Applied Mathematical Sciences, Vol. 6, 212, no. 63, 319-3117 Convergence Rates in Regularization for Nonlinear Ill-Posed Equations Involving m-accretive Mappings in Banach Spaces Nguyen Buong Vietnamese Academy of Science and Technology Institute of Information Technology 18, Hoang Quoc Viet, Hanoi, Vietnam nbuong@ioit.ac.vn Nguyen Thi Hong Phuong 9/5, Tran Quoc Hoan, Cau Giay, Ha Noi, Viet Nam Abstract. In this paper, based on a priory choice of regularization parameter, convergence rate of the regularized solution is established, for solving nonlinear ill-posed operator equations involving m-accretive mappings in Banach spaces without weak sequential continuous normalized duality mapping. Mathematics Subject Classification: 47J5, 47H9, 49J3 Keywords: Accretive and α-strong accretive operators, weakly, demi- continuous, strictly convex Banach space, Fréchet differentiable and Tikhonov regularization 1. INTRODUCTION AND PRELIMINARIES Let E be a real Banach space and E be its dual space. For the sake of simplicity, the norms of E and E are denoted by the symbol.. The symbol x, x denotes the value of x E at x E. A mapping J : E E is called a normalized duality mapping of E, if it satisfies the following condition: x, J(x) = x 2, J(x) = x, x E. Let A be an m-accretive and single-valued mapping on E, i.e., A : E E has the following properties:
311 Nguyen Buong and Nguyen Thi Hong Phuong (i) A(x) A(y), j(x y), x, y E, where j(x y) J(x y), and (ii) R(A + λi) = E for each λ >, where R(A) denotes the range of A and I is the identity operator in E. If there exists a positive constant α such that A(x) A(y), j(x y) α x y 2, x, y E, then A is said to be α-strongly accretive. When α =, A is called accretive. We are interested in solving the operator equation A(x) = f, f E, (1.1) where A is an m-accretive and single-valued mapping on E. Throught this paper, we assume that the set of all solutions of (1.1), denoted by S, is nonempty. Without additional conditions on the structure of A such as strongly or uniformly accretive property, equation (1.1) is, in general, an ill-posed problem. To solve (1.1), we have to use stable methods. A well known one is the Tikhonov regularization method. Its operator version for ill-posed equations involving accretive mapping has the form (see [1-9]) A(x) + α(x x + ) = f δ, f δ f δ, (1.2) where α > is the parameter of regularization, and x + E is a given initial guess. Since A is m-accretive, equation (1.2) has a unique solution x δ α for each fixed α > and δ >. Moreover, from (1.1), (1.2) and the accretive property of A it is easy to obtain the estimate x δ α y y x + + δ/α y S. (1.3) In [1] it was also shown that the function ρ(α) = α x δ α x + is continuous, monotone non-decreasing, and if A is continuous at x + then lim ρ(α) =, lim α ρ(α) = α + A(x+ ) f δ. Further, by the argument in [2], we can verify that if A(x + ) f δ > Kδ p, K > 2, < p 1, then there exists at least a value α = α(δ) such that A(x δ α(δ) ) f δ = Kδ p, and (K 1)δ p /α(δ) 2 y x +. Consequently, for the case < p < 1 we have δ/α(δ) 2 y x + δ 1 p /(K 1), as δ. Hence, if J is continuous and weak sequential continuous, then x δ α(δ) y S (see [1], [2], [8], [9]). Unfortunately, the class of infinite-dimensional proper Banach spaces having J with these properties is very small (only l p ). It is natural to ask when the algorithm (1.2) can be applied for other Banach spaces. In [3-7], we showed the strong convergence of regularized solutions x δ α to a solution of
Convergence rates in regularization 3111 (1.1) in Banach spaces without the weak sequential continuous property of J, when A is weakly continuous, A(x) A(y ) QA (y ) J(x y ) τ A(x) A(y ), y E, (1.4) where τ is some positive constant, A (x) denotes the Fréchet derivative of A at x E and Q is the normalized duality mapping of E, and there exists an element v E such that x + y = A (y )v. (1.5) In this paper, without the weak sequential continuous property of J and (1.4), we prove the strong convergence of the algorithm (1.2) in a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm and give an optimal order of convergence rates for the regularized solutions, when the regularization parameter is chosen a priory. We list some facts, that will be used to prove our results. Let E be a real normed linear space. Let S 1 () := {x E : x = 1}. The space E is said to have a Gâteaux differentiable norm (or to be smooth) if the limit x + ty x lim t t exists for each x, y S 1 (). The space E is said to have a uniformly Gâteaux differentiable norm if the limit is attained uniformly for x S 1 (). The space E is said to be strictly convex, iff for x, y S 1 () with x y, we have (1 λ)x + λy < 1 λ (, 1). It is well known (see, [1]) that if E is smooth, then the normalized duality mapping is single valued; and if the norm of E is uniformly Gâteaux differentiable, then the normalized duality mapping is norm to weak star uniformly continuous on every bounded subset of E. In the sequel, we shall denote the single valued normalized duality mapping by j. Let µ be a continuous linear functional on l and let (a 1, a 2,...) l. We write µ k (a k ) instead of µ((a 1, a 2,...)). We recall that µ is a Banach limit when µ satisfies µ = µ k (1) = 1 and µ k (a k+1 ) = µ k (a k ) for each (a 1, a 2,...) l. For a Banach limit µ, we know that lim inf a k µ k (a k ) lim sup a k k k for all (a 1, a 2,...) l. If a = (a 1, a 2,...) l, b = (b 1, b 2,...) l and a k c (respectively, a k b k ), as k, we have µ k (a k ) = µ(a) = c (respectively, µ k (a k ) = µ k (b k )).
