Approximate point spectrum of multiplication operator in $L^p$
$begingroup$
Let $M_phi f=phi f, ,, phiin L^infty(X,Omega,mu),,, fin L^p(X,Omega,mu),, 1leqslant p leqslant infty.$
I finded $$sigma(M_phi)=left{ lambda in mathbb{C} mid not exists epsilon>0 colon |lambda-f(x)|geqslant epsilon ,,text{a.e.} right}$$ and $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$
$sigma_{ap}(M_phi)=,?$
functional-analysis spectral-theory
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
$endgroup$
add a comment |
$begingroup$
Let $M_phi f=phi f, ,, phiin L^infty(X,Omega,mu),,, fin L^p(X,Omega,mu),, 1leqslant p leqslant infty.$
I finded $$sigma(M_phi)=left{ lambda in mathbb{C} mid not exists epsilon>0 colon |lambda-f(x)|geqslant epsilon ,,text{a.e.} right}$$ and $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$
$sigma_{ap}(M_phi)=,?$
functional-analysis spectral-theory
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
$endgroup$
add a comment |
$begingroup$
Let $M_phi f=phi f, ,, phiin L^infty(X,Omega,mu),,, fin L^p(X,Omega,mu),, 1leqslant p leqslant infty.$
I finded $$sigma(M_phi)=left{ lambda in mathbb{C} mid not exists epsilon>0 colon |lambda-f(x)|geqslant epsilon ,,text{a.e.} right}$$ and $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$
$sigma_{ap}(M_phi)=,?$
functional-analysis spectral-theory
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
$endgroup$
Let $M_phi f=phi f, ,, phiin L^infty(X,Omega,mu),,, fin L^p(X,Omega,mu),, 1leqslant p leqslant infty.$
I finded $$sigma(M_phi)=left{ lambda in mathbb{C} mid not exists epsilon>0 colon |lambda-f(x)|geqslant epsilon ,,text{a.e.} right}$$ and $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$
$sigma_{ap}(M_phi)=,?$
functional-analysis spectral-theory
functional-analysis spectral-theory
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
asked Dec 16 '18 at 22:31
QuantumDKQuantumDK
64
64
add a comment |
add a comment |
2 Answers
2
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$begingroup$
$M_phi$ is normal $Rightarrow sigma_{ap}=sigma$
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
$endgroup$
add a comment |
$begingroup$
$$
sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
$$
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$M_phi$ is normal $Rightarrow sigma_{ap}=sigma$
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
$endgroup$
add a comment |
$begingroup$
$M_phi$ is normal $Rightarrow sigma_{ap}=sigma$
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
$endgroup$
add a comment |
$begingroup$
$M_phi$ is normal $Rightarrow sigma_{ap}=sigma$
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
$endgroup$
$M_phi$ is normal $Rightarrow sigma_{ap}=sigma$
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
answered Dec 16 '18 at 22:54
QuantumDKQuantumDK
64
64
add a comment |
add a comment |
$begingroup$
$$
sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
$$
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
$$
sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
$$
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
$$
sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
$$
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
$$
sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
$$
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 16 at 6:17
DisintegratingByPartsDisintegratingByParts
59.8k42681
59.8k42681
add a comment |
add a comment |
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