 Solution

No. In the Lebesgue space $L^1([0,2\pi ])$, letFor every $g\in L^{\infty }([0,2\pi ])$, finiteness of the interval gives $g\in L^1([0,2\pi ])$, and the Riemann-Lebesgue lemma yieldsThe duality of Lp spaces therefore shows that $f_n\rightharpoonup 0$ in $L^1$. On the other hand,for every $n$. Thus weak convergence does not imply norm convergence in this $L^1$ space; the Weakly null sine sequence in L1 is the required counterexample.