Tauberian- and Convexity theorems for certain (N, p, q) –means

by

Rüdiger Kiesel and Ulrich Stadtmüller

1 Introduction

We consider throughout complex sequences (s_{n}) and discuss the relations of certain summability methods. We say a sequence (s_{n}) of complex numbers is summable to s by the

(i) Cesàro-method of order α > −1 , briefly s_{n} → s(C_{α}) , if

\frac{1}{\binom{n+α}{n}} ∑_{k=0}^{n} \binom{n − k + α − 1}{n − k}s_{k} → s (n → ∞);

(ii) Euler-method of order 0 < p ≤ 1 , briefly s_{n} → s(E_{p}) , if

∑_{k=0}^{n} \binom{n}{k}p^{k}(1 − p)^{n−k}s_{k} → s (n → ∞);

(iii) Abel-method, briefly s_{n} → s(A) , if

f(t) = (1 − t) ∑_{n=0}^{∞} s_{n}t^{n} exists for 0 < t < 1 and f(t) → s (t → 1−);

(iv) Borel-method, briefly s_{n} → s(B) , if

g(t) = e^{−t} ∑_{n=0}^{∞} \frac{s_{n}}{n!} t^{n} exists for t ∈ R I and g(t) → s (t → ∞).

The Cesàro- and Abel-method resp. the Euler- and Borel-method are known to be closely related, see [9,17,19]. Especially the following Abelian inclusions are well kown, see e.g. [9; Theorems 43,55,118,128]

for −1 < α ≤ β: s_{n} → s(C_{α}) ⇒ s_{n} → s(C_{β}) ⇒ s_{n} → s(A),

for 0 < p ≤ q ≤ 1 : s_{n} → s(E_{q}) ⇒ s_{n} → s(E_{p}) ⇒ s_{n} → s(B).

The following converse or Tauberian theorem for the Cesàro–Abel-case goes back to Littlewood [14] ( α, β ∈ N I ), and Anderson [1] ( α, β ≥ −1 ).

THEOREM TC 1

(i) Let −1 < α < β then

s_{n} → s(A) and s_{n} = O (1) (C_{α}) imply s_{n} → s(C_{β}) .

(ii) For −1 < α < δ ≤ β we have the so-called convexity-theorem

s_{n} → s(C_{β}) and s_{n} = O (1) (C_{α}) imply s_{n} → s(C_{δ}) .

Quite recently Boos and Tietz [4] proved that the situation is completely analogous for the Euler–Borel-case

THEOREM TC 2

(i) Let 0 < p < q ≤ 1 then

s_{n} → s(B) and s_{n} = O (1) (E_{q}) imply s_{n} → s(E_{p}) .

(ii) For 0 < p ≤ r < q ≤ 1 we have the convexity-theorem

s_{n} → s(E_{p}) and s_{n} = O (1) (E_{q}) imply s_{n} → s(E_{r}) .

Obviously part (ii) is in both cases a trivial consequence of the Abelian inclusion and part (i). The aim of this paper is to show that the above results are special cases of a more general setting.

For the following assume that (p_{n}) is a sequence of reals with the following properties:

(1.1)  \{ \begin{matrix}p_{0} > 0, p_{n} ≥ 0, n ∈ N I , such that the power series \\ p(t) = ∑_{n=0}^{∞} p_{n}t^{n} has radius of convergence R > 0. \end{matrix}

Since we can use p_{n}R^{n} as weights in case 0 < R < ∞, we only have to deal with the two cases R = 1 and R = ∞. Furthermore we define the α -th convolution p_{n}^{∗α} of a sequence (p_{n}) by

p_{n}^{∗1} := p_{n}, n = 0, 1, 2, … and p_{n}^{∗(α+1)} := ∑_{k=0}^{n} p_{n−k}^{∗α}p_{k}.

We now generalize the summability methods used in Theorems TC1 and TC2.

To this end we need a further sequence (q_{n}) of nonnegative reals, also satisfying (1.1), in general with a different radius of convergence R_{q} for the associated power series. We then say, that a sequence (s_{n}) is summable to s by the

(i) power series method of summability (P, p) , briefly s_{n} → s(P, p) , if

(1.2)  \{ \begin{matrix}p_{s}(t) = ∑_{n=0}^{∞} s_{n}p_{n}t^{n} converges for |t| < R \\ and if σ_{p}(t) = \frac{p_{s}(t)}{p(t)} → s as t → R − . \end{matrix}

(In case R = 1 we have the so-called (J_{p}) -methods, in case R = ∞ the (B_{p}) - methods).

