Existence of a $C_c^{infty}$ function $Phi$ s.t. $int nabla Phi neq 0.$
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Let $Omegasubseteq mathbb R^n$ be open , bounded and connected set with $mathcal{L^n}(Omega)>0$. Let $Omega=Omega_1cupOmega_2$ where $Omega_1capOmega_2=emptyset$ with $mathcal{L^n}(Omega_1)>0$ and $mathcal{L^n}(Omega_2)>0.$ Prove that $exists$ $Phiin C_c^{infty}(Omega)$ such that $displaystyle int_{Omega_1}nablaPhi(x)dxneq 0.$
Here $Phi$ is real valued function and $C_c^{infty}(Omega)$ consists of all $C^{infty}(Omega)$ functions which have compact support in $Omega.$
I tried to show by contradiction and wished to have $mathcal{L^n}(Omega_1)=0 $ but couldn't be able to do that.
Any help is appreciated. Thank you.
lebesgue-measure smooth-functions
edited Nov 21 at 5:00
Kemono Chen
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
add a comment |
up vote
0
down vote
favorite
Let $Omegasubseteq mathbb R^n$ be open , bounded and connected set with $mathcal{L^n}(Omega)>0$. Let $Omega=Omega_1cupOmega_2$ where $Omega_1capOmega_2=emptyset$ with $mathcal{L^n}(Omega_1)>0$ and $mathcal{L^n}(Omega_2)>0.$ Prove that $exists$ $Phiin C_c^{infty}(Omega)$ such that $displaystyle int_{Omega_1}nablaPhi(x)dxneq 0.$
Here $Phi$ is real valued function and $C_c^{infty}(Omega)$ consists of all $C^{infty}(Omega)$ functions which have compact support in $Omega.$
I tried to show by contradiction and wished to have $mathcal{L^n}(Omega_1)=0 $ but couldn't be able to do that.
Any help is appreciated. Thank you.
lebesgue-measure smooth-functions
edited Nov 21 at 5:00
Kemono Chen
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Omegasubseteq mathbb R^n$ be open , bounded and connected set with $mathcal{L^n}(Omega)>0$. Let $Omega=Omega_1cupOmega_2$ where $Omega_1capOmega_2=emptyset$ with $mathcal{L^n}(Omega_1)>0$ and $mathcal{L^n}(Omega_2)>0.$ Prove that $exists$ $Phiin C_c^{infty}(Omega)$ such that $displaystyle int_{Omega_1}nablaPhi(x)dxneq 0.$
Here $Phi$ is real valued function and $C_c^{infty}(Omega)$ consists of all $C^{infty}(Omega)$ functions which have compact support in $Omega.$
I tried to show by contradiction and wished to have $mathcal{L^n}(Omega_1)=0 $ but couldn't be able to do that.
Any help is appreciated. Thank you.
lebesgue-measure smooth-functions
edited Nov 21 at 5:00
Kemono Chen
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
Let $Omegasubseteq mathbb R^n$ be open , bounded and connected set with $mathcal{L^n}(Omega)>0$. Let $Omega=Omega_1cupOmega_2$ where $Omega_1capOmega_2=emptyset$ with $mathcal{L^n}(Omega_1)>0$ and $mathcal{L^n}(Omega_2)>0.$ Prove that $exists$ $Phiin C_c^{infty}(Omega)$ such that $displaystyle int_{Omega_1}nablaPhi(x)dxneq 0.$
Here $Phi$ is real valued function and $C_c^{infty}(Omega)$ consists of all $C^{infty}(Omega)$ functions which have compact support in $Omega.$
I tried to show by contradiction and wished to have $mathcal{L^n}(Omega_1)=0 $ but couldn't be able to do that.
