$H$ Hilbert space, $T$ symmetric bounded linear, when is $H=R(T) oplus N(T)$?












3












$begingroup$


I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?






functional-analysis operator-theory hilbert-spaces orthogonality direct-sum







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
    On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
    So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?






    functional-analysis operator-theory hilbert-spaces orthogonality direct-sum







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
      On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
      So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?






      functional-analysis operator-theory hilbert-spaces orthogonality direct-sum







      share|cite|improve this question











      $endgroup$




      I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
      On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
      So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?






      functional-analysis operator-theory hilbert-spaces orthogonality direct-sum




      functional-analysis operator-theory hilbert-spaces orthogonality direct-sum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 15:18









      Davide Giraudo

      125k16150261




      125k16150261










      asked Jan 8 at 14:26









      roi_saumonroi_saumon

      45628




      45628






















          1 Answer
          1






          active

          oldest

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          4












          $begingroup$

          The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.



          In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
          $$
          |Tx|geq c|x|
          $$
          for all $xin N(T)^perp$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
            $endgroup$
            – mechanodroid
            Jan 8 at 14:40










          • $begingroup$
            @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
            $endgroup$
            – roi_saumon
            Jan 8 at 14:48












          • $begingroup$
            @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
            $endgroup$
            – MaoWao
            Jan 8 at 14:50










          • $begingroup$
            @mechanodroid Not off the top of my head, no. I will think about it.
            $endgroup$
            – MaoWao
            Jan 8 at 14:51






          • 1




            $begingroup$
            @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
            $endgroup$
            – daw
            Jan 9 at 14:36













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          $begingroup$

          The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.



          In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
          $$
          |Tx|geq c|x|
          $$
          for all $xin N(T)^perp$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
            $endgroup$
            – mechanodroid
            Jan 8 at 14:40










          • $begingroup$
            @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
            $endgroup$
            – roi_saumon
            Jan 8 at 14:48












          • $begingroup$
            @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
            $endgroup$
            – MaoWao
            Jan 8 at 14:50










          • $begingroup$
            @mechanodroid Not off the top of my head, no. I will think about it.
            $endgroup$
            – MaoWao
            Jan 8 at 14:51






          • 1




            $begingroup$
            @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
            $endgroup$
            – daw
            Jan 9 at 14:36


















          4












          $begingroup$

          The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.



          In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
          $$
          |Tx|geq c|x|
          $$
          for all $xin N(T)^perp$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
            $endgroup$
            – mechanodroid
            Jan 8 at 14:40










          • $begingroup$
            @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
            $endgroup$
            – roi_saumon
            Jan 8 at 14:48












          • $begingroup$
            @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
            $endgroup$
            – MaoWao
            Jan 8 at 14:50










          • $begingroup$
            @mechanodroid Not off the top of my head, no. I will think about it.
            $endgroup$
            – MaoWao
            Jan 8 at 14:51






          • 1




            $begingroup$
            @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
            $endgroup$
            – daw
            Jan 9 at 14:36
















          4












          4








          4





          $begingroup$

          The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.



          In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
          $$
          |Tx|geq c|x|
          $$
          for all $xin N(T)^perp$.






          share|cite|improve this answer









          $endgroup$



          The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.



          In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
          $$
          |Tx|geq c|x|
          $$
          for all $xin N(T)^perp$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 14:35









          MaoWaoMaoWao

          2,543617




          2,543617












          • $begingroup$
            Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
            $endgroup$
            – mechanodroid
            Jan 8 at 14:40










          • $begingroup$
            @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
            $endgroup$
            – roi_saumon
            Jan 8 at 14:48












          • $begingroup$
            @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
            $endgroup$
            – MaoWao
            Jan 8 at 14:50










          • $begingroup$
            @mechanodroid Not off the top of my head, no. I will think about it.
            $endgroup$
            – MaoWao
            Jan 8 at 14:51






          • 1




            $begingroup$
            @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
            $endgroup$
            – daw
            Jan 9 at 14:36




















          • $begingroup$
            Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
            $endgroup$
            – mechanodroid
            Jan 8 at 14:40










          • $begingroup$
            @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
            $endgroup$
            – roi_saumon
            Jan 8 at 14:48












          • $begingroup$
            @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
            $endgroup$
            – MaoWao
            Jan 8 at 14:50










          • $begingroup$
            @mechanodroid Not off the top of my head, no. I will think about it.
            $endgroup$
            – MaoWao
            Jan 8 at 14:51






          • 1




            $begingroup$
            @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
            $endgroup$
            – daw
            Jan 9 at 14:36


















          $begingroup$
          Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
          $endgroup$
          – mechanodroid
          Jan 8 at 14:40




          $begingroup$
          Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
          $endgroup$
          – mechanodroid
          Jan 8 at 14:40












          $begingroup$
          @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
          $endgroup$
          – roi_saumon
          Jan 8 at 14:48






          $begingroup$
          @MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
          $endgroup$
          – roi_saumon
          Jan 8 at 14:48














          $begingroup$
          @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
          $endgroup$
          – MaoWao
          Jan 8 at 14:50




          $begingroup$
          @roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
          $endgroup$
          – MaoWao
          Jan 8 at 14:50












          $begingroup$
          @mechanodroid Not off the top of my head, no. I will think about it.
          $endgroup$
          – MaoWao
          Jan 8 at 14:51




          $begingroup$
          @mechanodroid Not off the top of my head, no. I will think about it.
          $endgroup$
          – MaoWao
          Jan 8 at 14:51




          1




          1




          $begingroup$
          @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
          $endgroup$
          – daw
          Jan 9 at 14:36






          $begingroup$
          @mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
          $endgroup$
          – daw
          Jan 9 at 14:36




















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