= Solution
<Weak convergence> means $\ell(u_j)\to\ell(u)$ for every <bounded linear functional> $\ell\in X^*$. <Norm convergence> implies this by $|\ell(u_j-u)|\leq\|\ell\|\|u_j-u\|$. The converse fails for coordinate unit vectors in $\ell^2$: their <norms> are one, but pairing with any square-summable <sequence> tends to zero, so they are weakly null.
For the <subsequence> assertion apply <Gram-Schmidt process> to a countable dense subset to choose a countable <orthonormal basis> $(e_m)$ of the <separable Hilbert space> and a bound $M$ for the <sequence>. Repeated <subsequence> selection and diagonalization make every coefficient $\langle u_{j_k},e_m\rangle$ converge to a scalar $a_m$. <Bessel inequality> gives $\sum_{m=1}^N|a_m|^2\leq M^2$ for every $N$, so $u=\sum_ma_me_m$ exists in $H$ with <norm> at most $M$. Pairings with finite <linear combinations> of the basis converge. Approximating any other test vector by such combinations and using the common <norm> bound proves \b[$u_{j_k}\rightharpoonup u$]. The finite-dimensional case uses the same argument with a finite basis.
Apply this to every <subsequence> of the bounded $L^2$ <sequence>. Any <weakly convergent> further <subsequence> has limit $v$ with $\int_a^bv=0$, since interval indicators are $L^2$ tests. The permitted fact gives $v=0$ almost everywhere. If the whole <sequence> failed to converge weakly to zero, some fixed test functional and <subsequence> would stay a positive distance from zero. Its <weakly convergent> further <subsequence> would then contradict the preceding conclusion. Thus \b[the entire <sequence> converges weakly to zero in $L^2(0,1)$].
Back to article page
