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I am trying to understand the proof of the following claim:

Let $f:A \subseteq \mathbb{S}^n \to \mathbb{S}^n$ be an $L$-Lipschitz* map (with $L <1$). Then $f(A)$ is contained in the interior of a hemisphere.

*The distance on $\mathbb{S}^n$ can be either the intrinsic one or the extrinsic (Euclidean) one, it does not matter.

In the standard proof I have found here (pg 7, lemma 2.8), the author shows that the convex hull of $f(A)$ in the unit ball of $\mathbb{R}^{n+1}$ does not contain the origin.

Why is this enough to deduce the claim?

Edit:

As commented by Jyrki Lahtonen, if a convex hull of a set $B$ is closed, then it is the intersection of all closed half-spaces containing $B$.

From this the following key-lemma follows:

Lemma: A closed convex set in $\mathbb{R}^n$ which does not contain the origin $\bar O$ is contained in a closed half-space which does not contain $\bar O$. In particular, $B$ is contained in the interior of a half-space whose boundary does contain $\bar O$.

Proof:

Suppose $B \subseteq \mathbb{R}^{n}$ is closed and convex.

$$\bar O \notin B=\operatorname{Conv}(B)=\cap_{\text{half-space } \, H, B \subseteq H} H$$ so there exist a closed half-space $H$ containing $B$ which does not contain the origin.

$H$ can be written in the form of $H:=\{ y \in \mathbb{R}^{n}| \langle y,v \rangle \ge c\}$ where $v \in \mathbb{R}^{n},c \in \mathbb{R}$.

$\bar O \notin H \Rightarrow c>0$. Thus,

$$ B \subseteq \{ y \in \mathbb{R}^{n}| \langle y,v \rangle \ge c > 0\} $$

is contained in the interior of the half-space $D=\{ y \in \mathbb{R}^{n}| \langle y,v \rangle \ge 0\}, \bar O \in \operatorname{bd} D$.


Note: The lemma is false if $B$ is not closed. Take for instance $$B=\{ (x,y)\in \mathbb{R}^2 \,| \, x \ge 0, y \ge 0 \} \setminus {(0,0)}$$

$B$ is convex, $\bar O \notin B$, but any closed half-space containing $B$, must contain $\bar O$. However, $B$ is contained in the interior of a half-space whose boundary does contain $\bar O$, the halh-space $\{ (x,y)\in \mathbb{R}^2 \,| \, x + y \ge 0 \}$.


The lemma implies the claim for when $f(A)$ is compact:

This still leaves open the question whether or not, the claim holds in more general case.

Proof:

First, observe that compactness of $f(A)$ implies closedness of $\operatorname{Conv}(f(A))$.

So, by the lemma there exist a closed half-space $H$ containing $f(A)$ which does not contain the origin.

$H$ can be written in the form of $H:=\{ y \in \mathbb{R}^{n+1}| \langle y,v \rangle \ge c > 0\}$ where $v \in \mathbb{R}^{n+1},c \in \mathbb{R}$.

Thus,

$$ f(A) \subseteq \{ y \in \mathbb{S}^n| \langle y,v \rangle \ge c > 0\} $$

is contained in the interior of a hemisphere $D=\{ y \in \mathbb{S}^n| \langle y,v \rangle \ge 0\} $, as we wanted.


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    $\begingroup$ Isn't the convex hull the intersection of the half-spaces containing the set $f(A)$? If the origin is not in the convex hull, then there exists a half-space containing $f(A)$ but not the origin, no? By a half-space I mean one side of any hyperplane - not necessarily thru the origin. IOW, the set of solutions of a linear inequality. $\endgroup$ – Jyrki Lahtonen Feb 26 '17 at 13:06
  • $\begingroup$ Thanks! It seems that you are essentially right. Convex hulls which are closed sets, indeed can be represented as intersection of all closed half-spaces containing the original set. So, it still remains to understand what are the most general conditions under which the theorem holds. I will edit the question to make this more clear. $\endgroup$ – Asaf Shachar Feb 26 '17 at 13:23
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Well, it turns out the theorem indeed holds in the general.

Let $f:A \subseteq \mathbb{S}^n \to \mathbb{S}^n$ be an $L$-Lipschitz* map (with $L <1$). Then $f(A)$ is contained in the interior of a hemisphere.

For closed subsets $A$, the proof is outlined in the question. If $A$ is not closed, one can always extend $f$ (uniquely) to be defined on the closure of $A$, while preserving the same Lipschitz constant (this is because the co-domain is complete).

Now apply the result to that extension.

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