3 Proof of Metric Space Carleson

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. We have that

\begin{equation} X=\bigcup _{R{\gt}0}B(o,R), \end{equation}

3.0.1

because every point of \(X\) has finite distance from \(o\).

Lemma 3.0.1 R truncation

For all integers \(R{\gt}0\)

\[ \int \mathbf{1}_{G\cap B(o,R)} \sup _{1/R{\lt}R_1{\lt}R_2{\lt}R}\sup _{{\vartheta }\in {\Theta }} \left| T_{R_1,R_2,{\vartheta }} f(x) \right|\, d\mu (x) \]

\begin{equation} \label{Rcut} \leq \frac{2^{450a^3}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}, \end{equation}

3.0.2

where

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

3.0.3

We first show how Lemma 3.0.1 implies Theorem 1.0.2. As \(R\) tends to \(\infty \), the integrand of the left-hand side of 3.0.2 grows monotonically toward the integrand of the left-hand side of 1.0.19 for all \(x\). By Lebesgue’s monotone convergence theorem, the left-hand side of 3.0.2 converges to the left-hand side of 1.0.19. This verifies Theorem 1.0.2.

It remains to prove Lemma 3.0.1. Fix an integer \(R{\gt}0\). By replacing \(G\) with \(G\cap B(o,R)\) if necessary, it suffices to show 3.0.2 under the assumption that \(G\) is contained in \(B(o,R)\). We make this assumption. For every \(x\in G\), the domain of integration in 3.0.3 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 3.0.2 under the assumption that \(F\) is contained in \(B(o,2R)\). We make this assumption.

Using the definition 2.0.5 of \(K_s\) and the partition of unity 2.0.4, we express 3.0.3 as the sum of

\begin{equation} \label{middles} {T}_{1,s_1,s_2,{\vartheta }}f(x):=\sum _{s_1 \le s\le s_2} \int K_s(x,y) f(y) e({\vartheta }(y)) \, \mathrm{d}\mu (y) \end{equation}

3.0.4

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({\vartheta }(y)) \, \mathrm{d}\mu (y), \end{equation}

3.0.5

where \(s_1\) is the smallest integer such that \(D^{s_1-2}R_2{\gt}\frac1{4D}\) and \(s_2\) is the largest integer such that \(D^{s_2+2}R_1{\lt}\frac12\). We restrict the summation index \(s\) by excluding summands with \(s{\lt}s_1-2\) or \(s{\gt}s_2+2\) because for these summands, the function \(K_s\) vanishes on the domain of integration. We also omit the restriction in the integral for the summands in 3.0.4 because in these summands, the support of \(K_s\) is contained in the set described by this restriction.

We apply the triangle inequality and estimate the versions of 3.0.2 separately with \(T_{R_1,R_2,{\vartheta }}\) replaced by 3.0.4 and by each summand of 3.0.5. To handle the case 3.0.4, we employ the following lemma. Here, we utilize the fact that if \(\frac1R\le R_1\le R_2\le R\), then \(s_1\) and \(s_2\) as in 3.0.4 are in an interval \([-S,S]\) for some sufficiently large \(S\) depending on \(R\).

Lemma 3.0.2 S truncation

For all integers \(S{\gt}0\)

\[ \int \mathbf{1}_{G}(x) \max _{-S{\lt}s_1\le s_2{\lt}S}\sup _{{\vartheta }\in {\Theta }} \left| T_{1, s_1,s_2,{\vartheta }} f(x) \right|\, d\mu (x) \]

\begin{equation} \label{Scut} \leq \frac{2^{446a^3}}{(q-1)^6} \mu (G)^{1 - \frac{1}{q}} \mu (F)^{\frac{1}{q}}, \end{equation}

3.0.6

where \(T_{1, s_1,s_2,{\vartheta }}\) is defined in 3.0.4.

