Norm estimate for a product of two orthogonal projectors Let $H$ denote a Hilbert space.
Consider two orthogonal projectors $\,P,Q\in\mathscr L(H)\,$ such that $H=\operatorname{Im}P\oplus\operatorname{Im}Q\,,$ that is


*

*both $\,\operatorname{Im}Q\,$ and $\,\operatorname{Im}P\,$ are closed subspaces of $H$,

*$\operatorname{Im}P+\operatorname{Im}Q=H$,

*$\{0\}=\operatorname{Im}P\cap\operatorname{Im}Q\,$.


It is not assumed that $\operatorname{Im}P\perp\operatorname{Im}Q$, or   equivalently $P+Q=\mathbb 1$.

Is it true then that $\|PQ\|<1$ ?

Note that $\|PQ\|=\|QP\|$ as the involution is isometric.
This is a follow-up  question to 
A "Crookedness criterion" for a pair of orthogonal projectors? .
Its answer shows that it is necessary to assume that 
$\,\operatorname{Im}P+\operatorname{Im}Q\,$ is closed in $H$.
 A: The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=\operatorname{Im}P\oplus\operatorname{Im}Q\,,$
let $Z\in\mathscr L(H)\,$ denote the projector onto the second summand. In general, this projection onto $\operatorname{Im}Q\,$ along $\operatorname{Im}P\,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $\operatorname{Im}Q\,$ nor $\operatorname{Im}P\,$ equals $\{0\}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$\|PQ\|\:\leqslant\:\sqrt{1-\|Z\|^{-2}}\;.$$
Proof:$\ $ If $x\in\operatorname{Im}Q\,$ is a unit vector, then $\,Zx=x\,$ and $\,ZPx=0$. Thus,
$$1=\|Zx-ZPx\|\:\leqslant\:\|Z\|\: \|(1-P)x\|\\[2ex]
\Longrightarrow\quad\frac1{\|Z\|^2} \:\leqslant\: \big\langle(1-P)x\mid (1-P)x\big\rangle
 = \big\langle(1-P)x\mid x\big\rangle
 = 1-\|Px\|^2\qquad\text{or}\quad\|Px\|\;\leqslant\; \sqrt{1-\|Z\|^{-2}}$$
The claim now follows from
$$\|PQ\|\:=\:\sup_{u\in H,\,\|u\|\leq 1}\|PQu\| 
 \:=\:\sup_{x\in\operatorname{Im}Q,\,\|x\|\leqslant 1}\|Px\|\:.$$
Remark:$\ $ Every non-zero projector satisfies $\|Z\|\geqslant1$. 
And $\|Z\|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
