If for some $ninmathbb{N}$, $T^{n+1}left( Xright) =T^{n}left( Xright)$ do we have $T^{n}left(Xright) $...
$begingroup$
Let $X$ be a Banach space, and let $T$ be a bounded operator on $X$ such
that for some $nin
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$, $T^{n+1}left( Xright) =T^{n}left( Xright) $.
Do we have $T^{n}left( Xright) $ closed ?
linear-algebra functional-analysis operator-theory banach-spaces
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space, and let $T$ be a bounded operator on $X$ such
that for some $nin
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$, $T^{n+1}left( Xright) =T^{n}left( Xright) $.
Do we have $T^{n}left( Xright) $ closed ?
linear-algebra functional-analysis operator-theory banach-spaces
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
$endgroup$
3
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
1
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
1
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17
add a comment |
$begingroup$
Let $X$ be a Banach space, and let $T$ be a bounded operator on $X$ such
that for some $nin
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$, $T^{n+1}left( Xright) =T^{n}left( Xright) $.
Do we have $T^{n}left( Xright) $ closed ?
linear-algebra functional-analysis operator-theory banach-spaces
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
$endgroup$
Let $X$ be a Banach space, and let $T$ be a bounded operator on $X$ such
that for some $nin
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$, $T^{n+1}left( Xright) =T^{n}left( Xright) $.
Do we have $T^{n}left( Xright) $ closed ?
linear-algebra functional-analysis operator-theory banach-spaces
linear-algebra functional-analysis operator-theory banach-spaces
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
asked Dec 23 '18 at 22:10
Djalal OunadjelaDjalal Ounadjela
30618
30618
3
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
1
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
1
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17
add a comment |
3
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
1
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
1
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17
3
3
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
1
1
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
1
1
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $implies$ Equivalent statement: Let us have a Banach space $X$ and some $Tin B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence ${x_i}$ in $T(X)$ there is a limit $xin X$, so $Tx_ito Txin T(X)$.
Proof: Equivalent statement $implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $overline{X}$ denote the completion of $X$. Define $Sin B(overline{X})$ by $S({x_i}):=lim Tx_i$. Then $S(S(overline{X}))=S(overline{X})=X$ and $X$ is not closed in $overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,|cdot|_p)$ of all finite sequences of real numbers.
Proof: Let $Tin B(d)$ be surjective. Then we have some sequence ${x_i}$ such that $Tx_i=e_i$. We have that $$y_i:=sum_{k=1}^ifrac{x_k}{|x_k|k^2}$$ is a Cauchy sequence, and ${Ty_i}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that ${Ty_i}$ is not convergent. I am still not sure whether there exists a surjective $Tin B(C[0,1],|cdot|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050743%2fif-for-some-n-in-mathbbn-tn1-left-x-right-tn-left-x-right-do%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
$begingroup$
This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $implies$ Equivalent statement: Let us have a Banach space $X$ and some $Tin B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence ${x_i}$ in $T(X)$ there is a limit $xin X$, so $Tx_ito Txin T(X)$.
Proof: Equivalent statement $implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $overline{X}$ denote the completion of $X$. Define $Sin B(overline{X})$ by $S({x_i}):=lim Tx_i$. Then $S(S(overline{X}))=S(overline{X})=X$ and $X$ is not closed in $overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,|cdot|_p)$ of all finite sequences of real numbers.
Proof: Let $Tin B(d)$ be surjective. Then we have some sequence ${x_i}$ such that $Tx_i=e_i$. We have that $$y_i:=sum_{k=1}^ifrac{x_k}{|x_k|k^2}$$ is a Cauchy sequence, and ${Ty_i}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that ${Ty_i}$ is not convergent. I am still not sure whether there exists a surjective $Tin B(C[0,1],|cdot|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
add a comment |
$begingroup$
This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $implies$ Equivalent statement: Let us have a Banach space $X$ and some $Tin B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence ${x_i}$ in $T(X)$ there is a limit $xin X$, so $Tx_ito Txin T(X)$.
Proof: Equivalent statement $implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $overline{X}$ denote the completion of $X$. Define $Sin B(overline{X})$ by $S({x_i}):=lim Tx_i$. Then $S(S(overline{X}))=S(overline{X})=X$ and $X$ is not closed in $overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,|cdot|_p)$ of all finite sequences of real numbers.
Proof: Let $Tin B(d)$ be surjective. Then we have some sequence ${x_i}$ such that $Tx_i=e_i$. We have that $$y_i:=sum_{k=1}^ifrac{x_k}{|x_k|k^2}$$ is a Cauchy sequence, and ${Ty_i}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that ${Ty_i}$ is not convergent. I am still not sure whether there exists a surjective $Tin B(C[0,1],|cdot|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
add a comment |
$begingroup$
This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $implies$ Equivalent statement: Let us have a Banach space $X$ and some $Tin B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence ${x_i}$ in $T(X)$ there is a limit $xin X$, so $Tx_ito Txin T(X)$.
Proof: Equivalent statement $implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $overline{X}$ denote the completion of $X$. Define $Sin B(overline{X})$ by $S({x_i}):=lim Tx_i$. Then $S(S(overline{X}))=S(overline{X})=X$ and $X$ is not closed in $overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,|cdot|_p)$ of all finite sequences of real numbers.
Proof: Let $Tin B(d)$ be surjective. Then we have some sequence ${x_i}$ such that $Tx_i=e_i$. We have that $$y_i:=sum_{k=1}^ifrac{x_k}{|x_k|k^2}$$ is a Cauchy sequence, and ${Ty_i}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that ${Ty_i}$ is not convergent. I am still not sure whether there exists a surjective $Tin B(C[0,1],|cdot|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
$endgroup$
This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $implies$ Equivalent statement: Let us have a Banach space $X$ and some $Tin B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence ${x_i}$ in $T(X)$ there is a limit $xin X$, so $Tx_ito Txin T(X)$.
Proof: Equivalent statement $implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $Tin B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $overline{X}$ denote the completion of $X$. Define $Sin B(overline{X})$ by $S({x_i}):=lim Tx_i$. Then $S(S(overline{X}))=S(overline{X})=X$ and $X$ is not closed in $overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,|cdot|_p)$ of all finite sequences of real numbers.
Proof: Let $Tin B(d)$ be surjective. Then we have some sequence ${x_i}$ such that $Tx_i=e_i$. We have that $$y_i:=sum_{k=1}^ifrac{x_k}{|x_k|k^2}$$ is a Cauchy sequence, and ${Ty_i}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that ${Ty_i}$ is not convergent. I am still not sure whether there exists a surjective $Tin B(C[0,1],|cdot|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
answered Dec 24 '18 at 15:09
SmileyCraftSmileyCraft
3,601517
3,601517
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
add a comment |
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 24 '18 at 18:58
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050743%2fif-for-some-n-in-mathbbn-tn1-left-x-right-tn-left-x-right-do%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
3
$begingroup$
You can get rid of the $n$. If $T^{n+1}(X)=T^n(X)$, then $T^{2n}(X)=T^n(X)$, so $(T^n)^2(X)=(T^n)^1(X)$. This is just an instance of the case $n=1$. Hence we just need to prove this for $n=1$.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:25
1
$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 22:27
1
$begingroup$
Is there a particular reason why you think that the image might be closed?
$endgroup$
– Nathanael Skrepek
Dec 23 '18 at 23:09
$begingroup$
Yes, because $Tleft( overline{T^{n}left( Xright) }right) subset T^{n}left( Xright) $
$endgroup$
– Djalal Ounadjela
Dec 23 '18 at 23:17