Convergence rates in regularization for a system of nonlinear ill-posed equations with m-accretive operators

Abstract

In this paper, we study an operator version of the modified Browder-Tikhonov regularization method for finding a common solution for a system of ill-posed operator equations involving m-accretive operators $A_i$, $i=0,\dots ,N$, in a reflexive Banach space. The convergence rates of the regularized solutions are estimated not only in the infinite-dimensional space, but also in connection with its finite-dimensional approximations without the weakly sequential continuity of the dual mapping.

MSC:47H17, 47H20.

1 Introduction

Let X be a real reflexive Banach space with the property of approximations and its dual space $X^{\ast }$ be strictly convex. The norms of X and $X^{\ast }$ are denoted by the symbol $\parallel \cdot \parallel$. We write $〈x,x^{\ast }〉$ instead of $x^{\ast }(x)$ for $x^{\ast }\in X^{\ast }$ and $x\in X$.

Definition 1.1 A Banach space X is said to be strictly convex if for $x,y\in S_X$ with $x\ne y$, then

$$\parallel (1-\lambda )x+\lambda y\parallel <1\mathrm{\forall }\lambda \in (0,1),$$

where $S_X$ is the unit sphere $S_X=\{x\in X:\parallel x\parallel =1\}$.

Definition 1.2 A mapping j from X onto $X^{\ast }$ is called the normalized dual mapping of X, if it satisfies the condition

$$〈x,j(x)〉={\parallel x\parallel }^2,\parallel j(x)\parallel =\parallel x\parallel \mathrm{\forall }x\in X.$$

It is well known that if $X^{\ast }$ is strictly convex then j is single-valued.

Definition 1.3 An operator A from X to X is said to be accretive, if

$$〈A(x)-A(y),j(x-y)〉\ge 0\mathrm{\forall }x,y\in D(A),$$

where $D(A)$ denotes the domain of A. An accretive operator A is said to be an m-accretive, if $\mathcal{R}(A+\lambda I)=X$ for $\lambda >0$ where $\mathcal{R}(A)$ and I denote the range of A and the identity mapping of X, respectively.

Definition 1.4 An operator A from X to X is said to be

• (i) demicontinuous if $x_n\to x$ in X implies $A(x_n)\rightharpoonup A(x)$,

• (ii) weakly continuous if $x_n\rightharpoonup x$ implies $A(x_n)\rightharpoonup A(x)$.

It is well known that if A is accretive, and is continuous, demicontinuous, or weakly continuous, then it is m-accretive [1–3].

Definition 1.5 A mapping A from X to X is called Fréchet differentiable at a point $x\in D(A)$, if

$$A(x+h)-A(x)=B(x)h+o(\parallel h\parallel )\mathrm{\forall }x+h\in D(A),$$

where $B(x)$ is a bounded linear mapping from X to X. And the Fréchet derivative of A at $x\in D(A)$ is denoted by $A^{\prime }(x)$.

Let ${\{A_i\}}_{i=0}^N$ be a family of $N+1$ accretive operators in X and satisfy one of the above mentioned three continuities.

Our problem is to find a common solution of the following operator equations:

$$A_i(x)=f_i,f_i\in \mathcal{R}(A_i),i=0,\dots ,N.$$

(1.1)

Set

$$S=\bigcap _{i=0}^N{S}_i,$$

where $S_i$ is the solution set of (1.1), that is, $S_i=\{x:A_i(x)=f_i\}$.

Suppose that $S\ne \mathrm{\varnothing }$.

For m-accretive operators, some results of the approximating solution for each equation in (1.1) under suitable different conditions are investigated in [4–10], and [11].

The system of equations (1.1) is ill-posed, because each one of the system is ill-posed. By ill-posedness, we mean that its solutions do not depend continuously on the data $(A_i,f_i)$. Therefore, we have to use the stable methods in order to solve the problem. Some stable methods of approximating solution for each equation in (1.1) with m-accretive operator are investigated in [12, 13], and [14] having the weakly sequentially continuous duality mapping j. In [15–19], the authors considered the modified Browder-Tikhonov regularization method with the regularization parameter choice without the property for j, for the case of demicontinuous or weakly continuous accretive operators $A_i$ satisfying the condition

$$\parallel A_i(y)-A_i\left(x_0^i\right)-j^{\ast }{A}_i^{\prime }\left(x_0^i\right)j(y-x_0^i)\parallel \le \tilde{\tau }\parallel y-x_0^i\parallel \parallel A_i^{\prime }(x_{\ast })j(y-x_0^i)\parallel$$

(1.2)

for y in some neighborhood of $S_i$, where $A_i^{\prime }(x_0^i)$ is the Fréchet derivative of $A_i$ at $x_0^i\in S_i$, $\tilde{\tau }$ is some positive constant, and $j^{\ast }$ is the normalized duality mapping of $X^{\ast }$.

