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?
This is a case of the Riemann–Lebesgue lemma.
If $f$ is has a continuous derivative and $f(1)=0$ then integration by parts yields $$ \int_0^1 f(x) \sin(kx) \,dx = f(1)(\sin k) - \int_0^1 f'(x) \frac 1 k \sin(kx) \, dx = \frac 1 k\int_0^1 f'(x) \sin(kx)\,dx $$ And taking absolute values on both sides and showing that the last expression goes to $0$ (since $|f'(x)\sin(kx)| \le \max\{ |f'(x)| : 0\le x\le 1\}$) will do it.
But what if $f(1)\ne0$ or $f$ does not have a continuous derivative? The linked Wikipedia article says one can approximate $f$ in $L^1$ with another function $g$ that has those propoerties.