8 Proof of the Hölder cancellative condition

We need the following auxiliary lemma. Recall that $τ = 1 / a$ .

Lemma 8.0.1 Lipschitz Holder approximation

✓

Let $z \in X$ and $R > 0$ . Let $φ : X \to C$ be a function supported in the ball $B := B (z, R)$ with finite norm $‖ φ ‖_{C^{τ} (B)}$ . Let $0 < t \leq 1$ . There exists a function $\tilde{φ} : X \to C$ , supported in $B (z, 2 R)$ , such that for every $x \in X$

| φ (x) - \tilde{φ} (x) | \leq t^{τ} ‖ φ ‖_{C^{τ} (B)}

8.0.1

and

‖ \tilde{φ} ‖_{Lip (B (z, 2 R))} \leq 2^{4 a} t^{- 1 - a} ‖ φ ‖_{C^{τ} (B)} .

8.0.2

Proof ▶

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

L (x, y) := max {0, 1 - \frac{ρ (x, y)}{t R}} .

8.0.3

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

y \in B (x, t R) .

8.0.4

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

| L (x, y) | \geq 2^{- 1} .

8.0.5

Hence

\int L (x, y) d μ (y) \geq 2^{- 1} μ (B (x, 2^{- 1} t R)) .

8.0.6

Let $n$ be the smallest integer so that

2^{n} t \geq 1 .

8.0.7

Iterating $n + 2$ times the doubling condition 1.0.5, we obtain

\int L (x, y) d μ (y) \geq 2^{- 1 - a (n + 2)} μ (B (x, 2 R)) .

8.0.8

Now define

\tilde{φ} (x) := {(\int L (x, y) d μ (y))}^{- 1} \int L (x, y) φ (y) d μ (y) .

Using that $φ$ is supported in $B (z, R)$ and 8.0.4, we have that $\tilde{φ}$ is supported in $B (z, 2 R)$ .

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

(\int L (x, y) d μ (y)) | φ (x) - \tilde{φ} (x) |

8.0.9

= | \int L (x, y) (φ (x) - φ (y)) d μ (y) | .

8.0.10

Using 8.0.4, we estimate the last display by

\leq \int_{B (x, t R)} L (x, y) | φ (x) - φ (y) | d μ (y) .

8.0.11

Using the definition of $‖ φ ‖_{C^{τ} (B)}$ , we estimate the last display further by

\leq (\int_{B (x, t R)} L (x, y) ρ (x, y)^{τ} d μ (y)) ‖ φ ‖_{C^{τ} (B)} R^{- τ} .

8.0.12

Using the condition on the domain of integration to estimate $ρ (x, y)$ by $t R$ and then expanding the domain by positivity of the integrand, we estimate this further by

\leq (\int L (x, y) d μ (y)) ‖ φ ‖_{C^{τ} (B)} t^{τ} .

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

| \int L (x, y) d μ (y) | | \tilde{φ} (x) | = | \int L (x, y) φ (y) d μ (y) |

8.0.14

\leq | \int L (x, y) d μ (y) | sup_{x^{'} \in X} | φ (x^{'}) | .

8.0.15

As $φ$ is supported on $B$ , dividing by the integral of $L$ , we obtain

| \tilde{φ} (x) | \leq sup_{x^{'} \in B} | φ (x^{'}) | \leq ‖ φ ‖_{C^{τ} (B)} .

8.0.16

If $ρ (x, x^{'}) \geq R$ , then we have by the triangle inequality

R \frac{| \tilde{φ} (x^{'}) - \tilde{φ} (x) |}{ρ (x, x^{'})} \leq 2 sup_{x^{″} \in X} | \tilde{φ} (x^{″}) | \leq 2 ‖ φ ‖_{C^{τ} (B)} .

8.0.17

Now assume $ρ (x, x^{'}) < R$ . 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) | \leq \frac{ρ (x, x^{'})}{t R} .

