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1 J. Math. Anal. Appl ) The equivalence between the convergences of Ishikawa and Mann iterations for an asymptotically nonexpansive in the intermediate sense and strongly successively pseudocontractive maps B.E. Rhoades a and Ştefan M. Şoltuz b,,1 a Department of Mathematics, Indiana University, Bloomington, IN , USA b T. Popoviciu Institute of Numerical Analysis, P.O. Box 68-1, 3400 Cluj-Napoca, Romania Received 8 June 2003 Submitted by Z.-J. Ruan Abstract The convergence of modified Mann iteration is equivalent to the convergence of modified Ishikawa iterations, when T is an asymptotically nonexpansive in the intermediate sense and strongly successively pseudocontractive map Elsevier Inc. All rights reserved. Keywords: Modified Mann iteration; Modified Ishikawa iteration; Asymptotically nonexpansive in the intermediate sense; Strongly successively pseudocontractive 1. Introduction Let X be a Banach space and let B be a nonempty subset of X, u 0,x 0 B be two arbitrary fixed points and T : B B be a map. Definition 1. The map T is said to be * Corresponding author. addresses: rhoades@indiana.edu B.E. Rhoades), soltuz@itwm.fhg.de, stefansoltuz@personal.ro Ş.M. Şoltuz). 1 Mailing address: Kurt Schumacher Str. 48, Ap. 38, Kaiserslautern, Germany X/$ see front matter 2003 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) i) asymptotically nonexpansive if there exists a sequence k n ) n, k n [1, ), n N, lim n k n = 1, such that T n x T n y k n x y, x,y B, n N; 1) ii) asymptotically nonexpansive in the intermediate sense if T is continuous for some m and lim sup sup T n x T n y x y ) 0; 2) n x,y B iii) strongly successively pseudocontractive if there exists k 0, 1) and n 0 N such that x y x y + t [ I T n ki)x I T n ki)y ], 3) for all x,y B, t>0andn n 0 ; iv) uniformly Lipschitzian if there exists L>0 such that T n x T n y L x y, x,y B, n N. 4) An example of an asymptotically nonexpansive in the intermediate sense which not continuous can be found in [1, Example 1.1, p. 456]. An asymptotically nonexpansive map is uniformly Lipschitzian for some L 1, i.e., L 1: T n x T n y L x y, x,y B, n N. It is clear now that ii) is weaker then i). Remark 2. An asymptotically nonexpansive map is asymptotically nonexpansive in the intermediate sense. The converse is not true. Setting n = n 0 := 1 in 3), we get the definition of a strongly pseudocontractive map. In Example 1.2 from [1], there is a map which is not strongly pseudocontractive but which is strongly successively pseudocontractive. We consider the following iteration, see [3]: u n+1 = 1 α n )u n + α n T n u n, n= 0, 1, 2,... 5) This iteration is known as modified Mann iteration. We consider the following iteration, known as modified Ishikawa iteration see [2]): x n+1 = 1 α n )x n + α n T n y n, y n = 1 β n )x n + β n T n x n, n= 0, 1, 2,... 6) The sequences {α n }, {β n } 0, 1) are such that lim α n = 0, lim β n = 0, α n =. 7) n n n=1 The sequence {α n } remains the same in both iterations. For β n = 0, n N, from 6) we get 5). We denote by FT) the set of fixed points of T. Replacing T n by T in 5) and 6) one obtains ordinary Mann and Ishikawa iteration. The aim of this note is to prove the equivalence between the convergences of the above two iterations when T is an asymptotically nonexpansive in the intermediate sense or strongly successively pseudocontractive map. The following lemma is from [7].

