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8
MohamedAkkouchi
Theorem1.Let(X,d)beametricspaceandT:X→Xbea
self-mappingsatisfyingtheinequality
(6)
d(Tx,Ty)≤φ(max{d(x,y),d(Tx,x),d(Ty,y),d(Tx,y),d(Ty,x)})
forallx,y∈X,whereφisagivenelementinA.
Supposethat(X,d)isT−orbitallycomplete.Then,wehave:
(i)ThasauniquefixedpointzinX.
(ii)ThefixedpointproblemofTiswell-posed.
(iii)Tiscontinuousatitsuniquefixedpoint.
Proof.(i)LetxobeanarbitrarypointinX.WeconsiderthePicard
sequenceassociatedtoxo.Thatisthesequence{xn}definedbyxn+1:=
Tnxo=T(xo),foreverynonnegativeintegern.
WestartbyshowingthatthePicardsequence{xn}isaCauchysequence.
Foreachnonnegativeintegern,weconsiderthesetOT(xo,n):={Tjxo:
0≤j≤n}.Weobservethat
diam(OT(xo))=sup{diam(OT(xo,n)):n≥0},
whereforanysubsetAofX,wedenotediam(A)tomeanthediameterofA.
By(i)ofLemma2,weknowthat
(7)
diam(OT(T
mxo,n))≤φm(diam(OT(xo,n+m)))
holdstrueforanypositiveintegersnandm.
Letn≥1.Forallintegersi,jsuchthat1≤i,j≤n,by(6),wehave
(8)
d(Tśx,Tjx)=d(T(Tś11x),T(Tj11x))
≤φ(max{d(Tś11x,Tj11x),d(Tś11x,Tśx),
d(Tj11x,Tjx),d(Tśx,Tj11x),d(Tjx,Tś11x)})
≤φ(diam(OT(xo,n))).
From(8),wededucethatthereexistsksuchthat1≤k≤nand
diam(OT(xo,n))=d(xo,T
kxo).
Then
diam(OT(xo,n))=d(xo,T
kxo)≤d(xo,Txo)+d(Txo,Tkxo)
≤d(xo,Txo)+diam(OT(Txo,n−1)).
Takingintoaccounttheinequality(i)ofLemma2,weobtainthat
diam(OT(xo,n))≤d(xo,Txo)+φ(diam(OT(xo,n))).