(ii) general Nörlund-means (N, p^{∗α}, q^{∗β}); α, β ∈ N I , briefly s_{n} → s (N, p^{∗α}, q^{∗β}), if

(1.3)  \{ \begin{matrix}\frac{1}{r_{n}} ∑_{k=0}^{n} p_{n−k}^{∗α}q_{k}^{∗β}s_{k} → s (n → ∞), where we suppose that \\ r_{n} := (p^{∗α} ∗ q^{∗β})_{n} = ∑_{k=0}^{n} p_{n−k}^{∗α}q_{k}^{∗β} > 0 for n = 0, 1, …. \end{matrix}

We require all methods to be regular. By Theorem 5 in [9], we have regularity for a power series method if and only if

(1.4) \{ \begin{matrix}(A) & P_{n} = ∑_{k=0}^{n} p_{k} → ∞, (n → ∞), in case R = 1, \\ (B) & p(t) is not a polynomial, i.e. \\ & p_{n} ≠ 0 for infinitely many n in case R = ∞. \end{matrix}

By Theorem 3 in [9] the general Nörlund mean (N, p^{∗α}, q^{∗β}) is regular if and only if

(1.5) \frac{p_{n−k}^{∗α}}{r_{n}} → 0 for any fixed k.

Remark 1: Important special cases are

(i) The Cesàro–Abel-methods: p_{n} = 1 : (P, p) = (A), (N, p^{∗α}, p) = (C_{α}), α ∈ N I .

(ii) The generalized Abel-method ( δ > 0 ):

p_{n} = \binom{n − 1 + δ}{n} : (P, p) = (A_{δ−1}), (N, p^{∗α}, 1) = (C_{αδ}), α ∈ N I .

(iii) The Euler-Borel-methods:

p_{n} = 1/n! : (P, p) = (B), (N, p^{∗α}, p) = (E_{\frac{1}{1+α}}), α ∈ N I . (We use the notation 1 for the sequence (1, 1, …) )

We now generalize the above results to our general setting, provided some regularity assumptions are satisfied.

2 Main results

In [10], Proposition 1, R.Kiesel showed that for α ≤ β, α, β ∈ N I the following inclusions hold true:

s_{n} → s(N, p^{∗α}, p) ⇒ s_{n} → s(N, p^{∗β}, p) ⇒ s_{n} → s(P, p),

provided that for all γ ∈ N I the methods (N, p^{∗γ}, p) are regular (for the second inclusion only the regularity of the (P, p) -method is needed.).

This is especially the case , if one of the following conditions is satisfied.

(2.1) \{ \begin{matrix}(A) & p_{n} ∼ n^{σ}L(n), & σ ≥ 0, n^{σ}L(n) is nondecreasing and L(.) is slowly \\ & & varying, see [3] §1.2 for the definition; \\ (B) & p_{n} ∼ exp\{−g(n)\}, & where g ∈ C_{2}[0, ∞), \\ & & with g′′(x) ↓ 0, x^{2}g′′(x) ↑ ∞ (x → ∞). \end{matrix}

Using the sequence of maximal weights” (Δ_{n}) defined by

(2.2) Δ_{n} = inf_{0<t<R} p(t)t^{−n},

we have in the above cases the following relationship

(2.3) Δ_{n} = \sqrt{2π}ϕ(n) p_{n} (n → ∞),

where ϕ(.) is a suitable, positive function.

For (x → ∞) we have in case (A) that \sqrt{2π}ϕ(x) ∼ Γ(σ + 1) (\frac{σ+1}{e})^{−σ−1} x and in

case (B) that ϕ(x) ∼ (g′′(x))^{− \frac{1}{2}} . Following [2,3 § 2.11] we call a function ψ : (0, ∞) → (0, ∞) self-neglecting if ψ satisfies ψ(x) = o (x) (x → ∞) , and if ψ(x + tψ(x))/ψ(x) → 1 (x → ∞) locally uniformly in t ∈ R I . Observe that g′′(x)^{− \frac{1}{2}} is self-neglecting because of (2.1) and since for e.g. t ≥ 0

(\frac{g′′(x + tg′′(x)^{−1/2})}{g′′(x)})^{− \frac{1}{2}} ≥ 1, and

(\frac{g′′(x + tg′′(x)^{−1/2})}{g′′(x)})^{− \frac{1}{2}} = (\frac{(x + tg′′(x)^{−1/2})^{2}g′′(x + tg′′(x)^{−1/2})}{x^{2}g′′(x)})^{− \frac{1}{2}} (1 + \frac{t}{\sqrt{x^{2}g′′(x)}})

≤ 1 + \frac{t}{\sqrt{x^{2}g′′(x)}} → 1 (x → ∞), locally uniformly in t.

