Proof Since $\mathbf {u} \not \in \{\mathbf {z}_1,\ldots , \mathbf {z}_k\}$ and $\{\mathbf {u}_m\}$ converges to $\mathbf {u}$ , we may assume that $\mathbf {u}_m \not \in \{\mathbf {z}_1,\ldots , \mathbf {z}_k\}$ for all m. Let $\mathbf {v}_m = \mathbf {u}_m + \big ( \sum _{i=1}^k t_{m,i} (\mathbf {z}_i - \mathbf {u}_m)^* \big )^*$ for some $t_{m,i} \ge 0$ with $\sum _{i=1}^k t_{m,i} =1$ . Without loss of generality, $\{t_{m,i}\}_m$ converges to $t_i \ge 0$ , and $\sum _{i=1}^k t_i = 1$ . So we can just take the limit to conclude. The converse is straightforward.

Proof Translations preserve generalized $(n -1)$ -spheres, so it is enough to prove that the transformation $T_{\mathbf {0}} : \mathbf {x} \mapsto \mathbf {x}^*$ preserves them. Any generalized $(n -1 )$ -sphere in $\hat {\mathbb {R}}^{n}$ is the set of points that satisfy the equation

(3.1) $$ \begin{align} \alpha \|\mathbf{x}\|^2 - 2 \langle \mathbf{x}, \mathbf{a} \rangle + \beta = 0, \end{align} $$

for some parameters $\alpha , \beta \in \mathbb {R}$ and $\mathbf {a} \in \mathbb {R}^{n}$ . By convention, $\infty $ satisfies this equation if and only if $\alpha = 0$ , and then the set is a hyperplane. When $\mathbf {x} \neq \mathbf {0}$ , dividing throughout by $\| \mathbf {x} \|^2$ , we get

(3.2) $$ \begin{align} \alpha - 2 \langle \mathbf{x}^* , \mathbf{a} \rangle + \beta \|\mathbf{x}^* \|^2 = 0, \end{align} $$

and this is the equation of the image of (3.1) under $T_{\mathbf {0}}$ . If $\mathbf {0}$ satisfies the original equation, then $\beta = 0$ , and (3.2) reduces to an equation of a hyperplane. In that case, $\infty $ also satisfies that equation.

Proof Any k-sphere in $\hat {\mathbb {R}}^{n}$ can be written as an intersection of a $(n-1)$ -sphere with $n - k -1$ distinct hyperplanes. Since $T_{\mathbf {u}}$ is bijective, these are sent to distinct hyperplanes and $(n -1)$ -spheres, and they intersect in a k-sphere.

Proof The Möbius transformation $T_{\mathbf {u}}$ sends the circle through $\mathbf {u}, \mathbf {z}_1, \mathbf {z}_2$ to the line through $\infty , T_{\mathbf {u}}(\mathbf {z}_1), T_{\mathbf {u}}(\mathbf {z}_2)$ . The arc of the circle lying between $\mathbf {z}_1, \mathbf {z}_2$ that does not include $\mathbf {u}$ gets sent to the segment between $T_{\mathbf {u}}(\mathbf {z}_1)$ and $T_{\mathbf {u}}(\mathbf {z}_2)$ not containing $\infty $ . Taking the inverse image of the points on this segment, the points on the arc can be written as $T_{\mathbf {u}} (tT_{\mathbf {u}}(\mathbf {z}_1) + (1-t) T_{\mathbf {u}}(\mathbf {z}_2) )$ for $0 \leq t \leq 1$ , which is exactly the parametrization in Definition 2.1.

Proof Let S be a $\mathbf {u}$ -convex set and $\mathbf {z}_1, \mathbf {z}_2 \in S$ . Since $T_{\mathbf {u}}$ sends $\mathrm {arc}_{\mathbf {u}} [\mathbf {z}_1, \mathbf {z}_2]$ to the line segment $\{ t T_{\mathbf {u}} (\mathbf {z}_1) + (1-t)T_{\mathbf {u}} (\mathbf {z}_2) : t \in [0,1] \}$ , we see that it is in $T_{\mathbf {u}}(S)$ . As $T_{\mathbf {u}}$ is a bijection, we see that the segment between any pair of points from $T_{\mathbf {u}} (S)$ is in $T_{\mathbf {u}} (S)$ , so it is convex. A similar argument shows the other half of the lemma.

Proof

1. (a) This part is straightforward:

   $$ \begin{align*} \|\mathbf{v} -\mathbf{u} \| &= \Big\| \Big(\sum _{i=1}^k t_i (\mathbf{z}_i - \mathbf{u} )^*\Big)^* \Big\| =\Big \|\sum _{i=1}^k t_i (\mathbf{z}_i - \mathbf{u})^*\Big\|^{-1}. \end{align*} $$

2. (b) This follows trivially from the fact that $(\mathbf {x}^*)^* = \mathbf {x}$ .

3. (c) We can rewrite the stated expression as

   $$ \begin{align*}0 = \sum _{i=1}^k \frac{t_i ( \|\mathbf{v} -\mathbf{u} \|^2 -\|\mathbf{z}_i - \mathbf{u}\|^2 - \|\mathbf{z}_i - \mathbf{v} \|^2)}{\|\mathbf{z}_i - \mathbf{u} \|^2}. \end{align*} $$
   Then, we expand the part in the parentheses in the numerator of each summand:
   $$ \begin{align*} \|\mathbf{v} - &\mathbf{u} \|^2 - \|\mathbf{z}_i - \mathbf{u} \|^2 - \|\mathbf{z}_i - \mathbf{v}\|^2\\ &= (\langle \mathbf{v}, \mathbf{v} \rangle - \langle \mathbf{v},\mathbf{u} \rangle - \langle \mathbf{u},\mathbf{v} \rangle + \langle \mathbf{u},\mathbf{u} \rangle) - (\langle \mathbf{z}_i , \mathbf{z}_i \rangle - \langle \mathbf{z}_i, \mathbf{u} \rangle - \langle \mathbf{u}, \mathbf{z}_i \rangle + \langle \mathbf{u},\mathbf{u} \rangle) \\ & \hspace{5.6cm} - (\langle \mathbf{z}_i , \mathbf{z}_i \rangle - \langle \mathbf{z}_i,\mathbf{v} \rangle - \langle \mathbf{v}, \mathbf{z}_i \rangle + \langle \mathbf{v} ,\mathbf{v} \rangle) \\ &= - \langle \mathbf{v},\mathbf{u} \rangle - \langle \mathbf{u},\mathbf{v} \rangle - \langle \mathbf{z}_i , \mathbf{z}_i \rangle + \langle \mathbf{z}_i, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{z}_i \rangle - \langle \mathbf{z}_i , \mathbf{z}_i \rangle + \langle \mathbf{z}_i,\mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{z}_i \rangle \\ &= \langle \mathbf{v} - \mathbf{u} , \mathbf{z}_i - \mathbf{u} \rangle + \langle \mathbf{z}_i - \mathbf{u} , \mathbf{v} - \mathbf{u} \rangle - 2 \langle \mathbf{z}_i - \mathbf{u} , \mathbf{z}_i - \mathbf{u} \rangle. \end{align*} $$
   Multiply the last expression by $t_i$ and divide it by $\|\mathbf {z}_i - \mathbf {u} \|^2$ . Then, sum over i from $1$ to k and use part (b) of the current lemma to obtain
   $$ \begin{align*} \Big\langle \mathbf{v} -\mathbf{u} , \sum _{i=1}^k t_i \frac{ \mathbf{z}_i - \mathbf{u} }{\|\mathbf{z}_i - \mathbf{u} \|^2} \Big\rangle &+\Big\langle \sum _{i=1}^k t_i\frac{ \mathbf{z}_i - \mathbf{u} }{\|\mathbf{z}_i - \mathbf{u} \|^2}, \mathbf{v} -\mathbf{u} \Big\rangle -2 \\ &= \langle \mathbf{v} -\mathbf{u} , (\mathbf{v} -\mathbf{u})^* \rangle +\langle (\mathbf{v} -\mathbf{u})^*, \mathbf{v} -\mathbf{u} \rangle -2 \\ &=0. \end{align*} $$

This completes the proof.

