Uniform convolution estimates for complex polynomial curves in ${\mathbb {C}}^3$

Abstract

We establish optimal (p, q) ranges for the weighted convolution operator associated with a complex polynomial curve. Establishing this estimate comes down to establishing a lower bound for the Jacobian of a mapping associated with the complex curve in question.

1 Introduction and statement of main theorems

Consider the complex polynomial curve $\varGamma (z):=\left( P_1(z),P_2(z),P_3(z)\right)$, for complex polynomials $P_i(z)$. Letting

$$\begin{aligned} L_\varGamma (z):=\left| \text {det}(\varGamma '(z),\varGamma ''(z),\varGamma '''(z))\right| ,\end{aligned}$$

(1.1)

and

$$\begin{aligned} \lambda _\varGamma (z):=\left| L_\varGamma (z)\right| ^\frac{1}{3},\end{aligned}$$

(1.2)

we define the weighted complex convolution operator associated with $\varGamma$ as:

$$\begin{aligned} \begin{aligned} T_\varGamma f({\textbf{z}}):=\int _{\mathbb {C}}f\left( {\textbf{z}}-\varGamma (w)\right) \lambda _\varGamma (w)\textrm{d}w, \end{aligned} \end{aligned}$$

(1.3)

In this paper, we establish a key bound for $T_\varGamma$:

Theorem 1

Let $\varGamma (z)$ be a complex polynomial curve in ${\mathbb {C}}^3$ of degree N, and $f\in L^{2,u}({\mathbb {C}})$. Then, for the complex convolution operator associated with $\varGamma$, $T_\varGamma$, there exists $C=C(N)$ such that:

$$\begin{aligned}\left| \left| T_\varGamma f\right| \right| _{L^{3,v}({\mathbb {C}})}\le C\left| \left| f\right| \right| _{L^{2,u}({\mathbb {C}})} \end{aligned}$$

for all $u\in [0,3]$, $v\in [2,\infty ]$, and $u<v$.

To establish this theorem, we will see that we first must establish the restricted weak-type estimate at $(p,q)=(2,3)$ and apply the argument from [11] to obtain this range of Lorentz estimates. Utilising this theorem, we can obtain more estimates that are uniform in $\varGamma$. This is done by interpolating between the estimate in Theorem 1 with its dual estimate, using the Marcinkiewicz interpolation theorem. This results in the fact that

$$\begin{aligned}||T_\varGamma f||_{L^{q_\theta }({\mathbb {C}}^3)}\le C||f||_{L^{p_\theta }({\mathbb {C}}^3)}, \end{aligned}$$

for $(p_\theta ,q_\theta )=(\frac{6}{3+\theta },\frac{6}{2+\theta })$, for $\theta \in (0,1)$, and some constant dependant only on N and $\theta$. Furthermore, if we replace $T_\varGamma$ with its local analogue, where the region of integration is restricted to some disc D, we can acquire a larger range of non-uniform estimates by interpolating with a trivial $L^\infty \rightarrow L^\infty$ estimate, and its dual.

Explicitly, letting ${\mathcal {R}}$ be the convex hull of the points $\left( 1,1\right)$, $\left( 0,0\right)$, $\left( p_0^{-1},q_0^{-1}\right)$, and $\left( p_1^{-1},q_1^{-1}\right)$. Then if $(p^{-1},q^{-1})\in {\mathcal {R}}$, then for all complex polynomial curves, $\varGamma$, there exists a constant $C=C(N,\varGamma ,D,p,q)$, such that for all functions $f\in L^p({\mathbb {C}}^3)$:

$$\begin{aligned} ||{ T_\varGamma f }||_{L^q({\mathbb {C}})}\le C||{f}||_{L^p({\mathbb {C}})}.\end{aligned}$$

Much work has been done on these kinds of estimates, and they are known when $T_\varGamma$ is replaced with its real analogue. In the real case, in arbitrary dimension, d, $\varGamma$ is a real polynomial curve, the integration is over some interval, and the weight function $\lambda _\varGamma$ is replaced with $\lambda _\varGamma (t):=L_\varGamma ^\frac{2}{d(d+1)}(t)$. Furthermore, our constant C must depend on the dimension we are in for these estimates to hold.

Usually, these estimates have been for a particular family of polynomial curves, or in a particular dimension. For example, in 1998, in [3], one of the early papers on this topic, Christ established this result for any moment curve in any dimension. That is to say he established it for $\varGamma (t)=\left( t,t^2\cdots t^n\right)$ in ${\mathbb {R}}^n$. It should be noted that for curves like this, we have $\lambda _\varGamma (t)=n!$, which simplifies the problem. This can be considered the most non-degenerate case, as all derivatives of $\varGamma$ are linearly independent. Later papers introduced the weight $\lambda _\varGamma$ in order to deal with problematic singular points of $\varGamma$, where the derivatives become linearly dependent. At such points, $\lambda _\varGamma$ vanishes, allowing us to mitigate the effect of such points during the integration.

Further estimates were established for other families of curves, but it was in 2002, in [9], that Oberlin established the real estimate for general polynomial curves in two dimensions, and in three dimensions, established it for curves of the form $\gamma (t)=\left( t,P_1(t),P_2(t)\right)$. In [6], this three-dimensional case was extended to general polynomial curves, and in [10], this was extended to general higher dimensions.

The first such estimates for surfaces instead of curves can be found in [8], in which Drury and Guo proved a class of estimates of this type, but instead of convolving with a measure arising from a polynomial curve, their estimates were pertaining to the case where $\varGamma$ was instead a two-dimensional surface in ${\mathbb {R}}^4$, of the form $(x,y,\phi _1(x,y),\phi _2(x,y))$, with certain conditions on $\phi _1$ and $\phi _2$.

In [4], Chung and Ham established similar results, considering surfaces that could be represented as complex curves in ${\mathbb {C}}^d$. They established these bounds for curves of the form $\left( z,z^2\cdots z^N\right)$ in ${\mathbb {C}}^d$, and $\left( z,z^2,\phi (z)\right)$ in ${\mathbb {C}}^3$, for analytic $\phi$. They do this by using a more powerful analogue to Lemma 4 in this paper, which is established in the complex Fourier restriction paper [1], involving a lower bound in terms of the arithmetic means of $\left\{ L_\varGamma (z_i)\right\}$, as opposed to the geometric mean. This method suffices for the complex curves in question in [4], but in general does not hold. In this paper, we do not utilise a lower bound in terms of the arithmetic mean, but instead adapt the geometric mean bound deriving from [7] into the complex case, and achieve the overall estimate using that.

Note that while the two-dimensional estimate has not been established for general complex polynomial curves, these estimates can be derived from the procedure described in our proof of Theorem 1, with simple modifications. The explicit description of this is omitted to aid the presentation of the three-dimensional case.

Before moving onto the process of proving Theorem 1, we first introduce some commonly accepted notation for these kinds of problems. We say that, for positive quantities A and B, we have $A\lesssim B$ if $A\le C B$, for some constant C. Furthermore, we say $A\sim B$ if $A\lesssim B\lesssim A$. Here, C, the implicit constant, can depend on will only depend on the degree of the curve, N

2 Decomposition of ${\mathbb {C}}$

To achieve the desired results, we first partition ${\mathbb {C}}$ into various convex sets. This is done by utilising two decompositions as described in [7]. In [7], these decompositions are described in relation to ${\mathbb {R}}$, but can be extended to ${\mathbb {C}}$. The first of these decompositions will be referred to as D1.

D1: Given a polynomial Q(z), we can decompose ${\mathbb {C}}$ into a bounded number of convex sets, $B_i$. On each of these $B_i$, we have $\left| Q(z)\right| \sim c(Q)\left| z-b_i\right| ^{k_i}$, for $b_i$ some root of Q(z). We will frequently omit the subscript of k and b for the sake of convenience.

To see how this decomposition functions, we first write ${\displaystyle Q(z)=A\prod \nolimits _{j=1}^{d'}(z-\eta _j)^{\alpha _j}}$, with $d'$ being the number of distinct roots of Q. We now decompose ${\mathbb {C}}$ into a union of $d'$ convex sets, given by the Voronoi diagram associated with $\left\{ \eta _1,\eta _2,\cdots ,\eta _{d'}\right\}$. On these sets, we will have our complex centres b be the root associated with this set. Formally, these sets are denoted $S(\eta _j)=\left\{ z\in {\mathbb {C}}\mid \left| z-\eta _j\right| \le \left| z-\eta _i\right| \text { for all }i\ne j \right\}$.

Each of these sets are further divided into finitely many sectors, centred at b. For $n\ge 1$ denote $\Delta ^n_\varepsilon (\eta _j)=\left\{ z-\eta _j=re^{i\theta }\in S(\eta _j)|(n-1)\varepsilon \le \theta \le n\varepsilon \right\}$. Here, $\varepsilon$ is chosen such that $\frac{2\pi }{\varepsilon }$ is an integer. Specifically, we need to take $\varepsilon \le \frac{\pi }{4}$, which is necessary to decompose the intersection of some annuli and these sectors into convex sets. While this is sufficient for D1 to function as described, we will see that we also need to choose $\varepsilon$ to be dependent on d, the degree of our polynomial curve. Thus, the choice of $\varepsilon$ is currently not specified.

Fixing one such $\Delta ^n_{\varepsilon }(\eta _j)=\Delta (\eta _j)$, we relabel $\eta _j={\hat{\eta }}_1$, and the remaining roots as ${\hat{\eta }}_k$ so that we have $\left| {\hat{\eta }}_1-{\hat{\eta }}_2\right| \le \left| {\hat{\eta }}_1-{\hat{\eta }}_3\right| \le \cdots \le \left| {\hat{\eta }}_1-{\hat{\eta }}_{d'}\right|$. That is to say we order them according to their distance from $\eta _j$ We let, for $k\ge 2$:

$$\begin{aligned} T^n_{k,j}=\left\{ z\in \Delta (\eta _j)\cap S(\eta _j)\bigm \vert \left| z-{\hat{\eta }}_1\right| \le \frac{1}{2}\left| {\hat{\eta }}_1-{\hat{\eta }}_k\right| \right\} . \end{aligned}$$

(2.1)

Labelling $T_{1,j}=\varnothing$, and $T_{d'+1,j}=\Delta (\eta _j)$, we see that $T_{1,j}\subseteq T_{2,j}\subseteq \cdots \subseteq T_{d',j}\subseteq T_{d'+1,j}=\Delta (\eta _j)$. Also note that if $z\in T^n_{k,j}$, then $\frac{1}{2}\left| {\hat{\eta }}_l-{\hat{\eta }}_1\right| \le \left| z-\eta _l\right| \le \frac{3}{2}\left| {\hat{\eta }}_l-{\hat{\eta }}_1\right|$, for all $l\ge k$, and if $z\notin T^n_{i,j}$, then $\left| z-{\hat{\eta }}_1\right|<\left| z-\eta _l\right| <3\left| z-\eta _1\right|$ for all $k\le i$.

Letting, for $1\le i\le d'$, $I^n_{i,j}=T^n_{i+1,j}\backslash T^n_{i,j}$, we can see that $\Delta ^n_\varepsilon (\eta _j)=\bigcup _{k=1}^{d'}I^n_{k,j}$, and on any fixed $I_{k,j}$, we have:

$$\begin{aligned} \left| Q(z)\right| =\left| A\right| \prod _{l=1}^{d'}\left| z-\eta _l\right| ^{\alpha _l}\sim \left| A\right| \left| z-{\hat{\eta }}_1\right| ^{{\hat{\alpha }}_1+{\hat{\alpha }}_2+\cdots +{\hat{\alpha }}_k}\prod _{l=k+1}^{d'}\left| {\hat{\eta }}_1-{\hat{\eta }}_l\right| ^{\hat{\alpha _l}}, \end{aligned}$$

Note here that ${\hat{\alpha }}_j$ denotes the multiplicity of ${\hat{\eta }}_j$.

So indeed, writing ${{\mathbb {C}}=\bigcup \nolimits _{l=1}^{\frac{2\pi }{\varepsilon }d'^2} I_l=\bigcup \nolimits _{j=1}^{d'}\bigcup _{n=1}^{\frac{2\pi }{\varepsilon }} \bigcup \nolimits _{i=1}^{d'} I^n_{i,j}}$, we have on $B_l=I^n_{i,j}$ that:

$$\begin{aligned} \begin{aligned} \left| Q(z)\right| \sim c(Q)\left| z-b_l\right| ^{k_l}, \end{aligned} \end{aligned}$$

(2.2)

where $b_l=b_{i,j,n}=\eta _j$, and $k_l=k_{i,j,n}={\hat{\alpha }}_1+{\hat{\alpha }}_2\cdots +{\hat{\alpha }}_i$.

