# Operator $T \colon L^p \to L^p$ is a conditional expectation

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.

The $p=1$ case:

Let $\cal G$ be as you have defined it.

By the theorem for $L^p$ duality, one has an isometric isomorphism $\psi: L^{\infty}(\mathcal B, \mu) \to L^1(\mathcal B, \mu)^*$ given by $\psi(g)(f):=\int_X fg d\mu$. Let $T^*: L^1(\mathcal B, \mu)^* \to L^1(\cal B, \mu)^*$ denote the canonical adjoint of $T$. Define $S:= \psi^{-1} \circ T^* \circ \psi: L^{\infty}(\mathcal B, \mu) \to L^{\infty}(\mathcal B, \mu)$. Note that $\|S\| \leq 1$ since $S$ is a composition of maps with operator norm $1$. Direct computation yields that for any $f \in L^1(\mathcal B, \mu)$ and $g\in L^{\infty}(\mathcal B, \mu)$, $$\int_X (Tf)g d\mu = \int_X f (Sg) d\mu$$ Next, notice that for any $f \in L^1(\mathcal B, \mu)$ and any $g \in L^{\infty}(\mathcal G, \mu)$ $$T(f)g=T(f)T(g)=T(fT(g))=T(fg)$$ and therefore $$\int_X T(f)g d\mu = \int_X T(fg)d\mu = \int_X fg \;S(1) d\mu$$ Define $h:=S(1)$. Letting $\;f=g=1$ gives that $\int_X h\;d\mu=1$. On the other hand, since $\|S\| \leq 1$ we see that that $\| h\|_{\infty} = \|S1 \|_{\infty} \leq \|1\|_{\infty}= 1$. Since $\|h\|_{\infty} \leq 1$ and $\int_X h\; d\mu=1$, it must hold true that $h=1$ almost surely. This means that for any $f \in L^1(\mathcal B, \mu)$ and any $g \in L^{\infty}(\mathcal G, \mu)$ $$\int_X (Tf)g d\mu = \int_X fg d\mu$$ which precisely means that $Tf=E[f|\mathcal G]$, once we show that $Tf$ is $\cal G$-measurable for all $f \in L^1(\cal B, \mu)$. To show this last part, just note that if $f\in L^{\infty}(\cal B, \mu)$, then it must be true that $Tf\in \cal C$, and thus by density of $L^{\infty}(\cal B, \mu)$ in $L^1(\cal B, \mu)$ it follows that any $Tf$ is an $L^1$-limit of functions in $\cal C$, hence is $\cal G$-measurable. This completes the proof.

The $p>1$ case:

If $p>1$, then all of the above duality computations still hold true with $L^1(\cal B, \mu)$ replaced by $L^p(\cal B, \mu)$, and $L^{\infty}(\cal B, \mu)$ replaced by $L^q(\cal B, \mu)$ where $q = \frac{p}{p-1}$. However, this time, we get the bound that $\|h \|_q\leq 1$ rather than the bound $\|h\|_{\infty}\leq 1$. This makes the story a little bit more complicated than the $p=1$ case, but we will now show that it is still true that $h=1$ a.s.

Since $\int_X h \; d\mu =1$, it follows that $\| h \|_1 \geq 1$. Thus monotonicity of $L^p$ norms gives us that $1 \leq \|h \|_1 \leq \|h\|_{(q+1)/2} \leq \|h\|_q \leq 1$, and hence $\|h\|_1 = \|h\|_{(q+1)/2}=\|h\|_q = 1$. Thus we get that $$\int_X |h|\big( |h|^{(q-1)/2}-1\big)^2 d\mu = \|h\|_q^q -2 \|h\|_{(q+1)/2}^{(q+1)/2}+ \|h\|_1 = 1-2+1=0$$ This in turn implies that $|h|\big( |h|^{(q-1)/2}-1\big)^2=0$ a.s, and hence $|h|$ takes values in $\{0,1\}$ almost surely. It follows that $h=1_E-1_F$ for some disjoint $E,F \in \cal B$. Then we get that $\mu(E)-\mu(F)=\int_X h\; d\mu=1$, and hence $\mu(E)=1$ and $\mu(F)=0$. Therefore $h=1$ a.s., which shows that $Tf=E[f|\mathcal G]$ and thus completes the proof.

