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In Narutaka Ozawa's solution over at Mathoverflow, the following result is implicitly used:

Let $M\subseteq B (H)$ be a von Neumann algebra, and let $\xi\in H$. If $u\in M$ is a partial isometry, then for $\epsilon>0$ we can find a partial isometry $v\in M$ with $\|(u-v)\xi\| <\epsilon$ and with $1-v^*v \sim 1-vv^*$.

Here $\sim$ is Murray-von Neumann equivalence. It's easy to see that if $1-v^*v \sim 1-vv^*$ then $v$ can be extended to a unitary in $M$. Why is this claimed result true?

If $M$ is finite, then as indicated in this answer we always have that if $p\sim q$ then $1-p\sim 1-q$ (see also, for example, Exercise 6.9.6 in Kadison and Ringrose, Volume 2). By definition $u^*u\sim uu^*$ so the claim follows with $v=u$. Indeed, this would also hold if $u^*u$ (equivalently, $uu^*$) were finite, even if $M$ were not.

Ozawa's wording suggests to me that in general, we might seek a projection $p\in M$ with $p\leq u^*u$ and set $v=up$. Then $v^*v = pu^*up = p$, and we want $\|u(1-p)\xi\|<\epsilon$, which seems plausible to achieve, perhaps? I have no clue how to get $1-v^*v \sim 1-vv^*$? Maybe instead a type-decomposition argument could work, but again I don't see how to get started.

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As stated, this is not true. Let $M=B(H)$, $\xi=e_1$, $u=S^*$ the adjoint of the unilateral shift. Then, for any unitary $v$, $$ \|(u-v)\xi\|=\|v\xi\|=1. $$


On the other hand, if I'm not wrong what Ozawa is claiming is that if $\|u\xi-\eta\|<\delta$, with $\|\xi\|=\|\eta\|=1$ and $u\in M$, then there exists $v\in M$, unitary, with $\|u\xi-v\xi\|<\varepsilon$. Below is what I could come up with, which no doubt is much less fancier than how Ozawa thought about it.

We may assume that $M$ has no finite-dimensional summand, as that case is easily dealt with. When $M$ has no finite-dimensional summand, the unitary group is sequentially wot dense in the unit ball (see Conway and Szücs, Indiana Univ. Math. J. 22 (1972/73), 763–768. Article and zbMath). So there exists a sequence $\{v_n\}\subset M$ with $v_n$ unitary and $v_n\to u$ wot. Then, noting that $\|u\xi\|>\|\eta\|-\delta=1-\delta$ \begin{align} \limsup_n\|u\xi-v_n\xi\|^2&=\limsup_n\|u\xi\|^2+\|\xi\|^2-2\operatorname{Re}\langle u\xi,v_n\xi\rangle\\[0.3cm] &\leq \limsup \delta+2-2\operatorname{Re}\langle u\xi,v_n\xi\rangle\\[0.3cm] &=\delta+2(1-\|u\xi\|)<\delta+2\delta=3\delta. \end{align} Choosing $n$ big enough, we can get a unitary $v=v_n$ such that $$ \|u\xi-v\xi\|<\sqrt{4\delta}=2\sqrt\delta. $$ Some can take $\varepsilon=2\sqrt\delta$.

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  • $\begingroup$ Great! Thanks not only for the counter-example, but especially for working out what I should have asked, and then answering that instead. $\endgroup$ Jan 15 at 9:55
  • $\begingroup$ I guess it's also worth stressing that this argument works for any contraction $u\in M$ while "my" application only needed it with $u$ a partial isometry; so this is very nice! $\endgroup$ Jan 15 at 10:10
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    $\begingroup$ Indeed. Many many years ago, the exercises in Kadison-Ringrose's chapter 5 about showing that in $B(H)$ the projections are wot-dense in the positive unit ball, and the unitaries in the full unit ball were eye-opening for me. $\endgroup$ Jan 15 at 12:38
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Here is a (sketch) proof of how to use Comparison Theory. We start with a partial isometry $u\in M$ and a vector $\xi\not=0$, and for $\epsilon>0$ we seek a partial isometry $v$ with $\|(u-v)\xi\| > (1-\epsilon) \|u\xi\|$. In fact, we will construct $v$ with $v^*v \leq u^*u$. First a couple of reductions. If $\xi'=u^*u\xi$ so $\|\xi'\| = \|u\xi\|$, and we can find $v$ with $\|(u-v)\xi'\| > (1-\epsilon)\|\xi'\|$ then as $v\xi = v(v^*v)\xi = v(v^*v)(u^*u)\xi = v(u^*u)\xi = v\xi'$, and $u\xi=u\xi'$, we have $\|(u-v)\xi\| = \|(u-v)\xi'\| > (1-\epsilon)\|\xi'\| = (1-\epsilon)\|u\xi\|$ as required. So wlog $u^*u\xi=\xi$ and $\xi$ is a unit vector. The problem also respects direct sums, so we may decompose $M$ as a direct sum and solve the problem in each part.

If $u^*u$ is finite, then as already observed in the OP, we're done. Otherwise $u^*u$ is infinite, so there is a central projection $z$ with $zu^*u$ finite and $(1-z)u^*u$ properly infinite. We have solved the problem in the summand $zM$, so we can focus on $(1-z)M$. So wlog assume that $e = u^*u$ is properly infinite.

By the "halving lemma" there is $g\leq e$ with $g\sim e$ and $e-g\sim e$. Notice that we can then further decompose (say) $e-g$, and so by induction, for any $n$ find orthogonal $g_1,\cdots,g_n$ with $\sum_i g_i=e$ and $g_i\sim e$ for each $i$. Notice also that if $f_1,f_2$ are orthogonal with $f_i\sim e$ for $i=1,2$ then $f_1\sim e \sim g$ and $f_2\sim e \sim e-g$ so $f_1+f_2\sim g+(e-g)=e$. An induction shows that if $f_1,\cdots,f_n$ are orthogonal with $f_i\sim e$ for each $i$ then $\sum_i f_i\sim e$. In particular, for any $i$, we find that $\sum_{i\not=j} g_j \sim e$.

Set $f=uu^*$. For any $i$, set $\hat g_i = e - \sum_{j\not=i} g_j$, and observe that $$ 1-\hat g_i = (1-e) + (e-\hat g_i) = (1-e) + g_i \sim (1-e)+e = 1. $$ Using the partial isometry $u$ which by definition gives $e\sim f$, we can transport the decomposition $e = \sum_j g_j$ to $f = \sum_j g_j'$ say. Set $\hat g_i' = f - \sum_{j\not=i} g'_j \sim \hat g_i$. So the same argument gives $1-\hat g_i' \sim 1$ and so $1-\hat g_i \sim 1-\hat g_i'$.

As $1 = \|\xi\|^2 = \|u^*u\xi\|^2 = \sum_j \|g_j\xi\|^2$ we see that for some $i$, we must have $\|g_i\xi\|^2 \leq 1/n$, and so $\|\hat g_i\xi\|^2 \geq 1-1/n$. Taking $n$ large enough and setting $v = u(\hat g_i^*\hat g_i)$ gives the required conclusion.

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    $\begingroup$ So this is not as general as Martin Argerami's answer, but we do prove a little more, namely that a "cut down" of $u$ may be extended to a unitary (rather than finding some unitary which does the job). $\endgroup$ Jan 16 at 11:42

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