$\newcommand{\ran}{\operatorname{ran}}$
Let $p$, $q$ be projections in a $C^*$-algebra $A$. If $\|p−q\|<1$ then $p$ and $q$ are MvN equivalent.
I'm tryind to understand the following proof:
Denote $\|p-q\|=1-\epsilon$ and $x=pq$.
So $p\geq xx^*=pqp=p-p(p-q)p\geq \epsilon p$. Simetrically, $q\geq x^*x\geq \epsilon q$.
Suppose our $C^*$ algebra $A$ is contained in $B(H)$. Let $x=v \sqrt{x^*x}$ be the polar decomposition. Denote $\sqrt{x^*x}=a$.
$v\in A$ , since $a$ is invertible as an element in $qAq$ and $v=x"a^{-1}"$ where $v=vq$ and $"a^{-1}"\in qAq$.
Now, $vv^*=p$ and $v^*v=q$.
First, I don't understand why $v\in A$ and the explanation given. If $a$ was invertible in $qAq$ it was also invertible in $A$, no? Moreover, do we need to think of the unit of $qAq$ as $q$?
Second, I have a feeling that the inequalities related to $xx^* =pqp$ and $x^*x=qpq$ can tell us something about the spectrum.
For example, if $p\in C^*(1,xx^*)$ then it is possible to look at Gelfand transform and maybe conclude something about $\sigma(xx^*)$. But, I don't know if $p$ is an element of this $C^*$ sub-algebra.
Now, I just want to be sure that the explanation I gave to $vv^*=p$ is right:
From the polar decomposition we know that $vv^*$ is a projection onto $\ran(v)=\overline{\ran(x)}=\ker(x^*)^\bot$. Let's try to understand $\ker(x^*)=\ker(qp)$.
Obviously, $\ker(p) \subseteq \ker(x^*)$.
On the other hand, Let $\xi \in \ker(x^*)$: $\langle p \xi, p \xi\rangle = \langle p \xi, \xi\rangle \leq {. 1 \over {\epsilon}} \langle xx^* \xi, \xi\rangle =0$. Thus, $p \xi =0$ and $\ker(x^*)=\ker(p)$. We can conclude now $\ran(v)=\ker(p)^\bot=\ran(p)$.
$p$ and $vv^*$ projections on the same space, so they're equal : $p=vv^*$. Similarily, I've shown $v^*v=q$. I'm worried that I complicated simple things, and not sure that's the right way to understand it. Thank you for your time.
Invertible in $qAq$ is not the same as invertible in $A$. For instance, $q$ is invertible in $qAq$ (it is the unit there). That $a$ is invertible in $qAq$ means that there exists $y\in qAq$ such that $ay=ay=q$. So, you have $x=va$; then $$ v=vq=vay=xy\in A. $$
About the spectrum, in general you cannot say anything: if $A=M_2(\mathbb C)$ and $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ \ q=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix}, $$ then $\sigma(xx^*)=\{0,t\}$. But, of course, $p\not\in C^*(1,xx^*)$.
Your argument for $p=vv^*$ is correct. In general, already by the polar decomposition you know that the range of $v$ agrees with the range of $xx^*$; and by the inequality $p\geq xx^*\geq \epsilon p$, you know that the range of $xx^*$ is $p$ (because $xx^*$ is invertible in $pAp$).
Finally, here is a shorter proof (and a few others in the other answers). Note that one proves the stronger result that $p$ and $q$ are unitarily equivalent.