409
2
2
2
1
1
1
1
1
1
n
n n
n
n n
n n
n n
n
n n
n n
n
n
n
n
n
n
n
n n
n n
n n
n
n n
n
n n
n
y p
T x
a x a v p
T x
a x a v
a p p a p
T x p
a x p a v p
x p
a x p a M
a x p a M
E
E
E
E
E
E
E
E
E
E
d
d
and
^
`
1
1
1
1
1
1
1
1
1
1
1
1
n
n n
n n n
n n
n n
n
n n
n n
n n
n
n
n
n
n
n
n
n n
n n
n n
n
n
n
n
n
n
n n
n
n
x p
T y
b y b u p
T y
b y b u
b p p b p
T y p
b y p b u p
y p
b y p b M
b y p b M
b
a x p a M b M
D
D
D
D
D
D
D
D
D
D
d
d
d
1 1
1
1
n
n n
n n
n
n n
n n n
n
n
n n
b a x p
b a M b M
a b a b x p a b a b M
n
n
n
x p a b M
d
,
n
n
x p d
2.3
where
n
n
n
d a b M
. Now by the assumptions that
1
n
n
a
f
f
¦
and
1
n
n
b
f
f
¦
, we have that
1
n
n
d
f
f
¦
.
Then Lemma 1.3 implies that
lim
n
n
x p
of
exists. From
2.3
and by induction, for
,
1
m n
t
and
1
2
p F T F T
, we have
1
n m
n m
n
i
i n
x
p x p
d
d
¦
.
2.4
From
2.3
and taking the infimum over
1
2
p F T F T
, we obtain
1
1
2
1
2
,
,
n
n
n
d x F T F T d x F T F T d
d
.
But, the assumption
1
2
liminf
,
0
n
n
d x F T F T
of
implies that there exists a subsequence of
^
`
1
2
,
n
d x F T F T
converging to zero. Therefore Lemma 1.3 tells us that
1
2
lim ,
0
n
n
d x F T F T
of
.
2.5
Next, we show that
^ `
n
x
is a Cauchy sequence in
X
. Let
0
H
!
From
2.5
and
1
n
n
d
f
f
¦
, there exist
1 2
,
N N N
such that
1
1
2
2
,
4
.
2
n
n
k
k m
n N d x F T F T
n m N d
H
H
!
t !
¦
Choose
^
`
1 2
max ,
N
N N
, there exists
0
n N
such that for
0
n n N
t !
, we have
1
2
,
,
4
n
d x F T F T
H
0
.
2
n
n n
d
H
f
¦
2.6
By the first inequality in
2.6
and the definition of infimum, there exists
0
1
2
p F T F T
such that
