# Newton-Kantorovich and Smale Uniform Type Convergence Theorem for a Deformed Newton Method in Banach Spaces

Authors: Rongfei Lin; Yueqing Zhao; Zdeněk Šmarda; Yasir Khan; Qingbiao Wu
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 1. Introduction

In this paper, we study the problem of approximating a unique solutionx* of a nonlinear operator equation
(1)F(x)=0,
where F is a Fréchet-differentiable operator defined on an open convex Ω of a Banach space X with values in a Banach space Y.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
(2)xn+1=xn-F′(xn)-1F(xn),(n≥0)(x0∈Ω).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
(3)∫xnx‍f′(t)dt≃(x-xn)f′(xn+x2).The deformed Newton method can be written as follows:(4)xn+1=xn-f(xn)f′(xn-f(xn)/2f′(xn)),
where f 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:
(5)yn=xn-F′(xn)-1F(xn),xn+1=xn-F′(xn+yn2)-1F(xn),
where F is defined on an open convex subset Ω of a Banach space X with values in a Banach space Y, F(x) has Fréchet derivatives in Ω, and F′(x)-1 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 x* of (1).


## 2. Main Results

Denoteg(t)=∫0t‍(t-u)L(u)du-t+η, u∈(0,R), η>0, and suppose L(u),L′(u) are the positive and nondecreasing continuous functions, limt→R+g(t)=g(R+)>0, ∫0R‍L(u)du>1, ∫0α‍L(u)du=1 for α∈(0,R), β=α-∫0α‍(α-u)L(u)du=∫0α‍uL(u)du.Assume that sequences{tn}, {sn} are generated by the following formulae [18]:
(6)sn=tn-g′(tn)-1g(tn),tn+1=tn-g′(tn+sn2)-1g(tn),t0=0.Firstly, we give some lemmas.Lemma 1.
Ifη≤β, then the function g(t) has two positive real roots r1, r2 (0<r1≤α≤r2<R).Proof.
Becauseg(0)=η>0,  g(R+)>0, and g′′(t)=L(t)>0, we know that g(t) is the convex function for t∈(0,R). Hence, α is a unique positive root of g′(t)=∫0t‍L(u)du-1. So, the necessary and sufficient condition that g(t) has two positive roots for t∈(0,R) is that the minimum of g(t) satisfies the condition g(α)≤0, which holds for η≤β. This completes the proof of Lemma 1.Lemma 2.
Suppose the sequences{tn}, {sn} are generated by (6). Then, for η≤β, the sequences {tn}, {sn} are increasing and converge to the minimum positive root of g(t), and
(7)0≤tn≤sn≤tn+1<r1.Proof.
Denote(8)U(x)=x-g(x)g′(x),V(x)=x-g(x)g′((x+U(x))/2).
On[0,r1), we know g(t)>0, g′(t)<0, g′′(t)>0, and g′′(t) is increasing.  Denoting y=(x+U(x))/2=x-g(x)/2g′(x), then
(9)U′(x)=g(x)g′′(x)g′(x)2>0,[g′(y)-g′(x)]=g′′(ξ)(y-x)=-g′′(ξ)g(x)2g′(x),ξ∈(x,y),V′(x)=1-(+(g(x)g′′(x)g′(x)2)g′(x)g′(y)-12g(x)g′′(y)×1+(g(x)g′′(x)g′(x)2)12)×(g′(y)2)-1=1g′(y)[g′(y)-g′(x)]+g(x)g′′(y)2g′(y)2+g(x)2g′′(x)g′′(y)2g′(x)2g′(y)2≥-g′′(ξ)g′(y)·g(x)2g′(x)+g(x)g′′(y)2g′(y)2=-g(x)g′′(ξ)2g′(y)2g′(x)[g′(y)-g′(x)]+g(x)g′′(y)-g(x)g′′(ξ)2g′(y)2=g(x)g′′(ξ)2g′(y)2g′(x)·g(x)g′′(ξ)2g′(x)+g(x)g′′(y)-g(x)g′′(ξ)2g′(y)2>0.
Therefore,U(x), V(x) are increasing on [0,r1]. Thus, for x∈[0,r1), U(x)<U(r1)=r1, V(x)<V(r1)=r1. Moreover,
(10)sn=U(tn),tn+1=V(tn),t0=0<r1;
hence we can easily prove Lemma 2 by the induction.
