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Why is $E$ discrete?
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I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.
I understand the proof, except for the very last sentence:
Consequently its set of zeros, precisely $E$, is discrete.
I understand that $Omega = { lambda in D : phi(
lambda)=0 }=E$ and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?
complex-analysis functional-analysis spectral-theory banach-algebras
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DJS
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up vote
1
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I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.
I understand the proof, except for the very last sentence:
Consequently its set of zeros, precisely $E$, is discrete.
I understand that $Omega = { lambda in D : phi(
lambda)=0 }=E$ and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?
complex-analysis functional-analysis spectral-theory banach-algebras
asked yesterday
DJS
2,340926
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.
I understand the proof, except for the very last sentence:
Consequently its set of zeros, precisely $E$, is discrete.
I understand that $Omega = { lambda in D : phi(
lambda)=0 }=E$ and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?
complex-analysis functional-analysis spectral-theory banach-algebras
asked yesterday
DJS
2,340926
I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.
I understand the proof, except for the very last sentence:
Consequently its set of zeros, precisely $E$, is discrete.
I understand that $Omega = { lambda in D : phi(
lambda)=0 }=E$ and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?
complex-analysis functional-analysis spectral-theory banach-algebras
complex-analysis functional-analysis spectral-theory banach-algebras
asked yesterday
DJS
2,340926
asked yesterday
DJS
2,340926
asked yesterday
DJS
2,340926
asked yesterday
DJS
2,340926
asked yesterday
DJS
2,340926
2,340926
add a comment |
add a comment |
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