In this section, we show that the boundary of \(\mathcal {A}_{3,n}\) is contained in \(\mathcal {B}^{+}_{n}\cup \mathcal {B}^{-}_{n}\) for \(n=4\) and conclude from this that the same statement holds for all \(n\ge 3\). We need to show that no point of
$$\begin{aligned} \left\{ k\begin{pmatrix}s\\ s^2\\ s^3\end{pmatrix}+\ell \begin{pmatrix}t\\ t^2\\ t^3\end{pmatrix}\,\Bigg | -1< s< t<1\right\} +a\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}+b\begin{pmatrix}-1\\ 1\\ -1\end{pmatrix} \end{aligned}$$
is contained in the boundary of \(\mathcal {A}_{3,4}\) for \((k,\ell ,a,b)=(1,1,1,1),(1,2,1,0)\), (2, 1, 0, 1), (2, 2, 0, 0). We start with the case \((k,\ell ,a,b)=(2,2,0,0)\).
Proposition 4.1
Take \(-1<s<t<1\). Then the point
$$\begin{aligned} p=2(s,s^2,s^3)+2(t,t^2,t^3) \end{aligned}$$
does not lie on the boundary of \(\mathcal {A}_{3,4}\).
Proof
Consider the system of equations
$$\begin{aligned} 2s+2t&=t_1+t_2+t_3+t_4,\\ 2s^2+2t^2&=t_1^2+t_2^2+t_3^2+t_4^2,\\ 2s^3+2t^3&=t_1^3+t_2^3+t_3^3+t_4^3 \end{aligned}$$
with the additional conditions that \(-1<t_1,t_2,t_3,t_4<1\) are pairwise distinct. If this system has a solution that satisfies the additional conditions, then the point p cannot lie on the boundary of \(\mathcal {A}_{3,4}\) by Proposition 3.1. It turns out that such a solution \((t_1,t_2,t_3,t_4)\) can even be found when we assume that \(t_1+t_2=t_3+t_4\). Indeed, let \(0\ne \alpha \ne \beta \ne 0\) be such that \(|\alpha |,|\beta |<1-\frac{1}{2}|s+t|\) and \(\alpha ^2+\beta ^2=\frac{1}{2}(s-t)^2\). Then
$$\begin{aligned} (t_1,t_2,t_3,t_4)=\left( \frac{s+t}{2}+\alpha ,\frac{s+t}{2}-\alpha ,\frac{s+t}{2}+\beta ,\frac{s+t}{2}-\beta \right) \end{aligned}$$
is a solution to the system equalities so that \(-1<t_1,t_2,t_3,t_4<1\) are pairwise distinct. One can check that \(|\alpha |,|\beta |<1-\frac{1}{2}|s+t|\) and \(\alpha ^2+\beta ^2=\frac{1}{2}(s-t)^2\) for \(\alpha ,\beta =\pm \frac{1}{2}(s-t)\). Here we use that \(|s-t|+|s+t|\le 2\cdot \max (|s|,|t|)<2\). It follows that for any point \((\alpha ,\beta )\) on the circle given by
$$\begin{aligned} \alpha ^2+\beta ^2=\frac{1}{2}(s-t)^2 \end{aligned}$$
that is sufficiently close to \((\frac{1}{2}(s-t),\frac{1}{2}(s-t))\) also satisfies these conditions. So to conclude the proof, we simply let \((\alpha ,\beta )\) be such a point with \(0\ne \alpha \ne \beta \ne 0\). \(\square \)
Next, we take care of the case \((k,\ell ,a,b)=(1,1,1,1)\).
Lemma 4.2
Take \(\delta >0\). Then there exists an \(\varepsilon >0\) such that
$$\begin{aligned} (s,s^2,s^3)+(t,t^2,t^3)+(1,1,1)-(0,0,\varepsilon ')\in \mathcal {A}_{3,3} \end{aligned}$$
for all \(-1<s<t<1\) and \(0\le \varepsilon '\le \varepsilon \) with \(1-t,t-s,s-(-1)\ge \delta \).
Proof
For \(0\le \lambda \ll 1\), consider \((t_1,t_2,t_3)=(s+\mu ,t+\lambda -\mu ,1-\lambda )\) where
$$\begin{aligned} \mu&=(\lambda +(t-s)-\sqrt{d})/2,\\ d&=(t-s)^2+2(t-s)\lambda +4(1-t)\lambda -3\lambda ^2. \end{aligned}$$
Assuming that \(d\ge 0\) and \(-1\le t_1,t_2,t_3\le 1\), one can check that
$$\begin{aligned} (s,s^2,s^3)+(t,t^2,t^3)+(1,1,1)-(0,0,\varepsilon '(\lambda ))=\sum _{i=1}^3(t_i,t_i^2,t_i^3)\in \mathcal {A}_{3,3} \end{aligned}$$
for \(\varepsilon '(\lambda )=3\lambda (1-t-\lambda )(1-s-\lambda )\). Note that \(\varepsilon '(0)=0\). So to prove the lemma, it suffices to find a \(c>0\) such that \(d\ge 0\) and \(-1\le t_1,t_2,t_3\le 1\) for all \(0\le \lambda \le c\) and \(\varepsilon '(c)>0\), because we can then take \(\varepsilon =\varepsilon '(c)\) and have \([0,\varepsilon ]\subseteq \varepsilon '([0,c])\).
Take \(c=\min (\delta ,\delta ^2)/1000>0\) and assume that \(0\le \lambda \le c\). Then we have
$$\begin{aligned} d\ge (t-s)^2+3\lambda (2\delta -\lambda )\ge (t-s)^2\ge 0 \end{aligned}$$
and hence \(\sqrt{d}\ge t-s\). We also have
$$\begin{aligned} \sqrt{d}-(t-s)=\frac{d-(t-s)^2}{\sqrt{d}+(t-s)}\le \frac{2(t-s)\lambda +4(1-t)\lambda -3\lambda ^2}{2(t-s)}\le \frac{12\lambda }{2\delta }\le \delta /100 \end{aligned}$$
since \(t-s,1-t\le 2\). Hence \(|\mu |\le \delta /100\) and therefore \(-1\le t_1,t_2,t_3\le 1\). We have
$$\begin{aligned} \varepsilon '(c)=3c(1-t-c)(1-s-c)\ge 3c\cdot \delta /3\cdot \delta /3=c\delta ^2/3>0. \end{aligned}$$
Hence the statement of the lemma holds. \(\square \)
Lemma 4.3
Take \(\delta >0\). Then there exists an \(\varepsilon >0\) such that
$$\begin{aligned} (s,s^2,s^3)+(t,t^2,t^3)+(-1,1,-1)+(0,0,\varepsilon ')\in \mathcal {A}_{3,3} \end{aligned}$$
for all \(-1<s<t<1\) and \(0\le \varepsilon '\le \varepsilon \) with \(1-t,t-s,s-(-1)\ge \delta \).
Proof
The proof is similar to the Proof of Lemma 4.2. For \(0\le \lambda \ll 1\), one considers \((t_1,t_2,t_3)=(s-\lambda +\mu ,t-\mu ,-1+\lambda )\) where
$$\begin{aligned} \mu&=(\lambda +(t-s)-\sqrt{d})/2,\\ d&=(t-s)^2+2(t-s)\lambda +4(s-(-1))\lambda -3\lambda ^2. \end{aligned}$$
Assuming that \(d\ge 0\) and \(-1\le t_1,t_2,t_3\le 1\), one can check that
$$\begin{aligned} (s,s^2,s^3)+(t,t^2,t^3)+(-1,1,-1)+(0,0,\varepsilon '(\lambda ))=\sum _{i=1}^3(t_i,t_i^2,t_i^3)\in \mathcal {A}_{3,3} \end{aligned}$$
for \(\varepsilon '(\lambda )=3\lambda (t-(-1)-\lambda )(s-(-1)-\lambda )\). The other details are left to the reader. \(\square \)
Proposition 4.4
Take \(-1<s<t<1\). Then the point
$$\begin{aligned} p=(s,s^2,s^3)+(t,t^2,t^3)+(1,1,1)+(-1,1,-1) \end{aligned}$$
does not lie on the boundary of \(\mathcal {A}_{3,4}\).
