LEAST SQUARES PROBLEM
SIMEON REICH AND HONG-KUN XU Received 3 December 2001
A constrained least squares problem in a Hilbert spaceH is considered. The standard Tikhonov regularization method is used. In the case where the set of the constraints is the nonempty intersection of a finite collection of closed convex subsets ofH, an iterative algorithm is designed. The resulting sequence is shown to converge strongly to the unique solution of the regularized problem. The net of the solutions to the regularized problems strongly converges to the minimum norm solution of the least squares problem if its solution set is nonempty.
1. Introduction
Consider the following constrained least squares problem in a Hilbert spaceH: minx∈C f(x) :=1
2Ax−b2, (1.1)
whereCis a nonempty closed convex subset ofH,A:H→His a bounded linear operator, andbis an element ofH.
LetSdenote the solution set of (1.1). It is known thatSis closed and convex, but it may well be empty.
Since problem (1.1) may be ill-posed, a regularization process is necessary in order to find a solution of (1.1) (if any). We will apply the standard Tikhonov regularization to (1.1); that is, we consider the following regularized problem (ε >0):
minx∈C fε(x) :=1
2Ax−b2+1
2εx2. (1.2)
Copyright©2003 Hindawi Publishing Corporation Abstract and Applied Analysis 2003:8 (2003) 503–512
2000 Mathematics Subject Classification: 90C55, 49M20, 65J15, 47H09 URL:http://dx.doi.org/10.1155/S1085337503212082
(The reader is referred to [6] by Engl et al. for regularization methods for inverse and ill-posed problems. See also Tikhonov and Arsenin [12] and Bj¨orck [3].)
Since fε is continuous, convex, and coercive (i.e., x → ∞ ⇒ fε(x)→ ∞), (1.2) has a unique solution which we denote byxε.
In this paper, we first discuss the asymptotic behavior of{xε}asε↓0 and then design an iterative algorithm which produces a sequence converging strongly to the unique solutionxεof (1.2) in the case whereCis the nonempty intersection of a finite collection of closed convex subsets ofH. Moreover, ifSis nonempty, then the net{xε}of the approximate solutions converges strongly to the mini- mum norm solution of (1.1).
2. Behavior of{xε}
LetC, f,fε,{xε}, andSbe given as in the introduction. It is the objective of this section to study the behavior of the net{xε}of the solutions of the regularized problem (1.2). The next result shows that as long as the solution setSof (1.1) is nonempty,{xε}behaves well and approaches the minimum norm solution of (1.1).
Theorem2.1. The behavior of{xε}is as follows:
(i)S= ∅if and only if{xε}is bounded asε↓0. In this case,{xε}converges strongly asε↓0to the minimum norm solution of (1.1);
(ii)S= ∅if and only if limε↓0xε = ∞.
Proof. To prove (i), first assume thatS= ∅. SinceSis closed and convex, there is a unique element inS, denoted byS◦, such that
S◦=minv:v∈S. (2.1)
Since fε(xε)≤ fε(S◦), we see that
εxε2−S◦2≤AS◦−b2−Axε−b2. (2.2)
But f(S◦)≤f(xε); that is,AS◦−b2≤ Axε−b2, it follows from (2.2) that
xε≤S◦. (2.3)
Hence,{xε}is bounded.
Next, assume that{xε}is bounded asε↓0. Assume thatεn↓0 andxεn→x¯ weakly. Clearly, ¯x∈C.
We claim that ¯x∈S. To see this, we note that the weak lower semicontinuity of f implies that for anyx∈C,
fx¯≤lim inf
n→∞ fxεn
≤lim sup
ε↓0 fxε
=lim sup
ε↓0 fε xε
≤lim sup
ε↓0 fε(x)= f(x). (2.4)
This shows that ¯x∈S.