3 268 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Lemma 3 [7]. Let {a n } be a nonnegative sequence which satisfies the following inequality a n+1 1 λ n )a n + δ n, 8) where λ n 0, 1), n N, n=1 λ n =, and δ n = oλ n ). Then lim n a n = The case of asymptotically nonexpansive in the intermediate sense Theorem 4. Let B be a closed convex bounded subset of an arbitrary Banach space X and {x n } and {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7).LetT : B B be an asymptotically nonexpansive in the intermediate sense and successively strongly pseudocontractive self-map of B. Put c n = max 0, sup x,y B T n x T n y x y )) 9) so that lim c n = 0. 10) n If u 0 = x 0 B, then the following two assertions are equivalent: i) Modified Mann iteration 5) converges to x FT). ii) Modified Ishikawa iteration 6) converges to x FT). Proof. If the modified Ishikawa iteration 6) converges to x, then it is clear that this x is a fixed point. Setting β n = 0, n N, in 6) we obtain the convergence of modified Mann iteration. Conversely, we shall prove that the convergence of modified Mann iteration implies the convergence of modified Ishikawa iteration. The proof is similar to the proof of Theorem 4 from [5]. From 6) we have x n = x n+1 + α n x n α n T n y n = 1 + α n )x n+1 + α n x n+1 α n T n x n+1 kα n x n+1 2α n x n+1 + kα n x n+1 + α n x n + α n T n x n+1 α n T n y n = 1 + α n )x n+1 + α n I T n ki)x n+1 2 k)α n x n+1 + α n x n + α n T n x n+1 T n y n ) = 1 + α n )x n+1 + α n I T n [ ki)x n+1 2 k)α n xn + α n T n y n x n ) ] + α n x n + α n T n x n+1 T n y n ) = 1 + α n )x n+1 + α n I T n ki)x n+1 1 k)α n x n + 2 k)αn 2 x n T n y n ) + α n T n x n+1 T n y n ). 11) Analogously, for 5) we get u n = 1 + α n )u n+1 + α n I T n ki)u n+1 1 k)α n u n + 2 k)α 2 n u n T n u n ) + α n T n u n+1 T n u n ). 12)

4 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Compute 11) 12) to obtain [ x n u n = 1 + α n )x n+1 u n+1 ) + α n I T n ki)x n+1 I T n ] ki)u n+1 1 k)α n x n u n ) + 2 k)αn 2 [ xn u n T n y n T n u n ) ] [ + α n T n x n+1 T n y n T n u n+1 T n u n ) ]. 13) Using the triangular inequality and 3) with x := x n+1, y := u n+1, t := α n /1 + α n ), x n u n 1 + α n ) x n+1 u n+1 ) + α n [ I T n ki)x n+1 I T n ] ki)u n α n 1 k)α n x n u n 2 k)αn 2 x n u n T n y n T n u n ) α n T n x n+1 T n y n T n u n+1 T n u n ) 1 + α n ) x n+1 u n+1 1 k)α n x n u n 2 k)αn 2 xn u n T n y n T n u n ) α n T n x n+1 T n y n T n u n+1 T n u n ). 14) Thus 1 + α n ) x n+1 u n+1 ) k)α n xn u n +2 k)αn 2 x n u n T n y n + T n u n + α n T n x n+1 T n u n+1 T n y n T n u n ) ) k)α n xn u n +2 k)αn 2 x n T n y n +2 k)αn 2 u n T n u n + α n T n u n+1 T n u n +α n T n x n+1 T n y n. 15) Using the facts that 1 + αn 2) 1 1and1 + αn 2) 1 1 α n + αn 2 we get x n+1 u n+1 ) k)α n 1 αn + αn) 2 xn u n { + α n 2 k)αn x n T n y n +2 k)α n u n T n u n + T n u n+1 T n u n + T n x n+1 T n y n } = ) k)α n 1 αn + αn) 2 xn u n +α n σ n, 16) where σ n := 2 k)α n x n T n y n +2 k)α n u n T n u n + T n u n+1 T n u n + T n x n+1 T n y n. 17) We have M := max { x 0, sup { T n x, x B, n N }} <. 18) The sequence { x n T n y n } is bounded because {T n y n } is in the bounded set B, and {x n } also is bounded by M. Supposing that x n M, a simple induction leads to x n+1 1 α n ) x n +α n M 1 α n )M + α n M = M. 19)

