Suppose that $f\in PC(2\pi)$ is such that $f'\in PC(2\pi)$ (if $f\in PC(2\pi)$ then $\int_{-\pi}^{\pi}f(x)dx=\int_c^{c+2\pi}f(x)dx \quad\forall c\in\mathbb{R}$). Then the Fourier series of $f$ converges uniformly absolutely to a continous function $g$.
My try:
I know that the coefficients of the Fourier series of $f$ can be written in terms of $f'$ Fourier coeficients. This is: $\forall n\in \mathbb{N}/\{0\}$, $a_n(f)=-\frac{b_n(f')}{n}\quad$ and $\quad b_n(f)=\frac{a_n(f')}{n}$.
$$S_n(f)(x)=\frac{a_0(f)}{2}+\sum_{k=1}^{n}\left[a_k(f) cos(kx)+b_k(f) sin(kx) \right]$$ $$= \frac{a_0(f)}{2}+\sum_{k=1}^{n}\left[-\frac{b_k(f') cos(kx)}{k}+\frac{a_k(f') sin(kx)}{k} \right]$$ $$= \frac{a_0(f)}{2}+\sum_{k=1}^{n}\left[\frac{a_k(f') sin(kx)}{k}-\frac{b_k(f') cos(kx)}{k} \right]$$
I'm not sure on how to prove that this series converges. I only know that by the Riemann Lebesgue Lemma the values of the series are zero when $n$ goes to infinity. Any suggestions would be great!