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Hilbert space, orthonormal basis

  1. Oct 27, 2013 #1
    My book says that "the countability of the ONS in a hilbert space H entails that H can be represented as closure of the span of countably many elements". I must admit my english is probably not that good. At least the above quote does not make sense to me. What is it trying to say?
    Previously it was talking about orthonormal bases in Hilbert spaces and the idea of maximiality:
    <g,e_k> = 0 for al k => g=0 (definition of maximality)
    Why is it we use this definition to characterize and orthonormal basis (e_k) and not that H=span(e_k) and how does it relate to the quote above?
     
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  3. Oct 27, 2013 #2

    Office_Shredder

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    If you just take a span of an orthonormal system, you are only allowed to have finite linear combinations. Usually H is the closure of the span - which means you can take infinite linear combinations as long as the sums converge.

    The quote by itself is a little odd as usually an orthonormal system just means you have a collection of vectors which are pairwise orthogonal and unit length - is the context that you have a countable orthonormal system which is maximal? Because then the quote makes more sense.
     
  4. Oct 28, 2013 #3
    it is. Can you elaborate? Im not sure I understand the idea of maximality. Why use this to define an ONB and not some kind of convergence condition on finite linear combinations?
     
  5. Oct 28, 2013 #4

    Office_Shredder

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    The point is that a maximal ONS has that every element of H is an infinite sum of your ONS elementals, which is a really nice thing to have. But we want to prove that we can do this without talking about the topology of the space, so it's really good to have a condition which is just a linear algebra condition.
     
  6. Oct 28, 2013 #5
    Not sure I understand yet:
    For an ONS is:
    ONS is maximal <=> for all g in H, g=Ʃ<g,e_k>e_k
    But not in general H=lim_n->∞[span{(e_k)}] - or does this last statement even make sense?
     
  7. Oct 28, 2013 #6

    Office_Shredder

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    By
    [tex] \lim_{n\to \infty} span( \{ e_1,....,e_n \}) [/tex]
    what you really want to write is
    [tex] \bigcup_{n=1}^{\infty} span(\{e_1,...,e_n \} ) [/tex].

    And H will not be equal to this in general. Let's take the canonical Hilbert space l2 of sequences [itex](a_1,...,a_n,...)[/itex] such
    [tex] \sum_{n=1}^{\infty} a_n^2 [/tex]
    converges.

    Then a maximal orthonormal system is
    [itex] e_1 = (1,0,0,....),\ e_2 = (0,1,0,....),\ e_3 = (0,0,1,0,0,...) [/itex] etc. There are countably many of these. The inner product of [itex] (a_1,a_2,...)[/itex] and [itex] (b_1,b_2,...)[/itex] is
    [tex] \sum_{n=1}^{\infty} a_n b_n [/tex].

    It should be clear that the en are an orthonormal system, and that <g,en> = 0 for all n implies that g = (0,0,0,...,0,...) is the zero vector. However almost none of the vectors in l2 are in
    [tex] \bigcup_{n=1}^{\infty} span( \{e_1,....,e_n \} ) [/tex]
    because to be in this set, you have to be a linear combination of finitely many of the en. So for example the vector (1,1/2,1/4,1/8,....) is in l2 but not in that above union, because it is writable as only an infinite linear combination of the ens.
     
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