full2010.pdf - page 446

408
Main Results
Let
X
be a
real Banach space and let
C
be a nonempty closed convex subset of
X
. For
1, 2,
i
let
:
i
T C C
o
be a quasi-nonexpansive mapping such that
1
2
F T F T
ˆ z ‡
in
C
. We are interested in
sequences in the following process. For
1
x C

and
1
n
t
, define the sequences
^ `
n
x
and
^ `
n
y
by
2
1
n
n n
n
n n
n n
y
T x
a x a v
E
E
1
1
1
n
n n
n
n n
n n
x
T y
b y b u
D
D
2.1
Where
^ `
n
a
,
^ `
n
b
,
^ `
n
D
and
^ `
n
E
are sequences in
> @
0,1
and
^ `
n
u
and
^ `
n
v
are sequences in
C
. Note that
since
C
is a nonempty convex subset of
X
and the sequences
^ `
n
x
and
^ `
n
y
are convex combinations of
elements in
C
, we conclude that
^ `
n
x
and
^ `
n
y
are sequences in
C
.
If
1
2
T T T
,
2.1
becomes
1
n
n n
n
n n
n n
y
Tx
a x a v
E
E
1
1
n
n n
n
n n
n n
x
Ty
b y b u
D
D
2.2
Theorem 2.1
Let
X
be a real Banach space and let
C
be a nonempty closed convex subset of
X
. For
1, 2,
i
let
:
i
T C C
o
be a quasi-nonexpansive mapping such that
1
2
F T F T
ˆ z ‡
in
C
. Let
^ `
n
a
,
^ `
n
b
,
^ `
n
D
and
^ `
n
E
be sequences in
> @
0,1
and
^ `
n
u
and
^ `
n
v
be sequences in
C
. Assume that
(i)
^
`
n
n
a
E
and
^
`
n
n
b
D
are sequences in
> @
0,1
and
1
n
n
a
f
f
¦
and
1
n
n
b
f
f
¦
;
(ii)
^ `
n
u
and
^ `
n
v
are bounded.
Then the iterative sequence
^ `
n
x
defined in
2.1
converges strongly to a common fixed point of
1
T
and
2
T
if
and only if
1
2
liminf
,
0
n
n
d x F T F T
of
ˆ
.
Proof.
We first prove the necessity.
Assume that
^ `
n
x
converges strongly to a common fixed point of
1
T
and
2
T
i.e. there exists
1
2
p F T F T
 ˆ
such that
lim
0.
n
n
x p
of
That is
liminf
0
n
n
x p
of
,
*
1
2
*
1
2
,
inf
n
n
n
p F T F T
d x F T F T
x p x p
 ˆ
ˆ
d
.
Taking limit infimum as
n
of
, we have
1
2
liminf
,
0
n
n
d x F T F T
of
ˆ
, as desired.
Now, we prove the sufficiency. Let
1
2
p F T F T
 ˆ
. By the boundedness of
^ `
n
u
and
^ `
n
v
,
we let
^
`
1
1
max sup
,sup
n
n
n
n
M
u p
v p
t
t
.
Since
:
i
T C C
o
is a quasi-nonexpansive mapping for
1, 2,
i
2.1
and by the triangle inequality,
we
have
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