This question arose from Durrett Theorem 4.1.4, as follows:
Conditional expectation is a contraction in $L^{p}$ space for $p\geq 1$.
This question has been addressed here: Contraction Property of Conditional Expectation, and it seems that a conditional expectation being a contraction in $L_{p}$ means that it satisfies $$\|\mathbb{E}(X|\mathcal{F}))\|_{p}\leq\|X\|_{p}.$$ but this definition seems not working in other cases. For instance, what it means by a random variable being a contraction in $L_{p}$?
From https://en.wikipedia.org/wiki/Contraction_(operator_theory), a bounded operator $T:X\longrightarrow Y$ between normed linear spaces $X$ and $Y$ means it satisfies $\|T\|_{Y}\leq 1$. So a random variable $X\in L_{p}(\Omega,\mathcal{F},\mathbb{P})$ being a contraction means it satisfies $\|X\|_{p}\leq 1$?
But from here:https://en.wikipedia.org/wiki/Contraction_mapping it seems that a map being contraction has another meaning.
Are these definitions equivalent or sort of related to each other or they are totally independent among each other?
The word contraction appears only once in Durrett, which is in this theorem. So I do need a general clarification of the definition of "contraction"
Please point it out if my question is not clear. Thank you so much!
Edit 1:
Thanks to "kimchi lover", now it is really clear. I edit this post to simply prove this statement and clarify everything for further readers.
[Contraction]. Given two normed linear space $V$ and $W$, a linear map $T:V\longrightarrow W$ is continuous if and only if there exists a real number $c$ such that $$\|T(v)\|_{W}\leq c\|v\|_{V},\ \text{for all}v\ \in V.$$
Then we define the operator norm of $T$ as $$\|T\|_{op}:=\inf\{c\geq 0:\|T(v)\|_{W}\leq c\|v\|_{V}\ \text{for all}v\ \in V\}.$$
An operator is said to be a contraction if $\|T\|_{op}\leq 1$.
[Refined Statement]. The operator of obtaining conditional expectation is a contraction in $L_{p}$ for $p\geq 1$
Proof: Let $(\Omega,\mathcal{F}_{0},\mathbb{P})$ be a probability space, $\mathcal{F}\subset\mathcal{F}_{0}$ a $\sigma-$algebra, we show that the operator $$T:L_{p}(\Omega,\mathcal{F},\mathbb{P})\longrightarrow L_{p}(\Omega,\mathcal{F},\mathbb{P})\ \text{defined by}$$ $$X\mapsto\mathbb{E}(X|\mathcal{F}),$$ is a contraction.
Firstly, the absolute value function $|\ \cdot\ |^{p}$ is convex for $p\geq 1$. Therefore, by Jensen we have $$\Big|\mathbb{E}(X|\mathcal{F})\Big|^{p}\leq\mathbb{E}(|X|^{p}|\mathcal{F}),$$ so taking expectation we have $$\mathbb{E}\Big|\mathbb{E}(X|\mathcal{F})\Big|^{p}\leq\mathbb{E}\Big(\mathbb{E}(|X|^{p}|\mathcal{F})\Big).$$
Now, $\mathcal{F}$ is a $\sigma-$algebra (in fact a sub-$\sigma-$algebra), so $\Omega\in\mathcal{F}$, then by the second criterion of conditional probability, we have $$\mathbb{E}\Big(\mathbb{E}(|X|^{p}|\mathcal{F})\Big)=\mathbb{E}|X|^{p}.$$
These together imply that $$\|\mathbb{E}(X|\mathcal{F})\|_{p}\leq\|X\|_{p},$$ which is saying that for all $X\in L_{p}(\Omega,\mathcal{F},\mathbb{P})$, $$\|T(X)\|_{p}\leq\|X\|_{p}.$$
Therefore, the operator norm of $T$, being the infimum, must be $\leq 1$. Thus, $T$ is a contraction.
Best Answer
Suppose $\mathcal F$ is a sub-algebra of $\sigma$. Then the operator $c_{\mathcal F}:L^p(\Omega,\sigma,P)\to L^p(\Omega,\sigma,P)$ given by $c_{\mathcal F}:X\mapsto E[X|\mathcal F]$ is a contraction in the usual sense.
Jensen's inequality shows $\|X\|_p \le \|c_{\mathcal F} X\|_p$, so the operator norm of $c_{\mathcal F}$ is $\le1$. But for $\mathcal F$-measurable rvs $X$ (including constant rvs) one has $c_{\mathcal F} X=X$, so the operator norm of $c_{\mathcal F}$is $1$.