I'm trying to fill in the details for the following inequality from a paper, but am thoroughly stumped.
Prelude
Let $f \in C_c^{\gamma}(\mathbb{R}^n)$ for some $\gamma \in (0,1)$ (that is, a compactly supported $\gamma$ Holder continuous function).
Define the Banach space $B = \{ X:\mathbb{R}^n \rightarrow \mathbb{R}^n | \; |X|_{1,\gamma} < \infty \}$ where the $|\cdot|_{1,\gamma}$ is a modified Holder norm
\begin{equation*} |X|_{1,\gamma} = | X(\mathbf{0})| + \|\nabla X \|_0 + |\nabla X|_{\gamma} \end{equation*} and $|\cdot|_{\gamma}$ is the usual Holder seminorm given by
\begin{equation*} |W|_{\gamma} = \sup_{x \neq y} \frac{|W(x)-W(y)|}{|x-y|^{\gamma}} \end{equation*}
Furthermore define the open subset $O_L \subset B$ by \begin{equation*} O_L = \left\{X \in B | \quad \inf_{\alpha} \{\det \nabla X(\alpha)\} > \frac{1}{L} \quad, |X|_{1,\gamma} < L \right\} \end{equation*}
I can show that $O_L$ is open in $B$ and that elements of $O_L$ are $C^1$ diffeomorphisms from $\mathbb{R}^n$ to $\mathbb{R}^n$.
So take $X \in O_L$ and $Y \in B$. Define $Z_{\epsilon} = X+ \epsilon Y$. (So by the openness of $O_L$, we have for small $\epsilon$ that $Z_{\epsilon} \in O_L$ and thus a diffeomorphism). We also have $Z_{\epsilon} \rightarrow X$ in $|\cdot|_{1,\gamma}$ as $\epsilon \rightarrow 0$.
Finally, I need to show \begin{equation*} |f(Z_{\epsilon}^{-1}(Z_{\epsilon}(x)-y))- f(X^{-1}(X(x)-y))| < C(\epsilon)|y|^{\gamma} \end{equation*} where $C(\epsilon) \rightarrow 0 $ as $\epsilon \rightarrow 0$.
Any thoughts?
Looks complicated at first, but if you work from the outside, the expression peels nicely. To begin with, the only thing we know about $f$ is its Hölder continuity. Therefore, our goal is really to prove $$ |Z_{\epsilon}^{-1}(Z_{\epsilon}(x)-y)- X^{-1}(X(x)-y)| \le C(\epsilon)|y| \tag1 $$ Since $Z_\epsilon$ is a diffeomorphism, we don't change the content of the left side of (1) by much if we apply it to both terms there. Then simplify:
$$ \begin{align} Z_{\epsilon}(x)-y&- Z_\epsilon(X^{-1}(X(x)-y))\\ &= X(x)+\epsilon Y(x) -y- X(X^{-1}(X(x)-y)) - \epsilon Y(X^{-1}(X(x)-y)) \\ &= X(x)+\epsilon Y(x) -y- X(x) +y - \epsilon Y(X^{-1}(X(x)-y)) \\ &= \epsilon(Y(x)-Y(X^{-1}(X(x)-y))) \end{align}\tag2$$ Since $Y$ is $C^{1}$,
$$ | \epsilon(Y(x)-Y(X^{-1}(X(x)-y))) |\le C\epsilon |x-X^{-1}(X(x)-y))| \tag3 $$ Finally, $X$ is also a diffeomorphism, which means we can compare $x-X^{-1}(X(x)-y))$ to $$ X(x)-X(X^{-1}(X(x)-y))) = X(x)- X(x)+y =y $$ and the fruit is completely peeled.