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.