Is the charateristic function $chi _{Omega }$ in the Sobolev space $W^{1,2}_{0}(Omega)$?
Given $Omega$ is a bounded, $C^1$ domain in $mathbb{R}^n$. $chi _{Omega }(x)$ is the characteristic function of $Omega$.
I have done the followings:
We can get $chi _{Omega }(x) in L^2(Omega)$ for all $n in mathbb{N}$ quite easily. The next thing is to show the existence of the weak derivative of $chi_{Omega}$.
For $n=1$, $Omega$ is an open interval. Let $phi in C_{c}^{infty }(Omega)$ be an arbitrary test function. We have
$int_{a}^{b}chi _{(a,b) }phi 'dx=int_{a}^{b}phi'dx =phi(b)-phi(a)=0$
So $int_{a}^{b}chi _{(a,b) }phi 'dx=-int_{a}^{b}0.phi dx$, and the weak derivative of $chi_{Omega}$ is $0$.
For $ngeq 2$, I get stuck and don't know if $chi_{Omega}$ has the weak derivative or not.
Thank you very much for your help.
functional-analysis lebesgue-integral sobolev-spaces
asked May 9 '16 at 14:10
user1210321
485
add a comment |
Given $Omega$ is a bounded, $C^1$ domain in $mathbb{R}^n$. $chi _{Omega }(x)$ is the characteristic function of $Omega$.
I have done the followings:
We can get $chi _{Omega }(x) in L^2(Omega)$ for all $n in mathbb{N}$ quite easily. The next thing is to show the existence of the weak derivative of $chi_{Omega}$.
For $n=1$, $Omega$ is an open interval. Let $phi in C_{c}^{infty }(Omega)$ be an arbitrary test function. We have
$int_{a}^{b}chi _{(a,b) }phi 'dx=int_{a}^{b}phi'dx =phi(b)-phi(a)=0$
So $int_{a}^{b}chi _{(a,b) }phi 'dx=-int_{a}^{b}0.phi dx$, and the weak derivative of $chi_{Omega}$ is $0$.
For $ngeq 2$, I get stuck and don't know if $chi_{Omega}$ has the weak derivative or not.
Thank you very much for your help.
functional-analysis lebesgue-integral sobolev-spaces
asked May 9 '16 at 14:10
user1210321
485
$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
1
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago
add a comment |
Given $Omega$ is a bounded, $C^1$ domain in $mathbb{R}^n$. $chi _{Omega }(x)$ is the characteristic function of $Omega$.
I have done the followings:
We can get $chi _{Omega }(x) in L^2(Omega)$ for all $n in mathbb{N}$ quite easily. The next thing is to show the existence of the weak derivative of $chi_{Omega}$.
For $n=1$, $Omega$ is an open interval. Let $phi in C_{c}^{infty }(Omega)$ be an arbitrary test function. We have
$int_{a}^{b}chi _{(a,b) }phi 'dx=int_{a}^{b}phi'dx =phi(b)-phi(a)=0$
So $int_{a}^{b}chi _{(a,b) }phi 'dx=-int_{a}^{b}0.phi dx$, and the weak derivative of $chi_{Omega}$ is $0$.
For $ngeq 2$, I get stuck and don't know if $chi_{Omega}$ has the weak derivative or not.
Thank you very much for your help.
functional-analysis lebesgue-integral sobolev-spaces
asked May 9 '16 at 14:10
user1210321
485
Given $Omega$ is a bounded, $C^1$ domain in $mathbb{R}^n$. $chi _{Omega }(x)$ is the characteristic function of $Omega$.
I have done the followings:
We can get $chi _{Omega }(x) in L^2(Omega)$ for all $n in mathbb{N}$ quite easily. The next thing is to show the existence of the weak derivative of $chi_{Omega}$.
For $n=1$, $Omega$ is an open interval. Let $phi in C_{c}^{infty }(Omega)$ be an arbitrary test function. We have
$int_{a}^{b}chi _{(a,b) }phi 'dx=int_{a}^{b}phi'dx =phi(b)-phi(a)=0$
So $int_{a}^{b}chi _{(a,b) }phi 'dx=-int_{a}^{b}0.phi dx$, and the weak derivative of $chi_{Omega}$ is $0$.
