Expanding the applicability of Lavrentiev regularization methods for ill-posed problems

  • Ioannis K Argyros1,

    Affiliated with

    • Yeol Je Cho2Email author and

      Affiliated with

      • Santhosh George3

        Affiliated with

        Boundary Value Problems20132013:114

        DOI: 10.1186/1687-2770-2013-114

        Received: 29 January 2013

        Accepted: 18 April 2013

        Published: 7 May 2013

        Abstract

        In this paper, we are concerned with the problem of approximating a solution of an ill-posed problem in a Hilbert space setting using the Lavrentiev regularization method and, in particular, expanding the applicability of this method by weakening the popular Lipschitz-type hypotheses considered in earlier studies such as (Bakushinskii and Smirnova in Numer. Funct. Anal. Optim. 26:35-48, 2005; Bakushinskii and Smirnova in Nonlinear Anal. 64:1255-1261, 2006; Bakushinskii and Smirnova in Numer. Funct. Anal. Optim. 28:13-25, 2007; Jin in Math. Comput. 69:1603-1623, 2000; Mahale and Nair in ANZIAM J. 51:191-217, 2009). Numerical examples are given to show that our convergence criteria are weaker and our error analysis tighter under less computational cost than the corresponding works given in (Bakushinskii and Smirnova in Numer. Funct. Anal. Optim. 26:35-48, 2005; Bakushinskii and Smirnova in Nonlinear Anal. 64:1255-1261, 2006; Bakushinskii and Smirnova in Numer. Funct. Anal. Optim. 28:13-25, 2007; Jin in Math. Comput. 69:1603-1623, 2000; Mahale and Nair in ANZIAM J. 51:191-217, 2009).

        MSC:65F22, 65J15, 65J22, 65M30, 47A52.

        Keywords

        Lavrentiev regularization method Hilbert space ill-posed problems stopping index Fréchet-derivative source function boundary value problem

        1 Introduction

        In this paper, we are interested in obtaining a stable approximate solution for a nonlinear ill-posed operator equation of the form
        F ( x ) = y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ1_HTML.gif
        (1.1)
        where F : D ( F ) X X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq1_HTML.gif is a monotone operator and X is a Hilbert space. We denote the inner product and the corresponding norm on a Hilbert space by , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq2_HTML.gif and http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq3_HTML.gif, respectively. Let U ( x , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq4_HTML.gif stand for the open ball in X with center x X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq5_HTML.gif and radius r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq6_HTML.gif. Note that F is a monotone operator if it satisfies the relation
        F ( x 1 ) F ( x 2 ) , x 1 x 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ2_HTML.gif
        (1.2)

        for all x 1 , x 2 D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq7_HTML.gif.

        We assume, throughout this paper, that y δ Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq8_HTML.gif is the available noisy data with
        y y δ δ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ3_HTML.gif
        (1.3)
        and (1.1) has a solution x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq9_HTML.gif. Since (1.1) is ill-posed, its solution need not depend continuously on the data, i.e., small perturbation in the data can cause large deviations in the solution. So, the regularization methods are used [18]. Since F is monotone, the Lavrentiev regularization is used to obtain a stable approximate solution of (1.1). In the Lavrentiev regularization, the approximate solution is obtained as a solution of the equation
        F ( x ) + α ( x x 0 ) = y δ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ4_HTML.gif
        (1.4)

        where α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq10_HTML.gif is the regularization parameter and x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq11_HTML.gif is an initial guess for the solution x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq9_HTML.gif.

        In [9], Bakushinskii and Smirnova proposed an iterative method
        x k + 1 δ = x k δ ( α k I + A k , δ ) 1 [ ( F ( x k δ ) y δ ) + α k ( x k δ x 0 ) ] , x 0 δ = x 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ5_HTML.gif
        (1.5)

        where A k , δ : = F ( x k δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq12_HTML.gif and ( α k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq13_HTML.gif is a sequence of positive real numbers satisfying α k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq14_HTML.gif as k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq15_HTML.gif. It is important to stop the iteration at an appropriate step, say k = k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq16_HTML.gif, and show that x k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq17_HTML.gif is well defined for 0 < k k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq18_HTML.gif and x k δ δ x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq19_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif (see [10]).

        In [9, 11, 12], Bakushinskii and Smirnova chose the stopping index k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq21_HTML.gif by requiring it to satisfy
        F ( x k δ δ ) y δ 2 τ δ < F ( x k δ ) y δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equa_HTML.gif
        for k = 0 , 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq22_HTML.gif and k δ 1 , τ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq23_HTML.gif. In fact, they showed that x k δ δ x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq19_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif under the following assumptions:
        1. (1)

          There exists L > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq24_HTML.gif such that F ( x ) F ( y ) L x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq25_HTML.gif for all x , y D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq26_HTML.gif;

           
        2. (2)
          There exists p > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq27_HTML.gif such that
          α k α k + 1 α k α k + 1 p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ6_HTML.gif
          (1.6)
           
        for all k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq28_HTML.gif;
        1. (3)
          ( 2 + L σ ) x 0 x ˆ t d σ 2 x 0 x ˆ t d α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq29_HTML.gif, where
          σ : = ( τ 1 ) 2 , t : = p α 0 + 1 , d = 2 ( t x 0 x ˆ + p σ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equb_HTML.gif
           

        However, no error estimate was given in [9] (see [10]).