In many papers, for each i, the regularized solution of (1.1) is constructed by the following operator equation:

$$A_i^h(x)+\alpha x=f_i^{\delta },$$

where $(A_i^h,f_i^{\delta })$ is the approximation of $(A_i,f_i)$ satisfying the conditions:

$$\parallel A_i^h(x)-A_i(x)\parallel \le hg(\parallel x\parallel ),\parallel f_i-f_i^{\delta }\parallel \le \delta ,h,\delta \to 0,$$

(1.3)

$g(t)$ is a nonnegative bounded (image of bounded set is bounded) real function, and $A_i^h$ is also accretive and the same continuity as $A_i$.

The system of equations (1.1) can be written in the form

$$\mathcal{A}(x)=f,$$

(1.4)

where $\mathcal{A}:\mathcal{X}\to \mathcal{X}:=X\times \cdots \times X$ is defined by $\mathcal{A}(x):=(A_0(x),\dots ,A_N(x))$, and $f:=(f_0,\dots ,f_N)$.

Note that (1.4) can be seen as a special case of (1.1) with $N=0$. However, one potential advantage of (1.1) over (1.4) can be that it might better reflect the structure of the underlying information $(f_0,\dots ,f_N)$ leading to the couplet system, than a plain concatenation into one single data element f could. In particular, the second advantage is that in estimating convergence rates of regularization solution, which is showed later, we need only the smooth property for one among $A_i$, while for (1.4) we need the property for $A_i$, $i=0,\dots ,N$.

When for each i, $A_i$ is the nonlinear Fréchet differentiable operator from the Hilbert space X to the Hilbert space $Y_i$ with derivative being uniformly bounded in a neighborhood of a common solution, a stable method for problem (1.1) is considered in [20].

In this paper, we show that a common solution of (1.1) involving m-accretive operators $A_i$, without the weakly sequentially continuous property of j, can be approximated by the modified Browder-Tikhonov regularization method which is described by the following operator equation:

$$A_0^h(x)+{\alpha }^{1+{\mu }_0}\sum _{i=1}^N(A_i^h(x)-f_i^{\delta })+\alpha x=f_0^{\delta },{\mu }_0\ge 0,$$

(1.5)

where $\alpha >0$ is a small regularization parameter. Since the operator

$$A_0^h+{\alpha }^{1+{\mu }_0}\sum _{i=1}^N(A_i^h-f_i^{\delta })$$

has the same properties as each $A_i^h$, it is also m-accretive. Therefore, (1.5) has a unique solution denoted by $x_{\alpha }^{\tau }$, $\tau =(\delta ,h)$, for every value $\alpha >0$.

In the following section, the convergence rates of the regularized solution $x_{\alpha }^{\tau }$ and its finite-dimensional approximations $x_{\alpha ,n}^{\tau }$ are established under an assumption similar to (1.2).

The symbols ‘→’ and ‘⇀’ denote strong and weak convergence, respectively, and the notation $a\sim b$ means that $a=o(b)$ and $b=o(a)$.

2 Main results

Assumption A There exists a constant ${\tau }_0>0$ such that

$$\parallel A_0(x)-A_0(x_0)-j^{\ast }{A}_0^{\prime }{(x_0)}^{\ast }j(x-x_0)\parallel \le {\tau }_0\parallel A_0(x)-A_0(x_0)\parallel ,\mathrm{\forall }x\in X.$$

Now, we are in a position to introduce the main theorem.

Theorem 2.1 Let X be a real reflexive Banach space with the property of approximations and its dual space $X^{\ast }$ be strictly convex. Let ${\{A_i\}}_{i=0}^N$ be a family of $N+1$ accretive operators in X and satisfy one of the above mentioned continuities. Assume that the following conditions hold:

• (i) $A_0$ is Fréchet differentiable at $x_0$ with Assumption A.

• (ii) There exists an element $z\in X$ such that

  $$A_0^{\prime }(x_0)z=-x_0.$$

• (iii) The parameter α is chosen such that $\alpha \sim {(\delta +h)}^{\mu }$, $0<\mu <1$.

