5 Proof of discrete Carleson

Let a grid structure $(D, c, s)$ and a tile structure $(P, I, Ω, Q)$ for this grid structure be given. In Section 5.1, we decompose the set $P$ of tiles into subsets. Each subset will be controlled by one of three methods. The guiding principle of the decomposition is to be able to apply the forest estimate of Proposition 2.0.4 to the final subsets defined in 5.1.23. This application is done in Section 5.4. The miscellaneous subsets along the construction of the forests will either be thrown into exceptional sets, which are defined and controlled in Section 5.2, or will be controlled by the antichain estimate of Proposition 2.0.3, which is done in Section 5.5. Section 5.3 contains some auxiliary lemmas needed for the proofs in Subsections 5.4-5.5.

5.1 Organisation of the tiles

In the following definitions, $k, n$ , and $j$ will be nonnegative integers. Define $C (G, k)$ to be the set of $I \in D$ such that there exists a $J \in D$ with $I \subset J$ and

μ (G \cap J) > 2^{- k - 1} μ (J),

5.1.1

but there does not exist a $J \in D$ with $I \subset J$ and

μ (G \cap J) > 2^{- k} μ (J) .

5.1.2

Let

P (k) = {p \in P : I (p) \in C (G, k)}

5.1.3

Define $M (k, n)$ to be the set of $p \in P (k)$ such that

μ (E_{1} (p)) > 2^{- n} μ (I (p))

5.1.4

and there does not exist $p^{'} \in P (k)$ with $p^{'} \neq p$ and $p \leq p^{'}$ such that

μ (E_{1} (p^{'})) > 2^{- n} μ (I (p^{'})) .

5.1.5

Define for a collection $P^{'} \subset P (k)$

{dens}_{k}^{'} (P^{'}) := sup_{p^{'} \in P^{'}} sup_{λ \geq 2} λ^{- a} sup_{p \in P (k) : λ p^{'} ≲ λ p} \frac{μ (E_{2} (λ, p))}{μ (I (p))} .

5.1.6

Sorting by density, we define

C (k, n) := {p \in P (k) : 2^{4 a} 2^{- n} < {dens}_{k}^{'} ({p}) \leq 2^{4 a} 2^{- n + 1}} .

5.1.7

Following Fefferman [ , we define for $p \in C (k, n)$

B (p) := {m \in M (k, n) : 100 p ≲ m}

5.1.8

and

C_{1} (k, n, j) := {p \in C (k, n) : 2^{j} \leq | B (p) | < 2^{j + 1}} .

5.1.9

and

L_{0} (k, n) := {p \in C (k, n) : | B (p) | < 1} .

5.1.10

Together with the following removal of minimal layers, the splitting into $C_{1} (k, n, j)$ will lead to a separation of trees. Define recursively for $0 \leq l \leq Z (n + 1)$

L_{1} (k, n, j, l)

5.1.11

to be the set of minimal elements with respect to $\leq$ in

C_{1} (k, n, j) ∖ ⋃_{0 \leq l^{'} < l} L_{1} (k, n, j, l^{'}) .

5.1.12

Define

C_{2} (k, n, j) := C_{1} (k, n, j) ∖ ⋃_{0 \leq l^{'} \leq Z (n + 1)} L_{1} (k, n, j, l^{'}) .

5.1.13

The remaining tile organization will be relative to prospective tree tops, which we define now. Define

U_{1} (k, n, j)

5.1.14

to be the set of all $u \in C_{1} (k, n, j)$ such that for all $p \in C_{1} (k, n, j)$ with $I (u)$ strictly contained in $I (p)$ we have $B_{u} (Q (u), 100) \cap B_{p} (Q (p), 100) = \emptyset$ .

We first remove the pairs that are outside the immediate reach of any of the prospective tree tops. Define

L_{2} (k, n, j)

5.1.15

to be the set of all $p \in C_{2} (k, n, j)$ such that there does not exist $u \in U_{1} (k, n, j)$ with $I (p) \neq I (u)$ and $2 p ≲ u$ . Define

C_{3} (k, n, j) := C_{2} (k, n, j) ∖ L_{2} (k, n, j) .

5.1.16

We next remove the maximal layers. Define recursively for $0 \leq l \leq Z (n + 1)$

L_{3} (k, n, j, l)

5.1.17

to be the set of all maximal elements with respect to $\leq$ in

C_{3} (k, n, j) ∖ ⋃_{0 \leq l^{'} < l} L_{3} (k, n, j, l^{'}) .

5.1.18

Define

C_{4} (k, n, j) := C_{3} (k, n, j) ∖ ⋃_{0 \leq l \leq Z (n + 1)} L_{3} (k, n, j, l) .

5.1.19

Finally, we remove the boundary pairs relative to the prospective tree tops. Define

L (u)

5.1.20

to be the set of all $I \in D$ with $I \subset I (u)$ and $s (I) = s (u) - Z (n + 1) - 1$ and

B (c (I), 8 D^{s (I)}) ⊄ I (u) .

5.1.21

Define

L_{4} (k, n, j)

5.1.22

to be the set of all $p \in C_{4} (k, n, j)$ such that there exists $u \in U_{1} (k, n, j)$ with $I (p) \subset ⋃ L (u)$ , and define

C_{5} (k, n, j) := C_{4} (k, n, j) ∖ L_{4} (k, n, j) .

5.1.23

We define three exceptional sets. The first exceptional set $G_{1}$ takes into account the ratio of the measures of $F$ and $G$ . Define $P_{F, G}$ to be the set of all $p \in P$ with

{dens}_{2} ({p}) > 2^{2 a + 5} \frac{μ (F)}{μ (G)} .

5.1.24

Define

G_{1} := ⋃_{p \in P_{F, G}} I (p) .

5.1.25

For an integer $λ \geq 0$ , define $A (λ, k, n)$ to be the set of all $x \in X$ such that

\sum_{p \in M (k, n)} 1_{I (p)} (x) > λ 2^{n + 1}

5.1.26

and define

G_{2} := ⋃_{k \geq 0} ⋃_{k \leq n} A (2 n + 6, k, n) .

5.1.27

Define

G_{3} := ⋃_{k \geq 0} ⋃_{n \geq k} ⋃_{0 \leq j \leq 2 n + 3} ⋃_{p \in L_{4} (k, n, j)} I (p) .

5.1.28

Define $G^{'} = G_{1} \cup G_{2} \cup G_{3}$ . The following bound of the measure of $G^{'}$ will be proven in Section 5.2.

Lemma 5.1.1 exceptional set

✓

We have

μ (G^{'}) \leq 2^{- 1} μ (G) .

5.1.29

In Section 5.4, we identify each set $C_{5} (k, n, j)$ outside $G^{'}$ as forest and use Proposition 2.0.4 to prove the following lemma.

Lemma 5.1.2 forest union

✓

Let

P_{1} = ⋃_{k \geq 0} ⋃_{n \geq k} ⋃_{0 \leq j \leq 2 n + 3} C_{5} (k, n, j)

5.1.30

For all $f : X \to C$ with $| f | \leq 1_{F}$ we have

\int_{G ∖ G^{'}} | \sum_{p \in P_{1}} T_{p} f | d μ \leq \frac{2^{433 a^{3}}}{(q - 1)^{4}} μ (G)^{1 - \frac{1}{q}} μ (F)^{\frac{1}{q}} .

5.1.31

In Section 5.5, we decompose the complement of the set of tiles in Lemma 5.1.2 and apply the antichain estimate of Proposition 2.0.3 to prove the following lemma.

Lemma 5.1.3 forest complement

✓

Let

P_{2} = P ∖ P_{1} .

5.1.32

For all $f : X \to C$ with $| f | \leq 1_{F}$ we have

\int_{G ∖ G^{'}} | \sum_{p \in P_{2}} T_{p} f | d μ \leq \frac{2^{153 a^{3}}}{(q - 1)^{5}} μ (G)^{1 - \frac{1}{q}} μ (F)^{\frac{1}{q}} .

5.1.33

Proof of Proposition 2.0.2 ▶

Proposition 2.0.2 follows by applying the triangle inequality to 2.0.22 according to the splitting in Lemma 5.1.2 and Lemma 5.1.3 and using both Lemmas as well as the bound on the set $G^{'}$ given by Lemma 5.1.1.

5.2 Proof of the Exceptional Sets Lemma

We prove separate bounds for $G_{1}$ , $G_{2}$ , and $G_{3}$ in Lemmas 5.2.1, 5.2.6, and 5.2.10. Adding up these bounds proves Lemma 5.1.1.

The bound for $G_{1}$ follows from the Vitali covering lemma, Proposition 2.0.6.

