# Limit almost everywhere of averages of uniformly bounded and integrable functions .

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 $$[0,1]$$.

$$(1)$$ $$\frac{1}{N^2}\sum_{n=1}^{N^2}f_n(x) \to 0$$ 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.

• how did you get $\frac{1}{N^2}\sum_{n=1}^{N^2} f_n(x) \to 0$ almost everywhere – mathworker21 Nov 29 '18 at 11:37
• I proved that $\sum_{n=1}^{\infty} \int_0^1|\frac{1}{n^2}\sum_{k=1}^{n^2}f_k(x)|^2dx< +\infty$..And then Beppo-Levi – Marios Gretsas Nov 29 '18 at 11:40

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 $$\frac{N-M^2}{M^2} \le \frac{2M+1}{M^2} \to 0$$.