Proof of the Riesz-Schauder Theorem (for compact operators) using the Analytical Fredholm Theorem












5














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







share|cite|improve this question



























    5














    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







    share|cite|improve this question

























      5












      5








      5


      4





      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







      share|cite|improve this question













      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






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      asked Dec 29 '14 at 17:08









      eleguitar

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      117114






















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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.






          share|cite|improve this answer





















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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.






            share|cite|improve this answer


























              0














              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.






              share|cite|improve this answer
























                0












                0








                0






                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 14 '17 at 10:56









                Mariah

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                    Musa

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                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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