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.

In this problem $(X,F,u)$ is a measure space, and let $f_{n}:X\rightarrow R $ be a sequence of measurable functions on it satisfying ()$$\int_{X}|f_{k}|^{2}du\leq M$$ for all k

()$$\int_{X}f_{k}f_{j}du=0$$ for all $j\neq k$,

where M is a finite constant independent of n, for each n=1,2,3, set $S_{n}=\sum_{k=1}^{n}f_{k}$ prove that $\lim_{n\rightarrow \infty} \frac{S_{n^{2}}}{n^{\alpha}}=0$,a.e for all $\alpha \geq 3/2$ . Here my idea is to show that $$\int_{X}\lim_{n\rightarrow \infty} \frac{s_{n^{2}}}{n^{\alpha}} du=\lim_{n\rightarrow \infty}\int_{X} \frac{s_{n^{2}}}{n^{\alpha}} du=0$$

For the second part we have $$\int_{X} \frac{s_{n^{2}}}{n^{\alpha}} du\leq \int_{X} |\frac{s_{n^{2}}}{n^{\alpha}} |du\leq \sqrt{\int_{X}\frac{n^{2}M}{n^{\alpha}}du}\sqrt{\int_{X}1du}$$ but here I dont know if X is finite, so I can not conclude the RHS go to 0 as n goes to infinity. I wonder if my idea is wrong here.

share|improve this question
2  
$\alpha\geqslant3/2$ or $\alpha\gt3/2$? –  Did Jan 1 '13 at 19:13

1 Answer 1

  • We have $\lVert S_{n^2}\rVert^2_{L^2}\leqslant Mn^2$ for all $n$ by the assumptions.
  • We have $\mu\left(\frac{|S_{n^2}|}{n^\alpha}>n^\beta\right)\leqslant Mn^{2(1-\alpha-\beta)}.$
  • To conclude by Borel-Cantelli's lemma, we just have to find $\beta<0$ such that $2(1-\alpha-\beta)<-1$.
share|improve this answer
2  
The introduction of $\beta$ is quite mysterious (and unneeded) but the technique which is delineated here, once repaired, yields the result for every $\alpha\gt\frac32$. –  Did Jan 1 '13 at 19:03
    
@Davide Thanks for your proof, its quite nice to use Chebsev to convert to the integral, while I still dont know why we need $2(1-\alpha-\beta)<-1$, can you explain that for me? –  user53800 Jan 1 '13 at 19:31
    
We want to ensure the convergence of $\sum_nn^{2(1-\alpha-\beta)}$. @did: I indeed need to think about the case $\alpha=3/2$. Do you have a shorter approach? –  Davide Giraudo Jan 1 '13 at 20:06

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.