Lemma 5.2.1 first exception

✓

We have

μ (G_{1}) \leq 2^{- 5} μ (G) .

5.2.1

Proof ▶

Let

K = 2^{2 a + 5} \frac{μ (F)}{μ (G)} .

For each $p \in P_{F, G}$ pick a $r (p) > 4 D^{s (p)}$ with

μ (F \cap B (c (p), r (p))) \geq K μ (B (c (p), r (p))) .

This ball exists by definition of $P_{F, G}$ and ${dens}_{2}$ . By applying Proposition 2.0.6 to the collection of balls

B = {B (c (p), r (p)) : p \in P_{F, G}}

and the function $u = 1_{F}$ , we obtain

μ (⋃ B) \leq 2^{2 a} K^{- 1} μ (F) .

We conclude with 2.0.10 and $r (p) > 4 D^{s (p)}$

μ (G_{1}) = μ (⋃_{p \in P_{F, G}} I (p)) \leq μ (⋃ B) \leq 2^{2 a} K^{- 1} μ (F) = 2^{- 5} μ (G) .

We turn to the bound of $G_{2}$ , which relies on the Dyadic Covering Lemma 5.2.2 and the John-Nirenberg Lemma 5.2.5 below.

Lemma 5.2.2 dense cover

✓

For each $k \geq 0$ , the union of all dyadic cubes in $C (G, k)$ has measure at most $2^{k + 1} μ (G)$ .

Proof ▶

The union of dyadic cubes in $C (G, k)$ is contained the union of elements of the set $M (k)$ of all dyadic cubes $J$ with $μ (G \cap J) > 2^{- k - 1} μ (J)$ . The union of elements in the set $M (k)$ is contained in the union of elements in the set $M^{*} (k)$ of maximal elements in $M (k)$ with respect to set inclusion. Hence

μ (⋃ C (G, k)) \leq μ (⋃ M^{*} (k)) \leq \sum_{J \in M^{*} (k)} μ (J)

5.2.2

Using the definition of $M (k)$ and then the pairwise disjointedness of elements in $M^{*} (k)$ , we estimate 5.2.2 by

\leq 2^{k + 1} \sum_{J \in M^{*} (k)} μ (J \cap G) \leq 2^{k + 1} μ (G) .

5.2.3

This proves the lemma.

Lemma 5.2.3 pairwise disjoint

✓

If $p, p^{'} \in M (k, n)$ and

E_{1} (p) \cap E_{1} (p^{'}) \neq \emptyset,

5.2.4

then $p = p^{'}$ .

Proof ▶

Let $p, p^{'}$ be as in the lemma. By definition of $E_{1}$ , we have $E_{1} (p) \subset I (p)$ and analogously for $p^{'}$ , we conclude from 5.2.4 that $I (p) \cap I (p^{'}) \neq \emptyset$ . Let without loss of generality $I (p)$ be maximal in ${I (p), I (p^{'})}$ , then $I (p^{'}) \subset I (p)$ . By 5.2.4, we conclude by definition of $E_{1}$ that $Ω (p) \cap Ω (p^{'}) \neq \emptyset$ . By 2.0.14 we conclude $Ω (p) \subset Ω (p^{'})$ . It follows that $p^{'} \leq p$ . By maximality 5.1.5 of $p^{'}$ , we have $p^{'} = p$ . This proves the lemma.

Lemma 5.2.4 dyadic union

✓

For each $x \in A (λ, k, n)$ , there is a dyadic cube $I$ that contains $x$ and is a subset of $A (λ, k, n)$ .

Proof ▶

Fix $k, n, λ, x$ as in the lemma such that $x \in A (λ, k, n)$ . Let $M$ be the set of dyadic cubes $I (p)$ with $p$ in $M (k, n)$ and $x \in I (p)$ . By definition of $A (λ, k, n)$ , the cardinality of $M$ is at least $1 + λ 2^{n + 1}$ . Let $I$ be a cube of smallest scale in $M$ . Then $I$ is contained in all cubes of $M$ . It follows that $I \subset A (λ, k, n)$ .

Lemma 5.2.5 John Nirenberg

✓

For all integers $k, n, λ \geq 0$ , we have

μ (A (λ, k, n)) \leq 2^{k + 1 - λ} μ (G) .

5.2.5

Proof ▶

Fix $k, n$ as in the lemma and suppress notation to write $A (λ)$ for $A (λ, k, n)$ . We prove the lemma by induction on $λ$ . For $λ = 0$ , we use that $A (λ)$ by definition of $M (k, n)$ is contained in the union of elements in $C (G, k)$ . Lemma 5.2.2 then completes the base of the induction.

Now assume that the statement of Lemma 5.2.5 is proven for some integer $λ \geq 0$ . The set $A (λ + 1)$ is contained in the set $A (λ)$ . Let $M$ be the set of dyadic cubes which are a subset of $A (λ)$ . By Lemma 5.2.4, the union of $M$ is $A (λ)$ . Let $M^{*}$ be the set of maximal dyadic cubes in $M$ .

Let $x \in A (λ + 1)$ and $L \in M^{*}$ such that $x \in L$ . Then by the dyadic property 2.0.8

\sum_{p \in M (k, n)} 1_{I (p)} (x) = \sum_{p \in M (k, n) : I (p) \subset L} 1_{I (p)} (x) + \sum_{p \in M (k, n) : L ⊊ I (p)} 1_{I (p)} (x) .

5.2.6

We now show

\sum_{p \in M (k, n) : I (p) \subset L} 1_{I (p)} (x) \geq 2^{n + 1} .

5.2.7

The left-hand side of 5.2.6 is strictly greater than $(λ + 1) 2^{n + 1}$ . If $L$ is the top cube the second sum $Q_{2}$ on the right-hand side of 5.2.6 is zero and 5.2.7 follows immediately. Otherwise consider the inclusion-minimal cube $\hat{L}$ with $L ⊊ \hat{L}$ ; all tiles $p$ over which $Q_{2}$ is summed over satisfy $\hat{L} \subset p$ , so $Q_{2}$ is constant for all $x \in \hat{L}$ . By maximality of $L$ , $Q_{2}$ is at most $λ 2^{n + 1}$ somewhere on $\hat{L}$ , thus on all of $\hat{L}$ and consequently also at $x$ . Rearranging the inequality yields 5.2.7.

By Lemma 5.2.3, we have

\sum_{p \in M (k, n) : I (p) \subset L} μ (E_{1} (p)) \leq μ (L) .

5.2.8

Multiplying by $2^{n}$ and applying 5.1.4, we obtain

\sum_{p \in M (k, n) : I (p) \subset L} μ (I (p)) \leq 2^{n} μ (L) .

5.2.9

We then have with 5.2.7 and 5.2.9

2^{n + 1} μ (A (λ + 1) \cap L) = \int_{A (λ + 1) \cap L} 2^{n + 1} d μ

5.2.10

\leq \int \sum_{p \in M (k, n) : I (p) \subset L} 1_{I (p)} d μ \leq 2^{n} μ (L) .

5.2.11

Hence

2 μ (A (λ + 1)) = 2 \sum_{L \in M^{*}} μ (A (λ + 1) \cap L) \leq \sum_{L \in M^{*}} μ (L) = μ (A (λ)) .

5.2.12

Using the induction hypothesis, this proves 5.2.5 for $λ + 1$ and completes the proof of the lemma.

Lemma 5.2.6 second exception

✓

We have

μ (G_{2}) \leq 2^{- 2} μ (G) .

5.2.13

Proof ▶

We use Lemma 5.2.5 and sum twice a geometric series to obtain

\sum_{0 \leq k} \sum_{k \leq n} μ (A (2 n + 6, k, n)) \leq \sum_{0 \leq k} \sum_{k \leq n} 2^{k - 5 - 2 n} μ (G)

5.2.14

\leq \sum_{0 \leq k} 2^{- k - 3} μ (G) \leq 2^{- 2} μ (G) .

5.2.15

This proves the lemma.

We turn to the set $G_{3}$ .

Lemma 5.2.7 top tiles

✓

We have

\sum_{m \in M (k, n)} μ (I (m)) \leq 2^{n + k + 3} μ (G) .

5.2.16

Proof ▶

We write the left-hand side of 5.2.16

\int \sum_{m \in M (k, n)} 1_{I (m)} (x) d μ (x) \leq 2^{n + 1} \sum_{λ = 0}^{| M |} μ (A (λ, k, n)) .

5.2.17

Using Lemma 5.2.5 and then summing a geometric series, we estimate this by

\leq 2^{n + 1} \sum_{λ = 0}^{| M |} 2^{k + 1 - λ} μ (G) \leq 2^{n + 1} 2^{k + 2} μ (G) .

