TITLE: A "Crookedness criterion" for a pair of orthogonal projectors?
QUESTION [3 upvotes]: If $P$ is an orthogonal projector on a Hilbert space $H$, then
$\,\operatorname{im}P=(\ker P)^\perp\subset H\,$ is a closed subspace, also called the support of $P$. And vice versa: Every closed subspace determines a  unique orthogonal projector, having the given subspace as support.
Let $Q$ be another orthogonal projector. Then we have

$\operatorname{im}P\perp\operatorname{im}Q\iff PQ=0\,,$
$\operatorname{im}P\subset\operatorname{im}Q\iff PQ=P\,.$

Both points are not hard to prove and are found in many textbooks.
My question is, if these can be condensed into one criterion as follows:

Does the equivalence
  $$\operatorname{Im}P\cap\operatorname{Im}Q=\{0\}\iff\|PQ\|<1\tag{1}$$
  hold?
  How can it be proved/disproved?

Note that the RHS condition of $(1)$ is also symmetric since 
$\|PQ\|=\|(QP)^*\|=\|QP\|$
by the orthogonality of the projectors and the isometry of the involution.
I did some textbook and online search but only came up with the "extreme" cases $\|\cdot\|\in\{0,1\}$ corresponding to the explicit points as of above.
Starting from the RHS and using as ansatz the definition of the operator norm was tried, but I couldn't see how to get through ...

The validity of $(1)$ would permit to define an angle between the subspaces  by $\,\cos\gamma= \|PQ\|\,$, thus measuring the "crookedness" between the projectors or their associated subspaces, respectively.
If $\,P=(\,\cdot\,| u_P)\,u_P\,$ and $\,Q=(\,\cdot\,|u_Q)\,u_Q\,$ are $1$-dimensional orthogonal projectors onto $\operatorname{span}\{u_P\}$ and $\operatorname{span}\{u_Q\}$, respectively, then
$$\|QP\| = \|PQ\| = |(u_P|u_Q)|$$
which fits well with a possible interpretation of $\|PQ\|$ as $\,\cos\gamma$.

REPLY [2 votes]: I think I found a counterexample in $\ell^2$.
Let $A:=\{x\in\ell^2 : x_{2n} = n x_{2n+1}\;\forall n\in\mathbb N\}$
and let $B:=\{x\in\ell^2 : x_{2n+1}=0\;\forall n\in\mathbb N\}$.
Then it can be seen that both $A$ and $B$ are closed.
Let $P$ be the orthogonal projector on $A$, and $Q$ the orthogonal projector on $B$.
Clearly, $\operatorname{im} P \cap \operatorname{im} Q=\{0\}$.
Let $n\in \mathbb N$. We consider the vector $e_{2n}\in\ell^2$.
It can be shown that
$$
Qe_{2n} =e_{2n}
$$
and
$$
Pe_{2n} =(1+\frac1{n^2})^{-1}\left( e_{2n} + \frac1n e_{2n+1}\right).
$$
Calculating the norm of the latter yields
$$
\|PQe_{2n}\|=\|Pe_{2n}\|=(1+\frac1{n^2})^{-1/2}\to 1
\qquad (n\to\infty)
$$
Thus $\|PQ\|=1$.
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