The theorem and the remark and an example on a remark are given below:
But I do not understand the example $f(x) = x$, specifically I do not understand:
1- what do the author mean by "$f(x) = x$ as a function in $PC(2\pi)$" and why this function has a discontinuity at $ x = (2k -1)\pi$, for all $k \in \mathbb{Z}.$
2- I also do not understand why this function is only piecewise differentiable in $(-\pi , \pi)$ and not differentiable? (by theway I know that differentiability implies piecewise differentiable ...... am I correct? )
Thank you.
The function $f(x)=x$ is not piecewise continuous, but if we define another function $g(x) = f(x)$ for $|x|< \pi$ and define $g$ so that it is $2 \pi$ periodic then we have a function that is equal to $f$ for $|x| < \pi$ and $g \in PC(2 \pi)$.
I have not defined $g$ at odd multiples of $\pi$, but there is no need to here as the Fourier series are 'insensitive' to changes on a set of measure zero.
The function $g$ is smooth (not just differentiable) in $(\pi,\pi)$, hence for $x \in (\pi,\pi)$ we have ${1 \over 2}( \lim_{t \uparrow x} g(t) + \lim_{t \downarrow x} g(t)) = g(x) = x$ and so the Fourier series of $g$ converges (pointwise) to the value $x$.
You can see that, similarly & unsurprisingly, the Fourier series converges to $0$ at odd multiples of $\pi$ (because $g$ is odd for $|x| < \pi$).
Note that for $|x| < \pi$ the convergence is not absolute.
Look at the excellent answer https://math.stackexchange.com/a/95155/27978 for more detail about the convergence.