Proof of the Riesz-Schauder Theorem (for compact operators) using the Analytical Fredholm Theorem
First of all sorry for my bad English, I'm an Italian student, hope to let you understand!
I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators.
Some infos in advance:
Let $B(mathbb{H})$ the Algebra of bounded operators in the Hilbert space $mathbb{H}$.
The resolvent of an operator $A$ is the subset of $mathbb{C}$, $rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} in B(mathbb{H})}}$.
The spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma(A)=mathbb{C}setminus rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} notin B(mathbb{H})}}$.
The point spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma_p (A)$={$lambda in mathbb{C}: Ay=lambda y$ has a solution $y ne 0$}, in other words $sigma_p (A)$ is the set of eigenvalues of $A$ and holds that $sigma_p (A) subseteq sigma(A)$.
For the proof of the Riesz-Schauder Theorem I will use the following:
Analytic Fredholm Theorem. Let $D$ an open connected subset of $mathbb{C}$. Let $f: D$ $rightarrow$ $B(mathbb{H})$ be an analytic operator-valued function such that $f(z)$ is compact for all $z in D$.
Then, either:
(a) $(I-f(z))^{-1}$ exists for no $z in D$
or
(b) $(I-f(z))^{-1}$ exists for all $z in Dsetminus S$, where $S$ is a discrete subset of $D$ (i.e. a set with has no limit points in $D$).
(Note that $I$ is the identity operator of $B(mathbb{H})$.)
Finally, the main theorem
Riesz-Schauder Theorem. The spectrum $sigma(A)$ of a compact operator $A$ is a discrete set having no limit points except (perhaps) $0$. $sigma(A)$ is finite or countable and in this case $0$ is the only limit point. Further, any nonzero $lambda in sigma(A)$ is an eigenvalue of finite multiplicity.
For the proof (as suggested here http://www.macs.hw.ac.uk/~hg94/pdst11/pdst11_AFTcompact.pdf ) , I will proceed like this:
Let $f: mathbb{C}$ $rightarrow$ $B(mathbb{H})$ such that $f(lambda)=lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $lambda in mathbb{C}$. Because $K(mathbb{H})$={A $in B(mathbb{H})$: $A$ is compact operator} is a $mathbb{C}$-subspace of $B(mathbb{H})$ then $f(lambda)$ is compact for all $lambda in mathbb{C}$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(lambda))^{-1}$ exists for no $z in mathbb{C}$
or
(b) $(I-f(lambda))^{-1}$ exists for all $z in mathbb{C}setminus S$, where $S$ is a discrete subset of $mathbb{C}$ with no limit points.
But, if $lambda=0$, then $(I-f(lambda))^{-1}=(I-0A)^{-1}=I$ exists, so option (b) is true.
Furthermore, from the Analytic Fredholm Theorem proof we know that $S=${$lambda in mathbb{C}$: $lambda A y=y$ has a solution $y ne 0$}, and it is easy to see that $0 notin S$.
We can observe that if $1/lambda notin S$, then
$$(A-lambda I)^{-1}=-1/lambda (I-1/lambda A)^{-1}$$
and so $lambda in rho(A)$.
But holds even the viceversa so, $1/lambda notin S Leftrightarrow lambda in rho(A)$.
Then we have $1/lambda in S Leftrightarrow lambda notin rho(A)$, thus
$$1/lambda in S Leftrightarrow lambda in sigma(A).$$
Here are my doubts. How can I prove that $sigma(A)$={$0$}$cup sigma_p (A)$?
Maybe it is possible to prove that $sigma(A)=Scup {0}$? In this way, it is easy to see that the only limit point is zero.
Instead, I've done something like this:
If $lambda in sigma(A)$ and $lambda ne 0$ then $1/lambda in S$. Thus $1/lambda A y =y$ has a solution $y ne 0$, and $lambda in sigma_p (A)$. So, because for a compact operator it holds that $0 in sigma(A)$, I can state that
$$sigma(A)=sigma_p (A) cup {0}.$$
But in this way I have some troubles attempting to prove that zero is the only limit point.
