Generalized convex perimeter of an n-d object

Problem #2 of http://www.math.dartmouth.edu/~pw/solutions.pdf states

At many train stations, post offices and courier services around the
world, the cost of sending a rectangular box is determined by the sum
of its dimensions; that is, length plus width plus height. Prove that
you can't "cheat" by packing a box into a cheaper box.

*** SPOILER WARNING ***

One could define a "convex perimeter" of any bounded set of points S
in the plane as p_2(S) = the perimeter of the convex hull of S.

Based on the elegant solution at the above site, one could define a
generalized convex perimeter of any bounded set of points S in R^n as
follows. Let S_r be the set of all points in R_n within distance r of
(a point in) S. Then we define p_n(S) as

p_n(S) = lim (r -> infinity) (Vol(S_r) - Vol(O_r))/r^(n-1),

where O is a set containing a single point (i.e., Vol(O_r) is the
volume of an n-ball with radius r). Note that this definition agrees
with the previous definition of p_2(S).

The definition makes it obvious (or, if not, see the site above) that
for any sets A and B, if A is contained in B, then p_n(A) <= p_n(B)
(let's say p_n "respects inclusion"). I call it the *convex*
generalized perimeter because p_n has the same value for S as for the
convex hull of S.

If S is a sphere, then p_3(S) is 4pi times its radius. If S is a
polyhedron, then p_3(S) is a weighted sum of the lengths of its edges,
where the weight of each edge is the supplement of the dihedral angle
at that edge (i.e., pi - the angle). Thus, one could pose the
problem, "prove that this funny weighted sum is between 4pi*r and
4pi*R where r and R are the radii of the inscribed and circumscribed
sphere, respectively."

Note that this generalized convex perimeter is a sort of dual to what
we might call the generalized convex surface area: i.e., let s_n(S)
be the (n-1)-d surface area of the convex hull of a point set S. Like
p_n, s_n respects inclusion (i.e., A being a subset of B implies
s_n(A) <= s_n(B)). This is obvious because the convex hull of a set S
has minimal (n-1)-d surface area among all sets containing S.

This stuff seems pretty nice - it must have been studied by someone.
Has anyone run across it?

Is there a nice integral expression for the p_n(S) when S is convex
and smooth?

For 1, n-1, and n dimensions, one could define the quantities p_n(S),
s_n(S), and v_n(S) (the volume of the convex hull of S), all of which
respect inclusion. Are there other such quantities for the dimensions
in between? Perhaps one can simply read off the expansion of Vol(S_r)
as a polynomial in r to get them, though I don't see how to prove they
respect inclusion. Note that p_2(S) = s_2(S), p_1(S) = v_1(S), and
p_0(S) = 0.

BTW, Problem #1 of http://www.math.dartmouth.edu/~pw/solutions.pdf is
also quite interesting. (It was the link to it, in another thread,
that led me to this problem in the first place.)

-Jim Ferry
Metron, Inc.
f rr @m tsc .c m
e y e i o

.

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