SupposeX and Y are the Banach spaces, Ω⊂X is an open convex subset, F:Ω⊂X→Y has the second-order Fréchet derivative, F′(x0)-1 exists for x0∈Ω, and the following conditions hold:
(11)∥F′(x0)-1F(x0)∥≤η,∥F′(x0)-1F′′(x0)∥≤L(0),∥F′(x0)-1(F′′(y)-F′′(x))∥≤∫ρ(x)ρ(x,y¯)L′(u)du,x,y∈Ω,ρ(x,y¯)<α,
where ρ(x)=∥x-x0∥ and ρ(x,y¯)=∥y-x∥+∥x-x0∥.Lemma 3.
SupposeF satisfies (11) and ∥x-x0∥<α. Then F′(x)-1 exists, and
(12)∥F′(x0)-1F′′(x)∥≤g′′(∥x-x0∥),∥F′(x)-1F′(x0)∥≤-1g′(∥x-x0∥).Proof.
Firstly, by the conditions (11), we know that
(13)∥F′(x0)-1F′′(x)∥≤∥F′(x0)-1F′′(x0)∥+∥F′(x0)-1F′′(x)-F′(x0)-1F′′(x0)∥≤L(0)+∫0∥x-x0∥L′(u)du=L(∥x-x0∥)=g′′(∥x-x0∥).
Secondly, we knowg′(t)<0 for t<α. Hence
(14)∥F′(x0)-1F′(x)-I∥=∥F′(x0)-1[F′(x)-F′(x0)]∥=∥F′(x0)-1∫01F′′(x0+t(x-x0))×(x-x0)dt∫01F′′(x0+t(x-x0))∥≤∫01g′′(t∥x-x0∥)∥x-x0∥dt=g′(∥x-x0∥)-g′(0)=g′(∥x-x0∥)+1<1.
By Banach Theorem, we knowF′(x)-1 exists, and
(15)∥F′(x)-1F′(x0)∥≤11-∥F′(x0)-1F′(x)-I∥=-1g′(∥x-x0∥).
This completes the proof of Lemma3.Lemma 4.
SupposeX and Y are Banach spaces, Ω is an open convex of the Banach space X, F:Ω⊂X→Y has the second-order Fréchet derivative, and the sequences {xn}, {yn} are generated by (5). Then, for any natural number n, the following formula holds:
(16)F(xn+1)=∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))(1-t)dt×(xn+1-yn)2+12∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))(1-t)dt×(xn+1-yn)(yn-xn)+12∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))(1-t)dt×(yn-xn)(xn+1-yn)+14∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))(1-t)dt×(yn-xn)2-14∫01‍F′′(xn+yn2-t(yn-xn2))(1-t)dt×(yn-xn)2.Proof.
By (5), we have
(17)F(xn+1)=F(xn+1)-F(xn+yn2)-F′(xn+yn2)(xn+1-xn+yn2)+F(xn+yn2)+F′(xn+yn2)×(xn+1-xn+yn2)=∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))(1-t)dt×(xn+1-xn+yn2)2+F(xn+yn2)+F′(xn+yn2)(xn+1-xn+yn2),F(xn+yn2)+F′(xn+yn2)(xn+1-xn+yn2)=F(xn+yn2)+F′(xn+yn2)×(xn+1-xn-yn-xn2)=F(xn+yn2)-F(xn)-F′(xn+yn2)yn-xn2=-14∫01‍F′′(xn+yn2-t(yn-xn2))(1-t)dt×(yn-xn)2.
Hence(18)F(xn+1)=∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))×(1-t)dt(xn+1-yn)2+12∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))×(1-t)dt(xn+1-yn)(yn-xn)+12∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))×(1-t)dt(yn-xn)(xn+1-yn)+14∫01‍F′′(xn+yn2+t(xn+1-xn+yn2))×(1-t)dt(yn-xn)2-14∫01‍F′′(xn+yn2-t(yn-xn2))(1-t)dt×(yn-xn)2.
This completes the proof of Lemma4.Theorem 5.
SupposeX and Y are Banach spaces, Ω⊂X is an open convex subset, F:Ω⊂X→Y satisfies condition (11), η≤β, and
(19)S(x0,r1)-={x∣∥x-x0∥≤r1,x∈X}⊂Ω.
Then the sequence {xn}n≥0 generated by (5) is well defined, xn∈S(x0,r1)-, and converges to the unique solution x* in S(x0,α) and
(20)∥xn-x*∥≤r1-tn.Proof.