3112 Nguyen Buong and Nguyen Thi Hong Phuong Lemma 1.1 [11]. Let C be a convex subset of a Banach space E whose norm is uniformly Gâteaux differentiable. Let {x k } be a bounded subset of E, let z be an element of C and µ be a Banach limit. Then, µ k x k z 2 = min u C µ k x k u 2 if and only if µ k u z, j(x k z) for all u C. For an m-accretive mapping A in E and any fixed element f E, we can define a mapping u = T f (x) by A f (u) + u = x, A f (.) = A(.) f, (1.6) for each x E. Since A f is also m-accretive, the existence of T f is asserted. It is not difficult to verify that T f has the following properties: (i) D(T f ) = E; (ii) T f is nonexpansive, i.e., T f x T f y x y ; (iii) F ix(t f ) = S where F ix(t f ) denotes the set of fixed points of T f, i.e., F ix(t f ) = {x E : x = T f (x)}. Recall that a mapping T in E is said to be pseudocontractive, if T x T y, j(x y) x y 2, for all x, y D(T ), the domain of T. Obviously, every nonexpansive mapping is pseudocontractive and if A is accretive, then T = I A is also pseudocontractive. Lemma 1.2 [9]. For any linear, bounded and accretive mapping F on a reflexive Banach space E, we have F (F + αi) 1 2 for each α >. 2. MAIN RESULTS Theorem 2.1. Let E be a real, reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm and let A be an m-accretive mapping on E. Then, for each α > and a fixed f E, the equation A(x) + α(x x + ) = f, (2.1) possesses a unique solution x α and, in addition, if the solution set of (1.1) S, then the net {x α } converges strongly to an element y E, solving the following variational inequality: Moreover, we have y S : y x +, j(y y) y S. (2.2) x δ α x α δ/α, where x δ α is the unique solution of (1.2), for any α > and f δ E.
Convergence rates in regularization 3113 Proof. Since A is m-accretive, equation (2.1) has a solution, denoted by x α, for each α >. This solution is unique, because the mapping A + α(i x + ) is α-strongly accretive. Next, we have, from (2.1), that A(x α ) A(y), j(x α y) + α x α x +, j(x α y) =, for any y S, and hence x α y 2 x + y, j(x α y) y S. (2.3) Therefore, x α y x + y. In means that {x α } is bounded. Again, from (2.1) it follows that A(x α ) f = α x α x + 2α x + y, and hence, lim A(x α) f =. (2.4) α Next, we consider the mapping T f := I A f. Clearly, p S iff p F ix(t f ). Moreover, we observe that the mapping 2I T f has a nonexpansive inverse, denoted by A. Indeed, since 2I T f = I + I T f = I + A f, from (1.6) it follows that A = T f = (I + A f ) 1, which is a single-valued nonexpansive mapping. So, we obtain F ix(a) = F ix(t f ) = S. Thus, from x δ α T f x δ α = (2I T f )x δ α x δ α = A(x δ α) f and it follows that A(2I T f )x δ α = (I + A f ) 1 (I + A f )(x δ α) = x δ α, x δ α Ax δ α = A(2I T f )x δ α Ax δ α (2I T f )x δ α x δ α = A(x δ α) f. This together with (2.4) implies that x α Ax α as α. Let {x k } be any subsequence of {x α } with α k as k. We consider the functional ϕ(x) = µ k x k x 2 for all x E. We see that ϕ(x) as x and ϕ is continuous and convex, so as E is reflexive, there exists ỹ E such that ϕ(ỹ) = min x E ϕ(x). Hence, the set C := {u E : ϕ(u) = min ϕ(x)}. x E It is easy to see that C is a bounded, closed, and convex subset of E. On the other hand, since x k Ax k, we have that ϕ(aỹ) = µ k x k Aỹ 2 = µ k Ax k Aỹ 2 µ k x k ỹ 2 = ϕ(ỹ),
