Let $\varphi\in C^{\infty}_0(\mathbb{R})$, $0<\alpha<1$, $\Omega\subset\mathbb{R}^d$ be a bounded domain.
Is it true that the map $\Phi:C^{0,\alpha}(\bar{\Omega})\to C^{0,\alpha}(\bar{\Omega})$ defined by \begin{eqnarray} \Phi(u):=\varphi\circ u \end{eqnarray} is continuous? Or are there any props similar to this?
This is true, and here is an approach: $$ \varphi\circ(u+h)-\varphi\circ u = h \int_0^1 \varphi'\circ(u+th)\,dt $$ On the right, the Hölder norm of the integral is controlled by the derivatives of $\varphi$ and the Hölder norm of $u+th$. It remains to use multiplicative property: $[uv]_\alpha\le C[u]_\alpha[v]_\alpha$. The latter holds because $$ |u(x)v(x)-u(y)v(y)|\le |u(x)| |v(x)-v(y)|+|u(x)-u(y)||v(y)| \le 2|x-y|^\alpha [u]_\alpha [v]_\alpha $$ and similarly for the supremum part of the $C^\alpha$ norm: $$ \sup_x|u(x)v(x)|\le \sup_x|u(x)| \sup_x|v(x)| \le [u]_\alpha [v]_\alpha $$