Any help is appreciated. Thank you.
lebesgue-measure smooth-functions
lebesgue-measure smooth-functions
edited Nov 21 at 5:00
Kemono Chen
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
edited Nov 21 at 5:00
Kemono Chen
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
edited Nov 21 at 5:00
Kemono Chen
1,618330
edited Nov 21 at 5:00
Kemono Chen
1,618330
edited Nov 21 at 5:00
Kemono Chen
1,618330
1,618330
asked Nov 21 at 4:02
nurun nesha
9692623
asked Nov 21 at 4:02
nurun nesha
9692623
asked Nov 21 at 4:02
nurun nesha
9692623
9692623
add a comment |
add a comment |
1 Answer
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1
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Hint: Assuming $Omega_1$ and $Omega_2$ are measurable, then there exists compact set $Fsubset Omega_1$ such that $mathcal{L}^n(F)+varepsilon=mathcal{L}^n(Omega_1)$. Likewise, there exists open set $Gsupset Omega_1$ such that $mathcal{L}^n(G)-varepsilon=mathcal{L}^n(Omega_1)$. Hence $mathcal{L}^n(GcapOmega_2)$ is small.
Then by smooth Urysohn's lemma there exists $Phi$ smooth such that $Phi equiv 1$ on $F$ and $Phiequiv 0$ on $G^c$.
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Assuming $Omega_1$ and $Omega_2$ are measurable, then there exists compact set $Fsubset Omega_1$ such that $mathcal{L}^n(F)+varepsilon=mathcal{L}^n(Omega_1)$. Likewise, there exists open set $Gsupset Omega_1$ such that $mathcal{L}^n(G)-varepsilon=mathcal{L}^n(Omega_1)$. Hence $mathcal{L}^n(GcapOmega_2)$ is small.
Then by smooth Urysohn's lemma there exists $Phi$ smooth such that $Phi equiv 1$ on $F$ and $Phiequiv 0$ on $G^c$.
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
add a comment |
up vote
1
down vote
accepted
Hint: Assuming $Omega_1$ and $Omega_2$ are measurable, then there exists compact set $Fsubset Omega_1$ such that $mathcal{L}^n(F)+varepsilon=mathcal{L}^n(Omega_1)$. Likewise, there exists open set $Gsupset Omega_1$ such that $mathcal{L}^n(G)-varepsilon=mathcal{L}^n(Omega_1)$. Hence $mathcal{L}^n(GcapOmega_2)$ is small.
Then by smooth Urysohn's lemma there exists $Phi$ smooth such that $Phi equiv 1$ on $F$ and $Phiequiv 0$ on $G^c$.
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Assuming $Omega_1$ and $Omega_2$ are measurable, then there exists compact set $Fsubset Omega_1$ such that $mathcal{L}^n(F)+varepsilon=mathcal{L}^n(Omega_1)$. Likewise, there exists open set $Gsupset Omega_1$ such that $mathcal{L}^n(G)-varepsilon=mathcal{L}^n(Omega_1)$. Hence $mathcal{L}^n(GcapOmega_2)$ is small.
Then by smooth Urysohn's lemma there exists $Phi$ smooth such that $Phi equiv 1$ on $F$ and $Phiequiv 0$ on $G^c$.
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
Hint: Assuming $Omega_1$ and $Omega_2$ are measurable, then there exists compact set $Fsubset Omega_1$ such that $mathcal{L}^n(F)+varepsilon=mathcal{L}^n(Omega_1)$. Likewise, there exists open set $Gsupset Omega_1$ such that $mathcal{L}^n(G)-varepsilon=mathcal{L}^n(Omega_1)$. Hence $mathcal{L}^n(GcapOmega_2)$ is small.
Then by smooth Urysohn's lemma there exists $Phi$ smooth such that $Phi equiv 1$ on $F$ and $Phiequiv 0$ on $G^c$.
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
edited Nov 21 at 5:46
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
answered Nov 21 at 5:20
Jacky Chong
17.3k21027
17.3k21027
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
add a comment |
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
If we take $Omega=(0,2)=(0,1]cup(1,2)=Omega_1 cup Omega_2$ then $G^c$ will not contain $Omega_2$
– nurun nesha
Nov 21 at 5:43
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
@nurunnesha Yes. I made the correction.
– Jacky Chong
Nov 21 at 5:46
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
But we need $int_{Omega_1} nabla Phi neq 0.$ So will we have to construct this $Phi$ on $Omega_1setminus F$ in such a way that $nabla Phi$ does not vanish there?
– nurun nesha
Nov 21 at 6:01
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
@nurunnesha Well. $Phi$ needs to transition from $1$ to $0$ so $nablaPhi$ will not be small. You could also make $F$ smaller.
– Jacky Chong
Nov 21 at 6:04
add a comment |
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