Proof By symmetry, we prove only the necessity. Without loss of generality, we may assume that none of $\mathbf {u} , \mathbf {v} ,\mathbf {z}_1 , \ldots , \mathbf {z}_k$ are $\infty $ . (Otherwise, pick a point $\mathbf {z} \not \in \{\mathbf {u}, \mathbf {v}, \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ , and apply $T_{\mathbf {z}}$ .) Let

$$ \begin{align*}\mathbf{v} = \mathbf{u} + \Big( \sum _{i=1}^k t_i (\mathbf{z}_i - \mathbf{u})^* \Big)^* ,\end{align*} $$

for some $t_i \ge 0$ , $\sum _{i=1}^k t_i =1$ . Observe that

$$ \begin{align*} \sum _{i=1}^k t_i \frac{ \mathbf{z}_i - \mathbf{v}}{\|\mathbf{z}_i - \mathbf{u}\|^2} &= \sum _{i=1}^k t_i (\mathbf{z}_i - \mathbf{u})^* + t_i \frac{\mathbf{u} - \mathbf{v}}{\|\mathbf{z}_i - \mathbf{u}\|^2}\\ &= (\mathbf{v}-\mathbf{u})^* + \sum _{i=1}^k t_i \frac{\mathbf{u} - \mathbf{v}}{\|\mathbf{z}_i - \mathbf{u}\|^2} \\ &= ( \mathbf{v} -\mathbf{u} ) \Big( \frac{1}{\|\mathbf{v} - \mathbf{u}\|^{2}} - \sum _{i=1}^k \frac{t_i}{\|\mathbf{z}_i - \mathbf{u}\|^{2}} \Big). \end{align*} $$

Thus, we continue

$$ \begin{align*} \Big( \sum _{i=1}^k t_i \frac{\mathbf{z}_i - \mathbf{v}}{\|\mathbf{z}_i - \mathbf{u}\|^2} \Big)^* &= (\mathbf{v} - \mathbf{u})^*\Big( \frac{1}{\|\mathbf{v} - \mathbf{u}\|^{2}} - \sum _{i=1}^k \frac{t_i}{\|\mathbf{z}_i - \mathbf{u}\|^{2}} \Big)^{-1}\\ &= (\mathbf{v} - \mathbf{u})^* \|\mathbf{v} - \mathbf{u}\|^2 \Big( 1 - \sum _{i=1}^k {t_i}\frac{\|\mathbf{v} - \mathbf{u}\|^2}{\|\mathbf{z}_i - \mathbf{u}\|^2} \Big)^{-1}\\ &= (\mathbf{u} - \mathbf{v})\Big( \sum _{i=1}^k t_i \frac{\|\mathbf{z}_i - \mathbf{v}\|^2}{\|\mathbf{z}_i - \mathbf{u}\|^2} \Big)^{-1}, \end{align*} $$

where in the last equality, we used part (c) of Lemma 4.1. Define

(4.1) $$ \begin{align} \mu_i := \frac{ {t_i \|\mathbf{z}_i - \mathbf{v}\|^2}\|\mathbf{z}_i - \mathbf{u} \|^{-2}}{\sum _{j=1}^k {t_j \|\mathbf{z}_j - \mathbf{v} \|^2}\|\mathbf{z}_j - \mathbf{u} \|^{-2}} \ge 0, \end{align} $$

and note that $\sum _{i=1}^k \mu _i =1$ . Using the above calculation in the second equality below, we get

$$ \begin{align*} \Big( \sum _{i=1}^k \mu_i (\mathbf{z}_i - \mathbf{v})^*\Big)^*&= \Big({\sum _{i=1}^k t_i\frac{ \|\mathbf{z}_i - \mathbf{v}\|^2}{\|\mathbf{z}_i - \mathbf{u}\|^2}} \Big)\Big( \sum _{i=1}^k t_i\frac{ \mathbf{z}_i - \mathbf{v}}{\|\mathbf{z}_i - \mathbf{u}\|^2} \Big)^*\\ &= \Big({\sum _{i=1}^k t_i \frac{ \|\mathbf{z}_i - \mathbf{v} \|^2}{\|\mathbf{z}_i - \mathbf{u}\|^2}} \Big) (\mathbf{u} - \mathbf{v})\Big( \sum _{i=1}^k t_i\frac{\|\mathbf{z}_i - \mathbf{v}\|^2}{\|\mathbf{z}_i - \mathbf{u}\|^2} \Big)^{-1}\\ &= \mathbf{u} -\mathbf{v}. \end{align*} $$

Adding $\mathbf {v}$ to the first and last terms of the above equalities, we conclude

$$ \begin{align*}\mathbf{u} = \mathbf{v} + \Big( \sum _{i=1}^k \mu_i (\mathbf{z}_i - \mathbf{v})^*\Big)^*, \end{align*} $$

and hence, $\mathbf {u} \in \mathrm {conv}_{\mathbf {v}} \{\mathbf {z}_1, \ldots , \mathbf {z}_k\}$ .

Proof To see necessity, fix an $i \in \{ 1 ,\ldots , k \}$ and assume

$$ \begin{align*}\mathbf{u} \in \mathrm{conv}_{\mathbf{z}_i} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_{i-1} , \mathbf{z}_{i+1}, \ldots , \mathbf{z}_k \}. \end{align*} $$

Since $\mathbf {u} \neq \mathbf {z}_i$ , there are $t_j \in [0,1]$ , for $j \in \{1 , \ldots , k \} \smallsetminus \{i \}$ , such that

$$ \begin{align*}\sum _{j \in \{1 , \ldots , k \}\smallsetminus \{i \}} t_j=1 \mbox{ and } \mathbf{u} = T_{\mathbf{z}_i} \big( \sum _{j \in \{1 , \ldots , k \} \smallsetminus \{i \}} t_j T_{\mathbf{z}_i} (\mathbf{z}_j)\big). \end{align*} $$

Moreover, since $\mathbf {u} \not \in \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}\smallsetminus \{\mathbf {z}_i \}$ , none of the $t_j$ can be $1$ . So, $\mathbf {u}$ is not a $\mathbf {z}_i$ -extreme point of $\mathrm {conv}_{\mathbf {z}_i} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_{i-1} , \mathbf {z}_{i+1}, \ldots , \mathbf {z}_k \}$ . Using Theorem 4.3, $\mathbf {z}_i \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_{i-1} , \mathbf {z}_{i+1}, \ldots , \mathbf {z}_k \}$ , and by definition, $\mathbf {z}_i$ is not a $\mathbf {u}$ -extreme point. The argument for the sufficiency is similar.

Proof If $\mathbf {v}$ is not a $\mathbf {u}$ -extreme point, then there are points $\mathbf {z}_1, \mathbf {z}_2 \in Z$ distinct from $\mathbf {v}$ such that $\mathbf {v} \in \mathrm {conv}_{\mathbf {u}} \{\mathbf {z}_1, \mathbf {z}_2 \}$ . Therefore, $\mathbf {u} \in \mathrm {conv}_{\mathbf {v}} \{ \mathbf {z}_1, \mathbf {z}_2 \} \subseteq \mathrm {conv}_{\mathbf {v}} (Z)$ . Conversely, if $\mathbf {u} \in \mathrm {conv}_{\mathbf {v}} (Z)$ , then by Carathéodory’s theorem, there are points $ \mathbf {z}_1, \ldots , \mathbf {z}_k \in Z$ , for some $k \le n+1$ , distinct from $\mathbf {u}$ and $\mathbf {v}$ , such that $\mathbf {u} \in \mathrm {conv}_{\mathbf {v}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_{k} \}$ . This implies that $\mathbf {v} \in \mathrm {conv}_{\mathbf {u}} \{\mathbf {z}_1 , \ldots , \mathbf {z}_{k} \}$ . Therefore, $\mathbf {v}$ cannot be $\mathbf {u}$ -extreme.

Proof Without loss of generality, we may assume $\mathbf {u} = \infty $ , so $\mathbf {v} , \mathbf {z}_1 , \ldots , \mathbf {z}_k \in \mathbb {R}^n$ . The assumption that the points are not on a circle now becomes that they are not on a line. We show that

$$ \begin{align*}\mathrm{conv}\{ \mathbf{z}_1, \ldots , \mathbf{z}_k\} \not\subseteq \mathrm{conv}_{\mathbf{v}} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \} \end{align*} $$

with the opposite non-inclusion being analogous. Since the points are not all on a circle, there are at least two distinct points in $\mathbf {z}_1 , \ldots , \mathbf {z}_k$ (i.e., $k \geq 2$ ). Since $\mathbf {v} \not \in \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ , we have that $\mathbf {v} \not \in \mathrm {conv}_{\mathbf {v}} \{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ . If $\mathbf {v} \in \mathrm {conv}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k\}$ , then we are done. Assume this is not the case. Then, there is a closed half-space H containing $\mathrm {conv}\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ having at least two points $\mathbf {z}_i , \mathbf {z}_j$ on the boundary and not containing $\mathbf {v}$ . (Note that if the points $\mathbf {v}, \mathbf {z}_1, \ldots , \mathbf {z}_k$ are on a line, such a half-space does not exist.) Let $\mathbf {z} \in \mathrm {conv}\{ \mathbf {z}_i , \mathbf {z}_j\} \smallsetminus \{ \mathbf {z}_1, \ldots , \mathbf {z}_k\}$ ; that is, $\mathbf {z} \in \mathrm {conv} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ . Since H is also $\mathbf {z}$ -convex, we get that $\mathbf {v} \not \in \mathrm {conv}_{\mathbf {z}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k\} \subseteq H$ . By Theorem 4.3, we obtain that ${\mathbf {z} \not \in \mathrm {conv}_{\mathbf {v}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}}$ , concluding the argument.

Proof Note that $T_{\mathbf {u}}(A)$ and $T_{\mathbf {u}}(B)$ are nonintersecting convex sets. The classical hyperplane separation theorem implies that there is a hyperplane H that separates $T_{\mathbf {u}}(A)$ and $T_{\mathbf {u}}(B)$ . Then, $T_{\mathbf {u}}(H)$ is an $(n-1)$ -sphere, having $\mathbf {u}$ on its boundary, separating A and B. Next, note that if $\mathbf {u} \not \in A \cup B$ , then $T_{\mathbf {u}} (A) , T_{\mathbf {u}} (B) \subseteq \mathbb {R}^n$ .

(1) If A is closed as a subset of $\hat {\mathbb {R}}^n$ , then it is compact. Since $\mathbf {u} \not \in A$ , $T_{\mathbf {u}} (A)$ is compact in $\mathbb {R}^n$ . Similarly, if B is closed in $\hat {\mathbb {R}}^n \smallsetminus \{\mathbf {u}\}$ , then $T_{\mathbf {u}} (B)$ is closed in $\mathbb {R}^n$ .

