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

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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.

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  • 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

up vote
5
down vote

favorite

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 $[0,1]$.

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.

share|cite|improve this question

  • 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

up vote
5
down vote

favorite

2

up vote
5
down vote

favorite

2
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 $[0,1]$.

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.

share|cite|improve this question

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.

real-analysis measure-theory lebesgue-measure

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asked Nov 29 at 11:34

Marios Gretsas

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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

  • 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

  • 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

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

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

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

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  • 1

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

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accepted

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

share|cite|improve this answer

  • 1

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

up vote
8
down vote

accepted

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

share|cite|improve this answer

  • 1

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

up vote
8
down vote

accepted

up vote
8
down vote

accepted

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

share|cite|improve this 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 $frac{N-M^2}{M^2} le frac{2M+1}{M^2} to 0$.

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edited Nov 29 at 12:17

answered Nov 29 at 11:44

mathworker21

8,2851827

8,2851827

  • 1

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

  • 1

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

1

1

+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

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