I have a function $f(x,y,z)$ with bounded first and second order partial derivatives on the compact set $[a,b]\times [a,b]\times[0,b-a]$ (where $a<b$), and $|\frac{\partial}{\partial z}f(x,x,z)_{z=0}| > 0$ for all $x$
Let $0<c<2$. I want to show that for any sufficiently small $\epsilon>0$ and for all $x\in [a,b]$ with $|x-y|\geq \epsilon$, I have $$ |f(x,y,|x-y|^c) - 0.5 f(x,y-\epsilon,|x-y+\epsilon|^c) - 0.5f(x,y+\epsilon,|x-y-\epsilon|^c)| \leq \begin{cases} M \frac{\epsilon^2}{|x-y|^{2-c}}, \text{ if } c \neq 1\\ M\epsilon^2, \text{ if } c=1 \end{cases} $$ for some constant $0<M<\infty$.
I think the assertion follows from a Taylor-expansion, but I am not able to show it.
Edit: The case $c=1$ is covered, see below. I am still having a lot of trouble showing the assertion for $c\neq 1$.
Edit 2: looking at my own answer, for $c\neq 1$, the problem narrows down to proof that for $|u|_{max}\geq |u|\geq \epsilon >0$ and $0<c<2$ we have:
$$||u+\epsilon|^c + |u-\epsilon|^c - 2|u|^c| \leq M_1 \frac{\epsilon^2}{|u|^{2-c}}.$$ for some constant $0<M_1<\infty$. Since the proof to this assertion seems to be more straightforward (and simpler) than the one of my original problem, I will prob. create another topic for this question.