(2) If A and B are both open, then so are the convex sets $T_{\mathbf {u}} (A)$ and $T_{\mathbf {u}} (B)$ .

In both cases, from ordinary convexity, we can find a separating hyperplane H such that $T_{\mathbf {u}}(A) \cap \partial H = \emptyset = T_{\mathbf {u}}(B) \cap \partial H$ . We conclude by setting $S:=T_{\mathbf {u}}(H)$ .

Proof Without loss of generality, we may assume $\mathbf {u} = \infty $ . If $\infty \not \in Z$ , then Z is a bounded closed set. So we can conclude that $\mathrm {conv}(Z)$ is closed. Otherwise, $\infty \in \mathrm {int}(Z)$ . Then, $\mathrm {conv}(Z) = \hat {\mathbb {R}}^n$ , so we are done.

Proof Note that Equation (5.1) is the same as saying that $\mathbf {0} \in \mathrm {conv}_{\mathbf {u}} \{\mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ . Assume that this is not the case. Since $ \mathbf {u} \not \in \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ , the set $\mathrm {conv}_{\mathbf {u}}\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ is closed. So, by Lemma 5.1, there is a spherical domain S, having $\mathbf {u}$ on its boundary, that strongly separates $\{ \mathbf {0} \}$ and $\mathrm {conv}_{\mathbf {u}} \{\mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ ; that is,

$$ \begin{align*}\partial S \cap \mathrm{conv}_{\mathbf{u}}\{ \mathbf{z}_1 , \ldots , \mathbf{z}_k\} = \emptyset = \partial S \cap \{ \mathbf{0} \}. \end{align*} $$

Let $\alpha \langle \mathbf {x} , \mathbf {x} \rangle + \langle \mathbf {x} , \mathbf {v} \rangle + \beta = 0$ be the equation of the boundary of the spherical domain S. Since it separates $\{ \mathbf {0} \}$ and $\mathrm {conv}_{\mathbf {u}} \{\mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ , both of them evaluate to different signs and neither of them are zero. In particular, since $\mathbf {0} \not \in \partial S$ , we have

$$ \begin{align*}\beta = \alpha \langle \mathbf{0} , \mathbf{0} \rangle + \langle \mathbf{0} , \mathbf{a} \rangle + \beta \neq 0. \end{align*} $$

If $\beta> 0$ , we are done; otherwise, take $-\mathbf {a}$ , $-\alpha $ , and $-\beta $ to get the inequalities in (5.2). Since $\mathbf {u} \in \partial S$ , we get the equality in (5.2).

Proof If $\mathbf {u} = - \sum _{i=1}^k t_i \mathbf {z}_i$ for $t_i \in [0, \infty )$ , then $\mathbf {u} \in -\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ , and so $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ is $\mathbf {u}$ -convex. (The argument for the latter is analogous to the one in Example 2.6.) Consequently, $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\} \cup \{ \infty \}$ is also $\mathbf {u}$ -convex. Thus, for any $\mathbf {v} \in \hat {\mathbb {R}}^n \smallsetminus \{ \infty , \mathbf {u} \}$ , either $\mathbf {v} \in \mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ or there exists a spherical domain S, having $\mathbf {u}$ on its boundary, separating $\mathbf {v}$ and $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}\cup \{ \infty \}$ . Again, since $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}\cup \{ \infty \}$ is closed, this domain can be chosen so that

$$ \begin{align*}\partial S \cap \big( \mathrm{cone}\{ \mathbf{z}_1 , \ldots , \mathbf{z}_k\}\cup \infty \big) = \emptyset = \partial S \cap \{ \mathbf{v} \}. \end{align*} $$

Let $\alpha \langle \mathbf {x} , \mathbf {x} \rangle + \langle \mathbf {x} , \mathbf {a} \rangle + \beta = 0$ be the equation of the boundary of the spherical domain S. Since $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ is an unbounded set, it must lie in an unbounded component of $\mathbb {R}^n \smallsetminus \partial S$ . If $\partial S$ happens to be a hyperplane, then $\alpha =0$ , and we can choose the signs of $\mathbf {a}$ and $\beta $ to satisfy (5.4). Otherwise, $\mathbb {R}^n \smallsetminus \partial S$ has only one unbounded component, and we can choose the coefficient $\alpha $ to be positive. Then, since $\mathrm {cone}\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ is an unbounded set, the quadratic term in $\alpha \langle \mathbf {z}_i , \mathbf {z}_i \rangle + \langle \mathbf {z}_i , \mathbf {a} \rangle + \beta $ determines the sign of the whole expression. The inequalities in (5.4) follow. Since $\mathbf {v}$ lies in the bounded component, we have $\alpha \langle \mathbf {v} , \mathbf {v} \rangle + \langle \mathbf {v} , \mathbf {a} \rangle + \beta < 0$ . Finally, the boundary of the spherical domain passes through $\mathbf {u}$ , giving the third statement in (5.4).

Proof The first condition is equivalent to $\mathbf {v} \in \mathrm {cone}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . Applying the transform $T_{\mathbf {u}}$ to this inclusion, it is equivalent to $T_{\mathbf {u}}(\mathbf {v}) \in \mathrm {cone} \{ T_{\mathbf {u}} (\mathbf {z}_1), \ldots , T_{\mathbf {u}}(\mathbf {z}_k ) \}$ . Assume this is not the case. Using the classical Farkas’ lemma, we get that there is a closed half-space H supporting $\mathrm {cone} \{ T_{\mathbf {u}} (\mathbf {z}_1), \ldots , T_{\mathbf {u}}(\mathbf {z}_k ) \}$ at $\mathbf {0}$ and separating it from $\{T_{\mathbf {u}}(\mathbf {v}) \}$ . Moreover, H is such that $T_{\mathbf {u}}(\mathbf {v})$ lies in the open complement of H. Apply $T_{\mathbf {u}}$ to the boundary hyperplane $\partial H$ , which contains $\infty $ by convention; see Example 3.6. We get that $S:= T_{\mathbf {u}}( \partial H)$ is a sphere separating $\{ \mathbf {v} \}$ and $\mathrm {cone}_{\mathbf {u}} \{\mathbf {z}_1, \ldots , \mathbf {z}_k \}$ such that $\mathbf {v} \not \in S $ and $\mathbf {u}, \mathbf {u} - \mathbf {u}^* \in S$ (since $\mathbf {u} = T_{\mathbf {u}}(\infty )$ and $ \mathbf {u} - \mathbf {u}^* = T_{\mathbf {u}}(\mathbf {0})$ ). Let $\alpha \langle \mathbf {z}, \mathbf {z} \rangle + \langle \mathbf {z}, \mathbf {a} \rangle + \beta = 0$ be the equation of S and choose the signs such that $\alpha \langle \mathbf {v}, \mathbf {v} \rangle + \langle {\mathbf {v}}, {\mathbf {v}} \rangle + \beta> 0$ . Since S is a separating sphere, we get that $\alpha \langle \mathbf {z}_i, \mathbf {z}_i \rangle + \langle \mathbf {z}_i, \mathbf {a} \rangle + \beta \leq 0$ for all $i = 1, \ldots , k$ . Moreover, since S passes through $\mathbf {u}, \mathbf {u} - \mathbf {u}^*$ , we get

(5.6) $$ \begin{align} \alpha \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{a} \rangle + \beta &= 0 \quad \mbox{ and } \\ \alpha \langle \mathbf{u} -\mathbf{u}^*, \mathbf{u} -\mathbf{u}^* \rangle + \langle \mathbf{u} - \mathbf{u}^*, \mathbf{a} \rangle + \beta &= 0. \nonumber \end{align} $$

Simplifying the second equation and using (5.6) gives

$$ \begin{align*} \alpha \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{a} \rangle + \beta -2 \alpha + \frac{\alpha}{\|\mathbf{u}\|^2} - \langle \mathbf{u}^*, \mathbf{a} \rangle &=0 \end{align*} $$

or $2\alpha \langle \mathbf {u}, \mathbf {u} \rangle + \langle \mathbf {u}, \mathbf {a} \rangle - \alpha = 0$ . Subtracting (5.6), one obtains

$$ \begin{align*}\alpha \langle \mathbf{u}, \mathbf{u} \rangle = \alpha + \beta, \end{align*} $$

therefore concluding the proof.

Proof Without loss of generality, we may assume $\mathbf {u} = \infty $ . Then, $\mathrm {conv}\{ \mathbf {z}_1 ,\ldots , \mathbf {z}_k \}$ is a convex polytope of full dimension and can be written as the intersection of supporting half-spaces corresponding to each of its maximal faces. Since the polytope $\mathrm {conv}\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ is of full dimension, there are clearly only finitely many half-spaces H, such that $\mbox {aff}\{\{\mathbf {z}_1 ,\ldots , \mathbf {z}_k \} \cap \partial H \}$ is a hyperplane. The second statement now follows from Remark 2.4.