8.0.18

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

(\int L (x, y) d μ (y)) | \tilde{φ} (x^{'}) - \tilde{φ} (x) | =

8.0.19

| \int L (x, y) \tilde{φ} (x^{'}) - L (x, y) \tilde{φ} (x) + L (x^{'}, y) \tilde{φ} (x^{'}) - L (x^{'}, y) \tilde{φ} (x^{'}) d μ (y) | .

8.0.20

Grouping the second and third and the first and fourth term, we obtain using the definition of $\tilde{φ}$ and Fubini,

\leq | \int (L (x^{'}, y) - L (x, y)) φ (y) d μ (y) |

8.0.21

+ | \int L (x, y) d μ (y) - \int L (x^{'}, y) d μ (y) | | \tilde{φ} (x^{'}) |

8.0.22

\leq 2 \int | L (x, y) - L (x^{'}, y) | d μ (y) ‖ φ ‖_{C^{τ} (B)},

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

\leq 2 \frac{ρ (x, x^{'})}{t R} μ (B (x, t R) \cup B (x^{'}, t R)) ‖ φ ‖_{C^{τ} (B)} .

8.0.24

Using $ρ (x, x^{'}) < R$ and the triangle inequality, we estimate the last display by

\leq 2 \frac{ρ (x, x^{'})}{t R} μ (B (x, 2 R)) ‖ φ ‖_{C^{τ} (B)} .

8.0.25

Dividing by the integral over $L$ and using 8.0.8 and 8.0.7, we obtain

\frac{R | \tilde{φ} (x^{'}) - \tilde{φ} (x) |}{ρ (x, x^{'})} \leq 2^{2 + a (n + 2)} t^{- 1} ‖ φ ‖_{C^{τ} (B)} \leq 2^{2 + 3 a} t^{- 1 - a} ‖ φ ‖_{C^{τ} (B)} .

8.0.26

Combining 8.0.17 and 8.0.26 using $a \geq 4$ and $t \leq 1$ 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 $R > 0$ and set $B = B (z, R)$ . Let $φ$ be given as in Proposition 2.0.5. Set

t := (1 + d_{B} (ϑ, θ))^{- \frac{τ}{2 + a}}

8.0.27

and define $\tilde{φ}$ as in Lemma 8.0.1. Let $ϑ$ and $θ$ be in $Θ$ . Then

| \int e (ϑ (x) - θ (x)) φ (x) d μ (x) |

8.0.28

\leq | \int e (ϑ (x) - θ (x)) \tilde{φ} (x) d μ (x) |

8.0.29

+ | \int e (ϑ (x) - θ (x)) (φ (x) - \tilde{φ} (x)) d μ (x) |

8.0.30

Using the cancellative condition 1.0.12 of $Θ$ on the ball $B (z, 2 R)$ , the term 8.0.29 is bounded above by

2^{a} μ (B (z, 2 R)) ‖ \tilde{φ} ‖_{Lip (B (z, 2 R))} (1 + d_{B (z, 2 R)} (ϑ, θ))^{- τ} .

8.0.31

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

2^{6 a} t^{- 1 - a} μ (B) ‖ φ ‖_{C^{τ} (B)} (1 + d_{B} (ϑ, θ))^{- τ} .

8.0.32

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

μ (B (z, 2 R)) t^{τ} ‖ φ ‖_{C^{τ} (B)} \leq 2^{a} μ (B) t^{τ} ‖ φ ‖_{C^{τ} (B)} .

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

2^{6 a} μ (B) ‖ φ ‖_{C^{τ} (B)} (1 + d_{B} (ϑ, θ))^{- \frac{τ}{2 + a}}

8.0.34

+ 2^{a} μ (B) ‖ φ ‖_{C^{τ} (B)} (1 + d_{B} (ϑ, θ))^{- \frac{τ^{2}}{2 + a}} .

8.0.35

\leq 2^{1 + 6 a} μ (B) ‖ φ ‖_{C^{τ} (B)} (1 + d_{B} (ϑ, θ))^{- \frac{τ^{2}}{2 + a}},

8.0.36

where we used $τ \leq 1$ . This completes the proof of Proposition 2.0.5. □