Prove the norm of operator derived from orthogonal projections is less or equal to 1 Suppose $H$ is a Hilbert space. We have two orthogonal projections $P_1: H\rightarrow A_1$ and $P_2: H\rightarrow A_2$. Here $A_1$ and $A_2$ are two closed subspace of $H$. Prove that 
$$
\lVert P_1 - P_2\rVert \le 1
$$
Any hints may help. Thank you.
 A: HINT: 
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $\|P-\frac{1}{2}I\|=\frac{1}{2}$ or $\|2P-I\|=1$, where $I$ is the identity operator on $H$.  To show this, note that $2P-I$ is a unitary.
A: For any $x\in H$ with $\|x\|=1$, we have for an orthogonal projection $P$ $$\langle Px,x\rangle=\langle P^*Px,x\rangle=\langle Px,Px\rangle=\|Px\|^2\leq \|Px\|^2+\|(I-P)x\|^2=\|x\|^2=1.$$
So 
$$
\langle (P_1-P_2)x,x\rangle=\langle P_1x,x\rangle-\langle P_2x,x\rangle\in[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$. 
For any selfadjoint operator $T$, we have $\|T\|=\sup\{|\langle Tx,x\rangle:\ \|x\|=1\}$. So 
$$
\|P_1-P_2\|=\sup\{|\langle (P_1-P_2)x,x\rangle|:\ \|x\|=1\}\leq1. 
$$
A: You can write $A_1 = B_1\oplus (A_1\cap A_2)$ and $A_2=B_2\oplus(A_1\cap A_2)$ where $\oplus$ is an orthogonal decomposition. Then
$$
     P_{A_1}=P_{B_1}+P_{A_1\cap A_2}\\
     P_{A_2}=P_{B_2}+P_{A_1\cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
     P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1\perp B_2$, it is easy to see that $\|P_{B_1}-P_{B_2}\|\le 1$ because
$$
    \|P_{B_1}x-P_{B_2}x\|^2=\|P_{B_1}x\|^2+\|P_{B_1}x\|^2=\|P_{B_1\oplus B_2}x\|^2\le \|x\|^2.
$$
