Interpolation inequality for Holder continuous functions. The 2019 Stack Overflow Developer Survey Results Are InProving an operator is compactfunctions in Holder spaceApproximate Holder continuous functions by smooth functions$f$ is a real function and it is $alpha$-Holder continuous with $alpha>1$. Is $f$ constant?Inclusion of Holder SpacesHolder norms inequalityInterpolation inequality involving Holder seminorms and Lebesgue normsHolder continuity EquivalenceDimension of Holder spaceClosure of Continuously DifferentiableFunctions in Holder Space
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Interpolation inequality for Holder continuous functions.
The 2019 Stack Overflow Developer Survey Results Are InProving an operator is compactfunctions in Holder spaceApproximate Holder continuous functions by smooth functions$f$ is a real function and it is $alpha$-Holder continuous with $alpha>1$. Is $f$ constant?Inclusion of Holder SpacesHolder norms inequalityInterpolation inequality involving Holder seminorms and Lebesgue normsHolder continuity EquivalenceDimension of Holder spaceClosure of Continuously DifferentiableFunctions in Holder Space
$begingroup$
Let $Omega$ be a bounded open connected set in $mathbbR^n$ with $C^1$ boundary and let $0<alpha<1$. Then there exists a real number $sigma_0>0$ and a dimensional constant $C>0$ such that $$||Du||_L^infty(Omega)leq sigma^alpha [|Du|]_alpha,Omega+fracCsigma||u||_L^infty(Omega)$$ and $$[u]_alpha,Omegaleq sigma[|Du|]_alpha,Omega+fracCsigma^alpha||u||_L^infty(Omega)$$ hold for all $0<sigma<sigma_0$ and for all $uin C^1,alpha(barOmega)$. Here $||u||_C^1,alpha=||u||_L^infty(Omega)+||Du||_L^infty(Omega)+[|Du|]_alpha$ and $[u]_alpha=sup_xneq yfracx-y$.
N.B. I have proved the above results for balls and then for domain with $C^2$ boundary. I cant proceed for $C^1$ boundary domain. Any help will be great.
real-analysis pde holder-spaces interpolation-theory
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a bounded open connected set in $mathbbR^n$ with $C^1$ boundary and let $0<alpha<1$. Then there exists a real number $sigma_0>0$ and a dimensional constant $C>0$ such that $$||Du||_L^infty(Omega)leq sigma^alpha [|Du|]_alpha,Omega+fracCsigma||u||_L^infty(Omega)$$ and $$[u]_alpha,Omegaleq sigma[|Du|]_alpha,Omega+fracCsigma^alpha||u||_L^infty(Omega)$$ hold for all $0<sigma<sigma_0$ and for all $uin C^1,alpha(barOmega)$. Here $||u||_C^1,alpha=||u||_L^infty(Omega)+||Du||_L^infty(Omega)+[|Du|]_alpha$ and $[u]_alpha=sup_xneq yfracx-y$.
N.B. I have proved the above results for balls and then for domain with $C^2$ boundary. I cant proceed for $C^1$ boundary domain. Any help will be great.
real-analysis pde holder-spaces interpolation-theory
$endgroup$
$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11
add a comment |
$begingroup$
Let $Omega$ be a bounded open connected set in $mathbbR^n$ with $C^1$ boundary and let $0<alpha<1$. Then there exists a real number $sigma_0>0$ and a dimensional constant $C>0$ such that $$||Du||_L^infty(Omega)leq sigma^alpha [|Du|]_alpha,Omega+fracCsigma||u||_L^infty(Omega)$$ and $$[u]_alpha,Omegaleq sigma[|Du|]_alpha,Omega+fracCsigma^alpha||u||_L^infty(Omega)$$ hold for all $0<sigma<sigma_0$ and for all $uin C^1,alpha(barOmega)$. Here $||u||_C^1,alpha=||u||_L^infty(Omega)+||Du||_L^infty(Omega)+[|Du|]_alpha$ and $[u]_alpha=sup_xneq yfracx-y$.
