Let $H$ be an infinitely dimensional Hilbert space, $\mathcal{M}$ be a non-degenerate von-Neumann sub-algebra of $B(H)$ and suppose $w$ is a proper isometry in $\mathcal{M}$. Put $q=ww^*$ and let $p$ be a projection in $\mathcal{M}$ such that the range of $p$ is infinitely dimensional, $pq\neq 0$ and $pq\neq p$. For each $n\in\mathbb{N}$, define $q_n = w^n q (w^n)^*$ and let $q_{\infty}$ be the $SOT$-limit of $\{q_n\}_{n\in\mathbb{N}}$ (which is obviously decreasing). My question is if there exists a sufficient condition for the following to be true:
Whenever there exists a proper subsequence $\{r_n\}_{n\in\mathbb{N}}$ of $\{q_n\}_{n\in\mathbb{N}}$ such that $r_n\perp p$ for each $n$, then $q_{\infty}\perp p$.
If $q_{\infty}$ can be approached by a proper subsequence, then the conclusion holds but I believe this is stronger than the one above. In $\ell^2$, it is very easy for $q_{\infty}$, the $SOT$-limit of $\{q_n\}$, to be $0$. I believe my statement above is false in general when $H = \ell^2$ but I am also trying to construct an example where $q_{\infty}\neq 0$. Any hints or references relevant to my question will be highly appreciated.
The statement holds in general, because the sequence is monotone.
Fix $x\in H$ and $\def\e{\varepsilon}\e>0$. Then there exists $n_0$ such that $\|(q_n-q_\infty)x\|<\e$ for all $n\geq n_0$. If we write the subsequence as $\{q_{n_k}\}$, then $n_k\geq k$ for all $k$. So for $k\geq n_0$ we have $n_k\geq n_0$ and so $\|(q_{n_k}-q_\infty)x\|<\e$.
Another consequence of the sequence being monotone is that if $r_n\perp p$ for all $n$, then $q_n\perp p$ for all $n$ bigger than the index of $r_1$. Because then $$ q_np=q_nr_1p=0. $$ To construct an example with $q_\infty\ne0$, take any example and now form $\widetilde{\mathcal M}=\mathbb C\oplus\mathcal M$ and $$ \tilde{q_n}=1\oplus q_n,\qquad\qquad \tilde w=1\oplus w,\qquad\qquad\tilde p=1\oplus p. $$