In light of the significance and the successful applications of the convexity of sets (in linear spaces) in many different mathematics branches, it is not unexpected for mathematicians to try to find its counterpart for sets in manifolds. Such an attempt started from the early 1940's with sets in $\mathbb{S}^{n-1}$, the unit sphere in the Euclidean space $\mathbb R^n$, since it is a typical manifold (see [2, 6, 9-11, 13] and the references therein).

Several different definitions of convex sets in $\mathbb{S}^{n-1}$ were proposed since 1940's. For instance, a set in $\mathbb S^{n-1}$ is called strongly convex if it contains no antipodal points ($x_1, x_2\in \mathbb S^{n-1}$ are called (a pair of) antipodal points if $x_2=-x_1$) and it contains, with each pair of its points, the shorter arc of the great circle determined by them (see [2]); weakly convex if it contains, with each pair of its points, the shorter arc or a semicircular arc of a great circle determined by them (see [10]); Robinson-convex if it contains, with each pair of its non-antipodal points, the shorter arc of the great circle determined by them (see [6]); Horn-convex if it contains, with each pair of its non-antipodal points, at least one of the great circle arcs determined by them (see [6]).

Clearly, every strongly convex set is weakly convex and the weak convexity implies the Robinson-convexity, and the Robinson-convexity implies the Horn-convexity (see [2]). Also, simple examples show that all the inclusion relations here are proper.

Although various convexities for sets in $\mathbb{S}^{n-1}$ were proposed for different purposes and a great progress in the research was made in the early stage (see [2, 6, 9, 10, 13]) and recently this topic regains the attentions of mathematicians (see [1, 3, 4, 7, 8] and the references therein), the progress in this area (in particular in the aspect of geometric invariants) seems not as fast as expected. One of the main obstacles of progressing is that all these definitions of convexity for sets in $\mathbb{S}^{n-1}$ are in geometric forms, which made it complicated and even impossible to formulate or demonstrate some conclusions in the high dimension cases. This situation can be seen from several recent important work: Lassak in 2015 defined the width of a strongly (or spherically) convex body (see below for definition) and studied its elementary properties (see [7]), and further investigated the properties of (spherically) reduced bodies on spheres (see [8]), but the arguments he presented were somehow complicated due to his pure geometric methods. Similarly, Vigodsky's early work on demonstrating an analogue of Carathéodory's theorem for strongly convex sets in $\mathbb S^{2}$ (see [13]) was also done by a geometric approach and thus hard to extend to the high dimension cases. Also, some authors have to deal with sets in $\mathbb{S}^{n-1}$ in the frame of Euclidean spaces (see [14]).

In this paper, taking strong convexity as an example, we present a pure analytic approach to defining the convexity for sets in $\mathbb S^{n-1}$ and studying the basic properties of strongly convex sets in $\mathbb S^{n-1}$. As an illustration, at the end we prove a structure theorem of Minkowski type for strongly convex sets in $\mathbb S^{n-1}$.

We work in the $n$-dimensional Euclidean space $\mathbb R^n$ with the classical inner product $\langle\cdot , \cdot\rangle$ and the induced norm $\parallel\cdot\parallel$. For any non-zero $u\in \mathbb{R}^{n}$, $H_u$ denotes the hyperplane $\{x\in\mathbb R^n\mid \langle u, x\rangle=0\}$, $H^{-}_{u}$ (or $H^{-}$ simply) denotes the open half space $\{x\in \mathbb{R}^{n}\mid \langle{u, x}\rangle<0\}$, and $\bar{H}^{-}_{u}$ (or $\bar{H}^{-}$ simply) denotes the closed half space $\{x\in \mathbb{R}^{n}\mid \langle{u, x}\rangle\leqslant 0\}$.

For non-empty $C\subset \mathbb R^n$, ${\rm co}(C)$ and ${\rm cone}(C)$ denote the convex hull, the convex conical hull of $C$, respectively. For other notation and terms refer to [5].