N.B. I have proved the above results for balls and then for domain with $C^2$ boundary. I cant proceed for $C^1$ boundary domain. Any help will be great.
real-analysis pde holder-spaces interpolation-theory
$endgroup$
Let $Omega$ be a bounded open connected set in $mathbbR^n$ with $C^1$ boundary and let $0<alpha<1$. Then there exists a real number $sigma_0>0$ and a dimensional constant $C>0$ such that $$||Du||_L^infty(Omega)leq sigma^alpha [|Du|]_alpha,Omega+fracCsigma||u||_L^infty(Omega)$$ and $$[u]_alpha,Omegaleq sigma[|Du|]_alpha,Omega+fracCsigma^alpha||u||_L^infty(Omega)$$ hold for all $0<sigma<sigma_0$ and for all $uin C^1,alpha(barOmega)$. Here $||u||_C^1,alpha=||u||_L^infty(Omega)+||Du||_L^infty(Omega)+[|Du|]_alpha$ and $[u]_alpha=sup_xneq yfracx-y$.
N.B. I have proved the above results for balls and then for domain with $C^2$ boundary. I cant proceed for $C^1$ boundary domain. Any help will be great.
real-analysis pde holder-spaces interpolation-theory
real-analysis pde holder-spaces interpolation-theory
asked Mar 24 at 5:43
mudokmudok
374315
374315
$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11
add a comment |
$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11
$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $Omega$ is bounded, its closure its compact. Also since $uin
C^1,alpha(overlineOmega)$, you have that $Du$ is continuous. Hence
there exists $x_0inoverlineOmega$ such that
$$
|Du(x_0)|=max_xinoverlineOmega|Du(x)|.
$$
In particular, $u$ is differentiable $x_0$. Thus,
$$
fracpartial upartialnu(x_0)=Du(x_0)cdotnu,
$$
where $fracpartial upartialnu$ is a directional derivative in an
admissible direction $nuinmathbbR^n$, with $|nu|=1$. Now since
$partialOmega$ is of class $C^1$, there is a cone $C$ such that every
point $xinoverlineOmega$ is the vertex of a cone $C_x$ congruent to $C$
and contained in $Omega$. Hence, if you consider the cone $C_x_0$, you
can find $n$ linearly independent directions $nu_1,ldots,nu_n$ such
that the segments $x_0+tnu_i$, $tinlbrack0,h]$ are contained in the
cone $C_x_0$, where $h>0$. If you consider the system of $n$ equations
$$
fracpartial upartialnu_i(x_0)=Du(x_0)cdotnu_i,
$$
in the $n$ unknowns $fracpartial upartial x_i(x_0)$, you have that
the determinant is different from zero since the vectors are linearly
independent. Hence, you can write
$$
fracpartial upartial x_i(x_0)=sum_j=1^nc_i,jfracpartial
upartialnu_j(x_0),
$$
where the numbers $c_i,j$ depent only on the directions $nu_1,ldots
,nu_n$.
Along each segment $S_i=x_0+tnu_i$, $tinlbrack0,h]$ you can apply
your inequality for $n=1$ to the function $g_i(t):=u(x_0+tnu_i)$,
$tinlbrack0,h]$.
Now you have to prove that the coefficients $c_i,j$ depend only on $Omega$.
I have to think about this, but if you rotate the cone, your new directions
are $Rnu_1,ldots,Rnu_n$, where $R$ is your rotation, so the determinant
should not change.
$endgroup$
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
Since $Omega$ is bounded, its closure its compact. Also since $uin
C^1,alpha(overlineOmega)$, you have that $Du$ is continuous. Hence
there exists $x_0inoverlineOmega$ such that
$$
|Du(x_0)|=max_xinoverlineOmega|Du(x)|.
$$
In particular, $u$ is differentiable $x_0$. Thus,
$$
fracpartial upartialnu(x_0)=Du(x_0)cdotnu,
$$
where $fracpartial upartialnu$ is a directional derivative in an
admissible direction $nuinmathbbR^n$, with $|nu|=1$. Now since
$partialOmega$ is of class $C^1$, there is a cone $C$ such that every
point $xinoverlineOmega$ is the vertex of a cone $C_x$ congruent to $C$
and contained in $Omega$. Hence, if you consider the cone $C_x_0$, you
can find $n$ linearly independent directions $nu_1,ldots,nu_n$ such
that the segments $x_0+tnu_i$, $tinlbrack0,h]$ are contained in the
cone $C_x_0$, where $h>0$. If you consider the system of $n$ equations
$$
fracpartial upartialnu_i(x_0)=Du(x_0)cdotnu_i,
$$
in the $n$ unknowns $fracpartial upartial x_i(x_0)$, you have that
the determinant is different from zero since the vectors are linearly
independent. Hence, you can write
$$
fracpartial upartial x_i(x_0)=sum_j=1^nc_i,jfracpartial
upartialnu_j(x_0),
$$
where the numbers $c_i,j$ depent only on the directions $nu_1,ldots
,nu_n$.