5.2.18

This proves the lemma.

Lemma 5.2.8 tree count

✓

Let $k, n, j \geq 0$ . We have for every $x \in X$

\sum_{u \in U_{1} (k, n, j)} 1_{I (u)} (x) \leq 2^{- j} 2^{9 a} \sum_{m \in M (k, n)} 1_{I (m)} (x)

5.2.19

Proof ▶

Let $x \in X$ . For each $u \in U_{1} (k, n, j)$ with $x \in I (u)$ , as $u \in C_{1} (k, n, j)$ , there are at least $2^{j}$ elements $m \in M (k, n)$ with $100 u ≲ m$ and in particular $x \in I (m)$ . Hence

1_{I (u)} (x) \leq 2^{- j} \sum_{m \in M (k, n) : 100 u ≲ m} 1_{I (m)} (x) .

5.2.20

Conversely, for each $m \in M (k, n)$ with $x \in I (m)$ , let $U (m)$ be the set of $u \in U_{1} (k, n, j)$ with $x \in I (u)$ and $100 u ≲ m$ . Summing 5.2.20 over $u$ and counting the pairs $(u, m)$ with $100 u ≲ m$ differently gives

\sum_{u \in U_{1} (k, n, j)} 1_{I (u)} (x) \leq 2^{- j} \sum_{m \in M (k, n)} \sum_{u \in U (m)} 1_{I (m)} (x) .

5.2.21

We estimate the number of elements in $U (m)$ . Let $u \in U (m)$ . Then by definition of $U (m)$

d_{u} (Q (u), Q (m)) \leq 100 .

5.2.22

If $u^{'}$ is a further element in $U (m)$ with $u \neq u^{'}$ , then

Q (m) \in B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100) .

5.2.23

By the last display and definition of $U_{1} (k, n, j)$ , none of $I (u)$ , $I (u^{'})$ is strictly contained in the other. As both contain $x$ , we have $I (u) = I (u^{'})$ . We then have $d_{u} = d_{u^{'}}$ .

By 2.0.15, the balls $B_{u} (Q (u), 0.2)$ and $B_{u} (Q (u^{'}), 0.2)$ are contained respectively in $Ω (u)$ and $Ω (u^{'})$ and thus are disjoint by 2.0.13. By 5.2.22 and the triangle inequality, both balls are contained in $B_{u} (Q (m), 100.2)$ .

By 1.0.11 applied nine times, there is a collection of at most $2^{9 a}$ balls of radius $0.2$ with respect to the metric $d_{u}$ which cover the ball $B_{u} (Q (m), 100.2)$ . Let $B^{'}$ be a ball in this cover. As the center of $B^{'}$ can be in at most one of the disjoint balls $B_{u} (Q (u), 0.2)$ and $B_{u} (Q (u^{'}), 0.2)$ , the ball $B^{'}$ can contain at most one of the points $Q (u)$ , $Q (u^{'})$ .

Hence the image of $U (m)$ under $Q$ has at most $2^{9 a}$ elements; since $Q$ is injective on $U (m)$ , the same is true of $U (m)$ . Inserting this into 5.2.21 proves the lemma.

Lemma 5.2.9 boundary exception

✓

Let $L (u)$ be as defined in 5.1.20. We have for each $u \in U_{1} (k, n, l)$ ,

μ (⋃_{I \in L (u)} I) \leq D^{1 - κ Z (n + 1)} μ (I (u)) .

5.2.24

Proof ▶

Let $u \in U_{1} (k, n, l)$ . Let $I \in L (u)$ . Then we have $s (I) = s (u) - Z (n + 1) - 1$ and $I \subset I (u)$ and $B (c (I), 8 D^{s (I)}) ⊄ I (u)$ . By 2.0.10, the set $I$ is contained in $B (c (I), 4 D^{s (I)})$ . By the triangle inequality, the set $I$ is contained in

X (u) := {x \in I (u) : ρ (x, X ∖ I (u)) \leq 12 D^{s (u) - Z (n + 1) - 1}} .

5.2.25

By the small boundary property 2.0.11, noting that

12 D^{s (u) - Z (n + 1) - 1} = 12 D^{s (I)} > D^{- S},

we have

μ (X (u)) \leq 2 \cdot (12 D^{- Z (n + 1) - 1})^{κ} μ (I (u)) .

Using $κ < 1$ and $D \geq 12$ , this proves the lemma.

Lemma 5.2.10 third exception

✓

We have

μ (G_{3}) \leq 2^{- 4} μ (G) .

5.2.26

Proof ▶

As each $p \in L_{4} (k, n, j)$ is contained in $\cup L (u)$ for some $u \in U_{1} (k, n, l)$ , we have

μ (⋃_{p \in L_{4} (k, n, j)} I (p)) \leq \sum_{u \in U_{1} (k, n, j)} μ (⋃_{I \in L (u)} I) .

5.2.27

Using Lemma 5.2.9 and then Lemma 5.2.8, we estimate this further by

\leq \sum_{u \in U_{1} (k, n, j)} D^{1 - κ Z (n + 1)} μ (I (u))

5.2.28

\leq 2^{100 a^{2} + 9 a + 1 - j} \sum_{m \in M (k, n)} D^{- κ Z (n + 1)} μ (I (m)) .

5.2.29

Using Lemma 5.2.7, we estimate this by

\leq 2^{100 a^{2} + 9 a + 1 - j} D^{- κ Z (n + 1)} 2^{n + k + 3} μ (G) .

5.2.30

Now we estimate $G_{3}$ defined in 5.1.28 by

μ (G_{3}) \leq \sum_{k \geq 0} \sum_{n \geq k} \sum_{0 \leq j \leq 2 n + 3} μ (⋃_{p \in L_{4} (k, n, j)} I (p))

5.2.31

\leq \sum_{k \geq 0} \sum_{n \geq k} \sum_{0 \leq j \leq 2 n + 3} 2^{100 a^{2} + 9 a + 3 + n + k - j} D^{- κ Z (n + 1)} μ (G)

5.2.32

Summing geometric series, using that $D^{κ Z} \geq 8$ by 2.0.1, 2.0.2 and 2.0.3, we estimate this by

\leq \sum_{k \geq 0} \sum_{n \geq k} 2^{100 a^{2} + 9 a + 4 + n + k} D^{- κ Z (n + 1)} μ (G)

5.2.33

= \sum_{k \geq 0} 2^{100 a^{2} + 9 a + 4 + 2 k} D^{- κ Z (k + 1)} \sum_{n \geq k} 2^{n - k} D^{- κ Z (n - k)} μ (G)

5.2.34

\leq \sum_{k \geq 0} 2^{100 a^{2} + 9 a + 5 + 2 k} D^{- κ Z (k + 1)} μ (G)

5.2.35

\leq 2^{100 a^{2} + 9 a + 6} D^{- κ Z} μ (G)

5.2.36

Using $D = 2^{100 a^{2}}$ and $a \geq 4$ and $κ Z \geq 2$ by 2.0.1 and 2.0.2 proves the lemma.

Proof of Lemma 5.1.1 ▶

5.3 Auxiliary lemmas

Before proving Lemma 5.1.2 and Lemma 5.1.3, we collect some useful properties of $≲$ .

Lemma 5.3.1 wiggle order 1

✓

If $n p ≲ m p^{'}$ and $n^{'} \geq n$ and $m \geq m^{'}$ then $n^{'} p ≲ m^{'} p^{'}$ .

Proof ▶

This follows immediately from the definition 2.0.24 of $≲$ and the two inclusions $B_{p} (Q (p), n) \subset B_{p} (Q (p), n^{'})$ and $B_{p^{'}} (Q (p^{'}), m^{'}) \subset B_{p^{'}} (Q (p^{'}), m)$ .

Lemma 5.3.2 wiggle order 2

✓

Let $n, m \geq 1$ and $k > 0$ . If $p, p^{'} \in P$ with $I (p) \neq I (p^{'})$ and

n p ≲ k p^{'}

5.3.1

then

(n + 2^{- 95 a} m) p ≲ m p^{'} .

5.3.2

Proof ▶

The assumption 5.3.1 together with the definition 2.0.24 of $≲$ implies that $I (p) ⊊ I (p^{'})$ . Let $ϑ \in B_{p^{'}} (Q (p^{'}), m)$ . Then we have by the triangle inequality

d_{p} (Q (p), ϑ) \leq d_{p} (Q (p), Q (p^{'})) + d_{p} (Q (p^{'}), ϑ)

The first summand is bounded by $n$ since

Q (p^{'}) \in B_{p^{'}} (Q (p^{'}), k) \subset B_{p} (Q (p), n),

using 2.0.24. For the second summand we use Lemma 2.1.2 to show that the sum is estimated by

n + 2^{- 95 a} d_{p^{'}} (Q (p^{'}), ϑ) < n + 2^{- 95 a} m .

Thus $B_{p^{'}} (Q (p^{'}), m) \subset B_{p} (Q (p), n + 2^{- 95 a} m)$ . Combined with $I (p) \subset I (p^{'})$ , this yields 5.3.2.

Lemma 5.3.3 wiggle order 3

✓

The following implications hold for all $q, q^{'} \in P$ :

q \leq q^{'} and λ \geq 1.1 ⟹ λ q ≲ λ q^{'},

5.3.3

10 q ≲ q^{'} and I (q) \neq I (q^{'}) ⟹ 100 q ≲ 100 q^{'},

5.3.4

2 q ≲ q^{'} and I (q) \neq I (q^{'}) ⟹ 4 q ≲ 500 q^{'} .

5.3.5

Proof ▶

5.3.4 and 5.3.5 are easy consequences of Lemma 5.3.1, Lemma 5.3.2 and the fact that $a \geq 4$ . For 5.3.3, if $I (q) = I (q^{'})$ then we get $q = q^{'}$ by 2.0.13 and 2.0.23. If $I (q) \neq I (q^{'})$ , then from 2.0.23, 2.0.24 and 2.0.15 it follows that $q ≲ 0.2 q^{'}$ , and 5.3.3 follows from an easy calculation using Lemma 5.3.2.

We call a collection $A$ of tiles convex if

p \leq p^{'} \leq p^{″} and p, p^{″} \in A ⟹ p^{'} \in A .

5.3.6

Lemma 5.3.4 P convex

✓

For each $k$ , the collection $P (k)$ is convex.

Proof ▶

Suppose that $p \leq p^{'} \leq p^{″}$ and $p, p^{″} \in P (k)$ . By 5.1.3 we have $I (p), I (p^{″}) \in C (G, k)$ , so there exists by 5.1.1 some $J \in D$ with

I (p^{'}) \subset I (p^{″}) \subset J

and $μ (G \cap J) > 2^{- k - 1} μ (J)$ . Thus 5.1.1 holds for $I (p^{'})$ . On the other hand, by 5.1.2, there exists no $J \in D$ with $I (p) \subset J$ and $μ (G \cap J) > 2^{- k} μ (J)$ . Since $I (p) \subset I (p^{'})$ , this implies that 5.1.2 holds for $I (p^{'})$ . Hence $I (p^{'}) \in C (G, k)$ , and therefore by 5.1.3 $p^{'} \in P (k)$ .

Lemma 5.3.5 C convex

✓

For each $k, n$ , the collection $C (k, n)$ is convex.

Proof ▶

Let $p \leq p^{'} \leq p^{″}$ with $p, p^{″} \in C (k, n)$ . Then, in particular, $p, p^{″} \in P (k)$ , so, by Lemma 5.3.4, $p^{'} \in P (k)$ . Next, we show that if $q \leq q^{'} \in P (k)$ then ${dens}_{k}^{'} ({q}) \geq {dens}_{k}^{'} ({q^{'}})$ . If $p \in P (k)$ and $λ \geq 2$ with $λ q^{'} ≲ λ p$ , then it follows from $q \leq q^{'}$ , 5.3.3 of Lemma 5.3.3 and transitivity of $≲$ that $λ q ≲ λ p$ . Thus the supremum in the definition 5.1.6 of ${dens}_{k}^{'} ({q})$ is over a superset of the set the supremum in the definition of ${dens}_{k}^{'} ({q^{'}})$ is taken over, which shows ${dens}_{k}^{'} ({q}) \geq {dens}_{k}^{'} ({q^{'}})$ . From $p^{'} \leq p^{″}$ , $p^{″} \in C (k, n)$ and 5.1.7 it then follows that

2^{4 a} 2^{- n} < {dens}_{k}^{'} ({p^{″}}) \leq {dens}_{k}^{'} ({p^{'}}) .

Similarly, it follows from $p \leq p^{'}$ , $p \in C (k, n)$ and 5.1.7 that

{dens}_{k}^{'} ({p^{'}}) \leq {dens}_{k}^{'} ({p}) \leq 2^{4 a} 2^{- n + 1} .

Thus $p^{'} \in C (k, n)$ .

Lemma 5.3.6 C1 convex

✓

For each $k, n, j$ , the collection $C_{1} (k, n, j)$ is convex.

Proof ▶

Let $p \leq p^{'} \leq p^{″}$ with $p, p^{″} \in C_{1} (k, n, j)$ . By Lemma 5.3.5 and the inclusion $C_{1} (k, n, j) \subset C (k, n)$ , which holds by definition 5.1.9, we have $p^{'} \in C (k, n)$ . By 5.3.3 and transitivity of $≲$ we have that $q \leq q^{'}$ and $100 q^{'} ≲ m$ imply $100 q ≲ m$ . So, by 5.1.8, $B (p^{″}) \subset B (p^{'}) \subset B (p)$ . Consequently, by 5.1.9

2^{j} \leq | B (p^{″}) | \leq | B (p^{'}) | \leq | B (p) | < 2^{j + 1},

thus $p^{'} \in C_{1} (k, n, j)$ .

Lemma 5.3.7 C2 convex

✓

For each $k, n, j$ , the collection $C_{2} (k, n, j)$ is convex.

Proof ▶

Let $p \leq p^{'} \leq p^{″}$ with $p, p^{″} \in C_{2} (k, n, j)$ . By 5.1.13, we have

C_{2} (k, n, j) \subset C_{1} (k, n, j) .

Combined with Lemma 5.3.6, it follows that $p^{'} \in C_{1} (k, n, j)$ . If $p = p^{'}$ the statement is trivially true, otherwise suppose that $p^{'} \notin C_{2} (k, n, j)$ . By 5.1.13, this implies that there exists $0 \leq l^{'} \leq Z (n + 1)$ with $p^{'} \in L_{1} (k, n, j, l^{'})$ . By the definition 5.1.11 of $L_{1} (k, n, j, l^{'})$ , this implies that $p^{'}$ is minimal with respect to $\leq$ in $C_{1} (k, n, j) ∖ ⋃_{l < l^{'}} L_{1} (k, n, j, l)$ . Since $p \leq p^{'}$ and $p \in C_{2} (k, n, j)$ , $p = p^{'}$ , a contradiction.

Lemma 5.3.8 C3 convex

✓

For each $k, n, j$ , the collection $C_{3} (k, n, j)$ is convex.

Proof ▶

Let $p \leq p^{'} \leq p^{″}$ with $p, p^{″} \in C_{3} (k, n, j)$ . By 5.1.16 and Lemma 5.3.7 it follows that $p^{'} \in C_{2} (k, n, j)$ . By 5.1.16 and 5.1.15, there exists $u \in U_{1} (k, n, j)$ with $2 p^{″} ≲ u$ and $I (p^{″}) ⊊ I (u)$ . From $p^{'} \leq p^{″}$ , 5.3.3 and transitivity of $≲$ we then have $2 p^{'} ≲ u$ and $I (p^{'}) ⊊ I (u)$ , so $p^{'} \in C_{3} (k, n, j)$ .

Lemma 5.3.9 C4 convex

✓

For each $k, n, j$ , the collection $C_{4} (k, n, j)$ is convex.

Proof ▶

The proof is entirely analogous to Lemma 5.3.7, substituting $C_{4}$ for $C_{2}$ , $C_{3}$ for $C_{1}$ and $p^{'} \leq p^{″}$ for $p \leq p^{'}$ .

Lemma 5.3.10 C5 convex

✓

For each $k, n, j$ , the collection $C_{5} (k, n, j)$ is convex.

Proof ▶

Let $p \leq p^{'} \leq p^{″}$ with $p, p^{″} \in C_{5} (k, n, j)$ . Then $p, p^{″} \in C_{4} (k, n, j)$ by 5.1.23, and thus by Lemma 5.3.9 $p^{'} \in C_{4} (k, n, j)$ . It suffices to show that if $p^{'} \in L_{4} (k, n, j)$ then $p \in L_{4} (k, n, j)$ by contraposition; this is true by 5.1.22 and $p \leq p^{'}$ .

Lemma 5.3.11 dens compare

✓

We have for every $k \geq 0$ and $P^{'} \subset P (k)$

{dens}_{1} (P^{'}) \leq {dens}_{k}^{'} (P^{'}) .

