Basic Topological Structures of Ordinary Differential Equations (Mathematics and Its Applications)

= {Xo:: a E A} of subsets of a topological space X converges with respect to

not converge to the set H in the sense of (7.1). This means that for some neighborhood OH of H in every set F k , k = 1,2, ... , there exists an element ak such that XO: k \ OH -# 0. The sequence {ak: k = 1,2, ... } is cofinal to the set

= {Xa: a E A} of subsets of a regular space X converge in the sense of (7.1) with respect to a directed set

~ H. Proof. Our aim will be achieved when we show that every point x E X \ H does not belong to the set lim top sup'!' .

Topological and metric spaces.

29

Take disjoint neighborhoods Ox and OH of the point x and of the set H. By (7.1) there exists an element F of the set l{J such that Xc>: ~ OH for of every index a E F. Therefore Xc>: n Ox = 0. So x 1. lim top sup'!' . The theorem is proved. • Theorem 7.2. Let = {Xc>:: a E A} be a generalized sequence of subsets of a compactum X. Then the sequence converges (with respect to l{J) in the sense of (7.1) to the set H = lim top sup'!' . Proof. Let OH be an arbitrary neighborhood of the set H in the compactum X. Then every point x of the set X \ OH does not belong to the set lim top sup'!' . Therefore there exist a neighborhood of it Ox and a set F(x) E l{J such that for every index a E F(x) the intersection Xc>: n Ox is empty. By Theorem 6.2 the closed set X \ OH is compact. Its cover {Ox: x E X \ OH} has a finite subcover {OXI, ... , Oxd. By virtue of the directed ness of the set l{J there exists an element F of it lying in the intersection of the sets F(XI), ... , F(xd. If a E F then Xc>: n OXi = 0 for every i = 1, ... , k. Therefore Xc>: n (X \ OH) ~ Xc>: n (U~=I OXi) = U~=I (Xc>: n OXi) = 0. So Xc>: ~ OH. The theorem is proved. • Thus in the case of a compact space the convergence of a generalized sequence with respect to l{J in the sense of (7.1) to a closed set H is equipotent to the inclusion lim top sup'!' ~ H. This means, in particular, that for a multi-valued mapping F : X ----t Y with closed values of a topological space X into a compactum Y the upper semicontinuity of the mapping F at a point x E X is equipotent with the inclusion lim tOpSUPt-+x F(t) ~ F(x). By analogy with a base and a sub-base at a point a family f3 of open subsets of a topological space X is called a base (respectively, a subbase) of neighborhoods of a set H ~ X if every element of f3 contains the set H and for each neighborhood OH of the set H in the space X there exists an element U of the family f3 such that U ~ OH (respectively, there are elements UI , . . . ,Uk of the family f3 such that UI n ... n Uk ~ OH). Lemma 7.3. Let = {Xc>:: a E A} be a generalized sequence of subsets of a topological space X. Let f3 be a sub-base of neighborhoods of a set H ~ X. The sequence converges to the set H with respect to a directed set l{J in the sense of (7.1) if and only if for every element U of the sub-base f3 there exists a set F E l{J such that Xc>: ~ U for every index aEF. Proof. Necessity follows immediately from definitions. Sufficiency. Take an arbitrary neighborhood OH of the set H in the space X. By the definition of a sub-base there are sets UI , ... ,Uk E f3 such that UI n· .. n Uk ~ 0 H. For every i = 1, ... , k fix an element Fi of the set l{J such that Xc>: ~ Ui for every index a E F i . By virtue of the directedness of the set l{J there exists element F of it lying in the intersection FI n ... n F k •

30

CHAPTER 1

For every index a E F we have: XCI. <;;;; UI n··· n Uk <;;;; OH. The lemma is proved. • This lemma implies: Theorem 7.3. Let F : X -+ Y be a multi-valued mapping of a topological space X into a topological space Y, f31 be a base at a point x EX, f32 be a sub-base of neighborhoods of the set F(x) in Y. The mapping F is upper semicontinuous at x if and only if for every V E f32 there exists U E f31 such that F(U) <;;;; V. • In the case of a single valued mapping the notion of the continuity at a point coincides with the notion of the upper semicontinuity at the point. Therefore all assertions about upper semicontinuous (at a point) mappings may be immediately applied to continuous single valued mappings. In the case of a metric space £-neighborhoods of a point constitute its base. Therefore from Theorem 7.3 we obtain Corollary. A mapping f : X -+ Y of a metric space (X, PI) into a metric space (Y, P2) is continuous at a point x E X if and only if for every £ > 0 there exists 8 > 0 such that if t E X and PI (x, t) < 8, then P2 (f (X ), f (t )) < £. • In this way we usually state the definition of the continuity at a point of a real function of a real argument (see in Example 4.1 the metric in the real line). A mapping of a topological space X into a topological space Y is called continuous if it is continuous at every point of the space X. A multi-valued mapping of a topological space X into a topological space Y is called upper semicontinuous if it is upper semicontinuous at every point of the space X. We say that a mapping f of a metric space (X, PI) into a metric space (Y, P2) satisfies Lipschitz condition with Lipschitz constant a ~ 0 if P2(f(s),f(t)):::;; apI(s,t) for every pair of points s,t of the space X. Every such mapping is continuous. This follows easily from the Corollary of Theorem 7.3. We put there 8 = £/(a + 1) for arbitrary £ > O. Example 7.2. Let d be a metric on the product of two real lines, which is mentioned in Example 4.5. Let the mapping

l
+ tl·ls -

tl :::;; 2als - tl·

Topological and metric spaces.

31

In particular, all functions in Examples 7.2-4 are continuous. A mapping of a metric space into a metric space is called contractive if it satisfies the Lipschitz condition with a constant being less than 1. A point x of a set X is called a fixed point of a mapping f : X ---t X if f(x) = x.

Theorem 7.4. Every contractive mapping of a complete metric into itself space possesses a fixed point and such a point is unique. Proof. Let a mapping f of a complete metric space (X, p) into itself satisfy the Lipschitz condition with a constant a < 1. Take an arbitrary point Xl of the space X. For k = 2, 3, . .. define sequentially the points Xk by the formula Xk = f(Xk-d· Because 0 < a < 1, a k - l ---t 0 as k ---t 00. Since P(Xb Xk+i)

:s; :s;

+ P(Xk+l, Xk+2) + ... + P(Xk+i-l, Xk+i) P(Xk, Xk+l) + ap(xk' xk+d + ... + a i - l P(Xk, xk+d P(Xk, xk+d

1 - ai

= --P(Xk, Xk+l)

:s;

I-a a --P(Xk-I,Xk) I-a

:s;

1

--P(Xk' xk+d I-a

:s; ... :s;

a k-

l

--P(XI,X2) I-a

for every k = 1,2, ... and i = 1,2, ... , the sequence {Xk: k = 1,2, ... } is fundamental. Since the space X is complete this sequence converges to a point X E X. The sequence {f(Xk): k = 1,2, ... } is obtained from the sequence {Xk: k = 1,2, ... } by rejection of the first member. Therefore the limits of these sequences coincide. By Corollary of Lemma 7.1 or by Lemma 7.2 f(x) = limk--+oo f(Xk) = limk--+oo Xk = x, i.e., x is a fixed point of the mapping f. We have proved the existence of a fixed point. Prove its uniqueness. Let t be a fixed point of the mapping f too. As p(x, t) = p(f(x), f(t)) :s; ap(x, t), where 0 :s; a < 1, then p(x, t) = 0 and x = t. The theorem is proved. • Theorem 7.5. Let F : X ---t Y be a multi-valued mapping of a topological space X into a topological space Y. Then the following conditions are equipotent: a) the mapping F is upper semicontinuous, b) for every open set V of the space Y the set {x: x E X, F(x) S;;; V} is open in X, c) for every closed set H of the space Y the set F-I(H) is closed in X. Proof. a{:::}b. This follows immediately from the definitions. b{:::}c. This follows from Theorem 3.3. The theorem is proved. • Theorem 7.6. Let a multi-valued mapping F : X ---t Y of a topological space X into a topological space Y be upper semicontinuous on every element of a finite closed cover of the space X. Then the mapping F is upper semicontinuous (on the entire space X).

32

CHAPTER 1

Proof. Let I be the finite closed cover of the space X. The words 'F is upper semicontinuous on H(E I)' mean that the mapping F IH is upper semicontinuous. Our assertion follows from Theorem 7.5c, Theorem 3.4, the obvious equality

F-1(M) =

U{(FIH)

-1

(M): H

EI} ,

and properties of the induced topology. The theorem is proved. • Theorem 7.7. Let a multi-valued mapping F : X -> Y of a topological space X into a topological space Y be upper semicontinuous on every element of an open cover of the space X. Then the mapping F is upper semicontinuous (on the entire space X). Proof. The proof is analogous to the proof of the previous theorem .• With reference to Theorems 7.6 and 7.7 it is necessary to notice that a restriction of an upper semicontinuous mapping to a subspace is upper semicontinuous with respect to the induced topology. In addition a composition of upper semicontinuous mappings is upper semicontinuous. This may be easily deduced right from definitions or with the help of Theorem 7.5. Theorem 7.8. Let F : X -> Y be an upper semicontinuous multi-valued mapping of a compact space X into a topological space Y. Let for each point x E X the set F(x) be compact. Then the set F(X) is compact. Proof. Let I be an arbitrary cover of the set F(X) by open subsets of the space Y. For every x E X fix a finite subfamily Ix of the cover I, which covers the set F(x). By Theorem 7.5b the set Ox = {t: t E X, F(t) ~ U is open in X. Evidently the set contains the point x. Thus we have an open cover {Ox: x E X} of the compact space X. Select its finite sub cover {OXl,'''' Oxd. As F(OXi) ~ U for every i = 1, ... , k, then F(X) = U~=lF(OXi) ~ U~=l (U'xJ. Hence the finite subfamily U~=llXi of I covers the set F(X). This gives what was required. • Naturally this result may be applied to single valued mappings too: A continuous image of a compact space is a compact space. We can use the latter observation to give an example of an upper semicontinuous multivalued mapping which is remote from continuous single valued mappings. Corollary. Let f : X -> Y be a continuous (single valued) mapping of a compact space X into a Hausdorff space Y. Then the (multi-valued) mapping F : Y -> X, F(y) = f-1(y), is upper semicontinuous. In order to obtain the corollary we need use Theorem 7.5c. Let H be an arbitrary closed subset of X. By Theorem 6.2 the set H is compact. By Theorem 7.8 the set F-1(H) = f(H) is compact too. We obtain from Theorem 6.1 the closedness of the set F-1(H), what was required. • Finish this section by a small remark. We will have an opportunity to use it below.

,X}

,X;

Topological and metric spaces.

33

Lemma 7.4. Let J and g be continuous (single valued) mappings oj a topological space X into a Hausdorff space Y. Then the set (oj points of coincidence of the mappings f and g) M = {x: x E X, J(x) = g(x)} is closed in X. Proof. Take an arbitrary point t E X \ M. Since the points f(t) and g(t) are different and the space Y is Hausdorff, there exist disjoint neighborhoods U and V of the points J(t) and get), respectively. By virtue of the continuity of the mappings f and g there exists a neighborhood at of the point t such that f(Ot) ~ U and g(Ot) ~ V. Evidently at n M = 0. Because of the arbitrariness of the point t E X \ M this implies the openness of the set X \ M. Hence the set M is closed. The lemma is proved. • 8. Zorn's lemma and well ordered sets A relation -< between elements of a set M is called a partial order or partial ordering on the set M, if for some couples x, y E M we say that x precedes y (or y succeeds x) with respect to -< and we write x -< y (or y »- x), moreover: if x -< y and y -< z, then necessarily x -< Z; x -< y and y -< x if and only if x = y. In this case the set M is called partially ordered. We say that a set N ~ M is linearly ordered by the relation -< if for every two its elements x, y: either x -< y, or y -< x. An element x of a partially ordered set M is called maximal (resp., minimal) if there is no element of the set M, which succeeds (resp., which precedes) x and is not equal to x. A subset N of a partially ordered set M is called bounded if there exists an element of the set M which succeeds all elements of the set N. The following assertion is one of fundamental facts of mathematics, see [MEl or [En, §I.4]. Lemma 8.1 (Zorn). If every linearly ordered subset of a partially ordered set M is bounded then the set M contains a maximal element. • The relation of the inclusion ~ is an example of a partial order on the set of subsets. In the connection with this partial order a linearly ordered subset N ~ M is called maximal (linearly ordered subset of the set M) if it is not contained in another linearly ordered subset (different from N) of the set M. Lemma 8.2. Let N be a linearly ordered subset of a partially ordered set M. Then there exists a maximal linearly ordered subset Nl of set M containing N: N ~ Nl ~ M.

CHAPTER 1

34

Proof. Consider the set X of all linearly ordered subsets of the set M containing N and the partial order ~ on X. Every linearly ordered subset a of the set X is bounded (recall that we consider the partial order ~: the element Ua of the set X succeeds all elements of the set a). We obtain what was required from Zorn's lemma. A linearly ordered set M is called well ordered if every nonempty subset N of it possesses a minimal element (with respect to the order) minN E N. Example 8.1. The segment [0,1] of the real line with its natural order is not well ordered: its subset (0,1] does not possess a minimal element (0 tJ. (0,1]). Theorem 8.1 (Zermelo). Every set may be made well ordered. • Proof. Let X be an arbitrary set. Consider a family r of all couples (M,,), where M is a subset of X and, is a well ordering on M. Let us partially order the set r: we put (Ml"d -< (M2' '2) if Ml ~ M2; the order is a restriction to Ml of the order ,2 and Xl -< X2 with respect to the order ,2 for every Xl E Ml and X2 E M2 \ M l · The set r is nonempty: take as M an arbitrary one point subset of the set X; we need not define the order, in this case. By Lemma 8.2 there exists a maximal linearly ordered subset ~ of the set r. Let Xl = U{M: (M,,) E ~ for some ,}. Introduce the following linear order on the set Xl: if Xl, X2 E Xl then Xl, X2 E M for some (M,,) E ~, and we put Xl -< X2 if this is true with respect to the order,. As orders, for (M,,) E ~ are compatible our definition of the relation Xl -< X2 does not depend on the choice of (M,,) E ~. Show that the set Xl is well ordered by this relation. Let a set N ~ Xl be nonempty. Take an arbitrary Xl E N. Fix an arbitrary set M :7 Xl being the first member in a couple (M,,) E ~. The set Nl = N n M is nonempty. Since the set M is well ordered by " there exists Xo = min N l . Then Xo -< X for every X E Nl by virtue of the inclusion Nl ~ M and Xo -< X for every X E N \ M by virtue of our definition of the partial order on ~. Show that Xl = X. If this is not true then the set X \ Xl is nonempty. Take arbitrary Xl E X \ Xl' Make the set M = Xl U {xd well ordered (by an order ,) by putting X -< Xl for every X E Xl' On the one hand (M,,) E ~. On the other hand M g Xl' This contradiction gives what was required. The theorem is proved. • A one to one mapping

,1

,1

Topological and metric spaces.

35

case two isomorphic partially ordered sets may be related by several isomorphisms (for instance, for the real line every mapping ip.x(x) = AX with ). > 0 is an isomorphism). For well ordered sets this is not so. Assertion 8.1. Let ipl,ip2 : (M1,')'d ---t (M2 ,')'2) be two isomorphisms of well ordered sets. Then ipl == ip2· Proof. Let M = {x: x E M 1 , ipl(X) = ip2(X)}. The set Mis nonempty: it contains min M. Our aim is to prove that M = M 1 . If this is not so then the set N = Ml \ M is nonempty. Since the mappings ipl and ip2 are isomorphisms, we have ipl(minN) = ip2(minN). Then minN E M, which contradicts the definition of N. The assertion is proved. • Equivalence classes of well ordered sets with respect to the relation introduced of isomorphicity are called ordinal number.. Evidently the cardinalities of all specimens of every such class are equal. This gives the possibility of speaking about finite ordinal numbers. Here the situation is simpler than in the general case: if the cardinalities of two finite well ordered sets coincide then they are isomorphic. The reader may prove it as a simple exercise. Infinite ordinal numbers are called transfinite. Here the situation is more complicated. Example 8.2. The set of positive integers N and the set N U {oo} with their natural orders are well ordered and countable, but they are not isomorphic. We make ordinal numbers well ordered when we put a -< (3 if for some Ml E a and M2 E (3 and for some x E M2 there exists an isomorphism ip : Ml ---t {t: t E M 2 , t -< x} (evidently an analogous isomorphism ip and x exist for every specimens Ml E a and M2 E (3). Show that this order is linear, i.e., every two ordinal numbers a and (3 are comparable. Fix Ml E a and M2 E (3 and denote by P the set of all of elements s E Ml satisfying the condition:

(8.1) there exists an element t E M2 such that the well ordered sets Nl = {x x E M 1 , X -< s} and N2 = {y: Y E M 2 , Y -< t} are isomorphic. Assertion 8.1 implies the existence of the unique isomorphism of the set P onto an initial part Q of the set N 2 . Let ip be the corresponding isomorphism. If one of the sets Ml \ P or M2 \ Q is empty then we have what was required. In the opposite case we extend the isomorphism ip to the set PU{min(Ml \P)} by putting ip(min(Ml \P)) = min(M2 \Q). This means that min(Ml \ P) E P. This is impossible. We have what was required. Evidently because of this observation ordinal numbers become well ordered.

36

CHAPTER 1

Below we will use remarks in this section only in cases of an extreme necessity. These results do not give explicit descriptions of the orders mentioned. By Theorem 8.1 the segment X = [0,1] may be well ordered. The reader can try to describe such an order in an explicit form. But what follows from the possibility of making the set X well ordered? The set X is uncountable. Therefore if M is a countable well ordered set then for some t E X the set M isomorphic to the set {y: y E X, Y -< t}. This means that for some t E X the set n of all at most countable ordinal numbers is isomorphic to the set {y: y E X, Y -< t}. The ordinal number corresponding to the set of positive integers N is denoted by Wo. The ordinal number corresponding to the set n is denoted by WI. Evidently the set n is uncountable; see an analogous situation in Example 8.l. If a is an arbitrary element of a well ordered set M then the element a + 1 = min{ t: t >- a, t i= a} of the set M succeeds directly a. In this case the element a of the set M precedes directly the element a + 1 in the meaning that there is no elements of the set M lying between a and a + l. An element of a well ordered set is called isolated if a directly preceding element exists. The opposite situation is possible too. We mean for some element {J of the set M there is no element a, which directly precedes the element {J, i.e., for every a -< {J, a i= {J, there exists an element I i= a, {J of the set M lying between a and {J: a -< I -< {J. In this case the ordinal number (transfinite) {J is called limit. Remark 8.1. If a transfinite {J < WI is limit then it is the limit of a monotonically increasing (with respect to the order) sequence {JI -< {J2 -< {J3 -< ... : The set B = {a: a -< {J, a i= {J} is countable and we can enumerate its elements: B = {ai: i = 1,2, ... } (we forget temporarily the order -<). The set B = {al} U {ai: i = 2,3, ... , al, ... , ai-l -< ad is infinite. The order -< coincides on B with the order due to the numeration. After an obvious renumbering of its elements (with preservation of the order, see §1) we obtain the needed sequence {JI -< {J2 -< {J3 -< ....

CHAPTER 2

SOME PROPERTIES OF TOPOLOGICAL, METRIC AND EUCLIDEAN SPACES

In this chapter we continue our acquaintance with topological and geometric notions which will be used in the later account. 1. Tychonoff topology of products

Let {Xc>: a E A} be a family of topological spaces. In §1.1 we have defined a product of sets. Our nearest aim here is the introduction of a topology on the set X = IT{Xc>: a E A}. Notice first that we can do it in various ways. We will define a so called Tychonoff topology of the product, which corresponds best to our aims. Let Ao be an arbitrary finite subset of the index set A. For every a E Ao fix an open subset Uc> of the space Xc> and put

oc> =

{Uc> X c>

if a E A o, if a E A \ Ao.

The set IT {Oc>: a E A} is called a Tychonoff neighborhood (of each of its points). If Mc> and M~ are subsets of the set Xc> for every a E A then

Therefore the intersection of every pair of Tychonoff neighborhoods is a Tychonoff neighborhood also. The set X is also a Tychonoff neighborhood. So the set of all Tychonoff neighborhoods satisfies conditions A and B of §1.2. Hence it is a base of an uniquely defined topology on the set X. The topology is called the Tychonofftopology of the product IT{Xc>: a E A}. It is easier to understand geometric particularities of a topological space when we have a simply described base of the space or bases at its points. In this connection let us notice that if for every a E A a base f3c> of the space Xc> (respectively, a base f3c> at a point Xc> E Xc» is fixed, then the family of all Tychonoff neighborhoods satisfying (with the above notation)

38

CHAPTER 2

the condition Uo. E f3o. is a base of the space X (respectively, a base at the point x = {xo.: a E A} EX). For every a E A let Yo. be a (nonempty) subspace the space Xo.. It is natural to consider the product Y = IT {Yo.: a E A} as a subset of the product X. So we have two topologies on the set Y. The first is the topology 71 which is induced on Y by the topology of the space X. The second is the Tychonoff topology 72 of the product IT {Yo.: a E A}. The formula (1.1) and the definition of the induced topology imply that if 0 is a Tychonoff neighborhood of the product X then the set 0 n Y is a Tychonoff neighborhood of the product Y. Every Tychonoff neighborhood in Y may be represented in the form of such an intersection. So the same base generates on Y both the topology 71 and the topology 72. Hence these topologies coincide. Let B be a subset of the index set A. The product X B = IT {Xo. : a E B} is called subproduct of the product IT {X a: a E A}. If the set B consists of one element a usually we speak not about a subproduct but about a factor Xo.. We may regard the subproduct X B as 'embedded' in the product. To define an 'embedding' for every a E A \ B we fix a point x~ E Xo.. For a point y = {Yo.: a E B} E X B we now put i(y) = {xo.: a E A}, where Xo. = Yo. for a E Band Xo. = x~ for a E A \ B. Our definition of an 'embedding' is not unique. It depends on the choice of the points x~ for a E A \ B. The mapping PB : X ----+ X B , PB({X",: a E A}) = {x",: a E B}, is called a projection. The projection is defined uniquely. For the projection of the product X = IT {X",: a E A} on the factor X{3 the preimage of an open subset U of the factor X{3 has the form of a Tychonoff neighborhood IT {O",: a E A}, where 0{3 = U and 00. = Xo. for a E A \ {f3}. Therefore Theorem 1.7.5b implies

Lemma 1.1. A projection of a product onto a factor is continuous . • Lemma 1.2. LetX = IT{X{3: f3 E B} be a product of topological spaces X{3, f3 E B, {xo. = {xo.{3: f3 E B}: a E A} be a generalized sequence of points of the space X. Let for every f3 E B the generalized sequence {xo.{3 : a E A} con verge with respect to a directed set

Some properties of topological, metric and Euclidean spaces.

39

mapping f is continuous if and only if for every a E A the 'coordinate' mapping fa = Paf is continuous. Proof. Necessity follows from Lemma 1.1 and from the fact that the composition of continuous mappings is continuous too. Sufficiency follows from Lemma 1.2. The theorem is proved. • When we compare Lemma 1.1 and Theorem 1.1, we obtain Corollary. A projection of a product of topological spaces onto every is subproduct is continuous. • It is natural to describe a mapping into a product by the definition of coordinate mappings (we do so, for instance, when we introduce a vector function with values in the three-dimensional space by writing separate formulae for the dependence on the argument of X-, Y- and z-coordinates). The inverse passage is possible too: Assume that for a E A a mapping fa : X --t Y a is fixed. The mapping f : X --t IT {Ya: a E A}, f(x) = {fa (X): a E A} is called a product (or a diagonal product) of the family {fa: a E A}. Theorem 1.1 implies: a diagonal product of continuous mappings is also a continuous mapping. Now let factors in the product be metric spaces. Does this cause any particularity in the questions under consideration? In particular, whether Tychonoff topology of the product is generated by a metric related in a natural way with metrics on factors? We will give an affirmative answer to the last question under some additional assumptions. Let diam M denote the diameter sup{p( s, t): s, t E M} of a subset M of a metric space (X, p): diam M = sup{p(s, t): s, t EM}. Consider the following situation:

(1.2) (Xb Pk), k

= 1,2, ... , is a sequence of metric spaces and diamXk

--t

O.

Under these assumptions define the distance d on the set X = IT {X k : k = 1,2, ... }. For x = {Xk: k = 1,2, ... } and Y = {Yk: k = 1,2, ... } put d(x, y) = SUP{Pk(Xk, yd: k = 1,2, ... }. Our assumptions imply that only a finite number of elements of the sequence {PdXk' yd: k = 1,2, ... } may be greater than 1. Therefore values of the function d are finite. The fulfilment of conditions 1 and 2 of the definition of a metric in §1.4 is obvious. Check the fulfilment of the triangle inequality. Let in addition Z = {Zk: k = 1,2, ... }. For every k = 1,2, ... Pk(Xk, Zk) ~ Pk(Xk, Yk)

+ pdYk, Zk)

~ d(x, y)

+ d(y, z).

Since the last expression does not depend on k, d(x, z)

= SUP{Pk(Xk, zd: k = 1,2, ... } ~ d(x, y) + d(y, z).

Which gives us what was required.

40

CHAPTER 2

Theorem 1.2. Under the condition (1.2) the Tychonoff topology of the product X = IT {X k: k = 1, 2, ... } is generated by the above metric d. Proof. I. Let a set U ~ X be open in the Tychonoff topology and x = {Xk: k = 1,2, ... } E U. By the definition of the Tychonoff topology there exists a number i = 1,2, ... and for every k = 1, ... , i there exists a neighborhood Vk of the point Xk in the space X k such that

I1{Vk : k=l, ... ,i}xI1{Xk : k=i+1,i+2, ... }~U. For every k = 1,2, ... , i fix a number [(k) > 0 such that OE(k)Xk ~ Vk. Put [ = min{E(l), ... , [(i)}. It remains to notice that

O"X ~ I1{O"Xk: k = 1, ... ,i} x I1{X k : k = i + 1,i + 2, ... } ~I1{Vk: k=l, ... ,i}xI1{Xk : k=i+1,i+2, ... }~U. In view of the arbitrariness of the point x E U the last fact means the openness of the set U with respect to the metric d. II. Take an arbitrary point x = {Xk: k = 1,2, ... } E X and an arbitrary number [ > O. Find a number i such that diam X k < [ for every k = i + 1, i + 2, .... For k = i + 1, i + 2, ... we have OcXk = X k. By virtue of our definition of the metric d on the product

O"X=I1{O"Xk: k=l, ... ,i}xI1{Xk : k=i+1,i+2, ... }, i.e., [-neighborhood of a point with respect to the metric d has the form of a Tychonoff neighborhood. So if a subset of the product is open with respect to the metric d, then it is open with respect to Tychonoff topology too. When we compare I and II we obtain that the Tychonoff topology on X coincides with the topology generated by the metric d. The theorem is proved. • In the next section we will show that the additional condition diam X k ---7 o for the existence of some metric on the product is inessential, but the above concrete description of the metric on the product will not appropriate. Every finite family of spaces may be extended to a countable family by taking as missing elements a one point space (a metric on it can be defined in a unique way). These additional elements change nothing in the consideration of the Tychonoff topology as well as in the introduction of the metric d. The condition diamX k ---7 0 is obviously fulfilled for the family completed in this way. Therefore the theorem proved may be automatically restated to a finite number of factors. In addition, we can propose to the reader as a simple exercise to revise the given reasoning, to reject conditions

Some properties of topological, metric and Euclidean spaces.

41

and remarks related with the infiniteness of the number of factors, and to give (an easier) direct proof of the existence of a metric on the product of a finite family of metric spaces. Example 1.1. The metric on the space JRn in Example 1.4.5 coincides with the metric introduced on this set as on the product of lines according to Theorem 1.2. Example 1.2. Let continuous real functions f and 9 be defined on a topological space X. By Theorem 1.1 the mapping : X --+ JR x JR, (x) = (f(x),g(x)), is continuous. Since the composition of continuous mappings is continuous, the remarks of Examples 1.7.2 and 1.1 imply the continuity of the function h( x) = f (x) + g( x). Example 1.3. The remarks of Example 1.7.4 and Theorem 1.7.7 imply the continuity of the function cp : JR --+ JR, cp( t) = e. Since

the remarks of the previous example imply the continuity of the mapping 'lj; : JR x JR --+ JR, 'lj;(x, y) = xy. The reasoning of the previous example now allows us to prove that under the same assumptions the function hI (x) = f(x)g(x) is continuous. Thus if a function can be constructed with help of the operations of addition and multiplication of continuous functions, then it is continuous. This remark and Lemma 1.1 imply the continuity of every mapping \]I : JRm --+ JRn with polynomials as coordinate functions (we consider the spaces with the metric of Example 1.4.5). Theorem 1.3. Let the hypotheses of Theorem 1.2 hold. Let the spaces Xb k = 1,2, ... , be compact. Then the product X is compact. Proof. Since by the previous theorem the space X is metric, to prove its compactness it is sufficient to check the fulfilment of condition (1.6.3). Let I be an arbitrary sequence of elements of the space X. We will choose sequentially subsequences of the sequence I' First we put 10 = I' Let the sequence Ik-I be constructed. Our next step is the construction of a sequence Ik. We do it as follows. Consider the sequence a of k-th coordinates of elements of Ik-I' Since the space X k is compact the sequence a contains a subsequence al converging to a point Yk E X k • We take as Ik the subsequence of the sequence Ik-I consisting of elements of Ik-I' the k-th coordinates of which belong to the sequence al' Now construct a subsequence 1* = {Xi; : j = 1,2, ... } of the sequence I' As Xit we take an arbitrary element of the sequence 11' Let elements Xi t , . . . ,Xi;_t be chosen. As Xi; we take an arbitrary element of the sequence Ij with the index > i j - I . Elements of 1* with indices > k belong to the sequence Ik' Therefore the k-th coordinates of the elements of 1* constitute

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42

a sequence converging to the point Yk. By Lemma 1.2 the sequence --y* converges to the point Y = {Yk: k = 1,2, ... }. The theorem is proved. • Theorem 1.4. Let X be a compactum and Y be an arbitrary topological space. Then the mapping F : Y -+ X x Y, F (y) = X x {y}, is upper semicontinuous. Proof. The proof use Theorem 1. 7.5. Let H be an arbitrary closed subset of the product X x Y and Y E Y\F- 1 (H). The last fact means that the intersection (X x {y} ) n H is empty: (X x {y }) n H = 0. Therefore for every x E X we can fix a Tychonoff neighborhood Ox x Vx of the point (x, y) in the product X x Y which does not meet the set H. The cover {Ox: x E X} of the compactum X contains a finite subcover {OX1,"" Oxd. Put Oy = n {Vx i l . . . , VXk } . Every point x E X belongs (at least) to one of the sets OXi' Therefore {x} x Oy <;;; OXi X VXi <;;; (X X Y) \ H. In view of the arbitrariness of the point x E X we have:

(X x Oy) n H = 0,

Oy n F- 1(H) =

0,

y ~ [F-l(H)].

Since the point y E Y \ F- 1 (H) was taken arbitrarily, the last fact means the closedness of the set F- 1 (H). The theorem is proved. • Corollary. The product of two compacta is a compactum. To obtain the corollary it is sufficient to remember Theorem 1.7.8. •

2. Continuous bijections. Homeomorphisms. Metrizability A continuous bijection is called an homeomorphism if its inverse mapping is continuous too. Let f : X -+ Y be an homeomorphism of a topological space X onto a topological space Y. We have one to one correspondence between elements of these two sets. Under these assumptions a set M <;;; X is open in the space X if (by virtue of the continuity of the mapping f) and only if (by virtue of the continuity of the mapping f-l) the set f(M) is open in the space Y (see Theorem 1.7.5). In other words, our one to one correspondence is extended to topologies of these spaces (with preservation of all relations of inclusion and belonging). This means that every fact which is stated in these terms and which is true for one of these homeomorphic spaces, is true for the other too. So the existence of an homeomorphism means a topological equivalence, an identity of topological properties of the spaces. For instance, if one of these spaces is Hausdorff, then the second one is Hausdorff too. If one of these spaces is compact, then the second one is compact too, etc .. We say that a space X homeomorphic to a space Y if there exists an homeomorphism of the space X onto the space Y. An identity mapping is an homeomorphism. Thus every space is homeomorphic to itself. Since a mapping that is inverse of an homeomorphism is an homeomorphism also,

Some properties of topological, metric and Euclidean spaces.

43

the relation of homeomorphism is symmetric: if a space X is homeomorphic to a space Y then the space Y is homeomorphic to the space X. Therefore usually we say more briefly that the spaces X and Yare homeomorphic. The composition of two homeomorphisms is an homeomorphism. Therefore (in view of the previous remark) if two spaces are homeomorphic to a third then they are homeomorphic to each other. A mapping f : X ---7 Y is called an embedding if when we consider it as a mapping of the space X onto the space f(X) it become an homeomorphism. This word (in inverted commas) was used in the previous section.

Example 2.1. Let real numbers a and b be different. Remarks at the end of the previous section imply the continuity of the function f(t) = (1 - t)a + tb and of its inverse function g(x) = (x - a)j(b - a). The segment [0, 1] is mapped by the function f onto the segment I with endpoints a and b (i.e., I = [a, b] if a < b, and I = [b, a] if a > b). The point goes into the point a and the point 1 goes into the point b. The mapping of the segment [0, 1] and its inverse mapping are the restrictions of the continuous mappings f and g. Thus they are continuous. They are homeomorphisms. We have proved: every (nondegenerate) segment of the real line is homeomorphic to the segment [0,1]. Corollary: Any two segments of the real line are homeomorphic.

°

Example 2.2. Every pair of intervals of the real line are homeomorphic. The proof for intervals with finite endpoints is analogous to the reasoning of the previous example. The function tan is an homeomorphism of the interval ( - ~, ~) onto the line ~ (the inverse function arctan is continuous too). The function f (t) = a + e t realizes an homeomorphism of the line onto the interval (a, 00). The function g(t) = a-e t realizes an homeomorphism ofthe line onto the interval (-00, a) (the inverse functions h±(t) = In(±(x - a)) are continuous too). Example 2.3. A segment is not homeomorphic to an interval: A segment is compact and an interval is not compact. Example 2.4. Affine transformations are continuous (see the end of the previous section). So affinely equivalent subspaces of the plane are homeomorphic. Thus any two triangles are homeomorphic, and the boundaries of any two triangles are homeomorphic. Example 2.5. Not every continuous bijection is an homeomorphism. Let (X, T) be an arbitrary topological space. Apart from the topology T let us consider the discrete topology Td on the set X. The identity mapping of the space (X, Td) onto the space (X, T) is continuous. This follows immediately from the definitions. It is an homeomorphism (i.e., its inverse mapping is continuous too) if and only if the topology T is discrete. Since not every topology is discrete, we obtain the required example.

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Theorem 2.1. A continuous bijection of a compact space onto a Hausdorff space is an homeomorphism. Proof. The proof reduces to referring to Corollary of Theorem 1. 7.S .• Often the presence of additional structures simplifies the investigation of topological spaces. We may see this in the previous account where we studied both a metric and a topology. Notice that in corresponding results as a rule we made conclusions about topological properties of spaces under consideration, i.e., the metric was not mentioned in the description of these properties. In such situations properties of any particular metric were not important and only the existence of a metric generating the topology in question played any role. A topological space (X, r) is called metrizable if there exists a metric on the set X generating the topology r. If there exists a one to one correspondence f between a set X and a metric space (Y, p), then we can define a metric d on the set X by putting d(s, t) = p(f(s), f(t)) for s, t EX. If in addition f is an homeomorphism of a topological space (X, r) onto the metric space (Y, p) then the metric d generates the topology r. Therefore a topological space which is homeomorphic to a metric space is metrizable. The inverse fact is obvious: a metrizable space is homeomorphic to a metric space. We speak here not about the existence of some metric on the given set. Such a metric always exists (see Examples 1.4.3 and 1.4.9, where a metric generating the discrete topology is pointed out). In Example 2.5 we have shown even that every space is the image of a discrete space under a continuous bijection. We discuss here the question of when, for a given topology, we can find a metric generating just this topology. So the discrete topology is metrizable (see Examples 1.4.3 and 1.4.9), although the initial definition of the discrete topology is not related to any metric. Example 2.6. The space in Example 1.6.3 is metrizable if and only if it is (at most) countable. In a metric space every point has an (at most) countable base, for instance the base consisting of its i-neighborhoods, k = 1,2, .... Thus every point may be represented as the intersection of an (at most) countable family of open sets. In our case the complement of every neighborhood of the point a (the notation of Example 1.6.3) is finite. Therefore the point a may be represented as an intersection of a countable number of open sets only in the case when the space X is countable. Consider the case of a countable space X. Enumerate elements of the set A : A = {ak: k = 1,2, ... }. Define the mapping f of the space X into the real line by putting f(a) = 0 and f(ad = for k = 1,2, .... The continuity of the mapping f follows easily from the definition of the topology of the space X. Evidently the mapping is injective. By Theorem 2.1 the space is homeomorphic to the subspace f(X) of the real line. The case of a finite space X is trivial. Such a space is discrete. Remark 2.1. Let for metrics PI and P2 on a set X there exist positive

i

Some properties of topological, metric and Euclidean spaces. numbers a and

!3

45

such that for every s, t E X

Then the topologies generated by the metrics PI and P2 coincide. In fact, the condition mentioned may be rewritten in the form of the system P2 ( s, t) ~ ~PI (s, t) { PI (s, t) ~ - P2 (s, t).

a

The first inequality of the system means the fulfilment of the Lipschitz condition for the identity mapping (X, PI) - t (X, P2). The second one means the fulfilment of the Lipschitz condition for the inverse mapping. Thus the identity mapping is an homeomorphism, as was required. In Example 1.4.2 we have introduced the Euclidean metric P in the space ]Rn. In Example 1.4.5 we have introduced on the same set the metric d. By virtue of Theorem 1.2 the metric d generates the Tychonoff topology in the space ]Rn when we consider it as the product of lines. These metrics are related by the inequality (2.1)

d(s, t) ~ p(s, t) ~ vnd(s, t).

Remark 2.1 implies the coincidence of the generated topologies, i.e.: Remark 2.2. The Euclidean topology on ]Rn coincides with the topology of the Tychonoff product. Remark 2.3. Let (X, PI) be a metric space, a > o. For s, t E X put P2(S,t) = min{PI(s,t),a}. Evidently the function P2 is a metric. The metrics PI and P2 generate the same topology on the set X: the family {{s: SEX, PI(S,t) < c}: 0 < c < a} is a base at the point t E X both for the topologies of the space (X, PI) and of the space (X, P2). By this remark the condition diamX k - t 0 in Theorems 1.2 and 1.3 is not very essential, i.e., we can change the metrics on the factors in a manner that the condition turns out to be fulfilled and the topologies are not changed. Therefore Theorem 1.2 implies: Theorem 2.2. A product of an {at most} countable family of metric spaces is metrizable. • Theorem 1.3 implies Theorem 2.3. A product of an {at most} countable family of metric compacta is a metric compact. • Example 2.7 'Hilbert cube'. Hilbert cube is the countable power of the segment. By Theorem 2.3 it is a metric compactum. Example 2.8. The space D consisting of the two points 0 and 1 (with the discrete topology) is called a two-point space. By Theorem 2.3

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46

its countable power DN is compact. Show that the space DN is homeomorphic to the Cantor perfect set. Use the description of the Cantor perfect set in Example 1.6.4. For k = 1,2, ... define the function ik in the following way. Let Sk-l([O,l]) be the union of disjoint segments [ai, bi ], i = 1, ... ,2 k- 1 (SO([O,l]) = [0,1]). Put the function fk equal to 0 on every set K n [ai, ai + t(b i - ai)] and equal to 1 on every set K n [ai + }(bi - a;), bd, i = 1, ... ,2 k - l . Each of these sets is closed in K. The function fk is constant on it. Hence the function is continuous there. Theorem 1.7.6 now implies the continuity of the function ik. Theorem 1.1 implies the continuity of the diagonal product f of mappings fk' k = 1,2, .... We propose that the reader, as a simple exercise, shows that the pre image of every point of the space K under the mapping f consists just of one point. Referring to Theorem 2.1 completes the reasoning. 3. Some properties of a metric Lemma 3.1. Let A be a nonempty subset of a metric space (X,p). Then Ip(x, A) - p(y, A)I ::::; p(x, y) for every points x, y of the space X Proof. I. Prove first that

p(p, A) ::::; p(p, q)

+ p(q, A)

for every points p, q EX. For tEA we have p(p, A) ::::; p(p, t) ::::; p(p, q) + p(q, t). When we pass to the lower bound with respect to tEA in the right hand side of this inequality we obtain the estimate needed. II. If we substitute in the inequality of! first p = x, q = y and next p = y, q = x, we obtain p(x, A) - p(y, A) ::::; p(x, y), p(y, A) - p(x, A) ::::; p(x, y). This gives what was required. • Corollary 1. Let (X, p) be a metric space. Let Sl, 82, t l , t2 EX. Then

because

+ (p(s2,td - P(S2,t2))1 P(S2' tdl + Ip(S2' td - P(S2' t 2)1

Ip(Sl,td - P(S2,t 2)1 = I(P(Sl,td - P(S2,t l )) ::::; Ip(Sl' t l ) ::::; P(Sl' S2) + P(tl' t2)'

Corollary 2. Let A be a nonempty subset of a metric space (X, p). Then the function cp(t) = p(t, A) is continuous on X. Because by Lemma 3.1 the function cp satisfies the Lipschitz condition with the constant 1. •

Some properties of topological, metric and Euclidean spaces.

