If $uin H^1_0(Omega)cap C(Omega)$ is it true that $uin H^1_0({u>0})$?
$begingroup$
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
asked Dec 19 '18 at 5:45
BobBob
1,5971725
$endgroup$
add a comment |
$begingroup$
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
asked Dec 19 '18 at 5:45
BobBob
1,5971725
$endgroup$
add a comment |
$begingroup$
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
asked Dec 19 '18 at 5:45
BobBob
1,5971725
$endgroup$
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
asked Dec 19 '18 at 5:45
BobBob
1,5971725
asked Dec 19 '18 at 5:45
BobBob
1,5971725
asked Dec 19 '18 at 5:45
BobBob
1,5971725
asked Dec 19 '18 at 5:45
BobBob
1,5971725
asked Dec 19 '18 at 5:45
BobBob
1,5971725
1,5971725
add a comment |
add a comment |
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 $u in H^1_0(D).$
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
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add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
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2 Answers
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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 $u in H^1_0(D).$
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
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 $u in H^1_0(D).$
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
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 $u in H^1_0(D).$
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
$endgroup$
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 $u in H^1_0(D).$
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
edited Dec 20 '18 at 9:32
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
answered Dec 19 '18 at 13:00
ktoiktoi
2,4061617
2,4061617
add a comment |
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
$endgroup$
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.
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
answered Dec 19 '18 at 6:52
BigbearZzzBigbearZzz
8,76121652
8,76121652
add a comment |
add a comment |
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