$\|P\|_2 = 1$ implies $P^* = P$ I try to proof the following: let $P$ be a projector, then if the 2-norm of $P$ equals 1, $P$ is an orthogonal projector. i.e.: $\|P\|_2 = 1 \Rightarrow P = P^*$.
Using the inproduct, we know $\langle Px, Px \rangle = \langle P^*Px, Px \rangle = \langle Px, P^2x \rangle$ etc, but I can't get to $P - P^* = 0$.
 A: Use $\mathcal{H}$ to denote the whole space. First show $x\in\textrm{ran}P$ iff $x=Px$. If $x=Px$, clearly $x\in\textrm{ran}P$. If $x\in\textrm{ran}P$, then there is $y\in\mathcal{H}$ such that $x=Py$. Since $P^2=P$, $Px=P^2y=Py=x$. 
Then prove that $\mathcal{H}=\textrm{ran}P\oplus \ker P$. If $x\in\textrm{ran}P\cap\ker P$, then $x=Px$ and $Px=0$, so $\textrm{ran}P\cap\ker P=\{0\}$. For every $x\in\mathcal{H}$,
$$x=Px+(x-Px).$$
where $Px\in\textrm{ran}P$ and $x-Px\in\ker P$. So $\mathcal{H}=\textrm{ran}P\oplus \ker P$.
Then prove $\textrm{ran}P\perp \ker P$ by contradiction. Suppose there exist nonzero $x\in\textrm{ran}P$ and $y\in\ker P$ such that $\langle x,y\rangle\neq0$. Note that $x,y$ should be linearly independent, since $\textrm{ran}P\cap\ker P=\{0\}$. Let $z=\|y\|^2x-\langle x,y\rangle y$, then $z\neq0$, and we can verify the following
\begin{align*}
\|Pz\|^2=&\|\|y\|^2Px-\langle x,y\rangle Py\|^2=\|\|y\|^2x\|^2=\|x\|^2\|y\|^4,
\\
\|z\|^2=&\langle \|y\|^2x-\langle x,y\rangle y,\|y\|^2x-\langle x,y\rangle y\rangle=\|x\|^2\|y\|^4-|\langle x,y\rangle|^2\|y\|^2.
\end{align*}
So $\|Pz\|>\|z\|$, which contradicts $\|P\|=1$.
Now we can prove $P^*=P$. Since $\mathcal{H}=\textrm{ran}P\oplus \ker P$ and $\textrm{ran}P\perp \ker P$, we have
$$\langle Px,y-Py\rangle=0,\quad\textrm{ for any }x,y\in\mathcal{H},$$
which means
$$\langle Px,y\rangle=\langle Px,Py\rangle,\quad\textrm{ for any }x,y\in\mathcal{H}.$$
Exchange $x$ and $y$ (since $x,y$ are arbitrary) and get
$$\langle Py,x\rangle=\langle Py,Px\rangle,\quad\textrm{ for any }x,y\in\mathcal{H}.$$
Combining the above two, we get
$$\langle Px,y\rangle=\langle x,Py\rangle,\quad\textrm{ for any }x,y\in\mathcal{H}.$$
which means $P^*=P$.
A: On the one hand if P is a projector you know that you have $P^2=P$ and then $||P||_2\leq ||P||_2^2$ and if $P$ is non zero then $||P||\geq 1$.
If $||P||_2=1$, $Im(P)\perp \ker P$ because for $x,y$ in $Im(P)$ and $\ker(P)$ you have ($Proj_y(x)$ is the orthogonal projection of $x$ on $y$ span
$$||x||^2=||Proj_y(x)||^2+||x-Proj_y(x)||^2\geq ||Proj_y(x)||^2+||P(x-Proj_y(x))||^2$$
And because $y\in\ker(P)$ $||P(x-Proj_y(x))||^2=||P(x)||^2=||x||^2$ and finally :
$$||Proj_y(x)||^2\leq 0$$
That gives you $||Proj_y(x)||=0$ and $x\perp y$
Then $\ker(P)\perp Im(P)$. Then you can show that this imply $P=P^*$ : for all $x,y$, $x=Px+z_x$ and $y=Py+z_y$ where $z_x,z_y$ are $x-Px$ and $y-Py$.
Because of $\ker(P)\perp Im(P)$ : $P_x\perp z_x$ and the same holds for $y$. Moreover $Px\perp z_y$ and $Py\perp z_x$. Then : 
$$\langle x,Py\rangle =\langle Px+z_x,y-z_y\rangle =\langle Px,y\rangle + \langle z_x,Py\rangle-\langle Px,z_y\rangle=\langle Px,y\rangle $$
