Let $f\in C^{0}[-L,L]$ be a function with period $2L$. And let be $\{a_{n}\}_{n\geq 0}$ and $\{b_{n}\}_{n\geq 1}$ their coefficients. Assume that
$$\sum n(|a_{n}|+|b_{n}|)<\infty$$
Prove that $f\in C^{1}(\mathbb{R})$ and that Fourier's series of $f'$ converges uniformly to $f'$.
This question is from my homework and I having a hard time with it. If you just prove that $f$ is differentiable is a good help.
The truncated Fourier series $$ S_n^f(x)=\frac{a_0}{2}+a_1\cos(x)+b_1\sin(x)+\cdots+a_n\cos(nx)+b_n\sin(nx) $$ converges uniformly by the Weierstrass M-test because, by assumption, one has the uniform estimate: $$ |S_n^f(x)| \le \frac{|a_0|}{2}+|a_1|+|b_1|+\cdots+|a_n|+|b|_n| $$ Similarly, the series of derivatives $(S_n^{f})'(x)$ also converges uniformly by the Weierstrass M-test. Then $S_n^f$ converges uniforly to $f$. If $(S_n^f)'$ converges uniformly to $g$, $g$ must be continuous, and the equation $$ S_n^f(x)=S_n^f(0)+\int_{0}^{x}(S_n^f)'(t)dt $$ has the limiting form $$ f(x)=f(0)+\int_{0}^{x}g(t)dt. $$ Hence $f$ is continuously differentiable, and the series $(S_n^f)'$ converges uniformly to $f'=g$.