Is a linear operator $M$ bounded given that $M^2$ is bounded?












2












$begingroup$



Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.




If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
What about the converse?



Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
$$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
which does not seem very fruitful.
I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.



I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.



Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?






functional-analysis operator-theory







share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.




    If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
    What about the converse?



    Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
    $$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
    which does not seem very fruitful.
    I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.



    I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.



    Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?






    functional-analysis operator-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$



      Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.




      If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
      What about the converse?



      Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
      $$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
      which does not seem very fruitful.
      I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.



      I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.



      Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?






      functional-analysis operator-theory







      share|cite|improve this question









      $endgroup$





      Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.




      If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
      What about the converse?



      Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
      $$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
      which does not seem very fruitful.
      I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.



      I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.



      Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?






      functional-analysis operator-theory




      functional-analysis operator-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 20:03









      MisterRiemannMisterRiemann

      5,8451624




      5,8451624






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Could one argue as follows:



          Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:



          $$My_n = nz_n,quad Mz_n = 0$$



          and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ${z_n}$ is not the complete Hamel basis
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:44










          • $begingroup$
            Yup. You are right.
            $endgroup$
            – Maksim
            Dec 6 '18 at 21:46










          • $begingroup$
            ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:47



















          1












          $begingroup$

          Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for a quick response. Could you provide an example of one such?
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:06






          • 1




            $begingroup$
            @MisterRiemann You may look up here, Example 3.3.
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 20:42










          • $begingroup$
            @A.Γ. Thank you.
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:44











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Could one argue as follows:



          Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:



          $$My_n = nz_n,quad Mz_n = 0$$



          and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ${z_n}$ is not the complete Hamel basis
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:44










          • $begingroup$
            Yup. You are right.
            $endgroup$
            – Maksim
            Dec 6 '18 at 21:46










          • $begingroup$
            ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:47
















          2












          $begingroup$

          Could one argue as follows:



          Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:



          $$My_n = nz_n,quad Mz_n = 0$$



          and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ${z_n}$ is not the complete Hamel basis
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:44










          • $begingroup$
            Yup. You are right.
            $endgroup$
            – Maksim
            Dec 6 '18 at 21:46










          • $begingroup$
            ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:47














          2












          2








          2





          $begingroup$

          Could one argue as follows:



          Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:



          $$My_n = nz_n,quad Mz_n = 0$$



          and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.






          share|cite|improve this answer











          $endgroup$



          Could one argue as follows:



          Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:



          $$My_n = nz_n,quad Mz_n = 0$$



          and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 20:44

























          answered Dec 6 '18 at 20:33









          Olof RubinOlof Rubin

          1,131316




          1,131316












          • $begingroup$
            ${z_n}$ is not the complete Hamel basis
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:44










          • $begingroup$
            Yup. You are right.
            $endgroup$
            – Maksim
            Dec 6 '18 at 21:46










          • $begingroup$
            ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:47


















          • $begingroup$
            ${z_n}$ is not the complete Hamel basis
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:44










          • $begingroup$
            Yup. You are right.
            $endgroup$
            – Maksim
            Dec 6 '18 at 21:46










          • $begingroup$
            ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
            $endgroup$
            – Olof Rubin
            Dec 6 '18 at 21:47
















          $begingroup$
          ${z_n}$ is not the complete Hamel basis
          $endgroup$
          – Olof Rubin
          Dec 6 '18 at 21:44




          $begingroup$
          ${z_n}$ is not the complete Hamel basis
          $endgroup$
          – Olof Rubin
          Dec 6 '18 at 21:44












          $begingroup$
          Yup. You are right.
          $endgroup$
          – Maksim
          Dec 6 '18 at 21:46




          $begingroup$
          Yup. You are right.
          $endgroup$
          – Maksim
          Dec 6 '18 at 21:46












          $begingroup$
          ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
          $endgroup$
          – Olof Rubin
          Dec 6 '18 at 21:47




          $begingroup$
          ..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
          $endgroup$
          – Olof Rubin
          Dec 6 '18 at 21:47











          1












          $begingroup$

          Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for a quick response. Could you provide an example of one such?
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:06






          • 1




            $begingroup$
            @MisterRiemann You may look up here, Example 3.3.
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 20:42










          • $begingroup$
            @A.Γ. Thank you.
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:44
















          1












          $begingroup$

          Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for a quick response. Could you provide an example of one such?
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:06






          • 1




            $begingroup$
            @MisterRiemann You may look up here, Example 3.3.
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 20:42










          • $begingroup$
            @A.Γ. Thank you.
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:44














          1












          1








          1





          $begingroup$

          Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.






          share|cite|improve this answer









          $endgroup$



          Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 20:05









          José Carlos SantosJosé Carlos Santos

          155k22124227




          155k22124227












          • $begingroup$
            Thanks for a quick response. Could you provide an example of one such?
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:06






          • 1




            $begingroup$
            @MisterRiemann You may look up here, Example 3.3.
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 20:42










          • $begingroup$
            @A.Γ. Thank you.
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:44


















          • $begingroup$
            Thanks for a quick response. Could you provide an example of one such?
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:06






          • 1




            $begingroup$
            @MisterRiemann You may look up here, Example 3.3.
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 20:42










          • $begingroup$
            @A.Γ. Thank you.
            $endgroup$
            – MisterRiemann
            Dec 6 '18 at 20:44
















          $begingroup$
          Thanks for a quick response. Could you provide an example of one such?
          $endgroup$
          – MisterRiemann
          Dec 6 '18 at 20:06




          $begingroup$
          Thanks for a quick response. Could you provide an example of one such?
          $endgroup$
          – MisterRiemann
          Dec 6 '18 at 20:06




          1




          1




          $begingroup$
          @MisterRiemann You may look up here, Example 3.3.
          $endgroup$
          – A.Γ.
          Dec 6 '18 at 20:42




          $begingroup$
          @MisterRiemann You may look up here, Example 3.3.
          $endgroup$
          – A.Γ.
          Dec 6 '18 at 20:42












          $begingroup$
          @A.Γ. Thank you.
          $endgroup$
          – MisterRiemann
          Dec 6 '18 at 20:44




          $begingroup$
          @A.Γ. Thank you.
          $endgroup$
          – MisterRiemann
          Dec 6 '18 at 20:44


















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