4 Combining the upper bounds
4.1 Preliminaries
Throughout this section, let \(\varepsilon {\gt}0\) be small but fixed. We now have everything in place to prove Theorem 1.4 for any
\[ \lambda \in (0,1+\delta -\varepsilon ). \]
and
\[ \delta \le 0.001-\varepsilon . \]
We shall write \(\delta \) as a symbol rather than its numerical value in order to clarify the argument. As will be clear from the proof, a somewhat larger value of \(\delta \) would also work.
Our goal is to prove that the upper bound in ?? holds for \(S^{*}_{\alpha ,\beta ,\gamma }(X)\), for any \(\alpha ,\beta ,\gamma \) such that
\begin{equation} \label{eq:goat} \alpha +\beta +\gamma \le \lambda {\lt} 1+\delta -\varepsilon . \end{equation}
4.1.1
The following result allows us to limit the range of \(\alpha ,\beta ,\gamma \) under consideration.
Proof
▶
This is an immediate consequence of 1.0.3 (with \(\varepsilon ^2/2\) in place of \(\varepsilon \)).
In the light of Proposition 2.6 (with \(\varepsilon ^2\) in place of \(\varepsilon \)), we wish to bound \(B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})\) for any pairwise coprime integers \(1\leq |c_1|,|c_2|,|c_3|\leq X^{\varepsilon ^2}\), any fixed \(d\geq 1\), and any choice of \(X_i,Y_i,Z_i\geq 1\), for \(1\leq i\leq d\) that satisfies ?? and ??. Moreover, \( \alpha ,\beta ,\gamma \) satisfy 4.1.1.
It will be convenient to define
Definition
4.2
Define \(a_i,b_i,c_i\in \mathbb {R}_{\geq 0}\) via
\[ X_i=X^{a_i}, \quad Y_i=X^{b_i}, \quad Z_i=X^{c_i}, \]
for \(1\leq i\leq d\), and \(a_i=b_i=c_i=0\) for \(i{\gt}d\).
Lemma
4.3
\begin{equation} \label{eq:oat1} \sum _{i\leq d} ia_i \leq 1, \quad \sum _{i\leq d} ib_i \leq 1, \quad 1-\varepsilon ^2\leq \sum _{i\leq d} ic_i \leq 1. \end{equation}
4.1.2
In particular, in the light of 4.1.1 and Proposition 4.1, we may henceforth assume that
Lemma
4.4
We have
\begin{equation} \label{eq:oat4} \sum _{i\leq d} (a_i+b_i)\geq 0.66-\varepsilon ^2,\quad \sum _{i\leq d} (a_i+c_i)\geq 0.66-\varepsilon ^2,\quad \sum _{i\leq d} (b_i+c_i) \geq 0.66-\varepsilon ^2 \end{equation}
4.1.3
and
\begin{equation} \label{eq:oat3} \sum _{i\leq d} (a_i+b_i+c_i) \leq 1+\delta -\varepsilon . \end{equation}
4.1.4
Proof
▶
Follows from 4.1.1 and Proposition 4.1.
Definition
4.5
It will be convenient to henceforth define
\begin{equation*} \nu =2\varepsilon ^2+\frac{\log B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})}{\log X}. \end{equation*}
Our goal is now to show that \(\nu \le 0.66\), since then ?? is a direct consequence of Proposition 2.6. This will then imply Theorem 1.4, via 2.0.1 and ??.
Before proceeding to the main tools that we shall use to estimate \(\nu \), we first show that we may assume that
\begin{equation*} 0.32 - \delta \leq \sum _{i\leq d} a_i ,~ \sum _{i\leq d} b_i ,~ \sum _{i\leq d} c_i \leq 0.34 + \delta - \frac{1}{2}\varepsilon , \end{equation*}
using the argument of Proposition 4.1.
Proposition
4.6
\begin{equation} \label{eq:oat5} 0.32 - \delta \leq \sum _{i\leq d} a_i ,~ \sum _{i\leq d} b_i ,~ \sum _{i\leq d} c_i \leq 0.34 + \delta - \frac{1}{2}\varepsilon , \end{equation}
4.1.5
Proof
▶
Indeed, suppose that \(\sum _{i\leq d} c_i{\gt} 0.34 + \delta -\varepsilon /2\). Then 4.1.4 implies that
\[ \sum _{i\leq d} (a_i + b_i){\lt} 0.66-\frac{1}{2}\varepsilon , \]
whence 1.0.3 yields \(\nu {\lt} 0.66\). This shows that we may suppose that the upper bound in 4.1.5 holds. Suppose next that \(\sum _{i\leq d} c_i{\lt} 0.32 - \delta \). Then, by the upper bound in 4.1.5, we have
\[ \sum _{i\leq d} (b_i + c_i){\lt} 0.66-\frac{1}{2}\varepsilon , \]
which is again found to be satisfactory, via 1.0.3.
Thus we may proceed under the assumption that the parameters \(a_i,b_i,c_i\) satisfy 4.1.4-4.1.5.
4.2 Summary of the main bounds
We now recast our bounds in Sections 3.1–3.3 in terms of an upper bound for \(\nu \), using the parameters \(a_i,b_i,c_i\). In all of the following bounds, we may freely permute the exponent vectors \((a_i),(b_i),(c_i)\).
Proposition
4.7
Fourier bound
\begin{align*} \nu & {\lt} \frac{1}{2}\Big(1+\delta +\sum _{i\leq d} \max (a_i,b_i) - \max _{m{\gt}1}(a_m,b_m)\Big). \end{align*}
Proof
▶
It follows from Proposition 3.1 that
\[ \nu \leq 3\varepsilon ^2+ \frac{1}{2}\sum _{i\leq d}\Big(a_i+b_i+c_i +\max (a_i,b_i) - \max _{m{\gt}1}b_m\Big). \]
Permuting variables, the claim now follows from 4.1.4.
