Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question on the definability of convolution

  1. Feb 19, 2010 #1
    In Apostol's "Mathematical Analysis", Page 328 (see the image below and the underlined sentence),
    f0n8gi.jpg
    why does the Lebesgue integral (41) exist for gif.gif ?[/URL]
    The definition of convolution is as follows:
    2b4rqr.jpg
    Thanks!
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 20, 2010 #2
    I'm not sure if I understand your question. For that special case, the Lebesgue integral doesn't have to exist for all [itex]x \in \left[a,b\right][/itex], and that is precisely what the author says.
     
  4. Feb 20, 2010 #3
    My question comes from the underlined sentence. That the convolution in (41) can be defined means the Lebesgue integral in (41) exists and the author asserts this integral exists for all x in [a,b] under the given condition. But I do not understand why this integral gif.latex?\int_0^x&space;f(t)g(x&space;-&space;t)dt.gif exists.
     

    Attached Files:

  5. Feb 22, 2010 #4
    What the author (Apostol) asserts is that, if f and g are Riemann integrable, then that convolution is well defined, for all [itex]x \in \left[a,b\right][/itex]. On the other hand, if the functions are Lebesgue integrable (and this include much less regular functions), then then their (Lebesgue) integral may not be defined. In fact, the author gives an example immediately after.
     
  6. Feb 22, 2010 #5
    My question is why this statement holds, not the statement on the other hand.
     
  7. Feb 22, 2010 #6
    Because the product of two Riemann integrable functions is also (Riemann) integrable. This is a consequence of, for example, theorem 7.48 of Apostol's book.
     
  8. Feb 22, 2010 #7
    But f and g are Riemann-integrable only on [a,b] under the given condition, which does not insure f(t) and g(x-t), and in turn their product, are Riemann-integrable on [0,x].
     
  9. Feb 22, 2010 #8
    Note that [itex]x \in \left[a,b\right][/itex], then for a<0, there is no problem, because [itex]\left[0,x\right]\subseteq \left[a,b\right][/itex]; if a>0, you may always assume that the functions are 0 outside the interval.
     
  10. Feb 22, 2010 #9
    Why? Why can we assume the functions are 0 outside the interval ([a,b] or [0,x])?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A question on the definability of convolution
Loading...