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What are the approximate eigenvalues of the right shift operator $R$ on $ell_infty$

3

$begingroup$

I have shown that the spectrum of $R=z$.

Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?

If they are not how do I find the rest?

Thanks.

functional-analysis operator-theory spectral-theory

share|cite|improve this question

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

$endgroup$

add a comment | 

3

$begingroup$

I have shown that the spectrum of $R=z$.

Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?

If they are not how do I find the rest?

Thanks.

functional-analysis operator-theory spectral-theory

share|cite|improve this question

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

$endgroup$

add a comment | 

$begingroup$

I have shown that the spectrum of $R=z$.

Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?

If they are not how do I find the rest?

Thanks.

functional-analysis operator-theory spectral-theory

share|cite|improve this question

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

$endgroup$

share|cite|improve this question

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

share|cite|improve this question

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

edited Mar 18 at 11:59

Andrews

1,2812422

edited Mar 18 at 11:59

Andrews

1,2812422

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

asked Mar 18 at 11:09

Jhon DoeJhon Doe

683414

1 Answer
1

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oldest

votes

2

$begingroup$

For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have

$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$

Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.

From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$

share|cite|improve this answer

answered Mar 18 at 12:18

FredFred

48.8k11849

$endgroup$

add a comment | 

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2

$begingroup$

For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have

$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$

Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.

From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$

share|cite|improve this answer

answered Mar 18 at 12:18

FredFred

48.8k11849

$endgroup$

add a comment | 

2

$begingroup$

For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have

$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$

Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.

From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$

share|cite|improve this answer

answered Mar 18 at 12:18

FredFred

48.8k11849

$endgroup$

add a comment | 

$begingroup$

For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have

$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$

Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.

From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$

share|cite|improve this answer

answered Mar 18 at 12:18

FredFred

48.8k11849

$endgroup$

share|cite|improve this answer

answered Mar 18 at 12:18

FredFred

48.8k11849

answered Mar 18 at 12:18

FredFred

48.8k11849

answered Mar 18 at 12:18

FredFred

48.8k11849