ITERATIVE SOLUTIONS OF NONLINEAR EQUATIONS WITH
φ-STRONGLY ACCRETIVE OPERATORS
SHIN MIN KANG, CHI FENG and ZEQING LIU
Abstract. Suppose that X is an arbitrary real Banach space and T : X → X is a Lipschitz
continuous φ-strongly accretive operator or uniformly continuous φ-strongly accretive operator.
We prove that under different conditions the three-step iteration methods with errors converge
strongly to the solution of the equation T x = f for a given f ∈ X.
1. Introduction
Let X be a real Banach space with norm k · k and dual X ∗ , and J denote the normalized
∗
duality mapping from X into 2X given by
J(x) = {f ∈ X ∗ : kf k2 = kxk2 = hx, f i},
JJ J
I II
Go back
x ∈ X,
where h·, ·i is the generalized duality pairing. In this paper, I denotes the identity operator
on X, R+ and δ(K) denote the set of nonnegative real numbers and the diameter of K for
any K ⊆ X, respectively. An operator T with domain D(T ) and range R(T ) in X is called
φ-strongly accretive if there exists a strictly increasing function φ : R+ → R+ with φ(0) = 0
such that for any x, y ∈ D(T ) there exists j(x − y) ∈ J(x − y) such that
(1.1)
hT x − T y, j(x − y)i ≥ φ(kx − yk)kx − yk.
Full Screen
Close
Quit
Received April 9, 2007.
2000 Mathematics Subject Classification. Primary 47H05, 47H10, 47H15.
Key words and phrases. φ-strongly accretive operators; three-step iteration method with errors; Banach
spaces.
If there exists a positive constant k > 0 such that (1.1) holds with φ(kx − yk) replaced
by kkx − yk, then T is called strongly accretive. The accretive operators were introduced
independently in 1967 by Browder [1] and Kato [8]. An early fundamental result in the
theory of accretive operator, due to Browder, states the initial value problem
du
+ T u = 0, u(0) = u0
dt
is solvable if T is locally Lipschitz and accretive on X. Martin [11] proved that if T : X → X
is strongly accretive and continuous, then T is subjective so that the equation
(1.2)
(1.3)
JJ J
I II
Go back
Tx = f
has a solution for any given f ∈ X. Using the Mann and Ishikawa iteration methods
with errors, Chang [3], Chidume [4], [5], Ding [7], Liu and Kang [10] and Osilike [12], [13]
obtained a few convergence theorems for Lipschitz φ-strongly accretive operators. Chang
[2] and Yin, Liu and Lee [16] also got some convergence theorems for uniformly continuous
φ-strongly accretive operators.
The purpose of this paper is to study the three-step iterative approximation of solution
to equation (1.3) in the case when T is a Lipschitz φ-strongly accretive operator and X is
a real Banach space. We also show that if T : X → X is a uniformly continuous φ-strongly
accretive operator, then the three-step iteration method with errors converges strongly to
the solution of equation (1.3). Our results generalize, improve the known results in [2]–[7],
[10], [12], [13] and [15].
Full Screen
2. Preliminaries
Close
Quit
The following Lemmas play a crucial role in the proofs of our main results.
Lemma 2.1 ([7]). Suppose that φ : R+ → R+ is a strictly increasing function with
∞
∞
∞
φ(0) = 0. Assume that {rn }∞
n=0 , {sn }n=0 , {kn }n=0 and {tn }n=0 are sequences of nonnegative numbers satisfying the following conditions:
∞
∞
∞
X
X
X
kn < ∞,
tn < ∞,
sn = ∞
(2.1)
n=0
n=0
n=0
and
(2.2)
rn+1 ≤ (1 + kn )rn − sn rn
φ(rn+1 )
+ tn
1 + rn+1 + φ(rn+1 )
for n ≥ 0.
Then limn→∞ rn = 0.
Lemma 2.2 ([10]). Suppose that X is an arbitrary Banach space and T : X → X is a
continuous φ-strongly accretive operator. Then the equation T x = f has a unique solution
for any f ∈ X.
∞
∞
Lemma 2.3 ([9]). Let {αn }∞
n=0 , {βn }n=0 and {γn }n=0 be three nonnegative real sequences
satisfying the inequality
JJ J
I II
Go back
Full Screen
Close
Quit
where {ωn }∞
n=0
limn→∞ αn = 0.
