Let \(s \in \mathbb {Z}\) with \(\underline{\sigma } ({\mathfrak u}, x) \le s \le \overline{\sigma } ({\mathfrak u}, x)\). By definition of \(\sigma \), there exists \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \({\mathrm{s}}({\mathfrak p}) = \underline{\sigma } ({\mathfrak u}, x)\) and \(x \in E({\mathfrak p})\), and there exists \({\mathfrak p}'' \in {\mathfrak T}({\mathfrak u})\) with \({\mathrm{s}}({\mathfrak p}'') = \overline{\sigma } ({\mathfrak u}, x)\) and \(x \in E({\mathfrak p}'') \subset {\mathcal{I}}({\mathfrak p}'')\). By property 2.0.7 of the dyadic grid, there exists a cube \(I \in \mathcal{D}\) of scale \(s\) with \(x \in I\). By property 2.0.13, there exists a tile \({\mathfrak p}' \in {\mathfrak P}(I)\) with \({Q}(x) \in {\Omega }({\mathfrak p}')\). By the dyadic property 2.0.8 we have \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p}'')\), and by 2.0.14, we have \({\Omega }({\mathfrak p}'') \subset {\Omega }({\mathfrak p}') \subset {\Omega }({\mathfrak p})\). Thus \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\), which gives with the convexity property 2.0.33 of \({\mathfrak T}({\mathfrak u})\) that \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u})\), so \(s \in \sigma ({\mathfrak u}, x)\).

Since \(\mathcal{J}(\mathfrak {S})\) is the set of inclusion maximal cubes in \(\mathcal{J}_0(\mathfrak {S})\), cubes in \(\mathcal{J}(\mathfrak {S})\) are pairwise disjoint by 2.0.8. The same applies to \(\mathcal{L}(\mathfrak {S})\).

If \(x \in \bigcup _{I \in \mathcal{D}} I\), then there exists by 2.0.7 a cube \(I \in \mathcal{D}\) with \(x \in I\) and \(s(I) = -S\). Then \(I \in \mathcal{J}_0(\mathfrak {S})\). There exists an inclusion maximal cube in \(\mathcal{J}_0(\mathfrak {S})\) containing \(I\). This cube contains \(x\) and is contained in \(\mathcal{J}(\mathfrak {S})\). This shows one inclusion in 7.1.1, the other one follows from \(\mathcal{J}(\mathfrak {S}) \subset \mathcal{D}\).

The proof of the two inclusions in 7.1.2 is similar.

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

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

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.

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

Let \({\mathfrak p}_s \in {\mathfrak T}({\mathfrak u})\) be a tile with \({\mathrm{s}}({\mathfrak p}_s) = s\) and \(x \in E({\mathfrak p}_s)\), and let \({\mathfrak p}'\) be a tile with \({\mathrm{s}}({\mathfrak p}') = \overline{\sigma }({\mathfrak u}, x)\) and \(x \in E({\mathfrak p}')\). Using the doubling property 1.0.8, the definition of \(d_{{\mathfrak p}}\) and Lemma 2.1.2, we can bound the previous display by

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

Using 2.1.3, it follows that

By 7.1.1, the collection \(\mathcal{J}\) is a partition of \(X\), so this is estimated by

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

Let \(J \in \mathcal{J}({\mathfrak T}({\mathfrak u}))\) with \(B(x, 0.5 D^s) \cap J \ne \emptyset \). By the triangle inequality and since \(x \in E({\mathfrak p}_s) \subset B({\mathrm{c}}({\mathfrak p}_s), 4D^{s})\), it follows that \(B({\mathrm{c}}({\mathfrak p}_s), 4.5D^s) \cap J \ne \emptyset \). If \(s(J) \ge s\) and \(s(J) {\gt} -S\), then it follows from the triangle inequality, 2.0.10 and 2.0.1 that \({\mathcal{I}}({\mathfrak p}_s) \subset B(c(J), 100 D^{s(J)+1})\), contradicting \(J \in \mathcal{J}(\mathfrak {T}({\mathfrak u}))\). Thus \(s(J) \le s - 1\) or \(s(J) = -S\). If \(s(J) = -S\) and \(s(J) {\gt} s - 1\), then \(s = -S\). Thus we always have \(s(J) \le s\). It then follows from the triangle inequality and 2.0.10 that \(J \subset B({\mathrm{c}}({\mathfrak p}_s), 16 D^s)\).

Thus we can continue our chain of estimates with

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

Since \(a \ge 4\), it follows that 7.1.6 is bounded by

Since \(L \in \mathcal{L}({\mathfrak T}({\mathfrak u}))\), we have \(s(L) \le {\mathrm{s}}({\mathfrak p})\) for all \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\). Since \(x\in L \cap {\mathcal{I}}({\mathfrak p}_s)\), it follows by 2.0.8 that \(L \subset {\mathcal{I}}({\mathfrak p}_s)\), in particular \(x' \in {\mathcal{I}}({\mathfrak p}_s) \subset B({\mathrm{c}}({\mathfrak p}_s), 16D^s)\). Thus

This completes the estimate for term 7.1.6.

