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










2












$begingroup$


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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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















2












$begingroup$


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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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













2












2








2


0



$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1





+100







$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1





+100







$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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















1





+100







$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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













1





+100







1





+100



1




+100



$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • $begingroup$
    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

















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