Orthographic projection on a closed subspace $E$ of $H$, Hilbert space

Question

Let $H$ a Hilbert space and $ P: H \rightarrow H$ a linear map. We have to show that $P$ is the orthographic projection on a closed linear subspace $E$ of $H$ if and only if $P^2=P$ and $\forall x \in H, \|P(x)\| \leq \|x\|$.

I rewrote the definition. $$\forall x \in H, \exists P(x) \in E, \|x-P(x)\| = \inf_{y\in E}\|x-y\|$$ and $\forall z \in E, (z-P(x),x-P(x)) \leq 0$.

How to show it ? I'm really "lost"... In more, they write that the norm of the projection is $1$. I really need help... Thank you in advance.

Answer 1

For the other direction, suppose that $P^2 = P$ and $\|Px\| \le \|x\|$ for all $x \in H$.

The second condition immediately gives that $P$ is bounded.

Note that $x \in \operatorname{Im}P \iff Px = x$. Namely, if $x \in \operatorname{Im}P$ then $\exists v \in H$ such that $x = Pv$. Therefore $$Px = P^2v = Pv = x$$ Conversely, if $x = Px$ then clearly $x \in \operatorname{Im} P$.

We can therefore conclude $\operatorname{Im}P\, \dot+ \ker P = H$, because clearly $\operatorname{Im}P\cap\ker P = \{0\}$ and $$x = \underbrace{Px}_{\in \operatorname{Im P}} + \underbrace{x-Px}_{\in \ker P}$$

for any $x \in H$.

$\operatorname{Im}P$ is a closed subspace of $H$ because $\operatorname{Im} P = (I-P)^{-1}(\{0\})$, where $I - P$ is continuous because $P$ is. Therefore, $P$ indeed projects to $\operatorname{Im} P$, i.e. if we have the unique decomposition $x = u + v$ where $u \in \operatorname{Im} P$ and $v \in \ker P$, then $Px = u$. All that it remains to show is that $P$ is an orthogonal projection, i.e. that $\operatorname{Im}P\perp\ker P$.

For that, recall that for $u, v \in H$ we have

$$u \perp v \iff \|u\| \le \|u + \lambda v\|, \,\text{for all scalars } \lambda$$

Now take $u \in \operatorname{Im} P$ and $v \in \ker P$.

We have

$$\|u\| = \|P(u + \lambda v)\| \stackrel{\text{assumption}}{\le} \|u + \lambda v\|$$

for all scalars $\lambda$, implying $u\perp v$.

We conclude $\operatorname{Im}P\perp\ker P$ so $\operatorname{Im}P\oplus \ker P = H$. Hence, $P$ is an orthogonal projection onto the closed subspace $\operatorname{Im} P$.

Answer 2

Here is one direction:

Suppose $P$ is the orthogonal projection. Show that if $x \in E$ then $Px=x$, from the definition that you wrote. This implies $P^2 = P$. To show the other claim, write $$\|x\|^2 = \|x - P(x) + P(x)\|^2 = \|x-P(x)\|^2 + \|P(x)\|^2 + 2 (x-P(x), P(x)) \overset{?}{\ge} \|P(x)\|^2$$ where you should figure out why each step holds. Thus $\|P(x)\| \le \|x\|$ which shows the norm of $P$ is $\le 1$. To show it equals $1$, pick some nonzero element $x \in E$ and recall what $P(x)$ is.