Proposition
4.8
Geometry bound
\begin{align*} \nu {\lt} \delta + \min _{I,I',I''\subset [d]} \left( \max \left( 1 \, ,\, \sum _{i\in I} ia_i +\sum _{i\in I'} ib_i + \sum _{i\in I''} ic_i\right) - \sum _{i\in I} a_i -\sum _{i\in I'} b_i - \sum _{i\in I''} c_i\right). \end{align*}
Proof
▶
Applying Proposition 3.2, we obtain
\[ \nu \leq 3\varepsilon ^2+ \min _{I,I',I''} \left( \sum _{i\notin I} a_i +\sum _{i\notin I'} b_i + \sum _{i\notin I''} c_i + \max \left( 0 \, ,\, \sum _{i\in I} ia_i +\sum _{i\in I'} ib_i + \sum _{i\in I''} ic_i- \sum _{i\in [d]} ic_i \right)\right) , \]
where the minimum runs over subsets \(I,I',I''\subset [d]\). Taking the lower bound \(\sum _{i\in [d]} ic_i\ge 1-\varepsilon ^2\), from 4.1.2, it follows that
\begin{equation} \nu \leq 4\varepsilon ^2+ \min _{I,I',I''} \left( \sum _{i\notin I} a_i +\sum _{i\notin I'} b_i + \sum _{i\notin I''} c_i+ \max \left( 0 \, ,\, \sum _{i\in I} ia_i +\sum _{i\in I'} ib_i + \sum _{i\in I''} ic_i-1\right)\right).\label{eq:Geoalt} \end{equation}
4.2.1
The proof now follows from 4.1.4.
Proposition
4.9
Determinant Bound
\begin{align*} \nu {\lt} \min _{p,q\ge 1} \left( 1+\delta - a_p - b_q +\min \left(\frac{a_p}{q}, \frac{b_q}{p}\right)\right). \end{align*}
Proof
▶
Proposition 3.14 implies that
\[ \nu \leq 3\varepsilon ^2+ \sum _{i\leq d} (a_i+b_i+c_i) -\max _{p,q\geq 1} \left(\min \left( \frac{a_p}{q}, \frac{b_q}{p}\right) -a_p-b_q\right). \]
The claimed bound now follows from 4.1.4.
Proposition
4.10
Thue bound
\begin{align*} \nu {\lt} 1 +\delta - \max _{p\ge 2}\sum _{p\mid i}(a_i+b_i). \end{align*}
Proof
▶
This easily follows from Proposition 3.15 and 4.1.4.
4.3 Completion of the upper bound for \(\nu \)
Assuming that \(\delta \leq 0.001\) and \(\varepsilon {\gt}0\) is sufficiently small, the remainder of this paper is devoted to a proof of the upper bound
\[ \nu \le 0.66, \]
for any choice of parameters \(a_i,b_i ,c_i\) satisfying the properties recorded in 4.1.2- 4.1.5. From this point onward, we will tacitly assume those properties hold.
Definition
4.11
It will be convenient to define constants \(\delta _a,\delta _b,\delta _c\) via
\begin{equation} \label{eq:ai1/3} \sum _{i\leq d} a_i \, = \frac{1}{3}-\delta _a, \quad \sum _{i\leq d} b_i \, = \frac{1}{3}-\delta _b, \quad \sum _{i\leq d} c_i = \frac{1}{3}-\delta _c, \end{equation}
4.3.1
together with
\begin{equation*} \delta _{ab}:=\delta _a+\delta _b,\quad \delta _{ac}:=\delta _a+\delta _c, \quad \delta _{bc}:=\delta _b+\delta _c, \end{equation*}
and \(\delta _s := \delta _a+\delta _b+\delta _c\).
Lemma
4.12
We have
\begin{equation} \label{eq:robin2} \delta _{ab}, \delta _{ac}, \delta _{bc} \leq 0.00\overline{6} + \varepsilon ^2, \end{equation}
4.3.2
\begin{equation} \label{eq:robin1} -0.00\bar6 - \delta \leq \delta _a,\delta _b,\delta _c \leq 0.01\bar3+\delta +\varepsilon , \end{equation}
4.3.3
and
\begin{equation} \label{eq:robin3} -\delta {\lt} \delta _s \leq 0.01+\varepsilon . \end{equation}
4.3.4
Proof
▶
It follows from 4.1.3 that
\begin{equation*} \delta _{ab}, \delta _{ac}, \delta _{bc} \leq 0.00\overline{6} + \varepsilon ^2, \end{equation*}
and from 4.1.5 that
\begin{equation*} -0.00\bar6 - \delta \leq \delta _a,\delta _b,\delta _c \leq 0.01\bar3+\delta +\varepsilon , \end{equation*}
and from 4.1.4 that \(1-\delta _s\leq 1+\delta . \) Moreover, 4.3.2 implies \(2\delta _s = \delta _{ab}+ \delta _{ac}+ \delta _{bc}\leq 0.02 + 3\varepsilon ^2\), so we must have
\begin{equation*} -\delta {\lt} \delta _s \leq 0.01+\varepsilon . \end{equation*}
Definition
4.13
Define \(s_i := a_i + b_i + c_i\).
Referring to 4.1.2, it will be convenient to record the inequalities
\begin{equation} \label{eq:ai} \sum _{i\ge 2} (i-1)a_i \le \frac{2}{3}+\delta _a, \quad \sum _{i\ge 3} (i-2)a_i\le \frac{1}{3}+a_1 + 2\delta _a, \quad \sum _{i\ge 4}(i-3)a_i \le 2a_1+a_2+3\delta _a, \end{equation}
4.3.5
that follow by subtracting. Similar relations hold for \(b_i\) and \(c_i\), and thus for \(s_i\).
