- Let \(f \in L ^ { 1 } ( \mathbb { R } ^ { d } )\) and \(M _ { f }\) denote the Hardy-Littlewood maximal function of \(f ;\) in other words,
\[ M _ { f } ( x ) = \operatorname* { s u p } _ { B } { \frac { 1 } { m ( B ) } } \int _ { B } | f ( y ) | d y , \quad x \in \mathbb { R } ^ { d } \]
where the supremum is taken over all balls containing the point x. Prove that
\[ m \big ( \{ x : M _ { f } ( x ) > \alpha \} \big ) \leq \frac { A } { \alpha } | | f | | _ { L ^ { 1 } ( \mathbb { R } ^ { d } ) } , \quad \forall \alpha > 0 \]
where A is a constant depending only on d and \(\begin{array} { r } { | | f | | _ { L ^ { 1 } ( \mathbb { R } ^ { d } ) } = \int _ { \mathbb { R } ^ { d } } | f ( x ) | d x . } \end{array}\)
Solution
1. Associate a good ball to every point of the superlevel set.
Fix \(\alpha>0\) and write \[ E_\alpha:=\{x\in\mathbb R^d:M_f(x)>\alpha\}. \] Take the balls in the definition to be open. Then \(E_\alpha\) is the union of all balls on which the average of \(|f|\) exceeds \(\alpha\): every point of such a ball has maximal function greater than \(\alpha\), and the converse follows from the definition of supremum. Thus \(E_\alpha\) is open, in particular measurable. If it is empty, the required inequality is immediate; otherwise proceed below. For every \(x\in E_\alpha\), by definition of \(M_f(x)\) there is a ball \(B_x\) containing \(x\) such that \[ \frac1{m(B_x)}\int_{B_x}|f(y)|\,dy>\alpha. \] Hence \[ \alpha\,m(B_x)<\int_{B_x}|f|\le \|f\|_1. \] In particular the radii of all such balls are uniformly bounded in terms of \(\alpha\), \(d\), and \(\|f\|_1\).
2. Apply the \(5r\) covering lemma.
Apply the \(5r\) covering lemma to the family \(\{B_x:x\in E_\alpha\}\). There is a finite or countable pairwise disjoint subfamily \(\{B_j\}\) such that \[ E_\alpha\subseteq\bigcup_j 5B_j, \] where \(5B_j\) denotes the ball with the same center and five times the radius.
Therefore \[ m(E_\alpha) \le \sum_j m(5B_j) =5^d\sum_j m(B_j). \]
3. Use disjointness and the defining average inequality.
For each selected ball, \[ \alpha\,m(B_j)<\int_{B_j}|f|. \] Since the \(B_j\) are pairwise disjoint, \[ \alpha\sum_j m(B_j) \le\sum_j\int_{B_j}|f| \le\int_{\mathbb R^d}|f| =\|f\|_1. \] Combining this with Step 2 gives \[ m(E_\alpha) \le \frac{5^d}{\alpha}\|f\|_1. \] Thus the desired estimate holds with, for example, \[ \boxed{A=5^d.} \]