3 Proof of the Lattice Point Theorem; Other Results 341m <n, such that|αjk|≤βfor allj,k, then the system of homogeneous linear
equations
Ax= 0has a solutionx=(ξk)in integers, not all 0, such that|ξk|≤ 1 +(nβ)m/(n−m)for allk.
Bombieri and Vaaler show that, ifAhas rankmand ifg>0 is the greatest common
divisor of allm×msubdeterminants ofA, then there aren−mlinearly independent
integral solutionsxj=(ξjk)(j= 1 ,...,n−m)such that
n∏−mj= 1‖xj‖≤[det(AAt)]^1 /^2 /g,where‖xj‖=maxk|ξjk|.
The second application, due to Gillet and Soul ́e (1991), may be regarded as an
arithmetic analogue of the Riemann–Roch theorem for function fields. Again letKbe
a compact symmetric convex subset ofRnwith nonempty interior and letμidenote
the infimum of allρ>0 such thatρKcontains at leastilinearly independent points
ofZn(i= 1 ,...,n).IfM(K)is the number of points ofZninK,andifhis the maxi-
mum number of linearly independent points ofZnin the interior ofK, then Gillet and
Soul ́e show thatμ 1 ···μh/M(K)is bounded above and below by positive constants,
which depend onnbut not onK.
A number of results in this section have dealt with compact symmetric convex sets
with nonempty interior. Since such sets may appear rather special, it should be pointed
out that they arise very naturally inconnection with normed vector spaces.
The vector spaceRnis said to benormedif with eachx∈Rnthere is associated a
real number|x|with the properties
(i)|x|≥0, with equality if and only ifx=O,
(ii)|x+y|≤|x|+|y|for allx,y∈Rn,
(iii)|αx|=|α||x|for allx∈Rnand allα∈R.
LetKdenote the set of allx∈Rnsuch that|x|≤1. ThenKis bounded, since all
norms on a finite-dimensional vector space are equivalent. In factKis compact, since
it follows from (ii) thatKis closed. MoreoverKis convex and symmetric, by (ii) and
(iii). Furthermore, by (i) and (iii),x/|x|∈Kfor each nonzerox ∈Rn. Hence the
interior ofKis nonempty and is actually the set of allx∈Rnsuch that|x|<1.
Conversely, letKbe a compact symmetric convex subset ofRnwith nonempty
interior. Then the origin is an interior point ofKand for each nonzerox∈Rnthere is a
uniqueρ>0suchthatρxis on the boundary ofK. If we put|x|=ρ−^1 ,and|O|=0,
then (i) obviously holds. Furthermore, since|−x|=|x|, it is easily seen that (iii) holds.
Finally, ify∈Rnand|y|=σ−^1 ,thenρx,σy∈Kand hence, sinceKis convex,
ρσ(ρ+σ)−^1 (x+y)=σ(ρ+σ)−^1 ρx+ρ(ρ+σ)−^1 σy∈K.Hence
|x+y|≤(ρ+σ)/ρσ=|x|+|y|.ThusRnis a normed vector space andKthe set of allx∈Rnsuch that|x|≤1.