Define a mapping
by
for all
. Then, for each
the following equation is obtained:
for all
and for all
, where
For each fixed
, one obtains from
that
for all
and
,
. Then, it follows from the above inequality that
for all
and
Therefore, we prove from this relation that, for any integers
,
for all
Since the right-hand side of the above inequality tends to 1 as
, the sequence
is
-Cauchy and thus converges in
. Hence, we may define a mapping
as
for all
and
In addition, we claim that the mapping
Q satisfies (
2). For this purpose, we calculate the following inequality:
for all
, where
. This means that
for all
. Hence, the mapping
Q satisfies (
2) and so
for all
. It follows that
for all
.
To prove the uniqueness, let
be another mapping satisfying (
2) and
for all
Thus, we have
for all
Taking the limit as
, then we conclude that
for all
Under the assumption that either
f is measurable or
is continuous in
for each fixed
, the quadratic mapping
Q satisfies
for all
and for all
by the same reasoning as the proof of [
25]. That is,
Q is
-quadratic. Let
. Putting
and
for all
in (
4) and dividing the resulting inequality by
we have
for all
Taking
and using the evenness of
Q, we obtain that
for all
and for each
The last relation is also true for
Now, let
a be a nonzero element in
and
K a positive integer greater than
Then, we have
By [
26] [Theorem 1], there exist three elements
such that
Thus, we calculate in conjunction with [
27] [Lemma 2.1] that
for all
and for all
Thus, the unique
-quadratic mapping
Q is also
-quadratic, as desired. This completes the proof. □