Our main results are contained in
Theorem 2.1.
Let
,
and let
; then
(1)
(2)
(3)if
, then
(4)if
, then
;
(5)if
, then 
Proof.
(
) Let
We prove that
If
has the series expansion
then
We use the fact that
for
and
these imply
for 
From (2.4) and (2.5), we deduce
By using the definition of
from this last inequality we, obtain (2.2) which implies
hence 
(
) 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.
Case 1.
Let
be defined by
and let
We have
or
and then
(see the definition of
).
Let now
Then, by a simple computation and by using the fact that
we obtain
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 
Case 2.
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), where
and
Then
which implies that
We have
where
is given by (2.17).
From
where
and
are given by (2.18) and (2.19), we obtain
If
and
, then
and if
then (2.24) also holds. By combining (2.24) with (2.23) and the definition of
, we obtain that