αn+1 ≤ (1 − ωn )αn + ωn βn + γn
for n ≥ 0,
P∞
P∞
⊂ [0, 1],
n=0 ωn = ∞, limn→∞ βn = 0 and
n=0 γn < ∞.
Then
3. Main Results
Theorem 3.1. Suppose that X is an arbitrary real Banach space and T : X → X is
∞
∞
a Lipschitz φ-strongly accretive operator. Assume that {un }∞
n=0 , {vn }n=0 , {wn }n=0 are se∞
∞
∞
quences in X and {an }n=0 , {bn }n=0 and {cn }n=0 are sequences in [0, 1] such that {kwn k}∞
n=0
is bounded and
∞
X
(3.1)
a2n < ∞,
n=0
∞
X
∞
X
an bn < ∞,
n=0
n=0
∞
X
(3.2)
kun k < ∞,
∞
X
kvn k < ∞,
n=0
an = ∞.
n=0
For any given f ∈ X, define S : X → X by Sx = f + x − T x for all x ∈ X. Then the
three-step iteration sequence with errors {xn }∞
n=0 defined for arbitrary x0 ∈ X by
zn = (1 − cn )xn + cn Sxn + wn ,
(3.3)
yn = (1 − bn )xn + bn Szn + vn ,
xn+1 = (1 − an )xn + an Syn + un ,
n≥0
converges strongly to the unique solution q of the equation T x = f . Moreover
kxn+1 − qk ≤ [1 + (3 + 3L3 + L4 )a2n + L(1 + L2 )an bn ]kxn − qk
− A(xn+1 , q)an kxn − qk + an bn L2 (3 + L)kwn k
(3.4)
JJ J
+ an L(3 + L)kvn k + (3 + L)kun k
I II
for n ≥ 0, where A(x, y) =
Go back
Full Screen
Close
Quit
φ(kx−yk)
1+kx−yk+φ(kx−yk)
∈ [0, 1) for x, y ∈ X.
Proof. It follows from Lemma 2.2 that the equation T x = f has a unique solution q ∈ X.
Let L0 denote the Lipschitz constant of T . From the definition of S we know that q is a
fixed point of S and S is also Lipschitz with constant L = 1 + L0 . Thus for any x, y ∈ X,
there exists j(x − y) ∈ J(x − y) such that
h(I − S)x − (I − S)y, j(x − y)i ≥ A(x, y)kx − yk2 .
This implies that
h(I − S − A(x, y))x − (I − S − A(x, y))y, j(x − y)i ≥ 0
and it follows from Lemma 1.1 of Kato [8] that
(3.5)
kx − yk ≤ kx − y + r[(I − S − A(x, y))x − (I − S − A(x, y))y]k
for x, y ∈ X and r > 0. From (3.3) we conclude that for each n ≥ 0
xn = xn+1 + an xn − an Syn − un
(3.6)
= (1 + an )xn+1 + an (I − S − A(xn+1 , q))xn+1 − (I − A(xn+1 , q))an xn
+ an (Sxn+1 − Syn ) + (2 − A(xn+1 , q))a2n (xn − Syn )
− [1 + (2 − A(xn+1 , q))an ]un
JJ J
I II
Go back
Full Screen
and
(3.7)
q = (1 + an )q + an (I − S − A(xn+1 , q))q − (I − A(xn+1 , q))an q.
