Let $u$ be a bounded function and let a closed set $E$ be given. The compliment of $E$ can be covered with a Whitney type covering $B\_i$ such that the following are satisfied: 


1) $E^c \subset \bigcup 4B\_i$


2) $16B\_i \cap E \neq \emptyset$


3) $\sum \chi\_{4B\_i}(x) \leq C(n)$


4) If $r\_i$ is the radius of $B\_i$, and if $B\_i \cap B\_j \neq \emptyset$, then $r\_i \leq 2 r\_j \leq 4r\_i$.


The rest of the properties of the standard Whitney covering also holds. 


I have the following uniform $C^{\alpha}$ bound: $$|u(x) - u(y)| \leq C |x-y|^{\alpha}$$ for all $x,y \in 8B\_i$ with the constant $C$ and $\alpha$ being independent of the Whitney covering. 


Question: How do I show that my function $u$ is Hölder continuous on $E^c$ with a uniform bound? 



Joining Hölder continuous functions on Whitney covering
**Your function need not be Hölder continuous.** Let $\Omega$ be the union of two exponential cusps with a common vertex and let $E$ be the complement of these cusps. Let $u=1$ in the upper cusp and $u=0$ in the lower cusp. Because the cusps are "sharp", if $B\_i$ is on one cusp, $8B\_i$ will not intersect the oper cusp so the function $u$ will satisfy the condition
$$
|u(x)-u(y)|\leq C|x-y|^\alpha,
\quad
x,y\in 8B\_i
$$
since the left hand side will be equal zero. However, $u$ is not Hölder continuous.