To reduce Lemma 3.0.1 to Lemma 3.0.2, we need estimates for the summands in 3.0.5. Using Lemma 2.1.3, we 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({\vartheta }(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 3.0.2, with \(T_{R_1,R_2,{\vartheta }}\) replaced by a summand of 3.0.5, can be estimated using Hölder’s inequality and Proposition 2.0.6 by

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

Applying the triangle inequality to estimate the left-hand side of 3.0.2 by contributions from the summands in 3.0.4 and 3.0.5, using Lemma 3.0.2 to control the first term, and the above to estimate the contribution from the four summands in 3.0.5, combined with \(a\geq 4\) and \(q {\lt} 2\), completes the reduction of Lemma 3.0.1 to Lemma 3.0.2.

It remains to prove Lemma 3.0.2. Fix \(S{\gt}0\).

Lemma 3.0.3 finitary S truncation

For all finite sets \(\tilde{{\Theta }}\subset {\Theta }\)

\[ \int \mathbf{1}_{G}(x) \max _{-S{\lt}s_1\le s_2{\lt}S}\sup _{{\vartheta }\in \tilde{{\Theta }}} \left| T_{1, s_1,s_2,{\vartheta }} f(x) \right|\, d\mu (x) \]

\begin{equation} \label{Sqcut} \leq \frac{2^{445a^3}}{(q-1)^6} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}\, . \end{equation}

3.0.9

We reduce Lemma 3.0.2 to Lemma 3.0.3. By the Lebesgue monotone convergence theorem, applied to an increasing sequence of finite sets \(\tilde{{\Theta }}\), inequality 3.0.9 continues to hold for countable \(\tilde{{\Theta }}\).

Let \(\epsilon =\frac{1}{2S+1}\). Pick some \({\vartheta }_0 \in {\Theta }\). For \(k \ge 0\), let the set \(\tilde{{\Theta }}_k\) be a subset of \(B_{B(o,2R)}({\vartheta }_0, k)\) of maximal size, such that for all \({\vartheta }, {\theta }\in \tilde{{\Theta }}_k\), it holds that \(d_{B(o, 2R)}({\vartheta }, {\theta }) \ge \epsilon \). Such a set exists, since by Lemma 2.1.1 there exists an upper bound for the size of such subsets in \(B_{B(o, 2R)}({\vartheta }_0, k)\). Define

\[ \tilde{{\Theta }} := \bigcup _{k \in \mathbb {N}} \tilde{{\Theta }}_k\, . \]

Then the set \(\tilde{{\Theta }}\) is at most countable, and it has the property that for any \({\theta }\in {\Theta }\), there exists \({\vartheta }\in \tilde{{\Theta }}\) with

\[ d_{B(o, 2R)}({\theta }, {\vartheta }) {\lt} \epsilon \, . \]

For every \({\vartheta }\in {\Theta }\), we have

\begin{equation} \left| T_{1, s_1,s_2,{\vartheta }} f(x)\right|= \left|\sum _{s_1 \le s\le s_2} \int K_s(x,y) f(y) e({\vartheta }(y)-{{\vartheta }(x)}) \, \mathrm{d}\mu (y)\right| \end{equation}

3.0.10

Moreover, there is a \(\tilde{{\vartheta }} \in \tilde{{\Theta }}\) with \(d_{B(o,2R)}({\vartheta },\tilde{{\vartheta }})\le \epsilon \). Hence,

\begin{align*} & \left| T_{1, s_1,s_2,{\vartheta }} f(x)\right|-\left|T_{1, s_1,s_2,\tilde{{\vartheta }}} f(x) \right|\\ & \leq \sum _{s_1 \le s\le s_2} \int |K_s(x,y)| \mathbf{1}_F(y) |e({\vartheta }(y)-{{\vartheta }(x)})-e({\tilde{{\vartheta }}(y)}-{\tilde{{\vartheta }}(x)})| \, \mathrm{d}\mu (y)\\ & \leq \sum _{s_1 \le s\le s_2} \epsilon \int |K_s(x,y)| \mathbf{1}_F(y) \, \mathrm{d}\mu (y) \end{align*}

Using Lemma 2.1.3, we can estimate the above expression by

\begin{align*} & \sum _{s_1 \le s\le s_2} \frac{2^{102 a^3}}{\mu (B(x, D^{s}))} \epsilon \int _{B(x, D^s)} \mathbf{1}_F(y) \, \mathrm{d}\mu (y)\\ & \leq (2S+1)\epsilon 2^{102 a^3} M\mathbf{1}_F(x)\le 2^{102 a^3}M\mathbf{1}_F(x) \end{align*}