Along each segment $S_i=x_0+tnu_i$, $tinlbrack0,h]$ you can apply
your inequality for $n=1$ to the function $g_i(t):=u(x_0+tnu_i)$,
$tinlbrack0,h]$.
Now you have to prove that the coefficients $c_i,j$ depend only on $Omega$.
I have to think about this, but if you rotate the cone, your new directions
are $Rnu_1,ldots,Rnu_n$, where $R$ is your rotation, so the determinant
should not change.
$endgroup$
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
|
show 1 more comment
$begingroup$
Since $Omega$ is bounded, its closure its compact. Also since $uin
C^1,alpha(overlineOmega)$, you have that $Du$ is continuous. Hence
there exists $x_0inoverlineOmega$ such that
$$
|Du(x_0)|=max_xinoverlineOmega|Du(x)|.
$$
In particular, $u$ is differentiable $x_0$. Thus,
$$
fracpartial upartialnu(x_0)=Du(x_0)cdotnu,
$$
where $fracpartial upartialnu$ is a directional derivative in an
admissible direction $nuinmathbbR^n$, with $|nu|=1$. Now since
$partialOmega$ is of class $C^1$, there is a cone $C$ such that every
point $xinoverlineOmega$ is the vertex of a cone $C_x$ congruent to $C$
and contained in $Omega$. Hence, if you consider the cone $C_x_0$, you
can find $n$ linearly independent directions $nu_1,ldots,nu_n$ such
that the segments $x_0+tnu_i$, $tinlbrack0,h]$ are contained in the
cone $C_x_0$, where $h>0$. If you consider the system of $n$ equations
$$
fracpartial upartialnu_i(x_0)=Du(x_0)cdotnu_i,
$$
in the $n$ unknowns $fracpartial upartial x_i(x_0)$, you have that
the determinant is different from zero since the vectors are linearly
independent. Hence, you can write
$$
fracpartial upartial x_i(x_0)=sum_j=1^nc_i,jfracpartial
upartialnu_j(x_0),
$$
where the numbers $c_i,j$ depent only on the directions $nu_1,ldots
,nu_n$.
Along each segment $S_i=x_0+tnu_i$, $tinlbrack0,h]$ you can apply
your inequality for $n=1$ to the function $g_i(t):=u(x_0+tnu_i)$,
$tinlbrack0,h]$.
Now you have to prove that the coefficients $c_i,j$ depend only on $Omega$.
I have to think about this, but if you rotate the cone, your new directions
are $Rnu_1,ldots,Rnu_n$, where $R$ is your rotation, so the determinant
should not change.
$endgroup$
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
|
show 1 more comment
$begingroup$
Since $Omega$ is bounded, its closure its compact. Also since $uin
C^1,alpha(overlineOmega)$, you have that $Du$ is continuous. Hence
there exists $x_0inoverlineOmega$ such that
$$
|Du(x_0)|=max_xinoverlineOmega|Du(x)|.
$$
In particular, $u$ is differentiable $x_0$. Thus,
$$
fracpartial upartialnu(x_0)=Du(x_0)cdotnu,
$$
where $fracpartial upartialnu$ is a directional derivative in an
admissible direction $nuinmathbbR^n$, with $|nu|=1$. Now since
$partialOmega$ is of class $C^1$, there is a cone $C$ such that every
point $xinoverlineOmega$ is the vertex of a cone $C_x$ congruent to $C$
and contained in $Omega$. Hence, if you consider the cone $C_x_0$, you
can find $n$ linearly independent directions $nu_1,ldots,nu_n$ such
that the segments $x_0+tnu_i$, $tinlbrack0,h]$ are contained in the
cone $C_x_0$, where $h>0$. If you consider the system of $n$ equations
$$
fracpartial upartialnu_i(x_0)=Du(x_0)cdotnu_i,
$$
in the $n$ unknowns $fracpartial upartial x_i(x_0)$, you have that
the determinant is different from zero since the vectors are linearly
independent. Hence, you can write
$$
fracpartial upartial x_i(x_0)=sum_j=1^nc_i,jfracpartial
upartialnu_j(x_0),
$$
where the numbers $c_i,j$ depent only on the directions $nu_1,ldots
,nu_n$.
