NonlinearAnaly:ds, Theory, Methods& Applications,Vol. 1 No. 3, pp. 235 ~47. PergamonPress, 1977. Printedin Great Britain.
CONTRACTIONS
AND
COMPLETELY
CONTINUOUS
MAPPINGS
A . MUKHERJEA* University of South Florida, Tampa, Florida 33620, U.S.A. (Received 30 July 1976) K e y words : Contractions, completely continuous mappings, Banach fixed point theorem, random operator, stochastic equation.
CONTRAC~ONS and completely continuous mappings are among the most widely studied mappings on metric spaces. In recent years, there have been numerous extensions of the Banach fixed point theorem for contractions and the Schauder fixed point theorem for completely continuous mappings and detailed studies of stochastic analogues of these theorems in the context of various random integral equations or certain other types of random operator equations (see [1], [2]). This article aims at surveying some of the most interesting extensions of the Banach fixed point theorem which have been obtained by various authors during the last decade, understanding some of the interesting variants of the contraction condition by means of several examples and finally, presenting a stochastic analogue of a fixed point theorem involving the sum of a nonlinear contraction and a completely continuous operator. While this article is partly a survey, most of the examples and some of the theorems (with proofs) are new ; also this article contains an up-to-date summary of recent results in the present context. 1
In what follows, let T be a mapping from a metric space (X, t0 into a metric space (Y, d). Note that for simplicity, we are using the same notation d for the metrics in X and E which, of course, may very well be different. Then T is said to be a contraction if there exists a real number k < 1 such that for all x, y e X, we have: d(Tx, Ty) ~< kd(x, y).
(1)
Then the following proposition is almost evident. PROPOSITION 1.1. Let X = Y and d be complete. Suppose T is continuous and there exists x o s X and a real number k < 1 satisfying the inequality: d(T" + i x0 ' T" + 2Xo) <~ k . d ( T " x o, T" + i Xo )
(2)
for all non-negative integers n. Then the sequence T " x o converges to a fixed point u of T as n ~ oo. An immediate consequence is: THE BANACH FIXED POINT THEOREM 1.2. Let T be a contraction from a complete metric spade X *The author is supported by a National Science Foundation grant MPS 74-06552-A01. 235
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A. MUKHERJEA
into itself. Then there exists a unique x o in X satisfying (i) Tx o = x o (ii) for any x in X, lim T " x = x o (iii) there exists an open set Vcontaining x o such that T " V --, x o as n ---, ~ , i.e. for any open set W containing x o, there exists a positive integer n(W) such that n > n(W) implies T " V ~ W. Recently, there have been numerous attempts to weaken the hypothesis of the Banach fixed point theorem, but at the same time retaining the convergence property of the successive iterates to the unique fixed point of the mapping. We present here some of these extensions. The following is a slightly simplified version of a theorem of Boyd and Wong. THEOREM 1.3. Let T be a mapping from a complete metric space X into itself. Suppose there exists a right continuous mapping f from R +, the non-negative reals into itself such that for all x, y i n X d(Tx, Ty) ~ f ( d ( x , y))
(3)
If f (t) < t for each t > 0, then Thas a unique fixed point u in X and for each x in X, lim T " x = u. Proof. Let x ~ X, x = T " x and d. = d(x., x + 1)" Then for n > 1, d = d ( T x , _ 1, T x ) <<.f ( d ~ _ 1) < d,,-1
so that the sequence d is decreasing. Let d = lim d,. Then d = 0, since the preceding inequality n~oo
implies that d <~ f ( d ) , which is less than d unless d = 0. Now we show that the sequence x is Cauchy. If not, then there exists ~ > 0 and for each positive integer k there exist positive integers nk and mk with k <~ m k < nk such that d(xm~, x , ) >i ~. With no loss of generality, we can also assume that nk is the smallest integer greater than mk satisfying the above inequality. Let tk = d(x,,~, x.~). Then e <~ t k <~ d(xm, x , , _ l) + d(x,,_ 1, x,~.) <<. e + d,,k- 1
and therefore, e ~< t k and lim t k = e. We have also e <. t k <. d ( x , . , x , . . + l ) + d(xmk+l,X,,~+l) + d ( x , , + l , x , , ) <~ d,.~ + f(tk) + d
~ f ( e ) a s k ~ oo.
