Finding the Spectrum of an element of $ell^infty$
$begingroup$
I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.
Am I missing something? Kindly Help!!!
Thanks & Regards in advance
c-star-algebras algebras
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
$endgroup$
add a comment |
$begingroup$
I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.
Am I missing something? Kindly Help!!!
Thanks & Regards in advance
c-star-algebras algebras
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
$endgroup$
add a comment |
$begingroup$
I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.
Am I missing something? Kindly Help!!!
Thanks & Regards in advance
c-star-algebras algebras
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
$endgroup$
I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.
How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.
Am I missing something? Kindly Help!!!
Thanks & Regards in advance
c-star-algebras algebras
c-star-algebras algebras
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
asked Jan 17 at 17:46
Devendra Singh RanaDevendra Singh Rana
7641416
7641416
add a comment |
add a comment |
1 Answer
1
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oldest
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The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.
For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
$endgroup$
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.
For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
$endgroup$
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
add a comment |
$begingroup$
The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.
For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
$endgroup$
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
add a comment |
$begingroup$
The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.
For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
$endgroup$
The spectrum of $finell^infty(S)$ definitely has to include (at least some) limit points of $f(S)$. Consider for example $S=mathbb N$, and $finell^infty(mathbb N)$ given by $f(n)=frac{1}{n}$. Then $0notin f(mathbb N)$, but $f$ is not invertible in $ell^infty(mathbb N)$, so $0$ must be in $sigma(f)$.
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $finell^infty(S)$. If $lambdanotinoverline{f(S)}$, then there is some $varepsilon>0$ such that $|f(s)-lambda|geqvarepsilon$ for all $sin S$. Then show that $g:Stomathbb C$ defined by $g(s)=frac{1}{f(s)-lambda}$ is in $ell^infty(S)$, and an inverse to $f-lambda$.
For the other direction, suppose $lambdain overline{f(S)}$. Then for each $varepsilon>0$, there is some $sin S$ such that $|f(s)-lambda|<varepsilon$. Then either $g$ as given above, is not well-defined, or has norm $>frac{1}{varepsilon}$, and since $varepsilon>0$ was arbitrary, we have $gnotinell^infty(S)$.
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
edited Jan 17 at 18:22
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
answered Jan 17 at 17:57
AweyganAweygan
14.2k21441
14.2k21441
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
add a comment |
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
1
1
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
If you want the details on how to prove the claim in this example, let me know and I will edit.
$endgroup$
– Aweygan
Jan 17 at 17:58
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
$begingroup$
A little hint of the proof will be helpful @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:15
1
1
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
@DevendraSinghRana I gave some details from which a proof can be constructed. Let me know if you need more details.
$endgroup$
– Aweygan
Jan 17 at 18:22
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
$begingroup$
This is actually enough, thanks @Aweygan
$endgroup$
– Devendra Singh Rana
Jan 17 at 18:34
1
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jan 17 at 18:34
add a comment |
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