A linear operator $P: H \to H$ on a Hilbert space $H$ is self-adjoint if for every $f, g \in H$, $\langle Pf, g\rangle=\langle f,Pg\rangle$ and is idempotent if for every $f \in H$, $P(P(f))=P(f)$ (i.e. $P^2=P$). Show that if $P$ is self-adjoint and idempotent then it is continuous with $\|P\|= 0$ or $1$ and that $Pf \perp (f-Pf)$ for any $f \in H$.
I'm not sure on things from Hilbert space other that I need = but I don't how to start or do this.
In situations like this (i.e., basic questions) it often pays just writing the definitions. To check that $P$ is bounded, you need to find a constant $c$ such that $\|Px\|\leq c\|x\|$ for all $x$. So we write $\|Px\|$ and we use what we know about $P$: $$ \|Px\|^2=\langle Px,Px\rangle=\langle P^*Px,x\rangle=\langle P^2x,x\rangle=\langle Px,x\rangle\leq \|Px\|\,\|x\|. $$ If $\|Px\|\ne0$, you get $\|Px\|\leq \|x\|$, showing that $P$ is bounded and $\|P\|\leq1$. For any $x$ in the range of $P$, you have $x=Py$ for some $y$, so $$ Px=P^2y=Py=x. $$ Then $\|Px\|=\|x\|$ for such $x$, which shows that $\|P\|\geq1$. You can also get $\|P\|\geq1$ from $\|P\|=\|P^2\|\leq\|P\|^2$.
To show the orthogonality condition, again you just write what you need: for two things to be orthogonal, you need their inner product to be zero. Now $$ \langle Px,y-Py\rangle=\langle P^2x,y-Py\rangle=\langle Px,P^*(y-Py)\rangle=\langle Px,P(y-Py)\rangle=0. $$