ON ANALYTIC CONTINUATION OF POWER

SERIES WITH THE CONDITION ON

THEIR GELFOND-LEONTEV DERIVATIVES

Myroslav M. Sheremeta

1.

For R ∈ [0, +∞] we denote by A(R) the class of power series f(z) = ∑_{k=0}^{∞} f_{k}z^{k} having radius of convergence R, and we say that f ∈ A^{+}(R) if f ∈ A(R) and f_{k} > 0 for all k ∈ ℤ_{+}. For f ∈ A(0) and l ∈ A^{+}(0) we put

D_{l}^{n}f(z) = ∑_{k=0}^{∞} \frac{l_{k}}{l_{k+n}} f_{k+n}z^{k}.

The operator D_{l}^{n} is called the Gelfond-Leontev operator. If l(z) ≡ e^{z},

then D_{l}^{n} = \frac{d^{n}}{dz^{n}} is the usual differential operator. We can assume that l_{0} = 1. We put

κ_{k} = \frac{l_{k}^{2}}{l_{k+1}l_{k−1}} , η_{k} = \frac{1}{k} ln \frac{1}{l_{k}} , ω_{k} = η_{k+1} − η_{k} (k ∈ ℕ).

It is easy to show [1] that if κ_{k} ↘ κ, then ω_{k} ↘ ω ≥ 0 as k → ∞.

Let 𝔑 be a class of increasing sequences (n_{p}) of nonnegative integers, n_{0} = 0, let Λ be the class of all positive sequences, and let Λ^{∗} = {λ ∈ Λ: ln λ_{k} ≤ a for every k ∈ ℕ and some a > 0}. We say that f ∈ A_{λ}(R) if f ∈ A(R) and |f_{k}| ≤ λ_{k}|f_{1}| for all k ∈ ℕ. In [1] the following theorem is proved.

The research described in this publication was supported partly by grants of the International Science Foundation N UCR000 and N SPU041078.

Theorem 0. Suppose that l ∈ A^{+}(∞) and κ_{k} ↘ κ as k → ∞. In order that for every (n_{p}) ∈ 𝔑, λ ∈ Λ^{∗}, and f ∈ A(0) the inclusion D_{l}^{n_{p}}f ∈ A_{λ}(0) for all p ∈ ℤ_{+} implies the inclusion f ∈ A(∞), it is necessary and sufficient that ω > 0.

From the condition l ∈ A^{+}(∞) and κ_{k} ↘ κ (k → ∞) it follows that ln(1/l_{k}) = O(k^{2}), k → ∞, and if ω > 0, then ln(1/l_{k}) ≥ ωk^{2} + O(k), k → ∞. Hence, if these three conditions hold, then ln(1/l_{k}) ≍ k^{2}.

The question arises whether the sufficiency in Theorem 0 remains valid if ln(1/l_{k}) increases faster than k^{2}. It is clear that in this case the sequence (κ_{k}) is not nonincreasing. On the other hand, if the sequence (κ_{k}) is oscillating, then the statement of Theorem 0 can be obtained in general [1]. Therefore, we examine the case when the sequence (κ_{k}) is decreasing. It is possible to show that in this case it follows from the condition l ∈ A^{+}(∞) that κ_{k} > 1 for all k ≥ k_{0}. For simplicity we assume that κ_{k} > 1 for all k ≥ 1.

We prove the following assertion.

Theorem 1. Suppose that l ∈ A^{+}(∞) and 1 < κ_{k} ↗ κ ≥ ∞ as k → ∞. Then for every (n_{p}) ∈ 𝔑, λ ∈ Λ^{∗}, and f ∈ A(0) the inclusion D_{l}^{n_{p}}f ∈ A_{λ}(0) for all p ∈ ℤ_{+} implies the inclusion f ∈ A(∞).

2.

For the proof of Theorem 1 we need some lemmas.

Lemma 1 ([1]). Let λ ∈ Λ, (n_{p}) ∈ 𝔑, and D_{l}^{n_{p}}f ∈ A_{λ}(0) for all p ∈ ℤ_{+}. Then for every p ∈ ℕ and k = 1, 2, …, n_{p+1}−n_{p}+1, the following inequality holds

|f_{n_{p}+k}| ≤ |f_{1}|l_{1}^{p}l_{n_{p}+k} \frac{λ_{k}}{l_{k}} ∏_{j=1}^{p} \frac{λ_{n_{j}−n_{j−1}+1}}{l_{n_{j}−n_{j−1}+1}} .

