Rudin functional Analysis chapter 10, exercise 20
This is the exercise:
Suppose $xin A$, $x_nin A$, and $lim x_n=x$. Suppose $Omega$ is
an open set in $mathbb C$ that contains a component of $sigma(x)$.
Prove that $sigma(x_n)$ intersects $Omega$ for all sufficiently
large $n$. (This strengthens Theorem 10.20.) Hint: If
$sigma(x)subsetOmegacupOmega_0$, where $Omega_0$ is an open
set disjoint from $Omega$, consider the function $f$ that is $1$ in
$Omega$, and $0$ in $Omega_0$.
According to Rudin's hint, we can define $fin
H(OmegacupOmega_0)$ as follows:
$$
f(lambda) = left {begin{array}{lll}
1 & lambdain Omega\
0 & lambdain Omega_0\
end{array}right.
$$
And then by symbolic calculus we have
$$
tilde{f}(x) = left {begin{array}{lll}
e & sigma(x)subset Omega\
0 & sigma(x)subset Omega_0\
end{array}right.
$$
For the $xin A$ that $x_nrightarrow x$, we have
$$tilde{f}(x)=tilde{f}(lim_{ntoinfty}x_n)=lim_{ntoinfty}tilde{f}(x_n),$$
meaning ${tilde{f}(x_n)}$ as a sequence in $A$ converges to
$tilde{f}(x)$. So for a large enough $n$;
$$sigmabig(tilde{f}(x_n)big)capsigmabig(tilde{f}(x)big)neqemptyset.$$
And by spectral mapping theorem;
$$fbig(sigma(x_n)big)cap fbig(sigma(x)big)neqemptyset,$$
$$fbig(sigma(x_n)big)cap {0,1}neqemptyset.$$
So we are forced to have only one of these three possibilities;
1) $fbig(sigma(x_n)big)={0,1}$
2) $fbig(sigma(x_n)big)={1}$
3) $fbig(sigma(x_n)big)={0}$
The third case never happens. Because if it happens by the
definition of $f$, we can conclude for any large $n$ that
$$sigma(x_n)subsetOmega_0,$$
in which by tending $n$ to infinity, we get the following
contradiction;
$$sigma(x)subsetOmega_0.$$
In the first and second cases, there will be at least one
$lambdain sigma(x_n)$ such that $f(lambda)=1$ which by
definition simply means $sigma(x_n)$ intersects $Omega$.
Is this true? Thank you for your help.
functional-analysis spectral-theory banach-algebras
edited Dec 5 '18 at 19:09
SmathFunity
53
asked Nov 27 '18 at 19:16
SsFf
385
add a comment |
This is the exercise:
Suppose $xin A$, $x_nin A$, and $lim x_n=x$. Suppose $Omega$ is
an open set in $mathbb C$ that contains a component of $sigma(x)$.
Prove that $sigma(x_n)$ intersects $Omega$ for all sufficiently
large $n$. (This strengthens Theorem 10.20.) Hint: If
$sigma(x)subsetOmegacupOmega_0$, where $Omega_0$ is an open
set disjoint from $Omega$, consider the function $f$ that is $1$ in
$Omega$, and $0$ in $Omega_0$.
According to Rudin's hint, we can define $fin
H(OmegacupOmega_0)$ as follows:
$$
f(lambda) = left {begin{array}{lll}
1 & lambdain Omega\
0 & lambdain Omega_0\
end{array}right.
$$
And then by symbolic calculus we have
$$
tilde{f}(x) = left {begin{array}{lll}
e & sigma(x)subset Omega\
0 & sigma(x)subset Omega_0\
end{array}right.
$$
For the $xin A$ that $x_nrightarrow x$, we have
$$tilde{f}(x)=tilde{f}(lim_{ntoinfty}x_n)=lim_{ntoinfty}tilde{f}(x_n),$$
meaning ${tilde{f}(x_n)}$ as a sequence in $A$ converges to
$tilde{f}(x)$. So for a large enough $n$;
$$sigmabig(tilde{f}(x_n)big)capsigmabig(tilde{f}(x)big)neqemptyset.$$
And by spectral mapping theorem;
$$fbig(sigma(x_n)big)cap fbig(sigma(x)big)neqemptyset,$$
$$fbig(sigma(x_n)big)cap {0,1}neqemptyset.$$
So we are forced to have only one of these three possibilities;
1) $fbig(sigma(x_n)big)={0,1}$
2) $fbig(sigma(x_n)big)={1}$
3) $fbig(sigma(x_n)big)={0}$
The third case never happens. Because if it happens by the
definition of $f$, we can conclude for any large $n$ that
$$sigma(x_n)subsetOmega_0,$$
in which by tending $n$ to infinity, we get the following
contradiction;
$$sigma(x)subsetOmega_0.$$
In the first and second cases, there will be at least one
$lambdain sigma(x_n)$ such that $f(lambda)=1$ which by
definition simply means $sigma(x_n)$ intersects $Omega$.
Is this true? Thank you for your help.
functional-analysis spectral-theory banach-algebras
edited Dec 5 '18 at 19:09
SmathFunity
53
asked Nov 27 '18 at 19:16
SsFf
385
What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23
add a comment |
This is the exercise:
Suppose $xin A$, $x_nin A$, and $lim x_n=x$. Suppose $Omega$ is
an open set in $mathbb C$ that contains a component of $sigma(x)$.
Prove that $sigma(x_n)$ intersects $Omega$ for all sufficiently
large $n$. (This strengthens Theorem 10.20.) Hint: If
$sigma(x)subsetOmegacupOmega_0$, where $Omega_0$ is an open
set disjoint from $Omega$, consider the function $f$ that is $1$ in
$Omega$, and $0$ in $Omega_0$.
