TITLE: Operator $T \colon L^p \to L^p$ is a conditional expectation
QUESTION [17 upvotes]: I'm trying to solve this problem:
Let $(X,\mathcal{B},\mu)$ a probability space and $T \colon L^p(\mu) \to L^p(\mu)$ a continuous linear operator ($1 \leq p < \infty$ ) with the following properties:
1) $||T||=1$.
2) $T(1) = 1$.
3) $\forall g \in L^\infty(\mu), f \in L^p(\mu) \colon \, T(gT(f))=T(f)T(g)$.
Then exists a sub $\sigma$-algebra $\mathcal{G} \subseteq \mathcal{B}$ such that $$ \forall f \in L^p(\mu) \colon \, T(f) = \mathbb{E}(f | \mathcal{G})$$ 

My attempt: 
We can define $$\mathcal{C}= \{g \in L^\infty \colon T(g) = g\}, \ \mathcal{G} = \sigma(\mathcal{C})$$ 
By the monotone class theorem is not to difficult to see that if $g \in L^\infty$ is $\mathcal{G}$-measurable then $T(g) = g$. Then I want to prove that for every $f \in L^\infty$ $T(f) = \mathbb{E}(f|\mathcal{G})$. 

$T(f) \in \mathcal{C}$ by property 2) then $T(f)$ is $\sigma(\mathcal{C})=\mathcal{G}$-measurable.
Let $g \in L^\infty$ and $\mathcal{G}$-measurable. Then $$\int_X T(f)g d\mu = \int_X T(fg) d\mu =\cdots $$

I don't know how to procede from here.
Any help will be appreciated.

REPLY [5 votes]: One possible approach can be based on Proposition I-2-13 in Neveu's book on discrete parameter martingales (available here or here). The Remark following the proof of this proposition explains that, if $T$ is a continuous linear operator on $L^p$ satisfying condition (3) above, then $T$ is of the form $T(f) = \mathbb{E}(uf | \mathcal{G})$ where the function $u$ satisfies $u\in L^\infty(\mathcal{B},\mu)$ if $p=1$, or $\mathbb{E}(|u|^q | \mathcal{G}) \in L^\infty(\mathcal{G},\mu)$ in the case $p>1$ (where $q = \frac{p}{p-1}$). In the case $p=1$, conditions (1) and (2) above imply that $\|u\|_\infty=1$ and $\int_X u d\mu=1$ (respectively), so that $u=1$ almost surely. In the case $p>1$, conditions (1) and (2) above imply that $\|\mathbb{E}(|u|^q | \mathcal{G})\|_\infty=1$ and $\int_X u d\mu=1$, from which we can conclude that $u\in L^q (\mathcal{B},\mu)$ with $\|u\|_q \leq 1$. When $p=q=2$, considering $\int_X (u-1)^2 d\mu$ shows that $u=1$ almost surely. For $p<2$, use monotonicity of $L^p$ norms to again conclude that $u=1$ almost surely. For $p>2$, $q\in(1,2)$ and we have $\|u\|_1\geq 1$ but $\|u\|_q \leq 1$, so $\|u\|_1=\|u\|_q =1$ by norm monotonicity. Since $\int_X u d\mu=1$, this implies that $u$ is positive. Now consider $\int_X u(u^\frac{q-1}{2} -1)^2 d\mu$, concluding that $u$ is of the form $1_E$. Then the fact that $\int_X u d\mu=1$ implies that $\mu(E)=1$, so $u=1$ almost surely.
Alternatively: Proposition I-2-14 in the book states that, if $p\geq 1$ but $p \neq 2$, then every idempotent linear contraction $T$ on $L^p$ such that $T(1)=1$ is necessarily a conditional expectation. Conditions (1)-(3) above ensure that the assumptions of this Proposition are satisfied, so $T$ can therefore be written as a conditional expectation. However, only a partial proof of the Proposition is given, for the case $p=1$. A full proof can be obtained by using Theorem 1 in Ando's paper on "Contractive Projections in $L_p$ Spaces" (ref 5 in the book), which states that if $T$ is a contractive idempotent in $L^p$ ($p>1, p\neq 2$) with $T(1)=1$ then $T$ is contractive in the $L^1$ norm. This results allows you to reduce the $p>1$ case to the $p=1$ case.