If $uin H^1_0(Omega)cap C(Omega)$ is it true that $uin H^1_0({u>0})$?












3












$begingroup$


Let $Omega$ be a non-empty open subset of $mathbb{R}^N$.



Let $H^1_0(Omega)$ be the closure in the $H^1(Omega)$ norm of the subspace $C^infty_c(Omega)$.



Let $uin C(Omega)cap H^1_0(Omega)$ such that $D:={xinOmega | u(x)>0}neqemptyset$.



Then $D$ is a non-empty open subset of $mathbb{R}^N$ and so it makes sense to talk about $H^1(D)$ and $H^1_0(D)$.




Is it true that $u|_Din H^1_0(D)$? I.e.: does there exist a sequence $(varphi_n)_{ninmathbb{N}}subset C^infty_c(D)$ such that $|u|_D-varphi_n|_{H^1(D)}to0,nrightarrowinfty$? If not, what about if $partialOmega$ is smooth or maybe if we require further regularity on $u$?







sobolev-spaces







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$endgroup$

















    3












    $begingroup$


    Let $Omega$ be a non-empty open subset of $mathbb{R}^N$.



    Let $H^1_0(Omega)$ be the closure in the $H^1(Omega)$ norm of the subspace $C^infty_c(Omega)$.



    Let $uin C(Omega)cap H^1_0(Omega)$ such that $D:={xinOmega | u(x)>0}neqemptyset$.



    Then $D$ is a non-empty open subset of $mathbb{R}^N$ and so it makes sense to talk about $H^1(D)$ and $H^1_0(D)$.




    Is it true that $u|_Din H^1_0(D)$? I.e.: does there exist a sequence $(varphi_n)_{ninmathbb{N}}subset C^infty_c(D)$ such that $|u|_D-varphi_n|_{H^1(D)}to0,nrightarrowinfty$? If not, what about if $partialOmega$ is smooth or maybe if we require further regularity on $u$?







    sobolev-spaces







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      3



      $begingroup$


      Let $Omega$ be a non-empty open subset of $mathbb{R}^N$.



      Let $H^1_0(Omega)$ be the closure in the $H^1(Omega)$ norm of the subspace $C^infty_c(Omega)$.



      Let $uin C(Omega)cap H^1_0(Omega)$ such that $D:={xinOmega | u(x)>0}neqemptyset$.



      Then $D$ is a non-empty open subset of $mathbb{R}^N$ and so it makes sense to talk about $H^1(D)$ and $H^1_0(D)$.




      Is it true that $u|_Din H^1_0(D)$? I.e.: does there exist a sequence $(varphi_n)_{ninmathbb{N}}subset C^infty_c(D)$ such that $|u|_D-varphi_n|_{H^1(D)}to0,nrightarrowinfty$? If not, what about if $partialOmega$ is smooth or maybe if we require further regularity on $u$?







      sobolev-spaces







      share|cite|improve this question









      $endgroup$




      Let $Omega$ be a non-empty open subset of $mathbb{R}^N$.



      Let $H^1_0(Omega)$ be the closure in the $H^1(Omega)$ norm of the subspace $C^infty_c(Omega)$.



      Let $uin C(Omega)cap H^1_0(Omega)$ such that $D:={xinOmega | u(x)>0}neqemptyset$.



      Then $D$ is a non-empty open subset of $mathbb{R}^N$ and so it makes sense to talk about $H^1(D)$ and $H^1_0(D)$.




      Is it true that $u|_Din H^1_0(D)$? I.e.: does there exist a sequence $(varphi_n)_{ninmathbb{N}}subset C^infty_c(D)$ such that $|u|_D-varphi_n|_{H^1(D)}to0,nrightarrowinfty$? If not, what about if $partialOmega$ is smooth or maybe if we require further regularity on $u$?







      sobolev-spaces




      sobolev-spaces






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 19 '18 at 5:45









      BobBob

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          2 Answers
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          $begingroup$

          Yes, this is true in general. The idea is to consider the sequence of functions,
          $$u_{varepsilon} = (u-varepsilon)_+ = max{u-varepsilon,0} in H^1_0(Omega)$$
          and approximate these by functions in $C^{infty}_c(D).$ If $Omega$ is bounded, the support of each $operatorname{supp} v_{varepsilon}$ is compactly contained in $Omega,$ so we can mollify each to obtain elements in $C^{infty}_c(D).$ The general case will require an additional cutoff argument.



