Let $\Omega$ be a bounded open connected set in $\mathbb{R}^n$ 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 $$||Du||_{L^\infty(\Omega)}\leq \sigma^\alpha [|Du|]_{\alpha,\Omega}+\frac{C}{\sigma}||u||_{L^\infty(\Omega)}$$ and $$[u]_{\alpha,\Omega}\leq \sigma[|Du|]_{\alpha,\Omega}+\frac{C}{\sigma^\alpha}||u||_{L^\infty(\Omega)}$$ hold for all $0<\sigma<\sigma_0$ and for all $u\in C^{1,\alpha}(\bar\Omega)$. Here $||u||_{C^{1,\alpha}}=||u||_{L^\infty(\Omega)}+||Du||_{L^\infty(\Omega)}+[|Du|]_{\alpha}$ and $[u]_\alpha=\sup_{x\neq y}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$.

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.

  • $\begingroup$ 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

Since $\Omega$ is bounded, its closure its compact. Also since $u\in C^{1,\alpha}(\overline{\Omega})$, you have that $Du$ is continuous. Hence there exists $x_{0}\in\overline{\Omega}$ such that $$ |Du(x_{0})|=\max_{x\in\overline{\Omega}}|Du(x)|. $$ In particular, $u$ is differentiable $x_{0}$. Thus, $$ \frac{\partial u}{\partial\nu}(x_{0})=Du(x_{0})\cdot\nu, $$ where $\frac{\partial u}{\partial\nu}$ is a directional derivative in an admissible direction $\nu\in\mathbb{R}^{n}$, with $|\nu|=1$. Now since $\partial\Omega$ is of class $C^{1}$, there is a cone $C$ such that every point $x\in\overline{\Omega}$ 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 $x_{0}+t\nu_{i}$, $t\in\lbrack0,h]$ are contained in the cone $C_{x_{0}}$, where $h>0$. If you consider the system of $n$ equations $$ \frac{\partial u}{\partial\nu_{i}}(x_{0})=Du(x_{0})\cdot\nu_{i}, $$ in the $n$ unknowns $\frac{\partial u}{\partial x_{i}}(x_{0})$, you have that the determinant is different from zero since the vectors are linearly independent. Hence, you can write $$ \frac{\partial u}{\partial x_{i}}(x_{0})=\sum_{j=1}^{n}c_{i,j}\frac{\partial u}{\partial\nu_{j}}(x_{0}), $$ where the numbers $c_{i,j}$ depent only on the directions $\nu_{1},\ldots ,\nu_{n}$.

Along each segment $S_{i}=\{x_{0}+t\nu_{i}$, $t\in\lbrack0,h]\}$ you can apply your inequality for $n=1$ to the function $g_{i}(t):=u(x_{0}+t\nu_{i})$, $t\in\lbrack0,h]$.

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 $R\nu_{1},\ldots,R\nu_{n}$, where $R$ is your rotation, so the determinant should not change.

  • $\begingroup$ 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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