$L^p-L^q$ estimates for heat equation - regularizing effect











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Where can I find a proof of the following estimate
$$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.










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    Where can I find a proof of the following estimate
    $$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
    where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.










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      Where can I find a proof of the following estimate
      $$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
      where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.










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      Where can I find a proof of the following estimate
      $$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
      where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.







      heat-equation regularity-theory-of-pdes semigroup-of-operators






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      edited Feb 29 '16 at 15:31

























      asked Feb 29 '16 at 15:30









      Reginaldo Demarque da Rocha

      856




      856






















          2 Answers
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          Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
          $$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
          Since
          $$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
          we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.



          We know that $S(t)v=N_tast v$. From Young's Inequality, we have
          $$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
          where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$



          Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
          $$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$



          Hence, we have the result.






          share|cite|improve this answer




























            up vote
            0
            down vote













            The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):



            Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.



            It follows from the Kato-Trotter product formula that
            $$
            0leq e^{tDelta-nchi}fleq e^{tDelta} f
            $$

            for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
            $$
            |e^{tDelta_Omega}f|leq e^{tDelta}|f|
            $$

            for all $fin L^2(Omega)$.



            This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.



            An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
            $$
            |f|_{2N/(N-2)}leq C|nabla f|_2,
            $$

            which can obviously be localized.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              up vote
              2
              down vote













              Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
              $$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
              Since
              $$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
              we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.



              We know that $S(t)v=N_tast v$. From Young's Inequality, we have
              $$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
              where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$



              Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
              $$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$



              Hence, we have the result.






              share|cite|improve this answer

























                up vote
                2
                down vote













                Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
                $$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
                Since
                $$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
                we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.



                We know that $S(t)v=N_tast v$. From Young's Inequality, we have
                $$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
                where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$



                Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
                $$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$



                Hence, we have the result.






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
                  $$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
                  Since
                  $$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
                  we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.



                  We know that $S(t)v=N_tast v$. From Young's Inequality, we have
                  $$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
                  where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$



                  Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
                  $$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$



                  Hence, we have the result.






                  share|cite|improve this answer












                  Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
                  $$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
                  Since
                  $$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
                  we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.



                  We know that $S(t)v=N_tast v$. From Young's Inequality, we have
                  $$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
                  where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$



                  Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
                  $$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$



                  Hence, we have the result.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 21 '16 at 10:04









                  Reginaldo Demarque da Rocha

                  856




                  856






















                      up vote
                      0
                      down vote













                      The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):



                      Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.



                      It follows from the Kato-Trotter product formula that
                      $$
                      0leq e^{tDelta-nchi}fleq e^{tDelta} f
                      $$

                      for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
                      $$
                      |e^{tDelta_Omega}f|leq e^{tDelta}|f|
                      $$

                      for all $fin L^2(Omega)$.



                      This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.



                      An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
                      $$
                      |f|_{2N/(N-2)}leq C|nabla f|_2,
                      $$

                      which can obviously be localized.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):



                        Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.



                        It follows from the Kato-Trotter product formula that
                        $$
                        0leq e^{tDelta-nchi}fleq e^{tDelta} f
                        $$

                        for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
                        $$
                        |e^{tDelta_Omega}f|leq e^{tDelta}|f|
                        $$

                        for all $fin L^2(Omega)$.



                        This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.



                        An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
                        $$
                        |f|_{2N/(N-2)}leq C|nabla f|_2,
                        $$

                        which can obviously be localized.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):



                          Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.



                          It follows from the Kato-Trotter product formula that
                          $$
                          0leq e^{tDelta-nchi}fleq e^{tDelta} f
                          $$

                          for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
                          $$
                          |e^{tDelta_Omega}f|leq e^{tDelta}|f|
                          $$

                          for all $fin L^2(Omega)$.



                          This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.



                          An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
                          $$
                          |f|_{2N/(N-2)}leq C|nabla f|_2,
                          $$

                          which can obviously be localized.






                          share|cite|improve this answer












                          The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):



                          Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.



                          It follows from the Kato-Trotter product formula that
                          $$
                          0leq e^{tDelta-nchi}fleq e^{tDelta} f
                          $$

                          for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
                          $$
                          |e^{tDelta_Omega}f|leq e^{tDelta}|f|
                          $$

                          for all $fin L^2(Omega)$.



                          This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.



                          An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
                          $$
                          |f|_{2N/(N-2)}leq C|nabla f|_2,
                          $$

                          which can obviously be localized.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 26 at 23:43









                          MaoWao

                          2,333616




                          2,333616






























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