Contraction Property of Conditional Expectation Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space with $\Omega$ finite. From Jensen's inequality
$$\big|\; \mathbb{E}[X|\mathcal{G}] \;\big|^{p} \;\leq\; \mathbb{E}\big[\, |X|^{p}\,|\,\mathcal{G}\,\big] \quad\mathbb{P}\text{-a.s.}$$
for $p\geq 1$, $\mathcal{G}\subset\mathcal{F}$ and $X:\Omega\rightarrow\mathbb{R}$. By taking expectations, this directly gives the famous contraction property
$$\big\|\; \mathbb{E}[X|\mathcal{G}] \;\big\|_{p} \;\leq\; \|\; X \;\|_{p}$$
for $p\geq 1$, $\mathcal{G}\subset\mathcal{F}$ and $X:\Omega\rightarrow\mathbb{R}$, where $\|X\|_{p}:=\mathbb{E}\big[\,|X|^{p}\,\big]$.
Is it possible to generalize this result to arbitrary norms $\|\cdot\|$ on $\mathbb{R}^{d}$, where $d=|\Omega|$? Can somebody contruct a counter example?
Thanks!

No. Conditional expectation may be thought of as an orthogonal projection (to the space of $\mathcal G$-measurable variables). And while orthogonal projection is a contraction with respect to the Euclidean norm, it may increase other norms.

Here is a particular example. Let  $\mathrm{P}(\omega_i) = 1/d$, $i=1,\dots,d$. Take $||X|| = \sum_{k=1}^d a_k|X(\omega_k)|$, where $0<a_1<a_2<\dots<a_d$. Then for $X = \mathbf{1}_{\omega_1}$ and trivial $\mathcal G$ we have $\mathrm{E}[X\mid\mathcal G]\equiv 1/d$, so the inequality you're interested in would read
$$
\frac1d \sum_{k=1}^d a_k \le a_1, 
$$
which is false.
