Let \({\mathfrak p},{\mathfrak p}'\) and \(x\) be given. Assume without loss of generality that \({\mathrm{s}}({\mathfrak p})\le {\mathrm{s}}({\mathfrak p}')\). As we have \(x\in E({\mathfrak p})\subset {\mathcal{I}}({\mathfrak p})\) and \(x\in E({\mathfrak p}')\subset {\mathcal{I}}({\mathfrak p}')\) by Definition 2.0.20, we conclude for \(i=1,2\) that \({Q}(x)\in {\Omega }({\mathfrak p})\) and \({Q}(x)\in {\Omega }({\mathfrak p}')\). By 2.0.14 we have \({\Omega }({\mathfrak p}')\subset {\Omega }({\mathfrak p})\). By Definition 2.0.23, we conclude \({\mathfrak p}\le {\mathfrak p}'\). As \(\mathfrak {A}\) is an antichain, we conclude \({\mathfrak p}={\mathfrak p}'\). 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 \(K_{{\mathrm{s}}({\mathfrak p})}(x,y)\neq 0\). 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

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

This together with \(a\ge 1\) proves the Lemma.

We have \(f=\mathbf{1}_Ff\). Using Hölder’s inequality, we obtain for each \(x\in B'\) and each \(B'\in \mathcal{B}\) using \(1{\lt}\tilde{q}\le 2\)

Taking the maximum over all \(B'\) containing \(x\), we obtain

We have with Proposition 2.0.6

Using \(1{\lt}\tilde{q}\le 2\) estimates the last display by

We obtain with Cauchy-Schwarz and then Lemma 6.1.2

With 6.1.14 and 6.1.16 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 6.1.21

with the adjoint operator

We have by expanding the square

We split the sum into the terms with \({\mathrm{s}}({\mathfrak p}')\le {\mathrm{s}}({\mathfrak p})\) and \({\mathrm{s}}({\mathfrak p}){\lt} {\mathrm{s}}({\mathfrak p}')\). Using the symmetry of each summand, we may switch \({\mathfrak p}\) and \({\mathfrak p}'\) in the second sum. Using further positivity of each summand to replace the condition \({\mathrm{s}}({\mathfrak p}'){\lt} {\mathrm{s}}({\mathfrak p})\) by \({\mathrm{s}}({\mathfrak p}')\le {\mathrm{s}}({\mathfrak p})\) in the second sum, we estimate 6.1.26 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 6.1.30, we estimate 6.1.27 by

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

Note that \(p\geq 4\) since \(a{\gt}4\). We estimate \(h({\mathfrak p})\) as defined in 6.1.32 with Hölder using \(|g|\le \mathbf{1}_G\) and \(E({\mathfrak p}')\subset B({\mathfrak p})\) by

Then we apply Lemma 6.1.6 to estimate this by

Let \(\mathcal{B}'\) 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 \(M_{\mathcal{B}',p'}\)

Hence we can estimate 6.1.34 by

With this estimate of \(h({\mathfrak p})\), using \(E({\mathfrak p})\subset B({\mathfrak p})\) by construction of \(B({\mathfrak p})\), we estimate 6.1.31 by

Using Lemma 6.1.1, the last display is observed to be

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

Using \(p{\gt}4\) and thus \(1{\lt}p'{\lt}\frac43\), we estimate the last display by

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

If \(\varphi (y)\) 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 \(y'\neq y\) additionally with 2.1.4

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

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

We have \({\mathcal{I}}({\mathfrak p}_1)\subset {\mathcal{I}}({\mathfrak p}_2)\) by 2.0.8. Using Lemma 2.1.2 it follows that

By the triangle inequality, we obtain from 6.2.11 and 6.2.12

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 6.2.13 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 6.2.10 follows with \(a\ge 1\).

Fix \({\mathfrak p}\) and \(y\) with 6.2.18. 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 \(K_{{\mathrm{s}}({\mathfrak p})}(x,y)\neq 0\), we have by 2.1.2 that

Now 6.2.19 follows by the triangle inequality.

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

Hence by the triangle inequality

Together with 6.3.7 and monotonicity 1.0.9 of \(d\)

Using the doubling property 1.0.10 \(5a+2\) times gives

Using \({\mathrm{s}}({\mathfrak p}){\lt}{\mathrm{s}}({\mathfrak p}')\) and \(D=2^{100a^2}\) and \(a\ge 4\) gives

With the triangle inequality and 6.3.5,

This shows

This implies 6.3.3 and completes the proof of the lemma.

