定理(英文描述) Let I be an interval of the real line and let f denote a real-valued, convex function defined on I. If x1,...,xn and y1,...,yn are numbers in I such that (x1,...,xn)majorizes(y1,...,yn), then f(x1)+⋯+f(xn)≥f(y1)+⋯+f(yn). (1)
Here majorization means that x1,...,xn and y1,...,yn satisfies x1≥x2≥⋯≥xn and y1≥y2≥⋯≥yn and we have the inequalities x1+⋯+xi≥y1+⋯+yi for all i∈1,...,n−1. and the equality x1+⋯+xn=y1+⋯+yn If f is a strictly convex function, then the inequality (1) holds with equality if and only if we have xi=yi for all i∈1,...,n.
证明: The inequality is written as f(a,x)+g(a,x)>0 where f(a,x)=−x1+a+(x−1)1+a+(x−1/2)1+a−(x−3/2)1+aand g(a,x)=(1/2−2a−1)((x−3/2)a−(x−1/2)a).Clearly, g(a,x)>0. It suffices to prove that f(a,x)≥0. Since x↦−x1+a is convex on (0,∞), and (x,x−3/2) majorizes (x−1/2,x−1), by using Karamata inequality, we have f(a,x)≥0. We are done. 2020年4月22日最后修改