I have the following problem. Upon staring at it, I think I've developed an intuition of how it should go but am unsure of the exact steps to follow.
Problem: Let $X\in L_2(\Omega, F,\mathbb{P})$ and $Y\in L_2(\Omega, G, \mathbb{P})$ where $G$ is a $\sigma$-field $\subseteq F$. Suppose that $Y=\mathbb{E}(X|G)$ and that $\lVert X\rVert _2 = \lVert Y \rVert_2$. Prove that in fact $X=Y$ almost surely.
Intuition: Through conditioning, $Y$ acts as a projection of $X$ onto the $G$ subspace. In laymen's terms, such a projection "reduces" or "coarsens" the original random variable, and hence we should expect its $L_2$ norm to decrease. However, that the $L_2$ norm remains the same shows that conditioning on $G$ has no effect and therefore $X=Y$ almost surely. Also, the $L_2$ norm amplifies the "coarsening" more than the $L_1$ norm, so it is possible that two random variables are equivalent in the latter but not in the former.
P.S. Apologies in advance if the title is wrong, as it is currently how I understand the problem. I will probably edit it to something more suitable later. Let me know if there is any better way to clarify what I've written above and I'll be glad to accomodate.
I'll give you two ways of seeing this, one from basic definitions and a faster way. $\|X\|_2 = \|Y\|_2$ is equivalent to $E[X^2] = E[Y^2]$. But one of the defining properties of the conditional expectation (well, I suppose that this depends on your prefered text) is that for any $Z \in L_2(\Omega, G, P)$, $Y = E[X \mid G]$ is the (up to null) unique random variable $L_2(\Omega, G , P)$ such that $$ E[XZ] = E[YZ]. $$ In particular, $$ E[XY] = E[Y^2]. $$ On the other hand, $Y = E[X \mid G]$ also minimizes $$ E[(X - Y)^2] = E[X^2 + Y^2 - 2XY] $$ so if $E[X^2] = E[E[X \mid G]^2]$, then $E[(X - Y)^2] = 0$ is possible, and thus it must be that $X = Y$ almost surely.
Alternatively, $$E[Y^2] = E[E[X \mid G]^2] \leq E[E[X^2 \mid G]] = E[X^2]$$ where we use Jensen and then the tower property; conditional Jensen's is an equality in this case only if $X$ is already $G$-measurable or degenerate, in which case $Y = X$ almost surely.