Let \({\vartheta },N,L\) be given and set

Let \({\mathfrak p}\in \mathfrak {A}'\). We have by Definition 2.0.28 using \(\lambda =2\) and the squeezing property 2.0.15

By the covering property 1.0.11, applied \(N+4\) times, there is a collection \({\Theta }'\) of at most \(2^{a(N+4)}\) elements such that

As each \({\mathcal{Q}}({\mathfrak p})\) with \({\mathfrak p}\in \mathfrak {A}_{{\vartheta },N}\) is contained in the left-hand-side of 6.3.19 by definition, it is in at least one \(B_{{\mathfrak p}}({\vartheta }', 0.2)\) with \({\vartheta }'\in {\Theta }'\).

For two different \({\mathfrak p},{\mathfrak p}'\in \mathfrak {A}'\), we have by 2.0.13 that \({\Omega }({\mathfrak p})\) and \({\Omega }({\mathfrak p}')\) are disjoint and thus by the squeezing property 2.0.15 we have for every \({\vartheta }'\in {\Theta }'\)

Hence at most one of \({\mathcal{Q}}({\mathfrak p})\) and \({\mathcal{Q}}({\mathfrak p})\) is in \(B_{{\mathfrak p}}({\vartheta }', 0.2)\). It follows that there are at most \(2^{a(N+4)}\) elements in \(\mathfrak {A}'\). Adding 6.3.18 over \(\mathfrak {A}'\) proves 6.3.16.

Let \({\mathfrak p}\) be any tile in \(\mathfrak {A}_{{\vartheta },N}\) with \({\mathrm{s}}({\mathfrak p}_{{\vartheta }}){\lt}{\mathrm{s}}({\mathfrak p})\). By definition of \(E\), the tile contributes zero to the sum on the left-hand side of 6.3.21 unless \({\mathcal{I}}({\mathfrak p})\cap {\mathcal{I}}({\mathfrak p}_{{\vartheta }}) \neq \emptyset \), which we may assume. With \({\mathrm{s}}({\mathfrak p}_{{\vartheta }}){\lt}{\mathrm{s}}({\mathfrak p})\) and the dyadic property 2.0.8 we conclude \({\mathcal{I}}({\mathfrak p}_{{\vartheta }})\subset {\mathcal{I}}({\mathfrak p})\). By the squeezing property 2.0.15, we conclude from \({\vartheta }\in {\Omega }({\mathfrak p}_{{\vartheta }})\) that

We conclude from \({\mathfrak p}\in \mathfrak {A}_{{\vartheta },N}\) that

With Lemma 6.3.1, we conclude

By Definition 2.0.27 of \(E_2\), we conclude

Using disjointedness of the various \(E({\mathfrak p})\) with \({\mathfrak p}\in \mathfrak {A}\) by Lemma 6.1.1, we obtain 6.3.21. This proves the lemma.

Fix \({\vartheta }\) and \(N\). Let \(\mathfrak {A}'\) be the set of \({\mathfrak p}\in \mathfrak {A}_{{\vartheta },N}\) such that \({\mathcal{I}}({\mathfrak p})\cap G\) is not empty.

Let \(\mathcal{L}\) be the collection of dyadic cubes \(I\in \mathcal{D}\) such that \(I\subset {\mathcal{I}}({\mathfrak p})\) for some \({\mathfrak p}\in \mathfrak {A}'\) and if \({\mathcal{I}}({\mathfrak p})\subset I\) for some \({\mathfrak p}\in \mathfrak {A}'\), then \({\mathrm{s}}({\mathfrak p})=-S\). By 2.0.7, for each \({\mathfrak p}\in \mathfrak {A}'\) and each \(x\in {\mathcal{I}}({\mathfrak p})\cap G\), there is \(I\in \mathcal{D}\) with \(s(I)=-S\) and \(x\in I\). By 2.0.8, we have \(I\subset {\mathcal{I}}({\mathfrak p})\). Hence

As each \(I\in \mathcal{L}\) satisfies \(I\subset {\mathcal{I}}({\mathfrak p})\) for some \({\mathfrak p}\) in \(\mathfrak {A’}\), we conclude

Let \(\mathcal{L}^*\) be the set of maximal elements in \(\mathcal{L}\) with respect to set inclusion. By 2.0.8, the elements in \(\mathcal{L}^*\) are pairwise disjoint and we have

Using the partition 6.3.29 into elements of \(\mathcal{L}\) in 6.3.30, it suffices to show for each \(L\in \mathcal{L}^*\)