47

Lemma 3.2. With the notation of Corollary 2 of Lemma 3.1 the set A is closed in X if and only if A = '11-1(0). Proof. The proof is obvious (see Remark 1.4.1). • By analogy with c-neighborhoods of points: for c > 0 and a nonempty subset M of a metric space (X, p) the set OoM = O(M,c) = {t: t E X, p(t,M) < c} is called c-neighborhood of the set M. Let also O,(M,c) = {t: t E X, p(t,M) ~ c} for c ~ O. Corollary 2 of Lemma 3.1 implies immediately: Lemma 3.3. With the above notation the set O(M, c) is open and the set O,(M, c) is closed in the space X. • The definition implies: Lemma 3.4. A point x of a metric space (X, p) belongs to the c-neighborhood of a nonempty subset A of the space X if and only if p(x,t) < c for a point tEA. In other words Oo(A) = U{Oot: tEA}. • Lemma 3.4 and the triangle inequality imply: Lemma 3.5. Let A and B be nonempty subsets of a metric space X. Let a and f3 be positive numbers and B ~ Oo:A. Then O{3B ~ Oo:+!3A. • Lemma 3.6. Let A and B be nonempty subsets of a metric space X and the set B be compact. Then there exists a positive number a such that B ~ Oo:A. Proof. The family, = {OoA: c > O} is an open cover of the space X. By virtue of the compactness of the set B the family , contains a finite subfamily {OOI A, ... , OOk A}, which cover B too. Evidently a = max {Cl, ... , Ck} satisfies the above conditions. The lemma is proved .• A subset M of a metric space X is called bounded if there are a point x E X and a number c > 0 such that M ~ Oox. Remark 3.1. The diameter of every bounded set is finite. Remark 3.2. If M is a subset of finite diameter of a metric space X and x is an arbitrary point of the space X, then M ~ Oox for some c > o. Proof of these assertions reduces to a simple verification with the help of the triangle inequality. They imply, in particular, that the conditions of the boundedness of the set and of the finiteness of its diameter are equipotent. On the other hand if we take a one point set as A in Lemma 3.6 we obtain: Lemma 3.7. Every compact subset of a metric space is bounded. • Lemma 3.5 implies: Remark 3.3. c-neighborhood (c > 0) of a bounded set is a bounded set too. Assertions of §1.6 and Lemma 3.7 imply: Remark 3.4. Let a real function f be defined and be continuous on a compact space X. Then -00 < inff(X) ~ supf(X) < 00 and {inff(X), sup f(X)} ~ f(X).

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As a particular case of Remark 3.4 we obtain: Remark 3.5. Let a positive function f be defined and be continuous on a compact space X. Then < inf f(X) ~ supf(X) < 00 and {inf f(X), sup f(X)} ~ f(X). A distance between nonclosed disjoint (nonempty) sets may be equal to zero. This is sufficiently obvious. We have already noticed that if a point x belongs to the closure of a set M, then p(x, M) = 0. The distance between two closed disjoint sets may be zero too, see below Example 4.2. However: Lemma 3.8. Let A and B be nonempty closed disjoint subsets of a metric space (X, p). Let the set B be compact. Then p(A, B) > and there exists a point b E B such that p(A, b) = p(A, B). Proof consists in referring to Corollary of Lemma 3.1 (with the notation of Lemma 3.1) and to Remark 3.5 (with f = 0. In other words, [-neighborhoods of a compact subset of a metric space constitute a base of its neighborhoods. • A topological space is called normal if it is Hausdorff and every pair of disjoint closed subsets of it possess disjoint neighborhoods. Evidently every normal space is regular. Lemma 3.9. Every metric space is normal. Proof. Let F1 and F2 be closed disjoint subsets of a metric space (X, p). The function g : X ~ ]R,

°

°

p(t, Fd - p(t, F 2 ) g () t = '-'------'----'--'------'p(t, Fd + p(t, F2)'

is continuous (see Corollary 2 of Lemma 3.1 and §1). The sets OF1 = g-1(( -00,0)) and OF2 = g-1((0, (0)) are disjoint neighborhoods of the sets F1 and F 2, respectively (see Theorem 1.7.5).

The lemma is proved.

•

4. Some properties of Euclidean, locally compact, and normed spaces

The metric of the Euclidean space ]Rn is defined in Example 1.4.2, see also Remark 2.2 and Example 1.1. Lemma 4.1. The set 0,(0, r), where r > 0, is compact. Proof. The inclusion 0,(0, r) ~ [-r, r]n (~ ]Rn) is obvious. The set [-r, r]n is compact by virtue of the compactness of the segment [-r, r] and Theorem 1.3. Lemma 3.3 and Theorem 1.6.2 imply what was required. The lemma is proved. •

Some properties of topological, metric and Euclidean spaces.

49

Theorem 4.1. A subset of an Euclidean space is compact if and only if it is closed and bounded at once. Proof. Sufficiency follows from remarks of the previous section, Lemma 4.1 and Theorem 1.6.2. Necessity follows from Theorem 1.6.1 and Lemma 3.7. The theorem is proved. • This theorem immediately implies that the closure of every bounded subset of an Euclidean space is compact, because this closure is a bounded set too. A topological space is called locally compact if every point possesses a neighborhood the closure of which (in the space under consideration) is compact. Evidently every compact space is locally compact. Every open subspace of a compactum is locally compact. The remark following Theorem 4.1 implies the local compactness of open subsets of the Euclidean space. So the following assertion is applicable directly to an arbitrary open subset U of the Euclidean space. Lemma 4.2. Let U be an arbitrary neighborhood of a nonempty compact subset B of a locally compact metric space X. Then there exists a number b > 0 such that the set 0f(B, b) (~O(B, b)) is compact and lies in U. Proof. Show first that there exists a neighborhood V of the set B such that the set [V] is compact and lies in U. According to the local compactness of the space X associate to every point b of the set B its neighborhood Wb such that the set [Wb ] is compact. According to the normalness of the space X (see Lemma 3.9) associate to each point b E X of the set Bits neighborhood Vb and a neighborhood Nb of the set Fb = X \ (Wb n U) such that Vb n Nb = 0. Then [Vb] n Fb = 0, [Vb] S;;; un W b. By virtue of the inclusion [Vb] S;;; Wb the set Vb is compact. The cover {Vb: b E B} of the compactum B has a finite subcover {Vb" ... , Vb.}. The set V = Vb, U·· ·UVb• lies in the compactum

that gives what was required. Our assertion now follows from Corollary of Lemma 3.8 (with V as the set U of the corollary) and from Lemma 3.3 .• Later we will use: Assertion 4.1. Let X be a locally compact metric space and M be its (at most) countable nonempty closed subset. Then set M has an isolated (in M) point. Proof. Take an arbitrary point p E M. By virtue of the local compactness of the space X for some c > 0 the set 0f(P,c) is compact. The set G = O(p, c) n M is open in the compactum K = O,(p, c) n M. It is (at most) countable. Enumerate its elements: G = {Xk: k = 1,2, ... }. Apply Baire's theorem 1.6.7 to the open subset G of the compact urn K. At least

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50

one of the sets {xd, k = 1,2, ... , has a nonempty interior. Let {xd be such a set. Since the set {xd consists of one point, the nonemptiness of its interior means that ({ xd) = {xd. Hence the set {xd is open (in K). Thus the point Xi is isolated (in G ~ K and hence it is isolated in M). The assertion is proved. • Corollary. Let U be an open subset of the space IRn. Let M be an (at most) countable nonempty closed (in U) subset of the set U. Then the set M has an isolated point (in M). • Let X and y be two arbitrary points of a vector space L. The set {tx + (1 - t)y: 0 ~ t ~ I} is called a segment (connecting the points x and y. The points x and yare called endpoints of the segment). A subset M of the space L is called convex, if for every two its points the set M contains the segment connecting the points. In order to investigate and to use properties of convex sets we will need the following notion. A function f : L ----t IR is called a linear functional, if f(u + v) = f(u) + f(v) and f(au) = af(u) for every elements u, v of the space L and for every number a. A very simple example of a linear functional on the space IR n is the function ai, i = 1, ... ,n, which associates to an element x = (Xl,' .. , x n ) of the space IRn its i-th coordinate Xi' This example may be generalized in the following way. Fix an arbitrary vector e = (al,"" an) E IRn. We define the linear functional

alxl

+ ... + anx n·

In fact, every linear functional f : IRn ----t IR may be represented in such a form. For i = 1, ... ,n let aj = f(O, . .. ,0,1,0, ... ,0), where the unity is at the i-th position. We obtain

This implies, in particular, the continuity of the functional f, i.e., every linear functional on the space IRn is continuous. When we introduce the notion of a metric space we take as a pattern subsets of the line, of the plane and of three-dimensional space. However not all properties of such sets have analogs in an arbitrary metric space. We are closer to the usual geometry when instead of a metric we consider a norm. A real function which is defined on a vector space L and which associates to a vector u E L a number denoted by lIull, is called a norm, if: 1) Ilull > 0 for every nonzero vector u E L; 2) Ilaull = lal . Ilull for every vector u ELand every number a; 3) Ilu + vii ~ lIull + Ilvll for every two vectors u, vEL. The number lIull is called the norm of the vector u. A pair consisting of a vector space and a norm on the space is called a normed space. Length of vectors in Euclidean space is a norm, i.e., it satisfies the listed conditions.

Some properties of topological, metric and Euclidean spaces.

51

In a normed space we introduce a metric as follows. For u, vEL we put p(u, v) = Ilu - vii. The verification that the function p is a metric is easy. Thus every normed space may be considered as a metric space. We can apply to normed spaces the terminology and theory elaborated for metric spaces. Lemma 4.3. Let M be a (nonempty) convex subset of a normed space. Then for every

f

>

°

the set OEM is convex.

Proof. Let p and q be arbitrary points of the set OeM. By Lemma 3.4 there are points x and y of the set M such that Ilx - pil < f and Ily - qll < f. Take now an arbitrary point a of the segment connecting the points p and q. The point a may be represented in the form tp + (1 - t)q, t E [0,1]. By virtue of the convexity of the set M the point b = tx + (1 - t)y belongs to the set M. On the other hand,

Iia - bll

+ (1 - t)(q - y)11 xii + (1 - t)llq - yll

= Ilt(p - x) ~ tllp -

< tf + (1 =

t)f

f.

Hence a E OEM. The lemma is proved. • Lemma 4.4. The intersection of every family of convex sets is a convex set.

Proof. The proof is obvious: A segment connecting two arbitrary points of the intersection lies in every element of the family. Therefore it lies in the intersection. The lemma is proved. • Lemma 4.5. The closure of a convex subset of a normed space is a convex set.

Proof. Let M be a (nonempty) convex subset of the normed space under consideration. The representation [M] =
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points of the stated form is convex. This may be easily proved by a direct verification. Thus the convex hull of a set M coincides with the set of all points of the form alxl + ... akXk, where Xl, ... ,Xk E M, al,· .. ak are non-negative number and al + ... + ak = 1. (Such linear combinations are called convex combinations.) Evidently the convex hull of a convex set coincides with the set. By virtue of Lemma 4.5 and the equality cc(M) = [c(M)] its closed convex hull coincides with its closure. Let A and B be subsets of a metric space (X, p). A point a E A is said to be nearest to the set B, if p(a, B) = p(A, B). The point b of the set B in Lemma 3.8 is nearest to the set A. Notice that in the definition we do not require the uniqueness of a nearest point. Lemma 4.6. Let A and B be nonempty closed subsets of the space JRn. Let the set B be compact. Then the set A has a point that is nearest to the set B. Proof. Take an arbitrary number c > p(A, B). The distance from points of the set A \ OcB to the set B is not less than c. Since p(A, B) < c, p(A, B) = p(A n [OcB], B). The set An [OcB] is compact. By Lemma 3.8 there exists a point a E An [OcB] such that p(a, B)

= p(A n [OcB], B) = p(A, B).

The lemma is proved. • Lemma 4.7. Let a segment I connect points p and q in the space JRn. Let X E JRn. Then the function r.p: I ~ JR, r.p(t) = p(t,x), takes its maximal value M at p or at q (or both at p and at q) and values of the function r.p at other points of the segment I are strongly less than M. Proof. Let 0 :::;; s :::;; 1 and t = sp + (1 - s)q = q + s(p - q). Then r.p(t) =

lit - xii

= V(q - x

= V(t - x, t - x)

+ s(p -

= V(q - x, q - x)

q), q - x

+ 2s(q -

+ s(p -

q))

X,p - q)

+ S2(p -

q,p - q).

The last subradical expression as a function on s has the form of a square trinomial. The coefficient in S2 is positive. Therefore the maximum of this expression occurs at one of the endpoints (or at both endpoints) of segment [0,1], i.e., for s = 0 or 1. Respectively, the maximal value of the function r.p occurs at t = P or at q. The lemma is proved. • Corollary. Let A be a non empty closed convex subset of the space JRn and x E ~n. Then there exists a unique point of set A being nearest to the point x.

Some properties of topological, metric and Euclidean spaces.

53

The existence of such a point is proved in Lemma 4.6. Let us prove its uniqueness. Assume that the set A has two different points p and q nearest to the point x. By Lemma 4.7 the segment connecting the points p and q (it lies in the set A) has a point that is nearer to the point x than the points p and q. This is impossible by the definition of a nearest point. • Example 4.1. Consider the segments A = {O}x[O, 1] and B = {1} x [0, 1] of the plane 1R2 (Figure 2.1). Every point of the set A is nearest to the set B and every point of the set B is nearest to the set A.

B

A

A Figure 2.1

Figure 2.2

Example 4.2. Consider the x-axis A and the set B = {(x, y) : x > 0, y ~ ~} (Figure 2.2) of the plane 1R2. Both of them are closed and convex. The distance between the sets A and B is equal to zero. Since they are disjoint the set A has no nearest point to the set B and the set B has no nearest point to the set A. Theorem 4.2. Let A and B be nonempty closed convex disjoint subsets of the space IRn. Let the set B be compact. Then there exist a linear functional f : IR n --+ IR and real numbers p < q such that f(A) ~ (-oo,p] and f(B) ~ [q, (0). Proof. Let a point a E A be nearest to the set B. Let a point b E B nearest to a (the existence of such points is guaranteed by Lemma 4.6). For u E IR n put f(u) = (u, b - a). 1. Let x E A. The point a E A is nearest to the set B. The segment with the endpoints a and x lies in A. Thus for s E (0,1]

II(sx + (1 ((a (a - b, a -

bll ~ p(sx + (1 - s)a, B) ~ p(a, B) = Iia - bll, II(a - b) + s(x - a)11 ~ Iia - bll, b) + s(x - a), (a - b) + s(x - a)) ~ (a - b, a - b), b) + 2s(a - b, x - a) + S2(X - a, x - a) ~ (a - b, a - b), 2(a - b, x - a) + s(x - a, x - a) ~ O.

- s)a) -

As it is true for every s E (0,1], then (a - b, x - a) f(x) ~ f(a).

~

0, (x, b- a)

~

(a, b- a),

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54

II. Let y E B. The point b E B is nearest to the point a. The segment with the endpoints band y lies in B. As in I for s E (0,1]

II(b 2(y - b, b -

+ s(y - b) II = IIsy + (1 a) + s(y - b, y - b) ~ 0, (y a)

s)b - all ~

lib -

all,

b, b - a) ~ 0, f(y) ~ f(b).

III. Since (b - a, b - a) > 0, f(b) > f(a). Put p = f(a) and q = f(b). By I and II all required conditions hold. The theorem is proved. • The just proved Theorem 4.2 is also a consequence of the Hahn-Banach theorem, see below §5.2. 5. Remarks on multi-valued mappings

Lemma 5.1. Let

final to

°

space X into a metric space Y and x EX. Then the following conditions are equipotent: a) the mapping F is upper semicontinuous at the point x and the set F(x) is compact;

Some properties of topological, metric and Euclidean spaces.

55

b) for every sequence Xk ---+ x of points of the space X every sequence of points Yk E F(Xk), k = 1,2, ... has a subsequence converging to a point of the set F(x). Proof. a~b. This follows from Lemma 5.1 because the point x possesses a countable base of neighborhoods, for instance, the base {O( x, 2-k) : k=1,2, ... }. b ~a. We can consider the stationary sequence Xk == x. When we apply the condition b to this sequence we obtain the compactness of the set F(x). The rest follows from Lemma 5.1. The theorem is proved. • Lemma 5.2. Let cp be a decreasing sequence Fl "2 F2 "2 F3 "2 ... of nonempty subsets of a set A. A generalized sequence IP = {Xa: a E A} of subsets of the space lRn converges with respect to cp to a nonempty convex compact set H ~ lRn in the sense of condition (1.7.1) if and only if H "2 n { cc (H U (u { X a: a E F})): F E cp}. Proof. Necessity. By Lemma 4.3 for every c > 0 the set O£H is convex. For some F E cp H U (U{Xa: a E F}) ~ O£H. By virtue of the convexity of the set O£H we have c(H U (U{Xa: a E F})) ~ O£H. Therefore cc(HU(U{Xa: a E F})) ~ [c(HU(U{Xa: a E F}))] ~ [OcH], n{cc(H U (U{Xa: a E F})): F E cp} ~ [O£H] (for every c > 0). This gives what was required. Sufficiency. Fix an arbitrary point x E H. Assume that the generalized sequence IP does not converge to the set H in the sense of (1. 7.1). By Corollary of Lemma 3.8 there exists a number c > 0 such that for every k = 1,2, ... there exists an index ak E Fk such that X ak \ O£H i 0. Fix a point Xk E X ak \ O£H. The segment connecting the points x and Xk lies in the set cc(H U (U{Xa: a E Fd)). Let tk = sup{s: S E [0,1]' SXk + (1 - s)x E O£H} (recall: the point x belongs and the point Xk does not belong to the set O£H). The point x~ = tkXk + (1 - tk)x belongs to the closure of the set O£H. It does not belong to this set itself. Hence the point lies in the boundary. The boundary of the set O£H is a closed bounded set. By Theoren_ 4.1 it is compact. Therefore the sequence {x~: k = 1,2, ... } has a limit point x* E O£H. We have x' ~ Hand X* E n{ cc(H U (U{ Xc> : a E F})): F E cp} at once. This contradicts our assumptions. Thus our assumption is false and the sequence IP converges to H with respect to cp in the sense of (1.7.1). The lemma is proved. • n Theorem 5.2. Let F : X ---+ lR be a multi-valued mapping of a metric space X into the space lRn , x EX. Let the set F (x) be nonempty compact and convex. Then the following conditions are equipotent: a) the mapping F is upper semicontinuous at the point x; b) F(x) ::2 n{ cc(F(O£x)): c > o}. Proof. The proof consists in referring to Lemma 5.2.

•

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56

Lemma 5.3. Let generalized sequences {D",: a E A} and {E", : a E A} of subsets of a normal space X converge in the sense of (1. 7.1) with respect to a directed set

n E",

~

OD

n OE

= (U U V)

n (U U W)

= U U (V

n W)

= UU0 =

u.

This gives what was required. The lemma is proved. • Lemma 5.3 implies immediately: Theorem 5.3. Let multi-valued mappings F, G, with closed values, of a topological space X into a normal space Y be upper semicontinuous (respectively, upper semicontinuous at a point x EX). Then the multi-valued mapping H : X ---+ Y, H(t) = F(t) n G(t), is upper semicontinuous (respectively, upper semicontinuous at the point x). • 6. Connectedness Every topological space has at least two sets which are both simultaneously open and closed. These are the set which coincides with the entire space and the empty set. A topological space X is called connected if it has no proper (i.e., different from X and 0) open-closed (i.e., both open and closed) subsets. In the opposite case, i.e., if (6.1) the space X has a proper open-closed subset, the space X is called disconnected. Denote the proper open-closed subset of the condition (6.1) by F and put G = X \ F. By Theorem 1.3.3 (6.1) implies the conditions (6.2) the space X may be represented as the union of two nonempty disjoint open subsets of it (namely, F and G), and (6.3) the space X may be represented as the union of two nonempty disjoint closed subsets of it

Some properties of topological, metric and Euclidean spaces.

57

(namely, F and G, moreover). By Theorem 1.3.3 condition (6.2) as well as (6.3) implies the fulfilment of condition (6.1). Thus conditions (6.1), (6.2) and (6.3) are equipotent. We have defined the connectedness and the disconnectedness of a topological space. As we usually do with all topological notions we allow one to speak about the connectedness and the disconnectedness of subsets and subspaces of topological spaces too. We then mean the induced topology. Example 6.1. A space consisting of one point is connected. Example 6.2. A segment [a, b], a < b, of the real line is a connected space. Assume the opposite. Then some proper subset M of it is openclosed. Take an arbitrary point t E [a, b] \ M. Denote by s a point of the set M nearest to the point t (see Lemma 3.8). The interval with the endpoints sand t lies in [a, b] \ M. Every neighborhood of the point s intersects the interval. Hence the set M is not open. This contradiction gives what was required. Example 6.3. The subset [0,1] U [2,3] of the real line is disconnected. It is sufficient to refer to (6.3). Lemma 6.1. Let a space X be represented in the form of the union of two disjoint open (respectively, closed) sets F and G. Let a set M ~ X intersect both F and G. Then the set M is disconnected. Proof. The sets MnF and MnG are nonempty and open (respectively, closed) in the subspace M. Their union coincides with M. They are disjoint. Referring to (6.2) (respectively, to (6.3)) completes the proof. • Corollary 1. Let a connected subset M of a topological space be covered by two disjoint open (respectively, closed) subsets. Then the set M lies • entirely in one of them. Corollary 2. Let a connected subset M of a topological space intersect an open-closed set F. Then M ~ F. • Theorem 6.1. Let a topological space X be the union of a family, of its connected subsets and 1= 0. Then the space X is connected. Proof. Fix an arbitrary point x En,. Assume that the space X is disconnected. Take its representation in the form of the union of open sets F and G according to (6.2). For definiteness let x E F. By Corollary 1 of ~ F Lemma 6.1 every element of the family ,lies in F. Therefore X = and G = 0. This contradicts the choice of G. The contradiction shows that Our assumption is false. The theorem is proved. • Example 6.4. A half interval (a, b] of the real line is connected: we apply Theorem 6.1 to the representation (a, b] = U{[c, b]: c E (a, b]}. Likewise a half interval [a, b) is connected. An interval (a, b) is connected: we apply Theorem 6.1 to the representation (a, b) = (a, c] U [c, b), where c is an arbitrary point of the interval (a, b).

n,

U,

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Theorem 6.2. Let M be a connected subset of a topological space. Then the set [M] is connected. Proof. Assume the opposite. Let [M] = FuG be a representation of the subspace [M] according to (6.2). The set M is dense in the space [M]. The sets F and G are nonempty and open in the space [M]. Thus M n F i= 0 and M n G i= 0. Now Lemma 6.1 implies the disconnectedness of the set M. This contradicts our initial assumption. So our assumption is false. The theorem is proved. • Theorem 6.3. Every connected subset of the real line is either a segment, either a half-interval, or an interval. (A one point subset of the line is regarded as a (degenerate) segment.) Proof. If the set is one point, it is a (degenerate) segment. Let a connected subset M of the real line contain more than one point. Show first that for every two points s < t of the set M the segment [s, t] lies in M. Assume the opposite, i.e., there exists a point c E [s, t] \ M. Then the sets M n (-00, c) and M n (c, (0) are nonempty. Lemma 6.1 implies the disconnectedness of the set M. This contradicts initial assumptions. Thus our additional assumption is false. So the interval (a, b), where a = inf M and b = sup M, lies in set M. The definition of a and b implies that the set M is either the segment [a, b], either the half interval [a, b), either the half interval (a, b], or the interval (a, b). All these sets are connected (see • Examples 6.1, 6.2 and 6.4). The theorem is proved. A connected metric compactum is called a continuum. Theorem 6.3 implies that a subset of the line is a continuum if and only if it is a segment (because in the list of Theorem 6.3 only segments are compact). Singletons are called trivial connected sets. Connected sets containing more than one point are called nontrivial. Theorem 6.4. Let h be an upper semicontinuous (multi-valued) mapping of a connected topological space X into a topological space Y. Let for every point x E X the set h(x) be nonempty and connected. Then the set h(X) is connected. Proof. Assume the opposite. The set h(X) may be represented in the form h(X) = F U G according to (6.2). Corollary 2 of Lemma 6.1 implies the equalities h-1(F) = {x: x E X, h(x) ~ F} and h-l(G) = {x : x E X, h(x) ~ G}. Therefore h-1(F) n h-l(G) = 0. These sets are nonempty. Their union coincides with X. Theorem 1.7.5 implies the openness of these sets. So we have the representation (6.2). Hence X is disconnected. This contradicts the initial assumption. So our additional assumption is false. The theorem is proved. • Corollary 1. The image of a connected topological space under a continuous (single valued) mapping is connected. •

Some properties of topological, metric and Euclidean spaces.

59

Example 6.5. A circle is connected. Consider for simplicity the circle x 2 + y2 = 1. To prove the connectedness it is sufficient to apply Corollary 1 to the mapping p : JR -. JR2, p(t) = (cos t, sin t). Corollary 2. Let h be a continuous mapping of a compactum X onto a connected Hausdorff space Y. Let the preimage of every point y E Y under the mapping h be connected. Then the compactum X is connected. (See Theorem 6.4 and Corollary of Theorem 1.7.8.) • Remark 6.1. Corollary 2 implies the connectedness of the product of two connected compacta. (We consider the projection of the product onto the factor.) However the requirement of the compactness of the factors in fact is superfluous: the product of two connected spaces is connected. To prove that we may consider the multi-valued mapping q being inverse to the projection p of the product onto a factor. In the general case the mapping q need not be upper semicontinuous. However, because the image under the mapping p of every open set is an open set replaces well the assumption of the upper semicontinuity of q. We repeat with corresponding small changes the proof of Theorem 6.4 to obtain what was required. A topological space is called locally connected if every its point possesses arbitrary small connected neighborhood (i.e., for every point x of the space in question and for every neighborhood Ox of this point there exists a connected neighborhood of x lying in Ox). Example 6.6. Not every locally connected space is connected. The space of Example 6.3 is locally connected and disconnected. Example 6.7. Not every connected space is locally connected. In order to construct a corresponding example, denote by lk, k = 1,2, ... , the segment connecting in the plane the points (0,1) and (2- k , 0) and put 10 = {O} x [0,1]. By Theorem 6.1 the set X = U{lk: k = 0,1,2, ... } (Figure 2.3) is connected. The point (0,0) does not possess a connected neighborhood (in the space X) lying in the neighborhood of the radius 1. Let x be an arbitrary point of a topological space X. Denote by ex the union of all connected subsets of the space X conFigure 2.3 taining the point x. The set ex is nonempty because it contains the point x. Theorem 6.1 implies its connectedness. Theorem 6.2 implies its closedness. The set ex is called a connected component (or, shortly, a component) of the point x in the space X. Theorem 6.1 implies that if two component intersect then they coincide. In particular,

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a point x belongs to the component of a point y if and only if the point y belongs to the component of the point x. Evidently: Theorem 6.5. A connected component of a locally connected space is an open set. • Theorem 6.6. Let, be an open cover of a topological space X consisting of pairwise disjoint nonempty connected sets. Then all connected components of the space X and only them are elements of ,. Proof. By Corollary 2 of Lemma 6.1 every component of the space X lies entirely in an element of the cover ,. It follows from the definition that this component coincides with the corresponding element of ,. By the same arguments every element of, is a component. The theorem is proved. • Theorem 6.7. Let U be an open subset of a locally connected space. Then the set U may be represented as the union of a family , of pairwise disjoint nonempty connected open subset and such a representation is umque. Proof. The subspace U is locally connected and Theorem 6.5 gives the needed representation. Theorem 6.6 implies its uniqueness. The theorem is • proved. Corollary. Let U be an open subset of a locally connected space with countable base. Then the set U may be represented as the union of an (at most) countable family of pairwise disjoint (nonempty) connected open sets and such a representation is unique. (See Corollary of Theorem 1.6.10.) • Example 6.8. Connected components of the Cantor perfect set (see Example 1.6.4) are singletons. A continuous mapping f of (every) segment [a, b] of the real line into a topological space X is called a path in the space X. The point f (a) is called the initial point or first endpoint and the point f (b) is called the end of the or second endpoint of the path f. They say that the path f connects the points f (a) and f (b). Every two nondegenerate segments are homeomorphic (see Example 2.1). Therefore when we have a path f we can take every other segment I of the real line and construct a path with the same endpoints and with the segment I as its domain of definition. The segment being the domain of definition of the path f is symmetric with respect to its middle point. When we use the symmetry in a analogous way we can construct a path having the point f (b) as its initial point and the point f (a) as the end. Lemma 6.2. Assume that there exist a path connecting points a and b and a path connecting points band c in a topological space X. Then there exists a path connecting the points a and c in the space X. Proof. The proof is obvious.

•

Some properties of topological, metric and Euclidean spaces.

61

A topological space is called arcwise connected if any two points of it can be connected by a path. A topological space is called locally arcwise connected if any point of it possesses an arbitrarily small arcwise connected neighborhood. Corollary 1 of Theorem 6.4 and Theorem 6.1 imply that every arcwise connected space is connected (recall: in Example 6.2 we have established the connectedness of a segment). So every locally arcwise connected space is locally connected. Let a space X be locally arcwise connected. For x E X denote by Lx the set of all points of the space X which can be connected to the point x by a path. If a point y belongs to the closure of the set Lx then (every) connected neighborhood Oy of it intersects the set Lx. Lemma 6.2 implies that the set Lx U Oy is arcwise connected. Hence it lies in Lx. In particular, Oy ~ Lx· Thus y E (Lx). In view of the arbitrariness of the point y E [Lx] this means that [Lx] ~ (Lx). Hence Lx = [Lx] = (Lx). So the set Lx is open-closed. By Corollary 1 of Lemma 6.1 the set Lx coincides with the connected component of the point x.

7. Plane regions An open connected subset of the space ]Rn is called a region. By the Corollary of Theorem 6.8 every open subset of the space ]Rn fall into the union of an (at most) countable family of pairwise disjoint regions. This concerns, in particular, the complement ]Rn \ X of an arbitrary compact subset X of the space ]Rn. Let , be the set of connected component of the set ]Rn \ X (we say that X splits the space ]Rn into the sets U E ,). By Lemma 3.7 and Remark 3.2, for some r > 0 the compact urn X lies in the ball 0(0, r). The set ]Rn \ 0, (0, r) ~ ]Rn \ X is (arcwise) connected. By Corollary 2 of Lemma 6.1 it lies in an element W oo(X) of the family,. All other elements of ,lie in the ball 0,(0, r). Hence they are bounded. Let Y be a Hausdorff space, a < b. A path f : [a, b] ~ Y is called simple if it is injective. A path f is called a loop, or a closed path, if f (a) = f (b). The loop (closed path) f is called simple if f (s) =f. f (t) for every pair of different points s, t E (a, b]. If f : [a, b] ~ Y is a simple path, then the mapping f is an embedding (see Theorem 2.1). A continuous mapping of a segment (i.e., a path), of a half interval or of an interval is called a curve. We say that a curve has no self-intersections if it is injective. A closed curve is a synonym for a closed path. A closed curve without self-intersections is a synonym for a simple loop. Let f : [a, b] ~ Y be a loop. The formula

27r(t - a) . 27r(t - a)) h() t = ( cos b -a ' SIn b -a

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defines a continuous mapping of the segment [a, bj onto the circle S of the center in the origin of coordinates and the radius 1. The continuity of the (single valued) mapping fh- 1 : S --+ f([a, b]) follows from the Corollary of Theorem 1.7.8 (because the composition of two upper semicontinuous mapping is upper semicontinuous). The constructed mapping fh- 1 is also called a closed path, or a loop. If f is a simple loop then the mapping fh- 1 is one to one. In this case by Theorem 2.1 the mapping fh- 1 is an homeomorphism. It is also called a simple closed path, or a simple loop. If f : [a, bj --+ Y is a simple path (respectively, a simple loop) then the set f([a, b]) is called a simple arc (respectively, a simple closed arc). So a simple arc is a subset homeomorphic to a segment. A simple closed arc is a subset homeomorphic to a circle. One of the corresponding implications is proved above; the converse (i.e., if a set is homeomorphic to a circle then it may be covered by a simple loop) is somewhat obvious. For brevity we will say that a simple loop bounds a region V ~ ]R2, if the corresponding simple closed arc is the boundary of the bounded region V. Points f (a) and f (b) are called an initial point and an end, respectively, of the simple arc f([a, b]). If the end of a simple arc coincides with the initial point of another and they do not have any other common points then their union is also a simple arc (with the initial point at the initial point of the first arc and with the end at the end of the second). If the end of a simple arc coincides with the initial point of a second and the end of the second arc coincides with the initial point of the first, and the arcs have no other common points, then their union is a simple closed arc. A simple arc does not split the plane (i.e., its complement is connected or, with the notation of the beginning of this section, the set , consists of the unique element Woo(X), where X denotes the simple arc in question) [Dij.

Theorem 7.1 (Jordan). A simple closed arc splits the plane into two regions and is the boundary of every of them. See [Dij.

•

Notice with reference to Theorem 7.1 that one of two components of the complement to a simple arc X is just the unbounded component Woo(X) and second component is a bounded set.

Theorem 7.2. Let a plane region U be bounded by a simple closed arc. Then there exists an homeomorphism g of the set [Uj onto the disc {(x,y): (x,y) E ]R2, x 2 + y :'( I} such that g(U) = {(x,y): (x,y) E 1l~.2, x 2 + y2 < I}. Proof. The proof of this assertion is based on methods of the theory of functions of a complex variable (Riemann theorem and the theorem on the correspondence of boundaries accompanying the Riemann theorem, see [Maj). •

Some properties of topological, metric and Euclidean spaces.

63

8. Degree of a mapping of a circle into itself The mapping p, p(t) = (cos t, sin t), maps the real line into the circle S ofthe radius 1 and the center in the origin of a rectangular system of coordinates of the plane, (see its graph in Figure 2.4). If we take an arbitrary intervall of the circle S (i.e., an arc without its endpoints) then its pre image under the mapping p falls into the union of pairwise disjoint intervals (0:+ 21fk, (3+ 21fk), where 0 < (3 -0: :( 21f and k = 0, ±1, ±2, ... , moreover the restriction of the mapping p to every of these intervals is an homeomorphism onto the intervall (Figure 2.4). We will consider the set e(X, JR.) (respectively, e(X, S)) of all upper semicontinuous mapping of a compactum X in the line JR. (respectively, in the circle S) with proper connected compact subset as values. By Theorem 6.3 values of mappings from e(X, JR.) are segments of the real line. The same theorem implies that values of mappings from e(X, S) are segments (i.e., arcs with their endpoints) of the circle S. A mapping 9 E e(X, JR.) is called an e-lifting of the mapping I E e(X, S), if 1= pg.

Figure 2.4

Let -00 < e < d < 00. When we consider mappings of a eompactum X, mappings from e(X x [e, d], JR.) and e(X x [e, d], S) are called e-homotopies. Mappings 11,12 E e(X, JR.) (respectively, f1, h E e(X, S)) are called e-homotopie if there exists an e-homotopy F E e(X x [e, d], JR.) (respectively, F E e(X x [e, d], S)) such that II (x) = F(x, e) and h(x) = F(x, d) for every x E X. All nondegenerate segments of the line are homeomorphic. So in this definition the segment [e, d] may be replaced by every other nondegenerate segment of the line, for instance, by the segment [0, 1]. Likewise in this definition the order of 11 and h is not essential. Every mapping 1 E e(X, IR) U e(X, S) is e-homotopic to itself (a corresponding e-homotopy may be defined by the formula F(x, t) = l(x)). If 11 is e-homotopic to 12 and is 12 e-homotopic to h, then 11 is e-homotopic 13 (a corresponding

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homotopy may be defined by the formula for x E X, t E [0,1], for x E X, t E [1,2]' where Fl is an e-homotopy from the mapping fl to the mapping hand F2 is an e-homotopy from f2 to h, see Theorem 1.7.6). Lemma 8.1. Let FE e(X x [c,d],S) and 9 E e(X,IR). Let F(x,c) = pg(x) for every point x E X. Then there exists an unique e-lifting G E e(X x [c, d], IR) of the e-homotopy F satisfying the condition

G(x, c) = g(x) for every point x E

x.

Proof. 1. Let us prove the uniqueness of the e-homotopy G. Let two e-homotopies G 1 and G 2 satisfy the condition imposed on the homotopy G and (x, s) E X x [c, d]. For every point t E [c,d] we have p(G 1 (x,t)) = F(x,t) = p(G 2 (x,t)). The lengths of the segments G 1 (x, t) and G 2 (x, t) are less than 271". Thus G 1 (x, t) = G 2 (x, t) + 271"k(t), where k(t) = 0, ±1, ±2, .... The upper semicontinuity of the mappings G 1 and G 2 implies the continuity of the function k. Values of the function k are integers. The set of integers does not contain nontrivial connected subsets. We obtain from the continuity of the function k and Corollary 1 of Theorem 6.4 that the function k is constant. Since k(c) = 0, k == 0. Hence G 1 (x, t) = G 2 (x, t). This gives what was required. II. Let us prove the existence of the e-homotopy G under the strong additional restriction F(X x [c, d]) i- S. Fix arbitrary points 0 E S\F(X x [c,d]) and a E p-l(O). The preimage of the set S\ {O} under the mapping p falls into the union of pairwise disjoint intervals I j = (a + 271"j, a + 271"(j + 1)), j = 0, ±1, ±2, .... The restriction of the mapping p to every such interval is an homeomorphism onto the set S \ {O}. Let qj : S \ {O} -Y I j denote the inverse homeomorphism. Since g(X) ~ IR\p-l(O), Corollary 1 of Lemma 6.1 and the membership 9 E e(X, IR) imply the representation of the compactum X as the union of its pairwise disjoint open subsets Xj = g-l(1j), j = 0, ±1, ±2, .... Since each of them is equal to the complement of the union of all the others, these sets are closed. Let G(x, t) = qjF(x, t) for x E Xj, where j = 0, ±1, ±2, .... By virtue of our remarks the mapping G satisfies all required conditions. III. Make weaker the restriction imposed in II and pass to the condition F(X x {t})

i- S

for every t E [c, d].

Some properties of topological, metric and Euclidean spaces.