          Let $chi_{1/varepsilon} in C^{infty}_c(B_{1+1/varepsilon})$ be a cutoff such that $chi_{1/varepsilon} equiv 1$ in $B_{1/varepsilon}$ and $|nablachi_{1/varepsilon}| leq 2$ everywhere. Put $v_{varepsilon} = u_{varepsilon}chi_{1/varepsilon},$ so one can check that $v_{varepsilon} rightarrow u$ in $H^1(D)$ as $varepsilon rightarrow 0.$ Now $K_{varepsilon} = operatorname{supp} v_{varepsilon} subset Omega$ is compact, by continuity of each $v_{varepsilon}$ and as $K_{varepsilon} cap partialOmega = emptyset.$ Hence the mollification $v_{varepsilon} ast eta_{delta}$ lies in $C^{infty}_c(Omega)$ provided $delta < delta_0(varepsilon).$ Taking $delta = delta_0(varepsilon)/2$ and letting $varepsilon rightarrow 0$ gives a sequence of $C^{infty}_c(D)$ functions converging to $u.$ Hence $u in H^1_0(D).$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            If the set $D$ is not too exotic then yes, we have $ulvert_D in H^1_0(D).$



            Assuming that $partial D$ is regular enough (says, of Lipschitz boundary), since $partial D$ is contained in the set $u^{-1}({0})$, we have $text{Tr}(u)=0$, where $text{Tr}:H^1(D)to L^2(partial D)$ is the trace operator.



            The set $partial D$ can be very irregular, however. This question on Mathoverflow discusses how bad can the zero set of a continuous (or even smooth) function can be. In this case I believe the space $H^1(D)$ itself would be pretty hard to describe.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

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              active

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              2












              $begingroup$

              Yes, this is true in general. The idea is to consider the sequence of functions,
              $$u_{varepsilon} = (u-varepsilon)_+ = max{u-varepsilon,0} in H^1_0(Omega)$$
              and approximate these by functions in $C^{infty}_c(D).$ If $Omega$ is bounded, the support of each $operatorname{supp} v_{varepsilon}$ is compactly contained in $Omega,$ so we can mollify each to obtain elements in $C^{infty}_c(D).$ The general case will require an additional cutoff argument.



              Let $chi_{1/varepsilon} in C^{infty}_c(B_{1+1/varepsilon})$ be a cutoff such that $chi_{1/varepsilon} equiv 1$ in $B_{1/varepsilon}$ and $|nablachi_{1/varepsilon}| leq 2$ everywhere. Put $v_{varepsilon} = u_{varepsilon}chi_{1/varepsilon},$ so one can check that $v_{varepsilon} rightarrow u$ in $H^1(D)$ as $varepsilon rightarrow 0.$ Now $K_{varepsilon} = operatorname{supp} v_{varepsilon} subset Omega$ is compact, by continuity of each $v_{varepsilon}$ and as $K_{varepsilon} cap partialOmega = emptyset.$ Hence the mollification $v_{varepsilon} ast eta_{delta}$ lies in $C^{infty}_c(Omega)$ provided $delta < delta_0(varepsilon).$ Taking $delta = delta_0(varepsilon)/2$ and letting $varepsilon rightarrow 0$ gives a sequence of $C^{infty}_c(D)$ functions converging to $u.$ Hence $u in H^1_0(D).$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Yes, this is true in general. The idea is to consider the sequence of functions,
                $$u_{varepsilon} = (u-varepsilon)_+ = max{u-varepsilon,0} in H^1_0(Omega)$$
                and approximate these by functions in $C^{infty}_c(D).$ If $Omega$ is bounded, the support of each $operatorname{supp} v_{varepsilon}$ is compactly contained in $Omega,$ so we can mollify each to obtain elements in $C^{infty}_c(D).$ The general case will require an additional cutoff argument.



                Let $chi_{1/varepsilon} in C^{infty}_c(B_{1+1/varepsilon})$ be a cutoff such that $chi_{1/varepsilon} equiv 1$ in $B_{1/varepsilon}$ and $|nablachi_{1/varepsilon}| leq 2$ everywhere. Put $v_{varepsilon} = u_{varepsilon}chi_{1/varepsilon},$ so one can check that $v_{varepsilon} rightarrow u$ in $H^1(D)$ as $varepsilon rightarrow 0.$ Now $K_{varepsilon} = operatorname{supp} v_{varepsilon} subset Omega$ is compact, by continuity of each $v_{varepsilon}$ and as $K_{varepsilon} cap partialOmega = emptyset.$ Hence the mollification $v_{varepsilon} ast eta_{delta}$ lies in $C^{infty}_c(Omega)$ provided $delta < delta_0(varepsilon).$ Taking $delta = delta_0(varepsilon)/2$ and letting $varepsilon rightarrow 0$ gives a sequence of $C^{infty}_c(D)$ functions converging to $u.$ Hence $u in H^1_0(D).$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Yes, this is true in general. The idea is to consider the sequence of functions,
                  $$u_{varepsilon} = (u-varepsilon)_+ = max{u-varepsilon,0} in H^1_0(Omega)$$
                  and approximate these by functions in $C^{infty}_c(D).$ If $Omega$ is bounded, the support of each $operatorname{supp} v_{varepsilon}$ is compactly contained in $Omega,$ so we can mollify each to obtain elements in $C^{infty}_c(D).$ The general case will require an additional cutoff argument.



