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

Confusion regarding the dominated convergence theorem

  1. Oct 12, 2011 #1
    So a well-known theorem from Lebesgue integration is the dominated convergence theorem. It talks about a sequence [itex]f_1,f_2,\ldots[/itex] of functions converging pointwise to a function [itex]f[/itex]. And if [itex]|f_n(x)| \leq g(x)[/itex] for an integrable function [itex]g[/itex], then we have [itex]\int f_n \to \int f[/itex].

    But what if we have a given function [itex]f_\epsilon[/itex] which depends on some parameter [itex]\epsilon[/itex], which we are taking to zero? Suppose you've shown that [itex]f_\epsilon \to 0[/itex] pointwise as [itex]\epsilon \to 0[/itex]; however, you know [itex]f_\epsilon[/itex] is NOT bounded in [itex]\epsilon[/itex] - if you take [itex]\epsilon[/itex] large enough, you can make [itex]\sup_{x\in \mathbb R} f_\epsilon(x)[/itex] arbitrarily large. But, of course, we don't WANT to do that - we want to take [itex]\epsilon[/itex] small. And suppose you know that [itex]\epsilon < \epsilon_0[/itex] means [itex]\sup f_{\epsilon} < \sup f_{\epsilon_0}[/itex]. My question is this: Can we still apply the DCT to show that [itex]\lim_{\epsilon \to 0} \int_{\mathbb R} f_\epsilon = 0[/itex]? If so, why?
     
  2. jcsd
  3. Feb 27, 2012 #2
    The answer is yes, but it needs non-trivial justifications
    Firstly because the DCT is stated for only a countable SEQUENCE of converging functions. Your epsilon seems to me to be a real parameter. But we can still apply DCT because EVERY countable subsequence you can get from the original epsilon sequence will converge to zero.
    Two, even though for large epsilon your function might be wild, remember that limits of sequences are not affected by the initial terms of the sequence. You only need demonstrate good behavior of your sequence in some little ball around 0, which you claim to have done, so you're fine.
    I hope this helps, even though its a couple of months late.
     
  4. Mar 4, 2012 #3
    This is what you want to prove:

    [tex]
    \forall \mathcal{E}>0\quad \exists \epsilon_0>0\quad\textrm{s.t.}\quad 0<\epsilon <\epsilon_0 \quad\implies\quad |\int f_{\epsilon}| < \mathcal{E}
    [/tex]

    This is what you know:

    For all sequences [itex]\epsilon_1,\epsilon_2,\epsilon_3,\ldots[/itex]
    [tex]
    \lim_{n\to\infty}\epsilon_n = 0\quad\implies\quad \lim_{n\to\infty}\int f_{\epsilon_n} = 0
    [/tex]

    In order to achieve what you want, you can make an anti-thesis, and use what you know. Assume that there exists [itex]\mathcal{E}>0[/itex], but does not exists [itex]\epsilon_0 > 0[/itex] such that the first implication would be satisfied. Then we obtain a sequence [itex]\epsilon_1,\epsilon_2,\epsilon_3,\ldots[/itex] such that

    [tex]
    \lim_{n\to\infty}\epsilon_n=0
    [/tex]

    and

    [tex]
    0<\epsilon < \epsilon_n \quad\nRightarrow\quad |\int f_{\epsilon}|<\mathcal{E}\quad\forall n=1,2,3,\ldots
    [/tex]

    Now we are close to a contradiction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confusion regarding the dominated convergence theorem
Loading...