Mixing
Bounded operators that are not closed.
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If a bounded operator, say $A:D(A)to X$, have $D(A)=X$ then it is closed.
Can anybody construct an example of a bounded linear operator, without resorting and restricting to $D(A)=X$, that is not closed?
Please give the examples including densely defined operators,
Thanks in advance.
functional-analysis operator-theory
$endgroup$
add a comment |
$begingroup$
If a bounded operator, say $A:D(A)to X$, have $D(A)=X$ then it is closed.
Can anybody construct an example of a bounded linear operator, without resorting and restricting to $D(A)=X$, that is not closed?
Please give the examples including densely defined operators,
Thanks in advance.
functional-analysis operator-theory
$endgroup$
$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
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@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54
add a comment |
$begingroup$
If a bounded operator, say $A:D(A)to X$, have $D(A)=X$ then it is closed.
Can anybody construct an example of a bounded linear operator, without resorting and restricting to $D(A)=X$, that is not closed?
Please give the examples including densely defined operators,
Thanks in advance.
functional-analysis operator-theory
$endgroup$
If a bounded operator, say $A:D(A)to X$, have $D(A)=X$ then it is closed.
Can anybody construct an example of a bounded linear operator, without resorting and restricting to $D(A)=X$, that is not closed?
Please give the examples including densely defined operators,
Thanks in advance.
functional-analysis operator-theory
functional-analysis operator-theory
asked Oct 1 '14 at 15:24
user180035
$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
$begingroup$
@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54
add a comment |
$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
$begingroup$
@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54
$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
$begingroup$
@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A (probably to) simple example is $T:Dto X:xmapsto 0$ for some unclosed dense $Dsubsetneqoverline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:Dto Y$ with $Dsubseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:overline{D}to Y$ that is indeed closed."
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
$endgroup$
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
add a comment |
$begingroup$
If $A : mathcal{D}(A) subseteq Xrightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : mathcal{D}(A)subseteq Xrightarrow X$ is the restriction of a bounded operator to a non-closed subspace $mathcal{D}(A)$ of $X$. Any non-closed subspace $mathcal{D}(A)$ and any $A in mathcal{L}(X)$ will work to produce a non-closed $A : mathcal{D}(A)subset Xrightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : mathcal{D}(A)subseteq Xrightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $hat{X}$, and then uniquely extending $A$ to a continuous linear operator $hat{A} : hat{X}rightarrow hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $hat{A} : hat{X}rightarrowhat{X}$ on a Banach space $hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $hat{A}X subseteq X$ and (b) a subspace $mathcal{D}(A)subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $hat{X}$.)
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A (probably to) simple example is $T:Dto X:xmapsto 0$ for some unclosed dense $Dsubsetneqoverline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:Dto Y$ with $Dsubseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:overline{D}to Y$ that is indeed closed."
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
$endgroup$
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
add a comment |
$begingroup$
A (probably to) simple example is $T:Dto X:xmapsto 0$ for some unclosed dense $Dsubsetneqoverline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:Dto Y$ with $Dsubseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:overline{D}to Y$ that is indeed closed."
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
$endgroup$
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
add a comment |
$begingroup$
A (probably to) simple example is $T:Dto X:xmapsto 0$ for some unclosed dense $Dsubsetneqoverline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:Dto Y$ with $Dsubseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:overline{D}to Y$ that is indeed closed."
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
$endgroup$
A (probably to) simple example is $T:Dto X:xmapsto 0$ for some unclosed dense $Dsubsetneqoverline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:Dto Y$ with $Dsubseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:overline{D}to Y$ that is indeed closed."
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
edited Oct 2 '14 at 2:46
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
answered Oct 2 '14 at 2:39
C-Star-PuppyC-Star-Puppy
8,22142162
8,22142162
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
add a comment |
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
A bounded $T : D subsetneq Xrightarrow X$ on a normed space $X$ is closable because the closure of its graph in $Xtimes X$ is still a graph as it cannot include $(0,y)$ unless $y=0$.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 12:55
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
@T.A.E.: Oh good point! :) So but then the domain of its closure is not necessarily the closure of the domain of itself?
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 13:30
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
That sounds correct.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 13:55
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@C-Star-Puppy: "A bounded operator T:D→Y with D⊆X and X not necessarily complete but Y complete, extends to a bounded operator TE:D→Y that is indeed closed." I couldn't find a reference with such a theorem. Could you suggest where I could find it, or how this statement could be proven?