        In [10], Mahale and Nair, motivated by the work of Qi-Nian Jin [13] for an iteratively regularized Gauss-Newton method, considered an alternate stopping criterion which not only ensures the convergence, but also derives an order optimal error estimate under a general source condition on x ˆ x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq30_HTML.gif. Moreover, the condition that they imposed on { α k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq31_HTML.gif is weaker than (1.6).

        In the present paper, we are motivated by [10]. In particular, we expand the applicability of the method (1.5) by weakening one of the major hypotheses in [10] (see Assumption 2.1(2) in the next section).

        In Section 2, we consider some basic assumptions required throughout the paper. Section 3 deals with the stopping rule and the result that establishes the existence of the stopping index. In Section 4, we prove results for the iterations based on the exact data and, in Section 5, the error analysis for the noisy data case is proved. The main order optimal result using the a posteriori stopping rule is provided in Section 6.

        2 Basic assumptions and some preliminary results

        We use the following assumptions to prove the results in this paper.

        Assumption 2.1
        1. (1)

          There exists r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq6_HTML.gif such that U ( x ˆ , r ) D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq32_HTML.gif and F : U ( x ˆ , r ) X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq33_HTML.gif is Fréchet differentiable.

           
        2. (2)
          There exists K 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq34_HTML.gif such that, for all u θ = u + θ ( x ˆ u ) U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq35_HTML.gif, θ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq36_HTML.gif and v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq37_HTML.gif, there exists an element, say ϕ ( x ˆ , u θ , v ) X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq38_HTML.gif, satisfying
          [ F ( x ˆ ) F ( u θ ) ] v = F ( u θ ) ϕ ( x ˆ , u θ , v ) , ϕ ( x ˆ , u θ , v ) K 0 v x ˆ u θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equc_HTML.gif
           
        for all u θ U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq39_HTML.gif and v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq37_HTML.gif.
        1. (3)

          ( F ( u ) + α I ) 1 F ( u θ ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq40_HTML.gif for all u θ U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq39_HTML.gif.

           
        2. (4)

          ( F ( u ) + α I ) 1 1 α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq41_HTML.gif for all u θ U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq39_HTML.gif.

           

        The condition (2) in Assumption 2.1 weakens the popular hypotheses given in [10, 14] and [15].

        Assumption 2.2 There exists a constant K > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq42_HTML.gif such that, for all x , y U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq43_HTML.gif and v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq37_HTML.gif, there exists an element denoted by P ( x , u , v ) X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq44_HTML.gif satisfying
        [ F ( x ) F ( u ) ] v = F ( u ) P ( x , u , v ) , P ( x , u , v ) K v x u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equd_HTML.gif

        Clearly, Assumption 2.2 implies Assumption 2.1(2) with K 0 = K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq45_HTML.gif, but not necessarily vice versa. Note that K 0 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq46_HTML.gif holds in general and K 0 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq47_HTML.gif can be arbitrarily large [1620]. Indeed, there are many classes of operators satisfying Assumption 2.1(2), but not Assumption 2.2 (see the numerical examples at the end of this study). Moreover, if K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq48_HTML.gif is sufficiently smaller than K, which can happen since K K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq49_HTML.gif can be arbitrarily large, then the results obtained in this study provide a tighter error analysis than the one in [10].

        Finally, note that the computation of constant K is more expensive than the computation of K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq50_HTML.gif.

        We need the auxiliary results based on Assumption 2.1.

        Proposition 2.3 For any u U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq51_HTML.gif and α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq52_HTML.gif,
        ( F ( u ) + α I ) 1 [ F ( x ˆ ) F ( u ) F ( u ) ] ( x ˆ u ) 3 K 0 2 x ˆ u 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Eque_HTML.gif
        Proof Using the fundamental theorem of integration, for any u U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq51_HTML.gif, we get
        F ( x ˆ ) F ( u ) = 0 1 F ( u + t ( x ˆ u ) ) ( x ˆ u ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equf_HTML.gif
        Hence, by Assumption 2.2,
        F ( x ˆ ) F ( u ) F ( u ) ( x ˆ u ) = 0 1 [ F ( u + t ( x ˆ u ) ) F ( x ˆ ) + F ( x ˆ ) F ( u ) ] ( x ˆ u ) d t = 0 1 F ( x ˆ ) [ ϕ ( u + t ( x ˆ u ) , x ˆ , x ˆ u ) ϕ ( u , x ˆ , x ˆ u ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equg_HTML.gif
        Then, by (2), (3) in Assumption 2.1 and the inequality ( F ( u ) + α I ) 1 F ( u θ ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq53_HTML.gif, we obtain in turn
        ( F ( u ) + α I ) 1 [ F ( x ˆ ) F ( u ) F ( u ) ] ( x ˆ u ) 0 1 ϕ ( u + t ( x ˆ u ) , x ˆ , x ˆ u ) + ϕ ( u , x ˆ , x ˆ u ) d t 0 1 K 0 x ˆ u 2 t d t + K 0 x ˆ u 2 3 K 0 2 x ˆ u 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equh_HTML.gif

