Sequence of functions having a convergent subsequence












0












$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49
















0












$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49














0












0








0


1



$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$




let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 '15 at 7:01









Mhenni Benghorbal

43.2k63675




43.2k63675










asked Jan 9 '15 at 5:47









LearnmoreLearnmore

17.8k324100




17.8k324100








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49














  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49








1




1




$begingroup$
What's your problem with the previous version
$endgroup$
– Learnmore
Jan 9 '15 at 5:48




$begingroup$
What's your problem with the previous version
$endgroup$
– Learnmore
Jan 9 '15 at 5:48












$begingroup$
I cant understand that.Please help me
$endgroup$
– Learnmore
Jan 9 '15 at 5:49




$begingroup$
I cant understand that.Please help me
$endgroup$
– Learnmore
Jan 9 '15 at 5:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35











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%2f1097339%2fsequence-of-functions-having-a-convergent-subsequence%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












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35
















1












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35














1












1








1





$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$



It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 7:34

























answered Jan 22 '15 at 20:20









Alex RavskyAlex Ravsky

41.5k32282




41.5k32282












  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35


















  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35
















$begingroup$
@ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
$endgroup$
– jasmine
Sep 14 '18 at 16:04






$begingroup$
@ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
$endgroup$
– jasmine
Sep 14 '18 at 16:04






1




1




$begingroup$
@jasmine Thanks for your remark. I corrected and extended my answer.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:35




$begingroup$
@jasmine Thanks for your remark. I corrected and extended my answer.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:35


















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%2f1097339%2fsequence-of-functions-having-a-convergent-subsequence%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more