I'm working on a problem that says: If $f \in C_c(\mathbb{R})$ is a compactly supported function that's Holder-continuous of degree $\alpha \in (0, 1)$ (i.e. there exists $C$ such that $|f(y) - f(x)| \leq |y - x |^\alpha$ for $|x - y| \leq 1$), then $$\left| \widehat{f}(\xi) \right| \leq C' \left| \xi \right|^{-\alpha}$$ for $|\xi| \geq 1$.
The hint says to look at $f_y(x) = f(x + y)$, and compute $\widehat{f}_y(\xi)$ in terms of $\widehat{f}(\xi)$. I did this and got that $\widehat{f}_y(\xi) = e^{i y \xi} \widehat{f}(\xi)$, but I don't know how to leverage this fact. I'm afraid this is as far as I've gotten.
Let $|\xi|\geq 1,$ and call $y=1/\xi.$ Note that $|y|\leq 1$. From your work, $$|\hat{f}(\xi)|=\frac{1}{|e^{iy\xi}-1|}|\hat{f}(\xi)-\widehat{f_y}(\xi)|=\frac{1}{|e^{i}-1|}|\hat{f}(\xi)-\widehat{f_y}(\xi)|.$$ Since $f$ is compactly-supported and $y$ is fixed, $f(x+y)-f(x)$ is also compactly supported, say from $-M$ to $M$. Using the Holder continuity condition, $$|\hat{f}(\xi)-\widehat{f_y}(\xi)|\leq \int\limits_{-\infty}^\infty |f(x)-f(x+y)||e^{-ix\xi}|\, dx\leq \int\limits_{-M}^MC |y|^{\alpha}\, dx=D|y|^{\alpha}=D|\xi|^{-\alpha}.$$ So, your $C'$ is $$C'=\frac{D}{|e^i-1|}=\frac{2MC}{|e^i-1|}.$$