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Inner product on C(R)

  1. Dec 18, 2008 #1
    My apologies in advance for asking what (to me) looks like an extremely stupid question, but I just can't figure it out.

    1. The problem statement, all variables and given/known data:
    Where is this an inner product:
    [tex]\int_{a}^{b}f(x)g(x) dx[/tex]

    a) on C[a,b]?
    b) on C(R)?

    The answer is that it is an inner product on a), but not on b) - apparently on b) the axiom of positivity fails. I do not understand how this is possible, since all functions that are C(R) are also C[a,b] - or have I just always misunderstood this notation? Does not C(R) mean "functions continuous on all of R"?

    2. Relevant equations

    This is what the answer key says: It fails on b) because [tex]\exists f \ne 0 : \Vert f \Vert^2 = 0[/tex]

    3. The attempt at a solution:

    has mainly consisted of trying (unsuccessfully) to work backwards towards a counterexample. I don't know how to do this in any other, more general way, since the whole idea seems illogical to me. Please enlighten me, somebody?
  2. jcsd
  3. Dec 18, 2008 #2
    The inner product of the function [tex] f(x) = \sqrt{x} [/tex] is zero and this f is not the zero function.
  4. Dec 18, 2008 #3


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    Any continuous function that is 0 on [a,b], but not anywhere else. sqrt(x) doesn't work for arbitrary a,b, but a simple piecewise function that looks similar to the absolute value function would.
  5. Dec 18, 2008 #4
    What are you talking about? The inproduct of sqrt(x) delivers x if you integrate this from -inf to +inf you'll get zero, right?
  6. Dec 18, 2008 #5


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    It's not from -inf to inf. It's from a to b. The reason the integral from a to b isn't an inner product on C(R) is because there is an f in C(R) such that it's integral from a to b is zero, but f is not zero everywhere.
  7. Dec 18, 2008 #6
    Yes you're right I didn't read well enough. The integral has bounds form a to b.
  8. Dec 19, 2008 #7
    I think this is occasion for a big *headdesk* (on my own behalf, obviously). How did I not see this? Thanks a lot. :)
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