I'm trying to solve the following question but I'm hopelessly lost. My ideas are detailed below.
For part (i), I can obviously get $$E((1- \pi(1))\pi(1)) = 0 \implies E(\pi^2(1)) = E(\pi(1)) $$ and we also know that $0 \leq E( (1-\pi(1)^2) = 1 - 2E(\pi(1)) + E(\pi^2(1)) = 1 - E(\pi(1) \implies E(\pi(1)) \leq 1$
I have absolutely no idea how to get a strict inequality on the lower bound at $0$ for part (i), which I know is equivalent to showing that $P(\pi(1) \ne 0) > 0$. I have tried everything I can think of, which is, truly disappointingly, absolutely useless. I know that since $1$ is NOT in the closure of $\mathcal{G}$, we must have that there exists $\epsilon > 0$ such that $E(1-g)^2 \ge \epsilon \quad \forall g \in \mathcal{G}$ and obviously $E(\pi(1)g) = 0$ as well.
I thought that maybe using the projection of $1$ onto the closure of $\mathcal{G}$ might be helpful, but this has also been useless for me.
I don't know how to do (ii) either, but I was thinking of trying to apply Holder's inequality - again this is (somewhat) useless: $$E(\pi(1) \pi(H))^2 \leq E(\pi^2(H)) E(\pi^2(1)) = E(\pi^2(H)) E(\pi(1)) $$ I don't know how to "get rid" of of the $\pi(1)$ in the LHS.
(iii) is also painful for me, because we know that $\pi(1) = \arg \min_{Z \in \mathcal{G}^\perp} E(Z-1)^2$ and therefore $$\text{Var}(\pi(1)/E(\pi(1)) = \frac{E(\pi^2(1)) - E(\pi(1))^2}{E(\pi(1))^2} = \frac{E(\pi(1))(1 - E(\pi(1)))}{E(\pi(1))^2} = \frac{1 - E(\pi(1))}{E(\pi(1))} \\ = \frac{E(1-\pi(1))^2}{E(\pi(1))} \leq \frac{E(Z-1)^2}{E(\pi(1))} = \text{Var(Z)}/E(\pi(1))\\ \forall Z \in \{D \in \mathcal{G}^\perp: E(D) = 1\}$$ but this does not give us what we want since $1/E(\pi(1)) \geq 1$. Please help if you can.
For part (i), you essentially said the answer yourself. Just note that $\pi_G(1) + \pi_{G^\perp}(1) = 1$, and you will see that if $\pi_{G^\perp}(1) = 0$ a.s., then $\pi_G(1) = 1$ which means $1$ was in $G$.
For part (ii), you again did the necessary calculation; just substitute $\pi(H)$ instead of $\pi(1)$ to get $E[(1 - \pi(1)) \pi(H)] = 0 $
For part (iii) you use part (ii), note that for any $D \in \mathcal{D}$, you can write the inequality in part (ii) as
$E[D^2] \geq \frac{1}{E[\pi(1)]}$ so you can get your desired inequality by subtracting 1 from both sides.