Close
Quit
It follows from (3.5)–(3.7) that
kxn − qk
= k(1 + an )xn+1 + an (I − S − A(xn+1 , q))xn+1 − (I − A(xn+1 , q))an xn
+ an (Sxn+1 − Syn ) + (2 − A(xn+1 , q))a2n (xn − Syn )
− [1 + (2 − A(xn+1 , q))an ]un − (1 + an )q − an (I − S − A(xn+1 , q))q
+ (I − A(xn+1 , q))an qk
an
≥ (1 + an )
xn+1 − q + 1 + an [(I − S − A(xn+1 , q))xn+1
− (I − S − A(xn+1 , q))q − an (1 − A(xn+1 , q))kxn − qk
− (2 − A(xn+1 , q))a2n kxn − Syn k − an kSxn+1 − Syn k
− [1 + (2 − A(xn+1 , q))an ]kun k
≥ (1 + an )kxn+1 − qk − an (1 − A(xn+1 , q))kxn − qk
− (2 − A(xn+1 , q))a2n kxn − Syn k − an kSxn+1 − Syn k
− [1 + (2 − A(xn+1 , q))an ]kun k,
which implies that
kxn+1 − qk
JJ J
I II
Go back
Full Screen
≤
(3.8)
1 + (1 − A(xn+1 , q))an
kxn − qk + (2 − A(xn+1 , q))a2n kxn − Syn k
1 + an
+ an kSxn+1 − Syn k + [1 + (2 − A(xn+1 , q))an ]kun k
≤ (1 − A(xn+1 , q)an + a2n )kxn − qk + 2a2n kxn − Syn k
+ an kSxn+1 − Syn k + (1 + 2an )kun k
Close
Quit
for n ≥ 0. By (3.3) we get that
kzn − qk ≤ (1 − cn )kxn − qk + cn kSxn − qk + kwn k
(3.9)
≤ (1 − cn )kxn − qk + Lcn kxn − qk + kwn k
≤ Lkxn − qk + kwn k,
(3.10)
kyn − qk ≤ (1 − bn )kxn − qk + bn kSzn − qk + kvn k
≤ (1 − bn )kxn − qk + Lbn kzn − qk + kvn k,
(3.11)
kxn − Szn k ≤ kxn − qk + kSzn − qk ≤ kxn − qk + Lkzn − qk,
(3.12)
kxn − yn k ≤ bn kxn − Szn k + kvn k
and
(3.13)
JJ J
I II
Go back
Full Screen
for n ≥ 0. From (3.9)–(3.13) we obtain that
(3.14)
kxn − Syn k ≤ (1 + L3 )kxn − qk + L2 bn kwn k + Lkvn k
and
kSxn+1 − Syn k ≤ (Lbn + L3 bn − Lan bn − L3 an bn + L3 an + L4 an )kxn − qk
Close
(3.15)
Quit
kSyn − yn k ≤ kSyn − qk + kyn − qk ≤ (1 + L)kyn − qk
+ (L2 bn + L3 an bn )kwn k + (L + L2 an )kvn k + Lkun k
for n ≥ 0. It follows from (3.8), (3.14) and (3.15) that
kxn+1 − qk ≤ [1 + (3 + 3L3 + L4 )a2n + L(1 + L2 )an bn ]kxn − qk
− A(xn+1 , q)an kxn − qk + an bn L2 (3 + L)kwn k
(3.16)
+ (3 + L)an kvn k + (3 + L)kun k
for n ≥ 0. Set
rn = kxn − qk,
kn = (3 + 3L3 + L4 )a2n + L(1 + L2 )an bn ,
sn = an ,
2
tn = an bn L (3 + L)kwn k + an L(3 + L)kvn k + (3 + L)kun k for n ≥ 0.
Then (3.16) yields that
(3.17)
rn+1 ≤ (1 + kn )rn − sn rn
φ(rn+1 )
+ tn
1 + rn+1 + φ(rn+1 )
for n ≥ 0.
It follows from (3.1), (3.2), (3.17) and Lemma 2.1 that rn → 0 as n → ∞. That is xn → q
as n → ∞. This completes the proof.
JJ J
I II
Go back
Full Screen
Close
Quit
Remark 3.2. Theorem 3.1 extends Theorem 5.2 of [3], Theorem 1 of [4], Theorem 2
of [5], Theorem 1 of [6], Theorem 3.1 of [10], Theorem 1 of [12], Theorem 1 of [13] and
Theorem 4.1 of [15].
∞
∞
∞
∞
∞
Theorem 3.3. Let X, {un }∞
n=0 , {vn }n=0 , {wn }n=0 , {an }n=0 , {bn }n=0 and {cn }n=0 be
as in Theorem 3.1 and T : D(T ) ⊂ X → X be a Lipschitz φ-strongly accretive operator.
Suppose that the equation T x = f has a solution q ∈ D(T ) for some f ∈ X. Assume that
∞
∞
the sequences {xn }∞
n=0 , {yn }n=0 and {zn }n=0 generated from an arbitrary x0 ∈ D(T ) by
∞
∞
(3.3) are contained in D(T ). Then {xn }n=0 , {yn }∞
n=0 and {zn }n=0 converge strongly to q
and satisfied (3.4).