Let \(s_1 = \underline{\sigma }({\mathfrak u}, x)\). By definition, there exists a tile \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \({\mathrm{s}}({\mathfrak p}) = s_1\) and \(x \in E({\mathfrak p})\). Then \(x \in {\mathcal{I}}({\mathfrak p}) \cap L\). By 2.0.8 and the definition of \(\mathcal{L}({\mathfrak T}({\mathfrak u}))\), it follows that \(L \subset {\mathcal{I}}({\mathfrak p})\), in particular \(x' \in {\mathcal{I}}({\mathfrak p})\), so \(x \in I_{s_1}(x')\). Next, let \(s_2 = \overline{\sigma }({\mathfrak u}, x)\) and let \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u})\) with \({\mathrm{s}}({\mathfrak p}') = s_2\) and \(x \in E({\mathfrak p}')\). Since \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u})\), we have \(4{\mathfrak p}' \lesssim {\mathfrak u}\). Since \({Q}(x) \in {\Omega }({\mathfrak p}')\), it follows that

Applying the doubling property 1.0.8 five times, we obtain

By the triangle inequality, we have \(B(x, D^{s_2}) \subset B(c({\mathfrak p}), 8 D^{s_2})\), so by 1.0.9

Finally, by applying 1.0.10 \(100a\) times, we obtain

Consequently, \(D^{s_2 - 1} {\lt} R_Q({\mathcal{Q}}({\mathfrak u}), x)\). The lemma now follows from the definition of \(T_{\mathcal{N}}\).

We have for \(J \in \mathcal{J}({\mathfrak T}({\mathfrak u}))\):

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

Suppose that \(s \in \sigma ({\mathfrak u}, x)\). If \(K_s(x,y) \ne 0\) for some \(y \in J \in \mathcal{J}({\mathfrak T}({\mathfrak u}))\) then, by 2.1.2, \(y \in B(x, 0.5 D^s) \cap J \ne \emptyset \). Let \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \({\mathrm{s}}({\mathfrak p}) = s\) and \(x \in E({\mathfrak p})\). Then \(B({\mathrm{c}}({\mathfrak p}_s), 4.5D^s) \cap J \ne \emptyset \) by the triangle inequality. If \(s(J) \ge s\) and \(s(J) {\gt} -S\), then it follows from the triangle inequality, 2.0.10 and 2.0.1 that \({\mathcal{I}}({\mathfrak p}) \subset B(c(J), 100 D^{s(J)+1})\), contradicting \(J \in \mathcal{J}(\mathfrak {T}({\mathfrak u}))\). Thus \(s(J) \le s - 1\) or \(s(J) = -S\). If \(s(J) = -S\) and \(s(J) {\gt} s - 1\), then \(s = -S\). So in both cases, \(s(J) \le s\). 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

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

Since \(a \ge 4\) we can continue our estimate with

This completes the proof.

Let \(L \in \mathcal{L}({\mathfrak T}({\mathfrak u}))\). We apply Lemma 7.1.3 to \(e({\mathcal{Q}}({\mathfrak u})) f\) to obtain for all \(y, x' \in L\)

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

Integrating this estimate yields

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

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

which by the above computation is bounded by

Applying Cauchy-Schwarz and Minkowski’s inequality, this is bounded by

By Proposition 2.0.6, Lemma 7.2.2 and Lemma 7.2.3, the second factor is at most

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

This completes the proof since \(a \ge 4\).

Fix \(s_1, s_2\). By 2.0.4 we have for all \(x \in (0, \infty )\)

Since \(\psi \) is supported in \([\frac{1}{4D}, \frac{1}{2}]\), 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

Since \(\psi \) is supported in \([\frac{1}{4D}, \frac{1}{2}]\), we further have

Finally, since \(\psi \ge 0\) and \(\sum _{s \in \mathbb {Z}} \psi (D^{-s}x) = 1\), we have for all \(x\)

Let \(x' \in I_{s_1}(x)\) and suppose that \(D^{s_2 - 1} \le R_Q({\vartheta }, x')\). By the triangle inequality and 2.0.10, it holds that \(\rho (x,x') \le 8D^{s_1}\). We have

The first term 7.2.3 is at most \(2T_{{Q}}^{\vartheta }f(x)\) , using with \(R_1:=8D^{s_1}\) and \(R_2:=\frac{1}{4}D^{s_2-1}\) and

the triangle inequality in the form

and making the second term larger by replacing the constraint \(\rho (x,x'){\lt}R_1\) by the constraint \(\rho (x,x'){\lt}R_2\).

