α n T z n ) q ( a 2 + b 2 ) ( ( 1 α n ) S x n + α n T z n ) q

( 1 α n ) S x n q + α n T z n q ( 1 α n ) ( a 1 x n q + b 1 x n q ) + α n ( a 2 z n q + b 2 z n q ) ( 1 α n ) ( a 1 + b 1 ) x n q + α n ( a 2 + b 2 ) z n q ( 1 α n ) x n q + α n z n q x n q α n x n q + α n z n q

x n + 1 q x n q α n = z n q x n q

and hence

x n + 1 q x n q x n + 1 q x n q α n = z n q x n q

which implies that x n + 1 q z n q (2.27)

Taking limit inferior in (2.27) we obtain

c lim inf n z n q (2.28)

From (2.20) and (2.28) we have

c = lim n z n q = lim n ( 1 β n ) x n + β n S x n q = lim n β n ( S x n q ) + ( 1 β n ) ( x n q ) (2.29)

Now from (2.24), (2.26), (2.29) and lemma (1.7), we have lim n S x n x n = 0 .

Now,

T x n q a 2 x n q + b 2 x n q x n q

lim sup n T x n q lim sup n x n q c (2.30)

Using the conditions of the lemma in (2.30), we can write

C = lim n β n ( T x n q ) + ( 1 β n ) ( x n q ) (2.31)

Using (2.24), (2.30), (2.31) along with the lemma (1.7), we have

lim n T x n x n = 0.

Theorem 2.11: Let H be a non-empty closed convex subset of a Banach space X satisfying Opial’s condition and S, T and { x n } n = 0 be same as defined in the lemma (2.10) .Then the sequence { x n } n = 0 converges weakly to some q F .

Proof: From lemma (2.10) we have, lim n T x n x n = 0 .

Since X is uniformly convex and hence it is reflexive so there exists a subsequence { x n m } of { x n } such that { x n m } converges weakly to some q 1 F . Since H is closed so q 1 H . Now we claim the weak convergence of { x n } to q 1 . Let it is not true, then there exists a subsequence of { x n i } of { x n } which converges weakly to q 2 and let q 1 q 2 . Also q 2 F . Now from lemma (2.9) lim n x n q 1 and lim n x n q 2 both exist. Using Opial’s condition we have,

lim n x n q 1 lim n x n m q 1 < lim n x n m q 2 = lim n x n q 2 = lim n x n i q 2 < lim n x n i q 1 lim n x n q 1

This is a contradiction, so we must have q 1 = q 2 . Thus the sequence { x n } n = 0 converges weakly to some q F .

Theorem 2.12: Let H be a non-empty closed compact subset of a Banach space X and S, T and { x n } n = 0 be same as defined in the lemma (2.10). Then the sequence { x n } n = 0 converges strongly to some q F .

Proof: Since H is compact and hence it is sequentially compact. So there exists a subsequence { x n i } of { x n } which converges to q H .

Now

x n i T q = x n i T x n i + T x n i T q x n i T x n i + a 2 x n i q + b 2 x n i q x n i T x n i + x n i q (2.32)

Taking limit n in (2.32) we have, T q = q that is q F . We have earlier proved that lim n x n q exists for q F . Hence the sequence { x n } n = 0 converges strongly to some q F .

In [8] it is proves that the K-iteration process converges faster than Picard-S, Thakur-New and Vatan two-step iterative process. Now we shall compare the rate of convergence the K-iteration process defined in [8] and our new modified K-iteration process for two mappings.

Table 1. Iterative values of K-iteration process and Modified K-iteration process.

Example 2.13: Let S , T : [ 0 , 3 ] [ 0 , 3 ] be two mappings defined by T ( x ) = x + 2 2 and s ( x ) = ( x + 2 ) 1 2 . Let α n , β n be the sequences defined by α n = β n = 1 4 . Let the initial approximation be x 0 = 2.25 . Clearly S, T has

unique common fixed point 2. The convergence pattern of K-iteration process and modified K-iteration process is shown in Table 1.

Clearly we can conclude from Table 1, that the modified K-iteration process has better rate of convergence than the k-iteration process.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Panwar, A. and Bhokal, R.P. (2019) Fixed Point Results for K-Iteration Using Non-Linear Type Mappings. Open Access Library Journal, 6: e5245. https://doi.org/10.4236/oalib.1105245

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