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§5. Quasilinear and Quasiquadratic Maps 133

(2) A function β : A × A → R is quasibilinear provided the pairs of functions

β(x + x , y) and β(x, y) + β(x , y) and β(x, y + y ) and β(x, y) + β(x, y ) are

equivalent functions A × A × A → R.
(3) A function q : A → R is quasiquadratic provided q(x, y) is quasibilinear
and q(x) = q(−x). Moreover, q is positive provided q(x) ≥ 0 for all x ∈ A. Again
q(x, y) = q(x + y) − q(x) − q(y).
We will construct a norm on a group from a quasiquadratic function. First we
need a characterization of quasibilinearity.
(5.5) Lemma. Let f : A → R be a function on an abelian groupA satisfying
f (x) = f (−x). Then f : A × A → R is quasibilinear (resp. bilinear) if and only
if the weak (resp. ordinary) parallelogram law holds, i.e., for (x, y) ∈ A × A

f (x + y) + f (x − y) ∼ 2 f (x) + 2 f (y)
(resp. f (x + y) + f (x − y) = 2 f (x) + 2 f (y)).

Proof. We form the following real valued function on A × A × A:

g(x, y, z) = f (x + y + z) − f (x + y) − f (x + z) − f (y + z)
+ f (x) + f (y) + f (z).
The function f is quasibilinear (resp. bilinear) if and only if g(x, y, z) is bounded
(resp. zero). The parallelogram law is easily seen to be equivalent to the relation
g(x, y, −z) ∼−g(x, y, z) (resp. g(x, y, −z) =−g(x, y, z)). Since f (x) = f (−x)
implies that g(−x, −y, −z) = g(x, y, z), and since g(x, y, z) is symmetric in
x, y, z, we see that the parallelogram law is equivalent to 2g(x, y, z) ∼ 0 (resp.
g(x, y, z) = 0). This proves the lemma.

In connection with Definition (5.4)(3) we recall that a function q : A → R is
quadratic provided q is bilinear on A × A.

(5.6) Proposition. A function q : A → R is quadratic if and only if q(x) =
q(−x), q(2x) = 4q(x), and q is quasiquadratic.
Proof. The direct implication is immediate. Conversely, suppose that q(x, y) and
q(x, −y) are bounded. Then the sum of the two

s(x, y) = q(x + y) + q(x − y) − 2q(x) − 2q(y)
is bounded in the absolute value by a constant A, and since the hypothesis implies
n
n
that 2 −2n s(2 x, 2 y) = s(x, y), we have zero in the limit as n →+∞, |s(x, y)|=
n
n
2 −2n |s(2 x, 2 y)|≤ 2 −2n A → 0. Hence the function s(x, y) = 0 for all (x, y) and
the parallelogram law holds. Thus by (5.5) the function q(x) is quadratic, and this
proves the proposition.
(5.7) Theorem. If q : A → R is a quadratic function satisfying q(x) = q(−x),
n
∗
∗
then q (x) = lim n→∞ 2 −2n q(2 x) exists and the function q (x) is quadratic.