Proof Without loss of generality, assume $\mathbf {u}_1 = \infty $ . If $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k\} \cup \{ \mathbf {u}_1 , \mathbf {u}_2\}$ lie on a hyperplane, the sets in (6.4) will also all lie in this hyperplane. So we can restrict to this hyperplane to assume that not all of $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k\} \cup \{ \mathbf {u}_1 , \mathbf {u}_2\}$ lie on a hyperplane. We prove the first equality; the second follows by symmetry. The containment $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}) \subseteq \mathrm {conv}_{\{\mathbf {u}_1 , \mathbf {u}_2\}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ follows from minimality in Definition 6.1. We want to show the opposite inclusion

$$ \begin{align*}\mathrm{conv}_{\{ \mathbf{u}_1 , \mathbf{u}_2 \}} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \} \subseteq \mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \}). \end{align*} $$

From Theorem 6.3, we see that $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}) = \bigcap _{i=1}^2 \bigcap _{S\in \mathcal {S}_i} S$ . All the domains S lying in either $\mathcal {S}_1$ or $\mathcal {S}_2$ are convex with respect to both $\mathbf {u}_1$ and $\mathbf {u}_2$ by definition. It follows that $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ is also convex with respect to both $\mathbf {u}_1$ and $\mathbf {u}_2$ . Again by minimality in Definition 6.1, we get that

$$ \begin{align*}\mathrm{conv}_{\{ \mathbf{u}_1 , \mathbf{u}_2 \}} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \} \subseteq \mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \}), \end{align*} $$

completing the proof.

Proof We prove the first equality; the other one follows by symmetry. The inclusion

$$ \begin{align*}\mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2}(Z)) \subseteq \mathrm{conv}_{\{\mathbf{u}_1 , \mathbf{u}_2\}}(Z) \end{align*} $$

follows by minimality. To see the other inclusion, it is sufficient to prove that $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2}(Z))$ is $\mathbf {u}_2$ -convex. Let $\mathbf {x}_1 , \mathbf {x}_2 \in \mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2}(Z))$ . Then, by repeated application of Carathéodory’s theorem, we get $\mathbf {x}_1 , \mathbf {x}_2 \in \mathrm {conv}_{\mathbf {u}_1}\{ \mathbf {v}_1 , \ldots , \mathbf {v}_\ell \}$ for some $\mathbf {v}_1 , \ldots , \mathbf {v}_\ell \in \mathrm {conv}_{\mathbf {u}_2}(Z)$ . By Carathéodory’s theorem again, we get that $\mathbf {x}_1 , \mathbf {x}_2 \in \mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2}\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \})$ for some $\mathbf {z}_1 , \ldots , \mathbf {z}_k \in Z$ , that may be assumed distinct. By Corollary 6.4, $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2}\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \})$ is $\mathbf {u}_2$ -convex, so

$$ \begin{align*}\mathrm{arc}_{\mathbf{u}_2}[\mathbf{x}_1 , \mathbf{x}_2] \subseteq \mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2}\{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \}) \subseteq \mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2}(Z)). \end{align*} $$

Therefore, the set $\mathrm {conv}_{\mathbf {u}_1}(\mathrm {conv}_{\mathbf {u}_2}(Z))$ is convex with respect to both $\mathbf {u}_1$ and $\mathbf {u}_2$ and contains Z. So, by minimality,

$$ \begin{align*}\mathrm{conv}_{\{\mathbf{u}_1 , \mathbf{u}_2\}}(Z) \subseteq \mathrm{conv}_{\mathbf{u}_1}(\mathrm{conv}_{\mathbf{u}_2}(Z)), \end{align*} $$

concluding the proof.

Proof As before, by minimality, we have the following inclusion:

$$ \begin{align*}\mathrm{conv}_{\mathbf{u}_m}(\mathrm{conv}_{\{\mathbf{u}_1, \ldots, \mathbf{u}_{m-1}\}}(Z)) \subseteq \mathrm{conv}_{\{\mathbf{u}_1, \ldots , \mathbf{u}_m\}}(Z). \end{align*} $$

So it is enough to prove that $\mathrm {conv}_{\mathbf {u}_m}(\mathrm {conv}_{\{\mathbf {u}_1, \ldots , \mathbf {u}_{m-1}\}}(Z)) $ is convex with respect to all of $\mathbf {u}_i$ , $i=1 , \ldots , m$ . We use induction on the number of poles, m. Note that the base case, $m=2$ , is Corollary 6.5. Assume the corollary to be true for $m-1$ poles. Then, by the induction hypothesis

$$ \begin{align*}\mbox{conv}_{ \{ \mathbf{u}_1, \ldots , \mathbf{u}_{m-1}\} } (Z) = \mbox{conv}_{\mathbf{u}_{m-1}}(\mbox{conv}_{ \{ \mathbf{u}_1, \ldots , \mathbf{u}_{m-2}\} } (Z)). \end{align*} $$

Taking the $\mathbf {u}_m$ convex hull and using Corollary 6.5, we get

$$ \begin{align*} \mbox{conv}_{\mathbf{u}_m} (\mbox{conv}_{ \{ \mathbf{u}_1, \ldots , \mathbf{u}_{m-1}\} } (Z)) &= \mbox{conv}_{\mathbf{u}_m}(\mbox{conv}_{\mathbf{u}_{m-1}}(\mbox{conv}_{ \{ \mathbf{u}_1, \ldots , \mathbf{u}_{m-2}\} } (Z))) \\ &= \mbox{conv}_{\mathbf{u}_{m-1}}(\mbox{conv}_{\mathbf{u}_m}(\mbox{conv}_{ \{ \mathbf{u}_1, \ldots , \mathbf{u}_{m-2}\} } (Z))). \end{align*} $$

By a similar reasoning, we may replace $\mathbf {u}_{m-1}$ by any other $\mathbf {u}_i$ for ${i = 1, \ldots , m-1}$ . Therefore, we conclude that $\mathrm {conv}_{\mathbf {u}_m}(\mathrm {conv}_{\{\mathbf {u}_1, \ldots , \mathbf {u}_{m-1}\}}(Z)) $ is indeed convex with respect to all the $\mathbf {u}_i$ ’s, so the corollary holds.

Proof If $\mathrm {conv}_{ U_m}(Z_k)$ is the entire $\hat {\mathbb {R}}^n$ , then the families $\mathcal {L}_i$ are empty and the theorem holds, so assume that this is not the case. The proof is by an induction on the dimension of the ambient space, n. When the dimension is $n = 2$ , the theorem is simply [Reference Sendov, Sendov and Wang11, Theorem 5.2]. Suppose that $n \ge 3$ and the result holds when the dimension of the space is $n-1$ or lower. This assumption means that the result holds in any subspace or affine space of $\hat {\mathbb {R}}^n$ of dimension $n-1$ or on any sphere in $\hat {\mathbb {R}}^n$ of dimension $n-1$ . To see the latter, simply send a point of the said sphere to $\infty $ using a Möbius transformation.

We begin with the following containments:

(6.7) $$ \begin{align} \bigcup_{\substack{S \in \mathcal{L}_j \\ j=1,\ldots, m}} \mathrm{conv}_{\partial S \cap U_m} (\partial S \cap Z_k ) \subseteq \mathrm{conv}_{ U_m }(Z_k) \subseteq \bigcap_{j=1}^{m} \bigcap_{S \in \mathcal{L}_j} S, \end{align} $$

where the inclusions follow by the minimality of the convex hulls. To conclude the proof, it is sufficient to show that

(6.8) $$ \begin{align} \bigcup_{\substack{S \in \mathcal{L}_j \\ j=1,\ldots, m}} \mathrm{conv}_{\partial S \cap U_m} (\partial S \cap Z_k ) = \partial \Big( \bigcap_{j=1}^{m} \bigcap_{S \in \mathcal{L}_j} S \Big). \end{align} $$

Indeed, assume (6.8) holds. Without loss of generality, assume that $\mathbf {u}_m = \infty $ or else apply a Möbius transformation to $\hat {\mathbb {R}}^n$ that sends $\mathbf {u}_m$ to $\infty $ . Then, $\bigcap _{j=1}^{m} \bigcap _{S \in \mathcal {L}_j} S$ is a closed convex set, and so it is equal to the convex hull of its boundary. Taking the convex hull of all parts in (6.7), we have

$$ \begin{align*} \mathrm{conv} &\Big(\bigcup_{\substack{S \in \mathcal{L}_j \\ j=1,\ldots, m}} \mathrm{conv}_{\partial S \cap U_m } (\partial S \cap Z_k ) \Big) \subseteq \mathrm{conv}_{ U_m }(Z_k) \subseteq \bigcap_{j=1}^{m} \bigcap_{S \in \mathcal{L}_j} S \\ &= \mathrm{conv} \Big( \partial \Big( \bigcap_{j=1}^{m} \bigcap_{S \in \mathcal{L}_j} S \Big) \Big) = \mathrm{conv} \Big(\bigcup_{\substack{S \in \mathcal{L}_j \\ j=1,\ldots, m}} \mathrm{conv}_{\partial S \cap U_m} (\partial S \cap Z_k ) \Big). \end{align*} $$

Thus, we have equalities throughout, and we are done.

For the remainder of the proof, we show (6.8). Since $ \bigcap _{j=1}^{m} \bigcap _{S \in \mathcal {L}_j} S$ is the intersection of spherical domains, the boundaries of such domains are the sole contributors to the boundary of $ \bigcap _{j=1}^{m} \bigcap _{S \in \mathcal {L}_j} S $ . To see that each domain S contributes exactly a piece of the form $\mathrm {conv}_{\partial S \cap U_m} (\partial S \cap Z_k)$ , fix some $S \in \mathcal {L}_i$ and restrict to its boundary $\partial S$ . Let

$$ \begin{align*}\{\mathbf{u}^{\prime}_1,\ldots, \mathbf{u}^{\prime}_{m'}\} := \partial S \cap U_m. \end{align*} $$

Recalling Definition 3.3, define the following families of spherical domains:

$$ \begin{align*} \mathcal{L}_j^{\prime} : = \big\{ S' \subset \partial S &: S' \mbox{ closed spherical domain, } \partial S \cap Z_k \subset S', \mathbf{u}_j^{\prime} \in \partial S', \\ &\partial S \cap U_m \subset \mathrm{cl}(S^{\prime c}) \mbox{ and } \mathrm{aff}_{\mathbf{u}_j^{\prime}} \{ \partial S \cap (Z_k \cup U_m) \cap \partial S' \} = \partial S' \big\}, \end{align*} $$

where in this definition, by $\partial S'$ , we understand the boundary of $S'$ relative to $\partial S$ , and by $S^{\prime c}$ , we understand the complement of $S'$ relative to $\partial S$ .

