The following is given: Let $f$ be a $2\pi$ periodic complex valued integrable function sucht that
$$[f]_{\dot{C}^{\alpha}}:=sup_{x,y\in[0,2\pi]}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}<\infty, \ \ \ \ \alpha\in(0,1].$$
I need to show that $$|\hat{f}(n)|\leq\frac{\pi^{\alpha}[f]_{\dot{C}^{\alpha}}}{2|n|^{\alpha}}, \ \ \ \ \forall \ n\neq0.$$
A hint I found is to first show
$$\hat{f}(n)=-\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+\frac{\pi}{n})e^{-inx}dx$$
and then write $\hat{f}(n)=\frac{\hat{f}(n)}{2}+\frac{\hat{f}(n)}{2}$.
But honestly I don't even understand the definition/notation of $[f]_{\dot{C}^{\alpha}}$.
Can anybody help with the first steps?
Assume we know that for a fixed $\alpha > 0$ : $$|f(x)-f(y)| < C_\alpha |x-y|^{\alpha}$$ Then $$|4\pi\hat{f}(n)| = |\int_{-\pi}^\pi (f(x)-f(x+\frac{\pi}n)) e^{-inx}dx| < \int_{-\pi}^\pi C_\alpha |x-(x+\frac{\pi}n)|^{\alpha} dx =2 C_\alpha \pi^{1+\alpha} |n|^{-\alpha}$$