TITLE: Show that $\lim_{k\to\infty} \int_0^1 f(x) \sin(kx) \, dx = 0$
QUESTION [7 upvotes]: I'm working through some old test problems in preparation for a Real Analysis exam. I got stuck on the following problem: 

Show that if $f$ is (Lebesgue) integrable on $[0,1]$, then $$\lim_{k\to\infty} \int_0^1 f(x) \sin(kx) \, dx = 0$$

It seems to me that the idea is to move the limit inside the integral. I know that $f(x)\sin(kx)$ is bounded by $f(x)$ which is integrable, so I thought maybe the bounded or dominated convergence theorems might be useful. However, I don't have a sequence of integrable functions converging point-wise to $0$. How can I proceed, or am I approaching this incorrectly?

REPLY [2 votes]: Assume function $f(x)$ has maximum $M$ and minimum $m$ in interval $[0,1].$ Since the integral exists so both $m$ and $M$ are real and bounded otherwise we will have counterexamples such as $f(x)={1\over x^2}$. Now we want to show that:
$$\lim_{k\to\infty} \int_0^1{f(x)\sin kx} \, dx=0$$
We try to bound the above integral. Since $m\le f(x)\le M$:
$$m\int_0^1 \sin k x \, dx \le \int_0^1f(x)\sin kx \,dx \le M \int_0^1\sin k x \, dx$$
or
$$m{1-\cos k \over k}\le \int_0^1 f(x)\sin k x \, dx \le M{1-\cos k \over k}$$
By tending $k\to\infty$ both bounds get zero and we obtain what we want.