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 \(\rho (x', y) \ge \frac{1}{4}D^{s_1 - 1}\), we have

Using \(D=2^{100a^2}\) and the doubling property 1.0.5 \(6 +100a^2\) times estimates the last display by

By the triangle inequality and 2.0.10, we have

Combining this with 7.2.10, we conclude that 7.2.4 is at most

For 7.2.5 we argue similarly. We have for all \(y\) with \(\rho (x', y) \ge \frac{1}{4}D^{s_2}\)

Using the doubling property 1.0.5 \(6 + 100a^2\) times estimates the last display by

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

Combining this, 7.2.5 is at most

Using \(a \ge 4\), taking a supremum over all \(x' \in I_{s_1}(x)\) and then a supremum over all \(-S \le s_1 {\lt} s_2 \le S\), we obtain

The lemma now follows from assumption 1.0.18, Proposition 2.0.6 and \(a \ge 4\).

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

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 \(\mathcal{C}\) is any finite collection of cubes \(J \in \mathcal{D}\) satisfying \(s(J) = s(I)\) and

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

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

Note that by definition, \(S_{1,{\mathfrak u}}f\) is a finite sum of indicator functions of cubes \(I \in \mathcal{D}\) 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 \(\bar g S_{1,{\mathfrak u}}f\) is integrable, and we have

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

By Lemma 7.2.4, there are at most \(2^{9a}\) cubes \(I\) at each scale with \(J \subset B(c(I), D^{s(I)})\). By convexity of \(t \mapsto D^t\) and since \(D \ge 2\), we have for all \(-1 \le t \le 0\)

so \((1 - D^{-1/a})^{-1} \le 2a \le 2^{a-1}\). Using this estimate for the sum of the geometric series, we conclude that 7.2.12 is at most

The collection \(\mathcal{J}\) is a partition of \(X\), so this equals

Using Cauchy-Schwarz and Proposition 2.0.6, we conclude

The lemma now follows by choosing \(g = S_{1,{\mathfrak u}}f\) and dividing on both sides by the finite \(\| S_{1,{\mathfrak u}}f\| _2\).

Denote

Then we have

By Lemma 7.2.1, this is bounded by

We bound the two factors separately. We have

By Cauchy-Schwarz and Lemma 7.3.2 this is at most

Since cubes \(L \in \mathcal{L}({\mathfrak T}({\mathfrak u}))\) are pairwise disjoint by Lemma 7.1.2, this is

Similarly, we have

By Cauchy-Schwarz, this is

Since cubes in \(\mathcal{J}({\mathfrak T}({\mathfrak u}))\) are pairwise disjoint by Lemma 7.1.2, this at most

Combining 7.3.4, 7.3.5 and 7.3.7 and using \(a \ge 4\) gives 7.3.1.

If \(f \le \mathbf{1}_F\) then \(f = f\mathbf{1}_F\), so

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

Combining this with 7.3.4, 7.3.6 and \(a \ge 4\) gives 7.3.2.

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

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

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

Thus \(L \cap E({\mathfrak q}) \subset E_2(9, {\mathfrak p})\). We obtain

By the definition of \(\operatorname{\operatorname {dens}}_1\), this is bounded by

Since \(a \ge 4\), 7.3.3 follows in this case.

Now suppose that for each \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \(L \cap E({\mathfrak p}) \ne \emptyset \), we have \({\mathrm{s}}({\mathfrak p}) {\gt} s(L)\). Since there exists at least one such \({\mathfrak p}\), there exists in particular at least one cube \(L'' \in \mathcal{D}\) with \(L \subset L''\) and \(s(L'') {\gt} s(L)\). By 2.0.7, there exists \(L' \in \mathcal{D}\) with \(L \subset L'\) and \(s(L') = s(L) + 1\). By the definition of \(\mathcal{L}({\mathfrak T}({\mathfrak u}))\) there exists a tile \({\mathfrak p}'' \in {\mathfrak T}({\mathfrak u})\) with \({\mathcal{I}}({\mathfrak p}'') \subset L'\). Let \({\mathfrak p}'\) be the unique tile such that \({\mathcal{I}}({\mathfrak p}') = L'\) and such that \(\Omega ({\mathfrak u}) \cap \Omega ({\mathfrak p}') \ne \emptyset \). Since by 2.0.32 \({\mathrm{s}}({\mathfrak p}') = s(L') \le {\mathrm{s}}({\mathfrak p}) {\lt} {\mathrm{s}}({\mathfrak u})\), we have by 2.0.8 and 2.0.14 that \(\Omega ({\mathfrak u}) \subset \Omega ({\mathfrak p}')\). Let \({\mathfrak q}\in {\mathfrak T}({\mathfrak u})\) with \(L \cap E({\mathfrak q}) \ne \emptyset \). As shown above, this implies \({\mathrm{s}}({\mathfrak q}) \ge s(L')\), so by 2.0.8 \(L' \subset {\mathcal{I}}({\mathfrak q})\). If \(q \in B_{{\mathfrak q}}({\mathcal{Q}}({\mathfrak q}), 1)\), then by a similar calculation as above, using the triangle inequality, Lemma 2.1.2 and 2.0.32, we obtain