Then, for $0<\delta +h<1$, we have

$$\parallel x_{\alpha }^{\tau }-x_0\parallel =o\left({(\delta +h)}^{\theta }\right),\theta =min\{1-{\mu }_{,}\mu /2\}.$$

Proof From the property of j, $A_i^h$, (1.1), (1.3), (1.5), and condition (ii), it follows that

$$\begin{array}{rl}{\parallel x_{\alpha }^{\tau }-x_0\parallel }^2=& 〈x_{\alpha }^{\tau }-x_0,j(x_{\alpha }^{\tau }-x_0)〉\\ =& \frac{1}{\alpha }〈f_0^{\delta }-A_0^h\left(x_{\alpha }^{\tau }\right)+{\alpha }^{1+{\mu }_0}\sum _{i=1}^N(f_i^{\delta }-A_i^h\left(x_{\alpha }^{\tau }\right))-\alpha x_0,j(x_{\alpha }^{\tau }-x_0)〉\\ \le & \frac{1+N{\alpha }^{1+{\mu }_0}}{\alpha }[\delta +hg(\parallel x_0\parallel )]\parallel x_{\alpha }^{\tau }-x_0\parallel \\ +〈z,A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha }^{\tau }-x_0)〉.\end{array}$$

(2.1)

Therefore, $\{x_{\alpha }^{\tau }\}$ is a bounded set. Since

$$〈z,A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha }^{\tau }-x_0)〉\le \parallel z\parallel \parallel A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha }^{\tau }-x_0)\parallel ,$$

by virtue of Assumption A, we have

$$\begin{array}{r}\parallel A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha }^{\tau }-x_0)\parallel \\ =\parallel j^{\ast }{A}_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha }^{\tau }-x_0)\parallel \\ \le ({\tau }_0+1)\parallel A_0\left(x_{\alpha }^{\tau }\right)-f_0\parallel \\ \le ({\tau }_0+1)[\parallel A_0^h\left(x_{\alpha }^{\tau }\right)-f_0^{\delta }\parallel +\delta +hg\left(\parallel x_{\alpha }^{\delta }\parallel \right)]\\ \le ({\tau }_0+1)[{\alpha }^{1+{\mu }_0}\sum _{i=1}^N\parallel A_i^h\left(x_{\alpha }^{\tau }\right)-f_i^{\delta }\parallel +\alpha \parallel x_{\alpha }^{\delta }\parallel +\delta +hg\left(\parallel x_{\alpha }^{\delta }\parallel \right)].\end{array}$$

Since $\alpha \sim {(\delta +h)}^{\mu }$, $0<\mu <1$, and $g(t)$ is a bounded function, from (2.1) and the last inequality, we obtain

$${\parallel x_{\alpha }^{\tau }-x_0\parallel }^2\le C_1{(\delta +h)}^{1-\mu }\parallel x_{\alpha }^{\tau }-x_0\parallel +C_2{(\delta +h)}^{\mu },0<\delta +h<1,$$

where $C_1$ and $C_2$ are positive constants. Now, by using the implication

$$a,b,c\ge 0,p>q,a^p\le b{a}^q+c⟹a^p=o(b^{p/(p-q)}+c),$$

we obtain

$$\parallel x_{\alpha }^{\tau }-x_0\parallel =o\left({(\delta +h)}^{\theta }\right),\theta =min\{1-{\mu }_{,}\mu /2\}.$$

This completes the proof. □

Now, we consider the problem of approximating (1.5) by the sequence of finite-dimensional problems

$$A_{0,n}^h(x)+{\alpha }^{1+{\mu }_0}\sum _{i=1}^N(A_{i,n}^h(x)-f_{i,n}^{\delta })+\alpha x=f_{0,n}^{\delta },x\in X_n,$$

(2.2)

where $f_{i,n}^{\delta }=P_n{f}_i^{\delta }$, $A_{i,n}^h=P_n{A}_i^h{P}_n$, $P_n$ is the linear projection from X onto $X_n$, $P_{n}x\to x$ for all $x\in X$, $\parallel P_n\parallel \le C_0$, $C_0$ is some positive constant, and $\{X_n\}$ is the sequence of finite-dimensional subspaces of X such that

$$X_1\subset X_2\subset \cdots \subset X.$$

It is easy to see that $A_{i,n}^h$ are also m-accretive. The aspects of existence and convergence of the solution $x_{\alpha ,n}^{\tau }$ of problem (2.2), as $n\to \mathrm{\infty }$, to the solution $x_{\alpha }^{\tau }$ of the operator equation (1.5) for each $\alpha >0$ has been studied in [21]. The question under which conditions the sequence $\{x_{\alpha ,n}^{\tau }\}$ converges to a solution $x_0$, as $\alpha ,\delta ,h\to 0$ and $n\to \mathrm{\infty }$, and the convergence rates of $\{x_{\alpha ,n}^{\tau }\}$ are subject of our further investigations.

