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.
Can I change the sum with the integral in this case?
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real-analysis
fourier-analysis
fourier-series
2 Answers
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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.
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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).
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0Ok thank you, do you know what theorem must I use to invert the sum with the integral? – 2017-02-12
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0It's called term by term integration. It basically states that if each term $f_k$ of the series is integrable over the interval of interest and the series converges uniformly to a function $f$ over the same interval, then the integral of the whole series over that interval equals the integral of $f$. Finally, it's not difficult to demonstrate that this means you can invert the operators, try doing it. – 2017-02-12