Because of this locally uniform convergence ϕ(.) is self-neglecting, too. We can now state our main theorem

THEOREM 1

Let α, β, γ, δ ∈ N I with α < δ ≤ β and assume that (p_{n}) satisfies (2.1). Then

(i) s_{n} → s(P, p^{∗γ}) and s_{n} = O (1) (N, p^{∗α}, p^{∗γ}) imply s_{n} → s(N, p^{∗β}, p^{∗γ}) . (ii) s_{n} → s(N, p^{∗β}, p^{∗γ}) and s_{n} = O (1) (N, p^{∗α}, p^{∗γ}) imply s_{n} → s(N, p^{∗δ}, p^{∗γ}) .

Remark 2: In case p_{n} ≡ 1, γ = 1 resp. p_{n} = 1/n!, γ = 1 Theorem 1 is Theorem TC1 resp. TC2 in the discrete index case.

In our paradigms Abel- and Borel-method we have the following relations of the methods (see [5]):

(i) Abel-case: (A_{α−1}) = (P, \binom{n+α−1}{n}) = (P, 1^{∗α}), α > 0 , then for μ > λ > −1 :

s_{n} → s(A_{μ}) ⇒ s_{n} → s(A_{λ}).

(ii) Borel-case: Since p_{n}^{∗α} = α^{n}/n! , we have

(B) = (P, 1/n!) ≈ (P, ((α^{n})/n!)) = (P, (1/n!)^{∗α}).

(Where we use ≈ to note that two methods are equivalent.)

So the question arises what the relation of (P, p^{∗α}) and (P, p^{∗β}) resp. (N, p, p^{∗α}) and (N, p, p^{∗β}) in the general case is. Unfortunatly we can only present answers to the question under additional assumptions.

PROPOSITION 1

Suppose α, β ∈ N I and the sequence (p_{n}) satisfies (1.1) with R = 1 or R = ∞ and p_{n}^{∗α} > 0. If we have furthermore that μ_{n} = (p_{n}^{∗β})/(p_{n}^{∗α}) is a totally monotone sequence, i.e.

(2.4) μ_{n} = ∫_{0}^{R} t^{n} dχ(t) < ∞

for all n = 0, 1, … with some (bounded) nondecreasing function χ, then we have

s_{n} → s(P, p^{∗α}) implies s_{n} → s(P, p^{∗β}).

This result can also be obtained using a theorem of Borwein in [5], but we are able to present a somewhat easier proof.

An answer to the question of inclusion in case of the (N, p, p^{∗α}) -means, was already given by Das [8], but again only under restricting additional assumptions.

PROPOSITION 2

Let α, β ∈ N I and (p_{n}) a sequence of strictly positive reals. If

(2.5) \frac{p_{n+1}}{p_{n}} ↑ 1 (n → ∞)

and if additionaly either

\frac{p_{n}^{∗β}}{p_{n}^{∗α}} ≥ \frac{p_{n+1}^{∗β}}{p_{n+1}^{∗α}} and (N, p, p^{∗β}) is regular,

or

\frac{p_{n}^{∗β}}{p_{n}^{∗α}} ≤ \frac{p_{n+1}^{∗β}}{p_{n+1}^{∗α}} and \frac{p_{n}^{∗β}p_{n}^{∗α+1}}{p_{n}^{∗β+1}p_{n}^{∗α}} = O (1) and (N, p, p^{∗α}) is regular.

Then (N, p, p^{∗α}) convergence implies (N, p, p^{∗β}) -convergence.

3 Auxiliary results

First we discuss the asymptotic properties of the (N, p, q) -means.

LEMMA 1

Assume that (p_{n}) satisfies (2.1).

(i) In case (A), i.e. p_{n} = n^{σ}L(n) , we have

p_{n}^{∗2} ∼ \{ \begin{matrix}n^{2σ+1}L^{2}(n)B(σ + 1, σ + 1), & if σ > −1, \\ L^{∗}(n)n^{−1}, & if σ = −1, \end{matrix}

with B(., .) denoting the beta-integral and L^{∗}(.) some slowly varying function.

(ii) In case (B), we have for any α ∈ N I

(3.1) p_{n}^{∗α} ∼ \sqrt{(2π)^{α−1}/α} ϕ(n/α)^{α−1} exp\{−αg(n/α)\} (n → ∞),

ϕ(.) as in (2.3).

Proof:

(i) are slight generalisations of Theorem 42 in [9] and Theorem 2.3.1 in chapter 5 of [20].

(ii) For α = 2 the result is contained in Proposition 3 of [10]. We use induction on α for the general case. By Definition we have

p_{n}^{∗(α+1)} = ∑_{ν=0}^{n} p_{ν}^{∗α}p_{n−ν}.

We define a function (3.2) ϵ(x) = x(x^{2}g′′(x))^{−1/4}. Then we can show that the essential part of the sum occurs for ν ∈ M(n) with

M(n) := \{ν : |ν − \frac{αn}{α + 1}| ≤ ϵ (\frac{αn}{α + 1})\} .

