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Domain of Sturm-Liouville operators

  1. Aug 31, 2013 #1
    Hello, folks,

    A Sturm-Liouville operator is typically defined not on the whole space of [itex]C^2[/itex] functions,
    but rather on some subspace described by boundary conditions. My question is: are those subspaces closed (hence complete, hence Hilbert) in [itex]L^2[/itex]? In case of an affirmative answer, how
    can one prove that?

    I've been searching for this answer in books, but apparently they don't even bother to consider
    that operators are defined in subspaces of [itex]L^2[/itex] and go on applying spectral theory to
    them as if they were defined on a Hilbert space.

    Please, let me know if the question was not clear.

    Thanks in advance.
     
  2. jcsd
  3. Aug 31, 2013 #2

    pasmith

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    The relevant Hilbert space is not the subspace of [itex]C^2[/itex] itself (which is not complete with respect to the [itex]L^2[/itex] norm anyway), but the completion of that subspace with respect to the [itex]L^2[/itex] norm.
     
  4. Aug 31, 2013 #3
    But how can we describe that completion?
     
  5. Sep 1, 2013 #4

    pasmith

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    As the set of equivalence classes of square-integrable but not necessarily continuous functions satisfying the given boundary conditions under the equivalence relation [itex]{\sim}[/itex], where [itex]f \sim g[/itex] if and only if the set [itex]\{ x : f(x) \neq g(x)\}[/itex] has measure zero. This construction is necessary to ensure that [itex]\left(\int f^2\,\mathrm{d}x\right)^{1/2}[/itex] remains a norm, since two functions which differ on a set of measure zero have the same integral. It also ensures that no two continuous functions are in the same equivalence class, since if two continuous functions differ at a point then they must differ on an open neighbourhood of that point, and open neighbourhoods have strictly positive measure.
     
  6. Sep 2, 2013 #5
    The completion detail is usually not described in textbooks. Do you have any idea why?
     
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