Proof. Let s \in \mathbb{Z} with \underline{\sigma} (\fu, x) \le s \le \overline{\sigma} (\fu, x). By definition of \sigma, there exists \fp \in \fT(\fu) with \ps(\fp) = \underline{\sigma} (\fu, x) and x \in E(\fp), and there exists \fp'' \in \fT(\fu) with \ps(\fp'') = \overline{\sigma} (\fu, x) and x \in E(\fp'') \subset \scI(\fp''). By property coverdyadic of the dyadic grid, there exists a cube I \in \mathcal{D} of scale s with x \in I. By property eq-dis-freq-cover, there exists a tile \fp' \in \fP(I) with \tQ(x) \in \fc(\fp'). By the dyadic property dyadicproperty we have \scI(\fp) \subset \scI(\fp') \subset \scI(\fp''), and by eq-freq-dyadic, we have \fc(\fp'') \subset \fc(\fp') \subset \fc(\fp). Thus \fp \le \fp' \le\fp'', which gives with the convexity property forest2 of \fT(\fu) that \fp' \in \fT(\fu), so s \in \sigma(\fu, x).

Proof. 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 dyadicproperty. The same applies to \mathcal{L}(\mathfrak{S}).

If x \in \bigcup_{I \in \mathcal{D}} I, then there exists by coverdyadic 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 eq-J-partition, the other one follows from \mathcal{J}(\mathfrak{S}) \subset \mathcal{D}.

The proof of the two inclusions in eq-L-partition is similar.

Proof. By definetp, if T_{\fp}[ e(-\fcc(\fu))f](x) \ne 0, then x \in E(\fp). Combining this with |e(\fcc(\fu)(x))| = 1, we obtain |\sum_{\fp \in \fT(\fu)} T_{\fp}[ e(-\fcc(\fu))f](x)| = \Bigg| \sum_{s \in \sigma(\fu, x)} \int e(-\fcc(\fu)(y) + \tQ(x)(y) + \fcc(\fu)(x) -\tQ(x)(x))\times K_s(x,y)f(y) \, \mathrm{d}\mu(y) \Bigg|. Using the triangle inequality, we bound this by the sum of three terms: \le \Bigg| \sum_{s \in \sigma(\fu, x)} \int (e(-\fcc(\fu)(y) + \tQ(x)(y) + \fcc(\fu)(x) -\tQ(x)(x))-1)\times K_s(x,y)f(y) \, \mathrm{d}\mu(y) \Bigg| + \Bigg| \sum_{s \in \sigma(\fu, x)} \int K_s(x,y) P_{\mathcal{J}(\fT(\fu))} f(y) \, \mathrm{d}\mu(y) \Bigg| + \Bigg| \sum_{s \in \sigma(\fu, x)} \int K_s(x,y) (f(y) - P_{\mathcal{J}(\fT(\fu))} f(y)) \, \mathrm{d}\mu(y) \Bigg|. The proof is completed using the bounds for these three terms proven respectively in Lemma 1.7.1.4, Lemma 1.7.1.5 and Lemma 1.7.1.6.

Proof. Let s \in \sigma(\fu,x). If x, y \in X are such that K_s(x,y)\neq 0, then, by supp-Ks, 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 osccontrol of the metrics d_B, it follows that |e(-\fcc(\fu)(y)+\tQ(x)(y)+\fcc(\fu)(x)-\tQ(x)(x))-1| \leq d_{B(x, 1/2 D^{s})}(\fcc(\fu), \tQ(x)). Let \fp_s \in \fT(\fu) be a tile with \ps(\fp_s) = s and x \in E(\fp_s), and let \fp' be a tile with \ps(\fp') = \overline{\sigma}(\fu, x) and x \in E(\fp'). Using the monotonicity property monotonedb, the doubling property firstdb repeatedly, the definition of d_{\fp} and Lemma 1.2.1.2, we can bound the previous display by d_{B(x, 4 D^{s})}(\fcc(\fu), \tQ(x)) \leq 2^{4a} d_{\fp_s}(\fcc(\fu), \tQ(x)) \le 2^{4a} 2^{s - \overline{\sigma}(\fu, x)} d_{\fp'}(\fcc(\fu), \tQ(x)). Since \fcc(\fu) \in B_{\fp'}(\fcc(\fp'), 4) by forest1 and \tQ(x) \in \Omega(\fp') \subset B_{\fp'}(\fcc(\fp'), 1) by eq-freq-comp-ball, this is estimated by \le 5 \cdot 2^{4a} 2^{s - \overline{\sigma}(\fu, x)}.

Using eq-Ks-size, it follows that eq-term-A is bounded by 5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\int_{B(x,0.5D^{s})}|f(y)|\,\mathrm{d}\mu(y). By eq-J-partition, the collection \mathcal{J} is a partition of \bigcup_{I \in \mathcal{D}} I, so this is estimated by 5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\sum_{\substack{J \in \mathcal{J}(\fT(\fu))\\J \cap B(x, 0.5D^s) \ne \emptyset} }\int_{J}|f(y)|\,\mathrm{d}\mu(y). This expression does not change if we replace |f| by P_{\mathcal{J}(\fT(\fu))}|f|.

Let J \in \mathcal{J}(\fT(\fu)) with B(x, 0.5 D^s) \cap J \ne \emptyset. By the triangle inequality and since x \in E(\fp_s) \subset B(\pc(\fp_s), 4D^{s}), it follows that B(\pc(\fp_s), 4.5D^s) \cap J \ne \emptyset. If s(J) \ge s and s(J) > -S, then it follows from the triangle inequality, eq-vol-sp-cube and defineD that \scI(\fp_s) \subset B(c(J), 100 D^{s(J)+1}), contradicting J \in \mathcal{J}(\mathfrak{T}(\fu)). Thus s(J) \le s - 1 or s(J) = -S. If s(J) = -S and s(J) > s - 1, then s = -S. Thus we always have s(J) \le s. It then follows from the triangle inequality and eq-vol-sp-cube that J \subset B(\pc(\fp_s), 16 D^s).

Thus we can continue our chain of estimates with 5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\int_{B(\pc(\fp_s),16 D^s)}P_{\mathcal{J}(\fT(\fu))}|f(y)|\,\mathrm{d}\mu(y). We have B(\pc(\fp_s), 16D^s)) \subset B(x, 32D^s), by eq-vol-sp-cube and the triangle inequality, since x \in \scI(\fp_s). Combining this with the doubling property doublingx, we obtain \mu(B(\pc(\fp_s), 16D^s)) \le 2^{5a} \mu(B(x, D^s)). Since a \ge 4, it follows that eq-term-A is bounded by 5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(\pc(\fp_s),16D^s))}\int_{B(\pc(\fp_s),16D^s)}P_{\mathcal{J}(\fT(\fu))}|f(y)|\,\mathrm{d}\mu(y).

Since L \in \mathcal{L}(\fT(\fu)) and x\in L \cap \scI(\fp_s), we have s(L) \le \ps(\fp_s). It follows by dyadicproperty that L \subset \scI(\fp_s), in particular x' \in \scI(\fp_s) \subset B(\pc(\fp_s), 16D^s). Thus \le 5\cdot 2^{104a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x') \le 10\cdot 2^{104a^3} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x'). This completes the estimate for term eq-term-A.

Next, let s_2 = \overline{\sigma}(\fu, x) and let \fp' \in \fT(\fu) with \ps(\fp') = s_2 and x \in E(\fp'). Since \fp' \in \fT(\fu), we have 4\fp' \lesssim \fu. Since \tQ(x) \in \fc(\fp'), it follows that d_{\fp}(\fcc(\fu), \tQ(x)) \le 5. Applying the doubling property firstdb five times, we obtain d_{B(c(\fp), 8D^{s_2})}(\fcc(\fu), \tQ(x)) \le 5 \cdot 2^{5a}. By the triangle inequality, we have B(x, D^{s_2}) \subset B(c(\fp), 8 D^{s_2}), so by monotonedb d_{B(x, D^{s_2})}(\fcc(\fu), \tQ(x)) \le 5 \cdot 2^{5a}. Finally, by applying seconddb 100a times, we obtain d_{B(x, D^{s_2-1})}(\fcc(\fu), \tQ(x)) \le 5 \cdot 2^{-95a} < 1. Consequently, D^{s_2 - 1} < R_Q(\fcc(\fu), x). The lemma now follows from the definition of T_{\mathcal{N}}.

Proof. We have for J \in \mathcal{J}(\fT(\fu)) \int_J K_{s}(x,y)(1 - P_{\mathcal{J}(\fT(\fu))})f(y) \, \mathrm{d}\mu(y) = \int_J \frac{1}{\mu(J)} \int_J K_s(x,y) - K_s(x,z) \, \mathrm{d}\mu(z) \,f(y) \, \mathrm{d}\mu(y). By eq-Ks-smooth and eq-vol-sp-cube, we have for y, z \in J |K_s(x,y) - K_s(x,z)| \le \frac{2^{127a^3}}{\mu(B(x, D^s))} \left(\frac{8 D^{s(J)}}{D^s}\right)^{1/a}.

Suppose that s \in \sigma(\fu, x). If K_s(x,y) \ne 0 for some y \in J \in \mathcal{J}(\fT(\fu)) then, by supp-Ks, y \in B(x, 0.5 D^s) \cap J \ne \emptyset. Let \fp \in \fT(\fu) with \ps(\fp) = s and x \in E(\fp). Then B(\pc(\fp_s), 4.5D^s) \cap J \ne \emptyset by the triangle inequality. If s(J) \ge s and s(J) > -S, then it follows from the triangle inequality, eq-vol-sp-cube and defineD that \scI(\fp) \subset B(c(J), 100 D^{s(J)+1}), contradicting J \in \mathcal{J}(\mathfrak{T}(\fu)). Thus s(J) \le s - 1 or s(J) = -S. If s(J) = -S and s(J) > s - 1, then s = -S. So in both cases, s(J) \le s. It then follows from the triangle inequality and eq-vol-sp-cube that J \subset B(x, 16 D^s).

Thus, we can estimate eq-term-C by 2^{127a^3 + 3/a}\sum_{\fp\in \mathfrak{T}}\frac{\mathbf{1}_{E(\fp)}(x)}{\mu(B(x,D^{\ps(\fp)}))}\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x, 16D^{\ps(\fp)})\\ s(J) \le \ps(\fp)}} D^{(s(J) - \ps(\fp))/a} \int_J |f|. = 2^{127a^3 + 3/a}\sum_{I \in \mathcal{D}} \sum_{\substack{\fp\in \mathfrak{T}\\ \scI(\fp) = I}}\frac{\mathbf{1}_{E(\fp)}(x)}{\mu(B(x, D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x, 16 D^{s(I)})\\ s(J) \le s(I)}} D^{(s(J) - s(I))/a} \int_J |f|. By eq-dis-freq-cover and defineep, the sets E(\fp) for tiles \fp with \scI(\fp) = I are pairwise disjoint. It follows from the definition of \mathcal{L}(\fT(\fu)) that x \in \scI(\fp) if and only if x' \in \scI(\fp), thus we can estimate the sum over \mathbf{1}_{E(\fp)}(x) by \mathbf{1}_{I}(x').

If x \in E(\fp) then in particular x \in \scI(\fp), so by eq-vol-sp-cube B(c(I),16D^{s(I)}) \subset B(x, 32D^{s(I)}). By the doubling property doublingx \mu(B(c(I), 16D^{s(I)})) \le 2^{5a} \mu(B(x, D^{s(I)})). Since a \ge 4 we can continue our estimate with \le 2^{128a^3}\sum_{I \in \mathcal{D}} \frac{\mathbf{1}_{I}(x')}{\mu(B(c(I), 16D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x, 16 D^{s(I)})\\ s(J) \le s(I)}} D^{(s(J) - s(I))/a} \int_J |f| = 2^{128a^3} S_{1,\fu} P_{\mathcal{J}(\fT(\fu))}|f|(x'). This completes the proof.

Proof of Lemma 1.7.2.1. Let L \in \mathcal{L}(\fT(\fu)). Let b(x') denote the right-hand side of Lemma 1.7.1.3. Apply this lemma to e(\fcc(\fu)) f, to obtain for all y, x' \in L \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \le b(x'). Hence, by taking an infimum, we have for y \in L \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \le \inf_{x' \in L} b(x'). Integrating this estimate yields \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y) \le \int_L |g(y)| \times \inf_{x' \in L} b(x') \, \mathrm{d}\mu(y) = \int_L P_{\mathcal{L}(\fT(\fu))}|g|(y) \times \inf_{x' \in L} b(x') \, \mathrm{d}\mu(y) \le \int_L P_{\mathcal{L}(\fT(\fu))}|g|(y) \times b(y) \, \mathrm{d}\mu(y). By definetp, we have T_{\fp} f = \mathbf{1}_{\scI(\fp)} T_{\fp} f for all \fp \in \fP, so \Bigg| \int \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \, \mathrm{d}\mu(y) \Bigg| = \Bigg| \int_{\bigcup_{\fp \in \fT(\fu)} \scI(\fp)} \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \, \mathrm{d}\mu(y) \Bigg|. Since \mathcal{L}(\fT(\fu)) partitions \bigcup_{\fp \in \fT(\fu)} \scI(\fp) by Lemma 1.7.1.2, we get from the triangle inequality \le \sum_{L \in \mathcal{L}(\fT(\fu))} \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y) which by the above computation is bounded by \sum_{L \in \mathcal{L}(\fT(\fu))} \int_L P_{\mathcal{L}(\fT(\fu))}|g|(y) \times b(y) \, \mathrm{d}\mu(y) = \int_X P_{\mathcal{L}(\fT(\fu))}|g|(y) \times b(y) \, \mathrm{d}\mu(y). Applying Cauchy-Schwarz, this is bounded by \|P_{\mathcal{L}(\fT(\fu))}|g|\|_2 \times \|b\|_2. By Minkowski's inequality, Theorem 1.2.6, Lemma 1.7.2.2 and Lemma 1.7.2.3, \|b\|_2 is at most 2^{129a^3} (2^{2a+1} + 2^{12a})\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 + 2^{103a^3} \|P_{\mathcal{J}(\fT(\fu))}[e(\fcc(\fu))f]\|_2. By the triangle inequality we have for all x \in X that |P_{\mathcal{J}(\fT(\fu))}[e(\fcc(\fu))f]|(x) \le P_{\mathcal{J}(\fT(\fu))}|f|(x), thus we can further estimate the above by (2^{129a^3} (2^{2a+1} + 2^{12a}) + 2^{103a^3}) \|P_{\mathcal{J}(\fT(\fu))}|f|\|_2. This completes the proof since a \ge 4.

Proof of Lemma 1.7.2.2. Fix s_1, s_2. By eq-psisum we have for all x \in (0, \infty) \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 1 - \sum_{s < s_1} \psi(D^{-s}x) - \sum_{s > s_1} \psi(D^{-s}x). 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 x \in [\frac{1}{2}D^{s_1-1}, \frac{1}{4} D^{s_2}] \implies \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 1. Since \psi is supported in [\frac{1}{4D}, \frac{1}{2}], we further have x \notin [\frac{1}{4}D^{s_1 - 1}, \frac{1}{2}D^{s_2}] \implies \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 0. Finally, since \psi \ge 0 and \sum_{s \in \mathbb{Z}} \psi(D^{-s}x) = 1, we have for all x 0 \le \sum_{s = s_1}^{s_2} \psi(D^{-s}x) \le 1.

Let x' \in I_{s_1}(x) and suppose that D^{s_2 - 1} \le R_Q(\mfa, x'). By the triangle inequality and eq-vol-sp-cube, it holds that \rho(x,x') \le 8D^{s_1}. We have \Bigg|\sum_{s = s_1}^{s_2} \int K_s(x',y) f(y) \, \mathrm{d}\mu(y)\Bigg| = \Bigg|\int \sum_{s = s_1}^{s_2} \psi(D^{-s}\rho(x',y)) K(x',y) f(y) \, \mathrm{d}\mu(y)\Bigg| \le \Bigg| \int_{8D^{s_1} < \rho(x',y) \le \frac{1}{4}D^{s_2-1}} K(x',y) f(y) \, \mathrm{d}\mu(y) \Bigg| + \int_{\frac{1}{4}D^{s_1-1} \le \rho(x',y) \le 8D^{s_1}} |K(x', y)| |f(y)| \, \mathrm{d}\mu(y) + \int_{\frac{1}{4}D^{s_2-1} \le \rho(x',y) \le \frac{1}{2}D^{s_2}} |K(x', y)| |f(y)| \, \mathrm{d}\mu(y).

The first term eq-sharp-trunc-term is at most 2T_{\tQ}^\mfa f(x), using with R_1 := 8D^{s_1}, R_2 := \frac{1}{4}D^{s_2-1} and R_1 < R_2 < R_{\tQ}(\mfa,x') the triangle inequality in the form \left|\int_{R_1 < \rho(x',y) \le R_2} K(x',y) f(y) \, \mathrm{d}\mu(y) \right| \le \left|\int_{R_1 < \rho(x',y) < R_{\tQ}(\mfa,x')} K(x',y) f(y) \, \mathrm{d}\mu(y) \right| + \left|\int_{R_2 < \rho(x',y) < R_{\tQ}(\mfa,x')} K(x',y) f(y) \, \mathrm{d}\mu(y) \right|.

The other two terms will be estimated by the finitary maximal function from Theorem 1.2.6. For the second term eq-lower-bound-term we use eqkernel-size which implies that for all y with \rho(x', y) \ge \frac{1}{4}D^{s_1 - 1}, we have |K(x', y)| \le \frac{2^{a^3}}{\mu(B(x', \frac{1}{4}D^{s_1 - 1}))}. Using D=2^{100a^2} and the doubling property doublingx 7 +100a^2 times estimates the last display by \frac{2^{7a+101a^3}}{\mu(B(x', 32D^{s_1}))}. By the triangle inequality and eq-vol-sp-cube, we have B(x', 8D^{s_1}) \subset B(c(I_{s_1}(x)), 16D^{s(I_{s_1}(x))}) \subset B(x', 32D^{s_1}). Combining this with pf-nontangential-operator-bound-imeq, we conclude that eq-lower-bound-term is at most 2^{7a + 101a^3} M_{\mathcal{B},1} f(x).

For eq-upper-bound-term we argue similarly. We have for all y with \rho(x', y) \ge \frac{1}{4}D^{s_2-1} |K(x', y)| \le \frac{2^{a^3}}{\mu(B(x', \frac{1}{4}D^{s_2-1}))}. Using the doubling property doublingx 7 + 100a^2 times estimates the last display by \frac{2^{7a + 101a^3}}{\mu(B(x', 32 D^{s_2}))}. Note that by dyadicproperty we have I_{s_1}(x) \subset I_{s_2}(x), in particular x' \in I_{s_2}(x). By the triangle inequality and eq-vol-sp-cube, we have B(x', 8D^{s_2}) \subset B(c(I_{s_2}(x)), 16D^{s(I_{s_2}(x))}) \subset B(x', 32D^{s_2}). Combining this, eq-upper-bound-term is at most 2^{7a+101a^3} M_{\mathcal{B},1} f(x).

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 < s_2 \le S, we obtain T_{\mathcal{N}} f(x) \le 2T_{\tQ}^\mfa f(x) + 2^{102a^3} M_{\mathcal{B},1} f(x). The lemma now follows from assumption linnontanbound, Theorem 1.2.6 and a \ge 4.