By induction, we can prove that the following formulae hold:(21)∥xn-x0∥≤tn;∥F′(xn)-1F′(x0)∥≤-g′(tn)-1;∥yn-xn∥≤sn-tn;∥yn-x0∥≤sn;∥F′(xn+yn2)-1F′(x0)∥≤-g′(tn+sn2)-1;∥xn+1-yn∥≤tn+1-sn;∥xn+1-xn∥≤tn+1-tn.
In fact, by Lemma2 we know tn<r1 for any natural number n. It is easy to prove that for n=0 the above formulae hold. Suppose the above formulae also hold for n>0. Then
(22)∥xn+1-x0∥≤∥xn+1-xn∥+∥xn-x0∥≤tn+1-tn+tn=tn+1.
By Lemma3, we get
(23)∥F′(xn+1)-1F′(x0)∥≤-g′(∥xn+1-x0∥)-1≤-g′(tn+1)-1.
By Lemmas 3 and 4 and the fact that -g′(t)-1, g′′(t) are positive and increasing on [0,α), we have
(24)∥F′(x0)-1[F′′(xn+yn2+t(xn+1-xn+yn2))-F′′(xn+yn2-t(yn-xn2))]∥≤∫0t∥xn+1-xn∥‍L′(u+∥xn+yn2-t(yn-xn2)-x0∥)du≤∫0t(tn+1-tn)‍L′(u+tn+sn2-tsn-tn2)du=L(tn+sn2+t(tn+1-tn+sn2))-L(tn+sn2-t(sn-tn2))=g′′(tn+sn2+t(tn+1-tn+sn2))-g′′(tn+sn2-t(sn-tn2)),∥F′(x0)-1F(xn+1)∥≤∫01‍g′′(tn+sn2+t(tn+1-tn+sn2))×(1-t)dt(tn+1-sn)2+12∫01‍g′′(tn+sn2+t(tn+1-tn+sn2))(1-t)dt×(tn+1-sn)(sn-tn)+12∫01‍g′′(tn+sn2+t(tn+1-tn+sn2))(1-t)dt×(sn-tn)(tn+1-sn)+14∫01‍g′′(tn+sn2+t(tn+1-tn+sn2))(1-t)dt×(sn-tn)2-14∫01‍g′′(tn+sn2-t(sn-tn2))(1-t)dt×(sn-tn)2=g(tn+1).
Hence we get(25)∥yn+1-xn+1∥=∥-F′(xn+1)-1F(xn+1)∥≤∥-F′(xn+1)-1F′(x0)∥×∥F′(x0)-1F(xn+1)∥≤-g′(tn+1)-1g(tn+1)=sn+1-tn+1,∥yn+1-x0∥≤∥yn+1-xn+1∥+∥xn+1-x0∥≤sn+1.
By Lemma3, we get
(26)∥F′(xn+1+yn+12)-1F′(x0)∥≤-g′(tn+1+sn+12)-1.
Moreover, we have(27)∥xn+2-yn+1∥=∥(xn+1+yn+12)-1F′(xn+1)-1F(xn+1)-F′(xn+1+yn+12)-1F(xn+1)∥=∥F′(xn+1+yn+12)-1[F′(xn+1+yn+12)-F′(xn+1)(xn+1+yn+12)]×F′(xn+1)-1F(xn+1)(xn+1+yn+12)∥=∥F′(xn+1+yn+12)-1F′(x0)F′(x0)-1×∫01‍F′′(xn+1+t2(yn+1-xn+1))dt×yn+1-xn+12F′(xn+1)-1×F′(x0)F′(x0)-1F(xn+1)F′(xn+1+yn+12)-1∥≤g′(tn+1+sn+12)-1×∫01‍g′′(tn+1+t2(sn+1-tn+1))dt×(sn+1-tn+1)2g′(tn+1)-1g(tn+1)≤g′(tn+1+sn+12)-1×[g′(tn+1+sn+12)-g′(tn+1)]×g′(tn+1)-1g(tn+1)=g′(tn+1)-1g(tn+1)-g′(tn+1+sn+12)-1×g(tn+1)=tn+2-sn+1,∥xn+2-xn+1∥≤∥xn+2-yn+1∥+∥yn+1-xn+1∥≤tn+2-tn+1.
Hence, the sequence{xn}n≥0 generated by (5) is well defined, xn∈S(x0,r1)¯, and {xn} converges to the solution x*∈S(x0,r1)¯ of (1).
