Big O proof of Fourier Coefficient Let $f(x)$ be a $2\pi$ periodic function on R. Assume that Hölder continuous:
$$\sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{-\alpha}} \leq C$$
for some constants $C$ and $\alpha \in \,]0,1]$.  Prove that
$$\hat{f}(n)= O(|n|^{-\alpha}).$$
I have been trying to work through using the corollary that states that if $f \in C^2$ on the circle then
$$\left|\hat{f} (n)\right| = O\left(\frac{1}{n^2}\right),$$
but I'm not reaching anywhere. Any help would be appreciated. Also if there is some general way of reaching the big-O status of Fourier coefficients please let me know. Would help me a great deal. I've searched through the postings but couldn't find a way to go about this problem. Thank you!
Ok so, the latest information I've been able to get is to reach a point in order to prove that $$\hat{f} (n) = -\frac{1}{2 \pi} \int\limits_{-\pi}^\pi f\left(x+\frac{\pi}{n}\right)\exp(-inx)dx ,$$
I am getting stuck after making the u-sub of $$\left(x+\frac{\pi}{n}\right) = u$$ and changing the limits of the integral for $\hat{f}(n)$to look like
$$\hat{f} (n) = \frac{1}{2 \pi} \int\limits_{-\pi+\frac{\pi}{n}}^{\pi+\frac{\pi}{n}} f(u)\exp(-inu) \exp(i \pi)du ,$$
Can someone at least help me with understanding how this integral would lead to the required form and I will be able to continue from there. I look forward to hearing something. Thanks!

By definition, we have
$$\hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x) e^{-inx}\,dx.\tag{1}$$
By periodicity,
$$\hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f\left(x+\frac{\pi}{n}\right) e^{-in(x+\pi/n)}\,dx = -\frac{1}{2\pi}\int_{-\pi}^\pi f\left(x+\frac{\pi}{n}\right)e^{-inx}\,dx.\tag{2}$$
Take the arithmetic mean of $(1)$ and $(2)$:
$$\hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi \frac{f(x) - f\left(x+\frac{\pi}{n}\right)}{2} e^{-inx}\,dx.\tag{3}$$
Taking the absolute value in $(3)$, the Hölder continuity yields the result.
