How are the two versions of the Lumer-Phillips theorem given in the books of Engel/Nagel and Pazy related? ...

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How are the two versions of the Lumer-Phillips theorem given in the books of Engel/Nagel and Pazy related?



The Next CEO of Stack OverflowNecessary and sufficient condition of being dissipativeShowing that an operator generates a contraction semigroupFind the iverse of the followning bounded operator?Proving the existence of a solution of the heat equation using semigroup methodsLumer-Phillips Theorem for non-contraction semigroups?If it holds that $forall f in mathcal{D}(A): T(t)f-f = int_0^t T(s) Af ds$ then $A$ is the generator of ${T(t)}$Proving density of $H^2(mathbb{R})$ in $L^2(mathbb{R})$ .Proving an operator is surjective for Lumer-Phillips Theorem application.Understanding a proof from Pazy's book on infinitesimal generatorsShowing that $lambda - (A + B)$ has dense range












2












$begingroup$


Let $E$ be a $mathbb R$-Banach space and $(mathcal D(A),A)$ be a densely-defined dissipative linear operator on $E$.



In the book of Engel and Nagel, I've found the following verison of the Lumer-Phillips theorem:




Version 1: The closure $(mathcal D(overline A),overline A)$ of $(mathcal D(A),A)$ is the generator of a contraction semigroup if and only if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some (and hence all) $lambda>0$.




Now, in the book of Pazy, I've found a different version:




Version 2: The range of $lambdaoperatorname{id}_{mathcal D(A)}-A$ is $E$ for some $lambda>0$ if and only if $(mathcal D(A),A)$ is the generator of a strongly continuous contraction semigroup.




Question 1: How are these two versions related? For example, if $(mathcal D(A),A)$ is the generator of a contraction semigroup, version 1 only yields that $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for all $lambda>0$. How do we see that these ranges are actually the whole space $E$? Clearly, the crucial thing must be that not only the closure $(mathcal D(overline A),overline A)$ but $(mathcal D(A),A)$ itself is the generator of a contraction semigroup.



Question 2: In version 1, there is no strong continuity claim on the semigroup. Is this a mistake in the book or are we not able to conclude the strong continuity of the contraction semigroup generated by $(mathcal D(overline A),overline A)$ if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some $lambda>0$?






functional-analysis operator-theory semigroup-of-operators







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $E$ be a $mathbb R$-Banach space and $(mathcal D(A),A)$ be a densely-defined dissipative linear operator on $E$.



    In the book of Engel and Nagel, I've found the following verison of the Lumer-Phillips theorem:




    Version 1: The closure $(mathcal D(overline A),overline A)$ of $(mathcal D(A),A)$ is the generator of a contraction semigroup if and only if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some (and hence all) $lambda>0$.




    Now, in the book of Pazy, I've found a different version:




    Version 2: The range of $lambdaoperatorname{id}_{mathcal D(A)}-A$ is $E$ for some $lambda>0$ if and only if $(mathcal D(A),A)$ is the generator of a strongly continuous contraction semigroup.




    Question 1: How are these two versions related? For example, if $(mathcal D(A),A)$ is the generator of a contraction semigroup, version 1 only yields that $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for all $lambda>0$. How do we see that these ranges are actually the whole space $E$? Clearly, the crucial thing must be that not only the closure $(mathcal D(overline A),overline A)$ but $(mathcal D(A),A)$ itself is the generator of a contraction semigroup.



    Question 2: In version 1, there is no strong continuity claim on the semigroup. Is this a mistake in the book or are we not able to conclude the strong continuity of the contraction semigroup generated by $(mathcal D(overline A),overline A)$ if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some $lambda>0$?






    functional-analysis operator-theory semigroup-of-operators







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $E$ be a $mathbb R$-Banach space and $(mathcal D(A),A)$ be a densely-defined dissipative linear operator on $E$.



      In the book of Engel and Nagel, I've found the following verison of the Lumer-Phillips theorem:




      Version 1: The closure $(mathcal D(overline A),overline A)$ of $(mathcal D(A),A)$ is the generator of a contraction semigroup if and only if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some (and hence all) $lambda>0$.




