I do not understand an argument (p. 58, l.2--3) regarding two "close" projections, in the proof of Theorem I.6.34, pp. 56--58, Kato's book "Perturbation Theory for Linear Operators".
The setting is in a Hilbert space $H$ (Chapter I is about finite dimensional spaces, but in the theorem below it does not seem the finiteness matters)
First, I introduce the setting.
The assumption:
Theorem 6.34. Let $P , Q$ be two orthogonal projections with $M = R(P)$, $N = R(Q)$ such that $\|(1-Q) P\|=\delta<1$.
In the proof we show
(6.55) $\ \|(1-P) Q_{0}\| \leq \delta=\|(1-Q) P\|$,
where $Q_0:H\to Im(Q|_{M})$ is an orthogonal projection.
Then, here is what I do not understand.
On p. 58 he writes
Applying the above result (6.55) to the pair $P,Q$ replaced by $Q_0,P$, we thus obtain $\left\|\left(1-Q_{0}\right) P\right\| \leqq\left\|(1-P) Q_{0}\right\|$.
But I do not know what this means. The final goal in this line is to show $\left\|\left(1-Q_{0}\right) P\right\| \leqq\left\|(1-P) Q_{0}\right\|$, so I tried to mimic the arguments to derive (6.55), but cannot figure out how.
Initially, Kato assumes that $P$ and $Q$ are orthogonal projectors satisfying $\,\|(1-Q)P\|<1$. He then proves $$\|(1-P)\,Q_0\|\:\leqslant\:\|(1-Q)P\,\|\tag{6.55}$$ where $Q_0$ is the orthogonal projector onto $N_0\!:=\,\operatorname{Im}\left(Q|_{\operatorname{Im}P}\right) \,=\,\operatorname{Im}(QP)\,$.
$(6.55)$ shows that $Q_0$ and $P$ also satisfy the initial assumptions, hence $$\|(1-Q_0)\,Z\|\:\leqslant\:\|(1-P)\,Q_0\|\tag{see page 58, line 3}$$ follows, this times with $Z$ being the orthogonal projector onto $$\operatorname{Im}\left(P|_{\operatorname{Im}Q_0}\right) \,=\,\operatorname{Im}(PQ_0)\,.$$ But in fact one has $Z=P\,$ because $\operatorname{Im}(PQ_0)=\operatorname{Im}(P)\,$. The preceding equality follows from section I.4.6 in Kato's book, which applies to the pair $P,Q_0\,$ as these are close by since $\|P-Q_0\|<1$ holds by $(6.58)$.