$X$ be real i.p.s. dim.>1 , if two closed balls,none of which is a subset of the other,intersect then do the boundaries of the balls intersect too?

Question

Let $X$ be a real inner product space of dimension more than $1$ , let $B[x;r] , B[y;s]$ be two closed

balls having non-empty intersection where none of the balls is a subset of the other , then is it true that $S[x;r] \cap S[y;s]$ is also non-empty ?

( where $S[x;r]:=\{a \in X : ||a-x||=r\}$ ) . The statement is trivially not true for dimension $1$ i.e. $\mathbb R$ . For $2$-dimensional Euclidean space , considering circles , it intuitively appears that the result is true , but I am unable to see any way of rigorously checking it . Please help . Thanks in advance

Answer 1

I try a solution:

a) Put $$A=\sqrt{\frac{ 4r^2\|x-y\|^2-(\|x-y\|^2+r^2-s^2)^2}{4\|x-y\|^2}}$$

$$B=\frac{\|x-y\|^2+r^2-s^2}{2\|x-y\|^2}$$ and $$z=x+B(y-x)+Au$$ with $u$ orthogonal to $x-y$, such that $\|u\|=1$.

My computations, if they are correct, show that we have $\|z-x\|=r^2$ and $\|z-y\|^2=s^2$. So $z$ belongs to both "circles".

There is a condition, namely that $4r^2\|x-y\|^2-(\|x-y\|^2+r^2-s^2)^2\geq 0$. This leads to ($\|x-y\|\geq r$ and $\|x-y\|\leq r+s$) or ($\|x-y\|\leq r$ and $\|x-y\|\geq r-s$). These are the traduction of your hypothesis of non-inclusion.

Answer 2

This actually works in arbitrary Banach spaces:

Note first that if $B[x,r] \cap B[y,s] \subseteq S[x,r] \cap S[y,s]$, then there's nothing to do. In particular, we can assume that $B[x,r] \cap B[y,s]$ contains an interior point of one (and, therefore, both) balls.

Let $B_{1}$ be the interior of $B[x,r]$ and $B_{2}$ be the interior of $B[y,s]$. Since $B_{1} \cap B_{2}$ is nonempty and neither is a subset of the other, $\partial B_{1} \cap B_{2}$ is nonempty (as is $\partial B_{2} \cap B_{1}$).

Let's focus on $\partial B_{1} \cap B_{2}$. Fix $w \in \partial B_{1} \setminus B_{2}$ and $z \in \partial B_{1} \cap B_{2}$. We know $z$ exists, what of $w$? Well, if $\partial B_{1} \subseteq B_{2}$, then for each $q \in B_{1}$, consider the line $\lambda q + (1 - \lambda) y$ for $\lambda \in \mathbb{R}$. As $\lambda$ goes to (positive) infinity, the distance from the point on the line to $x$ goes to infinity. However, at $\lambda = 1$, the distance to $x$ is bounded above by $r$. Thus, the line intersects $\partial B_{1}$ for some $\lambda > 1$. Now convexity of $B_{2}$ (and our assumption $\partial B_{1} \subseteq B_{2}$) implies $q \in B_{2}$. Therefore, we proved $\partial B_{1} \subseteq B_{2}$ implies $B_{1} \subseteq B_{2}$. Consequently, our hypothesis that neither ball is contained in the other implies $\partial B_{1} \setminus B_{2}$ must be non-empty, and, thus, I can choose $w$ as claimed above.

Consider the great circle on $S[x,r]$ containing $w$ and $z$. This makes sense in infinite dimensions: the arc of the circle from $w$ to $z$ is traced out by the mapping $$t \mapsto x + r \cdot \frac{(1 - \lambda)(w - x) + \lambda (z - x)}{\|(1 - \lambda)(w - x) + \lambda (z - x)\|}$$ for $t \in [0,1]$. Since this path is in $B_{2}^{c}$ when $\lambda = 0$ and in $B_{2}$ when $\lambda = 1$, we conclude that the path intersects $\partial B_{2}$ somewhere. (This follows from connectedness of $[0,1]$ and the fact that $B_{2}$ is closed.) This gives us our point in $\partial B_{1} \cap \partial B_{2} = S[x,r] \cap S[y,s]$.