65

For every t E [e, d] fix a point O(t) E S \ F(X x {t}). The set F-I(O(t)) does not intersect the set X x {t}. By Theorem 1.7.5 it is closed in the product X x [e, d]. By virtue of Theorem 1.4 for some t5(t) > 0 the set F-I(O(t)) lies outside the set X x 06( tl t. Hence F(X x 06(t)t) <;;; S\ {O(t)}. Let M denote the set of all m E [e, d] such that: for the e-homotopy F IXx[c,m] there exists an e-lifting G m satisfying all the above conditions but with the change of the segment [e, d] to the segment [e, m] <;;; [e, d]. The set M is nonempty because it contains the point e. The definition of the set M immediately implies that it has the form of a half interval [e, 'Y) or of a segment [e, 'Yl, where 'Y E [e, d]: if m E M and e ~ ml ~ m then ml E M, because we can take as G m, the restriction of the e-homotopy G m to the set X x [e, ml]' By I the mappings G m are defined uniquely. Take an arbitrary point tl of a (nonempty) set M n 06h)'Y' Let t2 be an arbitrary point of the set ['Y, d] n06h)'Y. We have F(X x [tl' t 2]) <;;; S\ {O("()}. The result of II is applicable to the e-homotopy F* = F IXX[t"t2] as F, to the mapping g* : X - t JR, g*(x) = G tl (x, h) as 9 and to the segment [tl, t 2 ] instead of [e, d]. Let G* be a e-lifting existing by II. Define the mapping

G(x t) = {GtG*(x,(x,t)t) l

t2'

t E [e, tIl, X, t E [tl, t 2l,

if x EX, if x E

on the set X x [e, t2]' The upper semieontinuity of the mapping G t2 follows from Theorem 1.7.6. Thus t2 E M. In view of the arbitrariness of the point t2 E [,,(, d] n 06h)'Y this implies the inclusion 06h)'Y <;;; M. The definition of'Y now implies that 'Y = d and M = [e, d]. So we have proved the existence of an e-lifting under the additional assumption. IV. Pass to the general case. Let x EX. By virtue of the upper semieontinuity of the mapping F (and Theorem 1.7.3) for an arbitrary t E [e, d] there exists a Tychonoff neighborhood 0; x Ot of the point (x, t) such that F( 0; x Ox) =f S. The open cover {Ot: t E [e, d]} of the segment [e, d] contains a finite sub cover {Otl , . .. ,Otd. Put Ox = n{ 0;1, ... ,O;k}. Every point t E [e, d] belongs at least to one of the sets Ot i . Therefore

(8.1) Apply Lemma 1.6.1 to the point x and to the set X \ Ox. Lemma 1.6.1 implies the existence of a neighborhood V x of the point x such that

66

[Vx]

CHAPTER 2 ~

Ox. By virtue of(8.1) the mappings FX

F I[Vxjx[c,dj andg = gl[V,j satisfy the condition imposed in II on F and g, respectively (with [Vx] as X). Let GX : [V x] X [c, d] - t ~ be an e-lifting of the mapping FX existing by II. If y E [Vxd n [VX2] and i E [c, d], then by I GXl(y, i) = G X2(y, i). Hence the formula G(y,i) = GX(y,i) for y E [Vx] uniquely defines the mapping G : X x [c, d] - t ~. Theorem 1.7.7 implies its upper semicontinuity. The fulfilment of all other conditions on G is obvious. The lemma is proved . • When the compactum X consists of one point we obtain: Corollary. Lei f E e([c, d]' S), I be a segment of the real line, p(1) = f(c). Then there exists a unique e-lifting h of the mapping f satisfying the condition h(c) = I. • Let us now define the notion which is mentioned in the title of this section. For every mapping g E e(S, S) we can apply Corollary of Lemma 8.1 to the mapping f = g . p I[O,2-11} [0,21f] - t S (we can take as I an arbitrary segment satisfying the condition p(I) = f(c)). If h is an e-lifting of the mapping f existing by Corollary of Lemma 8.1 then ph(O) = f(O) = f(21f) = ph(21f). Therefore the segment h(21f) has the form 1+ 21fk, where k = 0, ±1, ±2, .... The number k is called the degree of the mapping g. In the definition of the degree of a mapping we had an arbitrariness in the choice of the e-lifting h (or, to be precise, an arbitrariness in the choice of the segment 1). Every other e-lifting hl of the mapping f has the form hl (t) = h(i) + 21fi, i = 0, ±1, ±2, ... (this follows from the uniqueness of the e-lifting under restrictions of Corollary). Therefore hi (21f)

=

X

= h(21f) + 21fi = 1+ 21fk + 21fi = hl (21f) + 21fk,

i.e., the value of the degree of a mapping does not depend on the choice of its e-lifting. Every single valued continuous mapping 9 : S - t S belongs to e(S, S). Its degree corresponds to the number of entire rotations of the point g(i), when i runs the circle one time. The direction of the rotations defines the sign of the degree. Theorem 8.1. Mappings
h3 (c) = G(21f, c) = g(21f) = g(O)

+ 21fk = G(O, c) + 21fk = hl (c) + 21fk = h 2 ( c).

Some properties of topological, metric and Euclidean spaces.

67

The mappings h2 and h3 are e-liftings of the mapping a(t) = F(O, t). The uniqueness of such an e-lifting (see Corollary of Lemma 8.1) guarantees the coincidence of the mappings h2 and h 3. Thus

g*(27T) = G(27T, d) h3(d) = h2(d) = hl(d) + 27Tk = G(O, d) + 27Tk = g*(O) + 27Tk. =

Hence the degree of the mapping 1f2 is also equal to k. Sufficiency. Let k be the degree of the mappings Ifl and 1f2. Let 'l/Ji = lfiP l[o,27rr [0,27T] -+ 8 for i = 1,2. Let hI, h2 be e-liftings for Ifl and 1f2, respectively. Evidently the mapping H (s, t) = hI (s ) (1 - t) + h2 (s )t belongs to e([O, 27T] x [0,1]' JR) (the upper semicontinuity of the mapping H may be easily established, for instance, with the help of Theorem 5.1). The mapping F =pH belongs to e([0,27T] x [0,1],8). For every t E [0,1]

=

+ h2(27T)t) p«hl(O) + 27Tk)(1 - t) + (h 2(0) + 27Tk)t) p(h l (O)(1 - t) + h 2(0)t) + 27Tk) = p(h l (O)(1

=

F(O, t).

F(27T, t) = p(h l (27T)(1 - t) =

- t)

+ h2(0)t)

Therefore there exists an unique mapping G : 8 x [0,1] -+ 8 satisfying the condition F(x, t) = G(p(x), t) for every x E [0,27T] and t E [0,1]. Evidently G E e(8 x [0,1]' 8) (the upper semicontinuity of the mapping G follows easily from the upper semicontinuity of the mapping 1-1 : 8 x [0,1] -+ [0,27T] X [0,1], where 1 : [0,27T] X [0,1] -+ 8 x [0,1]' l(x, t) = (p(x), t), see Corollary of Theorem 1.7.8). It remains to notice that the e-homotopy G connects the mappings Ifl and 1f2. The theorem is proved. • A continuous (single valued) mapping F : X x [c, d] -+ Y of the product of a topological space X and a segment [c, d] of the real line into a topological space Y is called a homotopy connecting the mappings lc(x) = F(x, c) and Id(x) = F(x, d). Mappings are called homotopic if there exists a homotopy connecting them. Evidently a single valued e-homotopy is an homotopy. In the case of the single valuedness of the mappings Ifl and 1f2 in Theorem 8.1 the mapping G in the second part of its proof is single valued too. Therefore the last remark and Theorem 8.1 imply: Theorem 8.2. Continuous (single valued) mappings 1f1, 1f2 : 8 -+ 8 are homotopic (i.e., may be connected by a homotopy) if and only if their degrees coincide. • The formula x* (t) = x cos t - y sin t { y* (t) = x sin t + Y cos t

68

CHAPTER 2

defines a continuous mapping S x IR ----) S. For a fixed t E IR we have a rotation of the circle S about the angle t. Therefore the identity mapping S ----) S and the rotation of the circle S about an arbitrary angle are homotopic. This fact and Theorem 8.1 imply that in the definition of the degree of a mapping the segment [0,21f] may be changed to every other segment of the line of the length 21f. Mappings which are homotopic to constant mappings are called contractible. The degree of a constant mapping (of the circle into the circle) is equal to zero. Thus: Example 8.1. A continuous mapping of the circle into the circle is contractible if and only if its degree is equal to zero. Example 8.2. Identify the circle S with the circle {z: z E C, Izl = 1} in the complex plane C. It is not difficult to check that the degree of the mapping cp : S ----) S, cp(z) = zn, is equal to n. (Recall the Euler formula (cos cp + i sin cp) n = (cos ncp + i sin ncp )). Example 8.3. 'Fundamental Theorem of Algebra'. Every polynomial I(z) = zn+an_lzn-l + .. ·+alz+aO (n = 1,2, ... ) with complex coefficients has a complex radical. Let A = nmax{lan_ll, ... ,lall,laol,l}. Since An> lan_lIAn-l + ... + lallA + ao, the (continuous) function F(x, t) = xn + t(an_lx n- l + ... + alx + ao) does vanish for Ixl = A and 0 :::; t :::; 1. Thus the formula (z, t) = F(Az, t)(IF(Az, t)l)-l, where Izl = 1 and 0 :::; t :::; 1, defines an homotopy. It connects the functions cp(z) = zn and 'IjJ(z) = I(Az)(I/(Az)I)-l. We noticed in Example 8.2 that the degree of cp is equal to n. By Theorem 8.2 the degree of'IjJ is also equal to n. Assume now that the polynomial 1 has no radicals. Then the formula *(z, t) = f(tz)(lf(tz)l)-l, where Izl = 1 and 0 :::; t :::; A, defines an homotopy connecting the mapping 'IjJ and the constant mapping 'IjJ* == 1(0)(1/(0)1)-1. We noticed in Example 8.1 that the degree of a constant mapping is equal to zero. Therefore the result of the previous paragraph contradicts Theorem 8.2. Thus our assumption is false. This gives what was required. Lemma 8.2. Let mappings I, g E e(X, S) of a compactum X into the circle S satisfy the condition

I(x)

~

g(x) for every point x E X.

Then the mappings f and g are e-homotopic. Proof. We can define an e-homotopy connecting the mappings g by the formula

f(X) h(x, t) = { g(x)

for x E X,D:::; t < 1, for x EX, t = 1.

f and

Some properties of topological, metric and Euclidean spaces.

69

That it belongs to e(X x [0,1]' S) is obvious. • Corollary. Let mappings fI' h E e(X, S) of a compactum X into the circle S satisfy the condition:

for every x E X the set fl (x) U opposite points.

h (x)

does not contain diametrically

Then the mappings fl and f2 are e-homotopic. To obtain the corollary let us notice that under our assumptions for every x E X the set S \ (fl (x) U h (x)) consists of one or two arcs. But only one of them has a length greater than the length of half the circle. Denote by g(x) the complement to this arc. Evidently g E e(X, S). By Lemma 8.2 the mapping g is e-homotopic both to fl and f2' This gives what was required. • The next remark concerns the fulfilment of the condition: (8.2) every (single valued) continuous mapping f point (i.e., a point x E X such that f(x) = x).

:X

--*

X possesses a fixed

If a space X satisfies (8.2) and a space Y is homeomorphic to the space X then the space Y satisfies the condition (8.2) too. Theorem 8.3. Every continuous mapping of a closed disc into itself

possesses a fixed point. Proof. By virtue of the previous remark it is sufficient to consider the disc D = {( x, y) : (x, y) E lie, x 2 + y2 ~ 1} of the radius 1 and the center in the origin of the rectangular system of coordinates. Assume the opposite. Let a continuous mapping f : D --* D have no fixed points. For zED put u(z) = z - f(z) and v(z) = u(z) (1Iu(z)II)-I. Since (z, f(z)) ~ Izl . If(z)1 ~ 1,

(u(z), z) = (z - f(z), z) = (z, z) - (f(z), z)

~

0

for Ilzll = 1. Thus the points z and v(z) are not diametrically opposite. The degree of the identity mapping S --* S is equal to 1. By the Corollary of Lemma 8.2 and Theorem 8.1 the degree of the mapping 9 = v Is is equal to 1 too. The homotopy G(z, t) = v(tz), where z E S, 0 ~ t ~ 1, connects the mapping 9 and the constant mapping h == v(O). By Theorem 8.2 the degree of the mapping 9 is equal to zero. This contradicts the previous calculation. Thus our assumption is false. The theorem is proved. • The following more general assertions are true too.

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CHAPTER 2

Theorem 8.4 (Brouwer Fixed Point Theorem). Every continuous mapping of a closed ball of the Euclidean space (of arbitrary dimension) into itself possesses a fixed point. See [HW]. • Theorem 8.5 (Schauder Fixed Point Theorem). Every continuous mapping of a compact convex subset of a Banach space (see below §3.1) into itself possesses a fixed point. See [Ed]. • A mapping j* E e([c, d], S) satisfying the condition j*(c) = j*(d), is called an e-loop. By the Corollary of Lemma 8.1 the e-loop j* possesses an e-lifting h*, and as ph*(e) = j*(e) = j*(d) = ph'(d) then hOed) = h*(e) + 27fk for some k = 0, ±1, ±2, .... The number k is called the degree of the e-loop j*. As for the degree of a mapping of the circle the value of the degree of the e-loop j* does not depend on the choice of the e-lifting h'. With the notation of the definition of the degree of a mapping of the circle the degree of the mapping g is equal to the degree of the e-loop f. On the other hand, if a mapping q : [e, d] -+ S is defined by the formula _ ( 27f(t - c) . 27f(t - C») q (t ) - cos d ' sm d -c -e

then the degree ofthe e-loop j* is equal to the degree of the mapping j*q-l. The proof reduces to a simple comparison of definitions. The use of the notion of a e-loop simplify a little the calculation of the degree of mappings. It is based on the obvious Lemma 8.3. Let the degrees of e-loops fl E e([b, c], S), h E e([e, d], S) be equal, respectively, to kl and k 2. Let fl (c) = h (c). Then the degree of the e-loop f E e( [b, d]' S), for t E [b, e], for t E [e, d], is equal to kl + k 2 . Proof. Let hI E e([b, c], JR) be an e-lifting for fl' Fix an e-lifting h2 of e-loop h according to Corollary of Lemma 8.1 in order to satisfy the condition h2 (c) = hI (c). It remains to notice that the mapping h(t) = {hI (t) h2(t)

for t E [b, c], for t E [e, d],

is an e-lifting of the e-Ioop f and h(d) = h2(d) = h2(e) + 27fk2 = hl(e) + 27fk2 = hI(b) + 27fkl + 27fk2 = h(b) + 27f(kl + k2)' The lemma is proved .•

CHAPTER 3

SPACES OF MAPPINGS AND SPACES OF COMPACT SUBSETS

Usually in accounts of the theory of Ordinary Differential Equations in the description of topological properties of sets of solutions we use the notion of the uniform convergence of sequences of functions. We will discuss related topological structure in §§ 1 and 2. In fact its correspondence with this circle of questions is not very good because it deals with sets of functions with a fixed domain. In the definition of a solution of a differential equation we do not fix domains of definition. Functions with different domains are called solutions of the same equation. In order to describe topological properties of such sets of functions more completely, but rather briefly, we will introduce in §5 the space of partial mappings. In first few sections we consider some fundamental notions which are helpful in study of properties of the new space. See also [Ku1].

1. Metric and norm of uniform convergence Let X be an arbitrary set. Let (Y, p) be a metric space. We considered in Example 1.4.4 the space B(X, Y) of all bounded mappings of the set X into the space Y. The metric d of Example 1.4.4 is called the metric of uniform convergence. Lemma 1.1 Let (with the above notation) the space Y be complete. Then the space B(X, Y) is complete too. Proof. Let {Jk: k = 1,2, ... } be a fundamental sequence of elements of the space B(X, Y). By virtue of the estimate P(fi(X), fj(x)) ~ d(j;, fj) the sequence {Jk(X): k = 1,2, ... } is fundamental for every element x E X. By virtue of the completeness of the space Y the sequence converges to a point f (x). The verification of the facts of the function f just defined is bounded (and hence belongs to the space B(X, Y)) and that d(ik, J) ----+ 0 will complete the proof of the lemma. We leave it to the reader as a simple exercise. • Assume now in addition that X is a topological space. Denote the set of all bounded continuous mappings of the space X into the space Y by BG(X, Y).

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CHAPTER 3

Lemma 1.2. Under the conditions mentioned the set BC(X, Y) is closed in the space B(X, Y). Proof. The proof of this assertion reduces to the property that the limit of an uniformly convergent sequence of continuous functions is a continuous • function, see [AI]. Lemma 1.3. A closed subspace of a complete metric space is complete (with respect to the induced metric). Proof. The proof is obvious. Lemmas 1.1-3 immediately imply: Theorem 1.1. The space BC(X, Y) of all continuous bounded mappings of a topological space X into a complete metric space Y is complete (with respect to the metric of uniform convergence). • Denote by C(X, Y) the set of all continuous mappings of a topological space X into a topological space Y. (So, BC(X, Y) = B(X, Y)

n C(X, Y).)

If now a space X is compact and Y is a metric space then by Theorem 1.7.8 and Lemma 2.3.7 C(X, Y) = BC(X, Y). Thus in this case the metric of uniform convergence is defined on the entire set C(X, Y) and Theorem 1.1 implies Corollary. The space C(X, Y) of all continuous mappings of a compact space X into a complete metric space Y is complete (with respect to the metric of the uniform convergence). • Often we consider the space of mappings into a normed space. This situation is already met when we consider the space of real functions. In the corresponding assumptions the mapping space may be considered as a normed space, moreover its norm generates the metric of the uniform convergence. Let X be an arbitrary set and Y be a normed space. For an arbitrary mapping f : X --t Y let us put Ilfll = sup{lIf(x)lI: x E X}. A norm of an element of the space Y is nothing else than the distance from this element to the zero of the space Y. By virtue of the remarks in §2.3 the condition of the boundedness of a mapping f : X --t Y may be written as IIfll < 00. We can introduce in a natural way a linear structure on the space of mappings from X into Y. For two mappings f, g : X --t Y their sum is defined by the formula (J + g)(x) = f(x) + g(x). If a is an arbitrary number the mapping af is defined by the formula (aJ)(x) = a . f(x). With the notation described the function that associates the number IIfll to a mapping f E BC(X, Y) is a norm. The verification of this fact is not difficult. Thus the space BC(X, Y) turns out to be normed. A complete (in the sense of the metric related with the norm) normed space is called a Banach space. Theorem 1.1 implies:

Spaces of mappings and spaces of compact subsets.

73

Theorem 1.2. The space BC(X, Y) of all continuous bounded mappings of a topological space X into a Banach space Y is Banach. • As in the case of Theorem 1.1 we obtain Corollary. The space C(X, Y) of all continuous mappings of a compact space X into a Banach space Y is a Banach space. • We can introduce in a natural way the notion of a series Ul +U2 +U3 + ... and of their convergence as the existence of the limit of the sequence of partial sums Sk = UI + ... + Uk. The limit is called the sum of the series under consideration. Theorem 1.3. Let UI + U2 + U3 + . .. be series of elements of a Banach space Y. Let the numerical series IluI II + IIu2 II + IIu3 II + . .. converge. Then the series UI + U2 + U3 + ... converge too. Proof. By virtue of completeness of the space Y it is sufficient to show that the sequence of partial sums of the series UI + U2 + U3 + . .. is fundamental. It follows easily from the inequality

and from the convergence of the series IIuI II + II u 2 II + II u 3 II + ... (the convergence of the series implies that the sequence of partial sums is fundamental, see §1.6). The theorem is proved. • By virtue of Corollary of Theorem 1.2, Theorem 1.3 may be applied to series of continuous real functions on a compact space. We speak in this case about uniform convergence and about uniformly convergent series. 2. Compact open topology Consider now the possibility of introducing a topology on the set C(X, Y) in the general case, i.e., without the assumption of the previous section about the presence of a norm or a metric on the space Y. We can introduce a topology on the set C(X, Y) in different but quite natural ways. We will consider only one, but it corresponds the most to the circle of questions under consideration. Let X and Y be topological spaces. For K ~ X and U ~ Y denote O(K, U) = {f: f E C(X, Y), f(K) ~ U}. Denote by f3 the family of all sets of the form O(K, U), where K runs over the set of all closed compact subsets of the space X and U runs over the set of all open subsets of the space Y. The family f3 is a sub-base of a topology on the set C(X, Y). It is called the compact open topology on C(X, Y). Notice that our condition defines the topology uniquely. Theorem 2.1. Let X be a compactum and (Y, p) be a metric space. Then the compact open topology on the set C(X, Y) is generated by the metric of the uniform convergence.

74

CHAPTER 3

Proof. Denote by T the compact open topology on the set C(X, Y). Denote by d the metric of uniform convergence on the set C(X, Y). We can state our problem as follows: prove that the identity mapping h : (C(X, Y), d)

--t

(C(X, Y), T)

is an homeomorphism. We will do this with the help of Theorem 1.7.3. I. Let f E O(K, U), where K is a (nonempty) closed compact subset of the space X and U is a (nonempty) open subset ofthe space Y. By Theorem 1.7.8 the set f(K) is compact. By Corollary of Lemma 2.3.8 02cf(K) ~ U for some E > O. Let now g E O(f, E) (notice that in the first case the 2Eneighborhood of the set is taken in the space Y and in the second case the E-neighborhood of the point is taken in the space C(X, Y)). For every point t of the set K and for every point y of the set Y \ U

p(g(t), y)

~

p(f(t), y) - p(f(t), g(t))

~ 2E - E

= E.

In view of the arbitrariness of the point y E Y \ U this implies that E U (Figure 3.1). The point t E K is also chosen arbitrarily. Therefore g E O(K, U). We have established the continuity of the mapping h.

g(t)

Figure 3.1

II. Let f E C(X, Y) and E > O. Let Ox = !-I(Og;'3!(X)) for x E X. The open (see Theorem 1.7.5) cover {Ox: x E X} of the compact space X has a finite subcover {OXI,"" Oxd. By Lemma 2.3.3 and Theorem 1.7.5 the set Ki = !-I([Og/3!(Xi)]), i = 1, ... , k, is closed. Theorem 1.6.2 implies its compactness. Evidently Ki ;:::> OXi' Therefore the family {K I , ... ,Kd covers the space X.

Spaces of mappings and spaces of compact subsets.

75

Let U = n{O(Ki' 02f/3f(Xi)) : i = 1, ... , k}. Since f(Ki) = [O£/J!(Xi))] <;:;; 02£/3f(Xi), fEU. Take arbitrary 9 E U and x EX. For some i = 1, ... ,k we have x E K i ,

p(f(x),g(x)) ~ p(f(x), f(Xi))

E

2E

+ P(f(Xi),g(X)) < 3 +:3

= E

(Figure 3.2). This implies that the function 9 lies in an E-neighborhood of the function f (with respect to d). So we have established the continuity of the mapping h- 1 • The theorem is proved. •

Xi

Figure 3.2

Theorem 2.2. Let a locally compact space X have a countable base. Let Y be a metric space. Then the space C(X, Y) (with the compact open topology) is metrizable. Proof. Let f3 be (at most) a countable base of the space X. The set f30 of all elements of the base f3 with compact closures is (at most) countable: f30 = {Ui : i = 1,2, ... }. The local compactness of the space X implies that f30 is a base of X. For every k = 1,2, ... the set Mk = [U{Ui : i = 1, ... ,k}] is compact. By Theorem 2.1 the space C(Mk' Y) is metrizable. By Theorem 2.2.2 the space Z = IT {C(MkJ Y): k = 1,2, ... } is metrizable. The definition of the compact open topology implies the continuity of the mapping Pk : C(X, Y) - t C(Mk' Y), Pk(Z) = ZIMk ' k = 1,2, .... By Theorem 2.1.1 the diagonal product P : C(X, Y) - t Z of the mappings Pk, k = 1,2, ... , is continuous. The mappings P is injective. The theorem will be proved when we verify the continuity of the inverse mapping Q : P(C(X, Y)) - t C(X, Y). An arbitrary element of the above sub-base of the space C(X, Y) has the form O(K, U), where K is a closed compact subset of the space X and U is

CHAPTER 3

76

an open subset of the space Y. By virtue of Theorem 1.7.3 it is sufficient to prove the openness of the set Q-l (O(K, U)). The compactness of the set K and the equality UfJo = X imply that K ~ Mk for some k = 1,2, .... The set W = {h: h E C(Mk' Y), h(K) ~ U} is open in the space C(Mk' Y) by the definition of the compact open topology. We have

Q-'(O(K, U))

~

(It

C(M" Y) x W x

ilL i,

C(M V)) n P(C(X, V)).

This gives what was required. The theorem is proved. • The construction in the proof of the theorem allows us to characterize the compact open topology as the topology of the uniform convergence on compact subsets of the space X. In particular: Theorem 2.3. Under the hypotheses of Theorem 2.2 a sequence {Ii : i = 1,2, ... } ~ C(X, Y) converges (with respect to the compact open topology) to a function f E C(X, Y) if and only if for every closed compact subset K of the space X the sequence {Ii IK: i = 1,2, ... } converges uniformly to the function f IK. Proof. Necessity follows from Lemma 1.7.2 when we apply it to the (obviously continuous single valued) mapping associating to a function from C(X, Y) its restriction to the set K. Sufficiency follows from Lemma 2.1.2 when we apply it to the embedding P : C(X, Y) - t Z from the proof of Theorem 2.2. The theorem is proved. • Theorem 2.4. Let «I> : T x X - t Y be a continuous mapping of the product of topological spaces T and X into a topological space Y. Let 'Pt(x) = «I>(t, x) for t E T, x E X. Then the mapping F : T - t C(X, Y), F(t) = 'Pt, is continuous. Proof. Let to E T and F(t o) E O(K, U), where K is a closed compact subset of the space X and U is an open subset of the space Y. We have «I> ( {to} x K) = 'Pta (K) ~ U. By virtue of the continuity of the mapping «I> we can associate to every point x of the set K its neighborhood Ox and a neighborhood V(x) of the point to in a manner that «I>(V(x) x Ox) ~ U. The open cover {Ox: x E K} of the compact set K has a finite subcover {OXl, ... , Ox;}. Put V = V(xd n··· n V(Xi). Let t E V and x E K. For some j = 1, ... ,i the point x belongs to OXj. Therefore

In view of the arbitrariness of the point x E K this implies the inclusion 'Pt(K) ~ U. It means that F(t) = 'Pt E O(K, U). Referring to Theorem 1. 7.3 completes the proof. •

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77

3. Vietoris topology Let X be a topological space. Denote by exp X the set of all nonempty closed subsets of the space X. For U I , . .. ,Uk s:;; X put O(UI , ... ,Uk) = {F: FE expX, F s:;; Uf=IUi , F nUl i- 0, ... ,F n Uk i- 0}. Notice two particular cases. For U s:;; X O(U) = {F: F E expX, F s:;; U} and O(U, X) = {F: F E expX, F n U i- 0}. The following equality is obvious: (3.1) On the set exp X introduce a so called Vietoris topology. It is the smallest topology, in which all set of the forms O(U) and O(U, X) are open (i.e., sets of the form mentioned constitute a sub-base of the Vietoris topology). Here U runs over the topology of the space X. Without any detailed discussion let us notice that the openness of sets of the form O(U) corresponds to the notion of convergence in the sense of condition (1. 7.1), and the openness of sets of the form O(U, X) corresponds to the notion of convergence in the sense of the lower limit. By virtue of (3.1) sets of the form O(UI , · · · , Uk), are open in the Vietoris topology. They are called Vietoris neighborhoods (of their points). In fact, Vietoris neighborhoods constitute a base of the Vietoris topology because elements of the above sub-base (the sets O(U) and O(U, X)) are Vietoris neighborhoods too, and, on the other hand, if FE O(UI , . . . , Uk) n O(VI , ... , Vi) then FE O(WI ,· .. , W m ) s:;; O(UI , .. ·, Uk) n O(VI , . . .

,

Vi),

where {WI,"" W m } is the renumbered family {Ui n Vi: i = 1, ... , k, j = 1, ... ,I, F n Ui n Vi i- 0}. The set exp X with the Vietoris topology is called the space of (nonempty) closed subsets of the space X. We will not give an account here of the large ammount of theory related to the notion introduced, see [Ku1], [FF], [En]. We will restrict ourselves to some technical remarks. Let 9 : X -> Y be a continuous mapping of a topological space X into a topological space Y. It is naturally to define the mapping g* of the space exp X as its domain by putting g* (F) = g(F). However in this way we do not always obtain a mapping of the space exp X into the space exp Y because the image of a closed set need not to be closed. In order to pass over this case consider the set E(g*) of all closed subsets of the space X, images under the mapping 9 of which are closed in Y. Consider the set E(g*) with the topology induced by the Vietoris topology. Although the space E(g*) in the general case need not coincide with the space expX, often it turns out to be quite large. So, if the spaces X and Yare Hausdorff, by virtue of

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Theorems 1.6.1 and 1.7.8 the space E(g*) contains all (nonempty) compact subsets of the space X. Theorem 3.1. Let 9 be a continuous mapping of a topological space X into a topological space Y. Then the mapping g* : E(g*) --+ exp Y is continuous.

Proof. The proof reduces to a direct verification. Let F E E(g*). Let O(UI , ... , Uk) be an arbitrary Vietoris neighborhood of the point g*(F). For VI = g-I(UI ), ... , V k = g-I(Uk ) we have

Therefore F E o (v;. , ... , Vk). This gives what was required. The theorem is proved. • Theorem 3.2. Let X be a regular space. Let generalized sequences 8 = {D,,: a E A} ~ expX and c = {E,,: a E A} ~ expX converge (in the Vietoris topology) with respect to a directed set cp to D and E, respectively. Let D" ~ E" for every a E A. Then D ~ E. Proof. Take an arbitrary t E X \ E. The regularity of the space X implies the existence of disjoint neighborhoods Ot of the point t and OE of the set E in the space X. By virtue of the convergence of the sequence c to E with respect to cp there exists F E cp such that E" E O(OE) for every a E F. Hence D" ~ E" ~ OE, D" nOt = 0 and D" rf. O(Ot, X). By virtue of the convergence of the sequence 8 to D with respect to cp the open set O(Ot, X) cannot be a neighborhood of the point D, i.e., D n Ot = 0 and t rf. D. This gives what was required. The theorem is proved. • Theorem 3.3. Let X be a normal space. Then the set expc X of all connected closed subsets of the space X is closed in the space exp X. Proof. By Theorem 1.3.3 it is sufficient to prove that the set H = exp X \ expc X of all disconnected closed subset of X is open in the space exp X. Let Fo E H. The set Fo may be represented as the union of two nonempty closed disjoint subsets FI and F 2 . By virtue of the normality of the space X there are disjoint neighborhoods UI and U2 of the sets FI and F 2, respectively. The set O(UI , U2) is a neighborhood of the point Fo in the space exp X. Every F E 0 (U I , U2 ) is equal to the union of two nonempty disjoint closed subsets F \ U2 = F nUl and F \ U I = F n U2· Therefore FE H, O(UI , U2 ) ~ H. The theorem is proved. • Corollary. Let FI ;;2 F2 ;;2 F3 ;;2 ... be a non-increasing sequence of nonempty connected compact subsets of a metric space X. Then the set F = n{Fk: k = 1,2, ... } is connected.

To obtain the corollary it is sufficient to restrict ourselves to the case FI = X. We have F = lim tOPk--+oo Fk (see Example 1.5.3). By Corollary of Theorem 1.6.3 the set F is nonempty. Let U be an open subset of the space X. If F E O(U, X), then the definition of the lower limit implies

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79

that Fk E O(U, X), beginning with some k. If F E O(U), then by Theorem 1.7.2 Fk E O(U), beginning with some k. By Lemma 1.7.3 Fk --+ F in the Vietoris topology. This allows us to obtain the corollary. •

4. Hausdorff metric Let X be a Hausdorff space. By Theorem 1.6.1 every compact subset of the space X is closed, i.e., the set expc X of all nonempty compact subsets of the space X lies in exp X. The Vietoris topology was defined first on the set exp X. Now it induces a topology on the set expc X. The new topology is also called the Vietoris topology. Sets of the form Oc(U1, ... ,Uk ) = O(U1, ... ,Uk ) nexpcX, where U1"",Uk are open subsets of X (see the end of §1.3), constitute a base of the new topology. The set expc X with the mentioned topology is called the space of compact subsets of X. If X is a compact urn then by virtue of Theorems 1.6.1 and 1.6.2 expc X = exp X.

Figure 3.3

Figure 3.4

Let now (X, p) be a metric space. In this case consider one more possibility to introduce a topology on the set expc X. For F 1 , F2 E expc X put a(F1 ,F2) = inf{c: c > 0,F2 S;;; O[Fd, see Figures 3.3 and 3.4. Lemma 2.3.6 implies the nonemptiness of the set {c: c > 0, F2 S;;; O[Fd. The formula written defines a number a( Fl , F2)' The number is called the deflection of the set F2 from the set Fl' Notice properties of the deflection

a. For every (nonempty) compact subsets F 1 , F2 and F3 of the space X: 1) a(Fl,F2)~0; 2) a(F1 , F2) = if and only if F2 S;;; F 1 ; 3) a(Fl,F3):::; a(Fl,g) + a(F2,F3). Proof of these properties is simple. Notice only that in the case of the property 3) we need use Lemma 2.3.5. We have not mentioned the symmetry

°

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between properties of the deflection. The deflection does not possess this property and therefore it is not a metric. The Hausdorff distance between two compact subsets FI and F2 of the space X is defined as h(FI ,F2) = max{a(FI,F2),a(F2,Fd}. In contrast to the deflection the function h is symmetric (i.e.,

By virtue of the above properties of the deflection the function h is a metric on the set expc X (the Hausdorff metric). Theorem 4.1. The Vietoris topology on the space of compact subsets of a metric space is generated by the (corresponding) Hausdorff metric. Proof. Before this theorem is proved we must distinguish the space expc X of compact subsets of the metric space (X, p) with the Vietoris topology and the metric space H(X) with the same carrier set expc X and the Hausdorff metric h. Our aim is to prove that the identity mapping i : H(X) - t expc X is an homeomorphism (see §2.2). Since the bijectivity of this mapping is obvious we have to check the continuity of the mapping i and of its inverse mapping i-I. I. Let U be an open subset of the space X and Fo E Oc(U, X). Fix an arbitrary point x of the intersection Fo n U. Next, fix a number E > 0 such that O~x S;;; U (O~x denotes the E-neighborhood of the x point with respect to the metric p). If F E O~Fo (O~Fo denotes the E-neighborhood of the point Fo with respect to the metric h), then Fo S;;; O~F (the neighborhood with respect to the metric p). By Lemma 2.3.4 there exists a point t of the set F such that p(x, t) < E. SO t E O~x S;;; U. The intersection F n U contains the point t. Hence it is nonempty. Thus i(O~Fo) S;;; Oc(U,X). II. Let U be an open subset of the space X and Fo E 0 c(U). The condition Fo E Oc(U) may be rewritten as Fo S;;; U. By Corollary of Lemma 2.3.8 O,Fo S;;; U for some E > O. By the definition of the Hausdorff metric O~ Fo S;;; O(U), i.e., i(O~ Fo) S;;; O(U). III. The continuity of the mapping i follows from I, II and Theorem 1.7.3. IV. Let Fo E H(X) and E > o. The open cover {0,/4X: x E Fo} of the compactum Fo contains a finite sub cover , = {0,/4XI, ... , Oo/4xd. The following properties of the family, are essential for us:

and (4.2) if points s, t belong to an element of the family" then p( s, t) <

~.

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81

(the last property follows immediately from the triangle inequality). Lemma 2.3.4 and (4.2) imply immediately that if F I , F2 E O(Or/4XI, ... , Or/4Xk), then h(FI' F2) ::( c/2 < c. By virtue of (4.1) O(Or/4XI, ... ,Or/4Xk) ~ O~ Fo. This means the continuity of the mapping i-I (see Theorem 1.7.3). The theorem is proved. • The same topology on the set X may be generated by different metrics PI and P2· The metrics PI and P2 define (may be, different) Hausdorff metrics hI and h 2 • By virtue of Theorem 4.1 both Hausdorff metrics hI and h2 generate the same topology on the set expc X, namely just the Vietoris topology. The Vietoris topology may be defined directly by the topology of the space X without reference to the metric. The topology of a metric space may be characterized completely by the reserve of convergent sequences, see Theorem 1.5.3. Therefore the just worded independence of the topology of the space expc X on a concrete definition of a metric on the space X may be also deduced from the following assertion. Theorem 4.2. Let F, F 1 , F 2 , • .• be compact subsets of a metric space. The sequence {Fi: i = 1,2, ... } converges with respect to the Hausdorff metric to the compactum F if and only if simultaneously: 1) for every point x E F we can choose points Xi E Fi in a manner, that Xi

~

X,

2) every sequence Xk E Fik (il < i2 < ... ) has a subsequence {Xk : k E A} converging to a point of the set F. Proof. Necessity of condition 1. Denote by Xi a point of the set Fi nearest to the point X (see Lemma 2.3.8. and Figure 3.5). We have

F a· a .>

Figure 3.5

Figure 3.6

Necessity of condition 2. Denote byak a point of the compactum F nearest to the point Xk (Figure 3.6). By virtue of the compactness of the

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set F the sequence {ak k = 1,2, ... } has a subsequence {ak converging to a point a E F. Then for k E A

P(Xb a) ~ p(Xk, ad

+ p(ak' a)

~ a(F, FiJ

+ p(ak' a)

----t

k E A}

o.