Since $\partial S$ is a dimension $n -1 $ ambient space, the theorem holds by the induction hypothesis, so

(6.9) $$ \begin{align} \mathrm{conv}_{\partial S \cap U_m } (\partial S \cap Z_k) = \bigcap_{j=1}^{m'} \bigcap_{S' \in \mathcal{L}_j^{\prime}} S'. \end{align} $$

In the next two paragraphs, we explain how each domain $S' \in \mathcal {L}_j^{\prime }$ can be extended to a domain $S" \in \mathcal {L}_j$ , such that $S" \cap \partial S = S'$ . Without loss of generality, we may assume that both $\mathcal {L}_j^{\prime }$ and $\mathcal {L}_j$ correspond to the same pole $\mathbf {u}_j$ ; that is, $\mathbf {u}_j^{\prime } = \mathbf {u}_j$ for $j=1,\ldots , m'$ (or else we just relabel $\mathbf {u}_1,\ldots , \mathbf {u}_m$ so that $\mathbf {u}^{\prime }_1,\ldots , \mathbf {u}^{\prime }_{m'}$ are the first $m'$ of them).

If $\mathcal {L}_j^{\prime }$ are all empty, then $\bigcap _{j=1}^{m'} \bigcap _{S' \in \mathcal {L}_j^{\prime }} S' = \partial S$ . So, (6.9) becomes $\mathrm {conv}_{\partial S \cap U_m } (\partial S \cap Z_k) = \partial S$ , and then, (6.7) shows that

$$ \begin{align*}\partial S \subseteq \mathrm{conv}_{ U_m }(Z_k) \subseteq \bigcap_{j=1}^{m} \bigcap_{S \in \mathcal{L}_j} S. \end{align*} $$

In this case, since there is at least one point in $U_m \cup Z_k$ not lying on $\partial S$ , and $S \in \mathcal {L}_i$ implies that $\mathrm {aff}_{\mathbf {u}_i} \{ (Z_k \cup U_m) \cap \partial S \} = \partial S$ , we conclude that $\mathrm {conv}_{ U_m}(Z_k) = S$ . If $\mathcal {L}_j^{\prime }$ are not all empty, then fix $S' \in \mathcal {L}_j^{\prime }$ and assume that $\mathbf {u}_j = \infty $ or else apply a Möbius transformation to $\hat {\mathbb {R}}^n$ that sends $\mathbf {u}_j$ to $\infty $ . (Abusing notation, we keep the names S and $S'$ after that transformation.) Since $\mathbf {u}_j \in \partial S' \subset \partial S$ , $\partial S'$ becomes a hyperplane in $\partial S$ , which is our ambient space of dimension $n-1$ . (Note that $\partial S$ becomes a hyperplane in $\hat {\mathbb {R}}^n$ .) By definition, $S'$ contains the points $\partial S \cap Z_k$ and separates them from the points $\partial S \cap U_m $ , relative to $\partial S$ . Since $\partial S$ separates $Z_k$ and $U_m$ , the points $Z_k \setminus \partial S$ and $U_m \setminus \partial S$ are strictly on different sides of $\partial S$ . Thus, $S'$ can be extended to a half-space H in $\hat {\mathbb {R}}^n$ that contains the points $Z_k$ and separates them from the points $U_m$ . (To obtain this initial half space H, just rotate slightly the half-space S around $\partial S'$ .) That is, the half-space H is such that

$$ \begin{align*}U_m \subset \mbox{cl}(H^c) \mbox{ and } \mathrm{conv}_{ U_m}(Z_k)\subset H. \end{align*} $$

We show now that H can be chosen in such a way that

$$ \begin{align*}\mathrm{aff}_{\mathbf{u}_j} \{ (Z_k\cup U_m) \cap \partial H\} = \partial H. \end{align*} $$

In this way, we have $H \in \mathcal {L}_j$ , and by construction, $H \cap \partial S = S'$ .

If $\mathrm {dim\;} \mathrm {aff}_{\mathbf {u}_j} \{(Z_k\cup U_m) \cap \partial H\} = n-1$ , then we are done. So assume

(6.10) $$ \begin{align} \mathrm{dim\;} \mathrm{aff}_{\mathbf{u}_j} \{ (Z_k\cup U_m) \cap \partial H \} \le n-2. \end{align} $$

Since $ \partial S \cap (Z_k\cup U_m) \cap \partial S' \subseteq (Z_k\cup U_m) \cap \partial H$ , we have that

$$ \begin{align*}\partial S' = \mathrm{aff}_{\mathbf{u}_j} \{ \partial S \cap (Z_k\cup U_m) \cap \partial S' \} \subset \mathrm{aff}_{\mathbf{u}_j} \{(Z_k\cup U_m) \cap \partial H\}. \end{align*} $$

But the dimension of $\partial S'$ is $n-2$ , so we conclude that

$$ \begin{align*}\mathrm{aff}_{\mathbf{u}_j} \{(Z_k\cup U_m) \cap \partial H \} = \partial S', \end{align*} $$

and (6.10) holds with equality. Rotate $\partial H$ around $\partial S'$ until it hits a point in $(Z_k\cup U_m) \smallsetminus \partial S$ . (Note that the latter set difference is not empty since otherwise, $\mathrm {conv}_{ U_m }(Z_k) \subset \partial S$ , contradicting the assumption that the polar convex hull has nonempty interior.) It should be clear now that after the rotation, the dimension of $\mathrm {aff}_{\mathbf {u}_j} \{(Z_k\cup U_m) \cap \partial H\}$ is $n-1$ . Finally, let $S"$ be the inverse image of H under the Möbius transformation that sent $\mathbf {u}_j$ to $\infty $ .

By (6.9), we have that for each point $\mathbf {x} \in \partial S$ outside of $\mathrm {conv}_{\partial S \cap U_m } (\partial S \cap Z_k )$ , there is a spherical domain $S' \in \mathcal {L}_j^{\prime }$ , for some $j=1,\ldots , m$ , that excludes $\mathbf {x}$ , relative to $\partial S$ . By the above, $S'$ can be extended to an $S" \in \mathcal {L}_j$ that satisfies $S" \cap \partial S = S'$ , and hence, $S"$ excludes $\mathbf {x}$ .

Returning to (6.8), take a point $\mathbf {x} \in \partial \big ( \bigcap _{j=1}^{m} \bigcap _{S \in \mathcal {L}_j} S \big )$ . There is an $S \in \mathcal {L}_i$ for some $i=1,\ldots , m$ such that $\mathbf {x} \in \partial S$ . By the above observation, we need to have $\mathbf {x} \in \mathrm {conv}_{\partial S \cap U_m } (\partial S \cap Z_k )$ . So $\mathbf {x}$ belongs to the left-hand side of (6.8). Conversely, if $\mathbf {x}$ is in the left-hand side of (6.8), then $\mathbf {x} \in \partial S$ for some $S \in \mathcal {L}_i$ , $i=1,\ldots , m$ . By (6.7), $\mathbf {x} \in \bigcap _{j=1}^{m} \bigcap _{S \in \mathcal {L}_j} S$ , so $\mathbf {x}$ must be on the boundary of that intersection. The proof of (6.8) is completed.

Proof For any $S\in \mathcal {L}_i$ , with $|\partial S \cap Z_k| \geq 2$ , we have $U_m \subset \text {cl}(S^c)$ and $Z_k\cap S^c = \emptyset $ . So $\text {cl}(S^c)$ is convex with respect to all $\mathbf {z}_j$ , and we get

$$ \begin{align*}\mbox{conv}_{Z_k} (U_m) \subseteq \bigcap_{i=1}^m \bigcap_{\substack{S \in \mathcal{L}_i \\ |\partial S \cap Z_k| \geq 2}} \mathrm{cl}(S^c). \end{align*} $$

Next, suppose $\mathbf {v}$ does not belong to the right-hand side of (6.13). Then, $\mathbf {v} \in \mathrm {int} (S)$ for some $S \in \mathcal {L}_i$ with $|\partial S \cap Z_k| \geq 2$ , say $\mathbf {z}_1 , \mathbf {z}_2 \in \partial S$ . Then, $\mathrm {arc}_{\mathbf {v}} [\mathbf {z}_1 , \mathbf {z}_2] \not \subseteq S$ , so by (6.12), $\mathbf {v} \not \in \mathcal {P}(\mbox {conv}_{U_m}(Z_k))$ . Therefore,

$$ \begin{align*}\mathcal{P}(\mbox{conv}_{U_m}(Z_k)) \subseteq \bigcap_{i=1}^m \bigcap_{\substack{S \in \mathcal{L}_i \\ |\partial S \cap Z_k| \geq 2}} \mathrm{cl}(S^c). \end{align*} $$

Finally, for any $\mathbf {v}$ in the right-hand side of (6.13) and any $S\in \mathcal {L}_i$ with $ |\partial S \cap Z_k| \geq 2$ , we have $\mathbf {v} \not \in \mbox {int}(S)$ and so S is $\mathbf {v}$ -convex. Therefore, the right-hand side of (6.12) is $\mathbf {v}$ -convex, and so we conclude that $\mathbf {v} \in \mathcal {P}(\mbox {conv}_{U_m}(Z_k))$ . Thus,

$$ \begin{align*}\bigcap _{i=1}^m \bigcap _{\substack{S \in \mathcal{L}_i \\ |\partial S \cap Z_k| \geq 2}} \text{cl}(S^c) \subseteq \mathcal{P}(\mbox{conv}_{U_m}(Z_k)). \end{align*} $$

This completes the proof.

