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Why is $E$ open in $D$?
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I am reading through the proof of the following in Aupetit's A Primer on Spectral Theory.
Let $f$ be an analytic function from a domain $D$ of $mathbbC$ into a Banach algebra $A$. Suppose that $textSp(f(lambda)) subset mathbbR$ for all $lambda in D$, then $textSp(f(lambda))$ is constant on $D$.
I was able to prove that $E$ is closed in $D$, but I fail to see why $E$ is open. Can anyone please help explain to me why this is true?
complex-analysis functional-analysis spectral-theory banach-algebras
asked Jul 22 at 20:32
DJS
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I am reading through the proof of the following in Aupetit's A Primer on Spectral Theory.
Let $f$ be an analytic function from a domain $D$ of $mathbbC$ into a Banach algebra $A$. Suppose that $textSp(f(lambda)) subset mathbbR$ for all $lambda in D$, then $textSp(f(lambda))$ is constant on $D$.
I was able to prove that $E$ is closed in $D$, but I fail to see why $E$ is open. Can anyone please help explain to me why this is true?
complex-analysis functional-analysis spectral-theory banach-algebras
asked Jul 22 at 20:32
DJS
2,312924
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading through the proof of the following in Aupetit's A Primer on Spectral Theory.
Let $f$ be an analytic function from a domain $D$ of $mathbbC$ into a Banach algebra $A$. Suppose that $textSp(f(lambda)) subset mathbbR$ for all $lambda in D$, then $textSp(f(lambda))$ is constant on $D$.
I was able to prove that $E$ is closed in $D$, but I fail to see why $E$ is open. Can anyone please help explain to me why this is true?
complex-analysis functional-analysis spectral-theory banach-algebras
asked Jul 22 at 20:32
DJS
2,312924
I am reading through the proof of the following in Aupetit's A Primer on Spectral Theory.
Let $f$ be an analytic function from a domain $D$ of $mathbbC$ into a Banach algebra $A$. Suppose that $textSp(f(lambda)) subset mathbbR$ for all $lambda in D$, then $textSp(f(lambda))$ is constant on $D$.
I was able to prove that $E$ is closed in $D$, but I fail to see why $E$ is open. Can anyone please help explain to me why this is true?
complex-analysis functional-analysis spectral-theory banach-algebras
asked Jul 22 at 20:32
DJS
2,312924
asked Jul 22 at 20:32
DJS
2,312924
asked Jul 22 at 20:32
DJS
2,312924
asked Jul 22 at 20:32
DJS
2,312924
2,312924
add a comment |Â
add a comment |Â
1 Answer
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That is exactly what is shown in the part of the proof that precedes the definition of $E$: for some positive $delta$, the $delta$-neighborhood of $lambda_0$ is in $E$, as $Sp f(lambda)$ is constant there.
answered Jul 22 at 20:46
A. Pongrácz
2,054120
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That is exactly what is shown in the part of the proof that precedes the definition of $E$: for some positive $delta$, the $delta$-neighborhood of $lambda_0$ is in $E$, as $Sp f(lambda)$ is constant there.
answered Jul 22 at 20:46
A. Pongrácz
2,054120
add a comment |Â
up vote
1
down vote
accepted
That is exactly what is shown in the part of the proof that precedes the definition of $E$: for some positive $delta$, the $delta$-neighborhood of $lambda_0$ is in $E$, as $Sp f(lambda)$ is constant there.
answered Jul 22 at 20:46
A. Pongrácz
2,054120
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That is exactly what is shown in the part of the proof that precedes the definition of $E$: for some positive $delta$, the $delta$-neighborhood of $lambda_0$ is in $E$, as $Sp f(lambda)$ is constant there.
answered Jul 22 at 20:46
A. Pongrácz
2,054120
That is exactly what is shown in the part of the proof that precedes the definition of $E$: for some positive $delta$, the $delta$-neighborhood of $lambda_0$ is in $E$, as $Sp f(lambda)$ is constant there.
answered Jul 22 at 20:46
A. Pongrácz
2,054120
answered Jul 22 at 20:46
A. Pongrácz
2,054120
answered Jul 22 at 20:46
A. Pongrácz
2,054120
answered Jul 22 at 20:46
A. Pongrácz
2,054120
2,054120
add a comment |Â
add a comment |Â
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