Someone ask me the reason of the red one as below :
Let $f$ be defined on $(-\pi,\pi)$ by $f(\theta)=\theta$ . Then $f'$ is piecewise smooth with $f'(\theta)=1$ for $|\theta|<\pi$ and $f'(-\pi+)=f'(\pi-)=1~.$
Now the Fourier series of $f$ is $$\theta=S[f](\theta)=\sum_{n\ge1}\frac{2(-1)^{n+1}}{n}\sin(n\theta)=\sum_{n\ne0}\frac{(-1)^{n+1}}{in}e^{in\theta}.$$
But $$S[f'](\theta)=1\color{red}{\ne}\sum_{n\ne0}\frac{(-1)^{n+1}}{in}\big(e^{in\theta}\big)'=\sum_{n\ne0}\frac{(-1)^{n+1}}{in}\cdot ine^{in\theta}=\sum_{n\ne0}(-1)^{n+1}e^{in\theta}$$
I think the reason is the sum $\displaystyle\sum_{n\ne0}(-1)^{n+1}e^{in\theta}$ does not converge absolutely so , in general , we don't have this following $S[f'](\theta)=\displaystyle\sum_{n\ne0}(-1)^{n+1}e^{in\theta}$ .
Is the reason true for the red one , or , there is something that I make a mistake to understand ?
Any comment or valuable suggestion I will be grateful . Thanks for considering my request .
Usually it's the other way round, if $f_n$ converges to $f$ and somebody claims: "$f_n^\prime$ converges to $f^\prime$" there is the need to provide a proof for this (because it's not true in general). There is a well known theorem which tells you when you can expect it to be true (see here: https://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability), and, as you noticed, it is not applicable to your example.
It's not correct, however, to claim the result is not true because the prerequisites of the theorem are not met. The claim can be true without this.