Show that with $||e||=1$ and $F=\{e\}^\perp$ then $P_Fh=h - \langle h,e\rangle e $

Question

I am trying to solve the following homework: Let $e \in \mathcal{H}$ and $||e||=1$ and $F=\{e\}^\perp$

Show that F is a closed subspace and that $P_Fh=h - \langle h,e\rangle e $

The closedness seems be that because $e \subset \mathcal{H}$ then $e^\perp$ is closed due to some theorem in the textbook.

However for the projection I tried: $ \langle h,e\rangle e=h -P_Fh \in span(\{e\})$ because as I understand $\langle h,e\rangle e \in F^\perp=span(\{e\})$

This just doesn't seem like a solid proof. Furthermore it doesn't mention that $e$ is basis vector nor do I use that $||e||=1$

Any hint would be appreciated

Answer 1

I suppose $P_Fh$ is defined as the closest point in $F$ to $h$, which is well defined (infimum of distances) since $F$ is closed, by earlier part. Set $$ v = h - P_Fh$$ Suppose $v\not\in F^\perp$, then $c:=\langle v , e \rangle \neq 0$. Without loss, $c>0$. Now try to show that for $t\ll 1$, $P_F h + te $ is closer to $h$ than $P_F h$.

Bigger hint (Don't mouseover until you've tried)

$$ \|h - P_F h - t e \|^2 = \| v - t e\|^2 = \|v\|^2 - 2 ct+ t^2 \|e\|^2 $$

PS The term $h - \langle h,e\rangle e$ is not invariant in $\|e\|$. So $\|e\|=1$ is used to simplify this, otherwise you have to write $ h - \frac{\langle h,e\rangle e }{\|e\|^2 }$.