Let $\mathfrak{p},{\mathfrak{p}}^{\prime }$ and $x$ be given. Assume without loss of generality that $\mathrm{s}(\mathfrak{p})\le \mathrm{s}({\mathfrak{p}}^{\prime })$. As we have $x\in E(\mathfrak{p})\subset \mathcal{I}(\mathfrak{p})$ and $x\in E({\mathfrak{p}}^{\prime })\subset \mathcal{I}({\mathfrak{p}}^{\prime })$ by Definition 2.0.20, we conclude for $i=1,2$ that $Q(x)\in \mathrm{\Omega }(\mathfrak{p})$ and $Q(x)\in \mathrm{\Omega }({\mathfrak{p}}^{\prime })$. By 2.0.14 we have $\mathrm{\Omega }({\mathfrak{p}}^{\prime })\subset \mathrm{\Omega }(\mathfrak{p})$. By Definition 2.0.23, we conclude $\mathfrak{p}\le {\mathfrak{p}}^{\prime }$. As $\mathfrak{A}$ is an antichain, we conclude $\mathfrak{p}={\mathfrak{p}}^{\prime }$. This proves the lemma.

Fix $x\in X$. By Lemma 6.1.1, there is at most one $\mathfrak{p}\in \mathfrak{A}$ such that $T_{\mathfrak{p}}f(x)$ is not zero. If there is no such $\mathfrak{p}$, the estimate 6.1.2 follows.

Assume there is such a $\mathfrak{p}$. By definition of $T_{\mathfrak{p}}$ we have $x\in E(\mathfrak{p})\subset \mathcal{I}(\mathfrak{p})$ and by the squeezing property 2.0.10

Let $y\in X$ with $K_{\mathrm{s}(\mathfrak{p})}(x,y)\ne 0$. By Definition 2.0.5 of $K_{\mathrm{s}(\mathfrak{p})}$ we have

The triangle inequality with 6.1.3 and 6.1.4 implies

Using the kernel bound 1.0.14 and the lower bound in 2.1.2 we obtain

Using $D=2^{100a^2}$ and the doubling property 1.0.5 $5+100a^2$ times estimates the last display by

which, thanks to the closeness of the points $x$ and $\mathrm{c}(\mathfrak{p})$ shown in 6.1.3, is in turn bounded by

Using that $|e(\vartheta )|$ is bounded by $1$ for every $\vartheta \in \mathrm{\Theta }$, we estimate with the triangle inequality and the above information

This together with $a\ge 1$ proves the Lemma.

We have $f={\mathbf{1}}_{F}f$. Using Hölder’s inequality, we obtain for each $x\in B^{\prime }$ and each $B^{\prime }\in \mathcal{B}$ using $1<\tilde{q}\le 2$

Taking the maximum over all $B^{\prime }$ containing $x$, we obtain

We have with Proposition 2.0.6

Using $1<\tilde{q}\le 2$ estimates the last display by

We obtain with Cauchy-Schwarz and then Lemma 6.1.2

With 6.1.15 and 6.1.17 we can estimate the last display by

Using $a\ge 4$ and $(\tilde{q}-1{)}^{-1}=(q+1)/(q-1)\le 3(q-1{)}^{-1}$ proves the lemma.

We write for the expression inside the absolute values on the left-hand side of 6.1.22

with the adjoint operator

We have by expanding the square

We split the sum into the terms with $\mathrm{s}({\mathfrak{p}}^{\prime })\le \mathrm{s}(\mathfrak{p})$ and $\mathrm{s}(\mathfrak{p})<\mathrm{s}({\mathfrak{p}}^{\prime })$. Using the symmetry of each summand, we may switch $\mathfrak{p}$ and ${\mathfrak{p}}^{\prime }$ in the second sum. Using further positivity of each summand to replace the condition $\mathrm{s}({\mathfrak{p}}^{\prime })<\mathrm{s}(\mathfrak{p})$ by $\mathrm{s}({\mathfrak{p}}^{\prime })\le \mathrm{s}(\mathfrak{p})$ in the second sum, we estimate 6.1.27 by

Define for $\mathfrak{p}\in \mathfrak{P}$

and define

Note that by the squeezing property 2.0.10 and the doubling property 1.0.5 applied $6$ times we have

Using Lemma 6.1.5 and 6.1.31, we estimate 6.1.28 by

with $h(\mathfrak{p})$ defined as

Note that $p\ge 4$ since $a>4$. We estimate $h(\mathfrak{p})$ as defined in 6.1.33 with Hölder using $|g|\le {\mathbf{1}}_G$ and $E({\mathfrak{p}}^{\prime })\subset B(\mathfrak{p})$ by

Then we apply Lemma 6.1.6 to estimate this by

Let ${\mathcal{B}}^{\prime }$ be the collection of all balls $B(\mathfrak{p})$ with $\mathfrak{p}\in \mathfrak{A}$. Then for each $\mathfrak{p}\in \mathfrak{A}$ and $x\in B(\mathfrak{p})$ we have by definition 2.0.41 of $M_{{\mathcal{B}}^{\prime },p^{\prime }}$

