Generalization of Abel's theorem
K. V. Borozdin
Submitted 1961-01-01 | SovietRxiv: ru-196101.97922 | Translated from Russian

Abstract Generated abstract

This paper extends a theorem of A. G. Postnikov by proving two Abel-type limit theorems for weighted power series. The first shows that if a series converges and the function x f(x) is continuous and of bounded variation on [0,1], then the transformed series with terms a_i x^i f(x^i) tends to the ordinary sum multiplied by f(1). The second treats sequences whose partial sums have prescribed asymptotic growth of order n^\alpha, and proves a normalized limit expressed as an integral with kernel (ln 1/x)^{\alpha-1}. The proof uses decomposition of bounded-variation functions, Abel’s lemma, a Tauberian comparison with model coefficients, and a distribution-function argument based on Helly’s theorem.

Full Text

Mathematics

K. V. Borozdin

A Generalization of Abel’s Theorem

(Presented by Academician I. M. Vinogradov on XII 1, 1960)

In the present work two theorems are given which extend a theorem of A. G. Postnikov (¹).

Theorem 1. Let a convergent series \(\sum_{i=0}^{\infty} a_i = A\) be given, and let \(f(x)\) be such that the function \(x f(x)\) is a continuous function of bounded variation on \([0,1]\). Then

\[ \lim_{x\to 1-0}\sum_{i=0}^{\infty} a_i x^i f(x^i)=A f(1). \tag{1} \]

Proof. A continuous function with bounded variation is representable as the difference of two continuous nondecreasing functions. Therefore
\(x f(x)=g(x)-h(x)\), where the functions \(g(x)\) and \(h(x)\) on the interval \([0,1]\) are positive, continuous, and nondecreasing. The series \(\sum a_i\) is convergent; consequently, for any \(\varepsilon>0\) there exists an \(N\) such that, for \(N<n\le p\), one always has

\[ \left|\sum_{i=n}^{p} a_i\right|<\varepsilon . \]

The sequences \(\{g(x^i)\}\) and \(\{h(x^i)\}\) do not increase on the interval \([0,1]\). Applying Abel’s lemma to these two sequences (in turn) and to the series \(\sum a_i\) gives

\[ |a_n g(x^n)+a_{n+1}g(x^{n+1})+\cdots+a_p g(x^p)|<\varepsilon g(1), \]

\[ |a_n h(x^n)+\cdots+a_p h(x^p)|<\varepsilon h(1). \]

Consider the sum \(\sum_{i=n}^{p} a_i x^i f(x^i)\):

\[ |a_n x^n f(x^n)+\cdots+a_p x^p f(x^p)| \le |a_n g(x^n)+\cdots+a_p g(x^p)|+ \]

\[ \quad + |a_n h(x^n)+\cdots+a_p h(x^p)| <\varepsilon\bigl(g(1)+h(1)\bigr). \]

Thus, for the series of continuous functions \(\sum a_i x^i f(x^i)\) on the interval \([0,1]\), the conditions of the Cauchy criterion for uniform convergence are satisfied; consequently,

\[ \Phi(x)=\sum_{i=0}^{\infty} a_i x^i f(x^i) \]

is continuous on \([0,1]\). Therefore

\[ \lim_{x\to 1-0}\Phi(x)=\Phi(1)=A f(1), \]

which was required to prove.

Theorem 2. Let \(\alpha>0\) be given, and let

\[ \sum_{k=0}^{n} a_k \sim \frac{n^\alpha}{\Gamma(1+\alpha)}, \quad \text{where } n \to \infty . \tag{2} \]

Let a function \(f(x)\) be given such that, for some \(\beta \in [0,1)\), the expression \(x^\beta f(x)\) represents on the interval \([0,1]\) a continuous function of bounded variation. Then

\[ \lim_{x\to 1-0} (1-x)^\alpha \sum_{k=0}^{\infty} a_k x^k f(x^k) = \frac{1}{\Gamma(\alpha)} \int_{0}^{1} f(x)\left(\ln \frac{1}{x}\right)^{\alpha-1}\,dx . \tag{3} \]

In the proof of this theorem we shall need a lemma.