( Use techniques similar to those in the proof of Lemma 2 in [6], see also related calculations in [12,13] .) By the induction hypotheses we find

p_{n}^{∗(α+1)} ∼ ∑_{ν∈M(n)} \sqrt{(2π)^{α−1}/α} ϕ(ν/α)^{α−1} exp\{−αg(ν/α)\} exp\{−g(n − ν)\}.

We now use the asymptotics for p_{n} and the Taylor-expansion ( θ, ϑ ∈ (0, 1) ):

p_{n}^{∗(α+1)} ∼ ∑_{ν∈M(n)} \sqrt{\frac{1}{α} (\frac{2π}{g′′ (\frac{ν}{α})})^{α−1}}

exp \{−α ( g (\frac{n}{α + 1}) + g′ (\frac{n}{α + 1}) (\frac{ν}{α} − \frac{n}{α + 1})

+\frac{1}{2}g′′ (\frac{n}{α + 1} + θ (\frac{ν}{α} − \frac{n}{α + 1})) (\frac{ν}{α} − \frac{n}{α + 1})^{2})

−g (\frac{n}{α + 1}) − g′ (\frac{n}{α + 1}) (n − ν − \frac{n}{α + 1})

−\frac{1}{2}g′′ (\frac{n}{α + 1} + ϑ (n − ν − \frac{n}{α + 1})) (n − ν − \frac{n}{α + 1})^{2}\}

Now we use the basic inequality (13) in [6], namely

|\frac{g′′(t)}{g′′(x)} − 1| ≤ 4\frac{|t − x|}{x} for all sufficiently large t, x, if |t − x| ≤ x/4,

which is satisfied in our range M(n) , and the fact that ϵ(n)/n → 0 as n → ∞ to obtain

p_{n}^{∗(α+1)} ∼ \sqrt{\frac{1}{α} (\frac{2π}{g′′(\frac{n}{α+1})})^{α−1}} × exp \{−(α + 1)g (\frac{n}{α + 1})\}

× ∑_{ν∈M(n)} exp \{−\frac{(α + 1)}{2α}g′′ (\frac{n}{α + 1}) (ν − \frac{nα}{α + 1})^{2}\} (1 + o(1))

∼ \sqrt{\frac{(2π)^{α}}{α + 1}} ϕ (\frac{n}{α + 1})^{α} exp \{−(α + 1)g (\frac{n}{α + 1})\}

For the last step use the approximation of the sum with the integral of a Gaussian density with variance α/((α + 1)g′′(n/(α + 1))) .

Corollary

If (p_{n}) satisfies (2.1 B) and α, β ∈ N I , then we have for the entry a_{n,k} of the (N, p^{∗α}, p^{∗β}) - matrix the asymptotic relation

a_{n,k} ∼ \sqrt{\frac{α + β}{2παβ}} ϕ (\frac{n}{α + β} )^{−1} exp \{−\frac{α + β}{2αβ} (\frac{k − \frac{nβ}{α+β}}{ϕ (\frac{n}{α+β} )})^{2}\}

if |k − \frac{nβ}{α + β} | ≤ ϵ(n) with ϵ(.) as in (3.2) and furthermore

∑_{|k− \frac{nβ}{α+β} |>ϵ(n)} a_{n,k} → 0 (n → ∞).

Proof:

If |k − \frac{nβ}{α + β} | ≤ ϵ(n) then (θ, ξ ∈ (0, 1))

\frac{p_{n−k}^{∗α}p_{k}^{∗β}}{p_{n}^{∗(α+β)}} ∼ \begin{matrix}exp & \{−αg (\frac{n − k}{α} ) − βg (\frac{k}{β} )\} \\ exp & \{−(α + β)g (\frac{n}{α + β} )\} \end{matrix}\sqrt{\frac{(2π)^{α−1}(2π)^{β−1}g′′ (\frac{n}{α + β} )^{α+β−1} (α + β)}{(2π)^{α+β−1}g′′ (\frac{n − k}{α} )^{α−1} g′′ (\frac{k}{β} )^{β−1} αβ}}

∼ exp \{ − α ( g (\frac{n}{α + β} ) + g′ (\frac{n}{α + β} ) (\frac{n − k}{α} − \frac{n}{α + β} )

+\frac{1}{2}g′′ (\frac{n}{α + β} + θ (\frac{n − k}{α} − \frac{n}{α + β} )) (\frac{n − k}{α} − \frac{n}{α + β} )^{2})

−β ( g (\frac{n}{α + β} ) + g′ (\frac{n}{α + β} ) (\frac{k}{β} − \frac{n}{α + β} )

+\frac{1}{2}g′′ (\frac{n}{α + β} + ξ (\frac{k}{β} − \frac{n}{α + β} )) (\frac{k}{β} − \frac{n}{α + β} )^{2})\}

× exp \{(α + β)g (\frac{n}{α + β} )\} \sqrt{(1 + o(1))\frac{α + β}{2παβ} g′′ (\frac{n}{α + β} )}.