According to Rudin's hint, we can define $fin
H(OmegacupOmega_0)$ as follows:
$$
f(lambda) = left {begin{array}{lll}
1 & lambdain Omega\
0 & lambdain Omega_0\
end{array}right.
$$
And then by symbolic calculus we have
$$
tilde{f}(x) = left {begin{array}{lll}
e & sigma(x)subset Omega\
0 & sigma(x)subset Omega_0\
end{array}right.
$$
For the $xin A$ that $x_nrightarrow x$, we have
$$tilde{f}(x)=tilde{f}(lim_{ntoinfty}x_n)=lim_{ntoinfty}tilde{f}(x_n),$$
meaning ${tilde{f}(x_n)}$ as a sequence in $A$ converges to
$tilde{f}(x)$. So for a large enough $n$;
$$sigmabig(tilde{f}(x_n)big)capsigmabig(tilde{f}(x)big)neqemptyset.$$
And by spectral mapping theorem;
$$fbig(sigma(x_n)big)cap fbig(sigma(x)big)neqemptyset,$$
$$fbig(sigma(x_n)big)cap {0,1}neqemptyset.$$
So we are forced to have only one of these three possibilities;
1) $fbig(sigma(x_n)big)={0,1}$
2) $fbig(sigma(x_n)big)={1}$
3) $fbig(sigma(x_n)big)={0}$
The third case never happens. Because if it happens by the
definition of $f$, we can conclude for any large $n$ that
$$sigma(x_n)subsetOmega_0,$$
in which by tending $n$ to infinity, we get the following
contradiction;
$$sigma(x)subsetOmega_0.$$
In the first and second cases, there will be at least one
$lambdain sigma(x_n)$ such that $f(lambda)=1$ which by
definition simply means $sigma(x_n)$ intersects $Omega$.
Is this true? Thank you for your help.
functional-analysis spectral-theory banach-algebras
edited Dec 5 '18 at 19:09
SmathFunity
53
asked Nov 27 '18 at 19:16
SsFf
385
This is the exercise:
Suppose $xin A$, $x_nin A$, and $lim x_n=x$. Suppose $Omega$ is
an open set in $mathbb C$ that contains a component of $sigma(x)$.
Prove that $sigma(x_n)$ intersects $Omega$ for all sufficiently
large $n$. (This strengthens Theorem 10.20.) Hint: If
$sigma(x)subsetOmegacupOmega_0$, where $Omega_0$ is an open
set disjoint from $Omega$, consider the function $f$ that is $1$ in
$Omega$, and $0$ in $Omega_0$.
According to Rudin's hint, we can define $fin
H(OmegacupOmega_0)$ as follows:
$$
f(lambda) = left {begin{array}{lll}
1 & lambdain Omega\
0 & lambdain Omega_0\
end{array}right.
$$
And then by symbolic calculus we have
$$
tilde{f}(x) = left {begin{array}{lll}
e & sigma(x)subset Omega\
0 & sigma(x)subset Omega_0\
end{array}right.
$$
For the $xin A$ that $x_nrightarrow x$, we have
$$tilde{f}(x)=tilde{f}(lim_{ntoinfty}x_n)=lim_{ntoinfty}tilde{f}(x_n),$$
meaning ${tilde{f}(x_n)}$ as a sequence in $A$ converges to
$tilde{f}(x)$. So for a large enough $n$;
$$sigmabig(tilde{f}(x_n)big)capsigmabig(tilde{f}(x)big)neqemptyset.$$
And by spectral mapping theorem;
$$fbig(sigma(x_n)big)cap fbig(sigma(x)big)neqemptyset,$$
$$fbig(sigma(x_n)big)cap {0,1}neqemptyset.$$
So we are forced to have only one of these three possibilities;
1) $fbig(sigma(x_n)big)={0,1}$
2) $fbig(sigma(x_n)big)={1}$
3) $fbig(sigma(x_n)big)={0}$
The third case never happens. Because if it happens by the
definition of $f$, we can conclude for any large $n$ that
$$sigma(x_n)subsetOmega_0,$$
in which by tending $n$ to infinity, we get the following
contradiction;
$$sigma(x)subsetOmega_0.$$
In the first and second cases, there will be at least one
$lambdain sigma(x_n)$ such that $f(lambda)=1$ which by
definition simply means $sigma(x_n)$ intersects $Omega$.
Is this true? Thank you for your help.
functional-analysis spectral-theory banach-algebras
functional-analysis spectral-theory banach-algebras
edited Dec 5 '18 at 19:09
SmathFunity
53
asked Nov 27 '18 at 19:16
SsFf
385
edited Dec 5 '18 at 19:09
SmathFunity
53
asked Nov 27 '18 at 19:16
SsFf
385
edited Dec 5 '18 at 19:09
SmathFunity
53
edited Dec 5 '18 at 19:09
SmathFunity
53
edited Dec 5 '18 at 19:09
SmathFunity
53
53
asked Nov 27 '18 at 19:16
SsFf
385
asked Nov 27 '18 at 19:16
SsFf
385
asked Nov 27 '18 at 19:16
SsFf
385
385
What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23
add a comment |
What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23
What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23
add a comment |
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What is meant by $sigma(x)$?
– Olivier Moschetta
Nov 27 '18 at 19:18
It means the spectrum of $x$.
– SsFf
Nov 28 '18 at 10:58
Actually $sigma(x)={lambdain {mathbb C}: lambda e - x is not invertible}$.
– SsFf
Nov 28 '18 at 11:36
Can you also tell us what is $A$ (or what kind of objects are $x_n$ and $x$)?
– Joel Moreira
Dec 5 '18 at 20:39
They are unknown. The only thing we know is that we have an ordinary Banach algebra.
– SsFf
Dec 12 '18 at 20:23