Show that if $\|Q\|=1$, then $Q$ is the orthogonal projection of $H$ onto $R(Q)$ (The range of $Q$)

Question

Let $H$ be a Hilbert space, $Q \in B(H)$ idempotent. Show that if $\|Q\|=1$, then $Q$ is the orthogonal projection of $H$ onto $R(Q)$ (The range of $Q$). Hint: Show that $Q$ annihilates $R(Q)$ by considering $Qy+ty$ for $y \in$ (Orthogonal Complement of $R(Q)$ and $t \in \mathbb{R}$

I've already shown that since $Q$ is idempotent, it follows that $Qy=y$ for all $y \in R(Q)$. I've also shown that since $Q$ is idempotent, $R(Q)$ is closed and $\|Q\|\geq1$ if $Q \neq 0$.

I've been working on this one for awhile and fear that I have tunnel vision.. I would really appreciate some insight. I'm sure the hint is simple enough to use but I can't quite figure it out, or how it would help in showing that $Q$ is the orthogonal projection of $H$ onto $R(Q)$.

Answer 1

Hint: Let $y$ in the orthogonal of $R(Q)$, we have to show that $Q(y)=0$.

Since $\|Q\|=1$ and $Q^2=Q$, $(t+1)^2\|Q(y)\|^2=\|Q(Q(y)+ty)\|^2\leq \|Q(y)+ty\|^2=\|Q(y)\|^2+t^2\|y\|^2$, we deduce that

$(\|Q(y)\|^2-\|y\|^2)t^2+2t\|Q(y)\|^2\leq 0$ for every $t$, this implies that $Q(y)=0$.