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.