On Convergence a Variation of the Converse of Fabry Gap Theorem
Naser Abbasi, Molood Gorji^{*}
Department of Mathematics, Lorestan University, Khoramabad, Islamic Republic of Iran
Email address:
Naser Abbasi, Molood Gorji. On Convergence a Variation of the Converse of Fabry Gap Theorem. Science Journal of Applied Mathematics and Statistics. Vol. 3, No. 2, 2015, pp. 58-62. doi: 10.11648/j.sjams.20150302.15
Abstract: In this article we give a variation of the converse of Fabry Gap theorem concerning the location of singularities of Taylor-Dirichlet series, on the boundary of convergence. whose circle of convergence is the unit circle and for which the unit circle is not the natural boundary.
Keywords: Dirichlet Series, Entire Functions, Fabry Gap Theorem
1. Introduction
The gap theorem of Fabry states that if is a power series whose circle of convergence is the unit circle and then the unit circle is the natural boundary of .
Polya ([7]) proved the following converse of this result:
Letbe a sequence of integers for which ; then there exists a power series whose circle of convergence is the unit circle and for which the unit circle is not the natural boundary.
Based on a sequence as mentioned earlier, we construct a multiplicity sequence , that is, a sequence where, for one has , and eachappears times. For this sequence we prove that the infinite product which vanishes exactly on , satisfies every . That is, we have a sharp estimate for the th derivative function of evaluated on .
We assume that the reader is familiar with the theory of Entire Functions and the theory of Dirichlet series, as used in the books [1,5,8–11].
We note that other results concerning the location of singularities of Taylor–Dirichlet series have been derived by Blambert, Parvatham, and Berland (see [2–4]).
2. Auxiliary Results and Notions
In this section, we describe the definitions and also to express and prove the lemma, we need to prove the theorem.
Definition 2.1. We denote by the class of all sequenceswith distinct complex terms diverging to infinity, satisfying the following conditions: (see also [12])
(1) There is a constant so that for all.
(2)
(3) the .
Definition 2.2. Choose a sequence which belongs to the class. supposeandbe real positive numbers so that. We say that a sequence with complex terms, with the not necessarily in an increasing order, and , belongs to the class if for all we have and for all one of the following conditions holds:
(i)
(ii)
One observes that allows for the sequence to have coinciding terms. We may now rewrite in the form of a multiplicity sequencein the following way; first we split into groups of terms having the same modulus, and then within each group we order them by the size of their argument, beginning from smaller to larger. The arguments are taken in the interval . We shall say that is thereordering of .
Remark. We point out that the spacing condition of a sequence plays a very important role throughout the article.
Let be as in
.
One deduces that if then We also define to be the number of terms of and we shall refer to as the pseudo-multiplicity of .
In the lemma that follows, we get an upper bound for with respect to .
Lemma 2.3. There exist positive constants and so that for any n one has
Proof. First note that the relation
holds for all sinceConsider now any(). Then
It follows that Then one also gets
Finally, the spacing condition
yields that for any one has
Since is the number of terms of , then From the above equation it follows that there exists a positive so that Finally, the relationyields a positive so that
Lemma 2.4. For any one has
Proof. Let for some . From the previous lemma we know thatfor some . But the pseudo-multiplicity of is the multiplicityof . Thus, one obtains the relation .
Another important lemma, which is important for this paper can be stated as follows:
Lemma 2.5.Letbe a real positive sequence and letso that is real positive too, with its reordering. Then the regions of convergence of the three series as defined in
(1)
where is a polynomial with , and
(2)
are the same. For any point inside the open convex region, the three series converge absolutely. Similarly, if instead of a real sequencewe have a complex sequence .
Proof. We have to show that
is satisfied. First, note that from Lemma 2.4 one deduces that the right limit of
is valid. Thus, it remains to verify the left limit.
We claim that
This implies that
since , and we are done.
Let us justify our claim. It is obvious that . Assume that for some . We will prove that as well.
Note that there is at least one so that . If then .
Since , this implies that since for all , therefore . This means that , thus there is some with . It follows that
that is, which is false. Thus and this completes the proof.
3. Main Results
This section describes the main theorem of this paper, and it can be fixed by using proven methods Polya.
Theorem 3.1. suppose be a real positive sequence. Let so that is real positive too and letbe its reordering. Then any Taylor-Dirichlet series as in (1), satisfying
(3)
whose circle of convergence is the unit circle and for which the unit circle is not the natural boundary.
Proof. We follow on the lines of the proof of Theorem in [6].
Let and as defined in (1), (2) and
(4)
From Lemma 2.5, the regions of convergence of the three series are the same. Since theare real positive numbers, we consider the non-trivial case, that is when the three series converge in identical half-planes of the form . With no loss of generality we assume that the abscissa of convergence (ordinary and absolute) is the line .
In other words the relation
(5)
holds. Thus, all three series converge absolutely and uniformly in any half-plane . One also notes that from (3) we have
(6)
There clearly exist two sequences of integers and such that and the number of in is greater than . (The c's denote absolute positive constants.) The existence of these sequences is immediate from . Denote the in the intervals by . We clearly have . For construction of we shall use only the . Put We shall determine the so that the unit circle will be the circle of convergence and the pointwill be a regular point of . It will suffice to show that there exists a number , , such that the circle of convergence of
has radius greater than . We shall choose a sufficiently large integer. We have by the binomial expansion
We have to show that , for some choice of the with.
Let be a small but fixed number; we distinguish two cases.
In case does not lie in any of the intervals . Then we show that for every choice of the with .
This means that if is large enough and does not lie in then . Clearly
(7)
If we define
We find
(8)
By studying the quotient (8) we see that and by applying Stirling's formula
We note that as . It follows from (8) that there exists such that
(9)
(10)
and hence a simple calculation shows that there exists a such that
(11)
for not in .
Now clearly
where in the summation is extended over the and in over the . (By assumption does not lie in and in (7) theare all in ; thus if , and if , .) Thus from (9), (10) and (11) (by summing a geometric series)
or
which completes the proof.
In case
We write
where
indicates that the summation is extended only over those for which does not lie in , and in the summation is extended over the other .
We
and we can show that as before.
Now we show that we can choose the to be such as to make all the for equal to . Thus we must determine theso that
These are homogeneous equations for the . The number of these equations is less than for sufficiently small and the number of unknowns is greater than which is greater than the number of equations for large enough . Thus the system of equations always has a solution, and further we can suppose that the absolute value of the largest is . This will insure that the circle of convergence of will be the unit circle, which completes the proof.
4. Conclusion
In this study we examine a variation of the converse of Fabry Gap theorem.
Polya's result shows that in some sense Fabry's result is the best possible. Perhaps the elementary and direct proof that mentioned above might be of some interest.
To do this, a sequence with a series of new build and reordering the call, using the convergence of three series obtain upper and lower bounds. And using the Stirling's formula and we will achieve the desired result in this paper.
Acknowledgements
The author gratefully acknowledges the help of Prof. E. Zikkos to improve the original version of the paper.
References