$endgroup$
– Sanket_Diwale
Jan 4 at 13:59
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
$begingroup$
@Sanket_Diwale: Bounded, i.e. continuous, operators extend continuously to the completion (assuming that the target is complete), and a bounded operator is always closed, i.e. has a closed graph. The first statement and its proof can be found for example in Kreyszig, while the second is so trivial that it doesn't even require a mention, but probably you can find this in Kreyszig, too. Hope this helps ;)
$endgroup$
– C-Star-Puppy
Jan 6 at 14:17
add a comment |
$begingroup$
If $A : mathcal{D}(A) subseteq Xrightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : mathcal{D}(A)subseteq Xrightarrow X$ is the restriction of a bounded operator to a non-closed subspace $mathcal{D}(A)$ of $X$. Any non-closed subspace $mathcal{D}(A)$ and any $A in mathcal{L}(X)$ will work to produce a non-closed $A : mathcal{D}(A)subset Xrightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : mathcal{D}(A)subseteq Xrightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $hat{X}$, and then uniquely extending $A$ to a continuous linear operator $hat{A} : hat{X}rightarrow hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $hat{A} : hat{X}rightarrowhat{X}$ on a Banach space $hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $hat{A}X subseteq X$ and (b) a subspace $mathcal{D}(A)subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $hat{X}$.)
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
If $A : mathcal{D}(A) subseteq Xrightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : mathcal{D}(A)subseteq Xrightarrow X$ is the restriction of a bounded operator to a non-closed subspace $mathcal{D}(A)$ of $X$. Any non-closed subspace $mathcal{D}(A)$ and any $A in mathcal{L}(X)$ will work to produce a non-closed $A : mathcal{D}(A)subset Xrightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : mathcal{D}(A)subseteq Xrightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $hat{X}$, and then uniquely extending $A$ to a continuous linear operator $hat{A} : hat{X}rightarrow hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $hat{A} : hat{X}rightarrowhat{X}$ on a Banach space $hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $hat{A}X subseteq X$ and (b) a subspace $mathcal{D}(A)subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $hat{X}$.)
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
If $A : mathcal{D}(A) subseteq Xrightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : mathcal{D}(A)subseteq Xrightarrow X$ is the restriction of a bounded operator to a non-closed subspace $mathcal{D}(A)$ of $X$. Any non-closed subspace $mathcal{D}(A)$ and any $A in mathcal{L}(X)$ will work to produce a non-closed $A : mathcal{D}(A)subset Xrightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : mathcal{D}(A)subseteq Xrightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $hat{X}$, and then uniquely extending $A$ to a continuous linear operator $hat{A} : hat{X}rightarrow hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $hat{A} : hat{X}rightarrowhat{X}$ on a Banach space $hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $hat{A}X subseteq X$ and (b) a subspace $mathcal{D}(A)subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $hat{X}$.)
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
If $A : mathcal{D}(A) subseteq Xrightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : mathcal{D}(A)subseteq Xrightarrow X$ is the restriction of a bounded operator to a non-closed subspace $mathcal{D}(A)$ of $X$. Any non-closed subspace $mathcal{D}(A)$ and any $A in mathcal{L}(X)$ will work to produce a non-closed $A : mathcal{D}(A)subset Xrightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : mathcal{D}(A)subseteq Xrightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $hat{X}$, and then uniquely extending $A$ to a continuous linear operator $hat{A} : hat{X}rightarrow hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $hat{A} : hat{X}rightarrowhat{X}$ on a Banach space $hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $hat{A}X subseteq X$ and (b) a subspace $mathcal{D}(A)subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $hat{X}$.)
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Oct 2 '14 at 12:50
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
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$begingroup$
@T.A.E.: Actually that equivalence holds only if the codomain is complete. Moreover that extension also only exists if the codomain is complete.
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 0:11
$begingroup$
@Freeze_S : Yes, I retract that remark.
$endgroup$
– DisintegratingByParts
Oct 2 '14 at 1:51
$begingroup$
@T.A.E.: Well the problem is that most of the results really only hold for both the ambient spaces being Banach spaces even for the target space - except that little exception "closed domain operator bounded implies operator closed"...
$endgroup$
– C-Star-Puppy
Oct 2 '14 at 1:54