7 Proof of the Forest Operator Proposition

7.1 The pointwise tree estimate

Fix a forest $(U, T)$ . The main result of this subsection is Lemma 7.1.3, we begin this section with some definitions necessary to state the lemma.

For $u \in U$ and $x \in X$ , we define

σ (u, x) := {s (p) : p \in T (u), x \in E (p)} .

This is a subset of $Z \cap [- S, S]$ , so has a minimum and a maximum. We set

\overset{―}{σ} (u, x) := max σ (T (u), x)

\underset{―}{σ} (u, x) := min σ (T (u), x) .

Lemma 7.1.1 convex scales

✓

For each $u \in U$ , we have

σ (u, x) = Z \cap [\underset{―}{σ} (u, x), \overset{―}{σ} (u, x)] .

Proof ▶

Let $s \in Z$ with $\underset{―}{σ} (u, x) \leq s \leq \overset{―}{σ} (u, x)$ . By definition of $σ$ , there exists $p \in T (u)$ with $s (p) = \underset{―}{σ} (u, x)$ and $x \in E (p)$ , and there exists $p^{″} \in T (u)$ with $s (p^{″}) = \overset{―}{σ} (u, x)$ and $x \in E (p^{″}) \subset I (p^{″})$ . By property 2.0.7 of the dyadic grid, there exists a cube $I \in D$ of scale $s$ with $x \in I$ . By property 2.0.13, there exists a tile $p^{'} \in P (I)$ with $Q (x) \in Ω (p^{'})$ . By the dyadic property 2.0.8 we have $I (p) \subset I (p^{'}) \subset I (p^{″})$ , and by 2.0.14, we have $Ω (p^{″}) \subset Ω (p^{'}) \subset Ω (p)$ . Thus $p \leq p^{'} \leq p^{″}$ , which gives with the convexity property 2.0.33 of $T (u)$ that $p^{'} \in T (u)$ , so $s \in σ (u, x)$ .

For a nonempty collection of tiles $S \subset P$ we define

J_{0} (S)

to be the collection of all dyadic cubes $J \in D$ such that $s (J) = - S$ or

I (p) ⊄ B (c (J), 100 D^{s (J) + 1})

for all $p \in S$ . We define $J (S)$ to be the collection of inclusion maximal cubes in $J_{0} (S)$ .

We further define

L_{0} (S)

to be the collection of dyadic cubes $L \in D$ such that $s (L) = - S$ , or there exists $p \in S$ with $L \subset I (p)$ and there exists no $p \in S$ with $I (p) \subset L$ . We define $L (S)$ to be the collection of inclusion maximal cubes in $L_{0} (S)$ .

Lemma 7.1.2 dyadic partitions

✓

For each $S \subset P$ , we have

⋃_{I \in D} I = \dot{⋃_{J \in J (S)}} J

7.1.1

and

⋃_{I \in D} I = \dot{⋃_{L \in L (S)}} L .

7.1.2

Proof ▶

Since $J (S)$ is the set of inclusion maximal cubes in $J_{0} (S)$ , cubes in $J (S)$ are pairwise disjoint by 2.0.8. The same applies to $L (S)$ .

If $x \in ⋃_{I \in D} I$ , then there exists by 2.0.7 a cube $I \in D$ with $x \in I$ and $s (I) = - S$ . Then $I \in J_{0} (S)$ . There exists an inclusion maximal cube in $J_{0} (S)$ containing $I$ . This cube contains $x$ and is contained in $J (S)$ . This shows one inclusion in 7.1.1, the other one follows from $J (S) \subset D$ .

The proof of the two inclusions in 7.1.2 is similar.

For a finite collection of pairwise disjoint cubes $C$ , define the projection operator

P_{C} f (x) := \sum_{J \in C} 1_{J} (x) \frac{1}{μ (J)} \int_{J} f (y) d μ (y) .

Given a scale $- S \leq s \leq S$ and a point $x \in ⋃_{I \in D, s (I) = s} I$ , there exists a unique cube in $D$ of scale $s$ containing $x$ by 2.0.7. We denote it by $I_{s} (x)$ . Define for $ϑ \in Θ$ the nontangential maximal operator

T_{N}^{ϑ} f (x) := sup_{- S \leq s_{1} < S} sup_{x^{'} \in I_{s_{1}} (x)} sup_{\begin{matrix} s_{1} \leq s_{2} \leq S \\ D^{s_{2} - 1} \leq R_{Q} (ϑ, x^{'}) \end{matrix}} | \sum_{s = s_{1}}^{s_{2}} \int K_{s} (x^{'}, y) f (y) d μ (y) | .

7.1.3

Define for each $u \in U$ the auxiliary operator

S_{1, u} f (x)

:= \sum_{I \in D} 1_{I} (x) \sum_{\begin{matrix} J \in J (T (u)) \\ J \subset B (c (I), 16 D^{s (I)}) \\ s (J) \leq s (I) \end{matrix}} \frac{D^{(s (J) - s (I)) / a}}{μ (B (c (I), 16 D^{s (I)}))} \int_{J} | f (y) | d μ (y) .

7.1.4

Define also the collection of balls

B = {B (c (I), 2^{s} D^{s (I) + t}) : I \in D, 0 \leq s \leq S + 5, 0 \leq t \leq 2 S + 3} .

The following pointwise estimate for operators associated to sets $T (u)$ is the main result of this subsection.

Lemma 7.1.3 pointwise tree estimate

✓

Let $u \in U$ and $L \in L (T (u))$ . Let $x, x^{'} \in L$ . Then for all bounded functions $f$ with bounded support

| \sum_{p \in T (u)} T_{p} [e (- Q (u)) f] (x) |

\leq 2^{151 a^{3}} (M_{B, 1} + S_{1, u}) P_{J (T (u))} | f | (x^{'}) + | T_{N}^{Q (u)} P_{J (T (u))} f (x^{'}) |,

7.1.5

Proof ▶

By 2.0.21, if $T_{p} [e (- Q (u)) f] (x) \neq 0$ , then $x \in E (p)$ . Combining this with $| e (Q (u) (x)) | = 1$ , we obtain

| \sum_{p \in T (u)} T_{p} [e (- Q (u)) f] (x) |

\begin{matrix} = | \sum_{s \in σ (u, x)} \int e (- Q (u) (y) + Q (x) (y) + Q (u) (x) - Q (x) (x)) \times \\ K_{s} (x, y) f (y) d μ (y) | . \end{matrix}

Using the triangle inequality, we bound this by the sum of three terms:

\begin{matrix} \leq | \sum_{s \in σ (u, x)} \int (e (- Q (u) (y) + Q (x) (y) + Q (u) (x) - Q (x) (x)) - 1) \times \\ K_{s} (x, y) f (y) d μ (y) | \end{matrix}

+ | \sum_{s \in σ (u, x)} \int K_{s} (x, y) P_{J (T (u))} f (y) d μ (y) |

7.1.8

+ | \sum_{s \in σ (u, x)} \int K_{s} (x, y) (f (y) - P_{J (T (u))} f (y)) d μ (y) | .

7.1.9

The proof is completed using the bounds for these three terms proven in Lemma 7.1.4, Lemma 7.1.5 and Lemma 7.1.6.

Lemma 7.1.4 first tree pointwise

✓

For all $u \in U$ , all $L \in L (T (u))$ , all $x, x^{'} \in L$ and all bounded $f$ with bounded support, we have

(???) \leq 10 \cdot 2^{104 a^{3}} M_{B, 1} P_{J (T (u))} | f | (x^{'}) .

Proof ▶

Let $s \in σ (u, x)$ . If $x, y \in X$ are such that $K_{s} (x, y) \neq 0$ , then, by 2.1.2, we have $ρ (x, y) \leq 1 / 2 D^{s}$ . By $1$ -Lipschitz continuity of the function $t \mapsto \exp (i t) = e (t)$ and the property 1.0.7 of the metrics $d_{B}$ , it follows that

\begin{matrix} | e (- Q (u) (y) + Q (x) (y) + Q (u) (x) - Q (x) (x)) - 1 | \\ \leq d_{B (x, 1 / 2 D^{s})} (Q (u), Q (x)) . \end{matrix}

Let $p_{s} \in T (u)$ be a tile with $s (p_{s}) = s$ and $x \in E (p_{s})$ , and let $p^{'}$ be a tile with $s (p^{'}) = \overset{―}{σ} (u, x)$ and $x \in E (p^{'})$ . Using the monotonicity property 1.0.9, the doubling property 1.0.8 repeatedly, the definition of $d_{p}$ and Lemma 2.1.2, we can bound the previous display by

d_{B (x, 4 D^{s})} (Q (u), Q (x)) \leq 2^{4 a} d_{p_{s}} (Q (u), Q (x)) \leq 2^{4 a} 2^{s - \overset{―}{σ} (u, x)} d_{p^{'}} (Q (u), Q (x)) .

Since $Q (u) \in B_{p^{'}} (Q (p^{'}), 4)$ by 2.0.32 and $Q (x) \in Ω (p^{'}) \subset B_{p^{'}} (Q (p^{'}), 1)$ by 2.0.15, this is estimated by

\leq 5 \cdot 2^{4 a} 2^{s - \overset{―}{σ} (u, x)} .

Using 2.1.3, it follows that

(???) \leq 5 \cdot 2^{103 a^{3}} \sum_{s \in σ (x)} 2^{s - \overset{―}{σ} (u, x)} \frac{1}{μ (B (x, D^{s}))} \int_{B (x, 0.5 D^{s})} | f (y) | d μ (y) .

By 7.1.1, the collection $J$ is a partition of $⋃_{I \in D} I$ , so this is estimated by

5 \cdot 2^{103 a^{3}} \sum_{s \in σ (x)} 2^{s - \overset{―}{σ} (u, x)} \frac{1}{μ (B (x, D^{s}))} \sum_{\begin{matrix} J \in J (T (u)) \\ J \cap B (x, 0.5 D^{s}) \neq \emptyset \end{matrix}} \int_{J} | f (y) | d μ (y) .

This expression does not change if we replace $| f |$ by $P_{J (T (u))} | f |$ .

Let $J \in J (T (u))$ with $B (x, 0.5 D^{s}) \cap J \neq \emptyset$ . By the triangle inequality and since $x \in E (p_{s}) \subset B (c (p_{s}), 4 D^{s})$ , it follows that $B (c (p_{s}), 4.5 D^{s}) \cap J \neq \emptyset$ . If $s (J) \geq s$ and $s (J) > - S$ , then it follows from the triangle inequality, 2.0.10 and 2.0.1 that $I (p_{s}) \subset B (c (J), 100 D^{s (J) + 1})$ , contradicting $J \in J (T (u))$ . Thus $s (J) \leq s - 1$ or $s (J) = - S$ . If $s (J) = - S$ and $s (J) > s - 1$ , then $s = - S$ . Thus we always have $s (J) \leq s$ . It then follows from the triangle inequality and 2.0.10 that $J \subset B (c (p_{s}), 16 D^{s})$ .

Thus we can continue our chain of estimates with

5 \cdot 2^{103 a^{3}} \sum_{s \in σ (x)} 2^{s - \overset{―}{σ} (u, x)} \frac{1}{μ (B (x, D^{s}))} \int_{B (c (p_{s}), 16 D^{s})} P_{J (T (u))} | f (y) | d μ (y) .

We have $B (c (p_{s}), 16 D^{s})) \subset B (x, 32 D^{s})$ , by 2.0.10 and the triangle inequality, since $x \in I (p_{s})$ . Combining this with the doubling property 1.0.5, we obtain

μ (B (c (p_{s}), 16 D^{s})) \leq 2^{5 a} μ (B (x, D^{s})) .

Since $a \geq 4$ , it follows that 7.1.6 is bounded by

5 \cdot 2^{103 a^{3}} \sum_{s \in σ (x)} 2^{s - \overset{―}{σ} (u, x)} \frac{1}{μ (B (c (p_{s}), 16 D^{s}))} \int_{B (c (p_{s}), 16 D^{s})} P_{J (T (u))} | f (y) | d μ (y) .

Since $L \in L (T (u))$ and $x \in L \cap I (p_{s})$ , we have $s (L) \leq s (p_{s})$ . It follows by 2.0.8 that $L \subset I (p_{s})$ , in particular $x^{'} \in I (p_{s}) \subset B (c (p_{s}), 16 D^{s})$ . Thus

\leq 5 \cdot 2^{104 a^{3}} \sum_{s \in σ (x)} 2^{s - \overset{―}{σ} (u, x)} M_{B, 1} P_{J (T (u))} | f | (x^{'})

\leq 10 \cdot 2^{104 a^{3}} M_{B, 1} P_{J (T (u))} | f | (x^{'}) .

This completes the estimate for term 7.1.6.

Lemma 7.1.5 second tree pointwise

✓

For all $u \in U$ , all $L \in L (T (u))$ , all $x, x^{'} \in L$ and all bounded $f$ with bounded support, we have

| \sum_{s \in σ (u, x)} \int K_{s} (x, y) P_{J (T (u))} f (y) d μ (y) | \leq T_{N}^{Q (u)} P_{J (T (u))} f (x^{'}) .

Proof ▶

Let $s_{1} = \underset{―}{σ} (u, x)$ . By definition, there exists a tile $p \in T (u)$ with $s (p) = s_{1}$ and $x \in E (p)$ . Then $x \in I (p) \cap L$ . By 2.0.8 and the definition of $L (T (u))$ , it follows that $L \subset I (p)$ , in particular $x^{'} \in I (p)$ , so $x \in I_{s_{1}} (x^{'})$ . Next, let $s_{2} = \overset{―}{σ} (u, x)$ and let $p^{'} \in T (u)$ with $s (p^{'}) = s_{2}$ and $x \in E (p^{'})$ . Since $p^{'} \in T (u)$ , we have $4 p^{'} ≲ u$ . Since $Q (x) \in Ω (p^{'})$ , it follows that

d_{p} (Q (u), Q (x)) \leq 5 .

Applying the doubling property 1.0.8 five times, we obtain

d_{B (c (p), 8 D^{s_{2}})} (Q (u), Q (x)) \leq 5 \cdot 2^{5 a} .

By the triangle inequality, we have $B (x, D^{s_{2}}) \subset B (c (p), 8 D^{s_{2}})$ , so by 1.0.9

d_{B (x, D^{s_{2}})} (Q (u), Q (x)) \leq 5 \cdot 2^{5 a} .

