RHOADES Received 10 September 2003 Given a finite family of nonexpansive self-mappings of a Hilbert space, a particular qua-dratic functional, and a strongly positive selfadjoint bounded
Trang 1FOR A FAMILY OF NONEXPANSIVE MAPPINGS
IN HILBERT SPACE
B E RHOADES
Received 10 September 2003
Given a finite family of nonexpansive self-mappings of a Hilbert space, a particular qua-dratic functional, and a strongly positive selfadjoint bounded linear operator, Yamada
et al defined an iteration scheme which converges to the unique minimizer of the qua-dratic functional over the common fixed point set of the mappings In order to obtain their result, they needed to assume that the maps satisfy a commutative type condition
In this paper, we establish their conclusion without the assumption of any type of com-mutativity
Finding an optimal point in the intersectionF of the fixed point sets of a family of
nonexpansive maps is one that occurs frequently in various areas of mathematical sci-ences and engineering For example, the well-known convex feasibility problem reduces
to finding a point in the intersection of the fixed point sets of a family of nonexpan-sive maps (See, e.g., [3,4].) The problem of finding an optimal point that minimizes a given cost functionΘ : Ᏼ→RoverF is of wide interdisciplinary interest and practical
importance (See, e.g., [2,4,5,7,14].) A simple algorithmic solution to the problem of minimizing a quadratic function overF is of extreme value in many applications
includ-ing the set-theoretic signal estimation (See, e.g., [5,6,10,14].) The best approximation problem of finding the projectionP F(a) (in the norm induced by the inner product of Ᏼ) from any given point a in Ᏼ is the simplest case of our problem Some papers dealing
with this best approximation problem are [2,9,11]
LetᏴ be a Hilbert space, C a closed convex subset of Ᏼ, and T i, wherei =1, 2, ,N,
a finite family of nonexpansive self-maps ofC, with F : = ∩ n i =1Fix(T i)= ∅ Define a qua-dratic functionΘ : Ᏼ→Rby
Θ(u) : =1
whereb ∈ Ᏼ and A is a selfadjoint strongly positive operator We will also assume that
B : = I − A satisfies B < 1, although this is not restrictive, since µA is strongly positive
Copyright©2004 Hindawi Publishing Corporation
Fixed Point Theory and Applications 2004:2 (2004) 135–147
2000 Mathematics Subject Classification: 47H10
URL: http://dx.doi.org/10.1155/S1687182004309046
Trang 2with I − µA < 1 for any µ ∈(0, 2/ A ), and minimizingΘ(u): =(1/2) µAu,u − µb,u
overF is equivalent to the original minimization problem.
Yamada et al [13] show that there exists a unique minimizeru ∗of Θ over C if and
only ifu ∗satisfies
Au ∗ − b,u − u ∗
In their solution of this problem, Yamada et al [13] add the restriction that theT i
satisfy
Fix
T N ··· T1
=Fix
T1T N ··· T3T2
=Fix
T N −1T N −2··· T1T N
There are many nonexpansive maps, with a common fixed point set, that do not satisfy (3) For example, ifX =[0, 1] andT1andT2are defined byT1x = x/2 + 1/4 and T2x =
3x/4, then Fix(T1,T2)= {2/5 }, whereas Fix(T2,T1)= {3/10 }
In our solution, we are able to remove restriction (3) We will take advantage of the modified Wittmann iteration scheme developed by Atsushiba and Takahashi [1] Letα n1,α n2, ,α nN ∈(0, 1], n =1, 2, Given the mappings T1,T2, ,T N, one can define, for eachn, new mappings U1, ,U Nby
U n1 = α n1 T1+
1− α n1
I,
U n2 = α n2 T2U n1+
1− α n2
I,
U n,N −1 = α n,N −1T N −1U n,N −2+
1− α n,N −1
I,
W n:= U nN = α nN T N U n,N −1+
1− α nN
I.
(4)
From [1, Lemmas 3.1 and 3.2], if theT iare nonexpansive, so are theU ni, and both sets
of functions have the same fixed point set
The iteration scheme we will use is the following Letb ∈ C and choose any u0∈ C.
