3 Proof of Metric Space Carleson

In this section we prove Theorem 1.1.1 and Theorem 1.1.2.

Let \((X, \rho , \mu , a)\) be a doubling metric measure space and \(\Theta \) a cancellative, compatible collection of functions on \(X\). Let \(K\) be a one-sided Calderon-Zygmund kernel.

We begin by proving some continuity properties of the integrand in 1.1.14.

Lemma 3.0.1 int continuous

✓

Let \(f\) be a measurable function with \(|f| \le 1\). Then the function

\[ G: X \times \Theta \times (0,\infty ) \times (0, \infty ) \to \mathbb {C} \]

\[ G(x, {\vartheta }, R_1, R_2) := \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \]

is continuous in \({\vartheta }\) for fixed \(x, R_1, R_2\). It is right-continuous in \(R_1\) for fixed \(x, \vartheta , R_2\) and left-continuous in \(R_2\) for fixed \(x, \vartheta , R_1\). Finally, it is measurable in \(x\) and bounded for fixed \(\vartheta , R_1, R_2\).

Proof ▶

Measurability in \(x\) follows from joint measurability of

\[ K(x,y) \mathbf{1}_{B(x,R_2) \setminus B(x,R_1)}(y) \]

in \(x\) and \(y\) and (part of the proof of) Fubini’s theorem.

(Partial) continuity in \(R_1\) and \(R_2\) is also a standard lemma in measure theory. It follows for example by splitting the integrand as \(F_1 - F_{-1} + iF_i - iF_{-i}\) for positive, disjointly supported functions \(F_{-}\) and applying the monotone convergence theorem to each summand.

For continuity in \({\vartheta }\) note that

\[ |G(x, {\vartheta }, R_1, R_2) - G(x, {\vartheta }', R_1, R_2)| \]

\[ = \left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) (e({\vartheta }(y)) - e({\vartheta }'(y))) \, \mathrm{d}\mu (y) \right| \]

\[ \le \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} |K(x,y)| |f(y)| |e({\vartheta }(y) - {\vartheta }'(y) - {\vartheta }(o) + {\vartheta }'(o)) - 1| \, \mathrm{d}\mu (y). \]

By \(1\)-Lipschitz continuity of \(e^{ix}\), the property 1.1.4 of the metrics \(d\), the kernel upper bound 1.1.11 and \(|f| \le 1\) this is

\[ \le \mu (B(x, R_2)) \sup _{R_1 {\lt} \rho (x,y) {\lt} R_2} \frac{2^{a^3}}{V(x,y)} d_{B(x,\rho (o,x)+R_2)}(\vartheta , \vartheta '). \]

If \(R_1 {\lt} \rho (x,y) {\lt} R_2\) then there exists \(n\) with \(B(x,R_2) \subset B(x, 2^n \rho (x,y))\) and \(2^n \le 2 R_2/R_1\). Hence, by the doubling property 1.1.2,

\[ V(x,y) = \mu (B(x, \rho (x,y))) \ge 2^{-an} \mu (B(x, R_2)) \ge (2R_2 / R_1)^{-a} \mu (B(x, R_2)). \]

Hence

\[ |G(x, {\vartheta }, R_1, R_2) - G(x, {\vartheta }', R_1, R_2)| \le 2^{a^3} \Big(\frac{2R_2}{R_1}\Big)^a d_{B(x, \rho (o,x)+R_2)}({\vartheta }, {\vartheta }'). \]

Since the topology on \({\Theta }\) is the one induced by any of the metrics \(d_B\), continuity follows.

Finally, for boundedness as a function of \(x\) note that we also have by a similar computation using \(|e({\vartheta })|=1\)

\[ |G(x, {\vartheta }, R_1, R_2)| \le 2^{a^3} \Big(\frac{2R_2}{R_1}\Big)^a. \]

We now prove Theorem 1.1.1 using Theorem 1.1.2.

Proof of Theorem 1.1.1 ▶

Let Borel sets \(F\), \(G\) in \(X\) be given. Let a Borel function \(f: X \to \mathbb {C}\) with \(f \le \mathbf{1}_F\) be given.