Lemma. Let \(\alpha>0\) be given, and let \(0 \le \beta < 1\); let the function \(\varphi(x)\) be continuous on \([0,1]\). Then

\[ \lim_{x\to 1-0} (1-x)^\alpha \sum_{k=1}^{\infty} \frac{k^{\alpha-1}}{\Gamma(\alpha)} (x^{1-\beta})^k \varphi(x^k) = \frac{1}{\Gamma(\alpha)} \int_{0}^{1} x^{-\beta}\varphi(x) \left(\ln \frac{1}{x}\right)^{\alpha-1}\,dx . \tag{4} \]

Proof of the lemma. For \(|x|<1\) the expansion

\[ \frac{1}{(1-x)^\alpha} = 1+\alpha x+\frac{\alpha(\alpha+1)}{1\cdot 2}x^2+\cdots+ \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)\Gamma(n+1)}x^n+\cdots \tag{5} \]

is valid. It is easy to establish that, as \(n\to\infty\),

\[ \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)\Gamma(n+1)} \sim \frac{n^{\alpha-1}}{\Gamma(\alpha)} \left(1+O\left(\frac{1}{n}\right)\right). \tag{6} \]

From (6) and (5) it follows that, for the function
\(\Phi_\alpha(x)=\sum_{n=1}^{\infty}\frac{n^{\alpha-1}}{\Gamma(\alpha)}x^n\), where \(x\in[0,1]\), the relation

\[ \lim_{x\to 1-0} (1-x)^\alpha \Phi_\alpha(x)=1 \tag{7} \]

holds.

Consider the one-parameter family of distribution functions \(F_x(u)\), where the parameter \(x\in(0,1)\), defined as follows:

a) at the points \(u=x^n\), where \(n=1,2,\ldots\), \(F_x(u)\) has jumps equal to

\[ \frac{1}{\Phi_\alpha(x^{1-\beta})}\, \frac{n^{\alpha-1}}{\Gamma(\alpha)} (x^{1-\beta})^n; \]

b) at the remaining points of the interval \(u\in(-\infty,\infty)\) the function \(F_x(u)\) is constant and, in particular, \(F_x(0)=0\). The jumps of the function \(F_x(u)\) occur only on the interval \((0,1)\); their sum is

\[ \sum_{n=1}^{\infty} \frac{1}{\Phi_\alpha(x^{1-\beta})}\, \frac{n^{\alpha-1}}{\Gamma(\alpha)} (x^{1-\beta})^n =1. \]

We shall show that \(F_x(u)\to F(u)\), where \(F(u)\) is also a distribution function; since outside the interval \((0,1)\) all the \(F_x(u)\) coincide, only this interval needs to be investigated.

\[ F_x(u)= \sum_{x^n\le u} \frac{1}{\Phi_\alpha(x^{1-\beta})}\, \frac{n^{\alpha-1}}{\Gamma(\alpha)} (x^{1-\beta})^n = \frac{1}{\Phi_\alpha(x^{1-\beta})\Gamma(\alpha)} \sum_{n\ge \ln u/\ln x} n^{\alpha-1}(x^{1-\beta})^n . \tag{8} \]

It can be proved that, as \(x \to 1-0\),

\[ \sum_{n=[\ln u/\ln x+1]}^\infty n^{\alpha-1}\left(x^{1-\beta}\right)^n = \int_{\ln u/\ln x}^{\infty} t^{\alpha-1}\left(x^{1-\beta}\right)^t\,dt + O\left(\left(\ln \frac1x\right)^{1-\alpha}\right), \]

and, if one takes into account the well-known relations

\[ \ln \frac1x \sim 1-x,\qquad 1-x^{1-\beta}\sim (1-\beta)(1-x), \]

then (8) gives

\[ F(u)=\lim_{x\to 1-0}F_x(u) = \frac{(1-\beta)^\alpha}{\Gamma(\alpha)} \int_0^u v^{1-\beta}\left(\ln \frac1v\right)^{\alpha-1}\,dv . \tag{9} \]

It is obvious that, for the family \(F_x(u)\), the conditions of Helly’s second theorem (2) are satisfied and, consequently, the following relation holds:

\[ \lim_{x\to 1-0}\int_0^1 \varphi(u)\,dF_x(u) = \int_0^1 \varphi(u)\,dF(u). \tag{10} \]

Relation (4) is a consequence of relations (10), (8), and (9).

Proof of Theorem 2. The function \(\varphi(x)=x^\beta f(x)\), as a continuous function of bounded variation, can be represented in the form
\(\varphi(x)=\varphi_1(x)-\varphi_2(x)\), where \(\varphi_1(x)\) and \(\varphi_2(x)\) on the interval \([0,1]\) are continuous, nonnegative, and nondecreasing.

It is obvious that, for \(\alpha>0\) and \(n\to\infty\),

\[ \sum_{k=1}^n \frac{k^{\alpha-1}}{\Gamma(\alpha)} \sim \frac{n^\alpha}{\Gamma(\alpha+1)} . \tag{11} \]

Introduce the notation:

\[ s_n=\sum_{k=0}^n a_k,\qquad \psi_n=\sum_{k=1}^n \frac{k^{\alpha-1}}{\Gamma(\alpha)} . \tag{12} \]

From (2), (11), and (12) it follows that

\[ s_n\sim \psi_n . \tag{13} \]

Fix \(\varepsilon>0\); for \(n>N(\varepsilon)\), by virtue of (13), the following holds:

\[ |s_n-\psi_n|<\varepsilon\psi_n . \tag{14} \]

