OK, this may become a very long post... maybe we should switch to email? Or is anyone else following this discussion here on the forum?
First of all, I think the statement is true in both your Minkowski2.pdf (as I interpret it) and also in your Minkowski3.pdf. Note that I was not claiming B_kn = k*B_n to be true for all n. Rather, I said that there exists a certain n such that this holds for all k. In Minkowski3.pdf, you have drawn all B_n from B_1 to B_5. The relevant value for n here is n=2. And we have indeed B_4 = 2*B_2, as claimed.
However, your previous post did point out a problem in my proof. So I have been going back to the drawing board and thought about it all again. By now, I think I have a mathematically precise formulation of the statement as well as a rigorous proof. Here it comes.
The setting is the following: let us consider any periodic graph which is an infinite, connected and locally finite graph G=(V,E) together with an embedding of G into R^d, where d is arbitrary. The main assumption is that this embedding is periodic: there is a group Z^d acting as translations on R^d which maps the embedded graph to itself. Hence R^d decomposes into isomorphic unit cells of finite size, which are all translates of each other. In your example, the unit cell can be taken to be a hexagon made up out of 6 equilateral triangles.
Now fix any point of the graph as origin and take B_n to be the set of all vertices of the graph which can be reached from the origin by traversing at most n edges. So in contrast to my previous terminology, B_n is only a set of vertices, and not a polytope anymore; in particular, talking about "convexity" of B_n is meaningless. B_n is the set of points which can be reached in n time steps.
We are interested in how the shape of B_n / n depends on n. In particular, whether it is possible that this "velocity set" tends to a Euclidean ball as n --> oo.
*Claim:* The set
[math]
\lim_{n\rightarrow\infty}\frac{1}{n}B_n
[/math]
is a polytope. (More accurately: there is a polytope P such that the Hausdorff distance between P and B_n / n converges to 0 as n--> oo.)
*Proof:* For simplicity, let us consider first the case where every unit cell contains only one vertex of G. Then any vertex can be mapped into any other by a translation preserving the graph. In this case, I will now prove that the velocity polytope is precisely the convex hull
[math]
P = \mathrm{conv}(B_1)
[/math]
To see this, note that, as in the previous "proof", we get B_2 by translating a copy of B_1 to all the vertices of B_1. We obtain B_3 by translating copies of B_1 to all vertices of B_2. And so on. Hence,
[math]
B_n = \{x_1+\ldots+x_n\:|\:x_1,\ldots,x_n\in B_1\}
[/math]
Similarly,
[math]
\frac{1}{n}B_n=\left\{\frac{1}{n}x_1+\ldots+\frac{1}{n}x_n\:|\:x_1,\ldots,x_n\in B_1\right\}
[/math]
This clearly lies in the convex hull of B_1; morevoer, as n grows, we can approximate any point in the convex hull of B_1 by a point of this form. (Such a point is a convex combination of elements of B_1. Approximate the coefficients of this convex combination by rational numbers with denominator n.)
This proves the claim in the case that each unit cell contains exactly one vertex of the graph. The argument for the general case follows now. Define a an "admissible velocity" to be a vector
[math]
\vec{v}=\frac{\vec{s}}{t}
[/math]
where \vec{s} and t are given as follows: there needs to exist a path in the graph which begins at the origin x, ends at a vertex which is a translate Tx of x, and does not traverse any other translate of x, such that t is the number of edges in the path, and \vec{s} is the direction vector from x to Tx.
Then it is clear that there is only a finite number of admissible velocities. The convex hull of these velocities is a polytope Q. The claim now is that
[math]
Q=\lim_{n\rightarrow\infty}\frac{1}{n}B_n
[/math]
To see this, we prove the two inclusion separately. So why is the left-hand side contained in the right-hand side? The reason is that admissible velocities can be concatenated with each other any number of times, and this gives convex combinations of velocity vectors as above. Why is the right-hand side contained in the left-hand side? For any very long path which begins at the origin x, adding or removing a few edges does not change much, so we can assume that it ends at some translate of the origin Tx. But then we are back to a convex combination of velocity vectors.