Fix \(L\in \mathcal{L}^*\). By definition of \(L\), there exists an element \({\mathfrak p}'\in \mathfrak {A}'\) such that \(L\subset {\mathcal{I}}({\mathfrak p}')\). Pick such an element \({\mathfrak p}'\) in \(\mathfrak {A}\) with minimal \({\mathrm{s}}({\mathfrak p}')\). As \({\mathcal{I}}({\mathfrak p}')\not\subset L\) or \(s(L) = -S\) by definition of \(L\), we have with 2.0.8 that \(s(L){\lt} {\mathrm{s}}({\mathfrak p}')\) or \(s(L) = -S\). In particular \(s(L){\lt}S\), thus \(L \ne I_0\) and hence by 2.0.9 there exists a cube \(J \in \mathcal{D}\) with \(L \subsetneq J\). By 2.0.7, there is an \(L'\in \mathcal{D}\) with \(s(L')=s(L)+1\) and \(c(L)\in L'\). By 2.0.8, we have \(L\subset L'\).

We split the left-hand side of 6.3.30 as

We first estimate 6.3.31 with Lemma 6.3.2 by

We turn to 6.3.32. Consider the element \({\mathfrak p}'\in \mathfrak {A}'\) as above with \(L\subset {\mathcal{I}}({\mathfrak p}')\) and \(s(L){\lt}{\mathrm{s}}({\mathfrak p}')\). As \(L\subset L'\) and \(s(L')=s(L)+1\), we conclude with the dyadic property that \(L'\subset {\mathcal{I}}({\mathfrak p}')\). By maximality of \(L\), we have \(L'\not\in \mathcal{L}\). This together with the existence of the given \({\mathfrak p}'\in \mathfrak {A}\) with \(L'\subset {\mathcal{I}}({\mathfrak p}')\) shows by definition of \(\mathcal{L}\) that there exists \({\mathfrak p}''\in \mathfrak {A}'\) with \({\mathcal{I}}({\mathfrak p}'')\subset L'\).

By the covering property 2.0.13, there exists a unique \({\mathfrak p}_{{\vartheta }}\) with

such that \({\vartheta }\in {\Omega }({\mathfrak p}_{{\vartheta }})\). Note that

and as \({\mathfrak p}'' \in \mathfrak {A}_{{\vartheta },N}\) that

By Lemma 6.3.1, we conclude

As \({\mathfrak p}''\in \mathfrak {A}'\), we have by Definition 2.0.28 of \(\operatorname{\operatorname {dens}}_1\) that

Now let \({\mathfrak p}\) be any tile in the summation set in 6.3.32, that is, \({\mathfrak p}\in \mathfrak {A}'\) and \({\mathcal{I}}({\mathfrak p})\neq L'\). Then \({\mathcal{I}}({\mathfrak p})\cap L\neq \emptyset \). It follows by the dyadic property 2.0.8 and the definition of \(L\) that \(L\subset {\mathcal{I}}({\mathfrak p})\) and \(L\neq {\mathcal{I}}({\mathfrak p})\). By the dyadic property 2.0.8, we have \(s(L){\lt}{\mathrm{s}}({\mathfrak p})\) and thus \(s(L')\le {\mathrm{s}}({\mathfrak p})\). By the dyadic property 2.0.8 again, we have \(L'\subset {\mathcal{I}}({\mathfrak p})\). As \(L'\neq {\mathcal{I}}({\mathfrak p})\), we conclude \(s(L){\lt}{\mathrm{s}}({\mathfrak p})\). By Lemma 6.3.3, we can thus estimate 6.3.32 by

Using the decomposition into 6.3.31 and 6.3.32 and the estimates 6.3.33, 6.3.21, 6.3.37 we obtain the estimate

Using \(s(L')=s(L)+1\) and \(D=2^{100a^2}\) and the squeezing property 2.0.10 and the doubling property 1.0.5 \(100a^2+4\) times , we obtain

Inserting in 6.3.39 and using \(a\ge 4\) gives 6.3.30. This completes the proof of the lemma.

Using that \(\mathfrak {A}\) is the union of the \(\mathfrak {A}_{{\vartheta },N}\) with \(N\ge 0\), we estimate the left-hand side 6.1.45 with the triangle inequality by

We consider each individual term in this sum and estimate it’s \(p\)-th power. Using that for each \(x\in X\) by Lemma 6.3.4 there is at most one \({\mathfrak p}\in \mathfrak {A}\) with \(x\in E({\mathfrak p})\), we have

Using Lemma 6.3.4, we estimate the last display by

Using that with \(a\ge 4\) and since \(p = 4a^4\), we have

Hence we have for 6.3.46 the upper bound

Taking th \(p\)-th root and summing over \(N\ge 0\) gives for 6.3.41 the upper bound

Using that \(p = 4a^4\) and \(a \ge 4\), this proves the lemma.