Proposition
4.14
To show \(\nu \leq 0.66\), it suffices to assume
\begin{equation} \label{eq:Thue} a_j+b_j, \, a_j+c_j,\, b_j+c_j {\lt} 0.34+\delta , \end{equation}
4.3.6
for each \(j\geq 2\), and moreover,
\begin{equation} \label{eq:Thue2} a_2+a_4+b_2+b_4, \ a_2+a_4+c_2+c_4, b_2+b_4+c_2+c_4 {\lt} 0.34+\delta . \end{equation}
4.3.7
Hence
\begin{equation} \label{eq:sThue} s_5,s_3,s_2+s_4 {\lt} 0.51+\frac{3}{2}\delta . \end{equation}
4.3.8
Proof
▶
By the Thue bound, we have
\[ \nu {\lt} 1 +\delta - \max _{p\ge 2}\sum _{p\mid i} (a_i+b_i), \]
and similarly for \(a_i+c_i\) and \(b_i+c_i\). Now 4.3.6 follows by taking \(p = j\) and restricting the sum to \(i = j\), while 4.3.7 follows by taking \(p = 2\) and restricting \(i \leq 4\).
Finally, 4.3.6 and 4.3.7 imply
\begin{align} s_5,s_3,s_2+s_4 {\lt} \frac{3}{2}(0.34+\delta )\le 0.51+\frac{3}{2}\delta . \end{align}
Lemma
4.15
We may assume
\begin{equation} \label{eq:SpecialGeometry} s_1+s_2\leq 0.34 +\delta . \end{equation}
4.3.10
Proof
▶
If \(s_1+s_2{\gt} 0.34+\delta \) then the Geometry bound and 4.3.8 imply that
\begin{align*} \nu & \le \max \big(1,\, s_1+2s_2\big) -s_1-s_2 + \delta \\ & = \max (1-s_1-s_2, s_2) +\delta \\ & {\lt} \max \left(0.66, 0.51+3\delta \right) = 0.66. \end{align*}
Thus we may proceed under the premise that
\begin{equation*} s_1+s_2\leq 0.34 +\delta . \end{equation*}
Lemma
4.16
For any \(j\ge 3\), allow \(\tau _j\) to be an element
\begin{equation} \label{eq:taujdef} \tau _j\in \{ a_j,b_j,c_j, s_j, a_j+b_j,a_j+c_j,b_j+c_j\} . \end{equation}
4.3.11
Then
\begin{equation} \label{eq:Geotauj} \tau _j\in \left(0.34-s_1-s_2+\delta , \frac{0.66 -s_2-\delta }{j-1}\right) \Longrightarrow \nu {\lt} 0.66 \end{equation}
4.3.12
and
\begin{equation} \label{eq:Geotau3} \tau _3\in \left(0.34-s_1+\delta , 0.33-\frac{1}{2}\delta \right) \Longrightarrow \nu {\lt} 0.66. \end{equation}
4.3.13
Proof
▶
The Geometry bound gives
\begin{align*} \nu & \le \max \big(1, s_1+2s_2+j\tau _j\big)-s_1-s_2-\tau _j +\delta \\ & = \max \big(1-s_1-s_2-\tau _j, s_2+(j-1)\tau _j \big)+\delta . \end{align*}
Thus we have \(\nu {\lt} 0.66\) if \(\tau _j\in (0.34-s_1-s_2+\delta , \frac{0.66-s_2-\delta }{j-1})\). In particular when \(j=3\), we have \(\nu {\lt} 0.66\) if \(\tau _3\in (0.34-s_1-s_2+\delta , 0.33-\frac{1}{2}s_2-\frac{\delta }{2}\)). Similarly, by the Geometry bound, we have
\begin{align*} \nu & \le \max \big(1, s_1+3\tau _3\big)-s_1-\tau _3+\delta \\ & = \max \big(1-s_1-\tau _3, 2\tau _3\big)+\delta . \end{align*}
Thus \(\nu {\lt} 0.66\) if \(\tau _3\in (0.34-s_1+\delta , 0.33-\frac{\delta }{2})\).
Lemma
4.17
We have
\begin{equation} \label{eq:black1} a_3 \ge \frac{1}{3}-4\delta _a - 3a_1 - 2a_2, \end{equation}
4.3.14
\begin{equation*} a_3 \ge \frac{1}{3}-\frac{5}{2}\delta _a - 2a_1 - \frac{3}{2}a_2 - \frac{1}{2}a_4. \end{equation*}
Therefore
\begin{equation} s_3 \ge 1-4\delta _s - 3s_1 - 2s_2, \label{eq:s3{\gt}s12} \end{equation}
4.3.15
and
\begin{equation} s_3 \ge 1-\frac{5}{2}\delta _s - 2s_1 - \frac{3}{2}s_2 - \frac{1}{2}s_4. \label{eq:s3{\gt}s124} \end{equation}
4.3.16
Proof
▶
Note that 4.3.5 gives \(\sum _{i\ge 4}a_i \le \sum _{i\ge 4}(i-3)a_i \le 2a_1+a_2+3\delta _a\). Similarly, we have \(\sum _{i\ge 5}a_i \le \frac{1}{2}\sum _{i\ge 5}(i-3)a_i\le \frac{1}{2}(2a_1+a_2-a_4+3\delta _a)\). These imply that
\begin{equation*} a_3 = \frac{1}{3}-\delta _a - a_1-a_2 - \sum _{i\ge 4}a_i \ge \frac{1}{3}-4\delta _a - 3a_1 - 2a_2, \end{equation*}
and
\[ a_3 = \frac{1}{3}-\delta _a - a_1-a_2 - a_4 - \sum _{i\ge 5}a_i \ge \frac{1}{3}-\frac{5}{2}\delta _a - 2a_1 - \frac{3}{2}a_2 - \frac{1}{2}a_4. \]
Analogous bounds hold for \(b_3\) and \(c_3\), and so we obtain
\begin{align*} s_3 & \ge 1-4\delta _s - 3s_1 - 2s_2, \\ s_3 & \ge 1-\frac{5}{2}\delta _s - 2s_1 - \frac{3}{2}s_2 - \frac{1}{2}s_4. \end{align*}
We shall need to split the argument according to whether \(s_2{\lt}0.3\) or \(s_2\geq 0.3\). Without loss of generality, we shall assume that \(a_3\ge b_3\ge c_3\) in all that follows.
