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or there exists a positive integer s , such that
1
2
min
{ λ
z is , λ
z sj }=
(2.56)
Case 1 If for any k ,wehave
min
{ λ
z ik , λ
z kj }=
0
(2.57)
Then for any
λ ∈[
0
,
1
]
,ityields
max
{
v z ik ,
v z kj } >
1
λ
(2.58)
hence
mi k {
max
{
v z ik ,
v z kj }} ≥
v z ij >
1
λ
(2.59)
Considering the arbitrary of
λ
, when
λ
tends to be infinitely small, we get
ma k {
min
{ μ z ik z kj }} ≤
1
mi k {
max
{
v z ik ,
v z kj }} =
0
(2.60)
As a result,
ma k {
min
{ μ z ik z kj }} ≤ μ z ij
(2.61)
Case 2 If there exists a positive integer k 1 , such that
1
2
min
{ λ z ik 1 , λ z k 1 j }=
(2.62)
and for any k
=
k 1 ,let
min
{ λ z ik , λ z kj }=
0
(2.63)
Then according to Case 1, we have
k 1 {
{
v z ik ,
v z kj }} ≥
min
k
max
v z ij
(2.64)
=
max
k
k 1 {
min
{ μ z ik z kj }} ≤ μ z ij
(2.65)
=
and when k
=
k 1 , suppose that
min
{ μ z ik 1 z k 1 j }= μ z ik 1 z ik 1
z ij
(2.66)
Then let
λ = z ij + μ z ik 1 )/
2, and thus,
μ z ij λ μ z ik 1 . Accordingly,
z ik 1 = λ z k 1 j =
1
,
min
{ λ z ik 1 , λ z k 1 j }=
1
(2.67)
λ
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