\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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 22H / d / Solution

{c}
{title2=$\sin(nt)\rightharpoonup0$ 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 $\int_0^{2\pi}g(t)\sin(nt)\,dt\to0$. However,
$$
\lVert f_n\rVert_1=\int_0^{2\pi}|\sin(nt)|\,dt=4,
$$
so $L^1([0,2\pi])$ does not have the <Schur property>.

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