Lemma 2. If 1 < κ_{k} ↗ κ ≤ ∞ as k → ∞, then η_{k} ↗ ∞ and 0 < ω_{k} ↗ ω ≤ ∞ as k → ∞.

Indeed, l_{k}/l_{k+1} = \frac{1}{l_{1}} ∏_{j=1}^{k} κ_{j} and, therefore,

ln \frac{1}{l_{k}} = ∑_{j=1}^{k−1}(k − j) ln κ_{j} + k ln \frac{1}{l_{1}} , k ≥ 2.

Hence

η_{k} = ∑_{j=1}^{k−1}(1 − \frac{j}{k}) ln κ_{j} + ln \frac{1}{l_{1}} , k ≥ 2,

ON ANALYTIC CONTINUATION OF POWER SERIES 263

and

ω_{k} = \frac{1}{k(k + 1)} ∑_{j=1}^{k} jκ_{j}, k ≥ 1,

whence η_{k} ↗ ∞ (k → ∞) provided that l ∈ A^{+}(∞). Now

ω_{k+1} − ω_{k} = \frac{1}{k + 2}( ln κ_{k+1} − \frac{2}{k(k + 1)} ∑_{j=1} kj ln κ_{j})

≥ \frac{1}{k + 2} ln κ_{k+1}(1 − \frac{2}{k(k + 1)} ∑_{j=1}^{k} j) = 0,

whence ω_{k} ↗ ω ≤ ∞ (k → ∞). Recall that ω = ln κ. Lemma 3. Let

γ_{k} = γ_{k,n} = \frac{1}{n + k} \{ ln \frac{1}{l_{n+k}} − ln \frac{1}{l_{k}} + k ln \frac{1}{l_{1}} \}, n ∈ ℕ.

If 1 < κ_{k} ↗ κ ≤ ∞ (k → ∞), then γ_{k} ↗ γ ≤ ∞ (k → ∞).

Indeed, according to Lemma 2 we have

γ_{k+1} − γ_{k} = ω_{n+k} + \frac{k}{n + k} η_{k} − \frac{k + 1}{n + k + 1}η_{k+1} + \frac{n}{(n + k)(n + k + 1)} ln \frac{1}{l_{1}}

= ω_{n+k} − ω_{k} + \frac{n}{(n + k)(n + k + 1)}\{(n + k)ω_{k} − ∑_{j=1}^{k} ω_{j}\} > 0.

3.

Now we turn to the proof of Theorem 1. Since λ ∈ Λ^{∗} we have ln {λ_{k} ∏_{j=1}^{p} λ_{n_{j}−n_{j−1}+1}} ≤ 2a(k+n_{p}) for every p ∈ ℕ and k = 1, 2, …, n_{p}+ 1, and by Lemma 1

\frac{1}{n_{p} + k} ln \frac{1}{|f_{n_{p}+k}|} ≥ γ_{k,n_{p}} − \frac{1}{n_{p} + k} A_{p} + O(1), p → ∞,

where

A_{p} = ∑_{j=1}^{p} ln (\frac{1}{l_{n_{j}−n_{j−1}+1}} ).

Using Lemma 3 we get

(1) \frac{1}{n_{p} + k} ln \frac{1}{|f_{n_{p}+k}|} ≥ γ_{1,n_{p}} − \frac{1}{n_{p} + 1}A_{p} + \frac{(k − 1)A_{p}}{(n_{p} + 1)(n_{p} + k)} + O(1)

for all k = 1, 2, …, n_{p+1} − n_{p} + 1 as p → ∞. Since η_{k} ↗ ∞ (k → ∞), we have ln(1/l_{k}) ≥ 1 (k ≥ k_{0}), and therefore

A_{p} = ( ∑_{\begin{matrix}j=1 \\ n_{j}−n_{j−1}+1<k_{0} \end{matrix}}^{p} + ∑_{\begin{matrix}j=1 \\ n_{j}−n_{j−1}+1≥k_{0} \end{matrix}}^{p} ) ln \frac{1}{l_{n_{j}−n_{j−1}+1}} = B_{p} + C_{p},

where B_{p} = O(p), p → ∞, and C_{p} ≥ 0. Hence it follows that the third summand on the right side of (1) can be omitted, and for k = 1, 2, …, n_{p+1}− n_{p} + 1 we have

\frac{1}{n_{p} + k} ln \frac{1}{|f_{n_{p}+k}|} ≥ \frac{1}{n_{p} + 1}( ln \frac{1}{l_{n_{p}} + 1} − A_{p}) + O(1), p → ∞.

