Thus T(K)is a subset of a weakly compact set.
Hence by Theorem 6, Thas a fixed point.
Theorem 7. Let
K
be a compact subset of a Banach space
X
. Let
T
:
K→K
satisfies condition (C).
Let
{xn}
be a regular approximate fixed point sequence in
K
. Then
{xn}
converge strongly to a fixed
point of T.
Proof. Let {xn}be a regular approximate fixed point sequence in K.
Since Kis compact, there exist a subsequence {xnk}of {xn}and z∈Ksuch that xnk→z.
Therefore, A(K, {xnk})={z}.
Since
{xnk}
is an approximate fixed point sequence, by Lemma 3,
A
(
K, {xnk}
)is
T
-invariant. Hence
T(z)=z.
Since {xn}is regular, r(K, {xn})=r(K, {xnk})=0.
Since Kis compact and {xn}is an approximate fixed point sequence, A(K, {xn})is nonempty.
We know that if {xn}is regular, then for any subsequence {xnk},
A(K, {xn})⊆A(K, {xnk}).
Therefore, A(K, {xn})={z}. Hence xn→z.
The following example shows that even if
T
has fixed points, if
{xn}
is not regular, then
{xn}
need not
converge to a fixed point.
Example 2. Consider the compact set K=[−1,1] in Rand define T:K→Kas T(x)=x.
Clearly Tis a nonexpansive mapping and hence satisfies condition (C).
Consider xn=(−1)nf or all n ∈N
For all n∈N,||xn−Tx
n|| =0. Thus {xn}is an approximate fixed point sequence.
But {xn}does not converge to a fixed point.
This will not contradict the above theorem because {xn}is not regular.
Asymptotic radius of {xn}=r(K, {xn})=1and A(K, {xn})={0}.
Consider the subsequence {x2n}={1,1,1, ...}. Clearly x2n→1.
Therefore, A(K, {x2n})={1}and r(K, {x2n})=0.
Hence {xn}is not regular.
Theorem 8. Let
K
be a subset of a Banach space
X
and
T
:
K→K
satisfies condition (C). Suppose
T
(
K
)is contained in a compact subset of
K
and let
{xn}
be a regular approximate fixed point sequence
in Kwith nonempty asymptotic center. Then {xn}converge strongly to a fixed point of T.
Proof. Let
{xn}
be a regular approximate fixed point sequence in
K
with nonempty asymptotic
center.
Consider the sequence
{Tx
n}
in
T
(
K
). Since
T
(
K
)is compact, there exist a subsequence
{Tx
nk}
of
{Tx
n}and z∈T(K)such that Tx
nk→z.
Therefore, lim
n→∞||xnk−z|| ≤ lim
n→∞ ||xnk−Tx
nk|| =0.
Thus xnk→zand hence A(K, {xnk})={z}.
Since {xnk}is an approximate fixed point sequence, by Lemma 3, A(K, {xnk})is T-invariant.
Hence T(z)=z.
Since {xn}is regular, r(K, {xn})=r(K, {xnk})=0.
Also if {xn}is regular, then for any subsequence {xnk}, we have A(K, {xn})⊆A(K, {xnk}).
Thus we have A(K, {xn})={z}, which implies that xn→z.
3 CONCLUSIONS
In this paper, we have used the asymptotic center technique to establish the existence of fixed points
for Suzuki nonexpansive mappings in Banach spaces. We have shown that under certain condition, the
asymptotic radius and Chebyshev radius are equal. Using this result, we have established that every
https://doi.org/10.17993/3ctic.2022.112.15-24
approximate fixed point sequence is regular as well as uniform. The convergence of regular approximate
fixed point sequences to a fixed points of the Suzuki nonexpansive mapping is also established. A couple
of examples are given to illustrate the results.
ACKNOWLEDGMENT
The first author is highly grateful to University Grant Commission, India, for providing financial
support in the form of Junior Research fellowship.
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