Is this true that the sequences evenly sampled from functions in $L^p$ are in $l_p$?












0












$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12
















0












$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12














0












0








0





$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







share|cite|improve this question











$endgroup$




Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces




real-analysis sequences-and-series functional-analysis lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 20:37









Davide Giraudo

126k16150261




126k16150261










asked Oct 13 '16 at 10:33









bridgerbridger

503315




503315












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12


















  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12
















$begingroup$
$L_p$ consists of equivalence classes of a.e. equal functions...
$endgroup$
– user251257
Oct 13 '16 at 10:40




$begingroup$
$L_p$ consists of equivalence classes of a.e. equal functions...
$endgroup$
– user251257
Oct 13 '16 at 10:40




2




2




$begingroup$
No, "taking a sample" is unbounded operator in $L_p$.
$endgroup$
– A.Γ.
Oct 13 '16 at 10:41






$begingroup$
No, "taking a sample" is unbounded operator in $L_p$.
$endgroup$
– A.Γ.
Oct 13 '16 at 10:41














$begingroup$
@user251257 True but what does this imply ?
$endgroup$
– bridger
Oct 13 '16 at 11:04




$begingroup$
@user251257 True but what does this imply ?
$endgroup$
– bridger
Oct 13 '16 at 11:04












$begingroup$
@A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
$endgroup$
– bridger
Oct 13 '16 at 11:06




$begingroup$
@A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
$endgroup$
– bridger
Oct 13 '16 at 11:06




2




2




$begingroup$
@Hua It does not help. You may need $L_p$ bound on the derivative too.
$endgroup$
– A.Γ.
Oct 13 '16 at 11:12




$begingroup$
@Hua It does not help. You may need $L_p$ bound on the derivative too.
$endgroup$
– A.Γ.
Oct 13 '16 at 11:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30











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',
autoActivateHeartbeat: false,
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%2f1966625%2fis-this-true-that-the-sequences-evenly-sampled-from-functions-in-lp-are-in-l%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









1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30
















1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30














1












1








1





$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$



This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 13 '16 at 13:30









MaoWaoMaoWao

3,153617




3,153617












  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30


















  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30
















$begingroup$
Why could you exchange the sum and integral in the second equality ?
$endgroup$
– bridger
Oct 13 '16 at 14:08




$begingroup$
Why could you exchange the sum and integral in the second equality ?
$endgroup$
– bridger
Oct 13 '16 at 14:08












$begingroup$
That's an application of Fubini's theorem.
$endgroup$
– MaoWao
Oct 13 '16 at 16:19




$begingroup$
That's an application of Fubini's theorem.
$endgroup$
– MaoWao
Oct 13 '16 at 16:19












$begingroup$
Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
$endgroup$
– bridger
Oct 13 '16 at 16:24




$begingroup$
Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
$endgroup$
– bridger
Oct 13 '16 at 16:24




1




1




$begingroup$
There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
$endgroup$
– MaoWao
Oct 13 '16 at 16:30




$begingroup$
There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
$endgroup$
– MaoWao
Oct 13 '16 at 16:30


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1966625%2fis-this-true-that-the-sequences-evenly-sampled-from-functions-in-lp-are-in-l%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

How do I know what Microsoft account the skydrive app is syncing to?

Image
up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
Read more

When does type information flow backwards in C++?

Image
up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
Read more

Grease: Live!

Image
A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
Read more