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On the Structure of the Set of Subsequential Limit Points of a Sequence of Iterates
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
222, 297]304 Ž1998.
AY985944
NOTE On the Structure of the Set of Subsequential Limit Points of a Sequence of Iterates Zeqing Liu Department of Mathematics, Liaoning Normal Uni¨ ersity, Dalian, Liaoning 116029, People’s Republic of China Submitted by William F. Ames Received September 15, 1997
In this paper we obtain four results on the structure of the set of subsequential limit points of a sequence of iterates of a self-map of a metric space. Our results extend substantially the main results of Diaz and Metcalf, Maiti and Babu, and Park. Q 1998 Academic Press Key Words: subsequential limit point; closed set; connected set.
1. INTRODUCTION Diaz and Metcalf w2x initiated a study of the structure of the set of subsequential limit points of a sequence of iterates of a continuous self-map f of a metric space Ž X, d . satisfying the condition dŽ fx, F Ž f .. dŽ x, F Ž f .. for x / fx. Maiti and Babu w3x and Park w4x obtained similar results. In this paper we show that the results of Diaz and Metcalf w2x, Maiti and Babu w3x, and Park w4x may be derived under weaker conditions. In fact, we establish that discontinuous maps may also be accommodated in their works and we arrive at the same conclusions as in w2]4x about the set of subsequential limit points of a sequence of iterates of a self-map f in X satisfying the conditions dŽ f n x, F Ž f .. ª 0 as n ª `, or dŽ f n x, f nq1 x . ª 0 and dŽ f n x, LŽ x .. ª 0 as n ª `, or dŽ f n x, f nq1 x . ª 0 as n ª `. 297 0022-247Xr98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
298
NOTE
Let N be the set of positive integers, f a self-map of a metric space Ž X, d ., and F Ž f . the set of fixed points of f. For x g X and k g N, let O Ž x, f . s f n x : n g 0 4 j N 4 ,
O Ž x, f , k . s f n x : 0 F n F k 4 ,
let LŽ x . denote the set of subsequential limit points of the sequence f n x 4`ns 0 , and let O Ž x, f . denote the closure of O Ž x, f .. For A, B ; X, define d Ž x, A . s inf d Ž x, y . : y g A4 , d Ž A, B . s inf d Ž a, b . : a g A, b g B 4 and
d Ž A . s sup d Ž a, b . : a, b g A4 . Define T s t : t is a continuous map of X = X into 0, ` . that satisfies t Ž x, y . s 0 if and only if x s y 4 .
2. LEMMAS We need the following lemmas which are fundamental to our results. LEMMA 2.1. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. If LŽ x . is nonempty and f is continuous on O Ž x, f . for some x g X, then the following statements are equi¨ alent: Ž2.1. Ž2.2.
dŽ f n x, F Ž f .. ª 0 as n ª `; dŽ f n x, f nq1 x . ª 0 and dŽ f n x, LŽ x .. ª 0 as n ª `.
Proof. Ž2.1. « Ž2.2. We first show that f satisfies the following condition: Ž2.3. For every « ) 0, there exists d Ž « . ) 0 such that dŽ y, fz . - « whenever y g F Ž f ., z g O Ž x, f ., and dŽ y, z . - d Ž « .. If not, then there exist « ) 0 and sequences yn4ng N ; F Ž f . and z n4n g N ; O Ž x, f . such that both dŽ yn , z n . ª 0 as n ª ` and dŽ yn , fz n . G « for n g N. The compactness of F Ž f . implies that there exists a subsequence yn a4ag N of yn4ng N such that it converges to a point y in F Ž f .. Consequently, d Ž y, z n a . F d Ž y, yn a . q d Ž yn a , z n a . ª 0
as a ª `.
299
NOTE
Hence dŽ y, z n a . ª 0 as a ª `. Since f is continuous on O Ž x, f ., d Ž fz n a , y . s d Ž fz n a , fy . ª 0
as a ª `.
It follows that
« - d Ž yn a , fz n a . F d Ž yn a , y . q d Ž y, fz n a . ª 0
as a ª `.
This is a contradiction. We next show that dŽ f n x, f nq1 x . ª 0 as n ª `. Since F Ž f . is compact, there exists a sequence pn4ng N ; F Ž f . such that dŽ f n x, pn . s dŽ f n x, F Ž f .. for n g N. Let « ) 0. In view of Ž2.1. and Ž2.3., there exists m g N with dŽ f n x, pn . - min « , d Ž « .4 for n G m. Thus d Ž f n x, f nq 1 x . F d Ž f n x, pn . q d Ž f nq 1 x, pn . - 2 « for n G m. Consequently, d Ž f n x, f nq 1 x . ª 0
as n ª `.
We last show that dŽ f n x, LŽ x .. ª 0 as n ª `. Suppose that dŽ f n x, LŽ x .. ¢ 0 as n ª `. Then there exist « ) 0 and a subsequence f n a x 4ag N of f n x 4n g N such that dŽ f n a x, LŽ x .. G « for a g N. By the compactness of F Ž f ., we can find pn a g F Ž f . such that dŽ f n a x, F Ž f .. s dŽ f n a x, pn a . for a g N. The compactness of F Ž f . implies that there exist p g F Ž f . and a subsequence pm a4ag N of pn a4ag N satisfying pm a ª p as a ª `. It follows from Ž2.1. that d Ž f m a x, p . F d Ž f m a x, pm a . q d Ž pm a , p . ª 0
as a ª `,
which implies that p is in LŽ x .. Therefore
« F d Ž f m a x, L Ž x . . F d Ž f m a x, p . ª 0
as a ª `,
that is, « F 0, which is impossible. Ž2.2. « Ž2.1. Let p be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N such that f n a x ª p as a ª `. Since f is continuous on O Ž x, f ., d Ž f n a x, f n aq1 x . ª d Ž p, fp .
as a ª `.
It follows from Ž2.2. that p s fp. Hence LŽ x . ; F Ž f .. Consequently, d Ž f n x, F Ž f . . F d Ž f n x, L Ž x . . ª 0
as n ª `,
that is, dŽ f n x, F Ž f .. ª 0 as n ª `. This completes the proof.
300
NOTE
LEMMA 2.2. Let f be a self-map of a metric space Ž X, d . and let x g X. If O Ž x, f . is compact and f is continuous on O Ž x, f ., then the following statements are equi¨ alent: Ž2.4. dŽ f n x, f nq1 x . ª 0 as n ª `; Ž2.5. There exists t g T such that t Ž f n x, f nq1 x . ª 0 as n ª `; Ž2.6. B / LŽ x . ; F Ž f .; Ž2.7.
For e¨ ery t g T, t Ž f n x, f nq1 x . ª 0 as n ª `.
Proof. Ž2.4. « Ž2.5. and Ž2.7. « Ž2.4. are clear. Ž2.5. « Ž2.6. The compactness of O Ž x, f . implies LŽ x . / B. Let y be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N such that f n a x ª y as a ª `. The continuity of f and t implies that t Ž f n a x, f n aq1 x . ª t Ž y, fy . as a ª `. It follows from Ž2.5. that t Ž y, fy . s 0. Hence y s fy g F Ž f . and LŽ x . ; F Ž f .. Ž2.6. « Ž2.7. Suppose that Ž2.7. does not hold. Then there exists t g T such that t Ž f n x, f nq1 x . ¢ 0 as n ª `. Thus we can find an r ) 0 and a subsequence f n a x 4ag N of f n x 4n g N such that t Ž f n a x, f n aq1 x . G r for a g N. Since O Ž x, f . is compact, f n a x 4ag N has a subsequence, say f m a x 4ag N , converging to some n g LŽ x .. It is clear that t Ž f m a x, f m aq1 x . ª t Ž n , fn .
as a ª `.
It follows from Ž2.6. that 0 s t Ž n , fn . G r. This is a contradiction. This completes the proof.
3. RESULTS AND EXAMPLES THEOREM 3.1. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. If f is continuous on O Ž x, f . and satisfies Ž2.1. for some x g X, then LŽ x . is a closed and connected subset of F Ž f . and one of the following conditions holds: Ža. Žb. Žc.
LŽ x . s B; LŽ x . is singleton, and lim nª` f n x exists and belongs to F Ž f .; LŽ x . is uncountable, and it is contained in the boundary of F Ž f ..
Proof. If LŽ x . is empty, then there is nothing left to prove. Therefore throughout the rest of the argument we assume that LŽ x . is nonempty. Let p be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N
301
NOTE
such that f n a x ª p as a ª `. Since d Ž p, F Ž f . . s d lim f n a x, F Ž f . s lim d Ž f n a x, F Ž f . . s 0
ž
/
aª`
aª`
and F Ž f . is closed, then p belongs to F Ž f .. Hence LŽ x . ; F Ž f .. It is easy to see that LŽ x . is closed. To prove the connectedness of LŽ x ., we assume the contrary; that is, let LŽ x . s A j B, where A and B are both nonempty, closed, and disjoint. Note that F Ž f . is compact. Consequently, LŽ x ., A, and B are also compact and dŽ A, B . ) 0. Put 3m s dŽ A, B .. It follows from Lemma 2.1 that there exists k g N such that max dŽ f n x, f nq1 x ., dŽ f n x, LŽ x ..4 - m for n G k. The compactness of LŽ x . implies that there exists nn g LŽ x . such that dŽ f n x, LŽ x .. s dŽ f n x, nn .. If nn is in A, then d Ž f n x, L Ž x . . F d Ž f n x, A . F d Ž f n x, nn . s d Ž f n x, L Ž x . . - m for n G k. Consequently, either dŽ f n x, A. - m or dŽ f n x, B . - m for n G k. But both these inequalities cannot hold simultaneously for the same n because in that case 0 - 3m s d Ž A, b . F d Ž f n x, A . q d Ž f n x, B . - 2 m, which is absurd. The set of positive integers n G k, such that dŽ f n x, A. m, is not empty, because B / A ; LŽ x .. Similarly, the set of positive integers n G k, such that dŽ f n x, B . - m, is also not empty. It is easy to see that there exists n G k satisfying max dŽ f n x, A., dŽ f nq 1 x, B .4 - m. Thus 3m s d Ž A, B . F d Ž A, f n x . q d Ž f n x, f nq 1 x . q d Ž f nq 1 x, B . - 3m, which is impossible. Hence LŽ x . is connected. Theorem 1 of Berge w1, p. 96x ensures that LŽ x . is either a singleton or uncountable. In case LŽ x . is a singleton, Lemma 2.1 implies that Žb. holds. In case LŽ x . is uncountable, by using the method of Diaz and Metcalf w2, Theorem 2x we may similarly prove that Žc. holds also. This completes the proof. Remark 3.1. The main result of Diaz and Metcalf w2, Theorem 2x is a special case of Theorem 3.1. The following example reveals that our Theorem 3.1 is indeed a generalization of the result of Diaz and Metcalf. EXAMPLE 3.1. Let X s 0, 1, 3 4 j
½
1 4n
5 ½
: ngN j
4n 4n y 1
5
: ngN ,
302
NOTE
with the usual metric. Define a self-map f on X by f 0 s 0, f
1 4 Ž n q 1.
s
f 1 s 1, 1
f
1 4
f
and
4 Ž n q 2.
s 3,
f3s
4n 4n y 1
s
1 12
,
4 Ž n q 1. 4 Ž n q 1. y 1
for n g N. Then f is continuous, F Ž f . s 0, 14 , and dŽ f n Ž1r4., F Ž f .. ª 0 as n ª `; that is, the conditions of Theorem 3.1 are satisfied. But Theorem 2 of Diaz and Metcalf is not applicable since
ž
d f
1 4
, FŽ f . s 2 )
/
1 4
sd
ž
1 4
/
, FŽ f . .
The following result is an immediate consequence of Lemma 2.1 and Theorem 3.1. THEOREM 3.2. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. Assume that LŽ x . is nonempty and f is continuous on O Ž x, f . for some x g X. If f satisfies Ž2.2., then the conclusion of Theorem 3.1 holds. Using the method of Park w4, Theorem 1x, by Lemma 2.1 we immediately obtain the following result. THEOREM 3.3. Let f be a self-map of a metric space Ž X, d . and let O Ž x, f . be compact for some x g X. If f is continuous on O Ž x, f . and satisfies one of Ž2.4., Ž2.5., Ž2.6., and Ž2.7., then LŽ x . is a nonempty, closed, and connected subset of F Ž f . and either Žb. or Žc. holds. THEOREM 3.4. Let f be a self-map of a metric space Ž X, d . and x g X. Suppose that O Ž x, f . is compact and f is continuous on O Ž x, f .. If there exists k g N such that
Ž 3.1.
d Ž O Ž fy, f , k . . - d Ž O Ž y, f , k . .
for y g O Ž x, f . with d Ž O Ž y, f, k .. ) 0, then the conclusion of Theorem 3.1 holds. Proof. The compactness of O Ž x, f . ensures that LŽ x . is nonempty. Setting rn s d Ž O Ž f n x, f, k .., we have rnq1 F rn . Therefore rn ª r s inf rn : n g N 4 as n ª `. For any p g LŽ x ., there exists a subsequence f n a x 4ag N of f n x 4n g N such that f n a x ª p as a ª `. Since f is continuous on
303
NOTE
O Ž x, f ., then f n aqm x s f m f n a x ª f m p as a ª `. Consequently,
Ž 3.2.
r s lim d Ž O Ž f n a x, f , k . . s d Ž O Ž p, f , k . . ,
Ž 3.3.
r s lim d Ž O Ž f n aq1 x, f , k . . s d Ž O Ž fp, f , k . . .
aª`
aª`
We claim that r s 0. Otherwise r ) 0. It follows from Ž3.1., Ž3.2., and Ž3.3. that r s d Ž O Ž fp, f , k . . - d Ž O Ž p, f , k . . s r , which is impossible. Therefore r s 0 and p s fp g F Ž f ., that is, LŽ x . ; F Ž f .. Hence f satisfies Ž2.6.. Thus Theorem 3.4 follows from Theorem 3.3. This completes the proof. Remark 3.2. Theorem 1 of Park w4x follows from Theorem 3.3. The main result of Maiti and Babu w3, Theorem 3x and Theorem 2 of Park w4x also follow from Theorem 3.4. The following example shows Theorems 3.3 and 3.4 extend properly the results of Park and Maiti and Babu. EXAMPLE 3.2. Let X s 04 j
½
1 n
: n g N j w 2, 3 x ,
5
with the usual metric. Take the map f : X ª X such that f 0 s 0, f 3 s 1, fx s x for x g w2, 3. and f Ž1rn. s 1rŽ n q 1. for n g N. Obviously, O Ž 1, f . s 0 4 j
½
1 n
: ngN
5
is compact, f is continuous on O Ž1, f ., 1
d Ž f n1, f nq1 1 . s
nq1
y
1 nq2
ª0
as n ª `,
and
ž
d f
1 n
, FŽ f . s
/
1 nq1
-
1 n
sd
ž
1 n
, FŽ f .
/
for n g N. Thus we may invoke our Theorem 3 or Theorem 4 to show that LŽ1. is a nonempty, closed, and compact subset of F Ž f .. But we cannot invoke the results of Park w4x and Maiti and Babu w3x to show that LŽ1. is a nonempty, closed, and compact subset of F Ž f . because f is not continuous on X.
304
NOTE
ACKNOWLEDGMENT The author sincerely thanks the referee for his valuable suggestions.
REFERENCES 1. C. Berge, ‘‘Topological Spaces,’’ Oliver & Boyd, London, 1963. 2. J. B. Diaz and F. T. Metcalf, On the set of subsequential limit points of successive approximations, Trans. Amer. Math. Soc. 135 Ž1969., 459]485. 3. M. Maiti and A. C. Babu, On subsequential limit points of a sequence of iterates, Proc. Amer. Math. Soc. 82 Ž1981., 371]381. 4. S. Park, Remarks on subsequential limit points of a sequence of iterates, J. Korean Math. Soc. 19 Ž1982., 19]22.
222, 297]304 Ž1998.
AY985944
NOTE On the Structure of the Set of Subsequential Limit Points of a Sequence of Iterates Zeqing Liu Department of Mathematics, Liaoning Normal Uni¨ ersity, Dalian, Liaoning 116029, People’s Republic of China Submitted by William F. Ames Received September 15, 1997
In this paper we obtain four results on the structure of the set of subsequential limit points of a sequence of iterates of a self-map of a metric space. Our results extend substantially the main results of Diaz and Metcalf, Maiti and Babu, and Park. Q 1998 Academic Press Key Words: subsequential limit point; closed set; connected set.
1. INTRODUCTION Diaz and Metcalf w2x initiated a study of the structure of the set of subsequential limit points of a sequence of iterates of a continuous self-map f of a metric space Ž X, d . satisfying the condition dŽ fx, F Ž f .. dŽ x, F Ž f .. for x / fx. Maiti and Babu w3x and Park w4x obtained similar results. In this paper we show that the results of Diaz and Metcalf w2x, Maiti and Babu w3x, and Park w4x may be derived under weaker conditions. In fact, we establish that discontinuous maps may also be accommodated in their works and we arrive at the same conclusions as in w2]4x about the set of subsequential limit points of a sequence of iterates of a self-map f in X satisfying the conditions dŽ f n x, F Ž f .. ª 0 as n ª `, or dŽ f n x, f nq1 x . ª 0 and dŽ f n x, LŽ x .. ª 0 as n ª `, or dŽ f n x, f nq1 x . ª 0 as n ª `. 297 0022-247Xr98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
298
NOTE
Let N be the set of positive integers, f a self-map of a metric space Ž X, d ., and F Ž f . the set of fixed points of f. For x g X and k g N, let O Ž x, f . s f n x : n g 0 4 j N 4 ,
O Ž x, f , k . s f n x : 0 F n F k 4 ,
let LŽ x . denote the set of subsequential limit points of the sequence f n x 4`ns 0 , and let O Ž x, f . denote the closure of O Ž x, f .. For A, B ; X, define d Ž x, A . s inf d Ž x, y . : y g A4 , d Ž A, B . s inf d Ž a, b . : a g A, b g B 4 and
d Ž A . s sup d Ž a, b . : a, b g A4 . Define T s t : t is a continuous map of X = X into 0, ` . that satisfies t Ž x, y . s 0 if and only if x s y 4 .
2. LEMMAS We need the following lemmas which are fundamental to our results. LEMMA 2.1. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. If LŽ x . is nonempty and f is continuous on O Ž x, f . for some x g X, then the following statements are equi¨ alent: Ž2.1. Ž2.2.
dŽ f n x, F Ž f .. ª 0 as n ª `; dŽ f n x, f nq1 x . ª 0 and dŽ f n x, LŽ x .. ª 0 as n ª `.
Proof. Ž2.1. « Ž2.2. We first show that f satisfies the following condition: Ž2.3. For every « ) 0, there exists d Ž « . ) 0 such that dŽ y, fz . - « whenever y g F Ž f ., z g O Ž x, f ., and dŽ y, z . - d Ž « .. If not, then there exist « ) 0 and sequences yn4ng N ; F Ž f . and z n4n g N ; O Ž x, f . such that both dŽ yn , z n . ª 0 as n ª ` and dŽ yn , fz n . G « for n g N. The compactness of F Ž f . implies that there exists a subsequence yn a4ag N of yn4ng N such that it converges to a point y in F Ž f .. Consequently, d Ž y, z n a . F d Ž y, yn a . q d Ž yn a , z n a . ª 0
as a ª `.
299
NOTE
Hence dŽ y, z n a . ª 0 as a ª `. Since f is continuous on O Ž x, f ., d Ž fz n a , y . s d Ž fz n a , fy . ª 0
as a ª `.
It follows that
« - d Ž yn a , fz n a . F d Ž yn a , y . q d Ž y, fz n a . ª 0
as a ª `.
This is a contradiction. We next show that dŽ f n x, f nq1 x . ª 0 as n ª `. Since F Ž f . is compact, there exists a sequence pn4ng N ; F Ž f . such that dŽ f n x, pn . s dŽ f n x, F Ž f .. for n g N. Let « ) 0. In view of Ž2.1. and Ž2.3., there exists m g N with dŽ f n x, pn . - min « , d Ž « .4 for n G m. Thus d Ž f n x, f nq 1 x . F d Ž f n x, pn . q d Ž f nq 1 x, pn . - 2 « for n G m. Consequently, d Ž f n x, f nq 1 x . ª 0
as n ª `.
We last show that dŽ f n x, LŽ x .. ª 0 as n ª `. Suppose that dŽ f n x, LŽ x .. ¢ 0 as n ª `. Then there exist « ) 0 and a subsequence f n a x 4ag N of f n x 4n g N such that dŽ f n a x, LŽ x .. G « for a g N. By the compactness of F Ž f ., we can find pn a g F Ž f . such that dŽ f n a x, F Ž f .. s dŽ f n a x, pn a . for a g N. The compactness of F Ž f . implies that there exist p g F Ž f . and a subsequence pm a4ag N of pn a4ag N satisfying pm a ª p as a ª `. It follows from Ž2.1. that d Ž f m a x, p . F d Ž f m a x, pm a . q d Ž pm a , p . ª 0
as a ª `,
which implies that p is in LŽ x .. Therefore
« F d Ž f m a x, L Ž x . . F d Ž f m a x, p . ª 0
as a ª `,
that is, « F 0, which is impossible. Ž2.2. « Ž2.1. Let p be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N such that f n a x ª p as a ª `. Since f is continuous on O Ž x, f ., d Ž f n a x, f n aq1 x . ª d Ž p, fp .
as a ª `.
It follows from Ž2.2. that p s fp. Hence LŽ x . ; F Ž f .. Consequently, d Ž f n x, F Ž f . . F d Ž f n x, L Ž x . . ª 0
as n ª `,
that is, dŽ f n x, F Ž f .. ª 0 as n ª `. This completes the proof.
300
NOTE
LEMMA 2.2. Let f be a self-map of a metric space Ž X, d . and let x g X. If O Ž x, f . is compact and f is continuous on O Ž x, f ., then the following statements are equi¨ alent: Ž2.4. dŽ f n x, f nq1 x . ª 0 as n ª `; Ž2.5. There exists t g T such that t Ž f n x, f nq1 x . ª 0 as n ª `; Ž2.6. B / LŽ x . ; F Ž f .; Ž2.7.
For e¨ ery t g T, t Ž f n x, f nq1 x . ª 0 as n ª `.
Proof. Ž2.4. « Ž2.5. and Ž2.7. « Ž2.4. are clear. Ž2.5. « Ž2.6. The compactness of O Ž x, f . implies LŽ x . / B. Let y be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N such that f n a x ª y as a ª `. The continuity of f and t implies that t Ž f n a x, f n aq1 x . ª t Ž y, fy . as a ª `. It follows from Ž2.5. that t Ž y, fy . s 0. Hence y s fy g F Ž f . and LŽ x . ; F Ž f .. Ž2.6. « Ž2.7. Suppose that Ž2.7. does not hold. Then there exists t g T such that t Ž f n x, f nq1 x . ¢ 0 as n ª `. Thus we can find an r ) 0 and a subsequence f n a x 4ag N of f n x 4n g N such that t Ž f n a x, f n aq1 x . G r for a g N. Since O Ž x, f . is compact, f n a x 4ag N has a subsequence, say f m a x 4ag N , converging to some n g LŽ x .. It is clear that t Ž f m a x, f m aq1 x . ª t Ž n , fn .
as a ª `.
It follows from Ž2.6. that 0 s t Ž n , fn . G r. This is a contradiction. This completes the proof.
3. RESULTS AND EXAMPLES THEOREM 3.1. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. If f is continuous on O Ž x, f . and satisfies Ž2.1. for some x g X, then LŽ x . is a closed and connected subset of F Ž f . and one of the following conditions holds: Ža. Žb. Žc.
LŽ x . s B; LŽ x . is singleton, and lim nª` f n x exists and belongs to F Ž f .; LŽ x . is uncountable, and it is contained in the boundary of F Ž f ..
Proof. If LŽ x . is empty, then there is nothing left to prove. Therefore throughout the rest of the argument we assume that LŽ x . is nonempty. Let p be in LŽ x .. Then there exists a subsequence f n a x 4ag N of f n x 4ng N
301
NOTE
such that f n a x ª p as a ª `. Since d Ž p, F Ž f . . s d lim f n a x, F Ž f . s lim d Ž f n a x, F Ž f . . s 0
ž
/
aª`
aª`
and F Ž f . is closed, then p belongs to F Ž f .. Hence LŽ x . ; F Ž f .. It is easy to see that LŽ x . is closed. To prove the connectedness of LŽ x ., we assume the contrary; that is, let LŽ x . s A j B, where A and B are both nonempty, closed, and disjoint. Note that F Ž f . is compact. Consequently, LŽ x ., A, and B are also compact and dŽ A, B . ) 0. Put 3m s dŽ A, B .. It follows from Lemma 2.1 that there exists k g N such that max dŽ f n x, f nq1 x ., dŽ f n x, LŽ x ..4 - m for n G k. The compactness of LŽ x . implies that there exists nn g LŽ x . such that dŽ f n x, LŽ x .. s dŽ f n x, nn .. If nn is in A, then d Ž f n x, L Ž x . . F d Ž f n x, A . F d Ž f n x, nn . s d Ž f n x, L Ž x . . - m for n G k. Consequently, either dŽ f n x, A. - m or dŽ f n x, B . - m for n G k. But both these inequalities cannot hold simultaneously for the same n because in that case 0 - 3m s d Ž A, b . F d Ž f n x, A . q d Ž f n x, B . - 2 m, which is absurd. The set of positive integers n G k, such that dŽ f n x, A. m, is not empty, because B / A ; LŽ x .. Similarly, the set of positive integers n G k, such that dŽ f n x, B . - m, is also not empty. It is easy to see that there exists n G k satisfying max dŽ f n x, A., dŽ f nq 1 x, B .4 - m. Thus 3m s d Ž A, B . F d Ž A, f n x . q d Ž f n x, f nq 1 x . q d Ž f nq 1 x, B . - 3m, which is impossible. Hence LŽ x . is connected. Theorem 1 of Berge w1, p. 96x ensures that LŽ x . is either a singleton or uncountable. In case LŽ x . is a singleton, Lemma 2.1 implies that Žb. holds. In case LŽ x . is uncountable, by using the method of Diaz and Metcalf w2, Theorem 2x we may similarly prove that Žc. holds also. This completes the proof. Remark 3.1. The main result of Diaz and Metcalf w2, Theorem 2x is a special case of Theorem 3.1. The following example reveals that our Theorem 3.1 is indeed a generalization of the result of Diaz and Metcalf. EXAMPLE 3.1. Let X s 0, 1, 3 4 j
½
1 4n
5 ½
: ngN j
4n 4n y 1
5
: ngN ,
302
NOTE
with the usual metric. Define a self-map f on X by f 0 s 0, f
1 4 Ž n q 1.
s
f 1 s 1, 1
f
1 4
f
and
4 Ž n q 2.
s 3,
f3s
4n 4n y 1
s
1 12
,
4 Ž n q 1. 4 Ž n q 1. y 1
for n g N. Then f is continuous, F Ž f . s 0, 14 , and dŽ f n Ž1r4., F Ž f .. ª 0 as n ª `; that is, the conditions of Theorem 3.1 are satisfied. But Theorem 2 of Diaz and Metcalf is not applicable since
ž
d f
1 4
, FŽ f . s 2 )
/
1 4
sd
ž
1 4
/
, FŽ f . .
The following result is an immediate consequence of Lemma 2.1 and Theorem 3.1. THEOREM 3.2. Let f be a self-map of a metric space Ž X, d . and let F Ž f . be nonempty and compact. Assume that LŽ x . is nonempty and f is continuous on O Ž x, f . for some x g X. If f satisfies Ž2.2., then the conclusion of Theorem 3.1 holds. Using the method of Park w4, Theorem 1x, by Lemma 2.1 we immediately obtain the following result. THEOREM 3.3. Let f be a self-map of a metric space Ž X, d . and let O Ž x, f . be compact for some x g X. If f is continuous on O Ž x, f . and satisfies one of Ž2.4., Ž2.5., Ž2.6., and Ž2.7., then LŽ x . is a nonempty, closed, and connected subset of F Ž f . and either Žb. or Žc. holds. THEOREM 3.4. Let f be a self-map of a metric space Ž X, d . and x g X. Suppose that O Ž x, f . is compact and f is continuous on O Ž x, f .. If there exists k g N such that
Ž 3.1.
d Ž O Ž fy, f , k . . - d Ž O Ž y, f , k . .
for y g O Ž x, f . with d Ž O Ž y, f, k .. ) 0, then the conclusion of Theorem 3.1 holds. Proof. The compactness of O Ž x, f . ensures that LŽ x . is nonempty. Setting rn s d Ž O Ž f n x, f, k .., we have rnq1 F rn . Therefore rn ª r s inf rn : n g N 4 as n ª `. For any p g LŽ x ., there exists a subsequence f n a x 4ag N of f n x 4n g N such that f n a x ª p as a ª `. Since f is continuous on
303
NOTE
O Ž x, f ., then f n aqm x s f m f n a x ª f m p as a ª `. Consequently,
Ž 3.2.
r s lim d Ž O Ž f n a x, f , k . . s d Ž O Ž p, f , k . . ,
Ž 3.3.
r s lim d Ž O Ž f n aq1 x, f , k . . s d Ž O Ž fp, f , k . . .
aª`
aª`
We claim that r s 0. Otherwise r ) 0. It follows from Ž3.1., Ž3.2., and Ž3.3. that r s d Ž O Ž fp, f , k . . - d Ž O Ž p, f , k . . s r , which is impossible. Therefore r s 0 and p s fp g F Ž f ., that is, LŽ x . ; F Ž f .. Hence f satisfies Ž2.6.. Thus Theorem 3.4 follows from Theorem 3.3. This completes the proof. Remark 3.2. Theorem 1 of Park w4x follows from Theorem 3.3. The main result of Maiti and Babu w3, Theorem 3x and Theorem 2 of Park w4x also follow from Theorem 3.4. The following example shows Theorems 3.3 and 3.4 extend properly the results of Park and Maiti and Babu. EXAMPLE 3.2. Let X s 04 j
½
1 n
: n g N j w 2, 3 x ,
5
with the usual metric. Take the map f : X ª X such that f 0 s 0, f 3 s 1, fx s x for x g w2, 3. and f Ž1rn. s 1rŽ n q 1. for n g N. Obviously, O Ž 1, f . s 0 4 j
½
1 n
: ngN
5
is compact, f is continuous on O Ž1, f ., 1
d Ž f n1, f nq1 1 . s
nq1
y
1 nq2
ª0
as n ª `,
and
ž
d f
1 n
, FŽ f . s
/
1 nq1
-
1 n
sd
ž
1 n
, FŽ f .
/
for n g N. Thus we may invoke our Theorem 3 or Theorem 4 to show that LŽ1. is a nonempty, closed, and compact subset of F Ž f .. But we cannot invoke the results of Park w4x and Maiti and Babu w3x to show that LŽ1. is a nonempty, closed, and compact subset of F Ž f . because f is not continuous on X.
304
NOTE
ACKNOWLEDGMENT The author sincerely thanks the referee for his valuable suggestions.
REFERENCES 1. C. Berge, ‘‘Topological Spaces,’’ Oliver & Boyd, London, 1963. 2. J. B. Diaz and F. T. Metcalf, On the set of subsequential limit points of successive approximations, Trans. Amer. Math. Soc. 135 Ž1969., 459]485. 3. M. Maiti and A. C. Babu, On subsequential limit points of a sequence of iterates, Proc. Amer. Math. Soc. 82 Ž1981., 371]381. 4. S. Park, Remarks on subsequential limit points of a sequence of iterates, J. Korean Math. Soc. 19 Ž1982., 19]22.