Thus \(L \cap E({\mathfrak q}) \subset E_2(6, {\mathfrak p}')\). Since \({\mathcal{I}}({\mathfrak p}'') \subset {\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p})\) and \({\mathfrak p}'', {\mathfrak p}\in {\mathfrak T}({\mathfrak u})\), we have \({\mathfrak p}' \in {\mathfrak P}({\mathfrak T}({\mathfrak u}))\). We deduce using the definition 2.0.28 of \(\operatorname{\operatorname {dens}}_1\)

Using the doubling property 1.0.5, 2.0.10, and \(a \ge 4\) this is estimated by

This completes the proof.

Suppose first that there exists a tile \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \({\mathcal{I}}({\mathfrak p}) \subset B(c(J), 100 D^{s(J) + 1})\). By the definition of \(\mathcal{J}({\mathfrak T}({\mathfrak u}))\), this implies that \(s(J) = -S\), and in particular \({\mathrm{s}}({\mathfrak p}) \ge s(J)\). Using the triangle inequality and 2.0.10 it follows that \(J \subset B({\mathrm{c}}({\mathfrak p}), 200 D^{{\mathrm{s}}({\mathfrak p}) + 1})\). From the doubling property 1.0.5, \(D=2^{100a^2}\) and 2.0.10, we obtain

and hence

With the definition 2.0.29 of \(\operatorname{\operatorname {dens}}_2\) it follows that

completing the proof in this case.

Now suppose that there does not exist a tile \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) with \({\mathcal{I}}({\mathfrak p}) \subset B(c(J), 100 D^{s(J) + 1})\). If we had \({\mathrm{s}}({\mathfrak q}) \le s(J)\), then by 2.0.8 and 2.0.10 \({\mathcal{I}}({\mathfrak q}) \subset J \subset B(c(J), 100 D^{s(J) + 1})\), contradicting our assumption. Thus \({\mathrm{s}}({\mathfrak q}) {\gt} s(J)\). Then, by 2.0.7 and 2.0.8, there exists some cube \(J' \in \mathcal{D}\) with \(s(J') = s(J) + 1\) and \(J \subset J'\). By definition of \(\mathcal{J}({\mathfrak T}({\mathfrak u}))\) there exists some \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) such that \({\mathcal{I}}({\mathfrak p}) \subset B(c(J'), 100 D^{s(J') + 1})\). From the doubling property 1.0.5, \(D=2^{100a^2}\) and 2.0.10, we obtain

If \(J \subset B({\mathrm{c}}({\mathfrak p}), 4 D^{{\mathrm{s}}({\mathfrak p})})\), then we bound

and use the definition 2.0.29

From now on we assume \(J \not\subset B({\mathrm{c}}({\mathfrak p}), 4 D^{{\mathrm{s}}({\mathfrak p})})\). Since

we have by 2.0.10 and the triangle inequality

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

so from the doubling property 1.0.5

From here one completes the proof as in the other cases.

By 2.0.32, \(E({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u})\). Thus by 7.4.1

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

Thus

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

By Lemma 7.3.1, we have for all bounded \(f,g\) with bounded support that

Let \(f = \sum _{{\mathfrak p}\in {\mathfrak T}({\mathfrak u})} T_{{\mathfrak p}}^* g\). If \(g\) is bounded and has bounded support, then the same is true for \(f\). In particular \(\| f\| _2 {\lt} \infty \). Dividing 7.4.2 by \(\| f\| _2\) completes the proof.

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

By Lemma 7.4.1 and 2.0.8, the left hand side 7.4.3 is \(0\) unless \({\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak u}_2)\) or \({\mathcal{I}}({\mathfrak u}_2) \subset {\mathcal{I}}({\mathfrak u}_1)\). Without loss of generality we assume that \({\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak u}_2)\).

Define

Lemma 7.4.4 follows by combining the definition 2.0.3 of \(Z\) with the following two lemmas.

Suppose first that \({\mathfrak p}\in {\mathfrak T}({\mathfrak u}_1)\). Then \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak u}_2)\), by 2.0.32. Thus we have by the separation condition 2.0.36, 2.0.15, 2.0.32 and the triangle inequality

using that \(Z= 2^{12a}\ge 4\). Hence \({\mathfrak p}\in \mathfrak {S}\).

Suppose now that \({\mathfrak p}\in {\mathfrak T}({\mathfrak u}_2)\). If \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}_1)\), then the same argument as above with \({\mathfrak u}_1\) and \({\mathfrak u}_2\) swapped shows \({\mathfrak p}\in \mathfrak {S}\). If \({\mathcal{I}}({\mathfrak p}) \not\subset {\mathcal{I}}({\mathfrak u}_1)\) then, by 2.0.8, \({\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak p})\). Pick \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u}_1)\), we have \({\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak p})\). Hence, by Lemma 2.1.2 and the first paragraph

so \({\mathfrak p}\in \mathfrak {S}\).

By Lemma 7.1.2, it remains only to show that each \(J \in \mathcal{J}(\mathfrak {S})\) with \(J \cap {\mathcal{I}}({\mathfrak u}_1) \ne \emptyset \) is in \(\mathcal{J}'\). But if \(J \notin \mathcal{J}'\), then by 2.0.8 \({\mathcal{I}}({\mathfrak u}_1) \subsetneq J\). Pick \({\mathfrak p}\in {\mathfrak T}({\mathfrak u}_1) \subset \mathfrak {S}\). Then \({\mathcal{I}}({\mathfrak p}) \subsetneq J\). This contradicts the definition of \(\mathcal{J}(\mathfrak {S})\).

For each cube \(J \in \mathcal{J}\) let

and set

We define

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

Since \(\tilde\chi _J(z) \ge 4\) for all \(z \in B(c(J), 4) \supset J\) and by Lemma 7.5.1, we have that \(a(z) \ge 4\) for all \(z \in {\mathcal{I}}({\mathfrak u}_1)\). So we can estimate the above further by

If \(y' \notin B({\mathrm{c}}({\mathfrak p}), 8D^{{\mathrm{s}}({\mathfrak p})})\) then the second summand vanishes. Else, we can estimate the above, using also that \(|\tilde\chi _J(y')| \le 8\), by

By the triangle inequality, we have for all dyadic cubes \(I \in \mathcal{J}'\)

Using this above, we obtain

By Lemma 7.5.3, this is at most

By 2.0.10 and Lemma 7.5.1, the balls \(B(c(J'), \frac{1}{4} D^{s(J')})\) are pairwise disjoint, so by Lemma 7.5.3 the balls \(B(c(J'), \frac{1}{4} D^{s(J) - 1})\) are also 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

By the doubling property 1.0.5, we further have

for each such ball. Thus

Recalling that \(D=2^{100a^2}\), we obtain

Since \(a\ge 4\), 7.5.4 follows.

Suppose that \(s(J') {\lt} s(J) - 1\). Then \(s(J) {\gt} -S\). Thus, by the definition of \(\mathcal{J}'\) there exists no \({\mathfrak p}\in \mathfrak {S}\) with

Since \(s(J') {\lt} s(J)\) and \(J', J \subset {\mathcal{I}}({\mathfrak u}_1)\), we have \(J' \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). By 2.0.7, 2.0.8 there exists a cube \(J'' \in \mathcal{D}\) with \(J \subset J''\) and \(s(J'') = s(J') + 1\). By the definition of \(\mathcal{J}'\), there exists a tile \({\mathfrak p}\in \mathfrak {S}\) with

But by the triangle inequality and 2.0.1, we have

which contradicts 7.5.5 and 7.5.6.

By 7.4.1, we have

By the oscillation estimate 1.0.7, we have

Suppose that \(y, y' \in B({\mathrm{c}}({\mathfrak p}), 5D^{{\mathrm{s}}({\mathfrak p})})\), so that \(\rho (y,y') \le 10D^{{\mathrm{s}}({\mathfrak p})}\). Let \(k \in \mathbb {Z}\) be such that \(2^{ak}\rho (y,y') \le 10D^{{\mathrm{s}}({\mathfrak p})}\) but \(2^{a(k+1)} \rho (y,y') {\gt} 10D^{{\mathrm{s}}({\mathfrak p})}\). In particular, \(k \ge 0\). Then, using 1.0.8, we can bound 7.5.13 from above by

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

By definition of \(k\), we have

which gives

For all \(x \in {\mathcal{I}}({\mathfrak p})\), we have by 1.0.5 that

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

Since \(\rho (y,y') \le 10 D^{{\mathrm{s}}({\mathfrak p})}\), we conclude

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

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

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

It follows from the definition of \(\psi \) that

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

Combining the above with the previous estimate on \(\psi \), we get

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

Plugging this into 7.5.15 and using \(a \ge 4\), we get

which completes the proof of the lemma.

For the first estimate, assume that \({\mathrm{s}}({\mathfrak p}) {\lt} s(J)\), then in particular \({\mathrm{s}}({\mathfrak p}) \le {\mathrm{s}}({\mathfrak u}_1)\). Since \({\mathfrak p}\notin \mathfrak {S}\), we have by Lemma 7.4.7 that \({\mathcal{I}}({\mathfrak p}) \cap {\mathcal{I}}({\mathfrak u}_1) = \emptyset \). Since \(B\Big(c(J), \frac{1}{4} D^{s(J)}\Big) \subset {\mathcal{I}}(J) \subset {\mathcal{I}}({\mathfrak u}_1)\), this implies

On the other hand

by our assumption. Thus \(D^{{\mathrm{s}}({\mathfrak p})} \ge 64 D^{s(J)}\), which contradicts 2.0.1 and \(a \ge 4\).

For the second estimate, assume that \({\mathrm{s}}({\mathfrak p}) {\gt} s(J) + 3\). Since \(J \in \mathcal{J}'\), we have \(J \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). Thus there exists \(J' \in \mathcal{D}\) with \(J \subset J'\) and \(s(J') = s(J) + 1\), by 2.0.7 and 2.0.8. By definition of \(\mathcal{J}'\), there exists some \({\mathfrak p}' \in \mathfrak {S}\) such that \({\mathcal{I}}({\mathfrak p}') \subset B(c(J'), 100 D^{s(J) + 2})\). On the other hand, since \(B({\mathcal{I}}({\mathfrak p})) \cap B^\circ (J) \ne \emptyset \), by the triangle inequality it holds that

Using the definition of \(\mathfrak {S}\), we have

By 1.0.10, this is

and by 1.0.8 and the definition of \(\mathfrak {S}\)

This is a contradiction, the second estimate follows.

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

By Lemma 7.5.6, this is at most

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

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

Using 7.4.1, 2.1.3 and that \(a \ge 4\), we bound 7.5.16 by

For each \(I \in \mathcal{D}\), the sets \(E({\mathfrak p})\) for \({\mathfrak p}\in {\mathfrak P}\) with \({\mathcal{I}}({\mathfrak p}) = I\) are pairwise disjoint by 2.0.20 and 2.0.13. Further, if \(B({\mathcal{I}}({\mathfrak p})) \cap B^\circ (J) \ne \emptyset \) and \({\mathrm{s}}({\mathfrak p}) \ge s(J)\), then \(E({\mathfrak p}) \subset B(c(J), 32 D^{{\mathrm{s}}({\mathfrak p})})\). Thus the last display is bounded by

The lemma follows since \(a \ge 4\).

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

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

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

By Lemma 7.5.8, we have \({\mathrm{s}}({\mathfrak p}) \ge s(J)\) for all \({\mathfrak p}\) occurring in the sum. Further, for each \(s \ge s(J)\), the sets \(E({\mathfrak p})\) for \({\mathfrak p}\in {\mathfrak P}\) with \({\mathrm{s}}({\mathfrak p}) = s\) are pairwise disjoint by 2.0.20 and 2.0.13, and contained in \(B(c(J), 32D^{s})\) 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

By convexity of \(t \mapsto D^t\) and since \(D \ge 2\), we have for all \(-1 \le t \le 0\)

Since \(-1 \le -1/a {\lt}0\), it follows that

Estimate 7.5.18, and therefore the lemma, follow.

By Lemma 7.5.9

and by Lemma 7.5.7

This completes the proof.

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 \(B(J) \cap {\mathcal{I}}({\mathfrak u}_1)\). By 7.5.3 and Lemma 7.5.9, we have for all \(y \in B(J)\):

We have by the triangle inequality

As \(h_J\) is supported in \({\mathcal{I}}({\mathfrak u}_1)\), we can assume without loss of generality that \(y' \in {\mathcal{I}}({\mathfrak u}_1)\). If \(y \notin {\mathcal{I}}({\mathfrak u}_1)\), then 7.5.20 vanishes. If \(y \in {\mathcal{I}}({\mathfrak u}_1)\) then we have by 7.5.4, Lemma 7.5.9 and Lemma 7.5.10

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

By 7.5.3, and twice Lemma 7.5.9, we have

Using that \(\rho (y,y') \le 16D^{s(J)}\) and \(a \ge 4\), the lemma follows.

Since \(\emptyset \ne {\mathfrak T}({\mathfrak u}_1) \subset \mathfrak {S}\) by Lemma 7.4.7, there exists at least one tile \({\mathfrak p}\in \mathcal{S}\) with \({\mathcal{I}}({\mathfrak p}) \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). Thus \({\mathcal{I}}({\mathfrak u}_1) \notin \mathcal{J}'\), so \(J \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). Thus there exists a cube \(J' \in \mathcal{D}\) with \(J \subset J'\) and \(s(J') = s(J) + 1\), by 2.0.7 and 2.0.8. By definition of \(\mathcal{J'}\) and the triangle inequality, there exists \({\mathfrak p}\in \mathfrak {S}\) such that

Thus, by definition of \(\mathfrak {S}\):

By the doubling property 1.0.8, this is

which gives the lemma using \(a \ge 4\).

We have

By Lemma 7.4.1, the right hand side is supported in \({\mathcal{I}}({\mathfrak u}_1)\). Using 7.5.2 of Lemma 7.5.2 and the definition 7.5.7 of \(h_J\), we thus have

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

Using Lemma 7.5.4, Lemma 7.5.11 and \(a \ge 4\), we have that the above is bounded from above by

By the doubling property 1.0.5

thus

Summing over \(J \in \mathcal{J}'\), we obtain

Applying the Cauchy-Schwarz inequality, Lemma 7.4.5 follows.

