Vol. 80, No. 1, 1979

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Solution for an integral equation with continuous interval functions

John Albert Chatfield

Vol. 80 (1979), No. 1, 47–57
Abstract

Suppose R is the set of real numbers and all integrals are of the subdivision-refinement type. Suppose each of G and H is a function from R ×R to R and each of f and h is a function from R to R such that f(a) = h(a), dh is of bounded variation on [a,x], and axH2 = axG2 = 0 for x > a. The following two statements are equivalent:

(1) If x > a, then f is bounded on [a,x], axH exists, axG exists, (RL) ax(fG + fH) exists, and

                 ∫
x
f(x) = h(x)+ (RL ) a (fG + fH );

(2) If a p < q x, then each of p Πq(1 + H) and p Πq(1 G)1 exists and neither is zero,

   ∫ x
(R)   [⋅tΠx(1 +H )(1+ G)][(1− G )− 1]dh
a

exists, and

f(x) = f(a)a Πx(1 + H)(1 G)1
+ (R) ax[ t Πx(1 + H)(1 + G)][(1 G)1]dh.

Mathematical Subject Classification 2000
Primary: 45N05
Secondary: 26A42, 44A40
Milestones
Received: 16 July 1976
Published: 1 January 1979
Authors
John Albert Chatfield