Exercise 1.6.20 (Dyadic maximal inequality)

Question

Let $f:\mathbb{R}^d	o\mathbb{C}$ be an absolutely integrable function. Establish the dyadic Hardy-Littlewood maximal inequality
\begin{equation*}
m\left(\left\{x\in\mathbb{R}^d: \sup_{Q\mid x\in Q} \frac{1}{|Q|} \int_{Q} |f|\,\mathrm{d}m \geq \lambda ight\}ight) \leq \frac{1}{\lambda} \int_{\mathbb{R}^d} |f| \,\mathrm{d}m
\end{equation*}
where the supremum ranges over all dyadic cubes $Q$ that contain $x$.

Elu Thingol · Accepted Answer

\begin{hint*}
It is advised to read Lemma 1.2.11 from page 25 before starting with this proof.
\end{hint*}

It suffices to verify the claim with strict inequality,
\begin{equation*}
m\left(\left\{x\in\mathbb{R}^d: \sup_{Q\mid x\in Q} \frac{1}{|Q|} \int_{Q} |f|\,\mathrm{d}m > \lambda ight\}ight) \leq \frac{1}{\lambda} \int_{\mathbb{R}^d} |f| \,\mathrm{d}m,
\end{equation*}
as the non-strict case follows by perturbing $\lambda$ slightly and then taking limits.
ewline

Fix $f$ and $\lambda$. By \href{./exercise_1-2-15}{inner regularity}, it suffices to show that
$$m(K) \leq \frac{1}{\lambda} \int_{\mathbb{R}^d} |f| \,\mathrm{d}m$$
whenever $K$ is a compact set contained in
$$\left\{x\in\mathbb{R}^d: \sup_{Q\mid x\in Q} \frac{1}{|Q|} \int_{Q} |f|\,\mathrm{d}m > \lambda ight\}.$$
By construction, for every $x \in K$, there exists a dyadic cube $Q_x$ such that
\begin{equation} 
\frac{1}{|Q_x|} \int_{Q_x} |f|\,\mathrm{d}m > \lambda
\label{eq:cond}
\end{equation}
If we let $\mathcal{Q}$ be the collection of all the dyadic cubes $Q$ that correspond to \eqref{eq:cond} by the axiom of choice, we see that the union $\bigcup \mathcal{Q}$ of all these cubes covers $K$.\par
As there are only countably many dyadic cubes, $\mathcal{Q}$ is at most countable. Furthermore, let $\mathcal{Q}^*$ denote those cubes in $\mathcal{Q}$ which are maximal with respect to set inclusion, which means that they are not contained in any other cube in $\mathcal{Q}$. From the nesting property (and the fact that we have capped the maximum size of our cubes) we see that every cube in $\mathcal{Q}$ is contained in exactly one maximal cube in $\mathcal{Q}^*$, and that any two such maximal cubes in $\mathcal{Q}^*$ are almost disjoint. Thus, we see that $K$ is covered by the union $\bigcup \mathcal{Q}^*$ of almost disjoint cubes. As $\mathcal{Q}^*$ is at most countable, we have
$$m\left(\bigcup \mathcal{Q}^* ight)
\leq \sum_{Q \in \mathcal{Q}^*} m\left(Qight)
\leq \frac{1}{\lambda} \sum_{Q \in \mathcal{Q}^*} \int_{Q} |f|\,\mathrm{d}m 
\leq \frac{1}{\lambda} \int_{\mathbb{R}^d} |f|\,\mathrm{d}m.$$
Since $\mathcal{Q}^*$ covers $K$, we obtain the theorem assertion as desired.

Exercise 1.6.20 (Dyadic maximal inequality)

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Exercise 1.6.20 (Dyadic maximal inequality)

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