Is this linear mapping discontinuous?











up vote
1
down vote

favorite












Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







share|cite|improve this question
























  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25















up vote
1
down vote

favorite












Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







share|cite|improve this question
























  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







share|cite|improve this question















Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory




functional-analysis operator-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 7:52

























asked Nov 29 at 4:59









mathmax

1896




1896












  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25


















  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25
















For the first part consider the functions $frac {x^{n}} n$.
– Kavi Rama Murthy
Nov 29 at 5:25




For the first part consider the functions $frac {x^{n}} n$.
– Kavi Rama Murthy
Nov 29 at 5:25










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer





















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23











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%2f3018203%2fis-this-linear-mapping-discontinuous%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
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer





















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23















up vote
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer





















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23













up vote
1
down vote



accepted







up vote
1
down vote



accepted






For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer












For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 at 6:23









Martin Argerami

123k1176174




123k1176174












  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23


















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23
















Thanks for the answer
– mathmax
Nov 29 at 13:23




Thanks for the answer
– mathmax
Nov 29 at 13:23


















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%2f3018203%2fis-this-linear-mapping-discontinuous%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

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

Image
3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a
Read more

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
Read more

I want to find a topological embedding $f : X rightarrow Y$ and $g: Y rightarrow X$, yet $X$ is not...

Image
0 0 $begingroup$ This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this? general-topology connectedness share | cite | improve this question edited Dec 10 '18 at 3:46 user1101010 775 1 7 30 asked Sep 24 '13 at 23:15
Read more