Note that these sets are intersections of some convex sets, and annuli centred at $b_l$, and therefore are not necessarily convex. This can be remedied by “convexifying” our annuli sectors. This is done on any of these annuli sectors by constructing that tangent to the annuli which is perpendicular to the lower ray of the sector. Each annuli sector in the decomposition is now replaced with the trapezoid which has sides given by these tangents, and the two rays of the sector. More precisely, we replace the set

$$\begin{aligned}I_{k,j}^n=\left\{ z\in \Delta _\varepsilon ^n(\eta _j)\cap S(\eta _j)\biggm \vert \left| z-b\right| \in \left[ \frac{1}{2}\left| \eta _j-{\hat{\eta }}_i\right| ,\frac{1}{2}\left| \eta _j-{\hat{\eta }}_{i+1}\right| \right] \right\} ,\end{aligned}$$

with the set

$$\begin{aligned}{\tilde{I}}_{k,j}^n=\left\{ z\in \Delta _\varepsilon ^n(\eta _j)\cap S(\eta _j)\biggm \vert \text {Re}(e^{-(n-1)\varepsilon } (z-b))\in \left[ \frac{1}{2}\left| \eta _j-{\hat{\eta }}_i\right| ,\frac{1}{2}\left| \eta _j-{\hat{\eta }}_{i+1}\right| \right] \right\} ,\end{aligned}$$

where the relabeling of the ${\hat{\eta }}_i$ are done relative to $\eta _j$. Note that this set maintains the property that for all $z\in {\tilde{I}}_{k,j}^n$, (2.2) holds. This is because ${\tilde{I}}_{k,j}^n$ is contained in the annulus sector

$$\begin{aligned}J_{k,j}^n=\left\{ z\in \Delta _\varepsilon ^n(\eta _j)\biggm \vert \left| z-b\right| \in \left[ \frac{1}{2}\left| \eta _j-{\hat{\eta }}_i\right| ,\frac{\sqrt{1+(\tan \varepsilon )^2}}{2}\left| \eta _j-{\hat{\eta }}_{i+1}\right| \right] \right\} \end{aligned}$$

for which (2.2) holds for the same reasoning as for $I_{k,j}^n$, with slightly different constants. Furthermore, notice that ${\tilde{I}}_{k,j}^n$ are all trapezoids, with the exception of the sets ${\tilde{I}}_{k,1}^n$, which are right triangles. In either case, each ${\tilde{I}}_{k,j}^n$ is convex.

For the next decomposition, referred to as D2, we require a polynomial Q(z), and a centre, which is some complex number b.

D2: Given a polynomial Q(z), and complex number b, and some convex set J, then J can be decomposed into disjoint convex sets. These convex sets are of two types, either dyadic or gap. On dyadic sets, $\left| z-b\right| \sim c_1(Q)$, and on gap sets, $Q(z)\sim c_2(Q)\left| z-b\right| ^k$, for some positive integer k. Also, for $Q(z+b)=\Sigma c_iz^i$, if $c_j=0$, then there are no gap intervals such that $\left| Q(z)\right| \sim c_2(Q) \left| z-b\right| ^j$.

Note that, if we label the collection of dyadic sets as $\left\{ D_j\right\}$, and the gaps as $\left\{ G_j\right\}$we have:

• $D_j=\left\{ z\in {\mathbb {C}}|A^{-1}\left| z_j\right|<\left| z\right| <A\left| z_j\right| \right\}$,

• $G_j=\left\{ z\in {\mathbb {C}}|A\left| z_j\right|<\left| z\right| <A^{-1}\left| z_{j+1}\right| \right\}$,

for $A\in {\mathbb {R}}$, and $z_j$ being the roots of Q(z), ordered by magnitude.

D2 originates from [2] and describes a decomposition of ${\mathbb {R}}$. However, if we restrict ourselves to sectors of the form $\Delta ^n_\varepsilon (\eta _j)$, where $\eta _j$ are the roots of Q, and Voronoi sets $S(\eta _j)$, then the method can be extended to the complex case. Note again however that this will give us disjoint annuli in the complex case, so we repeat the thickening of these annuli, and converting them into corresponding convex sets as in D1.

For our purposes, we may assume an affine transformation has been applied to our curve $\varGamma$ to ensure that $C(Q)=c_1(Q)=c_2(Q)=1$, due to the affine invariance of the theorems addressed in this paper.

We also make use of the integral representation of the Jacobian in [7]. This states that, in 3-dimensions, for $L_3(z)=L_\varGamma (z)$, $L_2(x)=\text {det}\left( \begin{array}{cc} P_1'(z) &{} P_1''(z) \\ P_2'(z)&{} P_2''(z) \end{array}\right)$, and $L_1(z)=P_1'(z)$, then if we consider the mapping $\varPhi _\varGamma (z_1,z_2,z_3)=\varGamma (z_1)+\varGamma (z_2)+\varGamma (z_3)$, we have:

$$\begin{aligned} \begin{aligned} J_{\varPhi _\varGamma }(z_1,z_2,z_3)=\prod _{s=1}^3L_1(z_s)\int _{z_1}^{z_2}\int _{z_2}^{z_3}\prod _{s=1}^2\frac{L_2(w_s)}{L_1(w_s)^2}\int ^{w_2}_{w_1}\frac{L_1(y)L_3(y)}{L_2(y)^2}\,\textrm{d}y\,\textrm{d}w_2\,\textrm{d}w_1, \end{aligned}\nonumber \\ \end{aligned}$$

(2.3)

where $J_{\varPhi _\varGamma }$ is the real Jacobian of the mapping $\varPhi _\varGamma$, and the complex numbers appearing in the bounds of an integral represent that the integral in question is taken over the straight line joining those complex numbers. So in this case, $w_1$ is on the line joining $z_1$ to $z_2$, that $w_2$ is on the line joining $z_2$ to $z_3$, and y is on the line joining $w_1$ to $w_2$. Explicitly,

$$\begin{aligned}\int _{z_1}^{z_2}f(w)\textrm{d}w:=(z_2-z_1)\int _0^1 f(z_1+(z_2-z_1)t)\,\textrm{d}t.\end{aligned}$$

From here, we use D1 and D2 to decompose ${\mathbb {C}}$ into convex sets so that $L_1$, $L_2$, and $L_3$ can be approximated by either monomials or constants. This decomposition results in a bounded number of sets, which are divided into four different types, indexed by bitstrings of length 2. We index the degrees of the monomials that approximate the functions $L_i$ similarly. The degree associated with $L_3$ is indexed by a zero-length bitstring, and is denoted k. $L_1$ exponents are indexed by a bitstring of length one, given by the first digit of the bitstring associated with the set we’re on. So $\left| L_1(z)\right| \sim \left| z-b\right| ^{k_0}$ on $T_{00}$ and $T_{01}$ sets, and $\left| L_1(z)\right| \sim \left| z-b\right| ^{k_1}$ on $T_{10}$ and $T_{11}$ sets. Lastly the degree associated with $L_2$ is indexed by a bitstring of length 2, being identical to the bitstring of the set we’re on. So for example, $\left| L_2(z)\right| \sim \left| z-b\right| ^{k_{10}}$ on $T_{10}$ sets.

We are now ready to proceed with the decomposition.

First, we use D1 with respect to $L_3$, so that we get a finite number of convex sets such that $\left| L_3\right| \sim \left| z-b\right| ^k$ on each of the sets from this decomposition, with k and b depending on the sets. Next, we divide this family of sets into two types of sets, referred to as $T_0$ or $T_1$ sets. To do this, we first apply D2 to each set already obtained with respect to our centre from the first decomposition, and the polynomial $L_1$.

$T_0$ sets are those sets are gap annuli, on which $\left| L_1(z)\right| \sim \left| z-b\right| ^{k_0}$, and $\left| L_3(z)\right| \sim \left| z-b\right| ^k$.

To get the $T_1$ sets, we must further decompose the dyadic annuli. This is done by applying D1 with respect to $L_1$ on each of these annuli, which results in a family of sets on which $\left| L_1(z)\right| \sim \left| z-b'\right| ^{k_1}$, and $\left| L_3(z)\right| \sim 1$

We apply D2 again to the $T_0$ and $T_1$ sets, but this time, with respect to $L_2$, and the centre b, and $b'$, respectively. This divides the $T_0$ sets into two new families of sets, referred to as $T_{00}$ sets and $T_{01}$ sets.

$T_{00}$ sets are those on which $\left| L_2(z)\right| \sim \left| z-b\right| ^{k_{00}}$, while we still have $\left| L_1(z)\right| \sim \left| z-b\right| ^{k_0}$, and $\left| L_3(z)\right| \sim \left| z-b\right| ^k$

$T_{01}$ sets are again obtained by decomposing the dyadic annuli from the last D2 decomposition. We apply D1 to these annuli to obtain a family of convex sets on which $\left| L_3(z)\right| \sim \left| L_1(z)\right| \sim 1$, as $\left| z-b\right| \sim 1$, and $\left| L_2(z)\right| \sim \left| z-b''\right| ^{k_{01}}$

We decompose the $T_1$ sets precisely the same way to get families of sets named $T_{10}$ and $T_{11}$.

On $T_{10}$ sets, we have $\left| L_3\right| \sim 1$, $\left| L_1\right| \sim \left| z-b'\right| ^{k_1}$ and $\left| L_2(z)\right| \sim \left| z-b'\right| ^{k_{10}}$.

On $T_{11}$ sets, we have $\left| L_1(z)\right| \sim \left| L_3(z)\right| \sim 1$, and $\left| L_2(z)\right| \sim \left| z-b''\right| ^{k_{11}}$.

Note that the various b and k values depend on the specific set within these families that they belong to, as opposed to every set of type $T_{01}$ having the same $k_{01}$ value for example. The utility of dividing these sets into these types is that behaviour of the integrand that appears in (2.3) can be divided into four cases corresponding to these types. This will be shown at the end of the section, in (2.7) and the proceeding table.

Furthermore, we shall write $b'$ and $b''$ as just b for simplicity for the rest of this discussion. Note to make use of these estimates, we need the following lemma.

Lemma 1

For all $z_1,z_2,z_3$ in one of the sets in our decomposition, we have that

$$\begin{aligned}{} & {} |J_{\varPhi _\varGamma }(z_1,z_2,z_3)| \sim \prod _{s=1}^3\left| L_1(z_s)\right| \left| \int _{z_1}^{z_2}\left| \int _{z_2}^{z_3}\prod _{s=1}^2\frac{|L_2(w_s)|}{|L_1(w_s)|^2}\left| \int \right| ^{w_2}_{w_1}\right| \right| \\{} & {} \quad {{ \frac{|L_1(y)||L_3(y)|}{|L_2(y)|^2}\textrm{d}y\textrm{d}w_2}\textrm{d}w_1}. \end{aligned}$$

To see this, we first introduce the notion of a sector-contained function.

Definition 1

An $\varepsilon$-sector-contained function of (B, b) is, for $B\subset {\mathbb {C}}$ and $b\in {\mathbb {C}}$, is a function f such that for all $z\in B$, $f(z-b)\in \Delta _\varepsilon ^n(0)$ for some n, with notation as in the decomposition D1.

We will occasionally omit the reference to the set (B, b) when we refer to functions that satisfy this definition, understanding that a $\varepsilon$-sector-contained function refers to a $\varepsilon$-sector-contained function of (B, b), where B is any set in our decomposition, and b is centre associated with B.

Note that the reason we introduce the notion of these functions is because for f, an $\varepsilon$-sector-contained function, we have $\left| \int _u^v f(z-b)\textrm{d}z\right| \sim \left| \int _u^v \left| f(z-b)\right| \textrm{d}z\right|$ for u, v in any set of our decomposition, for $\varepsilon <\frac{\pi }{2}$.

To see this, note that we immediately have that $\left| \int _u^v f(z-b)\textrm{d}z\right| \le \left| \int _u^v \left| f(z-b)\right| \textrm{d}z\right|$, as

$$\begin{aligned}\left| \int _u^v f(z-b)\textrm{d}z\right|{} & {} =\left| (v-u){\int _0^1 f(z(t))\,\,\textrm{d}t}\right| \le \left| (v-u){\int _0^1\left| f(z(t)-b)\right| \,\,\textrm{d}t}\right| \\{} & {} =\left| \int _u^v \left| f(z-b)\right| \textrm{d}z\right| .\end{aligned}$$

To establish the second inequality, notice that by factoring out some complex number of unit length, we can assume that $f(z-b)$ is contained in a sector which has one ray in the direction of the positive real axis, and the second ray being given by the line containing the complex numbers with argument $\varepsilon$.

