0
$\begingroup$

Inspired by this question, I'm interested in knowing whether the following holds:

Suppose $f_n \in L^1 \cap L^\infty$. Suppose $f_n \to f$ in $L^1$ and $f_n \to g$ in $L^\infty$. Can we conclude that $f = g$ a.e.?

EDIT: I think I have a proof. Since $f_n \to f$ in $L^1$, there is a subsequence $f_{n_k} \to f$ a.e. Since $f_{n_k} \to g$ in $L^\infty$, there is a subsequence $f_{n_{m_k}} \to g$ a.e. But $f_{n_{m_k}}$ is a subsequence of $f_{n_k}$, and since limits are unique, we must have $f = g$ a.e.

Is this right? This seems to generalize nicely if it is. This shows that if you converge in $L^p$ and $L^q$, the limit function is unique.

  • 0
    Isn't $L^{\infty}$ contained in $L^{1}$? – 2017-01-10
  • 0
    No..... $f(x) = \log x$ shows that you can be in $L^p$ for any $p$ finite but not in $L^\infty$. – 2017-01-10
  • 0
    Yes, we can conclude that. A subsequence of $(f_n)$ converges almost everywhere to $f$. The full sequence converges almost everywhere to $g$. – 2017-01-10
  • 0
    @DanielFischer seems we typed that at the same time, though I passed to another subsequence which was completely unnecessary! Does the generalization hold with my proof though? – 2017-01-10
  • 0
    Yes. The same argument, $L^p$ convergence gives a subsequence converging almost everywhere to the limit function. – 2017-01-10
  • 0
    Do you mean, that $logx$ isn't essentially bounded? – 2017-01-10
  • 0
    Funnily enough no-one is referring to the measure space we are working on. I presume there might be some issues while working on general measure spaces – 2017-01-10

0 Answers 0