How to show weak convergence of $\mathrm{sign}(\sin(nx))$ as $n \to \infty$?

Question

How to show weak convergence of $\mathrm{sign}(\sin(nx)) \rightharpoonup 0$ in $L^2(0,2\pi)$ (or some $L^p$ space) as $n \to \infty$?

Without the sign function there, the proof uses I think Parseval's identity. I don't know how to tackle this.

Answer 1

Let $I:=(0, 2\pi)$, $f(x):=sign(\sin(x))$. $f$ is periodic with period $2\pi$. Note that $\int_I f(x)dx =0$. Define $$ g(x) = \int_0^x f(t) dt, $$ which implies continuity of $g$ and $|g(x)| \le \|f\|_{L^1(I)}$. Since $\int_I f(x)dx=0$, it follows that $g$ is $2\pi$-periodic, and the estimate $|g(x)| \le \|f\|_{L^1(I)}$ is valid for all $x>0$.

Take $0<a<b<2\pi$. Then with the substitution $t = nx$ $$ \int_a^b f(nx) dx = n^{-1} \int_{na}^{nb} f(t)dt = n^{-1} ( g(nb)-g(na)) \le 2 \|f\|_{L^1(I)} \cdot n^{-1} \to 0. $$ It shows that $\int_I \chi_{(a,b)}f_n(x)dx \to0$. Since characteristic functions of open intervalls are dense in $L^p(I)$ for $1\le p<\infty$, this implies $\int_I v(x)f_n(x)dx \to0$ for all $v\in L^p(I)$ for $1\le p<\infty$. Thus, $f_n \rightharpoonup 0$ in $L^q(I)$ for all $1<q<\infty$ and $f_n\rightharpoonup^*0$ in $L^\infty(I)=L^1(I)^*$.

This proof works for arbitrary periodic $f$ with integral mean zero: set $f_n(x):=f(nx)$ then $\|f_n\|_{L^q} = \|f\|_{L^q}$ and $f_n \rightharpoonup 0$ (or $\rightharpoonup^*0$) in $L^q$ as above.