3114 Nguyen Buong and Nguyen Thi Hong Phuong which implies that AC C, that is C is invariant under A. Now, we show that C contains a fixed point of A. Since E is a strictly convex and reflexive Banach space, any closed and convex subset in E is a Chebyshev set (see [12]). Then, for a point y F ix(a), there exists a unique ỹ C such that y ỹ = inf x C By virtue of y = Ay and Aỹ C, we have y x. y Aỹ = Ay Aỹ y ỹ, and hence Aỹ = ỹ. So, there exists a point ỹ F ix(a) C = S C. Now, from Lemma 2.1, we know that ỹ is a minimizer of ϕ(x) on E, if and only if µ k x ỹ, j(x k ỹ) x E. (2.5) By (2.3) with y = ỹ and taking x = x + in (2.5), we obtain that µ k x k ỹ 2 =. Hence, there exists a subsequence {x ki } of {x k } which strongly converges to ỹ as i. Again, from (2.3) and the norm to weak star continuous property of the normalized duality mapping j on bounded subsets of E, we obtain that y x +, j(ỹ y) y S. (2.6) Since y and ỹ belong to F ix(a), a closed and convex subset, by replacing y in (2.6) by sy + (1 s)ỹ for s (, 1), using the well-known property j(s(ỹ y)) = sj(ỹ y) for s >, dividing by s and taking s, we obtain ỹ x +, j(ỹ y) y S. The uniqueness of y in (2.2) guarantees that ỹ = y. So, all the net {x α } converges strongly to y as α. By using (1.3) and (2.1), we obtain A(x δ α) A(x α ), j(x δ α x α ) + α x δ α x α 2 = f δ f, j(x δ α x α ), that implies x δ α x α δ/α. This completes the proof. Theorem 2.2 Let E, A and f be as in Theorem 2.1 such that S. Assume that (1.5) and the following condition: the Fréchet derivative A (.) is locally Lipschitz continuous in a ball B r (y ) = {x E : x y x + y }, that is, there exists a Lipschitz constant L > such that A (x) A (z) L x z for all x, z B r (y ), hold. Then, for each α >, we have x α y 2(L v 2 + v )α.
Convergence rates in regularization 3115 Proof. Since A is accretive, A (y ) is also accretive. This implies that A (y ) is m-accretive. Put F = A (y ), R α = α(f + αi) 1 and B = F R α. Then, αf αb = αf αf R α = F (αi αr α ) = αf (I R α ) = αf [(F + αi) αi](f + αi) 1 = F B. Next, by putting z α = x α y, we obtain that implies A(x α ) A(y + Bv) + α(z α Bv) = A(y ) A(y + Bv) + α(z α Bv) αx α x + ) = A(y ) A(y + Bv) + α(x + y ) Bv) αbv = A(y ) A(y + Bv) + αf (v) αbv = A(y ) A(y + Bv) + F Bv, α x α Bv 2 A(y ) A(y + Bv) + F Bv, j(x α (y Bv)). Since A (x) A (z) L x z and x α, B r (y ), A(x) A(z) A (z) L x z L 2 x z 2, that implies z α Bv L 2α Bv 2 = L 2α F α(f + αi) 1 2 2L v 2. On the other hand, z α = x α y z α Bv + Bv 2(L v 2 + v )α. Theorem 2.3 Let E, A and f be as in Theorem 2.2. Assume that condition (1.5) and either the other condition in Theorem 2.2 or that there exists a constant k > such that k x + y < 1 and (A (x) A (y ))w = A (y )k(x, y, w), k(x, y, w) k w x y, x, w B r (y ), where r > r + δ/α, are satisfied. If α is chosen such that α = O( δ), then x δ α y O( δ).
3116 Nguyen Buong and Nguyen Thi Hong Phuong Proof. Obviously, when the conditions in Theorem 2.1 are satisfied, from Theorems 2.1 and 2.2 and (1.3) it follows that x δ α y δ α + 2(L v 2 + v )α. So, x δ α y O( δ), if α is chosen such that α = O( δ). Now, put x t = y +t(x α y ) with t (, 1). Clearly, x t y = t x α y x + y. Thus, x t B r. On the other hand, from (2.1) and A(x α ) A(y ) = A (y )(x α y ) + it implies that Therefore, = A (y )(x α y ) + α(x α x + ) = A (y )(x α y ) + α(x α y ) A (y )(x α y ) = α(x + y ) + Consequently, we have with Hence, x α y = α(f + αi) 1 F v (F + αi) 1 F v 2 v, (F + αi) 1 F k(x t, y, x α y ) dt 2 (A (x t ) A (y ))(x α y )dt A (y )k(x t, y, x α y )dt A (y )k(x t, y, x α y )dt. A (y )k(x t, y, x α y )dt. (F + αi) 1 F k(x t, y, x α y )dt k x t y x α y dt k x α y 2 k x + y x α y. (1 k x + y ) x α y 2α v. So, in this case, we also have x δ α y O( δ). This research is founded by Vietnamese National Foundation of Science and Technology Development under grant number 11.1-211.17.
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