 #17

    lavinia

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    The inner product allows the idea of projecting one function onto another.
    This can not be done for the other norms.

    There are many applications of this.

    - In probability, the inner product of two square summable (mean zero)random variables is their covariance. Orthogonal random variables are in most applications, independent.

    - a complete set of orthogonal norm 1functions can be used as a"basis" for the Hilbert space and the cosine of the angles of a function with this basis gives you a series approximation to the function as a linear combination of the basis functions(possibility infinite combination). An example is the Fourier series.
     
  19. Nov 8, 2014 #18

    WWGD

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    So you are saying that the set of finite sequences, or , equivalently, the subset S of l^2 of sequences that are eventually zero are a Pre-Hilbert space, which becomes a Hilbert space after it is completed metrically into l^2?

    I guess , e.g., we can have a sequence {a_n} with {a_n} zero after the n-th term, and this sequence would not
    converge to a sequence with finitely-many terms.

    Maybe I am missing something, but, what important property do we get in a Hilbert space that we do not get
    in either a Pre-Hilbert space, or in a normed inner-product space that is not a Pre-Hilbert space? Is it the
    projection operator that , given a subspace S and a point p, assigns to p a unique point p' so that ||p-p'||
    is minimal. Is there anything else, any other major property?
     
  20. Nov 8, 2014 #19

    WWGD

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    But there are inner-product spaces that are not Hilbert spaces. Do you mean the fact that the inner-product comes from the norm?

    And , is the existence of a maximal orthonormal set that becomes a Schauder basis unique to Hilbert spaces? Why is norm 1 special, can't we rescale any basis so that all basis vectors have norm 1? I think the Fourier series comes from the orthogonal projection of a function onto the span of the bases.

    Sorry if I am being dense, but I just don't get the points here.
     
  21. Nov 9, 2014 #20

    lavinia

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    - If the inner product space is complete then it is a Hilbert space by definition.

    -Completeness allows Cauchy sequences to converge.

    As has already been said, completeness together with the inner product leads to many results.

    For a Hilbert space with a countable complete basis, completeness guarantees that the Fourier series converges. This allows one to express a function in terms of an orthonomal basis.( In general this may be an infinite linear combination hence the need for completeness.)

    BTW: An orthonomal basis is used because convergence of the sum of the squares of Fourier coefficients easily implies that the Fourier series converges.

    If a countable basis is complete, I would imagine that it can always be adjusted to be orthonormal using the Gramm-Schmidt process inductively.
     
    Last edited: Nov 9, 2014
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