Sufficiency. I. Show that a(F, Fd ----t O. Assume the opposite. There exist a number E > 0 and an infinite set A <:;;; N such that a(F, F k ) ~ E for k E A (Figure 3.7). By virtue of the last inequality there exists a point

Figure 3.7

Figure 3.8

Xk E Fk such that p( x k, F) ~ E. By 2) there exists a subsequence {x k : Ad of the sequence {Xk : k E A} converging to a point x of the set F. Then P(Xk' F) ~ p(Xk, x) ----t 0 as k E Al and k ----t 00. This contradicts the choice of the point Xk: P(Xk' F) ~ E. II. Show that a(Fk, F) ----t o. Assume the opposite. That is, there exist a number E > 0 and an infinite set A <:;;; N such that a(Fk' F) ~ E for k E A (Figure 3.8). By virtue of last inequality there exists a point Xk E F such that P(Xk' Fd ~ E. By virtue of the compactness of the set F the sequence {Xk: k E A} has a subsequence {Xk: k E Ad converging to a point x E F. By 1) we can choose points ak E Fk in a manner that ak ----t x. Then p(ak, xd ~ p(ak' x) + p(x, xd ----t 0, P(Xk' Fd ~ P(Xk' ad ----t 0 as k E Al and k ----t 00. This contradicts the choice of the points Xk: P(Xk' Fd ~ E. III. The sufficiency follows from I and II. The lemma is proved. • Corollary. Let a sequence {FI' F2 , . . . } of compact subsets of a metric space converge with respect to the Hausdorff metric to a compactum F. Let OF be an arbitrary neighborhood of the compactum F. Then there exists an index io = 1,2, ... such that Fi <:;;; OF for i = io, io + 1, .... In the opposite case the set A = {i: i = 1,2, ... , Fqi \ OF -=1= 0} is infinite. For i E A fix a point Xi E Fi \ OF. The sequence {Xi: i = 1,2 ... } has no subsequence converging to a point of set A. This contradicts 2). (However, this corollary may be also deduced from the definition of the Vietoris topology and Theorem 4.1.) •

kE

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83

Theorem 4.3. The space of nonempty closed subsets of a metric compactum is compact. Proof. I. Let X denote the compact. Let {Fi: i = 1,2, ... } be a sequence of nonempty closed subsets of X. By Theorem 1.5.4 the sequence {Fi: i = 1,2, ... } has a subsequence {Fi: i E A} converging to a set F E expc X in the sense of the fulfilment of the equality

F = lim top inf Fi = lim top sup Fi . i-+oo

i-+oo

The compactness of X implies the nonemptiness of F = lim top sup F i . Let U be nonempty open subset of the compactum X. i--->oo II. Let F E 0 (U). By virtue of Theorem 1.7.2 Fi E 0 (U), beginning with some i = i 1 E A. III. Let F E 0 (U, X). Then Fi E 0 (U, X), beginning with some i = i 1 E A. This is a direct consequence of the definition of the lower limit. IV. Now II and III and Lemma 1.7.3 implies the convergence Fi - 7 F, i E A, with respect to the Vietoris topology. The theorem is proved. •

5. Space of partial mappings In this section we assume that (X, pd and (Y, P2) are metric spaces. Their product X x Y is also a metric space (see §2.1). Let P be the corresponding metric. Every continuous mapping of an arbitrary nonempty closed subspace of the space X into the space Y is called a partial mapping of the space X into the space Y, see also [Za], [Ku2] and [Ku3]. The set of all partial mappings of the space X into the space Y is denoted by Cv(X, Y). Thus Cv(X, Y) = U{C(F, Y): FE expX}. We will also consider partial mappings defined on compact subsets of the space X: Cvc(X, Y) = U{ C(F, Y): F E expc X}. Our next goal is the introduction of a topology and of a metric on the set Cvc(X, Y) which are suitable for the description of properties of solution sets of ordinary differential equations. Introduce a topology on the set Cvc(X, Y) in the following way. Consider the mapping Gr : Cvc(X, Y) - 7 expc(X x Y) of the set Cvc(X, Y) into the space expc(X x Y), associating its graph to a function. The mapping Gr is injective or in other words the mapping Gr : Cvc(X, Y) - 7 Gr (Cvc(X, Y)) is one to one. The set expc(X x Y) carries the Vietoris topology (generated by the corresponding Hausdorff distance). Recall that the topology does not depend on a concrete choice of the metrization of the product (see remarks of the previous section). The set Gr (Cvc(X, Y)) carries a topology and a metric induced from the space expc(X x Y). We define uniquely a

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topology and a metric on the set Cve(X, Y) when we declare the mapping Gr : Cve(X, Y) --. Gr (Cve(X, Y)) to be an isometry, i.e., we declare the distance between two elements of Cve(X, Y) to be equal to the Hausdorff distance between their graphs. The set Cve(X, Y) with the distance and topology so defined is called the space of partial mappings with compact domains of definition (of the metric space X into the metric space Y). Theorem 5.1. The mapping 1f : Cve(X, Y) --. eXPe X is continuous. Proof. The proof consists in referring to Theorem 3.1 and the continuity of the projection g : X x Y --. X. • For X = JR we have: Corollary. The mappings:

1fb : Cve(JR, Y) --. JR, 1fe : Cve(JR, Y) --. JR, 17r : Cve (JR, Y) --. JR,

1fb(Z) = inf1f(z), 1fe(z) = SUP1f(z), l7r(z) = 1fe(z) - 1fb(Z),

are continuous. • Let H E eXPe X. Denote C:C(X, Y) = {z: Z E Cve(X, Y), H ~ 1f(z)}. For Z E C:C(X, Y) put 0 the set A = {i : i E N, sup{P2(fi(t),f(t)): t E H} ~ c} is infinite. For every i E A fix a point ti E H, such that P2(fi(t i ), f(t i )) ~ E. The mapping cp : H x Y --. Y x Y, cp(t, y) = (f(t), y), is continuous. This follows from Theorem 2.1.1. The mapping 'ljJ : Y X Y --. JR, 'ljJ(x, y) = P2(X, y), is continuous. This follows from Corollary 1 of Lemma 2.3.1. Thus the function a : H x Y --. JR, a(t, y) = P2(f(t), y), is continuous. This implies the openness of the set

U = a- 1 ((-oo,c)) = {(t,y): t E H, y E Y, P2(f(t),y) < c}. in the product H x Y. By the Corollary of Lemma 2.3.8 there exists a number 8 > 0 such that

Spaces of mappings and spaces of compact subsets.

85

0 6 Gr(f) <:;; U u ((X \ H) x Y) = (X

x Y) \ ((H x Y) \ U) (Figure 3.9).

y

Then (ti' f(t i )) ~ 0 6 Gr(f) for every i E A. By virtue of condition 2) of .···~(H Theorem 4.2 and the closedness of the uset (XXY)\06 Gr(f) this contradicts the condition Gr(fi) --t Gr(f). Thus our assumption is false and the theo.:"."'" " : rem is proved. • Theorem 5.3. Under the hypo.X theses of Theorem 5.2 the identity mapping C H (X, Y) --t C(H, Y) is an Figure 3.9 homeomorphism. Proof. The continuity of the mapping follows from Theorem 5.2.

.J

liiii!ol,"~4aL""""";"<"4"'_'

Prove the continuity of the inverse mapping. Let h be the Hausdorff metric of the space Gr(CH(X, Y)). Let hI(f,g) = h(Gr(f),Gr(g)) be the corresponding metric on the space C H (X, Y). Let d be the metric of the uniform convergence in the space C(X, Y), see §1. With this notation we obtain what was required from the obvious estimate hI (f, g) ~ d(f, g) and remarks of §1. 7 about the Lipschitz condition. The theorem is proved. • Theorem 5.4. Let sequences

{Ji: i = 1,2, ... } <:;; Cvc(X, Y) and

{Di: i = 1,2, ... } <:;; expc X converge to f E CvJX, Y) and D E expc X, respectively. Let Di <:;; 7r(fi) (Figure 3.10) for every i = 1,2, .... Then D <:;; 7r(f) and filDi

--t

flD in the space Cvc(X, Y).

Figure 3.10

Proof. 1. The inclusion D <:;; 7r(f) follows from Theorem 5.1, Lemma 1.7.2 and Theorem 3.2.

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II. Show that the sequence Fi = Gr (JiI D) and the compactum F = Gr (JID) satisfy condition 1) of Theorem 4.2. Let x = (t, f(t)) E F. Since tED and Di ---+ D, by Theorem 4.2 we can fix points ti E Di in a manner that ti ---+ t. Our goal will be achieved w hen we prove the convergence (ti' fi(t i )) ---+ (t, f(t)). If we assume the opposite, we obtain that for some c > 0 the set A = {i: i E N, P((ti' fi(t i )), (t, f(t)))

Figure 3.11

~

c} is infinite. But

(Figure 3.11). Therefore the sequence {(ti, fi(t i )): i E A} contains a subsequence {(ti,fi(ti)): i E Ad converging to a point y = (8,1(8)) of the set Gr(f). By Lemmas 2.1.1 and 2.1.2 ti ---+ 8 for i E Al and i ---+ 00. Since ti ---+ t, t = 8. Therefore y = (t, f(t)). This contradicts the definition of the set A. III. Show now that the sequence Fi = Gr (JdD) and the compactum F = Gr (JI D) satisfy condition 2) of Theorem 4.2. Let a set A S;;; N be infinite. Let ti E Di and Xi = (ti' fi(t i )) for i E A. By virtue of the convergence Di ---+ D and Theorem 4.2 the sequence {t i : i E A} contains a subsequence {t i : i E Ad converging to a point t of the set D. Since (ti,fi(t i )) E Gr(fi) ---+ Gr(f), the sequence {(ti,fi(t;)): i E Ad contains a subsequence {( t i , fi (t i )): i E A 2 } converging to a point y = (8, f(8)) of the set Gr(f). By Lemmas 2.1.1 and 2.1.2 ti ---+ 8 as i E A 2 , i ---+ 00. Since ti ---+ t, 8 = tED. Therefore the subsequence {( t i , fi (t i )): i E A 2 } converges to the point y = (t, f(t)) of the set F = G(fID)' This is just what was • required. The theorem is proved. The position this theorem is well clarified by: Corollary. Let X be a compactum and Y be a Hausdorff space. Then the mapping : C(X, Y) x X ---+ Y, (z, x) = z(x), is continuous. To obtain the corollary let us notice that by virtue of Theorem 5.3, Theorem 5.4 implies the equality lim(s,t)-+(z,x) (8, t) = (z, x) for every point (z, x) of the product C(X, Y) x X (see also Lemma 1.7.2). It means just the continuity of the mapping at the point (z,x). •

Spaces of mappings and spaces of compact subsets.

Theorem 5.5. Lei {Ii: i = 1,2, ... } ~ CVC(X, Y), f E Cvc(X, Y), {Di i 1,2, ... } U {Ei i = 1,2, ... } ~ expc X. Lei D, E E expc X, Di U Ei ~ 7r(fi) (Figure 3.12) for every i = 1,2, ... , DUE ~ 7r(f), filDi - t fiD' hlEi - t fiE· Then

filDiUEi

-t

87

-E

fI DUE .

x

Proof. The theorem will Figure 3.12 be proved, when we check the fulfilment of conditions 1) and 2) of Theorem 4.2 for the sequence of the compacta Fi = Gr (JiIDiUE) and for the compactum F = Gr (JI DUE ). I. Take arbitrarily x = (t,f(t)) E F = Gr (JI DUE ). Let, tED for definiteness' sake. By virtue of the convergence filDi - t flD and by Theorem 4.2 we can fix points Xi E Gr (JiI D ) in a manner that Xi - t x. Thus Xi E Gr (JiI DiUE ) and Xi - t x. This means the fulfilment of condition 1). II. Let a set A ~ N be infinite, ti E Di U Ei and Xi = (t i , fi(t i )) for i E A.

Since A = Al U A 2, where

Al={i: iEA, tiEDi}, and A 2 ={i: iEA, tiEE;}, at least one of these two sets Al or A2 is infinite. Let the set Al be infinite for definiteness. By virtue of the convergence Gr (JiI D ) - t Gr (JI D ) and by Theorem 4.2 we can pass to a subsequence {(ti,fi(t i ))· i E B} of the sequence {(ti,fi(t;)): i E Ad converging to a point X of the set Gr (JI D ) ~ Gr (JIDUE). This means the fulfilment condition 2). The theorem is proved. •

6. Compactness in the space of partial mappings A subset Z of the set Cv(X, Y) of partial mappings of a metric space (X, pd into a metric space (Y, P2) is called equicontinuous, iffor every number c > 0 there exists a number 0 > 0 such that P2(Z(S), z(t)) < c for every mapping z E Z and for every two points s, t E 7r(z), which satisfy the condition Pl(S, t) < o. ~

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Theorem 6.1. Let (X, pd and (Y, P2) be metric spaces. Then every compact subset of the space Cvc(X, Y) is equicontinuous. Proof. Under these hypotheses the space Cvc(X, Y) is metrizable. Let Z be a compact subset of the space Cvc(X, Y). The definition of the Vietoris topology implies immediately that the mapping Gr (as a multi-valued mapping into the product X x Y) is upper semicontinuous. Theorem 1. 7.8 implies the compactness of the set E = U{Gr(z): z E Z}. Assume that the set Z is not equicontinuous. This means, in particular, that for some E > 0 for every k = 1,2, ... there exist a function Zk E Z and points Sk, tk E 7r(Zk) such that Pl(Skl td < 2- k and p2(zdsd, zdt k )) ~ E. By virtue of the compactness (and the metrizability) of the set Z we can assume in addition that the sequence {Zk: k = 1,2, ... } converges to a function Z E Z. By virtue of the compactness of the set E we can assume in addition that the sequences {( s k, Zk (s k)): k = 1, 2, ... } and {(t kl Zdtk)): k = 1,2, ... } converge, respectively, to points (s, yd and (t,Y2). Theorem 5.4 implies the equalities Yl = z(s) and Y2 = z(t). The estimate PI (s, t) ~ PI (s, sd + PI (Skl t k ) + PI (tk' t) follows from the triangle inequality. All terms in its right hand side tend to zero as k - t 00. Therefore PI(S,t) = O,s = t, z(s) = z(t). On the other hand,

When we pass to the limit as k - t 0 in the right hand side of the inequality we obtain the estimate P2(Z(S), z(t)) ~ E. Thus our assumption (that the set Z is not equicontinuous) has led to a contradiction. The theorem is proved. • The theorem proved remains valuable in the case when the set consists of one function only. Usually we use here another terminology and we call a function which constitutes an equicontinuous family uniformly continuous. Thus we obtain: Corollary. A continuous function defined on a metric compactum is uniformly continuous. • Theorem 6.2 (Arzela). Let K be a compact subset of the product X x Y of metric spaces (X,pd and (Y,P2). Let the sequence 0: = {Zk : k = 1,2, ... } ~ Cvc(X, Y) be equicontinuous. Let Gr(zk) ~ K for every k = 1,2, .... Then the sequence 0: has a convergent (in the space Cvc(X, Y)) subsequence. Proof. By virtue of Theorem 4.3 we can assume in addition the convergence of the sequence {Gr(zk): k = 1,2, ... } in the Vietoris topology to a point F of the space expc K. Denote by g the restriction to the set F of the projection X x Y - t X. Show that the mapping g : F - t X is injective. Assume the opposite, i.e.,

Spaces of mappings and spaces of compact subsets.

89

that there are two different points (X,YI) and (X,Y2) of the sets F with coinciding first coordinates. Put c = ~P2(YI' Y2) (> 0). By Theorem 4.2 we can choose points Sk, tk E 7r(zd for k = 1,2, ... in a manner that (Sk' ZdSk)) -- (x, yd and (tk' zk(td) -- (x, Y2)' By Lemma 2.1.2 Sk -- x and tk -- X. By virtue of equicontinuity of the sequence 0: there exists a number 8 > 0 such that p2(zds), zdt)) < c for every index k = 1,2, ... and every points s, t E 7r(zd satisfying the condition PI(S, t) < 8. Since Sk -- x and PI (s kl

x) <

8

2"'

beginning with some k. Hence:

(6.1) By the triangle inequality:

By Lemma 2.1.2 the first and third terms in the right hand side of (6.2) tend to zero as k -- 00. Because of (6.1)

Our assumption has led to a contradiction. So the mapping 9 is injective. This means that the set F is the graph of a mapping z : g(F) -- Y. By Theorem 2.2.1 the set g(F) is compact and the mapping h = g-l (inverse to 9 : F -- g(F)) is continuous. The mapping z may be represented in the form of the composition of the mapping h and the projection X x Y __ Y. This implies the continuity of z. Thus z E Cvc(X, Y). The convergence Gr( Zk) -- Gr( z) and the definition of the topology of the space of partial mappings imply the convergence Zk -- z. The theorem is proved. • 7. Space C.(M)

Let M be a subset of the product lR x X of the real line lR and of a metric space X. Denote by C.(M) the set {z: z E Cvc(lRxX), 7r(z) is connected, Gr(z) ~ M}. The topology induced on the set C.(M) by the space of partial mappings Cvc(lR, X) makes the set C.(M) a topological space. By the remarks of §5 this space is metrizable. L

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CHAPTER 3

By virtue of Theorems 2.6.3 and 3.3 the set of all segments is closed in the space expc JR. Therefore the continuity of the mapping 7f implies the closed ness of the subspace Cs(JR x X) in the space Cvc(JR, JRn). So Theorem 6.2 implies: Theorem 7.1. Let K be a compact subset of the product JR x X, a be an

equicontinuous sequence of elements of the set Cs(K). Then the sequence a has a convergent (in the space Cs(K)) subsequence. • In the rest part of this section we assume that X = JRn. Theorem 7.2. The mapping I: Cs(JR x JRn) --+ JRn,

J b

I(z) =

z(t)dt,

a

where [a, b] = 7f(z), is continuous. Proof. Take an arbitrary function Zo E Cs(JR x JRn) and an arbitrary number E > o. Let 7f(zo) = lao, bolo Let a number M > 0 be such that Im(zo) ~ 0(0, M) (see Theorem 1.7.8 and Lemma 2.3.6). By virtue of the continuity of the mappings 7f and 1m (see Theorem 3.1) there exists a neighborhood VI of the point Zo in the space C s (JR x JRn), for every element z of which -

Im(z) ~ 0(0, M), lao - al

E

E

< 8M and Ibo - bl < 8M'

where [a, b] = 7f(z). If ao = bo, then for z E VI

J

J

J

ao

a

a

bo

zo(t)dt -

Let ao

b

b

2E

z(t)dt :::;; 8M . M <

z(t)dt

E.

<

bo. Select points al and bl in order to satisfy the inequalities ao < al < bl < bo, lao - all < 8~' Ibo - bll < 8~· By virtue of the continuity of the mapping 7f there exists a neighborhood V2 of the point Zo in the space C s (JR x JRn) such that a < al and bl < b for every function z E V2 , where as above [a, b] = 7f(z). For z E VI n V2 we have a, b, b bo

j zo(t)dt - j z(t)dt :::;; j(zo(t) - z(t))dt a, a ao

j zo(t)dt b,

j zo(t)dt ao

a,

bo

+

+

+

j z(t)dt a

b

+

j z(t)dt b,

Spaces of mappings and spaces of compact subsets.

91

bt

~

j(zo(t) - z(t))dt

<

j (zo(t) - z(t))dt

bt

3c

+ 4'

Theorems 5.3 and 2.1 imply the existence of a neighborhood V3 of the point Zo in the space Cs(lR, IRn) such that Ilzo(t) - z(t)11 < 4(bl~a,) for every function z E V3 satisfying the condition 7r(z) 2 [aI, bd and for every point t E [aI, bd· Hence for z E VI n V2 n V3 bt

j(zo(t) - z(t))dt

~

al

bt

j Ilzo(t) - z(t)lldt al

III(zo) - I(Z)II <

c

3c

4+ 4

= c.

This means the continuity of the mapping I (at an arbitrary point Zo E Cs (IR x IRn) and hence on the entire space Cs (IR x IRn)). The theorem is proved. • Theorem 7.3. With the notation of Theorem 7.2 let 1

J(z) = b _ a I(z), if b =1= a and

J(z) = z(a),

= b. Then the mapping J : C s (IR x IRn) - t IRn is continuous. Proof. The continuity of the mapping J at a point Zo E Cs (IR x IRn), l,.(zo) > 0, follows immediately from Theorem 7.2, from the continuity of the mapping 1,. (see Corollary of Theorem 5.1) and of the division of vectors by nonzero numbers. Let 7r(zo) = {ao}. Let z E Cs(JR x JRn) and 7r(z) = la, b]. For a = b: if a

IIJ(zo) - J(z)11 = IIzo(ao) - z(a) II

~

h(Gr(zo), Gr(z)),

CHAPTER 3

92

where h denotes the Hausdorff metric generated by the Euclidean metric of the space IR x IRn = IRn+l. For a f- b:

IIJ(zo) - J(z)11

~ b~ a

b

J J

Ilzo(ao) - z(t)lldt

a

~ b~ a

b

h(Gr(zo), Gr(z))dt

a

= h(Gr(zo), Gr(z)).

These estimates imply the continuity of the mapping J at the point Zo. The theorem is proved. •

8. Alexander's lemma and its consequences Zorn's lemma 1.8.1 allows us to obtain results of wide generality. With the same success we may use assertions related to the axiom of choice and being equipotent to Zorn's lemma. However, t.he discussion of this question lies a long way from the framework of this book. In this section we will notice only possibilities for clarifying the meaning of methods used and some relations. We will generalize Theorems 2.2.2 and 4.3 with the help of Zorn's lemma. (In the main part of our account the theorems was not proved in their maximal generality because the corresponding methods are not usual in the circle of questions under consideration. It is sufficient to have versions of them obtained in the development of the technique of reasoning which will be necessary anywhere in the further account.) Lemma 8.1 (Alexander). Assume that a topological space X possesses a sub-base (3 such that: every cover of the space X by elements of (3 has a finite subcover. Then the space X is compact. Proof. Our aim is to show that

every open cover, of the space X has a finite sub cover. Assume the opposite. Let r denote the set of open covers space X without finite subcovers. Our assumption means that r is nonempty. Let

,0

E

r.

1. The set r is partially ordered by the relation ~. Lemma 1.8.2 implies the existence of its maximal linearly ordered subset N 3 {'o}.

Spaces of mappings and spaces of compact subsets.

93

II. If {'Yl, ... , Ik} ~ N then we can renumber the set 11, ... "k according to its order. That is, we can assume in addition the fulfilment of the condition 11 ~ ... ~ Ik' Then ubi,"" Id = Ik EN. III. Show that the cover I' = uN belongs to N. Assume the opposite, that is, that the cover I' has a finite subcover {U1 , ... , Ud. For every i = 1, ... ,k fix a cover Ii E N containing Ui . Since bl,' .. "d ~ N, by II u{ 11, ... "k} = Ik E N. By our choice the cover Ik has a finite sub cover {U1 , • •• ,Ud. This contradicts the condition Ik E N. IV. Show that the family f3' = I' n f3 covers the space X. If it is not so, there exists a point x E X \ (Uf3·). An element U of the cover I' contains the point x. Fix elements Vi,' .. , Vm of the sub-base f3 such that x E n {Vi, ... , Vm} ~ U. By virtue of the choice of the point x none of the sets Vi,' .. , Vm belongs to I'. By virtue of the maximality of the cover I' the cover {Vi} U I" i = 1, ... , m, has a finite sub cover

{Vi, W{, ... , W~(i)}' We have:

.......

and n{V1 , ... , Vm } ~ U. The finite subfamily {U} U {W/: j = 1, ... ,m, i = 1, ... ,n(j)} of I' covers the space X. This contradicts the condition 1* E f. Thus the additional assumption is false and the lemma is proved .• Theorem 8.1 (Tychonov). The product X = IT {Xa: a E A} of compact spaces X a , a E A, is compact. Proof. Sets of the form p;;l(U) (the notation of §2.1), where U runs over the topology of the space Xa and a runs over the set A, constitute a sub-base f3 of the Tychonoff topology of the product X = IT {Xa: a E A}. In order to refer to Alexander's lemma it is sufficient to check that an arbitrary cover of the space X by sets from f3 has a finite subcover. Assume the opposite. Let a cover I ~ f3 of the space X has no a finite subcover. For a E A denote by la the set of all elements of I which can be represented as p;;I (U) (just with this a). If for some a E A the family I~ = {U : U ~ X a , p;:;I(U) E la} covers the space X a , then by virtue of the compactness of the space Xa the cover I~ has a finite sub cover I~' This means that the cover I has a finite sub cover {p;;l(U): U E I~}' which contradicts the initial assumptions about the cover I' It remains to consider the case when for every a E A the family I~ does not cover Xa. In this case we have a possibility associate to every a E A a point Xa E Xa \ U/~, The point {xa: a E A} E X is not covered by,. By the initial assumption I covers X. Thus our additional assumption is false and the theorem is proved. •

94

CHAPTER 3

Theorem 8.2. If a space X is compact, then the space exp X is compact too.

Proof. Sets of the forms 0 (U) and 0 (U, X), where U runs over the topology of the space X, constitute a sub-base of the Viet oris topology of the space exp X. Let I be an arbitrary cover of the space X by elements of the mentioned subbase. Show that the cover I contains a finite subcover. Let Y = X\U where II is the set of all open subsets U of the space X such that 0 (U, X) E I. If Y = 0 then the family II covers the space X and by virtue of the compactness of X II has a finite subcover 12. Then exp X = U{ 0 (U, X) : U E 12}. So the cover I has a finite subcover. If Y =1= 0 then the point Y of the space exp X does not belongs to any set 0 (U, X) E I. Since I is a cover of the space exp X the point Y belongs to one of sets 0 (U) E I. Then Y ~ U. The cover II of the set X \ U contains a finite sub cover {U1 , ... , Urn}. It remains to notice that the family {O (U) ,0 (U1 , X) , ... ,0 (Urn' X)} covers the space exp X: if a point F E exp X does not belong to 0 (U1 , X) U ... U 0 (Urn' X), then F n (U: 1 Ui ) = 0 and therefore F ~ U, FE 0 (U). Referring to Alexander's lemma completes the proof. •

,1,

CHAPTER 4

DERIVATION AND INTEGRATION

Below we develop a theory directed to the study of properties of solutions of ordinary differential equations y' = f (t, y). It will be helpful in the investigation of equations with complicated discontinuities in their right hand sides. A natural approach to the definition of a solution of an equation is to consider a function as a solution, if at every point of its domain the function has a derivative satisfying the equation. This plan may be realized, for instance, when we study equations with continuous right hand sides. But there are many very simple cases when solutions in the sense stated are absent; however, there exist solutions in some near but different meaning. Moreover, these solutions correspond to the contents of the (physical, geometric, etc.) problem solved with the help of the equation under consideration. Our aim in this chapter is to introduce and to discuss a notion of the derivative which suits for large spectrum of such situations. A natural restriction is that if a problem may be solved with the help of classical methods then no new 'solution' appears on account of the play on definitions. As Integration theory in general and the theory of Measurable Mappings lie outside the framework of the theory of Ordinary Differential Equations, we will not try to give them here a complete account. We will follow the shortest way to a limited list of results and facts important for the understanding of the main contents of the book. It seems this is not a drawback. 1. Measure of a set on the line

Let M be a subset of the real line R When we cover the set M by an at most countable family of intervals 'Y = {(ak' bk ) k = 1,2, ... } then it is natural to say that b2 the 'length' of the set M 1---+-) --1(-+-)-+-)(---+does not exceed the sum d( 'Y) of the lengths (== total length) of intervals of the family 'Y, see FigFigure 4.1 ure 4.1. This argument pushes us to the consideration of the lower bound j.L*(M) of numbers db), t--(

96

CHAPTER 4

where, runs over the set H(M) of all at most countable families of intervals covering the set M. The word 'numbers' is used here conditionally: it may be, for all, E H(M) the sum d(,) is infinite. In addition, we may assume that the family, may contain intervals of infinite length. In these cases we write d(,) = 00. The number J.t*(M) (under the same reservation: it may be that J.t * (M) = (0) is called the outer measure of the set M. Lemma 1.1. If Ml ~ M2 (~~), then J.t*(Md ~ J.t*(M2). Proof. The proof is obvious: we obtain immediately what was required • from the inclusion H(M2 ) ~ H(Ml). Lemma 1.2. Let {Mk : k = 1,2, ... } be a family of subsets of the real line. Then

Proof. When the outer measure of one of the sets M k , k = 1,2, ... , is infinite or when the sum in the right hand side is infinite the proof is obvious. In what follows we assume the opposite. Take an arbitrary E > o. For every k = 1,2, ... fix a family of intervals such that U, "2 Mk and k d('d ~ J.t*(Mk) + 2- E. Such a family exists by the definition of the outer measure. For the family, = U{ ,k: k = 1,2, ... } we have

'k

U {Mk: k = 1, 2, ... } ~ d(,)

= 2:)d('d:

k

u"

= 1,2, ... }

~L{J.t*(Md+2-kc: k=1,2, ... } 00

k=l

=

L{J.t*(Md: k

= 1,2, ... } +E.

Hence J.t*(U{Mk: k = 1,2, ... }) ~ {J.t*(Md: k = 1,2, ... } + c.

Since this is true for every number c > 0 the lemma is proved. • The outer measure of the empty set is equal to zero. In Lemma 1.2 we did not assume the nonemptiness of the sets M k • Therefore the analog of the inequality just proved for a finite family of sets is its particular case. Lemma 1.3. Let a subset of the real line be represented as a union of a finite number of pairwise disjoint segments. Then the outer measure of the subset is equal to the total length of these segments. Proof. Let M = U{[ak, bkJ: k = 1, ... , s} be the set and its representation from the statement of the lemma. Notice that by virtue of

97

Derivation and integration.

Theorem 2.6.7 these segments, and only they, are connected components of the set M and such a representation is unique. Let d = 2:~=1 Ib k - ak I. The inequality J.L*(M) ~ d follows from the consideration of the families = {( ak - E, bk + E): k = 1, ... , s}, E > O. The lemma will be proved when we show that d(,) ~ d for every (at most countable) family, of intervals covering the set M. By virtue of the compactness of the set M it is sufficient to prove this assertion under the assumption of the finiteness of the cover ,. It will be done with the help of an induction on the number of elements of, = {U1 , . . . , Ud. We will do it not for a fixed set M, but for all such sets at once. For t = 1 the assertion is obvious. Let the assertion be proved for t = i - I ~ 1. Prove it for t = i. We do not lose any generality when we assume in addition that the segments [ak, bd are enumerated according to their position in the line: al ~ b1 < a2 ~ b2 < ... < as ~ bs . We can also assume that just the interval Ut = (p, q) contains the point bs (in the opposite case we renumber the family,). If p < al then as in

,t:

V1

Vt

V3 ~

~

~

( ( ) () ) ........ ".... ( -+ -t-+---t----1f-+-t-at

() ( )

}

b1 a2 '-..--'

V2 Figure 4.2

......

the case t = 1, the assertion is obvious. Consider the case al ~ P « bs ) (Figure 4.2). Let Ml = M n [al,pl and M2 = M n [p, bsl. Each of these sets may be also represented as the union of a finite number of pairwise disjoint segments. If d1 and d2 are the total lengths of the components of Ml and M 2, respectively, then d = d 1 + d 2. The family,1 = {U1 , •.• , Ut-d covers the set MI. By the inductive hypothesis d(,I) ~ d 1 • The length of the interval Ut is not less than d 2. So d(,) = d( '1) + d( {Ut }) ~ d 1 + d 2 = d. The lemma is proved. • Lemma 1.4. Let an (at most) countable family, = {Mk : k = 1,2, ... } consist of pairwise disjoint segments, intervals and half intervals of the real line and E > O. Then: a) if d(,) < 00, then there are segments II"'" I j lying in different elements of the family, such that: J.L*(I1 U··· U I j ) ~ d(,) - c and J.L*«U,) \ (U{Il,"" Ij}) < E; b) if d(,) = 00, then there are segments I 1 , ••• ,Ij lying in different elements of the family, such that J.L*(I1 U ... U I j ) ~ c .

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CHAPTER 4

Proof. Let ak ~ bk be endpoints of a segment, an interval or an half interval M k , k = 1,2, .... If ak = -00 or bk = 00 for some k, then we have b) and the validity of the assertion is obvious: the set Mk contains segments of arbitrarily large length. Below we assume that -00 < ak ~ bk < 00. For k = 1,2, ... denote by h the segment [ak + 2- k - 3 E, bk - 2- k - 3 E], if 2- k- 2E ~ bk - ak' If the last condition is not fulfiled we put h = 0. The length l(Id of the segment (or empty set) h is not less than the number max{bk - ak - 2- k- 2E,0}. For the calculus of d(,) we have the series L~=1 (b k - ad· For the calculus of the total length of the segments h, k = 1,2, ... we have the series L~=1 l(h). Let us estimate the difference of partial sums of these series: i

i

L

i

2)bk - ad l(h) k=1 k=1

~

L

Ibk - ak -l(h)1

k=1

~

i

L 2- k- 2

E

< ~.

k=1

Therefore the sums of these series, i.e., the limits of the sequences of its partial sums, differ at most by ! < ~, if db) < 00, and coincide if db) = 00. There exists a number j such that l(Id+· .. +l(Ij) > db) - ~ (> db) -E), for db) < 00 and l(Id + ... + l(Ij) > 10 for db) = 00. By Lemma 1.3 the last observation completes the proof in the case b). In the case a) the previous estimate proves the first part of the assertion. The second part of a) follows from the fact that difference (U,) \ (U{ II, ... ,Ij}) lies in the union of the sets [aI, al + 2- 4 10], [b 1- 2- 4 10, b1], ... , [aj, aj + 2- j - 3 E], [bj - 2- j - 3 E, bj ], [aj+l' bj+l], [aj+2' bj+2], .... Therefore Lemmas 1.1 and 1.3 and the inequality

imply the estimate

fL*((U,) \ (U{Il,'" ,Ij}) ~ 2- 3 10 + ... + 2- j - 2E + (bj+l - aj+l) + (bj+2 - aj+2)

<

J+

db) - (db) -

+ ...

~)

<

E.

The lemma is proved. • Lemma 1.5. Let G be an open subset of the real line and 10 > 0. Then there exists a closed subset H of the real line such that H ~ G and fL*(G \ H)

< E.

Proof. Theorems 2.8.8 and 2.8.3 guarantees the representation of an open subset of the line as the union of an at most countable family of disjoint intervals. Use Lemma 1.4 to find for every k = 0, ±1, ±2, ... a closed subset Mk of the line such that

99

Derivation and integration.

Mk t;;;; G n (k, k + 1) and J-L*((G Let M =

{a, ±1, ±2, ... } n G.

n (k, k + 1)) \ Md <

2- lkl - 2 E.

The set

H = M U (U{Mk: k = 0, ±1, ±2, ... }) is closed in the line. (Every point x E IR lies in an interval of the form (k, k + 2). If x rf. H then x rf. M U Mk U Mk+l. There exists a neighborhood Ox t;;;; (k, k + 2) of the point x without points of the set M U Mk U Mk+l. Since Ox n H = 0, x rf. [H]. In view of the arbitrariness of the point x E IR \ H this means the closedness of the set H.) By Lemma 1.2 00

00

k=-oo

k=O

The lemma is proved. • Lemma 1.6. Let F be a closed subset of the real line and 10 > 0. Then there exists a neighborhood OF of the set F such that J-L * (0 F \ F) < E. Proof. For G = IR \ F let us find H according to Lemma 1.5. When we take OF = IR \ H we obtain what was required because OF \ F = G \ H. The lemma is proved. • A subset M of the real line is called measurable if for every 10 > there are a closed subset F and an open subset G (Figure 4.3) of the line such that F t;;;; M t;;;; G and J-L * (G \ F) < E.

°

--+----+ G: f-

-)

+------ + F:

)-

.

H: Figure 4.3

Lemmas 1.5 and 1.6 imply that open and closed sets are measurable. Evidently the complement to a measurable set is measurable (see the proof of Lemma 1.6). If a set M t;;;; IR is measurable its outer measure J-L*(M) is called the measure of the set M and is denoted by J-L(M). Lemma 1. 7. The intersection of two measurable subsets of the real line is measurable. Proof. Let M i , i = 1,2, be measurable subsets of the line and c > 0. Take a closed Fi and an open G i sets such that Fi t;;;; Mi t;;;; G i and J-L(G i \ F i ) < ~. Evidently Fl nF2

C;;;;

Ml nM2 t;;;; G l nG 2

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CHAPTER 4

and By Lemma 1.2

/L*((GI n G 2 )

\

(FI n F2 ))

~ /L*(GI \ FI) + /L*(G 2 \ F2 ) < ~ + ~

=

E.

The lemma is proved. • Lemma 1.7 and the above remark about the measurability of a complement imply: Lemma 1.8. The difference of two measurable subsets of the real line is measurable. • In addition, MI U M2 = IR \ ((1R \ M I ) n (IR \ M 2)) and Lemma 1.7 and the same remark (or Lemma 1.8) imply the measurability of the union of two measurable set. An obvious induction on the number of elements gives: Lemma 1.9. The union of a finite family of measurable subsets of the real line is measurable. • Lemma 1.10. Let MI and M2 be disjoint closed subsets of the real line. Then ./L(MI U M 2 ) = /L(MI) + /L(M2 ).

Proof. The inequality /L(MI U M 2 ) ~ /L(MI ) + /L(M2 ) follows from Lemma 1.2. Our lemma will be proved when we establish the validity of the inverse inequality. Let, be an arbitrary cover of the set MI U M2 by intervals. By virtue of the normalness of the line there exist disjoint neighborhoods OMI and OM2 of the sets MI and M 2 , respectively. The sets Vi = (U,) n OMI and V2 = (U,) n OM2 are also disjoint neighborhoods of the sets MI and M 2 . The set of connected components of the set V = VI UV2 is split in two subset: components of the set Vi and components of the sets V2. By Lemmas 1.2-4 the measure of an open set is equal to the total length of its connected components. Therefore /L(V) = /L(Vd + /L(V2 ). By Lemma 1.2 /L(U,) ~ d(r). By Lemma 1.1 /L(V) ~ /L(U,) and

/L(Md

+ /-l(M2 )

~

/L(Vi)

+ /L(V2).

Thus

In view of the arbitrariness of the cover, we have

The lemma is proved.

•

101

Derivation and integration.

Theorem 1.1. Let measurable subsets M k , k = 1,2, ... , of the line be pairwise disjoint and L~l J.t(Md < 00. Then the set U{Mk : k = 1,2, ... } is measurable and 00

J.t(U{ M k : k

= 1,2, ... }) = L

J.t(Mk )·

k=l

Proof. Take an arbitrary c > O. According to the convergence of the series L~l J.t(Md fix an index j such that

Now Lemma 1.2 implies the existence of a family of intervals, such that

U{Mk: k = j

+ l,j + 2, ... } ~ U, and db) < ~.

For k = 1, ... , j fix a closed Fk and an open G k sets such that ~ Mk ~ G k and J.t(G k \ Fk ) < 2- k - 1 c. The sets F = FI U ... U Fj and G = G 1 U ... U Gj U (U,) satisfy the conditions

Fk

F ~ U{Mk: k = 1,2, ... } J.t(G \ F) ::::; J.t(G I \ Fd + ... + J.t(G j ::::;

c + ...

2- 2

+ 2-

In view of the arbitrariness of c

j - 1

c+

~ \

2- l

G, Fj ) c

+ J.t(U,)

< c.

> 0 this means the measurability of the

set

M = U{Mk: k = 1,2, ... }. By Lemmas 1.1 and 1.10 j

00

L J.t(Md k=l

c ::::;

L J.t(Mk) k=l

c

j

2 ::; L

J.t(Fk) = J.t(F)

k=l

::::; J.t(M) ::::; J.t(G) ::::; J.t(F) + J.t(G \ F) ::::; J.t(F)

k=l

k=l

In view of the arbitrariness c > 0 00

J.t(M) =

L

k=l

J.L(Mk).

+c

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The theorem is proved. • Lemma 1.11. A set M C ~ is measurable if and only if for every segment I ~ ~ the set M n I is measurable. Proof. Necessity follows from Lemma 1.7. Sufficiency. Take an arbitrary c > 0. For every k = 0, ±1, ±2, ... fix a closed Fk and an open G k subsets of the line such that

The set F = U{Fk: k = 0, ±1, ±2, ... } is closed. (Every point x E ~ lies in an interval (k, k + 2). If x f/. F, then x f/. Fk U Fk+I' The set Ox = (k, k+2)\(FkUFk+d is a neighborhood of the point x without points of the set F.) The set G = U{G k : k = 0, ±1, ±2, ... } is open and 00

F

~

M

~

J1.(G \ F) ~

G,

L k=-oo

00

2- lkl - 2 c

< L 2- k- I c = c. k=O

The lemma is proved. • Theorem 1.2. Let {Mk: k = 1,2, ... } be a family of measurable subsets of the real line. Then the set M = U{Mk: k = 1,2, ... } is measurable and

Proof. The proof use Lemma 1.11. Let [a, b] be an arbitrary segment of the line. We have the representation M n [a, b] = where

U"

, = {(Mk

n [a, b]) \

(U{MI , ... , Mk-d): k = 1,2, ... }.

The measurability of elements of the family , follows from Lemmas 1.8 and 1.9. By Theorem 1.1 and Lemmas 1.1 and 1.3 partial sums of the series l: {J1.(r) : r E ,} are bounded from above by the number b-a. Therefore the series converge. Theorem 1.1 implies the measurability of the set M n [a, b]. Referring to Lemma 1.11 completes the proof of the measurability of the set M. The validity of the inequality follows from Lemma 1.2. The theorem is proved. • Assertion 1.1. Under the hypotheses of Theorem 1.2 let MI ~ M2 ~ M3 ~ .... Then J1.(M) = .lim J1.(Mi)' '-+00 Proof. I. Lemma 1.1 implies that:

II. If limi-+oo J1.(Mi ) =

00,

then by I the fact in question is obvious.

103

Derivation and integration.

III. Let limi ..... oo J.t(Mi ) <

00.

By Theorem 1.1:

J.t(M) = J.t(M1 ) + J.t(M2 \ Md + J.t(M3 \ M 2 ) + ... = 1-+00 .lim (J.t(Md + J.t(M2 \ M 1 ) + ... + J.t(Mi \ Mi-d) = lim J.t(M;). 'l.~OO

The assertion is proved. • Theorem 1.3. Let {Mk: k = 1,2, ... } be a family of measurable subsets of the real line. Then the set n{Mk: k = 1,2, ... } is measurable. Proof. By virtue of the equality

n{Mk: k=1,2, ... }=~\(U{~\Mk: k=1,2, ... }) our assertion follows from Lemma 1.8 and Theorem 1.2. The theorem is proved. • Assertion 1.2. Let under the hypotheses of Theorem 1.3 J.t(Md < 00 and Ml ;2 M2 ;2 M3 ;2 . . .. Then

Proof.

