Cesaro average semigroup












1












$begingroup$


Let $(P_t)_{tgeq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,mu)$ with generator $(L,mathcal{D}(L))$. Given $finmathcal{D}(L^*)$, define the Cesaro average
$$
A_tf=frac{1}{t}int_0^tP_sf,ds,qquad t>0,
$$
where the integral is understood in the Bochner sense.



Why is $A_t finmathcal{D}(L)$ for all $t>0$ and
$$
LA_tf=frac1tint_0^tLP_sf,ds?
$$






hilbert-spaces semigroup-of-operators unbounded-operators







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $(P_t)_{tgeq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,mu)$ with generator $(L,mathcal{D}(L))$. Given $finmathcal{D}(L^*)$, define the Cesaro average
    $$
    A_tf=frac{1}{t}int_0^tP_sf,ds,qquad t>0,
    $$
    where the integral is understood in the Bochner sense.



    Why is $A_t finmathcal{D}(L)$ for all $t>0$ and
    $$
    LA_tf=frac1tint_0^tLP_sf,ds?
    $$






    hilbert-spaces semigroup-of-operators unbounded-operators







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $(P_t)_{tgeq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,mu)$ with generator $(L,mathcal{D}(L))$. Given $finmathcal{D}(L^*)$, define the Cesaro average
      $$
      A_tf=frac{1}{t}int_0^tP_sf,ds,qquad t>0,
      $$
      where the integral is understood in the Bochner sense.



      Why is $A_t finmathcal{D}(L)$ for all $t>0$ and
      $$
      LA_tf=frac1tint_0^tLP_sf,ds?
      $$






      hilbert-spaces semigroup-of-operators unbounded-operators







      share|cite|improve this question









      $endgroup$




      Let $(P_t)_{tgeq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,mu)$ with generator $(L,mathcal{D}(L))$. Given $finmathcal{D}(L^*)$, define the Cesaro average
      $$
      A_tf=frac{1}{t}int_0^tP_sf,ds,qquad t>0,
      $$
      where the integral is understood in the Bochner sense.



      Why is $A_t finmathcal{D}(L)$ for all $t>0$ and
      $$
      LA_tf=frac1tint_0^tLP_sf,ds?
      $$






      hilbert-spaces semigroup-of-operators unbounded-operators




      hilbert-spaces semigroup-of-operators unbounded-operators






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 26 at 15:56









      julianjulian

      342110




      342110






















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          $begingroup$

          We have $yin D(L)$ if the limit
          $$lim_{hto 0^+} frac{P_hy-y}{h}$$
          exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
          $$lim_{hto 0^+} frac{P_hA_tf-A_tf}{h}$$
          exists and is equal to
          $$frac1tint_0^tLP_sf,ds.$$
          For this, note that
          begin{align*}
          frac{P_hA_tf-A_tf}{h}&=frac{P_hfrac{1}{t}int_0^tP_sf,ds-frac{1}{t}int_0^tP_sf,ds}{h}\
          &=frac{1}{th}left(int_0^tP_{s+h}f,ds-int_0^tP_s f,dsright)\
          &=frac{1}{th}left(int_h^{t+h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
          &=frac{1}{th}left(int_0^{t+h}P_{tau}f,dtau-int_0^{h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
          &=frac{1}{th}left(int_t^{t+h}P_{tau}f,dtau-int_0^hP_tau f,dtauright)\
          &=frac{1}{t}left(frac{1}{h}int_t^{t+h}P_{tau}f,dtau-frac{1}{h}int_0^hP_tau f,dtauright)
          end{align*}
          Then, taking the limit as $hto0^+$, we conclude that
          $$LA_t f=frac{1}{t}left(P_tf-P_0fright)=frac{1}{t}int_0^tfrac{d}{ds}(P_s f),ds=frac{1}{t}int_0^tLP_s f,ds.$$
          because
          $$frac{1}{h}int_t^{t+h} P_sf,dsoverset{hto 0^+}{longrightarrow}P_t fquadtext{and}quad frac{d}{ds}(P_sf)=LP_sf.$$