        This completes the proof. □

        Proposition 2.4 For any u U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq51_HTML.gif and α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq52_HTML.gif,
        α ( F ( x ˆ ) + α I ) 1 ( F ( u ) + α I ) 1 K 0 x ˆ u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ7_HTML.gif
        (2.1)
        Proof Let T x ˆ , u = α ( ( F ( x ˆ ) + α I ) 1 ( F ( u ) + α I ) 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq54_HTML.gif for all v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq37_HTML.gif. Then we have, by Assumption 2.2,
        T x ˆ , u v = α ( F ( x ˆ ) + α I ) 1 ( F ( u ) F ( x ˆ ) ( F ( u ) + α I ) 1 ) v = ( F ( x ˆ ) + α I ) 1 F ( x ˆ ) ϕ ( u , x ˆ , α ( F ( u ) + α I ) 1 v ) K 0 x ˆ u v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equi_HTML.gif

        for all v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq37_HTML.gif. This completes the proof. □

        Assumption 2.5 There exists a continuous and strictly monotonically increasing function φ : ( 0 , a ] ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq55_HTML.gif with a F ( x ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq56_HTML.gif satisfying
        1. (1)

          lim λ 0 φ ( λ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq57_HTML.gif;

           
        2. (2)

          sup λ 0 α φ ( λ ) λ + α φ ( α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq58_HTML.gif for all α ( 0 , a ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq59_HTML.gif;

           
        3. (3)
          there exists v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq60_HTML.gif with v 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq61_HTML.gif such that
          x ˆ x 0 = φ ( F ( x ˆ ) ) v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ8_HTML.gif
          (2.2)
           

        Next, we assume a condition on the sequence { α k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq31_HTML.gif considered in (1.5).

        Assumption 2.6 ([10], Assumption 2.6)

        The sequence { α k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq62_HTML.gif of positive real numbers is such that
        1 α k α k + 1 μ , lim k 0 α k = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ9_HTML.gif
        (2.3)

        for a constant μ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq63_HTML.gif.

        Note that the condition (2.3) on { α k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq31_HTML.gif is weaker than (1.6) considered by Bakushinskii and Smirnova [9] (see [10]). In fact, if (1.6) is satisfied, then it also satisfies (2.3) with μ = p α 0 + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq64_HTML.gif, but the converse need not be true (see [10]). Further, note that for these choices of { α k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq31_HTML.gif, α k / α k + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq65_HTML.gif is bounded, whereas ( α k α k + 1 ) / α k α k + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq66_HTML.gif as k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq67_HTML.gif. (2) in Assumption 2.1 is used in the literature for regularization of many nonlinear ill-posed problems (see [4, 7, 8, 13, 21]).

        3 Stopping rule

        Let c 0 > 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq68_HTML.gif and choose k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq21_HTML.gif to be the first non-negative integer such that x k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq69_HTML.gif in (1.5) is defined for each k { 0 , 1 , 2 , , k δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq70_HTML.gif and
        α k δ ( A k δ δ + α k δ I ) 1 ( F ( x k δ δ ) y δ ) c 0 δ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ10_HTML.gif
        (3.1)
        In the following, we establish the existence of such a k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq21_HTML.gif. First, we consider the positive integer N N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq71_HTML.gif satisfying
        α N ( c 1 ) δ x 0 x ˆ < α k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ11_HTML.gif
        (3.2)

        for all k { 0 , 1 , , N 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq72_HTML.gif, where c > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq73_HTML.gif and α 0 > ( c 1 ) δ / x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq74_HTML.gif.

        The following technical lemma from [10] is used to prove some of the results of this paper.

        Lemma 3.1 ([10], Lemma 3.1)

        Let a > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq75_HTML.gif and b 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq76_HTML.gif be such that 4 a b 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq77_HTML.gif and θ : = ( 1 1 4 a b ) / 2 a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq78_HTML.gif. Let θ 1 , , θ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq79_HTML.gif be non-negative real numbers such that θ k + 1 a θ k 2 + b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq80_HTML.gif and θ 0 θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq81_HTML.gif. Then θ k θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq82_HTML.gif for all k = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq83_HTML.gif.

        The rest of the results in this paper can be proved along the same lines as those of the proof in [10]. In order for us to make the paper as self-contained as possible, we present the proof of one of them, and for the proof of the rest, we refer the reader to [10].

        Theorem 3.2 ([10], Theorem 3.2)

        Let (1.2), (1.3), (2.3) and Assumption  2.1 be satisfied. Let N be as in (3.2) for some c > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq73_HTML.gif and 6 c K 0 x 0 x ˆ / ( c 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq84_HTML.gif. Then x k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq69_HTML.gif is defined iteratively for each k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif and
        x k δ x ˆ 2 c x 0 x ˆ c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ12_HTML.gif
        (3.3)
        for all k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif. In particular, if r > 2 c x 0 x ˆ / ( c 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq86_HTML.gif, then x k δ B r ( x ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq87_HTML.gif for k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif. Moreover,
        α N ( A N δ + α N I ) 1 ( F ( x N δ ) y δ ) c 0 δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ13_HTML.gif
        (3.4)

        for c 0 : = 7 3 c + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq88_HTML.gif.

