I'm asked to find a Riemann integrable function $f$, that is not Lipschitz continuous but its Fourier series converges to $f$ uniformly.
I took a function $f(x)=\sqrt{|x|}$. It's not differentiable at $0$, and around zero it's derivative is not bounded. I tried to find Fourier coefficients: $A_n=\frac {1}{\pi}\int_{-\pi}^{\pi}\sqrt{|x|}\cos nx\,dx$, $B_n=\frac {1}{\pi}\int_{-\pi}^{\pi}\sqrt{|x|}\sin nx\,dx$ --> Equals zero, even func on symmetric interval.
My main problem is to find $A_n$. I tried to evaluate the integral but unsuccessfully.
You don't need an explicit form of the Fourier coefficients, a good upper estimate is enough. Using your example, integration by parts followed by substitution $u=nx$ gives $$ \begin{align*} A_n &= \frac2\pi \int_0^\pi \sqrt{x} \, \cos nx \, dx = \frac2\pi \left[ \sqrt{x} \frac{\sin nx}n \right]_0^\pi - \frac2\pi \int_0^\pi \frac{1}{2\sqrt{x}} \frac{\sin nx}{n} \, dx \\ &= -\frac{1}{\pi n} \int_0^\pi \frac{\sin nx}{\sqrt{x}} \, dx = -\frac{1}{\pi n} \int_0^{\pi n}\frac{\sin u}{\sqrt{u/n}} \, \frac{du}{n} = -\frac{1}{\pi n^{3/2}} \int_0^{\pi n} \frac{\sin u}{\sqrt{u}} \, du \end{align*} $$ Since $\int_0^\infty \frac{\sin u}{\sqrt{u}} \, du$ converges, there exists a constant $C$ such that $\left|\int_0^{\pi n} \frac{\sin u}{\sqrt{u}} \, du\right| \le C$ for all $n$, and so $|A_n| \le \frac{C}{\pi n^{3/2}}$, and thus $\sum_{n=0}^\infty |A_n| < \infty$, implying absolute and uniform convergence of the Fourier series.