Certainly I can state that either $sigma_p (A)$ is finite, or, on the contrary, because $sigma_p (A) subseteq sigma(A)$ and $sigma(A)$ is a limited set, then for the Bolzano - Weirstrass theorem $sigma_p (A)$ has a limit point. But $sigma(A)$ is norm-closed, so if $mu$ is a limit point for $sigma_p (A)$ then $mu in sigma(A)$. But how can I state that is $mu = 0$??
Thank you in advance for your time and sorry again for my English!
hilbert-spaces compactness analyticity compact-operators
asked Dec 29 '14 at 17:08
eleguitar
117114
add a comment |
First of all sorry for my bad English, I'm an Italian student, hope to let you understand!
I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators.
Some infos in advance:
Let $B(mathbb{H})$ the Algebra of bounded operators in the Hilbert space $mathbb{H}$.
The resolvent of an operator $A$ is the subset of $mathbb{C}$, $rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} in B(mathbb{H})}}$.
The spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma(A)=mathbb{C}setminus rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} notin B(mathbb{H})}}$.
The point spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma_p (A)$={$lambda in mathbb{C}: Ay=lambda y$ has a solution $y ne 0$}, in other words $sigma_p (A)$ is the set of eigenvalues of $A$ and holds that $sigma_p (A) subseteq sigma(A)$.
For the proof of the Riesz-Schauder Theorem I will use the following:
Analytic Fredholm Theorem. Let $D$ an open connected subset of $mathbb{C}$. Let $f: D$ $rightarrow$ $B(mathbb{H})$ be an analytic operator-valued function such that $f(z)$ is compact for all $z in D$.
Then, either:
(a) $(I-f(z))^{-1}$ exists for no $z in D$
or
(b) $(I-f(z))^{-1}$ exists for all $z in Dsetminus S$, where $S$ is a discrete subset of $D$ (i.e. a set with has no limit points in $D$).
(Note that $I$ is the identity operator of $B(mathbb{H})$.)
Finally, the main theorem
Riesz-Schauder Theorem. The spectrum $sigma(A)$ of a compact operator $A$ is a discrete set having no limit points except (perhaps) $0$. $sigma(A)$ is finite or countable and in this case $0$ is the only limit point. Further, any nonzero $lambda in sigma(A)$ is an eigenvalue of finite multiplicity.
For the proof (as suggested here http://www.macs.hw.ac.uk/~hg94/pdst11/pdst11_AFTcompact.pdf ) , I will proceed like this:
Let $f: mathbb{C}$ $rightarrow$ $B(mathbb{H})$ such that $f(lambda)=lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $lambda in mathbb{C}$. Because $K(mathbb{H})$={A $in B(mathbb{H})$: $A$ is compact operator} is a $mathbb{C}$-subspace of $B(mathbb{H})$ then $f(lambda)$ is compact for all $lambda in mathbb{C}$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(lambda))^{-1}$ exists for no $z in mathbb{C}$
or
(b) $(I-f(lambda))^{-1}$ exists for all $z in mathbb{C}setminus S$, where $S$ is a discrete subset of $mathbb{C}$ with no limit points.
But, if $lambda=0$, then $(I-f(lambda))^{-1}=(I-0A)^{-1}=I$ exists, so option (b) is true.
Furthermore, from the Analytic Fredholm Theorem proof we know that $S=${$lambda in mathbb{C}$: $lambda A y=y$ has a solution $y ne 0$}, and it is easy to see that $0 notin S$.
We can observe that if $1/lambda notin S$, then
$$(A-lambda I)^{-1}=-1/lambda (I-1/lambda A)^{-1}$$
and so $lambda in rho(A)$.
But holds even the viceversa so, $1/lambda notin S Leftrightarrow lambda in rho(A)$.
