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Page 104 Zhou et al. Intell Robot 2023;3(1):95-112 I http://dx.doi.org/10.20517/ir.2023.05

1
1
When − ( ) ≤ ( ) < 0 and −1 < < 0, ( ) = ( ) = − ( )| | , then
( ) ( )

( + 1) = ( ) ( ) − ( )| ( )| [ ( )] − ( ). (37)

1
.
Similar to the above proof process, we can obtain | ( + 1)| ≤ ( )
( )
1

From the above proof, it can be seen that | ( + 1)| ≤ ( ) is still satisfied with the conditions
( )
1 1
Φ ( ) 1− , −1 < < 0 and − ( ) ≤ ( ) < 0.
≥
( )
1 1 1
Case 2: If < Φ ( ) 1− , suppose ( ) = ( ) Φ ( ) 1− , 0 < | | < 1, then prove | ( + 1)| ≤
( )
1
Φ ( ) 1−
( ) . Submit ( ) to ( + 1), we can obtain that:

Φ ( ) 1− (38)
( + 1) = ( ) ( ) − ( ) ( ( ) ) − ( )

1 1
< Φ ( ) 1−
( )
1−
Φ ( )
⇒ <
( ) (39)
1− ( )
⇒ <
[ ( ) ]
1
Φ ( ) 1−
⇒ < ( ) .

When ( ) ≥ 1, ( ) ≥ ( ( ) ) , then, based on Lemma 2
( + 1)
1
Φ ( ) 1−
≤ ( ) ( ) − ( ) ( ( ) ) +

1
Φ ( ) 1−
≤ [ ( ) − ( ( ) ) + 1] ( ) (40)

1
Φ ( ) 1−
≤ ( ) ( )

1
Φ ( ) 1−
≤ ( )

( + 1)
1
Φ ( ) 1−
≥ ( ) ( ) − ( ) ( ( ) ) −

1
Φ ( ) 1−
≥ [ ( ) − ( ( ) ) − 1] ( ) (41)

1
Φ ( ) 1−
≥ − ( )

1
Φ ( ) 1−
≥ − ( ) ( ) ( ) .