# Is weak operator topology (WOT) limit of unitary operators isometry?

Let $(U_{\alpha})$ be net of unitary operator in $B\mathcal{(H)}$ s.t. $U_{\alpha} \xrightarrow {\text{WOT}}V$.Can we conclude that V is an isometry? If it be not true give a counter example.

Comments: I observe that if $U_{\alpha} \xrightarrow {\text{SOT}}V$ then V is necessarily isometry. But I could not able to justify the statement for weak operator topology case neither by proving nor by giving counter example.

Notations: $U_{\alpha} \xrightarrow {\text{WOT}}V$ means $\langle U_{\alpha}x,y\rangle\rightarrow\langle Vx,y\rangle$ for all x,y $\in\mathcal{H}$, $U_{\alpha} \xrightarrow {\text{SOT}}V$ means $\Vert U_{\alpha}x\Vert\rightarrow\Vert Vx\Vert$ for all x$\in \mathcal{H}$ and $\mathcal{H}$ is a Hilbert Space.

Any comment regarding proving the statement or giving counter example is highly appreciated.Thanks in advance.

• What if $H=\ell^2$ two-sided and $U_n=S^n$ where $S$ is the (right) shift $Sx_k=x_{k-1}$?
– A.Γ.
Commented Aug 25, 2018 at 9:32
• @A.Γ. It seems that your example works as $U_n$ being unitary is weak operator convergent to O operator but O is not isometry.Hope I am correct. Thank you for you answer.
– Piku
Commented Aug 25, 2018 at 10:32
• Yes, $U_n\to 0$ in WOT. Informally, for any given sequence almost all "energy" is located on a finite interval. When one sequence shifts wrt another, those intervals become not overlapping at the end so that the inner product is almost zero.
– A.Γ.
Commented Aug 25, 2018 at 11:15

The proof is a generalization of the idea, with matrices, that given any $$z\in\overline{\mathbb D}$$, there exists a unitary $$\begin{bmatrix} z&*\\ *&*\end{bmatrix}.$$ One of many possible choices is $$\begin{bmatrix}z&(1-|z|^2)^{1/2}\\(1-|z|^2)^{1/2}&-z\end{bmatrix},$$that we will mimic below.
Concretely, let $$x\in B(H)$$ be a contraction, i.e. $$\|x\|\leq 1$$. Neighbourhoods in the wot topology are of the form $$N=\{y:\ |\langle (x-y)\xi_j,\eta_j\rangle|\leq1:\ \xi_1,\ldots,\xi_m,\eta_1,\ldots,\eta_m\in H\}.$$ Fix one such $$N$$ and let $$L=\operatorname{span}\{\xi_1,\ldots,\xi_m,\eta_1,\ldots,\eta_m%,x\xi_1,\ldots,x\xi_m,x\eta_1,\ldots,x\eta_m \}.$$ Let $$p$$ be the orthogonal projection onto $$L$$. Since $$\dim L<\infty$$, there exists a subspace $$L'\subset L^\perp$$ with $$\dim L'=\dim L$$. Let $$q$$ be the orthogonal projection onto $$L'$$, and $$v$$ a partial isometry $$v:L'\to L$$, i.e. $$v^*v=q$$, $$vv^*=p$$. Now based on the decomposition $$H=L\oplus L'\oplus ( L\oplus L')^\perp$$, let $$u=pxp + (p-(pxp)^*pxp)^{1/2}v+ v^*(p-(pxp)^*pxp)^{1/2} - v^*pxpv+I-p-q.$$ In spirit, $$u=\begin{bmatrix} pxp & (p-(pxp)^*pxp)^{1/2}v&0\\ v^*(p-(pxp)^*pxp)^{1/2} &- v^*pxpv&0\\ 0&0&I\end{bmatrix}.$$ It is not hard to check that $$u$$ is a unitary. And $$\langle u\xi_j,\eta_j\rangle =\langle up\xi_j,p\eta_j\rangle =\langle pup\xi_j,\eta_j\rangle =\langle pxp\xi_j,\eta_j\rangle =\langle x\xi_j,\eta_j\rangle.$$ So $$u\in N$$. As we can do this for any neighbourhood $$N$$ of $$x$$, we can construct a net $$u_N$$ of unitaries with $$u_n\to x$$ wot.
This argument provides a way to show that the wot and sot topologies do not agree (and are, in fact, very different). The surprising thing, though, is that if $$\{u_j\}$$ are unitaries and $$u_j\xrightarrow{\ \rm wot\ }u$$ with $$u$$ unitary, then $$u_j\xrightarrow{\ \rm sot\ }u$$.