Mixing
A self-adjoint operator with essential spectrum={0} is compact
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Does every self adjoint operator (on a Hilbert space) with essential spectrum={0} is a compact operator ?
compactness spectral-theory compact-operators self-adjoint-operators
edited Feb 6 at 12:25
el_tenedor
2,297921
asked Feb 2 at 23:55
rihanirihani
132
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add a comment |
$begingroup$
Does every self adjoint operator (on a Hilbert space) with essential spectrum={0} is a compact operator ?
compactness spectral-theory compact-operators self-adjoint-operators
edited Feb 6 at 12:25
el_tenedor
2,297921
asked Feb 2 at 23:55
rihanirihani
132
$endgroup$
add a comment |
$begingroup$
Does every self adjoint operator (on a Hilbert space) with essential spectrum={0} is a compact operator ?
compactness spectral-theory compact-operators self-adjoint-operators
edited Feb 6 at 12:25
el_tenedor
2,297921
asked Feb 2 at 23:55
rihanirihani
132
$endgroup$
Does every self adjoint operator (on a Hilbert space) with essential spectrum={0} is a compact operator ?
compactness spectral-theory compact-operators self-adjoint-operators
compactness spectral-theory compact-operators self-adjoint-operators
edited Feb 6 at 12:25
el_tenedor
2,297921
asked Feb 2 at 23:55
rihanirihani
132
edited Feb 6 at 12:25
el_tenedor
2,297921
asked Feb 2 at 23:55
rihanirihani
132
edited Feb 6 at 12:25
el_tenedor
2,297921
edited Feb 6 at 12:25
el_tenedor
2,297921
edited Feb 6 at 12:25
el_tenedor
2,297921
2,297921
asked Feb 2 at 23:55
rihanirihani
132
asked Feb 2 at 23:55
rihanirihani
132
asked Feb 2 at 23:55
rihanirihani
132
132
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $ell^2$ given by
$$
D(T)={xiinell^2mid sum_k k^2 xi_{2k}^2<infty},,Txi=(0,xi_2,0,2xi_4,0,dots).
$$
On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $sigma(T)$).
Since $sigma_{mathrm{ess}}(T)={0}$, the set $sigma(T)setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $sigma(T)setminus{0}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
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add a comment |
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1 Answer
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1 Answer
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active
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active
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active
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votes
$begingroup$
If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $ell^2$ given by
$$
D(T)={xiinell^2mid sum_k k^2 xi_{2k}^2<infty},,Txi=(0,xi_2,0,2xi_4,0,dots).
$$
On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $sigma(T)$).
Since $sigma_{mathrm{ess}}(T)={0}$, the set $sigma(T)setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $sigma(T)setminus{0}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
$endgroup$
add a comment |
$begingroup$
If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $ell^2$ given by
$$
D(T)={xiinell^2mid sum_k k^2 xi_{2k}^2<infty},,Txi=(0,xi_2,0,2xi_4,0,dots).
$$
On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $sigma(T)$).
Since $sigma_{mathrm{ess}}(T)={0}$, the set $sigma(T)setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $sigma(T)setminus{0}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
$endgroup$
add a comment |
$begingroup$
If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $ell^2$ given by
$$
D(T)={xiinell^2mid sum_k k^2 xi_{2k}^2<infty},,Txi=(0,xi_2,0,2xi_4,0,dots).
$$
On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $sigma(T)$).
Since $sigma_{mathrm{ess}}(T)={0}$, the set $sigma(T)setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $sigma(T)setminus{0}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
$endgroup$
If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $ell^2$ given by
$$
D(T)={xiinell^2mid sum_k k^2 xi_{2k}^2<infty},,Txi=(0,xi_2,0,2xi_4,0,dots).
$$
On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $sigma(T)$).
Since $sigma_{mathrm{ess}}(T)={0}$, the set $sigma(T)setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $sigma(T)setminus{0}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
answered Feb 6 at 15:50
MaoWaoMaoWao
3,938618
3,938618
add a comment |
add a comment |
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