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Orthonormal system in Hilbert space

  1. Oct 22, 2013 #1
    Let H be a Hilbert space. Let F be a subset of H.
    F is dense in H if:
    <f,h>=0 for all f in F => h=0
    Now take an orthonormal set (ek) (this is a countable sequence indexed by k) in H. My book says that obviously:
    [itex]\bigcup[/itex]span(ek) is dense in H (the union runs over all k)
    =>
    g=Ʃ<g,ek>ek
    Now first of all:
    Why do we need the union of the spans in the above? What is the difference between the span of the sequence of countably many units vectors and the union of the spans?
    And then: Why is the above implication obvious?
     
    Last edited: Oct 22, 2013
  2. jcsd
  3. Oct 22, 2013 #2
    I don't know about the rest of your question, but think about two orthogonal unit vectors x and y in [itex]\mathbb{R}^2[/itex]. Span(x) and span(y) are both straight lines, so the union of span(x) and span(y) would be two crossed straight lines; it would look like an "X". On the other hand, span({x,y}) is the whole plane.
     
  4. Oct 22, 2013 #3
    yes, so why include the union if it is already included in span(e_k)
     
  5. Oct 22, 2013 #4
    The notation there isn't clear enough, but I think what it's naming is [tex]\bigcup_{n=1}^\infty \text{span}[\{e_k\}_{k=1}^n],[/tex] which some people would just call [itex]\text{span}[\{e_k\}_{k=1}^\infty][/itex]. The value of the former notation is that it reminds us that in a vector space the words "linear combination" mean (by definition) finite linear combination.
     
  6. Oct 22, 2013 #5

    mathman

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    That doesn't look right. F could be any subset that includes a basis. It doesn't have to be dense.
     
  7. Oct 22, 2013 #6
    ^ Good catch.

    Is it possible that the author has "taken the liberty" if misusing "a dense subset" to actually mean "a subset with dense span"?
     
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