Case 1: Assume \(s_2\ge 0.3\).
Lemma
4.18
If \(s_2 \geq 0.3\),
\begin{equation} \label{eq:s1} s_1 \leq 0.04+\delta \end{equation}
4.3.17
and \(s_4 {\lt} 0.21 +\frac{3}{2}\delta .\)
Proof
▶
Note that 4.3.10 gives
\begin{equation} s_1 \leq 0.34-s_2+\delta \le 0.04+\delta , \end{equation}
4.3.18
and 4.3.8 gives
\[ s_4 {\lt} 0.51-s_2 +\frac{3}{2}\delta \le 0.21 +\frac{3}{2}\delta . \]
We further split into subcases.
Subcase 1.1: Assume \(b_3\le 0.34-s_1-s_2+\delta \).
Lemma
4.19
\(\nu \leq 0.66\) in the case \(s_2 \geq 0.3\) and \(b_3\le 0.34-s_1-s_2+\delta \).
Proof
▶
Then
\begin{align*} b_3+c_3\le 2b_3 & \le 2(0.34-s_1-s_2+\delta )\\ & \le 0.68-2s_2 +2\delta \\ & \leq 0.33 -\frac{1}{2}s_2-\frac{1}{2}\delta , \end{align*}
for \(s_2\ge 0.3\) and \(\delta \leq 0.001\). Hence, in view of 4.3.13, we may assume \(b_3+c_3\le 0.34-s_1-s_2+\delta \). But then it follows from 4.3.16, 4.3.8 and 4.3.17 that
\begin{align*} a_3 = s_3 - (b_3 + c_3) & \ge 1 - \frac{5}{2}\delta _a- 2s_1 - \frac{3}{2}s_2 - \frac{1}{2}s_4 - (0.34-s_1-s_2+\delta )\\ & = 0.66 -\frac{5}{2}\delta _a - s_1 - \frac{1}{2}(s_2 + s_4)-\delta \\ & \geq 0.66 -\frac{5}{2}\delta _a - (0.04+\delta ) - \frac{1}{2}(0.51+\frac{3}{2}\delta )-\delta \\ & \ge 0.365 -\frac{5}{2}\delta _a -3\delta . \end{align*}
But 4.3.1 implies that \(\frac{1}{3}-\delta _a \ge a_3\geq 0.365 -\frac{5}{2}\delta _a -3\delta \). Thus 4.3.3 implies that
\[ 0.031\bar6{\lt}\frac{3}{2}\delta _a +3\delta \leq \frac{3}{2}(0.01\bar{3}+\delta +\varepsilon ) +3\delta \le 0.02+5\delta . \]
This contradicts our assumption \(\delta \leq 0.001\).
Subcase 1.2: Assume \(b_3{\gt} 0.34-s_1-s_2+\delta \).
Lemma
4.20
\(\nu \leq 0.66\) in the case \(s_2 \geq 0.3\) and \(b_3 {\gt} 0.34-s_1-s_2+\delta \).
Proof
▶
By 4.3.13 we may assume
\begin{equation} \label{eq:b3} b_3\geq 0.33 -\frac{1}{2}s_2-\frac{1}{2}\delta . \end{equation}
4.3.19
By permuting the variables in 4.3.5, we have
\[ \sum _{i\ge 4} (i-2)b_i\le \frac{1}{3}-b_3+b_1 + 2\delta _b. \]
We also have \(b_1\leq s_1\leq 0.34 -s_2+\delta \) by 4.3.10. Thus
\begin{align*} \sum _{i\geq 4}b_i \le \frac{1}{2}\left(\frac{1}{3}+b_1-b_3+2\delta _b\right) & \le \frac{1}{2}\Big(\frac{1}{3}+0.34 -s_2+\delta - (0.33 -\frac{s_2}{2} -\frac{\delta }{2}) +2\delta _b\Big)\\ & {\lt} 0.33-\frac{1}{2}s_2-\frac{\delta }{2}, \end{align*}
since 4.3.3 ensures that \(\delta _b\leq 0.01\bar3+\delta +\varepsilon \) (and we have \(\delta \leq 0.001\)). A fortiori the same bound holds for \(\sum _{i\geq 4}a_i\). Thus, in the light of 4.3.13, taking \(\varepsilon {\gt}0\) small we may assume that
\[ a_4,b_4,a_5,b_5,a_6,b_6\leq 0.34-s_1-s_2+\delta . \]
Now write \(M_i = \max (a_i,b_i)\) and \(m_i = \min (a_i,b_i)\), so that \(m_i+M_i=a_i+b_i\). By the Fourier bound we have
\begin{align*} \nu & {\lt} \frac{1}{2}\Big(1+\delta + \sum _{i\leq d} \max (a_i,b_i) - \max (a_2,b_2)\Big) = \frac{1}{2}\Big(1 +\delta + \sum _{i\neq 2}M_i\Big). \end{align*}
On using 4.3.1, this implies that
\begin{align*} 2\nu -1-\delta {\lt} \sum _{i\neq 2} M_i & \le \sum _{2\neq i \le 6}M_i + \sum _{i\ge 7} (a_i+b_i)\\ & = \sum _{2\neq i \le 6}M_i + \frac{2}{3} -\delta _{ab} - \sum _{i\le 6}(a_i+b_i)\\ & = \frac{2}{3} -\delta _{ab}-\sum _{2\neq i \le 6}m_i - (a_2+b_2). \end{align*}
Next we lower bound \(a_2+b_2\). To do this, we observe that by 4.3.5 we have