      Now, in the book of Pazy, I've found a different version:




      Version 2: The range of $lambdaoperatorname{id}_{mathcal D(A)}-A$ is $E$ for some $lambda>0$ if and only if $(mathcal D(A),A)$ is the generator of a strongly continuous contraction semigroup.




      Question 1: How are these two versions related? For example, if $(mathcal D(A),A)$ is the generator of a contraction semigroup, version 1 only yields that $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for all $lambda>0$. How do we see that these ranges are actually the whole space $E$? Clearly, the crucial thing must be that not only the closure $(mathcal D(overline A),overline A)$ but $(mathcal D(A),A)$ itself is the generator of a contraction semigroup.



      Question 2: In version 1, there is no strong continuity claim on the semigroup. Is this a mistake in the book or are we not able to conclude the strong continuity of the contraction semigroup generated by $(mathcal D(overline A),overline A)$ if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some $lambda>0$?






      functional-analysis operator-theory semigroup-of-operators







      share|cite|improve this question









      $endgroup$




      Let $E$ be a $mathbb R$-Banach space and $(mathcal D(A),A)$ be a densely-defined dissipative linear operator on $E$.



      In the book of Engel and Nagel, I've found the following verison of the Lumer-Phillips theorem:




      Version 1: The closure $(mathcal D(overline A),overline A)$ of $(mathcal D(A),A)$ is the generator of a contraction semigroup if and only if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some (and hence all) $lambda>0$.




      Now, in the book of Pazy, I've found a different version:




      Version 2: The range of $lambdaoperatorname{id}_{mathcal D(A)}-A$ is $E$ for some $lambda>0$ if and only if $(mathcal D(A),A)$ is the generator of a strongly continuous contraction semigroup.




      Question 1: How are these two versions related? For example, if $(mathcal D(A),A)$ is the generator of a contraction semigroup, version 1 only yields that $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for all $lambda>0$. How do we see that these ranges are actually the whole space $E$? Clearly, the crucial thing must be that not only the closure $(mathcal D(overline A),overline A)$ but $(mathcal D(A),A)$ itself is the generator of a contraction semigroup.



      Question 2: In version 1, there is no strong continuity claim on the semigroup. Is this a mistake in the book or are we not able to conclude the strong continuity of the contraction semigroup generated by $(mathcal D(overline A),overline A)$ if $lambdaoperatorname{id}_{mathcal D(A)}-A$ has dense range for some $lambda>0$?






      functional-analysis operator-theory semigroup-of-operators




      functional-analysis operator-theory semigroup-of-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 16 at 7:48









      0xbadf00d0xbadf00d

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          1 Answer
          1






          active

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          1












          $begingroup$


          1. If $A$ is a closed dissipative operator, then the range of $lambda-A$ is closed for all $lambda>0$. In fact, if $(lambda x_n-A x_n)$ is Cauchy, then
            $$
            |x_n-x_m|leq frac 1 lambda|(lambda-A)(x_n-x_m)|.
            $$
            Hence $(x_n)$ is Cauchy and thus $(Ax_n)$ is Cauchy as well. Since $A$ is closed, $lim_{ntoinfty}x_nin D(A)$ and
            $$
            lim_{ntoinfty}(lambda-A)x_n=(lambda-A)lim_{ntoinfty} x_nin mathrm{ran}(lambda-A).
            $$

          2. If a contraction semigroup has a densely defined generator, then it is strongly continuous. This result is also somewhere in the book of Engel an Nagel, but the argument is quite simple: Trajectories starting in the domain of the generator are clearly continuous, and the uniform bound $|T_t|leq 1$ allows one to extend this continuity to all trajectories starting in the closure of the domain of the generator.






          share|cite|improve this answer









          $endgroup$














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            1 Answer
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            active

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            1 Answer
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            active

