If for some $ninmathbb{N}$, $T^{n+1}left( Xright) =T^{n}left( Xright)$ do we have $T^{n}left(Xright) $...












7












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















7












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














7












7








7


4



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Dec 24 '18 at 18:58











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Dec 24 '18 at 18:58
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Dec 24 '18 at 18:58














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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