image of some ideal under the quotient map












1












$begingroup$


enter image description here!
I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.



Is my understanding correct?enter image description here






operator-theory operator-algebras c-star-algebras von-neumann-algebras







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$endgroup$

















    1












    $begingroup$


    enter image description here!
    I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.



    Is my understanding correct?enter image description here






    operator-theory operator-algebras c-star-algebras von-neumann-algebras







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      enter image description here!
      I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.



      Is my understanding correct?enter image description here






      operator-theory operator-algebras c-star-algebras von-neumann-algebras







      share|cite|improve this question









      $endgroup$




      enter image description here!
      I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.



      Is my understanding correct?enter image description here






      operator-theory operator-algebras c-star-algebras von-neumann-algebras




      operator-theory operator-algebras c-star-algebras von-neumann-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 16:16









      mathrookiemathrookie

      832512




      832512






















          1 Answer
          1






          active

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          1












          $begingroup$

          No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.



          For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
            $endgroup$
            – mathrookie
            Dec 7 '18 at 17:36












          • $begingroup$
            No, you don't. You are mixing the normalized and non-normalized cases.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 17:42










          • $begingroup$
            Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
            $endgroup$
            – mathrookie
            Dec 7 '18 at 18:46












          • $begingroup$
            Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 19:00










          • $begingroup$
            So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
            $endgroup$
            – mathrookie
            Dec 8 '18 at 9:44











          Your Answer





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          1 Answer
          1






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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.



          For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
            $endgroup$
            – mathrookie
            Dec 7 '18 at 17:36












          • $begingroup$
            No, you don't. You are mixing the normalized and non-normalized cases.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 17:42










          • $begingroup$
            Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
            $endgroup$
            – mathrookie
            Dec 7 '18 at 18:46












          • $begingroup$
            Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 19:00










          • $begingroup$
            So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
            $endgroup$
            – mathrookie
            Dec 8 '18 at 9:44
















          1












          $begingroup$

          No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.



          For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
            $endgroup$
            – mathrookie
            Dec 7 '18 at 17:36












          • $begingroup$
            No, you don't. You are mixing the normalized and non-normalized cases.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 17:42










          • $begingroup$
            Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
            $endgroup$
            – mathrookie
            Dec 7 '18 at 18:46












          • $begingroup$
            Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 19:00










          • $begingroup$
            So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
            $endgroup$
            – mathrookie
            Dec 8 '18 at 9:44














          1












          1








          1





          $begingroup$

          No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.



          For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.






          share|cite|improve this answer









          $endgroup$



          No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.



          For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 17:26









          Martin ArgeramiMartin Argerami

          125k1177177




          125k1177177












          • $begingroup$
            Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
            $endgroup$
            – mathrookie
            Dec 7 '18 at 17:36












          • $begingroup$
            No, you don't. You are mixing the normalized and non-normalized cases.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 17:42










          • $begingroup$
            Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
            $endgroup$
            – mathrookie
            Dec 7 '18 at 18:46












          • $begingroup$
            Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 19:00










          • $begingroup$
            So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
            $endgroup$
            – mathrookie
            Dec 8 '18 at 9:44


















          • $begingroup$
            Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
            $endgroup$
            – mathrookie
            Dec 7 '18 at 17:36












          • $begingroup$
            No, you don't. You are mixing the normalized and non-normalized cases.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 17:42










          • $begingroup$
            Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
            $endgroup$
            – mathrookie
            Dec 7 '18 at 18:46












          • $begingroup$
            Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
            $endgroup$
            – Martin Argerami
            Dec 7 '18 at 19:00










          • $begingroup$
            So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
            $endgroup$
            – mathrookie
            Dec 8 '18 at 9:44
















          $begingroup$
          Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
          $endgroup$
          – mathrookie
          Dec 7 '18 at 17:36






          $begingroup$
          Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
          $endgroup$
          – mathrookie
          Dec 7 '18 at 17:36














          $begingroup$
          No, you don't. You are mixing the normalized and non-normalized cases.
          $endgroup$
          – Martin Argerami
          Dec 7 '18 at 17:42




          $begingroup$
          No, you don't. You are mixing the normalized and non-normalized cases.
          $endgroup$
          – Martin Argerami
          Dec 7 '18 at 17:42












          $begingroup$
          Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
          $endgroup$
          – mathrookie
          Dec 7 '18 at 18:46






          $begingroup$
          Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
          $endgroup$
          – mathrookie
          Dec 7 '18 at 18:46














          $begingroup$
          Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
          $endgroup$
          – Martin Argerami
          Dec 7 '18 at 19:00




          $begingroup$
          Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
          $endgroup$
          – Martin Argerami
          Dec 7 '18 at 19:00












          $begingroup$
          So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
          $endgroup$
          – mathrookie
          Dec 8 '18 at 9:44




          $begingroup$
          So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
          $endgroup$
          – mathrookie
          Dec 8 '18 at 9:44


















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