This is a contradiction since f(e) < e. Therefore, the sequence x is Cauchy with some limit u in X. Clearly by the continuity of T, T(u) = u. The uniqueness of the fixed point follows immediately from the definition o f f Without right continuity of f above, Theorem 1.3 is, of course, false. For instance, let X = (x),x
= ~. (1 + I/j); then if Tx = x,+ 1, Tsatisfies (3) with f(1 + 1 / n ) = 1 + 1/(n + 1). j=l
Note that T has no fixed point and the function on f , if right continuous, has to satisfy f(1) = 1. The Boyd and Wong non-linear condition (3) has been replaced by a contraction type condition (but much weaker than the contraction condition) in another extension of Theorem 1.2 by V. M. Sehgal [3], later improved upon by Guseman [4]. The motivation of the Sehgal-Guseman result comes from the fact there there are mappings T which have at each point an iterate which is a
Contractions and completely continuous mappings
237
contraction and yet, none of the iterates of T is a contraction. F o r an example of such mappings, see [3]. The Sehgal-Guseman result can be stated as follows: THEOREM 1.4. Let T b e a mapping from a complete metric space X into itself. Suppose there exists B c X such that (i) T(B) ~ B; (ii) there is a real number k < 1 such that for each x e B, there is a positive integer n(x) such that all for all y e B, d(T"~)x, T"~)y) <~ k. d(x, y) and (4) (iii) for some z e B,
{T"z:n <~ 1} c B. Then there is a unique u E B such that Tu = u and for each x e B, lim T"x = u. n~oo
A different sort of extension of Theorem 1.2 is the following theorem of A. Meir and E. Keeler [5], which contains Theorem 1.3 properly. THEOREM 1.5. Let T be a m a p from a complete metric space X into itself satisfying the condition: Given e > 0, there exists ~ > 0 such that e ~< d(x, y) < e + ~ implies d(Tx, Ty) < e.
(5)
Then for any x e X, lim T"x exists and this limit is the unique fixed point of T. n~oo
Briefly, the proof of 1.5 runs as follows: Here T is contractive and c n = d(T"x, T"+lx) is a decreasing sequence with limit 0. If the sequence T"x is not Cauchy, then for infinitely m a n y n and m, d(T"x, T"x) is greater than 2 c (>0). F o r this e, let t3 be as in the hypothesis above. Let N be such that for n > N, c~ < 6'/3, where 6' = min {6, e}. Let n(> N) and k be such that d(T"x, T"+kx) > 2e. Since for all j with 1 <~j <~ k, ]d(Tnx, Tn+Jx) - d(T"x, T"+J+lx)l <~ c,,+j < 6'/3, it follows that for some j, 1 <~ j ~< k, we must have: e + 2t5'/3 < d(T'x, T"+Jx) < e + fi'. But for this j, d(T"x, T"÷Jx) <~ d(T"x, T"+lx) + d(T"+lx, T"+~+lx) + d(Tn+~x, T"+J+lx) < 6'/3 + e + 6'/3, a contradiction. Another interesting extension of 1.2 is a theorem of M. Edelstein for contractive mappings T (i.e. mappings satisfying d(Tx, Ty) < d(x, y) for all x, y ~ X and x ~ y), later improved upon by V. M. Sehgal [6] as follows: THEOREM 1.6. Let T be a continuous mapping from a complete metric space X into itself such that for all x, y e X with x :~ y,
d(Tx, Ty) < max {d(x, Tx), d(y, Ty), d(x, y)}.
(6)
Suppose for some x e X, the sequence (T"x) has a cluster point x 0. Then lim T"x = x 0 and x o is n~ct3
the unique fixed point of T. It is clear that a contractive T satisfies (6). But the converse need not be true. F o r instance, the mapping T: [0, 5] ~ [0, 5] defined by
Tx = x/2 for x in [0, 4]; --- - 2x + 10 for x in [4, 5] satisfies (6), though not contractive since T4 - T5 = 2.
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Our next fixed point theorem is another interesting extension of the Banach fixed point theorem. It involves mappings which are contractions in a different sense. To clarify this, let us consider the following variants of the contraction condition. In what follows, Tis a mapping from a complete metric space X into itself.