For $ngeq 2$, I get stuck and don't know if $chi_{Omega}$ has the weak derivative or not.
Thank you very much for your help.
functional-analysis lebesgue-integral sobolev-spaces
functional-analysis lebesgue-integral sobolev-spaces
asked May 9 '16 at 14:10
user1210321
485
asked May 9 '16 at 14:10
user1210321
485
asked May 9 '16 at 14:10
user1210321
485
asked May 9 '16 at 14:10
user1210321
485
asked May 9 '16 at 14:10
user1210321
485
485
$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
1
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago
add a comment |
$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
1
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago
$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
1
1
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago
add a comment |
1 Answer
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No, the function $chi_Omega$ is not in the Sobolev space $W_0^{1,2}(Omega)$
if the domain $Omega$ is bounded.
Recall that $W_0^{1,2}$ is defined as the closure of the subspace $C_c^infty(Omega)subset H^1(Omega)$.
Let us assume that $chi_Omegain W_0^{1,2}(Omega)$.
Then there exists a sequence $phi_nin C_c^infty(Omega)$
such that $phi_nto chi_Omega$ in $W_0^{1,2}(Omega)$.
It follows that
$$
|phi_n|_{L^p(Omega)}to |chi_Omega|_{L^p(Omega)}=|Omega|>0
$$
and
$$
|nabla phi_n|_{L^p(Omega)}to |nablachi_Omega|_{L^p(Omega)}=0.
$$
(Note that the function $chi_Omega$ has weak derivatives $0$
because it is equal to the classical derivative and $chi_Omega$ is constant.)
However, by the Poincaré inequality
we have
$$
|phi_n|_L^p(Omega) leq C|nablaphi_n|_{L^p(Omega)}
$$
for some constant $C>0$ (here it is useful that the domain $Omega$ is bounded).
This yields a contradiction to the convergencies that are mentioned above.
answered 18 hours ago
supinf
6,0791028
add a comment |
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1 Answer
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No, the function $chi_Omega$ is not in the Sobolev space $W_0^{1,2}(Omega)$
if the domain $Omega$ is bounded.
Recall that $W_0^{1,2}$ is defined as the closure of the subspace $C_c^infty(Omega)subset H^1(Omega)$.
Let us assume that $chi_Omegain W_0^{1,2}(Omega)$.
Then there exists a sequence $phi_nin C_c^infty(Omega)$
such that $phi_nto chi_Omega$ in $W_0^{1,2}(Omega)$.
It follows that
$$
|phi_n|_{L^p(Omega)}to |chi_Omega|_{L^p(Omega)}=|Omega|>0
$$
and
$$
|nabla phi_n|_{L^p(Omega)}to |nablachi_Omega|_{L^p(Omega)}=0.
$$
(Note that the function $chi_Omega$ has weak derivatives $0$
because it is equal to the classical derivative and $chi_Omega$ is constant.)
However, by the Poincaré inequality
we have
$$
|phi_n|_L^p(Omega) leq C|nablaphi_n|_{L^p(Omega)}
$$
for some constant $C>0$ (here it is useful that the domain $Omega$ is bounded).
This yields a contradiction to the convergencies that are mentioned above.
answered 18 hours ago
supinf
6,0791028
add a comment |
No, the function $chi_Omega$ is not in the Sobolev space $W_0^{1,2}(Omega)$
if the domain $Omega$ is bounded.
Recall that $W_0^{1,2}$ is defined as the closure of the subspace $C_c^infty(Omega)subset H^1(Omega)$.
Let us assume that $chi_Omegain W_0^{1,2}(Omega)$.
Then there exists a sequence $phi_nin C_c^infty(Omega)$
such that $phi_nto chi_Omega$ in $W_0^{1,2}(Omega)$.
It follows that
$$
|phi_n|_{L^p(Omega)}to |chi_Omega|_{L^p(Omega)}=|Omega|>0
$$
and
$$
|nabla phi_n|_{L^p(Omega)}to |nablachi_Omega|_{L^p(Omega)}=0.
$$
(Note that the function $chi_Omega$ has weak derivatives $0$
because it is equal to the classical derivative and $chi_Omega$ is constant.)
