Let $\{e_1, ... , e_n\}$ be a finite orthonormal system in an inner product space $(E, \langle \cdot , \cdot \rangle)$, let $F :=$ span$\{e_1, ... , e_n\}$ and let $P:E \to F$ be the orthogonal projection onto $F$. Show that the following assertions hold:
a). $PPf = Pf$ for all $f \in E$.
b). If $f,g \in E$ are such that $g \in F$ and $f-g \perp F$, then $g=Pf$.
c). Each $f \in E$ has a unique representation as a sum $f=u+v,$ where $u \in F$ and $v \in F^{\perp}$.
d). If $f\in E$ is such that $f \perp F^{\perp}$, then $f\in F$. Put differently: $(F^{\perp})^{\perp} = F$
e). Let $Qf := f-Pf, \quad f \in E$. Show that $QQf=Qf$ and $\| Qf \| \leq \|f\|$ for all $f \in E$.
I am working with the following definitions:
Two elements $f,g \in E$ are called orthogonal, written $f \perp g$, if $\langle f, g \rangle=0$.
Let $I$ be any nonempty index set. A collection of vectors $(e_j)_{e \in I}$ in an inner product space $E$ is called an orthonormal system if
$\langle e_i, e_j \rangle = \delta_{ij} := \left\{ \begin{array}{lr} 1 & : i=j,\\ 0 & : i \neq j \end{array} \right. $
The mapping $P: E \to E, \quad Pf = \sum_{j=1}^{n} \langle f,e_j \rangle e_j$ is called the orthogonal projection onto the subspace $F$.
I have been trying to use the given definitions to proceed, but I find that I am getting nowhere fast. As always, any assistance is greatly appreciated. Thanks in advance.
By the way, I am using the textbook Functional Analysis An Elementary Introduction by Markus Haase.
Hints:
(a) $Pe_j=e_j$ for all $j$ implies $Pf=f$ for $f\in F$.
(b) If $(f-g) \perp F$, then $(f-g,e_j)=0$ and $P(f-g)=0$ by definition. Also use (a)
(c) $f = Pf+(f-Pf)$. Let $u=Pf$ and $v=f-Pf$. If $u,u'\in F$, $v,v' \perp F$ with $u+v=u'+v'$ then $(u-u')=(v'-v)$ is orthogonal to itself.
(d) Use (c) to decompose $f=u+v$. Show $u \perp F^{\perp}$ so $f-u = v$ is orthogonal to itself.
(e) Use (a) and Pythogorean theorem: $\|x\|^{2}=\|x-Px\|^{2}+\|Px\|^{2}$.