Does positive part preserve Holder continuity?

Question

Let $u^+$ denote the positive part of the function $u$ on a bounded domain $\Omega.$

If $u \in C^{0,\alpha}(\bar \Omega)$, is also $u^+ \in C^{0,\alpha}(\bar \Omega)$?

Answer 1

Hint: $|u^+(x) - u^+(y)| \le |u(x) - u(y)|$.

Answer 2

Yes. Take $x,y \in \Omega$. If $u(x),u(y) > 0$ then $u^+(x) = u(x)$ and $u^+(y) = u(y)$ so that there is nothing to prove. If $u(x),u(y) \leq 0$, then $u^+(x) = u^+(y) = 0$, and again we are done. Assume hence that $u(x) < 0 < u(y)$. But then $$ \frac{|u^+(y) - u^+(x)|}{|x-y|^\alpha} = \frac{|u^+(y)|}{|x-y|^\alpha} = \frac{|u(y)|}{|x-y|^\alpha} \leq \frac{|u(y) - u(x)|}{|x-y|^\alpha}$$ as wanted.