The proof of Theorem 3.3 uses the same idea as that of Theorem 3.1. So we omit it.
Remark 3.4. Theorem 3.1 in [7] and Theorem 3.2 in [10] are special cases of our
Theorem 3.3.
Theorem 3.5. Suppose that X is an arbitrary real Banach space and T : X → X is a
uniformly continuous φ-strongly accretive operator, and the range of either (I − T ) or T is
bounded. For any f ∈ X, define S : X → X by Sx = f + x − T x for all x ∈ X and the
three-step iteration sequence with errors {xn }∞
n=0 by
x0 , u0 , v0 , w0 ∈ X,
zn = a00n xn + b00n Sxn + c00n wn ,
(3.18)
yn = a0n xn + b0n Szn + c0n vn ,
xn+1 = an xn + bn Syn + cn un ,
n ≥ 0,
∞
∞
∞
where {un }∞
n=0 , {vn }n=0 and {wn }n=0 are arbitrary bounded sequences in X and {an }n=0 ,
00 ∞
0 ∞
00 ∞
00 ∞
0 ∞
∞
∞
0 ∞
{bn }n=0 , {cn }n=0 , {an }n=0 , {bn }n=0 , {cn }n=0 , {an }n=0 , {bn }n=0 and {cn }n=0 are real sequences in [0, 1] satisfying the following conditions
JJ J
I II
(3.19)
Go back
Full Screen
Close
Quit
(3.20)
∞
X
n=0
an + bn + cn = 1,
a0n + b0n + c0n = 1,
a00n + b00n + c00n = 1,
bn + cn ∈ (0, 1),
bn = +∞,
n ≥ 0,
lim bn = lim b0n = lim c0n = lim
n→∞
n→∞
n→∞
n→∞
cn
= 0.
bn + cn
Then the sequence {xn }∞
n=0 converges strongly to the unique solution of the equation T x = f .
Proof. It follows from Lemma 2.2 that the equation T x = f has a unique solution q ∈ X.
By (1.2) we have
hT x − T y, j(x − y)i = h(I − S)x − (I − S)y, j(x − y)i ≥ A(x, y)kx − yk2 ,
where A(x, y) =
φ(kx−yk)
1+kx−yk+φ(kx−yk)
∈ [0, 1) for x, y ∈ X. This implies that
h(I − S − A(x, y))x − (I − S − A(x, y))y, j(x − y)i ≥ 0
for x, y ∈ X. It follows from Lemma 1.1 of Kato [8] that
(3.21)
kx − yk ≤ kx − y + r[(I − S − A(x, y))x − (I − S − A(x, y))y]k
for x, y ∈ X and r > 0. Now we show that R(S) is bounded. If R(I − T ) is bounded, then
kSx − Syk = k(I − T )x − (I − T )yk ≤ δ(R(I − T ))
for x, y ∈ X. If R(T ) is bounded, we get that
kSx − Syk = k(x − y) − (T x − T y)k
≤ φ−1 (kT x − T yk) + kT x − T yk
JJ J
≤ φ−1 (δ(R(T ))) + δ(R(T ))
I II
Go back
for x, y ∈ X. Hence R(S) is bounded. Put
dn = bn + cn ,
d0n = b0n + c0n ,
d00n = b00n + c00n
for n ≥ 0
Full Screen
and
Close
(3.22)
Quit
D = max{kx0 − qk,
sup{kx − qk : x ∈ {un , vn , wn , Sxn , Syn , Szn : n ≥ 0}}}.
By (3.18) and (3.22) we conclude that
max{kxn − qk, kyn − qk, kzn − qk} ≤ D
(3.23)
for n ≥ 0.
Using (3.18) we obtain that
(1 − dn )xn = xn+1 − dn Syn − cn (un − Syn )
(3.24)
= [1 − (1 − A(xn+1 , q))dn ]xn+1 + dn (I − S − A(xn+1 , q))xn+1
+ dn (Sxn+1 − Syn ) − cn (un − Syn ).
Note that
(3.25)
(1 − dn )q = [1 − (1 − A(xn+1 , q))dn ]q + dn (I − S − A(xn+1 , q))q.
