8 Proof of the Hölder cancellative condition

We need the following auxiliary lemma. Recall that \(\tau = 1/a\).

Lemma 8.0.1 Lipschitz Holder approximation

Let \(z\in X\) and \(R{\gt}0\). Let \(\varphi : X \to \mathbb {C}\) be a function supported in the ball \(B:=B(z,R)\) with finite norm \(\| \varphi \| _{C^\tau (B)}\). Let \(0{\lt}t \leq 1\). There exists a function \(\tilde\varphi : X \to \mathbb {C}\), supported in \(B(z,2R)\), such that for every \(x\in X\)

\begin{equation} \label{eq-firstt} |\varphi (x) - \tilde\varphi (x)| \leq t^{\tau } \| \varphi \| _{C^\tau (B)} \end{equation}

8.0.1

and

\begin{equation} \label{eq-secondt} \| \tilde\varphi \| _{\operatorname{\operatorname {Lip}}(B(z,2R))} \leq 2^{4a}t^{-1-a} \| \varphi \| _{C^{\tau }(B)}\, . \end{equation}

8.0.2

Proof ▶

Define for \(x,y\in X\) the Lipschitz and thus measurable function

\begin{equation} L(x,y) := \max \{ 0, 1 - \frac{\rho (x,y)}{tR}\} \, . \end{equation}

8.0.3

We have that \(L(x,y)\neq 0\) implies

\begin{equation} \label{eql01} y\in B(x, tR)\, . \end{equation}

8.0.4

We have for \(y\in B(x, 2^{-1}tR)\) that

\begin{equation} \label{eql30} |L(x,y)|\ge 2^{-1} \ . \end{equation}

8.0.5

Hence

\begin{equation} \int L(x,y) \, \mathrm{d}\mu (y)\ge 2^{-1}\mu (B(x, 2^{-1}tR))\, . \end{equation}

8.0.6

Let \(n\) be the smallest integer so that

\begin{equation} \label{2nt1} 2^nt\ge 1\, . \end{equation}

8.0.7

Iterating \(n+2\) times the doubling condition 1.0.5, we obtain

\begin{equation} \label{eql32} \int L(x,y) \, \mathrm{d}\mu (y)\ge 2^{-1-a(n+2)}\mu (B(x, 2R))\, . \end{equation}

8.0.8

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 8.0.4, we have that \(\tilde{\varphi }\) is supported in \(B(z,2R)\).

We prove 8.0.1. For any \(x\in X\), using that \(L\) is nonnegative,

\begin{equation} \label{eql1} \left(\int L(x,y) \, \mathrm{d}\mu (y)\right) |\varphi (x) - \tilde\varphi (x)| \end{equation}

8.0.9

\begin{equation} \label{eql2} = \left| \int L(x,y)(\varphi (x) - \varphi (y)) \, \mathrm{d}\mu (y)\right|\, . \end{equation}

8.0.10

Using 8.0.4, we estimate the last display by

\begin{equation} \label{eql3} \le \int _{B(x, tR)} L(x,y)|\varphi (x) - \varphi (y)| \, \mathrm{d}\mu (y)\, .\end{equation}

8.0.11

Using the definition of \(\| \varphi \| _{C^\tau (B)}\), we estimate the last display further by

\begin{equation} \label{eql4} \le \left(\int _{B(x, tR)} L(x,y) \rho (x,y)^\tau \, \mathrm{d}\mu (y) \right)\| \varphi \| _{C^\tau (B)}R^{-\tau }\, . \end{equation}

8.0.12

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

\begin{equation} \label{eql5} \le \left(\int L(x,y) \, \mathrm{d}\mu (y)\right) \| \varphi \| _{C^\tau (B)} t^{\tau } \, . \end{equation}

8.0.13

Dividing the string of inequalities from 8.0.9 to 8.0.13 by the positive integral of \(L\) proves 8.0.1.

We turn to 8.0.2. For every \(x\in X\), we have

\begin{equation} \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| \end{equation}

8.0.14

\begin{equation} \le \left|\int L(x,y) \, \mathrm{d}\mu (y)\right| \sup _{x'\in X} |{\varphi }(x')|\ . \end{equation}

8.0.15

As \(\varphi \) is supported on \(B\), dividing by the integral of \(L\), we obtain

\begin{equation} \label{eql42} |\tilde{\varphi }(x)|\le \sup _{x'\in B} |{\varphi }(x')|\le \| \varphi \| _{C^\tau (B)}\ . \end{equation}

8.0.16

If \(\rho (x,x')\ge R\), then we have by the triangle inequality

\begin{equation} \label{eql52} 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)}\, . \end{equation}

8.0.17

Now assume \(\rho (x,x'){\lt} R\). For \(y\in X\) we have by the triangle inequality and a two fold case distinction for the maximum in the definition of \(L\),

\begin{equation} \label{eql10} |L(x,y) - L(x',y)| \le \frac{\rho (x,x')}{tR}. \end{equation}

8.0.18

We compute with 8.0.18, first adding and subtracting a term in the integral,

\begin{equation} \left(\int L(x,y) \, \mathrm{d}\mu (y)\right) |\tilde{\varphi }(x') - \tilde\varphi (x)|= \end{equation}

8.0.19

\begin{equation} \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|\, . \end{equation}

8.0.20

Grouping the second and third and the first and fourth term, we obtain using the definition of \(\tilde\varphi \) and Fubini,

\begin{equation} \label{eql21} \le \left| \int (L(x',y)-L(x,y)) \varphi (y) \, \mathrm{d}\mu (y)\right| \end{equation}

