Compact operator on $L_2([0,1],m)$












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Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
begin{equation*}
T f(x)=x f(x) f in H, x in[0,1]
end{equation*}

Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.










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    0












    $begingroup$


    Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
    begin{equation*}
    T f(x)=x f(x) f in H, x in[0,1]
    end{equation*}

    Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
      begin{equation*}
      T f(x)=x f(x) f in H, x in[0,1]
      end{equation*}

      Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.










      share|cite|improve this question









      $endgroup$




      Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
      begin{equation*}
      T f(x)=x f(x) f in H, x in[0,1]
      end{equation*}

      Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.







      functional-analysis hilbert-spaces compact-operators






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      asked Jan 17 at 10:00









      user289143user289143

      913313




      913313






















          3 Answers
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          You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
          $$left( frac 1 2 - xright) f(x) = 1$$
          and see that no solution exists.
          But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$



              It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.



              Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.



              We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.



              Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
                $$left( frac 1 2 - xright) f(x) = 1$$
                and see that no solution exists.
                But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
                  $$left( frac 1 2 - xright) f(x) = 1$$
                  and see that no solution exists.
                  But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
                    $$left( frac 1 2 - xright) f(x) = 1$$
                    and see that no solution exists.
                    But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.






                    share|cite|improve this answer











                    $endgroup$



                    You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
                    $$left( frac 1 2 - xright) f(x) = 1$$
                    and see that no solution exists.
                    But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 17 at 11:22

























                    answered Jan 17 at 10:51









                    ShashiShashi

                    7,1781628




                    7,1781628























                        1












                        $begingroup$

                        It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.






                            share|cite|improve this answer









                            $endgroup$



                            It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 17 at 10:14









                            Kavi Rama MurthyKavi Rama Murthy

                            59.2k42161




                            59.2k42161























                                0












                                $begingroup$

                                Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$



                                It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.



                                Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.



                                We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.



                                Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$



                                  It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.



                                  Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.



                                  We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.



                                  Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$



                                    It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.



                                    Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.



                                    We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.



                                    Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$



                                    It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.



                                    Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.



                                    We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.



                                    Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 17 at 10:27









                                    MindlackMindlack

                                    3,86018




                                    3,86018






























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