Maximal function on small cubes

Question

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.

Answer 1

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 $Q^{-1/p}\lVert V\rVert_{p} = \lvert Av_{Q}(\left\lvert V\right\rvert^{p})\rvert^{1/p}$.