Show that convex hull of $\{x_j\}_{j =1}^N \subseteq \mathbb{S}^{n-1}$ meets $\mathbb{S}^{n-1}$ only in the $x_j$

Question

Let $x_1, \dots, x_N$ be a finite collection of distinct points on the unit sphere $\mathbb{S}^{n-1} \subseteq \mathbb{R}^n$. Consider the convex hull $C$ of these points, defined to be $$ C = \{ \sum^N_{j = 1} \lambda_j x_j : \lambda_j \ge 0, \, \sum^N_{j = 1} \lambda_j = 1 \} \subseteq \{x \in \mathbb{R}^n : \|x\| \le 1\}. $$

I would like to prove that $C \cap \mathbb{S}^{n-1} = \{x_1, \dots, x_N \}.$

When $N = 2$, there is a nice proof of this by finding the maxima the function $$ [0, 1] \ni t \mapsto \|tx_1 + (1-t)x_2 \|^2, $$ where $\| \cdot \|$ is the usual norm on $\mathbb{R}^n$. But I am unsure how to generalize to several points. Hints or solutions are greatly appreciated.

Answer 1

Let me slightly improve your case $N=2$. We will show that if $\lVert x\rVert=1$ and $\lVert y \rVert\leq 1$, then $\lVert tx+(1-t)y\rVert<1$ for $t \in (0, 1)$. In fact, $$ f(t)=\lVert tx+(1-t)y\rVert^2 =t^2+(1-t)^2\lVert y\rVert^2 + 2t(1-t)\langle x, y\rangle. $$ Just as in the post you quoted, the leading term is $t^2(1+\lVert y\rVert^2-2\langle x, y\rangle)$, which we can bound by Cauchy-Schwarz: $$ 1+\lVert y\rVert^2-2\langle x, y\rangle \geq 1-\lVert y\rVert^2\geq 0. $$ Therefore, if $\lVert y\rVert<1$, the map $f(t)$ is a quadratic function with positive leading term, and so its maximum is attained at the boundary. If we have equality, then it follows by the discussion in your linked answer.

This way, we can show inductively on $n\geq 2$ that $\lVert\sum_{j=1}^n \lambda_j x_j\rVert<1$ for $\lambda_j> 0, \sum_j \lambda_j=1$ (note the strict inequality!). The base case is already covered. For the induction step, write $$ \lambda_1 x_1+\dots+\lambda_{n+1}x_{n+1} = (1-\lambda_{n+1}) \left( \mu_1x_1+\dots+\mu_n x_n \right) + \lambda_{n+1}x_{n+1}, $$ where $\mu_i=\lambda_i/(1-\lambda_{n+1})$. The induction step shows that $\lVert \mu_1x_1+\dots+\mu_n x_n\rVert<1$, and so we conclude. This shows that the only elements in $C$ with norm 1 are the ones with $n=1$, as required

Answer 2

For higher dimensions you can argue that any two point $x_1, x_2$ span a plane $U$ which includes $0$ and intersects the unit sphere such that $$\mathbb{S}^n∩U=\mathbb{S}^2.$$

Furthermore

$$\lambda x_1 + (1-\lambda) x_2 \in \mathbb{S}^2, \qquad \forall\lambda\in[0,1].$$

We showed it for $N=2$ and it thus holds for $N>2$ as well.

Answer 3

Let $y \in C$. By Caratheodory's theorem there is an $m$-simplex P for some $m \in \{0,...,n\}$ with vertices in $\{x_1,...,x_N\}$ such that $y \in P$.

W.l.o.g. let $m$ be minimal. If $m = 0$, we have $y \in\{x_1,...,x_N\}$.

So let $m \geq 1$. Then we can further assume that $y \in C^{o}$, the open part of $C$, since the boundary is a union of $m-1$ simplices, but $m$ is minimal.

Because $C \in \bar B = \{x \in \mathbb{R}^{n} \ | \ ||x|| \leq 1\}$, it follows that $y \in C^{o} \subseteq (\bar B)^{o} = \{x \in \mathbb{R}^{n} \ | \ ||x|| < 1\}$. This means $y \notin \mathbb{S}^{n-1}$.

Answer 4

Here is another proof.

Suppose $\|x\|=1$, then let $H=\{ y \mid x^Ty = 1 \}$. Suppose $\|x'\| = 1$ and $x' \in H$. Then Cauchy Schwartz shows that $x' = x$. Note that if $x''$ is on the unit sphere, and $x'' \neq x$ then we must have $x^Tx'' < 1$.

Now let $x=\sum_{k=1}^n t_k x_k$ with the distinct $x_k$ on the unit circle, and the $t_k$ are strictly positive convex multipliers. Suppose $\|x\| = 1$, then $\sum_{k=1}^nt_k x^T x_k = 1$ and so $x^T x_k = 1$ for all $k$. In particular, this implies $n=1$ since the $x_k$ are presumed distinct, and so $x=x_1$.