Page - 274 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 407 Notall functionsϕ:R→ (0,∞), forwhichconditions (a1)and(a2)hold, satisfycondition(a3). Suchafunction isgivenbelow. Example2. Assumethat theunderlyingmeasureμ isσ-finite andnon-atomic. This is the case of theLebesgue measure. Letusconsider the function ϕ(u)= { e(u+1) 2/2, u≥0, e(u+1/2), u≤0, (3) which clearly is convex, and satisfies the limits limu→−∞ϕ(u) = 0 and limu→∞ϕ(u) =∞. Given any measurable functionu0: T→ (0,∞), wewill find ameasurable function c: T→Rwith ∫ T ϕ(c)dμ<∞, forwhichexpression (2) isnot satisfied. For eachm≥1,wedefine vm(t) := ( m log(2) u0(t) − u0(t) 2 −1 ) 1Em(t), whereEm={t∈T :mlog(2)u0(t) − u0(t) 2 −1>0}. Becausevm ↑∞,wecanfindasub-sequence{vmn} such that∫ Emn e(vmn+u0+1) 2/2dμ≥2n. According to (Lemma8.3 in [29]) , there exists a sub-sequencewk = vmnk and pairwise disjoint sets Ak⊆Emnk forwhich ∫ Ak e(wk+u0+1) 2/2dμ=1. Letusdefine c= c1T\A+∑∞k=1wk1Ak,whereA= ⋃∞ k=1Ak andc is anymeasurable functionsuch that ϕ(c(t))>0 for t∈T\Aand∫T\A ϕ(c)dμ<∞. Observing that e(wk(t)+u0(t)+1) 2/2=2mnke(wk(t)+1) 2/2, for t∈Ak, weget ∫ Ak e(wk+1) 2/2dμ= 1 2mnk , for everym≥1. Then,wecanwrite ∫ T ϕ(c)dμ= ∫ T\A ϕ(c)dμ+ ∞ ∑ k=1 ∫ Ak e(wk+1) 2/2dμ = ∫ T\A ϕ(c)dμ+ ∞ ∑ k=1 1 2mnk <∞. Ontheotherhand,∫ T ϕ(c+u0)dμ= ∫ T\A ϕ(c)dμ+ ∞ ∑ k=1 ∫ Ak e(u0+wk+1) 2/2dμ = ∫ T\A ϕ(c)dμ+ ∞ ∑ k=1 1=∞, whichshows that (2) isnot satisfied. 274