# 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?

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.
• Is the $V$ supposed to be $U$?