I would like to know whether for a bounded Lipschitz domain $\Omega \subset \mathbb{R}^n$ (in the weak Lipschitz, so a "Lipschitz manifold", sense, not necessarily a Lipschitz graph domain), there holds $$\bigl(C^\alpha(\overline\Omega),C^{1,\gamma}(\overline\Omega)\bigr)_{\theta,\infty} \hookrightarrow C^{0,1}(\overline\Omega) \quad \text{for some}~\theta < 1,$$
where $0 < \gamma \leq \alpha < 1$ for the case I am interested in.
- $C^\alpha(\overline\Omega)$ is the usual space of $\alpha$-Hölder continuous functions on $\Omega$,
- $C^{1,\gamma}(\overline\Omega)$ consists of continuously differentiable functions on $\Omega$ whose derivatives are bounded and $\gamma$-Hölder continuous.
For both spaces we identify the respective functions with their unique extension to the closure of $\Omega$.
It would be sufficient to have a continuous linear extension operator from $C^\alpha(\overline\Omega)$ to $\mathcal{C}^\alpha(\mathbb{R}^n)$ whose restriction to $C^{1,\gamma}(\overline\Omega)$ is continuous to $\mathcal{C}^{1+\gamma}(\mathbb{R}^n)$, because on Euclidean space the assertion would be true.
In fact, exactly such a property is claimed for the Whitney extension operator in Hölder Classes with Boundary Conditions as Interpolation Spaces by Acquistapace and Terreni (here), Proposition 1.4, however, without a precise proof, referring to the Singular Integrals book of Elias Stein, Chapter IV§2. I fail to understand how the restriction of the extension operator for the $C^\alpha$ scale to $C^{1,\gamma}$, resulting in a linear continuous operator, would work.
(The higher order Whitney operator is also defined for the $(1+\gamma)$-jet space which includes $C^{1+\gamma}(\overline\Omega)$ since a Lipschitz domain is quasiconvex, if I understand it right from this related question and the references there. A precise proof has eluded me so far, though.)
Any help or comments would be greatly appreciated.