A subset of the form $H^{-}\cap\mathbb{S}^{n-1}$ (resp. $\bar{H}^{-}\cap\mathbb{S}^{n-1}$) is call an open (resp. a closed) semi-sphere and a set of the form $H\cap \mathbb S^{n-1}$ is called a hypercircle. Generally, if $H_k\subset \mathbb R^n$ is a $k$-dimensional subspace ($1\leqslant k\leqslant n-1$), then $H_k\cap \mathbb S^{n-1}$ is called a $k$-circle. Thus a hypercircle is an $(n-1)$-circle and a 2-circle is a great circle named by other authors.

The main tool we used in our analytic approach is the radial function

$ \varphi:\mathbb{R}^{n}\longrightarrow\mathbb{S}^{n-1}\cup\{o\}, $

where $o$ stands for the origin of $\mathbb R^n$, defined by

$ \varphi(x):=\left\{\begin{array}{rl}\frac{x}{\|x\|}, &\;x\not=o, \\ o, &\;x=o.\end{array}\right. $

Observe that the radial function has the following properties

(1) $\varphi\circ\varphi=\varphi$;

(2) $\varphi({\lambda}x)=\varphi(x)$ for all $x\in \mathbb{R}^{n}$ and all $\lambda>0$;

(3) $\varphi(x)=x$ iff $x\in \mathbb{S}^{n-1}$ or $x=o$.

In terms of the radial function, we have the following definition.

Definition 2.1 For $x, y\in\mathbb{S}^{n-1}$ and $0\leqslant \lambda\leqslant 1$, we define

$ {\lambda}x+_{s}(1-{\lambda})y:=\varphi({\lambda}x+(1-{\lambda})y) $

called a spherically convex combination of $x, y$ ($s$-convex combination for brevity).

Generally, given $x_{1}, x_{2}, \cdots, x_{k}\in \mathbb{S}^{n-1}$, $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{k}\in [0, 1]$ with ${\sum\limits^{k}_{i=1}}\lambda_{i}=1$, we define their $s$-convex combination

$ (s){\sum\limits_{i=1}^{k}}{\lambda_{i}}x_{i}:=\varphi({\sum\limits_{i=1}^{k}}{\lambda_{i}}x_{i}). $

Natrually, when $k=2$, we write ${\lambda}x+_{s}(1-{\lambda})y$ instead of $(s)({\lambda}x+(1-{\lambda})y)$.

Remark 2.1 ⅰ) $x, y\in \mathbb S^{n-1}$ are non-antipodal points iff ${\lambda}x+_{s}(1-{\lambda})y\not=o$ or equivalently ${\lambda}x+(1-{\lambda})y\not=o$ for any $0\leqslant \lambda\leqslant 1$.

ⅱ) In terms of Definition 2.1, the short arc of the great circle determined by non-antipodal points $x, y\in\mathbb{S}^{n-1}$ is $\{\lambda x+_s(1-\lambda)y\mid 0\leqslant \lambda\leqslant 1\}$.

Now, we give the definition of strongly convex sets in $\mathbb S^{n-1}$ in an analytic form, and we adopt "spherically convex" instead of "strongly convex".

Definition 2.2 A set $C\subset \mathbb S^{n-1}$ is called spherically convex ($s$-convex for brevity$)$ if $\lambda x+_s(1-\lambda)y\in C$ for any $x, y\in C$ and $0\leqslant \lambda\leqslant 1$.

Remark 2.2 ⅰ) By Remark 2.1, an $s$-convex set contains no antipodal points for sure, and $C\subset \mathbb S^{n-1}$ is $s$-convex iff it contains, with each pair of its points, the short arc of the great circle determined by them.

ⅱ) It is easy to check that an open semi-sphere is $s$-convex while a closed semi-sphere, a $k$-circle and $\mathbb S^{n-1}$ are not since they all contain antipodal points.

The following simple property of $s$-convex sets will be needed later.