\[ 4\sum _{i\ge 7} a_i \le \sum _{i\ge 7}(i-3)a_i = 2a_1+a_2+3\delta _a -a_4-2a_5-3a_6, \]
whence
\begin{align*} \frac{1}{3} -\delta _a& = \sum _{i}a_i \le \sum _{i\le 6}a_i +\frac{1}{4}(2a_1+a_2+3\delta _a -a_4-2a_5-3a_6) = \frac{1}{4}\sum _{i\le 6}(7-i)a_i +\frac{3}{4}\delta _a. \end{align*}
Thus \(a_2 \ge \frac{4}{15} - \frac{1}{5}\sum _{2\neq i\le 6}(7-i)a_i-\frac{7}{5}\delta _a\), and similarly \(b_2 \ge \frac{4}{15} - \frac{1}{5}\sum _{2\neq i\le 6}(7-i)b_i-\frac{7}{5}\delta _b\). Since \(m_3=b_3\), it now follows that
\begin{align} \label{eq:2v-1.2} 2\nu -1-\delta & {\lt} \frac{2}{3}-\delta _{ab} -\sum _{2\neq i \le 6}m_i - \Big(\frac{8}{15} - \frac{1}{5}\sum _{2\neq i \le 6}(7-i)(a_i+b_i)-\frac{7}{5}\delta _{ab}\Big) \nonumber \\ & \le \frac{2}{15} +\frac{2}{5}\delta _{ab}+ \frac{1}{5}\Big(6M_1+m_1 + 4a_3-b_3 +3M_4+2M_5+ M_6\Big). \end{align}
Thus, using 4.3.19 and the bound \(a_3+b_3\leq 0.34+\delta \) coming from 4.3.6, we have
\begin{align*} 4a_3-b_3 & \le 4(0.34-b_3+\delta )-b_3 \\ & {\lt} 4(0.01+\frac{s_2}{2}+\frac{3\delta }{2}) - (0.33-\frac{s_2}{2}-\frac{\delta }{2})\\ & = \frac{5}{2}s_2-0.29+\frac{13\delta }{2}. \end{align*}
Also \(6M_1+m_1\le 6s_1\) and recall \(M_4,M_5,M_6 \le 0.34-s_1-s_2+\delta \). Hence plugging back into 4.3.20, we conclude
\begin{align*} 2\nu -1-\delta & {\lt} \frac{2}{15}+ \frac{2}{5}\delta _{ab} + \frac{1}{5}\Big(6s_1 + (\frac{5}{2}s_2-0.29+\frac{13\delta }{2})+6(0.34-s_1-s_2+\delta )\Big)\\ & {\lt} \frac{2}{15}+ \frac{2}{5}\delta _{ab} + \frac{1}{5}\Big(1.75-\frac{7}{2}s_2+13\delta \Big)\\ & \le 0.48\overline{3} - \frac{7}{10}s_2 +\frac{2}{5}\delta _{ab} +\frac{13}{5}\delta \\ & \le 0.48\overline{3} - \frac{7}{10}(0.3) +\frac{2}{5}(0.00\bar{6}+\varepsilon ^2) +\frac{13}{5}\delta {\lt} 0.279, \end{align*}
since \(s_2\geq 0.3\), \(\delta \le 0.001\), and 4.3.2 implies that \(\delta _{ab}\leq 0.00\bar{6}+\varepsilon ^2\). Hence \(\nu \le \frac{1.3}{2} = 0.65\), which is more than satisfactory.
Lemma
4.21
\(\nu \leq 0.66\) in the case \(s_2 \geq 0.3\).
Proof
▶
Immediate from Lemmas 4.19 and 4.20.
Case 2: Assume \(s_2{\lt} 0.3\).
Lemma
4.22
\begin{equation} \label{eq:2b3} b_3 {\lt} 0.17+\frac{\delta }{2}. \end{equation}
4.3.21
Proof
▶
It follows from 4.3.6 that
\begin{equation} 2b_3 \le a_3 + b_3 {\lt} 0.34+\delta , \end{equation}
4.3.22
whence \(b_3 {\lt} 0.17+\frac{\delta }{2}\).
Lemma
4.23
\begin{equation} \label{eq:Case2Basic2} a_3 \geq 0.32 - 4 \delta _s - s_1 - 2 \delta . \end{equation}
4.3.23
Proof
▶
We have \(0.17+\frac{\delta }{2} \leq 0.33-\frac{s_2}{2}-\frac{\delta }{2}\) since \(s_2{\lt}0.3\) and \(\delta \leq 0.001\). Thus, in view of 4.3.13, we may assume that \(b_3,c_3 \leq 0.34-s_1-s_2+ \delta \). Then 4.3.15 gives
\begin{align*} a_3 = s_3 - (b_3+c_3) & \ge 1-4\delta _s - 3s_1 - 2s_2 - 2(0.34 - s_1 - s_2+\delta ) \nonumber \\ & = 0.32-4\delta _s -s_1-2\delta . \end{align*}
We shall proceed by separately handing the subcases
\[ \mathbf{2.1}:\ a_3\geq 0.32 \qquad \qquad \qquad \mathbf{2.2}: b_3+c_3{\lt}0.33-\frac{s_2}{2}-\frac{\delta }{2}. \]
These will be instrumental to proving the following subcases
\begin{align*} \mathbf{2.3}& :\ 4s_1+3s_2 {\gt} 0.71, & \mathbf{2.5}& :\ 0.066\leq s_2\leq 0.204,\\ \mathbf{2.4}& :\ 4s_1+\ s_2 {\lt}0.4 , & \mathbf{2.6}& :\ 2s_1-s_2{\gt}0.025. \end{align*}
Handling these subcases will complete the proof.