In addition, suppose that j satisfies the following inequality:

$$\parallel j(x)-j(y)\parallel \le C(R){\parallel x-y\parallel }^{\nu },0<\nu <1,$$

(2.3)

where $C(R)$, $R>0$ is positive increasing function on $R=max\{\parallel x\parallel ,\parallel y\parallel \}$ (see [11]).

Set

$${\gamma }_n=\parallel (I-P_n)x_0\parallel .$$

Theorem 2.2 Let X be a real reflexive Banach space with the property of approximations and its dual space $X^{\ast }$ be strictly convex. Let ${\{A_i\}}_{i=0}^N$ be a family of $N+1$ accretive operators in X and satisfy one of the above mentioned continuities. Suppose that the following conditions hold:

• (i) $A_0$ is Fréchet differentiable with Assumption A and the derivative $A_0^{\prime }$ being uniformly bounded at $x_0$.

• (ii) There exists an element $z\in X$ such that

  $$A_0^{\prime }(x_0)z=-x_0.$$

• (iii) The parameter α is chosen such that $\alpha \sim {(\delta +h+{\gamma }_n)}^{\mu }$, $0<\mu <1$.

Then, for $0<\delta +h<1$, we have

$$\parallel x_{\alpha ,n}^{\tau }-x_0\parallel =o({(\delta +h+{\gamma }_n)}^{\theta }+{\gamma }_n^{\nu /2}),\theta =min\{1-{\mu }_{,}\mu /2\}.$$

Proof Set $x_0^n=P_n{x}_0$. From (2.2) and the property $j^n(x)=j(x)$ for all $x\in X_n$, where $j^n=P_n^{\ast }j{P}_n$ is the dual mapping of $X_n$ (see [13]), it follows that

$$\begin{array}{rl}{\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel }^2=& 〈x_{\alpha ,n}^{\tau }-x_0^n,j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ =& \frac{1}{\alpha }〈f_{0,n}^{\delta }-A_{0,n}^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ +〈-x_0^n,j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ +{\alpha }^{{\mu }_0}\sum _{i=1}^N〈f_{i,n}^{\delta }-A_{i,n}^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉.\end{array}$$

(2.4)

Clearly,

$$\begin{array}{r}〈f_{0,n}^{\delta }-A_{0,n}^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ =〈f_0^{\delta }-f_0+A_0(x_0)-A_0\left(x_0^n\right)+A_0\left(x_0^n\right)-A_0^h\left(x_0^n\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ +〈A_0^h\left(x_0^n\right)-A_0^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ \le [\delta +\parallel A_0(x_0)-A_0\left(x_0^n\right)\parallel +hg\left(\parallel x_0^n\parallel \right)]\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel .\end{array}$$

Due to condition (i) and $x_0^n\to x_0$ as $n\to \mathrm{\infty }$, we have

$$\parallel A_0\left(x_0^n\right)-A_0(x_0)\parallel \le C_0^{\prime }{\gamma }_n,$$

where $C_0^{\prime }$ is a positive constant such that

$$\parallel A_0^{\prime }(x_0)\parallel \le C_0^{\prime }$$

for x in a neighborhood of $x_0$. Thus, we have

$$〈f_{0,n}^{\delta }-A_{0,n}^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\le [\delta +hg\left(\parallel x_0^n\parallel \right)+C_0^{\prime }{\gamma }_n]\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel .$$

(2.5)

Each term of the sum in (2.4) is estimated as follows:

$$\begin{array}{r}〈f_{i,n}^{\delta }-A_{i,n}^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ =〈f_i^{\delta }-A_i^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ =〈f_i^{\delta }-A_i^h\left(x_0^n\right)+A_i^h\left(x_0^n\right)-A_i^h\left(x_{\alpha ,n}^{\tau }\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ \le 〈f_i^{\delta }-A_i^h\left(x_0^n\right),j^n(x_{\alpha ,n}^{\tau }-x_0^n)〉\\ \le [\delta +hg\left(\parallel x_0^n\parallel \right)+\parallel A_i(x_0)-A_i\left(x_0^n\right)\parallel ]\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel .\end{array}$$

(2.6)