5.3.7

Proof ▶

It suffices to show that for all $p^{'} \in P^{'}$ and $λ \geq 2$ and $p \in P (P^{'})$ with $λ p^{'} ≲ λ p$ we have

\frac{μ (E_{2} (λ, p))}{μ (I (p))} \leq sup_{p^{″} \in P (k) : λ p^{'} ≲ λ p^{″}} \frac{μ (E_{2} (λ, p^{″}))}{μ (I (p^{″}))} .

5.3.8

Let such $p^{'}$ , $λ$ , $p$ be given. It suffices to show that $p \in P (k)$ , that is, it satisfies 5.1.1 and 5.1.2.

We show 5.1.1. As $p \in P (P^{'})$ , there exists $p^{″} \in P^{'}$ with $I (p^{'}) \subset I (p^{″})$ . By assumption on $P^{'}$ , we have $p^{″} \in P (k)$ and there exists $J \in D$ with $I (p^{″}) \subset J$ and

μ (G \cap J) > 2^{- k - 1} μ (J) .

5.3.9

Then also $I (p^{'}) \subset J$ , which proves 5.1.1 for $p$ .

We show 5.1.2. Assume to get a contradiction that there exists $J \in D$ with $I (p) \subset J$ and

μ (G \cap J) > 2^{- k} μ (J) .

5.3.10

As $λ p^{'} ≲ λ p$ , we have $I (p^{'}) \subset I (p)$ , and therefore $I (p^{'}) \subset J$ . This contradicts $p^{'} \in P^{'} \subset P (k)$ . This proves 5.1.2 for $p$ .

Lemma 5.3.12 C dens1

✓

For each set $A \subset C (k, n)$ , we have

{dens}_{1} (A) \leq 2^{4 a} 2^{- n + 1} .

Proof ▶

We have by Lemma 5.3.11 that ${dens}_{1} (A) \leq {dens}_{k}^{'} (A)$ . Since $A \subset C (k, n)$ , it follows from monotonicity of suprema and the definition 5.1.6 that ${dens}_{k}^{'} (A) \leq {dens}_{k}^{'} (C (k, n)) .$ By 5.1.6 and 5.1.7, we have

{dens}_{k}^{'} (C (k, n)) = sup_{p \in C (k, n)} {dens}_{k}^{'} ({p}) \leq 2^{4 a} 2^{- n + 1} .

5.4 Proof of the Forest Union Lemma

Fix $k, n, j \geq 0$ . Define

C_{6} (k, n, j)

to be the set of all tiles $p \in C_{5} (k, n, j)$ such that $I (p) ⊄ G^{'}$ . The following chain of lemmas establishes that the set $C_{6} (k, n, j)$ can be written as a union of a small number of $n$ -forests.

For $u \in U_{1} (k, n, j)$ , define

T_{1} (u) := {p \in C_{1} (k, n, j) : I (p) \neq I (u), 2 p ≲ u} .

5.4.1

Define

U_{2} (k, n, j) := {u \in U_{1} (k, n, j) : T_{1} (u) \cap C_{6} (k, n, j) \neq \emptyset} .

5.4.2

Define a relation $\sim$ on $U_{2} (k, n, j)$ by setting $u \sim u^{'}$ for $u, u^{'} \in U_{2} (k, n, j)$ if $u = u^{'}$ or there exists $p$ in $T_{1} (u)$ with $10 p ≲ u^{'}$ .

Lemma 5.4.1 relation geometry

✓

If $u \sim u^{'}$ , then $I (u) = I (u^{'})$ and

B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100) \neq \emptyset .

Proof ▶

Let $u, u^{'} \in U_{2} (k, n, j)$ with $u \sim u^{'}$ . If $u = u^{'}$ then the conclusion of the Lemma clearly holds. Else, there exists $p \in C_{1} (k, n, j)$ such that $I (p) \neq I (u)$ and $2 p ≲ u$ and $10 p ≲ u^{'}$ . Using Lemma 5.3.1 and 5.3.4 of Lemma 5.3.3, we deduce that

100 p ≲ 100 u, 100 p ≲ 100 u^{'} .

5.4.3

Now suppose that $B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100) = \emptyset$ . Then we have $B (u) \cap B (u^{'}) = \emptyset$ , by the definition 5.1.8 of $B$ and the definition 2.0.24 of $≲$ , but also $B (u) \subset B (p)$ and $B (u^{'}) \subset B (p)$ , by 5.1.8, 2.0.24 and 5.4.3. Hence,

| B (p) | \geq | B (u) | + | B (u^{'}) | \geq 2^{j} + 2^{j} = 2^{j + 1},

which contradicts $p \in C_{1} (k, n, j)$ . Therefore we must have

B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100) \neq \emptyset .

It follows from $2 p ≲ u$ and $10 p ≲ u^{'}$ that $I (p) \subset I (u)$ and $I (p) \subset I (u^{'})$ . By 2.0.8, it follows that $I (u)$ and $I (u^{'})$ are nested. Combining this with the conclusion of the last paragraph and definition 5.1.14 of $U_{1} (k, n, j)$ , we obtain that $I (u) = I (u^{'})$ .

Lemma 5.4.2 equivalence relation

✓

For each $k, n, j$ , the relation $\sim$ on $U_{2} (k, n, j)$ is an equivalence relation.

Proof ▶

Reflexivity holds by definition. For transitivity, suppose that

u, u^{'}, u ” \in U_{1} (k, n, j)

and $u \sim u^{'}$ , $u^{'} \sim u^{″}$ . By Lemma 5.4.1, it follows that $I (u) = I (u^{'}) = I (u^{″})$ , that there exists

ϑ \in B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100)

and that there exists

θ \in B_{u^{'}} (Q (u^{'}), 100) \cap B_{u^{″}} (Q (u ”), 100) .

If $u = u^{'}$ , then $u \sim u^{″}$ holds by assumption. Else, there exists by the definition of $\sim$ some $p \in T_{1} (u)$ with $10 p ≲ u^{'}$ . Then we have $2 p ≲ u$ and $p \neq u$ by definition of $T_{1} (u)$ , so $4 p ≲ 500 u$ by 5.3.5. For $q \in B_{u^{″}} (Q (u^{″}), 1)$ it follows by the triangle inequality that

\begin{aligned} d_{u} (Q (u), q) & \leq d_{u} (Q (u), ϑ) + d_{u} (ϑ, Q (u^{'})) \\ + d_{u} (Q (u^{'}), θ) + d_{u} (θ, Q (u ”)) + d_{u} (Q (u ”), q) . \end{aligned}

Using 2.0.17 and the fact that $I (u) = I (u^{'}) = I (u^{″})$ this equals

\begin{aligned} d_{u} (Q (u), ϑ) + d_{u^{'}} (ϑ, Q (u^{'})) \\ + d_{u^{'}} (Q (u^{'}), θ) + d_{u^{″}} (θ, Q (u ”)) + d_{u^{″}} (Q (u ”), q) \\ < 100 + 100 + 100 + 100 + 1 < 500 . \end{aligned}

Since $4 p ≲ 500 u$ , it follows that $d_{p} (Q (p), q) < 4 < 10$ . We have shown that $B_{u^{″}} (Q (u^{″}), 1) \subset B_{p} (Q (p), 10)$ , combining this with $I (u^{″}) = I (u)$ gives $u \sim u^{″}$ .

For symmetry suppose that $u \sim u^{'}$ . By Lemma 5.4.1, it follows that $I (u) = I (u^{'})$ and that there exists $ϑ \in B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100)$ . Again, for $u = u^{'}$ symmetry is obvious, so suppose that $u \neq u^{'}$ . There exists $p \in T_{1} (u^{'})$ , which then satisfies $2 p ≲ u^{'}$ and $I (p) \neq I (u^{'})$ . By Lemma 5.3.1 and 5.3.5, it follows that

10 p ≲ 4 p ≲ 500 u^{'} .

5.4.4

If $q \in B_{u} (Q (u), 1)$ then we have from the triangle inequality and the fact that $I (u) = I (u^{'})$ :

\begin{aligned} d_{u^{'}} (Q (u^{'}), q) & \leq d_{u^{'}} (Q (u^{'}), ϑ) + d_{u^{'}} (ϑ, Q (u)) + d_{u^{'}} (Q (u), q) \\ = d_{u^{'}} (Q (u^{'}), ϑ) + d_{u} (ϑ, Q (u)) + d_{u} (Q (u), q) \\ < 100 + 100 + 1 < 500 . \end{aligned}

Combining this with 5.4.4 and 2.0.24, we get

B_{u} (Q (u), 1) \subset B_{p} (Q (p), 10) .

Since $2 p ≲ u^{'}$ , we have $I (p) \subset I (u^{'}) = I (u)$ . Thus, $10 p ≲ u$ which completes the proof of $u^{'} \sim u$ .

