= Solution
\b[No.] In the <Lebesgue space> $L^1([0,2\pi])$, let
$$
f_n(t)=\sin(nt).
$$
For 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> yields
$$
\int_0^{2\pi}f_n(t)g(t)\,dt\longrightarrow0.
$$
The <duality of Lp spaces> therefore shows that $f_n\rightharpoonup0$ in $L^1$. On the other hand,
$$
\lVert f_n\rVert_1
=\int_0^{2\pi}|\sin(nt)|\,dt=4
$$
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.
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