Example of $∫_Omega|f|^p$ uniformly bounded for $p<p_0$ but $∫_Omega|f|^{p_0}=infty$












0














I am seeking for an example of function $f$ that for $p<p_0$, $int_Omega|f|^p<M$ are uniformly bounded, but $int_Omega|f|^{p_0}=infty$.



Here $Omega$ is finite.



I tried the most usual example $int_0^1frac{1}{x^s}$ for $srightarrow1^-$ but $int_0^1frac{1}{x^s}=frac{1}{1-s}$ is not bounded.






real-analysis







share|cite|improve this question






















  • See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
    – Song
    Dec 29 '18 at 1:11


















0














I am seeking for an example of function $f$ that for $p<p_0$, $int_Omega|f|^p<M$ are uniformly bounded, but $int_Omega|f|^{p_0}=infty$.



Here $Omega$ is finite.



I tried the most usual example $int_0^1frac{1}{x^s}$ for $srightarrow1^-$ but $int_0^1frac{1}{x^s}=frac{1}{1-s}$ is not bounded.






real-analysis







share|cite|improve this question






















  • See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
    – Song
    Dec 29 '18 at 1:11
















0












0








0


1





I am seeking for an example of function $f$ that for $p<p_0$, $int_Omega|f|^p<M$ are uniformly bounded, but $int_Omega|f|^{p_0}=infty$.



Here $Omega$ is finite.



I tried the most usual example $int_0^1frac{1}{x^s}$ for $srightarrow1^-$ but $int_0^1frac{1}{x^s}=frac{1}{1-s}$ is not bounded.






real-analysis







share|cite|improve this question













I am seeking for an example of function $f$ that for $p<p_0$, $int_Omega|f|^p<M$ are uniformly bounded, but $int_Omega|f|^{p_0}=infty$.



Here $Omega$ is finite.



I tried the most usual example $int_0^1frac{1}{x^s}$ for $srightarrow1^-$ but $int_0^1frac{1}{x^s}=frac{1}{1-s}$ is not bounded.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 0:42









P.RP.R

184




184












  • See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
    – Song
    Dec 29 '18 at 1:11




















  • See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
    – Song
    Dec 29 '18 at 1:11


















See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
– Song
Dec 29 '18 at 1:11






See Fatou's lemma; en.m.wikipedia.org/wiki/Fatou%27s_lemma. You can drop the assumption $lvertOmegarvert<infty.$
– Song
Dec 29 '18 at 1:11












1 Answer
1






active

oldest

votes


















1














No such example exists. Since $Omega$ is finite (measure), we may restrict attention to the set where $|f| > 1$. On this set, $|f|^p$ increases to $|f|^{p_0}$ pointwise as $p uparrow p_0$. So by monotone convergence theorem, the integrals converge.






share|cite|improve this answer

















  • 1




    Thanks that's right. What if $p_0=infty$?
    – P.R
    Dec 29 '18 at 1:02










  • @P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
    – mathworker21
    Dec 29 '18 at 1:20












  • @P.R does that make sense?
    – mathworker21
    Dec 29 '18 at 8:31










  • Yeah. Thanks. I think it is right even with p -th root.
    – P.R
    Dec 30 '18 at 1:24











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%2f3055427%2fexample-of-%25e2%2588%25ab-omegafp-uniformly-bounded-for-pp-0-but-%25e2%2588%25ab-omegafp-0%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














No such example exists. Since $Omega$ is finite (measure), we may restrict attention to the set where $|f| > 1$. On this set, $|f|^p$ increases to $|f|^{p_0}$ pointwise as $p uparrow p_0$. So by monotone convergence theorem, the integrals converge.






share|cite|improve this answer

















  • 1




    Thanks that's right. What if $p_0=infty$?
    – P.R
    Dec 29 '18 at 1:02










  • @P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
    – mathworker21
    Dec 29 '18 at 1:20












  • @P.R does that make sense?
    – mathworker21
    Dec 29 '18 at 8:31










  • Yeah. Thanks. I think it is right even with p -th root.
    – P.R
    Dec 30 '18 at 1:24
















1














No such example exists. Since $Omega$ is finite (measure), we may restrict attention to the set where $|f| > 1$. On this set, $|f|^p$ increases to $|f|^{p_0}$ pointwise as $p uparrow p_0$. So by monotone convergence theorem, the integrals converge.






share|cite|improve this answer

















  • 1




    Thanks that's right. What if $p_0=infty$?
    – P.R
    Dec 29 '18 at 1:02










  • @P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
    – mathworker21
    Dec 29 '18 at 1:20












  • @P.R does that make sense?
    – mathworker21
    Dec 29 '18 at 8:31










  • Yeah. Thanks. I think it is right even with p -th root.
    – P.R
    Dec 30 '18 at 1:24














1












1








1






No such example exists. Since $Omega$ is finite (measure), we may restrict attention to the set where $|f| > 1$. On this set, $|f|^p$ increases to $|f|^{p_0}$ pointwise as $p uparrow p_0$. So by monotone convergence theorem, the integrals converge.






share|cite|improve this answer












No such example exists. Since $Omega$ is finite (measure), we may restrict attention to the set where $|f| > 1$. On this set, $|f|^p$ increases to $|f|^{p_0}$ pointwise as $p uparrow p_0$. So by monotone convergence theorem, the integrals converge.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 0:44









mathworker21mathworker21

8,8771928




8,8771928








  • 1




    Thanks that's right. What if $p_0=infty$?
    – P.R
    Dec 29 '18 at 1:02










  • @P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
    – mathworker21
    Dec 29 '18 at 1:20












  • @P.R does that make sense?
    – mathworker21
    Dec 29 '18 at 8:31










  • Yeah. Thanks. I think it is right even with p -th root.
    – P.R
    Dec 30 '18 at 1:24














  • 1




    Thanks that's right. What if $p_0=infty$?
    – P.R
    Dec 29 '18 at 1:02










  • @P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
    – mathworker21
    Dec 29 '18 at 1:20












  • @P.R does that make sense?
    – mathworker21
    Dec 29 '18 at 8:31










  • Yeah. Thanks. I think it is right even with p -th root.
    – P.R
    Dec 30 '18 at 1:24








1




1




Thanks that's right. What if $p_0=infty$?
– P.R
Dec 29 '18 at 1:02




Thanks that's right. What if $p_0=infty$?
– P.R
Dec 29 '18 at 1:02












@P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
– mathworker21
Dec 29 '18 at 1:20






@P.R I think $||f||_infty$ has to be at most $1$ in order for $sup_p int_Omega |f|^p$ to be finite (note that you don't have a $p$-th root). Indeed, if there were some $epsilon > 0$ so that ${|f| > 1+epsilon}$ had positive measure $mu$, then $int_Omega |f|^p ge (1+epsilon)^pmu$.
– mathworker21
Dec 29 '18 at 1:20














@P.R does that make sense?
– mathworker21
Dec 29 '18 at 8:31




@P.R does that make sense?
– mathworker21
Dec 29 '18 at 8:31












Yeah. Thanks. I think it is right even with p -th root.
– P.R
Dec 30 '18 at 1:24




Yeah. Thanks. I think it is right even with p -th root.
– P.R
Dec 30 '18 at 1:24


















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%2f3055427%2fexample-of-%25e2%2588%25ab-omegafp-uniformly-bounded-for-pp-0-but-%25e2%2588%25ab-omegafp-0%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

Questions related to Moebius Transform of Characteristic Function of the Primes

Image
2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
Read more

List of scandals in India

Image
List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more