Weak convergence in $L^1$ of $\sin(2\pi n x)$

Question

Is it possible to show that $\sin(2\pi n x)$ tends to $0$ weakly in $L^1([0,1])$ (with respect to the Lebesgue measure)? That is, to prove that $$ \int_{[0,1]}\sin(2\pi nx )g(x)\,\mathrm{d}{x} \xrightarrow{n\to\infty}0 $$ for any function $g\in L^\infty([0,1])$. I am working on different convergence results for $\sin(2\pi nx)$ but got stuck on this one. Any help is appreciated.

Answer 1

From Riemann-Lebesgue theorem we have if $g$ is integrable function then $$\lim_n \int e^{-inx}g(x)\,dx=0$$ In your situation you've got $g\in L^{\infty}[0,1]$. In particular this implies $$\int^1_0g(x)\,dx\leqslant \int^1_0|g(x)|\,dx\leqslant \sup_{x\in[0,1]}|g(x)|\int^1_0\,dx=\sup_{x\in[0,1]}|g(x)|<+\infty$$ hence $g\in L^1[0,1]$. By the theorem above we get $$\lim_n\int^1_0e^{-inx}g(x)\,dx=0\Rightarrow \lim_n\Im\Big(\int^1_0e^{-inx}g(x)\,dx\Big)=0\Rightarrow \lim_n\int^1_0\sin(nx)g(x)\,dx=0$$ For a proof of the Riemann-Lebesgue theorem you can refer to almost any standard book in real analysis.