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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.

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I think i got the case where $c=1$ covered: For some number $u$ we have $x=y+u$, where $\epsilon \leq |u| \leq max(|a-y|,|b-y|)$. We then have \begin{align*} f(x,y,|x−y|^c)−&0.5f(x,y−ϵ,|x−y+ϵ|^c)−0.5f(x,y+ϵ,|x−y−ϵ|^c)\\ &= f(y+u,y,|u|^c)−0.5f(y+u,y−ϵ,|u+ϵ|^c)−0.5f(y+u,y+ϵ,|u−ϵ|^c)\\ \end{align*} and it follows from Taylor-Expansions that: \begin{align*} f(y+u,y,|u|^c)&−0.5f(y+u,y−ϵ,|u+ϵ|^c)−0.5f(y+u,y+ϵ,|u−ϵ|^c)\\ &= f(y+u,y,|u|^c)\\ &-0.5(f(y+u,y,|u|^c)-\epsilon f_2(y+u,y,|u|^c)+(|u+\epsilon|^c-|u|^c)f_3(y+u,y,|u|^c) + R_1)\\ &-0.5(f(y+u,y,|u|^c)+\epsilon f_2(y+u,y,|u|^c) +(|u-\epsilon|^c-|u|^c)f_3(y+u,y,|u|^c) + R_2)\\ &= -0.5f_3(y+u,y,|u|^c)(|u+\epsilon|^c-|u|^c +|u-\epsilon|^c-|u|^c) - 0.5(R_1 + R_2) \end{align*} Now, for the case $c=1$ we have for $|u|\geq \epsilon$: $|u+\epsilon|^c-|u|^c +|u-\epsilon|^c-|u|^c= |u+\epsilon|-|u| +|u-\epsilon|-|u|=0$ and hence $$|-0.5f_3(y+u,y,|u|^c)(|u+\epsilon|^c-|u|^c +|u-\epsilon|^c-|u|^c) - 0.5(R_1 + R_2)| = 0.5 |(R_1+R_2)| \leq 0.5 (|R_1| +|R_2|). $$ But for some $\tilde{M}<\infty$ we have by the triangle inequality $$|R_1| \leq \tilde{M}(|\epsilon|^2 + ||u+\epsilon|-|u||^2) \leq \tilde{M}2 \epsilon^2,$$ while $$ |R_2| \leq \tilde{M}(|\epsilon|^2 + ||u-\epsilon|-|u||^2) \leq \tilde{M}2 \epsilon^2$$

and hence there exists a constant $M<\infty$ such that $$0.5 (|R_1| +|R_2|) \leq M\epsilon^2$$ which proves the statement for the case $c=1$. However I am still having a hard time showing the assertion for the case $c\neq 1$ and any help is greatly appreciated.


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