I'm trying to answer the next question of Kreyszig's Functional Analysis (Question 7 of section 9.6):
Let $(P_{n})$ be a monotone decreasing sequence of orthogonal projections on a Hilbert space $H.$ Suppose $P_{n}x\rightarrow Px$ for every $x\in H$ and $P$ is a projection defined on $H.$ Then $P$ projects $H$ onto $$P(H)=\bigcap_{n=1}^{\infty}P_{n}(H).$$
I'm trying to prove each set of the above equality is contained in the other.
So, because of $P_{n}\geq P_{n+1}$ for all $n\in\mathbb{N}$ then $P_{n}\geq P$ for all $n\in\mathbb{N}.$ The previous is equivalent to $P(H)\subset P_{n}(H)$ for all $n\in\mathbb{N}$ due to a theorem of equivalences in the book. Then $P(H)\subset\bigcap_{n=1}^{\infty}P_{n}(H).$
For the other direction, I was thinking in prove that $N(P)\subset N(P_{n})$ or all $n\in\mathbb{N}.$ If the behind holds then this equivalent to the desired. I'm stuck in this. I was trying using contradiction but I don't get any useful.
For the other hand, also I tried to use the hypotesis of strong convergence but I'm not sure how can use it, because if $x\in N(P)$ then $Px=0,$ and $P_{n}x\rightarrow 0,$ but is not clear that $P_{n}x\in N(P).$
Is there another way to prove this easier?
Any kind of help is thanked in advanced.
If $x\in \bigcap P_n (H) $ then for every $n,$ we have $$P_n (x) =x $$ but $P_n (x) \to P(x)$ hence $P(x)=x$ and therefore $x\in P(H).$