 Solution

Weak convergence means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. Norm convergence implies this by $|\ell (u_j-u)|\le \Vert \ell \Vert \Vert u_j-u\Vert$. 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 $⟨u_{j_k},e_m⟩$ converge to a scalar $a_m$. Bessel inequality gives ${\sum }_{m=1}^N|a_m{|}^2\le M^2$ for every $N$, so $u={\sum }_m{a}_m{e}_m$ 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 $u_{j_k}\rightharpoonup u$. 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 ${\int }_a^{b}v=0$, 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 the entire sequence converges weakly to zero in $L^2(0,1)$.