5 270 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Modified Mann iteration 5) converges, let x be that fixed point. Thus 0 u n T n u n T n x T n u n + u n x = T n x T n u n u n x ) + 2 u n x c n + 2 u n x 0 asn. 20) It is clear that { T n u n+1 T n u n } also converges to zero because 0 T n u n+1 T n u n T n u n+1 T n u n u n+1 u n + u n+1 u n 0 asn. 21) From 9) and 10) one obtains since T n x n+1 T n y n = [ T n y n T n x n+1 y n x n+1 ] + y n x n+1 c n + y n x n+1 0, as n, 22) y n x n+1 = β n x n + β n T n x n + α n x n α n T n y n 2β n M + 2α n M = 2Mα n + β n ) 0 asn. 23) The sequences {x n }, {T n x n } and {T n y n } are in the bounded set B, and bounded by M>0. The following inequality is, in fact, inequality 29) from [5]: k)αn ) 1 αn + α 2 n) = 1 kα n + kα 2 n + 1 k)α3 n 1 kα n + kα 2 n + 1 k)α2 n = 1 kα n + α 2 n. 24) The condition lim n α n = 0 implies the existence of a positive integer N such that for all n N α n k 2. 25) Substituting inequality 25) into 24) we get kαn ) ) 1 α n + αn 2 ) 1 kαn + αn 2 1 kα n + k 2 α n = 1 k 2 α n. 26) Relations 26) and 16) lead to x n+1 u n+1 1 k ) 2 α n x n u n +α n σ n, 27) where {σ n }is given by 17). From 22) we know that lim n σ n = 0. Denote a n := x n u n, λ n := k 2 α n, δ n := α n σ n = oλ n ). 28)

6 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Relations 28) and 27) lead to 8); using Lemma 3 we have lim n u n =0 n 29) to obtain 0 x n x x n u n + u n x 0 asn. 30) Hence lim n x n x =0. 3. The strongly successively pseudocontractive case 3.1. The Lipschitzian case Theorem 5. Let B be a closed convex without being necessarily bounded) subset of an arbitrary Banach space X and {x n }, {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7). Let T be a successively strongly pseudocontractive and uniformly Lipschitzian with L 1 self-map of B. If u 0 = x 0 B, then the following two assertions are equivalent: i) Modified Mann iteration 5) converges to x FT). ii) Modified Ishikawa iteration 6) converges to x FT). Proof. Supposing, again, that modified Ishikawa iteration converges, analogously as in the proof of Theorem 4 we obtain the convergence of modified Mann iteration. Conversely, supposing that modified Mann iteration converges, we will prove that modified Ishikawa iteration will converge. For that we need to evaluate x n u n. The map T is successively strongly pseudocontractive. Thus relations 11), 12), 14) 16), 24) hold: x n+1 u n+1 1 kα n + αn) 2 xn u n { + α n 2 k)αn x n T n y n +2 k)α n u n T n u n + T n u n+1 T n u n + T n x n+1 T n y n }. 31) We have x n T n y n x n u n + u n T n u n + T n u n T n y n x n u n + u n T n u n +L u n y n. 32) u n y n = 1 βn )u n x n ) + β n u n T n x n ) 1 β n ) x n u n +β n u n T n x n 1 β n ) x n u n +β n T n u n T n x n + u n T n u n ) 1 β n ) x n u n +β n L x n u n +β n u n T n u n = 1 β n + β n L) x n u n +β n u n T n u n L x n u n +β n u n T n u n, 33) because 1 L 1 β n + β n L L.

