Calderón–Zygmund lemma

Remarks

The proof is a stopping-time argument on dyadic cubes. Cut $\R^n$ into a grid of cubes so large that the average of $\abs{f}$ on each is at most $\alpha$. Divide each cube into $2^n$ subcubes by halving its sides, keep the subcubes on which the average exceeds $\alpha$, and continue dividing the others. A kept cube has a parent with average at most $\alpha$, which gives the bound $2^n\alpha$, and the Lebesgue differentiation theorem gives $\abs{f} \le \alpha$ almost everywhere on the points never kept.

Clause 3 gives $\abs{U} = \sum_k \abs{Q_k} \le \alpha^{-1}\norm{f}_{L^1}$. Setting $g = f$ on $F$ and $g$ equal to the average of $f$ on each $Q_k$, and $b = f - g$, one obtains the Calderón–Zygmund decomposition $f = g + b$. Here $\abs{g} \le 2^n\alpha$ almost everywhere, and $b$ has mean zero on each cube $Q_k$ and vanishes off $U$.

Example: for $n = 1$, $f = \mathbf{1}_{[0, 1]}$ and $\alpha = 1/2$, one may take $U = [0, 1]$ and $F = \R \setminus [0, 1]$; the average of $\abs{f}$ on $[0, 1]$ is $1$, which lies in $(1/2, 1]$.

Going deeper

For a student

The lemma, through the decomposition $f = g + b$, is the key step in proving that singular integral operators are of weak type $(1, 1)$. The good part $g$ is bounded and integrable, so it lies in $L^2$, where the operator is bounded. The bad part $b$ has mean zero on each cube, and the smoothness of the kernel controls the operator on it away from the cubes. The same stopping-time idea gives the weak type $(1, 1)$ bound for the Hardy–Littlewood maximal function and the John–Nirenberg inequality.