Finally, by applying 1.0.10 $100 a$ times, we obtain

d_{B (x, D^{s_{2} - 1})} (Q (u), Q (x)) \leq 5 \cdot 2^{- 95 a} < 1 .

Consequently, $D^{s_{2} - 1} < R_{Q} (Q (u), x)$ . The lemma now follows from the definition of $T_{N}$ .

Lemma 7.1.6 third tree pointwise

✓

For all $u \in U$ , all $L \in L (T (u))$ , all $x, x^{'} \in L$ and all bounded $f$ with bounded support, we have

| \sum_{s \in σ (u, x)} \int K_{s} (x, y) (f (y) - P_{J (T (u))} f (y)) d μ (y) |

\leq 2^{151 a^{3}} S_{1, u} P_{J (T (u))} | f | (x^{'}) .

Proof ▶

We have for $J \in J (T (u))$ :

\int_{J} K_{s} (x, y) (1 - P_{J (T (u))}) f (y) d μ (y)

= \int_{J} \frac{1}{μ (J)} \int_{J} K_{s} (x, y) - K_{s} (x, z) d μ (z) f (y) d μ (y) .

7.1.10

By 2.1.4 and 2.0.10, we have for $y, z \in J$

| K_{s} (x, y) - K_{s} (x, z) | \leq \frac{2^{150 a^{3}}}{μ (B (x, D^{s}))} {(\frac{8 D^{s (J)}}{D^{s}})}^{1 / a} .

Suppose that $s \in σ (u, x)$ . If $K_{s} (x, y) \neq 0$ for some $y \in J \in J (T (u))$ then, by 2.1.2, $y \in B (x, 0.5 D^{s}) \cap J \neq \emptyset$ . Let $p \in T (u)$ with $s (p) = s$ and $x \in E (p)$ . Then $B (c (p_{s}), 4.5 D^{s}) \cap J \neq \emptyset$ by the triangle inequality. If $s (J) \geq s$ and $s (J) > - S$ , then it follows from the triangle inequality, 2.0.10 and 2.0.1 that $I (p) \subset B (c (J), 100 D^{s (J) + 1})$ , contradicting $J \in J (T (u))$ . Thus $s (J) \leq s - 1$ or $s (J) = - S$ . If $s (J) = - S$ and $s (J) > s - 1$ , then $s = - S$ . So in both cases, $s (J) \leq s$ . It then follows from the triangle inequality and 2.0.10 that $J \subset B (x, 16 D^{s})$ .

Thus, we can estimate 7.1.9 by

2^{150 a^{3} + 3 / a} \sum_{p \in T} \frac{1_{E (p)} (x)}{μ (B (x, D^{s (p)}))} \sum_{\begin{matrix} J \in J (T (u)) \\ J \subset B (x, 16 D^{s (p)}) \\ s (J) \leq s (p) \end{matrix}} D^{(s (J) - s (p)) / a} \int_{J} | f | .

= 2^{150 a^{3} + 3 / a} \sum_{I \in D} \sum_{\begin{matrix} p \in T \\ I (p) = I \end{matrix}} \frac{1_{E (p)} (x)}{μ (B (x, D^{s (I)}))} \sum_{\begin{matrix} J \in J (T (u)) \\ J \subset B (x, 16 D^{s (I)}) \\ s (J) \leq s (I) \end{matrix}} D^{(s (J) - s (I)) / a} \int_{J} | f | .

By 2.0.13 and 2.0.20, the sets $E (p)$ for tiles $p$ with $I (p) = I$ are pairwise disjoint. It follows from the definition of $L (T (u))$ that $x \in I (p)$ if and only if $x^{'} \in I (p)$ , thus we can estimate the sum over $1_{E (p)} (x)$ by $1_{I} (x^{'})$ . If $x \in E (p)$ then in particular $x \in I (p)$ , so by 2.0.10 $B (c (I), 16 D^{s (I)}) \subset B (x, 32 D^{s (I)})$ . By the doubling property 1.0.5

μ (B (c (I), 16 D^{s (I)})) \leq 2^{5 a} μ (B (x, D^{s (I)})) .

Since $a \geq 4$ we can continue our estimate with

\leq 2^{151 a^{3}} \sum_{I \in D} \frac{1_{I} (x^{'})}{μ (B (c (I), 16 D^{s (I)}))} \sum_{\begin{matrix} J \in J (T (u)) \\ J \subset B (x, 16 D^{s (I)}) \\ s (J) \leq s (I) \end{matrix}} D^{(s (J) - s (I)) / a} \int_{J} | f |

= 2^{151 a^{3}} S_{1, u} P_{J (T (u))} | f | (x^{'}) .

This completes the proof.

7.2 An auxiliary $L^{2}$ tree estimate

In this subsection we prove the following estimate on $L^{2}$ for operators associated to trees.

Lemma 7.2.1 tree projection estimate

✓

Let $u \in U$ . Then we have for all $f, g$ bounded with bounded support

| \int_{X} \sum_{p \in T (u)} \bar{g} (y) T_{p} f (y) d μ (y) |

\leq 2^{152 a^{3}} ‖ P_{J (T (u))} | f | ‖_{2} ‖ P_{L (T (u))} | g | ‖_{2} .

7.2.1

Below, we deduce Lemma 7.2.1 from Lemma 7.1.3 and the following estimates for the operators in Lemma 7.1.3.

Lemma 7.2.2 nontangential operator bound

✓

For all bounded $f$ with bounded support and all $ϑ \in Θ$

‖ T_{N}^{ϑ} f ‖_{2} \leq 2^{102 a^{3}} ‖ f ‖_{2} .

Lemma 7.2.3 boundary operator bound

✓

For all $u \in U$ and all bounded functions $f$ with bounded support

‖ S_{1, u} f ‖_{2} \leq 2^{12 a} ‖ f ‖_{2} .

7.2.2

Proof of Lemma 7.2.1 ▶

Let $L \in L (T (u))$ . Let $b (x^{'})$ denote the right-hand side of Lemma 7.1.3. Apply this lemma to $e (Q (u)) f$ , to obtain for all $y, x^{'} \in L$

| \sum_{p \in T (u)} T_{p} f (y) | \leq b (x^{'}) .

Hence, by taking an infimum, we have for $y \in L$

| \sum_{p \in T (u)} T_{p} f (y) | \leq inf_{x^{'} \in L} b (x^{'}) .

Integrating this estimate yields

\int_{L} | g (y) | | \sum_{p \in T (u)} T_{p} f (y) | d μ (y)

\leq \int_{L} | g (y) | \times inf_{x^{'} \in L} b (x^{'}) d μ (y)

= \int_{L} P_{L (T (u))} | g | (y) \times inf_{x^{'} \in L} b (x^{'}) d μ (y)

\leq \int_{L} P_{L (T (u))} | g | (y) \times b (y) d μ (y)

By 2.0.21, we have $T_{p} f = 1_{I (p)} T_{p} f$ for all $p \in P$ , so

| \int \bar{g} (y) \sum_{p \in T (u)} T_{p} f (y) d μ (y) | = | \int_{⋃_{p \in T (u)} I (p)} \bar{g} (y) \sum_{p \in T (u)} T_{p} f (y) d μ (y) | .

Since $L (T (u))$ partitions $⋃_{p \in T (u)} I (p)$ by Lemma 7.1.2, we get from the triangle inequality

\leq \sum_{L \in L (T (u))} \int_{L} | g (y) | | \sum_{p \in T (u)} T_{p} f (y) | d μ (y)

which by the above computation is bounded by

\sum_{L \in L (T (u))} \int_{L} P_{L (T (u))} | g | (y) \times b (y) d μ (y)

= \int_{X} P_{L (T (u))} | g | (y) \times b (y) d μ (y)

Applying Cauchy-Schwarz, this is bounded by $‖ P_{L (T (u))} | g | ‖_{2} \times ‖ b ‖_{2}$ . By Minkowski’s inequality, Proposition 2.0.6, Lemma 7.2.2 and Lemma 7.2.3, $‖ b ‖_{2}$ is at most

2^{151 a^{3}} (2^{2 a + 1} + 2^{12 a}) ‖ P_{J (T (u))} | f | ‖_{2} + 2^{103 a^{3}} ‖ P_{J (T (u))} [e (Q (u)) f] ‖_{2} .

By the triangle inequality we have for all $x \in X$ that $| P_{J (T (u))} [e (Q (u)) f] | (x) \leq P_{J (T (u))} | f | (x)$ , thus we can further estimate the above by

(2^{151 a^{3}} (2^{2 a + 1} + 2^{12 a}) + 2^{103 a^{3}}) ‖ P_{J (T (u))} | f | ‖_{2} .

This completes the proof since $a \geq 4$ .

Now we prove the two auxiliary lemmas. We begin with the nontangential maximal operator $T_{N}$ .

Proof of Lemma 7.2.2 ▶

Fix $s_{1}, s_{2}$ . By 2.0.4 we have for all $x \in (0, \infty)$

\sum_{s = s_{1}}^{s_{2}} ψ (D^{- s} x) = 1 - \sum_{s < s_{1}} ψ (D^{- s} x) - \sum_{s > s_{1}} ψ (D^{- s} x) .

Since $ψ$ is supported in $[\frac{1}{4 D}, \frac{1}{2}]$ , the two sums on the right hand side are zero for all $x \in [\frac{1}{2} D^{s_{1} - 1}, \frac{1}{4} D^{s_{2} - 1}]$ , hence

x \in [\frac{1}{2} D^{s_{1} - 1}, \frac{1}{4} D^{s_{2}}] ⟹ \sum_{s = s_{1}}^{s_{2}} ψ (D^{- s} x) = 1 .

Since $ψ$ is supported in $[\frac{1}{4 D}, \frac{1}{2}]$ , we further have

x \notin [\frac{1}{4} D^{s_{1} - 1}, \frac{1}{2} D^{s_{2}}] ⟹ \sum_{s = s_{1}}^{s_{2}} ψ (D^{- s} x) = 0 .

Finally, since $ψ \geq 0$ and $\sum_{s \in Z} ψ (D^{- s} x) = 1$ , we have for all $x$

0 \leq \sum_{s = s_{1}}^{s_{2}} ψ (D^{- s} x) \leq 1 .

Let $x^{'} \in I_{s_{1}} (x)$ and suppose that $D^{s_{2} - 1} \leq R_{Q} (ϑ, x^{'})$ . By the triangle inequality and 2.0.10, it holds that $ρ (x, x^{'}) \leq 8 D^{s_{1}}$ . We have

| \sum_{s = s_{1}}^{s_{2}} \int K_{s} (x^{'}, y) f (y) d μ (y) |

= | \int \sum_{s = s_{1}}^{s_{2}} ψ (D^{- s} ρ (x^{'}, y)) K (x^{'}, y) f (y) d μ (y) |

\leq | \int_{8 D^{s_{1}} < ρ (x^{'}, y) < \frac{1}{4} D^{s_{2} - 1}} K (x^{'}, y) f (y) d μ (y) |

7.2.3

+ \int_{\frac{1}{4} D^{s_{1} - 1} \leq ρ (x^{'}, y) \leq 8 D^{s_{1}}} | K (x^{'}, y) | | f (y) | d μ (y)

7.2.4

+ \int_{\frac{1}{4} D^{s_{2} - 1} \leq ρ (x^{'}, y) \leq \frac{1}{2} D^{s_{2}}} | K (x^{'}, y) | | f (y) | d μ (y) .

7.2.5

The first term 7.2.3 is at most $T_{*} f (x)$ . The other two terms will be estimated by the finitary maximal function from Proposition 2.0.6. For the second term 7.2.4 we use 1.0.14 which implies that for all $y$ with $ρ (x^{'}, y) \geq \frac{1}{4} D^{s_{1} - 1}$ , we have

| K (x^{'}, y) | \leq \frac{2^{a^{3}}}{μ (B (x^{'}, \frac{1}{4} D^{s_{1} - 1}))} .

Using $D = 2^{100 a^{2}}$ and the doubling property 1.0.5 $7 + 100 a^{2}$ times estimates the last display by

\leq \frac{2^{7 a + 101 a^{3}}}{μ (B (x^{'}, 32 D^{s_{1}}))} .

7.2.6

By the triangle inequality and 2.0.10, we have

B (x^{'}, 8 D^{s_{1}}) \subset B (c (I_{s_{1}} (x)), 16 D^{s (I_{s_{1}} (x))}) \subset B (x^{'}, 32 D^{s_{1}}) .

Combining this with 7.2.6, we conclude that 7.2.4 is at most

2^{7 a + 101 a^{3}} M_{B, 1} f (x) .

For 7.2.5 we argue similarly. We have for all $y$ with $ρ (x^{'}, y) \geq \frac{1}{4} D^{s_{2} - 1}$

| K (x^{'}, y) | \leq \frac{2^{a^{3}}}{μ (B (x^{'}, \frac{1}{4} D^{s_{2} - 1}))} .

Using the doubling property 1.0.5 $7 + 100 a^{2}$ times estimates the last display by

\leq \frac{2^{7 a + 101 a^{3}}}{μ (B (x^{'}, 32 D^{s_{2}}))} .

7.2.7

Note that by 2.0.8 we have $I_{s_{1}} (x) \subset I_{s_{2}} (x)$ , in particular $x^{'} \in I_{s_{2}} (x)$ . By the triangle inequality and 2.0.10, we have

B (x^{'}, 8 D^{s_{2}}) \subset B (c (I_{s_{2}} (x)), 16 D^{s (I_{s_{2}} (x))}) \subset B (x^{'}, 32 D^{s_{2}}) .

Combining this, 7.2.5 is at most

2^{7 a + 101 a^{3}} M_{B, 1} f (x) .

Using $a \geq 4$ , taking a supremum over all $x^{'} \in I_{s_{1}} (x)$ and then a supremum over all $- S \leq s_{1} < s_{2} \leq S$ , we obtain

T_{N} f (x) \leq T_{*} f (x) + 2^{102 a^{3}} M_{B, 1} f (x) .

The lemma now follows from assumption 1.0.18, Proposition 2.0.6 and $a \geq 4$ .

We need the following lemma to prepare the $L^{2}$ -estimate for the auxiliary operators $S_{1, u}$ .

Lemma 7.2.4 boundary overlap

✓

For every cube $I \in D$ , there exist at most $2^{9 a}$ cubes $J \in D$ with $s (J) = s (I)$ and $B (c (I), 16 D^{s (I)}) \cap B (c (J), 16 D^{s (J)}) \neq \emptyset$ .

