On the equality problem of generalized Bajraktarevi\u0107 means

Abstract

The purpose of this paper is to investigate the equality problem of generalized Bajraktarević means, i.e., to solve the functional equation *f(-1)(p1(x1)f(x1)+⋯+pn(xn)f(xn)p1(x1)+⋯+pn(xn))=g(-1)(q1(x1)g(x1)+⋯+qn(xn)g(xn)q1(x1)+⋯+qn(xn)),\\documentclass[12pt]{minimal}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\begin{document}$$\\begin{aligned} f^{(-1)}\\bigg (\\frac{p_1(x_1)f(x_1)+\\dots +p_n(x_n)f(x_n)}{p_1(x_1)+\\dots +p_n(x_n)}\\bigg )=g^{(-1)}\\bigg (\\frac{q_1(x_1)g(x_1)+\\dots +q_n(x_n)g(x_n)}{q_1(x_1)+\\dots +q_n(x_n)}\\bigg ), \\end{aligned}$$\\end{document}which holds for all (x_1,dots ,x_n)in I^n, where nge 2, I is a nonempty open real interval, the unknown functions f,g:Irightarrow {mathbb {R}} are strictly monotone, f^{(-1)} and g^{(-1)} denote their generalized left inverses, respectively, and p=(p_1,dots ,p_n):Irightarrow {mathbb {R}}_{+}^n and q=(q_1,dots ,q_n):Irightarrow {mathbb {R}}_{+}^n are also unknown functions. This equality problem in the symmetric two-variable (i.e., when n=2) case was already investigated and solved under sixth-order regularity assumptions by Losonczi (Aequationes Math 58(3):223–241, 1999). In the nonsymmetric two-variable case, assuming the three times differentiability of f, g and the existence of iin {1,2} such that either p_i is twice continuously differentiable and p_{3-i} is continuous on I, or p_i is twice differentiable and p_{3-i} is once differentiable on I, we prove that (*) holds if and only if there exist four constants a,b,c,din {mathbb {R}} with adne bc such that cf+d>0,g=af+bcf+d,andqℓ=(cf+d)pℓ(ℓ∈{1,⋯,n}).\\documentclass[12pt]{minimal}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\begin{document}$$\\begin{aligned} cf+d>0,\\qquad g=\\frac{af+b}{cf+d},\\qquad \\text{ and }\\qquad q_\\ell =(cf+d)p_\\ell \\qquad (\\ell \\in \\{1,\\dots ,n\\}). \\end{aligned}$$\\end{document}In the case nge 3, we obtain the same conclusion with weaker regularity assumptions. Namely, we suppose that f and g are three times differentiable, p is continuous and there exist i,j,kin {1,dots ,n} with ine jne kne i such that p_i,p_j,p_k are differentiable.