Lemma
4.24
Assuming 4.27, 4.28, 4.29, 4.30, \(\nu \leq 0.66\) in the case \(s_2 {\lt} 0.3\).
Proof
▶
Indeed \(\mathbf{2.3}, \mathbf{2.4}, \mathbf{2.6}\) each define half-planes that cover \([0,1]^2\setminus T\), for the closed triangle \(T\) with vertices
\[ (s_1,s_2)\in \{ (0.06125,0.155), (0.0785,0.132), ( 0.0708\bar3, 0.11\bar6)\} . \]
But then \(\mathbf{2.5}\) covers \(T\). Hence subcases \(\mathbf{2.3}\)–\(\mathbf{2.6}\) will complete the proof of Case 2.
Subcase \(\bf {2.1}\): Assume \(a_3\geq 0.32\)
Lemma
4.25
\(\nu \leq 0.66\) in the case \(s_2{\lt}0.3\) and \(a_3\geq 0.32\).
Proof
▶
By 4.3.6 we have \(b_3,c_3\le 0.34+\delta -a_3\le 0.02+\delta \). Let \(m_i=\min (b_i,c_i)\), \(M_i=\max (b_i,c_i)\), and \(t_i=b_i+c_i=m_i+M_i\). If \(M :=\max _{i\ge 4}M_i {\gt} \frac{3}{4}(0.09)\), then using \(\delta \leq 0.001\) and the Determinant bound (with variables permuted) yields
\begin{align*} \nu \le 1 +\delta - a_3 - M + \min \left(\frac{M}{3}, \frac{a_3}{4}\right) \le 1+\delta - a_3 - \frac{2}{3}M & \le 1+\delta -0.32 - \frac{1}{2}(0.09) \le 0.636. \end{align*}
This is satisfactory. We may therefore assume that \(M_i\le \frac{3}{4}(0.09)\) for \(i\ge 4\). Then \(t_i\le 2M_i\le 0.135\) for \(i\ge 4\). Moreover, \(\sum _{i}(i-1)t_i \le \frac{4}{3}+\delta _{bc}\), by 4.1.2 and 4.3.1. Appealing to the Geometry bound in the form 4.2.1, we deduce that
\begin{align*} \nu \le \varepsilon + a_3+b_3 + m_4+\sum _{i\ge 5}t_i + \max \Big(0 , \sum _i is_i - 3(a_3+b_3) - 4m_4-\sum _{i\ge 5}it_i - 1\Big). \end{align*}
Thus we have \(\nu \le \max (\nu _1,\nu _2)+\varepsilon \), where
\begin{align*} \nu _1 := a_3+b_3 + m_4+\sum _{i\ge 5}t_i \qquad {\rm and}\qquad \nu _2 := \sum _i is_i -2(a_3+b_3) - 3m_4 - \sum _{i\ge 5}(i-1)t_i -1. \end{align*}
Using 4.3.5, we see that
\begin{align*} \nu _1 & \le a_3+b_3 +m_4+t_5+\frac{1}{5}\sum _{i\ge 6}(i-1)t_i\\ & \le a_3 + b_3 + \frac{1}{2}t_4+t_5+\frac{1}{5}\Big(\frac{4}{3}+\delta _{bc}-t_2-2t_3- 3t_4-4t_5\Big). \end{align*}
Using \(a_3+b_3\le 0.34+\delta \) (which follows from 4.3.21), \(\delta _{bc}{\lt}0.00\overline{6}+\varepsilon ^2\), and \(t_5\le 0.135\), we conclude that
\[ \nu _1 \le a_3+b_3 + \frac{1}{5}\Big(\frac{4}{3}+\delta _{bc}+t_5\Big) \le 0.34+ \delta + \frac{1}{5}\Big(\frac{4}{3}+0.00\overline{6}+\varepsilon ^2+0.135\Big) {\lt} 0.637. \]
Similarly, on recalling \(\sum _i i a_i \le 1\), we have \(\sum _i is_i -1 \le \sum _i it_i\), whence
\begin{align*} \nu _2 & \le \sum _i it_i - \sum _{i\ge 5}(i-1)t_i -2(a_3+b_3) - 3m_4\\ & = \sum _{i} t_i + \sum _{i\le 4}(i-1)t_i -2(a_3+b_3) - 3m_4\\ & = \frac{2}{3} -\delta _{bc}+ t_2 - 2(a_3-c_3) +3M_4, \end{align*}
by 4.3.1. Using \(t_2\le s_2{\lt} 0.3\), \(c_3\le 0.02+\delta \), and \(a_3\ge 0.32\) by assumption, we conclude that
\begin{align*} \nu _2 & {\lt} \frac{2}{3} -\delta _{bc} + 0.3 - 2(0.3-\delta ) + 3\cdot \frac{3}{4}(0.09) {\lt} 0.57-\delta _{bc}+2\delta . \end{align*}
Thus \(\nu _2{\lt}0.6\), since 4.3.3 implies that \(\delta _{bc}\geq -0.01\bar3-2\delta \), and \(\delta \leq 0.001\). Combining the bounds for \(\nu _1\) and \(\nu _2\), we conclude that \(\nu \le \max (\nu _1,\nu _2)+\varepsilon {\lt}0.64\), which suffices.
Subcase \(\mathbf{2.2}\): Assume \(b_3+c_3{\lt}0.33-\frac{s_2}{2}-\frac{\delta }{2}\).
Lemma
4.26
\(\nu \leq 0.66\) in the case \(s.2 {\lt} 0.3\) and \(b_3+c_3{\lt}0.33-\frac{s_2}{2}-\frac{\delta }{2}\).
