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.

I am trying to produce a sequence of sets $A_n \subseteq [0,1] $ such that their characteristic functions $\chi_{A_n}$ converge weakly in $L^2[0,1]$ to $\frac{1}{2}\chi_{[0,1]}$.

The sequence of sets $$A_n = \bigcup\limits_{k=0}^{2^{n-1} - 1} \left[ \frac{2k}{2^n}, \frac{2k+1}{2^n} \right]$$ seems like it should work to me, as their characteristic functions look like they will "average out" to $\frac{1}{2} \chi_{[0,1]}$ as needed. However, I'm having trouble completing the actual computation.

Let $g \in L^2[0,1]$, then we'd like to show that $$ \lim_{n \to \infty} \int_{[0,1]} \chi_{A_n} g(x) dx = \int_{[0,1]} \frac{1}{2}\chi_{[0,1]} g(x) dx = \frac{1}{2} \int_{[0,1]} g(x) dx $$ We have that $$ \int_{[0,1]} \chi_{A_n} g(x) dx = \sum\limits_{k=0}^{2^{n-1}-1} \int_{\left[ \frac{2k}{2^n}, \frac{2k+1}{2^n} \right] } \chi_{A_n} g(x) dx $$ Now I am stuck, as I don't see how to use a limit argument to show that this goes to the desired limit as $ n \to \infty$. Does anyone have any suggestions on how to proceed? Any help is appreciated! :)

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Suggestions:

  • First consider the case where $g$ is the characteristic function of an interval.
  • Generalize to the case where $g$ is a step function.
  • Use density of step functions in $L^2$.
share|improve this answer
    
Thank you! :) I was able to complete the argument this way. –  Nicole Jan 21 '13 at 21:23
  1. The sequence $\{\chi_{A_n}\}$ is bounded in $L^2$, and the polynomials are dense in $L^2[0,1]$, so it's enough to consider the case $g$ polynomial.
  2. By linearity, the case $g(x)=x^p$, $p\in\Bbb N$ is sufficient.
  3. The involved integral are explicitely computable. Then write $$(2k+1)^{p+1}-(2k)^{p+1}=\sum_{l=0}^p\binom{p+1}l(2k)^l,$$ and consider the term $l=p$ separately.
share|improve this answer
    
This was helpful, although I found Jonas Meyer's suggestion to be more straightforward for me. Thank you regardless! :) –  Nicole Jan 21 '13 at 21:23
    
It could be interesting to write both proofs in details. –  Davide Giraudo Jan 21 '13 at 21:32
    
Indeed, I will try to complete the problem in both ways, to make sure I understand it. –  Nicole Jan 21 '13 at 21:36

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.