        Proof We show (3.3) by induction. It is obvious that (3.3) holds for k = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq89_HTML.gif. Now, assume that (3.3) holds for some k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif. Then it follows from (1.5) that
        x k + 1 δ x ˆ = x k δ x ˆ ( A k δ + α k I ) 1 [ F ( x k δ ) y δ + α k ( x k δ x 0 ) ] = ( A k δ + α k I ) 1 ( ( A k δ + α k I ) ( x k δ x ˆ ) [ F ( x k δ ) y δ + α k ( x k δ x 0 ) ] ) = ( A k δ + α k I ) 1 [ A k δ ( x k δ x ˆ ) + y δ F ( x k δ ) + α k ( x 0 x ˆ ) ] = α k ( A k δ + α k I ) 1 ( x 0 x ˆ ) + ( A k δ + α k I ) 1 ( y δ y ) + ( A k δ + α k I ) 1 [ F ( x ˆ ) F ( x k δ ) + A k δ ( x k δ x ˆ ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ14_HTML.gif
        (3.5)
        Using (1.3), the estimates ( A k δ + α k I ) 1 1 / α k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq90_HTML.gif, ( A k δ + α k I ) 1 A k δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq91_HTML.gif and Proposition 2.3, we have
        α k ( A k δ + α k I ) 1 ( x 0 x ˆ ) + ( A k δ + α k I ) 1 ( y δ y ) x 0 x ˆ + δ α k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equj_HTML.gif
        and
        ( A k δ + α k I ) 1 [ F ( x ˆ ) F ( x k δ ) + A k δ ( x k δ x ˆ ) ] 3 K 0 2 x k δ x ˆ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equk_HTML.gif
        Thus we have
        x k + 1 δ x ˆ x 0 x ˆ + δ α k + 3 K 0 2 x k δ x ˆ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equl_HTML.gif
        But, by (3.2), δ α k x 0 x ˆ / ( c 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq92_HTML.gif and so
        x k + 1 δ x ˆ c x 0 x ˆ c 1 + 3 K 0 2 x k δ x ˆ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equm_HTML.gif
        which leads to the recurrence relation
        θ k + 1 a θ k 2 + b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equn_HTML.gif
        where
        θ k = x k δ x ˆ , a = 3 K 0 2 , b = c x 0 x ˆ c 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equo_HTML.gif
        From the hypothesis of the theorem, we have 4 a b = 6 c K 0 x 0 x ˆ c 1 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq93_HTML.gif. It is obvious that
        θ 0 x 0 x ˆ θ : = 1 1 4 a b 2 a = 2 b 1 + 1 4 a b 2 b = 2 c x 0 x ˆ c 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equp_HTML.gif
        Hence, by Lemma 3.1, we get
        x k δ x ˆ θ 2 c x 0 x ˆ c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ15_HTML.gif
        (3.6)

        for all k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif. In particular, if r > 2 c x 0 x ˆ / ( c 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq94_HTML.gif, then we have x k δ B r ( x ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq95_HTML.gif for all k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif.

        Next, let γ = α N ( A N δ + α N I ) 1 ( F ( x N δ ) y δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq96_HTML.gif. Then, using the estimates
        α N ( A N δ + α N I ) 1 1 , α N ( A N δ + α N I ) 1 A N δ α k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equq_HTML.gif
        and Proposition 2.3, we have
        γ δ + α N ( A N δ + α N I ) 1 ( F ( x N δ ) y + A N δ ( x N δ x ˆ ) A N δ ( x N δ x ˆ ) ) = δ + α N ( A N δ + α N I ) 1 [ F ( x N δ ) F ( x ˆ ) A N δ ( x N δ x ˆ ) + A N δ ( x N δ x ˆ ) ] δ + α N [ 3 K 0 x N δ x ˆ 2 2 + x N δ x ˆ ] δ + α N x N δ x ˆ [ 1 + 3 K 0 x N δ x ˆ 2 ] δ + 2 α N c x 0 δ x ˆ c 1 [ 1 + 3 K 0 c x 0 δ x ˆ c 1 ] δ + 2 c δ [ 1 + 1 6 ] ( 7 c 3 + 1 ) δ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ16_HTML.gif
        (3.7)

        Therefore, we have α N ( A N δ + α N I ) 1 ( F ( x N δ ) y δ ) c 0 δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq97_HTML.gif, where c 0 : = 7 3 c + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq88_HTML.gif. This completes the proof. □

        4 Error bound for the case of noise-free data

        Let
        x k + 1 = x k ( A k + α k I ) 1 [ F ( x k ) y + α k ( x k x 0 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ17_HTML.gif
        (4.1)

        for all k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq98_HTML.gif.

        We show that each x k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq17_HTML.gif is well defined and belongs to U ( x ˆ , r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq99_HTML.gif for r > 2 x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq100_HTML.gif. For this, we make use of the following lemma.

        Lemma 4.1 ([10], Lemma 4.1)

        Let Assumption  2.1 hold. Suppose that, for all k { 0 , 1 , , n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq101_HTML.gif, x k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq17_HTML.gif in (4.1) is well defined and ρ k : = α k ( A k + α k I ) 1 ( x 0 x ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq102_HTML.gif for some n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq103_HTML.gif. Then we have
        ρ k 3 K 0 x k x ˆ 2 2 x k + 1 x ˆ ρ k + 3 K 0 x k x ˆ 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ18_HTML.gif
        (4.2)

        for all k { 0 , 1 , , n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq104_HTML.gif.

