$\lim_{n\rightarrow \infty} \frac{S_{n^{2}}}{n^{\alpha}}=0$,a.e.

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.

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$\alpha\geqslant3/2$ or $\alpha\gt3/2$? – Did Jan 1 '13 at 19:13

• 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 $2(1-\alpha-\beta)<-1$.
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