[Acknowledgements to @S. Caterall for considering the function $|h| \big( |h|^{(q-1)/2}-1\big)^2 .\;\;$ ]

• Regarding your deleted answer here (math.stackexchange.com/questions/1504281/…): You can fix part of the argument by noting $\|S_n\|\leq n^{1/p}$ for$1\leq p\leq2$ . This follows from orthogonality for $p=2$, by the triangle inequality for $p=1$ and by interpolation in general. I will delete this comment tomorrow. Oct 30, 2015 at 6:01
• @PhoemueX: Thanks. I will think about it later on when I have the time. Oct 30, 2015 at 7:38

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.

• Here is the cited reference: books.google.de/… Oct 27, 2015 at 15:26
• Thanks for the link, I've added it to the text. Oct 27, 2015 at 16:18
• I think that your idea of considering the function $u \big( u^{(q-1)/2}-1\big)^2$ was rather ingenious. In fact it can be used to show that if $f$ is a function defined on some probability space such that $\|f\|_1=\|f\|_q=1$ for some $q>1$, then $f$ is of the form $1_E-1_F$ a.s. I hope you don't mind, but I incorporated this idea into my answer in order to complete it. Oct 28, 2015 at 21:49
• @Shalop Thanks! and I certainly don't mind you including this idea in your answer. Oct 28, 2015 at 22:06

This is meant to be a lengthy comment and not an actual answer (my actual answer is the other answer). Namely, I found some statements in the OP to be rather non-obvious and so I wanted to give a detailed proof for them.

The first quote:

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 $$T(g) = g$$.

We will show (more generally) that $$Tg=g$$ for all $$g \in L^p(\cal G, \mu)$$, where $$\cal G$$ is as defined above. It suffices to prove this in the special case that $$g=1_E$$ for some $$E \in \cal G$$, because then we can just approximate by simple functions and use that $$T$$ is continuous.

Let $$\mathcal E:= \{E \in \mathcal G : T(1_E) = 1_E\}$$. It is easy to see that $$\cal E$$ is a Dynkin system. But $$\cal E$$ is also closed under pairwise intersections, by condition (3). Thus $$\cal E$$ is a $$\sigma$$-algebra. Thus in order to show that $$\cal E = G$$, we just need to show that $$\cal E$$ contains a generating set for $$\cal G$$, i.e, all sets of the form $$f^{-1}(B)$$ with $$f \in \cal C$$ and $$B \subset \Bbb R$$ Borel.

Thus we just need to show that $$T(1_{f^{-1}(B)}) = 1_{f^{-1}(B)}$$ for all $$f \in \cal C$$ and all Borel $$B \subset \Bbb R$$. We will prove something stronger, namely, that if $$f \in \cal C$$ ,then for any bounded Borel measurable function $$h: \Bbb R \to \Bbb R$$, it holds true that $$T(h \circ f) = h \circ f$$.

