Closure of unitaries in the SOT Let $H$ be a Hilbert space. I am trying to find the closure of the collection of unitary operators $U(H)$ in $L(H)$ in SOT. I know the strong limit of unitary operators isnt unitary (Unitary shifts on $\ell^2$). I cannot find any information on this in the literature or google.
This isn't a homework problem or an exam.
This looks useful?
 A: The strong closure of the unitaries is the set of all isometries.  To see this suppose that  $\{u_i\}_i$ is a net of unitaries strongly
converging to an operator $v$.  Then, for every $x$ in $H$, one has that
$$
  \|v(x)\| = \lim_i  \|u_n(x)\| = \|x\|,
  $$
so $v$ is an isometry.
Conversely,  given any isometry $v$ in $B(H)$,  let $U$ be a neighborhood of $v$ in the strong
topology.  Then there exists some $\varepsilon >0$,  and vectors $x_1, x_2, \ldots , x_n$ in $H$,  such that
$$
  \{u\in  B(H): \|u(x_i)-v(x_i)\|<\varepsilon \}\subseteq U.
  $$
Let $H_1=\text{span}\{x_1, x_2, \cdots , x_n\}$ and $H_2=\text{span}\{v(x_1), v(x_2), \cdots , v(x_n)\}$.  Then evidently $v$ restricts
to a unitary operator from $H_1$ to $H_2$.  Observing that  $H_1^\perp$ and $H_2^\perp$ have the same dimension, there
exists a unitary operator $v'$ from $H_1^\perp$ to $H_2^\perp$.   The direct sum operator
$$
  u:= v|_{H_1}\oplus v': H_1\oplus H_1^\perp \to H_2\oplus H_2^\perp
  $$
is unitary and it belongs to $U$,  showing that $v$ lies in the closure of the set of unitary operators.