        Theorem 4.2 ([10], Theorem 4.2)

        Let Assumption  2.1 hold. If 6 K 0 x 0 x ˆ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq105_HTML.gif and r > 2 x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq100_HTML.gif, then, for all k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq106_HTML.gif, the iterates x k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq17_HTML.gif in (4.1) are well defined and
        x k x ˆ 2 x 0 x ˆ 1 + 1 6 K 0 x 0 x ˆ 2 x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ19_HTML.gif
        (4.3)

        for all k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq106_HTML.gif.

        Lemma 4.3 ([10], Lemma 4.3)

        Let Assumptions  2.1 and  2.6 hold and let r > 2 x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq100_HTML.gif. Assume that A η α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq107_HTML.gif and 4 μ ( 1 + η 1 ) K 0 x 0 x ˆ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq108_HTML.gif for some η with 0 < η < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq109_HTML.gif. Then, for all k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq106_HTML.gif, we have
        1 ( 1 + η ) μ x k x ˆ α k ( A k + α k I ) 1 ( x 0 x ˆ ) 1 1 η x k x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ20_HTML.gif
        (4.4)
        and
        1 η ( 1 + η ) μ x k x ˆ ( x k + 1 x ˆ ) ( 1 1 η + η ( 1 + η ) μ ) x k x ˆ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ21_HTML.gif
        (4.5)

        The following corollary follows from Lemma 4.3 by taking η = 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq110_HTML.gif. We show that this particular case of Lemma 4.3 is better suited for our later results.

        Corollary 4.4 ([10], Corollary 4.4)

        Let Assumptions  2.1 and  2.6 hold and let r > 2 x 0 x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq111_HTML.gif. Assume that A α 0 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq112_HTML.gif and 16 μ K 0 x 0 x ˆ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq113_HTML.gif. Then, for all k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq106_HTML.gif, we have
        3 4 μ x k x ˆ α k ( A + α k I ) 1 ( x 0 x ˆ ) 3 2 x k x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ22_HTML.gif
        (4.6)
        and
        x k x ˆ 2 μ ( x k + 1 x ˆ ) 2 x k x ˆ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equr_HTML.gif

        Theorem 4.5 ([10], Theorem 4.5)

        Let the assumptions of Lemma  4.3 hold. If x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq11_HTML.gif is chosen such that x 0 x ˆ N ( F ( x ˆ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq114_HTML.gif, then lim k x k = x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq115_HTML.gif.

        Lemma 4.6 ([10], Lemma 4.6)

        Let the assumptions of Lemma  4.3 hold for η satisfying
        ( 1 1 η ( 1 + η ) μ ) [ 1 + ( 2 μ 1 ) η + 2 μ ] + 2 η < 4 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ23_HTML.gif
        (4.7)
        Then, for all k , l N { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq116_HTML.gif with k l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq117_HTML.gif, we have
        x l x ˆ c η [ x k x ˆ + α l ( A + α l I ) 1 ( F ( x l ) y ) α k ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equs_HTML.gif
        where
        c η : = ( 1 b η ) 1 max { μ , 1 + ( 3 ε + 1 ) η 4 ( 1 η ) } , b η : = ( 3 ε + 1 ) η ( 1 η ) + 3 ε a 4 , ε : = 1 1 a a , a : = η ( 1 + η ) μ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equt_HTML.gif

        Remark 4.7 ([10], Remark 4.7)

        It can be seen that (4.7) is satisfied if η 1 / 3 + 1 / 24 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq118_HTML.gif.

        Now, if we take η = 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq110_HTML.gif, that is, K 0 x 0 x ˆ μ 1 / 16 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq119_HTML.gif in Lemma 4.6, then it takes the following form.

        Lemma 4.8 ([10], Lemma 4.8)

        Let the assumptions of Lemma  4.3 hold with η = 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq110_HTML.gif. Then, for all k l 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq120_HTML.gif, we have
        x l x ˆ c 1 / 3 [ x k x ˆ + α l ( A + α l I ) 1 ( F ( x l ) y ) α k ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equu_HTML.gif
        where
        c 1 / 3 = [ 1 8 μ + ( 8 μ + 1 ) 3 ε 16 μ ] 1 max { μ , 1 + 3 ε + 1 8 } , ε : = 4 μ 4 μ + 4 μ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equv_HTML.gif

        5 Error analysis with noisy data

        The first result in this section gives an error estimate for x k δ x k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq121_HTML.gif under Assumption 2.5, where k = 0 , 1 , 2 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq122_HTML.gif.

        Lemma 5.1 ([10], Lemma 5.1)

        Let Assumption  2.1 hold and let K 0 x 0 x ˆ 1 / m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq123_HTML.gif, where m > ( 7 + 73 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq124_HTML.gif, and N be the integer satisfying (3.2) with
        c > m 2 4 m 6 m 2 7 m 6 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equw_HTML.gif
        Then, for all k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif, we have
        x k δ x k δ ( 1 κ ) α k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ24_HTML.gif
        (5.1)
        where
        κ : = 1 m ( 4 + 3 c c 1 + 6 m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equx_HTML.gif

        If we take m = 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq125_HTML.gif in Lemma 5.1, then we get the following corollary as a particular case of Lemma 5.1. We make use of it in the following error analysis.

