Monotone sequence of orthogonal projections on a complex Hilbert space

Question

Suppose $P_n$ is a monotone sequence of orthogonal projections on a complex Hilbert space $\mathcal{H}$, i.e. $V_n= Im(P_n)$ is a decreasing or increasing sequence of subspaces and $P_n^\star=P_n$ and $P_n^2=P_n$ for all $n$. I want to prove that the sequence $\Vert P_n z \Vert$ is monotone for all $z\in \mathcal{H}$, but I can't quite see where to start. Somehow the $V_n$ being monotone should play into it, but I can't see how, since I can't find any obvious relations between elements of $V_n$ and $V_{n+1}$.

Answer 1

Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$.

In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$.

First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means that

$$\|P_{n+1}(z)\|^2=\|(P_{n+1}-P_n)(z)+P_n(z)\|^2={\|(P_{n+1}-P_n)(z)\|^2+\|P_n(z)\|^2}≥\|P_n(z)\|$$

And the sequence is increasing.

If $V_n$ is decreasing $(P_{n+1}-P_n)(z)$ is in $V_{n+1}^\perp$ and you get

$$\|P_n(z)\|^2=\|(P_{n}-P_{n+1})(z)+P_{n+1}(z)\|^2=\|(P_n-P_{n+1})(z)\|^2+\|P_{n+1}(z)\|^2≥\|P_{n+1}(z)\|^2$$

And the sequence is decreasing.

Answer 2

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$