                  Let $chi_{1/varepsilon} in C^{infty}_c(B_{1+1/varepsilon})$ be a cutoff such that $chi_{1/varepsilon} equiv 1$ in $B_{1/varepsilon}$ and $|nablachi_{1/varepsilon}| leq 2$ everywhere. Put $v_{varepsilon} = u_{varepsilon}chi_{1/varepsilon},$ so one can check that $v_{varepsilon} rightarrow u$ in $H^1(D)$ as $varepsilon rightarrow 0.$ Now $K_{varepsilon} = operatorname{supp} v_{varepsilon} subset Omega$ is compact, by continuity of each $v_{varepsilon}$ and as $K_{varepsilon} cap partialOmega = emptyset.$ Hence the mollification $v_{varepsilon} ast eta_{delta}$ lies in $C^{infty}_c(Omega)$ provided $delta < delta_0(varepsilon).$ Taking $delta = delta_0(varepsilon)/2$ and letting $varepsilon rightarrow 0$ gives a sequence of $C^{infty}_c(D)$ functions converging to $u.$ Hence $u in H^1_0(D).$






                  share|cite|improve this answer











                  $endgroup$



                  Yes, this is true in general. The idea is to consider the sequence of functions,
                  $$u_{varepsilon} = (u-varepsilon)_+ = max{u-varepsilon,0} in H^1_0(Omega)$$
                  and approximate these by functions in $C^{infty}_c(D).$ If $Omega$ is bounded, the support of each $operatorname{supp} v_{varepsilon}$ is compactly contained in $Omega,$ so we can mollify each to obtain elements in $C^{infty}_c(D).$ The general case will require an additional cutoff argument.



                  Let $chi_{1/varepsilon} in C^{infty}_c(B_{1+1/varepsilon})$ be a cutoff such that $chi_{1/varepsilon} equiv 1$ in $B_{1/varepsilon}$ and $|nablachi_{1/varepsilon}| leq 2$ everywhere. Put $v_{varepsilon} = u_{varepsilon}chi_{1/varepsilon},$ so one can check that $v_{varepsilon} rightarrow u$ in $H^1(D)$ as $varepsilon rightarrow 0.$ Now $K_{varepsilon} = operatorname{supp} v_{varepsilon} subset Omega$ is compact, by continuity of each $v_{varepsilon}$ and as $K_{varepsilon} cap partialOmega = emptyset.$ Hence the mollification $v_{varepsilon} ast eta_{delta}$ lies in $C^{infty}_c(Omega)$ provided $delta < delta_0(varepsilon).$ Taking $delta = delta_0(varepsilon)/2$ and letting $varepsilon rightarrow 0$ gives a sequence of $C^{infty}_c(D)$ functions converging to $u.$ Hence $u in H^1_0(D).$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 9:32

























                  answered Dec 19 '18 at 13:00









                  ktoiktoi

                  2,4161618




                  2,4161618























                      1












                      $begingroup$

                      If the set $D$ is not too exotic then yes, we have $ulvert_D in H^1_0(D).$



                      Assuming that $partial D$ is regular enough (says, of Lipschitz boundary), since $partial D$ is contained in the set $u^{-1}({0})$, we have $text{Tr}(u)=0$, where $text{Tr}:H^1(D)to L^2(partial D)$ is the trace operator.



                      The set $partial D$ can be very irregular, however. This question on Mathoverflow discusses how bad can the zero set of a continuous (or even smooth) function can be. In this case I believe the space $H^1(D)$ itself would be pretty hard to describe.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        If the set $D$ is not too exotic then yes, we have $ulvert_D in H^1_0(D).$



                        Assuming that $partial D$ is regular enough (says, of Lipschitz boundary), since $partial D$ is contained in the set $u^{-1}({0})$, we have $text{Tr}(u)=0$, where $text{Tr}:H^1(D)to L^2(partial D)$ is the trace operator.



                        The set $partial D$ can be very irregular, however. This question on Mathoverflow discusses how bad can the zero set of a continuous (or even smooth) function can be. In this case I believe the space $H^1(D)$ itself would be pretty hard to describe.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          If the set $D$ is not too exotic then yes, we have $ulvert_D in H^1_0(D).$



                          Assuming that $partial D$ is regular enough (says, of Lipschitz boundary), since $partial D$ is contained in the set $u^{-1}({0})$, we have $text{Tr}(u)=0$, where $text{Tr}:H^1(D)to L^2(partial D)$ is the trace operator.



                          The set $partial D$ can be very irregular, however. This question on Mathoverflow discusses how bad can the zero set of a continuous (or even smooth) function can be. In this case I believe the space $H^1(D)$ itself would be pretty hard to describe.






                          share|cite|improve this answer









                          $endgroup$



                          If the set $D$ is not too exotic then yes, we have $ulvert_D in H^1_0(D).$



                          Assuming that $partial D$ is regular enough (says, of Lipschitz boundary), since $partial D$ is contained in the set $u^{-1}({0})$, we have $text{Tr}(u)=0$, where $text{Tr}:H^1(D)to L^2(partial D)$ is the trace operator.



                          The set $partial D$ can be very irregular, however. This question on Mathoverflow discusses how bad can the zero set of a continuous (or even smooth) function can be. In this case I believe the space $H^1(D)$ itself would be pretty hard to describe.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 19 '18 at 6:52









                          BigbearZzzBigbearZzz

                          8,93521652




                          8,93521652






























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