Stationary point for convex difference measure

[noowutah]

Let $\mathbb{S}^n$ be a simplex in $\mathbb{R}^{n+1}$, so $\mathbb{S}^{n}=\{x\in\mathbb{R}^{n+1}|\sum{}x_{i}=1\}$. Let $D$ be a difference measure on $\mathbb{S}^{n}$ with $D(x,x)=0$ and $x=y$ for $D(x,y)=0$. $D$ is also smooth, so differentiable as much as we need.

Let (R) be a convexity requirement for $D$, corresponding to the intuition that on the periphery points are farther apart from each other than near the centre. $M$ is the midpoint of $\mathbb{S}^{n}$, so the coordinates of $M$ are all $m_{i}=1/(n+1)$. Let $d$ be the Euclidean distance function.

(R) if conditions (i)--(iii) are fulfilled for $X_{1},X_{2},X_{3},X_{4}$ then $D(X_{1},X_{2})^{2}>D(X_{3},X_{4})^{2}$.

Here are conditions (i)--(iii) (see the diagram for illustration in $\mathbb{S}^{1}$):

http://streetgreek.com/lpublic/various/horizon.png

(i) $X_{1},X_{2},X_{3},X_{4}$ are all on a straight line between $M$ and an arbitrary point $Z$ on the boundary of $\mathbb{S}^{n}$, so $v(OX_{i})=v(OM)+\lambda_{i}v(MZ)$ (where $v(OX)$ is the vector from the origin to $X$ and $\lambda_{i}\in[0,1]$).

(ii) $d(X_{1},X_{2})=d(X_{3},X_{4})$.

(iii) $\max\{\lambda_{i}\}=\lambda_{2}$, so the pair $X_{1},X_{2}$ is farther away from the midpoint than the pair $X_{3},X_{4}$.

I am using the squared difference in (R) because I am allowing $D$ to be negative-valued. Given only (R) can I show that $M$ is the only stationary point on $\mathbb{S}^{n}$ for $D$ such that

$$
\frac{\partial{}D}{\partial{}x_{i}}=0
$$

only at $M$? The idea would be to show that a non-zero partial derivative at $M$ is inconsistent with (R) and a zero-vector derivative anywhere else is inconsistent with (R) as well. I lack the mathematical sophistication for a proof. I don't even really know what methods to use and which tags to give to this post.

 

[noowutah]

Don't answer this question. It has a fatal flaw in its premises. I would delete it, but I can no longer edit it.