I am reading "The Uncertainty Principle" by Fefferman (Bull. AMS, 1983) and have some issues following the arguments. In Lemma $C$ we have the following setting:



> 
> Let $Q^0\subseteq \mathbb{R}^n$ be some dyadic cube, $V: \mathbb{R}^n \rightarrow (-\infty, 0]$ some function, $p>1$ and
> $$ V^+\_Q(x):= \sup\_{\_{Q' \text{ cube } \ : \ x\in Q' \subseteq Q}} \left( \frac{1}{\vert Q' \vert} \int\_{Q'} \vert V(y) \vert^p dy \right)^{1/p}. $$
> and we assume that there exists a constant $\gamma>0$ such that for every subcube $Q\subseteq Q^0$ we have
> $$ \left( \frac{1}{\vert Q\vert} \int\_Q \vert V\vert^p \right)^{1/p}\leq \gamma \text{diam}(Q)^{-2}.$$
> 
> 
> 


In the proof it is then claimed that there exists a constant $C\_p$ depending only on $p$ such that
$$ \frac{1}{\vert Q \vert} \int\_Q V\_Q^+ \leq C\_p \left( \frac{1}{\vert Q \vert} \int\_Q \vert V \vert^p \right)^{1/p}. $$


I'd appreciate any pointers on how to prove this (or a reference where to find such a proof).


This looks a bit like the boundedness of the maximal operator (restricted to our cube), but the powers look completely off.


**Added:** This claim does not appear in the statement of Lemma $C$, but in its proof. Namely it says "the maximal theorem and the hypothesis imply
$$\text{Av}\_Q V\_Q^+\leq C\_p (\text{Av}\_Q \vert V\vert^p)^{1/p} \leq C\_p \gamma (\text{diam}(Q)^{-2})."$$
Here $\text{Av}\_Q$ denotes the average over $Q$. It is the first inequality that is unclear to me.



Maximal function on small cubes
By Holder’s inequality


$$ \frac{1}{\lvert Q\rvert}\int\_{Q}V^{+}dm \leq \frac{1}{\lvert Q\rvert} \lvert Q\rvert^{1/q}\cdot \left(\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p} = \left(\lvert Q\rvert^{-1}\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p}.$$
The RHS term is $\lVert V^{+}\rVert\_{p}$ (up to the $Q$-normalization). Applying the strong maximal inequality of Hardy-Littlewood gives
$$ \left(\lvert Q\rvert^{-1}\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p} \leq C\_{p} Q^{-1/p}\left\lVert V\right\rVert\_{p},$$
where the integrals all happen inside $Q$, so the inequality follows as $Q^{-1/p}\lVert V\rVert\_{p} = \lvert Av\_{Q}(V)^{p}\rvert^{1/p}$.