Proof ▶

Suppose that $B (c (I), 16 D^{s (I)}) \cap B (c (J), 16 D^{s (J)}) \neq \emptyset$ and $s (I) = s (J)$ . Then $B (c (I), 33 D^{s (I)}) \subset B (c (J), 128 D^{s (J)})$ . Hence by the doubling property 1.0.5

2^{9 a} μ (B (c (J), \frac{1}{4} D^{s (J)})) \geq μ (B (c (I), 33 D^{s (I)})),

and by the triangle inequality, the ball $B (c (J), \frac{1}{4} D^{s (J)})$ is contained in $B (c (I), 33 D^{s (I)})$ .

If $C$ is any finite collection of cubes $J \in D$ satisfying $s (J) = s (I)$ and

B (c (I), 16 D^{s (I)}) \cap B (c (J), 16 D^{s (J)}) \neq \emptyset,

then it follows from 2.0.10 and pairwise disjointedness of cubes of the same scale 2.0.8 that the balls $B (c (J), \frac{1}{4} D^{s (J)})$ are pairwise disjoint. Hence

\begin{aligned} μ (B (c (I), 33 D^{s (I)})) & \geq \sum_{J \in C} μ (B (c (J), \frac{1}{4} D^{s (J)})) \\ \geq | C | 2^{- 9 a} μ (B (c (I), 33 D^{s (I)})) . \end{aligned}

Since $μ$ is doubling and $μ \neq 0$ , we have $μ (B (c (I), 33 D^{s (I)})) > 0$ . The lemma follows after dividing by $2^{- 9 a} μ (B (c (I), 33 D^{s (I)}))$ .

Now we can bound the operators $S_{1, u}$ .

Proof of Lemma 7.2.3 ▶

Note that by definition, $S_{1, u} f$ is a finite sum of indicator functions of cubes $I \in D$ for each locally integrable $f$ , and hence is bounded, has bounded support and is integrable. Let $g$ be another function with the same three properties. Then $\bar{g} S_{1, u} f$ is integrable, and we have

| \int \bar{g} (y) S_{1, u} f (y) d μ (y) |

Unknown environment 'bgroup'

Unknown environment 'bgroup'

Changing the order of summation and using $J \subset B (c (I), 16 D^{s (I)})$ to bound the first average integral by $M_{B, 1} | g | (y)$ for any $y \in J$ , we obtain

Unknown environment 'bgroup'

By Lemma 7.2.4, there are at most $2^{9 a}$ cubes $I$ at each scale with $J \subset B (c (I), 16 D^{s (I)})$ . Since $D^{- 1 / a} \leq \frac{1}{2}$ , $(1 - D^{- 1 / a})^{- 1} \leq 2$ . Using this estimate for the sum of the geometric series, we conclude that 7.2.8 is at most

2^{9 a + 1} \sum_{J \in J (T (u))} \int_{J} | f (y) | M_{B, 1} | g | (y) d μ (y) .

The collection $J$ is a partition of $⋃_{I \in D} I$ , so this equals

2^{9 a + 1} \int_{X} | f (y) | M_{B, 1} | g | (y) d μ (y) .

Using Cauchy-Schwarz and Proposition 2.0.6, we conclude

| \int \bar{g} S_{1, u} f d μ | \leq 2^{12 a} ‖ g ‖_{2} ‖ f ‖_{2} .

The lemma now follows by choosing $g = S_{1, u} f$ and dividing on both sides by the finite $‖ S_{1, u} f ‖_{2}$ .

7.3 The quantitative $L^{2}$ tree estimate

The main result of this subsection is the following quantitative bound for operators associated to trees, with decay in the densities ${dens}_{1}$ and ${dens}_{2}$ .

Lemma 7.3.1 densities tree bound

✓

Let $u \in U$ . Then for all bounded $f$ with bounded support and $g$ with $| g | \leq 1_{G}$ we have

| \int_{X} \bar{g} \sum_{p \in T (u)} T_{p} f d μ | \leq 2^{202.5 a^{3}} {dens}_{1} (T (u))^{1 / 2} ‖ f ‖_{2} ‖ g ‖_{2} .

7.3.1

If additionally $| f | \leq 1_{F}$ , then we have

| \int_{X} \bar{g} \sum_{p \in T (u)} T_{p} f d μ | \leq 2^{303 a^{3}} {dens}_{1} (T (u))^{1 / 2} {dens}_{2} (T (u))^{1 / 2} ‖ f ‖_{2} ‖ g ‖_{2} .

7.3.2

Below, we deduce this lemma from Lemma 7.2.1 and the following two estimates controlling the size of support of the operator and its adjoint.

Lemma 7.3.2 local dens1 tree bound

✓

Let $u \in U$ and $L \in L (T (u))$ . Then

μ (L \cap G \cap ⋃_{p \in T (u)} E (p)) \leq 2^{101 a^{3}} {dens}_{1} (T (u)) μ (L) .

7.3.3

Lemma 7.3.3 local dens2 tree bound

✓

Let $u \in U$ and $J \in J (T (u))$ . Then

μ (F \cap J) \leq 2^{201 a^{3}} {dens}_{2} (T (u)) μ (J) .

Proof of Lemma 7.3.1 ▶

Denote

E (u) = ⋃_{p \in T (u)} E (p) .

Then we have

| \int_{X} \bar{g} \sum_{p \in T (u)} T_{p} f d μ | = | \int_{X} \overset{―}{g 1_{E (u)}} \sum_{p \in T (u)} T_{p} (1_{I (u)} f) d μ | .

By Lemma 7.2.1, this is bounded by

\leq 2^{152 a^{3}} ‖ P_{J (T (u))} | f | ‖_{2} ‖ P_{L (T (u))} | 1_{E (u)} g | ‖_{2} .

7.3.4

We bound the two factors separately. We have

‖ P_{L (T (u))} | 1_{E (u)} g | ‖_{2} = {(\sum_{L \in L (T (u))} \frac{1}{μ (L)} {(\int_{L \cap E (u)} | g (y) | d μ (y))}^{2})}^{1 / 2} .

By Cauchy-Schwarz and Lemma 7.3.2 this is at most

\leq {(\sum_{L \in L (T (u))} 2^{101 a^{3}} {dens}_{1} (T (u)) \int_{L \cap E (u)} | g (y) |^{2} d μ (y))}^{1 / 2} .

Since cubes $L \in L (T (u))$ are pairwise disjoint by Lemma 7.1.2, this is

\leq 2^{50.5 a^{3}} {dens}_{1} (T (u))^{1 / 2} ‖ g ‖_{2} .

7.3.5

Similarly, we have

‖ P_{J (T (u))} | f | ‖_{2} = {(\sum_{J \in J (T (u))} \frac{1}{μ (J)} {(\int_{J} | f (y) | d μ (y))}^{2})}^{1 / 2} .

7.3.6

By Cauchy-Schwarz, this is

\leq {(\sum_{J \in J (T (u))} \int_{J} | f (y) |^{2} d μ (y))}^{1 / 2} .

Since cubes in $J (T (u))$ are pairwise disjoint by Lemma 7.1.2, this is at most

‖ f ‖_{2} .

7.3.7

Combining 7.3.4, 7.3.5 and 7.3.7 gives 7.3.1.

If $f \leq 1_{F}$ then $f = f 1_{F}$ , so

{(\sum_{J \in J (T (u))} \frac{1}{μ (J)} {(\int_{J} | f (y) | d μ (y))}^{2})}^{1 / 2}

= {(\sum_{J \in J (T (u))} \frac{1}{μ (J)} {(\int_{J \cap F} | f (y) | d μ (y))}^{2})}^{1 / 2} .

We estimate as before, using now Lemma 7.3.3 and Cauchy-Schwarz, and obtain that this is

\leq 2^{100.5 a^{3}} {dens}_{2} (T (u))^{1 / 2} ‖ f ‖_{2} .

Combining this with 7.3.4 and 7.3.5 gives 7.3.2.

Now we prove the two auxiliary estimates.

Proof of Lemma 7.3.2 ▶

If the set on the right hand side is empty, then 7.3.3 holds. If not, then there exists $p \in T (u)$ with $L \cap I (p) \neq \emptyset$ .

Suppose first that there exists such $p$ with $s (p) \leq s (L)$ . Then by 2.0.8 $I (p) \subset L$ , which gives by the definition of $L (T (u))$ that $s (L) = - S$ and hence $L = I (p)$ . Let $q \in T (u)$ with $E (q) \cap L \neq \emptyset$ . Since $s (L) = - S \leq s (q)$ it follows from 2.0.8 that $I (p) = L \subset I (q)$ . We have then by Lemma 2.1.2

\begin{aligned} d_{p} (Q (p), Q (q)) & \leq d_{p} (Q (p), Q (u)) + d_{p} (Q (q), Q (u)) \\ \leq d_{p} (Q (p), Q (u)) + d_{q} (Q (q), Q (u)) . \end{aligned}

Using that $p, q \in T (u)$ and 2.0.32, this is at most $8$ . Using again the triangle inequality and Lemma 2.1.2, we obtain that for each $q \in B_{q} (Q (q), 1)$

d_{p} (Q (p), q) \leq d_{p} (Q (p), Q (q)) + d_{q} (Q (q), q) \leq 9 .

Thus $L \cap G \cap E (q) \subset E_{2} (9, p)$ . We obtain

μ (L \cap G \cap ⋃_{q \in T (u)} E (q)) \leq μ (E_{2} (9, p)) .

By the definition of ${dens}_{1}$ , this is bounded by

9^{a} {dens}_{1} (T (u)) μ (I (p)) = 9^{a} {dens}_{1} (T (u)) μ (L) .

Since $a \geq 4$ , 7.3.3 follows in this case.

Now suppose that for each $p \in T (u)$ with $L \cap E (p) \neq \emptyset$ , we have $s (p) > s (L)$ . Since there exists at least one such $p$ , there exists in particular at least one cube $L^{″} \in D$ with $L \subset L^{″}$ and $s (L^{″}) > s (L)$ . By 2.0.7, there exists $L^{'} \in D$ with $L \subset L^{'}$ and $s (L^{'}) = s (L) + 1$ . By the definition of $L (T (u))$ there exists a tile $p^{″} \in T (u)$ with $I (p^{″}) \subset L^{'}$ .

It suffices to show that there exists a tile $p^{'} \in P (T (u))$ with $I (p^{'}) = L^{'}$ , $d_{p^{'}} (Q (p^{'}), Q (u)) < 4$ and $9 p^{″} ≲ 9 p^{'}$ . For then, let $q \in T (u)$ with $L \cap E (q) \neq \emptyset$ . As shown above, this implies $s (q) \geq s (L^{'})$ , so by 2.0.8 $L^{'} \subset I (q)$ . If $q \in B_{q} (Q (q), 1)$ , then by a similar calculation as above, using the triangle inequality, Lemma 2.1.2 and 2.0.32, we obtain

d_{p^{'}} (Q (p^{'}), q) \leq d_{p^{'}} (Q (p^{'}), Q (q)) + d_{q} (Q (q), q) \leq 9 .

Thus $L \cap G \cap E (q) \subset E_{2} (9, p^{'})$ . We deduce using the definition 2.0.28 of ${dens}_{1}$

μ (L \cap G \cap ⋃_{q \in T (u)} E (q)) \leq μ (E_{2} (9, p^{'})) \leq 9^{a} {dens}_{1} (T (u)) μ (L^{'}) .

Using the doubling property 1.0.5, 2.0.10, and $a \geq 4$ this is estimated by

9^{a} 2^{100 a^{3} + 5 a} {dens}_{1} (T (u)) μ (L) \leq 2^{101 a^{3}} {dens}_{1} (T (u)) μ (L) .

To show existence of $p^{'}$ with the given properties, if $I (p^{″}) = L^{'}$ we can take $p^{'} = p^{″}$ , which satisfies the distance property by 2.0.32 and the other properties trivially. Otherwise, let $p^{'}$ be the unique tile such that $I (p^{'}) = L^{'}$ and such that $Ω (u) \cap Ω (p^{'}) \neq \emptyset$ . Since $I (p^{'}) \subset I (p)$ and $p \in T (u)$ , we have $p^{'} \in P (T (u))$ . Since by 2.0.32 $s (p^{'}) = s (L^{'}) \leq s (p) < s (u)$ , we have by 2.0.8 and 2.0.14 that $Ω (u) \subset Ω (p^{'})$ , and hence the distance property. $9 p^{″} ≲ 9 p^{'}$ follows by the triangle inequality, 2.0.32, Lemma 2.1.2 and 2.0.15. This completes the proof.

Proof of Lemma 7.3.3 ▶

We prove the inequality with the constant $2^{201 a^{3}}$ replaced by $2^{200 a^{3} + 14 a}$ ; this is stronger because $a \geq 4$ . It suffices to show the existence of a tile $p \in T (u)$ and an $r \geq 4 D^{s (p)}$ such that $J \subset B (c (p), r)$ and $μ (B (c (p), r)) \leq 2^{200 a^{3} + 14 a} μ (J)$ , because then it follows from the definition 2.0.29 of ${dens}_{2}$ that

μ (F \cap J) \leq μ (F \cap B (c (p), r))

\leq {dens}_{2} (T (u)) μ (B (c (p), r)) \leq 2^{200 a^{3} + 14 a} {dens}_{2} (T (u)) μ (J) .

In particular, these criteria are satisfied, with $r = 4 D^{s (p)}$ , by any $p \in T (u)$ such that $J \subseteq B (c (p, 4 D^{s (p)}))$ and $μ (I (p)) \leq 2^{100 a^{3} + 10 a} μ (J)$ , because then by the doubling property 1.0.5,

μ (B (c (p), 4 D^{s (p)})) \leq 2^{4 a} μ (B (c (p), D^{s (p)} / 4))

\leq 2^{4 a} μ (I (p)) \leq 2^{100 a^{3} + 14 a} μ (J) .

Suppose first that $s (J) = S$ . Then $J = I_{0}$ , so 2.0.9 and the fact that $J \in J (T (u)) \subseteq J_{0} (T (u))$ imply that $s (J) = - S$ . Thus $S = 0$ . It follows that $J$ is the only dyadic cube, so any $p \in T (u)$ has $I (p) = J$ , and therefore satisfies $J \subseteq B (c (p, 4 D^{s (p)}))$ and $μ (I (p)) \leq 2^{100 a^{3} + 10 a} μ (J)$ .

