Let $\mathscr C^\alpha$ be a set of functions that satisfy $\alpha$- Holder condition, i.e., $$\mathscr C^\alpha:=\{f : \mathbb R\to \mathbb C \mid \exists M>0\ ;\ |f(x)-f(y)|\leqq M |x-y|^\alpha \ \mathrm{for \ all} \ x,y \in \mathbb R \}$$ for $\alpha >0.$
I want to consider whether [$f : \mathbb R \to \mathbb C$ is periodic and $C^1 \Rightarrow f\in \mathscr C^\alpha$ for all $\alpha >0$] holds or not.
For $0<\alpha \leqq 1.$
Since $f$ is periodic and continuous, $f$ is bounded and Lipschitz continuous.
So, there exists $L, C>0$ s.t. $|f(x)|\leqq C$ for all $x \in \mathbb R$ and $|f(x)-f(y)|\leqq L|x-y|$ for all $x,y \in \mathbb R$.
Let $M=\max\{2C, L\}$.
Fix $x,y\in \mathbb R$.
If $|x-y|\geqq 1$, then $|f(x)-f(y)|\leqq |f(x)|+|f(y)|\leqq 2C \leqq 2C|x-y|^\alpha \leqq M|x-y|^\alpha.$
If $|x-y|<1$, then $|f(x)-f(y)|\leqq L|x-y|\leqq L|x-y|^\alpha\leqq M|x-y|^\alpha.$
Thus $f \in \mathscr C^\alpha.$
But what about $\alpha >1$ ?
For $|x-y|\geqq 1,$ the same way can be applied, but for $|x-y|<1$, I couldn't show $|f(x)-f(y)|\leqq M|x-y|^\alpha.$
What can I do for $\alpha >1 $ ? Or, isn't the statement true for $\alpha>1$ ?