Weakly null sine sequence in L1 ($\sin (nt)\rightharpoonup 0$ in $L^1([0,2\pi ])$)

The functions $f_n(t)=\sin (nt)$ converge weakly to zero in $L^1([0,2\pi ])$. Indeed, every $g\in L^{\infty }$ also belongs to $L^1$ on this finite interval, and the Riemann-Lebesgue lemma gives ${\int }_0^{2\pi }g(t)\sin (nt)dt\to 0$. However,so $L^1([0,2\pi ])$ does not have the Schur property.