A characterization of convex bodies of constant width Let $C\in\mathbb{R}^n$ be a convex body. Assume that for all $x\in\text{bd}(C)$ (the boundary of $C$)
$$\max_{y\in C}||x-y||_2=D$$ where $D$ doesn't depend on $x$ 
($D$ is then necessarily the diameter of $C$). My question is : is $C$ a convex body of constant width (I know that the reverse is true) ? If the answer is negative, can you provide a counterexample ?
thanks for your help.
Elmerto
 A: Def : $x\in {\rm bd}\ (C)$ is a smooth point if
supporting plane at
$x$ for $C$ is unique.
i) There is no non-smooth point :
If ${\rm diam}\ C=d(x_0,y_0)$, then $D\geq d(x_0,y_0)$. Hence $D$ is
a diameter. By definition of $D$, for any $x\in {\rm bd}\ C$, there
is $y$ s.t. $d(x,y)=d(x_0,y_0)$.
If $S(a,R)$ is a $R$-sphere centered at $a$, then $y\in S(x,D)$ and
$x\in S(y,D)$. Here $C$ is in ${\rm conv}\ S(x,D)\bigcap {\rm conv}\
S(y,D)$. If $S_x,\ S_y$ are tangent plane of $S(y,D),\ S(x,D)$ at
$x,\ y$ respectively, then $C$ is between $S_x,\ S_y$.
Clearly, distance between two planes is $D$ so that $C$ has a
constant width.
ii) There is a non-smooth point :
For $v$ in $(n-1)$-dimensional unit sphere $ S^{n-1}$, if $H^v,\
H^{-v}$ are supporting planes for $C$ in direction $v$, and if they
have a distance $D$, then we are done.
If $x\in H^v\bigcap C$ is smooth, then by previous we are done.
So remaining case is as follows : $x,\ y$ are not smooth s.t. $x\in
H^v,\ y\in H^{-v}$ and $H^{\pm v}$ has a distance $<D$.
If $S_\epsilon$ is a geodesic closed ball of radius
$\frac{\pi}{2}-\epsilon$ in $S^{n-1}$, then define a finite cone
$C_\epsilon$ to be $[0, \frac{D}{2\sin\ \epsilon
}  )\cdot S_\epsilon$.
Hence $C^x_\epsilon$ is a cone whose vertex is $x$ so that $C$ is in $C':={\rm conv}\ C^x_\epsilon \bigcup C^y_\epsilon
\ \ast$. If $x',\ y'\in {\rm bd}\ C'$ are around $x,\
y$, then note that
$|x'-y'|<|x-y|=D\ \ast\ast$.
If $x_n\rightarrow x$, then assume that $d(x_n,y_n)=D$. If
$y_n\rightarrow y$, then by $\ast\ast$ it is a contradiction.
By compactness of $C$, we have a subsequence $y_n\rightarrow
y_\infty$. Clearly, $d(x,y_\infty)=D$.
From $\ast$, $d(y,y_\infty)\geq \sqrt{2}D$ so that it is a
contradiction.
