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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.
  • $\lim_{t \to \infty} |f(t)| =+\infty$.
  • $|f(0)| = |y|<r$.

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 $|y|<r$ 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?

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  • $\begingroup$ May I ask for the reason of the downvote? $\endgroup$
    – Akira
    May 23 at 8:58

1 Answer 1

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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.

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