This is the situation in three dimensions for a centrally symmetric surface $S$ with the induced Euclidean metric:
- For simply-connected $S$ (convexity is not required), if the diameter of
$S$ is attained, it is attained by a pair of antipodal points.
- This is not necessarily
true if $S$ is not simply connected.
Judging by the answers so far, these might be new results for non-convex $S$. So I'd better prove them.
Outline of Proof of 1 The idea is very simple. Let $X$ and $Y$ be any points in $S$, with antipodal points $X'$ and $Y'$. Let $\gamma_X$ and $\gamma_Y$ be rectifiable paths from $X$ to $X'$ and from $Y$ to $Y'$ respectively. Then we show that there exists a path from $X$ to $Y$ which is no longer than $\max(|\gamma_X|, |\gamma_Y|)$: if $P$ is a point of intersection of $\gamma_X$ and $\gamma_Y$, this will be either the path from $X$ to $P$ to $Y$, or the antipodal image of the path from $X'$ to $P$ to $Y'$. And if $\gamma_X$ doesn't intersect $\gamma_Y$, then the antipodal image of $\gamma_X$ must (because $S$ is simply connected).
Lemma Let $S$ be a simply-connected, centrally symmetric surface in $\mathbb R^3$. Let $X$ and $Y$ be points in $S$, with antipodal points $X'$ and $Y'$ respectively. Let $\gamma_X$ be any rectifiable path from $X$ to $X'$, and $\gamma_Y$ be any rectifiable path from $Y$ to $Y'$. Then there exists a rectifiable path $\gamma_{XY}$ from $X$ to $Y$ with $|\gamma_{XY}| \le \max(|\gamma_X|,|\gamma_Y|)$.
Proof of Lemma Let $\gamma'_X$ denote the antipodal image of $\gamma_X$. Then $\gamma_X+\gamma'_X$, the concatenation of $\gamma_X$ and $\gamma'_X$, is a closed curve; because $S$ is simply connected, this curve divides $S$ into two regions, which are antipodal images of each other. One of these two regions contains $Y$, and the other contains $Y'$; therefore $\gamma_Y$ crosses $\gamma_X+\gamma'_X$, at a point $P$. Without loss of generality, we may assume that $P$ lies on $\gamma_X$.
Now, letting $\gamma_{XP}$ denote the path along $\gamma_X$ from $X$ to $P$ etc., we have $|\gamma_X| = |\gamma_{XP}|+|\gamma_{PX'}|$ and $|\gamma_Y| = |\gamma_{YP}|+|\gamma_{PY'}|$. But then
$$\begin{align}
|\gamma_{XPY}|+|\gamma_{X'PY'}|
&= (|\gamma_{XP}|+|\gamma_{PY}|) + (|\gamma_{X'P}|+|\gamma_{PY'}|) \\
&= (|\gamma_{XP}|+|\gamma_{PX'}|) + (|\gamma_{YP}|+|\gamma_{PY'}|) \\
&= |\gamma_X|+|\gamma_Y| \\
&\le 2\max(|\gamma_X|,|\gamma_Y|)
\end{align}$$
So at least one of $|\gamma_{XPY}|, |\gamma_{X'PY'}|$ is $\le \max(|\gamma_X|,|\gamma_Y|)$. If it is $|\gamma_{XPY}|$, then take $\gamma_{XY} = \gamma_{XPY}$. If it is $|\gamma_{X'PY'}|$, then take $\gamma_{XY} = $ the antipodal image $\gamma'_{X'PY'}$. This proves the Lemma.
Proof of 1 Suppose the diameter $d$ of $S$ is attained by points $X,Y$: that is, all paths from $X$ to $Y$ have length at least $d$. Then $|\gamma_{XY}| \ge d$ for all paths $\gamma_{XY}$ from $X$ to $Y$. Hence, by the Lemma, if $\gamma_X$ and $\gamma_Y$ are paths from $X$ to $X'$ and from $Y$ to $Y'$, then one of $|\gamma_X|$ and $|\gamma_Y|$ is at least $d$. Hence the diameter is attained by antipodal points, either $(X,X')$ or $(Y,Y')$. $\square$
Proof of 2 We describe a centrally symmetric surface $S$, not simply connected, whose diameter is attained, but not by a pair of antipodal points.
Imagine a vertical hollow tube, of radius about 1 and height about 4. Its walls are thin. Half-way up, two long, sharp spikes protrude horizontally from it, one to the left and one to the right. They are conical, with a base radius of about $\frac12$, and length about $10$. (The exact numbers are not important; this is just to give you a picture.) These spikes are also hollow, with thin walls.
The two antipodal points that are furthest apart are the external tips of the spikes. But the diameter is attained by an external spike tip and the opposite, internal spike tip; the distance between these two points is greater, because any path between the two has to climb up over the lip of the tube to get from one to the other.
I hope this is clear enough; if not, I will try to draw some pictures.
