Clash Royale CLAN TAG#URR8PPP
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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.
add a comment 
5
down vote
favorite
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.

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

1I proved that $sum_{n=1}^{infty} int_0^1frac{1}{n^2}sum_{k=1}^{n^2}f_k(x)^2dx< +infty$..And then BeppoLevi
– Marios Gretsas
Nov 29 at 11:40
add a comment 
5
down vote
favorite
5
down vote
favorite
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.
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.

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

1I proved that $sum_{n=1}^{infty} int_0^1frac{1}{n^2}sum_{k=1}^{n^2}f_k(x)^2dx< +infty$..And then BeppoLevi
– Marios Gretsas
Nov 29 at 11:40
add a comment 

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

1I proved that $sum_{n=1}^{infty} int_0^1frac{1}{n^2}sum_{k=1}^{n^2}f_k(x)^2dx< +infty$..And then BeppoLevi
– Marios Gretsas
Nov 29 at 11:40
– mathworker21
Nov 29 at 11:37
– mathworker21
Nov 29 at 11:37
– Marios Gretsas
Nov 29 at 11:40
– Marios Gretsas
Nov 29 at 11:40
add a comment 
1 Answer
1
active
oldest
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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 $$frac{a_1+dots+a_N}{N} le frac{a_1+dots+a_N}{M^2} le frac{a_1+dots+a_{M^2}}{M^2}+frac{a_{M^2+1}+dots+a_N}{M^2}$$ $$ le frac{a_1+dots+a_{M^2}}{M^2}+frac{NM^2}{M^2}C.$$ Now just let $M to infty$, noting that $frac{NM^2}{M^2} le frac{2M+1}{M^2} to 0$.

1+1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18
add a comment 
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $$frac{a_1+dots+a_N}{N} le frac{a_1+dots+a_N}{M^2} le frac{a_1+dots+a_{M^2}}{M^2}+frac{a_{M^2+1}+dots+a_N}{M^2}$$ $$ le frac{a_1+dots+a_{M^2}}{M^2}+frac{NM^2}{M^2}C.$$ Now just let $M to infty$, noting that $frac{NM^2}{M^2} le frac{2M+1}{M^2} to 0$.

1+1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18
add a comment 
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 $$frac{a_1+dots+a_N}{N} le frac{a_1+dots+a_N}{M^2} le frac{a_1+dots+a_{M^2}}{M^2}+frac{a_{M^2+1}+dots+a_N}{M^2}$$ $$ le frac{a_1+dots+a_{M^2}}{M^2}+frac{NM^2}{M^2}C.$$ Now just let $M to infty$, noting that $frac{NM^2}{M^2} le frac{2M+1}{M^2} to 0$.

1+1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18
add a comment 
8
down vote
accepted
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 $$frac{a_1+dots+a_N}{N} le frac{a_1+dots+a_N}{M^2} le frac{a_1+dots+a_{M^2}}{M^2}+frac{a_{M^2+1}+dots+a_N}{M^2}$$ $$ le frac{a_1+dots+a_{M^2}}{M^2}+frac{NM^2}{M^2}C.$$ Now just let $M to infty$, noting that $frac{NM^2}{M^2} le frac{2M+1}{M^2} to 0$.
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 $$frac{a_1+dots+a_N}{N} le frac{a_1+dots+a_N}{M^2} le frac{a_1+dots+a_{M^2}}{M^2}+frac{a_{M^2+1}+dots+a_N}{M^2}$$ $$ le frac{a_1+dots+a_{M^2}}{M^2}+frac{NM^2}{M^2}C.$$ Now just let $M to infty$, noting that $frac{NM^2}{M^2} le frac{2M+1}{M^2} to 0$.

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

1+1 from me..very nice answer.
– Marios Gretsas
Nov 29 at 12:18
– Marios Gretsas
Nov 29 at 12:18
– Marios Gretsas
Nov 29 at 12:18
add a comment 
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– mathworker21
Nov 29 at 11:37
– Marios Gretsas
Nov 29 at 11:40