Now we prove the uniqueness. Supposey* is also a solution of (1) on S(x0,α). We know that g′(t)<0 for t∈[0,α). Then
(28)∥F′(x0)-1∫01F′(x*+t(y*-x*))dt-I∥≤∥F′(x0)-1∫01{F′[x*+t(y*-x*)]-F′(x0)}dt∥≤∥F′(x0)-1∫01∫01F′′(x0+s(x*-x0+t(y*-x*)))dsdt×(x*-x0+t(y*-x*))∫∫01F′′(x0+s(x*-x0+t(y*-x*))dsdt∥≤∫01∫01g′′(s∥x*-x0+t(y*-x*)∥)dsdt×∥x*-x0+t(y*-x*)∥=∫01g′(∥x*-x0+t(y*-x*)∥)dt-g′(0)=∫01g′(∥(1-t)(x*-x0)+t(y*-x0)∥)dt+1<1.
By Banach Theorem, we know the inverse of∫01F′[x*+t(y*-x*)]dt exists and
(29)0=F(y*)-F(x*)=∫01F′[x*+t(y*-x*)]dt(y*-x*);
hence we get y*=x*. This completes the proof of the uniqueness of the solution of (1).
Form>n, we know that
(30)∥xm-xn∥≤∥xm-xm-1∥+∥xm-1-xm-2∥+⋯+∥xn+1-xn∥≤tm-tn.
Whenm→∞, we get
(31)∥xn-x*∥≤r1-tn.
This completes the proof of Theorem5.Suppose thatL(u)=γ+Ku,u∈(0,+∞), γ,K>0. Then ∫ρ(x)ρ(x,y¯)‍L′(u)du=K∥x-y∥, g(t)=(1/6)Kt3+(1/2)γt2-t+ηα=2/(γ+γ2+2K), and β=α-(1/6)Kα3-(1/2)γα2=2(γ+2γ2+2K)/3(γ+γ2+2K)2.Corollary 6.
SupposeX and Y are the Banach spaces, Ω is an open convex subset of the Banach space X, F:Ω⊂X→Y has the second-order Fréchet derivative, F′(x0)-1 exists for x0∈Ω, and the following conditions hold:
(32)∥F′(x0)-1F(x0)∥≤η,∥F′(x0)-1F′′(x0)∥≤γ,∥F′(x0)-1(F′′(x)-F′′(y))∥≤K∥x-y∥x,y∈Ω,η≤2(γ+2γ2+2K)3(γ+γ2+2K)2,S(x0,r1)¯⊂Ω.
Then the sequence{xn}n≥0 generated by (5) is well defined, xn∈S(x0,r1)¯, and {xn} converges to the unique solution x* on S(x0,α) of (1), where r1≤r2 are two positive roots of g(t)=(1/6)Kt3+(1/2)γt2-t+η.SupposeL(u)=2γ(1-γu)-3, u∈(0,1/γ), g(t)=η-t+γt2/(1-γt), α=(1-2/2)(1/γ) and β=(3-22)/γ and for ∥x-x0∥<α, ∥F′(x0)-1F′′′(x)∥≤6γ2/(1-γ∥x-x0∥)4. Hence, for ∥x-x0∥+∥y-x∥<α, we get
(33)∥F′(x0)-1[F′′(y)-F′′(x)]∥=∥∫01F′(x0)-1F′′′(x+t(y-x))dt(y-x)∥≤∫016γ2[1-γ∥x-x0+t(y-x)∥]4dt∥y-x∥≤∫016γ2[1-γ(∥x-x0∥+t∥y-x∥)]4dt∥y-x∥=∫∥x-x0∥∥x-x0∥+∥y-x∥6γ2(1-γu)4u·=∫∥x-x0∥∥x-x0∥+∥y-x∥L′(u)du.Corollary 7 (see [10]).
SupposeX and Y are Banach spaces, Ω is an open convex subset of the Banach space X, F:Ω⊂X→Y has the third-order Fréchet derivative, F′(x0)-1 exists for x0∈Ω, and the following conditions hold:
(34)∥F′(x0)-1F(x0)∥≤η,∥F′(x0)-1F′′(x0)∥≤2γ,∥F′(x0)-1F′′′(x)∥≤6γ2(1-γ∥x-x0∥)4=g′′′(∥x-x0∥),x∈Ω,∥x-x0∥≤(1-12)1γ,ηγ≤3-22,S(x0,r1)-⊂Ω.
Then the sequence{xn}n≥0 generated by (5) is well defined, xn∈S(x0,r1)¯, and {xn} converges to the unique solution x* of (1) on S(x0,(1-1/2)(1/γ)), where
(35)r1=1+ηγ-(1+ηγ)2-8ηγ4γ,r2=1+ηγ+(1+ηγ)2-8ηγ4γ
are two positive roots of the equation g(t)=η-t+γt2/(1-γt).