We therefore have:

$$\begin{aligned} \begin{aligned} \left| \int _u^v f(z-b)\textrm{d}z\right|&= \left| (v-u)\int _0^1 \text {Re}(f(z(t)-b))+i\text {Im}(f(z(t)-b))\,\,\textrm{d}t\right| \\&\ge \left| (v-u)\int _0^1 \text {Re}(f(z(t)-b))\,\,\textrm{d}t\right| \\&\sim \left| \int _u^v\left| f(z-b)\right| \textrm{d}z\right| . \end{aligned} \end{aligned}$$

Therefore, to establish our lemma, we would like to show that each integrand is an $\varepsilon$-sector-contained function. However, because the argument of $(w_2-w_1)$ cannot be controlled, this is not quite true, so we instead show that the integrand is the product of some $\varepsilon$ contained function, and $(w_2-w_1)$, which we will see is sufficient to establish our lemma.

Lemma 2

Let $P(z)=\prod _{j=1}^d(z-\eta _j)$ be a polynomial. If $D_j(\eta _l)$ is the jth dyadic annulus of P(z) centred at $\eta _l$, then $D_j(\eta _l)\cap \Delta _\varepsilon (b)$ can be decomposed as $\bigcup _{i=1}^{C(d,{\hat{\varepsilon }})}B_{i,j,l,{\hat{\varepsilon }}}^P$ such that P(z) is an ${\hat{\varepsilon }}$-sector-contained function of $B_{i,j,l,{\hat{\varepsilon }}}^P$ for all i.

Proof of Lemma 2

In [4], it was shown in step 2 of lemma 4.2 that for a polynomial $P(z-b)=\prod _{j=1}^d (z-\eta _j)$, if we are in the jth dyadic annulus associated with P(z) and b, as in the terminology of D2, then we have, by factoring $P(z-b)$ as

$$\begin{aligned}P(z-b)=g_j(z)(-1)^{d-j}z^j\prod _{l=j+1}^d\eta _l,\end{aligned}$$

that on the jth gap or dyadic annulus, we can decompose these sets into a bounded number of sets so that $g_j(z)$ is contained in a sector of aperture less that $b_0$, for arbitrary $b_0$. Here, the number of sets are dependant on $b_0$. As $(-1)^{d-j}z^j\prod _{l=j+1}^d\eta _l$ will be contained in a sector of aperture $j\varepsilon$, we have that P(z) will be in a sector of aperture at most $j\varepsilon +b_0$ for z in any of our sets. Therefore, in our initial decomposition, we must ensure that we choose $\varepsilon$ and $b_0$ such that $d\varepsilon +b_0<{\hat{\varepsilon }}$ and our result holds. $\square$

Notice that both decomposition techniques D1 and D2 involve restricting our sets to $\Delta ^n_\varepsilon (\eta _j)\bigcap S(\eta _j)$, and on each of our sets we have applied D1 or D2 to each of $L_1,L_2$ and $L_3$. Therefore, we have no concern about utilising this result for any number of $L_1,L_2$ and $L_3$ simultaneously.

A similar result immediately follows for rational functions whose numerator and denominator satisfy the conditions of Lemma 2.

Corollary 1

Let $Q(z)=\frac{P_1(z)}{P_2(z)}$ for polynomials $P_1$ and $P_2$ with degrees $d_1$ and $d_2$. Then if $B_1$ and $B_2$ are two sets arising from the decomposition in Lemma 2 for a common centre b, being applied to $P_1$ and $P_2$, respectively, then Q(z) is a $(d_1+d_2){\hat{\varepsilon }}$ sector-contained function of $B_1\cap B_2$.

So therefore, for any set in our decomposition, we can decompose it further to ensure each of our $L_i$ are $(d_i\varepsilon +b_{0,i})$-sector-contained functions, for $d_i$ the degree of $L_i$. Therefore, we have that the rational function $\frac{L_1L_3}{L_2^2}$ is a $4(d+1)\varepsilon$-sector-contained function, and $\frac{L_2}{L_1^2}$ is a $3(d+1)\varepsilon$-sector-contained function where d is the maximum of the $d_i$ values.

Lastly, we wish to observe that $\int ^{w_2}_{w_1}f(y)\textrm{d}y$ is the product of a $\varepsilon$-sector-contained function and $(w_2-w_1)$.

Lemma 3

Suppose f(z) is a $\varepsilon$-sector-contained function. Then, there exists a $\varepsilon$-sector-contained function, F(z), such that $\int ^{w_2}_{w_1}f(y)\textrm{d}y=(w_2-w_1)F(w_1,w_2)$.

Proof of Lemma 3

Letting $\int ^{w_2}_{w_1}f(y)\textrm{d}y=(w_2-w_1)\text {lim}_{n\rightarrow \infty }\frac{1}{n}\Sigma _{j=1}^nf(y(\frac{j}{n}))$, for $y(t)=w_1+(w_2-w_1)t$. Now, as f(y) is in a sector for all y on this line segment, as B is convex, each $\frac{1}{n}\Sigma _{j=1}^nf(y(\frac{j}{n}))$ is in the sector, so therefore the limit of this sequence is in the sector.

Therefore, we have that $\int ^{w_2}_{w_1}f(y)\textrm{d}y=(w_2-w_1)F(w_1,w_2)$, where $F(w_1,w_2)=\int _0^1(f(y(t))\,\,\textrm{d}t$ is a $\varepsilon$-sector-contained function. $\square$

We can now proceed with establishing the main lemma of this section.

Proof of Lemma 1

Now note that we can write:

$$\begin{aligned} \begin{aligned} J_{\varPhi _\varGamma }(z_1,z_2,z_3)&=\prod _{s=1}^3L_1(z_s)\int _{z_1}^{z_2}\int _{z_2}^{z_3}\prod _{s=1}^2\frac{L_2(w_s)}{L_1(w_s)^2}\int ^{w_2}_{w_1}\frac{L_1(y)L_3(y)}{L_2(y)^2}\textrm{d}y\textrm{d}w_2\textrm{d}w_1\\&=\left( \prod _{s=1}^3L_1(z_s)\right) (z_2-z_1)(z_3-z_2) I\\ \end{aligned} \end{aligned}$$

where:

$$\begin{aligned} \begin{aligned} I:=\int _{0}^{1}\int _{0}^{1}\prod _{s=1}^2\frac{L_2(w_s)}{L_1(w_s)^2}(w_2-w_1)\int ^{1}_{0}\frac{L_1(y)L_3(y)}{L_2(y)^2}\,\,\textrm{d}t_3\,\,\textrm{d}t_2\,\,\textrm{d}t_1. \end{aligned} \end{aligned}$$

(2.4)

Here, note that $w_1=z_1+(z_2-z_1)t_1$, $w_2=z_2+(z_3-z_2)t_2$, and $y=w_1+(w_2-w_1)t_3$.

We now wish to bound the size of the integral I. Let $\alpha (t_1,t_2)$ and $\beta (t_1,t_2)$ be real valued functions such that $(w_2-w_1)=\alpha +i\beta$. Note that these polynomials, after our decomposition, are sector-contained functions, and the integral along the line from $w_1$ to $w_2$ of a sector-contained function is a sector-contained function multiplied by $(w_2-w_1)$. Therefore, the above integral can be written as:

$$\begin{aligned}I=\int _{0}^{1}\int _{0}^{1}\left( \alpha (t_1,t_2)+i\beta (t_1,t_2)\right) g(t_1,t_2)\,\,\textrm{d}t_2\,\,\textrm{d}t_1,\end{aligned}$$

where g is a $(7(d+1))\varepsilon$-sector-contained function. That is to say $g(t_1,t_2)=\xi (t_1,t_2)+i\eta (t_1,t_2)$, for real functions $\xi$ and $\eta$ with $|g(t_1,t_2)|\sim \xi (t_1,t_2)$. Note that we make the following assumptions about the points $z_1,z_2,z_3$. These assumptions can be made without loss of generality via renaming these points for the first two, and factoring out a unit complex number for the last. First, we assume that in the triangle $z_1,z_2,z_3$, the angle at $z_2$, which we label $\theta$, is the largest in the triangle. Next, we assume $|z_3-z_2|\ge |z_2-z_1|$. Lastly, we assume $z_2-z_1$ is a positive real number.

Note that this immediately gives that $\beta$ is single signed, and $\theta \in [\frac{\pi }{3},\pi ]$. We compute our estimate by splitting into cases when $\theta$ is acute, and when it is obtuse. We also define here the integral quantity

$$\begin{aligned}G:=\int _{0}^{1}\int _{0}^{1}\left| \alpha (t_1,t_2)+i\beta (t_1,t_2)\right| \left| g(t_1,t_2)\right| \,\,\textrm{d}t_2\,\,\textrm{d}t_1.\end{aligned}$$

The aim in the following sections is to show $\left| I\right| \gtrsim G$.

Case (i) If $\theta \in [ \dfrac{\pi }{3},\dfrac{\pi }{2}]$:

First, note that we have

$$\begin{aligned} \xi \sim |g(t_1,t_2)| =\prod _{s=1}^2\frac{|L_2(w_s(t_s))|}{|L_1(w_s(t_s))^2|}\left| \int ^{1}_{0}\frac{L_1(y(t_3))L_3(y(t_3))}{L_2(y(t_3))^2}\,\,\textrm{d}t_3\right| . \end{aligned}$$

Now, as the integrand here is a sector-contained function, we can move the absolute value signs inside the integral, yielding

$$\begin{aligned} \begin{aligned} |g(t_1,t_2)|&\sim \prod _{s=1}^2\frac{|L_2(w_s(t_s))|}{|L_1(w_s(t_s))^2|}\left| \int ^{1}_{0}\frac{|L_1(y(t_3))||L_3(y(t_3))|}{|L_2(y(t_3))^2|}\,\,\textrm{d}t_3\right| \\&\sim \prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int ^{1}_{0}|y(t_3)|^{\sigma _3}\,\,\textrm{d}t_3\right| . \end{aligned} \end{aligned}$$

(2.5)

Now, we let $t_2(t_1)$ is the smallest value of $t_2$ for which the complex number $(w_2-w_1)$ has imaginary part greater than half the absolute value of its real part. That is to say, $t_2(t_1)$ is the smallest value such that $\beta >\dfrac{|\alpha |}{2}$. Note that by our assumptions on $\theta$ and the relative lengths of our line segments, we have that $t_2(t_1)<\frac{1}{2}$ for all $t_1$.

From here we will proceed to estimate the $t_1$ integrand, by considering

$$\begin{aligned} \begin{aligned} G(t_1)&:=\int _{t_2(t_1)}^1|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&=\int _{2t_2(t_1)}^1|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&\qquad +\int _{t_2(t_1)}^{2t_2(t_1)}|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&:=G_1+G_2. \end{aligned} \end{aligned}$$

Now note that $L=L(t_1)=z_2-w_1$ is a positive real number, and for all the $t_2$ in the $G_2$ integral, we have $L\le |w_2-w_1|\le 2L$, and also that this relation holds for $t_2<t_2(t_1)$. Furthermore, we also note that the triangle with vertices $z_1,z_2,w_2(2t_2(0))$ is contained in the circle centred at $z_2$, with radius $\varepsilon |z_2|$. This can be seen by considering the case with $z_1,z_2$ are positive real numbers, $z_3$ is on the other ray of our sector, $\theta =\frac{\pi }{2}$, and $|z_2-z_1|=|z_3-z_2|$, as this arrangement maximises the distance from $z_2$ and $w_2(2t_2(0))$. Note for all points z in this circle, we have $|z|\sim |z_2|$. We therefore have:

$$\begin{aligned} \begin{aligned} G_2&\sim \int _{t_2(t_1)}^{2t_2(t_1)}|w_2-w_1||z_2|^{2\sigma _2+\sigma _3}\left| \int _0^1\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&\sim \int ^{t_2(t_1)}_0|w_2-w_1||z_2|^{2\sigma _2+\sigma _3}\left| \int _0^1\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&\sim \int ^{t_2(t_1)}_0|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\,\textrm{d}t_3\right| \,\,\textrm{d}t_2\\&:=G_3. \end{aligned} \end{aligned}$$

Thus, we can write:

$$\begin{aligned} \begin{aligned} G(t_1)&\sim G_1+G_2+G_3\\&=\int _0^1|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\textrm{d}t_3\right| \,\textrm{d}t_2. \end{aligned} \end{aligned}$$