Now |\frac{k}{β} − \frac{n}{α + β} | ≤ \frac{ϵ(n)}{β} and |\frac{n − k}{α} − \frac{n}{α + β} | ≤ \frac{ϵ(n)}{α}. Therefore we obtain the

desired result by the same calculations as used in Lemma 1. For the second part observe that

∑_{k=0}^{n} \frac{p_{n−k}^{∗α}p_{k}^{∗β}}{p_{n}^{∗(α+β)}} = 1 ∼ (1 + o (1)) ∑_{|k− \frac{nβ}{α+β} |≤ϵ(n)} exp\{…\} \sqrt{…}

We now give the asymptotics of the relevant power-series methods and show that for bounded sequences these methods are equivalent to certain generalized Valiron -type means, compare [6,11]

LEMMA 2

Assume that (p_{n}) satisfies (2.1 B). Then we have

(i) (p (exp \{g′ (\frac{x}{μ})\}))^{μ} ∼ (\sqrt{2πμ}ϕ (\frac{x}{μ}))^{μ} exp \{− (g (\frac{x}{μ}) − \frac{x}{μ}g′ (\frac{x}{μ}))\}

for x → R.

(ii) For bounded sequences (s_{n}) the following equivalence holds true

s_{n} → s(P, p^{∗μ}) ⇔ ∫_{−∞}^{∞} exp \{−\frac{1}{2μ} (\frac{x − t}{ϕ (\frac{x}{μ})})^{2}\} s(t)\frac{dt}{\sqrt{2πμ}ϕ (\frac{x}{μ})} → s,

for x → ∞, where s(t) = s_{[t]}, for t ≥ 0 and s(t) = 0 elsewhere.

Proof:

(i) follows directly from [12], Lemma 5, resp [13], Lemma 8, see also Lemma 2 in [6] (ii) In this case the calculations are similar to the calculations used in [6], Lemma 2 and [11],Theorem 2, so we only outline the major steps. We have by using Lemma 1 and part (i) (For the notation see (1.2)).

σ_{p^{∗μ}} (e^{g′(\frac{x}{μ})}) = \frac{(1 + o(1))}{\sqrt{2πμ}ϕ (\frac{x}{μ})} ∑_{n=0}^{∞} s_{n} exp \{−μg (\frac{n}{μ}) + ng′ (\frac{x}{μ}) + μg (\frac{x}{μ}) − xg′ (\frac{x}{μ})\}

= \frac{(1 + o(1))}{\sqrt{2πμ}ϕ (\frac{x}{μ})} ∑_{n=0}^{∞} s_{n} exp \{−\frac{μ}{2}g′′ (\frac{x}{μ} + θ (\frac{n}{μ} − \frac{x}{μ})) (\frac{n}{μ} − \frac{x}{μ})^{2}\}

= \frac{(1 + o(1))}{\sqrt{2πμ}ϕ (\frac{x}{μ})} ∑_{n=0}^{∞} s_{n} exp \{−\frac{1}{2μ}g′′ (\frac{x}{μ}) (\frac{n}{μ} − \frac{x}{μ})^{2}\}

= (1 + o(1)) ∫_{−∞}^{∞} exp \{−\frac{1}{2μ} (\frac{x − t}{ϕ (\frac{x}{μ})})^{2}\} s(t)\frac{dt}{\sqrt{2πμ}ϕ (\frac{x}{μ})}

Next we show that the (N, p^{∗α}, p^{∗β}) -means generalize some important properties of the Euler means.

First we consider the well known product-formula for the Euler-means

E_{α} ∘ E_{β} = E_{α+β}.

This becomes

LEMMA 3

Assume that (p_{n}) and (q_{n}) satisfy (1.1) (with possibly different radii of convergence) and let α, β, γ ∈ N I , α ≤ β.

(i) With r^{∗(α+β)} := p^{∗α} ∗ q^{∗β} , we have

(3.3) (N, p^{∗β}, q^{∗γ}) = (N, p^{∗(β−α)}, r^{∗(α+γ)}) ∘ (N, p^{∗α}, q^{∗γ})

resp. in case (p_{n}) = (q_{n})

(N, p^{∗β}, p^{∗γ}) = (N, p^{∗(β−α)}, p^{∗(α+γ)}) ∘ (N, p^{∗α}, p^{∗γ}).

(ii) If (N, p^{∗(β−α)}, r^{∗(α+γ)}) is regular, then

s_{n} → s(N, p^{∗α}, q^{∗γ}) implies s_{n} → s(N, p^{∗β}, q^{∗γ}) .