Proof Note that if Z is either $\hat {\mathbb {R}}^n$ , $\emptyset $ , or a singleton, then $\mathcal {P}(Z) = \hat {\mathbb {R}}^n$ , and so $\mathcal {P}(\mathcal {P}(Z)) = \hat {\mathbb {R}}^n$ . Similarly, if $\mathcal {P}(Z)$ is a singleton, then $\mathcal {P}(\mathcal {P}(Z)) = \hat {\mathbb {R}}^n$ .

Thus, we may assume that both Z and $\mathcal {P}(Z)$ contain at least two points. We need to show that if $\mathbf {z}_1 \in Z$ and if $\mathbf {u}_1, \mathbf {u}_2 \in \mathcal {P}(Z)$ , then $\mathrm {arc}_{\mathbf {z}_1}[ \mathbf {u}_1 , \mathbf {u}_2 ] \subseteq \mathcal {P}(Z)$ . Assume that the points $\mathbf {z}_1, \mathbf {u}_1$ , and $\mathbf {u}_2$ are distinct; otherwise, the inclusion $\mathrm {arc}_{\mathbf {z}_1}[ \mathbf {u}_1 , \mathbf {u}_2 ] \subseteq \mathcal {P}(Z)$ is trivial. In other words, one has to show that any $\mathbf {v} \in \mathrm {arc}_{\mathbf {z}_1}[\mathbf {u}_1, \mathbf {u}_2]$ is a pole for Z; that is, for any $\mathbf {z}_2, \mathbf {z}_3 \in Z$ , we have $\mathrm {arc}_{\mathbf {v}}[\mathbf {z}_2,\mathbf {z}_3] \subseteq Z$ . If $\mathbf {z}_2, \mathbf {z}_3$ happen to be on the circle determined by $\mathbf {z}_1, \mathbf {u}_1, \mathbf {u}_2$ , and $\mathbf {v}$ , then it is easy to see that $\mathrm {arc}_{\mathbf {v}}[\mathbf {z}_2,\mathbf {z}_3] \subseteq Z$ by considering several cases (we omit the details). Otherwise, by Remark 6.10, we may restrict to a smaller dimensional ambient space that has dimension at least two and where the set $\mbox {conv}_{\{\mathbf {u}_1, \mathbf {u}_2\}}\{ \mathbf {z}_1, \mathbf {z}_2, \mathbf {z}_3\}$ has nonempty interior. By Corollary 6.11, we have

$$ \begin{align*}\mathbf{v} \in \mathrm{arc}_{\mathbf{z}_1}[ \mathbf{u}_1, \mathbf{u}_2] \subseteq \mbox{conv}_{\{\mathbf{z}_1 ,\mathbf{z}_2 , \mathbf{z}_3\}}\{ \mathbf{u}_1, \mathbf{u}_2\} \subseteq \mathcal{P}(\mbox{conv}_{\{\mathbf{u}_1, \mathbf{u}_2\}}\{ \mathbf{z}_1, \mathbf{z}_2, \mathbf{z}_3\}). \end{align*} $$

This shows the first inclusion in

$$ \begin{align*}\mathrm{arc}_{\mathbf{v}} [\mathbf{z}_2, \mathbf{z}_3] \subseteq \mbox{conv}_{\{\mathbf{u}_1, \mathbf{u}_2\}}\{ \mathbf{z}_1, \mathbf{z}_2, \mathbf{z}_3\} \subseteq Z, \end{align*} $$

while the second inclusion follows since $\mathbf {u}_1, \mathbf {u}_2 \in \mathcal {P}(Z)$ . Finally, since $\mathbf {z}_2, \mathbf {z}_3 \in Z$ were arbitrary, we get that $\mathbf {v} \in \mathcal {P}(Z)$ .

Proof Clearly, the points $\sum _{i=1}^k \alpha _i \mathbf {z}_i,\sum _{i=1}^k \beta _i \mathbf {z}_i$ are in $\mbox {conv}\{\mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . For any $t \in [0,1]$ , we have

$$ \begin{align*} \mathbf{u} + \Big( t (\sum _{i=1}^k \alpha_i \mathbf{z}_i - \mathbf{u})^* + (1-t) (\sum _{i=1}^k \beta_i \mathbf{z}_i - \mathbf{u})^* \Big)^* &\in \mathrm{conv}_{\mathbf{u}}(\mathrm{conv} \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k \}) \\ &= \mathrm{conv}(\mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k \}). \end{align*} $$

By Carathéodory’s theorem, there are points $\mathbf {x}_1, \ldots , \mathbf {x}_n \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 ,\ldots , \mathbf {z}_k \}$ and parameters $\gamma _1, \ldots , \gamma _n \in [0,1]$ , such that $\sum _{i=1}^n \gamma _i = 1$ and

(6.16) $$ \begin{align} \mathbf{u} + \Big( t (\sum _{i=1}^k \alpha_i \mathbf{z}_i - \mathbf{u})^* + (1-t) (\sum _{i=1}^k \beta_i \mathbf{z}_i - \mathbf{u})^* \Big)^* = \sum _{i=1}^n \gamma_i \mathbf{x}_i. \end{align} $$

Since $\mathbf {x}_i \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 ,\ldots , \mathbf {z}_k \}$ , there must be parameters $\delta _{i,1}, \ldots , \delta _{i,k}$ such that $\sum _{j=1}^k \delta _{i,j} =1$ and

(6.17) $$ \begin{align} \mathbf{x}_i = \mathbf{u} + \Big( \sum _{j=1}^k \delta_{i,j} (\mathbf{z}_j - \mathbf{u})^* \Big)^*, \mbox{ for all } i=1,\ldots, k. \end{align} $$

Substituting the equations (6.17) back into (6.16) and simplifying gives the stated identity.

Proof of Lemma 2.1

By definition, we have

(7.1) $$ \begin{align} \mathbf{u} + \big( t (\mathbf{z}_1 - &\mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^* \big)^* \nonumber \\ &= \mathbf{u} + \frac{t (\mathbf{z}_1 - \mathbf{u})^* + (1-t)(\mathbf{z}_2 - \mathbf{u})^* }{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2} \nonumber \\ & =\mathbf{u} \Big( 1 - \frac{t / \|\mathbf{z}_1 - \mathbf{u}\|^{2} + (1-t) / \|\mathbf{z}_2 - \mathbf{u}\|^{2}}{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2}\Big) \\ &\hspace{0.5cm} + \frac{t\mathbf{z}_1 / \|\mathbf{z}_1 - \mathbf{u} \|^2 + (1-t)\mathbf{z}_2 / \|\mathbf{z}_2 - \mathbf{u} \|^2}{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2}. \nonumber \end{align} $$

We look at the two terms of the last displayed expression separately. For the first term, we have

$$ \begin{align*} \mathbf{u} \Big( 1 - &\frac{t / \|\mathbf{z}_1 - \mathbf{u}\|^{2} + (1-t) / \|\mathbf{z}_2 - \mathbf{u}\|^{2}}{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2}\Big)\\ & = \mathbf{u} \frac{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2 - t / \|\mathbf{z}_1 - \mathbf{u}\|^{2} - (1-t) / \|\mathbf{z}_2 - \mathbf{u}\|^{2}}{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2} \\ & = \frac{\mathbf{u}}{D} \big( \| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2 \|\mathbf{z}_1 - \mathbf{u}\|^{2} \|\mathbf{z}_2 - \mathbf{u}\|^{2} \\ & \hspace{6cm}- t \|\mathbf{z}_2 - \mathbf{u}\|^{2} - (1-t) \|\mathbf{z}_1 - \mathbf{u}\|^{2} \big), \end{align*} $$

where

$$ \begin{align*}D:= \| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2 \|\mathbf{z}_1 - \mathbf{u}\|^{2} \|\mathbf{z}_2 - \mathbf{u}\|^{2}. \end{align*} $$

In the numerator, expand $\| t (\mathbf {z}_1 - \mathbf {u})^* + (1-t) (\mathbf {z}_2 - \mathbf {u})^*\|^2$ as a dot product and multiply throughout by $ \|\mathbf {z}_1 - \mathbf {u}\|^{2} \|\mathbf {z}_2 - \mathbf {u}\|^{2}$ . After elementary simplifications, we arrive at

$$ \begin{align*} \frac{\mathbf{u}}{D} & \big( -t(1-t)\|\mathbf{z}_2 - \mathbf{u}\|^2 - t(1-t) \|\mathbf{z}_1 - \mathbf{u}\|^2 + t(1-t) \langle \mathbf{z}_1 - \mathbf{u}, \mathbf{z}_2 - \mathbf{u} \rangle \\ & \hspace{7cm}+ t(1-t) \langle \mathbf{z}_2 - \mathbf{u}, \mathbf{z}_1 - \mathbf{u} \rangle \big) \\ & = \frac{\mathbf{u}}{D}\big( -t(1-t)\|\mathbf{z}_1 - \mathbf{z}_2 \|^2 \big) \\ & = \mathbf{u} \frac{-t(1-t)\|\mathbf{z}_1 - \mathbf{z}_2 \|^2 \|\mathbf{z}_1 - \mathbf{u}\|^{2}\|\mathbf{z}_2 - \mathbf{u}\|^{2}} {\| t (\mathbf{z}_1 - \mathbf{u})\|\mathbf{z}_2 - \mathbf{u}\|^{2} + (1-t) (\mathbf{z}_2 - \mathbf{u})\|\mathbf{z}_1 - \mathbf{u}\|^{2}\|^2} \\ & = \mathbf{u} \frac{O(\|\mathbf{u} \|^4)}{\|\mathbf{u} \|^6}. \end{align*} $$