By virtue of the continuity of $A_i$, there exists a positive constant $C^{\prime }$ such that

$$\parallel A_i(x_0)-A_i\left(x_0^n\right)\parallel \le C^{\prime },i=1,\dots ,N.$$

From (2.4)-(2.6), we see that

$$\begin{array}{rl}{\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel }^2\le & [\frac{1}{\alpha }(\delta +h+C_0^{\prime }{\gamma }_n)+{\alpha }^{{\mu }_0}(\delta +hg\left(\parallel x_0^n\parallel \right)+C^{\prime })]\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel \\ +〈-x_0,j(x_{\alpha ,n}^{\tau }-x_0^n)〉.\end{array}$$

(2.7)

Consequently, $\{x_{\alpha ,n}^{\tau }\}$ is bounded as $\delta ,h,\alpha \to 0$ and $n\to \mathrm{\infty }$. Obviously, from (2.3), Assumption A, and condition (ii), it follows that

$$\begin{array}{rl}〈-x_0,j(x_{\alpha ,n}^{\tau }-x_0^n)〉=& 〈z,A_0^{\prime }{(x_0)}^{\ast }[j(x_{\alpha ,n}^{\tau }-x_0^n)-j(x_{\alpha ,n}^{\tau }-x_0)]〉+〈z,A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha ,n}^{\tau }-x_0)〉\\ \le & C(R_1)\parallel A_0^{\prime }{(x_0)}^{\ast }\parallel \parallel z\parallel {\gamma }_n^{\nu }+\parallel z\parallel \parallel A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha ,n}^{\tau }-x_0)\parallel ,\end{array}$$

where $R_1$ is a positive constant with $R_1\ge max\{\parallel x_0\parallel ,\parallel x_{\alpha ,n}^{\tau }\parallel \}$.

On the other hand,

$$\begin{array}{rl}\parallel A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha ,n}^{\tau }-x_0)\parallel & \le ({\tau }_0+1)\parallel A_0\left(x_{\alpha ,n}^{\tau }\right)-f_0\parallel \\ \le ({\tau }_0+1)[\parallel A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h\parallel +\delta +hg\left(\parallel x_{\alpha ,n}^{\tau }\parallel \right)].\end{array}$$

By virtue of the Hahn-Banach theorem, there exists an element $y^{\ast }\in X^{\ast }$ with $\parallel y^{\ast }\parallel =1$ such that

$$\parallel A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h\parallel =〈A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h,y^{\ast }〉.$$

Since

$$〈A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h,y^{\ast }〉=〈A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h,(I^{\ast }-P_n^{\ast })y^{\ast }〉+〈A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h,P_n^{\ast }{y}^{\ast }〉$$

and

$$\parallel (I^{\ast }-P_n^{\ast })y^{\ast }\parallel \le 1/2,$$

for sufficiently large n, where $I^{\ast }$ is the identity operator in $X^{\ast }$, we have

$$\parallel A_0^h\left(x_{\alpha ,n}^{\tau }\right)-f_0^h\parallel \le 2\parallel A_{0,n}^h\left(x_{\alpha ,n}^{\tau }\right)-f_{0,n}^h\parallel .$$

Therefore,

$$\begin{array}{rl}\parallel A_0^{\prime }{(x_0)}^{\ast }j(x_{\alpha ,n}^{\tau }-x_0)\parallel \le & ({\tau }_0+1)C_0[{\alpha }^{1+{\mu }_0}\sum _{i=1}^N\parallel A_i^h\left(x_{\alpha ,n}^{\tau }\right)-f_i^{\delta }\parallel \\ +\alpha \parallel x_{\alpha ,n}^{\tau }\parallel +\delta +hg\left(\parallel x_{\alpha ,n}^{\tau }\parallel \right)].\end{array}$$

Thus, (2.7) has the form

$${\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel }^2\le {\tilde{C}}_1{(\delta +h+{\gamma }_n)}^{1-\mu }\parallel x_{\alpha ,n}^{\tau }-x_0^n\parallel +{\tilde{C}}_2[{(\delta +h+{\gamma }_n)}^{\mu }+{\gamma }_n^{\nu }],$$

where ${\tilde{C}}_i>0$ ($i=1,2$). Consequently, we have

$$\parallel x_{\alpha ,n}^{\tau }-x_0\parallel =O({(\delta +h+{\gamma }_n)}^{\theta }+{\gamma }_n^{\nu /2}),\theta =min\{1-{\mu }_{,}\mu /2\}.$$

This completes the proof. □