Along each segment $S_i=x_0+tnu_i$, $tinlbrack0,h]$ you can apply
your inequality for $n=1$ to the function $g_i(t):=u(x_0+tnu_i)$,
$tinlbrack0,h]$.
Now you have to prove that the coefficients $c_i,j$ depend only on $Omega$.
I have to think about this, but if you rotate the cone, your new directions
are $Rnu_1,ldots,Rnu_n$, where $R$ is your rotation, so the determinant
should not change.
$endgroup$
Since $Omega$ is bounded, its closure its compact. Also since $uin
C^1,alpha(overlineOmega)$, you have that $Du$ is continuous. Hence
there exists $x_0inoverlineOmega$ such that
$$
|Du(x_0)|=max_xinoverlineOmega|Du(x)|.
$$
In particular, $u$ is differentiable $x_0$. Thus,
$$
fracpartial upartialnu(x_0)=Du(x_0)cdotnu,
$$
where $fracpartial upartialnu$ is a directional derivative in an
admissible direction $nuinmathbbR^n$, with $|nu|=1$. Now since
$partialOmega$ is of class $C^1$, there is a cone $C$ such that every
point $xinoverlineOmega$ is the vertex of a cone $C_x$ congruent to $C$
and contained in $Omega$. Hence, if you consider the cone $C_x_0$, you
can find $n$ linearly independent directions $nu_1,ldots,nu_n$ such
that the segments $x_0+tnu_i$, $tinlbrack0,h]$ are contained in the
cone $C_x_0$, where $h>0$. If you consider the system of $n$ equations
$$
fracpartial upartialnu_i(x_0)=Du(x_0)cdotnu_i,
$$
in the $n$ unknowns $fracpartial upartial x_i(x_0)$, you have that
the determinant is different from zero since the vectors are linearly
independent. Hence, you can write
$$
fracpartial upartial x_i(x_0)=sum_j=1^nc_i,jfracpartial
upartialnu_j(x_0),
$$
where the numbers $c_i,j$ depent only on the directions $nu_1,ldots
,nu_n$.
Along each segment $S_i=x_0+tnu_i$, $tinlbrack0,h]$ you can apply
your inequality for $n=1$ to the function $g_i(t):=u(x_0+tnu_i)$,
$tinlbrack0,h]$.
Now you have to prove that the coefficients $c_i,j$ depend only on $Omega$.
I have to think about this, but if you rotate the cone, your new directions
are $Rnu_1,ldots,Rnu_n$, where $R$ is your rotation, so the determinant
should not change.
answered Mar 30 at 15:30
Gio67Gio67
12.8k1627
12.8k1627
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
|
show 1 more comment
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
Very nice argument. What you essentially use is the interior cone property for $C^1$ domain. Which escaped my mind. I also think your $c_i,j$ dependence only on domain since it's come from the linear independent direction vectors depending on domain. Am I correct?
$endgroup$
– mudok
Mar 30 at 15:56
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
yes, you are correct
$endgroup$
– Gio67
Mar 30 at 16:08
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
one question : you say " a cone Cx congruent to C and contained in Ω" why congruent?
$endgroup$
– mudok
Mar 30 at 17:11
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
up to a translation and a rotation. And I should have said contained in the closure of $Omega$. You will need to rotate the cone.
$endgroup$
– Gio67
Mar 31 at 12:40
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
$begingroup$
$C^1$ boundary imply uniform cone property or only cone property?
$endgroup$
– mudok
Mar 31 at 13:04
|
show 1 more comment
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$begingroup$
It might help if you showed how you dealt with $C^2$ boundaries.
$endgroup$
– robjohn♦
Mar 28 at 15:20
$begingroup$
$C^2$ boundary have the interior ball property and i have the results for balls.
$endgroup$
– mudok
Mar 28 at 18:11