# Equality of two variable means revisited

Original Paper

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## Summary.

Let be given continuous functions such that where the integrals are Riemann – Stieltjes ones. From the mean value theorem it follows that
i.e.

*I*be an interval,$$ f,g:I \to \mathbb{R} $$

*g*(*x*)≠ 0 for*x*∈*I*and*h*(*x*) : =*f*(*x*)/*g*(*x*) (*x*∈*I*) is strictly monotonic (thus invertible) on*I*. Let further μ be an increasing non-constant function on [0, 1]. We consider the function$$ M_{{f,g,\mu }} (x,y): = h^{{ - 1}} {\left( {\frac{{{\int\limits_0^1 {f(tx + (1 - t)y)d\mu (t)} }}} {{{\int\limits_0^1 {g(tx + (1 - t)y)d\mu (t)} }}}} \right)}(x,y \in I) $$

$$ \min \{ x,y\} \leqslant M_{{f,g,\mu }} (x,y) \leqslant \max \{ x,y\} $$

*M*_{f, g, μ}is a mean on*I*. With suitable choice of μ we can get from it the quasi-arithmetic mean weighted by a weight function, and also the Cauchy or difference mean.The equality problem for these two classes of means has been solved in [22], [28]. The aim of this paper is twofold. First, we solve these equality problems in a unified way, second we get rid of the inconvenient conditions (vanishing or not vanishing of some functions) posed in the previous papers.

### Mathematics Subject Classification (2000).

Primary 39B22### Keywords.

Divided differences mean value functional equation## Preview

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## Copyright information

© Birkhäuser Verlag, Basel 2006