Lemma 2.9.11780F1882B8A5

Proof. We use the Calderón–Zygmund decomposition of $f$ at height $\lambda $ into a bounded part $g$, handled by the $L^2$ bound for $T_{ab}$, and parts $b_ Q$ of mean zero on disjoint cubes $Q$, handled by lemma 2.9.10. The dyadic cubes are the cubes $2^ k(m+[0,1)^ d)$ with $k\in \mathbb Z$ and $m\in \mathbb Z^ d$. Two dyadic cubes are either disjoint or nested, and each dyadic cube $Q$ lies in a unique dyadic cube of twice its side length, its parent, which has volume $2^ d|Q|$. Call a dyadic cube $Q$ selected if the average of $|f|$ over $Q$ exceeds $\lambda $, while the average over every strictly larger dyadic cube containing $Q$ does not. Every dyadic cube on which the average of $|f|$ exceeds $\lambda $ is contained in a selected cube, because the averages over its successive ancestors are at most $\| f\| _1$ divided by their volumes and therefore tend to zero. The selected cubes are pairwise disjoint. For each of them, the average over its parent is at most $\lambda $, and hence

Moreover, $|f|\leq \lambda $ almost everywhere outside the union of the selected cubes. Indeed, let $x$ be a point outside this union at which the conclusion of theorem 2.8.3 holds. Every dyadic cube containing $x$ has average of $|f|$ at most $\lambda $, since otherwise it would lie in a selected cube. The dyadic cube of side $2^ k$ containing $x$ lies in the cube $x+(-2^ k,2^ k)^ d$, whose volume is $2^ d$ times larger. Hence the average of $|f-f(x)|$ over it tends to zero as $k\to -\infty $, so its average of $|f|$ tends to $|f(x)|$, and therefore $|f(x)|\leq \lambda $.

For a selected cube $Q$, let $f_ Q=|Q|^{-1}\int _ Qf$ be the average of $f$. Define $g=f$ outside the selected cubes and $g=f_ Q$ on each selected cube $Q$, and put $b_ Q=(f-f_ Q)\mathbf1_ Q$. Each $b_ Q$ has integral zero, and

By the Cauchy–Schwarz inequality, $|Q||f_ Q|^2\leq \int _ Q|f|^2$, so $\| b_ Q\| _2\leq 2\| f\mathbf1_ Q\| _2$. Since the $b_ Q$ have disjoint supports, the series $\sum _ Qb_ Q$ converges to $f-g$ in $L^2$.

Let $c_ Q$ be the center of $Q$, and let $Q^*$ be the cube with center $c_ Q$ and $4\sqrt d$ times the side length of $Q$. If $x\notin Q^*$ and $y\in Q$, then $|x-c_ Q|\geq 2\operatorname {diam}Q$ and $|y-c_ Q|\leq \operatorname {diam}Q/2$, so every point of the segment from $x-c_ Q$ to $x-y$ has norm at least $|x-c_ Q|/2$. The function $b_ Q\in L^2$ vanishes outside the compact set $\overline Q$, which does not meet $\mathbb R^ d\setminus Q^*$, and it has integral zero. Therefore parts (1) and (2) of lemma 2.9.10 and the mean value theorem give, for almost every $x\notin Q^*$,

Integrating in polar coordinates about $c_ Q$ over the region $|x-c_ Q|\geq 2\operatorname {diam}Q$, which contains $\mathbb R^ d\setminus Q^*$, we obtain

Put $B^*=\bigcup _ QQ^*$. Then $|B^*|\leq (4\sqrt d)^ d\sum _ Q|Q|\leq C_ d\lambda ^{-1}\| f\| _1$. Enumerate the selected cubes, and let $S_ N$ be the partial sums of $\sum _ Qb_ Q$. Since $S_ N\to f-g$ in $L^2$, also $T_{ab}S_ N\to T_{ab}(f-g)$ in $L^2$, and a subsequence converges almost everywhere. Moreover $\int _{\mathbb R^ d\setminus B^*}|T_{ab}S_ N|\leq \sum _ Q\int _{\mathbb R^ d\setminus Q^*}|T_{ab}b_ Q|\leq C_ d\| f\| _1$, so Fatou's lemma gives

Finally, $T_{ab}f=T_{ab}g+T_{ab}(f-g)$, and the set where $|T_{ab}f|>\lambda $ is contained in the union of $B^*$, the set where $|T_{ab}g|>\lambda /2$, and the set of $x\notin B^*$ with $|T_{ab}(f-g)(x)|>\lambda /2$. By Chebyshev's inequality and the $L^2$ bound for $T_{ab}$, the second set has measure at most $4\lambda ^{-2}\| g\| _2^2\leq 2^{d+2}\lambda ^{-1}\| f\| _1$, and the third has measure at most $2\lambda ^{-1}\int _{\mathbb R^ d\setminus B^*}|T_{ab}(f-g)|\leq C_ d\lambda ^{-1}\| f\| _1$. Together with the bound for $|B^*|$, this proves the lemma. $\square$