J.t(M) = J.t(Md - J.t(M1 \ M) = J.t(Md - J.t(U{ Ml \ M i ~ J.t(Md - .lim J.t(M1

\

:

i = 1,2, ... })

Mi )

' ..... 00

= lim (J.t( Md - J.t( Ml \ M i )) ' ..... 00

The equality (*) is true by Assertion 1.1. The assertion is proved. • Example 1.1. A set M satisfying the condition J.t*(M) = 0 is called a set of measure zero. Every such a set is measurable (for the verification of the condition occurring in the definition of a measurable set we can take F = 0). Lemma 1.2 implies that the union of an (at most) countable family of sets of measure zero is a set of measure zero. In particular, the measure of an one point set is equal to zero. So the measure of every countable set is equal to zero (in particular, the set is measurable ). Example 1.2. Calculate the measure of the Cantor perfect set described in Example 1.6.4. Use the notation introduced there. For an arbitrary segment [a, b] we have J.t(S([a, b])) = ~J.t([a, b]). So J.t(Sk+l([O,l])) = ~J.t(Sk([O, 1])) and J.t(Sk([O,l])) = (~)k for every k = 1,2, .... By Lemma 1.1 J.t(K) ~ (~)k for every k = 1,2, .... Hence J.t(K) = O.

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Let M be a measurable subset of the line. A subset Mo of the set M is called a subset of full measure if the measure its complement in M is equal to zero: /l(M \ Mo) = O. The complement in the segment [0, 1J to the Cantor perfect set gives an example of a set of full measure. The set of rational numbers is countable. Hence its measure is equal to zero. Therefore the set of irrational numbers is a set of full measure on the line. rt is suitable also to use the following terminology. Let P be a property such that points of a measurable set M can possess or not possess it. We will say that the property P is fulfilled almost everywhere on the set M (or at almost all points of the set M) if the set of all points of the set M, in which the property P holds, is a subset (of the set M) of full measure. Non-measurable subsets exist. Here we will abstain from the construction of a corresponding example. 2. Measurable functions A function (a mapping) f defined on a measurable set M ~ ~ and taking values in a topological space Y is called measurable if for every open subset U of the space Y the set f-l(U) is measurable. By Theorem 1.3.3 and Lemma 1.8 a function f measurable if and only if for every closed subset H of the space Y the set f-l(H) is measurable. We say that a function f is measurable on a set M ~ 7r(J), if the function 11M is measurable. Example 2.1. A continuous function is measurable (see §1 and Theorem 1.7.5). Let functions fl : Ml --+ Y and 12 : M2 --+ Y be defined on measurable sets Ml and M2 and take values in a set Y. We say that the functions II and 12 coincide almost everywhere, if

We say that the functions II and f2 coincide almost everywhere on a measurable set M ~ ~ if the functions flIMn7l"(h) and f2IMn7l"(h) coincide almost everywhere. Lemma 2.1. Let functions fl and 12 be defined on measurable sets Ml and M 2, respectively, and take values in a topological space Y. Let they coincide almost everywhere and the function fl be measurable. Then the function 12 is measurable too. Proof. Let A = {t: t E M2 \ M o, h(t) E U}, Mo = {t: t E Ml n M 2, fl(t) = f2(t)}.

Derivation and integration.

105

The estimate J.t(A) ~ J.t(M2 \ Mo) = J.t(M2 ) - J.t(Mo) = 0 follows from Lemma 1.1 and Theorem 1.1. For every open set U of the space Y we have

Lemma 1.9 implies the measurability of the set f2- 1 (U). The lemma is proved. • Example 2.2. Define on the real line the function X ('Dirichlet function') in the following way. Let

X(t)={~

if t is irrational if t is rational.

The measure of the set of rational numbers is equal to zero. Hence the Dirichlet function coincides almost everywhere with the function identically equal to 1. Lemma 2.1 implies the measurability of the Dirichlet function. Theorem 2.1. Let continuous mappings fl and f2 of a segment [a, b], a < b, into a Hausdorff space Y coincide almost everywhere. Then the mappings fl and f2 coincide. Proof. Lemmas 1.3 and 1.1 and Theorem 1.1 imply that the segment [a, bJ itself is the unique closed subset of full measure of the segment. The assertion in question follows from Lemma 1.7.4. The theorem is proved . • Example 2.3. Let M = [0, IJ U {2}. The functions fl(t) == 1 and

h(t) =

{~

if t E [O,IJ if t = 2

are continuous. They coincide almost everywhere but they are different. Theorem 2.2. Let a function f be defined on a measurable set M ~ lR and take values in a topological space Y with a countable base. Then the following properties are equipotent: a) the function f is measurable; b) there exists a base of the space Y, for every element U of which the set f- 1 (U) is measurable; c) there exists a sub-base of the space Y, for every element U of which the set f- 1 (U) is measurable. Proof. a=? b =? c. This is obvious because every element of a base is an Open set and every base is a subbase. b =?a. Let f3 be a base occurring in condition b, V be an arbitrary open subset of the space Y. Denote by f30 the set of (all) elements of the base f3 lying in the set V. The family f30 covers the subspace V. Theorem 1.6.10 implies the existence of an (at most) countable subfamily f31 of f3o, which

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also covers the set V. Theorem 1.2 implies the measurability of the set f-l(V) = U{f-l(U): U E ,Bd. c =?b. Let ,B be a sub-base from c. By the definition of a sub-base the set ,Bo of intersections of all finite subfamilies of ,B is a base. Theorem 1.3 implies that for every element V = U1 n ... n Us of the base ,Bo (where U1 , ... ,Us E ,B) the set f-l(V) = f-1(Ud n ... n f-l(Us ) is measurable. The theorem is proved. • Theorem 2.3. Let a set M ~ IR be the union of an (at most) countable family of measurable sets {Mk : k = 1,2, ... }. Let a function f be defined on the set M and take values in a topological space Y. Let the function f be measurable on every set M k , k = 1,2, .... Then the function f zs measurable. Proof. For every open set U of the space Y we have f-l(U)

= U{f-l(U) n M k :

k

= 1,2, ... }.

Theorem 1.2 implies the measurability of the set f-l(U). The theorem is proved. • Corollary. Let a set M ~ IR be the union of an (at most) countable family of measurable sets {Mk: k = 1,2, ... }. Let a function f be defined on the set M and take values in a topological space Y. Let f be constant on every set MkI k = 1,2, .... Then the function f is measurable. • Theorem 2.4. Let measurable functions h, k = 1,2, ... , be defined almost everywhere on a measurable set M ~ IR and take values in a metric space (Y, p). Let the sequence {h: k = 1,2, ... } converge for almost all t E M to a function f (i.e., for almost all t E M we have fk(t) ---+ f(t)). Then the function f is measurable. Proof. The hypotheses and the conclusion of the theorem remain true under the change of the stated functions to functions which coincide with them almost everywhere (see Lemma 2.1). Thus it is sufficient to prove the theorem assuming in addition that the functions are defined on the entire set M and that the needed convergence is at every point of the set M. Let U be an arbitrary open subset of Y. If U = Y then the measurability of the set f-l(U) =M is obvious. In the opposite case by Corollary 2 of Lemma 2.3.1 the function cp(t) = p(t, Y \ U) is continuous. The sets Uk = {t: t E Y, cp(t) > 2- k }, k = 1,2, ... , are open. The sets Hk = {t : t E Y, cp(t) ~ 2- k }, k = 1,2, ... , are closed, U1 ~ HI ~ U2 ~ H2 ~ ... ~ U and U~lUk = U. If x E f-I(U) then f(x) E Urn for some m = 1,2, ... and Urn is a neighborhood of the point f(x). The convergence fk(x) ---+ f(x) implies the existence of a number n ~ 1,2, ... such that for every index k ~ n the point fk(x) belongs to Um. Therefore x E U~=I U~=l n":=nl;;l(Um ). )

Derivation and integration.

107

If x E M \ f- 1 (U) then for every index m = 1,2, ... the set Y \ Hm is a neighborhood of the point f(x). The convergence fk(X) - t f(x) implies the existence of a number n = 1,2, ... such that for every index k ~ n we have h(x) ¢ Hm ;2 Urn. Thus for every n = 1,2, ... the point x does not belong to the set n~nfk1(Urn). Hence x ¢ U~=l U~=l n'k=nfk- 1 (Um ). So f-1(U) = U~=l U~=1 n'k=nfk- 1(Um ). The assertion now follows from Theorems 1.2 and 1.3. The theorem is proved. •

Lemma 2.2. Let Yi and Y2 be topological spaces. Let a set M ~ ~ be measurable. Let a mapping f : M - t Yi be measurable and a mapping g: Y1 - t Y2 be continuous. Then the mapping gf : M - t Y2 is measurable. Proof. Let U be an arbitrary open subset of the space Y2 • By Theorem 1. 7.5 the set g-l (U) is open in Y1. The definition of a measurable mapping implies the measurability of the set (gf)-l(U) = f-1(g-1(U)). The lemma is proved. • Theorem 2.5. Let a function f be defined on a measurable set M ~ ~ and take values in the Euclidean space ~n: f (t) = (I1 (t), ... , f n (t)). The function f is measurable if and only if each function f1, . .. ,fn : M - t ~ is measurable. Proof. Necessity follows from Lemmas 2.1.1 and 2.2. Sufficiency. Let U be an arbitrary open subset of the line, i = 1, ... ,n. We have f- 1 (~x ... x ~ x U x ~ x ... x ~) = fi- 1 (U), where the factor U of the product is in the i-th position. Sets of the form ~ x ... x ~ x U x ~ x ... x ~ constitute a sub-base of the space ~n. Therefore the stated representation of the preimage of an element of this subbase, our assumptions about the functions f1, ... , fm and Theorem 2.2c imply the measurability of the function f. The theorem is proved. • This theorem, Lemma 2.2, and the continuity of operations of the addition and of the multiplication imply the following two assertion. Theorem 2.6. Let a set M ~ ~ and functions f, 9 : M - t ~n be measurable, A E R Then the functions f + 9 and Af are measurable too .• Theorem 2.7. Let a set M ~ ~ and functions f, 9 : M - t ~ be measurable. Then their product f 9 is a measurable function. • These assertions give a large reserve of measurable functions. The continuity of the function h(t) = t- 1 for t :I 0, the continuity of the multiplication, Lemma 2.2 and Theorem 2.5 imply the measurability of the fraction with a measurable function in the numerator and with a measurable nonvanishing function in the denominator. The function h(t) = It I is continuous. Lemma 2.2 implies that if a real function f(t) is measurable then the function If(t)1 is measurable too. Next, if f(t) and g(t) are measurable real functions then the last remark and

108

CHAPTER 4

Theorem 2.6 imply the measurability of the function.

'Pl(t)

= min{J(t),g(t)} =

f(t)

+ g(t) ~If(t) -

g(t)l,

'P2(t)

= max{J(t),g(t)} =

f(t)

+ g(t) + If(t) -

g(t)l.

2 In particular, for every measurable function f(t) its non-negative f+(t) = max{J(t), O} and non-positive f-(t) = min{J(t) , O} parts are measurable too. Let us finish this section by the following remark. Theorem 2.8. Let a real function f be defined on a measurable set M ~ lR. Then the following conditions are equipotent: a) the function f is measurable; b) for every number c E ~ the set {t: t E M, f(t) > c} is measurable; c) for every number c E ~ the set {t: t E M, f(t) ::::; c} is measurable; d) for every number c E ~ the set {t: t E M, f(t) < c} is measurable; e) for every number c E ~ the set {t: t E M, f(t) ~ c} is measurable. Proof. a =*b. This follows immediately from the definition of a measurable function and the openness of the subset (c, 00) of the line. b ¢::}c and d ¢::}e by virtue of the measurability of the complement to a measurable set (see Lemma 1.8). c =* d by virtue of the representation

{t: t E M, f(t) < c}

= U{{t:

t E M, f(t) ::::; c - 2- k }: k

= 1,2, ... }

and Theorem 1.2. e =* b may be proved analogously to the previous implication. Thus conditions b,c,d and e are equipotent. b,c=*a by Theorem 2.2c. The theorem is proved.

•

3. Derivation of non-decreasing functions Let the domain of a real function f contain a set M ~ lR. The function f is called: increasing on the set M if f(s) < f(t) for every two points s < t of the set M; non-decreasing on the set M if f(8) ::::; f(t) for every two points 8 < t of the set M; decreasing on the set M if f(8) > f(t) for every two points s < t of the set M; non-increasing on the set M if f(8) ~ f{t) for every two points 8 < t of the set M.

Derivation and integration.

109

All such functions are called monotone on the set M. Increasing or decreasing functions (on the set M) are also called strongly monotone (on the set M). Let a set M ~ lR lie in the domain of a real function I. A point s of the set M is called a right increasing point (of the function 1 on the set M, respectively, a left increasing point if on the right (respectively, on the left) of the point s there is a point t E M such that I(t) > I(s). If the domain of the function 1 coincides with the set M then usually we omit the mention of the set M in these terms. Lemma 3.1. Let a real function 1 be defined and be continuous on the segment [a, b], a < b, of the real line. Then the set S of right increasing points (respectively, left increasing) is open (in the segment [a, b]). If a < {3 are endpoints of a connected component of the set S then 1 (a) ::;;; 1 ({3) (respectively, 1 (a) ~ f ({3)). Proof. If s is a right increasing point then there exists a point t E (s, b] such that I(t) > f(s). By virtue of the continuity of the function f there exists a number c: > 0 such that f(OES) ~ (-oo,f(t)). Then OES ~ S. This means (in view of the arbitrariness of the point s E S) the openness Figure 4.4 of the set S. Let us prove the last assertion of the lemma. Take an arbitrary point s E (O'.,{3). The number r = sup/([s,b]) is greater than I(s). By Remark 2.3.4 r E I([s, b]). Let t = inf(j-l(r) n [s, b]). Then [s, t) ~ Sand t rf. S. Therefore {3 = t. In view of the arbitrariness of the point s E (a, {3) we have 1((O'.,{3)) ~ (-oo,r). The closed ness of the set (-oo,r] implies the inclusion 1([0'., {3]) ~ (-00, r]. Therefore f(O'.) ::;;; r = f(t) = 1({3)· The function g(x) = f( -x) is defined on the segment [-b, -a]. Left increasing points of the function f and right increasing points of the function g on the segment [-b, -a] are symmetric with respect to zero. Therefore the proved fact implies the assertion of the lemma for left increasing points. The lemma is proved. • Let us go on to the question about the existence of a derivative of a real function 1 defined on a segment of the real line. The derivative at a particular point need not exist. Corresponding examples are well known and we will not discuss them. But if a point x in question is not an endpoint of the segment then in every case we can define the lower )../ and the upper A/ limits of the quotient J(tt:~("') as the point t tends to the point x from the left and the lower )..r and the upper Ar limits of this quotient as the

110

CHAPTER 4

point t tends to the point x from the right, i.e.,

Al = lim inf f (t) - f (x) , t-+x, t<x t - x

f(t) - f(x) · A 1= 1Imsup ,

Ar = lim inf f (t) - f (x) , t-+x, t>x t - x

Ar = limsup f(t) - f(x). t-+x, t>x t- x

t- x

t-+x, t<x

Example 3.1. Consider the functions

f(x) = {xosin

() { M

9 x =

o

~

for x =/: 0, for x = 0,

sin ~ x

=/: 0,

for x

for x = 0,

on the segment [-1,1]. For the function f at the point x = 0 we have Al = Ar = -1, Al = Ar = 1 (Figure 4.5). For the function 9 at the same point we have Al = Ar = -00, Al = Ar = +00 (Figure 4.6).

! .

.

' HI

, "

+- - ; .

Figure 4.5

(, 'I

lil[!:: ; 1[. .~,

~ iV" ': J I! ..

\/

Figure 4.6

For the existence of the (finite) derivative of a function f at a point x the fulfilment of the condition -00 < Al = Al = Ar = Ar < 00 is necessary and sufficient. Lemma 3.2. Let a non-decreasing real function f be defined and be continuous on the segment [a, b], a < b, of the real line. Then the function f has the derivative at almost every point of the segment [a, bJ. Proof. I. Show first that almost everywhere Ar < 00. Let A be the set of (all) points of the segment [a, bJ, at which Ar = 00. For every point x of the set A and every positive integer k there exists a )

Derivation and integration.

111

point t E (x, b] such that

f(t) - f(x) > k. t-x This means that f(t) - kt > f(x) - kx, i.e., the point x is a right increasing point for the function 9k(S) = f(s) - ks. By Lemma 3.1 the set Ak ofright increasing points of the function 9k is open. The point b cannot be right increasing point. We obtain from Corollary of Theorem 2.6.7 and Theorem 2.6.3 that the set Ak may be represented as the union of an (at most) countable set 'Yk of its (pairwise disjoint) components. The components are either intervals or half intervals with the closed endpoint a (no more than one such components). By Lemma 3.1 for every element of the set 'Yk with endpoints a < 13 we have 9k(a) ~ 9k(f3), 13 - a ~ Hf(f3) - f(a)). In view of the monotonicity of the function f this implies that d("(k) ~ tU(b) - f(a)). So J.L*(A) ~ J.L*(Ak) ~ Hf(b)- f(a)). Since this is true for every k = 1,2, ... , J.L*(A) = O. This was required. II. Show that almost everywhere Ar ~ A/. At the points a and b one of these numbers is not defined. Denote by B the set of all points of the interval (a, b), at which the inequality under consideration is not fulfilled, i.e., we have there the inequality A/ < Ar • Denote by P the set of all couples p = (q, r) of rational numbers, in which the first member is strongly less than the second one: q < r. The set P is countable (see Example 1.1.2 and Corollary of Theorem 1.1. 7). The set B may be represented as the union of (countable number of) sets B p , p = (q, r) E P, where Bp is the set of points, at which A/ < q < r < Ar . Our aim will be achieved when we show that the outer measure of every set B p , PEP, is equal to zero. Assume the opposite, i.e., that J.L*(Bp) = C > 0 for some p = (q, r) E P. There exists a covering "I of the set Bp by intervals such that d(,,() < ~c (with the notation of §1). Let V = (a,f3) E "I, By = Bp n V and x E By. At the point x we have A/ < q. Thus there exists a point t E (a, x) such that J(tt~(x) < q and hence f(t) - qt > f(x) - qx, i.e., the point x is a left increasing point of the function f(s) - qs on the segment Iv = [a, 13]. By Lemma 3.1 the set Mv of (all) such points is open in the segment Iv. The left endpoint of the segment Iv cannot be an left increasing point. Corollary of Theorem 2.6.7 and Theorem 2.6.3 imply that the set Mv may be represented as the union of an (at most) countable set "Iv of its (pairwise disjoint) components. The components are either intervals or an half interval with the closed endpoint sup Iv. By Lemma 3.1 for every element U of the set "Iv with endpoints a' < 13' we have the inequality f(a') - qa' ~ f(f3') - qf3' or (3.1)

f(f3') - f(a')

~

q(f3' - a').

CHAPTER 4

112

By virtue of the inequality Ar > r for every point x E UnBir there exists a point t E (x, {3') such that f(tt:~(x) > r, i.e., f(x) - rx < f(t) - rt. Thus the point x is a right increasing point for function f (s) - r s on the segment [a', {3']. By Lemma 3.1 the set M*(c/,{3') of (all) such points is open in the segment [a', {3']. By Corollary of Theorem 2.6.7 and by Theorem 2.6.3 the set M*(a', {3') may be represented as the union of an (at most) countable set ,*(a',{3') of its (pairwise disjoint) components. The components are either intervals or an half interval with the closed endpoint, which coincides with a'. By Lemma 3.1 for every element of the set (a', {3') with endpoints a" < {3" we have the inequality f(a") - ra" ~ f({3") - r{3", i.e.,

,*

{3" - a"

~ ~(f({3") r

The monotonicity of the function

d(r*(a', {3'))

f(a")).

f implies that

~ ~(f({3') r

f(a')).

By (3.1) the total length of elements of the set ,~= U{r*(a',{3'):

(a',{3') E 'v, ( (a',{3] E 'v)}

does not exceed ;J.L(V). The family ,* = U{rir: V E ,} covers the set Bp. By Lemmas 1.2 and 1.1 (and the measurability of open sets)

The contradiction obtained (J.L*(Bp) < J.L*(Bp)) shows that our assumption is false. Hence J.L*(Bp) = O. III. When we go from the function f (x) to the function (x) = - f ( - x ) the notions of left and right change places. From this observation we obtain immediately from the previous reasoning that at almost every point of the segment [a, b] we have Al ~ Ar . When we compare the above inequalities with the obvious estimates Al ~ Al and Ar ~ Ar we obtain that 0 ~ Ar ~ Al ~ Al ~ Ar ~ Ar < 00 at almost every point of the segment [a, b]. So 0 ~ Ar = Al = Al = Ar < 00. Therefore at almost every point of the segment [a, b] the derivative exists. The lemma is proved. • It is natural to ask in what measure the derivative describes the behavior of a function. There exist situations being a long way from the ones which we met when we considered continuously differentiable functions. For the construction of a corresponding example we will need

r

Derivation and integration.

113

Lemma 3.3. Let M be a closed subset of the segment [a, b], a < b. Let a real function f be defined and be non-decreasing on the segment [a, b], be continuous on the set M and on the closure of every component of the set [a, b] \ M. Then the function f is continuous. Proof. The continuity of the function f at points of the set [a, b] \ M follows from the openness of the set [a, b] \ M in the segment [a, b] and hypotheses of the lemma. Thus we need to prove the continuity of the function f at an arbitrary point x of the set M. Take an arbitrary number c > O. By virtue of the continuity of the function f on the set M there exists a number 8 > 0 such that If(t) - f(x)1 < c for every t E (x - 8, x + 8) n M. If (x - 8, x) n M "# 0 then denote by c an arbitrary point of the set (x - 8, x) n M. If (x - 8, x) n M = 0 then by virtue of the continuity of the function f on the closure of the component of the set [a, b] \ M, which contains the interval (x - 8, x), there exists a number c E (x - 8, x) such that If(c) - f(x)1 < c. Thus in this case too we have fixed a point c. If (x, x + 8) n M "# 0 then denote by d an arbitrary point of the set (x,x + 8) n M. If (x,x + 8) n M = 0, then by virtue of the continuity of the function f on the closure of the component of the set [a, b] \ M, which contains the interval (x, x + 8), there exists a number d E (x, x + 8) such that If(d) - f(x)1 < c. Thus in this case too we have fixed a point d. The interval (c, d) is a neighborhood of the point x. By virtue of the monotonicity of the function f we have

f(( c, d))

~

[f( c), f(d)]

~

(f(x) - c, f(x)

+ c).

The lemma is proved. • Example 3.2. We will construct a continuous mapping of the segment [0, 1] onto itself, a so called 'Cantor stairs'. Use the notation of Examples 1.6.4 and 2.2.8. In discussion of properties of this function the result of Example 1.2 will be also essential: the measure of the Cantor perfect set is equal to zero. In Example 2.2.8 we have pointed an homeomorphism f : K --t DN. Define first the function h ('Cantor stairs') on the set K and next extend it to the entire segment [0, 1]. For a point x = {Xl, X2, ... } E DN put Sets VX1, ... ,Xk

=

{y

= {YI, Y2,'" }: Y E D N , YI = Xl,··· ,Yk =

= {Yl, Y2,' .. } E VX1, ... ,Xk' then h ( )1 ~ IXk+l - Yk+d + IXk+2 - Yk+21 + 0 Y '" 2k+1 2 + 2 ' ..

constitute a base at the point x. If Y

Ih o( x ) -

xd

k

~

111 + 2k+2 + ... = 2k '

2k+l

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114

This implies the continuity of the mapping ho and hence the continuity of the mapping h = hof. Let [e, d] be a component of the set Sk([O, 1]). Then (with the notation of Example 2.2.8) for every i = 1, ... , k the set fi([e, d]) consists of one point. Denote it by ai. The values of the mapping f at the points e and d may be written as

f(e) = {al, ... ,ak,O,O, ... }, Let el

f(d) = {al, ... , ab1, I, ... }.

= e + ~(d - e), d l = e + Hd - e). We have:

f(el) = {al, ... ,ak,O,I,I, ... }, Therefore

and (3.2)

h() = h(d ) = al el

1

2

+

...

ak

_1_ = h(e)

+ 2k + 2k+l

+ h(d) . 2

Thus values of the function h at the endpoints of a component (el, dd of the complement to the Cantor perfect set coincide. Extend the function h to the segment [0,1] and put h(x) = h(ed for every x E (el,dd. The equality (3.2) helps to imagine the graph of this function (Figure 4.7). The function h is monotone. Lemma 3.3 implies its continuity. The continuity, the equalities h(O) = and h(l) = 1, Corollary 1 of TheoFigure 4.7 rem 2.6.4, and Theorem 2.6.3 imply the coincidence h([O, 1]) = [0,1]. The complement of the Cantor perfect set in the segment [0,1] has the full measure. At every its point the derivative of the function h exists and is equal to zero. We have constructed the continuous non-decreasing nonconstant function, the derivative of which almost everywhere is equal to zero. Recall for comparison that if the derivative of a function exists and is equal to zero everywhere (but not 'almost everywhere') then the function

°

Derivation and integration.

115

is constant. It seems that Lemma 3.2 gives a possibility to introduce the notions of the integral and of the primitive for a new rather large class of functions. Example 3.2 shows that here we go away from relations between functions and derivatives to which we are accustomed in dealing with continuously differentiable functions. However in Lemma 3.2 we made a step in the direction needed. Our nearest aim will be to select a class of functions which have derivatives almost everywhere and which are uniquely determined by their derivatives up to constant terms. We will pass to it in the next section, but now Theorem 3.1. A convergent series of continuous non-decreasing functions f(x) = fl (x) + f2(X) + h(x) + ... on a segment [a, b], a < b, admits almost everywhere the termwise derivation:

f'(x)

= f~(x) + f~(x) + f~(x) + ....

Before to prove the theorem let us make some introductional remarks. As usual, the sum of the series means the limit of the sequence of the partial sums

as k -+ 00. In the statement of the theorem we did not define exactly the type of convergence of the series. Assume now the convergence of the series at points a and b only. By virtue of the monotonicity of the functions fk' k = 1,2, ... for every point t of the segment [a, b] and for every index i < j we have In view of the arbitrariness of the point t E [a, b] this implies that

where in the left hand side the norm of the uniform convergence appears. This inequality, the convergence of the sequences {s d a): k = 1, 2, ... } and {s k (b): k = 1, 2, ... }, and the fact that every convergent sequence is fundamental (see §1.6) imply that the sequence {Sk: k = 1,2, ... } is fundamental with respect to the norm of the uniform convergence. From Corollary of Theorem 3.1.2 we obtain the convergence of the sequence {Sk : k = 1,2, ... } to a continuous function f. This guarantees the summability of Our series with respect to the norm of the uniform convergence. The sum f of our series is non-decreasing. By Lemma 3.2 terms of the series and the sum have derivatives almost everywhere on the segment [a, b]. Proof of Theorem 3.1. Consider a subset M of full measure of the segment [a, bj, at points of which derivatives of all terms and of the sum of

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116

the series exist. For every point x E M and for every point t E [a, b] \ {x} we have =

t-x

t- x

Signs of terms of the numerator in the left hand side coincide with the sign of the denumerator. Thus for every number N

When we pass to the limit as t

--+

x we obtain

N

L f~(x) ~ J'(x). k=1

Since this is true for every N = 1,2, ... the nonnegativity of the terms implies the summability of the series of the derivatives and the estimate (3.3)

f{(x) +

f~(x)

+ f~(x) + ...

~

J'(x).

Now for every k = 1,2, ... fix a number i(k) = 1,2, ... such that SiCk) II ~ 2- k. The above remarks may be applied to the series

Ilf -

g(x) = gl (x) + g2(X) + g3(X) + ... , where gk(X) = f(x) - Si(k)(X) = fi(k)+l(X) + fi(k)+2(X) + .... Thus at almost every point x of the segment [a, b] we have

g'(x)

~ g~ (x)

+ g~(x) + g;(x) + ...

The convergence of a numerical series implies that their terms tend to zero. In our case this means that S~(k)(X) --+ f'(x) as k --+ 00. Now (3.3) and the nonnegativity of terms of the series in the left hand side of the inequality (3.3) imply that Si(X) --+ f'(x) as i --+ 00 for almost all x E [a, b]. The theorem is proved. • 4. Derivative with respect to a set Let the domain of a function f with values in the Euclidean space ~n (for n = 1 in the real line ~) contains a set M ~ R Let x be a non-isolated point of the set M. If the limit of the quotient I(tt:~("') as t E M, t --+ x exists, it is called a derivative of the function f with respect to the set M at the point x and is denoted by f~(x).

117

Derivation and integration.

Following the standard method we prove that if two functions have the derivatives with respect to a set M ~ IR at a point x E M then their sum, difference, and product of each of these functions by a number also have such a derivative and its value is equal, respectively, to the sum, the difference and the product by the number of the values of the derivatives of initial functions. A mapping f (t) = {fl (x), ... , f n (t)} of the set M into the space IR n has the derivative at a point x E M with respect to the set M if and only if the coordinate functions fl(X), ... , fn(x) have the corresponding derivatives (necessity follows from Lemmas 2.1.1 and 1.7.2, sufficiency follows from Lemma 2.1.2). In this case fk(x) = {(fl)~(X), ... , (fn)~(x)). Lemma 4.1. Let the domain of a function f with values in the Euclidean space IR n contain sets A ~ B (~ 1R). Let x be a non-isolated point of the set A (hence x is a non-isolated point of the set B too). Let the derivative f~(x) exist. Then the derivative f~(x) exists and f~(x) = f~(x). Proof. The proof is obvious. • Lemma 4.2. Let the domain of a real function f contain a (nonempty) compact set M ~ R Let the function f be continuous and non-decreasing on the set M. Then the function f has a derivative with respect to the set M at almost every point of M. Proof. It is sufficient to prove the assertion under the additional assumption 7r(f) = M. Let a = inf M and b = supM. The set L = [a, bj \ M is open in the segment [a, bj and does not contain the points a and b. Theorems 2.6.7 and 2.6.3 imply the representation of the set L as the union of a set 'Y of pairwise disjoint intervals. Endpoints of these intervals belong to M. Extend the function f to the entire segment [a, bj by putting f(t) = f(a)({3 - t) + f(f3)(t - a) (3-a

for t E [a,bj\M, where t E (a,{3) E 'Y, see Figure 4.8. By Lemma 3.3 the extended function f is continuous. Evidently it is non-decreasing. By Lemma 3.2 there exists a set Mo ~ [a, bj of measure zero such that for every point x of the set [a, bj \ Mo the derivative f'(x) exists. Lemma 4.1 implies the existence of the derivative fk(x) for every point x E M \ Mo. It remains to notice, that by Theorem 1.1 J.t(M \ Mo) = J.t(M). The lemma is proved. •

-M a

M

fJ Figure 4.8

M

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118

Lemma 4.3. Let the domain of a function f with values in the Euclidean space Rn contain a measurable set M ~ R Let the (finite) derivative f~ (x) exist at almost every point x of the set M. Then the function f~ is measurable. Proof. Let Mo be the set of all points x EM, at which the derivative f~(x) exists. If x E M o, then the quotient f(tt:~(x) tends to a finite limit, f(t) - f(x) = f(t) - f(x) (t - x) t-x

~

0

as t ~ x in the set M. Therefore the function f IM is continuous at x. Fix a point tik E (2- i k, 2- i (k+1))nM for i = 1,2, ... , k = 0, ±1, ±2, ... satisfying the condition (2- i k, 2- i (k + 1)) n M =I 0. The set Ml of all so fixed points is (at most) countable. Therefore IL(Md = O. For i = 1,2, ... , k = 0, ±1, ±2, ... and x E (2- i k, 2-i(k + 1)) n (Mo \ (Ml U Q)) put gi(X) = f(x) - f(t ik ). x - tik

The function gi is continuous on every set

i = 1,2, .... Therefore it is continuous (see Theorem 1.7.7). Thus the func-

tion is measurable (on Mo \ (Ml U Q)). For a fixed x E Mo \ (Ml U Q) by virtue of the existence of the derivative f~(x) the sequence gi(X) converges to f~(x). Theorem 2.4 implies the measurability of the function f~IMo\(MluQr Lemma 2.1 implies the measurability of the function f~· The lemma is proved. • 5. Absolutely continuous functions Let the domain of a function f with values in a metric space (X, p) contain a set M ~ R The function f is called absolutely continuous on the set M if for every number c > 0 there exists a number > 0 such that if

°

(*,0, M) a finite family,

= {(ai, (3i):

i

= 1, ... ,k} consists of pairwise

disjoint intervals with endpoints from the set M and with d(,) < 0, then E{p(f(ai),f({3;)): i

= 1, ... ,k}

~

c.

If the set M coincides with the domain of the function f then the function f is called absolutely continuous (without the mentioning of M). Lemma 5.1. Let a function f with values in a metric space (X, pd be absolutely continuous on a set M ~ R Let a mapp'ing 9 of the space X into

Derivation and integration.

119

a metric space (Y, P2) satisfy the Lipschitz condition. Then the function gf is absolutely continuous on M. Proof. Take an arbitrary number c > O. Let L (~ 0) be a Lipschitz constant for the mapping g. Select a number 0 > 0 such that as soon as a family'Y = {(ai, .Bi): i = 1, ... ,k} satisfies condition (*,0, M), then

For every such a family 'Y 'L3p2(gf(ai), gf(.Bi)): i = 1, ... ,k}

~ L 'L3P1 (f(ai), f(.Bi)): i = 1, ... , k}

~ L

Lc

+ 1 < c.

The lemma is proved. • Lemma 5.2. The diagonal product f of functions fj, j = 1, ... ,n, with values in metric spaces (Xj, pj) is absolutely continuous on a set M ~ IR (with respect to the metric on the product from Theorem 2.1.2) if and only if every function Ii is absolutely continuous on M. Proof. Necessity follows immediately from Lemma 5.1. Sufficiency. Let P be the metric on the product. Take an arbitrary number c > O. For every j = 1, ... ,n fix a number OJ > 0 in a manner that

for every family'Y = {(ai,.Bd: i = 1, ... , n} satisfying condition (*, OJ, M). If 0 = min{ 01,'" ,On} and a family 'Y = {(ai, .Bi): i = 1, ... ,k} satisfies condition (*,0, M), then L{p(f(ai),f(.Bi)): i = 1, ... ,k}

~L

{t

pj(Jj(ai), Ii (.Bi)) : i

= 1, ...

'k}

3=1

n

= LL{pj(Jj(ai),fj(.Bi)):

i

= 1, ... ,k}

i=1 ~

c n· n

The lemma is proved.

= c.

•

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CHAPTER 4

Lemma 5.3. A vector function f = {fl, ... , fn} with values in the Euclidean space IRn is absolutely continuous on a set M ~ IR if and only if every coordinate function fl, ... , fn is absolutely continuous on M. Proof. To be precise, in the statement we speak about the Euclidean metric of the space IRn. For the metric of Example 1.4.5 an analogous assertion follows immediately from Lemma 5.2. Because of the inequality (2.2.1) relating these two metrics the assertion in question follows from Lemma • 5.1. The lemma is proved. Lemma 5.4. Let the domains of functions f and 9 contain a set M ~ R Let the functions take values in the Euclidean space IRn and be absolutely continuous on the set M. Let>. be a real number. Then the functions f + 9 and >.f are absolutely continuous on the set M. Proof. Lemma 5.3 reduces the proof to the scalar case. Here we obtain what was required from Lemma 5.2, remarks of Examples 1.7.2 and 1.7.3 and Lemma 5.1. The lemma is proved. • Example 5.1. The identity mapping of a set M ~ IR into itself is absolutely continuous. This follows immediately from definitions. So Lemma 5.1 implies that if a function satisfies the Lipschitz condition then it is absolutely continuous. Assume that a function f is defined, has the continuous derivative on the segment [a, bj, and m = sup f'([a, b]). For every point s < t of the segment [a, bj there exists a point c E [s, tj such that f(t) - f(s) = f'(c)(t - s) (Lagrange formula). So If(t) - f(s)1 ~ mit - 81, i.e., the function f satisfies the Lipschitz condition. Hence it is absolutely continuous. Lemmas 5.4 and 5.3 extend immediately our list of examples of absolutely continuous functions. Every absolutely continuous function is uniformly continuous. We obtain it, when we take in (*,0, M) a family 'Y consisting of one element. Lemma 5.5. Let a function f with values in a metric space (X, p) be absolutely continuous on a set M ~ R Then for every number EO > 0 there exists a number > 0 such that if

°

(**,o,M) an (at most) countable family 'Y = {(ai,,Bi): i = 1,2, ... } consists of pairwise disjoint intervals with endpoints from the set M and with d(,) < 8, then 'L-{p(f(ai), f(,Bi)): i = 1,2, ... } ~ EO. Proof. For an arbitrary number EO > 0 find a number 8 > 0 according to the definition of the absolute continuity. If a family 'Y = {( ai, ,Bi) : i = 1,2, ... } is infinite and satisfies condition (**,8, M) then for every k = 1,2, .. .

Derivation and integration. When we pass to the limit as k

-t

00,

121

we obtain

(For a finite family, we obtain what was required directly from the definition.) The lemma is proved. • Theorem 5.1. Let a real function j be defined and be absolutely continuous on a segment [a, b]' a < b, of the real line. Then for every number c > 0, there exists a number 8 > 0 such that if M ~ [a, b] and J.L*(M) < 8, then J.L*(f(M)) :::;; c. Proof. I. For arbitrary c > 0 fix 8 according to Lemma 5.5. Let M ~ [a, b] and J.L*(M) < 8. II. Consider the particular case, where the set M is open in the segment [a, b]. By Corollary of Theorem 2.6.7 the set M may be represented as the union of an at most countable set, of its connected components. By Theorem 2.6.3 every element J of the set, is either an interval or a half interval or a segment. Let a(J) and (3(J) be its initial point and the end, s(J) = inf j([a(J), (3(J)]) and t(J) = sup j([a(J), (3(J)]) , see Figure 4.9. Remark 2.3.4 By s(J) = j(a'(J)) and t(J) t( J) = j ({3' (J)) for some cl(J), (3' (J) E [a( J), (3(J)] 1(J) (Figure 4.9). This means that under the action of j the image of the segs(J) ment I(J) with the endW(J) ~ a.'(J) . points a' (J) and (3' (J) co---------------a.(J)-;---~(J) incides with the image of the segment [a(J),{3(J)]. Figure 4.9 By the choice of 8 and by results of §1

J.L(f(M)) :::;; J.L(U{f([a(J),{3(J)]): J E ,}) = J.L(U{[s(J), t(J)]: J E ,}) :::;; 2)If(a'(J)) - j({3'(J))I: J E ,} :::;; c, because I(J) ~ [a(J), (3(J)] and

I)Ia'(J) - {3'(J)I: J E ,} :::;; 2)la(J) - {3(J)I: J E ,}

= J.L(M) < 8.

III. Let us go on to the general case. The definition of the outer measure implies the existence of an at most countable family of intervals, covering the set M with db) < o. By Theorem 1.2 the measure of the set G 1 =

U,

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122

is less than o. The measure of the (see Lemmas 1.7 and 1.1). The set definition of the induced topology. theorem is proved. Corollary. Let a real function f

set G 2 = G 1 n [a, bj is less than 0 too G 2 is open in the segment [a, bj by the By II JL(f(M)) ~ JL(f(G 2 )) ~ E. The •

be defined and be absolutely continuous on a segment [a, b], a < b, of the real line and M be a subset of the measure zero of the segment [a, bj. Then JL(f(M)) = o. • Lemma 5.6. Let a function f with values in a metric space (X,p) be defined and be continuous on a subset M of the real line R Let it be absolutely continuous on a set Mo ~ M. Then function f is absolutely continuous on the set Ml = [Moj n M. Proof. Take an arbitrary number E > O. According to the definition of the absolute continuity find 0 > O. Let a family 'Y = {(ai, ,8i) : i = 1, ... , k} satisfy condition (*,0, Md and s = db) « 0). For every n = 1,2, ... and i = 1, ... ,k fix points ar E O{ai n Mo and ,8': E 0e,8i n M o, where ~ = 2-nk(0 - s), such that intervals of the family 'Yn = {( ar,,8,:): i = 1, ... , k} are pairwise disjoint. We have

The family 'Yn satisfies condition (*,0, M). Thus z)p(f(a~),f(,8~)): i = 1, ... ,k} ~

E.