Hence we can estimate 6.1.35 by

With this estimate of $h(\mathfrak{p})$, using $E(\mathfrak{p})\subset B(\mathfrak{p})$ by construction of $B(\mathfrak{p})$, we estimate 6.1.32 by

Using Lemma 6.1.1, the last display is observed to be

Applying Cauchy-Schwarz and using Proposition 2.0.6 estimates the last display by

Using $p>4$ and thus $1<p^{\prime }<\frac{4}{3}$, we estimate the last display by

Now Lemma 6.1.4 follows by applying Cauchy-Schwarz on the left-hand side and using $a\ge 4$.

We have

Multiplying the $(2-q)$-th power of 6.1.11 and the $(q-1)$-th power of 6.1.22 and estimating gives after simplification of some factors

With the definition of $p$, this implies Proposition 2.0.3.

If $\phi (y)$ is not zero, then $K_{s_1}(x_1,y)$ is not zero and thus 2.1.2 gives 6.2.2.

We next have for $y$ with 2.1.3

and for $y^{\prime }\ne y$ additionally with 2.1.4

Adding the estimates 6.2.4 and 6.2.9 gives 6.2.3. This proves the lemma.

Fix $\mathfrak{p}$ and $y$ with 6.2.10. Then there exists $x\in E(\mathfrak{p})$ with

As $E(\mathfrak{p})\subset \mathcal{I}(\mathfrak{p})$ and by the squeezing property 2.0.10, we have

As $K_{\mathrm{s}(\mathfrak{p})}(x,y)\ne 0$, we have by 2.1.2 that

Now 6.2.11 follows by the triangle inequality.

Let $i\in \{1,2\}$. By Definition 2.0.20 of $E$, we have $Q(x_i)\in \mathrm{\Omega }({\mathfrak{p}}_i)$ With 2.0.15 we then conclude

We have by the triangle inequality and 2.0.10 that $\mathcal{I}({\mathfrak{p}}_1)\subset B(\mathrm{c}({\mathfrak{p}}_2),14D^{\mathrm{s}({\mathfrak{p}}_2)})$. Thus, using again 2.0.10 and the doubling property 1.0.8

By the triangle inequality, we obtain from 6.2.16 and 6.2.17

As $x_1\in \mathcal{I}({\mathfrak{p}}_1)$ by Definition 2.0.20 of $E$, we have by the squeezing property 2.0.10

and thus by 2.0.10 again and the triangle inequality

We thus estimate the right-hand side of 6.2.18 with monotonicity 1.0.9 of the metrics $d_B$ by

This is further estimated by applying the doubling property 1.0.8 three times by

Now 6.2.15 follows with $a\ge 1$.

We begin with 6.1.43. By Lemma 6.2.2, the left-hand side of 6.1.43 vanishes if $B(\mathrm{c}({\mathfrak{p}}_1),5D^{\mathrm{s}({\mathfrak{p}}_1)})\cap B(\mathrm{c}({\mathfrak{p}}_2),5D^{\mathrm{s}({\mathfrak{p}}_2)})=\mathrm{\varnothing }$. Thus we can assume for the remainder of the proof that

We expand the left-hand side of 6.1.43 as

By Fubini and the triangle inequality and the fact $|e(Q(x_i)(x_i))|=1$ for $i=1,2$, we can estimate 6.2.24 from above by

with

We estimate for fixed $x_1\in E({\mathfrak{p}}_1)$ and $x_2\in E({\mathfrak{p}}_2)$ the inner integral of 6.2.26 with Proposition 2.0.5. The function $\phi :={\phi }_{x_1,x_2}$ satisfies the assumptions of Proposition 2.0.5 with $z=x_1$ and $R=D^{s_1}$ by Lemma 6.2.1. We obtain with $B^{\prime }:=B(x_1,D^{\mathrm{s}({\mathfrak{p}}_1)})$,

Using 6.2.23, Lemma 6.2.3 and $a\ge 1$ estimates 6.2.28 by

As $x_2\in \mathcal{I}({\mathfrak{p}}_2)$ by Definition 2.0.20 of $E$, we have by 2.0.10

and thus by 2.0.10 again and the triangle inequality

Using three iterations of the doubling property 1.0.5 give

With $a\ge 1$ and 6.2.29 we conclude 6.1.43.

Now assume the left-hand side of 6.1.43 is not zero. There is a $y\in X$ with

By the triangle inequality and Lemma 6.2.2, we conclude

By the squeezing property 2.0.10 and the triangle inequality, we conclude

This completes the proof of Lemma 6.1.5.

By Lemma 2.1.2, we have

Together with 6.3.1 and the triangle inequality, we obtain

Now assume

By the doubling property 1.0.8, applied five times, we have

We have by the squeezing property 2.0.10