In von Neumann algebra, infimum of projections p and q is strong limit of (pq)^n

Question

Working through Vaughan Jones's Von Neumann Algebra Notes and am stuck on an exercise:

Exercise 3.4.3. [For $p, q$ projections (self-adjoint idempotent operators on a Hilbert space)] Show that $p \land q = \lim_{n \to \infty}^\text{strong} (pq)^n$.

If I understand correctly, this is equivalent to showing that where $H$ is a Hilbert space, $K_1, K_2 \subseteq H$ closed subspaces, $p$ the projection onto $K_1$, $q$ the projection onto $K_2$, and $r$ the projection onto $K_1 \cap K_2$, that for all $\xi \in H$, $$\lim_{n \to \infty} (pq)^n \xi = r\xi.$$

I'm not quite sure how to show this, although it's pretty trivial in the case that $\xi \in K_1 \cap K_2$ or $\xi \in K_1^\bot \cup K_2^\bot$.

Answer 1

It's easiest to work first with $(pqp)^n$. This sequence is easily seen to be monotone decreasing, so it has a sot limit $r$. As sot limits are multiplicative on bounded nets, one gets that $r^2=r$. So $r$ is a projection.

We have $$ rp=\lim_{\rm sot}(pqp)^np=\lim_{\rm sot}(pqp)^n=r. $$ Also, $$ r(1-q)r=\lim_{\rm sot}(pqp)^n(1-q)(pqp)^n=r-\lim_{\rm sot}(pqp)^nq(pqp)^n =r-\lim_{\rm sot}(pqp)^{2n+1}=r-r=0. $$ This implies that $(1-q)r=0$, that is $r=qr$. Thus $r=pr=qr$, so $r\leq p$ and $r\leq q$, and this gives $r\leq p\wedge q$.

Now if $x\in pH\cap qH$, we have $x=px=qx$ and so $(pqp)x=x$, which implies that $rx=x$. So $pH\cap qH\subset rH$, which says that $p\wedge q\leq r$.

So $r=p\wedge q$.

Finally, $$ \lim_{\rm sot}(pq)^n=\lim_{\rm sot}(pqp)^{n-1}q=rq=r. $$

Answer 2

This answer uses a bit more machinery than Martin Argerami's answer, but I think it is still instructive. Let $s = qpq$. It suffices to prove that $s^n\to r$ in the strong operator topology, because then for $x\in H$ we get $(pq)^nx = p(qpq)^{n-1}x\to prx=rx$ as $n\to\infty$. Note that $s$ is a positive operator of norm $\leq 1$. Let $f_n:[0,1]\to\Bbb R, f_n(t)=t^n$. Note that $f_n\to\chi_{\{1\}}$ (the indicator function of $\{1\}$) pointwise as $n\to\infty$ and $\lVert{f_n}\rVert$ is bounded. So by the strong operator continuity of the Borel functional calculus, we get $s^n=f_n(s)\to \chi_{\{1\}}(s)$. Now note that $\chi_{\{1\}}(s)$ is the spectral projection corresponding to $\{1\}\cap\sigma(s)\subset\sigma(s)$ and it is easy to see that $\ker(s-I)=K_1\cap K_2$, so $\chi_{\{1\}}(s)=P_{K_1\cap K_2}=r$ as we wanted to show.