Then we have $1/lambda in S Leftrightarrow lambda notin rho(A)$, thus
$$1/lambda in S Leftrightarrow lambda in sigma(A).$$
Here are my doubts. How can I prove that $sigma(A)$={$0$}$cup sigma_p (A)$?
Maybe it is possible to prove that $sigma(A)=Scup {0}$? In this way, it is easy to see that the only limit point is zero.
Instead, I've done something like this:
If $lambda in sigma(A)$ and $lambda ne 0$ then $1/lambda in S$. Thus $1/lambda A y =y$ has a solution $y ne 0$, and $lambda in sigma_p (A)$. So, because for a compact operator it holds that $0 in sigma(A)$, I can state that
$$sigma(A)=sigma_p (A) cup {0}.$$
But in this way I have some troubles attempting to prove that zero is the only limit point.
Certainly I can state that either $sigma_p (A)$ is finite, or, on the contrary, because $sigma_p (A) subseteq sigma(A)$ and $sigma(A)$ is a limited set, then for the Bolzano - Weirstrass theorem $sigma_p (A)$ has a limit point. But $sigma(A)$ is norm-closed, so if $mu$ is a limit point for $sigma_p (A)$ then $mu in sigma(A)$. But how can I state that is $mu = 0$??
Thank you in advance for your time and sorry again for my English!
hilbert-spaces compactness analyticity compact-operators
asked Dec 29 '14 at 17:08
eleguitar
117114
add a comment |
First of all sorry for my bad English, I'm an Italian student, hope to let you understand!
I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators.
Some infos in advance:
Let $B(mathbb{H})$ the Algebra of bounded operators in the Hilbert space $mathbb{H}$.
The resolvent of an operator $A$ is the subset of $mathbb{C}$, $rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} in B(mathbb{H})}}$.
The spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma(A)=mathbb{C}setminus rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} notin B(mathbb{H})}}$.
The point spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma_p (A)$={$lambda in mathbb{C}: Ay=lambda y$ has a solution $y ne 0$}, in other words $sigma_p (A)$ is the set of eigenvalues of $A$ and holds that $sigma_p (A) subseteq sigma(A)$.
For the proof of the Riesz-Schauder Theorem I will use the following:
Analytic Fredholm Theorem. Let $D$ an open connected subset of $mathbb{C}$. Let $f: D$ $rightarrow$ $B(mathbb{H})$ be an analytic operator-valued function such that $f(z)$ is compact for all $z in D$.
Then, either:
(a) $(I-f(z))^{-1}$ exists for no $z in D$
or
(b) $(I-f(z))^{-1}$ exists for all $z in Dsetminus S$, where $S$ is a discrete subset of $D$ (i.e. a set with has no limit points in $D$).
(Note that $I$ is the identity operator of $B(mathbb{H})$.)
Finally, the main theorem
Riesz-Schauder Theorem. The spectrum $sigma(A)$ of a compact operator $A$ is a discrete set having no limit points except (perhaps) $0$. $sigma(A)$ is finite or countable and in this case $0$ is the only limit point. Further, any nonzero $lambda in sigma(A)$ is an eigenvalue of finite multiplicity.
For the proof (as suggested here http://www.macs.hw.ac.uk/~hg94/pdst11/pdst11_AFTcompact.pdf ) , I will proceed like this:
Let $f: mathbb{C}$ $rightarrow$ $B(mathbb{H})$ such that $f(lambda)=lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $lambda in mathbb{C}$. Because $K(mathbb{H})$={A $in B(mathbb{H})$: $A$ is compact operator} is a $mathbb{C}$-subspace of $B(mathbb{H})$ then $f(lambda)$ is compact for all $lambda in mathbb{C}$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(lambda))^{-1}$ exists for no $z in mathbb{C}$
or
(b) $(I-f(lambda))^{-1}$ exists for all $z in mathbb{C}setminus S$, where $S$ is a discrete subset of $mathbb{C}$ with no limit points.
