# How many points does a line intersect a sphere in an infinite-dimensional normed vector space?

Let $$(E, |\cdot|)$$ be a n.v.s. We fix $$r>0$$ and $$x,y \in B(0, r)$$ such that $$x\neq y$$. Here $$B(0, r)$$ is the open ball centered at the origin and having radius $$r$$. The set of all points in the line though $$x$$ and $$y$$ is $$\{tx+(1-t)y \mid t \in \mathbb R\}.$$

Consider the map $$f: \mathbb R \to E, t \mapsto tx+(1-t)y$$. Then

• $$f$$ and thus $$|f|$$ are continuous.
• $$\lim_{t \to \infty} |f(t)| =+\infty$$.
• $$|f(0)| = |y|.

It follows that the set $$S := \{t\in \mathbb R \mid |f(t)|=r\}$$ is non-empty and bounded. If $$(E, \langle \cdot , \cdot \rangle)$$ is an inner product space, then \begin{align} |f(t)|=r &\iff |f(t)|^2=r^2 \\ &\iff |x|^2t^2 + 2\langle x,y \rangle t(1-t) + |y|^2(1-t)^2 = r^2 \\ &\iff |x-y|^2t^2+2(\langle x, y \rangle - |y|^2) t + |y|^2-r^2=0. \end{align}

We have $$\Delta = (\langle x, y \rangle - |y|^2)^2- |x-y|^2(|y|^2-r^2)>0$$ because $$|y| and $$|x-y|\neq 0$$. So $$S$$ has exactly $$2$$ elements in this case.

Does $$\operatorname{card} (S) =2$$ if $$E$$ is reflexive or uniformly convex?

• May I ask for the reason of the downvote? May 23 at 8:58

The following holds:

Every line in a normed space $$X$$ intersects the unit sphere $$S_X$$ at most twice if and only if $$X$$ is strictly convex.

There's no need for (local) uniform convexity here, and reflexivity is a distraction (e.g. consider finite-dimensional spaces with/without strict convexity).

Recall what it means for $$X$$ to be strictly convex:

$$(X, \| \cdot \|)$$ is strictly convex if and only if, for all distinct $$x, y \in S_X$$, $$\|x + y\| < 2$$.

It's not difficult to see that, if each line intersects at most twice with $$S_X$$, then $$X$$ is strictly convex. If we choose any distinct $$x, y \in S_X$$, and form a line between them (i.e. $$\{tx + (1 - t)y : t \in \Bbb{R}\}$$), then this line contains the midpoint $$\frac{x + y}{2}$$. As $$x$$ and $$y$$ are distinct, so too is $$\frac{x + y}{2}$$ distinct from the other two points, and hence, it cannot belong to $$S_X$$. We know by convexity of the norm that $$\left\|\frac{x + y}{2}\right\| \le \frac{\|x\| + \|y\|}{2} = 1,$$ but given that $$\frac{x + y}{2} \notin S_X$$, the above inequality is strict. Thus, $$\|x + y\| < 2$$, as required.

The converse is the slightly trickier direction. Suppose $$X$$ is strictly convex, and further, we have a line $$L = \{x + td : t \in \Bbb{R}\} \subseteq X$$ that intersects $$S_X$$ at least three times. Consider the function $$f : \Bbb{R} \to \Bbb{R}$$ defined by $$f(t) = \|x + td\|$$. Then, $$f$$ is convex, as \begin{align*} f(\lambda t + (1 - \lambda) s) &= \|x + (\lambda t + (1 - \lambda) s)d\| \\ &= \|(\lambda (x + td) + (1 - \lambda)(x + sd)\| \\ &\le \lambda \|x + td\| + (1 - \lambda)\|x + sd\| \\ &= \lambda f(t) + (1 - \lambda) f(s), \end{align*} for any $$s, t \in \Bbb{R}$$ and $$\lambda \in [0, 1]$$.

Now, $$f$$ achieves the value $$1$$ three times; let's name three of these solutions as $$t_1 < t_2 < t_3$$. Using the three slope lemma, we now show that $$f(t) = 1$$ for all $$t \in [t_1, t_3]$$. To show this, suppose that $$t \in (t_1, t_2)$$. Then the three slope lemma shows: $$\frac{f(t) - f(t_1)}{t - t_1} \le \frac{f(t_2) - f(t_1)}{t_2 - t_1} \le \frac{f(t_2) - f(t)}{t_2 - t},$$ which comes to: $$\frac{f(t) - 1}{t - t_1} \le 0 \le \frac{1 - f(t)}{t_2 - t},$$ i.e. $$f(t) \le 1$$. On the other hand, \begin{align*} &\frac{f(t_2) - f(t)}{t_2 - t} \le \frac{f(t_3) - f(t)}{t_3 - t} \le \frac{f(t_3) - f(t_2)}{t_3 - t_2} \\ \implies \; &\frac{1 - f(t)}{t_2 - t} \le \frac{1 - f(t)}{t_3 - t} \le 0 \\ \implies \; & f(t) \ge 1. \end{align*} So, $$f(t) = 1$$, for $$t \in (t_1, t_2)$$. A similar argument shows $$f(t) = t$$ for $$t \in (t_2, t_3)$$. Since $$f(t_1) = f(t_2) = f(t_3)$$, this shows $$f(t) = 1$$ for $$t \in [t_1, t_3]$$.

Either which way, this produces infinitely many points on $$L$$ that lie in $$S_X$$. If we consider $$x + t_1 d$$ and $$x + t_3 d$$, then their midpoint $$x + \frac{t_1 + t_3}{2} d$$ also lies in $$S_X$$, as $$\frac{t_1 + t_3}{2} \in [t_1, t_3]$$. This contradicts strict convexity.