Let H be a hilbert space. Let $P \in B(H,H)$, $\|P\| = 1, P^2 = P$. Then there exists unique closed subspace $L$ such that $P = P_L$.
I am thinking that $L = Im(P)$ then it is a subspace. I think the reason it is closed because P above is a open map? I am not sure why. I am not sure how to prove uniqueness either.
Let $y\in \overline L$. Then there exists a sequence $\{x_n\}$ such that $Px_n\to y$. Now, since $P $ is bounded, $$ Py=P(\lim Px_n)=\lim P^2x_n=\lim Px_n=y, $$ so $y=Py\in L$.
For the uniqueness, if $P=P_M$ then $P_L=P_M$, which implies that $M=L$. If you want to show this, let $m\in M$; then $$m=P_Mm=P_Lm\in L.$$So $M\subset L$. Similarly, $L\subset M$.