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







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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







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      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













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      edited Feb 6 at 12:25









      el_tenedor

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      asked Feb 2 at 23:55









      rihanirihani

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          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.






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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.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 6 at 15:50









                MaoWaoMaoWao

                3,938618




                3,938618






























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