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            active

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            1












            $begingroup$


            1. If $A$ is a closed dissipative operator, then the range of $lambda-A$ is closed for all $lambda>0$. In fact, if $(lambda x_n-A x_n)$ is Cauchy, then
              $$
              |x_n-x_m|leq frac 1 lambda|(lambda-A)(x_n-x_m)|.
              $$
              Hence $(x_n)$ is Cauchy and thus $(Ax_n)$ is Cauchy as well. Since $A$ is closed, $lim_{ntoinfty}x_nin D(A)$ and
              $$
              lim_{ntoinfty}(lambda-A)x_n=(lambda-A)lim_{ntoinfty} x_nin mathrm{ran}(lambda-A).
              $$

            2. If a contraction semigroup has a densely defined generator, then it is strongly continuous. This result is also somewhere in the book of Engel an Nagel, but the argument is quite simple: Trajectories starting in the domain of the generator are clearly continuous, and the uniform bound $|T_t|leq 1$ allows one to extend this continuity to all trajectories starting in the closure of the domain of the generator.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$


              1. If $A$ is a closed dissipative operator, then the range of $lambda-A$ is closed for all $lambda>0$. In fact, if $(lambda x_n-A x_n)$ is Cauchy, then
                $$
                |x_n-x_m|leq frac 1 lambda|(lambda-A)(x_n-x_m)|.
                $$
                Hence $(x_n)$ is Cauchy and thus $(Ax_n)$ is Cauchy as well. Since $A$ is closed, $lim_{ntoinfty}x_nin D(A)$ and
                $$
                lim_{ntoinfty}(lambda-A)x_n=(lambda-A)lim_{ntoinfty} x_nin mathrm{ran}(lambda-A).
                $$

              2. If a contraction semigroup has a densely defined generator, then it is strongly continuous. This result is also somewhere in the book of Engel an Nagel, but the argument is quite simple: Trajectories starting in the domain of the generator are clearly continuous, and the uniform bound $|T_t|leq 1$ allows one to extend this continuity to all trajectories starting in the closure of the domain of the generator.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$


                1. If $A$ is a closed dissipative operator, then the range of $lambda-A$ is closed for all $lambda>0$. In fact, if $(lambda x_n-A x_n)$ is Cauchy, then
                  $$
                  |x_n-x_m|leq frac 1 lambda|(lambda-A)(x_n-x_m)|.
                  $$
                  Hence $(x_n)$ is Cauchy and thus $(Ax_n)$ is Cauchy as well. Since $A$ is closed, $lim_{ntoinfty}x_nin D(A)$ and
                  $$
                  lim_{ntoinfty}(lambda-A)x_n=(lambda-A)lim_{ntoinfty} x_nin mathrm{ran}(lambda-A).
                  $$

                2. If a contraction semigroup has a densely defined generator, then it is strongly continuous. This result is also somewhere in the book of Engel an Nagel, but the argument is quite simple: Trajectories starting in the domain of the generator are clearly continuous, and the uniform bound $|T_t|leq 1$ allows one to extend this continuity to all trajectories starting in the closure of the domain of the generator.






                share|cite|improve this answer









                $endgroup$




                1. If $A$ is a closed dissipative operator, then the range of $lambda-A$ is closed for all $lambda>0$. In fact, if $(lambda x_n-A x_n)$ is Cauchy, then
                  $$
                  |x_n-x_m|leq frac 1 lambda|(lambda-A)(x_n-x_m)|.
                  $$
                  Hence $(x_n)$ is Cauchy and thus $(Ax_n)$ is Cauchy as well. Since $A$ is closed, $lim_{ntoinfty}x_nin D(A)$ and
                  $$
                  lim_{ntoinfty}(lambda-A)x_n=(lambda-A)lim_{ntoinfty} x_nin mathrm{ran}(lambda-A).
                  $$

                2. If a contraction semigroup has a densely defined generator, then it is strongly continuous. This result is also somewhere in the book of Engel an Nagel, but the argument is quite simple: Trajectories starting in the domain of the generator are clearly continuous, and the uniform bound $|T_t|leq 1$ allows one to extend this continuity to all trajectories starting in the closure of the domain of the generator.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 16 at 12:20









                MaoWaoMaoWao

                3,933618




                3,933618






























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