          share|cite|improve this answer









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            $begingroup$

            We have $yin D(L)$ if the limit
            $$lim_{hto 0^+} frac{P_hy-y}{h}$$
            exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
            $$lim_{hto 0^+} frac{P_hA_tf-A_tf}{h}$$
            exists and is equal to
            $$frac1tint_0^tLP_sf,ds.$$
            For this, note that
            begin{align*}
            frac{P_hA_tf-A_tf}{h}&=frac{P_hfrac{1}{t}int_0^tP_sf,ds-frac{1}{t}int_0^tP_sf,ds}{h}\
            &=frac{1}{th}left(int_0^tP_{s+h}f,ds-int_0^tP_s f,dsright)\
            &=frac{1}{th}left(int_h^{t+h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
            &=frac{1}{th}left(int_0^{t+h}P_{tau}f,dtau-int_0^{h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
            &=frac{1}{th}left(int_t^{t+h}P_{tau}f,dtau-int_0^hP_tau f,dtauright)\
            &=frac{1}{t}left(frac{1}{h}int_t^{t+h}P_{tau}f,dtau-frac{1}{h}int_0^hP_tau f,dtauright)
            end{align*}
            Then, taking the limit as $hto0^+$, we conclude that
            $$LA_t f=frac{1}{t}left(P_tf-P_0fright)=frac{1}{t}int_0^tfrac{d}{ds}(P_s f),ds=frac{1}{t}int_0^tLP_s f,ds.$$
            because
            $$frac{1}{h}int_t^{t+h} P_sf,dsoverset{hto 0^+}{longrightarrow}P_t fquadtext{and}quad frac{d}{ds}(P_sf)=LP_sf.$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We have $yin D(L)$ if the limit
              $$lim_{hto 0^+} frac{P_hy-y}{h}$$
              exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
              $$lim_{hto 0^+} frac{P_hA_tf-A_tf}{h}$$
              exists and is equal to
              $$frac1tint_0^tLP_sf,ds.$$
              For this, note that
              begin{align*}
              frac{P_hA_tf-A_tf}{h}&=frac{P_hfrac{1}{t}int_0^tP_sf,ds-frac{1}{t}int_0^tP_sf,ds}{h}\
              &=frac{1}{th}left(int_0^tP_{s+h}f,ds-int_0^tP_s f,dsright)\
              &=frac{1}{th}left(int_h^{t+h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
              &=frac{1}{th}left(int_0^{t+h}P_{tau}f,dtau-int_0^{h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
              &=frac{1}{th}left(int_t^{t+h}P_{tau}f,dtau-int_0^hP_tau f,dtauright)\
              &=frac{1}{t}left(frac{1}{h}int_t^{t+h}P_{tau}f,dtau-frac{1}{h}int_0^hP_tau f,dtauright)
              end{align*}
              Then, taking the limit as $hto0^+$, we conclude that
              $$LA_t f=frac{1}{t}left(P_tf-P_0fright)=frac{1}{t}int_0^tfrac{d}{ds}(P_s f),ds=frac{1}{t}int_0^tLP_s f,ds.$$
              because
              $$frac{1}{h}int_t^{t+h} P_sf,dsoverset{hto 0^+}{longrightarrow}P_t fquadtext{and}quad frac{d}{ds}(P_sf)=LP_sf.$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We have $yin D(L)$ if the limit
                $$lim_{hto 0^+} frac{P_hy-y}{h}$$
                exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
                $$lim_{hto 0^+} frac{P_hA_tf-A_tf}{h}$$
                exists and is equal to
                $$frac1tint_0^tLP_sf,ds.$$
                For this, note that
                begin{align*}
                frac{P_hA_tf-A_tf}{h}&=frac{P_hfrac{1}{t}int_0^tP_sf,ds-frac{1}{t}int_0^tP_sf,ds}{h}\
                &=frac{1}{th}left(int_0^tP_{s+h}f,ds-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_h^{t+h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_0^{t+h}P_{tau}f,dtau-int_0^{h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_t^{t+h}P_{tau}f,dtau-int_0^hP_tau f,dtauright)\
                &=frac{1}{t}left(frac{1}{h}int_t^{t+h}P_{tau}f,dtau-frac{1}{h}int_0^hP_tau f,dtauright)
                end{align*}
                Then, taking the limit as $hto0^+$, we conclude that
                $$LA_t f=frac{1}{t}left(P_tf-P_0fright)=frac{1}{t}int_0^tfrac{d}{ds}(P_s f),ds=frac{1}{t}int_0^tLP_s f,ds.$$
                because
                $$frac{1}{h}int_t^{t+h} P_sf,dsoverset{hto 0^+}{longrightarrow}P_t fquadtext{and}quad frac{d}{ds}(P_sf)=LP_sf.$$






                share|cite|improve this answer









                $endgroup$



                We have $yin D(L)$ if the limit
                $$lim_{hto 0^+} frac{P_hy-y}{h}$$
                exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
                $$lim_{hto 0^+} frac{P_hA_tf-A_tf}{h}$$
                exists and is equal to
                $$frac1tint_0^tLP_sf,ds.$$
                For this, note that
                begin{align*}
                frac{P_hA_tf-A_tf}{h}&=frac{P_hfrac{1}{t}int_0^tP_sf,ds-frac{1}{t}int_0^tP_sf,ds}{h}\
                &=frac{1}{th}left(int_0^tP_{s+h}f,ds-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_h^{t+h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_0^{t+h}P_{tau}f,dtau-int_0^{h}P_{tau}f,dtau-int_0^tP_s f,dsright)\
                &=frac{1}{th}left(int_t^{t+h}P_{tau}f,dtau-int_0^hP_tau f,dtauright)\
                &=frac{1}{t}left(frac{1}{h}int_t^{t+h}P_{tau}f,dtau-frac{1}{h}int_0^hP_tau f,dtauright)
                end{align*}
                Then, taking the limit as $hto0^+$, we conclude that
                $$LA_t f=frac{1}{t}left(P_tf-P_0fright)=frac{1}{t}int_0^tfrac{d}{ds}(P_s f),ds=frac{1}{t}int_0^tLP_s f,ds.$$
                because
                $$frac{1}{h}int_t^{t+h} P_sf,dsoverset{hto 0^+}{longrightarrow}P_t fquadtext{and}quad frac{d}{ds}(P_sf)=LP_sf.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 1:22









                PedroPedro

                10.8k23374




                10.8k23374






























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