However, by the Poincaré inequality
we have
$$
|phi_n|_L^p(Omega) leq C|nablaphi_n|_{L^p(Omega)}
$$
for some constant $C>0$ (here it is useful that the domain $Omega$ is bounded).
This yields a contradiction to the convergencies that are mentioned above.
answered 18 hours ago
supinf
6,0791028
add a comment |
No, the function $chi_Omega$ is not in the Sobolev space $W_0^{1,2}(Omega)$
if the domain $Omega$ is bounded.
Recall that $W_0^{1,2}$ is defined as the closure of the subspace $C_c^infty(Omega)subset H^1(Omega)$.
Let us assume that $chi_Omegain W_0^{1,2}(Omega)$.
Then there exists a sequence $phi_nin C_c^infty(Omega)$
such that $phi_nto chi_Omega$ in $W_0^{1,2}(Omega)$.
It follows that
$$
|phi_n|_{L^p(Omega)}to |chi_Omega|_{L^p(Omega)}=|Omega|>0
$$
and
$$
|nabla phi_n|_{L^p(Omega)}to |nablachi_Omega|_{L^p(Omega)}=0.
$$
(Note that the function $chi_Omega$ has weak derivatives $0$
because it is equal to the classical derivative and $chi_Omega$ is constant.)
However, by the Poincaré inequality
we have
$$
|phi_n|_L^p(Omega) leq C|nablaphi_n|_{L^p(Omega)}
$$
for some constant $C>0$ (here it is useful that the domain $Omega$ is bounded).
This yields a contradiction to the convergencies that are mentioned above.
answered 18 hours ago
supinf
6,0791028
No, the function $chi_Omega$ is not in the Sobolev space $W_0^{1,2}(Omega)$
if the domain $Omega$ is bounded.
Recall that $W_0^{1,2}$ is defined as the closure of the subspace $C_c^infty(Omega)subset H^1(Omega)$.
Let us assume that $chi_Omegain W_0^{1,2}(Omega)$.
Then there exists a sequence $phi_nin C_c^infty(Omega)$
such that $phi_nto chi_Omega$ in $W_0^{1,2}(Omega)$.
It follows that
$$
|phi_n|_{L^p(Omega)}to |chi_Omega|_{L^p(Omega)}=|Omega|>0
$$
and
$$
|nabla phi_n|_{L^p(Omega)}to |nablachi_Omega|_{L^p(Omega)}=0.
$$
(Note that the function $chi_Omega$ has weak derivatives $0$
because it is equal to the classical derivative and $chi_Omega$ is constant.)
However, by the Poincaré inequality
we have
$$
|phi_n|_L^p(Omega) leq C|nablaphi_n|_{L^p(Omega)}
$$
for some constant $C>0$ (here it is useful that the domain $Omega$ is bounded).
This yields a contradiction to the convergencies that are mentioned above.
answered 18 hours ago
supinf
6,0791028
answered 18 hours ago
supinf
6,0791028
answered 18 hours ago
supinf
6,0791028
answered 18 hours ago
supinf
6,0791028
6,0791028
add a comment |
add a comment |
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$chi_Omega$ is the same as just $1$ as far as $Omega$ is concerned, so its weak derivative is its strong derivative which is $0$. But it won't be $W^{1,p}_0$ because it isn't compactly supported.
– Ian
May 9 '16 at 14:20
1
@Ian: I didn't get why it's not compactly supported. I think $suppchi_{Omega}=overline{left { xinmathbb{R}^n: f(x)neq 0 right }}=overline{Omega}$, which is a compact set. Am I wrong somewhere?
– user1210321
May 9 '16 at 14:28
It means that it is compactly supported as a subset of $Omega$. $chi_Omega$ does not have that property, because its support inside $Omega$ is $Omega$ itself which is not compact.
– Ian
May 9 '16 at 14:29
OK thank you, I got it.
– user1210321
May 9 '16 at 14:31
@Ian it is misleading to say that it is not in $W_0^{1,2}$ because it is not compactly supported. Consider the function $x(x-1)$ which is in $W_0^{1,2}((0,1))$ but is not compactly supported in $(0,1)$.
– supinf
19 hours ago