In Vretblad's Fourier Analysis and its Applications, it says:
Suppose that $f \in C^1(\mathbb{T})$ , which means that both $f$ and its derivative $f'$ are continuous >on $\mathbb{T}$. We compute the Fourier coefficients of the derivative:
$$ \begin{aligned} c_n(f') &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} f'(t) \ e^{-int} \ dt \\ &= \frac{1}{2 \pi} \Big[ f(t) e^{-int} \Big]_{-\pi}^{\pi} - \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) (-in) e^{-int} dt \\ &= \frac{1}{2 \pi} \Big( f(\pi)(−1)^n − f(−\pi)(−1)^n \Big) + in c_n(f) \\ &= in c_n(f) \end{aligned}$$ The fact that $f$ is continuous on $\mathbb{T}$ implies that $f(−\pi) = f(\pi)$.
So my questions is: how do we know that $f$ being continuous on $\mathbb{T}$ implies that $f(−\pi) = f(\pi)$?
The book identifies a function that is $2\pi$-periodic as a function on the circle $\mathbf{T}$ (see Section 4.1 of the book):
So when $f$ is a function on the circle, it must be true that $$ f(-\pi)=f(-\pi+2\pi)=f(\pi). $$
"Continuity" is actually not quite relevant here, although the author mentioned it.