8.0.21

\begin{equation} \label{eql22} + \left| \int L(x,y) \, \mathrm{d}\mu (y)-\int L(x',y) \, \mathrm{d}\mu (y)\right||\tilde{\varphi }(x')| \end{equation}

8.0.22

\begin{equation} \label{eql23} \le 2 \int |L(x,y) -L(x',y)| \, \mathrm{d}\mu (y) \| \varphi \| _{C^\tau (B)}\, , \end{equation}

8.0.23

where in the last inequality we have used 8.0.16. Using further 8.0.18 and the support of \(L\), we estimate the last display by

\begin{equation} \label{eql224}\le 2 \frac{\rho (x,x')}{tR}\mu (B(x,tR)\cup B(x',tR)) \| \varphi \| _{C^\tau (B)}\, . \end{equation}

8.0.24

Using \(\rho (x,x'){\lt}R\) and the triangle inequality, we estimate the last display by

\begin{equation} \label{eql225}\le 2 \frac{\rho (x,x')}{tR} \mu (B(x,2R)) \| \varphi \| _{C^\tau (B)}\, . \end{equation}

8.0.25

Dividing by the integral over \(L\) and using 8.0.8 and 8.0.7, we obtain

\begin{equation} \label{eql226} \frac{R |\tilde{\varphi }(x') - \tilde\varphi (x)|}{\rho (x,x')} \le 2^{2+a(n+2)}t^{-1}\| \varphi \| _{C^\tau (B)} \le 2^{2+3a} t^{-1-a} \| \varphi \| _{C^\tau (B)}\, . \end{equation}

8.0.26

Combining 8.0.17 and 8.0.26 using \(a\ge 4\) and \(t\le 1\) and adding 8.0.16 proves 8.0.2 and completes the proof of Lemma 8.0.1.

We turn to the proof of Proposition 2.0.5.

Proof of Proposition 2.0.5 ▶

Let \(z\in X\) and \(R{\gt}0\) and set \(B=B(z,R)\). Let \(\varphi \) be given as in Proposition 2.0.5. Set

\begin{equation} \label{eql69} t:=(1+d_B({\vartheta },{\theta }))^{-\frac{\tau }{2+a}} \end{equation}

8.0.27

and define \(\tilde{\varphi }\) as in Lemma 8.0.1. Let \({\vartheta }\) and \({\theta }\) be in \({\Theta }\). Then

\begin{equation} \label{eql60} \left|\int e({\vartheta }(x)-{{\theta }(x)}) \varphi (x)\, \mathrm{d}\mu (x)\right| \end{equation}

8.0.28

\begin{equation} \label{eql61} \le \left|\int e({\vartheta }(x)-{{\theta }(x)}) \tilde{\varphi } (x)\, \mathrm{d}\mu (x)\right| \end{equation}

8.0.29

\begin{equation} \label{eql62} + \left|\int e({\vartheta }(x)-{{\theta }(x)}) (\varphi (x)-\tilde{\varphi }(x))\, \mathrm{d}\mu (x)\right| \end{equation}

8.0.30

Using the cancellative condition 1.0.12 of \({\Theta }\) on the ball \(B(z,2R)\), the term 8.0.29 is bounded above by

\begin{equation} \label{eql63} 2^a \mu (B(z,2R)) \| \tilde{\varphi }\| _{\operatorname{\operatorname {Lip}}(B(z,2R))} (1 + d_{B(z,2R)}({\vartheta },{\theta }))^{-\tau } \, . \end{equation}

8.0.31

Using the doubling condition 1.0.5, the inequality 8.0.2, and the estimate \(d_B\le d_{B(z,2R)}\) from the definition, we estimate 8.0.31 from above by

\begin{equation} \label{eql64} 2^{6a}t^{-1-a} \mu (B) \| {\varphi }\| _{C^\tau (B)} (1 + d_{B}({\vartheta },{\theta }))^{-\tau } \, . \end{equation}

8.0.32

The term 8.0.30 we estimate using 8.0.1 and that \({\vartheta }\) and \({\theta }\) are real and thus \(e({\vartheta })\) and \(e({\theta })\) bounded in absolute value by \(1\). We obtain for 8.0.30 with 1.0.5 the upper bound

\begin{equation} \label{eql65} \mu (B(z,2R)) t^{\tau } \| \varphi \| _{C^\tau (B)} \le 2^a \mu (B) t^{\tau } \| \varphi \| _{C^\tau (B)} \, . \end{equation}

8.0.33

Using the definition 8.0.27 of \(t\) and adding 8.0.32 and 8.0.33 estimates 8.0.28 from above by

\begin{equation} 2^{6a} \mu (B) \| {\varphi }\| _{C^\tau (B)} (1 + d_{B}({\vartheta },{\theta }))^{-\frac{\tau }{2+a}} \end{equation}

8.0.34

\begin{equation} + 2^a \mu (B) \| {\varphi }\| _{C^\tau (B)} (1 + d_{B}({\vartheta },{\theta }))^{-\frac{\tau ^2}{2+a}}\, . \end{equation}

8.0.35

\begin{equation} \label{eql66} \le 2^{1+6a} \mu (B) \| {\varphi }\| _{C^\tau (B)} (1 + d_{B}({\vartheta },{\theta }))^{-\frac{\tau ^2}{2+a}} \, , \end{equation}

8.0.36

where we used \(\tau \le 1\). This completes the proof of Proposition 2.0.5.