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8
AbdelkrimAlioucheandValeriuPopa
(01):Obviously.
u+u
2
(02):Let1<k<
}≤0.Then,u≤kt≤kamax{ujuj
1
a
,u≤ktand0(tjujujuju+uj0)=t−amax{uju,
u+u
2
}=kamax{uju}.Ifu>0
andu≥u,itfollowsthatu≤kau<uwhichisacontradictionandso
u≤hu,whereh=ka<1.Ifu=0,thereforeu≤hu.Similarly,u≤kt
and0(tjujujuj0ju+u)≤0impliesu≤hu.
Example3.0(t1jt2jt3jt4jt5jt6)=t1−amax{t2
2jt3t4jt5t6jt3t5jt4t6}
1
2,
0<a<
(01):Obviously.
√2
1
.
(02):Let1<k<
a√2
1
,u≤ktand0(tjujujuju+uj0)=t−
amax{u2juuju(u+u)}
1
2≤0.Then,u≤kt≤kamax{u2juuju(u+u)}
1
2.If
u>0andu≥u,itfollowsthatu≤ka√2u<uwhichisacontradiction
andsou≤hu,whereh=ka√2<1.Ifu=0,thereforeu≤hu.Similarly,
u≤ktand0(tjujujuj0ju+u)≤0impliesu≤hu.
Example4.0(t1jt2jt3jt4jt5jt6)=t2
1+
1+t5t6
t1
−at2
2−bt2
3−Ct2
4,a>0,
bjC≥0anda+b+C<1.
(01):Obviously.
(02):Let1<k<
√a+b+C
1
,u≤ktand0(tjujujuju+uj0)=t2+
t−au2−bu2−Cu2≤0.Then,t2≤au2+bu2+Cu2andu2≤k2t2≤
k2(au2+bu2+Cu2).Itfollowsthatu≤h1u,whereh1=kda+b
1−k2C
<1.
Similarly,u≤ktand0(tjujujuj0ju+u)≤0impliesu≤h2u,where
h2=kda+C
1−k2b
<1.Ifh=max{h1jh2},thenu≤hu.
Ct
p11
5
Example5.0(t1jt2jt3jt4jt5jt6)=t
t6},p≥2,0<a<1andC≥0.
p
1−max{at2t
p11
3
jat
p11
2
t4jat
p11
3
t4,
(01):Obviously.
(02):Let1<k<
√a
p
1
,u≤ktand0(tjujujuju+uj0)=tp−
max{aupjaup11u}≤0.Then,up≤kptp≤kpmax{aupjaup11u}.Ifu>0
andu≥u,itfollowsthatup≤akpup<upwhichisacontradictionandso
u≤hu,whereh=kp
√a<1.Ifu=0,thereforeu≤hu.Similarly,u≤kt
and0(tjujujuj0ju+u)≤0impliesu≤hu.
a)max{t2
Example6.0(t1jt2jt3jt4jt5jt6)=t1−b[amax{t2jt3jt4j
2jt3t4jt5t6j
1
2t3t6j
1
2t4t5}
1
2],0<b<1and0≤a<1.
t5+t6
2
}−(1−
