A counterexample of Banach Steinhaus Theorem












1














I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question
























  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36
















1














I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question
























  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36














1












1








1


2





I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question















I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces




functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 3:51

























asked Dec 9 at 3:04









Jeff

185




185












  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36


















  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36
















Please edit the question to include what you were reading this from.
– Shaun
Dec 9 at 3:11




Please edit the question to include what you were reading this from.
– Shaun
Dec 9 at 3:11












It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
– DanielWainfleet
Dec 9 at 3:44




It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
– DanielWainfleet
Dec 9 at 3:44












You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
– Jeff
Dec 9 at 3:59






You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
– Jeff
Dec 9 at 3:59






1




1




Isn't $k leq C$ for all $k$ a contradiction?
– Kavi Rama Murthy
Dec 9 at 4:54






Isn't $k leq C$ for all $k$ a contradiction?
– Kavi Rama Murthy
Dec 9 at 4:54














it is, but I already solve this problem with your help, thank you for helping me.
– Jeff
Dec 16 at 1:36




it is, but I already solve this problem with your help, thank you for helping me.
– Jeff
Dec 16 at 1:36










1 Answer
1






active

oldest

votes


















0














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37













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%2f3031953%2fa-counterexample-of-banach-steinhaus-theorem%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














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37


















0














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37
















0












0








0






The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer












The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 at 3:38









Riley

1395




1395












  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37




















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37


















Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
– Jeff
Dec 16 at 1:37






Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
– Jeff
Dec 16 at 1:37




















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%2f3031953%2fa-counterexample-of-banach-steinhaus-theorem%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

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more