Let \({\Theta }' \subset {\Theta }\) be a countable dense set. By Lemma 3.0.1 we have

\[ \sup _{{\vartheta }\in {\Theta }} \sup _{0 {\lt} R_1 {\lt} R_2}\left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \right| \]

\[ = \sup _{{\vartheta }\in {\Theta }'} \sup _{0 {\lt} R_1 {\lt} R_2, R_i \in \mathbb {Q}}\left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \right|. \]

Consider an enumeration of \({\Theta }'\) and let \({\Theta }_n\) be the finite set consisting of the first \(n+1\) functions in the enumeration. Then by the monotone convergence theorem

\[ \Big| \int _G Tf \, \mathrm{d}\mu \Big| \]

\[ = \int _G \sup _{{\vartheta }\in {\Theta }'} \sup _{0 {\lt} R_1 {\lt} R_2, R_i \in \mathbb {Q}}\left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \right| \, \mathrm{d}\mu (x) \]

\[ = \lim _{n \to \infty } \int _G \sup _{{\vartheta }\in {\Theta }_n} \sup _{0 {\lt} R_1 {\lt} R_2, R_i \in \mathbb {Q}}\left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \right| \, \mathrm{d}\mu (x). \]

For each \(n\), let \(Q_n(x)\) be the measurable function specifying the maximizer in the supremum in \({\vartheta }\) in the previous line. It is possible to construct such a measurable function of \(x\), because the function inside the supremum is measurable in \(x\), being a countable supremum of measurable functions, and because \({\Theta }_n\) is finite. (For example one may fix some order on the finite set \({\Theta }_n\) and pick the smallest maximizer with respect to that order. This is a measurable choice function.) Then the previous display becomes

\[ \le \lim _{n \to \infty } \int _G \sup _{0 {\lt} R_1 {\lt} R_2} \left| \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e(Q_n(x)(y)) \, \mathrm{d}\mu (y) \right| \, \mathrm{d}\mu (x). \]

\[ = \lim _{n \to \infty } \int _G T_{Q_n} f \, \mathrm{d}\mu . \]

It remains to verify the assumptions of Theorem 1.1.2, which when applied here completes the proof.

The assumptions of Theorem 1.1.1 and Theorem 1.1.2 are the same, with the exception of the assumption 1.1.20. The assumption 1.1.20 however is weaker than the assumption 1.1.15 of Theorem 1.1.1. Indeed, setting for fixed \(x, {\vartheta }\) the outer radius \(R_2\) in 1.1.13 to \(\min \{ R_2', R_Q({\vartheta }, x')\} \), where \(R_2'\) is the outer radius in 1.1.18, shows that 1.1.18 is smaller than or equal to 1.1.13. Thus we can apply Theorem 1.1.2, which completes the proof.

We continue with the proof of Theorem 1.1.2, via a series of reductions to simpler lemmas.

Let a measurable function \(Q\) with finite range be given.

Proof of Theorem 1.1.2 ▶

Let Borel sets \(F\), \(G\) in \(X\) with finite measure be given. Let a Borel function \(f: X \to \mathbb {C}\) with \(f \le \mathbf{1}_F\) be given.

For each \(0 {\lt} R_1,R_2,R\), we define \(T_{R_1,R_2,R}f\) as in 3.0.2. By Lemma 3.0.1, \(T_{R_1,R_2,R}f\) is measurable and bounded, and we clearly have for each \(x \in X\)

\[ T_Q f(x) = \lim _{n \to \infty } \sup _{2^{-n} {\lt} R_1 {\lt} R_2 {\lt} 2^n} T_{R_1, R_2, 2^n}f(x). \]

For each \(x\) and all \(f\), the functions \(\sup _{2^{-n} {\lt} R_1 {\lt} R_2 {\lt} 2^n} T_{R_1, R_2, 2^n} f(x)\) are measurable by Lemma 3.0.1 and form an increasing sequence in \(n\). By the monotone convergence theorem, the claimed estimate 1.1.21 then follows from Lemma 3.0.2.

