Proof. Define for x,y\in X the Lipschitz and thus measurable function L(x,y) := \max\{0, 1 - \frac{\rho(x,y)}{tR}\}. We have that L(x,y)\neq 0 implies y\in B(x, tR). We have for y\in B(x, 2^{-1}tR) that |L(x,y)|\ge 2^{-1}. Hence \int L(x,y) \, \mathrm{d}\mu(y)\ge 2^{-1}\mu(B(x, 2^{-1}tR)). Let n be the smallest integer so that 2^n t\ge 1. Iterating n+2 times the doubling condition doublingx, we obtain \int L(x,y) \, \mathrm{d}\mu(y)\ge 2^{-1-a(n+2)}\mu(B(x, 2R)).

Now define \tilde \varphi(x) := \left(\int L(x,y) \, \mathrm{d}\mu(y)\right)^{-1}\int L(x,y) \varphi(y) \, \mathrm{d}\mu(y). Using that \varphi is supported in B(z,R) and eql01, we have that \tilde{\varphi} is supported in B(z,2R).

We prove eq-firstt. For any x\in X, using that L is nonnegative, \left(\int L(x,y) \, \mathrm{d}\mu(y)\right) |\varphi(x) - \tilde \varphi(x)| = \left| \int L(x,y)(\varphi(x) - \varphi(y)) \, \mathrm{d}\mu(y)\right|. Using eql01, we estimate the last display by \le \int_{B(x, tR)} L(x,y)|\varphi(x) - \varphi(y)| \, \mathrm{d}\mu(y). We claim that in this integral, |\varphi(x) - \varphi(y)|\le \rho(x,y)^\tau \|\varphi\|_{C^\tau(B(z, 2R))}(2R)^{-\tau}. Indeed, if x or y does not belong to B(z, 2R), then the other point is not in B(z,R) as \rho(x,y)\le tR \le R. Therefore, \varphi(x)=\varphi(y)=0 since \varphi is supported in B (z, R). Otherwise, both points are in B(z, 2R), and the inequality follows from the definition of \|\varphi\|_{C^\tau(B(z, 2R))}.

Therefore, we can estimate the last display further by \le \left(\int_{B(x, tR)} L(x,y) \rho(x,y)^\tau \, \mathrm{d}\mu(y) \right)\|\varphi\|_{C^\tau(B(z, 2R))}(2R)^{-\tau}. Using the condition on the domain of integration to estimate \rho(x,y) by tR and then expanding the domain by positivity of the integrand, we estimate this further by \le \left(\int L(x,y) \, \mathrm{d}\mu(y)\right) \|\varphi\|_{C^\tau(B(z, 2R))} (t/2)^{\tau}. Dividing the string of inequalities from eql1 to eql5 by the positive integral of L proves eq-firstt.

We turn to eq-secondt. For every x\in X, we have \left|\int L(x,y) \, \mathrm{d}\mu(y)\right||\tilde{\varphi}(x)| =\left|\int L(x,y) {\varphi}(y)\, \mathrm{d}\mu(y)\right| \le \left|\int L(x,y) \, \mathrm{d}\mu(y)\right| \sup_{x'\in X} |{\varphi}(x')|. As \varphi is supported on B, dividing by the integral of L, we obtain |\tilde{\varphi}(x)|\le \sup_{x'\in B} |{\varphi}(x')|\le \|\varphi\|_{C^\tau(B(z, 2R))}. If \rho(x,x')\ge R, then we have by the triangle inequality R\frac{|\tilde{\varphi}(x') - \tilde \varphi(x)|}{\rho(x,x')} \le 2\sup_{x''\in X} |\tilde{\varphi}(x'')|\le 2\|\varphi\|_{C^\tau(B(z, 2R))}.

