Prove that the derivative of a continuous function $f:[0,2\pi]\to\mathbb{C}$ which has the fourier series expansion $f(x)=\sum_{n=-\infty}^\infty a_ne^{inx}$ satisfying $\sum |n||a_n|\leq 1$ is continuously differentiable.
My understanding is that for a function $f$ to be continuous on its domain we need to prove that $\lim_{x\to a^+} f(x)=\lim_{x\to a^-} f(x)=\lim_{x\to a} f(x)=f(a)$ for any element $a$ in its domain.
Taking the derivative $f'(x)=\sum_{n=-\infty}^\infty a_n.in.e^{inx}=g(x)$
For any $a\in[0,2\pi]$ $$ \lim_{x\to a}g(x)=\lim_{x\to a}\sum_{n=-\infty}^\infty a_n.in.e^{inx}=\sum_{n=-\infty}^\infty\lim_{x\to a} a_n.in.e^{inx}=\sum_{n=-\infty}^\infty a_n.in.e^{ina}=g(a) $$ Thereby proving $f'(x)$ is continuous and therefore continuously differentiable.
Is it sufficient to prove that the function is continuously differentiable ?
