I am a bit confused about the conditions for Dirichlet's theorem for the pointwise convergence of Fourier series. Despite the theorem only explicitly requiring that $f\in PC^1[-\pi,\pi]$ (piecewise continuous with a derivative that is piecewise continuous), it seems to me, from the two items that follow, that the theorem also requires that $f$ is periodic over $[-\pi,\pi]$ (denoted $f\in C_{per}[-\pi,\pi]$):
- The proof I have seen for the theorem is by induction over the number of discontinuity points, and begins with: "Let $f\in PC^1[-\pi,\pi]$. For $n=0$, $f\in C_{per}[-\pi,\pi] \cap PC^1[-\pi,\pi]$, and therefore...". I cannot see why $f \in PC^1[-\pi,\pi]$ would immediately mean it is also in $C_{per}[-\pi,\pi]$ unless it is an assumption.
- A solution for computing the following series: $$\sum_{n=1}^\infty \frac{\sin(n\pi\alpha)}{n} \qquad \alpha \in(0,1)$$first computed the Fourier series for the function: $f_\alpha(x):=\chi_{[0,\alpha]}(x)$ which is: $$ \alpha +\sum_{n=1}^\infty \frac{2}{n\pi}\sin(n\pi\alpha)\cos(n\pi x)$$ and then, wanting to plug in $x=0$, used the Dirichlet theorem, but stated "we now use the fact that the even expansion of $f_\alpha$ satisfies the conditions of the Dirichlet theorem, and therefore the Fourier series converges pointwise at $x=0$...". The only reasoning I can find for using the even expansion rather than $f_\alpha$ itself, is so that it is in $C_{per}[-1,1]$ (otherwise $f_\alpha$ is in $PC^1[-1,1]$ so why the even expansion?).
Despite these observations, the periodic part is not explicitly stated in the theorem. What am I missing?
Thanks.