University of Waterloo, Waterloo, ON, Canada N2L 3G1; and University of California, Irvine, CA, 92697-5100, USA.
Proc Natl Acad Sci U S A. 2002 Apr 16;99(8):5212-6. doi: 10.1073/pnas.082107799.
Two functional equations are considered that are motivated by three considerations: work in utility theory and psychophysics, questions concerning when pairs of degree 1 homogeneous functions can be homomorphic and calculating their homomorphisms, and the link of the latter questions to quasilinear mean values. The first equation is h(σ(y)x + [1 - σ(y)]y) = τ(y)h(x) + [1 - τ(y)]h(y) (x ≥ y ≥ 0), where h maps [0, ∞[into a subset of [0, ∞[and is strictly increasing and continuously differentiable; the functions σ and τ map [0, ∞[continuously into [0,1], σ(y) > 0 for y > 0 but σ is not 1 on]0, ∞[. The solutions are fully determined. (Recently Zsolt Páles has eliminated the differentiability assumption.) The second equation is h[y + f(x - y)] = h(y) + g[h(x) - h(y)] (x ≥ y ≥ 0), where h maps [0, ∞[onto a subinterval of positive length of [0, ∞[and is strictly increasing and twice continuously differentiable, f and g map [0, ∞[onto[0, ∞[and are twice differentiable, and either f"(0) ≠ 0 or g"(0) ≠ 0. The solutions are fully determined under these conditions. When f"(0) = g"(0) = 0 and h" is not identically zero, we determine the solutions under the added assumption of analyticity. It remains an open problem to find the solutions in the latter case under the assumption of only second order differentiability. A more general open problem is to eliminate all differentiability conditions for the second equation.
考虑了两个泛函方程,它们是由以下三个方面的考虑所驱动的:效用理论和心理物理学中的工作、关于一对次数为 1 的齐次函数何时可以同态以及计算它们的同态的问题,以及后一个问题与拟线性均值的联系。第一个方程是 h(σ(y)x + [1 - σ(y)]y) = τ(y)h(x) + [1 - τ(y)]h(y) (x ≥ y ≥ 0),其中 h 将 [0, ∞[映射到 [0, ∞[的一个子集,并且是严格递增和连续可微的;函数 σ 和 τ 将 [0, ∞[连续映射到 [0,1],σ(y) > 0 对于 y > 0,但 σ 不在]0, ∞[上是 1。解是完全确定的。(最近 Zsolt Páles 消除了可微性假设。)第二个方程是 h[y + f(x - y)] = h(y) + g[h(x) - h(y)] (x ≥ y ≥ 0),其中 h 将 [0, ∞[映射到 [0, ∞[的一个正长度子区间,并且是严格递增和二阶连续可微的,f 和 g 将 [0, ∞[映射到[0, ∞[并且是二阶可微的,并且 f"(0) ≠ 0 或 g"(0) ≠ 0。在这些条件下,解是完全确定的。当 f"(0) = g"(0) = 0 并且 h"不是恒等于零的时,我们在解析性假设下确定解。在仅二阶可微性假设下找到后一种情况下的解仍然是一个悬而未决的问题。一个更一般的未决问题是消除第二个方程的所有可微性条件。