Now we look at the second term in (7.1) and multiply its numerator and denominator by $\|\mathbf {u}\|^2$ . Then, it can be developed as

$$ \begin{align*} \frac{t\mathbf{z}_1 / \|\mathbf{z}_1 - \mathbf{u} \|^2 + (1-t)\mathbf{z}_2 / \|\mathbf{z}_2 - \mathbf{u} \|^2}{\| t (\mathbf{z}_1 - \mathbf{u})^* + (1-t) (\mathbf{z}_2 - \mathbf{u})^*\|^2} &=\frac{{t \mathbf{z}_1}{\Big\| \frac{\mathbf{z}_1}{\|\mathbf{u} \|} - \frac{\mathbf{u}}{\| \mathbf{u} \|} \Big\|^{-2}} + {(1-t) \mathbf{z}_2}{\Big\| \frac{\mathbf{z}_2}{\|\mathbf{u} \|} - \frac{\mathbf{u}}{\| \mathbf{u} \|} \Big\|^{-2}}}{\| t \big(\frac{\mathbf{z}_1}{\|\mathbf{u}\|} - \frac{\mathbf{u}}{\| \mathbf{u}\|}\big)^* + (1-t) \big(\frac{\mathbf{z}_2}{\|\mathbf{u}\|} - \frac{\mathbf{u}}{\| \mathbf{u}\|}\big)^*\|^2}. \end{align*} $$

Thus, taking the limit as $\|\mathbf {u}\| \rightarrow \infty $ , the first term in (7.1) converges to $0$ , while the second converges to $ t\mathbf {z}_1 + (1-t)\mathbf {z}_2$ . This completes the proof of the lemma.

Proof By symmetry, we may assume that $i=1$ . Since $\mathcal {L}_1 \subseteq \mathcal {S}_1$ , it is clear that

$$ \begin{align*}\bigcap _{S \in \mathcal{S}_1} S \subseteq \bigcap _{D \in \mathcal{L}_1} D. \end{align*} $$

To see the other containment, without loss of generality, assume $\mathbf {u}_1 = \infty $ . The set $\mathcal {L}_1$ consists of all supporting half-spaces that either correspond to the maximal faces of the polytope $\mathrm {conv}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ separating $\mathbf {u}_2$ and $\mathrm {conv}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ or those that correspond to the maximal faces of the cone

$$ \begin{align*}\mathbf{u}_2 + \mathrm{cone} \{ \mathbf{z}_1 - \mathbf{u}_2, \ldots, \mathbf{z}_k - \mathbf{u}_2 \}, \end{align*} $$

or both. Since $\mathcal {S}_1$ is the set of all half-spaces that separate $\mathbf {u}_2$ and $\mathrm {conv}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ , we have that $\bigcap _{D \in \mathcal {L}_1} D \subseteq S$ for all $S \in \mathcal {S}_1$ .

Proof of Theorem 6.3

Without loss of generality, we may assume $\mathbf {u}_2 = \infty $ and write $\mathbf {u}_1$ as $\mathbf {u}$ . In the proof, we need both families (6.1) and (6.2). So $\mathcal {S}_1$ , $\mathcal {L}_1$ correspond to $\mathbf {u}$ and $\mathcal {S}_2$ , $\mathcal {L}_2$ correspond to $\infty $ . It is clear that

$$ \begin{align*}\mathrm{conv}(\mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \}) \subseteq \bigcap_{i=1}^2 \bigcap_{S \in \mathcal{S}_i} S \end{align*} $$

because the right-hand side is convex with respect to both $\mathbf {u}$ and $\infty $ and contains $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . We aim to show the opposite inclusion.

$$ \begin{align*}\bigcap_{i=1}^2 \bigcap_{S \in \mathcal{S}_i} S \subseteq \mathrm{conv}(\mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \}). \end{align*} $$

If $\infty \in \mathrm {int}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ , then $\mathrm {conv}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}) = \hat {\mathbb {R}}^n$ , so the inclusion is trivial. Thus, we assume that $\infty \not \in \mathrm {int}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ and consider four cases based on whether $\infty $ and $\mathbf {u}$ are in $\{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ or not.

Case 1: Assume that $\infty , \mathbf {u} \not \in \{\mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ . We consider two sub-cases.

Case 1.a: If $\infty \not \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ , by Lemma 5.1, part (1), there is a spherical domain S, containing $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ and having $\mathbf {u}$ on its boundary, that strongly separates $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ and $\{\infty \}$ . Since $\infty $ is in $\mathrm {cl}(S^c)$ , S is convex with respect to both $\infty $ and $\mathbf {u}$ , we get that $S \in \mathcal {S}_1$ . So $\infty \not \in \bigcap _{i=1}^2 \bigcap _{S \in \mathcal {S}_i} S$ .

Since $\infty \not \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ , then $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ is closed and bounded. Thus, the set $\mathrm {conv}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ is closed and therefore also closed in $\hat {\mathbb {R}}^n \smallsetminus \{ \infty \}$ . So, by Lemma 5.1, part (1), for any $\mathbf {x} \not \in \mathrm {conv}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}) \cup \{ \infty \}$ , there is a spherical domain S, containing $\mathrm {conv} (\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ and having $\infty $ on its boundary, that strongly separates the sets $\{\mathbf {x}\}$ and $\mathrm {conv}(\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \})$ . We want to show that $\mathbf {x} \not \in \bigcap _{i=1}^2 \bigcap _{S \in \mathcal {S}_i} S$ .

Indeed, if $\mathbf {u} \in \mathrm {cl}(S^c)$ , then since S is convex with respect to both $\mathbf {u}$ and $\infty $ , we have $S \in \mathcal {S}_2$ . So, $\mathbf {x} \not \in \bigcap _{i=1}^2 \bigcap _{S \in \mathcal {S}_i} S$ . If $\mathbf {u} \not \in \mathrm {cl}(S^c)$ , then $\mathrm {cl}(S^c)$ is a closed $\mathbf {u}$ -convex spherical domain containing both $\mathbf {x}$ and $\infty $ and not containing $\mathbf {u}$ (in particular, $\mathbf {u}$ is not on the boundary of $\mathrm {cl}(S^c)$ ). Next, $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \} $ is a $\mathbf {u}$ -convex set that is closed in $\hat {\mathbb {R}}^n \smallsetminus \{\mathbf {u}\}$ . By Lemma 5.1, part (1), there is a spherical domain $S'$ , having $\mathbf {u}$ on its boundary, containing $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \} $ and strongly separating $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ and $\mathrm {cl} (S^c)$ . Since $\infty \in \mathrm {cl} (S^c)$ , we have that $S'$ does not contain $\infty $ and has $\mathbf {u}$ on its boundary. So, $S' \in \mathcal {S}_1$ , and since $\mathbf {x} \in \mathrm {cl} (S^c)$ , we get that $\mathbf {x} \not \in S'$ and conclude that $\mathbf {x} \not \in \bigcap _{i=1}^2 \bigcap _{S \in \mathcal {S}_i} S$ . This concludes the proof in the case when $\infty \not \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ .

Case 1.b: Suppose now $\infty \in \partial \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . Then, by Corollary 4.5, this is equivalent to $\mathbf {u} \in \partial \mathrm {conv}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . Every domain $S \in \mathcal {S}_1$ is $\mathbf {u}$ -convex and contains $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ by definition, so it contains $\mathrm {conv}_{\mathbf {u}}\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . This implies that S also contains $\infty $ , but by definition, $\infty \in \mathrm {cl}(S^c)$ , and therefore, $\infty \in \partial S$ . Similarly, $\mathbf {u} \in \partial S$ for all $S \in \mathcal {S}_2$ , so we conclude that $\mathcal {S}_1$ and $\mathcal {S}_2$ contain the same domains. Since a domain $S \in \mathcal {L}_1$ or $S \in \mathcal {L}_2$ is forced to have both $\mathbf {u}$ and $\infty $ on its boundary, they are all half-spaces. Therefore, for all $S \in \mathcal {L}_1 \cup \mathcal {L}_2$ , we get that

(7.3) $$ \begin{align} \mathrm{aff}_{\mathbf{u}} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} , \infty \} \cap \partial S \} &= \mathrm{aff}_{\infty} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} , \infty \} \cap \partial S \} \\ &= \mathrm{aff}_{\mathbf{u}} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} \} \cap \partial S \}. \nonumber \end{align} $$

To see the last equality, note that we have

$$ \begin{align*}\infty \in \partial \mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k \} \subseteq \mathrm{aff}_{\mathbf{u}} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u}\} \cap \partial S \}. \end{align*} $$