Now assume \rho(x,x')< R. For y\in X we have by the triangle inequality and a two fold case distinction for the maximum in the definition of L, |L(x,y) - L(x',y)| \le \frac{\rho(x,x')}{tR}. We compute with eql10, first adding and subtracting a term in the integral, \left(\int L(x,y) \, \mathrm{d}\mu(y)\right) |\tilde{\varphi}(x') - \tilde \varphi(x)|= \left|\int L(x,y) \tilde{\varphi}(x') -L(x,y) \tilde{\varphi}(x) +L(x',y) \tilde{\varphi}(x')- L(x',y) \tilde{\varphi}(x') \, \mathrm{d}\mu(y)\right|. Grouping the second and third and the first and fourth term, we obtain using the definition of \tilde \varphi and Fubini, \le \left| \int (L(x',y)-L(x,y)) \varphi(y) \, \mathrm{d}\mu(y)\right| + \left| \int L(x,y) \, \mathrm{d}\mu(y)-\int L(x',y) \, \mathrm{d}\mu(y)\right||\tilde{\varphi}(x')| \le 2 \int |L(x,y) -L(x',y)| \, \mathrm{d}\mu(y) \|\varphi\|_{C^\tau(B(z, 2R))}, where in the last inequality we have used eql42. Using further eql10 and the support of L, we estimate the last display by \le 2 \frac{\rho(x,x')} {tR}\mu(B(x,tR)\cup B(x',tR)) \|\varphi\|_{C^\tau(B(z, 2R))}. Using \rho(x,x')<R and the triangle inequality, we estimate the last display by \le 2 \frac{\rho(x,x')} {tR} \mu(B(x,2R)) \|\varphi\|_{C^\tau(B(z, 2R))}. Dividing by the integral over L and using eql32 and 2nt1, we obtain \frac {R |\tilde{\varphi}(x') - \tilde \varphi(x)|}{\rho(x,x')} \le 2^{2+a(n+2)}t^{-1}\|\varphi\|_{C^\tau(B(z, 2R))} \le 2^{2+3a} t^{-1-a} \|\varphi\|_{C^\tau(B(z, 2R))}. Combining eql52 and eql226 using a\ge 4 and t\le 1 and adding eql42 proves eq-secondt and completes the proof of Lemma 1.8.1.

Proof of Theorem 1.2.5. Let z\in X and R>0 and set B=B(z,R). Let \varphi be given as in Theorem 1.2.5. Set t:=(1+d_B(\mfa,\mfb))^{-\frac{\tau}{2+a}} and define \tilde{\varphi} as in Lemma 1.8.1. Let \mfa and \mfb be in \Mf. Then \left|\int e(\mfa(x)-{\mfb(x)}) \varphi (x)\, \mathrm{d}\mu(x)\right| \le \left|\int e(\mfa(x)-{\mfb(x)}) \tilde{\varphi} (x)\, \mathrm{d}\mu(x)\right| + \left|\int e(\mfa(x)-{\mfb(x)}) (\varphi (x)-\tilde{\varphi}(x))\, \mathrm{d}\mu(x)\right|.

Using the cancellative condition eq-vdc-cond of \Mf on the ball B(z,2R), the term eql61 is bounded above by 2^a \mu(B(z,2R)) \|\tilde{\varphi}\|_{\Lip(B(z,2R))} (1 + d_{B(z,2R)}(\mfa,\mfb))^{-\tau}. Using the doubling condition doublingx, the inequality eq-secondt, and the estimate d_B\le d_{B(z,2R)} from the definition, we estimate eql63 from above by 2^{6a}t^{-1-a} \mu(B) \|{\varphi}\|_{C^\tau(B)} (1 + d_{B}(\mfa,\mfb))^{-\tau}.

The term eql62 we estimate using eq-firstt and that \mfa and \mfb are real and thus e(\mfa) and e(\mfb) bounded in absolute value by 1. We obtain for eql62 with doublingx the upper bound 2^a \mu(B) t^{\tau} \|\varphi\|_{C^\tau(B)}.

Using the definition eql69 of t and adding eql64 and eql65 estimates eql60 from above by 2^{6a} \mu(B) \|{\varphi}\|_{C^\tau(B)} (1 + d_{B}(\mfa,\mfb))^{-\frac{\tau}{2+a}} + 2^a \mu(B) \|{\varphi}\|_{C^\tau(B)} (1 + d_{B}(\mfa,\mfb))^{-\frac{\tau^2}{2+a}}. This is \le 2^{1+6a} \mu(B) \|{\varphi}\|_{C^\tau(B)} (1 + d_{B}(\mfa,\mfb))^{-\frac{\tau^2}{2+a}}, where we used \tau\le 1. This completes the proof of Theorem 1.2.5.