Proof
Set \(\delta =\min (1-t,t-s,s-(-1))/2>0\) and let \(\varepsilon >0\) be the minimum of the two \(\varepsilon \)’s from Lemmas 4.2 and 4.3 . Write \(p=(x,y,z)\). The Jacobian of the map
$$\begin{aligned} (u,v)\mapsto (u,u^2)+(v,v^2)+(1,1)+(-1,1) \end{aligned}$$
is invertible at (s, t). It follows that all points in \(\mathbb {R}^2\) in a small neighborhood of (x, y) are of the form
$$\begin{aligned} (u,u^2)+(v,v^2)+(1,1)+(-1,1) \end{aligned}$$
with (u, v) in a small neighborhood of (s, t) by the Inverse Function Theorem. By shrinking this neighborhood, we may assume that \(-1<u<v<1\) and \(1-v,v-u,u-(-1)\ge \delta \). Lemmas 4.2 and 4.3 now tell us that
$$\begin{aligned} (u,u^2,u^3)+(v,v^2,v^3)+(1,1,1)+(-1,1,-1)+(0,0,\varepsilon ')\in \mathcal {A}_{3,4} \end{aligned}$$
for all (u, v) in the neighborhood and \(|\varepsilon '|<\varepsilon \). Hence p is in the interior of \(\mathcal {A}_{3,4}\). \(\square \)
Finally, we take care of the cases \((k,\ell ,a,b)=(1,2,1,0),(2,1,0,1)\).
Lemma 4.5
Take \(\delta >0\). Then there exists an \(\varepsilon >0\) such that
$$\begin{aligned} (s,s^2,s^3)+2(t,t^2,t^3)+(0,0,\varepsilon ')\in \mathcal {A}_{3,3} \end{aligned}$$
for all \(-1\le s<t<1\) and \(0\le \varepsilon '\le \varepsilon \) with \(1-t,t-s\ge \delta \).
Proof
The proof is similar to the Proof of Lemma 4.2. For \(0\le \lambda \ll 1\), one considers
$$\begin{aligned} (t_1,t_2,t_3)=(s+2\lambda ,t-\lambda +\mu ,t-\lambda -\mu ) \end{aligned}$$
where \(\mu =\sqrt{2(t-s)\lambda -3\lambda ^2}\) and finds that
$$\begin{aligned} (s,s^2,s^3)+2(t,t^2,t^3)+(0,0,\varepsilon '(\lambda ))=\sum _{i=1}^3(t_i,t_i^2,t_i^3)\in \mathcal {A}_{3,3}. \end{aligned}$$
for \(\varepsilon '(\lambda )=6(t-s)^2\lambda -24(t-s)\lambda ^2+24\lambda ^3\). The other details are left to the reader. \(\square \)
Lemma 4.6
Take \(\delta >0\). Then there exists an \(\varepsilon >0\) such that
$$\begin{aligned} 2(s,s^2,s^3)+(t,t^2,t^3)-(0,0,\varepsilon ')\in \mathcal {A}_{3,3} \end{aligned}$$
for all \(-1<s<t\le 1\) and \(0\le \varepsilon '\le \varepsilon \) with \(t-s,s-(-1)\ge \delta \).
