Best Constant in the Weak Type Inequality of the Dyadic Maximal Function
Let $\mathscr{D}$ be the collection of all dyadic cubes of $\mathbb{R}^n$. Let $\mu$ be a Borel measure on $\mathbb{R}^n$ such that $\mu(Q)<\infty$ for each $Q\in \mathscr{D}$. For any locally integrable function $f$, we define its average over a dyadic cube $Q$ to be
\[\langle f\rangle_Q:= \begin{cases} \frac{1}{\mu(Q)}\int_Q|f(y)|d\mu(y), & \mu(Q)\ne 0\\ 0, & \mu(Q)=0 \end{cases}.\]We define the dyadic maximal function of $f$ to be
\[M^\mu_d(f)(x):=\sup_{\mathscr{D}\ni Q\ni x}\langle |f|\rangle_Q.\]Question: What is the smallest constant $C>0$ such that the following inequality holds
\[\|M^\mu_d(f)\|_{L^{1,\infty}(\mu)}\leq C\|f\|_{L^1(\mu)}\]for all $f\in L^1(\mu)$? Equivalently, what is the operator norm $\parallel M_d^\mu\parallel_{L^1\rightarrow L^{1,\infty}}$? In particular, we are interested in the case that $\mu$ is the Lebesgue measure.
Answer: Suppose that the $\mu$-measure of each $d$-dimensional quadrant in $\mathbb{R}^d$ is infinite. Then $\parallel M_d^\mu\parallel_{L^1\rightarrow L^{1,\infty}}=1$.
The condition that each quadrant has infinite measure may seem weird, but it holds for the Lebesgue measure. Moreover, this property is needed for the next lemma.
Lemma: For each $\lambda>0$, there exists a collection of disjoint dyadic cubes ${Q_j}$ such that
\[\{M_d^\mu(f)>\lambda\}=\bigcup_j Q_j,\]where the cubes are maximal in the sense that for all dyadic cubes $Q\supsetneq Q_j$,
\[\langle |f|\rangle_Q\leq \lambda <\langle |f|\rangle_{Q_j}.\]Proof: By the definition of $M_d^\mu(f)$, for each point $x$ such that $M_d^\mu(f)(x)>\lambda$, we can find a dyadic cube $Q$ containing $x$ such that
\[\lambda <\langle |f|\rangle_{Q}.\]As the $\mu$-measure of each quadrant is infinite and $f\in L^1(\mu)$, $\langle \lvert f\rvert\rangle_{Q}\rightarrow 0$ as $Q$ gets larger and larger. Hence, eventually we will have $\langle \lvert f\rvert\rangle_{Q}\leq \lambda$, and a maximal dyadic cube $Q^x$ containing $x$ must exist. As a result,
\[\{M_d^\mu(f)>\lambda\}\subset \bigcup_x Q^x\]Notice that for every point $y\in Q^x$, $M_d^\mu(f)(y)\geq \langle \lvert f\rvert\rangle_{Q^x}>\lambda$. Therefore,
\[Q^x\subset \{M_d^\mu(f)>\lambda\}\]and
\[\{M_d^\mu(f)>\lambda\}=\bigcup_x Q^x\]Lastly, we show that if $Q_1$ and $Q_2$ are distinct maximal dyadic cubes, then they are disjoint. We know that two dyadic cubes are either disjoint or one of them is contained in the other. Assume WLOG $Q_1\subset Q_2$. As $Q_1\ne Q_2$, $Q_1\subsetneq Q_2$. By the maximality of $Q_1$ and $Q_2$, we have
\[\lambda <\langle |f|\rangle_{Q_2}\leq \lambda <\langle |f|\rangle_{Q_1}.\]This is absurd. $\square$
Remark: Since we only have countably many dyadic cubes, ${M_d^\mu(f)>\lambda}$ is a countable union of disjoint maximal dyadic cubes ${Q_j}$.
Now we are ready to prove that $\parallel M_d^\mu\parallel_{L^1\rightarrow L^{1,\infty}}=1$. Indeed, $\parallel M_d^\mu\parallel_{L^1\rightarrow L^{1,\infty}}\leq 1$ because for each $\lambda>0$, we have
\[\begin{align*} \lambda \mu(\{M_d^\mu(f)>\lambda\})&=\sum_j\lambda \mu(Q_j)\\ &\leq \sum_j \langle |f|\rangle_{Q_j}\mu(Q_j)\\ &\leq \sum_j\int_{Q_j}|f|d\mu\\ &=\int_{\bigcup_jQ_j}|f|d\mu\\ &\leq \|f\|_{L^1(\mu)}. \end{align*}\]This calculation implies that
\[\|M_d^\mu(f)\|_{L^{1,\infty}(\mu)}\leq \|f\|_{L^1(\mu)}.\]We also need to show that $1$ is the best constant. As each $d$-dimensional quadrant has infinite measure and $\mu$ is locally finite, there must exist some dyadic cube $\tilde{Q}$ such that $0<\mu(\tilde{Q})<\infty$. Consider the indicator function $g:=1_{\tilde{Q}}$. Let $C$ be any constant such that the inequality
\[\|M^\mu_d(f)\|_{L^{1,\infty}(\mu)}\leq C\|f\|_{L^1(\mu)}\]holds for all $f\in L^1(\mu)$. Observe that for all $0<\varepsilon<1$ and any point $x\in \tilde{Q}$, we have
\[M_d^\mu(g)(x)\geq \langle |g|\rangle_{\tilde{Q}}=1>1-\varepsilon.\]Thus, $\tilde{Q}\subset {M_d^\mu(g)>1-\varepsilon}$. We can then deduce that
\[\begin{align*} C\|g\|_{L^1(\mu)}\geq (1-\varepsilon)\mu(\{M_d^\mu(g)>1-\varepsilon\})&\geq (1-\varepsilon)\mu(\tilde{Q}),\\ C\mu(\tilde{Q})&\geq (1-\varepsilon)\mu(\tilde{Q}),\\ C&\geq 1-\varepsilon. \end{align*}\]As this inequality is true for all sufficiently small $\varepsilon$, we conclude that
\[C\geq 1\]and
\[\|M_d^\mu\|_{L^1\rightarrow L^{1,\infty}}=1.\]