Integrating both sides with respect to $t_1$ as $t_1$ goes from 0 to 1 yields

$$\begin{aligned}{} & {} \iint _{\left| \beta \right| \ge \frac{\left| \alpha \right| }{2}}\left| \alpha (t_1,t_2)+i\beta (t_1,t_2)\right| \left| g(t_1,t_2)\right| \\{} & {} \gtrsim \int _0^1\int _0^1|w_2(t_2)-w_1(t_1)|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\textrm{d}t_3\right| \,\textrm{d}t_2\,\textrm{d}t_1.\end{aligned}$$

Now, note that the left hand side of this inequality is similar in magnitude to $\iint _{\left| \beta \right| \ge \frac{\left| \alpha \right| }{2}}\left| \beta \right| \xi$, and the right hand side is precisely G. We therefore have that:

$$\begin{aligned} \begin{aligned} |I|&\ge \left| \text {Im}\left( \int _{0}^{1}\int _{0}^{1}(\alpha +i\beta )g(t_1,t_2)\,\textrm{d}t_2\,\textrm{d}t_1\right) \right| \\&=\left| \int _0^1\int _0^1(\xi \beta +\eta \alpha )\,\textrm{d}t_2\,\textrm{d}t_1\right| \\&\ge \int _0^1\int _0^1\xi \left| \beta \right| -\left| \eta \alpha \right| \,\textrm{d}t_2\,\textrm{d}t_1\\&\ge \int _0^1\int _0^1\xi (\left| \beta \right| -\varepsilon \left| \alpha \right| )\,\textrm{d}t_2\,\textrm{d}t_1\\&\gtrsim \iint _{\left| \beta \right| >\frac{|\alpha |}{2}}\left| \beta \right| \xi \,\textrm{d}t_2\,\textrm{d}t_1-\varepsilon G \\&\gtrsim G, \end{aligned} \end{aligned}$$

which establishes the lemma.

Case (ii) $\theta \in [\frac{\pi }{2},\pi ]$:

This case is simpler, as now we always have $\alpha >0$. Using the same notation as earlier, we have

$$\begin{aligned} \begin{aligned} |I|&=\left| \int _0^1\int _0^1(\alpha +i\beta )(\xi +i\eta )\,\textrm{d}t_2\,\textrm{d}t_1\right| \\&=\left| \int _0^1\int _0^1\alpha \xi -\beta \eta +i(\beta \xi +\alpha \eta )\,\textrm{d}t_2\,\textrm{d}t_1\right| . \end{aligned} \end{aligned}$$

(2.6)

Note that we also have, as $\beta$ is single signed, and $\alpha$ is positive,

$$\begin{aligned} \int _0^1\int _0^1\alpha \xi \,\textrm{d}t_2\,\textrm{d}t_1+ \left| \int _0^1\int _0^1\beta \xi \,\textrm{d}t_2\,\textrm{d}t_1\right| =\left| \int _0^1\int _0^1(\alpha +|\beta |)\xi \,\textrm{d}t_2\,\textrm{d}t_1\right| ,\end{aligned}$$

$$\begin{aligned}\sim \left| \int _0^1\int _0^1|w_2(t_2)-w_1(t_1)||g(t_1,t_2)|\,\textrm{d}t_2\,\textrm{d}t_1\right| =G.\end{aligned}$$

If $\left| \int _0^1\int _0^1\beta \xi \,\textrm{d}t_2\,\textrm{d}t_1\right| \ge \int _0^1\int _0^1\alpha \xi \,\textrm{d}t_2\,\textrm{d}t_1$, then we have:

$$\begin{aligned}|\text {Im}(I)|=\left| \int _0^1\int _0^1\beta \xi +\alpha \eta \,\textrm{d}t_2\,\textrm{d}t_1\right| \ge \left| \int _0^1\int _0^1\beta \xi \,\textrm{d}t_2\,\textrm{d}t_1\right| -\varepsilon \int _0^1\int _0^1\alpha \xi \,\textrm{d}t_2\,\textrm{d}t_1\\ \sim \left| \int _0^1\int _0^1\beta \xi \,\textrm{d}t_2\,\textrm{d}t_1\right| \gtrsim G.\end{aligned}$$

While if $\int _0^1\int _0^1\alpha \xi \,\textrm{d}t_2\,\textrm{d}t_1\ge \left| \int _0^1\int _0^1\beta \xi \,\textrm{d}t_2\,\textrm{d}t_1\right|$, then by the same line of reasoning we have

$$\begin{aligned}|\text {Re}(I)|=\left| \int _0^1\int _0^1\alpha \xi -\beta \eta \,\textrm{d}t_2\,\textrm{d}t_1\right| \gtrsim G.\end{aligned}$$

So indeed this gives us in either case that

$$\begin{aligned}|I|\gtrsim |\int _0^1\int _0^1|w_2-w_1|\prod _{s=1}^2|w_s(t_s)|^{\sigma _2}\left| \int _0^1|y(t_3)|^{\sigma _3}\,\textrm{d}t_3\right| \,\textrm{d}t_2\,\textrm{d}t_1|.\end{aligned}$$

as again can move the absolute value around the inner integral into it, as the integrand is a sector-contained function.

This completes the lemma, as in all cases, we indeed obtain:

$$\begin{aligned} |J_{\varPhi _\varGamma }(z_1,z_2,z_3)|\sim \prod _{s=1}^3|L_1(z_s)|\left| \int _{z_1}^{z_2}\left| \int _{z_2}^{z_3}\prod _{s=1}^2\frac{|L_2(w_s)|}{|L_1(w_s)^2|}\left| \int ^{w_2}_{w_1}\frac{|L_1(y)||L_3(y)|}{|L_2(y)^2|}\textrm{d}y\right| \textrm{d}w_2\right| \textrm{d}w_1\right| . \end{aligned}$$

Note that we can place absolute values around the $w_2$ integral as it is now clearly an integral of a real, positive integrand, and so we are done. $\square$

So therefore, we have, on any set in our decomposition:

$$\begin{aligned} \begin{aligned} \left| J_{\varPhi _\varGamma }(z_1,z_2,z_3)\right| \sim \prod _{s=1}^3\left| z_s-b\right| ^{\sigma _1}\left| \int _{z_1}^{z_2}\left| \int _{z_2}^{z_3}\prod _{s=1}^2\left| w_s-b\right| ^{\sigma _2}\left| \int ^{w_2}_{w_1}\left| y-b\right| ^{\sigma _3}\textrm{d}y\right| \textrm{d}w_2\right| \textrm{d}w_1\right| , \end{aligned}\nonumber \\ \end{aligned}$$

(2.7)

for some values of $\sigma _1,\sigma _2,\sigma _3$, depending on the type of set we are on. The explicit values of each $\sigma _i$ are given here:

	$\sigma _1$	$\sigma _2$	$\sigma _3$
$T_{00}$	$k_0$	$k_{00}-2k_0$	$k+k_0-2k_{00}$
$T_{01}$	0	$k_{01}$	$-2k_{01}$
$T_{10}$	$k_1$	$k_{10}-2k_1$	$k_1-2k_{10}$
$T_{11}$	0	$k_{11}$	$-2k_{11}$

3 The geometric inequality

In this section, we establish an important inequality which shall be used in both of our main theorems. This lemma is as follows:

Lemma 4

For $\left| J_{\varPhi _\varGamma }\right|$ yielding $\sigma =(\sigma _1,\sigma _2,\sigma _3)$ values as above after the described decomposition

$$\begin{aligned} \begin{aligned} \left| J_{\varPhi _\varGamma }(z_1,z_2,z_3)\right| \gtrsim \prod _{i=1}^3| L_\varGamma (z_i)|^\frac{1}{3}\prod _{1\le i<j\le 3}|z_j-z_i|, \end{aligned} \end{aligned}$$

(3.1)

for $\sigma _3\ne -1$, and $\sigma _2+\frac{\sigma _3}{2}<-2$ or $\ge 0$.

Furthermore, we can meet the conditions of this lemma, that is to say we can avoid certain values of $\sigma$, by applying an affine transformation to our curve $\varGamma$ before applying the decomposition D2. This is done to ensures that we avoid certain $\sigma$ values. We omit the details, which are exactly the same as in the real case, considered in [7]. Specifically, we ensure there is no $k_{10}$ value such that $k_{10}=\frac{-1-k_1}{2}$ on $T_{10}$ intervals, and no $k_{00}$ value such that $k_{00}=\frac{k+k_0+1}{2}$ on $T_{00}$ intervals. This ensures $\sigma _3\ne -1$. Similarly, we avoid $k_1=1$ on $T_{10}$ and any $k_0$ values on $T_{00}$ such that $k_0=\frac{k+i}{3}$ for $i=1,2,3,4$. Note that as $k_0$ is an integer, this involves excluding at most 2 values of $k_0$.

Note that since the conception of this paper, this inequality, and its higher dimensional analogues, have been established by J. de Dios Pont in [5]. The methods of this proof differ significantly from those presented here. The cited paper uses a more elegant, inductive argument to effectively reduce this lemma to the case where $\varGamma$ is a generalized moment curve, via establishing a pseudo-continuity between $\varGamma$ and $L_\varGamma$. Note that this continuity allows one to avoid a great deal of the messier computations that are to follow.

This limited version of the lemma presented here is included for completeness, while avoiding some of the more expansive arguments needed to establish the higher dimensional cases, which allows us to work with an explicit decomposition of ${\mathbb {C}}$ on which the lemma holds, rather than using the non-constructive methods in [5].

Proof of Lemma 4

To establish (3.2), we first seek to show that:

$$\begin{aligned} \begin{aligned} \left| J_{\varPhi _\varGamma }(z_1,z_2,z_3)\right| \gtrsim \prod _{i=1}^3\left| z_i\right| ^{\sigma _1+\frac{2\sigma _2}{3}+\frac{\sigma _3}{3}}\left| \int _{z_1}^{z_2}\int _{z_2}^{z_3}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_1\textrm{d}w_2\right| . \end{aligned} \end{aligned}$$

(3.2)

As calculation will yield that $\left| z_i\right| ^{\sigma _1+\frac{2\sigma _2}{3}+\frac{\sigma _3}{3}}\sim \left| L_\varGamma (z_i)\right| ^{\frac{1}{3}}$.

Firstly we wish to see that:

$$\begin{aligned} \begin{aligned} I_1:=\left| \int ^{w_2}_{w_1}\left| y-b\right| ^{\sigma _3}\textrm{d}y\right| \gtrsim \left| w_2-b\right| ^{\frac{\sigma _3}{2}}\left| w_1-b\right| ^{\frac{\sigma _3}{2}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| . \end{aligned} \end{aligned}$$

(3.3)

In proving (3.3), we may assume $\left| w_2\right| >\left| w_1\right|$, as if not, we can notice that

$\left| \int ^{w_2}_{w_1}\left| y-b\right| ^{\sigma _3}\textrm{d}y\right| =\left| -\int ^{w_1}_{w_2}\left| y-b\right| ^{\sigma _3}\textrm{d}y\right| =\left| \int ^{w_1}_{w_2}\left| y-b\right| ^{\sigma _3}\textrm{d}y\right|$.

So we may relabel to ensure that the larger of the two occurs at the upper bound of the integral.

Also notice that if $\left| w_1-b\right|<\left| w_2-b\right| <9\left| w_1-b\right|$, we have $\left| w_1-b\right| \sim \left| y-b\right| \sim \left| w_2-b\right|$ throughout this line integral, and our result immediately follows. We therefore only have to deal with the case in which $9\left| w_1-b\right| <\left| w_2-b\right|$.

Let:

• $L:=\left\{ y\mid 2\left| w_1-b\right|<\left| y-b\right| <3\left| w_1-b\right| \right\} .$

• $U:=\left\{ y \mid \frac{1}{3}\left| w_2-b\right|<\left| y-b\right| <\frac{1}{2}\left| w_2-b\right| \right\} .$

Note that after parameterizing our line integral, we can see that it is the integral of a purely positive quantity over a real interval, so we can delete portions of the line segment and ensure that this results in a lower bound.

So

$$\begin{aligned} \begin{aligned} I_1&\ge \left| \left[ \int _L+\int _U\right] {\left| y-b\right| ^{\sigma _3}}\textrm{d}y\right| \\&\sim \left| w_1-b\right| ^{\sigma _3+1}+\left| w_2-b\right| ^{\sigma _3+1}\\&=\left| w_2-b\right| \left( \left| w_2-b\right| ^{\sigma _3}+\frac{\left| w_1-b\right| ^{\sigma _3+1}}{\left| w_2-b\right| }\right) . \end{aligned} \end{aligned}$$

Note that $\left| w_2-b\right| \sim \left| w_2-w_1\right| =\left| \int _{w_1}^{w_2}\textrm{d}y\right|$, as $9\left| w_1-b\right| <\left| w_2-b\right|$.

