Property about orthogonal projections Let $H$ be a Hilbert space and $P_1,P_2$ orthogonal projections on the closed spaces $M_1$ and $M_2$.
I want to show that $M_1\subset M_2$ implies $\|P_1x\|\leq \|P_2x\|$
Some observations:
$M_2^\perp \subset M_1^\perp$, $P_1P_2=P_2P_1$
can someone give me a tip?
 A: The answer posted by Pp.. uses techniques worth knowning about, but there's a simpler way in this particular instance.  The point $P_1(x)$ is the point that minimizes $y\mapsto \|y-x\|$ subject to the constraint that $y\in M_1$.  If $M_2 \supseteq M_1$, then minimizing subject to the condition that $y\in M_2$ is minimizing with respect to a weaker condition; hence you get at least as small a minimum.  The idea is that there may or may not be some point in $y\in M_2$ for which $\|y-x\|$ is smaller than it is for all $y\in M_1$.  If there is none, then the minimum is what it was when one was restricted to $M_1$; otherwise it's smaller.  It can't get bigger.
A: Get an orthonomal basis $\{v_k\}_{k\in I}$ of $M_1$, extend it to an orthonormal basis $\{v_k\}_{k\in J}$, with $I\subset J$, of $M_2$, and express the projections in those bases.
You get $$P_1(x)=\sum_{k\in I}(x,v_k)v_k$$ and $$P_2(x)=\sum_{k\in J}(x,v_k)v_k$$
Compute their norms now.
$$\begin{align}\|P_1(x)\|^2&=\sum_{k\in I}|(x,v_k)|^2\\&\leq\sum_{k\in J}|(x,v_k)|^2=\|P_2(x)\|^2\end{align}$$