It remains to consider the case $s (J) < S$ . Then, by 2.0.7 and 2.0.8, there exists some cube $J^{'} \in D$ with $s (J^{'}) = s (J) + 1$ and $J \subset J^{'}$ . By definition of $J (T (u))$ there exists some $p \in T (u)$ such that $I (p) \subset B (c (J^{'}), 100 D^{s (J^{'}) + 1})$ .

Since $c (J) \in J \subset J^{'} \subset B (c (J^{'}), 4 D^{s (J^{'})})$ , the triangle inequality, $s (J^{'}) = s (J) + 1$ and $D = 2^{100 a^{2}}$ imply

B (c (J^{'}), 204 D^{s (J^{'}) + 1}) \subset B (c (J), 204 D^{s (J^{'}) + 1} + 4 D^{s (J^{'})}) \subset B (c (J), 2^{8} D^{s (J) + 2}) .

From the doubling property 1.0.5, $D = 2^{100 a^{2}}$ and 2.0.10, we obtain

μ (B (c (J^{'}), 204 D^{s (J^{'}) + 1})) \leq 2^{200 a^{3} + 10 a} μ (J) .

7.3.8

If $J \subset B (c (p), 4 D^{s (p)})$ , then we need only check that $μ (I (p)) \leq 2^{100 a^{3} + 10 a} μ (J)$ . This follows immediately from $I (p) \subset B (c (J^{'}), 100 D^{s (J^{'}) + 1})$ and 7.3.8.

From now on we assume $J ⊄ B (c (p), 4 D^{s (p)})$ . Since

c (p) \in I (p) \subset B (c (J^{'}), 100 D^{s (J^{'}) + 1}),

we have by 2.0.10 and the triangle inequality

J \subset J^{'} \subset B (c (J^{'}), 4 D^{s (J^{'})}) \subset B (c (p), 104 D^{s (J^{'}) + 1}) .

In particular this implies $104 D^{s (J^{'}) + 1} > 4 D^{s (p)}$ . By the triangle inequality we also have

B (c (p), 104 D^{s (J^{'}) + 1}) \subset B (c (J), 204 D^{s (J^{'}) + 1}),

so from 7.3.8,

μ (B (c (p), 104 D^{s (J^{'}) + 1})) \leq 2^{200 a^{3} + 10 a} μ (J),

which proves $p$ satisfies the needed criteria with $r = 104 D^{s (J^{'}) + 1}$ .

7.4 Almost orthogonality of separated trees

The main result of this subsection is the almost orthogonality estimate for operators associated to distinct trees in a forest in Lemma 7.4.4 below. We will deduce it from Lemmas 7.4.5 and 7.4.6, which are proven in Subsections 7.5 and 7.6, respectively. Before stating it, we introduce some relevant notation.

The adjoint of the operator $T_{p}$ defined in 2.0.21 is given by

T_{p}^{*} g (x) = \int_{E (p)} \overset{―}{K_{s (p)} (y, x)} e (- Q (y) (x) + Q (y) (y)) g (y) d μ (y) .

7.4.1

Lemma 7.4.1 adjoint tile support

✓

For each $p \in P$ , we have

T_{p}^{*} g = 1_{B (c (p), 5 D^{s (p)})} T_{p}^{*} 1_{I (p)} g .

For each $u \in U$ and each $p \in T (u)$ , we have

T_{p}^{*} g = 1_{I (u)} T_{p}^{*} 1_{I (u)} g .

Proof ▶

By 2.0.32, $E (p) \subset I (p) \subset I (u)$ . Thus by 7.4.1

T_{p}^{*} g (x) = T_{p}^{*} (1_{I (p)} g) (x)

= \int_{E (p)} \overset{―}{K_{s (p)} (y, x)} e (- Q (y) (x) + Q (y) (y)) 1_{I (p)} (y) g (y) d μ (y) .

If this integral is not $0$ , then there exists $y \in I (p)$ such that $K_{s (p)} (y, x) \neq 0$ . By 2.1.2, 2.0.10 and the triangle inequality, it follows that

x \in B (c (p), 5 D^{s (p)}) .

Thus

T_{p}^{*} g (x) = 1_{B (c (p), 5 D^{s (p)})} (x) T_{p}^{*} (1_{I (p)} g) (x) .

The second claimed equation follows now since $I (p) \subset I (u)$ and by 2.0.37 $B (c (p), 5 D^{s (p)}) \subset I (u)$ .

Lemma 7.4.2 adjoint tree estimate

✓

For all $g$ with $| g | \leq 1_{G}$ , we have that

{‖ \sum_{p \in T (u)} T_{p}^{*} g ‖}_{2} \leq 2^{155 a^{3}} {dens}_{1} (T (u))^{1 / 2} ‖ g ‖_{2} .

Proof ▶

By Lemma 7.3.1, we have for all bounded $f$ and $g$ with $| g | \leq 1_{G}$ that

| \int_{X} \overset{―}{\sum_{p \in T (u)} T_{p}^{*} g} f d μ | = | \int_{X} \overset{―}{g} \sum_{p \in T (u)} T_{p} f d μ |

\leq 2^{155 a^{3}} {dens}_{1} (T (u))^{1 / 2} ‖ g ‖_{2} ‖ f ‖_{2} .

7.4.2

Let $f = \sum_{p \in T (u)} T_{p}^{*} g$ . Since $| g | \leq 1_{G}$ , $f$ is bounded and has bounded support. In particular $‖ f ‖_{2} < \infty$ . Dividing 7.4.2 by $‖ f ‖_{2}$ completes the proof.

We define

S_{2, u} g := | \sum_{p \in T (u)} T_{p}^{*} g | + M_{B, 1} g + | g | .

Lemma 7.4.3 adjoint tree control

✓

We have for all $u \in U$ and $g$ with $| g | \leq 1_{G}$

‖ S_{2, u} g ‖_{2} \leq 2^{156 a^{3}} ‖ g ‖_{2} .

Proof ▶

This follows immediately from Minkowski’s inequality, Proposition 2.0.6 and Lemma 7.4.2, using that $a \geq 4$ .

Now we are ready to state the main result of this subsection.

Lemma 7.4.4 correlation separated trees

For any $u_{1} \neq u_{2} \in U$ and all bounded $g_{1}, g_{2}$ with bounded support, we have

| \int_{X} \sum_{p_{1} \in T (u_{1})} \sum_{p_{2} \in T (u_{2})} T_{p_{1}}^{*} g_{1} \overset{―}{T_{p_{2}}^{*} g_{2}} d μ |

7.4.3

\leq 2^{550 a^{3} - 3 n} \prod_{j = 1}^{2} ‖ S_{2, u_{j}} g_{j} ‖_{L^{2} (I (u_{1}) \cap I (u_{2}))} .

7.4.4

Proof of Lemma 7.4.4 ▶

By Lemma 7.4.1 and 2.0.8, the left hand side 7.4.3 is $0$ unless $I (u_{1}) \subset I (u_{2})$ or $I (u_{2}) \subset I (u_{1})$ . Without loss of generality we assume that $I (u_{1}) \subset I (u_{2})$ .

Define

S := {p \in T (u_{1}) \cup T (u_{2}) : d_{p} (Q (u_{1}), Q (u_{2})) \geq 2^{Z n / 2}} .

7.4.5

Lemma 7.4.4 follows by combining the definition 2.0.3 of $Z$ with the following two lemmas.

Lemma 7.4.5 correlation distant tree parts

We have for all $u_{1} \neq u_{2} \in U$ with $I (u_{1}) \subset I (u_{2})$ and all bounded $g_{1}, g_{2}$ with bounded support

| \int_{X} \sum_{p_{1} \in T (u_{1})} \sum_{p_{2} \in T (u_{2}) \cap S} T_{p_{1}}^{*} g_{1} \overset{―}{T_{p_{2}}^{*} g_{2}} d μ |

7.4.6

\leq 2^{541 a^{3}} 2^{- Z n / (4 a^{2} + 2 a^{3})} \prod_{j = 1}^{2} ‖ S_{2, u_{j}} g_{j} ‖_{L^{2} (I (u_{1}))} .

7.4.7

Lemma 7.4.6 correlation near tree parts

We have for all $u_{1} \neq u_{2} \in U$ with $I (u_{1}) \subset I (u_{2})$ and all bounded $g_{1}, g_{2}$ with bounded support

| \int_{X} \sum_{p_{1} \in T (u_{1})} \sum_{p_{2} \in T (u_{2}) ∖ S} T_{p_{1}}^{*} g_{1} \overset{―}{T_{p_{2}}^{*} g_{2}} d μ |

7.4.8

\leq 2^{222 a^{3}} 2^{- Z n 2^{- 10 a}} \prod_{j = 1}^{2} ‖ S_{2, u_{j}} g_{j} ‖_{L^{2} (I (u_{1}))} .

7.4.9

In the proofs of both lemmas, we will need the following observation.

Lemma 7.4.7 overlap implies distance

✓

Let $u_{1} \neq u_{2} \in U$ with $I (u_{1}) \subset I (u_{2})$ . If $p \in T (u_{1}) \cup T (u_{2})$ with $I (p) \cap I (u_{1}) \neq \emptyset$ , then $p \in S$ . In particular, we have $T (u_{1}) \subset S$ .

Proof ▶

Suppose first that $p \in T (u_{1})$ . Then $I (p) \subset I (u_{1}) \subset I (u_{2})$ , by 2.0.32. Thus we have by the separation condition 2.0.36, 2.0.15, 2.0.32 and the triangle inequality

\begin{aligned} d_{p} (Q (u_{1}), Q (u_{2})) & \geq d_{p} (Q (p), Q (u_{2})) - d_{p} (Q (p), Q (u_{1})) \\ \geq 2^{Z (n + 1)} - 4 \\ \geq 2^{Z n / 2}, \end{aligned}

using that $Z = 2^{12 a} \geq 4$ . Hence $p \in S$ .

Suppose now that $p \in T (u_{2})$ . If $I (p) \subset I (u_{1})$ , then the same argument as above with $u_{1}$ and $u_{2}$ swapped shows $p \in S$ . If $I (p) ⊄ I (u_{1})$ then, by 2.0.8, $I (u_{1}) \subset I (p)$ . Pick $p^{'} \in T (u_{1})$ , we have $I (p^{'}) \subset I (u_{1}) \subset I (p)$ . Hence, by Lemma 2.1.2 and the first paragraph

d_{p} (Q (u_{1}), Q (u_{2})) \geq d_{p^{'}} (Q (u_{1}), Q (u_{2})) \geq 2^{Z n},

so $p \in S$ .

To simplify the notation, we will write at various places throughout the proof of Lemmas 7.4.5 and 7.4.6 for a subset $C \subset P$

T_{C} f := \sum_{p \in C} T_{p} f, T_{C}^{*} g := \sum_{p \in C} T_{p}^{*} g .

7.5 Proof of the Tiles with large separation Lemma

Lemma 7.4.5 follows from the van der Corput estimate in Proposition 2.0.5. We apply this proposition in Section 7.5.3. To prepare this application, we first, in Section 7.5.1, construct a suitable partition of unity, and show then, in Section 7.5.2 the Hölder estimates needed to apply Proposition 2.0.5.

7.5.1 A partition of unity

Define

J^{'} = {J \in J (S) : J \subset I (u_{1})} .

Lemma 7.5.1 dyadic partition 1

✓

We have that

I (u_{1}) = \dot{⋃_{J \in J^{'}}} J .

Proof ▶

By Lemma 7.1.2, it remains only to show that each $J \in J (S)$ with $J \cap I (u_{1}) \neq \emptyset$ is in $J^{'}$ . But if $J \notin J^{'}$ , then by 2.0.8 $I (u_{1}) \subset J$ . Pick $p \in T (u_{1}) \subset S$ . Then $I (p) \subset J$ . This contradicts the definition of $J (S)$ .

For cubes $J \in D$ , denote

B (J) := B (c (J), 8 D^{s (J)}) .

7.5.1

The main result of this subsubsection is the following.

Lemma 7.5.2 Lipschitz partition unity

✓

There exists a family of functions $χ_{J}$ , $J \in J^{'}$ such that

1_{I (u_{1})} = \sum_{J \in J^{'}} χ_{J},

7.5.2

and for all $J \in J^{'}$ and all $y, y^{'} \in I (u_{1})$

0 \leq χ_{J} (y) \leq 1_{B (J)} (y),

7.5.3

| χ_{J} (y) - χ_{J} (y^{'}) | \leq 2^{226 a^{3}} \frac{ρ (y, y^{'})}{D^{s (J)}} .

7.5.4

In the proof, we will use the following auxiliary lemma.

Lemma 7.5.3 moderate scale change

✓

If $J, J^{'} \in J^{'}$ with

B (J) \cap B (J^{'}) \neq \emptyset,

then $| s (J) - s (J^{'}) | \leq 1$ .

Proof of Lemma 7.5.2 ▶

For each cube $J \in J$ let

{\tilde{χ}}_{J} (y) = 1_{I (u_{1})} (y) max {0, 8 - D^{- s (J)} ρ (y, c (J))},

and set

a (y) = \sum_{J \in J^{'}} {\tilde{χ}}_{J} (y) .

We define

χ_{J} (y) := \frac{{\tilde{χ}}_{J} (y)}{a (y)} .

Then, due to 2.0.37 and 7.5.1, the properties 7.5.2 and 7.5.3 are clearly true. Estimate 7.5.4 follows from 7.5.3 if $y, y^{'} \notin B (J)$ . Thus we can assume that $y \in B (J)$ . We have by the triangle inequality

| χ_{J} (y) - χ_{J} (y^{'}) | \leq \frac{| {\tilde{χ}}_{J} (y) - {\tilde{χ}}_{J} (y^{'}) |}{a (y)} + \frac{{\tilde{χ}}_{J} (y^{'}) | a (y) - a (y^{'}) |}{a (y) a (y^{'})}

Since ${\tilde{χ}}_{J} (z) \geq 4$ for all $z \in B (c (J), 4 D^{s (J)}) \supset J$ and by Lemma 7.5.1, we have that $a (z) \geq 4$ for all $z \in I (u_{1})$ . So we can estimate the above further by

\leq 2^{- 2} (| {\tilde{χ}}_{J} (y) - {\tilde{χ}}_{J} (y^{'}) | + {\tilde{χ}}_{J} (y^{'}) | a (y) - a (y^{'}) |) .