Therefore, we have to prove that

\frac{1}{n_{p} + 1}( ln \frac{1}{l_{n_{p}} + 1} − ∑_{j=1}^{p} \frac{1}{l_{n_{j}−n_{j−1}+1}} ) → +∞, p → ∞.

But

ln \frac{1}{l_{k}} = kη_{k} = k( ∑_{j=1}^{k−1} ω_{j} + ln \frac{1}{l_{1}} ).

Hence we have to prove that

(2) \frac{1}{n_{p} + 1}\{(n_{p} + 1) ∑_{m=1}^{n_{p}} ω_{m} − ∑_{j=1}^{p}(n_{j} − n_{j−1} + 1) ∑_{m=1}^{n_{j}−n_{j−1}+1} ω_{m}\} → +∞,

as p → ∞.

First let ω_{m} ↗ ω < ∞ (m → ∞). Then it is easy to show that the condition (2) is equivalent to the condition

\frac{1}{n_{p}} \{n_{p} ∑_{m=1}^{n_{p}} ω_{m} − ∑_{j=1}^{p}(n_{j} − n_{j−1}) ∑_{m=1}^{n_{j}−n_{j−1}} ω_{m}\} → +∞, p → ∞.

This condition holds provided that

∑_{m=1}^{n_{p}} ω_{m} − max { ∑_{m=1}^{n_{j}−n_{j−1}} ω_{m}: 1 ≤ j ≤ p} → +∞, p → ∞,

that is,

(3)  ∑_{m=1}^{n_{p}} ω_{m} − ∑_{m=1}^{n_{j}−n_{j−1}} ω_{m} = ∑_{m=n_{j}−n_{j−1}}^{n_{p}} ω_{m} → ∞, p → ∞,

where 1 ≤ j ≤ j_{p} ≤ p. Since ω_{1} > 0, ∑_{m=k+1}^{n} ω_{m} ≥ (n − k)ω_{1} and n_{p} − n_{j} +

n_{j−1} → ∞ for 1 ≤ j ≤ j_{p} ≤ p as p → ∞, we have (3), that is, Theorem 1 holds in the case ω < ∞.

Let now ω_{m} ↗ ∞ (m → ∞). Then \frac{1}{n_{p}+1} ∑_{j=1}^{n_{p}} ω_{m} → ∞ (p → ∞), and since

\frac{1}{n_{p} + 1}\{(n_{p} + 1) ∑_{m=1}^{n_{p}} ω_{m} − ∑_{j=1}^{p}(n_{j} − n_{j−1} + 1) ∑_{m=1}^{n_{j}−n_{j−1}+1} ω_{m}\}

= \frac{1}{n_{p} + 1} ∑_{m=1}^{n_{p}} ω_{m}+\frac{1}{n_{p} + 1} ∑_{j=1}^{p} \{(n_{j}−n_{j−1}+1) ∑_{m=n_{j}−n_{j−1}+2}^{n_{p}} ω_{m}− ∑_{m=1}^{n_{p}} ω_{m}\},

it is sufficient to prove that

(4)  (n_{j} − n_{j−1} + 1) ∑_{m=n_{j}−n_{j−1}}^{n_{p}} ω_{m} − ∑_{m=1}^{n_{p}} ω_{m} ≥ 0

for all p ≥ j ≥ j_{0}. Let j_{0} be such that n_{j−1} ≥ 4 for j ≥ j_{0}. Then 2 < n_{j} − n_{j−1} + 2 ≤ n_{p} − 2, and the inequality (4) will be proved if we show that

k ∑_{m=k+1}^{n} ω_{m} − ∑_{m=1}^{n} ω_{m} ≥ 0 (2 ≤ k ≤ n − 2).

The last inequality holds since ω_{k} is nondecreasing.

References

[1] M. M. Sheremeta, On power series with Gelfond-Leontev derivatives satisfying a special condition, Matematicheskaya Fizika, Analiz i Geometriya (to appear). (Russian)