Let $(\Omega,\mathcal F,\mu)$ be a probability space. It is well-known that if $\mathcal A$ is a sub-$\sigma$-algebra of $\mathcal F$, $p\geqslant 1$ and $X$ is an element of $\mathbb L^p$ which takes real values, then
$$\tag{1} \lVert\mathbb E[X\mid\mathcal A]\rVert_p\leqslant \lVert X\rVert_p.$$
We now consider for $p\gt1$ the weak $\mathbb L^p$-space, denoted by $\mathbb L^{p,\infty}$, which consists of all the real valued random variables such that
$\lVert X\rVert_{p,\infty}^*:= \left(\sup_{t\gt 0 }t^p\mu\{|X|\gt t\}\right)^{1/p} $ is finite. It is known that $\lVert \cdot \rVert_{p,\infty}^*$ is not a norm because the triangle inequality fails. However, defining $$\lVert X\rVert_{p,\infty}:=\sup\left\{\mu(A)^{1/p-1}\mathbb E[| X|\cdot\mathbb 1_{A} ], A\in\mathcal F,\mu(A)\gt 0 \right\},$$
then $\lVert X\rVert_{p,\infty}$ is a norm and
$$\tag{2} \lVert X\rVert_{p,\infty}^*\leqslant \lVert X\rVert_{p,\infty}\leqslant \frac p{p-1}\lVert X\rVert_{p,\infty}^*.$$
Questions:
- Does the inequality
$$\lVert\mathbb E[X\mid\mathcal A]\rVert_{p,\infty} \leqslant \lVert X\rVert_{p,\infty}$$
hold in general? - If not, is there a norm $N_p$ on $\mathbb L^{p,\infty}$ which is equivalent to $\lVert\cdot\rVert_{p,\infty} $ and for which the inequality
$$N_p\left(\mathbb E[X\mid\mathcal A]\right)\leqslant N_p(X) $$
takes place?
Motivations. In general, subadditive sequences have good properties. If $T\colon\Omega\to\Omega$ is a measure preserving map and $\mathcal M$ is a sub-$\sigma$-algebra of $\mathcal F$ such that $T \mathcal M\subset \mathcal M$, then defining $S_n(f):=\sum_{j=0}^{n-1} f\circ T^i$ ($f$ is $\mathcal M$-measurable), the sequence $\left(\lVert\mathbb E\left[S_n(f)\mid \mathcal M\right]\rVert_p\right)_{n\geqslant 1} $ is subadditive for any $p\geqslant 1$. A similar result for weak $\mathbb L^p$ norms would be nice.
For the first question, we can show that
$$t\mu\{\mathbb E[X\mid\mathcal A]>t\}^{1/p} \leqslant
\frac 1{\mu\{\mathbb E[X\mid\mathcal A]>t\}^{1-1/p}}\mathbb E[f\mathbb{1}\{\mathbb E[X\mid\mathcal A]>t\} ],$$
hence
$$\lVert \mathbb E[X\mid\mathcal A]\rVert_{p ,\infty}^*\leqslant \lVert f\rVert_{p,\infty},$$
and it follows from (2) that
$$\lVert \mathbb E[X\mid\mathcal A]\rVert_{p ,\infty}\leqslant \frac p{p-1} \lVert f\rVert_{p,\infty},$$
so the question is actually whether we can get rid of the factor $\frac p{p-1}$.
We can try to show this when $f$ is a simple function, but I fail to see how to apply efficiently convexity inequalities.
Best Answer
Yes. You can modify your definition of the norm as: $$\|X \|_{p,\infty}=\sup [(\int_\Omega \phi )^{1/p-1}\mathbb{E}(|X|\phi), \text{ } 0\leq \phi \leq 1] $$ Then $$\|\mathbb{E}(X|\mathcal{A}) \|_{p,\infty}=\sup [(\int_\Omega \mathbb{E}(\phi|\mathcal{A}))^{1/p-1}\mathbb{E}(|X|\mathbb{E}(\phi|\mathcal{A})), 0\leq \phi \leq 1] $$
And we can conclude because $0\leq \phi \leq 1 \Rightarrow 0\leq \mathbb{E}(\phi|\mathcal{A})\leq 1$