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
user1210321user1210321
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
user1210321user1210321
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
Jan 4 at 12:12
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
user1210321user1210321
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
user1210321user1210321
485
asked May 9 '16 at 14:10
user1210321user1210321
485
asked May 9 '16 at 14:10
user1210321user1210321
485
asked May 9 '16 at 14:10
user1210321user1210321
485
asked May 9 '16 at 14:10
user1210321user1210321
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
Jan 4 at 12:12
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
Jan 4 at 12:12
$chi_Omega$ 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
Jan 4 at 12:12
@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
Jan 4 at 12:12
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 Jan 4 at 13:36
supinfsupinf
6,0941028
add a comment |
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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 Jan 4 at 13:36
supinfsupinf
6,0941028
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 Jan 4 at 13:36
supinfsupinf
6,0941028
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 Jan 4 at 13:36
supinfsupinf
6,0941028
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 Jan 4 at 13:36
supinfsupinf
6,0941028
answered Jan 4 at 13:36
supinfsupinf
6,0941028
answered Jan 4 at 13:36
supinfsupinf
6,0941028
answered Jan 4 at 13:36
supinfsupinf
6,0941028
6,0941028
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
Jan 4 at 12:12