It follows from (3.21) and (3.23)–(3.25) that
(1 − dn )kxn − qk ≥ [1 − (1 − A(xn+1 , q))dn ]kxn+1 − q
dn
[(I − S − A(xn+1 , q))xn+1
1 − (1 − A(xn+1 , q))dn
− (I − S − A(xn+1 , q))q]k − dn kSxn+1 − Syn k − cn kun − Syn k
+
JJ J
≥ [1 − (1 − A(xn+1 , q))dn ]kxn+1 − qk − dn kSxn+1 − Syn k − 2Dcn .
I II
That is
Go back
1 − dn
kxn − qk
1 − (1 − A(xn+1 , q))dn
dn
2Dcn
+
kSxn+1 − Syn k +
1 − (1 − A(xn+1 , q))dn
1 − (1 − A(xn+1 , q))dn
≤ [1 − (1 − A(xn+1 , q))dn ]kxn − qk + M dn kSxn+1 − Syn k + M cn
kxn+1 − qk ≤
Full Screen
(3.26)
Close
Quit
for n ≥ 0, where M is some constant. In view of (3.18)–(3.20) we infer that
kxn+1 − yn k ≤ kxn+1 − xn k + kyn − xn k
≤ bn kSyn − xn k + cn kun − xn k + b0n kSzn − xn k + c0n kvn − xn k
≤ bn kSyn − xn k + cn kun − xn k + b0n kSzn − zn k + c0n kvn − xn k
+ b0n (b00n kSxn − xn k + c00n kwn − xn k)
≤ 2D(dn + d0n + b0n d00n ) → 0
as n → ∞. Since S is uniformly continuous, we have
(3.27)
kSxn+1 − Syn k → 0
as n → ∞.
Set inf{A(xn+1 , q) : n ≥ 0} = r. We claim that r = 0. If not, then r > 0. It is easy to check
that
kxn+1 − qk ≤ (1 − rdn )kxn − qk + M dn kSxn+1 − Syn k + M cn
for n ≥ 0.
Put
JJ J
Go back
Full Screen
Close
Quit
αn = kxn − qk,
cn = tn dn ,
I II
βn = M r
−1
ωn = rdn ,
(kSxn+1 − Syn k + tn ),
γn = 0 for n ≥ 0.
tn → 0 as n → ∞. It follows
(3.20), (3.27) and Lemma 2.3 that
(3.2) ensures that
P∞
Pfrom
∞
ωn ∈ (0, 1] with n=0 ωn = ∞, limn→∞ βn = 0, n=0 γn < ∞. So kxn − qk → 0 as n → ∞,
which means that r = 0. This is a contradiction. Thus r = 0 and there exists a subsequence
∞
{kxni +1 − qk}∞
i=0 of {kxn+1 − qk}n=0 satisfying
(3.28)
kxni +1 − qk → 0
as i → ∞.
From (3.28) and (3.29) we conclude that for given ε > 0 there exists a positive integer m
such that for n ≥ m,
kxnm +1 − qk < ε
(3.29)
and
(3.30)
cn
M kSxn+1 − Syn k + M
< min
dn
φ(ε)ε
1
ε,
.
2 1 + φ( 32 ε) + 32 ε
Now we claim that
kxnm +j − qk < ε for j ≥ 1.
(3.31)
In fact (3.29) means that (3.31) holds for j = 1. Assume that (3.31) holds for j = k. If
kxnm +k+1 − qk > ε, we get that
kxnm +k+1 − qk
JJ J
I II
(3.32)
Go back
Full Screen
≤ kxnm +k − qk + M dnm +k kSxnm +k+1 − Synm +k k + M cnm +k
1
φ(ε)ε
≤ ε + min
ε,
dnm +k
2 1 + φ( 32 ε) + 32 ε
3
≤ ε.
2
Note that φ(kxnm +k+1 − qk) > φ(ε). From (3.32) we get that
Close
(3.33)
Quit
A(xnm +k+1 , q) ≥
φ(ε)
.
1 + φ( 32 ε) + 32 ε
By virtue of (3.26) (3.30) and (3.33) we obtain that
kxnm +k+1 − qk
φ(ε)ε
≤ 1−
d
n +k kxnm +k − qk
1 + φ( 32 ε) + 32 ε m
+ M dnm +k kSxnm +k+1 − Synm +k k + M cnm +k
1
φ(ε)ε
φ(ε)ε
dnm +k
dn +k ε + min
ε,
≤ 1−
2 1 + φ( 32 ε) + 32 ε
1 + φ( 32 ε) + 32 ε m
≤ ε.
