10
$\begingroup$

Given a sequence $f_n \in L^p$ and $g \in L^p$, with $|f_n| \leq g$, I am trying to show that $f_n \to f$ in measure implies $f_n \to f$ in $L^p$.

Firstly, I know that if $f_n \to f$ in measure, then there is a subsequence $f_{n_i}$ such that $f_{n_i} \to f$ almost everywhere. Then I can use the dominated convergence theorem to show that $\lVert f_{n_i} - f_p\rVert \to 0$.

Now I am trying to show that $\lVert f_n - f\rVert_p \to 0$. My idea is to assume that $\lVert f_n - f\rVert_p \nrightarrow 0$ and then show that this contradicts the fact that $\lVert f_{n_i} - f\rVert_p \to 0$, but I am not sure of the details. Can anyone help me finish the argument?

  • 3
    In the title, you should add "with an hypothesis of domination". You are almost done: if $||f_n-f||_p$ doesn't converge to $0$, then we can find $\delta>0$ and a subsequence such that $||f_{n_k}-f||\geq \delta$. This subsequence still converges in measure to $f$, so by your previous argument we get a contradiction.2012-10-18
  • 0
    @DavideGiraudo Ok, got it. Thanks!2012-10-18
  • 1
    You can answer your own question (hence it won't remain unanswered), and you will have your homework done properly.2012-10-18
  • 0
    @DavideGiraudo : as one commenter makes note of below, we haven't used the boundedness of $f_n$ at all. Is this a problem?2017-08-08
  • 0
    Also, this question was featured on the UI-Urbana Champaign graduate qualifying exam from August 2015, if anyone cares to know.2017-08-08

1 Answers 1