Counterexample of non invertible operator












4














We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".



When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?






operator-theory operator-algebras







share|cite|improve this question






















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago
















4














We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".



When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?






operator-theory operator-algebras







share|cite|improve this question






















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago














4












4








4







We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".



When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?






operator-theory operator-algebras







share|cite|improve this question













We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".



When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?






operator-theory operator-algebras




operator-theory operator-algebras






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









sinbadh

6,312824




6,312824












  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago


















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago
















"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
5 hours ago




"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
5 hours ago










1 Answer
1






active

oldest

votes


















5














Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$



However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$



Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.






share|cite|improve this answer





















  • This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
    – Kavi Rama Murthy
    1 hour ago











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%2f3053246%2fcounterexample-of-non-invertible-operator%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









5














Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$



However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$



Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.






share|cite|improve this answer





















  • This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
    – Kavi Rama Murthy
    1 hour ago
















5














Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$



However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$



Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.






share|cite|improve this answer





















  • This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
    – Kavi Rama Murthy
    1 hour ago














5












5








5






Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$



However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$



Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.






share|cite|improve this answer












Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$



However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$



Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









mechanodroid

25.6k62245




25.6k62245












  • This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
    – Kavi Rama Murthy
    1 hour ago


















  • This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
    – Kavi Rama Murthy
    1 hour ago
















This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
1 hour ago




This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
1 hour ago


















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%2f3053246%2fcounterexample-of-non-invertible-operator%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

What visual should I use to simply compare current year value vs last year in Power BI desktop

Image
0 I have column SubToBind that shows current year ratio and column PY SubToBind which shows previous year ratio. How can I simply create a card that would be displaying them both in a nice way? Something like that: I cannot achieve that with regular card, I also tried Card with States by OKViz but also unable to see both values. In KPI visual - I dont like the word "Goal", it will confuse end users. Are there any cards or other visuals that would do the job? Thanks powerbi dax powerbi-embedded share | improve this question edited Jun 11 '18 at 20:11 Oleg ...
Read more

How to ignore python UserWarning in pytest?

Image
up vote 1 down vote favorite I am using openpyxl for parsing .xlsm files, and pytest for testing. When I open a file, I get the: OpenPyxl -> UserWarning: Data Validation extension is not supported and will be removed That is not really a problem bcs. program works and I can't change the .xlsm file to fix that. But... When I run pytest with something like: def test_wrong_file_format(): assert check_excel(open_excel('file.xlsm')) == True I will get the warning i mentioned altought check_excel(open_excel('file.xlsm')) returns True and the test should suceed... Is there a nice way to tell the pytest that "It's not a bug it's a feature" and tests should pass even when they get this warning? Is there other way than using something like: with pytest.warns(UserWarning): ...
Read more

Alexandru Averescu

Image
Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center ...
Read more