Seite - 259 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 277 Proof. Weprove (a). Let (φnk)kbeaconvergent subsequenceof (φk)kwhichconverges toφ∞. First, φ∞∈Φ0,becauseΦ0 is closedandthesubsequence (φnk) isasequenceofelementsofΦ0 (provedin Proposition1b). Letusnowshowthat thesubsequence (φnk+1)alsoconverges toφ∞.Wesimplyhave: ‖φnk+1−φ∞‖ ≤ ‖φnk−φ∞‖+‖φnk+1−φnk‖. Sinceφk+1−φk→0andφnk→φ∞,weconcludethatφnk+1→φ∞. By definition of φnk+1, it verifies the infimum in recurrence (11), so that the gradient of the optimizedfunction iszero: ∇Dˆϕ(pφnk+1,pφT)+∇Dψ(φnk+1,φnk)=0. Using the continuity assumptionsA1andACof thegradients, one canpass to the limitwith noproblem: ∇Dˆϕ(pφ∞,pφT)+∇Dψ(φ∞,φ∞)=0. However, thegradient∇Dψ(φ∞,φ∞)=0because (recall thatψ′(1)=0) foranyφ∈Φ ∇Dψ(φ,φ)= n ∑ i=1 ∫ X ∇hi(x|φ) hi(x|φ) ψ ′ ( hi(x|φ) hi(x|φ) ) hi(x|φ)dx= n ∑ i=1 ∫ X ∇hi(x|φ)ψ′(1)dx, which isequal tozerosinceψ′(1)=0. This implies that∇Dˆϕ(pφ∞,pφT)=0. Weprove(b).Weuseagainthedefinitionofthearginf.Astheoptimizedfunctionisnotnecessarily differentiableat thepointsof thesequence(φk)k, anecessaryconditionforφk+1 tobeaninfimumisthat 0belongs to thesubgradientof the functiononφk+1. SinceDψ(φ,φk) isassumedtobedifferentiable, theoptimalitycondition is translated into: −∇Dψ(φk+1,φk)∈∂Dˆϕ(pφk+1,pφT) ∀k. Since Dˆϕ(pφ,pφT) is continuous, then its subgradient isoutersemicontinuous (see [13]Chapter8, Proposition 7). We use the same arguments presented in (a) to conclude the existence of two subsequences (φnk)k and (φnk+1)k which converge to the same limit φ∞. By definition of outer semicontinuity,andsinceφnk+1→φ∞,wehave: limsup φnk+1→φ∞ ∂Dˆϕ(pφnk+1,pφT)⊂∂Dˆϕ(pφ∞,pφT). (12) Wewant toprove that0∈ limsup φnk+1→φ∞∂Dˆϕ(pφnk+1,pφT). Bydefinitionof the (outer) limsup (see [13]Chapter4,Definition1orChapter5B): limsup φ→φ∞ ∂Dˆϕ(pφ,pφT)= { u|∃φk→φ∞,∃uk→uwithuk∈∂Dˆϕ(pφk,pφT) } . Inourscenario,φ=φnk+1,φk=φnk+1,u=0anduk=∇1Dψ(φnk+1,φnk). Thecontinuityof∇1Dψ withrespect tobothargumentsandthefact that the twosubsequencesφnk+1 andφnk converge to the same limit, imply thatuk →∇1Dψ(φ∞,φ∞) = 0. Hence,u= 0∈ limsupφnk+1→φ∞∂Dˆϕ(pφnk+1,pφT). By inclusion(12),wegetourresult: 0∈∂Dˆϕ(pφ∞,pφT). Thisends theproof. Theassumption{φk+1−φk}→0used inProposition2 isnoteasy tobecheckedunlessonehasa close formulaofφk. Thefollowingpropositiongivesamethodtoprovesuchassumption. Thismethod 259