Mixing
Operator whose spectrum is given compact set
$begingroup$
Let $Asubset mathbb{C}$ be a compact subset.
Since $A$ is compact and metric space, it is separable, say $overline{lbrace a_nrbrace_{n=1}^infty}=A$.
Let $mathcal{l}^2(mathbb{Z})$ be the Hilbert space consisting of $L^2$-summable sequences and $lbrace e_nrbrace_{n=1}^infty$ be the canonical basis of $mathcal{l}^2(mathbb{Z})$.
Define an operator $Tcolonmathcal{l}^2(mathbb{Z})tomathcal{l}^2(mathbb{Z})$ by sending $e_n$ to $a_ne_n$. I want to prove that $A=sigma(T)$, where $sigma(T)$ is the spectrum of $T$.
What I can prove is that $A=overline{lbrace a_nrbrace_{n=1}^infty}subset sigma(T)$ because each $a_n$ is an eigenvalue of $T$ and $sigma(T)$ is closed. How can I prove the other inclusion, namely $sigma(T)subset A$?
functional-analysis
edited Mar 23 '11 at 16:44
AD.
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
$endgroup$
add a comment |
$begingroup$
Let $Asubset mathbb{C}$ be a compact subset.
Since $A$ is compact and metric space, it is separable, say $overline{lbrace a_nrbrace_{n=1}^infty}=A$.
Let $mathcal{l}^2(mathbb{Z})$ be the Hilbert space consisting of $L^2$-summable sequences and $lbrace e_nrbrace_{n=1}^infty$ be the canonical basis of $mathcal{l}^2(mathbb{Z})$.
Define an operator $Tcolonmathcal{l}^2(mathbb{Z})tomathcal{l}^2(mathbb{Z})$ by sending $e_n$ to $a_ne_n$. I want to prove that $A=sigma(T)$, where $sigma(T)$ is the spectrum of $T$.
What I can prove is that $A=overline{lbrace a_nrbrace_{n=1}^infty}subset sigma(T)$ because each $a_n$ is an eigenvalue of $T$ and $sigma(T)$ is closed. How can I prove the other inclusion, namely $sigma(T)subset A$?
functional-analysis
edited Mar 23 '11 at 16:44
AD.
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
$endgroup$
$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13
add a comment |
$begingroup$
Let $Asubset mathbb{C}$ be a compact subset.
Since $A$ is compact and metric space, it is separable, say $overline{lbrace a_nrbrace_{n=1}^infty}=A$.
Let $mathcal{l}^2(mathbb{Z})$ be the Hilbert space consisting of $L^2$-summable sequences and $lbrace e_nrbrace_{n=1}^infty$ be the canonical basis of $mathcal{l}^2(mathbb{Z})$.
Define an operator $Tcolonmathcal{l}^2(mathbb{Z})tomathcal{l}^2(mathbb{Z})$ by sending $e_n$ to $a_ne_n$. I want to prove that $A=sigma(T)$, where $sigma(T)$ is the spectrum of $T$.
What I can prove is that $A=overline{lbrace a_nrbrace_{n=1}^infty}subset sigma(T)$ because each $a_n$ is an eigenvalue of $T$ and $sigma(T)$ is closed. How can I prove the other inclusion, namely $sigma(T)subset A$?
functional-analysis
edited Mar 23 '11 at 16:44
AD.
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
$endgroup$
Let $Asubset mathbb{C}$ be a compact subset.
Since $A$ is compact and metric space, it is separable, say $overline{lbrace a_nrbrace_{n=1}^infty}=A$.
Let $mathcal{l}^2(mathbb{Z})$ be the Hilbert space consisting of $L^2$-summable sequences and $lbrace e_nrbrace_{n=1}^infty$ be the canonical basis of $mathcal{l}^2(mathbb{Z})$.
Define an operator $Tcolonmathcal{l}^2(mathbb{Z})tomathcal{l}^2(mathbb{Z})$ by sending $e_n$ to $a_ne_n$. I want to prove that $A=sigma(T)$, where $sigma(T)$ is the spectrum of $T$.
What I can prove is that $A=overline{lbrace a_nrbrace_{n=1}^infty}subset sigma(T)$ because each $a_n$ is an eigenvalue of $T$ and $sigma(T)$ is closed. How can I prove the other inclusion, namely $sigma(T)subset A$?
functional-analysis
functional-analysis
edited Mar 23 '11 at 16:44
AD.
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
edited Mar 23 '11 at 16:44
AD.
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
edited Mar 23 '11 at 16:44
AD.
8,76383163
edited Mar 23 '11 at 16:44
AD.
8,76383163
edited Mar 23 '11 at 16:44
AD.
8,76383163
8,76383163
asked Mar 23 '11 at 16:24
user8484user8484
351311
asked Mar 23 '11 at 16:24
user8484user8484
351311
asked Mar 23 '11 at 16:24
user8484user8484
351311
351311
$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13
add a comment |
$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13
$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13
$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: If $lambda$ in not is $A$, then there is $epsilon>0$ with $lvertlambda-a_nrvert>epsilon$ for all $n$. Use this to write down a bounded inverse for $T-lambdacdotmathrm{Id}$.
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
$endgroup$
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: If $lambda$ in not is $A$, then there is $epsilon>0$ with $lvertlambda-a_nrvert>epsilon$ for all $n$. Use this to write down a bounded inverse for $T-lambdacdotmathrm{Id}$.
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
$endgroup$
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
add a comment |
$begingroup$
Hint: If $lambda$ in not is $A$, then there is $epsilon>0$ with $lvertlambda-a_nrvert>epsilon$ for all $n$. Use this to write down a bounded inverse for $T-lambdacdotmathrm{Id}$.
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
$endgroup$
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
add a comment |
$begingroup$
Hint: If $lambda$ in not is $A$, then there is $epsilon>0$ with $lvertlambda-a_nrvert>epsilon$ for all $n$. Use this to write down a bounded inverse for $T-lambdacdotmathrm{Id}$.
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
$endgroup$
Hint: If $lambda$ in not is $A$, then there is $epsilon>0$ with $lvertlambda-a_nrvert>epsilon$ for all $n$. Use this to write down a bounded inverse for $T-lambdacdotmathrm{Id}$.
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
answered Mar 23 '11 at 16:33
RasmusRasmus
14.3k14479
14.3k14479
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
add a comment |
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
$begingroup$
I think that now I can solve it. Thank you.
$endgroup$
– user8484
Mar 23 '11 at 16:34
6
6
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
$begingroup$
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one.
$endgroup$
– Rasmus
Mar 23 '11 at 16:44
add a comment |
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$begingroup$
I'm interested in where is the question from and its background.
$endgroup$
– Jack
Apr 26 '11 at 22:13