# Joining Hölder continuous functions on Whitney covering

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?

• Do you know for sure that there is such a bound?
– SBK
Feb 12, 2020 at 9:24
• In my very specific situation, I get such a bound.
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.