Condition (K). There is a constant c < ½ such that for all x, y in X d(Tx, Ty) ~< c. [d(x, Tx) + d(y, Ty)]. Condition (K 0. There is a constant c < ½ such that for all x, y in X d(Tx, Ty) <~ c. [d(x, ry) + d(y, Tx)]. Condition (K2). There exist non-negative constants a, b and c such that 2a + 2b + c < 1 and for all x, y in X d(Tx, Ty) <%a. [d(x, rx) + d(y, Ty)] + b. [d(x, Ty) + d(y, Tx)] + cd(x, y). Condition (K) was first studied by R. Kannan in 1968. Since then, the other two conditions have been studied independently by various mathematicians. It is clear that the first two conditions imply the third. But none of these conditions, though seemingly contractions of some kind, are contractions. They need not even imply continuity. The following examples clarify this and their relationships to one another. 1.7. Examples of mappings T satisfying the conditions (K), (K 0 or (K2)
Example 1. An example of a discontinuous T satisfying condition (K), but not condition (K1) is the following: Define Tx = - x / 2 for x in ( - 1, 1) and T1 = T ( - 1) = 0. Then T: [ - 1, 1] ~ [ - 1, 1] and is discontinuous. T does not satisfy condition (KI) since for x in (0, 1), ITx - T ( - x ) I = x = Ix -
T(-x) I +
I-x
- Tx I.
That T satisfies condition (K) with c in (1/3, 1/2) is clear from the following three cases:
Casel.-1
1.
Here ITx - Ty I = [(x - y)/21
<= c[3lxl/2 + 31y[/2]
= c[[x - Tx[ + [y - Ty[]. Case 2. x = l o r x = - l and y in ( - 1 , 1 ) . This case is trivial to verify, as is the next case. Case 3. x = l a n d y =
-1.
Example 2. An example of a discontinuous 7" satisfying condition (K1), but not condition (K) is the following: Define Tx = x/2 if x ~ [0, 1) and T1 = 0. Then T: [0, 1] --, [0, 1].
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Contractions and completely continuous mappings
This Tdoes not satisfy condition (K) since for x in (0, 1),
TO[ = x/2 = I x -
ITx-
Txl + I O -
T01.
But T satisfies condition (KI) with c in (1/3, 1/2) as shown below. Suppose 0 ~< y ~< x ~< 1. Case 1. y < x / 2 <-_ x < 1. Then we have c[[x -
T y [ + lY -
T x l ] = c [ x - y/2 + x / 2 - y] = 3c(x - y)/2 > [ T x -
T y I.
Case 2. x / 2 ~ y __
c[I x - Ty I + [Y - T x I1 = c [ x - y/2 + y - x/21 = (c/2)[x + y] > 3 c x / 4 > (x - y)/2 = I T x -
Tyl.
Case 3. x = 1 and y in [0, 1). In this case, we have c [ I x - T y [ + ]y -
T x ] ] = c[1 - y/2 + Yl > 3cy/2
->lTy -
Tx[.
E x a m p l e 3. The mapping T x = x / 2 from [ - 1, 11 into [ - 1, 11 is the example of a contraction which does not satisfy condition (K1). The proof of this follows as in that of Example 1. E x a m p l e 4. The mapping T x = x / 2 from [0, 11 into [0, 11 is the example of a contraction which does not satisfy condition (K). The proof of this is trivial. E x a m p l e 5. An example of a continuous T which satisfies condition (K), but satisfies neither condition (K1) nor the contraction condition is the following. Consider the mapping T: [ - 5, 51 ~ [ - 5, 5] defined by T x = - x/2, i f x e [ - 4 , 4 1 ;
= 2x - 10, if x e [4, 51; = 2x + 10, i f x e [ - 5 ,
-41 .
One can verify that T satisfies condition (K) with c in (2/5, 1/2). The details being messy are omitted. Since T5 - T4 = 2, T is not a contraction. Also T does not satisfy condition (K~) since if x = - 4 a n d y = 4 , then
[Tx
-
Ty I =
4 = Ix -
Ty I +
lY -
Tx[.