Choose a set $U_{3} (k, n, j)$ of representatives for the equivalence classes of $\sim$ in $U_{2} (k, n, j)$ . Define for each $u \in U_{3} (k, n, j)$

T_{2} (u) := ⋃_{u \sim u^{'}} T_{1} (u^{'}) \cap C_{6} (k, n, j) .

5.4.5

Lemma 5.4.3 C6 forest

✓

We have

C_{6} (k, n, j) = ⋃_{u \in U_{3} (k, n, j)} T_{2} (u) .

5.4.6

Proof ▶

Let $p \in C_{6} (k, n, j)$ . By 5.1.19 and 5.1.23, we have $p \in C_{3} (k, n, j)$ . By 5.1.15 and 5.1.16, there exists $u \in U_{1} (k, n, j)$ with $2 p ≲ u$ and $I (p) \neq I (u)$ , that is, with $p \in T_{1} (u)$ . Then $T_{1} (u)$ is clearly nonempty, so $u \in U_{2} (k, n, j)$ . By the definition of $U_{3} (k, n, j)$ , there exists $u^{'} \in U_{3} (k, n, j)$ with $u \sim u^{'}$ . By 5.4.5, we have $p \in T_{2} (u^{'})$ .

Lemma 5.4.4 forest geometry

✓

For each $u \in U_{3} (k, n, j)$ , the set $T_{2} (u)$ satisfies 2.0.32.

Proof ▶

Let $p \in T_{2} (u)$ . By 5.4.5, there exists $u^{'} \sim u$ with $p \in T_{1} (u^{'})$ . Then we have $2 p ≲ u^{'}$ and $I (p) \neq I (u^{'})$ , so by 5.3.5 $4 p ≲ 500 u^{'}$ . Further, by Lemma 5.4.1, we have that $I (u^{'}) = I (u)$ and there exists $ϑ \in B_{u^{'}} (Q (u^{'}), 100) \cap B_{u} (Q (u), 100)$ . Let $θ \in B_{u} (Q (u), 1)$ . Using the triangle inequality and the fact that $I (u^{'}) = I (u)$ , we obtain

\begin{aligned} d_{u^{'}} (Q (u^{'}), θ) & \leq d_{u^{'}} (Q (u^{'}), ϑ) + d_{u^{'}} (Q (u), ϑ) + d_{u^{'}} (Q (u), θ) \\ = d_{u^{'}} (Q (u^{'}), ϑ) + d_{u} (Q (u), ϑ) + d_{u} (Q (u), θ) \\ < 100 + 100 + 1 < 500 . \end{aligned}

Combining this with $4 p ≲ 500 u^{'}$ , we obtain

B_{u} (Q (u), 1) \subset B_{u^{'}} (Q (u^{'}), 500) \subset B_{p} (Q (p), 4) .

Together with $I (p) \subset I (u^{'}) = I (u)$ , this gives $4 p ≲ u$ , which is 2.0.32.

Lemma 5.4.5 forest convex

✓

For each $u \in U_{3} (k, n, j)$ , the set $T_{2} (u)$ satisfies the convexity condition 2.0.33.

Proof ▶

Let $p, p^{″} \in T_{2} (u)$ and $p^{'} \in P$ with $p \leq p^{'} \leq p^{″}$ . By 5.4.5 we have $p, p^{″} \in C_{6} (k, n, j) \subset C_{5} (k, n, j)$ . By Lemma 5.3.10, we have $p^{'} \in C_{5} (k, n, j)$ . Since $p \in C_{6} (k, n, j)$ we have $I (p) ⊄ G^{'}$ , so $I (p^{'}) ⊄ G^{'}$ and therefore also $p^{'} \in C_{6} (k, n, j)$ .

By 5.4.5 there exists $u^{'} \in U_{2} (k, n, j)$ with $p^{″} \in T_{1} (u^{'})$ and hence $2 p^{″} ≲ u^{'}$ and $I (p^{″}) \neq I (u^{'})$ . Together this implies $I (p^{″}) ⊊ I (u^{'})$ . With the inclusion $I (p^{'}) \subset I (p^{″})$ from $p^{'} \leq p^{″}$ , it follows that $I (p^{'}) ⊊ I (u^{'})$ and hence $I (p^{'}) \neq I (u^{'})$ . By 5.3.3 and transitivity of $≲$ we further have $2 p^{'} ≲ u^{'}$ , so $p^{'} \in T_{1} (u^{'})$ . It follows that $p^{'} \in T_{2} (u)$ , which shows 2.0.33.

Lemma 5.4.6 forest separation

✓

For each $u, u^{'} \in U_{3} (k, n, j)$ with $u \neq u^{'}$ and each $p \in T_{2} (u)$ with $I (p) \subset I (u^{'})$ we have

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

5.4.7

Proof ▶

By the definition 5.1.13 of $C_{2} (k, n, j)$ , there exists a tile $p^{'} \in C_{1} (k, n, j)$ with $p^{'} \leq p$ and $s (p^{'}) \leq s (p) - Z (n + 1)$ . By Lemma 2.1.2 we have

d_{p} (Q (p), Q (u^{'})) \geq 2^{95 a Z (n + 1)} d_{p^{'}} (Q (p), Q (u^{'})) .

By 5.3.3 we have $2 p^{'} ≲ 2 p$ , so by transitivity of $≲$ there exists $v \sim u$ with $2 p^{'} ≲ v$ and $I (p^{'}) \neq I (v)$ . Since $u, u^{'}$ are not equivalent under $\sim$ , we have $v ≁ u^{'}$ , thus $10 p^{'} ≴ u^{'}$ . This implies that there exists $q \in B_{u^{'}} (Q (u^{'}), 1) ∖ B_{p^{'}} (Q (p^{'}), 10)$ .

From $p^{'} \leq p$ , $I (p^{'}) \subset I (p) \subset I (u^{'})$ and Lemma 2.1.2 it then follows that

\begin{aligned} d_{p^{'}} (Q (p), Q (u^{'})) \\ \geq - d_{p^{'}} (Q (p), Q (p^{'})) + d_{p^{'}} (Q (p^{'}), q) - d_{p^{'}} (q, Q (u^{'})) \\ \geq - d_{p^{'}} (Q (p), Q (p^{'})) + d_{p^{'}} (Q (p^{'}), q) - d_{u^{'}} (q, Q (u^{'})) \\ > - 1 + 10 - 1 = 8 . \end{aligned}

The lemma follows by combining the two displays with the fact that $95 a \geq 1$ .

Lemma 5.4.7 forest inner

✓

For each $u \in U_{3} (k, n, j)$ and each $p \in T_{2} (u)$ we have

B (c (p), 8 D^{s (p)}) \subset I (u) .

5.4.8

Proof ▶

Let $p \in T_{2} (u)$ . Then $p \in C_{4} (k, n, j)$ , hence there exists a chain

p \leq p_{Z (n + 1)} \leq \dots \leq p_{0}

of distinct tiles in $C_{3} (n, k, j)$ . We pick such a chain and set $q = p_{0}$ . Then we have from distinctness of the tiles in the chain that $s (p) \leq s (q) - Z (n + 1)$ . By 5.1.16 there exists $u^{″} \in U_{1} (k, n, j)$ with $2 q ≲ u^{″}$ and $s (q) < s (u^{″})$ . Then we have in particular by Lemma 5.3.1 that $10 p ≲ u^{″}$ . Let $u^{'} \sim u$ be such that $p \in T_{1} (u^{'})$ . By the definition of $\sim$ , it follows that $u^{'} \sim u^{″}$ . By transitivity of $\sim$ , we have $u \sim u^{″}$ . By Lemma 5.4.1, we have $\sc (u^{″}) = \sc (u)$ , hence $s (q) < s (u)$ and $s (p) \leq s (q) - Z (n + 1) \leq s (u) - Z (n + 1) - 1$ .

Thus, there exists some cube $I \in D$ with $s (I) = s (u) - Z (n + 1) - 1$ and $I \subset I (u)$ and $I (p) \subset I$ . Since $p \in C_{5} (k, n, j)$ , we have that $I \notin L (u)$ , so $B (c (I), 8 D^{s (I)}) \subset I (u)$ . By the triangle inequality, 2.0.1 and $a \geq 4$ , the same then holds for the subcube $I (p) \subset I$ .

Lemma 5.4.8 forest stacking

✓

It holds for $k \leq n$ that

\sum_{u \in U_{3} (k, n, j)} 1_{I (u)} \leq (4 n + 12) 2^{n} .

