Property of conditional expectation operator in $L^1$.

Question

Let $\mathscr{G}\subset\mathscr{F}$ be two $\sigma$-algebras. It's easy to see that the conditional expectation operator $$E[\,\cdot \mid\mathscr{G}]\in \mathscr{L}(L^1(\mathscr{F}))$$ satisfies $\|E[\,\cdot \mid\mathscr{G}]\|_{\mathscr{L}(L^1)}\le 1$. Then for $f\in L^1(\mathscr{F})$, if $$\|E[f \mid\mathscr{G}]\|_{L^1}=||f||_{L^1}$$ do we have $f\in L^1(\mathscr{G})$? What about in $L^p(\mathscr{F})$, for $p\neq 2$?

Answer 1

If $f$ is integrable and non-negative, we have $$\left\lVert\mathbb E\left[f\mid\mathcal G\right]\right\rVert_1=\mathbb E\left[\mathbb E\left[f\mid\mathcal G\right]\right]=\mathbb E\left[f\right]=\lVert f\rVert_1.$$ In general, if $\left\lVert\mathbb E\left[f\mid\mathcal G\right]\right\rVert_1=\lVert f\rVert_1$, then we can say that $$\lVert f\rVert_1 =\left\lVert\mathbb E\left[f\mid\mathcal G\right]\right\rVert_1 \leqslant \left\lVert\mathbb E\left[|f|\mid\mathcal G\right]\right\rVert_1=\lVert f\rVert_1.$$ Since $\left|\mathbb E\left[f\mid\mathcal G\right]\right|\leqslant \mathbb E\left[|f|\mid\mathcal G\right]$, we deduce (from the fact $g\geqslant 0, \int g=0\Rightarrow g=0$ a.s.) that $$\left|\mathbb E\left[f\mid\mathcal G\right]\right|= \mathbb E\left[|f|\mid\mathcal G\right].$$ This does not mean that $f$ is non negative or $\mathcal G$-measurable. But this holds if $f=hg$, where $h$ is non-negative and independent of $\mathcal G$ and $g$ is $\mathcal G$-measurable.