Is this statement correct? Local Elliptic Regularity
$begingroup$
(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.
real-analysis functional-analysis pde elliptic-operators
$endgroup$
add a comment |
$begingroup$
(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.
real-analysis functional-analysis pde elliptic-operators
$endgroup$
add a comment |
$begingroup$
(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.
real-analysis functional-analysis pde elliptic-operators
$endgroup$
(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
real-analysis functional-analysis pde elliptic-operators
edited Dec 26 '18 at 0:25
CL.
asked Dec 25 '18 at 0:33
CL.CL.
2,3082925
2,3082925
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
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.
$endgroup$
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
add a comment |
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$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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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