If $y^{'} \notin B (c (p), 8 D^{s (p)})$ then the second summand vanishes. Else, we can estimate the above, using also that $| {\tilde{χ}}_{J} (y^{'}) | \leq 8$ , by

\leq 2^{- 2} | {\tilde{χ}}_{J} (y) - {\tilde{χ}}_{J} (y^{'}) | + 2 \sum_{\begin{matrix} J^{'} \in J^{'} \\ B (J^{'}) \cap B (J) \neq \emptyset \end{matrix}} | {\tilde{χ}}_{J^{'}} (y) - {\tilde{χ}}_{J^{'}} (y^{'}) | .

By the triangle inequality, we have for all dyadic cubes $I$

| {\tilde{χ}}_{I} (y) - {\tilde{χ}}_{I} (y^{'}) | \leq ρ (y, y^{'}) D^{- s (I)} .

Using this above, we obtain

| χ_{J} (y) - χ_{J} (y^{'}) | \leq ρ (y, y^{'}) (\frac{1}{4} D^{- s (J)} + 2 \sum_{\begin{matrix} J^{'} \in J^{'} \\ B (J^{'}) \cap B (J) \neq \emptyset \end{matrix}} D^{- s (J^{'})}) .

By Lemma 7.5.3, this is at most

\frac{ρ (y, y^{'})}{D^{s (J)}} (\frac{1}{4} + 2 D | {J^{'} \in J^{'} : B (J^{'}) \cap B (J) \neq \emptyset} |) .

By 2.0.10 and Lemma 7.5.1, the balls $B (c (J^{'}), \frac{1}{4} D^{s (J^{'})})$ are pairwise disjoint. By the triangle inequality and Lemma 7.5.3, each such ball for $J^{'}$ in the set of the last display is contained in

B (c (J), 9 D^{s (J) + 1}) .

By the doubling property 1.0.5, we further have

μ (B (c (J), 9 D^{s (J) + 1})) \leq 2^{200 a^{3} + 7 a} μ (B (c (J^{'}), \frac{1}{4} D^{s (J^{'})}))

for each such ball. Thus

| {J^{'} \in J^{'} : B (J^{'}) \cap B (J) \neq \emptyset} | \leq 2^{200 a^{3} + 7 a} .

Recalling that $D = 2^{100 a^{2}}$ , we obtain

\frac{1}{4} + 2 D | {J^{'} \in J^{'} : B (J^{'}) \cap B (J) \neq \emptyset} | \leq 2^{200 a^{3} + 100 a^{2} + 7 a + 2} .

Since $a \geq 4$ , 7.5.4 follows.

Proof of Lemma 7.5.3 ▶

Suppose that $s (J^{'}) < s (J) - 1$ . Then $s (J) > - S$ . Thus, by the definition of $J^{'}$ there exists no $p \in S$ with

I (p) \subset B (c (J), 100 D^{s (J) + 1}) .

7.5.5

Since $s (J^{'}) < s (J)$ , there exists a cube $J^{″} \in D$ with $J \subset J^{″}$ and $s (J^{″}) = s (J^{'}) + 1$ . By the definition of $J^{'}$ , there exists a tile $p \in S$ with

I (p) \subset B (c (J^{″}), 100 D^{s (J^{'}) + 2}) .

7.5.6

But by the triangle inequality and 2.0.1, we have

B (c (J^{″}), 100 D^{s (J^{'}) + 2}) \subset B (c (J), 100 D^{s (J) + 1}),

which contradicts 7.5.5 and 7.5.6.

7.5.2 Hölder estimates for adjoint tree operators

Let $g_{1}, g_{2} : X \to C$ be bounded with bounded support. Define for $J \in J^{'}$

h_{J} (y) := χ_{J} (y) \cdot (e (Q (u_{1}) (y)) T_{T (u_{1})}^{*} g_{1} (y)) \cdot \overset{―}{(e (Q (u_{2}) (y)) T_{T (u_{2}) \cap S}^{*} g_{2} (y))} .

7.5.7

The main result of this subsubsection is the following $τ$ -Hölder estimate for $h_{J}$ , where $τ = 1 / a$ .

Lemma 7.5.4 Holder correlation tree

We have for all $J \in J^{'}$ that

‖ h_{J} ‖_{C^{τ} (B (c (J), 8 D^{s (J)}))} \leq 2^{535 a^{3}} \prod_{j = 1, 2} (inf_{B (c (J), \frac{1}{8} D^{s (J)})} | T_{T (u_{j})}^{*} g_{j} | + inf_{J} M_{B, 1} | g_{j} |) .

7.5.8

We will prove this lemma at the end of this section, after establishing several auxiliary results.

We begin with the following Hölder continuity estimate for adjoints of operators associated to tiles.

Lemma 7.5.5 Holder correlation tile

✓

Let $u \in U$ and $p \in T (u)$ . Then for all $y, y^{'} \in X$ and all bounded $g$ with bounded support, we have

| e (Q (u) (y)) T_{p}^{*} g (y) - e (Q (u) (y^{'})) T_{p}^{*} g (y^{'}) |

\leq \frac{2^{151 a^{3}}}{μ (B (c (p), 4 D^{s (p)}))} {(\frac{ρ (y, y^{'})}{D^{s (p)}})}^{1 / a} \int_{E (p)} | g (x) | d μ (x) .

7.5.9

Proof ▶

By 7.4.1, we have

| e (Q (u) (y)) T_{p}^{*} g (y) - e (Q (u) (y^{'})) T_{p}^{*} g (y^{'}) |

\begin{matrix} = | \int_{E (p)} e (Q (x) (x) - Q (x) (y) + Q (u) (y)) \overset{―}{K_{s (p)} (x, y)} g (x) \\ - e (Q (x) (x) - Q (x) (y^{'}) + Q (u) (y^{'})) \overset{―}{K_{s (p)} (x, y^{'})} g (x) d μ (x) | \end{matrix}

\begin{matrix} \leq \int_{E (p)} | g (x) | | e (Q (x) (y) - Q (x) (y^{'}) - Q (u) (y) + Q (u) (y^{'})) \overset{―}{K_{s (p)} (x, y)} \\ - \overset{―}{K_{s (p)} (x, y^{'})} | d μ (x) \end{matrix}

\begin{matrix} \leq \int_{E (p)} | g (x) | | e (- Q (x) (y) + Q (x) (y^{'}) + Q (u) (y) - Q (u) (y^{'})) - 1 | \\ \times | \overset{―}{K_{s (p)} (x, y)} | d μ (x) \end{matrix}

+ \int_{E (p)} | g (x) | | \overset{―}{K_{s (p)} (x, y)} - \overset{―}{K_{s (p)} (x, y^{'})} | d μ (x) .

7.5.12

By the oscillation estimate 1.0.7, we have

| - Q (x) (y) + Q (x) (y^{'}) + Q (u) (y) - Q (u) (y^{'}) |

\leq d_{B (y, 1.6 ρ (y, y^{'}))} (Q (x), Q (u)) .

7.5.13

Suppose that $y, y^{'} \in B (c (p), 5 D^{s (p)})$ , so that $ρ (y, y^{'}) \leq 10 D^{s (p)}$ . Let $k \in Z$ be such that $2^{a k} ρ (y, y^{'}) \leq 10 D^{s (p)}$ but $2^{a (k + 1)} ρ (y, y^{'}) > 10 D^{s (p)}$ . In particular, $k \geq 0$ . Then, using 1.0.10 followed by 1.0.8, we can bound 7.5.13 from above by

2^{- k} d_{B (c (p), 16 D^{s (p)})} (Q (x), Q (u)) \leq 2^{6 a - k} d_{p} (Q (x), Q (u)) .

Since $x \in E (p)$ we have $Q (x) \in Ω (p) \subset B_{p} (Q (p), 1)$ , and since $p \in T (u)$ we have $Q (u) \in B_{p} (Q (p), 4)$ , so this is estimated by

\leq 5 \cdot 2^{6 a - k} .

By definition of $k$ , we have

- k < 1 - \frac{1}{a} \log_{2} (\frac{10 D^{s (p)}}{ρ (y, y^{'})}),

which gives

| - Q (x) (y) + Q (x) (y^{'}) + Q (u) (y) - Q (u) (y^{'}) | \leq 10 \cdot 2^{6 a} {(\frac{ρ (y, y^{'})}{10 D^{s (p)}})}^{1 / a} .

7.5.14

For all $x \in I (p)$ , we have by 1.0.5 that

μ (B (x, D^{s (p)})) \geq 2^{- 3 a} μ (B (c (p), 4 D^{s (p)})) .

Combining the above with 2.1.3, 2.1.4 and 7.5.14, we obtain

(???) + (???) \leq \frac{2^{3 a}}{μ (B (c (p), 4 D^{s (p)}))} \int_{E (p)} | g (x) | d μ (x) \times

(2^{102 a^{3}} \cdot 10 \cdot 2^{6 a} {(\frac{ρ (y, y^{'})}{D^{s (p)}})}^{1 / a} + 2^{150 a^{3}} {(\frac{ρ (y, y^{'})}{D^{s (p)}})}^{1 / a})

Since $ρ (y, y^{'}) \leq 10 D^{s (p)}$ , we conclude

(???) + (???) \leq \frac{2^{151 a^{3}}}{μ (B (c (p), 4 D^{s (p)}))} {(\frac{ρ (y, y^{'})}{D^{s (p)}})}^{1 / a} \int_{E (p)} | g (x) | d μ (x) .

Next, if $y, y^{'} \notin B (c (p), 5 D^{s (p)})$ , then $T_{p}^{*} g (y) = T_{p}^{*} g (y^{'}) = 0$ , by Lemma 7.4.1. Then 7.5.9 holds.

Finally, if $y \in B (c (p), 5 D^{s (p)})$ and $y^{'} \notin B (c (p), 5 D^{s (p)})$ , then

| e (Q (u) (y)) T_{p}^{*} g (y) - e (Q (u) (y^{'})) T_{p}^{*} g (y^{'}) | = | T_{p}^{*} g (y) |

\leq \int_{E (p)} | K_{s (p)} (x, y) | | g (x) | d μ (x) .

By the same argument used to prove 2.1.6, this is bounded by

\leq 2^{102 a^{3}} \int_{E (p)} \frac{1}{μ (B (x, D^{s}))} ψ (D^{- s} ρ (x, y)) | g (x) | d μ (x) .

7.5.15

It follows from the definition of $ψ$ that

ψ (x) \leq max {0, (2 - 4 x)^{1 / a}} .

Now for all $x \in E (p)$ , it follows by the triangle inequality and 2.0.10 that

\begin{matrix} 2 - 4 D^{- s (p)} ρ (x, y) \leq 2 - 4 D^{- s (p)} ρ (y, c (p)) + 4 D^{- s (p)} ρ (x, c (p)) \\ \leq 18 - 4 D^{- s (p)} ρ (y, c (p)) \leq 4 D^{- s (p)} ρ (y, y^{'}) - 2 . \end{matrix}

Combining the above with the previous estimate on $ψ$ , we get

ψ (D^{- s (p)} ρ (x, y)) \leq 4 (D^{- s (p)} ρ (y, y^{'}))^{1 / a} .

Further, we obtain from the doubling property 1.0.5 and 2.0.10 that

μ (B (x, D^{s (p)})) \geq 2^{- 3 a} μ (B (c (p), 4 D^{s (p)})) .

Plugging this into 7.5.15 and using $a \geq 4$ , we get

| T_{p}^{*} g (y) | \leq \frac{2^{103 a^{3}}}{μ (B (c (p), 4 D^{s (p)}))} {(\frac{ρ (y, y^{'})}{D^{s (p)}})}^{1 / a} \int_{E (p)} | g (x) | d μ (y),

which completes the proof of the lemma.

Recall that

B (J) := B (c (J), 8 D^{s (J)}) .

We also denote

B^{\circ} (J) := B (c (J), \frac{1}{8} D^{s (J)}) .

Lemma 7.5.6 limited scale impact

✓

Let $p \in T (u_{2}) ∖ S$ , $J \in J^{'}$ and suppose that

B (I (p)) \cap B^{\circ} (J) \neq \emptyset .

Then

s (J) \leq s (p) \leq s (J) + 3 .

Proof ▶

For the first estimate, assume that $s (p) < s (J)$ , then in particular $s (p) \leq s (u_{1})$ . Since $p \notin S$ , we have by Lemma 7.4.7 that $I (p) \cap I (u_{1}) = \emptyset$ . Since $B (c (J), \frac{1}{4} D^{s (J)}) \subset I (J) \subset I (u_{1})$ , this implies

ρ (c (J), c (p)) \geq \frac{1}{4} D^{s (J)} .

On the other hand

ρ (c (J), c (p)) \leq \frac{1}{8} D^{s (J)} + 8 D^{s (p)},

by our assumption. Thus $D^{s (p)} \geq 64^{- 1} D^{s (J)}$ , which contradicts 2.0.1 and $a \geq 4$ .

For the second estimate, assume that $s (p) > s (J) + 3$ . Since $J \in J^{'}$ , we have $J ⊊ I (u_{1})$ . Thus there exists $J^{'} \in D$ with $J \subset J^{'}$ and $s (J^{'}) = s (J) + 1$ , by 2.0.7 and 2.0.8. By definition of $J^{'}$ , there exists some $p^{'} \in S$ such that $I (p^{'}) \subset B (c (J^{'}), 100 D^{s (J) + 2})$ . On the other hand, since $B (I (p)) \cap B^{\circ} (J) \neq \emptyset$ , by the triangle inequality it holds that

B (c (J^{'}), 100 D^{s (J) + 3}) \subset B (c (p), 10 D^{s (p)}) .

Using the definition of $S$ , we have

2^{Z n / 2} \leq d_{p^{'}} (Q (u_{1}), Q (u_{2})) \leq d_{B (c (J^{'}), 100 D^{s (J) + 2})} (Q (u_{1}), Q (u_{2})) .

By 1.0.10, this is

\leq 2^{- 100 a} d_{B (c (J^{'}), 100 D^{s (J) + 3})} (Q (u_{1}), Q (u_{2}))

\leq 2^{- 100 a} d_{B (c (p), 10 D^{s (p)})} (Q (u_{1}), Q (u_{2})),

and by 1.0.8 and the definition of $S$

\leq 2^{- 94 a} d_{p} (Q (u_{1}), Q (u_{2})) \leq 2^{- 94 a} 2^{Z n / 2} .

This is a contradiction, the second estimate follows.

