Approximate point spectrum of multiplication operator in $L^p$












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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 $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$



$sigma_{ap}(M_phi)=,?$










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    0












    $begingroup$


    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 $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$



    $sigma_{ap}(M_phi)=,?$










    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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 $$sigma_p(M_phi)={ lambda in mathbb{C} colon muleft( {xin X colon f(x)=lambda } right)>0 }.$$



      $sigma_{ap}(M_phi)=,?$










      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 16 '18 at 22:31









      QuantumDKQuantumDK

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          $begingroup$

          $M_phi$ is normal $Rightarrow sigma_{ap}=sigma$






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          $endgroup$





















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            $begingroup$

            $$
            sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
            $$






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              2 Answers
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              0












              $begingroup$

              $M_phi$ is normal $Rightarrow sigma_{ap}=sigma$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $M_phi$ is normal $Rightarrow sigma_{ap}=sigma$






                share|cite|improve this answer









                $endgroup$
















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                  $begingroup$

                  $M_phi$ is normal $Rightarrow sigma_{ap}=sigma$






                  share|cite|improve this answer









                  $endgroup$



                  $M_phi$ is normal $Rightarrow sigma_{ap}=sigma$







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Dec 16 '18 at 22:54









                  QuantumDKQuantumDK

                  64




                  64























                      0












                      $begingroup$

                      $$
                      sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $$
                        sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          0












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                          0





                          $begingroup$

                          $$
                          sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          $$
                          sigma_{ap}(M_phi) = { lambdainmathbb{C} : mu{ nu : 0 <|phi(nu)-lambda| < epsilon} >0;; forall epsilon > 0 }.
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 16 at 6:17









                          DisintegratingByPartsDisintegratingByParts

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                          59.3k42580






























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