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

Riemann lebesgue lemma. wikipedia. 2010-06-26

  1. Jun 26, 2010 #1

    Have I made a mistake when it looks to me that the Wikipedia proof on Riemann-Lebesgue lemma looks like nonsense?

    How are you supposed to use dominated convergence theorem there?
  2. jcsd
  3. Jun 26, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    f(x) eitx is dominated by f(x)?
  4. Jun 26, 2010 #3


    User Avatar
    Science Advisor

    I haven't analyzed it in detail, but I think it means that any integrable function can be shown to be dominated by a step function (over a finite interval).
  5. Jun 26, 2010 #4


    User Avatar
    Science Advisor

    Could you tell us why you think it looks like "non-sense"?
  6. Jul 2, 2010 #5
    I have read one proof for this theorem from a book, and I understood it. Now I see another proof in Wikipedia, I'm unable to understand it, and the author of this piece of wiki-info has not given any sources. That's the origin of my doubts.

    If you first prove the result for step functions, and them mention the use of dominated convergence theorem, it looks like that you want to approximate some integrable function with step functions and move a limit from inside the integral to the outside. Like this:

    \lim_{t\to\infty} \int e^{itx} f(x) dx = \lim_{t\to\infty}\lim_{n\to\infty} \int e^{itx} f_n(x)dx

    But what is this good for? It looks like an example of a case where the author justifies some simple step and then completely omits the justification for more difficult step.

    If [itex]\epsilon > 0 [/itex] and an integrable function [itex]f[/itex] are given, it would make lot more sense to find a step function [itex]f_n[/itex] such that [itex]\|f -f_n\|_1 < \frac{\epsilon}{2}[/itex], and then choose [itex]T>0[/itex] such that

    \big| \int e^{itx} f(x)dx\big| \leq \int |f(x) - f_n(x)| dx \;+\; \big|\int e^{itx} f_n(x) dx\big| < \epsilon

    for all [itex]t > T[/itex]. But you don't need dominated convergence for this direction of proof.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook