Proof. Let \fp,\fp' and x be given. Assume without loss of generality that \ps(\fp)\le \ps(\fp'). As we have x\in E(\fp)\subset \scI(\fp) and x\in E(\fp')\subset \scI(\fp') by definition defineep, we conclude for i=1,2 that \tQ(x)\in\fc(\fp) and \tQ(x)\in\fc(\fp'). By eq-freq-dyadic we have \fc(\fp')\subset \fc(\fp). By definition straightorder, we conclude \fp\le \fp'. As \mathfrak{A} is an antichain, we conclude \fp=\fp'. This proves the lemma.

Proof. Fix x\in X. By Lemma 1.6.1.1, there is at most one \fp \in \mathfrak{A} such that T_{\fp} f(x) is not zero. If there is no such \fp, the estimate hlmbound follows.

Assume there is such a \fp. By definition of T_{\fp} we have x\in E(\fp)\subset \scI(\fp) and by the squeezing property eq-vol-sp-cube \rho(x, \pc(\fp))\le 4D^{\ps(\fp)}.

Let y\in X with K_{\ps(\fp)}(x,y)\neq 0. By definition defks of K_{\ps(\fp)} we have \frac{1}{4} D^{\ps(\fp)-1} \leq \rho(x,y) \leq \frac{1}{2} D^{\ps(\fp)}. The triangle inequality with eqtttt0 and supp-Ks1 implies \rho(\pc(\fp),y) < 8D^{\ps(\fp)}.

Using the kernel bound eqkernel-size and the lower bound in supp-Ks we obtain |K_{\ps(\fp)}(x,y)|\le \frac{2^{a^3}}{\mu(B(x,\frac 14 D^{{\ps(\fp)}-1}))}. Using D=2^{100a^2} and the doubling property doublingx 5 +100a^2 times estimates the last display by \frac{2^{5a+101a^3}}{\mu(B(x, 8D^{\ps(\fp)}))}, which, thanks to the closeness of the points x and \pc(\fp) shown in eqtttt0, is in turn bounded by \frac{2^{6a+101a^3}}{\mu(B(\pc(\fp), 8D^{\ps(\fp)}))}.

Using that |e(\mfa)| is bounded by 1 for every \mfa\in \Mf, we estimate with the triangle inequality and the above information | T_{\fp} f(x)| \le \frac{2^{6a+101 a^3}}{\mu(B(\pc(\fp), 8D^{\ps(\fp)}))} \int _{\mu(B(\pc(\fp), 8D^{\ps(\fp)}))} |f(y)|\, dy This together with a\ge 4 proves the lemma.

Proof. We have f=\mathbf{1}_Ff. Using Holder's inequality, we obtain for each x\in B' and each B'\in \mathcal{B} using 1<\tilde{q}\le 2 \frac 1{\mu(B')}\int_{B'} |f(y)|\, d\mu(y) \le \left(\frac 1{\mu(B')}\int_{B'} |f(y)|^{\frac {2{\tilde{q}}}{3\tilde{q}-2}}\, d\mu(y)\right)^{\frac 32-\frac 1{\tilde{q}}} \left(\frac 1{\mu(B')}\int_{B'} \mathbf{1}_F(y)\, d\mu(y)\right)^{\frac 1{\tilde{q}}-\frac 12} \le \left(M_{\mathcal{B}} (|f|^{\frac {2{\tilde{q}}}{3{\tilde{q}}-2}})(x)\right)^{\frac 32-\frac 1{\tilde{q}}} \dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12}. Taking the maximum over all B' containing x, we obtain M_{\mathcal{B}}|f|\le M_{\mathcal{B},\frac {2{\tilde{q}}}{3{\tilde{q}}-2} } |f| \dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12}.

We have with Theorem 1.2.6 \left\|M_{\mathcal{B}, \frac {2{\tilde{q}}}{3{\tilde{q}}-2}} f\right\|_2\le 2^{2a}(3\tilde{q}-2)(2\tilde{q}-2)^{-1}\|f\|_2. Using 1<\tilde{q}\le 2 estimates the last display by 2^{2a+2} (\tilde{q}-1)^{-1} \|f\|_2. We obtain with Cauchy-Schwarz and then Lemma 1.6.1.2 |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)| \le \|g\|_2 \Big\| \sum_{\fp \in \mathfrak{A}} T_{\fp} f \Big\|_2 \le 2^{102a^3}\|g\|_2 \| M_{\mathcal{B}}f \|_2. With eqttt1 and eqttt2 we can estimate the last display by \le 2^{102a^3+2a+2}(\tilde{q}-1)^{-1} \|g\|_2 \|f\|_2\dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12} Using a\ge 4 and (\tilde q - 1)^{-1} = (q+1)/(q-1) \le 3(q-1)^{-1} proves the lemma.

Proof. We write for the expression inside the absolute values on the left-hand side of eqttt3 \sum_{\fp \in \mathfrak{A}}\iint \overline{g(x)} \mathbf{1}_{E(\fp)}(x) {K_{\ps(\fp)}(x,y)}e(\tQ(x)(y) - \tQ(x)(x)) f(y)\, d\mu(y)\,d\mu(x) =\int \sum_{\fp \in \mathfrak{A}} \overline{T_{\fp} ^*g(y)} f(y)\, d\mu(y) with the adjoint operator T_{\fp}^*g(y)=\int_{E(\fp)} \overline{K_{\ps(\fp)}(x,y)}e(-\tQ(x)(y)+ \tQ(x)(x))g(x)\, d\mu(x). We have by expanding the square \int \Big|\sum_{\fp\in \mathfrak{A}}T^*_{\fp}g(y)\Big|^2\, d\mu(y)= \int \left(\sum_{\fp\in \mathfrak{A}} T^*_{\fp}g(y)\right) \left(\sum_{\fp'\in \mathfrak{A}}\overline{T^*_{\fp'}g(y)}\right)\, d\mu(y) \le \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}} \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|. We split the sum into the terms with \ps(\fp')\le \ps(\fp) and \ps(\fp)< \ps(\fp'). Using the symmetry of each summand, we may switch \fp and \fp' in the second sum. Using further positivity of each summand to replace the condition \ps(\fp')< \ps(\fp) by \ps(\fp')\le \ps(\fp) in the second sum, we estimate eqtts1 by \le2 \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}: \ps(\fp')\le \ps(\fp)} \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|.

Define for \fp\in \fP B(\fp):=B(\pc(\fp), 14D^{\ps(\fp)}) and define \mathfrak{A}(\fp):=\{\fp'\in\mathfrak{A}: \ps(\fp')\leq \ps(\fp) \land \scI(\fp') \subset B(\fp)\}. Note that by the squeezing property eq-vol-sp-cube and the doubling property doublingx applied 6 times we have \mu(B(\fp))\le 2^{6a} \mu(\scI(\fp)). Using Lemma 1.6.1.5 and eqttt4, we estimate eqtts2 by \le 2^{232a^3+6a+1} \sum_{\fp\in \mathfrak{A}} \int_{E(\fp)}|g|(y) h(\fp)\, d\mu(y) with h(\fp) defined as \frac 1{\mu(B(\fp))}\int \sum_{\fp'\in \mathfrak{A}(\fp)} {(1+d_{\fp'}(\fcc(\fp'), \fcc(\fp))^{-1/(2a^2+a^3)}}(\mathbf{1}_{E(\fp')}|g|)(y')\, d\mu(y').

Note that p>4 since a\ge 4. Let p' be the dual exponent of p, satisfying 1/p+1/p'=1. We estimate h(\fp) as defined in def-hp with Holder using |g|\le \mathbf{1}_G and E(\fp')\subset B(\fp) by \frac{\|g\mathbf{1}_{B(\fp)}\|_{p'}}{\mu(B(\fp))} \Big\|\sum_{\fp\in\mathfrak{A}(\fp)}(1+d_{\fp}(\fcc(\fp), \fcc(\fp'))^{-1/(2a^2+a^3)}\mathbf{1}_{E(\fp)}\mathbf{1}_G\Big\|_{p}. Then we apply Lemma 1.6.1.6 to estimate this by \le 2^{5a} \frac{\|g\mathbf{1}_{B(\fp)}\|_{p'}}{\mu(B(\fp))} \dens_1(\mathfrak{A})^{\frac 1p}\mu(B(\fp))^{\frac 1p}. Let \mathcal{B}' be the collection of all balls B(\fp) with \fp\in \mathfrak{A}. Then for each \fp\in \mathfrak{A} and x\in B(\fp) we have by definition def-hlm of M_{\mathcal{B}',p'} \|g\mathbf{1}_{B(\fp)}\|_{p'}\le \mu(B(\fp))^{\frac 1{p'}} M_{\mathcal{B}',p'}g(x). Hence we can estimate eqttt5 by \le 2^{5a} (M_{\mathcal{B}', p'}g(x)) \dens_1(\mathfrak{A})^{\frac 1p}.

With this estimate of h(\fp), using E(\fp)\subset B(\fp) by construction of B(\fp), we estimate eqtts3 by \le 2^{232a^3+11a + 1} { \dens_1(\mathfrak{A})^{\frac 1p}}\sum_{\fp\in \mathfrak{A}} \int_{E(\fp)}|g|(y)M_{\mathcal{B}', p'}g(y) \, dy. Using Lemma 1.6.1.1, the last display is observed to be \le 2^{232a^3+11a + 1} {\dens_1(\mathfrak{A})^{\frac 1p}} \int |g|(y)(M_{\mathcal{B}', p'}g)(y) \, dy. Applying Cauchy-Schwarz and using Theorem 1.2.6 and 1<p'<\frac 32 estimates the last display by \le 2^{232a^3+11a + 1} \dens_1(\mathfrak{A})^{\frac 1p} \|g\|_2 \|M_{\mathcal{B}', p'} g\|_2 \le 2^{232a^3+12a+3} \dens_1(\mathfrak{A})^{\frac 1p}\|g\|_2 ^2. Now Lemma 1.6.1.4 follows by applying Cauchy-Schwarz on the left-hand side and using a\ge 4.

Proof of Theorem 1.2.3. We have \left(\frac 1{\tilde{q}} -\frac 12\right) (2-q)= \frac 1q -\frac 12. Multiplying the (2-q)-th power of eqttt9 and the (q-1)-th power of eqttt3 and estimating gives after simplification of some factors \Big|\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)\Big| \le 2^{117a^3}(q-1)^{-1} \dens_1(\mathfrak{A})^{\frac {q-1}{2p}}\dens_2(\mathfrak{A})^{\frac 1{q}-\frac 12} \|f\|_2\|g\|_2. With the definition of p, this implies Theorem 1.2.3.

Proof. If \varphi(y) is not zero, then K_{s_1}(x_1, y) is not zero and thus supp-Ks gives eqt10.

We next have for y with eq-Ks-size |\varphi(y)|\le \frac{2^{204 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))} and for y'\neq y additionally with eq-Ks-smooth |\varphi(y)-\varphi(y')| \le |K_{s_1}(x_1,y)-K_{s_1}(x_1,y')|| K_{s_2}(x_2, y)| +|K_{s_1}(x_1, y')| |K_{s_2}(x_2, y) - K_{s_2}(x_2, y')| \le \frac{2^{229 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))} \left(\left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{1/a}+ \left(\frac{ \rho(y,y')}{D^{s_2}}\right)^{1/a}\right) \le \frac{2^{230 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))} \left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{1/a}. Adding the estimates suppart and holderpart gives eqt11. This proves the lemma.

Proof. Fix \fp and y with tstargnot0. Then there exists x\in E(\fp) with \overline{K_{\ps(\fp)}(x,y)}e(-\tQ(x)(y) +\tQ(x)(x))g(x) \neq 0. As E(\fp)\subset \scI(\fp) and by the squeezing property eq-vol-sp-cube, we have \rho(x,\pc(\fp)) < 4D^{\ps(\fp)}. As K_{\ps(\fp)}(x,y)\neq 0, we have by supp-Ks that \rho(x,y)\le \frac 12 D^{\ps(\fp)}. Now ynotfar follows by the triangle inequality.

Proof. Let i\in \{1,2\}. By definition defineep of E, we have \tQ(x_i)\in \fc(\fp_i). With eq-freq-comp-ball we then conclude d_{\fp_i}(\tQ(x_i),\fcc(\fp_i)) < 1. We have by the triangle inequality and eq-vol-sp-cube that \scI(\fp_1) \subset B(\pc(\fp_2),14D^{\ps(\fp_2)}). Thus, using again eq-vol-sp-cube and the doubling property firstdb d_{\fp_1}(\tQ(x_2), \fcc(\fp_2)) \le 2^{6a} d_{\fp_2}(\tQ(x_2), \fcc(\fp_2)) \le 2^{6a}. By the triangle inequality, we obtain from dponetwo and tgeo0.5 1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2))\le 2 + 2^{6a} +d_{\fp_1}(\tQ(x_1), \tQ(x_2)). As x_1\in \scI(\fp_1) by definition defineep of E, we have by the squeezing property eq-vol-sp-cube d(x_1,\pc(\fp_1))\le 4D^{\ps(\fp_1)} and thus by eq-vol-sp-cube again and the triangle inequality \scI(\fp_1)\subset B(x_1,8D^{\ps(\fp_1)}). We thus estimate the right-hand side of tgeo1 with monotonicity monotonedb of the metrics d_B by \le 2 + 2^{6a} + d_{B(x_1,8D^{\ps(\fp_1)})}(\tQ(x_1), \tQ(x_2)). This is further estimated by applying the doubling property firstdb three times by \le 2 + 2^{6a} +2^{3a}d_{B_1(x_1, D^{\ps(\fp_1)})}(\tQ(x_1), \tQ(x_2)). Now tgeo follows with a\ge 4.

