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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?

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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.

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  • $\begingroup$ Is the $V$ supposed to be $U$? $\endgroup$
    – user515599
    Dec 27, 2020 at 16:07
  • $\begingroup$ Yes! Thanks for pointing it out. I should be OK now. $\endgroup$
    – Ruy
    Dec 27, 2020 at 16:43

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