By Lemma 7.1.2, it remains only to show that each \(J \in \mathcal{J}({\mathfrak T}({\mathfrak u}_1))\) with \(J \cap {\mathcal{I}}({\mathfrak u}_1) \ne \emptyset \) is in \(\mathcal{J}'\). But if \(J \notin \mathcal{J}'\), then by 2.0.8 \({\mathcal{I}}({\mathfrak u}_1) \subsetneq J\). Pick \({\mathfrak p}\in {\mathfrak T}({\mathfrak u}_1)\). Then \({\mathcal{I}}({\mathfrak p}) \subsetneq J\). This contradicts the definition of \(\mathcal{J}({\mathfrak T}({\mathfrak u}_1))\).

By Lemma 7.2.1 and Lemma 7.4.1, we have

It follows from the definition of the projection operator \(P\) and Jensen’s inequality that

Since cubes in \(\mathcal{J}'\) are pairwise disjoint and by Lemma 7.6.1, a cube \(J \in \mathcal{J}'\) intersect \({\mathcal{I}}({\mathfrak u}_1)\) if and only if \(J \in \mathcal{J}'\). Thus

Combining this with Lemma 7.6.2, the definition 2.0.2 and \(a \ge 4\) proves the lemma.

Suppose that \({\mathrm{s}}({\mathfrak p}) {\gt} s(J) + 2 -\frac{Zn}{202a^3} =: s(J) - s_1\). Then, we have

There exists a tile \({\mathfrak q}\in {\mathfrak T}({\mathfrak u}_1)\). By 2.0.32, it satisfies \({\mathcal{I}}({\mathfrak q}) \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). Thus \({\mathcal{I}}({\mathfrak u}_1) \notin \mathcal{J}'\). It follows that \(J \subsetneq {\mathcal{I}}({\mathfrak u}_1)\). By 2.0.7 and 2.0.8, there exists a cube \(J' \in \mathcal{D}\) with \(J \subset J'\) and \(s(J') = s(J) + 1\). By definition of \(\mathcal{J}'\), there exists a tile \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u}_1)\) with

By the triangle inequality, the definition 2.0.1 and \(a \ge 4\), we have

Since \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u}_1)\) and \({\mathcal{I}}({\mathfrak u}_1) \subset {\mathcal{I}}({\mathfrak u}_2)\), we have by 2.0.36

Hence, by 2.0.32, the triangle inequality and using that by 2.0.3 \(Z \ge 2\)

It follows that

Using 1.0.8, we obtain

Since \({\mathfrak p}' \notin \mathfrak {S}\) this is bounded by

Thus

contradicting the definition of \(s_1\).

Since \(J \in \mathcal{J}'\) we have \(J \subset {\mathcal{I}}({\mathfrak u}_1)\). Thus, if \(B(I) \cap J \ne \emptyset \) then

Furthermore, for each \(s\) the balls \(B(I)\) with \(s(I) = s\) have bounded overlap: Consider the collection \(\mathcal{D}_{s,x}\) of all \(I \in \mathcal{D}\) with \(x \in B(I)\) and \(s(I) = s\). By 2.0.10 and 2.0.8, the balls \(B(c(I), \frac{1}{4} D^{s(I)})\), \(I \in \mathcal{D}_{s,x}\) 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

for each \(I \in \mathcal{D}_{s,x}\). Thus

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

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

Expanding the definition of \(P_{\mathcal{J}'}\), we have

We split the innermost sum according to the scale of the tile \({\mathfrak p}\), and then apply the triangle inequality and Minkowski’s inequality:

By Lemma 7.4.1, the integral in the last display is \(0\) if \(J \cap B({\mathcal{I}}({\mathfrak p})) = \emptyset \). By Lemma 7.6.3, it follows with \(s_1 := \frac{Zn}{202a^3} - 2\):

We have by Lemma 7.4.1 and 2.1.3

If \(x \in E({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak p})\), then we have by 2.0.10 that

Using the doubling property 1.0.5, it follows that

Thus, using also \(a \ge 4\)

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

By Lemma 7.4.7, we have \({\mathcal{I}}({\mathfrak p}) \cap {\mathcal{I}}({\mathfrak u}_1) = \emptyset \) for all \({\mathfrak p}\in {\mathfrak T}({\mathfrak u}_2) \setminus \mathfrak {S}\). Thus we can estimate 7.6.3 by

which is by Cauchy-Schwarz at most

Using Lemma 7.6.4, we bound 7.6.4 by

and, since dyadic cubes in \(\mathcal{J}'\) form a partition of \({\mathcal{I}}({\mathfrak u}_1)\) by Lemma 7.6.1, \(\kappa \le 1\) by 2.0.2, and \(a \ge 4\)

By convexity of \(t \mapsto D^t\) and since \(D \ge 2\), we have for all \(-1 \le t \le 0\)

Using this for \(t = -\kappa /2\) and using that \(s_1 = \frac{Zn}{202a^3} - 2\) and the definitions 2.0.1 and 2.0.2 of \(\kappa \) and \(D\)

Using the definition 2.0.2 of \(\kappa \) and \(a \ge 4\), the lemma follows.