Lemma 7.5.7 local tree control

✓

For all $J \in J^{'}$ and all bounded $g$ with bounded support

sup_{B^{\circ} (J)} | T_{T (u_{2}) ∖ S}^{*} g | \leq 2^{104 a^{3}} inf_{J} M_{B, 1} | g |

Proof ▶

By the triangle inequality and since $T_{p}^{*} g = 1_{B (c (p), 5 D^{s (p)})} T_{p}^{*} g$ , we have

sup_{B^{\circ} (J)} | T_{T (u_{2}) ∖ S}^{*} g | \leq sup_{B^{\circ} (J)} \sum_{\begin{matrix} p \in T (u_{2}) ∖ S \\ B (I (p)) \cap B^{\circ} (J) \neq \emptyset \end{matrix}} | T_{p}^{*} g | .

By Lemma 7.5.6, this is at most

\sum_{s = s (J)}^{s (J) + 3} \sum_{\begin{matrix} p \in P, s (p) = s \\ B (I (p)) \cap B^{\circ} (J) \neq \emptyset \end{matrix}} sup_{B^{\circ} (J)} | T_{p}^{*} g | .

7.5.16

If $x \in E (p)$ and $B (I (p)) \cap B^{\circ} (J) \neq \emptyset$ , then

B (c (J), 16 D^{s (p)}) \subset B (x, 32 D^{s (p)}),

by 2.0.10 and the triangle inequality. Using the doubling property 1.0.5, it follows that

μ (B (x, D^{s (p)})) \geq 2^{- 5 a} μ (B (c (J), 16 D^{s (p)})) .

Using 7.4.1, 2.1.3 and that $a \geq 4$ , we bound 7.5.16 by

2^{103 a^{3}} \sum_{s = s (J)}^{s (J) + 3} \sum_{\begin{matrix} p \in P, s (p) = s \\ B (I (p)) \cap B^{\circ} (J) \neq \emptyset \end{matrix}} \frac{1}{μ (B (c (J), 16 D^{s})} \int_{E (p)} | g | d μ .

For each $I \in D$ , the sets $E (p)$ for $p \in P$ with $I (p) = I$ are pairwise disjoint by 2.0.20 and 2.0.13. Further, if $B (I (p)) \cap B^{\circ} (J) \neq \emptyset$ and $s (p) \geq s (J)$ , then $E (p) \subset B (c (J), 16 D^{s (p)})$ . Thus the last display is bounded by

2^{103 a^{3}} \sum_{s = s (J)}^{s (J) + 3} \frac{1}{μ (B (c (J), 16 D^{s}))} \int_{B (c (J), 16 D^{s})} | g | d μ .

\leq inf_{x^{'} \in J} 2^{103 a^{3} + 2} M_{B, 1} | g | .

The lemma follows since $a \geq 4$ .

Lemma 7.5.8 scales impacting interval

✓

Let $C = T (u_{1})$ or $C = T (u_{2}) \cap S$ . Then for each $J \in J^{'}$ and $p \in C$ with $B (I (p)) \cap B (J) \neq \emptyset$ , we have $s (p) \geq s (J)$ .

Proof ▶

By Lemma 7.4.7, we have that in both cases, $C \subset S$ . If $p \in C$ with $B (I (p)) \cap B (J) \neq \emptyset$ and $s (p) < s (J)$ , then $I (p) \subset B (c (J), 100 D^{s (J) + 1})$ . Since $p \in S$ , it follows from the definition of $J^{'}$ that $s (J) = - S$ , which contradicts $s (p) < s (J)$ .

Lemma 7.5.9 global tree control 1

✓

Let $C_{1} = T (u_{1})$ and $C_{2} = T (u_{2}) \cap S$ . Then for $i = 1, 2$ and each $J \in J^{'}$ and all bounded $g$ with bounded support, we have

\begin{array}{r} sup_{B (J)} | T_{C_{i}}^{*} g | \leq inf_{B^{\circ} (J)} | T_{C_{i}}^{*} g | + 2^{151 a^{3} + 4 a + 2} inf_{J} M_{B, 1} | g | \end{array}

and for all $y, y^{'} \in B (J)$

| e (Q (u_{i}) (y)) T_{C_{i}}^{*} g (y) - e (Q (u_{i}) (y^{'})) T_{C_{i}}^{*} g (y^{'}) |

\leq 2^{151 a^{3} + 4 a + 1} {(\frac{ρ (y, y^{'})}{D^{s (J)}})}^{1 / a} inf_{J} M_{B, 1} | g | .

7.5.18

Proof ▶

Note that 7.5.17 follows from 7.5.18, since for $y^{'} \in B^{\circ} (J)$ , by the triangle inequality,

{(\frac{ρ (y, y^{'})}{D^{s (J)}})}^{1 / a} \leq (8 + \frac{1}{8})^{1 / a} \leq 2.

By the triangle inequality, Lemma 7.4.1 and Lemma 7.5.5, we have for all $y, y^{'} \in B (J)$

| e (Q (u_{i}) (y)) T_{C_{i}}^{*} g (y) - e (Q (u_{i}) (y^{'})) T_{C_{i}}^{*} g (y^{'}) |

7.5.19

\leq \sum_{\begin{matrix} p \in C_{i} \\ B (I (p)) \cap B (J) \neq \emptyset \end{matrix}} | e (Q (u_{i}) (y)) T_{p}^{*} g (y) - e (Q (u_{i}) (y^{'})) T_{p}^{*} g (y^{'}) |

\leq 2^{151 a^{3}} ρ (y, y^{'})^{1 / a} \sum_{\begin{matrix} p \in C_{i} \\ B (I (p)) \cap B (J) \neq \emptyset \end{matrix}} \frac{D^{- s (p) / a}}{μ (B (c (p), 4 D^{s (p)}))} \int_{E (p)} | g | d μ .

By Lemma 7.5.8, we have $s (p) \geq s (J)$ for all $p$ occurring in the sum. Further, for each $s \geq s (J)$ , the sets $E (p)$ for $p \in P$ with $s (p) = s$ are pairwise disjoint by 2.0.20 and 2.0.13, and contained in $B (c (J), 32 D^{s})$ by 2.0.10 and the triangle inequality. Using also the doubling estimate 1.0.5, we obtain that the expression in the last display can be estimated by

2^{151 a^{3}} ρ (y, y^{'})^{1 / a} \sum_{S \geq s \geq s (J)} D^{- s / a} \frac{2^{4 a}}{μ (B (c (J), 32 D^{s}))} \int_{B (c (J), 32 D^{s})} | g | d μ

\leq 2^{151 a^{3} + 4 a} {(\frac{ρ (y, y^{'})}{D^{s (J)}})}^{1 / a} \sum_{S \geq s \geq s (J)} D^{(s (J) - s) / a} inf_{J} M_{B, 1} | g | .

Since $D^{- 1 / a} \leq \frac{1}{2}$ , we have

\sum_{S \geq s \geq s (J)} D^{(s (J) - s) / a} \leq 2.

Estimate 7.5.18, and therefore the lemma, follow.

Lemma 7.5.10 global tree control 2

We have for all $J \in J^{'}$ and all bounded $g$ with bounded support

sup_{B (J)} | T_{T (u_{2}) \cap S}^{*} g | \leq inf_{B^{\circ} (J)} | T_{T (u_{2})}^{*} g | + 2^{155 a^{3}} inf_{J} M_{B, 1} | g | .

Proof ▶

By Lemma 7.5.9

sup_{B (J)} | T_{T (u_{2}) \cap S}^{*} g | \leq inf_{B^{\circ} (J)} | T_{T (u_{2}) \cap S}^{*} g | + 2^{154 a^{3}} inf_{J} M_{B, 1} | g |

\leq inf_{B^{\circ} (J)} | T_{T (u_{2})}^{*} g | + sup_{B^{\circ} (J)} | T_{T (u_{2}) ∖ S}^{*} g | + 2^{154 a^{3}} inf_{J} M_{B, 1} | g |,

and by Lemma 7.5.7

\leq inf_{B^{\circ} (J)} | T_{T (u_{2})}^{*} g | + (2^{104 a^{3}} + 2^{154 a^{3}}) inf_{J} M_{B, 1} | g | .

This completes the proof.

Proof of Lemma 7.5.4 ▶

Let $P$ be the product on the right hand side of 7.5.8, and $h_{J}$ as defined in 7.5.7.

By 7.5.3 and Lemma 7.4.1, the function $h_{J}$ is supported in $B (J) \cap I (u_{1})$ . By 7.5.3 and Lemma 7.5.9, we have for all $y \in B (J)$ :

| h_{J} (y) | \leq 2^{308 a^{3}} P .

We have by the triangle inequality

\begin{aligned} | h_{J} (y) - h_{J} (y^{'}) | \\ \leq | χ_{J} (y) - χ_{J} (y^{'}) | | T_{T (u_{1})}^{*} g_{1} (y) | | T_{T (u_{2}) \cap S}^{*} g_{2} (y) | \\ + | χ_{J} (y^{'}) | | T_{T (u_{1})}^{*} g_{1} (y) - T_{T (u_{1})}^{*} g_{1} (y^{'}) | | T_{T (u_{2}) \cap S}^{*} g_{2} (y) | \\ + | χ_{J} (y^{'}) | | T_{T (u_{1})}^{*} g_{1} (y^{'}) | | T_{T (u_{2}) \cap S}^{*} g_{2} (y) - T_{T (u_{2}) \cap S}^{*} g_{2} (y^{'}) | . \end{aligned}

As $h_{J}$ is supported in $I (u_{1})$ , we can assume without loss of generality that $y^{'} \in I (u_{1})$ . If $y \notin I (u_{1})$ , then 7.5.20 vanishes. If $y \in I (u_{1})$ then we have by 7.5.4, Lemma 7.5.9 and Lemma 7.5.10

(???) \leq 2^{534 a^{3}} \frac{ρ (y, y^{'})}{D^{s (J)}} P,

where $P$ denotes the product on the right hand side of 7.5.8.

By 7.5.3, Lemma 7.5.9 and Lemma 7.5.10, we have

(???) \leq 2^{310 a^{3}} P .

By 7.5.3, and twice Lemma 7.5.9, we have

(???) \leq 2^{308 a^{3}} P .

Using that $ρ (y, y^{'}) \leq 16 D^{s (J)}$ and $a \geq 4$ , the lemma follows.

7.5.3 The van der Corput estimate

Lemma 7.5.11 lower oscillation bound

✓

For all $J \in J^{'}$ , we have that

d_{B (J)} (Q (u_{1}), Q (u_{2})) \geq 2^{- 201 a^{3}} 2^{Z n / 2} .

Proof ▶

Since $\emptyset \neq T (u_{1}) \subset S$ by Lemma 7.4.7, there exists at least one tile $p \in S$ with $I (p) ⊊ I (u_{1})$ . Thus $I (u_{1}) \notin J^{'}$ , so $J ⊊ I (u_{1})$ . Thus there exists a cube $J^{'} \in D$ with $J \subset J^{'}$ and $s (J^{'}) = s (J) + 1$ , by 2.0.7 and 2.0.8. By definition of $J^{'}$ and the triangle inequality, there exists $p \in S$ such that

I (p) \subset B (c (J^{'}), 100 D^{s (J^{'}) + 1}) \subset B (c (J), 128 D^{s (J) + 2}) .

Thus, by definition of $S$ :

\begin{array}{r} 2^{Z n / 2} \leq d_{p} (Q (u_{1}), Q (u_{2})) \leq d_{B (c (J), 128 D^{s (J) + 2})} (Q (u_{1}), Q (u_{2})) . \end{array}

By the doubling property 1.0.8, this is

\leq 2^{200 a^{3} + 4 a} d_{B (J)} (Q (u_{1}), Q (u_{2})),

which gives the lemma using $a \geq 4$ .

Now we are ready to prove Lemma 7.4.5.

Proof of Lemma 7.4.5 ▶

We have

(???) = | \int_{X} T_{T (u_{1})}^{*} g_{1} \overset{―}{T_{T (u_{2}) \cap S}^{*} g_{2}} | .

By Lemma 7.4.1, the right hand side is supported in $I (u_{1})$ . Using 7.5.2 of Lemma 7.5.2 and the definition 7.5.7 of $h_{J}$ , we thus have

\leq \sum_{J \in J^{'}} | \int_{B (J)} e (Q (u_{2}) (y) - Q (u_{1}) (y)) h_{J} (y) d μ (y) | .

Using Proposition 2.0.5 with the ball $B (J)$ , we bound this by

\leq 2^{8 a} \sum_{J \in J^{'}} μ (B (J)) ‖ h_{J} ‖_{C^{τ} (B (J))} (1 + d_{B (J)} (Q (u_{1}), Q (u_{1})))^{- 1 / (2 a^{2} + a^{3})} .

Using Lemma 7.5.4, Lemma 7.5.11 and $a \geq 4$ , we have that the above is bounded from above by

\begin{matrix} \leq 2^{540 a^{3}} 2^{- Z n / (4 a^{2} + 2 a^{3})} \sum_{J \in J^{'}} μ (B (J)) \\ \times \prod_{j = 1}^{2} (inf_{B^{\circ} (J)} | T_{T (u_{j})}^{*} g_{j} | + inf_{J} M_{B, 1} g_{j}) . \end{matrix}

By the doubling property 1.0.5

μ (B (J)) \leq 2^{6 a} μ (B^{\circ} (J)),

thus

μ (B (J)) \prod_{j = 1}^{2} (inf_{B^{\circ} (J)} | T_{T (u_{j})}^{*} g_{j} | + inf_{J} M_{B, 1} g_{j})

\leq 2^{6 a} \int_{B^{\circ} (J)} \prod_{j = 1}^{2} (| T_{T (u_{j})}^{*} g_{j} | (x) + M_{B, 1} g_{j} (x)) d μ (x)

\leq 2^{6 a} \int_{J} \prod_{j = 1}^{2} (| T_{T (u_{j})}^{*} g_{j} | (x) + M_{B, 1} g_{j} (x)) d μ (x) .

Summing over $J \in J^{'}$ , we obtain

(???) \leq 2^{541 a^{3}} 2^{- Z n / (4 a^{2} + 2 a^{3})} \int_{X} \prod_{j = 1}^{2} (| T_{T (u_{j})}^{*} g_{j} | (x) + M_{B, 1} g_{j} (x)) d μ (x) .

Applying the Cauchy-Schwarz inequality, Lemma 7.4.5 follows.

7.6 Proof of The Remaining Tiles Lemma

We define

J^{'} = {J \in J (T (u_{1})) : J \subset I (u_{1})},

note that this is different from the $J^{'}$ defined in the previous subsection.

