If $\mathcal H$ is a Hilbert space and $U_n \in B(\mathcal H)$ is a strong-operator convergent sequence of unitary operators, say $U_n\rightarrow U$, is it true that $U$ is unitary? More explicitly, assume $\|(U_n-U)x\|\rightarrow 0$ as $n \rightarrow \infty$ for all $x\in \mathcal H$ and $U_n U_n^*=1= U_n^*U_n$, is it then true that $UU^*=1= U^*U$?
It is clear that $U$ is an isometry, i.e. $\|Ux\|=\|x\|$ for all $x\in\mathcal H$, so it suffices to show that $U$ is onto, or that $U^*$ is an isometry too. Clearly, the assertion then holds true in the finite dimensional setting.
It is clear that $U$ is unitary if the convergence happens in norm, since this implies that $U_n\rightarrow U$ and $U_n^*\rightarrow U^*$ in the strong operator topology, ensuring that $U$ and $U^*$ are isometries (and therefore that $U$ is unitary).
Also, it is clear that $\langle U_n^* x,y\rangle\rightarrow\langle U^* x,y\rangle$. It follows that if $U_n^*$ converges in the strong operator topology, then $U_n^*\rightarrow U^*$, again finishing the argument.
In order to find a counter example one needs to find a unitary sequence $U_n$ which converges in the strong operator topology but does not converge in norm, while $U_n^*$ converges weakly but not strongly. $U=\lim_{n\rightarrow\infty} U_n$ cannot be onto (but must be an isometry) and $U^*=\lim_{n\rightarrow\infty} U_n^*$ cannot be an isometry.
Any help or hint towards a proof or counter example will be much appreciated!
Best Answer
$U$ need not be unitary. Let $U_n$ be the map on $\ell^2$ that sends $e_j\mapsto e_{j+1}$ for $j=1,\ldots, n$, $e_{n+1}\mapsto e_1$, and $U_n$ is the identity on $L(\{e_j: j>n+1\})$.
Then $U_n$ is unitary (it sends an ONB to an ONB again), but $U=S$ (the shift) is only an isometry because $e_1$ is not in the range.