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 andthese 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), whereand 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