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. 
 A: 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. 
