Our main results are contained in
Let , and let ; then
(3)if , then
(4)if , then ;
(5)if , then
() Let We prove that
If has the series expansion
We use the fact that for andthese imply
From (2.4) and (2.5), we deduce
By using the definition of from this last inequality we, obtain (2.2) which implies
() Let be in Then (2.7) holds and, by using (2.3), this is equivalent to
For if , from (2.8) we obtain
which is equivalent to
Then multiplying the relation last inequality with we obtain
() if is a real positive number, then the definitions of and are equivalent, hence By using () and () from this theorem, we obtain ().
() We have the following two cases.
Let be defined by
and let We have
and then(see the definition of ).
Then, by a simple computation and by using the fact that
where, , and
For we, have where is the disc with the center
and the radius
We have where and we deduce that for all does not hold.
We have obtained that for but and in this case
We consider the function defined by (2.11) for In this case, the inequality (2.13) holds too and this implies that
We also obtain that like in Case 1.
() Let be given by (2.11), whereand Then
which implies that
where is given by (2.17).
From where and are given by (2.18) and (2.19), we obtain
If and, then
then (2.24) also holds. By combining (2.24) with (2.23) and the definition of , we obtain that