Was wondering about this as I brushed my teeth this morning. I have a differentiable function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ that has bounded and $\gamma$-Holder continuous derivatives. Can I prove that $\exists \; C >0$ such that
\begin{equation*} |f(x+h)-f(x)| \leq C|h|^{1+\gamma} \quad \forall x,h \in \mathbb{R}^n \quad? \end{equation*}
On
As Thomas said, the inequality does not hold as stated. But it can be modified to become correct (and useful). In order to have a super-linear decay of the remainder, you need to subtract the linear part of $f$, not only the constant $f(x)$. That is, $$|f(x+h)-f(x)-f'(x)h|\le C|h|^{1+\gamma}\tag1$$ To prove (1), let $g(h)=f(x+h)-f(x)-f'(x)h$ and observe that $g(0)=g'(0)=0$. Since $g'$ is Hölder continuous, integration along a line segment yields $$|g(h)|\le |h|\int_0^1 |g'(th)|\,dt \le |h|\int_0^1 C|h|^\gamma t^\gamma\,dt = \frac{C|h|^{1+\gamma}}{1+\gamma} $$
Brushing for a little longer, one can prove that a $C^{m,\gamma}$-smooth function satisfies the following: for every $x$ there is a polynomial $P_x$ of degree $m$ such that
$$|f(x+h)-P_x(h)|\le C|h|^{m+\gamma} $$
No, functions fulfilling your estimate are constant, as can be seen by dividing by $|h|$ and letting $h \to 0$.