1. ## Fourier Series problem

2. Originally Posted by silversand
Extend g by periodicity to a 2π-periodic function on $\mathbb{R}$, and calculate its Fourier coefficients:

\begin{aligned}\hat{g}(k) = \frac1{2\pi}\int_0^{2\pi}\!\!\!g(\tau)e^{-ik\tau}d\tau &= \frac1{2\pi}\int_0^{2\pi}\sum_{n=-\infty}^\infty F(2n\pi+\tau)e^{-ik\tau}d\tau \\ &= \frac1{2\pi}\sum_{n=-\infty}^\infty \int_0^{2\pi}\!\!\!F(2n\pi+\tau)e^{-ik\tau}d\tau \\ &= \frac1{2\pi}\sum_{n=-\infty}^\infty\int_{2n\pi}^{2(n+1)\pi}\!\!\!F(\sig ma) e^{-ik\sigma} d\sigma \end{aligned}
${\color{white}\hat{g}(k) = \frac1{2\pi}\int_0^{2\pi}\!\!\!g(\tau)e^{-ik\tau}d\tau} = \frac1{2\pi}\int_{-\infty}^\infty F(\sigma)e^{-ik\sigma} d\sigma = \tfrac1{2\pi}\tilde{F}(k).$

(in the second line, the exchange of limiting operations is justified by the uniform convergence of the series. In the next line, I made the substitution $\sigma = 2n\pi+\tau$ and used the fact that $e^{2nk\pi i} = 1$.)

I now want to complete the proof by saying that g is the sum of its Fourier series: $g(\tau) = \sum_{k=-\infty}^\infty \hat{g}(k)e^{ik\tau} = \frac1{2\pi}\sum_{k=-\infty}^\infty \tilde{F}(k)e^{ik\tau}$. To justify that, I need to quote some theorem saying that g is the sum of its Fourier series. The only such result that I know says that this holds if g is continuous. But the given information does not seem to ensure that g is continuous. Perhaps we need to be told that F is continuous? That would imply that g is continuous, because of the uniform convergence.