7 272 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Substituting 33) into 32) we get x n T n y n u n x n + u n T n u n + L L x n u n +β n u n T n u n ) 1 + L 2 ) x n u n +1 + Lβ n ) u n T n u n. 34) Now T n x n+1 T n y n L x n+1 y n =L 1 αn )x n + α n T n y n y n = L 1 αn )x n y n ) + α n T n y n y n ) L 1 α n ) x n y n +α n T n y n y n ). 35) Using 33), T n y n y n T n y n T n u n + T n u n u n + u n y n 1 + L) u n y n + T n u n u n 1 + L) L u n x n +β n T n u n u n ) + T n u n u n = 1 + L)L x n u n + [ 1 + L)β n + 1 ] T n u n u n. 36) x n y n = x n 1 β n )x n β n T n x n = β n x n T n x n [ β n xn u n + T n u n u n + T n x n T n u n ] β n 1 + L) xn u n + T n u n u n ). 37) Substituting 36) and 37) in 35) one obtains T n x n+1 T n y n L [ 1 α n ) x n y n +α n T n y n y n ] L {1 α n ) β n 1 + L) un x n + T n u n u n )) + α n 1 + L)L xn u n + [ 1 + L)β n + 1 ] T n u n u n )} = 1 α n )β n 1 + L)L x n u n +L1 α n )β n T n u n u n + α n 1 + L)L 2 x n u n + α n L [ 1 + L)β n + 1 ] T n u n u n = L1 α n )β n 1 + L) + α n 1 + L)L 2) x n u n + β n L1 α n ) + α n L [ 1 + L)β n + 1 ]) T n u n u n. 38) Replacing 38) and 32) in 31) we get x n+1 u n+1 1 kα n + 2αn) 2 xn u n + 2 k)αn L 2 ) x n u n +1 + β n L) u n T n u n ) + 2 k)αn 2 u n T n u n +α n T n u n+1 T n u n + α n L1 αn )β n 1 + L) + α n 1 + L)L 2) x n u n + α n βn L1 α n ) + α n L [ 1 + L)β n + 1 ]) u n T n u n

8 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) = { 1 kαn + 2α 2 n) + 2 k)α 2 n 1 + L 2 ) + α n L1 + L) 1 α n )β n + α n L )} x n u n { + 2 k)αn β [ nl) + α n βn L1 α n ) + α n L [ 1 + L)β n + 1 ]]} u n T n u n + α n T n u n+1 T n u n. 39) Formula 30) from [5] with M = k)1 + L 2 ) + L L) leads us to 1 kα n ) + 2αn k)α2 n 1 + L2 ) + α n L1 + L) 1 α n )β n + α n L ) 1 kα n + α n 2αn + 2 k)α n 1 + L 2 ) + L1 + L) 1 α n )β n + α n L )) 1 kα n + α n Mα n + β n ) 1 kα n + α n k1 k) = 1 k 2 α n, 40) for all n sufficiently large, since lim n α n + β n ) = 0. Relations 40) and 39) lead to x n+1 u n+1 1 k 2 α n ) x n u n + α n {[2 k)α n 2 + β n L) + [ β n L1 α n ) + α n L [ 1 + L)β n + 1 ]]] } u n T n u n + T n u n+1 T n u n = 1 k 2 α n ) x n u n +α n ɛ n, [ ɛ n := 2 k)α n 2 + β n L) + [ β n L1 α n ) + α n L [ 1 + L)β n + 1 ]]] u n T n u n + T n u n+1 T n u n. 41) Supposing that lim n u n x =0, with Tx = x, we have lim n u n T n u n =0 because 0 u n T n u n u n x + T n x T n u n u n x +L u n x =1 + L) u n x 0 asn. 42) It is clear that if lim n u n T n u n =0, then lim n T n u n+1 T n u n =0. Denote by a n := x n u n, λ n := k 2 α n, δ n := α n ɛ n = oλ n ). 43) Supposing that modified Mann iteration converges, i.e., lim n u n = x, we get from 42) lim u n+1 T n u n =0 n and lim n T n u n =0 n

