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(I asked this question at math stackexchange 4 months ago, but received no answers)

Let $\{e_{kj}\}$ be the canonical matrix units in $B(H)$, with $H$ separable. Define projections $q_k$ by $$ q_k=\sum_{n=1}^ke_{nn}. $$ Let $\{p_1,p_2,\ldots\}\subset B(H)$ be a sequence of orthogonal projections in $B(H)$ with the property that $q_kp_hq_k=q_kp_kq_k$ whenever $h\geq k$ (i.e. the sequence "fixes" the upper left corner as the index grows).

Question: Does the sequence $\{p_k\}$ converge strongly?

(my gut feeling is that it should, but after a while thinking about it I couldn't get neither a proof nor a counterexample; it is easy to show that the sequence converges weakly so it would be enough to prove that the limit is a projection, but I got nowhere through this route either)

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1 Answer 1

up vote 7 down vote accepted

$Q_k$ is just the orthogonal projection to the linear span $L_k$ of $e_1,\dots,e_k$, right? Let $u_m$ be some sequence of unit vectors. Let $P_mx=(x,u_m)u_m$. Now, $Q_kP_mQ_k x=(Q_kx,u_m)Q_ku_m=(x,Q_ku_m)Q_ku_m$, so the condition is that $Q_ku_m$ stabilize for each $k$. Let's stabilize them to $\frac 12e_1$ (just add some tails far away to get norm $1$). Take $v=e_1$ as the test vector for strong convergence. Then $P_m v=\frac 12 u_m$. Now, the tails are wild and carry a lot of norm, so, alas, $u_m$ do not converge to anything in norm...

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Nice example. Thanks! –  Martin Argerami Oct 26 '12 at 1:40

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