Since

- p(f(ai), f(,8i)) I ~Ip(f(a~), f(,8~)) - p(f(a~), f(,8i)) I + Ip(f(an, f(,8i)) - p(f(ai), f(,8d) I ~p(f(,8~), f(,8i)) + p(f(a~), f(ai))

Ip(f(a~), f(,8~))

(the last passage is made by Lemma 2.3.1), z)lp(f(a~), f(,8~)) - p(f(ai), f(,8i)) I : i = 1, ... , k} ~ L {p(f(a~), f(ai)): i = 1, ... ,k}

+ L{p(f(,8~),f(,8d): Terms in the right hand side tend to zero as n

L

-+ 00.

Therefore

= 1, ... ,k} {p(f(ai), f(,8i)): i = 1, ... ,k}.

L)p(f(ar), f(,8':)): i -+

i = 1, ... ,k}.

123

Derivation and integration.

When we pass here to the limit, we obtain what was required:

z)p(f(ad,f(,8i)): i = 1, ... ,k}:S; c. The lemma is proved. • Lemma 5.7. Let a function f with values in a metric space X be absolutely continuous on a set M ~ JR. and MI ~ M. Then the function f is absolutely continuous on the set MI' Proof. The proof is obvious. •

6. Derivation of absolutely continuous functions Lemma 6.1. Let a real function f be defined and be absolutely continuous on a (nonempty) compact subset M of the real line JR.. Then there are absolutely continuous and non-decreasing functions g and h defined on the set M such that f = g - h. Proof. 1. For points a < ,8 of the set M denote by V(a,,8) the upper bound of sums

where a = Xo :S; ... :S; Xk+l = ,8 and {Xl,'" , xd ~ M. For now we do not assert that this upper bound is finite. II. Let a :S; ,8 :S; , be points of the set M. Consider an arbitrary sum

where a = xo:S; ... :S; Xk+l =, and {xI, ... ,xd such that X j :S; ,8 :S; X j+l' Then

s:S; (l)lf(xi+l) - f(xi)l: i = O, ... ,j - I}

+ (If(xj+d - f(,8)1 + 2: {If(XHI) :S; V(a,,B) + V(,8,,).

~

M. Fixj

+ If(,8)

= O, ... ,k

- f(xj)l)

f(xi)l: i = j

+ 1, ... , k})

When we pass to the upper bound in the left hand side of this inequality we obtain (6.1)

V(a,,) :S; V(a,,B)

On the other hand, for every c >

V(,8,,) we can fix points a

= Xo

+ V(,8, ,).

°

by the definition of V(a,,8) and

~ ... :S; Xj+l

= ,8

~ Xj+2 ~ ... ~ Xk+l

of the set M such that

L {If(xHd L {If(xHd -

f(xi)l: i = 0, ... ,j} ~ V(a,,8) - c,

!(xi)l: i = j + 1, ... , k} ~ V(,8,,) - c.

=,

CHAPTER 4

124 Therefore

By the definition of V(a,,)

V(a,,)

~

V(a,.8)

+ V(.8,,)

- 2£.

In view of the arbitrariness in the choice of £ > 0 (6.2)

V(a,,)

~

V(a,.B)

+ V(.8, ,).

When we compare (6.1) and (6.2), we obtain

V(a,,) = V(a,.8)

+ V(.8, ,).

III. For an arbitrary number £ > 0 take a number Ii > 0 according to the condition which occurs in the definition of the absolute continuity. Show that as soon as a family, = {( ai, .8i): i = 1, ... , k} satisfies the condition (*, Ii, M) of §5, then

Assume the opposite, i.e., assume the existence of a family , = {( ai,.8d: i = 1, ... ,k} satisfying condition (*, Ii, M) such that the number 'fJ = L {V(ai' .8i) : i = 1, ... , k} - c is positive. By the definition of V(ai' .8i) for every i = 1, ... ,k we can choose points

of the set M in a manner that

The family U{{(Xi,/,Xi,l+d: (*, Ii, M) and the estimate

1 = O, ... ,j(i)}:

~ {If(Xi,I+l) - f(Xi,/)I: 1 = 0, ... ,j(i), i

i = 1, ... ,k} satisfies

=

1, ... , k} > c.

This contradicts the choice of Ii. Thus our assumption is false. This gives what was required. IV. Let a = inf M. For t E M put g(t) = V(a, t). It follows from II and from the nonnegativity of values of V(a, .8), that the function 9 is non-decreasing. Show that for every t E M the value g(t) is finite. The set )

Derivation and integration.

125

Mo = {t: t E M, g(t) < oo} is nonempty because it contains the point a: g(a) = O. Let to = sup Mo.

By virtue of the monotonicity of the function g we have Mn[a, to) ~ Mo. For c = 1 find a number 6 > 0 according to III. Next, fix an arbitrary point s E (to-~, to+~)nMo. For every point t E (s, to+~)nM (and, in particular, for t = to) g(t) = g(s) + V(s, t) ~ g(s) + 1 < 00. Therefore t E Mo. The set Mo is open in the subspace M. If to =1= sup M, then the set M \ Mo is nonempty. In this case by virtue of the openness of the set Mo the point tl = inf(M \ Mo) is different from the point to and (to, td n M = 0. The last equality implies that V(to, td = If(td - f(to)l. So

Hence tl E Mo. This contradicts the definition of the point t 1 . Thus our assumption is false and to = sup M. This means the finiteness of values of the function g. V. The absolute continuity of the function g follows from II and III. VI. The absolute continuity of the functions f and g implies the absolute continuity of the function h = g - f. If points a < (3 belong to the set M, then the definition V( a, (3) implies immediately that If((3) - f(a)1 ~ V(a, (3).

Therefore h((3) - h(a) = (g((3) - g(a)) - (f((3) - f(a)) ~

V(a, (3) - If((3) - f(a)1 ~ 0,

i.e., the function h is non-decreasing. The lemma is proved. • Theorem 6.1. Let a function f with values in the space IR n be defined and be absolutely continuous on a compact set M ~ R Then almost everywhere on the set M the function f has the derivative.

Proof. Lemma 5.2 reduces the proof to the scalar case. Here the assertion follows from Lemmas 6.1 and 4.2. The theorem is proved. • 7. Lebesgue integral Lemma 7.1. Let a real function f be defined and be continuous on a segment [a, b], a < b, of the real line. Let it be absolutely continuous on a closed set M ~ [a, bj and at almost every point of the set M its derivative with respect to the set M be positive. Let the function f be increasing on every component of the set [a, bj \ M. Then the function f is non-decreasing. Proof. If a point x E M \ {b} is not the left endpoint of a component of the complement to the set M (but almost every point of the set M satisfies

126

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this condition), the derivative Jk(x) exists and is positive, then there exists a point t E M n (x, b) such that

J(t) - J(x) > t-x

:........:...-'-----=--..:........:...

o.

Hence J(t) - J(x) > 0, i.e., the point x is a right increasing point of the function J. Every point of the complement to the set M (except the point b, if it belongs to the complement) is also a right increasing point. This follows immediately from the hypotheses of the lemma. Thus the set 8 of right increasing points of the function J has the full measure on the segment [a, bj. By Lemma 3.1 the set 8 is open. For endpoints Q < f3 of every its component we have J(Q) ~ J(f3). Enumerate elements of the set 'Y of connected components of the set 8 and for k = 1,2, ... denote by 'Yk the set of components with numbers 1, ... ,k. Elements of the set 'Yk are pairwise disjoint. Therefore their positions in the segment [a, bj are ordered: if we take two its different elements then one of them lies entirely on the left of the another element. Enumerate anew elements of the set 'Yk in order to guarantee that every element with smaller number be placed in the left of every element with larger number: 'Yk = {II, ... ,ld. Introduce the numbers

+ ... + (J(f3~) - J(Q~)), J(f3f)) + ... + (J(Q~) - J(f3LI))'

Pk = (J(f3f) - J(Q~)) qk = (J(Q~) where Q~

< f3; are endpoints of Ii. Since J.t(U'Yk) ~ f3~ - Q~ ~ b - a b- a

f3t -

Q~

---t

= 1'(8) = J.t(U'Y) = "{J.t(l): L....J

and

I E 'Y}

= k--+oo lim J.t(U'Yk),

b - a. The inequalities

a ~ ... ~ Q~ ~ Q~ ~ f3i ~ f3~ ~ ... ~ b, imply that Q~ ---t a and f3t ---t b. Since Pk + qk = J(f3t) - J(Q~), we obtain the convergence Pk + qk ---t J(b) - J(a) as k ---t 00. The endpoints of the intervals (f3f, Q~), ... ,(f3LI' Q~) belong to the set M. Their total length (i.e., by virtue of Theorem 1.1, the measure of their union) is ~ (b - a) - J.t(u'Yd. Since the set 8 has the full measure on the segment [a, b] the total length tends to zero as k ---t 00. Now the absolute continuity of the function J on the set M implies that

Iqkl ~ If(Q~) - f(f3f)1

+ ... + If(Q~)

as k ---t 00. On the other hand, Pk f(b) - f(a) ~ 0, i.e., f(b) ~ f(a).

~

o.

- f(f3:-I)1

---t

0

When we compare, we obtain

Derivation and integration.

127

If we take two arbitrary points 8 < t of the segment [a, b]' then we see that the hypotheses of the lemma hold for the segment [8, t], the function fl[s,t] and the set M n [8, tj. By the just proved assertion f(t) ~ f(8). The lemma is proved. • We can pass immediately to a more precise result. Lemma 7.2. Let a real function f be defined and be continuous on the segment [a, b], a < b, of the real line, be absolutely continuous on the closed set M ~ [a, bj. Let at almost every point of the set M its derivative with respect to the set M be non-negative. Let the function f be non-decreasing on every component of the set [a, b] \ M. Then the function f is non-decreasing. Proof. For every e > 0 the function f(x) + eX satisfies the hypotheses of the previous lemma. Apply it. When we pass to the limit as c ~ 0 we obtain what was required. The lemma is proved. • Corollary. Let a real function be defined and be absolutely continuous on the segment [a, bj. Let at almost every point of the segment [a, bj its derivative be non-negative. Then the function f is non-decreasing. • The following version of Lemma 7.2 will be helpful too. Proposition 7.1. Let a real function f be defined and be absolutely continuous on a compact subset M ~ [a, bj of the real line. Let at almost every point of the set M its derivative with respect to the set M be non-negative. Let f(a) ~ f(fJ) for every component (a, f3) of the set [inf M, sup Mj \ M. Then the function f is non-decreasing on the set M. Proof. Extend the function f on the segment [inf M, sup Mj and define it at a point t of a component (a, fJ) of the set [inf M, sup Mj \ M by the formula

f(t)

=

t-a fJ-t fJ _ a f (f3) - fJ _ af(a).

The so extended function f satisfies the hypotheses of Lemma 7.2 (see • Lemma 3.3). This gives what was required. Lemma 7.3. Let a function f with values in the Euclidean space ~n be defined and be absolutely continuous on a segment [a, b], a < b, of the real line. Let at almost every point of this segment its derivative be equal to zero. Then the function f is constant . Proof. In the scalar case Corollary of Lemma 7.2 implies that both functions f and - f are non-decreasing. This may be only when the function f is constant. In the vector case the hypotheses of the lemma imply that almost everywhere derivatives of the coordinate functions are equal to zero. The previous remark implies that these functions are constant. Thus the function f is constant too. The lemma is proved. •

128

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Example 3.2 ('Cantor stairs') shows that in Lemma 7.3 we cannot omit the requirement of the absolute continuity of the function f without its compensation by any other restrictions. Let a function f be defined almost everywhere on the segment [a, b]' a < b, of the real line and take values in the space IRn. The function f is called Lebesgue integrable if there exists an absolutely continuous function F : [a, b] - t IRn such that F'(t) = f(t) for almost every t E [a, b]. The function F is called a primitive (or an indefinite integraQ of the function f. As in the case of the Riemmann integral (Le., of the integral of a continuous function whose theory is usually accounted at first steps of the course of Mathematical Analysis) the primitive is defined up to an additive constant: if F and G are two primitives of a function f on [a, b], then for almost all x E [a, b] we have (F(x) - G(x))' = f(x) - f(x) = O. By Lemmas 7.3 the difference F - G is constant. On the other hand, if F is a primitive of f on [a,b] and c E IRn , then (F(x) + c)' = F'(x) at every x E [a,b]' at which the derivative F'(x) exists. Thus F + c is also a primitive to f. By virtue of the last remark for every points s, t of the segment [a, b] the difference F(t) - F(8) does not depend on the choice of the primitive F. This vector (number in the scalar case) is called a (definite) integral of the function f from 8 to t (or over the segment [8, t], when 8 ~ t). It is denoted by t

J

f(x)dx.

8

Evidently t

J

J

8

t

f(x)dx = -

8

f(x)dx.

The derivation is made in each coordinate independently. Thus the finding of a primitive, which is a vector function, and calculation of a definite integral, which is a vector, may be also made in each coordinate independently. If a function f is integrable on a segment [a, b], then it is integrable on every smaller segment I, i.e., the function fll is integrable. If F be a primitive of the function f then the function FII is a primitive of the function fl/' On the other hand, if 8 E [a, bJ and a function f is integrable on the segments [a,8J and [8, bJ then it is integrable on the segment [a, bJ. Proof of this assertion may be proposed to the reader as an easy exercise. J

129

Derivation and integration.

f

If s, t and u are three arbitrary points of the segment [a, b], a function is defined and Lebesgue integrable on [a, bj, then u

Jf(x)dx = F(u) - F(s) 8

= (F(t) - F(s)) t

=

+ (F(u)

- F(t»

u

J

f(x)dx

+

J

f(x)dx,

t

8

where F is an arbitrary primitive of the function f. Let F be a primitive of a function f on a segment [a, bj and s E [a, bj. For t E [a, bj put t

G(t) =

J

f(x)dx = F(t) - F(s).

s

The function G differs from the primitive F by a constant -F(s). Because of above remarks it is a primitive for the function f on the segment [a, bj (moreover G(s) = 0), i.e., a definite integral when we consider it as a function on its upper limit of integration (i.e., of the number t in the above formula), is a primitive. Let a function f be defined on a interval or on a half interval J. The function f a is called locally integrable if it is integrable on every segment lying in J. For every s E J the formula t

G(t) =

Jf(x)dx s

defines a primitive of the function f on J (i.e., for every segment I ~ J the function Gil is a primitive of the function fII). For a function f being locally integrable on a half interval [a, (0) the limit s

8li.~

Jf(x)dx a

(under the natural assumption that it exists) is denoted by

J 00

f(x)dx.

a

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CHAPTER 4

Analogously we define the value of b

J

f(x) dx

-00

Put also

a

J

J

00

f(x)dx =

-00

J 00

f(x)dx

+

f(x)dx,

a

-00

where a E (-00,00) (evidently the sum in the right hand side does not depend on the choice of a). In these formulae we can use every other suitable symbol instead of x to denote of the independent variable. Let M be a subset of the set X. The function

X(X)={~

if x E M, if x E X \ M.

is called a characteristic function of the set M (in X). Let M be a (measurable) subset of the real line. We regard the symbol

J

f(x)dx

M

as equipotent to the symbol 00

J

f(x)X(x)dx,

-00

where X is a characteristic function of the set M (in 1R). Example 7.1. Every continuous function is Lebesgue integrable (see Example 5.1). Its Lebesgue integral coincides with its Riemann integral. Example 7.2. Let -00 < a < (3 < 00. The characteristic function of a segment, of an interval or of a half interval with endpoints a, (3 is Lebesgue integrable: the function

F(x) = {x ~ a (3-a

if x < a, if a ~ x ~ (3, if x > (3

is its primitive (the absolute continuity of the function F may be easy checked directly).

131

Derivation and integration.

Theorem 7.1. Let functions f and g with values in the space ~n be defined almost everywhere on a segment [a, b], a < b, of the real line. Let them be Lebesgue integrable. Let>. be an arbitrary real number. Then the functions f + g and >.g (defined almost everywhere on the segment [a, bj) are Lebesgue integrable and b

b

+ g(x))dx

jU(x)

=

a

b

j f(x)dx a

+j

g(x)dx,

a

b

b

j >.f(x)dx = >. j f(x)dx. a

a

Proof. The integrability follows from Lemma 5.4. The equality may be • easily checked directly. The theorem is proved. Lemma 7.4. Let a non-negative function f be defined (almost everywhere) and be integrable on a segment [a, b], a < b, of the real line. Then b

j f(x)dx

~ O.

a

Proof. By Corollary of Lemma 7.2 (each) primitive of the function f is a non-decreasing function, that immediately gives what was required. The lemma is proved. • Theorem 7.2. Let non-negative real functions
j(
+ ... +
a

for every k

= 1,2, ... ). Then for almost all t

converges, the function
f

=
L f
boob

a

E [a, bj the series


k=l

a

CHAPTER 4

132

Proof. By Corollary of Lemma 7.2 each function t

Fk(t) =

J

'h(x)dx

a

is non-decreasing. By remarks preceding the proof of Theorem 3.1 the series FI + F2 + F3 + ... converges uniformly. Theorem 3.1 will imply our assertion, when we show that the sum F of the series is absolutely continuous (measurability of derivative is established in Lemma 4.3). Take an arbitrary number c > O. By virtue of the uniform convergence of the series FI + F2 + F3 + . .. there exists a number k such that

IIF - (FI

+ ... + Fk)1I

= IIPHI

c

+ FH2 + FH3 + ···11 < "2.

According to the condition, which occurs in the definition of the absolute continuity, find a number 8 > 0 for the function FI + ... + Fk (its absolute continuity follows from Lemma 5.4) and for the number ~ (as c in the condition). For every family, = {(aI, ,6d, ... ,(aj, ,6j)} of pairwise disjoint intervals with db) ~ 8 we have

+ ... + Fk(ai))) - (F(,6i) - (FI (,6i) + ... + Fk(,6i)))I: i = 1, ... ,j} {1(Fk+l (ai) + FH2 (ai) + FH3 (ai) + ... ) - (FHI (,6i) + Fk+2(,6i) + Fk+3(,6i) + ···)1: i =

L {I(F(ai) - (FI (ai) = L

=

L {L {F/(,6i) -

1, ... ,j}

F/(ai): i = 1, ... ,j} : 1 = k + 1, k + 2, k + 3, ... }

~

L {F/(b) -

=

F(b) - (FI(b)

F/(a): 1 = k + 1, k + 2, k + 3, ... } c

+ ... + Fk(b)) < "2.

Therefore

L{IP(ai) - F(,6i)l: i = 1, ... ,j}

+ ... + Fk(ai))) - (F(,6i) - (FI (,6i) + ... + Fk(,6i))) I : {I(FI (ai) + ... + Fk(ai))

~ L{(IP(ai) - (FI(a;)

+L

- (F1 (,6i) ~

c

c

2" + 2" = c.

+ ... + Fk(,6i))l:

i

i = 1, ... ,j}

= 1, ... ,j}

Derivation and integration.

133

This proves the absolute continuity of the function F. The theorem is proved. • Theorem 7.3. Let real functions h, k = 1,2, ... }, be defined almost everywhere on a segment [a, b], a < b, and be Lebesgue integrable. Let their integrals are bounded in totality. Let the sequence Uk: k = 1,2, ... } converge from below to the function f (Le., for almost all x E [a, bJ

f1{X) ~ f2{X) ~ iJ{x) ~ ...

Figure 4.10

and f{x) = limk ..... oo fk{X)) (see Figure 4.10). Then the function f is integrable and b

f

b

f{x)dx =

1~~

a

f

fk{x)dx.

a

Proof. We apply the previous theorem to the series composed by the functions f1' h - fll iJ - f2' f4 - iJ,···· The theorem is proved. • Example 7.3. The characteristic function of an open subset U of a segment [a, b], a < b, may be represented as the sum of the series composed by characteristic functions of its component (see Example 7.2). The integrals of partial sums of this series are bounded from above by the number b - a. Theorem 7.2 implies the integrability of the characteristic function Xu of the set U. Theorems 7.2 and 1.1 imply the equality b

f

Xu{x)dx = J.L{U).

a

So Theorem 7.1 implies the integrability of the characteristic function XM of every closed set M ~ [a, bJ (by virtue of the representation M = [a, bj \ U). By Theorems 7.1 and 1.1 b

f

XM{x)dx

= J.L{M).

a

Example 7.4. Let M be an arbitrary measurable subset of a segment

[a, bj, a < b. The definition of a measurable set implies the existence of a

134

CHAPTER 4

sequence MI ~ M2 ~ M3 ~ ... of subsets of the set M, which are closed in the segment [a, bj and the measure of the union of which is equal to the measure of the set M. The integrals of the characteristic functions XMi are bounded from above by the number b - a. When we apply Theorem 7.3 to the sequences of the characteristic functions, we obtain the integrability of the function XM. By Theorems 7.2, 7.1 and 1.1

!

b

!

boob

XM(x)dx =

a

XM1(X)dx

+ L ! XMk+l\Mk(X)dx

a

=

k=1 a

! b

t; 00

XM1(X)dx+

(

! b

XMk+l(X)dx-

!

b )

XMk(X)dx

00

= J.l(Md

+ L(J.l(Mk+l) -

J.l(Mk ))

k=1 00

k=1

Lemma 7.5. Let real functions f and fo be defined (almost everywhere) and be integrable on a segment [a, b], a < b, of the real line. Let almost everywhere the function fo majorize the function f (i.e., f(x) :::;; fo(x) for almost all x E [a, b]). Then b

! a

b

f(x)dx :::;;

!

fo(x)dx.

a

Proof. Theorem 7.1 and Lemma 7.4 imply:

!

b

a

!

b

fo(x)dx -

a

b

f(x)dx = !Uo(x) - f(x))dx

~ o.

a

The lemma is proved. • Example 7.5. Let a non-negative function f be defined almost everywhere on a segment [a, bj, a < b. Let it have the form mentioned in the Corollary of Theorem 2.3. Let an integrable function fo (defined almost everywhere on the segment [a, b]) majorize almost everywhere the function f. The function f may be represented as the sum of the series alXI + a2X2 + a3X3 + ... , where Xk is the characteristic function of the set Mk and the coefficient ak is non-negative. By Lemma 7.5 the integrals of the partial sums of the mentioned series are bounded in totality by the j

Derivation and integration.

135

integral of the function fo. Theorem 7.2 implies the integrability of the function f. Example 7.6. Let a non-negative measurable function f be defined (almost everywhere) on the segment [a, bj, a < b. Let a Lebesgue integrable real function fo (defined also almost everywhere on the segment [a, b]) majorize almost everywhere the function f. For every pair of positive integers i,j put Mii = {x: x E [a, b], 2- i (j - 1) ~ f(x) < 2- i j}. For i = 1,2, ... define the function fi by the condition

These functions are studied in Example 7.5. Their integrals are bounded from above by the integral of the function fo (see Lemma 7.5). Theorem 7.3 implies the integrability of the function f. Lemma 7.5 implies the inequality b

J

f(x)dx

a

b

~

J

fo(x)dx.

a

Example 7.7. Let a real function f be defined (almost everywhere) on the segment [a, b], a < b, and be Lebesgue integrable. Let f+ be its non-negative part (see §2). When we apply Lemma 6.1 to (every) primitive of the function f by virtue of the nonnegativity of the derivative of a nondecreasing function (everywhere, where the derivative exists) we obtain a representation of the function f as the difference of two non-negative integrable functions: f = fl - f2' Evidently f+(x) ~ fl(X) for almost all x E [a, bj. By remarks of Example 7.6 the function f+ is integrable. Theorem 7.1 implies the integrability of the functions f- (see §2) and If I· If g is one more integrable function, then the functions min{f(x), g(x)} and max{f(x),g(x)} are integrable (see a corresponding representation in §2). The examples mentioned give a rather large reserve of integrable real functions and the remark following the definition of the Lebesgue integral gives a large reserve of integrable vector functions. Lemma 7.6. Let a measurable vector function f with values in the space ]R.n and a non-negative function fo be defined (almost everywhere) on a segment [a, bj. Let IIf(x)1I ~ fo(x) for almost all x E [a, bj. Let the function fo be Lebesgue integrable on the segment [a, bj. Then the function f is also Lebesgue integrable on the segment [a, bj and b

J

f(x)dx

a

~

b

J

fo(x)dx.

a

136

CHAPTER 4

Proof. By our remarks the question about the integrability reduces to the scalar case. By virtue of the reasoning of Example 7.7 in the scalar case the non-negative and nonpositive parts of the function fare integrable. This implies the integrability of the function f. To prove the inequality fix a rectangular system of coordinates in the space ~n in order to make one of its axis co-directed with the vector F(b) - F(a), where F is an arbitrary primitive of the function f (if F(b) - F(a) = 0, then our assertion is obvious). Let fi be the corresponding coordinate function of the function f and ft be its nonnegative part. We have:

J

10

-"-'-- .....

-/0

Figure 4.11

b

IIF(b) - F(a)11 =

J b

fi(X)dx

~

a

ft(x)dx

a

J b

~

fo(x)dx.

a

The lemma is proved. • Theorem 7.4 (Lebesgue). Let functions fk' k = 1,2, ... , f with values in ~n be defined almost everywhere on a segment [a, b], a < b, of the real line. Let the functions fk' k = 1,2, ... , be almost everywhere majorized in the norm by an integrable scalar function fo (i.e., Ilfk(X)11 ~ fo(x) for almost all x E [a, b] and for every k = 1,2, ... ). Let fk(X) ~ f(x) for almost every x E [a, b]. Then the function f is integrable and b

J

f(x)dx =

a

b

l~~

J

h(x)dx.

a

Proof. The integrability of the function f follows from Theorem 2.4 and Lemma 7.6. By virtue of our remarks and Lemma 2.1.2 the question about the limit passage under the integral reduces to the scalar case. In this latter case for k = 1,2, ... consider the functions

137

Derivation and integration. and

cp;;(x) = infUdx), fk+l (x), ... } = lim min{h(x), ... ,fk+p(X)}. p-+oo

Theorem 2.4 and Lemma 7.6 imply the Lebesgue integrability of these functions. Their integrals are majorized in the norm by the integral of the function fo. For almost all x E [a,b] the sequences {cpt(x): k = 1,2, ... } and {cp;; (x): k = 1, 2, ... } converge monotonically to f (x). By Theorem 7.3 b

(7.1)

J b

J

f(x)dx = lim

k-+oo

CPt (x)dx = lim

k-+oo

a

a

J b

cp;;(x)dx.

a

For almost all x E [a, b] we have:

cp;;(x) ~ fk(X) ~ cpt(x). By Lemma 7.5: b

b

J

cp;;(x)dx

~

a

b

J

fk(X)dx

~

a

f

cpt(x)dx.

a

Now (7.1) gives the limit passage in question. The theorem is proved. • Lemma 7.7 (Fatou). Let non-negative functions fk' k = 1,2, ... , be defined almost everywhere on a segment [a, b], a < b, of the real line and be Lebesgue integrable. Let their integrals (over the segment [a, b]) are not greater than m < 00. Let a function f be defined almost everywhere on the segment [a,b] and fk(X) -+ f(x) for almost every x E [a,b]. Then the function f is Lebesgue integrable and b

o~

f

f(x)dx

~ m.

a

Proof. For k = 1,2, ... define the function

CPk(X) = inf{h(x), fk+l(X), ... } = lim minUdx), ... , fk+p(x)}. p-+oo

Theorem 2.6 and Lemma 7.6 imply the Lebesgue integrability of these functions. For almost every x E [a, b] the sequence {CPk(X): k = 1,2, ... } Converges from below to f (x). Since

o~

b

J

CPk{x)dx

a

~

b

J

h(x)dx

a

~ m,

CHAPTER 4

138

for every k = 1,2, ... (see Lemma 7.5), Theorem 7.3 and Lemma 7.5 imply our assertion. The lemma is proved. •

8. Density points and approximate derivatives Let M be an arbitrary measurable subset of a segment [a, b], a < b, X be its characteristic function and H be a primitive of the function x. A point x E [a, b] is called a density point of the set M if the derivative of the function H at the point x exists and is equal to 1. The definition of a primitive and the integrability of the characteristic function of a measurable set (see Example 7.4) imply that almost every point of the set M is its density point and the measure of the set of density points of the set M lying outside M is equal to zero. Fix a point x E [a, b]. For arbitrary real number e denote by /::"0 the segment with endpoints x and x + e. Since the integral of the characteristic function of a (bounded) set is equal to the measure of this set (see Example 7.4), the derivative of the function H at the point x is calculated as the limit of the expressions

as e ~ O. Thus a point x is a density point of the set M if and only if one of the following two equipotent conditions lim JL(/::"o n M) = 1 0-+0

lei

holds. The following fact is essential for us. Lemma 8.1. Let x be a density point of two measurable sets M I , M2 (lying in a segment [a, b], a < b, of the real line). Then the point x a density point of the set MI n M 2 • Proof. For every real number e

Since x is a density point both of the sets MI and M 2 , the terms of the last expression tend to zero as c ~ O. This gives what was required. The lemma is proved. •

Derivation and integration.

139

Lemma 8.2. Let sets M 1 , M2 (lying in a segment [a, b], a < b, 01 the real line) be measurable. Let x E Ml n M2 be a density point lor each 01 them. Let a lunction I : [a, bJ --t IRn have the derivatives at the point x with respect to each 01 these sets. Then the derivatives 01 the lunction I at the point x with respect to the sets Ml and M2 coincide. Proof. By Lemma 8.1 x is a density point of the set M = Ml n M2 too. A density point is always a limit point of the corresponding set. Lemma 4.1 implies the existence of the derivative I~(x) and the equalities I~(x) = I~I (x) and I~(x) = 1~2(X). This gives what was required. The lemma is proved. • A vector v E IRn is called an approximate derivative of a function I : [a, bJ --t IRn at a point x E [a, b], a < b, if there exists a measurable set M ~ [a, bJ such that x EM, the point x is a density point of M and the vector v is equal to the derivative of the function I at the point x with respect to the set M. (When I is a scalar function values of its approximate derivative are numbers.) By Lemma 8.2 the approximate derivative (if it exists) is defined uniquely. In what follows we use the notation of the usual derivative to denote the approximate derivative. We omit the word 'approximate', when this does not lead to misunderstandings. 9. Generalized absolutely continuous functions and the Denjoy integral We say that a function I (defined on a subset M of the real line) is generalized absolutely continuous if it is continuous and the set M may be represented as the union of an at most countable number of subsets, on each of which the function I is absolutely continuous. By Lemma 5.6 we can assume in addition that the subsets in this definition are closed. Lemma 9.1. Let I = 11 U 12 be segments 01 the real line. Let a lunction f with values in the space IRn be defined on the segment I. Let the functions fill and fl12 be generalized absolutely continuous. Then the function f is generalized absolutely continuous too. Proof. The proof is obvious. • Lemma 9.2. Let functions I and 9 with values in the space IRn be defined and be generalized absolutely continuous on the subset M of the real line. Let A be a real number. Then the functions f + g and AI are generalized absolutely continuous. Proof. Let M = u{Ml: i = 1,2, ... } = u{Ml: i = 1,2, ... }. Let the function f be absolutely continuous on every set Ml, i = 1,2, .... Let the function g be absolutely continuous on every set Ml, i = 1,2, .... By

140

CHAPTER 4

Lemma 5.4 the function f+g is absolutely continuous on every set MlnMJ, i,j = 1,2, .... The equality M = u{Ml n MJ: i,j = 1,2, ... } implies the generalized absolute continuity of the function f + g. The generalized absolute continuity of the function )..f follows from Lemma 5.4 directly. The lemma is proved. • By Theorem 6.1 (and Lemma 5.7) a generalized absolutely continuous function has an approximate derivative almost everywhere. By Lemma 4.3 and Theorem 2.3 an approximate derivative is a measurable function. By Lemma 5.3 a vector function with values in the space JRn is generalized absolutely continuous if and only if its coordinate functions are generalized absolutely continuous. By remarks of §4 coordinates of the approximate derivative of such a vector function are approximate derivatives of its corresponding coordinate functions. Lemma 9.3. Let the approximate derivative of a generalized absolutely continuous function f : [a, bj -+ JR, a < b, be non-negative at almost every point of the segment [a, bj. Then the function f is non-decreasing. Proof. Denote by , the set of all connected open subsets of the segment [a, bj, on which the function f is non-decreasing. Let U = U,. If the intersection of two elements of, is nonempty then their union belongs also to ,. So every connected component of U belongs to " Le., the function f is non-decreasing on every of them. The lemma will be proved when we check that [a, bj = U. Assume the opposite, Le., that the set M = [a, bj \ U is nonempty. Let the segment [a, bj be represented as the union of its closed subsets M i , i = 1,2, ... , on every of which the function f is absolutely continuous. We have M = U{M n M i : i = 1,2, ... }. The set M is compact (see Theorem 1.6.2). Hence it is complete (Theorem 1.6.6). Apply Corollary of Baire theorem 1.6.7. By virtue of the definition of the induced topology we obtain the existence of a nonempty interval (a, (3) ~ [a, bj and of a number i = 1,2, ... such that 0 i= (a, /3) n M ~ Mi. At almost every point of the set Mi the approximate derivative coincides with the derivative with respect to this set. When we apply Lemma 7.2 to the function fl[o:,,6] we obtain: (a, /3) E f. This contradicts the nonemptiness of the set (a, /3) n M. Thus our assumption is false and the lemma is proved. • Lemma 9.4. Let the approximate derivative of a generalized absolutely continuous function f : [a, b] -+ JRn, a < b, is equal to zero almost everywhere. Then the function f is constant. Proof. The proof (by virtue of the previous lemma) repeats the proof of Lemma 7.3. • Let a function f be defined almost everywhere either on a segment or on a half interval or on an interval I of the real line and take values in the space JRn. The function f is called Denjoy integrable if there exists a generalized absolutely continuous function F : I -+ JRn such that F'(t) = J(t) for almost i

Derivation and integration.

141

all tEl. The function F (as in the cases of the Riemann and Lebesgue integrals) is called a primitive or an indefinite integral of the function f. As in the cases of the Riemann and Lebesgue integrals (and by analogous arguments, see §7 and Lemma 9.4) a primitive F is defined up to an additive constant and for s, tEl the difference F(t) - F(s) does not depend on the choice of the primitive F. It is called a (definite) integral of the function f from s to t and it is denoted by t

J

f(x)dx.

s

If u is an arbitrary third point of I then t

u

J

f(x)dx

=

u

J

f(x)dx

+

J

f(x)dx.

s s t

In this case too, when we consider the integral as a function of the upper limit of the integration the integral is one of primitives, namely it is the primitive whose value at the lower limit of the integration is zero. Example 9.1. Every Lebesgue integrable function is Denjoy integrable and the notions of primitives and integrals for it in both sense coincide. Example 9.2. Define on the segment [-1, 0] a function f in the following way. Let f(O) = o. For k = 1,2, ... and for an arbitrary point x of the half interval [_2- k +1, _2- k ) put (Figure 4.12) f(x) = (_2)k Ik. The function f is Lebesgue integrable on every segment [-I,e], where -1 < e < O. We can consider its primitive t

F(t) =

J

f(x)dx.

-1

Since IF( _2- k +1) - F( -2- k )1 = 11k --t 0, in view of the alternation of sign of the function on adjacent half interval [_2-k+l, _2- k ) the limit of the function F as t --t 0 exists.

Figure 4.12

142

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Extend the function F by the continuity on the entire segment [-I,Oj. Keep its old notation. On every segment [-1, _2- k j, k = 1,2, ... , the function F is a primitive of a Lebesgue integrable function. Thus it is absolutely continuous there. It is also absolutely continuous on the one point set {O}. Thus the function F is generalized absolutely continuous (on the segment [-1,0]). Hence the function f is Denjoy integrable (on the segment [-1,0]) and F is its Denjoy primitive. The function f is not Lebesgue integrable: If we assume such an integrability, we obtain that the function If I must be integrable too. By virtue of the continuity of its primitive the (finite) limit o

lim! If(x)1 dx.

0 ..... 0

-1

must exist. A direct calculation shows that this limit is infinite: 2k

!

If(x)ldx

1

1

= 1 + 2" + ... + k

--+ 00.

-1

Lemma 9.5. The derivative of a non-decreasing continuous real function (defined on a segment [a, b], a < b, of the real line) is Lebesgue integrable. Proof. The hypotheses of Lemma 3.2 hold for the domain M = [a, bj of the function F in question. Lemma 3.2 implies that the derivative f of the function F is defined almost everywhere on the segment [a, bj. Extend the function F on the entire line by putting F(t) = F(a) for t < a and F(t) = F(b) for t > b. Next, put

fork=I,2, ... andtE[a,bj. The definition of a derivative and its existence for the function F almost everywhere on the segment [a, b]) imply that

for almost all t E [a, bJ. The non-decreasing of the function F implies that the functions

143

Derivation and integration. Do it:

i~.(x)dx ~ CT ~ CT ~

2'

F(x)dx -

2'

F(x)dx -

i 7'

F(X)dX)

F(X)dX)

« 00).

F(b) - F(a)

The lemma is proyed. • Remark 9.1. Return to last inequality of the proof of Lemma 9.5. For every k = 1,2, ...

J b

q,k(x)dx

~

F(b) - F(a).

a

When we pass to the limit as k

---t

J

00 we obtain

b

f(x)dx

~

F(b) - F(a).

a

The example of the 'Cantor stairs' shows that the inequality may be strong. Theorem 9.1. A non-negative Denjoy integrable function is Lebesgue integrable. Proof. Let a non-negative function f be defined almost everywhere on the segment la, bj, a < b. Let F be its Denjoy primitive. By Lemma 9.3 the function F is non-decreasing. By Lemma 3.2 at almost every point t of the segment la, bj the function F has a derivative g(t) (in usual sense). By Lemma 8.2 for almost all tEla, bj we have the coincidence f(t) = g(t). By Lemma 9.5 the function g is Lebesgue integrable. Hence the function f is Lebesgue integrable too. The theorem is proved. • The following assertion is analogous to Lemma 7.5. Lemma 9.6. Let real functions f and fo be defined (almost everywhere) and be Denjoy integrable on a segment la, b], a < b, of the real line. Let almost everywhere the function fo majorize the function f. Then b

J

f{t)dt

a

~

b

J

fo(t)dt.

a

•

CHAPTER 4

144

10. Mean Value theorem Lemma 10.1. Let a function f be defined almost everywhere on a segment [a, b] of the real line and take values in the space ~n. Let it be Denjoy integrable (respectively, Lebesgue integrable). Let g : ~n - t ~ be a linear functional. Then the function gf is Denjoy integrable (respectively, Lebesgue integrable) on the segment [a, b] and 9

(!

f(t)dt)

~

i

g(J(t))dt.

Proof. Let F be an arbitrary primitive of the function f. Let f(t) = (f1 (t), ... , fn(t)) and F(t) = (F1 (t), ... Fn(t)). Let g(u) = /-t1X1 + ... + /-tnXn for u = (Xl"'" Xn). We have Ig(u)1 :::;; M sup{lxd, .. ·, Ixnl}, where M = 1/-t11 + ... + l/-tnl. Thus Ig(u)1 :::;; Mllull, i.e., the functional g satisfies the Lipschitz condition. Lemma 5.1 implies the generalized absolute continuity (respectively, absolute continuity) of the function H = gF. For almost all s E [a, b]

H'(s) = (/-t1F1 + ... + /-tnFn)'(s) = /-t1 F; (s) + ... + /-tnF~ (s ) = /-td1(S) + ... + /-tnfn(s) = g(f(s)). Thus b

J

g(f(t))dt = H(b) - H(a) = g(F(b)) - g(F(a))

a

~ g(F(b) -

F(a))

~9

(!

f(t)dt) .