The right-hand sides of (3) and (4) are identical (since \(\varphi(x)=x^\beta f(x)\)); let us prove the equivalence of the left-hand sides as \(x\to 1-0\). Denote

\[ W[\varphi_i(x)] = \left| (1-x)^\alpha \sum_{k=0}^\infty a_k\left(x^{1-\beta}\right)^k\varphi_i(x^k) - \right. \]

\[ \left. - (1-x)^\alpha \sum_{k=1}^\infty \frac{k^{\alpha-1}}{\Gamma(\alpha)} \left(x^{1-\beta}\right)^k\varphi_i(x^k) \right|; \]

then

\[ W[\varphi(x)]\leq W[\varphi_1(x)]+W[\varphi_2(x)], \tag{15} \]

\[ W[\varphi_1(x)] = (1-x)^\alpha \left| s_0\varphi_1(1) + \sum_{k=1}^\infty (s_k-s_{k-1})\left(x^{1-\beta}\right)^k\varphi_1(x^k) - \right. \]

\[ \left. - \sum_{k=1}^\infty (\psi_k-\psi_{k-1})\left(x^{1-\beta}\right)^k\varphi_1(x^k) \right| = (1-x)^\alpha \left| \sum_{k=0}^\infty (s_k-\psi_k)\left\{\left(x^{1-\beta}\right)^k\varphi_1(x^k)-\right. \]

\[ -\left(x^{1-\beta}\right)^{k+1}\varphi_1\left(x^{k+1}\right)\right\} \leq (1-x)^\alpha \left|\sum_{k=0}^{N}(s_k-\psi_k)\left\{\left(x^{1-\beta}\right)^k\varphi_1\left(x^k\right)-\right.\right. \]

\[ \left.\left.-\left(x^{1-\beta}\right)^{k+1}\varphi_1\left(x^{k+1}\right)\right\}\right| +\varepsilon(1-x)^\alpha \sum_{N+1}^{\infty}\psi_k \left|\left(x^{1-\beta}\right)^k\varphi_1\left(x^k\right) -\left(x^{1-\beta}\right)^{k+1}\varphi_1\left(x^{k+1}\right)\right|. \]

On the interval \([0,1]\) the function \(\varphi_1(x)\) is nonincreasing. Therefore

\[ W[\varphi_1(x)] < (1-x)^\alpha \left|\sum_{k=0}^{N}(s_k-\psi_k)\left\{\left(x^{1-\beta}\right)^k\varphi_1\left(x^k\right) -\left(x^{1-\beta}\right)^{k+1}\varphi_1\left(x^{k+1}\right)\right\}\right|+ \]

\[ +\varepsilon(1-x)^\alpha\psi_{N+1}\left(x^{1-\beta}\right)^{N+1}\varphi_1\left(x^{N+1}\right) +\varepsilon(1-x)^\alpha\sum_{k=1}^{\infty} \frac{k^{\alpha-1}}{\Gamma(\alpha)} \left(x^{1-\beta}\right)^k\varphi_1\left(x^k\right). \]

Here the first two terms on the right-hand side tend to \(0\) as \(x\to 1\), and to the third we apply the lemma (formula (4)):

\[ \varlimsup_{x\to 1-0} W[\varphi_1(x)] \leq \frac{\varepsilon}{\Gamma(\alpha)} \int_{0}^{1} x^{-\beta}\varphi_1(x)\left(\ln \frac{1}{x}\right)^{\alpha-1}\,dx =\varepsilon M_1. \]

An analogous inequality is established also for \(W[\varphi_2(x)]\). Denote \(M=M_1+M_2\); then \(\varlimsup_{x\to 1-0} W[\varphi(x)]\leq \varepsilon M\), where \(M\) is constant and \(\varepsilon\) is arbitrarily small. Consequently,

\[ \lim_{x\to 1-0}(1-x)^\alpha\sum_{k=0}^{\infty}a_k x^k f\left(x^k\right) = \lim_{x\to 1-0}(1-x)^\alpha\sum_{k=1}^{\infty} \frac{k^{\alpha-1}}{\Gamma(\alpha)} \left(x^{1-\beta}\right)^k\varphi\left(x^k\right) = \]

\[ =\frac{1}{\Gamma(\alpha)} \int_{0}^{1} f(x)\left(\ln \frac{1}{x}\right)^{\alpha-1}\,dx, \]

which was required to be proved.

Steklov Mathematical Institute
Academy of Sciences of the USSR

Received
30 XI 1960

REFERENCES

¹ A. G. Postnikov, DAN, 96, No. 5, 913 (1954). ² V. I. Glivenko, The Stieltjes Integral, Moscow–Leningrad, 1936.

Submission history

Generalization of Abel's theorem