Define{ u n }by
u n+1 = λ n b +I − λ n AW n u n, (5)
where theW nare the self-maps ofC generated by (4)
Theorem 1 Let T i:Ᏼ→ Ᏼ (i =1, ,N) be nonexpansive with nonempty common fixed point set F = ∅ Assume that { λ n } and { α ni } satisfy
(i) 0≤ λ n ≤ 1,
(ii) limλ n = 0,
(iii)
n ≥1λ n = ∞ ,
Trang 3n ≥1| λ n − λ n −1| < ∞ ,
(v)
n ≥1| α ni − α n −1,i | < ∞ for each i =1, 2, ,N.
Then, for any point u0∈ Ᏼ, the sequence { u n } generated by ( 5 ) converges strongly to the unique minimizer u ∗ of the function Θ of ( 1 ) over F.
Proof From [15],u ∗exists and is unique We will first assume that
u0∈ C u ∗:=
x ∈Ᏼ|x − u ∗ ≤b − Au ∗
1− B
whereA and B are as previously defined.
For anyx ∈Ᏼ and 0≤ λ ≤1, define
Then, for anyy ∈ Ᏼ, since W is nonexpansive,
T λ(x) − T λ(y) = (I − λA)W(x) − W(y) 1− λ1− B x − y (8) Also, sinceu ∗ ∈ F,
T λ
u ∗
Thus,
T λ(x) − u ∗ ≤ T λ(x) − T λ
u ∗+T λ
u ∗
− u ∗
≤ 1− λ1− B x − u ∗+λ
1− B b − Au ∗
1− B
≤x − Au ∗
1− B
(10)
If, in (7), we make the substitutionλ = λ n,T λ(x) = u n+1, andW(x) = W n u n, then it follows from (9) and (10) thatu nandW n u nbelong toC u ∗ for eachn Thus, { u n }and
{ W n u n }are bounded Since B < 1, { BW n u n }is also bounded
LetK denote the diameter of C u ∗
We may write (5) in the form
u n+1 = λ n b +I − λ n(I − B)W n u n
= λ n b +I − λ n
We will first show that
limu n+1 − u n =0. (12)
Trang 4Using (11), since eachW nis nonexpansive and B < 1,
u n+1 − u n
=λ n b +
1− λ n
W n u n+λ n BW n u n − λ n −1b
−1− λ n −1
W n −1u n −1− λ n −1BW n −1u n −1
≤ λ n − λ n −1 b +
1− λ nW n u n − W n −1u n
+ λ n − λ n −1 W n −1u n −1+
1− λ nW n −1u n − W n −1u n −1
+λ n B W n u n − W n −1u n+λ n B W n −1u n − W n −1u n −1
+ λ n − λ n −1 BW n −1u n −1
≤3 λ n − λ n −1 K +
1− λ n+λ n B
× W n u n − W n −1u n+
1− λ n+λ n B W n −1u n − W n −1u n −1 .
(13)
From (4), sinceT NandU n −1,N −1are nonexpansive,
W n u n − W n −1u n
=α nN T N U n,N −1u n+
1− α nN
u n − α n −1,N T N U n −1,N −1u n −1− α n −1,N
u n
≤ α nN − α n −1,N u n+α nN T N U n,N −1u n − α n −1,N T N U n −1,N −1u n
≤ α nN − α n −1,N u n+α nN
T N U n,N −1u n − T N U n −1,N −1u n
+ α nN − α n −1,N T N U n −1,N −1u n
≤ α nN − α n −1,N u n+α nNU n,N −1u n − U n −1,N −1u n+ α nN − α n −1,N K
≤2K α nN − α n −1,N +α nNU n,N −1u n − U n −1,N −1u n.
(14)
Again, from (4),
U n,N −1u n − U n −1,N −1u n
=α n,N −1T N −1U n,N −2u n+
1− α n,N −1
u n
− α n −1,N −1T N −1U n −1,N −2u n −1− α n −1,N −1u n
≤ α n,N −1− α n −1,N −1 u n
+α n,N −1T N −1U n,N −2u n − α n −1,N −1T N −1U n −1,N −2u n
≤ α n,N −1− α n −1,N −1 u n
+α n,N −1 T N −1U n,N −2u n − T N −1U n −1,N −2u n
+ α n,N −1− α n −1,N −1 K
≤2K α n,N −1− α n −1,N −1 +α n,N −1U n,N −2u n − U n −1,N −2u n
≤2K α n,N −1− α n −1,N −1 +U n,N −2u n − U n −1,N −2u n.
