Why we assume closed graph when we discuss the spectrum of an unbounded operator?
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This might be a duplicate, but I searched for it and did't find any discussion about it. If there is can anyone give me a link?
In my textbook it says that it restricts the discussion of the spectrum of unbounded operators to those with closed graph, but it doesn't illustrate why. I am not so clear about this requirement. Can anyone give an example of what will happen if it doesn't have a closed graph?
functional-analysis spectral-theory
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
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add a comment |
$begingroup$
This might be a duplicate, but I searched for it and did't find any discussion about it. If there is can anyone give me a link?
In my textbook it says that it restricts the discussion of the spectrum of unbounded operators to those with closed graph, but it doesn't illustrate why. I am not so clear about this requirement. Can anyone give an example of what will happen if it doesn't have a closed graph?
functional-analysis spectral-theory
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
$endgroup$
add a comment |
$begingroup$
This might be a duplicate, but I searched for it and did't find any discussion about it. If there is can anyone give me a link?
In my textbook it says that it restricts the discussion of the spectrum of unbounded operators to those with closed graph, but it doesn't illustrate why. I am not so clear about this requirement. Can anyone give an example of what will happen if it doesn't have a closed graph?
functional-analysis spectral-theory
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
$endgroup$
This might be a duplicate, but I searched for it and did't find any discussion about it. If there is can anyone give me a link?
In my textbook it says that it restricts the discussion of the spectrum of unbounded operators to those with closed graph, but it doesn't illustrate why. I am not so clear about this requirement. Can anyone give an example of what will happen if it doesn't have a closed graph?
functional-analysis spectral-theory
functional-analysis spectral-theory
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
asked Dec 1 '18 at 23:12
ApocalypseApocalypse
1378
1378
add a comment |
add a comment |
2 Answers
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The original study of the resolvent operator was made by Fredholm in latter part of the 1800's. He was studying the eigenvalue/eigenfunction problem for symmetric differential operators. Fredholm's "resolvent" operator $(L-lambda I)^{-1}$ was constructed through an argument involving discretization of the PDE and limits of determinants of matrices. Fredholm found singularities in $lambda$ at the eigenvalues of the PDE, and the residues at poles of that resolvent were projections onto the eigenspaces associated with the eigenvalue. This remarkable achievement led to a general study of the resolvent operator. The set of singularities of $lambdamapsto(L-lambda I)^{-1}$ was referred to as spectrum, because of its similarity to the line spectrum observed for atoms. We now know that there are strong connections between the two through Quantum Mechanics, but that was not known at the time; Quantum Mechanics did not yet exist. Fredholm was the first to define an abstract linear operator $L$, to study the resolvent of $L$, and to study orthogonal expansions of a selfadjoint operator $L$ through its resolvent and the resides of that resolvent.
Riesz abstracted Fredholm's techniques--which were quite elaborate and rather opaque--to come up with a theory of compact operators whose basic presentation has not changed much in the century since. Spectral theory was set up to deal with the singularities of the resolvent $(L-lambda I)^{-1}$, which was assumed to be a bounded operator. The underlying space was assumed to be complete. A bounded inverse $(L-lambda I)^{-1}$ that is defined on a full Banach space forces $L$ to have a closed graph. So Spectral Theory required from the start that $L$ be closed. This was necessary in order to consider a classical spectral analysis through the resolvent.
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
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If $A: D(A) to X$ has a closed graph and if $A - lambda$ happens to be injective and surjective, then the inverse operator $(A-lambda)^{-1}$ is immediately bounded. This follows from the closed graph theorem. This is probably why your textbook restricts to this case.
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
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$begingroup$
So it is making the resolvent have better property?
