Solution
= Solution
Yes. If a subsequence converges weakly to $\nu$, bounded continuity of $e^{iuy}$ gives
$$
\widehat\nu(u)=\lim_k\psi_{n(k)}(u)=\psi(u).
$$
The <uniqueness theorem for characteristic functions> makes $\nu$ unique. Tightness implies that every subsequence has a further weakly convergent subsequence, and every such limit is $\nu$. This subsequence criterion proves that the entire sequence converges weakly to $\nu$.