Sequences of orthogonal projection in bounded Hilbert space

functional-analysishilbert-spacesprojection

Let {$P_i$} $\subset$ B(H) (bounded Hilbert space) be the projections. Suppose that:

  1. If $P_1<P_2<…$ and there exists a projection P $\subset$ B(H)such that $\lim_{i \to \infty}P_i\zeta=P\zeta$ for all $\zeta \in H$. In this case if we set $M=\overline{[\bigcup P_iH]}$, then $P=P_M$

  2. If $P_1>P_2>…$ and there exists a projection P $\subset$ B(H)such that $\lim_{i \to \infty}P_i\zeta=P\zeta$ for all $\zeta \in H$. In this case if we set $M=\overline{[\bigcap P_iH]}$, then $P=P_M$

I have no idea how to proceed on this problem, so I hope to receive some help. Any ideas or clues are appreciated. Thank you.

Best Answer

I will do that for an increasing/ decreasing net of projections. For this approach one needs to know that for $p,q$ projections on a Hilbert space we have that $p\leq q$ iff $pq=p$ iff $p(H)\subset q(H)$.

Since $p_\lambda(H)\subset M:=\overline{\bigcup_\lambda p_\lambda(H)}$ (it is evident that $M$ is a closed subspace), if $p$ is the projection of $H$ onto $M$, we have that $p$ is an upper bound for the increasing net $(p_\lambda)$, i.e. $p_\lambda\leq p$ for all $\lambda$. By Vigier's theorem, $(p_\lambda)$ converges strongly to a projection $q$ that also satisfies $q\leq p$, since $p$ is an upper bound for the net $(p_\lambda)$ (the strong limit obtained by Vigier's theorem is the least upper bound). We will show that $q=p$, so it suffices to show that $p\leq q$. But if $q$ is the projection onto the subspace $N$, then we have that $p_\lambda(H)\subset N$ for all $\lambda$, thus $\bigcup_\lambda p_\lambda(H)\subset N$ and since $N$ is closed we have $M\subset N$, hence $p\leq q$.

Let $M:=\bigcap_\lambda p_\lambda(H)$ and $p$ be the projection onto $M$. Then since $M\subset p_\lambda(H)$ for all $\lambda$ it is $p\leq p_\lambda$ for all $\lambda$. Thus $(p_\lambda)$ is decreasing and bounded below, so, by Vigier's theorem, it converges strongly to some $q\in B(H)$ and as we saw $q$ is a projection and it also satisfies $p\leq q$. To show that $q\leq p$, note that if $q$ is the projection onto $N$, then since $q\leq p_\lambda$ for all $\lambda$ we have $N\subset p_\lambda(H)$ for all $\lambda$, ths $N\leq\bigcap_\lambda p_\lambda(H)=M$, so, $q\leq p$. So it is $q=p$ and we are done.

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