Proof
The proof is similar to the Proof of Lemma 4.2. For \(0\le \lambda \ll 1\), one considers
$$\begin{aligned} (t_1,t_2,t_3)=(s+\lambda +\mu ,s+\lambda -\mu ,t-2\lambda ) \end{aligned}$$
where \(\mu =\sqrt{2(t-s)\lambda -3\lambda ^2}\) and finds that
$$\begin{aligned} 2(s,s^2,s^3)+(t,t^2,t^3)-(0,0,\varepsilon '(\lambda ))=\sum _{i=1}^3(t_i,t_i^2,t_i^3)\in \mathcal {A}_{3,3}. \end{aligned}$$
for \(\varepsilon '(\lambda )=6(t-s)^2\lambda -24(t-s)\lambda ^2+24\lambda ^3\). The other details are left to the reader. \(\square \)
Proposition 4.7
Take \(-1<s<t<1\). Then the point
$$\begin{aligned} p=(s,s^2,s^3)+2(t,t^2,t^3)+(1,1,1) \end{aligned}$$
does not lie on the boundary of \(\mathcal {A}_{3,4}\).
Proof
The proof is similar to the proof of Proposition 4.4. For (u, v) in a small neighborhood of (s, t), we find points in \(\mathcal {A}_{3,4}\) above
$$\begin{aligned} (u,u^2,u^3)+2(v,v^2,v^3)+(1,1,1) \end{aligned}$$
using Lemma 4.5 and we find points below using Lemma 4.6. Note here that in the latter case the role of the pair (s, t) from Lemma 4.6 is played by (v, 1). \(\square \)
Proposition 4.8
Take \(-1<s<t<1\). Then the point
$$\begin{aligned} p=2(s,s^2,s^3)+(t,t^2,t^3)+(-1,1,-1) \end{aligned}$$
does not lie on the boundary of \(\mathcal {A}_{3,4}\).
Proof
The proof is similar to the proof of Proposition 4.7. For (u, v) in a small neighborhood of (s, t), we find points in \(\mathcal {A}_{3,4}\) above
$$\begin{aligned} 2(u,u^2,u^3)+(v,v^2,v^3)+(-1,1,-1) \end{aligned}$$
using Lemma 4.5 and we find points below using Lemma 4.6. Note here that in the former case the role of the pair (s, t) from Lemma 4.5 is played by \((-1,u)\). \(\square \)
By combining Propositions 3.1, 4.1, 4.4, 4.7 and 4.8 , we see that the boundary of \(\mathcal {A}_{3,n}\) is contained in \(\mathcal {B}^{+}_{n}\cup \mathcal {B}^{-}_{n}\) for \(n=4\). We now use this knowledge to prove the same for \(n>4\).
Theorem 4.9
The boundary of \(\mathcal {A}_{3,n}\) is contained in the union of \(\mathcal {B}^{+}_{n}\) and \(\mathcal {B}^{-}_{n}\).
Proof
For \(n=3\), this follows directly from Proposition 3.1. For \(n=4\), we additionally use Propositions 4.1, 4.4, 4.7 and 4.8 . For \(n>4\), we need to show that points of the form
$$\begin{aligned} k\begin{pmatrix}s\\ s^2\\ s^3\end{pmatrix}+\ell \begin{pmatrix}t\\ t^2\\ t^3\end{pmatrix}+a\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}+b\begin{pmatrix}-1\\ 1\\ -1\end{pmatrix} \end{aligned}$$
with \(-1< s< t<1\), \(k,\ell \ge 1\) and \(a,b\ge 0\) are not on the boundary of \(\mathcal {A}_{3,n}\) when one of the following conditions holds:
-
(1)
\(k,\ell \ge 2\),
-
(2)
\(k=\ell =1\) and \(a,b>0\),
-
(3)
\(k=1\), \(\ell >1\) and \(a>0\),
-
(4)
\(k>1\), \(\ell =1\) and \(b>0\).
This is done by combining Lemma 3.2 with Propositions 4.1, 4.4, 4.7 and 4.8 . \(\square \)