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

Clash Royale CLAN TAG#URR8PPP

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 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 at 11:40

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 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 at 11:40

2

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.

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.

real-analysis measure-theory lebesgue-measure

Marios Gretsas

8,42511437

8,42511437

• how did you get $frac{1}{N^2}sum_{n=1}^{N^2} f_n(x) to 0$ almost everywhere
– mathworker21
Nov 29 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 at 11:40

• how did you get $frac{1}{N^2}sum_{n=1}^{N^2} f_n(x) to 0$ almost everywhere
– mathworker21
Nov 29 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 at 11:40

how did you get $frac{1}{N^2}sum_{n=1}^{N^2} f_n(x) to 0$ almost everywhere
– mathworker21
Nov 29 at 11:37

how did you get $frac{1}{N^2}sum_{n=1}^{N^2} f_n(x) to 0$ almost everywhere
– mathworker21
Nov 29 at 11:37

1

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 at 11:40

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 at 11:40

active

oldest

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

• +1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18

active

oldest

active

oldest

active

oldest

active

oldest

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

• +1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18

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

• +1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18

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

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

edited Nov 29 at 12:17

mathworker21

8,2851827

8,2851827

• +1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18

• +1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18

1

– Marios Gretsas
Nov 29 at 12:18

– Marios Gretsas
Nov 29 at 12:18

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