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Zhou et al. Intell Robot 2023;3(1):95-112 I http://dx.doi.org/10.20517/ir.2023.05 Page 103

1 1 1
Case1: Consider ≥ Φ ( ) 1− ,suppose ( ) = ( ) , 0 < | | < 1,showthat | ( + 1)| ≤
( ) ( )
1
. | ( )| ≤ , we have
( )
( )

( + 1) = ( ) ( ) ( ) − ( )| ( )| − ( )
(32)

≤ ( ) ( ) − ( ( ) ) + [ ( )] +

( + 1) = ( ) ( ) ( ) − ( )| ( )| − ( ) (33)

≥ ( ) ( ) − ( ( ) ) + [ ( )] − .
1 1
Since ≥ Φ ( ) 1− , then
( )
1 1
Φ ( ) 1−
≥
( )
1−
Φ ( )
⇒ ≥
( ) (34)
1−
⇒ [ ( ) ( )] ≥ ( )
1

⇒ ≤ ( ) ( ) .
( )
Since 0 < < 1, lead to ( ) ≥ 1 or 0 < ( ) < 1. When ( ) ≥ 1, we have

( + 1) ≤ ( ) ( ) − ( ( ) ) +
1

≤ ( ) ( )
( )
1 (35)

≤ ( )
( )
1

≤ ( )
( )

( + 1) ≥ ( ) ( ) − ( ( ) ) −
1

≥ ( ) ( ) − [( ( ) ) + 1] ( )
( )
1

≥ [ ( ) − ( ( ) ) − 1] ( ) (36)
( )
1

≥ − ( )
( )
1

≥ − ( ) .
( )
We consider the situation that 0 < ( ) < 1, according to Lemma 3, we can also obtain | ( + 1)| ≤
1
.
( )
( )
1
1 1 1−
, when Φ ( ) and 0 < < 1,
From the above proof, we have | ( + 1)| ≤ ( ) ≥
( ) ( )
0 < ( ) ≤ ( ) .