Proof: (ii) is a trivial consequence of (i). To prove (i) observe that

(p^{∗(β−α)} ∗ r^{∗(α+γ)})_{n} = (p^{∗β} ∗ q^{∗γ})_{n}

and

\begin{matrix}∑_{k=0}^{n} p_{n−k}^{∗(β−α)}r_{k}^{∗(α+γ)} \frac{1}{r_{k}^{∗(α+γ)}} ∑_{ν=0}^{k} p_{k−ν}^{∗α}q_{ν}^{∗γ}s_{ν} \\ = ∑_{ν=0}^{n} q_{ν}^{∗γ}s_{ν} ∑_{k=0}^{n−ν} p_{n−ν−k}^{∗(β−α)}p_{k}^{∗α} = ∑_{ν=0}^{n} p_{n−ν}^{∗β}q_{ν}^{∗γ}s_{ν}. \end{matrix}

Now s_{n} → s (N, p^{∗(β−α)}, r^{∗(α+γ)}) ∘ (N, p^{∗α}, q^{∗γ}) means that

\frac{1}{(p^{∗(β−α)} ∗ r^{∗(α+γ)})_{n}} ∑_{k=0}^{n} p_{n−k}^{∗(β−α)}r_{k}^{∗(α+γ)} \frac{1}{r_{k}^{∗(α+γ)}} ∑_{ν=0}^{k} p_{k−ν}^{∗α}q_{ν}^{∗γ}s_{ν} → s (n → ∞),

but by the above identities this is the same as

\frac{1}{(p^{∗β} ∗ q^{∗γ})_{n}} ∑_{ν=0}^{n} p_{n−ν}^{∗β}q_{ν}^{∗γ}s_{ν} → s (n → ∞),

which is (N, p^{∗β}, q^{∗γ}) convergence.

A classical result of Knopp [9, Theorem 149] gives a connection between Cesáro convergence with speed and Euler convergence. We generalize this for general (p_{n}) with an additional condition on the sequence (s_{n}) . (In [10, Theorem 2] this generalization is given with an additional condition on the (p_{n}) , but without conditions on the (s_{n}) .)

LEMMA 4

Let (p_{n}) be a sequence of weights satisfying (2.1B) and ϕ(.) as in (2.3). Furthermore assume that s_{n} = O (1) . Then

\frac{1}{n + 1} ∑_{k=0}^{n}(s_{k} + ϵ_{k}) = s + o (\frac{ϕ(n)}{n}) , (n → ∞), with some nullsequence (ϵ_{n})

implies s_{n} → s(N, p^{∗α}, p^{∗β}) for every α, β ∈ N I .

Proof: Since s_{n} = O (1) we can use the asymptotic weights computed in the Corollary to Lemma 1 in the (N, p^{∗α}, p^{∗β}) method. By inclusion we have only to show the implication for the (N, p, p^{∗β}) method. Because of regularity and linearity we can suppose s = 0 and omit the convergent sequence (ϵ_{k}) . Thus the hypothesis becomes

∑_{k=0}^{n} s_{k} = o (ϕ(n)) (n → ∞).

For given ϵ > 0 we can find a N ∈ N I such that for n ≥ l ≥ m ≥ N

| ∑_{k=m}^{l} s_{k}| ≤ ϵϕ(l) ≤ ϵϕ(n),

using also the monotonicity of ϕ(.) . By the Corollary to Lemma 1 and since s_{n} = O (1) we have for the (N, p, p^{∗β}) -transform t_{n}

t_{n} = \sqrt{\frac{β + 1}{2πβ}}ϕ (\frac{n}{β + 1})^{−1} ∑_{|k− \frac{nβ}{β+1} |≤ϵ(n)} exp \{−\frac{β + 1}{2β} (\frac{k − \frac{nβ}{β+1}}{ϕ (\frac{n}{β+1})})^{2}\} s_{k} + o (1) ,

with a function ϵ(.) as in (3.2). So the weights are piecewise monotonic and the maximal weight is for k = \frac{nβ}{β+1} . We therefore split the sum in two parts, namely

t_{n} = ∑_{\frac{nβ}{β+1} −ϵ(n)≤k< \frac{nβ}{β+1}} … + ∑_{\frac{nβ}{β+1} ≤k≤ \frac{nβ}{β+1} +ϵ(n)} … + o (1) .

Using Abels partial summantion and the monotonicity of the weights we find that each of the two sums is bounded by ϵ\frac{ϕ(n)}{ϕ(n/β+1)} . Since ϕ(n/γ) = O (ϕ(n)) for any fixed γ > 0 , we obtain the desired result.

Cesàro-convergence with speed is also connected to the methods of moving-averages by the following

PROPOSITION 3

The following statements are equivalent for a self-neglecting function ϕ(.)