We estimate the left-hand-side of 3.0.6 by the sum of left-hand-side of 3.0.9 and

\begin{align*} & \int \mathbf{1}_{G}(x) \max _{-S{\lt}s_1\le s_2{\lt}S}\sup _{{\vartheta }\in {{\Theta }}} \inf _{\tilde{{\vartheta }}\in \tilde{{\Theta }}} (| T_{1, s_1,s_2,{\vartheta }}f(x)|-|T_{1, s_1,s_2,\tilde{{\vartheta }}}f(x) |)\, d\mu (x)\, ,\\ \end{align*}

which, as we have just shown, is estimated by

\begin{align*} & 2^{102 a^3}\int \mathbf{1}_{G}(x) M\mathbf{1}_{F}(x)\, d\mu (x). \end{align*}

By Hölder’s inequality and Proposition 2.0.6 (more precisely, 2.0.46 with \(p=q\)), the above is no greater than

\begin{align*} & \frac{2^{102 a^3+4a}q}{q-1} \mu (G)^{1-\frac{1}{q}} \mu (F)^{\frac{1}{q}}. \end{align*}

Combining this with Lemma 3.0.3 and the fact that

\begin{align*} \frac{2^{102 a^3+4a}q}{q-1}\leq \frac{2^{445a^3}}{(q-1)^6} \end{align*}

proves Lemma 3.0.2.

It remains to prove Lemma 3.0.3. Fix a finite set \(\tilde{{\Theta }}\).

Lemma 3.0.4 linearized truncation

Let \(\sigma _1,\sigma _2\colon X\to \mathbb {Z}\) be measurable functions with finite range \([-S,S]\) and \(\sigma _1\leq \sigma _2\). Let \({Q}\colon X\to {\tilde{{\Theta }}}\) be a measurable function. Then we have

\begin{equation} \label{Sqlin} \int \mathbf{1}_{G}(x) \left| {T}_{2,\sigma _1,\sigma _2,{Q}}f(x)\right|\, d\mu (x) \le \frac{2^{445a^3}}{(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,{Q}}f(x)=\sum _{\sigma _1(x) \le s\le \sigma _2(x)} \int K_s(x,y) f(y) e({Q}(x)(y) - {Q}(x)(x)) \, \mathrm{d}\mu (y)\, . \end{equation}

3.0.12

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

\[ \max _{s'\leq s_2{\lt}S}\max _{{\vartheta }\in \tilde{{\Theta }}} | T_{1, s',s_2,{\vartheta }} f(x)| = \max _{-S{\lt}s_1\le s_2{\lt}S}\max _{{\vartheta }\in \tilde{{\Theta }}} | {T}_{1, s_1,s_2,{\vartheta }} f(x)|:={T}_{1, x}. \]

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

\[ \max _{{\vartheta }\in \tilde{{\Theta }}} | {T}_{1, {\sigma }_1(x), s'',{\vartheta }} f(x)| = {T}_{1,x}\, . \]

Finally, choose a total order of the finite set \(\tilde{{\Theta }}\) and let \({Q}(x)\) be the minimal element \({\vartheta }\) with respect to this order such that

\[ | {T}_{1, {\sigma }_1(x),{\sigma }_2(x),{\vartheta }} f(x)| = {T}_{1,x}\, . \]

With these choices, and noting that

\begin{equation*} {T}_{1, {\sigma }_1(x),{\sigma }_2(x),{Q}(x)} f(x)={T}_{2, {\sigma }_1,{\sigma }_2,{Q}} f(x), \end{equation*}

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

It remains to prove 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 \(n\ge 0\), \(\mu (G_{n})\le 2^{-n} \mu (G)\) and

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

\begin{equation} \le \frac{2^{440a^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, {Q}}f(x) \right|\, d\mu (x) \le \frac{2^{445a^3}}{(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, {Q}}f(x) \right|\, d\mu (x) \le \frac{2^{445a^3}}{(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 1.0.2.