Is this statement correct? Local Elliptic Regularity












2












$begingroup$



(The local regularity theorem) Let $P$ be a differential operator of order $k$ that is elliptic over $bar{U}, U subseteq Bbb R^n$ relatively comapct. Let $k,l$ be integers, $f in W^l$ and $u in W^r$. Assume that $Pu=f$, this equation takes place in $W^{r-k}$. Then for each function $mu in C_c^infty(U)$, $mu u in W^{l+k}$.






The spaces where $W^l$ is the completion of the Schwartz space $S(Bbb R^n)$ under Sobolev $l$-norm.
$$ ||f||_l^2 := int (1+|xi|^2)^l |hat{f} (xi)|^2 , dxi $$
For an open set $U subseteq Bbb R^n$, we define $W^l(U)$ to be the completion of $C_c^infty(U)$ under the Sobolev $l$-norm.





What confuses me is the appearance of the integer $r$. Is $r$ supposed to equal to $l$?



Is the statement true even when $k=0$?





Original source maybe hepful: page 46, Theorem 3.5.1.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    (The local regularity theorem) Let $P$ be a differential operator of order $k$ that is elliptic over $bar{U}, U subseteq Bbb R^n$ relatively comapct. Let $k,l$ be integers, $f in W^l$ and $u in W^r$. Assume that $Pu=f$, this equation takes place in $W^{r-k}$. Then for each function $mu in C_c^infty(U)$, $mu u in W^{l+k}$.






    The spaces where $W^l$ is the completion of the Schwartz space $S(Bbb R^n)$ under Sobolev $l$-norm.
    $$ ||f||_l^2 := int (1+|xi|^2)^l |hat{f} (xi)|^2 , dxi $$
    For an open set $U subseteq Bbb R^n$, we define $W^l(U)$ to be the completion of $C_c^infty(U)$ under the Sobolev $l$-norm.





    What confuses me is the appearance of the integer $r$. Is $r$ supposed to equal to $l$?



    Is the statement true even when $k=0$?





    Original source maybe hepful: page 46, Theorem 3.5.1.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      (The local regularity theorem) Let $P$ be a differential operator of order $k$ that is elliptic over $bar{U}, U subseteq Bbb R^n$ relatively comapct. Let $k,l$ be integers, $f in W^l$ and $u in W^r$. Assume that $Pu=f$, this equation takes place in $W^{r-k}$. Then for each function $mu in C_c^infty(U)$, $mu u in W^{l+k}$.






      The spaces where $W^l$ is the completion of the Schwartz space $S(Bbb R^n)$ under Sobolev $l$-norm.
      $$ ||f||_l^2 := int (1+|xi|^2)^l |hat{f} (xi)|^2 , dxi $$
      For an open set $U subseteq Bbb R^n$, we define $W^l(U)$ to be the completion of $C_c^infty(U)$ under the Sobolev $l$-norm.





      What confuses me is the appearance of the integer $r$. Is $r$ supposed to equal to $l$?



      Is the statement true even when $k=0$?





      Original source maybe hepful: page 46, Theorem 3.5.1.










      share|cite|improve this question











      $endgroup$





      (The local regularity theorem) Let $P$ be a differential operator of order $k$ that is elliptic over $bar{U}, U subseteq Bbb R^n$ relatively comapct. Let $k,l$ be integers, $f in W^l$ and $u in W^r$. Assume that $Pu=f$, this equation takes place in $W^{r-k}$. Then for each function $mu in C_c^infty(U)$, $mu u in W^{l+k}$.






      The spaces where $W^l$ is the completion of the Schwartz space $S(Bbb R^n)$ under Sobolev $l$-norm.
      $$ ||f||_l^2 := int (1+|xi|^2)^l |hat{f} (xi)|^2 , dxi $$
      For an open set $U subseteq Bbb R^n$, we define $W^l(U)$ to be the completion of $C_c^infty(U)$ under the Sobolev $l$-norm.





      What confuses me is the appearance of the integer $r$. Is $r$ supposed to equal to $l$?



      Is the statement true even when $k=0$?





      Original source maybe hepful: page 46, Theorem 3.5.1.







      real-analysis functional-analysis pde elliptic-operators






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 0:25







      CL.

















      asked Dec 25 '18 at 0:33









      CL.CL.

      2,3082925




      2,3082925






















          1 Answer
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          active

          oldest

          votes


















          3












          $begingroup$

          Depending on how you interpret the expression $Pu$, the prerequisite $uin W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:



          If $u in mathscr{D}'(U)$ and $Pu in W^l(U)$, then $mu uin W^{l+k}(U)$ for all $muin C_c^infty(U)$.



          Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $rge k$ and $uin W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.



