Show that Fourier coefficients approach zero uniformly

Question

Let $f(t)$, $g(t)$ be piecewise continuous functons on $[-\pi,\pi]$, periodically continued on $\mathbb R$. I want to show that $$ a_n(x) = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x+t)g(t) \sin(nt) \, \mathrm dt $$ approaches zero as $n \to \infty$ uniformly on $[-\pi,\pi]$. It can be done using Parseval identity and Dini's rule, but maybe there is some easier way that doesn't use such machinery?

Answer 1

Since $$a_n(x)= \frac{1}{\pi}\int_{-\pi-\pi/n}^{\pi-\pi/n} f(x+t+\pi/n)g(t+\pi/n) \sin (nt+\pi) \,dt$$ adding this to the original formula for $a_n(x)$ yields $$\begin{split} 2\pi a_n(x) &= \int_{-\pi}^{\pi-\pi/n}\left\{ f(x+t)g(t)- f(x+t+\pi/n)g(t+\pi/n)\right\} \sin (nt) \,dt \\ &+ \int_{-\pi-\pi/n}^{-\pi}f(x+t+\pi/n)g(t+\pi/n) \sin (nt+\pi) \\ & + \int_{\pi-\pi/n}^\pi f(x+t)g(t) \sin (nt+\pi) \,dt \end{split} $$ The latter two integrals can be estimated by $(\pi/n)\max|f|\max|g|$. The first one is a bit tedious but also elementary: set aside small intervals (length $\pi/n$ or so) around discontinuities of $f$ and $g$, and on the rest estimate $f(x+t)g(t)- f(x+t+\pi/n)g(t+\pi/n)$ using the uniform continuity of the product of two uniformly continuous, bounded functions.

Here I'm interpreting "piecewise continuity" in the sense that I'm used to: the existence of a partition such that the function is uniformly continuous on each open subinterval in the partition, i.e., it has one-sided limits at the partition points.