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











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













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






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






active

oldest

votes

















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018513%2flimit-almost-everywhere-of-averages-of-uniformly-bounded-and-integrable-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








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










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

8,2601827




8,2601827








  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018513%2flimit-almost-everywhere-of-averages-of-uniformly-bounded-and-integrable-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

AnyDesk - Fatal Program Failure

How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

QoS: MAC-Priority for clients behind a repeater