9 274 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) because T is uniformly Lipschitzian. Thus lim n ɛ n = 0, which means that δ n = oλ n ). Relations 43) and Lemma 3 lead us to lim n u n =0. n 44) The inequality 0 x n x x n u n + u n x 0 asn 45) leads us to conclusion that lim n x n = x The non-lipschitzian case Let X be a real Banach space, B be a nonempty subset of X and T : B B. The map J : X 2 X given by Jx := {f X : x,f = x 2, f = x }, x X, is called the normalized duality mapping. The Hahn Banach theorem assures that Jx, x X. It is an easy task to see that jx),y x y, x,y X, jx) Jx). In [1, Lemma 2.1, p. 459] it is shown that the definition of successively strongly pseudocontractive map is equivalent to the following definition: Definition 6. T is successively strongly pseudocontractive map if there exists k 0, 1) and a jx y) Jx y) such that T n x T n y,jx y) k x y 2, x,y B. 46) We need the following lemma from [4]. Lemma 7 [4]. If X is a real Banach space, then the following relation is true: x + y 2 x y,jx + y), x,y X, jx+ y) Jx + y). 47) We are able now to prove the following result: Theorem 8. Let X be a real Banach space with dual uniformly convex and B a nonempty, closed, convex, bounded subset of X. LetT : B B be a successively strongly pseudocontractive operator and {x n }, {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7). Then for u 0 = x 0 B the following assertions are equivalent: i) Modified Mann iteration 5) converges to the fixed point of T. ii) Modified Ishikawa iteration 6)convergesto the fixed pointof T. Proof. The proof is similar to the proof of the main result from [6]. If either 5) or 6) converges to a point x, then x is a fixed point for T. Using 5), 6), 47) with x := 1 α n )x n u n ), y := α n T n y n T n u n ) observe that x + y = x n+1 u n+1 ) and 46) we get x n+1 u n+1 2 = 1 αn )x n u n ) + α n T n y n T n u n ) 2 1 α n ) 2 x n u n 2 + 2α n T n y n T n u n,jx n+1 u n+1 )

10 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) = 1 α n ) 2 x n u n 2 + 2α n T n y n T n u n,jx n+1 u n+1 ) Jy n u n ) + 2α n T n y n T n u n,jy n u n ) 1 α n ) 2 x n u n 2 + 2α n k y n u n 2 + 2α n T n y n T n u n,jx n+1 u n+1 ) Jy n u n ) 1 α n ) 2 x n u n 2 + 2α n k y n u n 2 + 2α n T n y n T n u n Jxn+1 u n+1 ) Jy n u n ) 1 α n ) 2 x n u n 2 + 2α n k y n u n 2 + 2α n M Jxn+1 1 u n+1 ) Jy n u n ), 48) for some M 1 > 0. Observe that { T n y n T n u n } is bounded. We prove that Jx n+1 u n+1 ) Jy n u n ) 0 asn. 49) If the dual is uniformly convex, then J is single map and uniformly continuous on every bounded set. To prove 49) it is sufficient to see that xn+1 u n+1 ) y n u n ) = xn+1 y n ) u n+1 u n ) = α n x n + α n T n y n + β n x n β n T n x n + α n u n α n T n u n α n xn + T n y n + u n + T n u n ) + β n xn + T n x n ) α n + β n )M 0 asn, 50) where M = sup n x n + T n y n + u n + T n u n ), x n + T n x n )) <. The sequences {u n }, {x n }, {T n x n }, {T n u n } and {T n y n } are bounded, being in the bounded set B. Hence one can see that the M above is finite and 49) holds. We define σ n := 2α n M Jxn+1 1 u n+1 ) Jy n u n ). 51) Again, using 6) and 47) with x := 1 β n )x n u n ), y := β n T n x n u n ) observe that x + y = y n u n )weget y n u n 2 = 1 βn )x n u n ) + β n T n x n u n ) 2 1 β n ) 2 x n u n 2 + 2β n T n x n u n,jy n u n ) x n u n 2 + β n M 2. 52) The last inequality is true because { T n x n u n,jy n u n ) } is bounded, with a constant M 2 > 0. Replacing 51) and 52) in 48), we obtain x n+1 u n α n ) 2 x n u n 2 + 2α n k x n u n 2 + σ n + α n 2k)β n M 2 = 1 21 k)α n + αn) 2 xn u n 2 + oα n ). 53) The condition lim n α n = 0 implies the existence of an n 0 such that for all n n 0 we have α n 1 k). 54)