(i) \frac{1}{n + 1} ∑_{k=0}^{n}(s_{k} + ϵ_{k}) = s + o (\frac{ϕ(n)}{n}) (n → ∞) for some ϵ_{n} → 0.

(ii) \frac{1}{uϕ(n)} ∑_{n≤k<n+uϕ(n)} s_{k} → s, ∀u > 0, (n → ∞).

For the proof see [2], for notation and properties of self-neglecting functions consult [3, §2.11].

In the Euler–Borel case we have the identity (B) ∘ (E_{p}) ≈ (B) . A similar identity can be obtained in the general case. For a related calculation compare [7].

LEMMA 5

Assume that (p_{n}) and (q_{n}) satisfy (1.1) with the same radius of convergence R and let α, β ∈ N I then

s_{n} → s(P, q^{∗β}) ⇔ s_{n} → s(P, r^{∗(α+β)}) ∘ (N, p^{∗α}, q^{∗β}).

Proof:

s_{n} → s(P, q^{∗β}) means that \frac{∑_{n=0}^{∞} s_{n}q_{n}^{∗β}x^{n}}{(q(x))^{β}} → s (x → R) ,

and s_{n} → s(P, r^{∗(α+β)}) ∘ (N, p^{∗α}, q^{∗β}) means that

\frac{∑_{n=0}^{∞} (∑_{k=0}^{n} p_{n−k}^{∗α}s_{k}q_{k}^{∗β}) x^{n}}{(p(x))^{α}(q(x))^{β}} → s (x → R).

But

\frac{∑_{n=0}^{∞} s_{n}q_{n}^{∗β}x^{n}}{(q(x))^{β}} = \frac{∑_{n=0}^{∞} (∑_{k=0}^{n} p_{n−k}^{∗α}s_{k}q_{k}^{∗β}) x^{n}}{(p(x))^{α}(q(x))^{β}} ,

and this proves the proposition.

Using Borweins Theorem, i.e. Proposition 1, we obtain

Corollary

If the assumptions of Lemma 5 hold true and if \frac{q_{n}^{∗β}}{r_{n}^{∗(α+β)}} is a totally monotone

sequence, then s_{n} → s (P, q^{∗β}) ⇒ s_{n} → s (P, q^{∗β}) ∘ (N, p^{∗α}, q^{∗β}).

Generalizing Theorem 1 in [10] slightly we obtain the following Tauberian theorem

THEOREM 2

Assume that (p_{n}) satisfies (1.1) and (2.1 B). Then we have under the Tauberian condition s_{n} = O (1) that for any γ ∈ N I

s_{n} → s(P, p^{∗γ}) implies s_{n} → s(N, p^{∗α}, p^{∗β})

for all α, β ∈ N I .

Remark 3:

(i) Under (2.1) (N, p^{∗α}, p^{∗β}) is regular for all α, β ∈ N I .

(ii) s_{n} → s(N, p^{∗α}, p^{∗β}) implies always s_{n} → s(P, p^{∗β}), since

σ_{p^{∗β}}(t) = \frac{∑_{n=0}^{∞} s_{n}p_{n}^{∗β}x^{n}}{(p(x))^{β}} = \frac{∑_{n=0}^{∞} p_{n}^{∗(α+β)} \frac{1}{p_{n}^{∗(α+β)}} (∑_{k=0}^{n} p_{n−k}^{∗α}p_{k}^{∗β}s_{k}) x^{n}}{(p(x))^{α}(p(x))^{β}} ,

and since (P, p^{∗(α+β)}) is regular, the Abelian conclusion follows.

Proof: By Lemma 3 (ii), it is sufficient to consider α = 1 . Define s(u) = s_{[u]} if u ≥ 0 and s(u) = 0 if u < 0 and K(x) = 1/\sqrt{2π} exp\{−x^{2}/2\}.

Since s_{n} = O (1) we have by Lemma 2 (ii), that s_{n} → s(P, p^{∗γ}) implies

(3.4)  lim_{x→∞} ∫_{−∞}^{∞} K (\frac{x − t}{\sqrt{γ}ϕ (\frac{x}{γ})}) s(t) \frac{dt}{\sqrt{γ}ϕ(x/γ)} = s

The conditions of Theorem 1 of [15], i.e. K(x) ∈ L^{1}(−∞, ∞) , the Fourier- transform of K is nonvanishing for any real argument and ϕ(.) is self-neglecting, are trivially satisfied. It follows now from that theorem that if we choose ϵ > 0 and define

H(x) = \{ \begin{matrix}\frac{1}{ϵ}, \\ 0, \end{matrix} if \begin{matrix}x ∈ (−ϵ, 0), \\ x ∉ (−ϵ, 0), \end{matrix}

that

lim_{x→∞} ∫_{−∞}^{∞} H (\frac{x − t}{\sqrt{γ}ϕ(x)}) s(t)\frac{dt}{\sqrt{γ}ϕ(x/γ)} = lim_{x→∞} \frac{1}{ϵ\sqrt{γ}ϕ(x/γ)} ∑_{x≤k<x+ϵ\sqrt{γ}ϕ(x/γ)} s_{k} = s.