Proof
▶
Then by 4.3.13 we may assume \(\tau _3=b_3+c_3{\lt}0.34-s_1-s_2+\delta \). By 4.3.16 we have
\begin{align*} a_3 = s_3 - (b_3+c_3) & {\gt} 1 - \frac{5}{2}\delta _s - 2s_1 - \frac{3}{2}s_2 - \frac{1}{2}s_4 - (0.34-s_1-s_2+ \delta )\\ & = 0.66- \frac{5}{2}\delta _s - s_1 - \frac{1}{2}(s_2+s_4)-\delta . \end{align*}
It follows from 4.3.4 that \(\delta _s\leq 0.01+\varepsilon \) and from 4.3.8 that \(s_2+s_4{\lt}0.51+3\delta /2\). Hence
\begin{align*} a_3 & {\gt} 0.66- \frac{5}{2}(0.01+\varepsilon ) - s_1 - \frac{1}{2}(0.51+\frac{3\delta }{2}) - \delta \ge 0.38-s_1- 3\delta . \end{align*}
Since \(\delta \le 0.001\), we see that \(a_3{\gt} 0.34-s_1+ \delta \). Thus it follows from 4.3.13 that we may assume \(\tau _3=a_3{\gt}0.33-\frac{\delta }{2}\geq 0.32\). Hence Subcase \(\mathbf{2.1}\) completes the proof.
Subcase \(\mathbf{2.3}\): Assume \(4s_1+3s_2{\gt}0.71\).
Lemma
4.27
\(\nu \leq 0.66\) in the case \(s.2 {\lt} 0.3\) and \(4s_1+3s_2{\gt}0.71\).
Proof
▶
Then the inequalities \(b_3,c_3 \le 0.34-s_1-s_2+ \delta \) give
\[ b_3+c_3{\lt}0.68-2(s_1+s_2)+2\delta {\lt} 0.325 -\frac{s_2}{2}+2\delta . \]
Since \(\delta \leq 0.001\), we see that \(b_3+c_3{\lt}0.33-\frac{s_2}{2}-\frac{\delta }{2}\). Hence Subcase \(\mathbf{2.2}\) completes the proof.
Subcase \(\mathbf{2.4}\): Assume \(4s_1+s_2{\lt}0.4\).
Lemma
4.28
\(\nu \leq 0.66\) in the case \(s.2 {\lt} 0.3\) and \(4s_1+s_2{\lt}0.4\).
Proof
▶
In this case, 4.3.6 and 4.3.23 give
\[ b_3,c_3 \le 0.34-a_3+\delta \le 0.34 - (0.32-4\delta _s-s_1-2\delta ) +\delta = 0.02+4\delta _s+s_1+3\delta . \]
In view of 4.3.4 and our assumption \(4s_1+s_2{\lt}0.4\), we deduce that
\begin{align*} b_3+c_3 \le 0.12+8\varepsilon + 2s_1+6\delta {\lt} 0.32-\frac{s_2}{2}+6\delta +8\varepsilon . \end{align*}
Since \(\delta \leq 0.001\), we have \(b_3+c_3 \le 0.33-\frac{s_2}{2}-\frac{\delta }{2}\). Hence Subcase \(\mathbf{2.2}\) completes the proof.
Subcase \(\mathbf{2.5}\): Assume \(0.066\leq s_2\leq 0.204\).
Lemma
4.29
\(\nu \leq 0.66\) in the case \(s.2 {\lt} 0.3\) and \(0.066\leq s_2\leq 0.204\).
Proof
▶
It follows from 4.3.4 and 4.3.23 that
\[ a_3 \geq 0.32-4\delta _s-s_1 -2\delta \geq 0.28-4\varepsilon -s_1 -2\delta . \]
Thus \(a_3{\gt}0.34-s_1-s_2 +\delta \), since \(s_2\geq 0.066 \ge 0.062+4\varepsilon +3\delta \) and \(\delta \leq 0.001\). It now follows from 4.3.13 that we may assume \(\tau _3=a_3 \geq 0.33 -\frac{s_2}{2}- \frac{\delta }{2}\). Thus 4.3.6 gives \(b_3,c_3 \le 0.34-a_3 +\delta {\lt} 0.01 +\frac{s_2}{2}+ \frac{3\delta }{2}\), which in turn gives \(b_3+c_3 {\lt} 0.02+s_2 +3\delta \). Since \(s_2\leq 0.204\) and \(\delta \leq 0.001\), we deduce that \(b_3+c_3{\lt}0.33-\frac{s_2}{2}-\frac{\delta }{2}\). Hence Subcase \(\mathbf{2.2}\) completes the proof.
Subcase \(\mathbf{2.6}\): Assume \(2s_1-s_2{\gt}0.025\).
Lemma
4.30
\(\nu \leq 0.66\) in the case \(s_2{\lt}0.3\) and \(2s_1-s_2{\gt}0.025\).
Proof
▶
In this case we note that the intervals in 4.3.13 overlap, since \(\delta \leq 0.001\). Hence for any \(\tau _3\) belonging to the set 4.3.11, we have
\begin{equation} \label{eq:tau3S6} \tau _3\in \left(0.34-s_1-s_2+ \delta , \; 0.33-\frac{\delta }{2}\right) \Longrightarrow \nu {\lt} 0.66. \end{equation}
4.3.23
Furthermore, in the light of Subcases \(\mathbf{S_3}\) and \(\mathbf{S_4}\), we may assume that \(4s_1+3s_2\leq 0.71\) and \(4s_1+s_2\geq 0.4\). In particular, these imply that \(s_1\leq \frac{0.71}{4}\leq 0.1775\) and \(s_2 \leq \frac{1}{3}(0.71 - 4s_1) \leq \frac{1}{3}(0.71-0.4+s_2)\), so that \(s_2\leq 0.155\). Then, on appealing to Subcase \(\mathbf{S}_5\), we may assume that \(s_2{\lt} 0.066\). Similarly, it follows from Subcase \(\mathbf{S}_1\) that we may also assume \(a_3{\lt}0.32\). Thus 4.3.23 and the bound \(\delta \leq 0.001\) imply that we may assume \(a_3{\lt} 0.34-s_1-s_2+ \delta \).