So $I_1\gtrsim \left| \int _{w_1}^{w_2}\textrm{d}y\right| \text {max}\left( \left| w_2-b\right| ^{\sigma _3},\frac{\left| w_1-b\right| ^{\sigma _3+1}}{\left| w_2-b\right| }\right)$.

If $\sigma _3$ is positive, we have

$$\begin{aligned}I_1\gtrsim \left| \int _{w_1}^{w_2}\textrm{d}y\right| \left| w_2-b\right| ^{\sigma _3}>\left| \int _{w_1}^{w_2}\textrm{d}y\right| \left| w_2-b\right| ^{\frac{\sigma _3}{2}}\left| w_1-b\right| ^{\frac{\sigma _3}{2}}.\end{aligned}$$

If $\sigma _3$ is negative, then, as $\sigma _3$ is an integer which is not equal to -1, $\sigma _3+1$ is negative also, so

$$\begin{aligned}I_1\gtrsim \left| \int _{w_1}^{w_2}\textrm{d}y\right| \frac{\left| w_1-b\right| ^{\sigma _3+1}}{\left| w_2-b\right| }>\left| \int _{w_1}^{w_2}\textrm{d}y\right| \left| w_2-b\right| ^{\frac{\sigma _3}{2}}\left| w_1-b\right| ^{\frac{\sigma _3}{2}}.\end{aligned}$$

As this holds for all $\sigma _3\ne -1$, we are done. With this, we can conclude that

$$\begin{aligned} \left| J_{\varPhi _\varGamma }(z_1,z_2,z_3)\right| \gtrsim \prod _{s=1}^3\left| z_s-b\right| ^{\sigma _1}\left| \int _{z_1}^{z_2}\int _{z_2}^{z_3}\prod _{s=1}^2\left| w_s-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\textrm{d}w_1\right| . \end{aligned}$$

We now seek to establish a similar inequality for

$$\begin{aligned}I_2=\left| \int _{z_2}^{z_{3}}\left| w_2-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| .\end{aligned}$$

We want to see that $I_2\gtrsim \text {max}\left( \left| z_{3}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}},\frac{\left| z_{2}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}}{\left| z_{3}-b\right| ^{2}}\right) \left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right|$, for $\sigma _2+\frac{\sigma _3}{2}<-2$ or $\ge 0$.

We assume that $\left| z_2-b\right| <\left| z_{3}-b\right|$, and relabel if we are not in such a case. We also notice that the inequality is immediate if we are in the case that $\left| z_2-b\right|<\left| z_{3}-b\right| <9\left| z_2-b\right|$, so we therefore assume that $9\left| z_2-b\right| <\left| z_{3}-b\right|$, and again restrict the integral to L and U, which now represent the sets:

• $L:=\left\{ y\mid 2\left| z_2-b\right|<\left| y-b\right| <3\left| z_2-b\right| \right\}$.

• $U:=\left\{ y \mid \frac{1}{3}\left| z_{3}-b\right|<\left| y-b\right| <\frac{1}{2}\left| z_{3}-b\right| \right\}$.

So we have $I_2\ge \left| \left[ \int _L+\int _U\right] \left| w_2-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\left| w_2-w_1\right| \textrm{d}w_2\right|$. Identically to last time, we have $\left| w_2-w_1\right| \sim \left| w_2-b\right|$, so we can write:

$$\begin{aligned} \begin{aligned} I_2&\gtrsim \left| \left[ \int _L+\int _U\right] \left| w_2-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+1}\textrm{d}w_2\right| \\&\sim \left| z_2-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}+\left| z_3-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}\\&=\left| z_3-b\right| ^2\left( \frac{\left| z_2-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}}{\left| z_3-b\right| ^2}+\left| z_3-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\right) \\&\gtrsim \text {max}\left( \left| z_{3}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}},\frac{\left| z_{2}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}}{\left| z_{3}-b\right| ^{2}}\right) \left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\mathrm{\textrm{d}}y\right| \textrm{d}w_2\right| . \end{aligned} \end{aligned}$$

Now, if $\sigma _2+\frac{\sigma _3}{2}>0$, we have:

$$\begin{aligned} \begin{aligned} I_2&= \left| z_{3}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| \\&\ge \left| z_{3}-b\right| ^{\frac{2\sigma _2}{3}+\frac{\sigma _3}{3}}\left| z_{2}-b\right| ^{\frac{\sigma _2}{3}+\frac{\sigma _3}{6}}\left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| . \end{aligned} \end{aligned}$$

If instead, $\sigma _2+\frac{\sigma _3}{2}<-2$, we have

$$\begin{aligned} \begin{aligned} I_2&\gtrsim \frac{\left| z_{2}-b\right| ^{\sigma _2+\frac{\sigma _3}{2}+2}}{\left| z_{3}-b\right| ^{2}}\left| \int _{z_2}^{z_{3}}\left| w_2-w_1\right| \textrm{d}w_2\right| \\&>\left| z_{3}-b\right| ^{\frac{2\sigma _2}{3}+\frac{\sigma _3}{3}}\left| z_{2}-b\right| ^{\frac{\sigma _2}{3}+\frac{\sigma _3}{6}}\left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| , \end{aligned} \end{aligned}$$

as $\sigma _2+\frac{\sigma _3}{2}+2<0$

A similar analysis yields that for

$$\begin{aligned}I_3=\left| \int _{z_1}^{z_{2}}\left| w_1-b\right| ^{\sigma _2+\frac{\sigma _3}{2}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_1\right| ,\end{aligned}$$

we have

$$\begin{aligned} I_3\gtrsim \left| z_{1}-b\right| ^{\frac{2\sigma _2}{3}+\frac{\sigma _3}{3}}\left| z_{2}-b\right| ^{\frac{\sigma _2}{3}+\frac{\sigma _3}{6}}\left| \int _{z_2}^{z_{3}}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| ,\end{aligned}$$

for the same constraints on our $\sigma$ values as previously outlined. So we therefore indeed have (3.2).

From here, we seek to resolve the triple integral which remains in (3.2) and establish that:

$$\begin{aligned} \begin{aligned} \left| \int _{z_1}^{z_2}\left| \int _{z_2}^{z_3}\left| \int _{w_1}^{w_2}\textrm{d}y\right| \textrm{d}w_2\right| \textrm{d}w_1\right| \gtrsim \left| z_3-z_1\right| \left| z_3-z_2\right| \left| z_2-z_1\right| . \end{aligned} \end{aligned}$$

(3.4)

Clearly, the inner most integral is given by $\left| \int ^{w_2}_{w_1}\textrm{d}y\right| =\left| w_2-w_1\right|$.

Consider $J=J(w_1)=\left| \int _{z_2}^{z_{3}}\left| w_2-w_1\right| \textrm{d}w_2\right|$. If $z_3-z_2=\left| z_3-z_2\right| e^{i\phi }$, and $t=e^{-i\phi }(w_2-z_2)\in {\mathbb {R}}$, we have

$$\begin{aligned}J=\int _0^{\left| z_3-z_2\right| }\left| t-w_1'\right| =\int _0^{\left| z_3-z_2\right| }\sqrt{(t-\text {Re}(w_1'))^2+\text {Im}(w_1')^2}\,\textrm{d}t:=\int _0^{\left| z_3-z_2\right| }g_1(t)\,\textrm{d}t,\end{aligned}$$

where $w_1'=e^{-i\phi }(w_1-z_2)$.

To bound this from below, we introduce a new function $f_1(t)=|{t-\widehat{w_1'}(t)}|$, where

Note that, for all values of $t\in [0,|z_3-z_2|]$, we have that $\text {Re}(w_1')=\text {Re}(\widehat{w_1'}(t))$, and that $|\text {Im}(w_1'(t))|>|\text {Im}(\widehat{w_1'}(t))|$. We also have $\widehat{w_1'}(0)=\widehat{w_1'}(|z_3-z_2|)=w_1'$.

Clearly $g_1(t)\ge f_1(t)$ for all t in the interval $[0,|z_3-z_2|]$, as t is real, and $\widehat{w_1'}(t)$ is $w_1'$ with a smaller imaginary part. Note that computation shows that $f_1(t)$ is given by:

So $J=\int _0^{\left| z_3-z_2\right| }g_1(t)\,\textrm{d}t>\int _0^{\left| z_3-z_2\right| }f_1(t)\,\textrm{d}t$. Note that $g_1(t)$ increases as $\left| t-\text {Re}(w_1')\right|$ increases, $g_1(0)=\left| z_2-w_1\right| =f_1(0)$ and $g_1(\left| z_3-z_2\right| )=\left| z_3-w_1\right| =f_1(\left| z_3-z_2\right| )$.

Therefore if $\text {Re}(w_1')<\frac{\left| z_3-z_2\right| }{2}$, we have $\left| z_3-w_1\right| >\left| z_2-w_1\right|$.

Alternatively if $\text {Re}(w_1')>\frac{\left| z_3-z_2\right| }{2}$, then we have $\left| z_2-w_1\right| >\left| z_3-w_1\right|$.

Now, suppose $\text {Re}(w_1')\le 0$. Then $\left| z_3-w_1\right| >\left| z_2-w_1\right|$, and we have:

$$\begin{aligned} \begin{aligned} J&>\int _0^{\left| z_3-z_2\right| }f_1(t)\,\textrm{d}t\\&=\int _0^{\left| z_3-z_2\right| } \frac{\left| z_3-w_1\right| }{\left| z_3-z_2\right| -\text {Re}(w_1')}(t-\text {Re}(w_1'))\,\textrm{d}t\\&= \frac{\left| z_3-w_1\right| }{\left| z_3-z_2\right| -\text {Re}(w_1')}\left[ \frac{t(t-\text {Re}(w_1'))}{2}\right] ^{\left| z_3-z_2\right| }_0\\&\sim \left| z_3-z_2\right| \left| z_3-w_1\right| . \end{aligned} \end{aligned}$$

A symmetric argument gives that if $\text {Re}(w_1')>\left| z_3-z_2\right|$ then $\left| z_2-w_1\right| >\left| z_3-w_1\right|$, and $J\gtrsim \left| z_3-z_2\right| \left| z_2-w_1\right|$.

Lastly, if $\text {Re}(w_1')\in \left[ 0,\left| z_3-z_2\right| \right]$, then:

$$\begin{aligned} \begin{aligned} J&>\int _0^{\text {Re}(w_1')}\frac{\left| z_2-w_1\right| }{\text {Re}(w_1')}(\text {Re}(w_1')-t)\,\textrm{d}t+\int _{\text {Re}(w_1')}^{\left| z_3-z_2\right| }\frac{\left| z_3-w_1\right| }{\left| z_3-z_2\right| -\text {Re}(w_1')}(t-\text {Re}(w_1'))\,\textrm{d}t\\&=\frac{1}{2}\left( \left| z_2-w_1\right| \text {Re}(w_1')+\left| z_3-w_1\right| (\left| z_3-z_2\right| -\text {Re}(w_1') )\right) \\&\sim \left| z_3-z_2\right| \text {max}\left( \left| z_3-w_1\right| ,\left| z_2-w_1\right| \right) . \end{aligned} \end{aligned}$$

So indeed $J\gtrsim \left| z_3-z_2\right| \text {max}(\left| z_3-w_1\right| ,\left| z_2-w_1\right| )$ in each of these cases. Applying this to the triple integral on the left of (3.4), we get

$$\begin{aligned}\left| \int _{z_1}^{z_2}\int _{z_2}^{z_{3}}\left| w_2-w_1\right| \textrm{d}w_2\textrm{d}w_1\right| \gtrsim \left| z_3-z_2\right| \left| \int _{z_1}^{z_2}\left| z_3-w_1\right| \textrm{d}w_1\right| .\end{aligned}$$

As we can go through an identical procedure for $J':=\int _{z_1}^{z_2}\left| z_3-w_1\right| \textrm{d}w_1$, this indeed establishes (3.4). This in conjunction with (3.2) yields the proof of Lemma 4. $\square$

4 Application to the convolution operator

We will now use the work from sections 2 and 3 to establish Theorem 1. Before that however, we must use them to establish the restricted weak-type estimate at (2, 3).