But, if $lambda=0$, then $(I-f(lambda))^{-1}=(I-0A)^{-1}=I$ exists, so option (b) is true.
Furthermore, from the Analytic Fredholm Theorem proof we know that $S=${$lambda in mathbb{C}$: $lambda A y=y$ has a solution $y ne 0$}, and it is easy to see that $0 notin S$.
We can observe that if $1/lambda notin S$, then
$$(A-lambda I)^{-1}=-1/lambda (I-1/lambda A)^{-1}$$
and so $lambda in rho(A)$.
But holds even the viceversa so, $1/lambda notin S Leftrightarrow lambda in rho(A)$.
Then we have $1/lambda in S Leftrightarrow lambda notin rho(A)$, thus
$$1/lambda in S Leftrightarrow lambda in sigma(A).$$
Here are my doubts. How can I prove that $sigma(A)$={$0$}$cup sigma_p (A)$?
Maybe it is possible to prove that $sigma(A)=Scup {0}$? In this way, it is easy to see that the only limit point is zero.
Instead, I've done something like this:
If $lambda in sigma(A)$ and $lambda ne 0$ then $1/lambda in S$. Thus $1/lambda A y =y$ has a solution $y ne 0$, and $lambda in sigma_p (A)$. So, because for a compact operator it holds that $0 in sigma(A)$, I can state that
$$sigma(A)=sigma_p (A) cup {0}.$$
But in this way I have some troubles attempting to prove that zero is the only limit point.
Certainly I can state that either $sigma_p (A)$ is finite, or, on the contrary, because $sigma_p (A) subseteq sigma(A)$ and $sigma(A)$ is a limited set, then for the Bolzano - Weirstrass theorem $sigma_p (A)$ has a limit point. But $sigma(A)$ is norm-closed, so if $mu$ is a limit point for $sigma_p (A)$ then $mu in sigma(A)$. But how can I state that is $mu = 0$??
Thank you in advance for your time and sorry again for my English!
hilbert-spaces compactness analyticity compact-operators
asked Dec 29 '14 at 17:08
eleguitar
117114
First of all sorry for my bad English, I'm an Italian student, hope to let you understand!
I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators.
Some infos in advance:
Let $B(mathbb{H})$ the Algebra of bounded operators in the Hilbert space $mathbb{H}$.
The resolvent of an operator $A$ is the subset of $mathbb{C}$, $rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} in B(mathbb{H})}}$.
The spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma(A)=mathbb{C}setminus rho(A)={{lambda in mathbb{C}: (A-lambda I)^{-1} notin B(mathbb{H})}}$.
The point spectrum of an operator $A$ is the subset of $mathbb{C}$, $sigma_p (A)$={$lambda in mathbb{C}: Ay=lambda y$ has a solution $y ne 0$}, in other words $sigma_p (A)$ is the set of eigenvalues of $A$ and holds that $sigma_p (A) subseteq sigma(A)$.
For the proof of the Riesz-Schauder Theorem I will use the following:
Analytic Fredholm Theorem. Let $D$ an open connected subset of $mathbb{C}$. Let $f: D$ $rightarrow$ $B(mathbb{H})$ be an analytic operator-valued function such that $f(z)$ is compact for all $z in D$.
Then, either:
(a) $(I-f(z))^{-1}$ exists for no $z in D$
or
(b) $(I-f(z))^{-1}$ exists for all $z in Dsetminus S$, where $S$ is a discrete subset of $D$ (i.e. a set with has no limit points in $D$).
(Note that $I$ is the identity operator of $B(mathbb{H})$.)
Finally, the main theorem
Riesz-Schauder Theorem. The spectrum $sigma(A)$ of a compact operator $A$ is a discrete set having no limit points except (perhaps) $0$. $sigma(A)$ is finite or countable and in this case $0$ is the only limit point. Further, any nonzero $lambda in sigma(A)$ is an eigenvalue of finite multiplicity.
