Now I want to figure out the following problem: Suppose that $f\in L^1(-\pi,\pi)$. Define $F(x)$ as \begin{align} F(x) = \int_{-\pi}^x (f(t)-c_0(f)) dt\,,~x\in[-\pi,\pi] \end{align} where $c_n(f)$ is \begin{align} c_n(f) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x) e^{-inx}dx\,. \end{align} Then evaluate $c_0(F)$.
I tried to solve this problem as follows. Using Fubini's theorem, we have \begin{align} 2\pi c_0(F) &= \int_{-\pi}^\pi F(x) dx = \int_{-\pi}^\pi \int_{-\pi}^x (f(t)-c_0(f)) dt dx \\ &= \int_{-\pi}^\pi \int_{t}^\pi dx (f(t)-c_0(f)) dt = \int_{-\pi}^\pi (\pi-t)(f(t)-c_0(f)) dt\,. \end{align} Since we have $\int_{-\pi}^\pi (f(t)-c_0(f))dt=0$, then \begin{align} \int_{-\pi}^\pi (\pi-t)(f(t)-c_0(f))dt=\int_{-\pi}^\pi -t(f(t)-c_0(f))dt=\int_{-\pi}^\pi -tf(t)dt = 2\pi c_0(F)\,. \end{align} After progressing this point, it is currently stuck. What can I do to get better results here?
Now I have solved the OP using Conrad's suggestion. Since $F\in W^1_1(-\pi,\pi)$ clearly and $F(\pi)=F(-\pi)=0$, hence it is periodic, therefore we can say that the Fourier series of $F$ converges to $F$ pointwisely, i.e., \begin{align} F(x)=c_0(F)+\sum\limits_{n\neq 0}\frac{c_n(f)}{in}e^{inx}\,. \end{align} Then, we have \begin{align} F(x)-F(-\pi)=\sum\limits_{n\neq0}\frac{c_n(f)}{in}(e^{inx}-e^{-in\pi})\,. \end{align} Integrate both sides from $-\pi$ to $\pi$ yields, \begin{align} 2\pi c_0(F) &= \sum\limits_{n\neq0}\frac{c_n(f)}{in}\int_{-\pi}^\pi (e^{inx} - e^{-in\pi}) dx = 0\,, \end{align} since $\int_{-\pi}^\pi e^{inx}dx=0$ for $n\neq0$ and $\int_{-\pi}^\pi e^{-in\pi} dx = (-1)^n2\pi$; hence the sum of all the pairs $-n$ and $n$ vanish because of the term $(in)^{-1}$.