Proposition 2.1 Let $C\subset \mathbb S^{n-1}$ be an $s$-convex set. Then we have

$ {\rm cone}(C)=\bigcup\limits_{t\geqslant 0}tC=\{t x\mid x\in C, t\geqslant 0\}. $

Moreover, ${\rm cone}(C)$ is closed iff $C$ is closed $($so compact$)$.

Proof For the first conclusion, since $\bigcup\limits_{t\geqslant 0}t C$ is clearly a cone, we need only to show that it is convex. More precisely, we need only to show $x_1+x_2\in \bigcup\limits_{t\geqslant 0}t C$ whenever $x_1, x_2\in \bigcup\limits_{t\geqslant 0}t C$. Without loss of generality, we may assume that both $x_1$ and $x_2$ are non-zero. Let $x_1=t_1x'_1$ and $x_2=t_2x'_2$ for some $x'_1, x'_2\in C$ and $t_1, t_2>0$. Thus by the $s$-convexity of $C$,

$ x:=\frac {x_1+x_2}{\| x_1+x_2\|}=\frac{t_1x'_1+t_2x'_2}{\parallel t_1x'_1+t_2x'_2 \parallel} =\frac{\frac{t_1}{t_1+t_2}x'_1+\frac {t_2}{t_1+t_2}x'_2}{\parallel \frac{t_1}{t_1+t_2}x'_1+\frac {t_2}{t_1+t_2}x'_2 \parallel} =\frac{t_1}{t_1+t_2}x'_1+_s\frac {t_2}{t_1+t_2}x'_2\in C, $

where we used the fact that $\frac{t_1}{t_1+t_2}+\frac {t_2}{t_1+t_2}=1$, and so $x_1+x_2=\| x_1+x_2\| x\in \bigcup\limits_{t\geqslant 0}t C$.

The argument for the second conclusion is standard and straightforward.

In this section, we study basic properties of $s$-convex sets and in turn define the spherically convex hull ($s$-convex hull for brevity) of a set in $\mathbb S^{n-1}$. We start with the following simple property.

Proposition 3.1 ⅰ) If $\{C_i\}_{i\in\Lambda}$ is a family of $s$-convex sets in $\mathbb S^{n-1}$ with $C:=\bigcap\limits_{i\in\Lambda} C_i\not=\emptyset$, then $C$ is $s$-convex.

ⅱ) If $C$ is $s$-convex and $\mathbb S_k:=H_k\cap \mathbb S^{n-1}$ is a $k$-circle ($2\leqslant k\leqslant n-1$), then $C\cap \mathbb S_k$ is $s$-convex provided that the intersection is not empty.

Proof ⅰ) For any $x, y\in C$, we have $x, y\in C_i$ for each $i\in \Lambda$. So $x+_sy\in C_i$ for each $i\in \Lambda$ by the $s$-convexity of $C_i$ and in turn $x+_sy\in C$.

ⅱ) For any $x, y\in C\cap \mathbb S_k$, we have $x+_sy\in C$ by the $s$-convexity of $C$, $x+_sy=\frac{x+y}{\|x+y\|}\in H_k$ and $x+_sy\in \mathbb S^{n-1}$ by the definition. So $x+_sy\in C\cap \mathbb S_k$.

Next is an analogue of the conclusion for convex sets in $\mathbb R^n$, which is not easy to formulate and prove in a geometric approach.

Theorem 3.1 If $C\subset \mathbb{S}^{n-1}$ is $s$-convex, then for any $x_{1}, x_{2}, \cdots, x_{k}\in C$ and ${\lambda}_{1}, {\lambda}_{2}, \cdots, {\lambda}_{k}$ $\in [0, 1]$ with ${\sum\limits^{k}_{i=1}}\lambda_{i}=1$, $k=1, 2, \cdots$, we have $(s)\sum\limits^{k}_{i=1}{\lambda}_{i}x_{i}\in C$. In particular, $o\notin {\rm co}(C)$.

