Uniform convergence doesn't preserve improper integrals












0














Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







share|cite|improve this question






















  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15
















0














Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







share|cite|improve this question






















  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15














0












0








0







Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







share|cite|improve this question













Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence




real-analysis improper-integrals uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 17:57









9Sp

244




244












  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15


















  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15
















The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
– Federico
Nov 26 at 18:06




The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
– Federico
Nov 26 at 18:06












Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
– Federico
Nov 26 at 18:07




Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
– Federico
Nov 26 at 18:07












You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
– Mark Viola
Nov 26 at 18:08






You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
– Mark Viola
Nov 26 at 18:08














If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
– Federico
Nov 26 at 18:08




If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
– Federico
Nov 26 at 18:08












I haven't studied measure yet, but will look into it! thanks.
– 9Sp
Nov 26 at 18:15




I haven't studied measure yet, but will look into it! thanks.
– 9Sp
Nov 26 at 18:15










1 Answer
1






active

oldest

votes


















0














If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20











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%2f3014671%2funiform-convergence-doesnt-preserve-improper-integrals%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









0














If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20
















0














If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20














0












0








0






If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer












If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 18:15









José Carlos Santos

149k22117219




149k22117219












  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20


















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20
















If the improper integral converges uniformly, then the interchange is permissible.
– Mark Viola
Nov 26 at 18:20




If the improper integral converges uniformly, then the interchange is permissible.
– Mark Viola
Nov 26 at 18:20


















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%2f3014671%2funiform-convergence-doesnt-preserve-improper-integrals%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more