Next, we prove, assuming thatS= ∅, that{xε}converges strongly asε↓0 to S◦, the minimum norm solution of (1.1). Letvbe any weak limit point of{xε} whenε↓0. Then, we have a sequenceεn↓0 such thatxεn→vweakly. Since the norm ofHis weakly lower semicontinuous, it follows from (2.3) that
v ≤S◦. (2.5)
Due to the uniqueness of the minimum norm element, this implies thatv=S◦, which in turn implies thatS◦ is the only weak limit point of{xε}. Therefore, again, due to the weak lower semicontinuity of the norm and (2.3), we have
w−lim
ε↓0xε=S◦, lim
ε↓0
xε=S◦. (2.6)
This implies thats−limε↓0xε=S◦.
In order to prove (ii), we first note that ifS= ∅, then by (i){xε}is bounded.
So, it remains to show thatS= ∅implies limε↓0
xε= ∞. (2.7)
Suppose on the contrary that (2.7) does not hold. Then we can find a sequence {εn} ↓0 and some ˜x∈Csuch that
xεn−→x˜ weakly. (2.8)
Repeating the first half of the proof of part (i), we obtain that
fx˜≤ f(x), x∈C. (2.9)
That is, ˜x∈SandSis nonempty. The contradiction we have reached completes
the proof.
Remark 2.2. Lety∈Hand letxε(y) denote the unique solution of the regular- ized optimization problem
minx∈C fε(x) :=1
2Ax−b2+1
2εx−y2. (2.10)
LetPS be the nearest point projection fromH ontoS. Then, we can prove the following results:
(i)S= ∅ if and only if{xε(y)}is bounded as ε↓0. In this case, {xε(y)} converges strongly asε↓0 to the solution of (1.1) closest toy(i.e.,PSy);
(ii)S= ∅if and only if limε↓0xε(y) = ∞. 3. Iterative algorithm
In this section, we will introduce an iterative algorithm which produces a se- quence that converges strongly to the unique solution of the regularized prob- lem (1.2) in the case where the setCof constraints is the intersection of finitely many closed convex subsets ofH. We need some lemmas.Lemma 3.1is elemen- tary (see [14] for a proof).Lemma 3.2 is an easy exercise in an inner product space.Lemma 3.3is well known in the fixed-point theory of nonexpansive map- pings (see, e.g., Goebel and Kirk [7]).Lemma 3.4 may be known. A proof is, however, included.
Lemma3.1. Let{sn}be a sequence of nonnegative numbers satisfying sn+1≤
1−αn
sn+αnβn, n≥0, (3.1) where{αn}and{βn}are sequences of real numbers such that
(i){αn} ⊂[0,1]and∞n=0αn= ∞or equivalently, ∞
n=0
1−αn :=lim
n→∞
n k=0
1−αk
=0; (3.2)
(ii) lim supn→∞βn≤0, ornαnβnis convergent.
Thenlimn→∞sn=0.
Lemma3.2. LetH be a Hilbert space. Thenx+y2≤ x2+ 2y,x+y, for all x, y∈H.
Lemma3.3. LetDbe a closed convex subset ofH andT:D→Ha nonexpansive mapping with a fixed point. If{xn}is a sequence inDsuch thatxn→xweakly and (I−T)xn→ystrongly, then(I−T)x=y. (HereIis the identity operator onH.) Lemma3.4 (the optimality condition for (1.2)). The pointxε∈Csolves (1.2) if and only if
x−xε,A∗b−
A∗A+εIxε ≤0, x∈C, (3.3) whereA∗is the adjoint ofA.
Proof. Assume thatxε∈Cis the minimizer of fεoverC. We have, for anyx∈C,
0≤lim
t↓0
1 t
fε
xε+tx−xε
−fε xε
=lim
t↓0
Axε−b,Ax−xε +εxε,x−xε
+t 2
Ax−xε2+εx−xε2
=
A∗Axε−b+εxε,x−xε
=
A∗A+εIxε−A∗b,x−xε .
(3.4)
Assume next that (3.3) holds. Calculations yield the derivative f(x)=
A∗A+εIx−A∗b. (3.5)
Hence, by the subdifferential inequality, we obtain, for allx∈C, f(x)≥fxε
+fxε ,x−xε
=fxε
+A∗A+εIxε−A∗b,x−xε ≥ fxε
. (3.6)
This shows thatxεis indeed a minimizer of fεoverC.
We now assume thatS= ∅and look for the minimum norm solution of (1.1) in the case where
C= N i=1
Ci, (3.7)
whereN≥1 is an integer and{C1,...,CN}is a finite collection of closed convex subsets of a Hilbert spaceH. We assume that for each i,Ciis the fixed-point set of a nonexpansive mapping Ti (i.e.,Ci= {x∈H:Tix=x}). Recall that a mappingT:H→H is nonexpansive ifTx−T y ≤ x−yfor all x, y∈H.
It is well known that the (nearest point) projectionPDfrom H onto a closed convex subsetDofHis nonexpansive. Note that eachCiis the fixed-point set of the projectionPCi.
We now introduce the iterative algorithm. Let a sequence{αn}of real num- bers in the interval [0,1] be given. Forε >0, we put
B≡Bε:=A∗A+εI, u:=A∗b. (3.8)
Starting with an arbitrary initial guessx0∈H, we define the sequence{xn} recursively by the following algorithm:
x1=
I−α1BT1x0+α1u, x2=
I−α2BT2x1+α2u, ...
xN=
I−αNBTNxN−1+αNu, xN+1=
I−αN+1BT1xN+αN+1u, ...
(3.9)
We can rewrite the algorithm in a more compact form xn+1=
I−αn+1BTn+1xn+αn+1u, n≥0, (3.10) whereTn=TnmodNand the mod function takes values in{1,2,...,N}.
Remark 3.5. The algorithm is a modification of an iteration scheme proposed by Halpern [8] and developed by Lions [9], Wittmann [13], Reich [10, 11], Bauschke [1], Xu [14,15], and others. The algorithm and its variants apply to the convex feasibility problem (CFP). (See [2,4] for some of the recent develop- ments regarding the CFP.)
We now prove the main result of the paper. Some related work can also be found in a paper by Deutsch and Yamada [5].
Theorem3.6. Assume that (1)αn→0;
(2)nαn= ∞;
(3)αn/αn+N→1orn|αn−αn+N|<∞; (4)C=N
i=1F(Ti)=F(T1T2···TN)=F(TNT1···TN−1)= ··· =F(T2T3
···TNT1).
Then, the sequence(xn)generated by algorithm (3.10) converges in norm to the unique solutionxεof (1.2).
Proof. We will divide the proof into several steps.
(1) First, we show that the sequence{xn}is bounded. Indeed, sinceBis self- adjoint, we have that for any 0< α <(A2+ε)−1,I−αBis positive. Hence,
I−αB = sup
x=1
I−αBx,x ≤1−αε. (3.11)
(Without loss of generality, we may assume throughout the proof that αn<
(A2+ε)−1for alln.) It follows from (3.11) that forp∈C, xn+1−p=I−αn+1BTn+1xn−p+αn+1(u−Bp)
≤
1−εαn+1xn−p+αn+1u−Bp. (3.12) Hence, by induction, we obtain
xn−p≤maxx0−p,u−Bp ε
, n≥0. (3.13)
(2) Next, we note thatxn+1−Tn+1xn →0. This follows from assumption (1), step (1), and the fact thatxn+1−Tn+1xn =αn+1u−BTn+1xn.
(3) Now, we prove thatxn+N−xn →0. Indeed, we have (note thatTn+N= Tn)
xn+N−xn=I−αn+NBTn+Nxn+N−1−Tnxn−1
+αn+N−αn
u−BTnxn−1
≤
1−εαn+Nxn+N−1−xn−1+Mαn+N−αn
=
1−εαn+Nxn+N−1−xn−1+εαn+Nβn,
(3.14)
where M is a constant such that u−BTnxn−1 ≤M for all n and βn:= Mε−1α−n+1N|αn+N−αn|. ApplyingLemma 3.1together with assumptions (2) and (3), we obtain thatxn+N−xn →0.
(4) Next, we observe thatxn−Tn+N···Tn+1xn s
−
→0.
Indeed, noting that eachTnis nonexpansive and using step (2), we observe that
xn+N−Tn+Nxn+N−1 s
−→0, Tn+Nxn+N−1−Tn+NTn+N−1xn+N−2 s
−→0, ... Tn+N···Tn+2xn+1−Tn+N···Tn+1xn−→s 0.
(3.15)
Adding up and using step (3), we see thatxn−Tn+N···Tn+1xn s
−→0.
(5) Now, we show that lim supn→∞u−Bxε,xn−xε ≤0, where xε is the unique solution of (1.2).
Take a subsequence{xnj}of{xn}such that lim sup
n→∞
u−Bxε,xn−xε =lim
j→∞
u−Bxε,xnj−xε . (3.16)
Since{xn}is bounded, we may also assume that there exists some ˜x∈H such thatxnj
−→w x. Since the pool of mappings is finite, we may further assume (pass-˜ ing to a further subsequence if necessary) that, for somei∈ {1,2,...,N},
Tnj≡Ti, ∀j≥1. (3.17)
It follows from step (4) that
xnj−Ti+N···Ti+1xnj−→s 0. (3.18) Lemma 3.3then ensures that the weak limit ˜xof{xnj}is a fixed point of the map- pingTi+N···Ti+1. This together with assumption (4) implies that ˜x∈F(Ti+N
···Ti+1)=C. Therefore, we have, by (3.16) andLemma 3.4, that lim supn→∞u− Bxε,xn−xε = u−Bxε,x˜−xε ≤0.
(6) Finally, we claim thatxn s
−
→xε. Indeed, we can write xn+1−xε=
I−αn+1BTn+1xn−xε +αn+1
u−Bxε
. (3.19)
ApplyLemma 3.2to get
xn+1−xε2≤I−αn+1BTn+1xn−xε2+ 2αn+1
u−Bxε,xn+1−xε
≤
1−εαn+1xn−xε2+ 2αn+1
u−Bxε,xn+1−xε . (3.20) UsingLemma 3.1and step (5), we now conclude thatxn−xε2→0.
Remark 3.7. Assumption (4) inTheorem 3.6is automatically satisfied if eachTi
isfirmly nonexpansive(i.e.,Tisatisfies the property:x−y,Tix−Tiy ≥ Tix− Tiy2, forx, y∈H). Since a projection is firmly nonexpansive, we have the fol- lowing consequence ofTheorem 3.6.
Corollary3.8. Assume that (1)αn→0;
(2)nαn= ∞;
(3)αn/αn+N→1orn|αn−αn+N|<∞.
Letx0∈Hbe chosen arbitrarily and let{xn}be generated by the iterative algorithm xn+1=
I−αn+1BPn+1xn+αn+1u, n≥0, (3.21) wherePk=PCk(1≤k≤N). Then,{xn}converges in norm to the unique solution xεof (1.2).
Remark 3.9. We have used two steps to find a solution of problem (1.1); that is, we first find the unique solutionxεof the regularized problem (1.2) and then ap- proximatexεvia the iterative algorithm (3.10). It is not clear if one can combine the two steps into one. In other words, if one defines a sequence{xn}by
xn+1=
I−αn+1Bεn
Tn+1xn+αn+1u, n≥0, (3.22) where{εn} ⊂(0,1) andεn→0 as n→ ∞, does the sequence {xn} defined by (3.22) converge to the minimum norm solution of problem (1.1) (if the solution setSof (1.1) is nonempty)?
Acknowledgment
The first author was partially supported by the Israel Science Foundation founded by the Israel Academy of Sciences and Humanities (Grant 592/00), by the Fund for the Promotion of Research at the Technion, and by the Technion VPR Fund.
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Simeon Reich: Department of Mathematics, The Technion - Israel Institute of Technol- ogy, 32000 Haifa, Israel
E-mail address:[email protected]
Hong-Kun Xu: Department of Mathematics, University of Durban-Westville, Private Bag X54001, Durban 4000, South Africa
E-mail address:[email protected]
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