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.






real-analysis measure-theory lebesgue-measure







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













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







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.






real-analysis measure-theory lebesgue-measure







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




real-analysis measure-theory lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 at 11:34









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








  • 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
1






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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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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