Let $V$ be a real inner product space and let $v\in V$ be a unit vector. Consider the set $S_\epsilon:=\{w\in V: ||w||=1 \text{ and } \langle v,w\rangle\geq 1-\epsilon\}$ for small $\epsilon >0$. I want to know in general what is the diameter of this set. In two dimension I have drawn the following so that my calculation becomes clear:
I want to first compute the length of the straight line between $v$ and $w$. Assume $v$ and $w$ are unit vectors such that $\langle v,w\rangle=\cos (\theta)=1-\epsilon$. Then it follows that $\theta=O(\sqrt{\epsilon})$. We have that this straight line between $v$ and $w$ has length $2\sin(\theta)=O(\sqrt{\epsilon})$. The only thing left is to find the Jacobian of the diffeomorphism that maps this straight line on the circle between $v$ and $w$.
I do not know much about this. So I would like to ask the following:
- Does this Jacobian do much to the $O(\sqrt{\epsilon})$?
- If so what is this Jacobian (asymptotic value is enough)?
You may assume ${\bf v}={\bf e}_1$. Then $S_\epsilon$ is a spherical cap: $$S_\epsilon=\bigl\{{\bf x}\in V\,\bigm|\,|{\bf x}|=1, \ x_1\geq 1-\epsilon\bigr\}\ .$$ The base of this cap is a ball of codimension $1$ and radius $\sqrt{1-(1-\epsilon)^2}=\sqrt{2\epsilon-\epsilon^2}$. I claim that the diameter of $S_\epsilon$ is given by $${\rm diam}(S_\epsilon)=2\sqrt{2\epsilon-\epsilon^2}\ .$$ Proof. Let ${\bf x}=(x_1,{\bf x}')$, ${\bf y}=(y_1,{\bf y}')\in S_\epsilon$, $1-\epsilon\leq \ x_1\leq y_1$. Then $|{\bf x}-{\bf y}|$ is enlarged if ${\bf y}'$ is turned opposite to ${\bf x}'$ in the orthogonal complement of ${\bf e}_1$ . We now choose ${\bf e}_2$ parallel to ${\bf x}'$, so that for all practical purposes $${\bf x}=(x_1,\sqrt{1-x_1^2}),\qquad{\bf y}=(y_1,-\sqrt{1-y_1^2})\ .$$ The claim now follows from inspection of a simple 2D figure.