5.4.9

Proof ▶

Suppose that a point $x$ is contained in more than $(4 n + 12) 2^{n}$ cubes $I (u)$ with $u \in U_{3} (k, n, j)$ . Since $U_{3} (k, n, j) \subset C_{1} (k, n, j)$ for each such $u$ , there exists $m \in M (k, n)$ such that $100 u ≲ m$ . We fix such an $m (u) := m$ for each $u$ , and claim that the map $u \mapsto m (u)$ is injective. Indeed, assume for $u \neq u^{'}$ there is $m \in M (k, n)$ such that $100 u ≲ m$ and $100 u^{'} ≲ m$ . By 2.0.8, either $I (u) \subset I (u^{'})$ or $I (u^{'}) \subset I (u)$ . By 5.1.14, $B_{u} (Q (u), 100) \cap B_{u^{'}} (Q (u^{'}), 100) = \emptyset$ . This contradicts $Ω (m)$ being contained in both sets by 2.0.15. Thus $x$ is contained in more than $(4 n + 12) 2^{n}$ cubes $I (m)$ , $m \in M (k, n)$ . Consequently, we have by 5.1.26 that $x \in A (2 n + 6, k, n) \subset G_{2}$ . Let $I (u)$ be an inclusion minimal cube among the $I (u^{'}), u^{'} \in U_{3} (k, n, j)$ with $x \in I (u)$ . By the dyadic property 2.0.8, we have $I (u) \subset I (u^{'})$ for all cubes $I (u^{'})$ containing $x$ . Thus

I (u) \subset {y : \sum_{u \in U_{3} (k, n, j)} 1_{I (u)} (y) > 1 + (4 n + 12) 2^{n}} \subset G_{2} .

Thus $T_{1} (u) \cap C_{6} (k, n, j) = \emptyset$ . This contradicts $u \in U_{2} (k, n, j)$ .

We now turn to the proof of Lemma 5.1.2.

Proof of Lemma 5.1.2 ▶

We first fix $k, n, j$ . By 2.0.21 and 2.0.20, we have that $1_{I (p)} T_{p} f (x) = T_{p} f (x)$ and hence $1_{G ∖ G^{'}} T_{p} f (x) = 0$ for all $p \in C_{5} (k, n, j) ∖ C_{6} (k, n, j)$ . Thus it suffices to estimate the contribution of the sets $C_{6} (k, n, j)$ . By Lemma 5.4.8, we can decompose $U_{3} (k, n, j)$ as a disjoint union of at most $4 n + 12$ collections $U_{4} (k, n, j, l)$ , $1 \leq l \leq 4 n + 12$ , each satisfying

\sum_{u \in U_{4} (k, n, j, l)} 1_{I (u)} \leq 2^{n} .

By Lemmas 5.4.4, 5.4.5, 5.4.6, 5.4.7 and 5.3.12, the pairs

(U_{4} (k, n, j, l), T_{2} |_{U_{4} (k, n, j, l)})

are $n$ -forests for each $k, n, j, l$ , and by Lemma 5.4.3, we have

C_{6} (k, n, j) = ⋃_{l = 1}^{4 n + 12} ⋃_{u \in U_{4} (k, n, j, l)} T_{2} (u) .

Since $I (p) ⊄ G_{1}$ for all $p \in C_{6} (k, n, j)$ , we have $C_{6} (k, n, j) \cap P_{F, G} = \emptyset$ and hence

{dens}_{2} (⋃_{u \in U_{4} (k, n, j, l)} T_{2} (u)) \leq 2^{2 a + 5} \frac{μ (F)}{μ (G)} .

Using the triangle inequality according to the splitting by $k, n, j$ and $l$ in 5.1.31 and applying Proposition 2.0.4 to each term, we obtain the estimate

\sum_{k \geq 0} \sum_{n \geq k} (2 n + 3) (4 n + 12) 2^{432 a^{3}} 2^{- (1 - \frac{1}{q}) n} (2^{2 a + 5} \frac{μ (F)}{μ (G)})^{\frac{1}{q} - \frac{1}{2}} ‖ f ‖_{2} ‖ 1_{G ∖ G^{'}} ‖_{2}

for the left hand side of 5.1.31. Since $| f | \leq 1_{F}$ , we have $‖ f ‖_{2} \leq μ (F)^{1 / 2}$ , and we have $‖ 1_{G ∖ G^{'}} ‖_{2} \leq μ (G)^{1 / 2}$ . We get a bound

2^{432 a^{3}} μ (F)^{\frac{1}{q}} μ (G)^{1 - \frac{1}{q}} 2^{2 a + 5} \sum_{k \geq 0} \sum_{n \geq k} (2 n + 3) (4 n + 12) 2^{- (1 - \frac{1}{q}) n} .

Interchanging the order of summation, the last factor equals

2^{2 a + 5} \sum_{n \geq 0} (2 n + 3) (4 n + 12) (n + 1) 2^{- \frac{q - 1}{q} n} .

Up to an explicit constant, the sum is bounded by $\sum n^{3} 2^{- \frac{q - 1}{q} n}$ , which is at most some constant times $1 / (q - 1)^{4}$ by comparing to an integral. Since $a \geq 4$ , this is overall bounded by $2^{a^{3}} / (q - 1)^{4}$ , which completes the proof of the lemma.

5.5 Proof of the Forest Complement Lemma

Define $P_{G ∖ G^{'}}$ to be the set of all $p \in P$ such that $μ (I (p) \cap (G ∖ G^{'})) > 0$ .

Lemma 5.5.1 antichain decomposition

✓

We have that

\begin{aligned} P_{2} \cap P_{G ∖ G^{'}} \\ = ⋃_{k \geq 0} ⋃_{n \geq k} L_{0} (k, n) \cap P_{G ∖ G^{'}} \\ \cup ⋃_{k \geq 0} ⋃_{n \geq k} ⋃_{0 \leq j \leq 2 n + 3} L_{2} (k, n, j) \cap P_{G ∖ G^{'}} \\ \cup ⋃_{k \geq 0} ⋃_{n \geq k} ⋃_{0 \leq j \leq 2 n + 3} ⋃_{0 \leq l \leq Z (n + 1)} L_{1} (k, n, j, l) \cap P_{G ∖ G^{'}} \\ \cup ⋃_{k \geq 0} ⋃_{n \geq k} ⋃_{0 \leq j \leq 2 n + 3} ⋃_{0 \leq l \leq Z (n + 1)} L_{3} (k, n, j, l) \cap P_{G ∖ G^{'}} . \end{aligned}

Proof ▶

Let $p \in P_{2} \cap P_{G ∖ G^{'}}$ . Clearly, for every cube $J = I (p)$ with $p \in P_{G ∖ G^{'}}$ there exists some $k \geq 0$ such that 5.1.1 holds, and for no cube $J \in D$ and no $k < 0$ does 5.1.2 hold. Thus $p \in P (k)$ for some $k \geq 0$ .

Next, since $E_{2} (λ, p^{'}) \subset I (p^{'}) \cap G$ for every $λ \geq 2$ and every tile $p^{'} \in P (k)$ with $λ p ≲ λ p^{'}$ , it follows from 5.1.2 that $μ (E_{2} (λ, p^{'})) \leq 2^{- k} μ (I (p^{'}))$ for every such $p^{'}$ , so ${dens}_{k}^{'} ({p}) \leq 2^{- k}$ . Combining this with $a \geq 0$ , it follows from 5.1.7 that there exists $n \geq k$ with $p \in C (k, n)$ .

Since $p \in P_{G ∖ G^{'}}$ , we have in particular $I (p) ⊄ A (2 n + 6, k, n)$ , so there exist at most $1 + (4 n + 12) 2^{n} < 2^{2 n + 4}$ tiles $m \in M (k, n)$ with $p \leq m$ . It follows that $p \in L_{0} (k, n)$ or $p \in C_{1} (k, n, j)$ for some $1 \leq j \leq 2 n + 3$ . In the former case we are done, in the latter case the equality to be shown follows from the definitions of the collections $C_{i}$ and $L_{i}$ .

Lemma 5.5.2 L0 antichain

✓

We have that

L_{0} (k, n) = \dot{⋃_{0 \leq l < n}} L_{0} (k, n, l),

where each $L_{0} (k, n, l)$ is an antichain.

Proof ▶

It suffices to show that $L_{0} (k, n)$ contains no chain of length $n + 1$ . Suppose that we had such a chain $p_{0} \leq p_{1} \leq \dots \leq p_{n}$ with $p_{i} \neq p_{i + 1}$ for $i = 0, \dots, n - 1$ . By 5.1.7, we have that ${dens}_{k}^{'} ({p_{n}}) > 2^{- n}$ . Thus, by 5.1.6, there exists $p^{'} \in P (k)$ and $λ \geq 2$ with $λ p_{n} \leq λ p^{'}$ and

\frac{μ (E_{2} (λ, p^{'}))}{μ (I (p^{'}))} > λ^{a} 2^{4 a} 2^{- n} .

