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What is 'completeness' (function space)

  1. Jan 19, 2010 #1
    given a set of functions that depend on a parameter lambda [tex] f(\lambda x) [/tex] , how can be proved or what does it mean that this set of functions is COMPLETE in [tex] L^{p} (a,b) [/tex] do the functions [tex] f(\lambda x) [/tex] need to form an orthogonal basis or it is enough that for diffrent values of lambda ,there is a linear independence.
     
  2. jcsd
  3. Jan 19, 2010 #2
    I'm not sure about sets which are not orthogonal, but I think that completness generally says that there is no other non-trivial vector which is linearly independent of the set vectors.
     
  4. Jan 19, 2010 #3

    Landau

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    In this context, completeness usually means 'maximal spanning', i.e. every other element of L^p(a,b) should be expressible as (not necessarily finite) linear combination of these 'functions'. It is an extension of the concept of 'basis' from finite-dimensional linear algebra to infinite dimensions. The orthogonal (or even orthonormal) requirement usually is explicitly added, as in "complete orthonormal set".

    See e.g. here.
     
  5. Jan 21, 2010 #4
    Picky aside: orthogonality isn't well-defined when p != 2.
     
  6. Jan 21, 2010 #5
    The L^p(a,b) is a metric space. The space of functions is complete if each Cauchy sequence converges in the L^p metric to another function in the space.
     
  7. Jan 22, 2010 #6

    Landau

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    @wofsy: that's a different type of completeness; here we're talking about completeness of a set of vectors in L^p, not of the (metric) space L^p itself!
    See the link in my previous post.
     
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