Lemma 5

$T_\varGamma$ is of restricted weak-type at (2, 3). That is to say for measurable subsets of ${\mathbb {C}}^3$, E and F, there exists $C=C(N)$ such that

$$\begin{aligned}\langle T_\varGamma \chi _E,\chi _F\rangle \le C\left| E\right| ^{\frac{1}{2}}\left| F\right| ^{\frac{2}{3}}.\end{aligned}$$

Before proving this lemma, we first reiterate why this lemma implies that we obtain uniform estimates for $T_\varGamma$ along the “critical line” of the convolution problem, not including its endpoints, and non-uniform estimates in the rest of the region ${\mathcal {R}}$. To see this, first notice that we need only establish that the operator T is of restricted weak-type at $(p,q)=(2,3)$, as the operator is “almost” self dual, we can use duality properties to conclude that we also have the bound corresponding to $(p,q)=(\frac{3}{2},2)$. Furthermore, following from the Marcinkiewicz interpolation theorem, this lemma establishes the strong type estimate for all $(p_\theta ,q_\theta )=(\frac{6}{3+\theta },\frac{6}{2+\theta })$, for $\theta \in (0,1)$. Pairing this with the fact that the local bound at $(p,q)=(\infty ,\infty )$ follows trivially from the fact that $T_\varGamma f$ is bounded whenever f is bounded, we see that, with this lemma, and utilising duality properties and interpolation, we acquire strong type estimates for all $(p,q)\in {\mathcal {R}}$, except for at $(p_0,q_0)$, and $(p_1,q_1)$, where we have restricted weak-type estimates.

Note also, the bound we will establish at $\left( 2,3\right)$ is uniform, as opposed to the trivial bound at (0, 0). Therefore, we have uniform estimates along the line segment joining the vertices $\left( \frac{1}{2},\frac{1}{3}\right)$ and $\left( \frac{2}{3},\frac{1}{2}\right)$, and local estimates, with implicit constants dependent on the disc and the curve, at every other (p, q) pair in ${\mathcal {R}}$.

Proof of Lemma 5

Throughout this proof, we will denote the measure $\left| z\right| ^\frac{k}{3}\textrm{d}z=\textrm{d}\sigma (z)$, where k is such that $\lambda _\varGamma (z)\sim \left| z\right| ^{\frac{k}{3}}$. We also denote two quantities, $\alpha$ and $\beta$, where, for measurable sets E and F, they are given by the equations:

$$\begin{aligned} \begin{aligned} \alpha \left| F\right| =\langle T{_\varGamma } \chi _{_E}, \chi _{_F}\rangle =\beta \left| E\right| . \end{aligned} \end{aligned}$$

Here, $\alpha$ and $\beta$ can be interpreted as the average value of $T_\varGamma \chi _E$ on F, and the average value of $T^*_\varGamma \chi _F$ on E, respectively.

Furthermore, we let $\gamma =\max (\alpha ,\beta )$, $\nu =\frac{3}{k+6}$, and denote the ball $B_x:=\left\{ z\in {\mathbb {C}}\,|\,\left| z\right| <c_\nu x^\nu \right\}$, for $c_\nu =(8\pi )^{-\nu }$.

An easy computation shows:

$$\begin{aligned} \begin{aligned} \sigma (B_x)&=\int _{B_x}\left| z\right| ^\frac{k}{3}\textrm{d}z\\&=2\pi \int _0^{(8\pi )^{-\nu }x^\nu }r^{\frac{k+3}{3}}\textrm{d}r\\&=\frac{x\nu }{4}\\&\le \frac{x}{8}. \end{aligned} \end{aligned}$$

Notice that if we replace $T_\varGamma$ with ${\widetilde{T}}f(z):=\int _{\Delta {\setminus } B_\gamma }f(z-\varGamma (w))\textrm{d}\sigma (w)$, and if $\gamma =\alpha$ we can notice that:

$$\begin{aligned} \begin{aligned} \langle {\widetilde{T}}\chi _{E},\chi _F\rangle&\ge \langle T_\varGamma \chi _{E},\chi _F\rangle -\left| F\right| \frac{\alpha }{2}\\&\ge \frac{\alpha }{2}\left| F\right| =\frac{\beta }{2}\left| E\right| , \end{aligned} \end{aligned}$$

Similarly, if $\gamma =\beta$:

$$\begin{aligned} \begin{aligned} \langle {\widetilde{T}}\chi _{E},\chi _F\rangle&=\langle \chi _{E},{\widetilde{T}}^*\chi _F\rangle \\&\ge \langle \chi _{E},T^*_\varGamma \chi _F\rangle -\left| E\right| \frac{\beta }{2}\\&\frac{\beta }{2}\left| E\right| \ge \frac{\alpha }{2}\left| F\right| , \end{aligned} \end{aligned}$$

Therefore, we can introduce the refined set $E_1:=\left\{ y\in E\,|\,{\widetilde{T}}^*\chi _F(y)\ge \frac{\beta }{2}\right\}$ and have it be non-empty.

Note that:

$$\begin{aligned}\langle {\widetilde{T}}\chi _{E_1},\chi _F \rangle =\langle {\widetilde{T}}\chi _{E},\chi _F \rangle -\langle {\widetilde{T}}\chi _{E\setminus E_1},\chi _F \rangle \ge \alpha \left| F\right| -\frac{\beta }{2}\left| E\right| =\frac{\alpha }{2}\left| F\right| .\end{aligned}$$

That is to say that the average value of ${\widetilde{T}}\chi _{E_1}$ on F is greater than $\frac{\alpha }{2}$. We therefore can introduce the set $F_1:=\left\{ x\in F\,|\,{\widetilde{T}}\chi _{E_1}(x)\ge \frac{\alpha }{4}\right\}$, and this will also be non-empty.

Now, if we take $z_0\in F_1$, we can define the set

$$\begin{aligned}P_1=P_{1,z_0}:=\left\{ z_1\in \Delta ^n_\varepsilon \setminus B_\gamma \,|\,z_0-\varGamma (z_1)\in E_1\right\} .\end{aligned}$$

Notice that $\sigma (P_1)=\int _{\Delta ^n_\varepsilon {\setminus } B_\gamma }\chi _{E_1}(z_0-\varGamma (z_1))\left| z\right| ^{\frac{k}{3}}\textrm{d}z\sim {\widetilde{T}}_\varGamma \chi _{E_1}(z_0)\gtrsim \alpha$. Now to each $z_1\in P_1$ associate a set:

$$\begin{aligned}P_2=P_{2,z_0,z_1}=\left\{ z_2\in \Delta ^n_\varepsilon \setminus B_\gamma \,|\,z_0-\varGamma (z_1)+\varGamma (z_2)\in F_1\right\} , \end{aligned}$$

which has the property $\sigma (P_2)\sim {\widetilde{T}}_\varGamma ^*\chi _{F_1}(z_0-\varGamma (z_1))\gtrsim \beta$, from the construction of $P_1$.

Lastly, letting:

$$\begin{aligned}P_3=P_{2,z_0,z_1,z_2}=\left\{ z_3\in \Delta ^n_\varepsilon \setminus B_\gamma \,|\,z_0-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3)\in E_1\right\} , \end{aligned}$$

We again see that $\sigma (P_3)\gtrsim \alpha$, by the same logic as for $P_1$.

Now, letting $H(z_1,z_2,z_3):=-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3)$, and $\varPhi =P_1\times P_2\times P_3$, we can notice $H(\varPhi )\subset E$. Thus, we obtain the lower bound for $\left| E\right|$ via:

$$\begin{aligned} \begin{aligned} \left| E\right|&\gtrsim \iiint _\varPhi \left| J_{\mathbb {R}}H(z_1,z_2,z_3)\right| \textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&= \iiint _\varPhi \left| J_{\mathbb {C}}H(z_1,z_2,z_3)\right| ^2\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^{\frac{2k}{3}}\prod _{1 \le i< j \le 3}\left| z_i-z_j\right| ^2\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&=\int _{P_1}\int _{P_2}\int _{P_3}\prod _{i=1}^3\left| z_i\right| ^{\frac{k}{3}}\prod _{1 \le i < j \le 3}\left| z_i-z_j\right| ^2\textrm{d}\sigma (z_3)\textrm{d}\sigma (z_2)\textrm{d}\sigma (z_1).\\ \end{aligned} \end{aligned}$$

Also notice that we have the property that on $P_i$, $\left| z_i\right| \ge c_\nu \gamma ^\nu$. From here, we aim to acquire sets $P_i'\subset P_i$, such that $\sigma (P_i')\gtrsim \alpha$ for $i=1,3$, or $\gtrsim \beta$ for $i=2$, but we also want to establish lower bounds for the separations of the variables.

Specifically, we wish to construct $P_i'$, for $z_i\in P_i'$

• $\left| z_3-z_i\right| \gtrsim \alpha ^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}$,

• $\left| z_2-z_1\right| \gtrsim \beta ^{\frac{1}{2}}\left| z_1\right| ^{-\frac{k}{6}}$.

To establish these bounds, we can notice the condition that the inequality $\left| z\right| \gtrsim x^\nu$ implies $\left| z\right| \gtrsim x^{\frac{1}{2}}\left| z\right| ^{-\frac{k}{6}}$, which is obtained by raising either side of the inequality to $\frac{1}{2\nu }$, and multiplying across by $\left| z\right| ^{-\frac{k}{6}}$.

In the first case, we must consider the balls $B_x(w):=\left\{ z \in {\mathbb {C}}\,|\,\left| z-w\right| <c_0 x^\frac{1}{2}\left| w\right| ^{-\frac{k}{6}}\right\}$ for some $c_0$ to be chosen. The aim here will be to remove $B_\alpha (z_i)$ from $P_3$, so we would like to show that its $\sigma$-measure is a small fraction of $\alpha$.

Now, if $z\in B_\alpha (z_i)$, we have $\left| z-z_i\right|<c_0 \alpha ^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}<c_0\left| z_i\right|$, which implies $\left| z\right| \lesssim c_0\left| z_i\right|$. We can now estimate $\sigma (B_\alpha (z_i))$ via:

$$\begin{aligned} \begin{aligned} \sigma (B_\alpha (z_i))&=\int _{B_\alpha (z_i)}\left| z\right| ^{\frac{k}{3}}\textrm{d}z\\&\lesssim c_0^2\alpha \left| z_i\right| ^{-\frac{k}{3}}\times c_0^\frac{k}{3}\left| z_i\right| ^{\frac{k}{3}}\\&\le c_0^{2+\frac{k}{3}}\alpha . \end{aligned} \end{aligned}$$

So for sufficiently small $c_0$, $B_\alpha (z_i)$, can be removed from $P_3$ to ensure the first bullet point holds.

The second bullet point similarly holds via the same argument, by replacing $\alpha$ with $\beta$, $z_3$ with $z_2$, and $P_3$ with $P_2$.

Finally, we need to equally redistribute the exponent of the monomial on the right hand side of this inequality. Specifically, we wish to show that $\left| z_3-z_i\right| \gtrsim \alpha ^{\frac{1}{2}}\left| z_3z_i\right| ^\frac{-k}{12}$, and $\left| z_2-z_1\right| \gtrsim \beta ^{\frac{1}{2}}\left| z_2z_1\right| ^\frac{-k}{12}$.

Clearly if $\left| z_i\right| \sim \left| z_j\right|$ these follow immediately, so we need only show it in the case $2\left| z_3\right| <\left| z_1\right|$, or $2\left| z_1\right| <\left| z_3\right|$, and similarly for the inequality involving $z_2$ and $z_1$.

