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CONVERGENCEOFCONZE-LESIGNEAVERAGES
BERNARDHOSTANDBRYNAKRA
Abstract.Westudytheconvergenceof1
NPf1(Ta1nx)f2(Ta2nx)f3(Ta3nx),
1.Introduction
form lim N!11 NN1X n=0f
1(Ta1nx)f2(Ta2nx):::f`(Ta`nx);(1)
f andConzeandLesigne[5]. lim N!11 NN1X n=0f
1(Tnx)f2(Tn2x):
generalexpressions.
InSection4.3,weshow:
1
2BERNARDHOSTANDBRYNAKRA
distinctintegersandf1;f2;f32L1().Then lim N!11 NN1X n=03 Y i=1f i(Tainx) existsinL2().
ThisiscarriedoutinSection2.
Section4.4,wegivetheactualformula.
2.Reductiontoasimplersystem
usingergodicdecomposition.
2.1.Characteristicfactors.
a withE(fijY)=0,thelimit lim N!11 NN1X n=0` Y i=1f iTain existsinL2()andisequalto0. lim N!1 1 NN1X n=0` Y i=1f iTain1NN1X n=0` Y i=1E(fiTainjY)! =0 inL2().
CONVERGENCEOFCONZE-LESIGNEAVERAGES3
Yhasasimpleform.
wewrite~fthefunctiononZdenedby~f=E(fjZ). anddeduced: L 1() lim N!11 NN1X n=0f
1(Tb1nx)f2(Tb2nx)
existsinL2()andequalsZ Z ~f1(z+b1)~f2(z+b2)dm(); wherez=(x). theclosedsubgroup
Z=(z+a1t;z+a2t;z+a3t):z;t2Z
ByTheorem2.1,forf1;f2;f32L1(),
lim N!11 NN1X n=0Z X3 Y i=1f i(Tainx)dm(x)=Z 3Y i=1~ fi(z+ai)dm(z)dm() Z Z3 Y i=1~ fi(zi)d~m(~z):(2)
Writing
~=(a1;a2;a3); thetransformation~Sisgivenby
S~z=~z+~:
Weiss[7].)
4BERNARDHOSTANDBRYNAKRA
Usingthis,werewriteEquation(2)andhave
lim N!11 NN1X n=0Z X3 Y i=1f i(Tainx)d(x)=Z X3 Y i=1f i(xi)d~(x1;x2;x3): B
IfH=L,wesaythatXisagroupextensionofY.
alsoergodic. 1 NN1X n=03 Y i=1f i(Taix) convergeto0inL2(). u schemesfa1;a2;a3g.
Weiss[7]forthedetails.
Thusinordertoprovetheexistenceof
lim N!11 NN1X n=03 Y i=1f i(Tainx) extension
CONVERGENCEOFCONZE-LESIGNEAVERAGES5
followingdiagramexplainstheserelations: X bZ=ZL=K ZL=X1=Z1L1 Z Z1 facts.
Z f(z+as)!s(z)f(z)j2dm(z)<2;(5) forZandksubstitutedfor. K jfj(z+k)fj(z)j2dmK(z)Furthermore,foralls2K,Z K jfj(z+s)fj(z)!s(z+j)j2dmK(z)Bythersthypothesis,wehave k2X j=0Z K jcj+1 j+1(z)cj j(z)j2dmK(z) Z K jc0
0(z+k)ck1
k1(z)j2dmK(z)10BERNARDHOSTANDBRYNAKRA forsomeconstantC.Thus j+1= jmodK?for0j2bZwith j= modK?forallj.Moreoverwehavejcj+1cjjThereexistsacomplexnumberwithk=1andj ()jItisnowimmediatethattheanefunction!=c satiseskf!k2Settinggi(z)=fi(z+bis) fi(z),usingthebound(7)againweget Z
theexistenceinL2()of lim N!11 NN1X n=03 Y i=1f i(Tainx):(14) modiedsystemdescribedinSection2.10. f i(z;g)=wi(z)i(g)(15) M ?.Then lim N!11 NN1X n=03 Y i=1f i(Tainx) existsinL2()andequals0. Z M3 Y i=1f i(zi;gi+ui)dmM(u1;u2;u3)= 3 Y i=1w i(zi)i(gi)Z M3 Y i=1 i(ui)dmM(u1;u2;u3)=0 forall(u1;u2;u3)2M,
CONVERGENCEOFCONZE-LESIGNEAVERAGES13
i=1f i(zi;gi)isor- L
2(m)suchthat
n(z)=3Y i=1 i(nai)(z) foralln2Z.
Wenotethatsisofmodulusoneforallt.
For1i3andk2Nwewrite
f k;i(z)=i(nkai)(z):
WehavetoprovethatQ3
i=1fk;i(z)convergesinL2(Z)ask!1. ~~(~z)=3Y i=1 i(ai)(zi) ofmodulus1onZZsuchthatforallk2N 3 Y i=1f k;i(z+ait)=3Y i=1 i(nkai)(z+ait)=b(z;t+nk) b(z;t) continuousway, 3Y i=1f k;i(z+ait)!b(z;t+s) b(z;t) inL2(ZZ)ask!1.Moreover,fori=1;2;3andk2N, f k;i(z+) f k;i(z+) fk;i(z)!i(z+ais)(z) inL2(Z)ask!1.
14BERNARDHOSTANDBRYNAKRA
1 NN1X i=13 Y i=1f i(Tainx)=3Y i=1 i(g)1NN1X n=03 Y i=1w i(z+nai)i(nai)(z):