Idempotent linear operators are projections

Question

I'm working on this problem:

Suppose $V$ is a Hilbert space and $P : V \to V$ is a linear map such that $P^2= P$ and $\Vert Pf \Vert \leq \Vert f \Vert$ for every $f \in V$. Prove that there exists a closed subspace $U$ of $V$ such that $P = P_U$.

I did do some searching online and I know that $P$ is an idempotent matrix and I know that $U$ is supposed to be $\text{range}P$, which I can show is closed. However, I'm having a hard time showing $P_{\text{range}P} = P$. I know that I can decompose $f = g + h$, where $g \in \text{range}P$ and $h$ is in the orthogonal complement of $\text{range}P$. I'm not sure how to proceed.

Most of the online searches require that $P$ be self-adjoint, but this question doesn't have this condition.

Can I get some help?

Answer 1

Let $U$ be the image of $P$ and let $W$ be the image of $1-P$. It suffices to show that $U$ and $W$ are orthogonal to each other, since then $$v=Pv+(1-P)v$$ must be the orthogonal decomposition of $v$ with respect to $U$ and $W$ and so $P$ is the projection onto $U$.

So, suppose $U$ and $W$ are not orthogonal; say $u\in U$ and $w\in W$ with $\langle u,w\rangle\neq 0$. The idea is to now consider linear combinations of $u$ and $w$: if $v=au+w$ for some scalar $a$, then $$\|v\|^2=|a|^2\|u\|^2+\|w\|^2+2\operatorname{Re}(a\langle u,w\rangle)$$ and $$\|Pv\|^2=|a|^2\|u\|^2$$ (here we use the fact that $P(1-P)=0$ so $Pw=0$ since $w\in W$). To get a contradiction, you now just have to choose $a$ such that $2\operatorname{Re}(a\langle u,w\rangle)<-\|w\|^2$, which is possible since $\langle u,w\rangle\neq 0$.

Answer 2

We have $U = range(P) = \{x : Px = x\} = \{x : Px - x = 0\}$ which is clearly closed and a subspace. Recall that we may view $V$ as the direct sum of $U$ and $U^\bot$. Thus, it suffices to check that $P$ acts on both $U$ and $U^\bot$ as a projection operator. Clearly, $Px = P_U x$ whenever $x \in U$. Suppose now that we have $x \in U^\bot$ and that $Px \neq 0$. Let $w$ be the norm of $Px.$ Then consider positive $\epsilon$; then $P(\epsilon x + Px) = (1 + \epsilon) Px$, with a norm of $w (1 + \epsilon) \leq \sqrt{\epsilon^2 |x|^2 + w^2}$. But clearly for sufficiently small $\epsilon$, this is impossible. Then $Px = 0$.