Now I'm really stuck on proving the following.
Let $f(x)=\vert x \vert^\alpha$, $x\in[-\pi,\pi]$, $0<\alpha<1$. Show that for $0<c_1<c_2$, as $n\rightarrow\infty$, $$ c_1\vert n \vert^{-1-\alpha} \leq \vert c_n(f) \vert \leq c_2\vert n \vert^{-1-\alpha}\,, $$ where $c_n(f)$ is the $n$-th Fourier coefficient of $f$. I have shown that $\vert x \vert^{\alpha}$ satisfies the Holder condition and proceeded as below. For $n>0$,
\begin{align} \vert c_n(f) \vert &= \frac{1}{2\pi}\left\vert \int_{-\pi}^\pi f(x) e^{-inx} dx \right\vert \\ &= \frac{1}{2\pi} \left\vert \sum_{j=0}^{n-1} \int_{-\pi+\frac{2\pi j}{n}}^{-\pi+\frac{2\pi(j+1)}{n}} \left(f(x)-f(-\pi+\frac{2\pi j}{n})\right)e^{-inx} dx \right\vert \\ &\leq \frac{1}{2\pi} \sum_{j=0}^{n-1} \int_{-\pi+\frac{2\pi j}{n}}^{-\pi+\frac{2\pi(j+1)}{n}} \left\vert f(x)-f(-\pi+\frac{2\pi j}{n}) \right\vert dx \\ &\leq \frac{1}{2\pi} \sum_{j=0}^{n-1}\frac{2\pi}{n} \left(\frac{2\pi}{n}\right)^\alpha\,, \end{align} which concludes only $\mathcal{O}(n^{-\alpha})$. How can I reduce the upper bound in terms of $n^{-1-\alpha}$?
Note that $f$ is even, so it has a Fourier cosine series and for $n \ge 1$ we have $$\pi c_n=\int_0^{1/n}x^{\alpha}\cos nxdx+\int_{1/n}^{\pi}x^{\alpha}\cos nxdx$$
By the second mean value theorem for integrals $$\int_0^{1/n}x^{\alpha}\cos nx dx=(1/n)^{\alpha}\int_0^{x_0}\cos nx dx =n^{-1-\alpha}\sin nx_0$$ Integrating by parts we have $$\int_{1/n}^{\pi}x^{\alpha}\cos nxdx=n^{-1-\alpha}\sin 1-\alpha/n \int_{1/n}^{\pi}x^{\alpha-1}\sin nx dx$$ and applying one more time the second mean value theorem, we have $$\frac{\alpha}{n}\int_{1/n}^{\pi}x^{\alpha-1}\sin nx =\frac{\alpha}{n}(1/n)^{\alpha-1}\int_{1/n}^{\pi} \sin nxdx=O(n^{-1-\alpha})$$ so the upper bound $|c_n|=O(n^{-1-\alpha})$ follows
For the lower bound one integrates by parts to get $$\int_0^{\pi}x^{\alpha}\cos nx dx=-\frac{\alpha}{n}\int_0^{\pi}x^{\alpha-1}\sin nx dx$$ and then splitting the integrals in $\pi/n$ chunks where $\sin nx$ has constant sign, grouping them in $(+-)+(+-)+...$, changing variables in the negative integral and noting that each such pair then is positive since $x^{\alpha-1}$ is decreasing, it is enough to look at $$\int_0^{\pi/n}(x^{\alpha-1}-(x+\pi/n)^{\alpha-1})\sin nx dx$$ and show it is $>>n^{-\alpha-1}$ as we add to it nonnegative terms; that follows using $\sin x \ge cx, 0 \le x \le \pi/2$, restricting to the integral from $0$ to $\pi/(2n)$ and computing the integrals after the majorization $\sin nx \ge c nx$