This is an exercise I encountered in my functional analysis class.
Let $H$ be a separable Hilbert space, and we take a family of bounded, linear operators on $H$ denoted $P_{\Omega}$ where the $\Omega$ indices are subsets of the real line $\mathbb{R}$. These satisfy the following two properties
a. Each $P_{\Omega}$ is an orthogonal projection, meaning $P^2_{\Omega}=P_{\Omega}$ and $P^*_{\Omega} = P_{\Omega}$.
b. If $\Omega = \bigcup_{n=1}^{\infty} \Omega_n$ with $\Omega_n \cap \Omega_m =\emptyset $ for $m \neq n$, then we have the strong operator convergence
$$
P_{\Omega} = \text{s-lim}_{N \to \infty} \left( \sum_{n=1}^N P_{\Omega_n} \right)
$$
We are asked to prove from properties a and b that $$P_{\Omega_1}P_{\Omega_2}=P_{\Omega_1 \cap \Omega_2}$$
To give some context, these operators are supposed to model spectral projections, but the instructions of the question specifically ask us not to look at these operators as spectral projections, but only operators that satisfy the above requirements. We are supposed to think about this abstractly using general operator theory (we are supposed to solve the question not using that these operators are supposed to be spectral projections). I do not see how to do this, I do not see how the operators being orthogonal projections and the strong convergence together imply what we are supposed to prove. I would appreciate any help with this and I thank all helpers.
Best Answer
First I will prove that if $\Omega_1 \cap \Omega_2 = \emptyset$ then $P_{\Omega_1}$ and $P_{\Omega_2}$ have orthogonal ranges. Indeed, suppose $x \in \operatorname{Ran} P_{\Omega_1}$. Then by the second assumption $$P_\Omega x = (P_{\Omega_1} + P_{\Omega_2}) x = x + P_{\Omega_2}x = (1-P_{\Omega_2})x + 2P_{\Omega_2}x$$ where $\Omega = \Omega_1 \cup \Omega_2$. We then have $$\|x\|^2 \geq \|P_{\Omega}x\|^2 = \|(1-P_{\Omega_2})x + 2P_{\Omega_2}x\|^2 = \|(1-P_{\Omega_2})x\|^2 + 4 \|P_{\Omega_2}x\|^2 = \|x\|^2 + 3\|P_{\Omega_2} x\|^2$$ by exploiting orthogonality of the ranges of $(1-P_{\Omega_2})$ and $P_{\Omega_2}$. Therefore we must have $P_{\Omega_2}x = 0$ which is what we wanted to prove. Therefore we have that $P_{\Omega_1}P_{\Omega_2} = 0 = P_{\Omega_2}P_{\Omega_1}$. Then, for general $\Omega_1, \Omega_2$ we can write $$P_{\Omega_1} P_{\Omega_2} = (P_{\Omega_1 \setminus \Omega_1 \cap \Omega_2} + P_{\Omega_1 \cap \Omega_2})(P_{\Omega_2 \setminus \Omega_1 \cap \Omega_2} + P_{\Omega_1 \cap \Omega_2}) = P_{\Omega_1 \cap \Omega_2}^2 = P_{\Omega_1 \cap \Omega_2}$$ using the fact that most of the products appearing in the second term are indexed by pairs of disjoint sets and thus don't contribute to the resulting sum.