“bounding” an unbounded operator












8












$begingroup$


I was wondering if, given a certain unbounded operator on a Hilbert space, it can (naively speaking) be "cutted" (or "bounded") by certain projections.



So, thinking about this in a more sensible way, I have the following question:



Let $T$ be a unbounded (densely defined) self-adjoint (positive) operator on $mathcal{H}$, and let ${P_lambda}_{lambda > 0}$ the spectral family of $T$. Then we can look at the following: $mathcal{H}_lambda:= P_{(lambda^{-1},lambda)}mathcal{H}$, where $P_{(lambda^{-1},lambda)}$ correspond to the Borel functional calculus on $T$ of the characteristic function on the interval $(lambda^{-1},lambda)$.



Is it true that, on $mathcal{H}_lambda$, $||Tx||leq lambda||x||$ ?






functional-analysis operator-theory spectral-theory unbounded-operators







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$endgroup$

















    8












    $begingroup$


    I was wondering if, given a certain unbounded operator on a Hilbert space, it can (naively speaking) be "cutted" (or "bounded") by certain projections.



    So, thinking about this in a more sensible way, I have the following question:



    Let $T$ be a unbounded (densely defined) self-adjoint (positive) operator on $mathcal{H}$, and let ${P_lambda}_{lambda > 0}$ the spectral family of $T$. Then we can look at the following: $mathcal{H}_lambda:= P_{(lambda^{-1},lambda)}mathcal{H}$, where $P_{(lambda^{-1},lambda)}$ correspond to the Borel functional calculus on $T$ of the characteristic function on the interval $(lambda^{-1},lambda)$.



    Is it true that, on $mathcal{H}_lambda$, $||Tx||leq lambda||x||$ ?






    functional-analysis operator-theory spectral-theory unbounded-operators







    share|cite|improve this question









    $endgroup$















      8












      8








      8





      $begingroup$


      I was wondering if, given a certain unbounded operator on a Hilbert space, it can (naively speaking) be "cutted" (or "bounded") by certain projections.



      So, thinking about this in a more sensible way, I have the following question:



      Let $T$ be a unbounded (densely defined) self-adjoint (positive) operator on $mathcal{H}$, and let ${P_lambda}_{lambda > 0}$ the spectral family of $T$. Then we can look at the following: $mathcal{H}_lambda:= P_{(lambda^{-1},lambda)}mathcal{H}$, where $P_{(lambda^{-1},lambda)}$ correspond to the Borel functional calculus on $T$ of the characteristic function on the interval $(lambda^{-1},lambda)$.



      Is it true that, on $mathcal{H}_lambda$, $||Tx||leq lambda||x||$ ?






      functional-analysis operator-theory spectral-theory unbounded-operators







      share|cite|improve this question









      $endgroup$




      I was wondering if, given a certain unbounded operator on a Hilbert space, it can (naively speaking) be "cutted" (or "bounded") by certain projections.



      So, thinking about this in a more sensible way, I have the following question:



      Let $T$ be a unbounded (densely defined) self-adjoint (positive) operator on $mathcal{H}$, and let ${P_lambda}_{lambda > 0}$ the spectral family of $T$. Then we can look at the following: $mathcal{H}_lambda:= P_{(lambda^{-1},lambda)}mathcal{H}$, where $P_{(lambda^{-1},lambda)}$ correspond to the Borel functional calculus on $T$ of the characteristic function on the interval $(lambda^{-1},lambda)$.



      Is it true that, on $mathcal{H}_lambda$, $||Tx||leq lambda||x||$ ?






      functional-analysis operator-theory spectral-theory unbounded-operators




      functional-analysis operator-theory spectral-theory unbounded-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 18 at 0:16









      HaroldFHaroldF

      591416




      591416






















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          Yes, of course.



          The functional calculus preserves positivity. The function $(lambda-t),1_{(0,lambda)}(t)$ is non-negative, so the operator $(lambda,I-T),P_{(0,lambda)}$ is positive. That is, $T,P_{(0,lambda)}leqlambda,P_{(0,lambda)}$. In particular, $|T,P_{(0,lambda)}|leqlambda$.






          share|cite|improve this answer









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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Yes, of course.



            The functional calculus preserves positivity. The function $(lambda-t),1_{(0,lambda)}(t)$ is non-negative, so the operator $(lambda,I-T),P_{(0,lambda)}$ is positive. That is, $T,P_{(0,lambda)}leqlambda,P_{(0,lambda)}$. In particular, $|T,P_{(0,lambda)}|leqlambda$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes, of course.



              The functional calculus preserves positivity. The function $(lambda-t),1_{(0,lambda)}(t)$ is non-negative, so the operator $(lambda,I-T),P_{(0,lambda)}$ is positive. That is, $T,P_{(0,lambda)}leqlambda,P_{(0,lambda)}$. In particular, $|T,P_{(0,lambda)}|leqlambda$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes, of course.



                The functional calculus preserves positivity. The function $(lambda-t),1_{(0,lambda)}(t)$ is non-negative, so the operator $(lambda,I-T),P_{(0,lambda)}$ is positive. That is, $T,P_{(0,lambda)}leqlambda,P_{(0,lambda)}$. In particular, $|T,P_{(0,lambda)}|leqlambda$.






                share|cite|improve this answer









                $endgroup$



                Yes, of course.



                The functional calculus preserves positivity. The function $(lambda-t),1_{(0,lambda)}(t)$ is non-negative, so the operator $(lambda,I-T),P_{(0,lambda)}$ is positive. That is, $T,P_{(0,lambda)}leqlambda,P_{(0,lambda)}$. In particular, $|T,P_{(0,lambda)}|leqlambda$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 18 at 3:47









                Martin ArgeramiMartin Argerami

                127k1182183




                127k1182183






























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