Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there some version of DCT for $p=\infty$. That is, is it true that if there is a sequence of measurable functions defined on an open set $\Omega$ in $\mathbb{R}^n$, $f_n$ converging pointwise to a function $f$ such that there exists $g\in L^{\infty}(\Omega)$ with $|f_n(x)|\leq|g(x)|$, then $f_n,f\in L^{\infty}(\Omega)$ and $||f_n-f||_{\infty}\to 0$?

If this is indeed true, could someone point out the proof?

share|improve this question
1  
It's not true. Consider the sequence $(f_n)$ defined by $f_n=\chi_{(0,1/n)}$ in $L^\infty(0,1)$. –  David Mitra Sep 11 '12 at 21:19

2 Answers 2

up vote 1 down vote accepted

In $\Bbb R$, this version doesn't work: take $f_n:=\chi_{(n,n+1)}$. We have $|f_n|\leq 1=:g\in L^{\infty}$ and $\lVert f_n-f\rVert_{\infty}=1$ for all $n$, with $f=0$.

Actually, the aim of dominated convergence is to switch a limit with an integral. When we work with $\lVert\cdot\rVert_{\infty}$, we don't use integrals (only properties of the measure).

The domination, for a sequence in $L^{\infty}$, means that the sequence of the norms is bounded, and pointwise convergence doesn't improve the things.

share|improve this answer
    
do you mean $f=0$? –  Vivek Sep 11 '12 at 21:31
    
@Vivek Yes, it was a typo. Thanks! –  Davide Giraudo Sep 11 '12 at 21:36

No, it's false.

Take $$f_n = \begin{cases} 1 : & n < x < n+1 \\ 0 : &\textrm{otherwise}\end{cases}$$

The sequence $f_n$ converges pointwise to $0$, and is dominated by $g(x) \equiv 1 \in L^\infty$, but $||f_n - f||_\infty \equiv 1$ (so does not go to $0$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.