5.5.6

Let $O$ be the set of all $p^{″} \in P (k)$ such that we have $I (p^{″}) = I (p^{'})$ and $B_{p^{'}} (Q (p^{'}), λ) \cap Ω (p^{″}) \neq \emptyset$ . We now show that

| O | \leq 2^{4 a} λ^{a} .

5.5.7

The balls $B_{p^{'}} (Q (p^{″}), 0.2)$ , $p^{″} \in O$ are disjoint by 2.0.15, and by the triangle inequality contained in $B_{p^{'}} (Q (p^{'}), λ + 1.2)$ . By assumption 1.0.11 on $Θ$ , this ball can be covered with

2^{a ⌈ \log_{2} (λ + 1.2) + \log_{2} (5) ⌉} \leq 2^{a (\log_{2} (λ) + 4)} = 2^{4 a} λ^{a}

many $d_{p^{'}}$ -balls of radius $0.2$ . Here we have used that for $λ \geq 2$

⌈ \log_{2} (λ + 1.2) + \log_{2} (5) ⌉ \leq 1 + \log_{2} (1.6 λ) + \log_{2} (5) = 4 + \log_{2} (λ) .

By the triangle inequality, each such ball contains at most one $Q (p^{″})$ , and each $Q (p^{″})$ is contained in one of the balls. Thus we get 5.5.7.

By 2.0.26 and 2.0.27 we have $E_{2} (λ, p^{'}) \subset ⋃_{p^{″} \in O} E_{1} (p^{″})$ , thus

2^{4 a} λ^{a} 2^{- n} < \sum_{p^{″} \in O} \frac{μ (E_{1} (p^{″}))}{μ (I (p^{″}))} .

Hence there exists a tile $p^{″} \in O$ with

μ (E_{1} (p ”)) \geq 2^{- n} μ (I (p^{'})) .

By the definition 5.1.5 of $M (k, n)$ , there exists a tile $m \in M (k, n)$ with $p^{'} \leq m$ . From 5.5.6, the inclusion $E_{2} (λ, p^{'}) \subset I (p^{'})$ and $a \geq 1$ we obtain

2^{n} \geq 2^{4 a} λ^{a} \geq λ .

From the triangle inequality, Lemma 2.1.2 and $a \geq 1$ , we now obtain for all $ϑ \in B_{m} (Q (m), 1)$ that

\begin{aligned} d_{p_{0}} (Q (p_{0}), ϑ) \\ \leq d_{p_{0}} (Q (p_{0}), Q (p_{n})) + d_{p_{0}} (Q (p_{n}), Q (p^{'})) + d_{p_{0}} (Q (p^{'}), Q (p ”)) \\ + d_{p_{0}} (Q (p ”), Q (m)) + d_{p_{0}} (Q (m), ϑ) \\ \leq 1 + 2^{- 95 a n} (d_{p_{n}} (Q (p_{n}), Q (p^{'})) + d_{p^{'}} (Q (p^{'}), Q (p ”)) \\ + d_{p^{″}} (Q (p ”), Q (m)) + d_{m} (Q (m), ϑ)) \\ \leq 1 + 2^{- 95 a n} (λ + (λ + 1) + 1 + 1) \leq 100 . \end{aligned}

Thus, by 2.0.23, $100 p_{0} ≲ m$ , a contradiction to $p_{0} \notin C (k, n)$ .

Lemma 5.5.3 L2 antichain

✓

Each of the sets $L_{2} (k, n, j)$ is an antichain.

Proof ▶

Suppose that there are $p_{0}, p_{1} \in L_{2} (k, n, j)$ with $p_{0} \neq p_{1}$ and $p_{0} \leq p_{1}$ . By Lemma 5.3.1 and Lemma 5.3.2, it follows that $2 p_{0} ≲ 200 p_{1}$ . Since $L_{2} (k, n, j)$ is finite, there exists a maximal $l \geq 1$ such that there exists a chain $2 p_{0} ≲ 200 p_{1} ≲ \dots ≲ 200 p_{l}$ with all $p_{i}$ in $C_{1} (k, n, j)$ and $p_{i} \neq p_{i + 1}$ for $i = 0, \dots, l - 1$ . If we have $p_{l} \in U_{1} (k, n, j)$ , then it follows from $2 p_{0} ≲ 200 p_{l} ≲ p_{l}$ and 5.1.15 that $p_{0} \notin L_{2} (k, n, j)$ , a contradiction. Thus, by the definition 5.1.14 of $U_{1} (k, n, j)$ , there exists $p_{l + 1} \in C_{1} (k, n, j)$ with $I (p_{l}) ⊊ I (p_{l + 1})$ and $ϑ \in B_{p_{l}} (Q (p_{l}), 100) \cap B_{p_{l + 1}} (Q (p_{l + 1}), 100)$ . Using the triangle inequality and Lemma 2.1.2, one deduces that $200 p_{l} ≲ 200 p_{l + 1}$ . This contradicts maximality of $l$ .

Lemma 5.5.4 L1 L3 antichain

✓

Each of the sets $L_{1} (k, n, j, l)$ and $L_{3} (k, n, j, l)$ is an antichain.

Proof ▶

By its definition 5.1.11, each set $L_{1} (k, n, j, l)$ is a set of minimal elements in some set of tiles with respect to $\leq$ . If there were distinct $p, q \in L_{1} (k, n, j, l)$ with $p \leq q$ , then $q$ would not be minimal. Hence such $p, q$ do not exist. Similarly, by 5.1.17, each set $L_{3} (k, n, j, l)$ is a set of maximal elements in some set of tiles with respect to $\leq$ . If there were distinct $p, q \in L_{3} (k, n, j, l)$ with $p \leq q$ , then $p$ would not be maximal.

We now turn to the proof of Lemma 5.1.3.

Proof of Lemma 5.1.3 ▶

If $p \notin P_{G ∖ G^{'}}$ , then $μ (I (p) \cap (G ∖ G^{'})) = 0$ . By 2.0.21 and 2.0.26, it follows that $1_{G ∖ G^{'}} T_{p} f (x) = 0$ for almost every $x$ . We thus have, almost everywhere,

1_{G ∖ G^{'}} \sum_{p \in P_{2}} T_{p} f (x) = 1_{G ∖ G^{'}} \sum_{p \in P_{2} \cap P_{G ∖ G^{'}}} T_{p} f (x) .

Let $L (k, n)$ denote any of the terms $L_{i} (k, n, j, l) \cap P_{G ∖ G^{'}}$ on the right hand side of 5.5.1, where the indices $j, l$ may be void. Then $L (k, n)$ is an antichain, by Lemmas 5.5.2,5.5.3, 5.5.4. Further, we have

{dens}_{1} (L (k, n)) \leq 2^{4 a + 1 - n}

by Lemma 5.3.12, and we have

{dens}_{2} (L (k, n)) \leq 2^{2 a + 5} \frac{μ (F)}{μ (G)},

since

L (k, n) \cap P_{F, G} \subset P_{G ∖ G^{'}} \cap P_{F, G} = \emptyset .

Applying now the triangle inequality according to the decomposition in Lemma 5.5.1, and then applying Proposition 2.0.3 to each term, we obtain the estimate

\begin{matrix} \leq \sum_{k \geq 0} \sum_{n \geq k} (n + (2 n + 4) + 2 (2 n + 4) (1 + Z (n + 1))) \\ \times 2^{150 a^{3}} (q - 1)^{- 1} (2^{4 a + 1 - n})^{\frac{q - 1}{8 a^{4}}} (2^{2 a + 5} \frac{μ (F)}{μ (G)})^{\frac{1}{q} - \frac{1}{2}} ‖ f ‖_{2} ‖ 1_{G ∖ G^{'}} ‖_{2} . \end{matrix}

Because $| f | \leq 1_{F}$ , we have $‖ f ‖_{2} \leq μ (F)^{1 / 2}$ , and we have $‖ 1_{G ∖ G^{'}} ‖_{2} \leq μ (G)^{1 / 2}$ . Using this and 2.0.3, we bound

\leq 2^{151 a^{3}} (q - 1)^{- 1} μ (F)^{\frac{1}{q}} μ (G)^{\frac{1}{q^{'}}} \sum_{k \geq 0} \sum_{n \geq k} n^{2} 2^{- n \frac{q - 1}{8 a^{4}}} .

The last sum equals, by changing the order of summation,

\sum_{n \geq 0} n^{2} (n + 1) 2^{- n \frac{q - 1}{8 a^{4}}} \leq \frac{2^{2 a^{3}}}{(q - 1)^{4}} .

This completes the proof. □