Joining Hölder continuous functions on Whitney covering

Question

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?

Answer 1

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.