Asymptotic behaviour of Fourier coefficients

Question

Let $f: [0,1]\rightarrow \mathbb R$ be continuous.

Then $\lim_{n\rightarrow \infty}\int_{0}^{1} f(x)\sin nx dx =0$

I wasn't able to solve it. The best upper bound I arrive at is $ \frac{2M}{\pi}$, Is there any other way?

Answer 1

Here is a trick that I learned in a harmonic analysis class: Let $M \geq 0$ be a bound of $f$. Then by using the substitution $x \mapsto x+\frac{\pi}{n}$,

\begin{align*} \left| 2\int_{0}^{1} f(x) \sin(nx) \, \mathrm{d}x \right| &= \left| \int_{0}^{1} f(x) \sin(nx) \, \mathrm{d}x - \int_{-\frac{\pi}{n}}^{1-\frac{\pi}{n}} f\left(x+\frac{\pi}{n}\right)\sin(nx) \, \mathrm{d}x \right| \\ &\leq \frac{2\pi M}{n} + \int_{0}^{1-\frac{\pi}{n}} \left| f(x) - f\left(x+\frac{\pi}{n}\right)\right| \, \mathrm{d}x. \end{align*}

Now use the fact that $f$ is uniformly continuous on $[0, 1]$ to show that this bound converges to $0$ as $n\to\infty$.