Can I change the sum with the integral in this case?

Question

Let $f : \mathbb{R} \to \mathbb{C}$ be a function $2 \pi$-periodic such that $\int_{- \pi}^{\pi} |f| d x < \infty$. We suppose that exists a sequence ${\{c_k\}}_{k = - \infty}^{\infty}$ of complex numbers such that $$ f(x) = \sum_{k = - \infty}^{\infty} c_k e^{i k x} $$ for all $x \in \mathbb{R}$. Is $$ \int_{- \pi}^{\pi} \left(\sum_{k = - \infty}^{\infty} c_k e^{i (k - j) x}\right) d x = \sum_{k = - \infty}^{\infty} \left(\int_{- \pi}^{\pi} c_k e^{i (k - j) x} d x\right) $$ true for all $j = \ldots , - 2 , - 1 , 0 , 1 , 2 , \ldots$? Thank you very much. I think it is false in general but I am not sure.

Answer 1

You should be fine if $\sup_{x \in [-\pi,\pi]} f(x)<\infty$, by the bounded convergence theorem.

If you're interested in cases where $f$ diverges, you should just make sure that it is bounded by an integrable function. Namely, if there exists some $g \in L^1[-\pi,\pi]$ that bounds $f$ (i.e: $|f(x)| \leq g(x)$ for all $x$) then the dominated convergence theorem applies.

Answer 2

You are supposing that $f$ is expandable into Fourier series, so the series obliviously converges uniformly to $f$ at least in $(-\pi,\pi)$. For this reason the integrand $\sum_{k = - \infty}^{\infty} c_k e^{i (k - j) x}$ does converge uniformly over the integration interval and as a consequence of that, you can invert the integral and the sum operators (Term by term integration).