]> No Title

a a a AND and and and and and and and, A. Apply apply assumptionbe be be between

Jrase consisting contained contrary

defined diagonal

equal equality every every existence existence exists

family following For For for for for

have have

If If If implies in including is It

JENKINS J. justifyK.

Lemma Lemma Let let limit limit limits: limits.

module much

ε(b)<ξ(b), ξ(b)>Reζn+η.

bbα0bγJμjn, ξ2γjċ-n=1ε>0, γj-1γ(b)γ(c)Rj, b<b, ξ2>ξ*α0=-,x0>-, α0>-. α0>-. γ(b). j=1, ..., n,

V(c*), γ1, ..., γn, γn+1=γ(b)

ιr(a0, b)-π-1ξ(b)μ(a0, b)-π-1ξ(b), μ(b, b)μ¯(b, b)π-1(ξ(b)-ξ(b))+2

u(a0, b)-1πξ(b)μ(a0, c*)-1πξn(c*)+2-nf(η3),

lim(μ(a0, b)-1πξ(b))=limb(μ(a0, b)-1πξκ(b)).

varlimsup(μ(a0, b)-1πξn(b))=varlimsupb(μ(a0, b)-1πξ(b))

u(a0, b)-μ(a0, b)=μ(b, b)μ¯(b, b)π-1(ξ(b)-ξ(b)).

varliminf(μ(a0, b)-1πξn(b))=varliminfb(μ(a0, b)-1πξ(b)),

varlimsup(μ(a0, b)-1πξ(b))varliminfb(μ(a0, b)-1πξ(b)).

varlimsup(μ(a0, b)-1πξ(b)-2)varliminfb(μ(a0, b)-1πξm(b)).

u¯(c*, b)=j=1n+1μj1π(ξ(b)-ξ(c*))+2-j=1nf(ξj-ξj),


obtain obtain of of of of of of of OIlIAWA

16'. 17. 18. 3, 3, 48 (il) (22) (22) (24), (27) (27) (27)

Proof prove


3atisfy show show so so such suffices sufficient

take that that that that, The the the the the the the the the the the the the thererherefore therefore this those to to to to to to to trivially


We We we we which which will with with