Let $D\subset\mathbb{R}$. We say that $f:D\to \mathbb{R}$ is Holder of order $\alpha>1$ if for the largest integer $l<\alpha$ (i.e. $\alpha-l\in (0,1)$:
$f$ is $l$ times differentiable, and there exists a $K\in\mathbb{R}$ such that for all $x,y\in D$: $|f^{(l)}(x)-f^{(l)}(y)|<K|x-y|^{\alpha-l}$. Show that if $f$ is Holder of order $\alpha>1$, then $f'$ is Holder of order $\alpha-1$.
By definition $f'$ is $l-1$ times differentiable. Notice that $l-1$ is the biggest integer that is strictly smaller then $\alpha-1$. Meaning that we can write the inequality as:
For $x,y\in D$ we must have that there exist an $M\in\mathbb{R}$ such that:
$|f^{(l-1)}(x)-f^{(l-1)}(y)|<M|x-y|^{(\alpha-1)-(l-1)}(=M|x-y|^{\alpha-l})$.
But I can't see how this inequality holds.
Any help is greatly appreciated!