The first equality in (7.3) shows that the families $\mathcal {L}_1$ and $\mathcal {L}_2$ are equal. By assumption, $\mathbf {z}_1 , \ldots , \mathbf {z}_k ,\mathbf {u}, \infty $ are not all on a hyperplane. So, using Lemma 7.1, we have

$$ \begin{align*}\bigcap _{i=1}^2 \bigcap _{S \in \mathcal{S}_i} S = \bigcap _{i=1}^2 \bigcap _{S \in \mathcal{L}_i} S = \bigcap _{S \in \mathcal{L}_1} S. \end{align*} $$

Let $\mathcal {L}$ be the family of spherical domains described in Lemma 6.1. Then, the second equality in (7.3) shows that $\mathcal {L}_1 \subseteq \mathcal {L}$ . Note that if $S \in \mathcal {L}$ and $\infty \in \partial S$ , then $S \in \mathcal {L}_1$ . Therefore, if $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ , then $\infty \not \in \partial S$ ; in other words, $\partial S$ is a bounded set. Any spherical domain $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ contains $\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ and has $\mathbf {u}$ on its boundary, and hence S contains $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ . Thus, S is unbounded since $\infty \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ , by Corollary 4.4. Since $\mathcal {L} \smallsetminus \mathcal {L}_1$ is a finite set, there is an $R> 0$ such that $D(\mathbf {0}; R)^c \subset S \smallsetminus \{\mathbf {u}\}$ for all $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ . (Here, $D(\mathbf {0}; R)$ is the open ball with center $\mathbf {0}$ and radius R.) Thus, we have

$$ \begin{align*}D(\mathbf{0}; R)^c \subseteq \bigcap_{S \in \mathcal{L}\smallsetminus \mathcal{L}_1 } S \smallsetminus \{ \mathbf{u} \}. \end{align*} $$

Therefore,

$$ \begin{align*} \bigcap_{S \in \mathcal{L}_1} S \cap D(\mathbf{0}; R)^c &\subseteq \Big( \bigcap_{S \in \mathcal{L}_1} S\Big) \cap \Big( \bigcap_{S \in \mathcal{L} \smallsetminus \mathcal{L}_1} S \smallsetminus \{ \mathbf{u} \}\Big) \\ & = \bigcap_{S \in \mathcal{L}} S \smallsetminus \{ \mathbf{u} \} = \mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \}, \end{align*} $$

where in the last equality, we used Lemma 6.1. Since we are in a case where $\mathbf {u} \not \in \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ and $\mathbf {u} \in \partial \mathrm {conv} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ , the boundary of the intersection of half-spaces, $\bigcap _{S \in \mathcal {L}_1} S$ , contains at least a line, so

$$ \begin{align*}\mathrm{conv}\Big( \bigcap_{S \in \mathcal{L}_1} S \cap D(\mathbf{0}; R)^c \Big) = \bigcap_{S \in \mathcal{L}_1} S , \end{align*} $$

and by Lemma 7.1, we have

$$ \begin{align*}\bigcap _{i=1}^2 \bigcap _{S \in \mathcal{S}_i} S = \bigcap _{i=1}^2 \bigcap _{S \in \mathcal{L}_i} S = \bigcap_{S \in \mathcal{L}_1} S \subseteq \mathrm{conv}( \mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \}). \end{align*} $$

Case 2: Assume that $\infty \in \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\}$ , but $\mathbf {u} \not \in \{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ . By definition,

$$ \begin{align*}\mathrm{conv} \{ \mathbf{z}_1, \ldots , \mathbf{z}_k\} = \mathrm{conv} \{ \mathbf{z}_i : \mathbf{z}_i \neq \infty, i=1,\ldots, k \} \cup\{\infty\}. \end{align*} $$

If $\mathbf {u} \in \mathrm {conv} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k\}$ , then $\mathbf {u} \in \mathrm {conv} \{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1,\ldots , k \}$ . So we have $ \infty \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1 , \ldots , k \}$ , and $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1 , \ldots , k \} = \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ . This, of course, implies that

$$ \begin{align*}\mathrm{conv} \big( \mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_i : \mathbf{z}_i \neq \infty, i=1 , \ldots , k \} \big) = \mathrm{conv} \big(\mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \}\big). \end{align*} $$

Therefore, any spherical domain S, convex with respect to both $\mathbf {u}$ and $\infty $ , that contains $\{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1 , \ldots , k \}$ is forced to contain $\infty $ because it contains $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1 , \ldots , k \}$ . Thus, $\infty \in \partial S$ because S is convex. This implies that the families $\mathcal {S}_1, \mathcal {S}_2, \mathcal {L}_1$ , and $\mathcal {L}_2$ corresponding to the sets $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k \}$ and $\{ \mathbf {z}_i : \mathbf {z}_i \neq \infty , i=1 , \ldots , k \}$ are the same. So, we can consider the set $\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k\} \smallsetminus \{\infty \}$ and argue as in Case 1.

Therefore, we may assume that $\mathbf {u} \not \in \mathrm {conv} \{ \mathbf {z}_1, \ldots , \mathbf {z}_k\}$ . If $S \in \mathcal {L}_1$ , then by definition, $\infty \in \mathrm {cl}(S^c)$ , but the premise of the current case implies that $\infty \in S$ , so we need to have $\infty \in \partial S$ . Since in addition, $\mathbf {u} \in \partial S$ , one can see that

(7.4) $$ \begin{align} \mathrm{aff}_{\mathbf{u}} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} , \infty \} \cap \partial S \} &= \mathrm{aff}_{\infty} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} , \infty \} \cap \partial S \} \\ &= \mathrm{aff}_{\mathbf{u}} \{ \{ \mathbf{z}_1 ,\ldots , \mathbf{z}_k, \mathbf{u} \} \cap \partial S \}. \nonumber \end{align} $$

The first equality shows that $\mathcal {L}_1 \subseteq \mathcal {L}_2$ . Therefore, we have

(7.5) $$ \begin{align} \bigcap_{i=1}^2 \bigcap _{S\in \mathcal{L}_i} S = \bigcap _{S\in \mathcal{L}_2} S. \end{align} $$

The family $\mathcal {L}_1$ is the set of all half-spaces that support the maximal faces of the cone $\mathbf {u} + \mathrm {cone}\{ \mathbf {z}_1 - \mathbf {u} , \ldots , \mathbf {z}_k - \mathbf {u}\}$ . (Note that the latter cone has a nonempty interior, or else $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k , \mathbf {u}, \infty \}$ lie on a $(n - 1)\mbox {-sphere}$ , contradicting our assumption.) That is,

(7.6) $$ \begin{align} \bigcap _{S\in \mathcal{L}_1} S &= \mathbf{u} + \mathrm{cone}\{ \mathbf{z}_1 - \mathbf{u} , \ldots , \mathbf{z}_k - \mathbf{u}\} \nonumber \\ &= \{ \mathbf{u} + t(\mathbf{z}-\mathbf{u}) : t \ge 0, \mathbf{z} \in \mathrm{conv}(\{ \mathbf{z}_1, \ldots , \mathbf{z}_k\} \smallsetminus \{ \infty \}) \} \cup \{\infty\}. \end{align} $$

A domain $S \in \mathcal {L}_2$ can be of two types: either $\mathbf {u} \in \partial S$ or $\mathbf {u} \in S^c$ . In the first case, S is a half-space that supports a maximal face of the cone $\mathbf {u} + \mathrm {cone}\{ \mathbf {z}_1 - \mathbf {u} , \ldots , \mathbf {z}_k - \mathbf {u}\}$ , while in the second case, S is a half-space that supports a maximal face of the polytope $\mathrm {conv}(\{ \mathbf {z}_1, \ldots , \mathbf {z}_k\} \smallsetminus \{ \infty \})$ that separates $\mathbf {u}$ from $\{ \mathbf {z}_1, \ldots , \mathbf {z}_k\}$ . Thus,

(7.7) $$ \begin{align} \bigcap _{S\in \mathcal{L}_2} S = \{ \mathbf{z} + t(\mathbf{z}-\mathbf{u}) : t \ge 0, \mathbf{z} \in \mathrm{conv}(\{ \mathbf{z}_1, \ldots , \mathbf{z}_k\} \smallsetminus \{ \infty \}) \} \cup \{\infty\}. \end{align} $$

We need to prove that

$$ \begin{align*}\bigcap _{S\in \mathcal{L}_2} S \subseteq \mathrm{conv}( \mathrm{conv}_{\mathbf{u}} \{ \mathbf{z}_1 , \ldots , \mathbf{z}_k \}). \end{align*} $$

Let again $\mathcal {L}$ be the family as described in Lemma 6.1. Then, the second equality in (7.4) shows that $\mathcal {L}_1 \subseteq \mathcal {L}$ . Note that if $S \in \mathcal {L}$ and $\infty \in \partial S$ , then $S \in \mathcal {L}_1$ . Therefore, if $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ , then $\infty \not \in \partial S$ ; in other words, $\partial S$ is a bounded set. Any spherical domain $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ contains $\{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ and has $\mathbf {u}$ on its boundary; hence, S contains $\mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ . Thus, S is unbounded since $\infty \in \mathrm {conv}_{\mathbf {u}} \{ \mathbf {z}_1 , \ldots , \mathbf {z}_k \}$ by Corollary 4.4. Since $\mathcal {L} \smallsetminus \mathcal {L}_1$ is a finite set, there is an $R> 0$ such that $D(\mathbf {0}; R)^c \subset S \smallsetminus \{\mathbf {u}\}$ for all $S \in \mathcal {L} \smallsetminus \mathcal {L}_1$ . We get that