Lemma 7.6.1 dyadic partition 2

✓

We have

I (u_{1}) = \dot{⋃_{J \in J^{'}}} J .

Proof ▶

By Lemma 7.1.2, it remains only to show that each $J \in J (T (u_{1}))$ with $J \cap I (u_{1}) \neq \emptyset$ is in $J^{'}$ . But if $J \notin J^{'}$ , then by 2.0.8 $I (u_{1}) ⊊ J$ . Pick $p \in T (u_{1})$ . Then $I (p) ⊊ J$ . This contradicts the definition of $J (T (u_{1}))$ .

Lemma 7.4.6 follows from the following key estimate.

Lemma 7.6.2 bound for tree projection

We have

‖ P_{J^{'}} | T_{T (u_{2}) ∖ S}^{*} g_{2} | ‖_{2} \leq 2^{118 a^{3}} 2^{- \frac{100}{202 a} Z n κ} ‖ 1_{I (u_{1})} M_{B, 1} | g_{2} | ‖_{2}

We prove this lemma below. First, we deduce Lemma 7.4.6.

Proof of Lemma 7.4.6 ▶

By Lemma 7.2.1 and Lemma 7.4.1, we have

\begin{array}{r} (???) \leq 2^{104 a^{3}} ‖ P_{L (T (u_{1}))} | g_{1} 1_{I (u_{1})} | ‖_{2} ‖ 1_{I (u_{1})} P_{J (T (u_{1}))} | T_{T (u_{2}) ∖ S}^{*} g_{2} | ‖_{2} . \end{array}

It follows from the definition of the projection operator $P$ and Jensen’s inequality that

‖ P_{L (T (u_{1}))} | g_{1} 1_{I (u_{1})} | ‖_{2} \leq ‖ g_{1} 1_{I (u_{1})} ‖_{2} .

Since cubes in $J^{'}$ are pairwise disjoint and by Lemma 7.6.1, a cube $J \in J^{'}$ intersect $I (u_{1})$ if and only if $J \in J^{'}$ . Thus

1_{I (u_{1})} P_{J (T (u_{1}))} | T_{T (u_{2}) ∖ S}^{*} g_{2} | = P_{J^{'}} | T_{T (u_{2}) ∖ S}^{*} g_{2} | .

Combining this with Lemma 7.6.2, the definition 2.0.2 and $a \geq 4$ proves the lemma.

We need two more auxiliary lemmas before we prove Lemma 7.6.2.

Lemma 7.6.3 thin scale impact

✓

If $p \in T (u_{2}) ∖ S$ and $J \in J^{'}$ with $B (I (p)) \cap B (J) \neq \emptyset$ , then

s (p) \leq s (J) + 2 - \frac{Z n}{202 a^{3}} .

Proof ▶

Suppose that $s (p) > s (J) + 2 - \frac{Z n}{202 a^{3}} =: s (J) - s_{1}$ . Then, we have $s_{1} + 2 \geq 0$ so

ρ (c (p), c (J)) \leq 8 D^{s (J)} + 8 D^{s (p)} \leq 16 D^{s (p) + s_{1} + 2} .

There exists a tile $q \in T (u_{1})$ . By 2.0.32, it satisfies $I (q) ⊊ I (u_{1})$ . Thus $I (u_{1}) \notin J^{'}$ . It follows that $J ⊊ I (u_{1})$ . By 2.0.7 and 2.0.8, there exists a cube $J^{'} \in D$ with $J \subset J^{'}$ and $s (J^{'}) = s (J) + 1$ . By definition of $J^{'}$ , there exists a tile $p^{'} \in T (u_{1})$ with

I (p^{'}) \subset B (c (J^{'}), 100 D^{s (J^{'}) + 1}) .

By the triangle inequality, the definition 2.0.1 and $a \geq 4$ , we have

B (c (J^{'}), 100 D^{s (J^{'}) + 1}) \subset B (c (p), 128 D^{s (p) + s_{1} + 2}) .

Since $p^{'} \in T (u_{1})$ and $I (u_{1}) \subset I (u_{2})$ , we have by 2.0.36

d_{p^{'}} (Q (p^{'}), Q (u_{2})) > 2^{Z (n + 1)} .

Hence, by 2.0.32, the triangle inequality and using that by 2.0.3 $Z (n + 1) = 2^{12 a} (n + 1) \geq 3$

d_{p^{'}} (Q (u_{1}), Q (u_{2})) > 2^{Z (n + 1)} - 4 \geq 2^{Z (n + 1) - 1} .

It follows that

2^{Z (n + 1) - 1} \leq d_{p^{'}} (Q (u_{1}), Q (u_{2})) \leq d_{B (c (p), 128 D^{s (p) + s_{1} + 2})} (Q (u_{1}), Q (u_{2})) .

Using 1.0.8, we obtain

\leq 2^{9 a + 100 a^{3} (s_{1} + 3)} d_{p} (Q (u_{1}), Q (u_{2})) .

Since $p^{'} \notin S$ this is bounded by

\leq 2^{9 a + 100 a^{3} (s_{1} + 3)} 2^{Z n / 2} .

Thus

Z n / 2 + Z - 1 \leq 9 a + 100 a^{3} (s_{1} + 3),

contradicting the definition of $s_{1}$ .

Lemma 7.6.4 square function count

✓

For each $J \in J^{'}$ and all $s \geq 0$ , we have

\frac{1}{μ (J)} \int_{J} (\sum_{\begin{matrix} I \in D, s (I) = s (J) - s \\ I \cap I (u_{1}) = \emptyset \\ J \cap B (I) \neq \emptyset \end{matrix}} 1_{B (I)})^{2} d μ \leq 2^{14 a + 1} (8 D^{- s})^{κ} .

Proof ▶

Since $J \in J^{'}$ we have $J \subset I (u_{1})$ . Thus, if $B (I) \cap J \neq \emptyset$ then

B (I) \cap J \subset {x \in J : ρ (x, X ∖ J) \leq 8 D^{s (I)}} .

7.6.1

Furthermore, for each $s$ the balls $B (I)$ with $s (I) = s$ have bounded overlap: Consider the collection $D_{s, x}$ of all $I \in D$ with $x \in B (I)$ and $s (I) = s$ . By 2.0.10 and 2.0.8, the balls $B (c (I), \frac{1}{4} D^{s (I)})$ , $I \in D_{s, x}$ are disjoint, and by the triangle inequality, they are contained in $B (x, 9 D^{s})$ . By the doubling property 1.0.5, we have

μ (B (x, 9 D^{s})) \leq 2^{7 a} μ (B (c (I), \frac{1}{4} D^{s (I)}))

for each $I \in D_{s, x}$ . Thus

μ (B (x, 9 D^{s})) \geq \sum_{I \in D_{s, x}} μ (B (c (I), \frac{1}{4} D^{s (I)})) \geq 2^{- 7 a} | D_{s, x} | μ (B (x, 9 D^{s})) .

Dividing by the positive $μ (B (x, 9 D^{s}))$ , we obtain that for each $x$

(\sum_{\begin{matrix} I \in D, s (I) = s (J) - s \\ I \cap I (u_{1}) = \emptyset \\ J \cap B (I) \neq \emptyset \end{matrix}} 1_{B (I)} (x))^{2} \leq | D_{s (J) - s, x} |^{2} \leq 2^{14 a} .

7.6.2

Combining 7.6.1, 7.6.2 and the small boundary property 2.0.11, noting that $8 D^{s (I)} = 8 D^{- s} D^{s (J)}$ , the lemma follows.

Proof of Lemma 7.6.2 ▶

Expanding the definition of $P_{J^{'}}$ , we have

‖ P_{J^{'}} | T_{T (u_{2}) ∖ S}^{*} g_{2} | ‖_{2}

= {(\sum_{J \in J^{'}} \frac{1}{μ (J)} {| \int_{J} \sum_{p \in T (u_{2}) ∖ S} T_{p}^{*} g_{2} d μ (y) |}^{2})}^{1 / 2} .

We split the innermost sum according to the scale of the tile $p$ , and then apply the triangle inequality and Minkowski’s inequality:

\leq \sum_{s = - S}^{S} {(\sum_{J \in J^{'}} \frac{1}{μ (J)} {| \int_{J} \sum_{\begin{matrix} p \in T (u_{2}) ∖ S \\ s (p) = s \end{matrix}} T_{p}^{*} g_{2} d μ (y) |}^{2})}^{1 / 2} .

By Lemma 7.4.1, the integral in the last display is $0$ if $J \cap B (I (p)) = \emptyset$ . By Lemma 7.6.3, it follows with $s_{1} := \frac{Z n}{202 a^{3}} - 2$ :

= \sum_{s = s_{1}}^{s_{1} + 2 S} (\sum_{J \in J^{'}} \frac{1}{μ (J)} | \int_{J} \sum_{\begin{matrix} p \in T (u_{2}) ∖ S \\ s (p) = s (J) - s \\ J \cap B (I (p)) \neq \emptyset \end{matrix}} T_{p}^{*} g_{2} d μ (y) |^{2})^{1 / 2} .

7.6.3

We have by Lemma 7.4.1 and 2.1.3

\int | T_{p}^{*} g_{2} | (y) d μ (y)

\leq 2^{102 a^{3}} \int_{B (I (p))} \int \frac{1}{μ (B (x, D^{s (p)}))} 1_{E (p)} (x) | g_{2} | (x) d μ (x) d μ (y) .

If $x \in E (p) \subset I (p)$ , then we have by 2.0.10 that

B (c (p), 4 D^{s (p)}) \subset B (I (p)) .

Using the doubling property 1.0.5, it follows that

μ (B (c (p), 4 D^{s (p)})) \leq 2^{3 a} μ (B (x, D^{s})) .

Thus, using also $a \geq 4$

\int | T_{p}^{*} g_{2} | (y) d μ (y)

\leq 2^{103 a^{3}} \int_{B (I (p))} \int \frac{1}{μ (B (c (p), 4 D^{s (p)}))} 1_{E (p)} (x) | g_{2} | (x) d μ (x) d μ (y) .

Since for each $I \in D$ the sets $E (p), p \in P (I)$ are disjoint, it follows that

| \int_{J} \sum_{\begin{matrix} p \in T (u_{2}) ∖ S \\ I (p) = I \\ J \cap B (I (p)) \neq \emptyset \end{matrix}} T_{p}^{*} g_{2} d μ |

\leq 2^{103 a^{3}} \int_{J} 1_{B (I)} \frac{1}{μ (B (c (p), 4 D^{s (p)}))} \int_{B (c (p), 4 D^{s (p)})} | g_{2} | (x) d μ (x)

\leq 2^{103 a^{3}} \int_{J} M_{B, 1} | g_{2} | (y) 1_{B (I)} (y) d μ (y) .

By Lemma 7.4.7, we have $I (p) \cap I (u_{1}) = \emptyset$ for all $p \in T (u_{2}) ∖ S$ . Thus we can estimate 7.6.3 by

2^{103 a^{3}} \sum_{s = s_{1}}^{s_{1} + 2 S} (\sum_{J \in J^{'}} \frac{1}{μ (J)} | \int_{J} \sum_{\begin{matrix} I \in D, s (I) = s (J) - s \\ I \cap I (u_{1}) = \emptyset \\ J \cap B (I) \neq \emptyset \end{matrix}} M_{B, 1} | g_{2} | 1_{B (I)} d μ |^{2})^{\frac{1}{2}},

which is by Cauchy-Schwarz at most

2^{103 a^{3}} \sum_{s = s_{1}}^{s_{1} + 2 S} (\sum_{J \in J^{'}} \int_{J} (M_{B, 1} | g_{2} |)^{2} \frac{1}{μ (J)} \int_{J} (\sum_{\begin{matrix} I \in D, s (I) = s (J) - s \\ I \cap I (u_{1}) = \emptyset \\ J \cap B (I) \neq \emptyset \end{matrix}} 1_{B (I)})^{2} d μ)^{\frac{1}{2}} .

7.6.4

Using Lemma 7.6.4, we bound 7.6.4 by

2^{103 a^{3}} \sum_{s = s_{1}}^{s_{1} + 2 S} {(\sum_{J \in J^{'}} \int_{J} (M_{B, 1} | g_{2} |)^{2} 2^{104 a^{2}} (8 D^{- s})^{κ})}^{\frac{1}{2}},

and, since dyadic cubes in $J^{'}$ form a partition of $I (u_{1})$ by Lemma 7.6.1, $κ \leq 1$ by 2.0.2, and $a \geq 4$

\leq 2^{116 a^{3}} \sum_{s = s_{1}}^{s_{1} + 2 S} D^{- s κ / 2} ‖ 1_{I (u_{1})} M_{B, 1} | g_{2} | ‖_{2}

\leq 2^{116 a^{3}} D^{- s_{1} κ / 2} \frac{1}{1 - D^{- κ / 2}} ‖ 1_{I (u_{1})} M_{B, 1} | g_{2} | ‖_{2} .

By convexity of $t \mapsto D^{t}$ and since $D \geq 2$ , we have for all $- 1 \leq t \leq 0$

D^{t} \leq 1 + t (1 - \frac{1}{D}) \leq 1 + \frac{1}{2} t .

Using this for $t = - κ / 2$ and using that $s_{1} = \frac{Z n}{202 a^{3}} - 2$ and the definitions 2.0.1 and 2.0.2 of $κ$ and $D$

\leq 2^{116 a^{3}} 2^{- 100 a^{2} (\frac{Z n}{202 a^{3}} - 2) \frac{κ}{2}} \frac{2}{κ} ‖ 1_{I (u_{1})} M_{B, 1} | g_{2} | ‖_{2}

\leq \frac{2^{117 a^{3} + 1}}{κ} 2^{- \frac{100}{202 a} Z n κ} ‖ 1_{I (u_{1})} M_{B, 1} | g_{2} | ‖_{2} .

Using the definition 2.0.2 of $κ$ and $a \geq 4$ , the lemma follows.

7.7 Forests

In this subsection, we complete the proof of Proposition 2.0.4 from the results of the previous subsections.

Define an $n$ -row to be an $n$ -forest $(U, T)$ , i.e. satisfying conditions 2.0.32 - 2.0.37, such that in addition the sets $I (u), u \in U$ are pairwise disjoint.

Lemma 7.7.1 forest row decomposition

✓

Let $(U, T)$ be an $n$ -forest. Then there exists a decomposition

U = \dot{⋃_{1 \leq j \leq 2^{n}}} U_{j}

such that for all $j = 1, \dots, 2^{n}$ the pair $(U_{j}, T |_{U_{j}})$ is an $n$ -row.