        Corollary 5.2 ([10], Corollary 5.2)

        Let Assumption  2.1 hold and let 16 K 0 x 0 x ˆ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq126_HTML.gif. Let N be the integer defined by (3.2) with c > 13 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq127_HTML.gif. Then, for all k { 0 , 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq85_HTML.gif, we have
        x k δ x k δ ( 1 κ ) α k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equy_HTML.gif
        where
        κ : = 31 c 19 32 ( c 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equz_HTML.gif

        Lemma 5.3 ([10], Lemma 5.3)

        Let the assumptions of Lemma  5.1 hold. Then we have
        α k ( A + α k δ I ) 1 ( F ( x k δ ) y ) c 1 δ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equaa_HTML.gif
        Moreover, if k δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq128_HTML.gif, then, for all 0 k < k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq129_HTML.gif, we have
        α k ( A + α k I ) 1 ( F ( x k ) y ) c 2 δ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equab_HTML.gif
        where
        c 1 = ( 1 + 2 c K 0 x 0 x ˆ c 1 ) ( c 0 + 2 κ 1 κ + 3 K 0 μ x 0 x ˆ 2 ( 1 κ ) 2 ( c 1 ) ) , c 2 = c 0 ( ( 2 κ ) ( 1 κ ) ) ( 3 K 0 x 0 x ˆ / 2 ( 1 κ ) 2 ( c 1 ) ) 1 + 2 ( c K 0 x 0 x ˆ / ( c 1 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equac_HTML.gif

        with c 0 = 7 3 c + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq130_HTML.gif and κ as in Lemma  5.1.

        Theorem 5.4 ([10], Theorem 5.4)

        Let Assumptions  2.1 and  2.6 hold. If 16 k μ x 0 x ˆ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq131_HTML.gif and the integer k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq21_HTML.gif is chosen according to stopping rule (3.1) with c 0 > 94 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq132_HTML.gif, then we have
        x k δ δ x ˆ ξ inf { x k x ˆ + δ α k : k 0 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ25_HTML.gif
        (5.2)

        where ξ = max { 2 μ ϱ , c 1 / 3 c 1 + 1 1 κ , c } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq133_HTML.gif, ϱ : = 1 + μ ( 1 + 3 K 0 x 0 x ˆ ) c 2 ( 1 κ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq134_HTML.gif with c 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq135_HTML.gif and κ as in Lemma  4.8 and Corollary  5.2, respectively, and c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq136_HTML.gif, c 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq137_HTML.gif as in Lemma  5.3.

        6 Order optimal result with an a posteriori stopping rule

        In this section, we show the convergence x k δ δ x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq19_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq138_HTML.gif and also give an optimal error estimate for x k δ δ x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq139_HTML.gif.

        Theorem 6.1 ([10], Theorem 6.1)

        Let the assumptions of Theorem  5.4 hold and let k δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq21_HTML.gif be the integer chosen by (3.1). If x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq11_HTML.gif is chosen such that x 0 x ˆ N ( F ( x ˆ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq114_HTML.gif, then we have lim δ 0 x k δ δ = x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq140_HTML.gif. Moreover, if Assumption  2.5 is satisfied, then we have
        x k δ δ x ˆ ξ μ ψ 1 ( δ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equad_HTML.gif

        where ξ : = 8 μ ξ / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq141_HTML.gif with ξ as in Theorem  5.4 and ψ : ( 0 , φ ( a ) ] ( 0 , a φ ( a ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq142_HTML.gif is defined as ψ ( λ ) : = λ φ 1 ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq143_HTML.gif, λ ( 0 , φ ( a ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq144_HTML.gif.

        Proof From (4.6) and (5.2), we get
        x k δ δ x ˆ ξ inf { α k ( A + α k I ) 1 ( x 0 x ˆ ) + δ α k : k = 0 , 1 , } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ26_HTML.gif
        (6.1)
        where ξ = 4 μ 3 max { 2 μ ( 1 + μ ( 1 + 3 k 0 x 0 x ˆ ) c 2 ( 1 κ ) ) , c 1 / 3 c 1 + 1 1 κ , c } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq145_HTML.gif. Now, we choose an integer m δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq146_HTML.gif such that m δ = max { k : α k δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq147_HTML.gif. Then we have
        x k δ δ x ˆ ξ inf { α m δ ( A + α m δ I ) 1 ( x 0 x ˆ ) + δ α m δ : k = 0 , 1 , } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ27_HTML.gif
        (6.2)
        Note that δ α m δ δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq148_HTML.gif, so δ α m δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq149_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif. Therefore by (6.2) to show that x k δ δ x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq19_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif, it is enough to prove that α m δ ( A + α m δ I ) 1 ( x 0 x ˆ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq150_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif. Observe that for w R ( F ( x ˆ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq151_HTML.gif, i.e., w = F ( x ˆ ) u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq152_HTML.gif for some u D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq153_HTML.gif, we have α m δ ( A + α m δ I ) 1 w α m δ u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq154_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif. Now since R ( F ( x ˆ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq155_HTML.gif is a dense subset of N ( F ( x ˆ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq156_HTML.gif, it follows that α m δ ( A + α m δ I ) 1 ( x 0 x ˆ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq157_HTML.gif as δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq20_HTML.gif. Using Assumption 2.5, we get that
        α k ( A + α k I ) 1 ( x 0 x ˆ ) φ ( α k ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ28_HTML.gif
        (6.3)
        So, by (6.2) and (6.3), we obtain that
        x k δ δ x ˆ ξ inf { φ ( α k ) + δ α k : k = 0 , 1 , } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ29_HTML.gif
        (6.4)
        Choose k ˆ δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq158_HTML.gif such that
        φ ( α k ˆ δ ) α k ˆ δ δ < φ ( α k ) α k for  k = 0 , 1 , , k δ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ30_HTML.gif
        (6.5)
        This also implies that
        ψ ( φ ( α k ˆ δ ) ) δ < ψ ( φ ( α k ) ) for  k = 0 , 1 , , k δ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ31_HTML.gif
        (6.6)