E x a m p l e 6. The following is an example of a discontinuous T which satisfies neither condition (K) nor condition (K1), but does satisfy condition (K2). Consider the mapping T: [ - 2 , 11 ~ [ - 2 , 11 defined by T x = x / 2 , for x in [0, 1);
-
1/8, for x = 1;
= - x / 2 , for x in [ - 2, 0).
We now show that this T meets the demands in the example. T does not satisfy conditions (K~)
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A. MUKHERJEA
or (K) since
I T ( - 2 ) - T(1)[ = I + 1/8 > (1/2)(15/8) = 1/2[[-2 - Til)[ + I1 - T ( - 2 ) [ ] and for x in (0, 1),
]Tx-
V0[ = I x -
Vx[ + [ 0 -
V0[.
T h a t T satisfies condition (K2) a b o v e with a = c = 1/5 and b = 12/65 follows from the following five cases.
Casel. O< y<
x < 1.
Here (1/5)[Ix - Tx[ + I)' - Ty[ + Ix - y[] + (12/65)[[x - Ty[ + lY - Tx[] = il/5)[(x + y)/2 + ix - 3') + (12/13)ix - y/2) + i12/13)[y - x/21]
= ix - y)/2 + (1/5)[(20/13)y - i l / 1 3 ) x + (12/13)[y - x/2[] > [ T x - Ty[, since the expression within the parenthesis is easily verified to be non-negative. Case 2. - 2 =< x , y =< O. Here (1/5)[Ix - Tx[ + lY - Ty[ + Ix - Yl]
= (1/5)[3/2)(1x [ + [y[) + Ix - Yl] _-> Ix - y1/2 = [ T x - Ty[. Therefore, in this case T satisfies condition (K2).
Case 3 . 0 < y < 1, x = 1. In this case, after s o m e c o m p u t a t i o n s we have:
(1/5)[Ix - Tx[ + [y - Zy[ + Ix - Y[ + (12/13)(1x - Ty[ + [y - Tx[)]
=
329 52--6
Y > ~v + 1 = 13o =
Tx-
Ty I.
Case 4. - 2 =< y =< O, x = 1. H e r e we have: (1/5)[Ix - T x I + [y - Zy[ + Ix - y[] = [y[/2 + ~1 7 >= - y2 +
1 =[Tx-
Therefore, T satisfies condition iK2) in this case.
Case 5. - 2 <~ y <.N O <~ x < 1. Here [ T x - Ty[ = Ix + y 1/2. Suppose t = (1/15)[Ix - Tx[ + [y - Yy[ + Ix - Yl + (12/13)([x Then if - y / > x, it is clear that t >>. i l / 5 ) [ , 3 y / 2
+ (x-
y)] >>. Ix +
y[/2
Ty[ + [y - Tx[)].
T y I.
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241
If --y < X, then we have:
t =(1/5) [ -x
2
"~Y+x-y+(12/13)
>i (1/5)[5x/2 +
( x +y~ +x~ - y) 1
513,1/2] >>-Ix + yl/2.
This completes the demonstration of the Example. Now, we state and prove a fixed point theorem for mappings satisfying the condition (K2). This theorem has been obtained independently by many authors including A. W. Goodman and this author and can be extended (see [7] and [8]) easily by replacing the condition (K2) by the condition : There is a real number c < 1 such that for all x, y e X,
d(Tx, Ty) <~c. max {d(x, Tx), d(y, Ty), d(x, Ty), d(y, Tx), d(x, y)}.
(7)
THEOREM 1.8. Let T be a mapping from a complete metric space X into itself satisfying condition (K2). Then for any x in X, lim T"x exists as n tends to infinity and this limit is the unique fixed point of T.
Proof. For all x, y in X, T satisfies the inequality: d(Tx, Ty) <=aid(x, Tx) + d(y, Ty)] + b[d(x, Ty) + d(y, Tx)] + cd(x, y), where 2a + 2b + c < 1.
