Hardy–Littlewood maximal inequality

Remarks

Part 1 says that $M$ is of weak type $(1, 1)$. Its proof uses the Vitali covering lemma. Each point of a compact subset $K$ of $\set{Mf > \lambda}$ is the centre of a ball $B$ with $\int_B \abs{f} > \lambda \abs{B}$. The covering lemma selects finitely many disjoint such balls $B_1, \dots, B_m$ whose dilates $3B_k$ cover $K$, so $\abs{K} \le 3^n \sum_k \abs{B_k} \le 3^n \lambda^{-1} \norm{f}_1$. For $p = \infty$, part 2 is the trivial bound $Mf \le \norm{f}_\infty$, and for $1 < p < \infty$ it follows from part 1 by the Marcinkiewicz interpolation theorem.

The constant in part 2 grows like $1/(p - 1)$ as $p \to 1$, and part 2 fails for $p = 1$. For example, $f = \mathbf{1}_{[0, 1]}$ on $\R$ has $Mf(x) = 1/(2x)$ for $x > 1$, which is not integrable. In fact $Mf$ is integrable only when $f = 0$ almost everywhere.

Going deeper

For a student

The inequality is the main tool for proving that averages converge almost everywhere. Combined with density of continuous functions it gives the Lebesgue differentiation theorem: the averages of $f$ over balls shrinking to $x$ tend to $f(x)$ for almost every $x$. Many operators in harmonic analysis, such as convolutions with radially decreasing integrable kernels, are bounded pointwise by a multiple of $Mf$, so they inherit its $L^p$ bounds. The weighted version of the inequality leads to the Muckenhoupt $A_p$ condition.