Given $\sigma$-algebras $G \subset F$ and a random variable $X \in L^2(\Omega,F,P)$, I aim to show the following inequality: $$ \text{Var}(X | G) \leq E[(X - \zeta)^2 | G] \text{ almost surely,} $$ for all $\zeta \in L^2(\Omega,G,P)$. Conditional variance is defined as Var$(X|G) := E[(X - E[X|G])^2 | G]$.
I have so far been able to show that Var$(X | G) = E[X^2|G] - E[X | G]^2$ and $E[(X-\zeta)^2 | G] = E(X^2| G) - 2\zeta E[X|G] + \zeta^2$ using definitions of conditional expectation. I am now trying to show that $E[X|G]^2 \geq 2\zeta E[X|G] - \zeta^2$ or some identical rearrangement, but I am stuck. Is this the right way to go about this? Is there some conditional expectation trickery which gets us closer to the answer? In addition, why is the fact that we are in $L^2$ relevant?
Conceptually I understand what is going on; essentially $E[X|G]$ is the "closest" $G$-measureable random variable to $X$, in that the variance is minimized. Any suggestions would be appreciated.
It's relevant that $X$ is in $L^2$ since otherwise, its conditional variance might not be defined.
Now, as for showing that the conditional expectation is the $G$-measurable minimizer of the conditional second moments, simply note that, using your computations,
\begin{align} \mathbb{E}[(X-\zeta)^2|G]-\mathbb{E}[(X-\mathbb{E}[X|G])^2|G] &=\zeta^2-\mathbb{E}[X|G]^2-2\mathbb{E}[X|G](\zeta-\mathbb{E}[X|G])\\ &=(\zeta-\mathbb{E}[X|G])(\zeta+\mathbb{E}[X|G]-2\mathbb{E}[X|G])\\ &= (\zeta-\mathbb{E}[X|G])^2\geq 0, \end{align} et voila.