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– Apocalypse
Dec 2 '18 at 0:02
add a comment |
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2 Answers
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active
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2 Answers
2
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votes
$begingroup$
The original study of the resolvent operator was made by Fredholm in latter part of the 1800's. He was studying the eigenvalue/eigenfunction problem for symmetric differential operators. Fredholm's "resolvent" operator $(L-lambda I)^{-1}$ was constructed through an argument involving discretization of the PDE and limits of determinants of matrices. Fredholm found singularities in $lambda$ at the eigenvalues of the PDE, and the residues at poles of that resolvent were projections onto the eigenspaces associated with the eigenvalue. This remarkable achievement led to a general study of the resolvent operator. The set of singularities of $lambdamapsto(L-lambda I)^{-1}$ was referred to as spectrum, because of its similarity to the line spectrum observed for atoms. We now know that there are strong connections between the two through Quantum Mechanics, but that was not known at the time; Quantum Mechanics did not yet exist. Fredholm was the first to define an abstract linear operator $L$, to study the resolvent of $L$, and to study orthogonal expansions of a selfadjoint operator $L$ through its resolvent and the resides of that resolvent.
Riesz abstracted Fredholm's techniques--which were quite elaborate and rather opaque--to come up with a theory of compact operators whose basic presentation has not changed much in the century since. Spectral theory was set up to deal with the singularities of the resolvent $(L-lambda I)^{-1}$, which was assumed to be a bounded operator. The underlying space was assumed to be complete. A bounded inverse $(L-lambda I)^{-1}$ that is defined on a full Banach space forces $L$ to have a closed graph. So Spectral Theory required from the start that $L$ be closed. This was necessary in order to consider a classical spectral analysis through the resolvent.
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
$endgroup$
add a comment |
$begingroup$
The original study of the resolvent operator was made by Fredholm in latter part of the 1800's. He was studying the eigenvalue/eigenfunction problem for symmetric differential operators. Fredholm's "resolvent" operator $(L-lambda I)^{-1}$ was constructed through an argument involving discretization of the PDE and limits of determinants of matrices. Fredholm found singularities in $lambda$ at the eigenvalues of the PDE, and the residues at poles of that resolvent were projections onto the eigenspaces associated with the eigenvalue. This remarkable achievement led to a general study of the resolvent operator. The set of singularities of $lambdamapsto(L-lambda I)^{-1}$ was referred to as spectrum, because of its similarity to the line spectrum observed for atoms. We now know that there are strong connections between the two through Quantum Mechanics, but that was not known at the time; Quantum Mechanics did not yet exist. Fredholm was the first to define an abstract linear operator $L$, to study the resolvent of $L$, and to study orthogonal expansions of a selfadjoint operator $L$ through its resolvent and the resides of that resolvent.
Riesz abstracted Fredholm's techniques--which were quite elaborate and rather opaque--to come up with a theory of compact operators whose basic presentation has not changed much in the century since. Spectral theory was set up to deal with the singularities of the resolvent $(L-lambda I)^{-1}$, which was assumed to be a bounded operator. The underlying space was assumed to be complete. A bounded inverse $(L-lambda I)^{-1}$ that is defined on a full Banach space forces $L$ to have a closed graph. So Spectral Theory required from the start that $L$ be closed. This was necessary in order to consider a classical spectral analysis through the resolvent.
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
$endgroup$
add a comment |
$begingroup$
The original study of the resolvent operator was made by Fredholm in latter part of the 1800's. He was studying the eigenvalue/eigenfunction problem for symmetric differential operators. Fredholm's "resolvent" operator $(L-lambda I)^{-1}$ was constructed through an argument involving discretization of the PDE and limits of determinants of matrices. Fredholm found singularities in $lambda$ at the eigenvalues of the PDE, and the residues at poles of that resolvent were projections onto the eigenspaces associated with the eigenvalue. This remarkable achievement led to a general study of the resolvent operator. The set of singularities of $lambdamapsto(L-lambda I)^{-1}$ was referred to as spectrum, because of its similarity to the line spectrum observed for atoms. We now know that there are strong connections between the two through Quantum Mechanics, but that was not known at the time; Quantum Mechanics did not yet exist. Fredholm was the first to define an abstract linear operator $L$, to study the resolvent of $L$, and to study orthogonal expansions of a selfadjoint operator $L$ through its resolvent and the resides of that resolvent.