Lemma 3.0.2 R truncation

✓

Let \(F\), \(G\) be Borel sets in \(X\). Let \(f:X\to {\mathbb {C}}\) be a Borel function with \(|f|\le 1_F\). Then for all \(R\in 2^{\mathbb {N}}\) we have

\begin{equation} \label{Rcut} \int \mathbf{1}_G \sup _{1/R{\lt}R_1{\lt}R_2{\lt}R} |T_{R_1,R_2,R} f(x)|\, d\mu (x) \le \frac{2^{443a^3}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}, \end{equation}

3.0.1

where

\begin{equation} \label{TRR} T_{R_1,R_2,R} f(x)= \mathbf{1}_{B(o,R)}(x) \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu (y) . \end{equation}

3.0.2

Proof ▶

Let \(F,G,f\) as in the lemma be given. Let \(R\in 2^{\mathbb {N}}\) be given. By replacing \(G\) with \(G\cap B(o,R)\) if necessary, a replacement that does not change the conclusion of the Lemma 3.0.2, it suffices to show 3.0.1 under the assumption that \(G\) is contained in \(B(o,R)\) and thus bounded. We make this assumption. For every \(x\in G\) and \(R_2 {\lt} R\), the domain of integration in 3.0.2 is contained in \(B(o,2R)\). By replacing \(F\) with \(F\cap B(o,2R)\) if necessary, and correspondingly restricting \(f\) to \(B(o, 2R)\), it suffices to show 3.0.1 under the assumption that \(F\) is contained in \(B(o,2R)\) and thus bounded. We make this assumption.

Using the definition 2.0.5 of \(K_s\) and the partition of unity 2.0.4, we observe that for fixed \(R_1{\lt}R_2\) we have

\begin{equation} \label{KKs} K(x,y)\mathbf{1}_{R_1{\lt}\rho (x,y){\lt}R_2}=\sum _{s=s_1-2}^{s_2+2} K_s(x,y) \mathbf{1}_{R_1{\lt}\rho (x,y){\lt}R_2}, \end{equation}

3.0.3

where \(s_1=\lfloor \log _D 2R_1\rfloor +3\) and \(s_2=\lceil \log _D 4R_2\rceil -2\). To obtain the identity 3.0.3, we have used that on the set where \(R_1{\lt}\rho (x,y){\lt}R_2\) the kernels \(K_s\) vanish unless \(s\) is inside the interval of summation in 3.0.3. Similarly, we observe

\begin{equation} \label{KsrhoKs} \sum _{s=s_1}^{s_2} K_s(x,y) \mathbf{1}_{R_1{\lt}\rho (x,y){\lt}R_2} = \sum _{s=s_1}^{s_2} K_s(x,y), \end{equation}

3.0.4

because on the support of \(K_s\) with \(s_1\le s\le s_2\) we have necessarily \(R_1{\lt}\rho (x,y){\lt}R_2\). We thus express 3.0.2 as the sum of

\begin{equation} \label{middles} T_{s_1,s_2}f(x):=\sum _{s_1 \le s\le s_2} \int K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu (y) \end{equation}

3.0.5

and

\begin{equation} \label{boundarys} \sum _{s=s_1-2,s_1-1, s_2+1, s_2+2} \int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu (y), \end{equation}

3.0.6

We apply the triangle inequality and estimate 3.0.1 separately with \(T_{R_1,R_2,R}\) replaced by 3.0.5 and by 3.0.6. To handle the case 3.0.5, we employ Lemma 3.0.3. Here, we utilize the fact that if \(\frac1R\le R_1\le R_2\le R\), then \(s_1\) and \(s_2\) as in 3.0.5 are in an interval \([-S,S]\) for some sufficiently large \(S\) depending on \(R\). To handle the case 3.0.6, we use the triangle inequality and the properties 2.1.2, 2.1.3 of \(K_s\) and \(|f| \le \mathbf{1}_F\) to obtain for arbitrary \(s\) the inequality

\begin{multline} \left|\int _{R_1 {\lt} \rho (x,y) {\lt} R_2} K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu (y)\right|\\ \leq \frac{2^{102 a^3}}{\mu (B(x, D^{s}))} \int _{B(x, D^s)} \mathbf{1}_F(y) \, \mathrm{d}\mu (y) \leq 2^{102 a^3} M\mathbf{1}_F(x), \end{multline}

where \(M\mathbf{1}_F\) is as defined in Proposition 2.0.6. Now the left-hand side of 3.0.1, with \(T_{R_1,R_2,R}\) replaced by 3.0.6, can be estimated using Hölder’s inequality and Proposition 2.0.6 by

\[ 2^{102a^3+2}\int \mathbf{1}_{G}(x) M\mathbf{1}_F(x)\, d\mu (x) \le \frac{2^{102a^3+4a+3}q}{q-1}\mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}\, . \]

Summing the contributions from 3.0.5 and 3.0.6 completes the proof.

Lemma 3.0.3 S truncation

✓

Let \(F\), \(G\) be bounded Borel sets in \(X\). Let \(f:X\to {\mathbb {C}}\) be a Borel function with \(|f|\le 1_F\). Then for all \(S\in {\mathbb {Z}}\) we have

\begin{equation} \label{Scut} \int \mathbf{1}_G(x) \sup _{-S\le s_1\le s_2\le S} |T_{s_1,s_2} f(x)|\, d\mu (x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}, \end{equation}

3.0.9

where

\begin{equation} \label{Tss} T_{s_1,s_2} f(x) = \sum _{s_1\le s \le s_2} \int _X K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu (y). \end{equation}

3.0.10

Proof of Lemma 3.0.3 ▶

We reduce Lemma 3.0.3 to Lemma 3.0.4. For each \(x\), let \(\sigma _1(x)\) be the minimal element \(s'\in [-S,S]\) such that

\[ \max _{s'\le s_2\le S} |T_{s',s_2} f(x)| = \max _{-S\le s_1\le s_2\le S} |T_{s_1,s_2} f(x)| =: T_{1,x}. \]

Similarly, let \({\sigma }_2(x)\) be the minimal element \(s''\in [-S,S]\) such that

\[ |T_{\sigma _2(x), s''} f(x)| = T_{1,x}\, . \]

With these choices, and noting that with the definition of \(T_{2, \sigma _1, \sigma _2}\) from 3.0.12

\begin{equation*} T_{\sigma _1(x),\sigma _2(x)} f(x)=T_{2,\sigma _1,\sigma _2} f(x), \end{equation*}

we conclude that the left-hand side of 3.0.9 and 3.0.11 are equal. Therefore, Lemma 3.0.3 follows from Lemma 3.0.4.

Lemma 3.0.4 linearized truncation

✓

Let \(\sigma _1,\sigma _2\colon X\to \mathbb {Z}\) be measurable functions with finite range and \(\sigma _1\leq \sigma _2\). Then we have

\begin{equation} \label{Sqlin} \int \mathbf{1}_{G}(x) \left| {T}_{2,\sigma _1,\sigma _2}f(x)\right|\, d\mu (x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}, \end{equation}

3.0.11

with

\begin{equation} \label{middles1} {T}_{2,\sigma _1,\sigma _2}f(x)=\sum _{\sigma _1(x) \le s\le \sigma _2(x)} \int K_s(x,y) f(y) e({Q}(x)(y)) \, \mathrm{d}\mu (y)\, . \end{equation}

3.0.12

Proof of Lemma 3.0.4 ▶

Fix \(\sigma _1\), \(\sigma _2\) and \({Q}\) as in the lemma. Applying Proposition 2.0.1 recursively, we obtain a sequence of sets \(G_n\) with \(G_0=G\) and, for each \(n\ge 0\), \(G_{n+1} \subset G_n\), \(\mu (G_{n})\le 2^{-n} \mu (G)\) and

\begin{equation*} \int \mathbf{1}_{G_{n}\setminus G_{n+1}}(x) \left| {T}_{2,\sigma _1,\sigma _2} f(x) \right|\, d\mu (x) \end{equation*}

\begin{equation} \le \frac{2^{442a^3}}{(q-1)^5} \mu (G_n)^{1 - \frac{1}{q}} \mu (F)^{\frac{1}{q}}. \end{equation}

3.0.13

Adding the first \(n\) of these inequalities, we obtain by bounding a geometric series

\begin{equation} \label{Sqcut2} \int \mathbf{1}_{G\setminus G_{n}}(x) \left| {T}_{2,\sigma _1,\sigma _2}f(x) \right|\, d\mu (x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}. \end{equation}

3.0.14

As the integrand is non-negative and non-decreasing in \(n\), we obtain by the monotone convergence theorem

\begin{equation} \label{Sqcut3} \int \mathbf{1}_{G}(x) \left| {T}_{2,\sigma _1,\sigma _2}f(x) \right|\, d\mu (x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}. \end{equation}

3.0.15

This completes the proof of Lemma 3.0.4 and thus Theorem 1.1.1.