        From (6.4), x k δ δ x ˆ ξ { φ ( α k ˆ δ ) + δ α k ˆ δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq159_HTML.gif. Now, using (6.5) and (6.6), we get x k δ δ x ˆ 2 ξ δ α k ˆ δ 2 ξ μ δ α k ˆ δ 1 2 ξ μ ψ 1 ( δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq160_HTML.gif. This completes the proof. □

        7 Numerical examples

        We provide two numerical examples, where K 0 < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq161_HTML.gif.

        Example 7.1 Let X = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq162_HTML.gif, D ( F ) = U ( 0 , 1 ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq163_HTML.gif, x ˆ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq164_HTML.gif and define a function F on D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq165_HTML.gif by
        F ( x ) = e x 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ32_HTML.gif
        (7.1)
        Then, using (7.1) and Assumptions 2.1(2) and 2.2, we get
        K 0 = e 1 < K = e . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equae_HTML.gif
        Example 7.2 Let X = C ( [ 0 , 1 ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq166_HTML.gif (: the space of continuous functions defined on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq167_HTML.gif equipped with the max norm) and D ( F ) = U ( 0 , 1 ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq163_HTML.gif. Define an operator F on D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq165_HTML.gif by
        F ( h ) ( x ) = h ( x ) 5 0 1 x θ h ( θ ) 3 d θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ33_HTML.gif
        (7.2)
        Then the Fréchet-derivative is given by
        F ( h [ u ] ) ( x ) = u ( x ) 15 0 1 x θ h ( θ ) 2 u ( θ ) d θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ34_HTML.gif
        (7.3)

        for all u D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq153_HTML.gif. Using (7.2), (7.3), Assumptions 2.1(2), 2.2 for x ˆ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq164_HTML.gif, we get K 0 = 7.5 < K = 15 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq168_HTML.gif.

        Next, we provide an example where K K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq49_HTML.gif can be arbitrarily large.

        Example 7.3 Let X = D ( F ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq169_HTML.gif, x ˆ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq164_HTML.gif and define a function F on D ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq165_HTML.gif by
        F ( x ) = d 0 x d 1 sin 1 + d 1 sin e d 2 x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ35_HTML.gif
        (7.4)

        where d 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq170_HTML.gif, d 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq171_HTML.gif and d 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq172_HTML.gif are the given parameters. Note that F ( x ˆ ) = F ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq173_HTML.gif. Then it can easily be seen that, for d 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq172_HTML.gif sufficiently large and d 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq171_HTML.gif sufficiently small, K K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq49_HTML.gif can be arbitrarily large.

        We now present two examples where Assumption 2.2 is not satisfied, but Assumption 2.1(2) is satisfied.

        Example 7.4 Let X = D ( F ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq169_HTML.gif, x ˆ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq164_HTML.gif and define a function F on D by
        F ( x ) = x 1 + 1 i 1 + 1 i + c 1 x c 1 i i + 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ36_HTML.gif
        (7.5)
        where c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq136_HTML.gif is a real parameter and i > 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq174_HTML.gif is an integer. Then F ( x ) = x 1 / i + c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq175_HTML.gif is not Lipschitz on D. Hence Assumption 2.2 is not satisfied. However, the central Lipschitz condition in Assumption 2.2(2) holds for K 0 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq176_HTML.gif. We also have that F ( x ˆ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq177_HTML.gif. Indeed, we have
        F ( x ) F ( x ˆ ) = | x 1 / i x ˆ 1 / i | = | x x ˆ | x ˆ i 1 i + + x i 1 i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equaf_HTML.gif
        and so
        F ( x ) F ( x ˆ ) K 0 | x x ˆ | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equag_HTML.gif
        Example 7.5 We consider the integral equation
        u ( s ) = f ( s ) + λ a b G ( s , t ) u ( t ) 1 + 1 / n d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ37_HTML.gif
        (7.6)

        for all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq178_HTML.gif, where f is a given continuous function satisfying f ( s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq179_HTML.gif for all s [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq180_HTML.gif, λ is a real number and the kernel G is continuous and positive in [ a , b ] × [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq181_HTML.gif.

