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?

  • $\begingroup$ Do you know for sure that there is such a bound? $\endgroup$
    – SBK
    Feb 12, 2020 at 9:24
  • $\begingroup$ In my very specific situation, I get such a bound. $\endgroup$
    – Adi
    Feb 17, 2020 at 8:00

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.