Let $f_n :[0,1] \to \Bbb{R}$ a sequence of uniformly bounded measurable functions with the property: $$\int_0^1f_n(x)f_m(x)dx=0,\forall m \neq n$$
Prove that $\frac{1}{N}\sum_{n=1}^Nf_n(x) \to 0$ almost everywhere in $[0,1]$.
I have already proven that:
$(1)$ $\frac{1}{N^2}\sum_{n=1}^{N^2}f_n(x) \to 0$ almost everywhere.
$(2)$ $\frac{1}{N+N^2}\sum_{n=1}^{N+N^2}f_n(x) \to 0$ almost everywhere.
$(3)$ $\frac{1}{N}\sum_{n=1}^{N}f_n(x)-\frac{1}{N}\sum_{n=1}^{N}f_{n+m}(x) \to 0$ almost everywhere, $\forall m \in \Bbb{N}$
Because of the fact that we have already a subsequence converging to $0$ almost everywhere i tried to show that $\frac{1}{N}\sum_{n=1}^{N}f_n(x)$ is a Cauchy sequence for almost every $x$.
But i did not manage anything.
Can someone give me a hint to solve this?
I do not want a full solution.
Thank you in advance.
Best Answer
The conditions (1) and that the $f_n$'s are uniformly bounded are sufficient.
That is, let $(a_n)_{n \ge 1}$ be any sequence of reals such that $|a_n| \le C$ for each $n$ and $\frac{1}{N^2}\sum_{n \le N^2} a_n \to 0$. Then $\frac{1}{N}\sum_{n \le N} a_n \to 0$. The reason this is true is that the squares occur frequently enough in the integers.
Suppose $M^2 \le N \le (M+1)^2$. Then $$|\frac{a_1+\dots+a_N}{N}| \le |\frac{a_1+\dots+a_N}{M^2}| \le |\frac{a_1+\dots+a_{M^2}}{M^2}|+|\frac{a_{M^2+1}+\dots+a_N}{M^2}|$$ $$ \le |\frac{a_1+\dots+a_{M^2}}{M^2}|+\frac{N-M^2}{M^2}C.$$ Now just let $M \to \infty$, noting that $\frac{N-M^2}{M^2} \le \frac{2M+1}{M^2} \to 0$.