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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! :)

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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 $L^2$.
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  • $\begingroup$ Thank you! :) I was able to complete the argument this way. $\endgroup$ – Nicole Jan 21 '13 at 21:23
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  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.
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  • $\begingroup$ This was helpful, although I found Jonas Meyer's suggestion to be more straightforward for me. Thank you regardless! :) $\endgroup$ – Nicole Jan 21 '13 at 21:23
  • $\begingroup$ It could be interesting to write both proofs in details. $\endgroup$ – Davide Giraudo Jan 21 '13 at 21:32
  • $\begingroup$ Indeed, I will try to complete the problem in both ways, to make sure I understand it. $\endgroup$ – Nicole Jan 21 '13 at 21:36

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