If we also had \(b_3+c_3{\lt}0.34-s_1-s_2+\delta \), then we would have \(s_3 = a_3 + b_3 +c_3 {\lt} 0.68-2s_1-2s_2+2\delta \). Combining this with 4.3.15, we would then conclude that
\[ 0.68-2s_1-2s_2+2\delta {\gt} s_3 \ge 1-4\delta _s-3s_1-2s_2, \]
which implies that \(s_1{\gt}0.32-4\delta _s-2\delta \). Recalling 4.3.4 and the inequalities \(s_1\leq 0.1775 \) and \(\delta \leq 0.001\), this is a contradiction. Hence we may assume that \(b_3+c_3\ge 0.34-s_1-s_2+ \delta \), and by 4.3.23, we may assume \(\tau _3=b_3+c_3\geq 0.33-\frac{\delta }{2}\), so \(b_3{\gt}0.165-\frac{\delta }{4}{\gt} 0.164\). Thus we have \(a_3,b_3\in [0.164, 0.341-s_1-s_2]\), since \(\delta \leq 0.001\). In particular the interval is nontrivial, so \(s_1+s_2\le 0.1775\).
Letting \(M_i = \max (a_i, b_i)\) and \(m_i = \min (a_i, b_i)\), it follows from the Fourier bound that
\begin{align*} \nu & {\lt} \frac{1}{2}\Big(1+\delta +\sum _{i}\max (a_i, b_i) - \max (a_3,b_3)\Big) = \frac{1}{2}\Big(1+\delta +\sum _{i\neq 3}M_i\Big). \end{align*}
It follows from 4.3.1 that \(\sum _i (M_i+m_i)=\frac{2}{3}-\delta _{ab}\), and so
\begin{align*} 2\nu -1 -\delta \le \sum _{i\neq 3}M_i & \le M_1+M_2 + \sum _{i\ge 4}(M_i+m_i)\\ & = M_1+M_2 + \Big(\frac{2}{3} - \delta _{ab} - \sum _{i\le 3}(M_i+m_i)\Big)\\ & \le \frac{2}{3} - \delta _{ab} - m_1 - m_2 - m_3 - M_3. \end{align*}
By 4.3.14 we have \(3a_1 + 2a_2 + a_3 \ge \frac{1}{3}-4\delta _a\), and similarly, \(3b_1 + 2b_2 +b_3 \ge \frac{1}{3}-4\delta _b\). Thus \(m_1 \ge \frac{1}{3}(\frac{1}{3}-2M_2 - M_3-4\max (\delta _a,\delta _b))\). This together with the bounds \(\delta \leq 0.001\) and 4.3.3 lead to the upper bound
\begin{align*} 2\nu -1 -\delta & \le \frac{2}{3} -\delta _{ab} + \frac{1}{3}(2M_2 + M_3-\frac{1}{3}+4\max (\delta _a,\delta _b)) - m_2 - m_3 - M_3 \\ & \le \frac{5}{9} + \frac{2}{3}M_2 +\frac{1}{3}\max (\delta _a,\delta _b)- \min (\delta _a,\delta _b) - m_3 - \frac{2}{3}M_3 \\ & \le \frac{5}{9} + \frac{2}{3}M_2+ \frac{1}{3}(0.01\bar{3}+\delta +\varepsilon ) + (0.00\bar{6}+\delta ) - m_3 - \frac{2}{3}M_3\\ & \le 0.568+\frac{2}{3}(M_2-M_3)- m_3, \end{align*}
using \(\delta _a,\delta _b\in [-0.00\bar{6}-\delta ,\, 0.01\bar{3}+\delta +\varepsilon ]\) and \(b_3 + c_3 \ge 0.33 - \frac{\delta }{2}\). Thus we have
\begin{align} \label{eq:nuleqM2} \nu & \le 0.784 + \frac{1}{3}\big(\max (a_2,b_2)-\max (a_3,b_3)\big) - \frac{1}{2}\min (a_3,b_3) \nonumber \\ & \le 0.784 + \frac{1}{3}\big(\max (a_2,b_2)-a_3\big) - \frac{1}{2}b_3 \end{align}
Next, if \(a_2{\lt}b_2\), let \(e_i := a_i\) for all \(i\ge 1\), otherwise let \(e_i := b_i\) for all \(i\ge 1\). In particular, note \(e_2 = \min (a_2,b_2)\) and \(e_3 \ge \min (a_3,b_3)\ge c_3\). By a similar argument with \((a_i,b_i)_i\) replaced by \((e_i,c_i)_i\), we have
\begin{align*} \nu & \le 0.784 + \frac{1}{3}\big(\max (e_2,c_2)-\max (e_3,c_3)\big) - \frac{1}{2}\min (e_3,c_3)\\ & \le 0.784 + \frac{1}{3}\big(\max (e_2,c_2)-e_3\big) - \frac{1}{2}c_3. \end{align*}
Averaging the bound with 4.3.24, we obtain
\begin{align} \nu & \le 0.784 + \frac{1}{6}\big(\max (a_2,b_2)+\max (e_2,c_2)-a_3-e_3\big) - \frac{1}{4}(b_3+c_3) \nonumber \\ & \le 0.784 + \frac{1}{6}s_2 - \frac{5}{12}(b_3+c_3) \nonumber \\ & \le 0.784 + \frac{1}{6}(0.066) - \frac{5}{12}(0.33 - \tfrac {\delta }{2}) \le 0.658. \end{align}
Here we used \(\max (a_2,b_2)+\max (e_2,c_2) = \max \big(a_2+b_2, \, \max (a_2,b_2) + c_2\big) \le s_2\) and \(a_3+e_3 \ge b_3+c_3 \ge 0.33 - \frac{\delta }{2}\).
This completes the proof.
Proof
▶
Immediate from Lemmas 4.21 and 4.24.