I'm using Dirac notation here.
Given only that a projection operator is defined by the property $P=P^2$, prove that $P$ is a positive operator on the Hilbert Space, i.e. $ \langle v|P|v\rangle \geq 0 \quad \forall\ |v\rangle \in H$.
I'm attempting to prove this using contradiction:
Let there be a vector $|v\rangle$ such that $\langle v| P |v \rangle < 0$ $$ \begin{aligned} \langle v| P |v \rangle &= \langle v| PP |v\rangle \\ &= \langle P^\dagger |v\rangle, P|v\rangle\rangle \end{aligned} $$
If I can prove that $P$ is Hermitian, then that would lead to the contradiction, but I'm not sure how to get there.
This is false. Consider $L^{2}([0,1])$. Take any function $g$ in this space such that $\int_E g=1$ but $\int_0^{1} g<0$ where $E=(0,\frac 1 2)$. Let $Pf=(\int_E f) g$. I will let you verify that $P^{2}=P$ but $ \langle Pf, f \rangle<0$ when $f$ is the constant function $1$.