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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.

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    $\begingroup$ What if $H=\ell^2$ two-sided and $U_n=S^n$ where $S$ is the (right) shift $Sx_k=x_{k-1}$? $\endgroup$
    – A.Γ.
    Commented Aug 25, 2018 at 9:32
  • $\begingroup$ @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. $\endgroup$
    – Piku
    Commented Aug 25, 2018 at 10:32
  • $\begingroup$ 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. $\endgroup$
    – A.Γ.
    Commented Aug 25, 2018 at 11:15

1 Answer 1

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The answer is a resounding no. In fact, the wot-closure of the set of unitaries is the whole unit ball.

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 $u_j\xrightarrow{\ \rm sot\ }u$.

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