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Let $S$ be the unit hypersphere in $\mathbb{R}^n$, and let $A_1$ and $A_2$ be two disjoint spherical caps of $S.$ The spherical caps include all their boundary points except the ones on the hyperplane that cuts them off. See figure demonstrating $S\setminus A_1$).

I want to show that: $$conv(S\setminus A_1) \cap conv(S\setminus A_2) = conv((S\setminus A_1) \cap (S \setminus A_2)).$$

The following direction is easy: $$conv((S \setminus A_1) \cap (S \setminus A_2)) \subseteq conv(S \setminus A_1) \cap conv(S \setminus A_2).$$ We can show it directly from the definition of convexity as follows.

Let $p \in conv((S \setminus A_1) \cap (S \setminus A_2))$, then $\exists \lambda: 0 \leq \lambda \leq 1$, and $\exists p_1, p_2 \in (S \setminus A_1) \cap (S \setminus A_2)$ s.t. $p=\lambda p_1 + (1-\lambda)p_2$. Then we have $p_1 \in S \setminus A_1, p_2 \in S \setminus A_1$, which implies $p \in conv(S \setminus A_1)$. Similarly, we have $p_1 \in S \setminus A_2$, and $p_2 \in S \setminus A_2$, which implies $p \in conv(S \setminus A_2)$, which completes one direction of the proof.

Even though the other direction also seems very intuitive, I am not able to move forward yet. Note that, I haven't used the disjointness of $A_1$ and $A_2$ in the previous direction, but it is easy to see that the other direction:

$$conv(S \setminus A_1) \cap conv(S \setminus A_2) \subseteq conv((S \setminus A_1) \cap (S \setminus A_2)),$$

does not hold when $A_1$ and $A_2$ are not disjoint spherical caps.

Any ideas are appreciated!

Edit: I would like a technique that works for an arbitrary number of spherical cap $A_i$'s, not just two.

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    Are the caps supposed to be closed? – 2017-02-03
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    This is a good clarification question, thank you! They include all their boundary points except the ones on the hyperplane that cuts them off! – 2017-02-03
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    Just added a figure to clarify! – 2017-02-03
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    So $S \setminus A_1$ is closed? I would say that for $n \ge 3$ the disjointness of $A_1$ and $A_2$ is not needed. – 2017-02-03
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    Yes $S\setminus A_1$ is closed. It is definitely needed in $n=2$, can you explain you intuition on why we don't need it for $n \geq 3$? – 2017-02-03

1 Answers 1

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You always find vectors $n_1, n_2 \in \mathbb{R}^n$ and $\alpha_1, \alpha_2 \in \mathbb{R}$, such that $$S \setminus A_i = \{x \in S \mid n_i^\top x \le \alpha_i \}.$$

It is easy to see that $$conv(S \setminus A_i) = \{x \in B \mid n_i^\top x \le \alpha_i \},$$ with $B = \{x \mid \|x\|\le1\}$. Hence, we have to check $$ M:=\{x \in B \mid \forall i \in \{1,2\}: n_i^\top x \le \alpha_i \} \subset conv(\{x \in S \mid \forall i \in \{1,2\}: n_i^\top x \le \alpha_i \})=:N.$$

Let us first assume that $n \ge 3$. Then, there is some vector $n \ne 0$ with $n^\top n_i = 0$ for $i =1,2$. Hence, an arbitrary $x \in M$ can be written as $$x = \lambda \, (x + \beta_1 \, n) + (1-\lambda) \, (x - \beta_2 \, n)$$ with $\lambda \in [0,1]$ and $\beta_1, \beta_2 > 0$ chosen such that $\| x \pm \beta_i\,n\| = 1$. This shows the assertion.

Let us now assume that $n = 2$. Then, by disjointness of the caps, the set $N$ contains the two segments $\{x \in B \mid n_i^\top x = \alpha_i\}$, $i = 1,2$. Then, an arbitrary $x$ can be written as $$x = \lambda \, (\beta_1 \, x) + (1-\lambda) \, (\beta_2 \, x)$$ with $\lambda \in [0,1]$ and $\beta_1 > 0$ and $\beta_2 < 0$ such that $\beta_i \, x$ either has norm $1$ (and satisfies $n_i^\top (\beta_i\,x) \le \alpha_i$) or belongs to one of the above segments.