For the proof (as suggested here http://www.macs.hw.ac.uk/~hg94/pdst11/pdst11_AFTcompact.pdf ) , I will proceed like this:
Let $f: mathbb{C}$ $rightarrow$ $B(mathbb{H})$ such that $f(lambda)=lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $lambda in mathbb{C}$. Because $K(mathbb{H})$={A $in B(mathbb{H})$: $A$ is compact operator} is a $mathbb{C}$-subspace of $B(mathbb{H})$ then $f(lambda)$ is compact for all $lambda in mathbb{C}$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(lambda))^{-1}$ exists for no $z in mathbb{C}$
or
(b) $(I-f(lambda))^{-1}$ exists for all $z in mathbb{C}setminus S$, where $S$ is a discrete subset of $mathbb{C}$ with no limit points.
But, if $lambda=0$, then $(I-f(lambda))^{-1}=(I-0A)^{-1}=I$ exists, so option (b) is true.
Furthermore, from the Analytic Fredholm Theorem proof we know that $S=${$lambda in mathbb{C}$: $lambda A y=y$ has a solution $y ne 0$}, and it is easy to see that $0 notin S$.
We can observe that if $1/lambda notin S$, then
$$(A-lambda I)^{-1}=-1/lambda (I-1/lambda A)^{-1}$$
and so $lambda in rho(A)$.
But holds even the viceversa so, $1/lambda notin S Leftrightarrow lambda in rho(A)$.
Then we have $1/lambda in S Leftrightarrow lambda notin rho(A)$, thus
$$1/lambda in S Leftrightarrow lambda in sigma(A).$$
Here are my doubts. How can I prove that $sigma(A)$={$0$}$cup sigma_p (A)$?
Maybe it is possible to prove that $sigma(A)=Scup {0}$? In this way, it is easy to see that the only limit point is zero.
Instead, I've done something like this:
If $lambda in sigma(A)$ and $lambda ne 0$ then $1/lambda in S$. Thus $1/lambda A y =y$ has a solution $y ne 0$, and $lambda in sigma_p (A)$. So, because for a compact operator it holds that $0 in sigma(A)$, I can state that
$$sigma(A)=sigma_p (A) cup {0}.$$
But in this way I have some troubles attempting to prove that zero is the only limit point.
Certainly I can state that either $sigma_p (A)$ is finite, or, on the contrary, because $sigma_p (A) subseteq sigma(A)$ and $sigma(A)$ is a limited set, then for the Bolzano - Weirstrass theorem $sigma_p (A)$ has a limit point. But $sigma(A)$ is norm-closed, so if $mu$ is a limit point for $sigma_p (A)$ then $mu in sigma(A)$. But how can I state that is $mu = 0$??
Thank you in advance for your time and sorry again for my English!
hilbert-spaces compactness analyticity compact-operators
hilbert-spaces compactness analyticity compact-operators
asked Dec 29 '14 at 17:08
eleguitar
117114
asked Dec 29 '14 at 17:08
eleguitar
117114
asked Dec 29 '14 at 17:08
eleguitar
117114
asked Dec 29 '14 at 17:08
eleguitar
117114
asked Dec 29 '14 at 17:08
eleguitar
117114
117114
add a comment |
add a comment |
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http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.
answered Sep 14 '17 at 10:56
Mariah
1,282518
add a comment |
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http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.
answered Sep 14 '17 at 10:56
Mariah
1,282518
add a comment |
http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.
answered Sep 14 '17 at 10:56
Mariah
1,282518
add a comment |
http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.
answered Sep 14 '17 at 10:56
Mariah
1,282518
http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.
answered Sep 14 '17 at 10:56
Mariah
1,282518
answered Sep 14 '17 at 10:56
Mariah
1,282518
answered Sep 14 '17 at 10:56
Mariah
1,282518
answered Sep 14 '17 at 10:56
Mariah
1,282518
1,282518
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