        For example, when G ( s , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq182_HTML.gif is the Green kernel, the corresponding integral equation is equivalent to the boundary value problem
        u = λ u 1 + 1 / n , u ( a ) = f ( a ) , u ( b ) = f ( b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equah_HTML.gif
        These types of problems have been considered in [1620]. The equation of the form (7.6) generalizes the equation of the form
        u ( s ) = a b G ( s , t ) u ( t ) n d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ38_HTML.gif
        (7.7)
        which was studied in [1620]. Instead of (7.6), we can try to solve the equation F ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq183_HTML.gif, where
        F : Ω C [ a , b ] C [ a , b ] , Ω = { u C [ a , b ] : u ( s ) 0 , s [ a , b ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equai_HTML.gif
        and
        F ( u ) ( s ) = u ( s ) f ( s ) λ a b G ( s , t ) u ( t ) 1 + 1 / n d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equaj_HTML.gif
        The norm we consider is the max-norm. The derivative F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq184_HTML.gif is given by
        F ( u ) v ( s ) = v ( s ) λ ( 1 + 1 n ) a b G ( s , t ) u ( t ) 1 / n v ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equak_HTML.gif
        for all v Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq185_HTML.gif. First of all, we notice that F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq184_HTML.gif does not satisfy the Lipschitz-type condition in Ω. Let us consider, for instance, [ a , b ] = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq186_HTML.gif, G ( s , t ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq187_HTML.gif and y ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq188_HTML.gif. Then we have F ( y ) v ( s ) = v ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq189_HTML.gif and
        F ( x ) F ( y ) = | λ | ( 1 + 1 n ) a b x ( t ) 1 / n d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equal_HTML.gif
        If F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq184_HTML.gif were the Lipschitz function, then we had
        F ( x ) F ( y ) L 1 x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equam_HTML.gif
        or, equivalently, the inequality
        0 1 x ( t ) 1 / n d t L 2 max x [ 0 , 1 ] x ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equ39_HTML.gif
        (7.8)
        would hold for all x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq190_HTML.gif and for a constant L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq191_HTML.gif. But this is not true. Consider, for example, the function
        x j ( t ) = t j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equan_HTML.gif
        for all j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq192_HTML.gif and t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq193_HTML.gif. If these are substituted into (7.7), then we have
        1 j 1 / n ( 1 + 1 / n ) L 2 j j 1 1 / n L 2 ( 1 + 1 / n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equao_HTML.gif
        for all j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq192_HTML.gif. This inequality is not true when j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq194_HTML.gif. Therefore, Assumption 2.2 is not satisfied in this case. However, Assumption 2.1(2) holds. To show this, suppose that x ˆ ( t ) = f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq195_HTML.gif and γ = min s [ a , b ] f ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq196_HTML.gif. Then, for all v Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq185_HTML.gif, we have
        [ F ( x ) F ( x ˆ ) ] v = | λ | ( 1 + 1 n ) max s [ a , b ] | a b G ( s , t ) ( x ( t ) 1 / n f ( t ) 1 / n ) v ( t ) d t | | λ | ( 1 + 1 n ) max s [ a , b ] G n ( s , t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equap_HTML.gif
        where G n ( s , t ) = G ( s , t ) | x ( t ) f ( t ) | x ( t ) ( n 1 ) / n + x ( t ) ( n 2 ) / n f ( t ) 1 / n + + f ( t ) ( n 1 ) / n v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq197_HTML.gif. Hence it follows that
        [ F ( x ) F ( x ˆ ) ] v = | λ | ( 1 + 1 / n ) γ ( n 1 ) / n max s [ a , b ] a b G ( s , t ) d t x x ˆ K 0 x x ˆ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equaq_HTML.gif

        where K 0 = | λ | ( 1 + 1 / n ) γ ( n 1 ) / n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq198_HTML.gif and N = max s [ a , b ] a b G ( s , t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq199_HTML.gif. Then Assumption 2.1(2) holds for sufficiently small λ.

        In the following remarks, we compare our results with the corresponding ones in [10].

        Remark 7.6 Note that the results in [10] were shown using Assumption 2.2, whereas we used weaker Assumption 2.1(2) in this paper. Next, our result, Proposition 2.3, was shown with 3 K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq200_HTML.gif replacing K. Therefore, if 3 K 0 < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq201_HTML.gif (see Example 7.3), then our result is tighter. Proposition 2.4 was shown with K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq50_HTML.gif replacing K. Then, if K 0 < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq161_HTML.gif, then our result is tighter. Theorem 3.2 was shown with 6 K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq202_HTML.gif replacing 2K. Hence, if 3 K 0 < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq201_HTML.gif, our result is tighter. Similar favorable to us observations are made for Lemma 4.1, Theorem 4.2 and the rest of the results in [10].

        Remark 7.7 The results obtained here can also be realized for the operators F satisfying an autonomous differential equation of the form
        F ( x ) = P ( F ( x ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_Equar_HTML.gif

        where P : X X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq203_HTML.gif is a known continuous operator. Since F ( x ˆ ) = P ( F ( x ˆ ) ) = P ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq204_HTML.gif, we can compute K 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq50_HTML.gif in Assumption 2.1(2) without actually knowing x ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq9_HTML.gif. Returning back to Example 7.1, we see that we can set P ( x ) = x + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-114/MediaObjects/13661_2013_Article_355_IEq205_HTML.gif.

        Declarations

        Acknowledgements

        Dedicated to Professor Hari M Srivastava.

        This paper was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170).

        Authors’ Affiliations

        (1)
        Department of Mathematical Sciences, Cameron University
        (2)
        Department of Mathematics Education and the RINS, Gyeongsang National University
        (3)
        Department of Mathematical and Computational Sciences, National Institute of Technology Karnataka

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