11 276 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) Substituting 54) into 53), we obtain 1 21 k)α n + αn k)α n + 1 k)α n = 1 1 k)α n. Thus, from 53) x n+1 u n+1 2 ) 1 1 k)α n xn u n 2 + oα n ). 55) Define a n := x n u n 2, λ n := 1 k)α n 0, 1). Then Lemma 3 implies that lim n a n = lim n x n u n 2 = 0, i.e., lim x n u n =0. 56) n Suppose that modified Mann iteration converges, i.e., lim n u n = x. The inequality 0 x x n u n x + x n u n 57) and 56) imply that lim n x n = x. Analogously lim n x n = x implies lim n u n = x. 4. The equivalence between T-stability Let FT):= {x X: x = Tx )}, x FT). Consider ε n := xn+1 1 α n )x n α n T n y n, 58) δ n := un+1 1 α n )u n α n T n u n. 59) Definition 9. If lim n ε n = 0 respectively lim n δ n = 0) implies that lim n x n = x respectively lim n u n = x ), then 6) respectively 5)) is said to be T-stable. It is obvious if we take the limit in 6), respectively 5). Remark 10. Let X be a normed space with B a nonempty, convex, closed, and bounded subset. Let T : B B be a map. If the modified Mann respectively Ishikawa) iteration converges, then lim n δ n = 0 respectively lim n ε n = 0). The remark holds without the boundeness assumption of B, when the map T is uniformly Lipschitzian. Proof. Let lim n u n = x. Then from 59) we have 0 δ n u n+1 u n +α n u n T n u n u n+1 x + u n x +α n u n x +α n x T n u n 0 asn. We are able now to prove the following result: Theorem 11. Let B be a closed convex bounded subset of an arbitrary Banach space X and {x n } and {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7). Let T be an asymptotically nonexpansive in the intermediate sense and successively strongly pseudocontractive self-map of B. Let {c n } be as in 9) satisfying lim n c n = 0. If u 0 = x 0 B, then the following two assertions are equivalent:

12 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) i) Modified Ishikawa iteration 6) is T -stable. ii) Modified Mann iteration 5) is T -stable. Proof. From Definition 9 we know that the equivalence i) ii) means that lim n ε n = 0 lim n δ n = 0. The implication lim n ε n = 0 lim n δ n = 0 is obvious by setting β n = 0 in 6). Conversely, suppose that 5) is T-stable. Using Definition 9, again, we get lim δ n = 0 lim u n = x. 60) n n Theorem 4 assures that lim n u n = x lim n x n = x. Using Remark 10 we have lim n ε n = 0. Thus we get lim n δ n = 0 lim n ε n = 0. Similarly one can prove the following result. Theorem 12. Let B beaclosedconvexwithout being necessarily bounded) subset of an arbitrary Banach space X and {x n }, {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7). Let T be a successively strongly pseudocontractive and uniformly Lipschitzian with L 1 self-map of B. If u 0 = x 0 B, then the following two assertions are equivalent: i) Modified Ishikawa iteration 6) is T -stable. ii) Modified Mann iteration 5) is T -stable. Also the following results holds using Theorem 8: Theorem 13. Let X be a real Banach space with dual uniformly convex and B a nonempty, closed, convex, bounded subset of X. LetT : B B be a successively strongly pseudocontractive operator and {x n }, {u n } defined by 6) and 5) with {α n }, {β n } 0, 1) satisfying 7). Then for u 0 = x 0 B the following assertions are equivalent: i) Modified Ishikawa iteration 6) is T -stable. ii) Modified Mann iteration 5) is T -stable. Our theorems are also true for set-valued mappings, if such maps admit appropriate single-valued selections. References [1] Z. Liu, J.K. Kim, K.H. Kim, Convergence theorems and stability problems of the modified Ishikawa iterative sequences for strictly successively hemicontractive mappings, Bull. Korean Math. Soc ) [2] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc ) [3] W.R. Mann, Mean value in iteration, Proc. Amer. Math. Soc ) [4] C. Morales, J.S. Jung, Convergence of paths for pseudocontractive mappings in Banach spaces, Proc. Amer. Math. Soc )

13 278 B.E. Rhoades, Ş.M. Şoltuz / J. Math. Anal. Appl ) [5] B.E. Rhoades, Ş.M. Şoltuz, On the equivalence of Mann and Ishikawa iteration methods, Internat. J. Math. Math. Sci ) [6] B.E. Rhoades, Ş.M. Şoltuz, The equivalence of Mann iteration and Ishikawa iteration for non-lipschitzian operators, Internat. J. Math. Math. Sci ) [7] X. Weng, Fixed point iteration for local strictly pseudocontractive mapping, Proc. Amer. Math. Soc )

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