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Inner Product Space

  1. Oct 8, 2011 #1
    Hi everyone!

    I would like to ask how would you verify if functions form an inner product space? For example, if one has functions of the form q(x)e^-(x^2/2) where q(x) is a polynomial of degree < N in x, on the interval -∞ < x < ∞. Also, how would you specify the dimension of the space, if it exists?

    Thank you!
     
  2. jcsd
  3. Oct 8, 2011 #2

    mathman

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    The dimension is N. It forms an inner product space, since the integral of the product of any two such functions is finite (the exponential term insures this).
     
    Last edited: Oct 8, 2011
  4. Oct 9, 2011 #3
    Check using the definition of an inner product would be my initial suggestion.

    1) [itex](u|u)\geq 0[/itex] and [itex]0[/itex] iff [itex]u = 0[/itex]
    2) [itex](\alpha u+ \beta v|w) = \alpha (u|w) + \beta (v|w) [/itex] Aka that it is linear

    If this holds true [itex]\forall u,v,w[/itex] then the inner product is defined for the said space and is thus a inner product space (given of course that it is in a vector space to begin with)
     
  5. Oct 9, 2011 #4

    Fredrik

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    Note that vector spaces whose elements are functions usually fail the condition [itex]\langle f,f\rangle=0\Rightarrow 0[/itex]. (Define f by f(x)=1 when x=0 and f(x)=0 otherwise; then [itex]\langle f,f\rangle=0[/itex] but f≠0). Such a space is sometimes called a semi-inner product space.
     
  6. Oct 9, 2011 #5

    mathman

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    Since f = 0 almost everywhere, this will be an inner product space. Moreover it is complete, so it is L2.
     
  7. Oct 9, 2011 #6

    Fredrik

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    By definition of inner product, it's not. But you can define an equivalence relation by saying that f~g if f=g almost everywhere, and then define an inner product on the set of equivalence classes by [itex]\langle [f],[g]\rangle=\langle f,g\rangle[/itex]. That's an inner product on the left and a semi-inner product on the right.
     
  8. Oct 14, 2011 #7
    Thank you so much for the insights! All of your comments gave me an idea on how to attack the problem! Thanks once again!
     
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