          To illustrate the difference between the two formulations, consider the following example: Let
          $$uin L^1_{loc}(U) text{ with }int_U u Delta varphi = 0 text{ for all } varphi in C_c^infty(U). tag{$star$}$$
          Now your formulation of elliptic regularity gives the following: If we additionally assume that $u in W^2(U)$, then we can integrate by parts and see that $Delta u = 0$, which lies in $W^l(U)$ for all $lge 0$. Since $Delta$ is elliptic, this implies that $mu uin W^{l+2}(U)$ for all $l$ and thus $u$ is smooth.
          However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($star$) is enough to conclude that $u$ is smooth.



          Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $ain C^infty(U)$ and ellipticity means that $a(x)neq 0$ for all $xin U$. In particular $mu u = (mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $mu a^{-1}in C_c^infty(U)$ certainly yields a function in $W^l(U)$ again.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
            $endgroup$
            – ktoi
            Dec 25 '18 at 12:31












          • $begingroup$
            Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
            $endgroup$
            – CL.
            Dec 26 '18 at 0:25






          • 1




            $begingroup$
            Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:52








          • 1




            $begingroup$
            As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:56














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          1 Answer
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          active

          oldest

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          3












          $begingroup$

          Depending on how you interpret the expression $Pu$, the prerequisite $uin W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:



          If $u in mathscr{D}'(U)$ and $Pu in W^l(U)$, then $mu uin W^{l+k}(U)$ for all $muin C_c^infty(U)$.



          Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $rge k$ and $uin W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.



          To illustrate the difference between the two formulations, consider the following example: Let
          $$uin L^1_{loc}(U) text{ with }int_U u Delta varphi = 0 text{ for all } varphi in C_c^infty(U). tag{$star$}$$
          Now your formulation of elliptic regularity gives the following: If we additionally assume that $u in W^2(U)$, then we can integrate by parts and see that $Delta u = 0$, which lies in $W^l(U)$ for all $lge 0$. Since $Delta$ is elliptic, this implies that $mu uin W^{l+2}(U)$ for all $l$ and thus $u$ is smooth.
          However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($star$) is enough to conclude that $u$ is smooth.



          Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $ain C^infty(U)$ and ellipticity means that $a(x)neq 0$ for all $xin U$. In particular $mu u = (mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $mu a^{-1}in C_c^infty(U)$ certainly yields a function in $W^l(U)$ again.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
            $endgroup$
            – ktoi
            Dec 25 '18 at 12:31












          • $begingroup$
            Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
            $endgroup$
            – CL.
            Dec 26 '18 at 0:25






          • 1




            $begingroup$
            Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:52








          • 1




            $begingroup$
            As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:56


















          3












          $begingroup$

          Depending on how you interpret the expression $Pu$, the prerequisite $uin W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:



          If $u in mathscr{D}'(U)$ and $Pu in W^l(U)$, then $mu uin W^{l+k}(U)$ for all $muin C_c^infty(U)$.



          Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $rge k$ and $uin W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.



          To illustrate the difference between the two formulations, consider the following example: Let
          $$uin L^1_{loc}(U) text{ with }int_U u Delta varphi = 0 text{ for all } varphi in C_c^infty(U). tag{$star$}$$
          Now your formulation of elliptic regularity gives the following: If we additionally assume that $u in W^2(U)$, then we can integrate by parts and see that $Delta u = 0$, which lies in $W^l(U)$ for all $lge 0$. Since $Delta$ is elliptic, this implies that $mu uin W^{l+2}(U)$ for all $l$ and thus $u$ is smooth.
          However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($star$) is enough to conclude that $u$ is smooth.



          Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $ain C^infty(U)$ and ellipticity means that $a(x)neq 0$ for all $xin U$. In particular $mu u = (mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $mu a^{-1}in C_c^infty(U)$ certainly yields a function in $W^l(U)$ again.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
            $endgroup$
            – ktoi
            Dec 25 '18 at 12:31












          • $begingroup$
            Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
            $endgroup$
            – CL.
            Dec 26 '18 at 0:25






          • 1




            $begingroup$
            Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:52








          • 1




            $begingroup$
            As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:56
















          3












          3








          3





          $begingroup$

          Depending on how you interpret the expression $Pu$, the prerequisite $uin W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:



          If $u in mathscr{D}'(U)$ and $Pu in W^l(U)$, then $mu uin W^{l+k}(U)$ for all $muin C_c^infty(U)$.



          Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $rge k$ and $uin W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.



          To illustrate the difference between the two formulations, consider the following example: Let
          $$uin L^1_{loc}(U) text{ with }int_U u Delta varphi = 0 text{ for all } varphi in C_c^infty(U). tag{$star$}$$
          Now your formulation of elliptic regularity gives the following: If we additionally assume that $u in W^2(U)$, then we can integrate by parts and see that $Delta u = 0$, which lies in $W^l(U)$ for all $lge 0$. Since $Delta$ is elliptic, this implies that $mu uin W^{l+2}(U)$ for all $l$ and thus $u$ is smooth.
          However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($star$) is enough to conclude that $u$ is smooth.



          Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $ain C^infty(U)$ and ellipticity means that $a(x)neq 0$ for all $xin U$. In particular $mu u = (mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $mu a^{-1}in C_c^infty(U)$ certainly yields a function in $W^l(U)$ again.






          share|cite|improve this answer









          $endgroup$



          Depending on how you interpret the expression $Pu$, the prerequisite $uin W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:



          If $u in mathscr{D}'(U)$ and $Pu in W^l(U)$, then $mu uin W^{l+k}(U)$ for all $muin C_c^infty(U)$.



          Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $rge k$ and $uin W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.



          To illustrate the difference between the two formulations, consider the following example: Let
          $$uin L^1_{loc}(U) text{ with }int_U u Delta varphi = 0 text{ for all } varphi in C_c^infty(U). tag{$star$}$$
          Now your formulation of elliptic regularity gives the following: If we additionally assume that $u in W^2(U)$, then we can integrate by parts and see that $Delta u = 0$, which lies in $W^l(U)$ for all $lge 0$. Since $Delta$ is elliptic, this implies that $mu uin W^{l+2}(U)$ for all $l$ and thus $u$ is smooth.
          However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($star$) is enough to conclude that $u$ is smooth.



          Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $ain C^infty(U)$ and ellipticity means that $a(x)neq 0$ for all $xin U$. In particular $mu u = (mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $mu a^{-1}in C_c^infty(U)$ certainly yields a function in $W^l(U)$ again.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 11:28









          Jan BohrJan Bohr

          3,3071521




          3,3071521








          • 1




            $begingroup$
            To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
            $endgroup$
            – ktoi
            Dec 25 '18 at 12:31












          • $begingroup$
            Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
            $endgroup$
            – CL.
            Dec 26 '18 at 0:25






          • 1




            $begingroup$
            Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:52








          • 1




            $begingroup$
            As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:56
















          • 1




            $begingroup$
            To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
            $endgroup$
            – ktoi
            Dec 25 '18 at 12:31












          • $begingroup$
            Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
            $endgroup$
            – CL.
            Dec 26 '18 at 0:25






          • 1




            $begingroup$
            Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:52








          • 1




            $begingroup$
            As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
            $endgroup$
            – Jan Bohr
            Dec 26 '18 at 8:56










          1




          1




          $begingroup$
          To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
          $endgroup$
          – ktoi
          Dec 25 '18 at 12:31






          $begingroup$
          To add, if $u in mathscr{D}'(U)$ and $V subsetsubset U,$ then the restriction $u|_V in W^r(V)$ for some $r in mathbb R$ (see e.g. corollary 6.8 of Folland's 'Partial Differential Equations'). By taking a compact exhaustion of $U,$ you can obtain this stronger form.
          $endgroup$
          – ktoi
          Dec 25 '18 at 12:31














          $begingroup$
          Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
          $endgroup$
          – CL.
          Dec 26 '18 at 0:25




          $begingroup$
          Thanks a lot, where is a good reference for this result (Folland's ?) . Also, the definition that we take for $W^r(U)$ is as edited. The proof I linked proves the cases when $k ge 1$. It seems to me that your argument does not require $ u in W^r(U)$, but $u in W^r$ is sufficient. Am I right?
          $endgroup$
          – CL.
          Dec 26 '18 at 0:25




          1




          1




          $begingroup$
          Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
          $endgroup$
          – Jan Bohr
          Dec 26 '18 at 8:52






          $begingroup$
          Be aware that this is not the standard definition of Sobolev spaces (in case $Uneq mathbb{R}^n$). Taking the closure of $C_c^infty(U)$ in the $W^k$-norm will only yield functions whose derivatives up to order $k-1$ vanish at the boundary of $U$. The space of those functions is usually denoted $W_0^k(U)$, whereas $W^k(U)$ is reserved for all Sobolev functions, without any boundary restrictions.
          $endgroup$
          – Jan Bohr
          Dec 26 '18 at 8:52






          1




          1




          $begingroup$
          As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
          $endgroup$
          – Jan Bohr
          Dec 26 '18 at 8:56






          $begingroup$
          As for references for elliptic regularity, you can also have a look at Evan's PDE book or Shubin's book on $Psi$DOs. They both generalise different aspects: Evans allows for $L^p$-bases Sobolev spaces and DO with non-smooth coefficients. Shubin works on manifolds, treats $Psi$DOs of all orders, but only treats smooth coefficients and $L^2$-bases spaces.
          $endgroup$
          – Jan Bohr
          Dec 26 '18 at 8:56




















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