(15)
Trang 5U n,N −1u n − U n −1,N −1u n
≤2K α n,N −1− α n −1,N −1 + 2K α n,N −2− α n −1,N −2
+U n,N −3u n − U n −1,N −3u n
≤2K N −1
i =2
α ni − α n −1,i +U n1 u n − U n −1,1u n
=α n1 T1u n+
1− α n1
u n − α n −1,1T1u n −1− α n −1,1
u n
+ 2K N −
1
i =2
α ni − α n −1,i
≤ α n1 − α n −1,1 u n+α n1 T1u n − α n −1,1T1u n
+ 2K N −1
i =2
α ni − α n −1,i
≤2K N −
1
i =1
α ni − α n −1,i .
(16)
Substituting (16) into (14),
W n u n − W n −1u n ≤2K α nN − α n −1,N + 2α nN K N −1
i =1
α ni − α n −1,i
≤2K N
i =1
Using (17) in (13),
u n+1 − u n ≤ 1− λ n
1− B u n − u n −1+ 3K λ n − λ n −1 + 2
1− λ n
1− B K N
i =1
α ni − α n −1,i . (18) Thus, since 0< 1 − λ n(1− B )< 1 for all n,
u n+m+1 − u n+m ≤ n+m
i = m
1− λ i
1− B u i+1 − u i
+ 3K
n+m
i = m
λ i − λ i −1 + 2K n+m
i = m
N
j =1
α ij − α i −1,j . (19)
Trang 6From (iii), since the product diverges to zero,
lim sup
n
u n+1 − u n =lim sup
n
u n+m+1 − u m+n
≤2K ∞
i = m
λ i − λ i −1 + 2K ∞
i = m
N
j =1
α ij − α i −1,j . (20) Therefore, taking the lim supmof both sides and using (iv) and (v),
lim sup
n
and (12) is satisfied
Now, for any nonexpansive self-mapT of C u ∗, defineG t:C u ∗ → C u ∗ by
G t(x) = tb + (1 − t)TG t(x) + tBTG t(x) (22) for eacht ∈(0, 1] Using an argument similar to the proof of [8, Theorem 12.2, page 45], we will now show that ifT has a fixed point, then, for each x in C u ∗, the strong limitt →0G t(x) exists and is a fixed point of T.
Definey(t) = G t(x) and let w be a fixed point of T:
y(t) − w = t(b − w) + (1 − t)T y(t) − w+tBT y(t). (23) SinceT is nonexpansive,
y(t) − w ≤ t b − w + (1− t)T y(t) − w+t B T y(t)
≤ t b − w + (1− t)y(t) − w+t B T y(t),
ty(t) − w ≤ t b − w +t B T y(t) − w+t B w , (24) or
y(t) − w ≤ b − w + B y(t) − w+ B w , (25) which, since B < 1, yields
y(t) − w ≤ 1
1− B b − w + B w , (26) andy(t) remains bounded as t →0
Also,
BT y(t)<T y(t) ≤ T y(t) − Tw+ w ≤y(t) − w+ w , (27) and bothBT y(t) and T y(t) remain bounded as t →0
Hence,
y(t) − T y(t) = tb − T y(t) + BT y(t) −→0 ast −→0. (28)
Trang 7Definey n = y(t n) and lett n →0 Letµ nbe a Banach limit and f : C u ∗ →R+defined by
f (z) = µ n
y n − z 2
Since f is continuous and convex, f (z) → ∞as z → ∞ SinceᏴ is reflexive, f attains
it infimum overC u ∗
LetM be the set of minimizers of f over C u ∗ Ifu ∈ C u ∗, then
f (Tu) = µ n
y n − Tu 2
= µ n
T y n − Tu 2
≤ µ n
y n − u 2
= f (u). (30) Therefore,M is invariant under T Since it is also bounded, closed, and convex, it must
contain a fixed point ofT Denote this fixed point by v Then,
y n − T y n,y n − v=y n − v, y n − v+
v − T y n,y n − v
=y n − v 2
+
Tv − T y n,y n − v. (31)
But
Tv − T y n,y n − v ≤ Tv − T y ny n − v ≤ y n − v 2
so that
Since
y n = t n b +1− t n
T y n+t n BT y n,
y n − b =1− t n
T y n − b+t n BT y n
=1− t n
T y n − y n
+
1− t n
y n − b+t n BT y n,
(34)
thus,
ty n − b=1− t n
T y n − y n
or
y n − b − Bv =1− t n
t n
T y n − y n
Therefore, from (33),
y n − b − Bv, y n − v
=1− t n
t n
T y n − y n,y n − v+
BT y n − Bv, y n − v
≤BT y n − Bv, y n − v.
(37)
Trang 8For anyz ∈ C u ∗,
y n − v 2
=y n −
1− t n
v − t n z + t n(z − v) − t n b − t n Bv + t n(b + Bv) 2
≥y n −
1− t n
v − t n b − t n Bv 2
+ 2t n
z − v + b + Bv, y n −1− t n
v − t n z − t n b − t n Bv.
(38)
Let > 0 be given Since Ᏼ is uniformly smooth, there exists a t0> 0 such that, for all
t n ≤ t0,
z − v + b + Bv,
y n − v−y n −1− t n
v − t n z − t n b − t n Bv < . (39) Thus, from (38),
z − v + b + Bv, y n − v
< +
z − v + b + Bv, y n −1− t n
v − t n z − t n b − t n B
< + 1
2t
y n − v 2
−y n −
1− t n
v − t n b − t n Bv 2
.
(40)
Since the Gateaux derivative exists inᏴ, we obtain
µ n
Settingz = θ in (41) and adding (37) and (41) yields
µ n
y n − v, y n − v≤ µ n
or
µ n
y n − v 2
≤ µ nBT y n − Bvy n − v
≤ µ n
B T y n − Tvy n − v
≤ B µ n
y n − v 2
.
(43)
Therefore,µ n y n − v 2=0 Thus, there is a subsequence of{ y n }converging strongly
tov Suppose that lim k →∞ y(t n k)= v1and limk →∞ y(t m k)= v2 From (37), we have
v1− b − Bv2,v1− v2
≤BTv1− Bv2,v1− v2
,
v2− b − Bv1,v2− v1
≤BTv2− Bv1,v2− v1
Adding these inequalitites, we obtain
v1− BTv1+BTv2− v2,v1− v2
or
v1− v2,v1− v2
≤BTv1− BTv2,v1− v2
Trang 9
that is,
v1− v2 2
≤BTv2− BTv1v1− v2
≤ B Tv2− Tv1v1− v2
≤ B v2− v1 2
,
(47)
which, since B < 1, implies that v1= v2, and thus limy n = v.
Now, settingz = θ in (41), we obtain
µ n
or
µ n
which, from (2), implies thatv = u ∗
Letu nkdenote the unique element ofC u ∗such that
u nk =1k b+
1−1k
W n u nk+1
From what we have just proved, limk u nk → u ∗ Using (11),
u n+1 − W n+1 u n+1,k
=λ n b +
1− λ n+λ n BW n u n − W n+1 u n+1,k
≤ λ nb − W n+1 u n+1,k+
1− λ nW n u n − W n+1 u n+1,k
+λ n B W n u n − W n+1 u n+1,k+λ nBW n+1 u n+1,k
< 3Kλ n+
1− λ n+λ n B W n u n − W n u nk+W n u nk − W n u n+1,k
+W n u n+1,k − W n+1 u n+1,k
≤3Kλ n+
1− λ n+λ n B u n − u nk+u nk − u n+1,k
+W n u n+1,k − W n+1 u n+1,k .
(51)
As in (17),
W n u n+1,k − W n+1 u n+1,k ≤2K N
i =1
α n+1,i − α ni . (52) From the definition ofu nk,
u nk =1
k b+
1−1
k
W n u nk+1
k BW n u nk,
u n+1,k =1k b+
1−1k
W n+1 u n+1,k+1
k BW n+1 u n+1,k,
u n+1,k − u nk =
1−1
k
W n+1 u n+1,k − W n u nk
+1
k B
W n+1 u n+1,k − W n u nk
.
(53)
Trang 10Therefore, sinceW n+1is nonexpansive,
u n+1,k − u nk ≤1−1k+1k B
W n+1 u n+1,k − W n u nk
≤1−1
k+
1
k B W n+1 u n+1,k − W n+1 u nk+W n+1 u nk − W n u nk
≤
1−1k+
1
k B u n+1,k − u nk+W n+1 u nk − W n u nk .
(54) Thus, using (17),
1− B
k u n+1,k − u nk ≤k −1 + B
N
i =1
α n+1,i − α n,i (55) or
u n+1,k − u nk ≤k −1 + B
1− B 2K N
i =1
α n+1,i − α n,i . (56)
Substituting (56) and (52) into (51) yields
u n+1 − W n+1 u n+1,k
≤3Kλ n+
1− λ n+λ n B u n − u nk+ k
1− B
2K N
i =1
α n+1,i − α n,i . (57) Thus, using (iii) and (v), we have
µ n
u n − W n u nk 2
= µ n
u n+1 − W n+1 u n+1,k 2
≤ µ n
u n − u nk 2
From (53),
u nk − u n =1
k
b − u n +
1−1
k
W n u nk − u n
+1
Hence,
1− k1
u n − W n u nk
= u n − u nk −1k
u n − b+1
1−1
k
2
u n − W n u nk 2
≥u n − u nk 2
−2
k
u n − b − BW n u nk,u n − u nk
=
1−2k
u n − u nk 2
−2k
u nk − b − BW n u nk,u n − u nk
.
(61)
Trang 11Therefore, using (58) and (61),
1−1k
2
µ nu n − u nk 2
≥
1− k1
2
µ nu n − W n u nk 2
≥
1− k2
µ nu n − u nk 2
−2
k µ n
u nk − b − BW n u nk,u n − u nk
,
(62)
which implies that
1
2k µ n
u n − u nk 2
≥ µ n
b − u nk+BW n u nk,u n − u nk
Since limk u nk → u ∗, independent ofn, it follows that
0≥ µ n
b − u ∗+Bu ∗,u n − u ∗
= µ n
b −(I − B)u ∗,u n − u ∗
= µ n
b − Au ∗,u n − u ∗
.
(64)
From (12),
lim b − u ∗,u n+1 − u ∗
−b − u ∗,u n − u ∗ =0. (65)
We need the following result from [12] IfA is a real number and { a1,a2, } ∈ ∞such thatµ n { a n } ≤ a for all Banach limits µ nand lim supn(a n+1 − a n)≤0, then lim supn a n ≤ a.
Consequently,
lim sup
n
b − u ∗,u n − u ∗
Sinceu ∗ ∈ F,
W n u n − u ∗ = W n u n − W n u ∗ ≤ u n − u ∗. (67) From (11),
u n+1 − u ∗ = λ n
b − u ∗ +
1− λ n
W n u n − u ∗
+λ n BW n u n
= λ n
b − u ∗ +
1− λ n+λ n BW n u n − u ∗
+λ n Bu ∗ (68) or
1− λ n+λ n BW n u n − u ∗
= u n+1 − u ∗ − λ n
b − u ∗
− λ n Bu ∗ (69) Therefore,
1− λ n+λ n B 2W n u n − u ∗ 2
≥u n+1 − u ∗ 2
−2λ n
b − u ∗+Bu ∗,u n+1 − u ∗
, (70)
Trang 12which implies that
u n+1 − u ∗ 2
≤1− λ n+λ n B 2W n u n − u ∗ 2
+ 2λ n
b − u ∗,u n+1 − u ∗ + 2λ n
Bu ∗,u n+1 − u ∗
From (ii) and the boundedness ofC u ∗, there exists a positive integerN such that, for
alln ≥ N,
λ n
b − u ∗,u n+1 − u ∗
≤
4, λ n
Bu ∗,u n+1 − u ∗
≤
Therefore, forn ≥ N,
u n+1 − u ∗ 2
≤1− λ n+λ n B 2u n − u ∗ 2
+
2+
2,
u n+m − u ∗ 2
≤
n+m−1
i = m
1− λ i+λ i B 2
u m − u ∗ 2
+
n+m−1
i = m
1− λ i+λ i B
2
.
(73)
Using (iii),
lim sup
n
u n − u ∗ 2
=lim sup
n
u n+m − u ∗ 2
Thus,{ u n }converges strongly tou ∗
Now letu0∈Ᏼ Let{ s n }be another sequence generated by (11) for somes0∈ C u ∗ Then, by what we have just proved, lims n = u ∗ SinceW nis nonexpansive for eachn,
u n+1 − s n+1 = λ n b +1− λ n AW n u n − λ n b −1− λ n AW n s n
≤1− λ n A
W n u n − W n s n
≤1− λ n+λ n B W n u n − W n s n
≤1− λ n+λ n B u n − s n.
(75)
By induction,
u n − s n ≤ u0− s0 n
k =1
1− λ k
1− B . (76)
Therefore, using (iii), lim u n − s n =0 and u n − u ∗ ≤ u n − s n + s n − u ∗ so that
... n+1− α nN
u n − α n −1,N T N... −1u n − T N U n −1,N −1u n
+... −1,N −1T N −1U n −1,N −2u