Number Theory: An Introduction to Mathematics

3 Proof of the Lattice Point Theorem; Other Results 335

xj−xk=zj−zk∈Λ( 1 ≤j,k≤m+ 1 ).

Suppose next thatSis compact andλ(S)=md(Λ).Let{εv}be a decreasing
sequence of positive numbers such thatεv→0asv→∞,andletSvdenote the set of
all points ofRndistant at mostεvfromS.ThenSvis compact,λ(Sv)>λ(S)and

S 1 ⊃S 2 ⊃···, S=∩

v

Sv.

By what we have already proved, there existm+1 distinct pointsx 1 (v),...,xm(v+) 1

ofSvsuch thatx(jv)−xk(v)∈Λfor allj,k.SinceSv ⊆S 1 andS 1 is compact, by

restricting attention to a subsequence we may assume thatx(jv)→xjasv→∞(j=

1 ,...,m+ 1 ).Thenxj∈Sandx(jv)−x(kv)→xj−xk.Sincex(jv)−xk(v)∈Λ,this

is only possible ifxj−xk =x(jv)−x(kv)for all largev. Hencex 1 ,...,xm+ 1 are
distinct.

Siegel (1935) has given an analytic formula which underlies Proposition 9 and
enables it to be generalized. Although we will make no use of it, this formula will now
be established. For notational simplicity we restrict attention to the (self-dual) lattice
Λ=Zn.

Proposition 10IfΨ :Rn→Cis a bounded measurable function which vanishes
outside some compact set, then

∫

Rn

Ψ(x)φ(x)dx=

∑

w∈Zn

∣

∣

∣

∣

∫

Rn

Ψ(x)e−^2 πiw

tx

dx

∣

∣

∣

∣

2

,

where

φ(x)=

∑

z∈Zn

Ψ(x+z).

Proof SinceΨvanishes outside a compact set, there exists a finite setT⊆Znsuch
thatΨ(x+z)=0forallx ∈Rnifz ∈Zn\T. Thus the sum definingφ(x)has
only finitely many nonzero terms andφalso is a bounded measurable function which
vanishes outside some compact set.
If we write

x=(ξ 1 ,...,ξn), z=(ζ 1 ,...,ζn),

then the sum definingφ(x)is unaltered by the substitutionζj→ζj+1 and henceφ
has period 1 in each of the variablesξj(j= 1 ,...,n).LetΠdenote the fundamental
parallelotope

Π={x=(ξ 1 ,...,ξn)∈Rn:0≤ξj≤1forj= 1 ,...,n}.

Since the functionse^2 πiw
tx
(w∈Zn)are an orthogonal basis forL^2 (Π),Parseval’s
equality (Chapter I,§10) holds: