Today, my differential equations professor said:
A piecewise continuous function on $[a,b]$ is only discontinuous at finitely many points in $[a,b]$, and, furthermore, at any discontinuity point, both one-sided limits exist.
If a periodic function $f$ and its derivative $f'$ are both piecewise continuous, then $f$'s Fourier series converges pointwise to
$x \mapsto \begin{cases} f(x) & f \text{ continuous at } x \\ \dfrac{f(x^+) + f(x^-)}2 & f \text{ discontinuous at } x \end{cases}$
Where $f(x^-)$ and $f(x^+)$ are the left-sided and right-sided limits of $f$ at $x$.
Wouldn't it have been easier to say this instead?
- A piecewise continuous function on $[a,b]$ has all left-sided limits in $(a,b]$ and all right-sided limits in $[a,b)$.
- If a periodic function $f$ and its derivative $f'$ are both piecewise continuous, then $f$'s Fourier series converges pointwise to $x \mapsto \frac{f(x^+) + f(x^-)}2$. There's no need to case-analyze whether $f$ is continuous at $x$.
Or is my version subtly wrong in some way?
Regarding the first point, there are monotone functions defined on a compact interval with countably many jump discontinuities. So requiring the function to have only finitely many discontinuities is an additional assumption that your definition would not capture.
Regarding the Fourier series, I would say that it is "reassuring" to know that the Fourier series converges to the function value at all except finitely many points. Also, this is usually the important part of the theorem that is most useful in application. Yes, your statement is logically equivalent, but it takes more work to extract the useful and reassuring part.