TITLE: When two projections in a C*-algebra are "almost" Murray-von Neumann equivalent, they are equivalent
QUESTION [14 upvotes]: Let $A$ be a C*-algebra and $p,q \in A$ be projections. Assume there is an element $a\in A$ such that $\|aa^*-p\|<\frac{1}{4}$ and $\|a^*a-q\|<\frac{1}{4}$. Then there is a partial isometry $v$ with $vv^*=p$ and $v^*v=q$.
This is considered obvious in the paper I'm reading. I've tried to mimic the proof of the following proposition. 

Let $p$, $q$ be projections in a C*-algebra A. If $\|p-q\|<1$ then p and q are unitarily 
  equivalent. 

Which involves using a suitable expression ($1-p-q+2pq$) and polar decomposition. But it's not going anywhere.

REPLY [3 votes]: Here is my own attempt on the problem, which turns out to be essentially San’s argument above (I only realized this when writing my response). I would therefore like to focus on demystifying the continuous functional calculus that takes place in the argument and on making portions of it more transparent. All credit is due to San, of course, for posting the first solution. My only goal here is to explain my thought processes to show that the problem yields to a systematic attack and is therefore not exceedingly difficult.

We will prove the stronger version of the problem as presented in San’s answer. We will also relax the condition that $ A $ is unital.
As all projections have norm $ 1 $, playing around with the Triangle Inequality yields
$$
\| a a^{*} \| = \| a^{*} a \| < \frac{3}{2}.
$$
As $ a a^{*},a^{*} a \in A_{\geq 0} $, we thus have
$$
\sigma(a a^{*}),\sigma(a^{*} a) \subseteq \left[ 0,\dfrac{3}{2} \right).
$$
Sam has already shown that $ \dfrac{1}{2} \notin \sigma(a a^{*}) $ and $ \dfrac{1}{2} \notin \sigma(a^{*} a) $, under the extra assumption that $ A $ is unital. However, we do not have to require that $ A $ is unital, for once we have argued in $ A^{\sim} $ to get $ \dfrac{1}{2} \notin \sigma(a a^{*}) $ and $ \dfrac{1}{2} \notin \sigma(a^{*} a) $, we can switch back to $ A $. The remainder of the argument does not rely on the existence of a unit. However, when applying the continuous functional calculus in the absence of a unit, any function introduced must satisfy the property that its value at $ 0 $ is $ 0 $.
By a standard Banach-algebra result, we get
$$
\sigma(a a^{*}) \cup \{ 0 \} = \sigma(a^{*} a) \cup \{ 0 \}.
$$
We can thus find an $ r \in \left[ 0,\dfrac{1}{2} \right) $ and an $ s \in \left( \dfrac{1}{2},\dfrac{3}{2} \right) $ such that
$$
          \sigma(a a^{*}),\sigma(a^{*} a)
\subseteq D \stackrel{\text{df}}{=} [0,r] \cup \left[ s,\dfrac{3}{2} \right].
$$
Then the function $ f: D \to \Bbb{R} $ defined by
$$
\forall x \in D: \quad
f(x) \stackrel{\text{df}}{=}
\begin{cases}
0 & \text{if $ x \in [0,r] $}, \\
1 & \text{if $ x \in \left[ s,\dfrac{3}{2} \right] $},
\end{cases}
$$
is continuous on both $ \sigma(a a^{*}) $ and $ \sigma(a^{*} a) $. Applying the continuous functional calculus, we have well-defined projections $ f(a a^{*}) $ and $ f(a^{*} a) $, because $ f = f^{*} = f^{2} $. These are called spectral projections, and they exist for self-adjoint elements that have a gap in their spectrum.
Now, $ f(a a^{*}) $ and $ f(a^{*} a) $ are Murray-von Neumann equivalent. To prove this, we want to exploit the fact that $ f(a a^{*}) a = a f(a^{*} a) $, as already explained in this other post. Suppose that there is a positive continuous function $ g: D \to \Bbb{R} $ such that $ f(x) = x [g(x)]^{2} $ for all $ x \in D $. If we let
$$
b \stackrel{\text{df}}{=} g(a a^{*}) a,
$$
then
\begin{align}
    b b^{*}
& = [g(a a^{*}) a] [a^{*} \overline{g}(a a^{*})] \\
& = [g(a a^{*}) a] [a^{*} g(a a^{*})] \quad (\text{As $ g $ is positive.}) \\
& = g(a a^{*}) (a a^{*}) g(a a^{*}) \\
& = f(a a^{*}). \quad (\text{As $ g(x) x g(x) = f(x) $ for all $ x \in D $.})
\end{align}
Similarly,
\begin{align}
    b^{*} b
& = [a^{*} \overline{g}(a a^{*})] [g(a a^{*}) a] \\
& = [a^{*} g(a a^{*})] [g(a a^{*}) a] \quad (\text{As $ g $ is positive.}) \\
& = [g(a^{*} a) a^{*}] [a g(a^{*} a)] \\
& = g(a^{*} a) (a^{*} a) g(a^{*} a) \\
& = f(a^{*} a). \quad (\text{As $ g(x) x g(x) = f(x) $ for all $ x \in D $.})
\end{align}
Of course, we still have to find $ g $, and the most natural (in fact, only) choice is
$$
\forall x \in D: \quad
g(x) \stackrel{\text{df}}{=}
\begin{cases}
0                   & \text{if $ x \in [0,r] $}, \\
\dfrac{1}{\sqrt{x}} & \text{if $ x \in \left[ s,\dfrac{3}{2} \right] $},
\end{cases}
$$
which is the same function exhibited in San’s answer.
Observe next that $ \| p - f(a a^{*}) \| < 1 $. This is because
\begin{align}
       \| p - f(a a^{*}) \|
& \leq \| p - a a^{*} \| + \| a a^{*} - f(a a^{*}) \| \\
& \leq \| p - a a^{*} \| + \| \text{id}|_{D} - f \|_{\infty} \quad
       (\text{By the continuous functional calculus.}) \\
& <    \frac{1}{2} + \frac{1}{2} \quad
       (\text{Compare the graphs of $ \text{id} $ and $ f $.}) \\
& =    1.
\end{align}
Similarly, $ \| q - f(a^{*} a) \| < 1 $.
Projections that are within distance $ 1 $ of each other are homotopy-equivalent (this is folklore). It is also not hard to show that homotopy equivalence implies Murray-von Neumann equivalence. Therefore,
$$
p \sim f(a a^{*}) \sim f(a^{*} a) \sim q.
$$