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134 6. Proof of Mordell’s Finite Generation Theorem

Proof. By (5.5) the function q is quasiquadraticif and only if the weak parallelogram
law holds q(x + y) + q(x − y) ∼ 2q(x) + 2q(y). Setting x = y we have q(2x) ∼
n
4q(x),thatis, |q(2x) − 4q(x)|≤ A for positive constant A. Replacing x by 2 x,we
obtain the inequality
n
|2 −2(n+1) q(2 n+1 x) − 2 −2n q(2 x)|= 2 −2n A
which leads to the following estimate for all n and p
4A
−2n
−2(n+p)
n+p
n
−2n
|2 q(2 x) − 2 q(2 x)|= 2 · .
3
∗
Thus the sequence defining q (x) is Cauchy, and, therefore, it is convergent. The last
∗
assertion follows now from the previous proposition since the condition q (2x) =
∗
∗
∗
4q (x) follows from the defining limit and the conditions that q (x) = q (−x) and
q (x) be quasiquadraticare preserved in the limit. This proves the theorem.
∗
Another important situation where an equivalence can be modified to become an
equality is contained in the next proposition.
(5.8) Proposition. Let f : X → X be a function, let d > 1, and let h : X → R
n
be a function such that h f ∼ d · h. Then the limit h f (x) = lim n→∞ d −n · h( f (x))
exists uniformly on X and:
2
2
(1) h f ( f (x)) = dh f (x) and |h f (x) − h(x)|≤ cd /(d − 1) where
|h( f (x)) − d · h(x)|≤ c
for all x ∈ X.
(2) If h is proper, then h f is proper.

(3) If g : X → X with hg ∼ d · h for a constant d ,thenh f g ∼ d · h f holds on X.
Proof. From the inequality |h( f (x) − d · h(x)|≤ c,wehave
n
|d −n−1 h( f n+1 (x) − d −n h( f (x))|≤ cd −n−1 ,
and, therefore, we have the estimate
n

c 1
n+p
−(n+p)
−n
n
|d h( f (x)) − d h( f (x))|≤ .
1 − (1/d) d
n
Thus d −n h( f (x)) is a Cauchy sequence, and it converges uniformly to h f .
As for assertion (1), the relation h f ( f (x)) = d · h f (x) follows since the two
sides are rearrangements of the same limit. The second part results from the above
estimate by letting p →∞ and taking n = 0. Assertion (2) follows from (1) and
(5.3). As for (3), we calculate the difference

h f (g(x)) − d · h f (x)

= [h f (g(x)) − h(g(x))] + [h(g(x)) − d · h(x)] + [d · h(x) − h f (x)],
and see that each of the three terms is bounded. This proves the proposition.