Proof It is trivial for $k=1, 2$ by Definition 2.2. Suppose the conclusion holds for all $1\leqslant j\leqslant k-1$ (which implies, in particular, that $o\notin {\rm co}\{x_1, x_2, \cdots, x_j\}$ for any $x_1, x_2, \cdots, x_j\in C$ and $\mu_1, \mu_2, \cdots, \mu_j>0$ with $\sum\limits_{i=1}^j\mu_i=1$). Now for $x_{1}, x_{2}, \cdots, x_{k}\in C$ and ${\lambda}_{i}\geqslant 0$, ${\sum\limits^{k}_{i=1}}\lambda_{i}=1$ (we may clearly assume $\lambda_i>0$ for all $i$), we denote $z=\sum\limits^{k-1}_{i=1}\frac{{\lambda}_{i}}{\sum\limits^{k-1}_{j=1}{\lambda}_{j}}x_{i}$. Clearly, $z\neq o$ and ${\varphi}(z)=\frac{z}{{\parallel}z{\parallel}}\in C$ by the induction hypothesis.

Thus we have, by Definition 2.1 and the properties of $\varphi$,

$ \begin{equation}\notag \begin{array}{rl} (s)\sum\limits^{k}_{i=1}{\lambda}_{i}x_{i}&={\varphi}(\sum\limits^{k}_{i=1}{\lambda}_{i}x_{i}) ={\varphi}(\sum\limits^{k-1}_{i=1}\frac{{\lambda}_{i}}{\sum\limits^{k-1}_{j=1}{\lambda}_{j}}x_{i}+\frac{{\lambda}_{k}}{\sum\limits^{k-1}_{j=1}{\lambda}_{j}}x_{k})\\ &={\varphi}(z+\frac{{\lambda}_{k}}{\sum\limits^{k-1}_{j=1}\lambda_{j}}x_k)={\varphi}(\frac{z}{{\parallel}z{\parallel}}+\frac{{\lambda}_{k}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}\lambda_j}x_{k})\\ &={\varphi}((\sum\limits^{k-1}_{j=1}\lambda_j)\frac{z}{{\parallel}z{\parallel}}+\frac{{\lambda}_{k}}{{\parallel}z{\parallel}}x_{k})={\varphi}({\parallel}z{\parallel}(\sum\limits^{k-1}_{j=1}{\lambda}_{j})\frac{z}{{\parallel}z{\parallel}}+{\lambda}_{k}x_{k})\\ &={\varphi}(\frac{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j} +\lambda_{k}}\frac{z}{{\parallel}z{\parallel}}+\frac{{\lambda}_{k}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda_{j}+{\lambda}_{k}}}x_{k})\\ &=\frac{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j} +\lambda_{k}}\frac{z}{{\parallel}z{\parallel}}+_s\frac{{\lambda}_{k}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda_{j}+{\lambda}_{k}}}x_{k}\in C, \end{array} \end{equation} $

where we used the fact that $\frac{z}{{\parallel}z{\parallel}}$, $x_{k}\in C$ and

$ \frac{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda}_{j} +\lambda_{k}}+\frac{{\lambda}_{k}}{{\parallel}z{\parallel}\sum\limits^{k-1}_{j=1}{\lambda_{j}+{\lambda}_{k}}}=1. $

Now, $o\notin {\rm co}(C)$ is trivial.

Next, we consider the spherically convex hull of a set in $\mathbb S^{n-1}$, i.e., the smallest $s$-convex set containing a set. Since if $C\subset \mathbb S^{n-1}$ is $s$-convex, then $o\notin {\rm co}(C)$ by the definition and in turn $o\notin {\rm co}(S)$ if $S\subset C$, so we see that a set $S\subset \mathbb S^{n-1}$ has a spherically convex hull iff $o\notin {\rm co}(S)$.

Definition 3.1 Given $S\subset \mathbb{S}^{n-1}$ with $o\notin {\rm co}(S)$, we define its spherically convex hull ${\rm Sco}(S)$ ($s$-convex hull for brevity) by

$ {\rm Sco}(S):=\varphi({\rm co}(S))=\{{(s){\sum\limits^{k}_{i=1}}{\lambda_{i}}x_{i}\mid}x_{i}\in S, {\lambda_{i}}\geqslant 0, {\sum\limits^{k}_{i=1}}\lambda_{i}=1, k=1, 2, \cdots\}. $

Theorem 3.2 Let $S\subset \mathbb{S}^{n-1}$ with $o\notin {\rm co}(S)$. Then

ⅰ) ${\rm Sco}(S)$ is $s$-convex.

ⅱ) ${\rm Sco}(S)=\cap\{C\subset\mathbb S^{n-1} \mid C\supset S \ {\rm and }\ C $ is $s-{\rm convex}\}$, i.e., ${\rm Sco}(S)$ is the smallest $s$-convex set containing $S$.

ⅲ) ${\rm Sco}(S)={\rm cone}(S)\cap\mathbb S^{n-1}$.

Proof ⅰ) Let $x:=(s){\sum\limits^{k}_{i=1}}{\lambda_{i}}x_{i}, y:=(s){\sum\limits^{l}_{j=1}}{\mu_{i}}y_{j}\in {\rm Sco}(S)$ and $0<\lambda<1$. Then by the properties of $\varphi$ (write $a:=\parallel \sum\limits^{k}_{i=1}{\lambda_{i}}x_{i}\parallel, \;b:=\parallel\sum\limits^{l}_{j=1}{\mu_{j}}y_{j}\parallel$),

$ \begin{array}{rl}&\lambda x+_s(1-\lambda)y=\varphi\left(\lambda x+(1-\lambda)y\right)=\varphi\left(\frac \lambda a\sum\limits^{k}_{i=1}{\lambda_i}x_i+\frac{(1-\lambda)}b\sum\limits^{l}_{j=1}{\mu_{i}}y_{j}\right)\\ =&\varphi\left(\lambda b\sum\limits^{k}_{i=1}{\lambda_i}x_i+(1-\lambda)a\sum\limits^{l}_{j=1}{\mu_{i}}y_{j}\right)=\varphi\left(\sum\limits^{k}_{i=1}\frac{\lambda b\lambda_i}{\lambda b+(1-\lambda)a}x_i+\sum\limits^{l}_{j=1}\frac {(1-\lambda)a\mu_{i}}{\lambda b+(1-\lambda)a}y_{j}\right)\\ =&(s)\left(\sum\limits^{k}_{i=1}\frac{\lambda b\lambda_i}{\lambda b+(1-\lambda)a}x_i+\sum\limits^{l}_{j=1}\frac {(1-\lambda)a\mu_{i}}{\lambda b+(1-\lambda)a}y_{j}\right)\in {\rm Sco}(S), \end{array} $

where we used the definition of ${\rm Sco}(S)$ and the fact that

$ \frac{\lambda b\lambda_i}{\lambda b+(1-\lambda)a}\geqslant 0, \; \frac {(1-\lambda)a\mu_{i}}{\lambda b+(1-\lambda)a}\geqslant 0 $

and

$ \sum\limits^{k}_{i=1}\frac{\lambda b\lambda_i}{\lambda b+(1-\lambda)a}+\sum\limits^{l}_{j=1}\frac {(1-\lambda)a\mu_{i}}{\lambda b+(1-\lambda)a}=1. $

ⅱ) Denote $D:=\cap\{C\subset\mathbb S^{n-1} \mid C\supset S \ {\rm and }\ C $ is $s-{\rm convex}\}$ ($D$ is well-defined since ${\rm Sco}(S)\supset S$ and ${\rm Sco}(S)$ is $s$-convex by ⅰ)). Then $D\subset {\rm Sco}(S)$ clearly since ${\rm Sco}(S)\supset S$ and it is $s$-convex by ⅰ) above. Conversely, since for each $s$-convex $C\supset S$, ${\rm Sco}(S)\subset C$ by Theorem 3.1, we have ${\rm Sco}(S)\subset D.$ Therefore ${\rm Sco}(S)=D$.