Proof of Lemma 1.6.1.5. We begin with eq-basic-TT*-est. By Lemma 1.6.2.2, the left-hand side of eq-basic-TT*-est vanishes if B(\pc(\fp'),5D^{\ps(\fp')}) \cap B(\pc(\fp),5D^{\ps(\fp)}) = \emptyset. Thus we can assume for the remainder of the proof that B(\pc(\fp'),5D^{\ps(\fp')}) \cap B(\pc(\fp),5D^{\ps(\fp)}) \neq \emptyset. We expand the left-hand side of eq-basic-TT*-est as \left|\int \int_{E(\fp')} \overline{K_{\ps(\fp')}(x_1,y)}e(-\tQ(x_1)(y)+ \tQ(x_1)(x_1))g(x_1)\, d\mu(x_1) \right. \times \left.\int_{E(\fp)} {K_{\ps(\fp)}(x_2,y)}e(\tQ(x_2)(y) -\tQ(x_2)(x_2))\overline{g(x_2)}\, d\mu(x_2)\, d\mu(y)\right|. By Fubini and the triangle inequality and the fact |e(\tQ(x_i)(x_i))|=1 for i=1,2, we can estimate tstartstar and tstartstar' from above by \int_{E(\fp')} \int_{E(\fp)} {\bf I}(x_1, x_2)\, d\mu(x_1)d\mu(x_2), with {\bf I}(x_1, x_2):= \left|\int e(-\tQ(x_1)(y)+\tQ(x_2)(y))\varphi_{x_1,x_2}(y) d\mu(y) \, g(x_1)g(x_2)\right|.

We estimate for fixed x_1\in E(\fp') and x_2\in E(\fp) the inner integral of eqa1 with Theorem 1.2.5. The function \varphi:=\varphi_{x_1,x_2} satisfies the assumptions of Theorem 1.2.5 with z = x_1 and R = D^{s_1} by Lemma 1.6.2.1. We obtain with B':= B(x_1, D^{\ps(\fp')}) {\bf I}(x_1, x_2) \le 2^{8a} \mu(B') \|{\varphi}\|_{C^\tau(B')} (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-1/(2a^2+a^3)}|g(x_1)g(x_2)| \le \frac{2^{231a^3+8a}} {\mu(B(x_2, D^{\ps(\fp)}))} (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-1/(2a^2+a^3)}. Using intersec5B, Lemma 1.6.2.3 and a\ge 1 estimates eqa1.5 by \le \frac{2^{231a^3 + 8a + 1}} {\mu(B(x_2, D^{\ps(\fp)}))} (1+d_{\fp'}(\fcc(\fp'), \fcc(\fp)))^{-1/(2a^2+a^3)}|g(x_1)g(x_2)|. As x_2\in \scI(\fp) by definition defineep of E, we have by eq-vol-sp-cube \rho(x_2,\pc(\fp)) < 4D^{\ps(\fp)} and thus by eq-vol-sp-cube again and the triangle inequality \scI(\fp)\subset B(x_2,8D^{\ps(\fp)}). Using three iterations of the doubling property doublingx give \mu(\scI(\fp))\le 2^{3a}\mu(B(x_2,D^{\ps(\fp)})). With a\ge 4 and eqa2 we conclude eq-basic-TT*-est.

Now assume the left-hand side of eq-basic-TT*-est is not zero. There is a y\in X with T^*_{\fp'}g(y)\overline{T^*_{\fp}g(y)}\neq 0 By the triangle inequality and Lemma 1.6.2.2, we conclude \rho(\pc(\fp),\pc(\fp'))\le \rho(\pc(\fp),y) +\rho(\pc(\fp'),y) \le 5D^{\ps(\fp)}+5D^{\ps(\fp')}\le 10 D^{\ps(\fp)}. By the squeezing property eq-vol-sp-cube and the triangle inequality, we conclude \scI(\fp') \subset B(\pc(\fp), 14D^{\ps(\fp)}). This completes the proof of Lemma 1.6.1.5.