For each $t\in[0,1]$, evaluation $\delta_t:f\mapsto f(t)$ is a <bounded linear functional> of <norm> one. Thus <weak convergence> to zero gives $f_n(t)\to0$ at every $t$. The set $\{f_n:n\ge1\}$ is a <weakly bounded set>, since every scalar sequence $\ell(f_n)$ converges, and part (a) supplies a uniform bound $\|f_n\|_\infty\le M$.

The constant $M$ is integrable for <Lebesgue measure> on $[0,1]$. Applying the <dominated convergence theorem> to $|f_n|$ gives the <weakly null continuous functions converge in L1> conclusion
$$
\boxed{\|f_n\|_{L^1[0,1]}=\int_0^1|f_n(t)|\,dt\longrightarrow0.}
$$
Pointwise convergence alone would not provide the needed uniform dominating function; it is the weak boundedness argument that supplies it.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 106 / 2 / b / i / Solution

{title2=$f_n\rightharpoonup0\text{ in }C[0,1]\Longrightarrow\|f_n\|_1\to0$}

<Weak convergence> to zero gives pointwise convergence by the continuous evaluation functionals and uniform boundedness in the <supremum norm> by the <weakly bounded set> criterion. The <dominated convergence theorem> for Lebesgue measure then proves the displayed implication. The finite measure of the interval and uniform norm bound are the relevant hypotheses.

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