That is
ε < kxnm +k+1 − qk ≤ ε,
which is a contradiction. Hence kxnm +k+1 − qk ≤ ε. By induction (3.29) holds for j ≥ 1.
Thus (3.31) yields that xn → q as n → ∞. This completes the proof.
JJ J
I II
Go back
Remark 3.6. Theorem 3.5 extends and improves Theorem 3.4 in [2] and Theorem 3.1
in [16].
Acknowledgement. This work was supported by the Science Research Foundation of
Educational Department of Liaoning Province (20060467).
Full Screen
Close
Quit
1. Browder F. E., Nonlinear mappings of nonexpansive and accretive type in Banach spaces, Bull. Amer.
Math. Soc. 73 (1967), 875–882.
2. Chang S. S., Some problems and results in the study of nonlinear analysis, Nonlinear Anal. 30 (1997),
4197–4208.
JJ J
I II
Go back
Full Screen
Close
Quit
3. Chang S. S., ChoY. J., Lee B. S. and S. M. Kang, Iterative approximations of fixed points and solutions
for strongly accretive and strongly pseudo-contractive mappings in Banach spaces, J. Math. Anal. Appl.
224 (1998) 149–165.
4. Chidume C. E., An iterative process for nonlinear Lipschitzian strongly accretive mapping in Lp spaces,
J. Math. Anal. Appl. 151 (1990), 453–461.
5.
, Iterative solution of nonlinear equations with strongly accretive operators, J. Math. Anal. Appl.
192 (1995), 502–518.
6. Deng L., On Chidume’s open questions, J. Math. Anal. Appl. 174 (1993), 441–449.
7. Ding X. P., Iterative process with errors to nonlinear φ-strongly accretive operator equations in arbitrary
Banach spaces, Computers Math. Applic. 33 (1997), 75–82.
8. Kato T., Nonlinear semigroups and evolution equations, J. Math. Soc. Japan 19 (1967), 508–520.
9. Liu L. S., Ishikawa and Mann iterative process with errors for nonlinear strongly accretive mapping in
Banach spaces, J. Math. Anal. Appl. 194 (1995), 114–125.
10. Liu Z. and Kang S. M. Convergence theorems for φ-strongly accretive and φ-hemicontractive operators,
J. Math. Anal. Appl. 253 (2001), 35–49.
11. Martin R. H., Jr. A global existence theorem for autonomous differential equations in Banach spaces,
Proc. Amer. Math. Soc. 26 (1970), 307–314.
12. Osilike M. O., Iterative solution of nonlinear equations of the φ-strongly accretive type, J. Math. Anal.
Appl. 200 (1996), 259–271.
13.
, Ishikawa and Mann iteration methods with errors for nonlinear equations of the accretive type,
J. Math. Anal. Appl. 213 (1997), 91–105.
14.
, Iterative solution of nonlinear φ-strongly accretive operator equations in arbitrary Banach
spaces, Nonlinear Anal. TMA 36 (1999), 1–9.
15. Tan K. K. and Xu H. K., Iterative solutions to nonlinear equations of strongly accretive operators in
Banach spaces, J. Math. Anal. Appl. 178 (1993), 9–21.
16. Yin Q., Liu Z. and Lee B. S., Iterative solutions of nonlinear equations with φ-strongly accretive operators, Nonlinear Anal. Forum 5 (2000), 87–89.
Shin Min Kang, Department of Mathematics and the Research Institute of Natural Science, Gyeongsang
National University, Jinju 660-701, Korea, e-mail: smkang@nongae.gsnu.ac.kr
Chi Feng, Department of Science, Dalian Fisheries College, Dalian, Liaoning, 116023, People’s Republic of
China, e-mail: windmill-1129@163.com
Zeqing Liu, Department of Mathematics, Liaoning Normal University, P.O. Box 200, Dalian, Liaoning,
116029, People’s Republic of China, e-mail: zeqingliu@sina.com.cn
JJ J
I II
Go back
Full Screen
Close
Quit