# Interpolation inequality for Holder continuous functions.

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.

• It might help if you showed how you dealt with $C^2$ boundaries. – robjohn Mar 28 at 15:20
• $C^2$ boundary have the interior ball property and i have the results for balls. – 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.
• 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? – mudok Mar 30 at 15:56
• 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. – Gio67 Mar 31 at 12:40
• $C^1$ boundary imply uniform cone property or only cone property? – mudok Mar 31 at 13:04