Do we have the range projection $r(TP)\downarrow 0$ as $P\downarrow 0$?

Question

Let $T$ be a closed (possibly unbounded) operator on a Hilbert space $\mathcal{H}$ and $P$ be a projection on $\mathcal{H}$. Do we have the range projection $r(TP)\downarrow 0$ as $P\downarrow 0$?

Answer 1

If $T$ is bounded. Then T is a linear combination of 4 unitary operators (see Page 209, Reed, Simon, Vol1 of Methods of mathematical physics. Functional analysis). Now, we consider unitary operator $U$. It is easy to see that $r(UP) = UP U^*$, which decreasing to $0$ as $P\downarrow 0$. So, $r(TP)\downarrow 0$ as $P\downarrow 0$.

Now, if $T$ is unbounded. Note that $r(E^{|T|}[n,\infty)|T|) \ge r(E^{|T|}[n,\infty)|T|P)$, we have \begin{align*} \langle r(|T| P)x,x\rangle &= \langle r(E^{|T|}[0,n)|T| P)x,x\rangle + \langle r(E^{|T|}[n,\infty)|T| P)x,x\rangle\\ &\le \langle r(E^{|T|}[0,n)|T| P)x,x\rangle + \langle r(E^{|T|}[n,\infty)|T|)x,x\rangle.\end{align*} When $n$ is large enough, the right hand side is decreasing to $0$ if $E^{|T|}[n,\infty)\downarrow 0$(I know a very large amount of operators satisfying this property). Also, for every $n$, $r(E^{|T|}[0,n)|T| P) \downarrow 0$ as $P\downarrow 0$. So, $r(|T|P)\downarrow 0$ and therefore, $r(TP)\downarrow 0$.