ⅲ) Since ${\rm cone}(S)=\bigcup\limits_{t\geqslant 0}t{\rm co}(S)$, ${\rm Sco}(S)\subset {\rm cone}(S)\cap\mathbb{S}^{n-1}$ clearly by the definition of ${\rm Sco}(S)$. Conversely, if $x\in {\rm cone}(S)\cap\mathbb{S}^{n-1}$, then $x=\sum\limits^{k}_{i=1}{\lambda}_{i}x_{i}$ with $x_{i}\in S, $ ${\lambda}_{i}\geqslant 0$ and ${\parallel}x{\parallel}=1$ (so $\sum\limits_{i=1}^k\lambda_i\not=0$). Thus

$ x=\varphi(x)=\varphi(\frac x{\sum\limits_{j=1}^k\lambda_j}) =(s)\sum\limits^{k}_{i=1}\frac{\lambda_i}{\sum\limits_{j=1}^k\lambda_j}x_{i}\in {\rm Sco}(S) $

by the definition of ${\rm Sco}(S)$ again. Hence ${\rm cone}(S)\cap\mathbb{S}^{n-1}\subset {\rm Sco}(S)$ and so ${\rm Sco}(S)={\rm cone}(S)\cap\mathbb S^{n-1}$.

In this section, we establish a theorem of Minkowski type for $s$-convex sets, which states that a closed $s$-convex set can be expressed as the $s$-convex hull of its $s$-extreme points (see below for definition). We start with some necessary notation and terms.

For an $s$-convex $C\subset \mathbb S^{n-1}$, we denote ${\rm dim}C={\rm dim}({\rm cone}(C))-1$ (notice that ${\rm cone}(C)$ is convex), called the dimension of $C$, ${\rm rbd}(C):= {\rm rbd}({\rm cone}(C))\cap {\rm cl}C$, where "rbd" in the right-handed side denotes the relative boundary, called the relative boundary of $C$ and ${\rm ri}(C):= {\rm ri}({\rm cone}(C))\cap C={\rm ri}({\rm cone}C)\cap \mathbb{S}^{n-1}$ (the latter equality can be checked by Proposition 2.1), where "ri" in the right-handed side denotes the relative interior, called the relative interior of $C$. It is easy to show that if $C\subset \mathbb S^{n-1}$ is $s$-convex, then so is ${\rm ri}(C)$ and that if $C$ is closed and $s$-convex, then $C={\rm rbd}(C)\cup {\rm ri}(C)$.

Definition 4.1 Let $C\subset \mathbb S^{n-1}$ be a closed $s$-convex set. A point $x\in C$ is called an $s$-extreme point of $C$ if $x=\frac 12 x_1+_s\frac 12 x_2$ for some $x_1, x_2\in C$, then $x_1=x_2$. The set of $s$-extreme points of $C$ is denoted by ${\rm Sext}(C)$.

Remark 4.1 It is easy to check that if $x$ is an $s$-extreme point of $C$, then $x\in {\rm rbd}(C)$.

Proposition 4.1 If $C\subset \mathbb S^{n-1}$ is a closed and $s$-convex set, then ${\rm Sext}(C)\not=\emptyset$.

Proof Observe first that ${\rm cone}(C)$ is closed by Proposition 2.1. Now we show that ${\rm cone}(C)$ is also line-free: Suppose there is a line $l:=\{x_0+\mu u\mid \mu\in\mathbb R\}\subset {\rm cone}(C)$ for some $x_0\in \mathbb R^n$ and $u\in\mathbb S^{n-1}$. Then $x_0+\frac 1tu, x_0-\frac 1tu\in {\rm cone}(C)$ for all $t>0$ and in turn $tx_0+u=t(x_0+\frac 1tu), tx_0-u=t(x_0-\frac 1tu)\in {\rm cone}(C)$ for all $t>0$ since ${\rm cone}(C)$ is a cone. Thus, by the closedness of ${\rm cone}(C)$, we have

$ u=\lim\limits_{t\to 0^+}(tx_0+u)\in {\rm cone}(C){\rm and } -u=\lim\limits_{t\to 0^+}(tx_0-u)\in {\rm cone}(C), $

which together with $\|u\|=\|-u\|=1$ leads to (the antipodal points) $u, -u\in C$ by Proposition 2.1, a contradiction.