Define recursively \({\mathfrak U}_j\) to be a maximal disjoint set of tiles \({\mathfrak u}\) in

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

Now we show by induction on \(j\) that each point is contained in at most \(2^n - j\) cubes \({\mathcal{I}}({\mathfrak u})\) with \({\mathfrak u}\in {\mathfrak U}\setminus \bigcup _{j' \le j} {\mathfrak U}_{j'}\). This implies that \(\bigcup _{j = 1}^{2^n} {\mathfrak U}_j = {\mathfrak U}\), 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 \({\mathcal{I}}({\mathfrak u})\) with \({\mathfrak u}\in {\mathfrak U}\setminus \bigcup _{j' {\lt} j} {\mathfrak U}_{j'}\), then it is contained in a maximal such cube. Thus it is contained in a cube in \({\mathcal{I}}({\mathfrak u})\) with \({\mathfrak u}\in {\mathfrak U}_j\). Thus the number \({\mathfrak u}\in {\mathfrak U}\setminus \bigcup _{j' \le j} {\mathfrak U}_{j'}\) with \(x\in {\mathcal{I}}({\mathfrak u})\) is zero, or is less than the number of \({\mathfrak u}\in {\mathfrak U}\setminus \bigcup _{j' \le j-1} {\mathfrak U}_{j'}\) with \(x \in {\mathcal{I}}({\mathfrak u})\) by at least one.

By Lemma 7.3.1 and the density assumption 2.0.35, we have for each \({\mathfrak u}\in {\mathfrak U}\) and all bounded \(f\) of bounded support that

and

Since for each \(j\) the top cubes \({\mathcal{I}}({\mathfrak u})\), \({\mathfrak u}\in {\mathfrak U}_j\) are disjoint, we further have for all bounded \(g\) of bounded support by Lemma 7.4.1

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

Again by disjointedness of the cubes \({\mathcal{I}}({\mathfrak u})\), this is estimated by

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

To save some space we will write for subsets \({\mathfrak C}\subset {\mathfrak P}\)

We have by Lemma 7.4.1 and the triangle inequality that

By Lemma 7.4.4, this is bounded by

We apply the Cauchy-Schwarz inequality in the form

to the outer two sums:

By pairwise disjointedness of the sets \({\mathcal{I}}({\mathfrak u})\) for \({\mathfrak u}\in {\mathfrak U}_j\) and of the sets \({\mathcal{I}}({\mathfrak u}')\) for \({\mathfrak u}' \in {\mathfrak U}_{j'}\), we have

Arguing similar for \(g_2\), we can estimate 7.7.5 to be

The lemma now follows from Lemma 7.4.3.

Suppose that \({\mathfrak p}\in {\mathfrak T}({\mathfrak u})\) and \({\mathfrak p}' \in {\mathfrak T}({\mathfrak u}')\) with \({\mathfrak u}\ne {\mathfrak u}'\) and \(x \in E({\mathfrak p}) \cap E({\mathfrak p}')\). Suppose without loss of generality that \({\mathrm{s}}({\mathfrak p}) \le {\mathrm{s}}({\mathfrak p}')\). Then \(x \in {\mathcal{I}}({\mathfrak p}) \cap {\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak u}')\). By 2.0.8 it follows that \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}')\). By 2.0.36, it follows that

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

Since \(Z \ge 3\) by 2.0.3, it follows that \({\mathcal{Q}}({\mathfrak p}') \notin B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 1)\), so \(\Omega ({\mathfrak p}') \not\subset \Omega ({\mathfrak p})\) by 2.0.15. Hence, by 2.0.14, \(\Omega ({\mathfrak p}) \cap \Omega ({\mathfrak p}') = \emptyset \). But if \(x \in E({\mathfrak p}) \cap E({\mathfrak p}')\) then \(Q(x) \in \Omega ({\mathfrak p}) \cap \Omega ({\mathfrak p}')\). This is a contradiction, and the lemma follows.

To save some space, we will write

By 7.4.1, we have for each \(j\)

Hence, by Lemma 7.7.1,

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:

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

and by disjointedness of the sets \(E_j\), this is at most

Taking adjoints and square roots, it follows that for all \(f\)

On the other hand, we have by disjointedness of the sets \(E_j\) from Lemma 7.7.4

If \(|f| \le \mathbf{1}_F\) then we obtain from Lemma 7.7.2 and taking square roots that

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.