So let $$f \in \cal C$$, let $$h$$ be as stated, and let $$\epsilon>0$$. Since $$f \in L^{\infty}(\cal B, \mu)$$, there exists some compact interval $$J$$ such that $$\mu(f \notin J)=0$$. Since continuous functions are dense in $$L^p(J, \;f_*\mu)$$, it follows that there exists a continuous function $$h_1$$ on $$J$$ such that $$\| h-h_1 \|_{L^p(f_*\mu)}<\epsilon/4$$. Similarly, there exists a polynomial $$P$$ on $$J$$ such that $$\| h_1-P \|_{L^p(f_*\mu)}<\epsilon/4$$. It follows that $$\| h \circ f - P \circ f \|_{L^p(\mu)} = \|h-P \|_{L^p(f_*\mu)} < \epsilon/2$$. Since $$T$$ has operator norm $$1$$, it follows that $$\| T(h \circ f) - T(P \circ f) \|_{L^p(\mu)} < \epsilon/2$$. But since $$P$$ was assumed to be a polynomial, it follows that $$T(P \circ f) = P \circ f$$, simply by repeatedly applying (3) and using the fact that $$f \in \cal C$$. Hence we see that $$\| T(h \circ f) - h \circ f \|_{L^p(\mu)} \leq \| T(h \circ f) - T(P \circ f) \|_{L^p(\mu)} + \| P \circ f - h \circ f \|_{L^p(\mu)} < \epsilon/2+\epsilon/2=\epsilon$$ Letting $$\epsilon \downarrow 0$$, we get that $$T(h \circ f) = h \circ f$$.

Another nontrivial quote is the following:

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).

A priori, it is not necessarily true that if $$f \in L^{\infty}(\cal B, \mu)$$ then $$Tf \in L^{\infty}(\cal B, \mu)$$. We will prove this.

Let $$f \in L^{\infty}(\cal B, \mu)$$. Define a sequence as $$f_1:=f$$ and $$f_{n+1}=f\cdot Tf_n$$. It is clear that $$f_n \in L^p(\cal B, \mu)$$ for all $$n$$. Moreover, we get the relation that $$\|f_{n+1}\|_p = \|f \cdot Tf_n \|_p \leq \|f\|_{\infty} \cdot \|Tf_n\|_p \leq \|f\|_{\infty} \cdot \|f_n\|_p$$. Thus by induction we see that $$\| f_n \|_p \leq \|f\|_{\infty}^{n-1} \cdot \|f_1\|_p \leq \|f\|_{\infty}^n$$.

But it is also true that $$Tf_{n+1}=T(f\cdot Tf_n) = Tf \cdot Tf_n$$, and thus by induction we see that $$Tf_n = (Tf)^n$$ for all $$n$$. Hence we see that for all $$n \in \Bbb N$$: $$\|Tf\|_{np}^n = \|(Tf)^n\|_p = \|Tf_n\|_p \leq \|f_n\|_p \leq \|f\|_{\infty}^n$$ Taking $$n^{th}$$ roots, we see that $$\|Tf\|_{np}\leq \|f\|_{\infty}$$ for all $$n$$. Letting $$n \to \infty$$, we get that $$\|Tf\|_{\infty} \leq \|f\|_{\infty} < \infty$$, so that $$Tf \in L^{\infty}(\cal B, \mu)$$.

• Why did you change $L^ \infty$ by $L^ p$ in the OP's quote? Oct 26, 2015 at 13:55
• Oh yeah. Fixed it. Oct 26, 2015 at 15:59
• An alternative (probably closer to what the OP intended) is to use Dynkins multiplicative system theorem about which I recently learned here (math.stackexchange.com/a/47521/151552). Let $E=\{g \in L^\infty \mid Tg=g\}$ and note that this contains the constants, is a vector space and closed under "bounded pointwise convergence". Finally, $E$ contains the multiplicative (by property (3)) system $\mathcal{C}$, so that the theorem above implies that $E$ contains all $\sigma(\mathcal{C})$ measurable bounded functions. Lifting to unbounded functions is straightforward. Oct 26, 2015 at 21:44
• @PhoemueX: That's really cool. I didn't know about it, and it does simplify the argument. On an unrelated note, I wonder if it can be used to show that characteristic functions uniquely determine the distribution of a r.v? Oct 26, 2015 at 22:02
• @PhoemueX: Just to be clear, the argument given in the above proof (i.e, polynomial approximation of measurable functions) is basically the same argument used to prove Dynkin's multiplicative system theorem, so really they are the same proof disguised as different ones. Nevertheless it's good to have the most abstract form of the theorem at one's disposal. Oct 26, 2015 at 22:55