Since ${\rm cone}(C)$ is closed and line-free, it has at least one extreme ray by Theorem 1.4.3 in [12]. Let $L$ be an extreme ray of ${\rm cone}(C)$ and let $x=L\cap C$. Then $L=\{tx\mid t\geqslant 0\}$. Now, if $x=\frac 12x_1+_s\frac 12 x_2=(\frac 12x_1+\frac 12 x_2)/\|\frac 12x_1+\frac 12 x_2\|$ for some $x_1, x_2\in C$, then $\frac 12x_1+\frac 12 x_2=\|\frac 12x_1+\frac 12 x_2\|x\in L$. So $x_1, x_2\in L$ since $L$ is an extreme ray and in turn $x_1, x_2\in L\cap C$ which leads to $x_1=x_2=x$ clearly. Thus, $x$ is an $s$-extreme point.

The following is a theorem of Minkowski type for closed $s$-convex sets in $\mathbb S^{n-1}$, which, again, it is hard to prove in geometric methods.

Theorem 4.1 Let $C\subset\mathbb S^{n-1}$ be a closed and $s$-convex set. Then $C={\rm Sco}\left( {\rm Sext}(C)\right)$.

Proof Since ${\rm cone}(C)$ is closed by Proposition 2.1 and line-free (see the argument for Proposition 4.1), we have ${\rm cone}(C)={\rm co}({\rm extr}({\rm cone}(C)))$ by Theorem 1.4.3 in [12] again, where ${\rm extr}({\rm cone}(C))$ denotes the union of extreme rays of ${\rm cone}(C)$. Thus, if $x\in C\subset {\rm cone}(C)$, then $x=\sum\limits_{i=1}^m\lambda_ix_i$ with non-zero $x_i\in L_i$ for some extreme ray $L_i$ and $\lambda_i>0, \;\sum\limits_{i=1}^m\lambda_i=1$. Hence, by the properties of $\varphi$,

$ \begin{array}{rl}x&=\varphi(x)=\varphi(\sum\limits_{i=1}^m\lambda_ix_i)=\varphi(\sum\limits_{i=1}^m\lambda_i\|x_i\|\frac{x_i}{\|x_i\|})\\ &=\varphi(\sum\limits_{i=1}^m\frac{\lambda_i\|x_i\|}{\sum\limits_{j=1}^m\lambda_j\|x_j\|}\frac{x_i}{\|x_i\|}) =(s)(\sum\limits_{i=1}^m\frac{\lambda_i\|x_i\|}{\sum\limits_{j=1}^m\lambda_j\|x_j\|}\frac{x_i}{\|x_i\|})\in {\rm Sco}\left( {\rm Sext}(C)\right), \end{array} $

where we used the fact that all $\frac{x_i}{\|x_i\|}=L_i\cap C$ are in ${\rm Sext}(C)$ (see the argument for Proposition 4.1), and $\sum\limits_{i=1}^m\frac{\lambda_i\|x_i\|}{\sum\limits_{j=1}^m\lambda_j\|x_j\|}=1$.

Final Remark From Theorem 3.1 and Theorem 4.1, one can see that our analytic approach indeed makes it possible to formulate and demonstrate some conclusions which can hardly be formulated and proved by pure geometric methods. It is expectable that, with this analytic approach, more conclusions will be established and more concepts, such as the meaningful geometric invariants for $s$-convex sets etc., will be proposed later. We leave these topics to other papers.