(8)
For each positive integer n and x in X, we replace x by T ~- ix and y by T"x in (8). Then using the triangular inequality, we have:
d(T~x, T"+ix) _<- a[d(T"-lx, T~x) + d(T"x, T~+lx)] + b[d(r"-lx, r"+lx) + d(rnx, Tnx)] + cd(r"-lx, r'x) <=a[d(r"-ix, r"x) + d(r~x, T n+ix)] + bid(T"-ix, r"x) + d(r~x, T "+ ix)] + cd(r"-ix, rnx). Simplifying, we have:
d(T"x, T "+ ix) _< kd(T ~- ix, T"x) a+b+c where k - 1 - a - b' which is less than 1. Repeating the above process n times, we have'
d(r"x, r n+ix) ~ k"d(x, rx) and therefore, for each positive integer m we can write" m--1
d(T"x, Tn+mx) <= ~ d(T"~ix, T"~i+lx) i=O m-1
< ~. k"+~. d(x. rx). i=0
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A. MUKr~RJEA
Thus the sequence T~x is a Cauchy sequence and since X is complete, the sequence has a limit ~. To prove that T~ = ~, we use (8) and find that
d(~, T~ )<~d(~, T~x) + d(T~x, T~) <~ d(~, T~x) + a[a(T ~- ix, T~x) + d(~, T~)] + b[d(T ~- Ix, T~) + d(~, T~x) + ccl(T~- ix, ~). Noting that d(T ~- ix, T¢) <~ d(T ~- ix, ~) + d(¢, T~), we see from above that (1 - a - b) d(¢, T~) can be made as small as we please. Hence ~ = T~. To prove the uniqueness of the fixed point, let T~ = ~ and T r / = :7. Then from (8), writing x = ~ and y = r/, we have
d(~, tl) = d(r~, T~l) < (2b + c) d(~, t/). Since 2b + c is less than 1, ~ = t/. With reference to the condition (K), we remark here that a mapping T: [a, b] -. [a, b] satisfying the condition I T x - Ty I <~ ½[Ix - Tx I + lY - Ty I] for all x, y e [a, b] (8') has a unique fixed point ifa and b are in T([a, hi). The fixed point is the mid-point ~ a + b). This result, of course, is no longer true if condition (8') is replaced by:
I T x - Ty[ ~ ½l'[x - Ty[ + [y - T x 1] for all x, y • [a, b].
(9)
It is also an interesting fact that a (K2)-type condition can force a sequence of mappings T from a complete metric space into itself having a common fixed point. This is a result ofB. N. Roy extended by R. Iseki [9]. The condition is: for each pair T~, Tj, there are non-negative constants a, b and c (depending upon i and j) such that for all x, y in X
d(Tix, Ty) <~ a[d(x, T~x) + d(y, Ty)] + bl,d(x, Tff) + d(y, T~x)] + cd(x, y)
(10)
where2a+2b+c< 1. The way to get a c o m m o n fixed point is to choose any x o • X and then set x n = T x _ ~, n = 1, 2 . . . . ; the sequence (xn) then turns out to be Cauchy and its limit is the unique common fixed point of the T~'s. This result can be extended to an arbitrary family of mappings. Our next theorem assumes a condition on T which is weaker than condition (K2). TBEORE~ 1.9. Let T be a mapping of a complete metric space X into itself such that the following conditions hold: (i) the map x ~ d(x, Tx) is lower semicontinuous; (ii) there exists a sequence (x,) in X such that d(x,, T x ) goes to 0 as n tends to infinity; (iii) there exist nonnegative constants A, B, C, D, and E such that C + D + E < 1 and for all x, y i n X d(Tx, Ty) < Ad(x, Tx) + Bd(y, Ty) + Cd(x, Ty) + Dd(y, Tx) + Ed(x, y). (11) Then T has a unique fixed point which is the limit of the sequence x . Moreover, this result need not hold if any one of the conditions (i), (ii) or (iii) is dropped.
Proof. First, we show that the sequence (x,) is Cauchy. Using (10), we have:
Contractions and completelycontinuous mappings
243
d(x n, Txn÷p) < Ad(x,, Tx~) + Bd(x ÷p, T x +p) + Cd(x,, Tx,+p) + Dd(x~+p, T x ) + Ed(x n, Xn+p) < (A + C + E)d(x,, Tx,) + (B + D + E)d(x,+~, Tx,+p) + (C + D + E)d(Tx,, Tx,,+p). Since C + D + E < 1, it is clear that the sequence (Tx~) is Cauchy by condition (ii). This means that the sequence ( x ) is Cauchy and therefore, has a limit ~. By assumptions (i) and (ii), it follows that T~ = ~. The uniqueness of the fixed point follows easily. Now to see that none of the conditions (i), (ii), or (iii) can be dropped, it suffices to examine the following examples. Let X be the set of real numbers with the absolute value metric. 1. Consider T x = [xl/2, for x non-zero and = 1, for x = 0. Then T satisfies conditions (ii) and (iii), but not (i) and T has no fixed point. 2. Consider Tx = - 1, for x nonnegative and = 1, for x negative. Then T has conditions (i) and (iii), but not (ii) and T has no fixed point. 3. Consider T x = x for all x. Then T has conditions (i) and (ii), but not (iii) and T has uncountably many fixed points. This completes the proof of the theorem. The concept of contraction has also been extended in yet another direction, by that of diminishing orbital diameter, introduced by W. A. Kirk [10]. Let T be a mapping from a complete metric space X into itself. Let O(x) denote the orbit o f x e X, that is the set {T"x: n >1 O, T°x = x} and let p(x) = {sup d(z, w) : z, w ~ O(x)} = the diameter of the orbit ofx. Then the mapping T is said to be of diminishing orbital diameter ifp(x) < ~ for all x ~ X and p(x) > 0 implies lim p(T"x) < p(x). I1~00
For example, if X = {(x, y): x t> 0, y I> 0} with the usual 'distance' metric and if T: X ~ X be defined by T(x, y) = (x, xZ), then T has diminishing orbital diameter and the function p(x) is a continuous function on X. The following is a result ofW. A. Kirk as extended by Sehgal [6]. THEOREM 1.10. Let T: X ~ X, X a complete metric space. Suppose that T is continuous and has diminishing orbital diameter. Ifp(x) is continuous on X, then each limit point ~ in X of the sequence (T~x) for x e X, is a fixed point of T and ~ = lim T~x, n-~oo
Finally, in this section we will consider a converse of the Banach fixed point theorem: Does every continuous mapping T from a complete metric space into itself for which the conclusions of Theorem 1.2 hold have to be a contraction with respect to an equivalent metric, that is a metric generating the same topology as the original metric? Usefulness of and interest in such converses are mainly because of the fact that contractions are almost ideal for the study of some iterative numerical process. L. Janos [11] was perhaps the first to prove a converse of Theorem 1.2 for compact metric spaces. F o r general complete metric spaces, the result was proven by P. R. Meyers [12]. He proved : THEOREM 1.1 1. Let T be a continuous mapping from a complete metric space X into itself such that the conclusions (i), (ii) and (iii) of Theorem 1.2 hold.Then for each real number k in (0, 1), there is an equivalent metric d k such that for all x, y e X,
dk(TX, Ty) <~ k dk(X, y). The following corollary follows easily.
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A. MUKHERJEA
COROLLARY1.12. Let T be a continuous mapping from a complete metric space X into itself with a unique fixed point ~, such that for each x ~ X, lira T"x = 4. Then if~ has an open neighbourhood n~oo
with compact closure, the conclusion of Theorem 1.11 holds. To prove this corollary, let U((), V(O be an open neighborhood of ~ and K = V--~ be compact. For each x e K, lim T"x = ~ and therefore, there is a smallest non-negative integer n(x) such n-.-~ oo
that n >>,n(x) ~ T"x e U((). It is sufficient to show that n(U) = {sup n(x): x e K} is finite. Ifnot, then there exist x i ~ K such that n(xi) > i. By compactness of K, there is a subsequence x~/-~ y e K. Since T"tY~xij ~ T"tr)y a s j -~ 0o, it is clear that n(xi/) <~ n(y) for sufficiently largej, contradicting that n(x~) > i V i. The following example will now demonstrate that Corollary 1.12 may be false without the assumption that ~ has an open neighborhood with compact closure.
Example 1.13. Consider on the plane infin~ely many lines L~, L 2, L 3. . . . all passing through a single point ~ (as in the picture). Let X = w L~. We define a metric on X as : i
I
d(P1, P2) = the distance P1P2, if P~, P2 are on the same line and d(P 1, Q) = the distance Q~ + the distance P ~ , if P~, Q are on different lines. Define. T x = n/(n + 1)x, ifx ~ L,. Here by x, we mean the point Q ~ L, whose distance from ~ is x. Then it is clear that ~ is the unique fixed point of T and for each x ~ X, lim T"x = ~. It is also clear that T cannot be a contraction with respect to an equivalent metric. 2. In this section, X will always denote a Banach space. Then a mapping T: X --* X is called completely continuous if T is continuous and maps bounded sets into compact sets. The well known fixed point theorem of Schauder [13] states that every continuous mapping T of a compact convex subset E (of X) into itself has a fixed point. Tychonoff [14] later extended this result to the case when X is replaced by a general locally convex topological vector space. To deal with the analytical question of measurability of the solution of a stochastic equation,it is often necessary and important to derive stochastic analogues of the fixed point theorems available in the deterministic (or non-stochastic) literature. For a useful account of these and other relevant material, see [ 1] and [2]. To be more specific about the stochastic analogues, let us define a random operator T: f~ x X --* X, where (fL Z, P) is some probability measure space and X a separable (separability is assumed for technical simplicity) Banach space, to be a mapping such that for each x ~ X, T(tn}[x] = T(~o, x) is a random variable from t) into X, i.e. {o~: T(co)[x] ~ B} e Z for every Borel set B = X. Then the following stochastic analogue of Theorem 1.2 is immediate.
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THEOREM2.1. Suppose T is a random operator defined as above. Suppose there is a non-negative real-valued function k(o9): f~ ~ R such that 0 < k(o)) < 1 for each ~oe f~, and II T(to)Ux] -- T(~,)[y] II ~< k(o~)[Ix -- y l]
(12)
for all, x, y e X. Then there exists a unique random variable ¢(~o)such that T(~o)[~(~o)] = ~(t~;~. The proof of this theorem is immediate since for any x • X, lim T"(ug)[x] exists and is the unique fixed point of the operator T(og). Many consequences of the above theorem in the context of random integral equations have been studied by Tsokos and his co-workers (see [2]). A stochastic analogue of Schauder's Theorem is less simple since the fixed point in this theorem need not be unique, and consequently, finding a measurable fixed point for the random operator becomes a problem. However, one can get around this difficulty as is shown by the following theorem due to Bharucha-Reid and the author. THEOREM 2.2. Let T be a random operator such that for each 09 • t), T(to): E ~ E is continuous, where E is a compact convex subset of X. Then there exists a random variable ¢(~o): f~ --. E such that T(~o)[~(t~)] = ~(e~) for almost all ~.
Proof.
Let A(to) = {x e E I T(to) x = x}. Then by Schauder's Theorem, for each to the set A(to) is non-empty. Furthermore, for any closed subset F of E
{tolA(to) c~ F is non-empty} = {~olA(to) x = x for some x in F} oo
oo
n=l
i=1
=
{tol II T(to)
-
l[ <
1/,}
where the xi's form a dense sequence in F. It is therefore clear that the set {wlA(to) c~ F is nonempty} is measurable for every closed subset F of E. To prove the theorem, it is sufficient to find an E-valued random variable ~(to) such that ~(to) e A(to). It is known [151 that we can associate with the space E a sequence of triples (C., p., qb) (n = a non-negative integer) such that (i) (ii) (iii) (iv)
C.'s are all countable and p. maps C + 1 onto C ; • maps C, into a class of non-empty closed subsets of E of diameter < 2-"; E = {~o(C)[C e Co} ; for each n and for each c in C , ~n(C) = k.) {f~)n + 1 (Ct) l C' • p~ 1@)}.
We now assume with no loss of generality that C o and each p~- i (c) with c in C. are naturally linearly ordered such that only finitely many elements can precede any element in this order. We now use an idea of Castaing [16]. For each n, we intend to find a suitable partition of f~. We proceed inductively as follows: For each c in Co, we define f2c by ¢oe f~c if and only if A(to) r~ ¢I)o(C) is non-empty and A(to) n q)o(C') is empty for c' e C o, c' < c. Then the f~c's are pairwise disjoint measurable sets with union f~. Suppose now that we have found a partition off~ corresponding to the elements of C k. To do this for Ck+ 1, we define for c in Ck+ 1 the set f~ by to e f~ if and only if to e f~p~¢~and A(to) c~ ~k+ 1(c) is non-empty, hut A(to) ~ ~,+ 1@') is empty for c' in p~-l(Pk(C)) and c' < c. Now we define for each positive n: ¢,(to) e ¢},(c) if and only if to e f~cwhere the f~'s are members of
246
A. MUKHERJEA
the partitition of f~ corresponding to the elements of C . Then each ~.(o9) is measurable and I(¢,(~o) - ¢. + 1(o~)II -<- 2-",
d(¢.(o,), A(o,)) -5_ 2-".
Therefore if ¢(o~) = lira ~,(o), then ~(co)e A(~o)and the theorem follows. [Note that in this theorem E could be assumed to be a compact subset of a metric space instead ofa Banach space if each A(~o) was guaranteed to be non-empty by some other conditions.] It is now easy to obtain a more general theorem by using an extension of a theorem of Krasnoselskii [17] by M. Z. Nashed and J. S. Wong [18]. The Nashed and Wong Theorem reads as follows: THEOREM2.3. Let X be a Banach space and S be a closed, bounded and convex subset of X. Suppose A, B are mappings from S into X such that A x + B y e S for all x, y e S. Then if B is completely continuous and A is a non-linear contraction, i.e. for all x, y • X and some real-valued continuous function f satisfying f ( t ) < t for t > 0, []ax-
Ay]l <~
f(llx
- yll),
(*)
the equation A x + B x = x has a solution in S. The proof of this theorem is not difficult. Because of (,), it easily follows that (I - A)- 1 exists and is continuous; also since for every y • S, there exists x (depending on y) such that A x + B y = x, the mapping (I - A)- 1B is well-defined and completely continuous as a mapping from S into A. Consequently, Schauder's fixed-point Theorem applies on (I - A)- 1B and therefore, there exists x such that x = (I - A ) - I B x or A x + B x = x. Finally, as a consequence of Theorems 2.2 and 2.3, we have easily the following stochastic analogue of Theorem 2.3. THEOREM 2.4. Let E be a compact, convex subset of a Banach space X, and A(og) and B(~o) be continuous random operators on X mapping E into itself. Suppose that (i) for all x, y e E, ~0 e f L A(o~)[x] + B(og)[y] e E and (ii) for all x, y e E and co e f~, IIA(og)[x] - A(co)[y] II ~< f(l! x - y II) where f is a non-negative real valued continuous function such that f ( t ) < t for t > 0. Then there exists a random variable ¢(o9): f~ ~ E such that A(0~)[¢(o9)] + B(ca)[¢(o~)] = ¢(co) for almost all ~oefL This theorem follows immediately if one observes that for each co e f L [1 - A(co)] - 1 B(~o) is a well-defined completely continuous mapping from E into E; moreover, [1 - Aim)I-1 B(en) is a random operator. Therefore, by Theorem 2.2, there exists a random variable ¢(09): f~ -+ E such that for almost all ~o, [ I -- A ( o J ) ] - 1 3(oJ)[~(o))] = ¢(o)) or
A(co)[¢(o9)] + B(og)[~(co)] = ~(o9). Finally, we must mention Nadler's criterion for ITC. We say that ITC holds for a metric space (X, d) if for every contractive mapping T on X having a fixed point u, the implication that lirn T"(x) = u is valid for all x in X. In Can. math. Bull. 15(3), 1972, Nadler established the validity
Contractions and completely continuous mappings
247
of ITC for the class of locally compact connected metric spaces (including all finitedimensional Banach spaces), whereas ITC is shown to be not valid for any infinite dimensional Hilbert space or the Ip spaces (p/> 1) or the space c o. Nadler also showed the necessity of local compactness as well as connectedness for his result. Acknowledgement--The author appreciates assistance from Profs. Bharucha-Reid and Sehgal for many discussions on the subject and Dr. Nakassis for Example 1.13. The author is especially thankful to Prof. A. W. Goodman for initiating some of the discussions in this paper.
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