Abstract

Newton-Kantorovich and Smale uniform type of convergence theorem of a deformed Newton method having the third-order convergence is established in a Banach space for solving nonlinear equations. The error estimate is determined to demonstrate the efficiency of our approach. The obtained results are illustrated with three examples.

1. Introduction

In this paper, we study the problem of approximating a unique solution of a nonlinear operator equation where is a Fréchet-differentiable operator defined on an open convex of a Banach space with values in a Banach space .

There are many iterative methods (see [1–3]), which have been used for finding a solution of (1). For example, the well-known iterative method for solving (1) is Newton's method defined by

Under the appropriate assumptions, Newton's method is the second-order convergence. Kantorovich (see [4]) presented the famous convergence result regarding a solution of (1). Many Newton-Kantorovich type of convergence theorems were given in papers [5–11]. Frontini and Sormani (see [12]) presented a new deformed Newton method with

The deformed Newton method can be written as follows: where is a real or a complex function. In papers [13–17], the local convergence theorem has been established and the deformed method in a real or a complex space was discussed.

In the paper, we generalize the deformed Newton method [18] in a Banach space. The deformed Newton method [18] is shown as follows: where is defined on an open convex subset of a Banach space with values in a Banach space , has Fréchet derivatives in , and exists.

We establish Newton-Kantorovich and Smale uniform type convergence theorem (see [18]) for the deformed Newton method with the third-order in a Banach space with new sufficient conditions for the existence of a well-defined sequence which converges to a unique solution of (1).

2. Main Results

Denote , , , and suppose are the positive and nondecreasing continuous functions, , , for , .

Assume that sequences , are generated by the following formulae [18]:

Firstly, we give some lemmas.

Lemma 1. If , then the function has two positive real roots , ().

Proof. Because ,  , and , we know that is the convex function for . Hence, is a unique positive root of . So, the necessary and sufficient condition that has two positive roots for is that the minimum of satisfies the condition , which holds for . This completes the proof of Lemma 1.

Lemma 2. Suppose the sequences , are generated by (6). Then, for , the sequences , are increasing and converge to the minimum positive root of , and

Proof. Denote
On , we know , , , and is increasing. Denoting , then
Therefore, , are increasing on . Thus, for , , . Moreover, hence we can easily prove Lemma 2 by the induction.
Suppose and are the Banach spaces, is an open convex subset, has the second-order Fréchet derivative, exists for , and the following conditions hold: where and .

Lemma 3. Suppose satisfies (11) and . Then exists, and

Proof. Firstly, by the conditions (11), we know that
Secondly, we know for . Hence
By Banach Theorem, we know exists, and
This completes the proof of Lemma 3.

Lemma 4. Suppose and are Banach spaces, is an open convex of the Banach space , has the second-order Fréchet derivative, and the sequences , are generated by (5). Then, for any natural number , the following formula holds:

Proof. By (5), we have
Hence
This completes the proof of Lemma 4.

Theorem 5. Suppose and are Banach spaces, is an open convex subset, satisfies condition (11), , and Then the sequence generated by (5) is well defined, , and converges to the unique solution in and

Proof. By induction, we can prove that the following formulae hold:
In fact, by Lemma 2 we know for any natural number . It is easy to prove that for the above formulae hold. Suppose the above formulae also hold for . Then
By Lemma 3, we get By Lemmas 3 and 4 and the fact that , are positive and increasing on , we have
Hence we get
By Lemma 3, we get
Moreover, we have
Hence, the sequence generated by (5) is well defined, , and converges to the solution of (1).
Now we prove the uniqueness. Suppose is also a solution of (1) on . We know that for . Then
By Banach Theorem, we know the inverse of exists and hence we get . This completes the proof of the uniqueness of the solution of (1).
For , we know that
When , we get
This completes the proof of Theorem 5.