Riesz abstracted Fredholm's techniques--which were quite elaborate and rather opaque--to come up with a theory of compact operators whose basic presentation has not changed much in the century since. Spectral theory was set up to deal with the singularities of the resolvent $(L-lambda I)^{-1}$, which was assumed to be a bounded operator. The underlying space was assumed to be complete. A bounded inverse $(L-lambda I)^{-1}$ that is defined on a full Banach space forces $L$ to have a closed graph. So Spectral Theory required from the start that $L$ be closed. This was necessary in order to consider a classical spectral analysis through the resolvent.
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
$endgroup$
The original study of the resolvent operator was made by Fredholm in latter part of the 1800's. He was studying the eigenvalue/eigenfunction problem for symmetric differential operators. Fredholm's "resolvent" operator $(L-lambda I)^{-1}$ was constructed through an argument involving discretization of the PDE and limits of determinants of matrices. Fredholm found singularities in $lambda$ at the eigenvalues of the PDE, and the residues at poles of that resolvent were projections onto the eigenspaces associated with the eigenvalue. This remarkable achievement led to a general study of the resolvent operator. The set of singularities of $lambdamapsto(L-lambda I)^{-1}$ was referred to as spectrum, because of its similarity to the line spectrum observed for atoms. We now know that there are strong connections between the two through Quantum Mechanics, but that was not known at the time; Quantum Mechanics did not yet exist. Fredholm was the first to define an abstract linear operator $L$, to study the resolvent of $L$, and to study orthogonal expansions of a selfadjoint operator $L$ through its resolvent and the resides of that resolvent.
Riesz abstracted Fredholm's techniques--which were quite elaborate and rather opaque--to come up with a theory of compact operators whose basic presentation has not changed much in the century since. Spectral theory was set up to deal with the singularities of the resolvent $(L-lambda I)^{-1}$, which was assumed to be a bounded operator. The underlying space was assumed to be complete. A bounded inverse $(L-lambda I)^{-1}$ that is defined on a full Banach space forces $L$ to have a closed graph. So Spectral Theory required from the start that $L$ be closed. This was necessary in order to consider a classical spectral analysis through the resolvent.
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
edited Dec 2 '18 at 21:00
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
answered Dec 2 '18 at 1:32
DisintegratingByPartsDisintegratingByParts
58.8k42579
58.8k42579
add a comment |
add a comment |
$begingroup$
If $A: D(A) to X$ has a closed graph and if $A - lambda$ happens to be injective and surjective, then the inverse operator $(A-lambda)^{-1}$ is immediately bounded. This follows from the closed graph theorem. This is probably why your textbook restricts to this case.
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
$endgroup$
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
add a comment |
$begingroup$
If $A: D(A) to X$ has a closed graph and if $A - lambda$ happens to be injective and surjective, then the inverse operator $(A-lambda)^{-1}$ is immediately bounded. This follows from the closed graph theorem. This is probably why your textbook restricts to this case.
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
$endgroup$
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
add a comment |
$begingroup$
If $A: D(A) to X$ has a closed graph and if $A - lambda$ happens to be injective and surjective, then the inverse operator $(A-lambda)^{-1}$ is immediately bounded. This follows from the closed graph theorem. This is probably why your textbook restricts to this case.
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
$endgroup$
If $A: D(A) to X$ has a closed graph and if $A - lambda$ happens to be injective and surjective, then the inverse operator $(A-lambda)^{-1}$ is immediately bounded. This follows from the closed graph theorem. This is probably why your textbook restricts to this case.
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
answered Dec 1 '18 at 23:51
Bartosz MalmanBartosz Malman
8161620
8161620
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
add a comment |
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
$begingroup$
So it is making the resolvent have better property?
$endgroup$
– Apocalypse
Dec 2 '18 at 0:02
add a comment |
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