Show that $\|(pqp)^{-\frac{1}{2}}pq-pq\|<1$ for projections $p,q$ with $\|p-q\|<1$ in a $C^*$-Algebra

Question

Let $p,q$ be projections in a (unital) $C^*$-Algebra with $\|p-q\|<1$.

Definition of $(pqp)^{-\frac{1}{2}}$: Let $pAp$ be the $C^*$-Algebra of elements $pap$, $a\in A$ with unit $p$. Then $pqp$ is invertible in $pAp$ because $$ \|pqp-p\| = \|p(q-p)p\| \le \|q-p\| < 1 $$ Because $pqp$ is also selfadjoint, its spectrum is real. $pqp$ is positive, because by assuming $p,q$ are projections in a Hilbert space (via an isometric representation of the $C^*$-algebra), it follows from $q\ge 0$ that $$ \langle pqp(v),v\rangle = \langle qp(v), p(v)\rangle \ge 0$$ So $(pqp)^{-\frac{1}{2}}$ ist well defined using functional calculus in $pAp$.

$\newcommand\id{\operatorname{id}}$ Claim is true for 1-dimensional projections. Let $H$ be a Hilbert space, $n_1,n_2\in H$ unit vectors and $p$ be the projection onto $\mathbb{C}n_1$, $q$ the projection onto $\mathbb{C}n_2$. Let $\alpha:=|\langle n_1,n_2\rangle|$, then $pqp|_{p(H)} = \alpha^2\id_{p(H)}$. So $$ (pqp)^{-\frac{1}{2}}|_{p(H)} = (pqp|_{p(H)})^{-\frac{1}{2}} = \alpha^{-1} \id_{p(H)} $$ So \begin{align*} \|(pqp)^{-\frac{1}{2}}pq(v)-pq(v)\| &= \|\alpha^{-1}pq(v)-pq(v)\| \\ &= |\alpha^{-1}-1|\cdot\|pq(v)\| \\ &= |\alpha^{-1}-1|\cdot\alpha\|q(v)\| \\ &\le |\alpha^{-1}-1|\cdot \alpha \|v\| \end{align*} So $\|(pqp)^{-\frac{1}{2}}pq-pq\|\le|\alpha^{-1}-1|\cdot\alpha<1$.

How can I show this for arbitrary projections in a Hilbert space?

Answer 1

We first need the following result:

Claim: if $0\leq x\leq I$ and $\|I-x\|<1$, then $\|I-x^{1/2}\|<1$.

Indeed, note first that $0\leq x\leq I$ implies $0\leq x^{1/2}\leq I$. If $\|I-x^{1/2}\|=1$, since $I-x^{1/2}\geq0$, this means that $1\in\sigma(I-x^{1/2})$, so $0\in\sigma(-x^{1/2})$ and so $x$ would not be invertible, contradicting $\|I-x\|<1$.

Now we compute: using $\|y\|^2=\|yy^*\|$, \begin{align} \|(pqp)^{-1/2}pq-pq\|^2&=\|((pqp)^{-1/2}pq-pq)(qp(pqp)^{-1/2}-qp)\|\\ \ \\ &=\|(pqp)^{-1/2}pqp(pqp)^{-1/2}+pqp-2\,\text{Re}\,pqp(pqp)^{-1/2}\|\\ \ \\ &=\|p+pqp-2\,\text{Re}\,(pqp)^{1/2}\|\\ \ \\ &=\|p+pqp-2\,(pqp)^{1/2}\|\\ \ \\ &=\|(p-(pqp)^{1/2})^2\|\\ \ \\ &=\|(p-(pqp)^{1/2})\|^2\\ \ \\ &<1 \end{align} by the Claim, since $\|p-pqp\|<1$.