## 3. Numerical Examples

In this section, we apply the convergence theorem and show three numerical examples.Example 1.
Consider the equation(36)F(x)=16x3+16x2-56x+13=0.
We choose the initial pointx0=0, Ω=[-1,1]; then
(37)η=|F′(0)-1F(0)|=25,γ=|F′(0)-1F′′(0)|=25,K=65,2(γ+2γ2+2K)3(γ+γ2+2K)2=35>η.
Hence, by Corollary6, the sequence {xn}n≥0 generated by (5) is well defined, and {xn} converges to the solution x* of (36).
Now, we will analyze errors∥xn-x*∥ by Corollary 6 (see Table 1). In this case, we take x0=0; then r1=0.462598422⋯.Table 1
Error results for Corollary6  (∥xn-x*∥≤r1-tn).
Step
r
1
-
t
n
Step
r
1
-
t
n
k
=
1
1.616985
×
10
-
2
k
=
2
2.236349
×
10
-
6
k
=
3
6.225929
×
10
-
18
k
=
4
1.343387
×
10
-
52
k
=
5
1.349560
×
10
-
156
k
=
6
1.368249
×
10
-
468Example 2.
Consider the system of equation [18] F(u,v)=0, where
(38)F(u,v)=(uv-1,uv+u-2v)T.
Then, we have(39)F′(u,v)=(vuv+1u-2),F′(u,v)-1=-1u+2v(u-2-u-v-1v),F′′(u,v)=(01100110).
We choosex0=(u0,v0)=(1.75,1.75) and Ω={x∣∥x-x0∥≤1.75}. We take the max-norm in R2 and the norm ∥A∥=max{|a11|+|a12|,|a21|+|a22|} for A=(a11a12a21a22). Define the norm of a bilinear operator B on R2 by
(40)∥B∥=sup∥u∥=1maxi∑j=12|∑k=12bijkuk|,
where u=(u1,u2)T and
(41)B=(b111b112b121b122b211b212b221b222).
Then we get the following results:(42)η=∥F′(x0)-1F(x0)∥=914,γ=∥F′(x0)-1F′′(x0)∥=1621,K=0,2(γ+2γ2+2K)3(γ+γ2+2K)2>η.
This means that the hypotheses of Corollary6 are satisfied.
Now, we will analyze errors∥xn-x*∥ by Corollary 6 (see Table 2). In this case, we take x0=(u0,v0)=(1.75,1.75); then r1=1.125.Table 2
Error results for Corollary6  (∥xn-x*∥≤r1-tn).
Step
r
1
-
t
n
Step
r
1
-
t
n
k
=
1
2.736486
×
10
-
1
k
=
2
3.044252
×
10
-
2
k
=
3
1.588069
×
10
-
4
k
=
4
2.844419
×
10
-
11
k
=
5
1.636509
×
10
-
30
k
=
6
3.116680
×
10
-
92Example 3.
Consider the following integral equations:(43)x(s)=1+14x(s)∫01‍ss+tx(t)dt
and the space X=C[0,1] with the norm
(44)∥x∥=max0≤s≤1|x(s)|.
This equation arises in the theory of radiative transfer and neutron transport and in the kinetic theory of gases. Define the operatorF on X by
(45)F(x)=14x(s)∫01‍ss+tx(t)dt-x(s)+1.
Then, forx0=1, we obtain
(46)η=∥F′(x0)-1F(x0)∥=0.2652,2γ=∥F′(x0)-1F′′(x0)∥=1.5304×2·14max0≤s≤1|∫01ss+tdt|=1.5304×ln22=0.5303,ηγ=0.07032<3-22,∥F′(x0)-1F′′′(x)∥=0<6γ2(1-γ∥x-x0∥)4.
This means that the hypotheses of Corollary 7 are satisfied.
Now, we will analyze errors∥xn-x*∥ by Corollary 7 (see Table 3). In this case, we take x0=1; then r1=0.289222⋯.Table 3
Error results for Corollary7  (∥xn-x*∥≤r1-tn).
Step
r
1
-
t
n
Step
r
1
-
t
n
k
=
1
2.764303
×
10
-
3
k
=
2
4.099223
×
10
-
9
k
=
3
1.344301
×
10
-
26
k
=
4
4.741124
×
10
-
79
k
=
5
2.079868
×
10
-
236
k
=
6
<1.0 × 10−500



---

*Source: 923898-2013-12-31.xml*