To see this, notice that the separation ensures

$$\begin{aligned}\left| z_3-z_i\right| \gtrsim \left| z_3\right| =\left| z_3\right| ^{1+\frac{k}{12}}\left| z_i\right| ^{\frac{k}{12}}\left| z_iz_3\right| ^{-\frac{k}{12}}\gtrsim \gamma ^{1+\frac{k}{6}}\left| z_iz_3\right| ^{-\frac{k}{12}}\ge \alpha ^{\frac{1}{2}}\left| z_iz_3\right| ^{-\frac{k}{12}},\end{aligned}$$

so we are done. An identical argument works for $\left| z_2-z_1\right|$. Now we can return to our lower bound for $\left| E\right|$, with these new estimates.

$$\begin{aligned} \begin{aligned} \left| E\right|&\gtrsim \int _{P'_1}\int _{P'_2}\int _{P'_3}\prod _{i=1}^3\left| z_i\right| ^{\frac{k}{3}}\prod _{1 \le i < j \le 3}\left| z_i-z_j\right| ^2\textrm{d}\sigma (z_3)\textrm{d}\sigma (z_2)\textrm{d}\sigma (z_1)\\&\gtrsim \alpha ^2\beta \int _{P'_1}\int _{P'_2}\int _{P'_3}\prod _{i=1}^3\left| z_i\right| ^{\frac{k}{3}}\left( \left| z_1z_2\right| \left| z_2z_3\right| \left| z_1z_3\right| \right) ^{-\frac{k}{6}}\textrm{d}\sigma (z_3)\textrm{d}\sigma (z_2)\textrm{d}\sigma (z_1)\\&=\alpha ^2\beta \sigma (P_1')\sigma (P_2')\sigma (P_3')\\&\gtrsim \alpha ^4\beta ^2, \end{aligned} \end{aligned}$$

and after substituting in $\alpha =\frac{\langle T{_\varGamma } \chi _{_E}, \chi _{_F}\rangle }{\left| F\right| }$, and $\beta =\frac{\langle T{_\varGamma } \chi _{_E}, \chi _{_F}\rangle }{\left| E\right| }$, we see this implies Lemma 5. $\square$

Now, we want to follow-up from this proof and establish Theorem 1. The method demonstrated here follows almost identically to the method described in [4], which in turn was based on the work done in the real case of [11], and is included here for the sake of completion. To establish this theorem, we need only establish the following two lemmas, which via the reasoning in the appendix from [11], imply the theorem in question. Note that the argument presented in that appendix relies upon the restricted weak-type estimate already established, so we could not have simply started with an attempt to establish this range of Lorentz estimates.

Lemma 6

Let $E_1$, $E_2$, $F\subset {\mathbb {C}}^3$ be measurable sets with finite measure. If we have $\alpha _1$ and $\alpha _2$ such that $T_\varGamma \chi _{E_i}(x)\ge \alpha _i$ for $x\in F$, and $\alpha _2\ge \alpha _1$, then we have the following lower bound for the measure of $E_2$:

$$\begin{aligned}\left| E_2\right| \gtrsim \alpha _1^{2-2\nu }\alpha _2^{2+2\nu }\beta ^2,\end{aligned}$$

for $\nu =\frac{3}{k+6}$, and $\beta =\alpha _1\frac{\left| F\right| }{\left| E_1\right| }$.

Lemma 7

Let $F_1$, $F_2$, $E\subset {\mathbb {C}}^3$ be measurable sets with finite measure. If we have $\beta _1$ and $\beta _2$ such that $T^*_\varGamma \chi _{F_i}(y)\ge \beta _i$ for $y\in E$, and $\beta _2\ge \beta _1$, then we have the following lower bound for the measure of $F_2$:

$$\begin{aligned}\left| F_2\right| \gtrsim \alpha ^{3}\beta _1\beta _2^2,\end{aligned}$$

for $\alpha =\beta _1\frac{\left| E\right| }{\left| F_1\right| }$.

For both of these proofs, we will frequently reuse the notation from Lemma 5, such as the measure $\sigma$ being given by $d\sigma (z)=\left| z\right| ^{\frac{k}{3}}\textrm{d}z$.

Proof of Lemma 6

First, we wish to construct 3 sets, $P_1$, $P_2$, $P_3\subset \Delta ^n_\varepsilon$, with the following properties:

• $\sigma (P_1)\gtrsim \alpha _1$, and if $z_1\in P_1$, then $\left| z_1\right| \ge c_\nu \max {(\alpha _1,\beta )}^\nu$,

• $\sigma (P_2)\gtrsim \beta$, and if $z_2\in P_2$, then $\left| z_2\right| \ge c_\nu \max {(\alpha _1,\beta )}^\nu$,

• $\sigma (P_3)\gtrsim \alpha _2$, and if $z_3\in P_3$, then $\left| z_3\right| \ge c_\nu '\alpha _2^\nu$.

To ease notation, we define $\gamma :=\max {(\alpha _1,\beta )}$. Furthermore, we denote, as before, $B_x$ to be the open ball of radius $\left( \frac{x}{8\pi }\right) ^\nu$.

We therefore again have:$\sigma (B_x)=\int _{B_x}\left| z\right| ^\frac{k}{3}\textrm{d}z\le \frac{x}{2}.$

Now, letting ${\widetilde{T}}f(z):=\int _{\Delta {\setminus } B_\gamma }f(z-\varGamma (w))\textrm{d}\sigma (w)$, we can notice that:

$$\begin{aligned} \begin{aligned} \langle {\widetilde{T}}\chi _{E_1},\chi _F\rangle&\ge \langle T_\varGamma \chi _{E_1},\chi _F\rangle -\left| F\right| \frac{\gamma }{2}\\&\ge \frac{\alpha _1}{2}\left| F\right| =\frac{\beta }{2}\left| E_1\right| , \end{aligned} \end{aligned}$$

irrespective of which value $\gamma$ takes on.

This shows that the set $F^1:=\left\{ x\in F\,|\,{\widetilde{T}}\chi _{E_1}(x)\ge \frac{\alpha _1}{4}\right\}$ is non-empty. Furthermore:

$$\begin{aligned} \begin{aligned} \langle \chi _{E_1},{\widetilde{T}}^*\chi _{F^1}\rangle&=\langle \chi _{E_1},{\widetilde{T}}^*\chi _{F^1}\rangle -\langle {\widetilde{T}}\chi _{E_1\setminus E_1^1},\chi _{F^1}\rangle \\&\ge \frac{\alpha _1}{2}\left| F\right| -\frac{\alpha _1}{4}\left| F\right| \\&=\frac{\beta }{4}\left| E_1\right| =\frac{\alpha _1}{4}\left| F\right| . \end{aligned} \end{aligned}$$

Meaning we can define the non-empty set $E_1^1:=\left\{ y\in E_1\,|\, {\widetilde{T}}^*\chi _{F^1}(y)\ge \frac{\beta }{8}\right\}$, and by a similar argument, a non-empty set $F^2:=\left\{ x\in F^1\,|\,{\widetilde{T}}\chi _{E_1^1}(x)\ge \frac{\alpha _1}{16}\right\}$.

We are now ready to define $P_1,P_2,P_3$. First fix $x_0\in F^2$, and define

$$\begin{aligned} \begin{aligned} P_1:=&\left\{ z_1\in \Delta ^n_\varepsilon \setminus B_\gamma \,|\, x_0-\varGamma (z_1)\in E_1^1\right\} ,\\ P_2=P_{2,z_1}:=&\left\{ z_2\in \Delta ^n_\varepsilon \setminus B_\gamma \,|\, x_0-\varGamma (z_1)+\varGamma (z_2)\in F^1\right\} ,\\ P_3=P_{3,z_1,z_2}:=&\left\{ z_3\in \Delta ^n_\varepsilon \setminus B_{\alpha _2} \,|\, x_0-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3)\in E_2\right\} , \end{aligned} \end{aligned}$$

where it is understood that $z_i\in P_i$ when we are defining $P_j$, for $j>i$.

Notice that we have $\sigma (P_1)={\widetilde{T}}^*\chi _{F^1}(x_0)\ge \frac{\alpha _1}{16}$. Similarly $\sigma (P_2)={\widetilde{T}}\chi _{E_1^1}(x_0+\varGamma (z_1))\ge \frac{\beta }{8}$.

For $P_3$, notice that:

$$\begin{aligned} \begin{aligned} \sigma (P_3)&=\int _{\Delta ^n_\varepsilon \setminus B_{\alpha _2}}\chi _{E_2}(x_0-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3))\textrm{d}\sigma (z_3)\\&\ge \int _{\Delta ^n_\varepsilon }\chi _{E_2}(x_0-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3))\textrm{d}\sigma (z_3)-\sigma (B_{\alpha _2})\\&\ge \frac{\alpha _2}{2}. \end{aligned} \end{aligned}$$

Thus, as $P_1,\,P_2,\,$ and $P_3$ are contained within sectors with the appropriate balls removed from them, these sets indeed satisfy the conditions we initially laid out. From here, we must now construct subsets $P_i'\subset P_i$, for each i, satisfying the following separation conditions:

• if $z_2\in P_2'$, and $z_1\in P_1'$, then $\left| z_2-z_1\right| \ge c\beta ^\frac{1}{2}\left| z_1\right| ^{-\frac{k}{6}}$,

• if $z_3\in P_3'$, and $z_i\in P_i'$ then:

  • if $\left| z_i\right| \le \frac{1}{2}c_\nu \alpha _2^\nu$, then $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_3\right| ^{-\frac{k}{6}}$.

  • if $\left| z_i\right| \ge \frac{1}{2} c_\nu \alpha _2^\nu$, then $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}$.

where $c_\nu$ is as in the previous section, and c is some small constant. We also need to retain the property than $\sigma (P_1)\gtrsim \alpha _1$, $\sigma (P_2)\gtrsim \beta$, and $\sigma (P_3)\gtrsim \alpha _2$.

To show the cases in the second bullet point, we denote $B_x(w):=\Big \{z \in \Delta \,|\,\left| z-w\right| <c_0 x^\frac{1}{2}\left| w\right| ^{-\frac{k}{6}}\Big \}$, with $c_0$ to be specified. Also we must note that, for any $\alpha$, the condition $\left| z\right| \ge c_\nu \alpha ^\nu$ is equivalent to $\left| z\right| \ge c_\nu ^\frac{1}{2\nu }\alpha ^\frac{1}{2}\left| z\right| ^{-\frac{k}{6}}$, which is acquired by raising each side of the inequality to the power $\frac{1}{2\nu }$. We now prove the first of the two cases in this bullet point.

(1) if $\left| z_i\right| \le \frac{1}{2}c_\nu \alpha _2^\nu$, then $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}$:

This immediately follows from the fact that no element of $B_{\alpha _2}$ was in $P_3$, and $z_i\in B_{\frac{\alpha _2}{2}}$ which implies the separation required such that:

$$\begin{aligned}\left| z_3-z_i\right| \ge \frac{\left| z_3\right| }{2}\ge \frac{1}{2} c_\nu ^\frac{1}{2\nu }\alpha ^\frac{1}{2}\left| z_3\right| ^{-\frac{k}{6}}.\end{aligned}$$

(2) if $\left| z_i\right| \ge \frac{1}{2} c_\nu \alpha _2^\nu$, then $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}$:

In this case, we must first consider the sigma-measure of $B_{\alpha _2}(z_i)$ and show that it is a small multiple of $\alpha _2$, as we will wish to remove it from $P_3$, and ensure separation between the variables.

We will need to utilise the fact that if $z\in B_{\alpha _2}(z_i)$, it follows that:

$$\begin{aligned}\left| z-z_i\right|<c_0 \alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}<2c_0c_\nu \left| z_i\right| .\end{aligned}$$

which implies $\left| z\right| <\left( 1+2c_0c_\nu ^{-\frac{1}{2\nu }}\right) \left| z_i\right|$. From here, we can bound the size of $B_{\alpha _2}(z_i)$ via:

$$\begin{aligned} \begin{aligned} \sigma (B_{\alpha _2}(z_i))&=\int _{B_{\alpha _2}(z_i)}\left| z\right| ^\frac{k}{3} \textrm{d}z\\&\le \pi \left( c_0\alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}\right) ^2\times \left| \left( 1+2c_0c_\nu ^{-\frac{1}{2\nu }}\right) \left| z_i\right| \right| ^{\frac{k}{3}}\\&=\pi c_0^2\left( 1+2c_0c_\nu ^{-\frac{1}{2\nu }}\right) ^{\frac{k}{3}}\alpha _2=c_1\alpha _2. \end{aligned} \end{aligned}$$

Thus, we can choose $c_0$ sufficiently small, say to ensure $\sigma (B_{\alpha _2}(z_i))\ge \frac{\alpha _2}{4}$, and have the set $P_3'=P_3\setminus B_{\alpha _2}(z_i)$ satisfy the condition that $\sigma (P_3')\gtrsim \alpha _2$.

The first bullet point follows in an identical argument to this, by replacing $\alpha _2$ with $\alpha _1$. As both $P_1$ and $P_2$ contain no elements of $B_\gamma$, we know $\left| z_1\right| \ge c_\nu \frac{1}{2} \gamma ^\nu \ge c_\nu \frac{1}{2} \beta ^\nu$. Thus, we can ensure that, for some sufficiently small value of $c_0$, setting $P_2'=P_2\setminus B_{\beta }(z_1)$, yields $\sigma (P_2)\gtrsim \beta$. Lastly, leaving $P_1'=P_1$, we have our 3 further refined sets.