Proof ▶

Define recursively $U_{j}$ to be a maximal disjoint set of tiles $u$ in

U ∖ ⋃_{j^{'} < j} U_{j^{'}}

with inclusion maximal $I (u)$ . Properties 2.0.32, -2.0.37 for $(U_{j}, T |_{U_{k}})$ follow immediately from the corresponding properties for $(U, T)$ , and the cubes $I (u), u \in U_{j}$ are disjoint by definition. The collections $U_{j}$ are also disjoint by definition.

Now we show by induction on $j$ that each point is contained in at most $2^{n} - j$ cubes $I (u)$ with $u \in U ∖ ⋃_{j^{'} \leq j} U_{j^{'}}$ . This implies that $⋃_{j = 1}^{2^{n}} U_{j} = U$ , which completes the proof of the Lemma. For $j = 0$ each point is contained in at most $2^{n}$ cubes by 2.0.34. For larger $j$ , if $x$ is contained in any cube $I (u)$ with $u \in U ∖ ⋃_{j^{'} < j} U_{j^{'}}$ , then it is contained in a maximal such cube. Thus it is contained in a cube in $I (u)$ with $u \in U_{j}$ . Thus the number $u \in U ∖ ⋃_{j^{'} \leq j} U_{j^{'}}$ with $x \in I (u)$ is zero, or is less than the number of $u \in U ∖ ⋃_{j^{'} \leq j - 1} U_{j^{'}}$ with $x \in I (u)$ by at least one.

We pick a decomposition of the forest $(U, T)$ into $2^{n}$ $n$ -rows

(U_{j}, T_{j}) := (U_{j}, T |_{U_{j}})

as in Lemma 7.7.1.

Lemma 7.7.2 row bound

For each $1 \leq j \leq 2^{n}$ and each bounded $g$ with bounded support with $| g | \leq 1_{G}$ , we have

{‖ \sum_{u \in U_{j}} \sum_{p \in T (u)} T_{p}^{*} g ‖}_{2} \leq 2^{156 a^{3}} 2^{- n / 2} ‖ g ‖_{2}

7.7.1

and

{‖ \sum_{u \in U_{j}} \sum_{p \in T (u)} 1_{F} T_{p}^{*} g ‖}_{2} \leq 2^{257 a^{3}} 2^{- n / 2} {dens}_{2} (⋃_{u \in U} T (u))^{1 / 2} ‖ g ‖_{2} .

7.7.2

Proof ▶

By Lemma 7.3.1 and the density assumption 2.0.35, we have for each $u \in U$ and all bounded $f$ of bounded support that

{‖ 1_{G} \sum_{p \in T (u)} T_{p} f ‖}_{2} \leq 2^{155 a^{3}} 2^{(4 a + 1 - n) / 2} ‖ f ‖_{2}

7.7.3

and

{‖ 1_{G} \sum_{p \in T (u)} T_{p} 1_{F} f ‖}_{2} \leq 2^{256 a^{3}} 2^{(4 a + 1 - n) / 2} {dens}_{2} (T (u))^{1 / 2} ‖ f ‖_{2} .

7.7.4

Since for each $j$ the top cubes $I (u)$ , $u \in U_{j}$ are disjoint, we further have for all bounded $g$ of bounded support and $| g | \leq 1_{G}$ by Lemma 7.4.1

{‖ 1_{F} \sum_{u \in U_{j}} \sum_{p \in T (u)} T_{p}^{*} g ‖}_{2}^{2} = {‖ 1_{F} \sum_{u \in U_{j}} \sum_{p \in T (u)} 1_{I (u)} T_{p}^{*} 1_{I (u)} g ‖}_{2}^{2}

= \sum_{u \in U_{j}} \int_{I (u)} {| 1_{F} \sum_{p \in T (u)} T_{p}^{*} 1_{I (u)} g |}^{2} d μ \leq \sum_{u \in U_{j}} {‖ \sum_{p \in T (u)} 1_{F} T_{p}^{*} 1_{G} 1_{I (u)} g ‖}_{2}^{2} .

Applying the estimate for the adjoint operator following from equation 7.7.4, we obtain

\leq 2^{256 a^{3}} 2^{(4 a + 1 - n) / 2} max_{u \in U_{j}} {dens}_{2} (T (u))^{1 / 2} \sum_{u \in U_{j}} {‖ 1_{I (u)} g ‖}_{2}^{2} .

Again by disjointedness of the cubes $I (u)$ , this is estimated by

2^{256 a^{3}} 2^{(4 a + 1 - n) / 2} max_{u \in U_{j}} {dens}_{2} (T (u))^{1 / 2} ‖ g ‖_{2}^{2} .

Thus, 7.7.2 follows, since $a \geq 4$ . The proof of 7.7.1 from 7.7.3 is the same up to replacing $F$ by $X$ .

Lemma 7.7.3 row correlation

For all $1 \leq j < j^{'} \leq 2^{n}$ and for all $g_{1}, g_{2}$ with $| g_{i} | \leq 1_{G}$ , it holds that

| \int \sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} \sum_{p \in T_{j} (u)} \sum_{p^{'} \in T_{j^{'}} (u^{'})} T_{p}^{*} g_{1} \overset{―}{T_{p^{'}}^{*} g_{2}} d μ | \leq 2^{862 a^{3} - 3 n} ‖ g_{1} ‖_{2} ‖ g_{2} ‖_{2} .

Proof ▶

To save some space we will write for subsets $C \subset P$

T_{C}^{*} = \sum_{p \in C} T_{p}^{*} .

We have by Lemma 7.4.1 and the triangle inequality that

| \int \sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} \sum_{p \in T_{j} (u)} \sum_{p^{'} \in T_{j^{'}} (u^{'})} T_{p}^{*} g_{1} \overset{―}{T_{p^{'}}^{*} g_{2}} d μ |

\leq \sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} | \int T_{T_{j} (u)}^{*} (1_{I (u)} g_{1}) \overset{―}{T_{T_{j^{'}} (u^{'})}^{*} (1_{I (u^{'})} g_{2})} d μ | .

By Lemma 7.4.4, this is bounded by

2^{550 a^{3} - 3 n} \sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} ‖ S_{2, u} g_{1} ‖_{L^{2} (I (u^{'}) \cap I (u)} ‖ S_{2, u^{'}} g_{2} ‖_{L^{2} (I (u^{'}) \cap I (u))} .

7.7.5

We apply the Cauchy-Schwarz inequality in the form

\sum_{i \in M} a_{i} b_{i} \leq (\sum_{i \in M} a_{i}^{2})^{1 / 2} (\sum_{i \in M} b_{i}^{2})^{1 / 2}

to the outer two sums:

\leq 2^{550 a^{3} - 3 n} {(\sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} ‖ S_{2, u} g_{1} ‖_{L^{2} (I (u^{'}) \cap I (u))}^{2})}^{1 / 2}

{(\sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} ‖ S_{2, u^{'}} g_{2} ‖_{L^{2} (I (u^{'}) \cap I (u))}^{2})}^{1 / 2} .

By pairwise disjointedness of the sets $I (u)$ for $u \in U_{j}$ and of the sets $I (u^{'})$ for $u^{'} \in U_{j^{'}}$ , we have

\sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} ‖ S_{2, u} g_{1} ‖_{L^{2} (I (u^{'}) \cap I (u))}^{2} = \sum_{u \in U_{j}} \sum_{u^{'} \in U_{j^{'}}} \int_{I (u) \cap I (u^{'})} | S_{2, u} g_{1} (y) |^{2} d μ (y)

\leq \int_{X} | S_{2, u} g_{1} (y) |^{2} d μ (y) = ‖ S_{2, u} g_{1} ‖_{2}^{2} .

Arguing similar for $g_{2}$ , we can estimate 7.7.5 to be

\leq 2^{550 a^{3} - 3 n} ‖ S_{2, u} g_{1} ‖_{2} ‖ S_{2, u^{'}} g_{2} ‖_{2} .

The lemma now follows from Lemma 7.4.3.

Define for $1 \leq j \leq 2^{n}$

E_{j} := ⋃_{u \in U_{j}} ⋃_{p \in T (u)} E (p) .

Lemma 7.7.4 disjoint row support

✓

The sets $E_{j}$ , $1 \leq j \leq 2^{n}$ are pairwise disjoint.

Proof ▶

Suppose that $p \in T (u)$ and $p^{'} \in T (u^{'})$ with $u \neq u^{'}$ and $x \in E (p) \cap E (p^{'})$ . Suppose without loss of generality that $s (p) \leq s (p^{'})$ . Then $x \in I (p) \cap I (p^{'}) \subset I (u^{'})$ . By 2.0.8 it follows that $I (p) \subset I (u^{'})$ . By 2.0.36, it follows that

d_{p} (Q (p), Q (u^{'})) > 2^{Z (n + 1)} .

By the triangle inequality. Lemma 2.1.2 and 2.0.32 it follows that

\begin{aligned} d_{p} (Q (p), Q (p^{'})) & \geq d_{p} (Q (p), Q (u^{'})) - d_{p} (Q (p^{'}), Q (u^{'})) \\ > 2^{Z (n + 1)} - d_{p^{'}} (Q (p^{'}), Q (u^{'})) \\ \geq 2^{Z (n + 1)} - 4 . \end{aligned}

Since $Z \geq 3$ by 2.0.3, it follows that $Q (p^{'}) \notin B_{p} (Q (p), 1)$ , so $Ω (p^{'}) ⊄ Ω (p)$ by 2.0.15. Hence, by 2.0.14, $Ω (p) \cap Ω (p^{'}) = \emptyset$ . But if $x \in E (p) \cap E (p^{'})$ then $Q (x) \in Ω (p) \cap Ω (p^{'})$ . This is a contradiction, and the lemma follows.

Now we prove Proposition 2.0.4.

Proof of Proposition 2.0.4 ▶

To save some space, we will write

T_{R_{j}}^{*} = \sum_{u \in U_{j}} \sum_{p \in T (u)} T_{p}^{*} .

By 7.4.1, we have for each $j$

T_{R_{j}}^{*} g = \sum_{u \in U_{j}} \sum_{p \in T (u)} T_{p}^{*} g = \sum_{u \in U_{j}} \sum_{p \in T (u)} T_{p}^{*} 1_{E_{j}} g = T_{R_{j}}^{*} 1_{E_{j}} g .

Hence, by Lemma 7.7.1,

{‖ \sum_{u \in U} \sum_{p \in T (u)} T_{p}^{*} g ‖}_{2}^{2} = {‖ \sum_{j = 1}^{2^{n}} T_{R_{j}}^{*} g ‖}_{2}^{2} = {‖ \sum_{j = 1}^{2^{n}} T_{R_{j}}^{*} 1_{E_{j}} g ‖}_{2}^{2}

= \int_{X} {| \sum_{j = 1}^{2^{n}} T_{R_{j}}^{*} 1_{E_{j}} g |}^{2} d μ

= \sum_{j = 1}^{2^{n}} \int_{X} | T_{R_{j}}^{*} 1_{E_{j}} g |^{2} + \sum_{j = 1}^{2^{n}} \sum_{\begin{matrix} j^{'} = 1 \\ j^{'} \neq j \end{matrix}}^{2^{n}} \int_{X} \overset{―}{T_{R_{j}}^{*} 1_{E_{j}} g} T_{R_{j^{'}}}^{*} 1_{E_{j^{'}}} g d μ .

We use Lemma 7.7.2 to estimate each term in the first sum, and Lemma 7.7.3 to bound each term in the second sum:

\leq 2^{312 a^{3} - n} \sum_{j = 1}^{2^{n}} ‖ 1_{E_{j}} g ‖_{2}^{2} + 2^{862 a^{3} - 3 n} \sum_{j = 1}^{2^{n}} \sum_{j^{'} = 1}^{2^{n}} ‖ 1_{E_{j}} g ‖_{2} ‖ 1_{E_{j^{'}}} g ‖_{2} .

By Cauchy-Schwarz in the second two sums, this is at most

2^{862 a^{3}} (2^{- n} + 2^{n} 2^{- 3 n}) \sum_{j = 1}^{n} ‖ 1_{E_{j}} g ‖_{2}^{2},

and by disjointedness of the sets $E_{j}$ , this is at most

2^{863 a^{3} - n} ‖ g ‖_{2}^{2} .

Taking adjoints and square roots, it follows that for all $f$

{‖ \sum_{u \in U} \sum_{p \in T (u)} T_{p} f ‖}_{2} \leq 2^{432 a^{3} - \frac{n}{2}} ‖ f ‖_{2} .

7.7.6

On the other hand, we have by disjointedness of the sets $E_{j}$ from Lemma 7.7.4

{‖ \sum_{u \in U} \sum_{p \in T (u)} T_{p} f ‖}_{2}^{2} = {‖ \sum_{j = 1}^{2^{n}} 1_{E_{j}} T_{R_{j}} f ‖}_{2}^{2} = \sum_{j = 1}^{2^{n}} ‖ 1_{E_{j}} T_{R_{j}} f ‖_{2}^{2} .

If $| f | \leq 1_{F}$ then we obtain from Lemma 7.7.2 and taking square roots that

\leq 2^{257 a^{3}} {dens}_{2} (⋃_{u \in U} T (u))^{\frac{1}{2}} 2^{- \frac{n}{2}} (\sum_{j = 1}^{2^{n}} ‖ f ‖_{2}^{2})^{\frac{1}{2}}

= 2^{257 a^{3}} {dens}_{2} (⋃_{u \in U} T (u))^{\frac{1}{2}} ‖ f ‖_{2} .

7.7.7

Proposition 2.0.4 follows by taking the product of the $(2 - \frac{2}{q})$ -th power of 7.7.6 and the $(\frac{2}{q} - 1)$ -st power of 7.7.7. □

7 Proof of the Forest Operator Proposition

7.1 The pointwise tree estimate

7.2 An auxiliary L2 tree estimate

7.3 The quantitative L2 tree estimate

7.4 Almost orthogonality of separated trees

7.5 Proof of the Tiles with large separation Lemma

7.5.1 A partition of unity

7.5.2 Hölder estimates for adjoint tree operators

7.5.3 The van der Corput estimate

7.6 Proof of The Remaining Tiles Lemma

7.7 Forests

7.2 An auxiliary $L^{2}$ tree estimate

7.3 The quantitative $L^{2}$ tree estimate