Because ϕ(.) is self-neglecting and ϕ(x/γ) = O (ϕ(x)) , for any fixed γ > 0 , we obtain by Proposition 3 , that

\frac{1}{n + 1} ∑_{k=0}^{n}(s_{k} + ϵ_{k}) = s + o (\frac{ϕ(n)}{n}) ,

which in turn by Lemma 4 implies that s_{n} → s(N, p, p^{∗β}) .

4 Proofs

Proof of Theorem 1: Part (i): By Lemma 5:

s_{n} → s(P, p^{∗γ}) ⇔ s_{n} → s(P, p^{∗(α+γ)}) ∘ (N, p^{∗α}, p^{∗γ}).

In case (A) We apply Karamatas Tauberian theorem (observe Lemma 1) (see [2, Theorem 1.7.6 , 18]) and obtain

s_{n} → s(N, 1, p^{∗(α+γ)}) ∘ (N, p^{∗α}, p^{∗γ}).

Since s_{n} = O (1) (N, p^{∗α}, p^{∗γ}) we can use the asymptotic weights and assume w.l.o.g that p_{n}^{∗(β−α)} is nondecreasing and by Theorem 3 in Das [8] we get

s_{n} → s(N, p^{∗(β−α)}, p^{∗(α+γ)}) ∘ (N, p^{∗α}, p^{∗γ}),

which by Lemma 3(i) implies our result. In case (B) Since s_{n} = O (1) (N, p^{∗α}, p^{∗γ}) we can use Theorem 2 to obtain directly

s_{n} → s(N, p^{∗(β−α)}, p^{∗(α+γ)}) ∘ (N, p^{∗α}, p^{∗γ}).

The last step is as above.

Part (ii) is directly implied by part (i) by the Abelian inclusion.

Remark 4: Boos/Tietz [4] gave an alternative proof of Theorem 1 in the Borel-case. The basic steps are as follows ( α = γ = 1, β = 2 )

(i) s_{n} → s(P, p) ⇒ s_{n} → s(P, p^{∗3})(N, p^{∗2}, p)

(ii) (N, p^{∗2}, p) = (N, p, p^{∗2})(N, p, p) .

Hence if (∗) ((N, p, p^{∗2})x)_{n} − ((N, p, p^{∗2})x)_{n−1} = O (1/ϕ(n)) for bounded sequences (x_{n}) , one can use the O-Tauberian theorems in [12,13] to conclude

(iii) s_{n} → s(N, p^{∗2}, p) .

The statement (∗) in (ii) is true for some special cases, like p_{n} = 1/n! , but has not been obtained in general so far.

Proof of Proposition 1: Observe that e.g. in case R = ∞

σ_{p^{∗β}}(x) = \frac{∑_{n=0}^{∞} s_{n} \frac{p_{n}^{∗β}}{p_{n}^{∗α}} p_{n}^{∗α}x^{n}}{(p(x))^{β}} = ∫_{0}^{∞} \frac{p(xt)^{α}}{p(x)^{β}} σ_{p^{∗α}}(xt)dχ(t) = L (σ_{p^{∗α}}(.), x) .

The interchange of integral and sum is allowed because of the absolute convergence for x > 0 . We now follow the arguments in an unpublished paper by A. Jakimovski (oral communication, see also [16] for details.) L(f,x) is a positive linear operator on a linear space of real functions in C[0, ∞) with the properties:

(i) There exists e(t) > 0, e(t) → 1, t → ∞ such that L(e(.), x) → 1, x → ∞, namely e(t) = σ_{p^{∗α}}(t) with the sequence (s_{n}) chosen to be (1, 1, …) .

(ii) There exists some e_{0}(t) > 0 such that L(e_{0}(.), x) → 0, x → ∞, namely e_{0}(t) = σ_{p^{∗α}}(t) = p_{0}^{∗α}/p(t)^{α}, with the sequence (s_{n}) chosen to be (1, 0, 0, …) .

From (i) and the assumptions we find

|f(t) − se(t)| < ϵ/2 ≤ ϵe(t), for t ≥ t_{0}(ϵ),

and by (ii)

|f(t) − se(t)| ≤ M ≤ \frac{M}{m} e_{0}(t), t ∈ [0, t_{0}(ϵ)],

with suitable M, m. Hence for t ≥ 0 :

|f(t) − se(t)| ≤ ϵe(t) + \frac{M}{m} e_{0}(t).

Since L is linear and positive we obtain that L(f(.), x) → s if f(x) → s, which yields the desired result.

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