We need one final improvement on these separations that for $z_i\in P_i'$, we have:

• $\left| z_2-z_1\right| \gtrsim \beta ^{\frac{1}{2}}\left| z_2z_1\right| ^{-\frac{k}{12}}$,

• $\left| z_3-z_i\right| \gtrsim \alpha _2^{\frac{1+2\nu }{4}}\alpha _1^{\frac{1-2\nu }{4}}\left| z_3z_i\right| ^{-\frac{k}{12}}$.

Notice that if $\frac{\left| z_1\right| }{2}<\left| z_2\right| <2\left| z_1\right|$, the first estimate follows from the fact that $\left| z_2-z_1\right| \ge c\beta ^\frac{1}{2}\left| z_1\right| ^{-\frac{k}{6}}\sim \beta ^{\frac{1}{2}}\left| z_2z_1\right| ^{-\frac{k}{12}}$.

Similarly the second estimate follows in the case that $\frac{\left| z_i\right| }{2}<\left| z_3\right| <2\left| z_i\right|$, as either $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_i\right| ^{-\frac{k}{6}}$, or $\left| z_3-z_i\right| \ge c \alpha _2^\frac{1}{2}\left| z_3\right| ^{-\frac{k}{6}}$.

Therefore, we need only consider the case where there is sufficient separation of the variables.

First, for the first bullet point, assume $2\left| z_1\right| <\left| z_2\right|$, or $2\left| z_2\right| <\left| z_1\right|$. In either case, again using the fact that we have removed $B_\gamma$ from $P_1'$ and $P_2'$, we have:

$$\begin{aligned} \begin{aligned} \left| z_2-z_1\right|&\gtrsim \left| z_2\right| \\&=\left| z_2\right| ^{1+\frac{k}{12}}\left| z_1\right| ^{\frac{k}{12}}\left| z_1z_2\right| ^{-\frac{k}{12}}\\&\ge (c_\nu \gamma ^\nu )^{1+\frac{k}{12}}(c_\nu \gamma ^\nu )^{\frac{k}{12}}\left| z_1z_2\right| ^{-\frac{k}{12}}\\&\ge \beta ^{\frac{1}{2}}\left| z_1z_2\right| ^{-\frac{k}{12}}. \end{aligned} \end{aligned}$$

For the second bullet point that is to establish that $\left| z_3-z_i\right| \gtrsim \alpha _2^{\frac{1+2\nu }{4}}\alpha _1^{\frac{1-2\nu }{4}}\left| z_3z_i\right| ^{-\frac{k}{12}}$, again, we only need to deal with the case when $2\left| z_i\right| <\left| z_3\right|$ or $2\left| z_3\right| <\left| z_i\right|$. Using the same logic as last time, except utilising the fact that $\left| z_3\right| \gtrsim \alpha _2^\nu$ as well, we have:

$$\begin{aligned} \begin{aligned} \left| z_3-z_i\right|&\gtrsim \left| z_3\right| \\&\ge (c_\nu \alpha _2^\nu )^{1+\frac{k}{12}}(c_\nu \gamma ^\nu )^{\frac{k}{12}}\left| z_1z_2\right| ^{-\frac{k}{12}}\\&\ge \alpha _2^{\frac{1+2\nu }{4}}\alpha _1^{\frac{1-2\nu }{4}}\left| z_1z_2\right| ^{-\frac{k}{12}}. \end{aligned} \end{aligned}$$

With this, we can now finish the proof of this lemma, as, recalling that the image of $\varPhi =P_1'\times P_2'\times P_3'$ under the mapping $(z_1,z_2,z_3)\rightarrow H(z_1,z_2,z_3)=x_0-\varGamma (z_1)+\varGamma (z_2)-\varGamma (z_3)$ is contained in $E_2$, we have:

$$\begin{aligned} \begin{aligned} \left| E_2\right|&\gtrsim \iiint _\varPhi \left| J_{\mathbb {R}}H(z_1,z_2,z_3)\right| \textrm{d}z_1\textrm{d}z_2\textrm{d}z_3\\&=\iiint _\varPhi \left| J_{\mathbb {C}}H(z_1,z_2,z_3)\right| ^2\textrm{d}z_1\textrm{d}z_2\textrm{d}z_3\\&\gtrsim \iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3}\prod _{1\le i<j\le 3}\left| z_i-z_j\right| ^2\textrm{d}z_1\textrm{d}z_2\textrm{d}z_3\\&\gtrsim \beta \alpha _1^{1-2\nu }\alpha _2^{1+2\nu }\iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3} \left| z_1z_2z_3\right| ^{-\frac{k}{3}}\textrm{d}z_1\textrm{d}z_2\textrm{d}z_3 \\&\gtrsim \beta \alpha _1^{1-2\nu }\alpha _2^{1+2\nu } \sigma (P_1')\sigma (P_2')\sigma (P_3')\\&=\beta ^2\alpha _1^{2-2\nu }\alpha _2^{2+2\nu }, \end{aligned} \end{aligned}$$

so we are done. $\square$

Proof of Lemma 7

To prove the second lemma, we initially follow the same process as with the previous lemma, except with $\beta _i$ replacing $\alpha _i$, $\alpha$ replacing $\beta$, $F_i$ replacing $E_i$, and E replacing F. This results in a collection of 3 sets, $Q_1'$,$Q_2'$ and $Q_3'$, such that, for $z_i\in Q_i'$, and $\nu =\frac{3}{k+6}$, and $\gamma =\max (\beta _1,\alpha )$:

• $\sigma (Q_1')\gtrsim \beta _1$, and $\left| z_1\right| \ge c_\nu \gamma ^\nu$,

• $\sigma (Q_2')\gtrsim \alpha$ and $\left| z_2\right| \ge c_\nu \gamma ^\nu$,

• $\sigma (Q_3')\gtrsim \beta _2$ and $\left| z_3\right| \ge c_\nu \beta _2^\nu$.

As well as the following separations of our variables:

• $\left| z_2-z_1\right| \gtrsim \alpha ^{\frac{1}{2}}\left| z_2z_1\right| ^{-\frac{k}{12}}$,

• if $\left| z_i\right| <\frac{1}{2}c_\nu \beta _2^\nu$, then $\left| z_3-z_i\right| \gtrsim \beta _2^{\frac{1}{2}}\left| z_3\right| ^{-\frac{k}{6}}$,

• if $\left| z_i\right| \ge \frac{1}{2}c_\nu \beta _2^\nu$, then $\left| z_3-z_i\right| \gtrsim \beta _2^{\frac{1}{2}}\left| z_i\right| ^{-\frac{k}{6}}$.

If we continued following the previous proof from here we’d obtain $\left| F_2\right| \gtrsim \beta _1^{2-2\nu }\beta _2^{2+2\nu }\alpha ^2$. However, this would not establish our lemma.

We can however note that this is indeed sufficient if $\beta \gtrsim \alpha _1$, say $2\beta >\alpha _1$, as we now have $\left| F_2\right| \gtrsim \beta _1^{2-2\nu }\beta _2^{2+2\nu }\alpha ^2\gtrsim \beta _1^{2-2\nu }\beta _2^{2+2\nu }\alpha ^2\left( \frac{\alpha }{\beta _1}\right) \left( \frac{\beta _1}{\beta _2}\right) ^{2\nu }=\beta _1\beta _2^2\alpha ^3$.

Therefore, we need only establish Lemma 7 when $2\beta _1\le \alpha$. To deal with this case, we define $B_{\delta ,\alpha }(z_i)=\left\{ z\in \Delta \,|\,\left| z-z_i\right| <\delta \alpha ^{\frac{1}{2}}\left| z_1z_2\right| ^{-\frac{k}{12}}\right\}$. If we set $\delta =\min \left\{ \frac{c}{3},\frac{c_\nu ^{\frac{k}{6}+1}}{2}\right\}$, where c is the implicit constant in the inequality $\left| z_2-z_1\right| \gtrsim \alpha ^{\frac{1}{2}}\left| z_2z_1\right| ^{-\frac{k}{12}}$, we ensure the balls $B_{\delta ,\alpha }(z_1)$, and $B_{\delta ,\alpha }(z_2)$ are disjoint. We prove this lemma by considering whether $z_3$ is in one of these balls, or if it is neither.

First, if $z_3\in B_{\delta ,\alpha }(z_1)$, and therefore $z_3\notin B_{\delta ,\alpha }(z_2)$, we have the estimate $\left| z_3-z_1\right|<\delta \alpha ^{\frac{1}{2}}\left| z_1z_2\right| ^{-\frac{k}{12}}<\delta \alpha ^{\frac{1}{2}}\left| c_\nu \alpha ^\nu \right| ^{-\frac{k}{6}}<\delta c_\nu ^{-\frac{k}{6}}\alpha ^{\nu }<\delta c_\nu ^{-\frac{k}{6}-1}\left| z_1\right| \le \frac{\left| z_1\right| }{2}$.

We can therefore conclude $\left| z_3\right| \sim \left| z_1\right|$.

As $z_3\notin B_{\delta ,\alpha }(z_2)$, we have $\left| z_3-z_2\right| >\delta \alpha ^{\frac{1}{2}}\left| z_1z_2\right| ^{-\frac{k}{12}}\sim \alpha ^{\frac{1}{2}}\left| z_3z_2\right| ^{-\frac{k}{12}}.$

Furthermore, from the properties of the $Q_i'$, due to similarity in size of $z_1$ and $z_3$ we have that $\left| z_3-z_1\right| \gtrsim \beta _2^{\frac{1}{2}}\left| z_1z_3\right| ^{-\frac{k}{12}}$. We can therefore substitute these inequality into our lower bound for the measure of $F_2$ to get:

$$\begin{aligned} \begin{aligned} \left| F_2\right|&\gtrsim \iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3}\prod _{1\le i<j\le 3}\left| z_j-z_i\right| ^2\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \beta _2\alpha ^2\iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3} \left| z_1z_2z_3\right| ^{-\frac{k}{3}}\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \beta _1\beta _2^2\alpha ^3, \end{aligned} \end{aligned}$$

which establishes the lemma. An identical argument yields this result for $z_3\in B_{\delta ,\alpha }(z_2)$.

We now only have to deal with the case where $z_3\notin B_{\delta ,\alpha }(z_i)$ for $i=1,2$. In this situation, we need to further divide this case, based on the relative sizes of $z_1$ and $z_3$.

If $\left| z_3\right|<\left| z_1\right| <2\left| z_3\right|$, or $\left| z_1\right|<\left| z_3\right| <2\left| z_1\right|$, then we have that $\left| z_3-z_2\right| \gtrsim \alpha ^\frac{1}{2}\left| z_3z_2\right| ^{-\frac{k}{12}}$, as $\left| z_1\right| \sim \left| z_3\right|$ and $z_3\notin B_{\delta ,\alpha }(z_2)$. Similarly, from the initial list of properties of our $Q_i$, and as $\left| z_1\right| \sim \left| z_3\right|$, we have $\left| z_3-z_1\right| \gtrsim \beta _2^{\frac{1}{2}}\left| z_3z_1\right| ^{-\frac{k}{12}}$.

This is enough again to immediately observe $\left| F_2\right| \gtrsim \alpha ^2\beta _2\sigma (Q_1')\sigma (Q_2')\sigma (Q_3')\gtrsim \alpha ^3\beta _2^2\beta _1$, by the standard argument used up to here.

The final cases are $\left| z_3\right| >2\left| z_1\right|$ or $\left| z_1\right| >2\left| z_3\right|$. If this is the case, we obtain, using the fact that $\left| z_3-z_1\right| \gtrsim \left| z_3\right|$, that $z_3\notin B_{\delta ,\alpha }(z_2)$, and that $\left| z_3\right| \gtrsim \beta _2^\nu$, to obtain:

$$\begin{aligned} \begin{aligned} \left| F_2\right|&\gtrsim \iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3}\prod _{1\le i<j\le 3}\left| z_j-z_i\right| ^2\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \alpha ^2\iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{2k}{3} \left| z_1z_2\right| ^{-\frac{k}{3}}\left| z_3\right| ^{2}\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \alpha ^2\iiint _\varPhi \prod _{i=1}^3\left| z_i\right| ^\frac{k}{3}\left| z_3\right| ^{\frac{1}{\nu }}\textrm{d}z_3\textrm{d}z_2\textrm{d}z_1\\&\gtrsim \alpha ^2\beta _2\sigma (Q_1')\sigma (Q_2')\sigma (Q_3')\\&\gtrsim \alpha ^3\beta _2^2\beta _1. \end{aligned} \end{aligned}$$

Therefore, the lemma has been established in all cases, so we are done. $\square$