I want to prove that
If $E_1$ and $E_2$ are orthogonal projections, then $E_2 \ge E_1$ if and only if $||E_2f|| \ge ||E_1f||$ for all $f \in \mathcal{H}$ where $\mathcal{H}$ is the Hilbert space. We denote $E_2 \ge E_1$ to mean that $E_2 – E_1$ is a positive operator.
($\Rightarrow$): if $E_2 \ge E_1$, then using Cauchy Schwarz, I get
\begin{align*}
&\langle (E_2 – E_1) f, f \rangle = \langle E_2f, f\rangle – \langle E_1f, f\rangle \ge 0\\
&\langle E_2f, f\rangle \ge \langle E_1f, f\rangle\\
&||E_2f||\; ||f|| \ge \langle E_1f, f\rangle\\
\end{align*}
I'm not sure how to make the right hand side into $||E_1f||\;||f||$.
($\Leftarrow$): if $||E_2f|| \ge ||E_1f||$ then
\begin{align*}
0 &\le ||E_2f|| – ||E_1f|| \\
&\le ||E_2f – E_1f||\\
&= ||(E_2 – E_1) f||\\
&= \sqrt{\langle (E_2-E_1)f, (E_2-E_1)f\rangle}\\
&= \sqrt{\langle E_2f, E_2f \rangle – \langle E_1f, E_2f\rangle + \langle E_1f, E_1f\rangle – \langle E_2f, E_1f \rangle}\\
&= \sqrt{||E_2f||^2 + ||E_1f||^2 – \langle E_1f, E_2f\rangle- \langle E_2f, E_1f \rangle}
\end{align*}
Again, I'm not sure where to go from here.
Any suggestions or hints would be appreciated.
Best Answer
I am going to use a bit different notation. We need the following lemma:
We only prove the direction $pq=p\implies p\leq q$, this is all we are going to use.
If $p=pq$, then $p=p^*=qp$, so $p=p^2=qp^2q=qpq$. Now since $p$ is a projection we have that $0\leq p\leq1$, so $0\leq qpq\leq q1q=q^2=q$, so $p\leq q$.
Now for our problem: assume that $p\leq q$. Then, if $\xi\in H$, we have that $\|q\xi\|^2-\|p\xi\|^2=\langle q\xi,\xi\rangle-\langle p\xi,\xi\rangle=\langle(q-p)\xi,\xi\rangle\geq0$, so $\|q\xi\|\geq\|p\xi\|$ as we wanted.
Conversely, assume that $\|p\xi\|\leq\|q\xi\|$ for all $\xi\in H$. Then, if $\xi\in H$, consider $\xi'=(1-q)\xi$. Since $\|p\xi'\|\leq\|q\xi'\|$, we have that $0\leq\|p(1-q)\xi\|\leq\|q(1-q)\xi\|=\|(q-q^2)\xi\|=\|(q-q)\xi\|=0$. Since this is true for any $\xi\in H$, we have that $p(1-q)=0$, so $p=pq$. Using our lemma, we conclude that $p\leq q$.