Given that \(f\) is Reimann Integrable and given a Fourier Series on \([-\pi,\pi]\), then
$$
\begin{align*}
f(\theta) &\sim \sum_{n\in\mathbb{Z}} \widehat{f}(n)e^{in\theta}
\end{align*}
$$
Then
- If the sum of the Fourier coefficients is absolutely summable so \( \sum_{n \in \mathbb{Z}} |\widehat f(n)| < \infty \), then
\[
g(\theta) := \sum_{n\in\mathbb Z}\widehat f(n)e^{in\theta}
\]
converges absolutely and uniformly to \(g\) (Proof here). Moreover, recall that the partial sums
\[
S_N(\theta)=\sum_{n=-N}^{N}\widehat f(n)e^{in\theta}
\]
is is a finite sum of continuous exponential functions. Since a uniform limit of continuous functions is continuous, then \(g\) is continuous.
Next, we want to verify that the coefficients are matching. That is \(\widehat f(n) = \widehat g(n)\). Expand the definition of \(\widehat g(n)\) \[ \begin{aligned} \widehat{g}(n) &= \frac{1}{2\pi}\int_0^{2\pi}g(x)e^{-inx}\,dx \\ &= \frac{1}{2\pi}\int_0^{2\pi} \left(\sum_{k\in\mathbb{Z}}\widehat{f}(k)e^{ikx}\right) e^{-inx}\,dx \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sum_{k\in\mathbb Z} \widehat f(k)e^{ikx}e^{-inx}\,dx \\ \end{aligned} \] Then because \(S_N\to g\) uniformly, interchange the sum and and the integral as follows \[ \begin{aligned} \widehat{g}(n) &= \frac{1}{2\pi} \sum_{k\in\mathbb Z} \left( \widehat f(k) \int_0^{2\pi}e^{ikx}e^{-inx}\,dx. \right) \\ &= \frac{1}{2\pi} \sum_{k\in\mathbb{Z}} \left( \widehat{f}(k) \int_0^{2\pi}e^{i(k-n)x}\,dx \right) \end{aligned} \] Note that every term is zero except the term where \(n=k\). Hence when \(k=n\) \[ \begin{aligned} \widehat{g}(n) &= \frac{1}{2\pi} \widehat{f}(n) \int_0^{2\pi}e^{i0x}\,dx \\ &= \frac{1}{2\pi} \widehat{f}(n) (2\pi) = \widehat{f}(n) \\ \end{aligned} \] Hence we see that \(\widehat g(n)=\widehat f(n)\) for every \(n\in\mathbb Z\). Now comes the Uniqueness Theorem for Fourier series which says that if \(\widehat f(n) = 0\) for all \(n \in \mathbb{Z}\), then the function itself must vanish so \(f(x) = 0\) at every point \(x\) where \(f\) is continuous. Applying it to \(f-g\) gives \[ f-g=0 \] at every point where \(f\) is continuous. Consequently, \(f\) and \(g\) can differ only on the measure zero set where \(f\) is discontinuous. Therefore, \(f-g=0\) almost everywhere. Hence \[ f=g\quad\text{almost everywhere} \] - Additionally if \(f\) is continuous, we get that \(f=g\) at every point.
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi