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On lower semicontinuity of some set-valued iteration semigroups
Nonlinear Analysis 71 (2009) 5644–5654
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On lower semicontinuity of some set-valued iteration semigroups Grażyna Łydzińska Institute of Mathematics, Silesian University, Bankowa 14, PL-40-007 Katowice, Poland
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Article history: Received 4 February 2009 Accepted 29 April 2009
We study a lower semicontinuity of set-valued iteration semigroups which are the counterparts of the fundamental form of continuous iteration semigroups of single-valued functions on an interval. © 2009 Elsevier Ltd. All rights reserved.
MSC: 39B12 39B52 26E25 54C60 26A18 Keywords: Iteration theory Set-valued function Iteration semigroup Lower semicontinuity
1. Introduction It is well known (cf. [1, Chap. IX, Sec. 1], [2, Theorems 5.1–8.1], [3, p. 98–99], [4, Chap. I, Sec. 1.7], also [5, Theorem 1]) that the formula f (t , x) = α −1 α(x) + t
yields the fundamental form of iteration semigroups, i.e. functions satisfying the translation equation f (s + t , x) = f t , f (s, x) .
More generally we have the following observation. Remark 1. Let α be a bijection mapping a set X onto an interval I ⊂ R with the right endpoint q ∈ (−∞, +∞]. Assume that q ∈ I whenever q is finite. Then the function f : (0, ∞) × X → X , given by f (t , x) = α −1 min{α(x) + t , q} ,
(1)
is an iteration semigroup. For the first time the above remark was made by M.C. Zdun (cf. [2, Theorems 5.1–8.1]). As he proved there, (1) with homeomorphic α is a general form of the so-called continuous iteration semigroups on an interval (cf. also [5, Theorem 1]). In the present paper we resign the bijectivity of α and consider a counterpart of (1) which is a set-valued function F : (0, ∞)× X → 2X given by (A) (see Section 2); we study the lower semicontinuity of the multifunctions F (t , ·) and F (·, x).
E-mail address: [email protected]. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.04.051
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The obtained results are the first step in studying the lower semicontinuity of, considered in [6–8], set-valued counterpart of (1) given by the formula
F (t , x) = A− A(x) + min t , q − inf A(x)
where A : X → 2R and A− denotes the lower preimage under A defined below. 2. Preliminaries Let X , Y be non-empty sets, and F : X → 2Y . Put F (D) :=
[
F (x)
for every D ⊂ X
x∈D
and F − (B) := {x ∈ X : F (x) ∩ B 6= ∅} for every B ⊂ Y . The following remark is clear. Remark 2. For every family {Bt : t ∈ T } of subsets of Y and for every F : X → 2Y we have
! F
−
[
=
Bt
t ∈T
[
F − (Bt ).
t ∈T
Assume that X and Y are topological spaces and F : X → 2Y . We say that F is lower semicontinuous if for every open set U ⊂ Y , the set F − (U ) is open in X . Fix an arbitrary set X and an arbitrary, not necessarily bijective, function A : X → R. Put S := A(X )
and q := sup S .
Throughout this paper we always assume that S is an interval. Now put
F (t , x) := A−1 {A(x)} + min t , q − A(x) ,
(A)
where A−1 (V ) stands for the usual preimage of V ⊂ R under A. The next observation follows immediately from the definition of F . Lemma 1 (See Also [7, Lemma 3]). Let F : (0, ∞) × X → 2X be given by (A) and let t ∈ (0, ∞) and x ∈ X . If t < q − A(x) then
F (t , x) = A−1 {A(x)} + t 6= ∅ and if t ≥ q − A(x) then F (t , x) =
A−1 ({q}),
∅
if q ∈ S , otherwise.
Theorem 1 (See [8, Theorem 3]). Every F : (0, ∞) × X → 2X , given by (A), is a set-valued iteration semigroup: F t , F (s, x) = F (s + t , x)
for every x ∈ X and s, t ∈ (0, ∞). We say that an iteration semigroup F : (0, ∞) × X → 2X is lower semicontinuous if for every x ∈ X multifunction F (·, x) is lower semicontinuous. Remark 3. Let F : (0, ∞) × X → 2X be given by (A), t ∈ (0, ∞) and V ⊂ X . If q 6∈ A(V ) then
F (t , ·)− (V ) = A−1 A(V ) − t , and if q ∈ A(V ) then F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 [q − t , q]
= A−1 A(V ) − t ∪ A−1 (q − t , q] .
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Proof. First assume that q 6∈ A(V ). Then by Lemma 1, F (t , ·)− (V ) = x ∈ X : t < q − A(x) and
= = = =
F (t , x) ∩ V 6= ∅
∪ x ∈ X : t > q − A(x) and F (t , x) ∩ V 6= ∅ x ∈ X : t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅ x ∈ X : t < q − A(x) and A(x) + t ∈ A(V ) x ∈ X : A(x) + t ∈ A(V ) A−1 A(V ) − t .
Assume now that q ∈ A(V ). Due to Lemma 1 we have F (t , ·)− (V ) = x ∈ X : t < q − A(x) and
A−1 {A(x)} + t ∩ V 6= ∅
∪ x ∈ X : t > q − A(x) and A−1 {q} ∩ V 6= ∅ h i = A−1 (−∞, q − t ) ∩ A−1 A(V ) − t ∪ A−1 [q − t , q] = A−1 A(V ) − t ∪ A−1 [q − t , q] . Since q − t ∈ A(V ) − t we get also
F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 (q − t , q] . Remark 4. Let F : (0, ∞) × X → 2X be given by (A), x ∈ X and V ⊂ X . If q 6∈ A(V ) then F (·, x)− (V ) = (0, ∞) ∩ A(V ) − A(x) ,
and if q ∈ A(V ) then
(0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ = (0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ ,
F (·, x)− (V ) =
when A(x) < q and F (·, x)− (V ) = (0, ∞), when A(x) = q. Proof. First assume that q 6∈ A(V ). Then, by Lemma 1, F (·, x)− (V ) = t ∈ (0, ∞): t < q − A(x) and F (t , x) ∩ V 6= ∅
= = = =
∪ t ∈ (0, ∞): t > q − A(x) and F (t , x) ∩ V 6= ∅ t ∈ (0, ∞): t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅ t ∈ (0, ∞): t < q − A(x) and A(x) + t ∈ A(V ) t ∈ (0, ∞): A(x) + t ∈ A(V ) (0, ∞) ∩ A(V ) − A(x) .
And now assume that q ∈ A(V ). Consider the case when A(x) < q. Then F (·, x)− (V ) = t ∈ (0, ∞): t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅
∪ t ∈ (0, ∞): t > q − A(x) and A−1 {q} ∩ V 6= ∅ = t ∈ (0, q − A(x)): A(x) + t ∈ A(V ) ∪ q − A(x), ∞ = 0, q − A(x) ∩ A(V ) − A(x) ∪ q − A(x), ∞ h i = (0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ .
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Since q − A(x) ∈ A(V ) − A(x) we get also
F (·, x)− (V ) = (0, ∞) ∩ A(V ) − A(x)
∪ q − A(x), ∞ .
Now consider the case when q ∈ A(V ) and A(x) = q. Then
t ∈ (0, ∞): t < q − A(x) and F (t , x) ∩ V 6= ∅ = ∅.
On the other hand, we have
t ∈ (0, ∞): t ≥ q − A(x) and F (t , x) ∩ V 6= ∅ = t ∈ (0, ∞): A−1 {q} ∩ V 6= ∅ = (0, ∞).
Hence F (·, x)− (V ) = (0, ∞) which completes the proof.
Now we remind the definition and a certain property of a locally connected space. A topological space X is said to be locally connected if for every point x ∈ X and for each of its neighbourhood U ⊂ X there exists a connected neighbourhood V ⊂ X of x such that V ⊂ U. The following well-known fact can be found e.g. in [9]. Fact 1. A topological space X is locally connected if and only if every connected component of every open subset of X is open. 3. Lower semicontinuity Theorem 2. Let X be a topological space, A : X → R a continuous function and let F : (0, ∞) × X → 2X be given by (A). If F (·, x) is lower semicontinuous for every x ∈ X then F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Proof. Assume that for every x ∈ X the multifunction F (·, x) is lower semicontinuous. Fix t ∈ (0, ∞) and an open set V ⊂ X . We show that F (t , ·)− (V ) is open in X . Notice that in the case when A(V ) is open in R we have q 6∈ A(V ), whence, by Remark 3 and the continuity of A, the required condition holds. Thus assume that A(V ) is not open in R, so A(V ) \ int A(V ) 6= ∅.
(2)
We show that A(V ) \ int A(V ) ⊂ {inf S , q}.
(3)
Suppose that there exists a y ∈ A(V ) \ int A(V ) such that inf S < y < q. Therefore we can find an x ∈ X such that inf S 6 A(x) < y < q. Thus 0 < y − A(x) ∈ A(V ) − A(x), y − A(x) 6∈ int A(V ) − A(x)
and y − A(x) < q − A(x). Consequently, due to Remark 4, the set F (·, x)− (V ) is not open, which contradicts our assumption. Thus (3) is satisfied. At first consider the case q 6∈ A(V ). Then, using (2) and (3), we obtain A(V ) \ int A(V ) = {inf S }, whence A(V ) = int A(V ) ∪ {inf S } and A−1 A(V ) − t = A−1
int A(V ) ∪ {inf S } − t
= A−1
int A(V ) − t ∪ {inf S − t }
= A−1 int A(V ) − t . Therefore, by Remark 3 and the continuity of A, the set F (t , ·)− (V ) is open in X . Now assume that q ∈ A(V ). If A is constant, then A(x) = q for every x ∈ X and, by Remark 3, F (t , ·)− (V ) = A−1 {q} − t ∪ A−1 [q − t , q] =
= A−1 [q − t , q] = A−1 [q − t , q] ∩ S = A−1 {q} = X ,
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so F (t , ·)− (V ) is open. Assume that A is not constant. Since q ∈ A(V ), by Remark 4, we get that (0, ∞) ∩ A(V ) − A(x)
∪
q − A(x), ∞ is open in R for every x ∈ X . Then there exists ε ∈ (0, ∞) such that (q − ε, q] ⊂ A(V ). Therefore int A(V ) − t ∪ [q − t , q + ε) is open in R. Moreover, due to (3), A−1 A(V ) − t ⊂ A−1 int A(V ) ∪ {inf S } ∪ {q} − t = A−1 int A(V ) − t ∪ {q} − t ⊂ ⊂ A−1 A(V ) − t .
Thus, according to Remark 3, we see that F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 (q − t , q]
= A−1 int A(V ) − t ∪ [q − t , q] = A−1 int A(V ) − t ∪ [q − t , q + ε) is an open subset of X .
In the following theorem we present a necessary condition for a lower semicontinuity of F (t , ·) for every t ∈ (0, ∞), under the assumption that X is an arcwise connected space. Theorem 3. Let X be an arcwise connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Assume that F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Then
(S)
A|U ≡ inf S
or
max A(U ) = q
for every open set U ⊂ X with sup A(U ) ∈ A(U ), and
(I)
min A(V ) = inf S
for every open set V ⊂ X with inf A(V ) ∈ A(V ). Proof. At first we show that condition (I) holds. Let V ⊂ X be an open set such that y0 := inf A(V ) ∈ A(V )
(4)
and let x0 ∈ V satisfy the equality A(x0 ) = y0 . Suppose on the contrary that y0 = min A(V ) > inf S . We can find an x1 ∈ X for which A(x1 ) < y0 . Fix ε ∈ (0, 1) such that y0 −ε > A(x1 ). Since X is connected and A is continuous, there exists an xε ∈ X such that A(xε ) = y0 − ε. Let p : [0, 1] → X be a continuous function with p(0) = xε
and
p(1) = x1 ,
and let K := p [0, 1] ∩ A−1 {y0 } − ε .
Of course xε ∈ K , so K 6= ∅. Moreover, x1 6∈ K and K is compact. Let t0 := max t ∈ [0, 1]: p(t ) ∈ K ,
x := p(t0 ) and P := F (ε, ·)− (V ). By the assumption the set P is open in X . Notice that A(x) = y0 − ε ∈ A(V ) − ε , so, according to Remark 3, x ∈ P and, by the Darboux property of A ◦ p, the definition of t0 and by the inequality (A ◦ p)(1) = A(x1 ) < y0 − ε , we have
(A ◦ p)(t ) < y0 − ε
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for every t ∈ (t0 , 1]. Therefore, due to (4), for every t ∈ (t0 , 1] we get A p(t ) 6∈ A(V ) − ε and A p(t ) 6∈ (q − ε, q]. Then, by Remark 3, x ∈ P \ int P, which contradicts the openness of P and completes the proof of (I). Now we pass to the proof of (S). Let U ⊂ X be an open set such that y0 := sup A(U ) ∈ A(U ).
(5)
Assume that q 6= max A(U ). Then q 6∈ A(U ) and y0 < q. Take x0 ∈ U such that A(x0 ) = y0 . At first we show that A |U is constant. Suppose that this is not true. Therefore there exists an x1 ∈ U such that A(x1 ) 6= y0 . Then A(x1 ) < y0 = A(x0 ). Let ε := A(x0 ) − A(x1 ), p : [0, 1] → X be a continuous function such that p(0) = x1
and
p(1) = x0
and let K := p [0, 1] ∩ A−1 {y0 } − ε .
Of course x1 ∈ K , so K 6= ∅. Moreover K is compact and x0 6∈ K . Let t0 := max {t ∈ [0, 1]: p(t ) ∈ K }, x := p(t0 ) and P := A−1 (A(U ) − ε). Obviously, by Remark 3, the set P is open in X . Notice that x ∈ P and, due to the Darboux property of A ◦ p, the definition of t0 and the inequality (A ◦ p)(1) = A(x0 ) > y0 − ε , we get
(A ◦ p)(t ) > y0 − ε for every t ∈ (t0 , 1]. Then, according to (5), we have x ∈ P \ int P, which contradicts the openness of P. So we have shown that A |U ≡ const. As follows from the example below the assumption of the arcwise connectedness is essential for the validity of Theorem 3. Example 1. Let X be any set of the cardinality c, endowed with the discrete topology, and let A : X → R be an arbitrary function such that S is an interval and card S > 1. Clearly A is continuous and F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). On the other hand take any x ∈ X with inf S < A(x) < q. Obviously V := {x} is open in X . Since A |V takes the smallest and the greatest values, which are not equal to q and inf S, conditions (S) and (I) do not hold. Theorem 4. Let X be a locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). If conditions (S) and (I) hold then F (·, x) is lower semicontinuous for every x ∈ X and F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Proof. Assume conditions (S) and (I). We show that F (·, x) is lower semicontinuous for every x ∈ X . Fix x ∈ X and an open set V ⊂ X . We prove that T := F (·, x)− (V ) is open in (0, ∞). Since X is locally connected we can assume that V is connected (see Fact 1 and Remark 2). Therefore A(V ) is an interval. If A(V ) is open in R then q 6∈ A(V ), so, due to Remark 4, the set T is open. Thus consider the case when A(V ) is not open in R. Since A(V ) is an interval, we have sup A(V ) ∈ A(V )
(6)
inf A(V ) ∈ A(V ).
(7)
or
At first assume that inclusion (6) is satisfied. Then, by (S), we get A|V ≡ inf S
(8)
max A(V ) = q.
(9)
or
Assume that (9) holds. Then A(V ) = inf A(V ), q
or A(V ) = inf A(V ), q .
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If the first of the above possibilities holds, then, by Remark 4,
(0, ∞) ∩ inf A(V ) − A(x), q − A(x) ∪ q − A(x), ∞ = (0, ∞) ∩ inf A(V ), ∞ , so T is open in (0, ∞). If A(V ) = inf A(V ), q , then, due to (I), we have A(V ) = inf S , q , T =
whence, by Remark 4, T = (0, ∞). Now assume that condition (9) does not hold. Then q 6∈ A(V ) and equality (8) is satisfied. Then A(V ) = {inf S } and, consequently, Remark 4 gives A(V ) − A(x) ⊂ (−∞, 0]
T = ∅.
and
Now we pass to case (7). We may assume that (6) does not hold. Therefore A(V ) = inf A(V ), sup A(V ) .
According to condition (I) we get A(V ) = inf S , sup A(V ) .
Hence, by Remark 4, T = (0, ∞) ∩ A(V ) − A(x)
= (0, ∞) ∩
inf S , sup A(V ) − A(x)
= (0, ∞) ∩ inf S − A(x), sup A(V ) − A(x) . Therefore either T = 0, sup A(V ) − A(x)
if A(x) < sup A(V ),
or T =∅
otherwise.
Thus T is open in (0, ∞). The lower semicontinuity of F (t , ·) for every t ∈ (0, ∞) follows immediately from Theorem 2.
The next example shows that the assumption of the local connectedness of X in Theorem 4 is essential. Example 2. Let X be the sum of J = [0, 1] × {0} and the semilines (x, y) ∈ R2 : x = r and y 6 0 for every r ∈ [0, 1] ∩ Q. In the set X we consider the metric generated by the Euclidean metric on the plane. X is an arcwise connected (so connected) space, but it is not locally connected in the points (x, y) for y < 0. Let A : X → R be given by
A(x, y) = x. Obviously A is continuous. If U ⊂ X is an open set such that sup A(U ) ∈ A(U ) then 1 ∈ A(U ), so max A(U ) = 1 = q. Thus condition (S) holds. Notice also that (I) is satisfied. Let V be the open ball centered in 1/2, −1 , with the radius 1/4. In our case q ∈ [0, 1] = S, but q 6∈ 1/4, 3/4 ∩ Q = A(V ). Take t = 1/4. We have F 1/4, ·
−
(V ) = A−1 A(V ) − 1/4 = A−1 1/4, 3/4 ∩ Q − 1/4 o n = A−1 (0, 1/2) ∩ Q = (x, y) ∈ X : x ∈ 0, 1/2 ∩ Q . −
Since the points (x, y), where x ∈ 0, 1/2 ∩ Q and y = 0, do not belong to int F 1/4, · (V ), this set is not open in X . Therefore F (t , ·) is not lower semicontinuous. By Theorem 2 we get also that F (·, x) is not lower semicontinuous for some x ∈ X . A similar result can be obtained in the cases when J = [0, 1)×{0} (then q 6∈ S) or J = (x, y) ∈ R2 : x > 0 and y = 0 (then q = ∞).
The next theorem follows immediately from Theorems 2–4. Theorem 5. Let X be an arcwise connected and locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Then the following conditions are pairwise equivalent: (i) F (·, x) is lower semicontinuous for every x ∈ X ; (ii) F (t , ·) is lower semicontinuous for every t ∈ (0, ∞); (iii) conditions (S) and (I) hold.
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We can say much more in the case when A is defined on an interval. Theorem 6. Let I ⊂ R be an interval, A : I → R be a continuous function, and let F : (0, ∞) × I → 2I be given by (A). Assume that q 6∈ S. Then the interval I is not compact and the following conditions are pairwise equivalent: (i) F (·, x) is lower semicontinuous for every x ∈ I; (ii) F (t , ·) is lower semicontinuous for every t ∈ (0, ∞); (iii) the condition (U) there exist a, b ∈ R such that inf I 6 a 6 b 6 sup I, if inf I ∈ I then a = inf I , if sup I ∈ I then b = sup I and is strictly decreasing ,
A |I ∩(−∞,a) A |I ∩[a,b]
is constant ,
A |I ∩(b,∞)
is strictly increasing
holds; (iv) for every x ∈ I and t ∈ (0, ∞) card F (t , x) 6 2 and, when I is closed from one side, card F (t , x) 6 1. Before the proof of Theorem 6 we will present the following lemma. Lemma 2. Let I ⊂ R be an interval and A : I → R be a continuous function. Then (i) if I is open then (U) is equivalent to the condition: (P1) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) > A(x2 ), or A(x1 ) = A(x2 ) 6 A(x3 ), or A(x1 ) < A(x2 ) < A(x3 ); (ii) if I is closed from the left then (U) is equivalent to the condition: (P2) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) = A(x2 ) = A(x3 ), or A(x1 ) = A(x2 ) < A(x3 ), or A(x1 ) < A(x2 ) < A(x3 ); (iii) if I is closed from the right then (U) is equivalent to the condition: (P3) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) > A(x2 ) > A(x3 ), or A(x1 ) > A(x2 ) = A(x3 ), or A(x1 ) = A(x2 ) = A(x3 ).
(10) (11) (12)
Proof. (i) Assume that I is open and condition (U) holds. Fix the points x1 , x2 , x3 ∈ I for which x1 < x2 < x3 . Studying the form of the function A assumed in (U) it is easy to notice that one of conditions (10)–(12) is satisfied. Therefore we pass to the proof of the opposite implication. Assume (P1). Let C := y ∈ I: A(x) > A(y)
for x ∈ (inf I , y) ,
D := y ∈ I: A(y) < A(z )
for z ∈ (y, sup I )
and
a :=
inf I sup C
if C = ∅, , if C = 6 ∅
b :=
sup I inf D
if D = ∅, if D = 6 ∅.
Obviously inf I 6 a and b 6 sup I. Notice also that if u ∈ C and v ∈ D then u ≤ v . Consequently, a ≤ b.
(13)
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Assume that I ∩ (−∞, a) 6= ∅. Then a > inf I and C 6= ∅. Fix x1 , x2 ∈ I ∩ (−∞, a) = (inf I , a) such that x1 < x2 and suppose that A(x1 ) 6 A(x2 ).
(14)
By the definition of a there exists an x3 ∈ (x2 , a] ∩ I such that x3 ∈ C . Thus A(x2 ) > A(x3 ), which, together with inequality (14), contradicts condition (P1). Thus A is strictly decreasing in the interval I ∩ (−∞, a). Similarly we can prove that A is strictly increasing in I ∩ (b, ∞). Now we will show that A |I ∩[a,b] is constant. Assume that a < b and fix x1 , x2 ∈ (a, b) such that x1 < x2 . Suppose that A(x1 ) 6= A(x2 ), for instance A(x1 ) < A(x2 ).
(15)
Since x2 < b we have x2 6∈ D, and thus there exists an x3 ∈ (x2 , sup I ) for which A(x2 ) > A(x3 ), which, due to (15), contradicts assumption (P1). Similarly, in the case A(x1 ) > A(x2 ) there exists a point x0 ∈ (inf I , x1 ) such that A(x0 ) 6 A(x1 ), contrary to (P1). Thus A |I ∩[a,b] is constant. (ii) Let I = [c , d), c < d where c ∈ R and d ∈ R. As earlier it is easy to notice that in this case the condition (U) implies (P2). Now assume that condition (P2) holds. Let D be given by (13) and
a := inf I
and
b :=
sup I inf D
if D = ∅, if D = 6 ∅.
Obviously a 6 b. Assume that I ∩ (b, ∞) 6= ∅. Fix x1 , x2 ∈ I ∩ (b, ∞) = (b, d), x1 < x2 and suppose that A(x1 ) > A(x2 ).
(16)
According to the definition of the element b there exists an x0 ∈ [b, x1 ) ∩ I such that x0 ∈ D, and thus A(x0 ) < A(x1 ). Hence by (16) we get a contradiction with condition (P2). Thus A is strictly increasing in the interval I ∩ (b, ∞). Now assume that a < b. We show that A |(a,b) is constant, which due to the continuity of A will complete the proof of (ii). Suppose that there exist x1 , x2 ∈ (a, b), x1 < x2 for which A(x1 ) 6= A(x2 ). By the fact that x2 6∈ D we can find a point x3 ∈ (x2 , sup I ) such that A(x2 ) > A(x3 ), which contradicts condition (P2). The proof of assertion (iii) is similar.
Now we pass to the proof of Theorem 6. Proof of Theorem 6. At first we show that in the considered case condition (iii) and the conjunction of (S) and (I) are equivalent which, by Theorem 5, will complete the proof of the pairwise equivalence of conditions (i), (ii) and (iii). Studying the form of the function A assumed in (U) it is easy to observe that (iii) implies conditions (S) and (I). Passing to the proof of the opposite implication suppose that condition (iii) does not hold. At first consider the case when the interval I is open. Then, by Lemma 2, there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and either A(x1 ) < A(x2 )
and
A(x2 ) > A(x3 ),
(17)
or A(x1 ) = A(x2 ) > A(x3 ),
(18)
A(x1 ) < A(x2 ) = A(x3 ).
(19)
or
Let U := (x1 , x3 ). Since A is continuous, each of the above conditions implies that sup A(U ) ∈ A(U ) and max A(U ) > A(x2 ). Thus A|U 6≡ inf S, whence, since q 6∈ S, condition (S) is not satisfied.
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Now let I = [c , d), where c ∈ R and d ∈ R, c < d. Due to Lemma 2 condition (P2) does not hold, whence there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and one of conditions (17)–(19) or A(x1 ) > A(x2 )
(20)
is satisfied. Let U1 := [c , x3 ) and U2 := [c , x2 ). Of course U1 and U2 are open in I. Since A is continuous, each of conditions (17)–(19) implies that sup A(U1 ) ∈ A(U1 ), max A(U1 ) > A(x2 )
and
A|U1 6≡ inf S .
Moreover, if (20) holds then sup A(U2 ) ∈ A(U2 ), max A(U2 ) > A(x1 )
and
A|U2 6≡ inf S .
Thus (S) is not satisfied. When I is closed from the right the proof is similar. Now we will prove the equivalence of the conditions (iii) and (iv). First observe that, by Lemma 1, for every t ∈ (0, ∞) and x ∈ I we have card F (t , x) = card PA(x)+t , where Pk denotes the set consisted of the common points of a line y = k, k ∈ R and the graph of A. Obviously A(x) + t > inf S. Thus, looking at the form of A we observe that (iii) ⇒ (iv). Now assume that (iii) does not hold. We will apply Lemma 2 and in each of its case (i)–(iii) we will find a number k > inf S for which card Pk is greater than the cardinality of F (t , x) allowed by (iv); as stated in the preceding paragraph, this will complete the proof. At first consider the case when I is an open interval. Then, due to Lemma 2, there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and one of conditions (17)–(19) holds. Since q 6∈ S, there exists an x4 ∈ I \ (x1 , x3 ) for which A(x4 ) > max A [x1 , x3 ] . Assume that (17) is satisfied. Let
k ∈ max A(x1 ), A(x3 ) , A(x2 ) .
Obviously k > inf S. Hence, by the continuity of A and the choice of x4 , we get card Pk > 3. Now assume condition (18). If A is constant in the interval [x1 , x2 ] then let k = A(x1 ) and observe that k > inf S and Pk is infinite. Consider the case when A takes a locally maximum at a point x5 ∈ (x1 , x2 ) and A(x5 ) > A(x1 ). Let k ∈ A(x1 ), A(x5 ) .
Obviously k > inf S. Thus, by the continuity of A and the property of x4 , card Pk ≥ 3. Now assume that A has a locally minimum at a point x5 ∈ (x1 , x2 ) and A(x5 ) < A(x1 ). Take
k ∈ max A(x3 ), A(x5 ) , A(x1 ) .
Of course k > inf S and, as A is continuous, card Pk > 3. If we assume condition (19), then the proof is similar. Consider the case when I = [c , d), where c ∈ R and d ∈ R, c < d. Then, by Lemma 2, there exist x1 , x2 , x3 ∈ I, x1 < x2 < x3 , satisfying one of conditions (17)–(20). Since q 6∈ S, there exists an x4 ∈ (x3 , d) such that A(x4 ) > max A [x1 , x3 ] . Observe that if (18) or (19) holds then, taking k = A(x2 ), we obtain k > inf S and card Pk > 2. Assuming condition (17) let
k ∈ max A(x1 ), A(x3 ) , A(x2 ) .
Then k > inf S and, by the continuity of A, we get Pk > 2. Finally, if (20) holds then, by the choice of x4 , we get card Pk > 2 for any k ∈ A(x2 ), A(x1 ) . In the case when I is closed from the right the proof is similar. In particular, Theorem 6 shows that in the case when q 6∈ S the generators of lower semicontinuous iteration semigroups of the form (A) are not far from the bijections and those iteration semigroups are close to single-valued functions. In higher dimensions the situation seems to be more complicated. In the following example we show that if q 6∈ S then a lower semicontinuous iteration semigroup not always has such small values as in condition (iv) of Theorem 6. Example 3. Let X = R2 and let A : X → R be defined by A(x, y) = x2 + y2 . 2
Obviously S = [0, ∞) and q = ∞. Consider function F : (0, ∞) × R2 → 2R given by (A). Then, by Lemma 1,
F t , (x, y) = A−1 {A(x, y)} + t = A−1 ({x2 + y2 } + t ) p = T (0, 0), x2 + y2 + t , where T ((0, 0), r ) stands for the circle centered at (0, 0) and with the radius r. By Theorem 1 the multifunction F is an iteration semigroup. Of course A is continuous and the conditions (S) and (I) hold. Therefore, due to Theorem 5, F (t , ·) is lower semicontinuous for every t ∈ (0, ∞) and F (·, x) is lower semicontinuous for every x ∈ X .
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The next example shows that the assumption ‘‘q 6∈ S’’ is essential for the validity of Theorem 6. Example 4. Let A : R → R be given by A(x) = sin x. Then S = [−1, 1]. Of course the conditions (S) and (I) hold. Let F : (0, ∞) × R → 2R be defined by (A). Due to Theorem 5 the multifunction F (t , ·) is lower semicontinuous for every t ∈ (0, ∞) and F (·, x) is lower semicontinuous for every x ∈ X . Nevertheless condition (iii) fails to be true. As follows from Example 1 the assumption of the arcwise connectedness is essential for the validity of Theorem 3 and, consequently, Theorem 5. Nevertheless, without any assumption imposed on the topological space X , the lower semicontinuity of the multifunctions F (·, x), x ∈ X , implies conditions (S) and (I). Theorem 7. Suppose that X is a topological space. Let A : X → R and F : (0, ∞) × X → 2X be given by (A). If F (·, x) is lower semicontinuous for every x ∈ X then conditions (S) and (I) hold. Proof. At first suppose that condition (S) does not hold, so q 6∈ A(U ) and A |U 6≡ inf S
(21)
for some open set U ⊂ X with sup A(U ) ∈ A(U ). Let y = max A(U ). Then, by (21), we have inf S < y. Choose an x0 ∈ X in such a manner that A(x0 ) < y.
(22)
Thus 0 < y − A(x0 ) ∈ A(U ) − A(x0 ) whence, due to Remark 4 and the inclusion y ∈ A(U ) \ int A(U ), we obtain that the set F (·, x0 )− (U ) is not open in (0, ∞), which contradicts the assumption. Now suppose that V ⊂ X is an open set such that A |V takes the smallest value and min A(V ) > inf S .
(23)
Let y = min A(V ). According to (23) we have inf S < y. Take an x0 ∈ X satisfying (22). By Remark 4 and the inequality y 6 q we get 0 < y − A(x0 ) = min F (·, x0 )− (V ). Hence F (·, x0 )− (V ) is not open in (0, ∞), which contradicts the assumption of the lower semicontinuity of F (·, x0 ).
As a direct consequence of Theorems 7 and 4 we have the following characterization of the lower semicontinuity of the multifunctions of form (A). Theorem 8. Let X be a locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Then F (·, x) is lower semicontinuous for every x ∈ X if and only if conditions (S) and (I) hold. Remark 5. Observe that the space X considered in Example 1 is locally connected. Therefore, as follows from Theorem 8, Example 1 yields a multifunction F of form (A) for which the multifunctions F (t , ·), t ∈ (0, ∞), are lower semicontinuous, whereas F (·, x), x ∈ X , are not. Acknowledgment This research was supported by the Silesian University Mathematics Department (Iterative Functional Equations and Real Analysis program). References [1] [2] [3] [4] [5] [6] [7] [8] [9]
M. Kuczma, Functional equations in a single variable, Monografie Mat. 46, PWN, Warszawa 1968. M.C. Zdun, Continuous and differentiable iteration semigroups, Pr. Nauk. Uniw. Śl. Katow. 308 (1979). Gy. Targonski, Topics in iteration theory, Vandenhoeck and Ruprecht, Göttingen, 1981. M. Kuczma, B. Choczewski, R. Ger, Iterative functional equations, in: Encyclopedia of Math. and Its Applications, vol. 32, Cambridge University Press, 1990. G. Guzik, W. Jarczyk, J. Matkowski, Cocycles of continuous iteration semigroups, Bull. Polish Acad. Sci. Math. 51 (2003) 187–197. G. Łydzińska, Iteration semigroups for which expansion implies collapse, Aeqationes Math. 70 (2005) 247–253. G. Łydzińska, On collapsing iteration semigroups of set-valued functions, Publ. Math. Debrecen 64 (2004) 285–298. G. Łydzińska, On expanding iteration semigroups of set-valued functions, Math. Pannon. 15 (2004) 55–64. K. Kuratowski, Introduction to set theory and topology, Pergamon Press, London, 1961.
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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
On lower semicontinuity of some set-valued iteration semigroups Grażyna Łydzińska Institute of Mathematics, Silesian University, Bankowa 14, PL-40-007 Katowice, Poland
article
abstract
info
Article history: Received 4 February 2009 Accepted 29 April 2009
We study a lower semicontinuity of set-valued iteration semigroups which are the counterparts of the fundamental form of continuous iteration semigroups of single-valued functions on an interval. © 2009 Elsevier Ltd. All rights reserved.
MSC: 39B12 39B52 26E25 54C60 26A18 Keywords: Iteration theory Set-valued function Iteration semigroup Lower semicontinuity
1. Introduction It is well known (cf. [1, Chap. IX, Sec. 1], [2, Theorems 5.1–8.1], [3, p. 98–99], [4, Chap. I, Sec. 1.7], also [5, Theorem 1]) that the formula f (t , x) = α −1 α(x) + t
yields the fundamental form of iteration semigroups, i.e. functions satisfying the translation equation f (s + t , x) = f t , f (s, x) .
More generally we have the following observation. Remark 1. Let α be a bijection mapping a set X onto an interval I ⊂ R with the right endpoint q ∈ (−∞, +∞]. Assume that q ∈ I whenever q is finite. Then the function f : (0, ∞) × X → X , given by f (t , x) = α −1 min{α(x) + t , q} ,
(1)
is an iteration semigroup. For the first time the above remark was made by M.C. Zdun (cf. [2, Theorems 5.1–8.1]). As he proved there, (1) with homeomorphic α is a general form of the so-called continuous iteration semigroups on an interval (cf. also [5, Theorem 1]). In the present paper we resign the bijectivity of α and consider a counterpart of (1) which is a set-valued function F : (0, ∞)× X → 2X given by (A) (see Section 2); we study the lower semicontinuity of the multifunctions F (t , ·) and F (·, x).
E-mail address: [email protected]. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.04.051
G. Łydzińska / Nonlinear Analysis 71 (2009) 5644–5654
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The obtained results are the first step in studying the lower semicontinuity of, considered in [6–8], set-valued counterpart of (1) given by the formula
F (t , x) = A− A(x) + min t , q − inf A(x)
where A : X → 2R and A− denotes the lower preimage under A defined below. 2. Preliminaries Let X , Y be non-empty sets, and F : X → 2Y . Put F (D) :=
[
F (x)
for every D ⊂ X
x∈D
and F − (B) := {x ∈ X : F (x) ∩ B 6= ∅} for every B ⊂ Y . The following remark is clear. Remark 2. For every family {Bt : t ∈ T } of subsets of Y and for every F : X → 2Y we have
! F
−
[
=
Bt
t ∈T
[
F − (Bt ).
t ∈T
Assume that X and Y are topological spaces and F : X → 2Y . We say that F is lower semicontinuous if for every open set U ⊂ Y , the set F − (U ) is open in X . Fix an arbitrary set X and an arbitrary, not necessarily bijective, function A : X → R. Put S := A(X )
and q := sup S .
Throughout this paper we always assume that S is an interval. Now put
F (t , x) := A−1 {A(x)} + min t , q − A(x) ,
(A)
where A−1 (V ) stands for the usual preimage of V ⊂ R under A. The next observation follows immediately from the definition of F . Lemma 1 (See Also [7, Lemma 3]). Let F : (0, ∞) × X → 2X be given by (A) and let t ∈ (0, ∞) and x ∈ X . If t < q − A(x) then
F (t , x) = A−1 {A(x)} + t 6= ∅ and if t ≥ q − A(x) then F (t , x) =
A−1 ({q}),
∅
if q ∈ S , otherwise.
Theorem 1 (See [8, Theorem 3]). Every F : (0, ∞) × X → 2X , given by (A), is a set-valued iteration semigroup: F t , F (s, x) = F (s + t , x)
for every x ∈ X and s, t ∈ (0, ∞). We say that an iteration semigroup F : (0, ∞) × X → 2X is lower semicontinuous if for every x ∈ X multifunction F (·, x) is lower semicontinuous. Remark 3. Let F : (0, ∞) × X → 2X be given by (A), t ∈ (0, ∞) and V ⊂ X . If q 6∈ A(V ) then
F (t , ·)− (V ) = A−1 A(V ) − t , and if q ∈ A(V ) then F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 [q − t , q]
= A−1 A(V ) − t ∪ A−1 (q − t , q] .
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Proof. First assume that q 6∈ A(V ). Then by Lemma 1, F (t , ·)− (V ) = x ∈ X : t < q − A(x) and
= = = =
F (t , x) ∩ V 6= ∅
∪ x ∈ X : t > q − A(x) and F (t , x) ∩ V 6= ∅ x ∈ X : t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅ x ∈ X : t < q − A(x) and A(x) + t ∈ A(V ) x ∈ X : A(x) + t ∈ A(V ) A−1 A(V ) − t .
Assume now that q ∈ A(V ). Due to Lemma 1 we have F (t , ·)− (V ) = x ∈ X : t < q − A(x) and
A−1 {A(x)} + t ∩ V 6= ∅
∪ x ∈ X : t > q − A(x) and A−1 {q} ∩ V 6= ∅ h i = A−1 (−∞, q − t ) ∩ A−1 A(V ) − t ∪ A−1 [q − t , q] = A−1 A(V ) − t ∪ A−1 [q − t , q] . Since q − t ∈ A(V ) − t we get also
F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 (q − t , q] . Remark 4. Let F : (0, ∞) × X → 2X be given by (A), x ∈ X and V ⊂ X . If q 6∈ A(V ) then F (·, x)− (V ) = (0, ∞) ∩ A(V ) − A(x) ,
and if q ∈ A(V ) then
(0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ = (0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ ,
F (·, x)− (V ) =
when A(x) < q and F (·, x)− (V ) = (0, ∞), when A(x) = q. Proof. First assume that q 6∈ A(V ). Then, by Lemma 1, F (·, x)− (V ) = t ∈ (0, ∞): t < q − A(x) and F (t , x) ∩ V 6= ∅
= = = =
∪ t ∈ (0, ∞): t > q − A(x) and F (t , x) ∩ V 6= ∅ t ∈ (0, ∞): t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅ t ∈ (0, ∞): t < q − A(x) and A(x) + t ∈ A(V ) t ∈ (0, ∞): A(x) + t ∈ A(V ) (0, ∞) ∩ A(V ) − A(x) .
And now assume that q ∈ A(V ). Consider the case when A(x) < q. Then F (·, x)− (V ) = t ∈ (0, ∞): t < q − A(x) and A−1 {A(x)} + t ∩ V 6= ∅
∪ t ∈ (0, ∞): t > q − A(x) and A−1 {q} ∩ V 6= ∅ = t ∈ (0, q − A(x)): A(x) + t ∈ A(V ) ∪ q − A(x), ∞ = 0, q − A(x) ∩ A(V ) − A(x) ∪ q − A(x), ∞ h i = (0, ∞) ∩ A(V ) − A(x) ∪ q − A(x), ∞ .
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Since q − A(x) ∈ A(V ) − A(x) we get also
F (·, x)− (V ) = (0, ∞) ∩ A(V ) − A(x)
∪ q − A(x), ∞ .
Now consider the case when q ∈ A(V ) and A(x) = q. Then
t ∈ (0, ∞): t < q − A(x) and F (t , x) ∩ V 6= ∅ = ∅.
On the other hand, we have
t ∈ (0, ∞): t ≥ q − A(x) and F (t , x) ∩ V 6= ∅ = t ∈ (0, ∞): A−1 {q} ∩ V 6= ∅ = (0, ∞).
Hence F (·, x)− (V ) = (0, ∞) which completes the proof.
Now we remind the definition and a certain property of a locally connected space. A topological space X is said to be locally connected if for every point x ∈ X and for each of its neighbourhood U ⊂ X there exists a connected neighbourhood V ⊂ X of x such that V ⊂ U. The following well-known fact can be found e.g. in [9]. Fact 1. A topological space X is locally connected if and only if every connected component of every open subset of X is open. 3. Lower semicontinuity Theorem 2. Let X be a topological space, A : X → R a continuous function and let F : (0, ∞) × X → 2X be given by (A). If F (·, x) is lower semicontinuous for every x ∈ X then F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Proof. Assume that for every x ∈ X the multifunction F (·, x) is lower semicontinuous. Fix t ∈ (0, ∞) and an open set V ⊂ X . We show that F (t , ·)− (V ) is open in X . Notice that in the case when A(V ) is open in R we have q 6∈ A(V ), whence, by Remark 3 and the continuity of A, the required condition holds. Thus assume that A(V ) is not open in R, so A(V ) \ int A(V ) 6= ∅.
(2)
We show that A(V ) \ int A(V ) ⊂ {inf S , q}.
(3)
Suppose that there exists a y ∈ A(V ) \ int A(V ) such that inf S < y < q. Therefore we can find an x ∈ X such that inf S 6 A(x) < y < q. Thus 0 < y − A(x) ∈ A(V ) − A(x), y − A(x) 6∈ int A(V ) − A(x)
and y − A(x) < q − A(x). Consequently, due to Remark 4, the set F (·, x)− (V ) is not open, which contradicts our assumption. Thus (3) is satisfied. At first consider the case q 6∈ A(V ). Then, using (2) and (3), we obtain A(V ) \ int A(V ) = {inf S }, whence A(V ) = int A(V ) ∪ {inf S } and A−1 A(V ) − t = A−1
int A(V ) ∪ {inf S } − t
= A−1
int A(V ) − t ∪ {inf S − t }
= A−1 int A(V ) − t . Therefore, by Remark 3 and the continuity of A, the set F (t , ·)− (V ) is open in X . Now assume that q ∈ A(V ). If A is constant, then A(x) = q for every x ∈ X and, by Remark 3, F (t , ·)− (V ) = A−1 {q} − t ∪ A−1 [q − t , q] =
= A−1 [q − t , q] = A−1 [q − t , q] ∩ S = A−1 {q} = X ,
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so F (t , ·)− (V ) is open. Assume that A is not constant. Since q ∈ A(V ), by Remark 4, we get that (0, ∞) ∩ A(V ) − A(x)
∪
q − A(x), ∞ is open in R for every x ∈ X . Then there exists ε ∈ (0, ∞) such that (q − ε, q] ⊂ A(V ). Therefore int A(V ) − t ∪ [q − t , q + ε) is open in R. Moreover, due to (3), A−1 A(V ) − t ⊂ A−1 int A(V ) ∪ {inf S } ∪ {q} − t = A−1 int A(V ) − t ∪ {q} − t ⊂ ⊂ A−1 A(V ) − t .
Thus, according to Remark 3, we see that F (t , ·)− (V ) = A−1 A(V ) − t ∪ A−1 (q − t , q]
= A−1 int A(V ) − t ∪ [q − t , q] = A−1 int A(V ) − t ∪ [q − t , q + ε) is an open subset of X .
In the following theorem we present a necessary condition for a lower semicontinuity of F (t , ·) for every t ∈ (0, ∞), under the assumption that X is an arcwise connected space. Theorem 3. Let X be an arcwise connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Assume that F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Then
(S)
A|U ≡ inf S
or
max A(U ) = q
for every open set U ⊂ X with sup A(U ) ∈ A(U ), and
(I)
min A(V ) = inf S
for every open set V ⊂ X with inf A(V ) ∈ A(V ). Proof. At first we show that condition (I) holds. Let V ⊂ X be an open set such that y0 := inf A(V ) ∈ A(V )
(4)
and let x0 ∈ V satisfy the equality A(x0 ) = y0 . Suppose on the contrary that y0 = min A(V ) > inf S . We can find an x1 ∈ X for which A(x1 ) < y0 . Fix ε ∈ (0, 1) such that y0 −ε > A(x1 ). Since X is connected and A is continuous, there exists an xε ∈ X such that A(xε ) = y0 − ε. Let p : [0, 1] → X be a continuous function with p(0) = xε
and
p(1) = x1 ,
and let K := p [0, 1] ∩ A−1 {y0 } − ε .
Of course xε ∈ K , so K 6= ∅. Moreover, x1 6∈ K and K is compact. Let t0 := max t ∈ [0, 1]: p(t ) ∈ K ,
x := p(t0 ) and P := F (ε, ·)− (V ). By the assumption the set P is open in X . Notice that A(x) = y0 − ε ∈ A(V ) − ε , so, according to Remark 3, x ∈ P and, by the Darboux property of A ◦ p, the definition of t0 and by the inequality (A ◦ p)(1) = A(x1 ) < y0 − ε , we have
(A ◦ p)(t ) < y0 − ε
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for every t ∈ (t0 , 1]. Therefore, due to (4), for every t ∈ (t0 , 1] we get A p(t ) 6∈ A(V ) − ε and A p(t ) 6∈ (q − ε, q]. Then, by Remark 3, x ∈ P \ int P, which contradicts the openness of P and completes the proof of (I). Now we pass to the proof of (S). Let U ⊂ X be an open set such that y0 := sup A(U ) ∈ A(U ).
(5)
Assume that q 6= max A(U ). Then q 6∈ A(U ) and y0 < q. Take x0 ∈ U such that A(x0 ) = y0 . At first we show that A |U is constant. Suppose that this is not true. Therefore there exists an x1 ∈ U such that A(x1 ) 6= y0 . Then A(x1 ) < y0 = A(x0 ). Let ε := A(x0 ) − A(x1 ), p : [0, 1] → X be a continuous function such that p(0) = x1
and
p(1) = x0
and let K := p [0, 1] ∩ A−1 {y0 } − ε .
Of course x1 ∈ K , so K 6= ∅. Moreover K is compact and x0 6∈ K . Let t0 := max {t ∈ [0, 1]: p(t ) ∈ K }, x := p(t0 ) and P := A−1 (A(U ) − ε). Obviously, by Remark 3, the set P is open in X . Notice that x ∈ P and, due to the Darboux property of A ◦ p, the definition of t0 and the inequality (A ◦ p)(1) = A(x0 ) > y0 − ε , we get
(A ◦ p)(t ) > y0 − ε for every t ∈ (t0 , 1]. Then, according to (5), we have x ∈ P \ int P, which contradicts the openness of P. So we have shown that A |U ≡ const. As follows from the example below the assumption of the arcwise connectedness is essential for the validity of Theorem 3. Example 1. Let X be any set of the cardinality c, endowed with the discrete topology, and let A : X → R be an arbitrary function such that S is an interval and card S > 1. Clearly A is continuous and F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). On the other hand take any x ∈ X with inf S < A(x) < q. Obviously V := {x} is open in X . Since A |V takes the smallest and the greatest values, which are not equal to q and inf S, conditions (S) and (I) do not hold. Theorem 4. Let X be a locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). If conditions (S) and (I) hold then F (·, x) is lower semicontinuous for every x ∈ X and F (t , ·) is lower semicontinuous for every t ∈ (0, ∞). Proof. Assume conditions (S) and (I). We show that F (·, x) is lower semicontinuous for every x ∈ X . Fix x ∈ X and an open set V ⊂ X . We prove that T := F (·, x)− (V ) is open in (0, ∞). Since X is locally connected we can assume that V is connected (see Fact 1 and Remark 2). Therefore A(V ) is an interval. If A(V ) is open in R then q 6∈ A(V ), so, due to Remark 4, the set T is open. Thus consider the case when A(V ) is not open in R. Since A(V ) is an interval, we have sup A(V ) ∈ A(V )
(6)
inf A(V ) ∈ A(V ).
(7)
or
At first assume that inclusion (6) is satisfied. Then, by (S), we get A|V ≡ inf S
(8)
max A(V ) = q.
(9)
or
Assume that (9) holds. Then A(V ) = inf A(V ), q
or A(V ) = inf A(V ), q .
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If the first of the above possibilities holds, then, by Remark 4,
(0, ∞) ∩ inf A(V ) − A(x), q − A(x) ∪ q − A(x), ∞ = (0, ∞) ∩ inf A(V ), ∞ , so T is open in (0, ∞). If A(V ) = inf A(V ), q , then, due to (I), we have A(V ) = inf S , q , T =
whence, by Remark 4, T = (0, ∞). Now assume that condition (9) does not hold. Then q 6∈ A(V ) and equality (8) is satisfied. Then A(V ) = {inf S } and, consequently, Remark 4 gives A(V ) − A(x) ⊂ (−∞, 0]
T = ∅.
and
Now we pass to case (7). We may assume that (6) does not hold. Therefore A(V ) = inf A(V ), sup A(V ) .
According to condition (I) we get A(V ) = inf S , sup A(V ) .
Hence, by Remark 4, T = (0, ∞) ∩ A(V ) − A(x)
= (0, ∞) ∩
inf S , sup A(V ) − A(x)
= (0, ∞) ∩ inf S − A(x), sup A(V ) − A(x) . Therefore either T = 0, sup A(V ) − A(x)
if A(x) < sup A(V ),
or T =∅
otherwise.
Thus T is open in (0, ∞). The lower semicontinuity of F (t , ·) for every t ∈ (0, ∞) follows immediately from Theorem 2.
The next example shows that the assumption of the local connectedness of X in Theorem 4 is essential. Example 2. Let X be the sum of J = [0, 1] × {0} and the semilines (x, y) ∈ R2 : x = r and y 6 0 for every r ∈ [0, 1] ∩ Q. In the set X we consider the metric generated by the Euclidean metric on the plane. X is an arcwise connected (so connected) space, but it is not locally connected in the points (x, y) for y < 0. Let A : X → R be given by
A(x, y) = x. Obviously A is continuous. If U ⊂ X is an open set such that sup A(U ) ∈ A(U ) then 1 ∈ A(U ), so max A(U ) = 1 = q. Thus condition (S) holds. Notice also that (I) is satisfied. Let V be the open ball centered in 1/2, −1 , with the radius 1/4. In our case q ∈ [0, 1] = S, but q 6∈ 1/4, 3/4 ∩ Q = A(V ). Take t = 1/4. We have F 1/4, ·
−
(V ) = A−1 A(V ) − 1/4 = A−1 1/4, 3/4 ∩ Q − 1/4 o n = A−1 (0, 1/2) ∩ Q = (x, y) ∈ X : x ∈ 0, 1/2 ∩ Q . −
Since the points (x, y), where x ∈ 0, 1/2 ∩ Q and y = 0, do not belong to int F 1/4, · (V ), this set is not open in X . Therefore F (t , ·) is not lower semicontinuous. By Theorem 2 we get also that F (·, x) is not lower semicontinuous for some x ∈ X . A similar result can be obtained in the cases when J = [0, 1)×{0} (then q 6∈ S) or J = (x, y) ∈ R2 : x > 0 and y = 0 (then q = ∞).
The next theorem follows immediately from Theorems 2–4. Theorem 5. Let X be an arcwise connected and locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Then the following conditions are pairwise equivalent: (i) F (·, x) is lower semicontinuous for every x ∈ X ; (ii) F (t , ·) is lower semicontinuous for every t ∈ (0, ∞); (iii) conditions (S) and (I) hold.
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We can say much more in the case when A is defined on an interval. Theorem 6. Let I ⊂ R be an interval, A : I → R be a continuous function, and let F : (0, ∞) × I → 2I be given by (A). Assume that q 6∈ S. Then the interval I is not compact and the following conditions are pairwise equivalent: (i) F (·, x) is lower semicontinuous for every x ∈ I; (ii) F (t , ·) is lower semicontinuous for every t ∈ (0, ∞); (iii) the condition (U) there exist a, b ∈ R such that inf I 6 a 6 b 6 sup I, if inf I ∈ I then a = inf I , if sup I ∈ I then b = sup I and is strictly decreasing ,
A |I ∩(−∞,a) A |I ∩[a,b]
is constant ,
A |I ∩(b,∞)
is strictly increasing
holds; (iv) for every x ∈ I and t ∈ (0, ∞) card F (t , x) 6 2 and, when I is closed from one side, card F (t , x) 6 1. Before the proof of Theorem 6 we will present the following lemma. Lemma 2. Let I ⊂ R be an interval and A : I → R be a continuous function. Then (i) if I is open then (U) is equivalent to the condition: (P1) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) > A(x2 ), or A(x1 ) = A(x2 ) 6 A(x3 ), or A(x1 ) < A(x2 ) < A(x3 ); (ii) if I is closed from the left then (U) is equivalent to the condition: (P2) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) = A(x2 ) = A(x3 ), or A(x1 ) = A(x2 ) < A(x3 ), or A(x1 ) < A(x2 ) < A(x3 ); (iii) if I is closed from the right then (U) is equivalent to the condition: (P3) for every x1 , x2 , x3 ∈ I with x1 < x2 < x3 we have either A(x1 ) > A(x2 ) > A(x3 ), or A(x1 ) > A(x2 ) = A(x3 ), or A(x1 ) = A(x2 ) = A(x3 ).
(10) (11) (12)
Proof. (i) Assume that I is open and condition (U) holds. Fix the points x1 , x2 , x3 ∈ I for which x1 < x2 < x3 . Studying the form of the function A assumed in (U) it is easy to notice that one of conditions (10)–(12) is satisfied. Therefore we pass to the proof of the opposite implication. Assume (P1). Let C := y ∈ I: A(x) > A(y)
for x ∈ (inf I , y) ,
D := y ∈ I: A(y) < A(z )
for z ∈ (y, sup I )
and
a :=
inf I sup C
if C = ∅, , if C = 6 ∅
b :=
sup I inf D
if D = ∅, if D = 6 ∅.
Obviously inf I 6 a and b 6 sup I. Notice also that if u ∈ C and v ∈ D then u ≤ v . Consequently, a ≤ b.
(13)
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Assume that I ∩ (−∞, a) 6= ∅. Then a > inf I and C 6= ∅. Fix x1 , x2 ∈ I ∩ (−∞, a) = (inf I , a) such that x1 < x2 and suppose that A(x1 ) 6 A(x2 ).
(14)
By the definition of a there exists an x3 ∈ (x2 , a] ∩ I such that x3 ∈ C . Thus A(x2 ) > A(x3 ), which, together with inequality (14), contradicts condition (P1). Thus A is strictly decreasing in the interval I ∩ (−∞, a). Similarly we can prove that A is strictly increasing in I ∩ (b, ∞). Now we will show that A |I ∩[a,b] is constant. Assume that a < b and fix x1 , x2 ∈ (a, b) such that x1 < x2 . Suppose that A(x1 ) 6= A(x2 ), for instance A(x1 ) < A(x2 ).
(15)
Since x2 < b we have x2 6∈ D, and thus there exists an x3 ∈ (x2 , sup I ) for which A(x2 ) > A(x3 ), which, due to (15), contradicts assumption (P1). Similarly, in the case A(x1 ) > A(x2 ) there exists a point x0 ∈ (inf I , x1 ) such that A(x0 ) 6 A(x1 ), contrary to (P1). Thus A |I ∩[a,b] is constant. (ii) Let I = [c , d), c < d where c ∈ R and d ∈ R. As earlier it is easy to notice that in this case the condition (U) implies (P2). Now assume that condition (P2) holds. Let D be given by (13) and
a := inf I
and
b :=
sup I inf D
if D = ∅, if D = 6 ∅.
Obviously a 6 b. Assume that I ∩ (b, ∞) 6= ∅. Fix x1 , x2 ∈ I ∩ (b, ∞) = (b, d), x1 < x2 and suppose that A(x1 ) > A(x2 ).
(16)
According to the definition of the element b there exists an x0 ∈ [b, x1 ) ∩ I such that x0 ∈ D, and thus A(x0 ) < A(x1 ). Hence by (16) we get a contradiction with condition (P2). Thus A is strictly increasing in the interval I ∩ (b, ∞). Now assume that a < b. We show that A |(a,b) is constant, which due to the continuity of A will complete the proof of (ii). Suppose that there exist x1 , x2 ∈ (a, b), x1 < x2 for which A(x1 ) 6= A(x2 ). By the fact that x2 6∈ D we can find a point x3 ∈ (x2 , sup I ) such that A(x2 ) > A(x3 ), which contradicts condition (P2). The proof of assertion (iii) is similar.
Now we pass to the proof of Theorem 6. Proof of Theorem 6. At first we show that in the considered case condition (iii) and the conjunction of (S) and (I) are equivalent which, by Theorem 5, will complete the proof of the pairwise equivalence of conditions (i), (ii) and (iii). Studying the form of the function A assumed in (U) it is easy to observe that (iii) implies conditions (S) and (I). Passing to the proof of the opposite implication suppose that condition (iii) does not hold. At first consider the case when the interval I is open. Then, by Lemma 2, there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and either A(x1 ) < A(x2 )
and
A(x2 ) > A(x3 ),
(17)
or A(x1 ) = A(x2 ) > A(x3 ),
(18)
A(x1 ) < A(x2 ) = A(x3 ).
(19)
or
Let U := (x1 , x3 ). Since A is continuous, each of the above conditions implies that sup A(U ) ∈ A(U ) and max A(U ) > A(x2 ). Thus A|U 6≡ inf S, whence, since q 6∈ S, condition (S) is not satisfied.
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Now let I = [c , d), where c ∈ R and d ∈ R, c < d. Due to Lemma 2 condition (P2) does not hold, whence there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and one of conditions (17)–(19) or A(x1 ) > A(x2 )
(20)
is satisfied. Let U1 := [c , x3 ) and U2 := [c , x2 ). Of course U1 and U2 are open in I. Since A is continuous, each of conditions (17)–(19) implies that sup A(U1 ) ∈ A(U1 ), max A(U1 ) > A(x2 )
and
A|U1 6≡ inf S .
Moreover, if (20) holds then sup A(U2 ) ∈ A(U2 ), max A(U2 ) > A(x1 )
and
A|U2 6≡ inf S .
Thus (S) is not satisfied. When I is closed from the right the proof is similar. Now we will prove the equivalence of the conditions (iii) and (iv). First observe that, by Lemma 1, for every t ∈ (0, ∞) and x ∈ I we have card F (t , x) = card PA(x)+t , where Pk denotes the set consisted of the common points of a line y = k, k ∈ R and the graph of A. Obviously A(x) + t > inf S. Thus, looking at the form of A we observe that (iii) ⇒ (iv). Now assume that (iii) does not hold. We will apply Lemma 2 and in each of its case (i)–(iii) we will find a number k > inf S for which card Pk is greater than the cardinality of F (t , x) allowed by (iv); as stated in the preceding paragraph, this will complete the proof. At first consider the case when I is an open interval. Then, due to Lemma 2, there exist x1 , x2 , x3 ∈ I such that x1 < x2 < x3 and one of conditions (17)–(19) holds. Since q 6∈ S, there exists an x4 ∈ I \ (x1 , x3 ) for which A(x4 ) > max A [x1 , x3 ] . Assume that (17) is satisfied. Let
k ∈ max A(x1 ), A(x3 ) , A(x2 ) .
Obviously k > inf S. Hence, by the continuity of A and the choice of x4 , we get card Pk > 3. Now assume condition (18). If A is constant in the interval [x1 , x2 ] then let k = A(x1 ) and observe that k > inf S and Pk is infinite. Consider the case when A takes a locally maximum at a point x5 ∈ (x1 , x2 ) and A(x5 ) > A(x1 ). Let k ∈ A(x1 ), A(x5 ) .
Obviously k > inf S. Thus, by the continuity of A and the property of x4 , card Pk ≥ 3. Now assume that A has a locally minimum at a point x5 ∈ (x1 , x2 ) and A(x5 ) < A(x1 ). Take
k ∈ max A(x3 ), A(x5 ) , A(x1 ) .
Of course k > inf S and, as A is continuous, card Pk > 3. If we assume condition (19), then the proof is similar. Consider the case when I = [c , d), where c ∈ R and d ∈ R, c < d. Then, by Lemma 2, there exist x1 , x2 , x3 ∈ I, x1 < x2 < x3 , satisfying one of conditions (17)–(20). Since q 6∈ S, there exists an x4 ∈ (x3 , d) such that A(x4 ) > max A [x1 , x3 ] . Observe that if (18) or (19) holds then, taking k = A(x2 ), we obtain k > inf S and card Pk > 2. Assuming condition (17) let
k ∈ max A(x1 ), A(x3 ) , A(x2 ) .
Then k > inf S and, by the continuity of A, we get Pk > 2. Finally, if (20) holds then, by the choice of x4 , we get card Pk > 2 for any k ∈ A(x2 ), A(x1 ) . In the case when I is closed from the right the proof is similar. In particular, Theorem 6 shows that in the case when q 6∈ S the generators of lower semicontinuous iteration semigroups of the form (A) are not far from the bijections and those iteration semigroups are close to single-valued functions. In higher dimensions the situation seems to be more complicated. In the following example we show that if q 6∈ S then a lower semicontinuous iteration semigroup not always has such small values as in condition (iv) of Theorem 6. Example 3. Let X = R2 and let A : X → R be defined by A(x, y) = x2 + y2 . 2
Obviously S = [0, ∞) and q = ∞. Consider function F : (0, ∞) × R2 → 2R given by (A). Then, by Lemma 1,
F t , (x, y) = A−1 {A(x, y)} + t = A−1 ({x2 + y2 } + t ) p = T (0, 0), x2 + y2 + t , where T ((0, 0), r ) stands for the circle centered at (0, 0) and with the radius r. By Theorem 1 the multifunction F is an iteration semigroup. Of course A is continuous and the conditions (S) and (I) hold. Therefore, due to Theorem 5, F (t , ·) is lower semicontinuous for every t ∈ (0, ∞) and F (·, x) is lower semicontinuous for every x ∈ X .
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The next example shows that the assumption ‘‘q 6∈ S’’ is essential for the validity of Theorem 6. Example 4. Let A : R → R be given by A(x) = sin x. Then S = [−1, 1]. Of course the conditions (S) and (I) hold. Let F : (0, ∞) × R → 2R be defined by (A). Due to Theorem 5 the multifunction F (t , ·) is lower semicontinuous for every t ∈ (0, ∞) and F (·, x) is lower semicontinuous for every x ∈ X . Nevertheless condition (iii) fails to be true. As follows from Example 1 the assumption of the arcwise connectedness is essential for the validity of Theorem 3 and, consequently, Theorem 5. Nevertheless, without any assumption imposed on the topological space X , the lower semicontinuity of the multifunctions F (·, x), x ∈ X , implies conditions (S) and (I). Theorem 7. Suppose that X is a topological space. Let A : X → R and F : (0, ∞) × X → 2X be given by (A). If F (·, x) is lower semicontinuous for every x ∈ X then conditions (S) and (I) hold. Proof. At first suppose that condition (S) does not hold, so q 6∈ A(U ) and A |U 6≡ inf S
(21)
for some open set U ⊂ X with sup A(U ) ∈ A(U ). Let y = max A(U ). Then, by (21), we have inf S < y. Choose an x0 ∈ X in such a manner that A(x0 ) < y.
(22)
Thus 0 < y − A(x0 ) ∈ A(U ) − A(x0 ) whence, due to Remark 4 and the inclusion y ∈ A(U ) \ int A(U ), we obtain that the set F (·, x0 )− (U ) is not open in (0, ∞), which contradicts the assumption. Now suppose that V ⊂ X is an open set such that A |V takes the smallest value and min A(V ) > inf S .
(23)
Let y = min A(V ). According to (23) we have inf S < y. Take an x0 ∈ X satisfying (22). By Remark 4 and the inequality y 6 q we get 0 < y − A(x0 ) = min F (·, x0 )− (V ). Hence F (·, x0 )− (V ) is not open in (0, ∞), which contradicts the assumption of the lower semicontinuity of F (·, x0 ).
As a direct consequence of Theorems 7 and 4 we have the following characterization of the lower semicontinuity of the multifunctions of form (A). Theorem 8. Let X be a locally connected space, A : X → R be a continuous function and let F : (0, ∞) × X → 2X be given by (A). Then F (·, x) is lower semicontinuous for every x ∈ X if and only if conditions (S) and (I) hold. Remark 5. Observe that the space X considered in Example 1 is locally connected. Therefore, as follows from Theorem 8, Example 1 yields a multifunction F of form (A) for which the multifunctions F (t , ·), t ∈ (0, ∞), are lower semicontinuous, whereas F (·, x), x ∈ X , are not. Acknowledgment This research was supported by the Silesian University Mathematics Department (Iterative Functional Equations and Real Analysis program). References [1] [2] [3] [4] [5] [6] [7] [8] [9]
M. Kuczma, Functional equations in a single variable, Monografie Mat. 46, PWN, Warszawa 1968. M.C. Zdun, Continuous and differentiable iteration semigroups, Pr. Nauk. Uniw. Śl. Katow. 308 (1979). Gy. Targonski, Topics in iteration theory, Vandenhoeck and Ruprecht, Göttingen, 1981. M. Kuczma, B. Choczewski, R. Ger, Iterative functional equations, in: Encyclopedia of Math. and Its Applications, vol. 32, Cambridge University Press, 1990. G. Guzik, W. Jarczyk, J. Matkowski, Cocycles of continuous iteration semigroups, Bull. Polish Acad. Sci. Math. 51 (2003) 187–197. G. Łydzińska, Iteration semigroups for which expansion implies collapse, Aeqationes Math. 70 (2005) 247–253. G. Łydzińska, On collapsing iteration semigroups of set-valued functions, Publ. Math. Debrecen 64 (2004) 285–298. G. Łydzińska, On expanding iteration semigroups of set-valued functions, Math. Pannon. 15 (2004) 55–64. K. Kuratowski, Introduction to set theory and topology, Pergamon Press, London, 1961.