Give an example showing that this fails if \(g\) is not bounded.
Show that if \(f\in C^1\) then \(f\) is equal to its Fourier series.
The first part refers to the identity \(\int \hat f\,g = \int f\,\hat g\), which holds for \(f, g \in L^1\) by Fubini’s theorem.
Solution
On \(\RR\) use \(\hat f(\xi) = \int f(x)e^{-ix\xi}\,dx\); on the circle use \(\hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx\).
1. For \(f(x) = e^{-|x|}\) and \(g(\xi) = 1 + \xi^2\), \(f \in L^1(\RR)\) and \(\int \hat f\,g = \infty\), while \(\hat g\) is not defined as a Lebesgue integral.
1.1. \(\hat f(\xi) = \frac{2}{1 + \xi^2}\).
\(\int_0^\infty e^{-x}e^{-ix\xi}\,dx = \frac{1}{1+i\xi}\) and \(\int_{-\infty}^0 e^{x}e^{-ix\xi}\,dx = \frac{1}{1-i\xi}\), and their sum is \(\frac{2}{1+\xi^2}\).
1.2. Q.E.D.
By step 1, \(\hat f\,g \equiv 2\), whose integral over \(\RR\) is infinite. Since \(g \notin L^1(\RR)\), the integral \(\int g(\xi)e^{-ix\xi}\,d\xi\) does not converge absolutely for any \(x\), so the right side \(\int f\,\hat g\) is not defined.
2. If \(f\) is \(2\pi\)-periodic and \(C^1\), then \(\sum_n \hat f(n)e^{inx}\) converges absolutely and uniformly to \(f\).
2.1. For \(n \ne 0\), \(\hat f(n) = \frac{1}{in}\widehat{f'}(n)\).
Integration by parts gives \(\hat f(n) = \frac{1}{2\pi}\Big[\frac{f(x)e^{-inx}}{-in}\Big]_0^{2\pi} + \frac{1}{in}\cdot\frac{1}{2\pi}\int_0^{2\pi} f'(x)e^{-inx}\,dx\), and the boundary term vanishes by \(2\pi\)-periodicity of \(f\).
2.2. \(\sum_{n \in \ZZ}|\hat f(n)| < \infty\).
By step 1 and the Cauchy–Schwarz inequality, \(\sum_{n\ne0}|\hat f(n)| = \sum_{n\ne0}\frac{|\widehat{f'}(n)|}{|n|} \le \Big(\sum_{n\ne0}|\widehat{f'}(n)|^2\Big)^{1/2}\Big(\sum_{n\ne0}\frac{1}{n^2}\Big)^{1/2}\). The first factor is finite by Bessel’s inequality, since \(f'\) is continuous and hence in \(L^2\).
2.3. The series \(\sum_n \hat f(n)e^{inx}\) converges uniformly to a continuous function \(g\) with \(\hat g(n) = \hat f(n)\) for all \(n\).
Uniform convergence follows from step 2 and the Weierstrass M-test, since \(|e^{inx}| = 1\). The uniform limit of the continuous partial sums \(S_N\) is continuous, and uniform convergence lets the limit pass under the integral, so \(\hat g(n) = \lim_N \widehat{S_N}(n) = \hat f(n)\).
2.4. \(g = f\).
2.3. By Fejér’s theorem \(h \ast F_N \to h\) uniformly, where \(F_N\) is the Fejér kernel, and \(h \ast F_N = \sum_{|k|\le N}\big(1 - \frac{|k|}{N+1}\big)\hat h(k)e^{ikx} = 0\). So \(h = 0\).
2.5. Q.E.D.
Steps <2>2–<2>4.