The lemma is proved. • Theorem 10.1. Let a function f be defined on a subset of full measure of a segment [a, b], a < b, of the real line and take values in the space ~n. Let it be Denjoy integrable on the segment [a, b]. Then

b~ a

b

J

f(t)dt E cc(f(M)).

a

Proof. Assume the opposite. By Theorem 2.4.2 there exist a linear functional g : ~n - t ~ and numbers p < q such that (10.1)

g(cc(f(M)))

~

(-oo,pj,

Derivation and integration.

145

9(b~a! !(t)dt) ~q

(10.2) By (10.1) g(f(M))

~

(-oo,p], i.e., g(f(t))

~

p for all t E M. By Lemma

9.6

J b

g(f(t))dt

(10.3)

~ p(b -

a).

a

By (10.2) and Lemma 10.1

b~ a

b

J

~ q,

g(f(t))dt

a

J b

g(f(t))dt

~

q(b - a).

a

This estimate and (10.3) imply the inequality q(b - a) ~ p(b - a). So q ~ p. This contradicts the choice of the numbers p and q. Thus our assumption is false. The theorem is proved. • Notice that in the case of a continuous real function f defined on the entire segment [a, b] the proved inclusion turns up to a little weakened version of the classical Lagrange formula

b~ a

b

J

f(t)dt = f(e) for some e E [a, b].

a

11. Change of variable in an integral and derivation of a composite function Fix notation. Let -00 < a < b < 00, -00 < e < d < 00. Let f: [a, b] --+ [e, d] be a non-decreasing absolutely continuous function. Lemma 11.1. Let a set A ~ [a, b] be measurable and B = f(A). Then the set B is measurable too and p,(B) = fA f'(t) dt. Proof. 1. If A is either an interval (p, q), either a half interval [p, q) or (p, q], or a segment [p, q] then our assertion is obvious: q

p,(B)

= f(q) - f(p) =

J

f'(t) dt.

p

146

CHAPTER 4

II. An arbitrary open set A ~ [a, b] may be represented as the union of the family 'Y of its connected components. By I and Theorems 1.1 and 7.3 (see Example 7.3)

p.(B) = p.(U{J(r): f E 'Y}) ="L{p.(f(f)): fE'Y} =

L ([ f(t) dt:

=

J

r

E'Y}

f'(t) dt.

A

la, b] \ A is open. By II and

III. In the case of a closed set A the set C = Theorem 1.1

J

J

J

a

C

A

b

p.(B)

= (d - c) - p.(f(C)) =

f'(t) dt -

f'(t) dt =

IV. Let us go on to the general case. For arbitrary c that

la, b]

f'(t) dt =

"L {J(sup J) -

> 0 fix 0 > 0 such

and p.(M)

< 0,

f(inf J): J E 'Y}

< c,

if M is an open subset of the segment

J

f'(t) dt.

then

M

where 'Y is the set of all connected component of the set M. This may be done by virtue of the absolute continuity of the function f. Let A be an arbitrary measurable set. By virtue of the measurability of the set A we can find a closed and an open sets Hand G such that

H ~ A~ G The set f(H)

~

and

p.(G \ H) <

o.

B is closed. The set f(G) 2 B is measurable. By II

p.(f(G) \ f(H))

~ p.(f(G \

H)) =

J

f'(t) dt < c.

G\H

By virtue of the measurability of the set f(G) there exists an open set G l ;2 f(G) such that p.(G l \ f(Gd) < ~. Then (11.1)

p.(G l

\

f(H» ~ p.(G l

\

f(G»

+ p.(f(G) \ i

f(H)) <

c

c

2" + 2" = c.

147

Derivation and integration.

The closedness of the set f(H), the openness of the set G I , (11.1) and the arbitrariness in the choice of e imply the measurability of the set B. We have

/-l(J(H))

=

J

J'(t) dt

~

J

J'(t) dt

~

J

J'(t) dt

= /-l(J(G))

H A G

and

/-l(J(H)) ~ /-l(B) ~ /-l(J(G)). So

J

J'(t)dt - /-l(B) <

~ < e.

A

In view of the arbitrariness in the choice of e the last estimate implies the equality

J

J'(t)dt = /-l(B).

A

The lemma is proved. • Lemma 11.2. Let B ~ [c, d]' /-l(B) = 0 and A = f-I(B). Then the measure of the set {t: tEA, the derivative l' (t) exists and l' (t) =1= O} is equal to zero. Proof. The condition /-l(B) = 0 implies the existence of a sequence of open sets G I = [c, dJ ;2 G 2 ;2 G 3 ;2 ... ;2 B such that /-l(G i ) -+ O. Let G = n{Gi : i = 1,2, ... }. For every i = 1,2, ... the set Hi = f-I(G i ) is open in the segment [a, b] and the set (11.2) is the preimage of the set G. By (11.2) and Theorem 1.3 the set H is measurable. By Lemma 11.1

J

J'(t) dt = /-l(G) = O.

H

So 1'(t) = 0 for almost all t E H. Now the inclusion A ~ H implies What was required. The lemma is proved. • Lemma 11.3. Let a function g : [c, d] -+ IRn be Lebesgue integrable. Then there exists a sequence gk : [c, d] -+ IRn, k = 1,2, ... , of continuous functions converging to the function g almost everywhere.

148

CHAPTER 4

IT outside the segment [c, d]. Put

Proof. Extend the function 9 by

Now the derivation of the integral with respect to its upper limit of inte• gration gives what was required. The lemma is proved.

Lemma 11.4. Let a real function 9 : [c, d] --+ lR be measurable and Ig(x) I ~ M for all x E [c, d]. Then the sequence {gk: k = 1,2, ... } in Lemma 11.3 may be chosen in such a manner that Igk(X)1 ~ M for all x E [c, d] and k = 1,2, .... Proof. Let {g,i;: k Put

= 1,2, ... } be sequence pointed in Lemma 11.3. if gic(x) if - M if g,i;(x)

~ ~ ~

M, g,i;(x) -M.

~

M,

We obtain the sequence with the needed properties. The lemma is proved.

•

Theorem 11.1. Let a function 9 : [c, d] --+ lRn be Lebesgue integrable on the segment [c, d] = f([a, b]). Then the function g(f(t» . f'(t) is Lebesgue integrable and d

b

J

J

e

a

g(x) dx =

g(f(t» . f'(t) dt.

Proof. The integration is made in each coordinate independently. Thus the general vector case reduces to the scalar one. In what follows we assume that n = 1. 1. Consider the case of a continuous function g. Let (11.3) Xo = c < Xl < ... < Xp-l < xp = d, mk = infg([xk_I,Xk]), Mk = sUpg([Xk_I,Xk]) for k = 1, .. . p, tk = j-I(Xk) for k = 0,1, ... p,

149

Derivation and integration. By Theorem 7.5

L{mk(Xk - xk-d: k = 1, .. . p}

7

= L{mk

k= 1, ... P}

f'(t) dt:

tk-l

J b

~

(11.4)

g(f(t)) . f'(t) dt

a

~L

7

{Mk

f'(t) dt: k = 1, .. .

p}

tk-l

Let 8 = SUp{ Xl - XO, ... ,Xp - Xp-l}. By Theorem 7.4 the end sides in (11.4) tend to fed g(x) dx as 8 --t 0 in (11.3). This gives what was required. II. Consider the case of a measurable function g. Assume that the function 9 is bounded in modulus by a number M: Ig(x)1 ~ M for all X E [c, d]. Fix an arbitrary sequence of continuous functions {gk: k = 1,2, ... } in a manner that the absolute values of the functions are bounded by the number M and the sequence converges almost everywhere to the function g, see Lemma 11.4. By I d

J

g(x) dx =

e

d

kl~~

b

J

gdx) dx =

kl~~

J

gk(f(t)) . f'(t) dt.

a

c

The expressions in the integral in the last side is almost everywhere bounded from above by the Lebesgue integrable function M f'(t). It remains to show that almost everywhere
gk(f(t)) . f'(t)

--t

g(f(t)) . f'(t).

The derivative f'(t) exists on the set M ~ [a, bj of the measure b - a. The condition gk(X) --t g(x) may be not fulfilled on the set N ~ [c, dj of measure zero. At points of the set M n j-l([C, dj \ N) = M \ j-l(N) we have the needed convergence. We will get what was required when we show that the convergence (11.5) takes place for almost all t E M n j-l(N). By Lemma 11.2 f'(t) = 0 for almost all t E j-l(N). This implies immediately what was required.

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150

III. Consider the case of a non-negative integrable function g. Omit the assumption of II about its boundedness. Let gk(X) = min{g(x), k}. We have gk(X) - t g(x) for every x E [c, dJ. Hence the convergence (11.5) takes place at every point t E [a, b], at which the derivative f'(t) exists. By II d

d

! 9(X) dx

b

= k-+oo lim !9k(X)

e

dx

= k-+oo lim !9k(J(t». J'(t)

dt.

a

e

Values of the integrals in the last side constitute a monotonically increasing g(x) dx. sequence bounded from above by the number Theorem 1.7.3 implies that the function g(J(t» . f'(t) is integrable and

t

!

d

!

b

g(x) dx

=

g(J(t» . J'(t) dt.

a

e

IV. An arbitrary function 9 may be represented as the difference of two non-negative integrable functions: 9 = gl - g2. When we apply III to each of the functions gl and g2, we obtain d

! e

d

g(x) dx =

!

d

91(X) dx -

!

e

g2(X) dx

e b

b

= j gl(J(t))· J'(t) dt - j g2(J(t» . J'(t) dt a

a b

= j(91(J(t» - g2(J(t»)· J'(t) dt a b

= j g(J(t» . J'(t) dt. a

The theorem is proved. • Corollary. Let under the hypotheses of Theorem 11.1 G be a primitive of the function g. Then for almost all t E [a, bJ (11.6)

(G(J(t»)' = g(J(t)) . J'(t).

This may be obtained by the derivation of the expression t

G(f(t» - G(f(a» =

Jg(f(s» . f'(s) ds a

151

Derivation and integration.

with respect to t. • Remark 11.1. Let a function G be absolutely continuous on the closed subset M of the segment [c,d] and Ml = f-l(M}. Then the function Gf is absolutely continuous on the set MI' The proof of this fact is quite obvious. The last observation implies immediately that if the function G is generalized absolutely continuous, then the function G f is also generalized absolutely continuous. Assertion 11.1. Let a function g be Denjoy integrable on the segment [c, d] = f([a, b]) and G be a primitive of the function g. Then for almost all t E [a, b] condition (11.6) holds. Proof. The function G is generalized absolutely continuous. Thus the segment [c, d] may be represented as the union of countable number of closed subsets M i , i = 1,2, ... , on each of which the function G is absolutely continuous. Fix i = 1,2, .... Let 'Y be a family of all connected components of the complement [c, d] \ Mi' The function G(t) {

H(t) =

t-

in~f

G(supf)

supf - mff (Figure 4.13)

+

supf -: t G(infr) supf - mff

for t E f E 'Y

is absolutely continuous. It is a primitive of the function

'-v-'

r

Figure 4.13

Figure 4.14

get) h(t) = {

G(supr) - G(inff) supf - inff

for

t E f E 'Y.

(Figure 4.14). By Remark 11.1 at almost every point t of the set D = f-l(Mi ) the derivative (H(f(t)))'v with respect to set D exists. It is equal to the derivative (G(f(t)))'v. Almost all points of the set Dare

CHAPTER 4

152

density points of this set. At these points the equality goes into the equality of the corresponding approximate derivatives. By Corollary of Theorem 11.1 and by Lemma 11.2

(G(f(t)))'

= (H(f(t)))' = h(f(t)) . f'(t) = g(J(t)) . f'(t).

Since this is true for every i = 1,2, ... the assertion is proved.

•

12. Measurable multi-valued mappings Let a multi-valued mapping F be defined on a measurable subset M of the real line and take values in a topological space Y. The mapping F is called measurable if for every open subset U of the space Y the set F-l(U) = {t: t E M, F(t) n U 1= 0} is measurable. In the case of a single valued mapping this notion of the measurability coincides with this one of

§2. Theorem 12.1. Let a multi-valued mapping F be defined on a measurable subset M of the real line and take compact values in a metric space Y. Then the following conditions are equipotent: a) the mapping F is measurable; b) for every closed subset H of the space Y the set F- 1 (H) is measurable; c) for every open subset U of the space Y the set {t: t E M, F(t) ~ U} is measurable; d) for every closed subset H of the space Y the set {t: t E M, F(t) ~ H} is measurable. If Y is the Euclidean space then the listed conditions are equipotent to the conditions: e) for every open and convex subset U of the space Y the set F- 1 (U) is measurable; f) for every open ball U of the space Y the set F-l(U) is measurable. Proof. a:::}b. By virtue of the compactness of values of the mapping F and Lemma 2.3.8 we have F-1(H) = n'f:lF-1(Ol/kH). Now a) and Theorem 1.3 imply the measurability of the set F-l(H). b:::} a. Let U be an open subset of the space Y. For k = 1,2,... put Hk = Y \ Ol/k(Y \ U). The set H k, k = 1,2, ... , is closed, U = U'f:lHk and F-l(U) = U{F-l(Hd: k = 1,2, ... }. Now b) and Theorem 1.1.2 implies the measurability of the set F-l(U). a ¢:>c and b ¢:>d. It is obvious because

{t: tEM, F(t)c;H}=M\F-1(Y\H). a:::}e:::}j. It is obvious. j:::}a. Open balls constitute a base of the space ~n. Therefore every open subset U of the space ~n may be represented as the union of a family

Derivation and integration.

153

'Y of open balls. By Theorem 1.6.10 the presence of a countable base of the space ~n allows to assume that the family 'Y is (at most) countable: 'Y = {Vk : k = 1,2, ... }. The equality

and Theorem 1.1.2 imply the measurability of the set F-I(U). The theorem is proved. • Theorem 12.2. Let a set M ~ ~ be represented as the union of an {at most} countable family of measurable sets {Mk: k = 1,2, ... }. Let a multi-valued mapping F be defined on the set M, take values in a topological space Y and be measurable on every set Mk {i. e., the mapping F IMk , k = 1,2, ... , be measurable}. Then the mapping F is measurable. Proof. The proof repeats the proof of Theorem 2.3 (the assertion in question is a direct generalization of the theorem). • Lemma 12.1. Let Fk : M - t Y, k = 1,2, ... , be measurable multivalued mappings of a measurable subset M of the real line in a metric space Y. Let their values be compact, FI(t) ;2 F 2 (t) ;2 F3(t) ;2 ... and F(t) = n{Fk(t): k = 1,2, ... } for every point t E M. Then the mapping F is measurable. Proof. Let H be an arbitrary closed subset of the space Y. By Corollary of Theorem 1.6.3 the compactness of values of the mappings under consideration implies the equality F-I(H) = n{Fk-I{H): k = 1,2, ... }. Theorems 1.3 and 1.12b imply the measurability of the set F- I (H). By Theorem 1.12b the mapping F is measurable. The lemma is proved. • Lemma 12.2. Let a multi-valued mapping F be defined on a measurable subset M of the real line, be measurable and take values in a metric space Y. Let HI be a closed subset of the space Y and FI (t) = F{t) n HI for t EM. Then the mapping FI : M - t Y is measurable. Proof. For every closed subset H of the space Y the set F1-l{H) = F-I{H n Hd is measurable by Theorem 12.1b. One more application of this theorem gives the measurability of the mapping F I . The lemma is proved. • Theorem 12.3. Let M be a measurable subset of the real line. Let Y be a metric space with a countable base. Let values of a (multi-valued) mapping F : M - t Y be nonempty and compact. Then there exists a single valued measurable mapping f : M - t Y such that f{t) E F{t) for every point

tEM. Proof. Construct a sequence {Fk: k = 0, 1,2, ... } of multi-valued measurable mappings. Put Fo = F. Let the mappings Fk - 1 , k = 1,2, ... , be constructed. Construct now the mapping Fk in the following way. By Theorem 1.6.10 the open cover {O(y, 21k): y E Y} of the space Y has an

CHAPTER 4

154

(at most) countable subcover {O(Yj, 21k): j = 1,2, ... }. Let

Ml = Fk-!l M2 = Mj =

([O(Yl' 21k)]) ,

21k)]) \ M1, ... , 1k)]) \ (Ml U··· U Mj-d,··· Fk-!l ([O(Yj, 2

F;;!l

([0(Y2'

.

The measurability of these sets follows from Theorem 12.1b and results of §1. For t E M j put Fk(t) = Fk-1(t) n [O(Yj, A)]. Since M = U{Mj : j = 1,2, ... } and the sets M j , j = 1,2, ... , are pairwise disjoint, the mapping Fk : M - t Y is defined correctly. For every t E M the set Fk(t) is nonempty and compact. If points x, Y belong to Fk(t), then p(x, y) ~ i.e., diamFk(t) ~ By Lemma 12.2 and Theorem 12.2 the mapping Fk is measurable. For t E M put F(t) = n{Fk(t): k = 1,2, ... }. By Lemma 12.1 the mapping F is measurable. By virtue of the compactness of the sets Fk(t), k = 1,2, ... , and Corollary of Theorem 1.6.3 the set F(t) is nonempty. Since

h

i.

diamF(t)

~ diamFk(t) ~ ~

for every k = 1,2, ... the set F(t) is one point. Denote it by f(t). The mapping f is constructed. The theorem is proved. • We will need the following additional assertions. Lemma 12.3. Let a multi-valued mapping F be defined on a measurable subset M of the real line, take values in the Euclidean space ~n and be measurable. Let a non-negative measurable function cp be defined on the set M. For t E M let G(t) = Of(F(t),cp(t)). Then the mapping G is measurable. Proof. The proof use Theorem 12.1f. Take an arbitrary point x E ~n and an arbitrary number c > O. The set

n OEX =J 0} F(t) n OE+CP(t)X =J 0}

G-1(OEX) = {t: t E M, G(t) = {t: t E M,

= U{ {t: t E M, F(t)

n Orx =J 0}

n{t: tEM, cp(t»r-c}: rEQ, r>O} is measurable by virtue of the measurability of the mapping F, of the measurability of the function cp and by Theorems 2.8, 1.2 and 1.3. Now our • assertion follows from Theorem 12.1. The lemma is proved. Lemma 12.4. Let measurable multi-valued mappings F, G: M - t JRn be defined on a measurable subset M of the real line JR. Let their values be

155

Derivation and integration.

closed. Let H{t) = F(t) n G(t) for t E M. Then the mapping H : M --+ ]Rn is measurable. Proof. 1. Let U be an arbitrary open subset of the space ]Rn. By virtue of the measurability of the mappings F and G and Theorem 1.3 the set {t: t

M, F(t) n U 1= 0, G(t) n U 1= 0} ={t: tEM, F(t)nU1=0}n{t: tEM, G(t)nU1=0} E

is measurable. II. Let 13 be an arbitrary (at most) countable family of open subsets of the space ]Rn. By I and Theorem 1.2 the set

u{ {t: t

E

M, F(t) n U 1= 0, G(t) n U 1= 0}: U E f3}

is measurable. III. Let {13k: k = 1,2, ... } be an (at most) countable family of (at most) countable families of open subsets of the space ]Rn. By II and Theorem 1.3 the set

n{u{{ t: t

E

M, F(t) n U 1=

0,

G(t) n U 1=

0}:

U

E

f3d: k

= 1,2, ... }

is measurable. IV. Let K be an arbitrary compact subset of ]Rn. Show that the set

M*={t: tEM, H(t)nK1=0} is measurable. Let 13 be an arbitrary countable base of the space ]Rn. For k = 1, 2, ... let 13k = {U: U E 13, Un K 1= 0, diamU < 2- k}. Evidently M* ~ M**, where M** = n{u{ {t: t E M, F(t) n U 1= 0, G(t) n U 1= 0}: U E f3d: k = 1,2, ... }. Lemma 2.3.8 implies the inverse inclusion M** ~ M*. Thus M* = M**. By III the set M* is measurable. V. Let x be an arbitrary point of the space ]Rn and c > O. The representation O(x, c) = U {Of (x, c(k;

1)) : k

= 2,3, ... }

implies the equality H-l(O(X, c»

= {t: t

E

= U {{ t:

M, H(t) n O(x, c) 1= t E M,

H(t) n Of (x,

0}

c(k-l») k 1= 0 } :

k

= 1,2, ..

CHAPTER 4

156

By virtue of the compactness of the sets Of (x,c(k -l)/k), k = 1,2, ... , IV and Theorem 1.2 the set H- 1 (0(x, c)) is measurable. Referring to Theorem 12.1 completes the proof. The lemma is proved. •

13. Multi-valued mappings defined on products Consider the following situation:

(13.1) I = [a, bj, a < b, is a segment of the real line; K is a metric compact; Y is a metric space with a countable base; F : I x K -+ Y is a multi-valued mapping with nonempty compact values; for every tEl the mapping F(t, x) is upper semicontinuous in x; for every x E K the mapping F( t, x) is measurable in t.

With the notation of (13.1) for A

~

K and B

~

Y let us put

M*(A,B)={t: tEl, F({t}xA)~B}.

We will also need the following strengthening of the last condition in (13.1): (13.2) for every open subset U of the compactum K and for every open subset V of the space Y the set M*(U, V) is measurable.

Lemma 13.1. With the notation of (13.1) for every closed subset H of the compactum K and for every open subset V of the space Y we have M*(H, V) = U'f:1M*(OtH, V). Proof. For every k = 1,2, ... we have M*(OtH, V) ~ M*{H, V) because H ~ 0tH. Thus U'f:1M*(OtH, V) ~ M*(H, V). The problem reduces to the proof of the inclusion M*(H, V) ~ Uk'=l M*(OtH, V). Assume the opposite. Let t E M*(H, V) \ U'f:1M*(OtH, V). Since t ~ U'f:1M*(OtH, V) for every k = 1,2, ... there are a point Xk E 0tH and a point Yk E F(t, xd \ By virtue of the compactness of the space K the sequence {Xk: k = 1,2, ... } has a subsequence {Xk: k E A} converging to a point x* E K. Since Xk E 0tH for k = 1,2, ... , x* E H (see Corollary of Lemma 2.3.8). The existence of the sequences {Xk : k E A} and {Yk: k E A}, Theorem 2.5.1 and the inclusion F(t,x) ~ F({t} x H) ~ V imply that the mapping F(t, x) is not upper semicontinuous in x at the point (t, x*) (because the sequence {Yk: k E A} ~ K \ V has no a subsequence converging to a point of F(t,x». This contradicts (13.1). Thus our assumption is false. The lemma is proved. •

v.

Derivation and integration.

157

Lemma 13.2. Under the hypotheses of (13.1) condition (13.2) is equipotent to the condition: (13.3) for every closed subset H of the compactum K and for every open subset V of the space Y the set M*(H, V) is measurable.

Proof. Necessity follows from Lemma 13.1 and Theorem 1.2. Sufficiency. Let U be an open subset of the compactum K and U For k = 1,2, ... put Hk = K \ 0t(K \ U). Since U = Uk:lHk,

=1=

K.

The measurability of the set M*(U, V) follows from (13.3) and Theorem 1.3. Lemma is proved. • Theorem 13.1. Let conditions (13.1) and (13.2) hold. Let X be a closed subset of the product I x K. Let G(t) = F(( {t} x K) n X) for t E I. Then the mapping G : I --t Y is measurable . Proof. I. Let the set X be the product of compacta II and K l . In this case our assertion follows from Lemma 13.2 and Theorem 12.1c. II. Let the set X be the union of the products I j x K j , j = 1, ... , s, of compacta I j and K j . Let a mapping G j be constructed according to the statement of the theorem by the set I j x K j • By I the mapping G j is measurable. Therefore for every open subset V of the space Y the set Gjl (V) is measurable. The equality G- l (V) = Uj=l Gjl (V) implies the measurability of the set G-l(V). Hence the mapping G is measurable. III. Let us go on to the general case. Let k = 1,2, .... For every point x E X find its Tychonoffneighborhood Tx x Ux such that [Tx x Uxl ~ 0tX. The constructed open cover {Tx x Ux : x E X} of the compactum X contains a finite subcover {TXl x UXl , • • • ,Tx. x Ux.}. Put

The set X k satisfies the hypotheses of II. The corresponding mapping G k is measurable. By Theorem 1.7.8 values of the mapping G k are compact. Since X ~ n~lXk ~ n~lOtX = X (i.e., X = nk:lxd, the upper semicontinuity of the mapping F(t, x) in x implies: G(t) = nk:lGk(t). By Lemma 12.1 the mapping G is measurable. • The theorem is proved. Let us highlight a particular case of the theorem proved. Corollary. Let conditions (13.1) and (13.2) hold. Let f be a continuous (single valued) mapping of the segment I in the compactum K.

158

CHAPTER 4

Let the (multi-valued) mapping G : I - t Y be defined by the formula G(t) = F(t, f(t». Then the mapping G is measurable. • Theorem 13.2. Let condition (13.1) hold. Then condition (13.2) is equipotent to the condition:

(13.4) for every number € > 0 there exists a closed subset H of the segment [a, b] such that f..L(H) ~ b-a-€ and the mapping FIHxK is upper semicontinuous.

a Figure 4.15

Proof. Necessity. Fix a countable base a = {Am: m = 1,2, ... } of the compactum K and a countable base f3* of the space Y. The family f3 of the unions of all possible finite subfamily of the base f3* is countable: f3 = {Bn: n = 1,2, ... }. By (13.2) the set Mmn = M*(Am' B n ), m, n = 1,2, ... , is measurable. Therefore there are closed Pmn and open Qmn subsets of the segment [a, b] such that Pmn ~ Mmn ~ Qmn and f..L(Wmn ) < 2- m- n€, where W mn = Qmn \ Pmn- The set [a, b] \ W mn may be represented as the union of two disjoint closed subsets Pmn and [a, b] \ Qmn of the segment [a, b]. Both these sets are closed in the subspace [a, b] \ Wmn- Since they are complements in [a, b] \ W mn of each other they are open in [a, b] \ W mn. By Theorem 1.2 the measure of the set W = U{Wmn: m, n = 1,2, ... } is less than €. Put H = [a,b] \ W. Take an arbitrary open subset V of the space Y. Our aim is to show that the set G v = {(t,x): (t,x) E H x K, F(t,x) ~ V} is open in the product H x K (Figure 4.15). Consider first the particular case V = Bn- If (t* , x*) E G Bn then by virtue of the upper semicontinuity of the mapping F(t, x) in x there exists an element Am 3 x* of the base a such that F( {t*} x Am) ~ B n , i.e., t* E Mmn- In view of the arbitrariness of the point (t*,x*) E G Bn this means that G Bn = U{(Mmn \ W) x Am: m = 1,2, ... }. Hence the set G Bn is open in the product H x K. In the general case for every point (t*, y*) E G v in view of the compactness of the set F(t*, x*) there are elements Vi, ... , Vs of the base f3* such that F( t*, x*) ~ Vi U ... U Vs ~ V. The set Vi U ... U Vs ~ V belongs to the family f3, i.e., G v = U{ G B : B E {3, B ~ V}. This implies the openness of the set G v in the product HxK. Sufficiency. By Lemma 13.2 it is sufficient to show that (13.4) implies (13.3). Let Kl be a closed subset of the compactum K, V be an open {

Derivation and integration.

159

subset of the space Y. For every j = 1,2, ... fix a closed subset H j of the segment [a, b] such that {t(Hj) ~ b-a- j and the mapping FIHj xK is upper semicontinuous. The last condition implies the openness of the set G = {(t, y): (t, y) E H j xK, F(t, y) ~ V} in the product H j x K. Theorem 1.6.2 implies the compactness of the set (Hj x K) \ G. Lemma 2.1.1 and Theorem 1.7.8 and 1.6.1 imply the closed neSS of the set P = {t: t E H j , ({t} x Kd\G =1= 0}. By Theorem 1.4.2 the set H j \P = M(Kl' V)nHj is open in the space H j . Lemma 1.7 implies its measurability. By Theorem 1.2 the set Q = M(Kl' V)n(U{Hj: j = 1,2, ... }) is measurable. We have M(Kl' V)\ Q ~ nk:l ([a, b] \ Hj ). By Lemma 1.1 and Theorem 1.1 the measure of the last set is equal to zero. Theorem 1.1 implies the measurability of the set M(K1 , V). The theorem is proved. • Lemma 13.3. Let condition (13.1) hold. Let:

(13.5) for every point t of the segment [a, b] and for every open subset V of the space Y the set N = {x: x E K, F(t,x)nV =1= 0} lie in the closure of its interior: N ~ [(N)]. Then condition (13.2) holds too. Proof. I. Let U be an open subset of the compactum K, V be an open subset of the space Y. Fix a countable everywhere dense subset A of U ~ K. Evidently the set M = n{M*({x}, V): x E A} contains the set M*(U, V). By Theorem 1.3 the set M is measurable. Show that M ~ M*(U, [V]). Assume the opposite, i.e., the existence of a point t* E M \ M*(U, [V]). Then there exists a point x* E U such that F(t*, x*) \ [V] =1= 0. By virtue of the density of the set A in U and of (13.5) F(t*,x) \ [V] =1= 0 for a point x E A. This contradicts the choice of the point t* EM. Hence our assumption is false and the inclusion in question is true. Thus M*(U, V) ~ M ~ M*(U, [V]). II. Let H be a closed subset of the compactum K and V be an open SUbset of the space Y. By Lemma 13.1 M*(H, V) = nk:lM*(OtH, V). By I for every k = 1,2, ... there exists a measurable set Mk such that M*(OtH, V) ~ Mk ~ M*(OtH, [V]) (~ M*(H, [V])). By Theorem 1.2 the set Mo = Uk:lMk is measurable. By Lemma 13.1 M.*(H, V) ~ Mo ~ M*(H, [V]). III. Let H be a closed subset of the compactum K and V be an open SUbset of the space Y.

CHAPTER 4

160

For k = 1,2, ... put Vk = Y\ [Ot(Y\ V)]. The sets Vk , k are open and Vi ~ [Vd ~ V2 ~ ... ~ V (therefore M*(H, Vd

~

M*(H, [Vi])

~

M*(H, V2 )

~

•••

~

= 1,2, ... ,

M*(H, V)),

Uk:l Vk = V (see, for instance, Corollary 2 of Lemma 2.3.1 and Lemma 2.3.2). By Theorem 1.7.8 for every t E [a, b] the set F( {t} x H) is compact. Therefore M*(H, V) = Uk:1M*(H, Vk ).

By II for every k = 1,2, ... we can fix a measurable set Mk such that M*(H, Vk) ~ Mk ~ M*(H, [Vk]). Hence M*(H, V) = Uk:lMk' By Theorem 1.2 the set M*(H, V) is measurable. We have checked the fulfilment of the condition (13.3). Now our assertion follows from Lemma 13.2. The lemma is proved. • Theorem 13.3. Let condition (13.1) hold. Let the mapping F be single valued. Then condition (13.2) holds. Proof. Notice that in the case of a single valued mapping the condition of the upper semicontinuity of the mapping F in the second argument goes into the condition of the continuity and the fulfilment of (13.5) become obvious. By virtue of this observation our assertion follows from Lemma • 13.3. The theorem is proved. Theorems 13.3 and 13.2 imply: Theorem 13.4 (Scorza-Dragoni). Let:

(13.6) I = [a, b], a < b, be a segment of the real line; K be a metric compact; Y be a metric space with a countable base; f : I x K -+ Y be a (single valued) mapping; for every tEl the mapping f (t, x) be continuous in x; for every x E K the mapping f (t, x) be measurable in t.

Then:

(13.7) for every number £ > 0 there exists a closed subset H of the segment [a, b] such that J.t(H) ~ b - a - £ and the mapping fl HxK is continuous . • Theorem 13.5 (A strengthening of the Egorov theorem). Let condition (13.6) hold. Let every function Ii, i = 1,2, ... , satisfy the condition imposed on the function f in (13.6). For almost every t E [a, b] let the sequence of functions gH x) = fi (t, x) converge uniformly to the function lex) = f(t, x). Then for every £ > 0 there exists a closed set H ~ [a, b] such that J.L(H) ~ b - a - £, the functions fil HxK ' i = 1,2, ... , and fl HxK are continuous and the sequence {fiIHxK: i = 1,2, ... } converges uniformly to the function fl HxK '

Derivation and integration.

161

Proof. I. We can change values of the functions fi' i = 1,2, ... , f on a set A x K, where the measure of the set A ~ [a, bj is equal to zero, in order to guarantee the fulfilment of the condition: for every t E [a, b] the sequence of functions {gf : i uniformly to the function l.

= 1,2, ... } converges

II. Denote by Ko the compactum K x ({O} U {2- i Define the function: F(t

x) = {fi(t, y, x) ,y, f(t,y)

if if

x = 2- i i x = 0

i = 1,2, ... }).

= 1,2, ... ,

on the product [a, bj x K o. The hypotheses of Theorem 13.4 hold. This implies that we can fix a closed set Hl ~ [a, b] such that p,(Hd ~ b - a - ~ and the function FIHl xKo is continuous. There exists an open set U 2 A of the measure < ~. The measure of the set H = Hl \ U is greater than b - a-c. Values of the functions fi' i = 1,2, ... , f were not changed on the set H x K. The function FI HxKo is continuous. By Theorems 3.2.4 and 3.2.3 the last fact implies the uniform convergence fi IH x K -+ f IH x K· The theorem is proved. • As a corollary of Theorem 13.5 we obtain: Theorem 13.6 (Egorov). Let real functions h i = 0,1,2, ... , be defined and be measurable on a segment [a, b], -00 < a < b < 00, of the real line IR. Let fi(t) -+ fo(t) for almost all t E [a, b]. Then for every c > 0 there exists a closed subset H of the segment [a, b] such that p,(H) ~ b - a - c, the functions filH' i = 0,1,2, ... , are continuous and the sequence {fiI H : i = 1,2, ... } converges uniformly to the function foln- • When fi == f Theorem 13.6 implies: Theorem 13.7 (Luzin). Let a real function f be defined and be measurable on a segment [a, b]' a < b, of the real line. Then for every number c > 0 there exists a closed subset H of the segment [a, b] such that p,{H) ~ b-a-c and the function flH is continuous. • Remark 13.1. Let us strengthen Lemma 11.3 from the point of view of results of this section. Assume that the conditions (13.1-2) hold for Y = IRn. Following Theorem 13.2 fix a sequence {a, b} ~ Hl ~ H2 ~ ... of closed subsets of the segment [a, b] such that P,{Hi) -+ b - a as i -+ 00 ~nd the mapping F IHi X K is upper semi continuous (see also Theorem 1. 7.6 In connection with the construction of a non-decreasing sequence of sets). Define a mapping Fi : [a, b] x K -+ IRn. Let it coincide with F on the set

162

CHAPTER 4

Hi X K: Fi IHi X K = F IHi X K' Let (c, d) be a connected component of the set [a, b] \ Hi' For t E (c, d) and x E K put t-c Fi(t,x) = d _ cF(d,x)

d-t

+ d _ cF(c,x).

The upper semicontinuity of the mapping Fi may be easily established with the using of Theorem 2.5.1. The convergence of the sequence of the mappings {Fi: i = 1,2, ... } to the mapping F has a rather strong character: for almost all t E [a,b] we have Fi(t,x) = f(t,x) for all x E K, beginning with some i = io. If the mapping F is single valued then the mappings Fi are single valued too. If values of the mapping F are convex, then values of the mappings Fi are convex too. If values of the mapping F are bounded from above by a number M, that is lIuli ~ M for every vector u E F([a, b] x K), then the analogous condition is true for the mappings Fi .

CHAPTER 5

WEAK TOPOLOGY ON THE SPACE L1 AND DERIVATION OF CONVERGENT SEQUENCES

In this chapter we collect the material in Functional Analysis and in the theory of Functions of Real Variable which is important for our Cauchy problem theory for differential equations and inclusions. In particular, we obtain the possibility of investigating of equations with complicated discontinuities in their right hand sides in space variables. Assertions of the functional analysis cited in this chapter will be helpful in our methods of proving the convergence of sequences of solution spaces of equations. In particular, they give the possibility of studying the mentioned class of equations. But they may be applied with advantage in the investigation of equations with continuous right hand sides too. We will meet such situations later. We do not aim here to give an exhaustive account of the stated fields of mathematics, but in the framework of our needs the corresponding material is given in entirety. The Hahn-Banach theorem and a simple version of the Dunford-Pettis criterion of the weak compactness are key points in this chapters (see §8 here and [Ed]). 1. Continuous linear functionals

We will restrict ourselves to the consideration of vector spaces over the field of real numbers JR. Let L be a normed space (see §2.4). Let D = {u: u E L, Ilull ~ I} be the closed unit ball of the space L. Assertion 1.1. Let f : L --t JR be a linear functional. Then the following conditions are equipotent: a) the functional f is continuous; b) the functional f is continuous at the point 0; c) sup{lf(u)l: u E D} < 00. Proof. a=>b. It is obvious. b=>a. Take arbitrary u ELand c > O. By b) there exists a number 6 > 0 such that. if IIwll < 6, then If(w)1 < c. If now v E 06U, then the

164

CHAPTER 5

vector w = v - u belongs to 0 6 0. So

If(v) - f(u)1

= If(v -

u)1

= If(w)1 < c.

b=;.c. By b) there exists a number c > 0 such that if If(w)1 < 1. If now u E D, then

Therefore

c

2"lf(u)1 < 1,

Ilwll <

c, then

2 If(u)1 < -.

c By virtue of the arbitrariness of u E D this implies that

sup{lf(u)l: u E D}

~

2 -. c

c=;.b. Let m = sup{lf(u)l: u ED}. For arbitrary c > 0 consider

8 = 2~. If now u E 0 6 0, then

2m

-u c

E

D,

2m

-If(u)1 c

~ m,

If(u)1

~

c -2 < c.

The assertion is proved. We noticed in §2.4 that every linear functional on the space

• ~n

is con-

tinuous. The set L * of all continuous linear functionals on the space L is a vector space. Assertion 1.1 allows to introduce on L* the function II . II : L* ~ [0, (0), lIull = sup{lf(u)1 :u ED}. We use for the function the symbol of the norm. Show that it is in fact a norm. If f E L* and flD == 0, then f == o. This implies that Ilfll = 0 ifand only if f == o. Thus the new function II . II satisfies condition 1) of the definition of a norm in §2.4. The fulfilment of condition 2) is obvious. Show the fulfilment of condition 3). Let f, gEL and u E D. We have

If(u)

+ g(u)1

~ If(u)1

+ Ig(u)1

~

Ilfll + IIgll·

Since it holds for every vector u E D,

Ilf + gil

= sup{lf(u)

+ g(u)l:

u

E

IIfll + IIgll· If(u)1 ~ Ilfll . Ilull·

D} ~

Assertion 1.2. Let u ELand f E L*. Then Proof. For u = 0 the estimate in question is obvious. Let u # O. The vector v = II~II belongs to D. Therefore

If(v)1

~

Ilfll,

If(u)1

-W~lIfli.

Weak topology on the space L1 and derivation of convergent sequences. 165 This gives what was required. • Besides the topology generated by the mentioned norm the space L * carries a so called 'weak' topology. The 'weak' topology is generated by the 'embedding' i : L* ~ II~.L: each linear functional f E L* is a function f : L ~ JR and so it represents an element of the Tychonoff product JRL. For the normed space L * we will keep the notation L *. Denote the space L * with the weak topology by L:. Associate to an arbitrary vector u E L the linear functional CPu : L * ~ JR, CPu(J) = f(u). It is continuous: (1.1)

if f E L* and

Ilfll

~ 1, then

ICPu(J)1

=

If(u)1 ~

IIfll·llull

~

Ilull·

Remark 1.1. For u E L the mapping CPu : L* ~ JR coincides with the composition of the mapping i and of the projection of the product JRL on the corresponding factor. This implies the continuity of the mapping i (see Theorem 2.1.1). The mapping : L ~ (L*)*, (u) = CPu, is linear:

CPo:u+(3v(J)

= f(au + /3v) = af(u) + /3f(v) = acpu(J) + /3CPv(J).

It is continuous: by virtue of the linearity of and (1.1)

1I(u) - (v) II

=

11(u - v)11

~

Ilu - vii.

In the next section we will continue the investigation of the mapping . Remark 1.2. Often the mapping is an isomorphism. In this case space L is called reflexive. But this need not take place. Assertion 1.3. The space L* is complete. (Independently of the completeness of the space L or its absence.) Proof. I. For every fixed point u E D the mapping BC(D, JR) ~ JR (see notation in §3.1), which associates to a function f E BC(D, JR) its value f(u), is continuous because If(u) - g(u)1 ~ Ilf - gil. II. Because of I for every a E [0, 1J and u, v E D the set

{J: f E BC(D, JR), f(au

+ (1 -

a)v) = af(u)

+ (1 -

a)f(v)}

is closed in the space BC(D, JR). III. The closedness of the set

A

= {J: f

f(au + (1 - a)v) = af(u) for every a E [O,lJ and u, v E D}

E BC(D, JR),

+ (1 -

a)f(v)

in the space BC(D, JR) follows from II and from Theorem 1.3.4. IV. The mapping L* ~ A, which associates to a functional f E L* its restriction fl D , is an isometry.

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CHAPTER 5

V. Now III, IV, Lemma 3.1.3 and Theorem 3.1.2 imply what was re• quired. The assertion is proved.

2. Hahn-Banach theorem Theorem 2.1 (Hahn and Banach). Let L be a vector space. Let a function p : L --t [0,00) satisfy the conditions:

(2.1) p(u+v)

~p(u)

+p(v) for allu,v E L;

(2.2) p(au) = ap(u) for all u E L, a

~

O.

Let a linear functional fo : Lo --t IR be defined on a vector subspace Lo of the space L and satisfy the condition

(2.3) fo(u)

~

p(u) for all u E Lo.

Then there exists a linear functional f : L --t IR such that fiLa = 10 and -pC -u) ~ feu) ~ p(u) for all u E L. Proof. I. Use Lemma 1.8.2. Denote by A the set of all pairs (M, cp), where M 2 Lo is a vector subspace of the space L, cP : M --t IR is a linear functional, cplLo = 10 and cp(u) ~ p(u) for all u E M. The set A is partially ordered by the rule: (MI' cpd -< (M2' CP2), if MI ~ M2 and CP21 M1 = CPl' II. Show that if (MI' cpd E A and MI i= L then there exists an element (M2' CP2) >- (MI' cpd, M2 i= M I, of the set A. Since MI i= L, there exists a vector Wo E L \ MI' The set M2 = {u + AWo: u E M I , A E 1R} is a vector subspace of the space L. Since Wo E L \ M I , the representation of every element of the set M2 as a sum u + AWo is unique. Take an arbitrary c E lR. Consider the functional 'l/Jc(u + AWo) = CPI(U) + AC. on the space M2· Our nearest aim is to find a value of c such that the pair (M2' 'l/Jc) belongs to A. For WI, W2 E MI we have CPI(wd

+ CPI(W2) = CPI(WI + W2)

~ P(WI

+ W2)

+ (W2 + wo)) wo) + P(W2 + wo).

= P((WI - wo) ~

P(WI -

Therefore CPI(wd -P(WI -wo) ~ P(W2+ WO) -CPI(W2)' Thus the numbers = inf{p(w + wo) - CPI(W) : wE Md are related by the inequality ql ~ q2' ql

= SUp{CPI(W) - pew - wo): wE Md and q2

Weak topology on the space Ll and derivation of convergent sequences. 167

Take now as C an arbitrary number of the segment [ql, q2]. For every vector u + AWo E M2 with u E Ml and A =J 0 we have

For A > 0 the right hand inequality implies:

CA ~ p(u

+ AWo) -

CPl(U),

'l/Jc(u

+ AWo) =

CPl(U)

+ AC ~ p(u + AWo)·

For A < 0, when we multiply the left-hand inequality by -A, we obtain: CPl (u) - p(u + AWo) ~ -AC,

'l/Jc(u + AWo) = CPl (u)

+ AC ~ p(u + AWo)·

For A = 0

Thus (M2' 'l/Jc) E A. III. By Lemma 1.8.2 there exists a maximal linearly ordered (by the relation -< of I) set B S;; A. Its maximality means that the set B is not a proper subset of an other linearly ordered subset of the set A. Denote by M(3 the first element of the pair (3 = (M, cp) E B (i.e., the subspace M S;; L). Denote by CP(3 the second one (i.e., the functional cp). The set if = U{M(3 : (3 E B} is a vector subspace of the space L. We define uniquely the linear functional rp on if by the condition rpiM/3 = CP(3. Since functionals CP(3 satisfy (2.3), this condition is satisfied by the functional rp too. Thus (if, rp) E A. The pair (if, rp) is not less than every element of the set B (in the sense of the order of I). By virtue of the maximality of the set B the pair (if, rp) belongs to B. If if =J L then by II there exists an element of the set B following (ii, rp). This is impossible by the definition of (if, rp). Thus if = L. For U E if = L we have rp(u) = -rp( -u) ~ -p( -u). The theorem is proved. • Remark 2.1. The norm II . II satisfies (2.1 - 2). For every m > 0 the /unction p(u) = mllull satisfies (2.1 - 2). By Remark 2.1 Theorem 2.1 and Assertion 1.1 imply Theorem 2.2. Let Lo be a vector subspace of a normed space L. Let fo : Lo - t IR be a continuous linear functional. Then there exists a contin• uous linear functional f : L - t IR extending the functional fo. Theorem 2.3. Let L be a vector space. Let a function p : L - t [0,(0) satisfy conditions (2.1- 2) and Uo E L. Then there exists a linear functional f : L - t IR such that

168

CHAPTER 5

and -p( -u)

~

f(u)

~

p(u)

for every vector u E L. Proof. Define on M = {Auo : ). E JR} an (obviously, continuous) functional f 0 ().uo) = ).p( uo). Theorem 2.1 implies the existence of its extension f to L. The extension f satisfies the estimates -p( -u) ~ f(u) ~ p(u). The theorem is proved. • When we take the norm as the function p in Theorem 2.3, we obtain, in particular (with the notation of Remark 1.1), that for every vector u E L there exists a linear functional f E L* such that f(u) :f. O. This implies: Assertion 2.1. If u ELand u :f. 0, then (u) :f. O. • Moreover, the functional f mentioned in the proof of Theorem 2.3 satisfies conditions

f(uo) = Iluo II,

(2.4) (2.5)

If(u)1 ~

It follows from (2.4-5) that

Ilull for of all u IIfll

E L.

= 1. By (2.4)

Therefore By (1.1) this means that

11(uo)11

=

Iluoll·

Thus the mapping : L ---t (L*)* keeps the norm. So it is isometric. So the mapping . : L ---t (L*):, which coincides with as the mapping of the sets, is continuous. The bijection . gives the possibility of defining a new topology on the set L (denote the corresponding topological vector space by L.):

a set U ~ L. is open in the space L. if and only if the set .(U) is open in the space 8(L) ~ (L*):. The new introduced topology on the vector space L is called the weak topology of the normed space L. By virtue of Remark 1.1 and the isometry of the mapping the identity mapping L ---t L8 is continuous.

Weak topology on the space £1 and derivation of convergent sequences. 169 3. Convex sets and linear functionals Let A :3 0 be a convex open subset of a normed space L. Then G"O ~ A for some c > O. Therefore: for every vector U E L there exists a number A > 0 such that A-lU E A (it is sufficient to take A = ~, see E Figure 5.1).

AA

In other words the set AA = {.Aw: w E A} contains the vector u. So the set Figure 5.1 Su = {.A: A ~ 0, u E AA} is nonempty. The Minkowski functional of the set A is defined by the formula JLA(U) = inf suo By the remarks just made 0 ~ JLA(U) < 00. Assertion 3.1. Under the above assumptions: 1) JLA(U + v) ~ JLA(U) + JLA(V) for all u, VEL; 2) JLA(Au) = AJLA(U) for all U ELand A ~ 0; 3) A = {u: U E L, JLA(U) < 1}. Proof. I. Take arbitrary numbers a > JLA(U) and b > JLA(V). Then a- 1 u, b- 1 v E A. By virtue of the convexity of the set A the vector u +V _ -a- (-1) a U a+b a+b

b (b- 1 v ) +--

a+b

belongs to A. Therefore a + bE su+v. This means that a + b ~ JLA(U + v). By virtue of the arbitrariness in the choice of a and b this implies 1). II. The property 2) follows trivially from the obvious equality SAU = As u . III. The continuity of the operation of the multiplication of a vector by a number and of the openness of the set A implies 3). The assertion is • proved. Lemma 3.1. Let f : L ~ ~ be a continuous linear functional and f t= O. Let A be an open subset of the space L. Then the set f(A) is open in R Proof. Take an arbitrary x E f(A). Let U E A and f(u) = x. The condition f t= 0 implies the existence of a vector v satisfying the inequality f(v) > o. The continuity of the operations of the addition of vectors and of the multiplication of a vector by a number, the openness of the set A imply the existence of a number c > 0 such that U + .xv E A for

170

CHAPTER 5

all ). E (-e, e). It remains to notice that

f(A) ;;2 {f(u

+ ).v):

). E (-e, e)} = (x - ef(v), x

+ ef(v».

The lemma is proved. • Theorem 3.1. Let A and B be nonempty convex disjoint subsets of a normed space L. Let the set A be open. Then there exist a continuous linear functional f : L ---t lR and a number 'Y E lR such that

(3.1) f(u) < 'Y :::; f(v) for all u E A and v E B (Figure 5.2).

Proof. Fix an arbitrary couple of vectors Ua E A and Va E B. The set

+ Va Ua + Va

C= A- B = {u U

= u{ {u u E

A}:

B

Ua

V -

E A,

[

V

E B}

V -

Ua

V

A

:

+ Va:

E B}

is open in the space L and contains O. So

Figure 5.2

(3.2) for some e > O. The convexity of the sets A and B implies the convexity of the set C. Since AnB = 0 the vector Wa = Va -Ua does not belong to C. Therefore (3.3) For U E L put p( u) = ing the condition

(3.4)

::/::;>. Construct a functional f : L

---t

lR satisfy-

-JLe( -u) ~ -p( -u) ~ f(u) ~ p(u) ~ JLc{u) for all u E L

according to Theorem 2.3. By (3.2)

Weak topology on the space L1 and derivation of convergent sequences. 171 for every vector u E L. Therefore u E latter estimate and (3.4) imply: f(u) ~ c- 1 11ull

c 1 11ull· C,

f (u - v

Therefore f(u) - f(v)

(3.5)

f(u)

<

+ wo)

+ f(wo) = f(v)

~

f.

/La( u - v

f(u) - f(v)

for all

u EA

c 1 11ull. The

u E L.

for all

This means the continuity of the functional For arbitrary u E A and v E B we have

/La(u) ~

+ wo) < 1. + 1 < 1, and

v EB

So'Y < 00 for 'Y = sup{f(u): u E A}. The openness of the set A and Lemma 3.1 imply the openness of the set f(A). Therefore'Y > f(u) for every vector u E A. This observation and (3.5) imply the fulfilment of (3.1). The theorem is proved. • Theorem 3.2. Let A and B be nonempty convex disjoint closed subsets of a normed space L. Let the set A be compact. Then there exist a continuous linear functional f : L ---* ~ and numbers p < q satisfying the condition

(3.6)

f(u)

f(v) for all u E A and v E B.

Proof. By Lemma 2.3.8 the distance c = p(A, B) is positive. By Lemma 2.4.3 the set DcA is convex. By the choice of c we have DcAnB = 0. When we apply Theorem 3.1 to the couple consisting of the sets DcA and B, we obtain the existence of a continuous linear functional f : L ---* ~ and a number'Y satisfying (3.6). The set f(A) is compact. Therefore the number p = sup f(A) belongs to f(A). Hence p < 'Y. It remains to put q = 'Y. The theorem is proved. • 4. Weakly convergent sequences We say that a sequence a = {Uk: k = 1,2, ... } of elements of a normed space L converges weakly to u E L if the sequence a converges to u with respect to weak topology of the space L. By virtue of the definition of the weak topology a sequence {Uk : k = 1,2, ... } ~ L converges weakly to u E L if and only if f(uk) ---* f(u) for every functional f E L * . Remark 4.1. If a sequence a = {Uk: k = 1,2, ... } ~ L converges weakly to u E L, then u E cc(a). To prove it, consider a pair consisting of the one point set A = {u} and the closed convex set B = cc(a). If we

172

CHAPTER 5

assume that our assertion is false then An B = 0. By Theorem 3.2 there exists a continuous linear functional f : L - t IR such that f(u) < inf f(B). Then f(u) # lim f(ud (~f(B)). k->oo

This contradicts the initial assumption. Remark 4.2. Under the assumptions of Remark 4.1 we can fix positive integers kl < k2 < ... in such a manner that the vector u is a limit (with respect to the norm of the space L) of the sequence of vectors Vi E c( {Uk., ... , Uk.+l-d). The mentioned vectors may be pointed sequentially. Since u E cc(a) = [c(a)], Remark 3.1 implies the existence of a vector VI E c(a) n 0 1 u. For some kl we have VI E c( {Ul, . .. , Ukl-l}). So we have done the first step of our inductive construction. Assume now that the vectors VI, ... , Vi and the numbers kl' ... ' k i are fixed. Since u E CC({Uk" Uk.+!, ... }), as in the first step there exists a vector 'liHI E c( {Uk., Uk.+!, ... }) n O2 -, U. Moreover, Vi+! E c( {Uk., ... , Uk.+l- 1 }) for some kHI > k i . Since the sequence {Vi: i = 1,2, ... } satisfies the condition Vi E O2 -.+1 U, we have the convergence Vi - t U with respect to the norm. Assertion 4.1. Let {Uk: k = 1,2, ... } ~ Land

(4.1) for every continuous linear functional f : L {If(udl: k = 1,2, ... } be bounded.

Then sup{llukll: k Proof. Let

-t

IR the sequence

= 1,2, ... } < 00.

Mi=n{{f: fEL*, If(udl=:;;i}: k=1,2, ... }. 1. Because of (4.1) L*=U{Mi: i=1,2, ... }. The inequality in the definition of the set Mi is non-strong. Therefore the set Mi is closed in the space L *. Assertion 1.3 and Baire theorem 1.6.7 imply that for some j = 1,2, ... the interior (Mj ) of the set M j is nonempty. Therefore for some point g E M j there exists a number E E (0,1) such that

Oog

~

Mj

.

If now h E L* and Ilhll < k = 1,2, ... we have:

E

then g - h E M j

.

Therefore for every

II. By Theorem 2.3 for every k = 1,2, ... we can fix a functional fk E L* in a manner that fk(Uk) = IIUkll and IIfkll =:;; 1.

Weak topology on the space L1 and derivation of convergent sequences. 173 By I (with h =

~fk)

for every k = 1,2, .... The assertion is proved. Assertion 4.2. Let a sequence {Uk: k = 1,2, ... } to a vector u E L. Then

~

• L converge weakly

Proof. Assume in addition the existence of a limit lim Iluk II (if the k ..... oo limit does not exist we choose a corresponding subsequence). Consider a continuous linear functional f E L* such that f(u) = Ilull and Ilfll ::;;; 1. Such a functional exists by Theorem 2.3 (with the norm 11·11 as the function p). Then The assertion is proved. • Theorem 4.1. A sequence {Uk: k = 1,2, ... } ~ L converges weakly to a vector u E L if and only if the following two conditions hold at once: 1) sup{lIukll: k = 1,2, ... } < 00; 2) there exists a dense subset M of the space L * such that lim f(Uk) = f(u) for every functional f E M. k-+oo

Proof. The necessity of condition 1) follows from Assertion 4.1. The necessity of condition 2) is obvious. Let us prove the sufficiency of these conditions. Let f E L, m = sup{ Iluk II: k = 1,2, ... } and € > O. Consider an arbitrary functional 9 E Oef nM. We have:

If(ud - f(u)1 ::;;; If(Uk) - g(uk)1 + Ig(ud - g(u)1 ::;;; €llukll + Ig(ud - g(u)1 + €lIull, lim sup If(Uk) - f(u)1 ~ 2€m.

+ Ig(u)

- f(u)1

k ..... oo

By virtue of the arbitrariness in the choice of € > 0 the last estimate implies the equality lim sup If(Uk)- f(u)1 = O. Therefore lim f(Uk) = f(u). k-+oo

The theorem is proved.

k ..... oo

•

5. Spaces Ll and Loo Let n = 1,2, ... and -00 < a < b < 00. The space Ll = Ll (la, b]) mentioned in the title of the section consists of Lebesgue integrable functions

174

CHAPTER 5

I : [a, b] ~ ]R.n; moreover, if two such functions coincide almost everywhere we mean that they represent the same element of the space L 1 . The Lebesgue integrability of the function I and remarks of §4.7 imply the integrability of coordinate functions Ii, i = 1, ... ,n, of the function I. This implies the integrability of the function g(x) = max{IIl(x)I,···, IIn(x)I}. The inequality (2.2.1) and Lemma 4.7.6 imply that: (5.1)

the function 11111 Lebesgue integrable.

Vice versa, a measurable function I : [a, b] ~ ]R. satisfying (5.1) belongs to the space Ll by Lebesgue theorem 4.7.4. The structure of a vector space on the set Ll may be defined in a natural way and the function b

11111 =

J

III(t)1I dt

a

is a norm on the space L 1 • Introduce in the consideration the function 1111100 = inf{m : m ~ 0, /-l({t: t E [a,b], II(t)1 > m}) = O}. For some part of elements I of the space Ll 1111100 = 00. The set Loo = {f: I ELI, 1111100 < oo} is a vector subspace of the space L 1 . The function 1111100 is a norm on Loo. This norm is defined independently on the main norm II I II of the space L 1 • The topology generated by the norm 1111100 need not coincide with the topology generated on Loo by the norm 11111. On the other hand, (5.2)

11111 ~ (b - a)IIIlIoo.

Therefore we obtain the continuity of the identity mapping Loo ~ L 1 • Here we consider the space Loo with the norm 11·1100 and we consider the space Ll with the norm II . II ( (5.2) implies the satisfaction of Lipschitz condition). Theorem 5.1. The space Ll is complete. Proof. Let Ib 12, ... be a fundamental sequence of elements of the space L 1 • Fix indices i(l) < i(2) < ... such that and IIIi(k) - Ijll < 2- k for j ~ i(k). Pass to the subsequence CPk = Ii(k)l k = 1,2, .... In particular,

(5.3) Because of (5.3) the partial sums IIcp2 - CPIII + IIcp3 -CP211 + ... + Ilcpk -CPk-III of the series IIcp2 - cpIiI + IIcp3 - CP211 + ... are bounded in totality by the number ~ + ~ + ... = l. By Theorem 4.7.2 the series (5.4)

Weak topology on the space Ll and derivation of convergent sequences. 175 converges almost everywhere on the segment [a, b] to a (non-negative) function 1jJ(x). At every point of convergence of the series (5.4) the series

(x) = (CP2(X) - CPl(X))

(5.5)

+ (CP3(X)

- CP2(X))

+ ...

converges too; moreover 11(x)11 ~ 1jJ(x). Theorem 4.7.4 implies the Lebesgue integrability of the function (x). Let f(x) = CPl(X)+(x). For m = 1,2, ... put l(m) = max{k: k = 1,2, ... , i(k) ~ m}. Then

II -

IIf - fmll ~

(5.6)

CPI{m) II

+ Ilfi{l{m)) - fmll·

By virtue of the estimates

Ilf - CPkll = II + CP1 - CPkll = II(CPk+l ~ 2- k + 2- k- 1 + ... = 2- k+l

CPk) --t

--t

- cpk+d

+ .. ·11

0,

Ilfi{l{m)) - fmll ~ 2- i {l{m)) and (5.6) we have the convergence fm The theorem is proved.

+ (CPk+2 --t

0,

f in the metric of the space L 1 • •

6. Common representation of a linear functional on the space L1 Let

-00 < a < b < 00, n = 1,2, .... Let cP E Loo and e= Ilcplloo. Then b

(6.1)

J

b

(u(s), cp(s)) ds

a

~

JlIu(s)ll· Ilcp(s)11

b

ds

J

~ e lIu(s)11

a

ds =

ellull

a

for every function u ELI, where (.,.) in the first integral denotes the scalar product. Thus the formula

J b

f(u) =

(u(s), cp(s)) ds

a

defines a continuous linear functional on the space L 1 • Let us show that really this is a common representation of a continuous linear functional on the space L 1 • For x E [a, b] denote by ax the characteristic function of the subset [a, x] of the segment [a, b]: _ {

ax ( t ) -

1

o

if if

t E [a, xj, t E (x,bj.

CHAPTER 5

176

Remark 6.1. Let e be an arbitrary unit vector of the space JRn. Let the mapping [a, b] - Ll associate to a number x E [a, b] the function axe. The mapping is continuous. It is even isometric: b

Ilaxe - ayell =

y

J

J

a

x

lax(t) - ay(t)1 dt =

1 dt = Ix - YI·

Let f : Ll - JR be a continuous linear functional. Let e be an arbitrary unit vector of the space JRn. Remark 6.1 implies the continuity of the function rp(x) = f(axe). Let (6.2) , = {(ai, bi ): i of the segment [a, b],

=

1, ... , k} be a family of pairwise disjoint intervals a ~ al < bl ~ a2 < ... ~ ak < bk ~ b.

For i = 1, ... ,k put

O"(i) = {

1 -1

if if

f(abie) - f(aaie) ~ 0, f(abie) - f(aaie) < o.

Define the function f3 : [a, b] - JR,

f3(t) = {

o 0"( i)

U"

if t rJ. if t E (ai, bi ),

i = 1, ... , k.

The function f3 admits the representation

(in the sense of the space L l , so we do not distinguish functions coinciding almost everywhere). Therefore (6.3)

f(f3e) = O"(l)(f(ah e) - f(aal e)) + ... + l7{k)(f(ah e ) - f(aa. e )) = If(ab1e) - f(aale)1 + ... + If(ab. e ) - f(aake)l·

Assume now that (6.4) for i = 1,2, ... a family condition (6.2)

'i of intervals of the segment [a, b] satisfies

The sequence of corresponding functions f3i converges in the space Ll to the function [J == o. By virtue of the continuity of the functional f

Weak topology on the space L1 and derivation of convergent sequences. 177

f(f3ie) --t 0 as i --t 00. By (6.3) this means the absolute continuity of the function cpo Remark 6.2. Assume that U is the characteristic function of a set E ~ [a, b], e is an arbitrary unit vector ofthe space ~n, bi: i = 1,2, ... } is a sequence of families of intervals of the segment [a, bj satisfying (6.4), (6.5) p,( Gi

E \

E)

~ --t

Gi = Uri 0 as i --t

for every i = 1,2, ... , G1 ~ G 2 ~ G 3 ~ ••• and Ui is the characteristic function of the set Gi .

00,

The sequence of the functions Ui converges in the space L1 to the function u. By virtue of the continuity of the functional f

f( Ui e ) --t f( ue)

(6.6)

as

i

--t

00.

We have

f(Ui e ) = (f(ab 1 e ) - f(aal e » + ... + (f(ahe) - f(aak e » = (cp( ab 1 ) - cp( aal » + ... + (cp( abk) - cp( aak» (we keep the notation of the previous remark). Thus

J

f(Ui e ) =

cp'(s) ds.

Gi

By Theorem 4.7.3 and by (6.5-6)

J

J

E

E

cp'(s) ds =

cp'(s) ds

= .lim

'1.--+00

J

+ i~~

J

cp'(s) ds

Gi\E

cp'(s) ds = '1.--+00 .lim f(Ui e ) = f(ue).

Gi

Let e1, ... ,en be a basis of the space ~n, CPm(x) = f(axe m), cp = {CP1, ... , CPn}. By our remarks the function cp is absolutely continuous. Show that cp' E Loo and IIcp'lloo ~ Ilfll. Assume the opposite. Then the measure of the set M = {t: t E [a,bj, IIcp'(t) II > IIfll} is positive. Let A be the family of all closed balls S of the space ~n satisfying the conditions: all coordinates of the center of the ball S are rational; the radius of the ball S is rational; the ball S lies entirely outside the closed ball of the radius IIfll with the center at the origin of coordinates. The family A is countable: A = {Si: i = 1,2, ... }. We have M = U{Mi: i = 1,2, ... }, where Mi = {t: t E [a, bj, cp'(t) E Sd. By Theorem 4.1.2 JL(Mi ) =F 0 for some i = 1,2, ....

178

CHAPTER 5

Let e = {a1,'''' an} be the unit vector co-directed with the position vector of the center of the ball Si' Let u be the characteristic function of the set E = Mi. By Remark 6.2

f(ue) = ar!(ue1) + ...

+ anf(uen )

= a1 j cp~(s)ds + ... + an j cp~(s)ds E

E

(a1cp~ (s) + ... + ancp~(s))ds

= j E

= j(e,cp'(s))ds E

> Ilfll . Ji,(E) = Ilfll . Ilull = Ilfll ·lIuell· This contradicts Assertion 1.2. To complete the reasoning consider the continuous functional b

(6.7)

g(u) = f(u) - j(cp'(s),u(s)) ds. a

By Remark 6.2 g(u) = 0 for every product u of the characteristic function of an arbitrary measurable subset of the segment [a, b] by an arbitrary vector. By virtue of the linearity of the functional 9 we have g( v) = 0 for every sum v of such products. Such sums constitute an everywhere dense subset of the space L1 (see Example 4.7.6). Therefore the continuity of the functional 9 implies that 9 = O. In other words, b

f(u)

= j(cp'(s),u(s)) ds a

for every function u E L 1 • Theorem 6.1. A sequence {h: k = 1,2, ... } ~ L1 converges weakly to a function fELl if and only if the following two conditions hold at once: (6.8)

sup{llfkll: k = 1,2, ... }

< 00,

and (6.9)

I

D

D~Ia,bl·

h(s) ds

~

I

D

f(s) ds as k

~ 00

for every measurable set

Weak topology on the space Ll and derivation of convergent sequences. 179 Proof. Theorem 4.1 and our remarks imply that the convergence of the sequence Uk: k = 1,2, ... } to the function f is equipotent to the fulfilment at once of conditions (6.8) and b

(6.10)

b

j(h(s),u(s)) ds

~

a

jU(s),u(s)) ds a

as k ~ 00 for every sum u E Leo of products of characteristic functions by vectors. The equivalence of (6.9) and (6.10) is obvious (note that b

j h(s) ds = j h(s)XD(s) ds). D

a

•

The theorem is proved. 7. Space of measurable sets Let

-00

< a < b < 00.

Denote by Mo(a, b) the set of all measurable subset of the segment [a, b]. We regard sets A, B E Mo (a, b) equivalent if the measure of the set AAB = (A \ B) U (B \ A) is equal to zero. Notice that b

(7.1)

Jl(AAB)

= J IXA(S) - XB(s)1 ds = IIXA - XBII· a

Let the mapping <1>0 : Mo ~ Ll associate to a set its characteristic function. The images of sets A and B under the mapping <1>0 coincide if and only if the set A and B are equivalent. Therefore the mapping <1>0 generates the mapping <1> : M ~ Ll of the quotient set M of the set Mo with respect to the above equivalentness. Moreover (7.1) implies that the fUnction p(A, B) = Jl(AAB) is a metric on the set M and the mapping <1> turns out to be an isometry. Assertion 7.1. The metric space (M, p) is complete. Proof. Let a = {Ak: k = 1,2, ... } be an arbitrary fundamental sequence of elements of the space M. It is sufficient to prove the existence of a limit for a subsequence of the sequence a. Thus without loss of generality We can assume in addition that p(Ak' A,) < 2- k for every I > k. Let Bk = U{Ai: i = k, k + 1, ... }.

CHAPTER 5

180

peAk. B k ) ~ jl(U{Ai+1 \ Ai: i = k, k ~

2- k

+ 1, ... })

+ 2- k - 1 + ... = 2- k +l.

Therefore it is sufficient to prove the existence of a limit of the sequence (3 = {B k : k = 1,2, ... }. This is obvious: the set n(3 is the limit of the sequence (3. The assertion is proved. •

8. Weak compactness in the space L1 Keep the notation of the previous sections. Our aim in this section is to understand, when (8.1) every subsequence of a (the) sequence Uk: k = 1,2, ... } ~ L1 has a subsequence converging weakly to a function fELl' A sequence Uk : k = 1,2, ... } satisfying condition (8.1) is called relatively weakly compact (see §1.6). By Theorem 6.1 the fulfilment of condition (6.8) is necessary for the fulfilment of condition (8.1). In order to point a second necessary condition

consider following situation: (8.2) {Mk : k = 1,2, ... } is a sequence of subsets of the segment [a, b] and jl(Mk ) - t 0 as k - t 00.

k

Lemma 8.1. Let n = 1, condition (8.2) hold and the sequence converge weakly to the function f == O. Then

Uk

= 1,2, ... }

(8.3)

J

h(s) ds

-t

0 as k

-t

00.

Mk

Proof. 1. With the notation of §7 the set S =
J b

GJ(u)

=

f(s)u(s) ds

a

is continuous.

~ -t

L1 is complete. JR,

Weak topology on the space L1 and derivation of convergent sequences. 181 For

f

E

Loo this is obvious:

100f(u) - O!f(v)1

~

b

J

If(s)I·lu(s) - v(s)1 ds

~ Ilflloo ·llu - vii·

a

In the general case fix a sequence Uk: k = 1,2, ... } to the function f in the space L 1 • Then for every function u E S

~

L oo , converging

b

100fk(u) - O!f(u)1

~

J

IJk(S) - f(s)l· u(s) ds

~ Ilh - fll·

a

So we have the uniform convergence O!h ~ O!f. This implies what was required. II. Take an arbitrary c > O. For l = 1,2, ... put

By virtue of the weak convergence of the sequence Uk: k = 1,2, ... } to the function f == 0 we have b

O!fk (u)

=

J

h(s)u(s) ds

~0

a

for every u E S. Thus S = U{S/: l = 1,2, ... }. The sets S/ are closed in the space S (the functions O! h are continuous and the inequality in the definition of S/ is non-strong). By Baire's theorem 1.6.7 for some m = 1,2, ... the set (Sm) is nonempty. Therefore there are A E Sm and 6 > 0 such that the 6-neighborhood U of the point A in the space S lies in the set Sm. If BE U and n = m, m + 1, ... , then

III. If with the notation of II C ~ A and p,(C) ~ 6, then

IlxA -

J

fn(s) ds

C

=

J

XA\clI ~

fn(s) ds -

A

8,

J A\C

fn(s) ds,

182

CHAPTER 5

Jf (

C

~ "'8

8) ds

n

C

C

+ -8

= -

4·

C

IV. With the notation of III put C+ = {s: C _ = {s: sEC, f n(s) < O}. When we apply III to these sets we obtain

J

Ifn(s)1 ds

C

J

=

SEC, fn(s)

J

fn(s) ds -

fn(s) ds

~

O} and

~ ~ + ~ = ~.

c_

C+

V. If with the notations of II C ~ S \ A and /-l(C) ~ 15, then IlxA - XA\cll ~ 15,

J

J

fn(s) ds =

J

fn(s) ds -

fn(s) ds,

AuC

C

A

J

f n ( s) ds ~

C

C

C

8 + 8 = "4.

C

Now when we repeat the reasoning IV, we obtain

J

Ifn ( S ) I ds ~

C

C

"4 + "4

C

2·

=

C

VI. Let a set C

~

[a, b] be measurable, /-l( C)

J

Ifn(s)1 ds =

J

Ifn(s)1 ds

CnA

C

~

J

15 and n ~ m. Then

Ifn(s)1 ds

C\A

-C + -C =

'" 2

+

~

2

c.

The last estimate implies (8.3). The lemma is proved. Corollary. Let condition (8.2) hold. Let a sequence converge weakly to the function f == O. Then

J

fk(S) ds

---t

0 as

k

Uk:

• k = 1,2, ... }

---t 00.

Mk

To obtain the corollary notice that if g E (JRn)*, then the sequence {g(h): k = 1,2, ... } converges weakly to the function h == O. By Lemma 8.1

Weak topology on the space Ll and derivation of convergent sequences. 183

By virtue of the arbitrariness of 9 E (IR.n)* we have

• Assertion 8.1. Let conditions (8.1- 2) hold. Then condition (8.3) holds too.

Proof. Assume the opposite, i.e., for some

> 0 let the set

E

be infinite. By (8.1) the sequence Uk: k E A} has a subsequence Uk: k E Ad converging weakly to a function fELl. Since XMk --+ 0 almost everywhere, fXMk --+ 0 almost everywhere. Theorem 4.7.4 implies the convergence b

J

(S.4)

f(s) ds =

J

f(S)XMk(S) ds

--+

0

as

k

--+ 00

a

(we can take the function If I as a majorant). When we apply Lemma 8.1 to the sequence obtain

J

(S.5)

(Jk(S) - f(s)) ds

--+

0

Uk - f

as k E AI, k

k E

Ad,

we

--+ 00.

Mk By (S.4-f))

J

f k(s) ds =

Mk

as k

J

(Jk(s) - f (s )) ds

Mk

+

J

f (s) ds

--+

0

Mk

k --+ 00. So our assumption is false and the assertion is proved. We say that a sequence of functions 'Pk : [a, b] --+ ~n, k equi-absolutely continuous if: E AI,

=

• 1,2, ... , is

(8.6) for every E > 0 there exists 8 > 0 such that if a family {(ai, bi ) : i == 1, ... ,p} of pairwise disjoint intervals of the segment [a, b] satisfies

184

CHAPTER 5

the condition E{lbi - ail: i = 1, ... ,p} < 8 then E{II
-


i = 1, ... ,p}

Remark 8.1. If condition (8.6) holds then: (8.6') for every I:: > 0 there exists 8 > 0 such that if a family {(ai, bi ) : i = 1,2, ... } of pairwise disjoint intervals of the segment [a, bj satisfies the condition E{lbi - ail: i = 1,2, ... } < 8, then E{II
(8.6/1) for every J.L(M) < 8, then

I::

> 0 there exists a 8 > 0 such that if M

J
ds

C IR and

< I::

M

for every k

= 1,2, ....

To deduce (8.6/1) from Remark 8.1 fix 8 according to condition (8.6'), but for 1::/2 instead of 1::. By (8.6'): for every open set M k = 1,2, ...

~

[a, bJ of the measure < 8 and for every

J
< 1::/2.

M

Fix of a decreasing sequence of open sets

Weak topology on the space L1 and derivation of convergent sequences. 185 such that

Then for every k

f cp~(t)dt f cp~(t)dt + l~~ f i~~ f cp~(t)dt

M

M

cp~(t)dt

Mi\M

=

Mi

c ~-
[a, b]

--t

IRn , k = 1,2, ... ,

be equi-absolutely continuous and converge uniformly on the segment [a, b] to a function CPo. Then the function CPo is absolutely continuous. Proof. Take an arbitrary rJ > O. Fix 8 > 0 according to (8.6) for C = rJ/2. Let 'Y = {(ai,b i ): i = 1, ... ,p} be a family of pairwise disjoint intervals of the segment [a, b] with d(r) < 8. When we pass to the limit as k --t 00 in the last inequality of (8.6), we obtain: E{IICPo(bi ) - cpo(ai)11 : i = 1, ... ,p} ~ rJ/2 < rJ. This is what was required. The lemma is proved. • Lemma 8.3. Let t

CPi (t) =

J fi (s)

ds for

a

i == 0,1,2, ... and the sequence {CPi: i = 1,2, ... } converge uniformly to a /unction CPo.

Let conditions (6.8), (8.6) hold. Then

{8.8} condition {8.3} holds for every sequence of measurable sets {Mk k == 1,2, ... } satisfying {8.2}. and the sequence a converges weakly to the function 10 = cp~. Proof. By Theorem 6.1 and Assertion 8.1 it is sufficient to prove the fulfilment of {6.9}.

CHAPTER 5

186

1. For D = (c, d) ~ [a, bj (in (6.9) ) the convergence in (6.9) takes place by virtue of the convergence d

/ h(s) ds = 'Pk(d) - 'Pk(C)

~ 'Po(d)

d

- 'Po(c) = /10(s) ds.

c

c

II. By I in the case, when the set D may be represented as the union of finite number of intervals, the corresponding convergence takes place too. III. Let D be an arbitrary open subset of the segment [a, bj, 'Y = {G i : i = 1,2, ... } be the family of (all) its connected components. Take an arbitrary c > o. Lemma 8.2 implies the equi-absolute continuity of the sequence {'Pi: i = 0,1,2, ... }. Remarks 8.1-2 imply that for 'f/ = ~ (8.9) there exists 0

J.L(M) < 0, then

> 0 such that if the set M ~ [a, bj is measurable and every i = 0,1,2, ....

II J 1i(S) dsll < 1J for M

(This implies the fulfilment of (8.8).) Since J.L(D) = J.L(U'Y) ::;; b - a, there exists an index j = 1,2, ... such that J.L(U{ Gj , Gj+I, ... }) < o. Let H = G I U··· U G j - 1 . By II there exists an index l = 1,2, ... such that

/1k(S) ds - /10(s) ds < H

Thus for k

~

i

for

k

~ l.

H

l

/ 1k(S) ds - / 10(s) ds D

D

( /1k(S) ds H

+ /

1k(S) dS) - (/10(S) ds

D\H

: ; J1k(S) ds - J10(s) ds H

H

H

+

+ /

10(s) dS)

D\H

J 10(s) D\H

ds

Weak topology on the space Ll and derivation of convergent sequences. 187 By virtue of the arbitrariness in the choice of c this means the convergence

1 h{s) ds

-+

D

110{s) ds. D

IV. Let D be an arbitrary measurable subset of the segment [a, bJ. Take an arbitrary c > O. Fix 8 according to (8.9). By virtue of the measurability of the set D there exists a open subset H 2 D of the segment [a, bJ such that J.L{H \ D) < 8. By III there exists an index l = 1,2, ... such that

11k{s) ds - 1 10 {s) ds < H

Thus for k

~

i

for

k

~ l.

H

l

11k{S) ds - 1 10 {s) ds D

D

1

( 1 h{s) ds H

~

10{s) dS) - (110{s) ds -

H\D

1

1k{S) ds - 110{s) ds

H

c ~ ...., 3

H

+

H

c

1

10{s) dS)

H\D

1 10{s) ds H\D

c

+ -3 + -3 = c .

By virtue of the arbitrariness in the choice of c this means the convergence

1 1k{S) ds D

-+

110{s) ds. D

The lemma is proved. • Remark 8.3. Condition (8.8) implies that for every 'TJ > 0 condition (8.9) holds. If we take in (8.9) an open set (it falls into the union of the family'Y of its connected components) we obtain the fulfilment of (8.6) (see the notation in (8.7)).

CHAPTER 5

188

Assertion 8.2. Let a = Uk: k = 1,2, ... } ~ L 1 • Let conditions (6.8) and (8.9) hold. Then the sequence a satisfies condition (8.1). Proof. Let t


J

fi(s) ds for i = 1,2, ....

a

For arbitrary 'f/ > 0 fix 8 > 0 according to (8.9), see Remark 8.3. In particular, if c < d are points of the segment [a, b] and d - c < 8, then lI 0 according to (8.9) for 'f/ = 1. Choose points

to

= a < tl < ... < tj = b

in a manner that ti - t i- l < 8 for all i = 1, ... ,j. Take an arbitrary k = 1,2, .... For i = 1, ... ,j put Mi = {t: t E [ti-l, til, fk{t) ~ O} and Ni = [ti-l, til \ Mi· Then

11M =

!

1/.(t)l dt =

t (j

I.(t) dt -

!

I,(t) dt) :;; 2j.

The last estimate does not depend on k. The theorem is proved. • We have obtained a working (that is, applicable in real situations) test of the weak relative compactness in the space L 1 . Remark 8.4. Let a sequence of functions Uk: k = 1,2, ... } satisfy (8.8). Then the sequence of the functions {Ihl: k = 1,2, ... } satisfies (8.8) too. To prove it with the notation of (8.2) put M; = {t: t E Mk, fi{t) ~ O} and Mk* = Mk \ M;. Then

J

Ifk(S)1 ds =

~

J

fk(S) ds -

~

J

fk(S) ds.

~.

Both terms in the right hand side of the equality tend to zero. This gives what was required.

Weak topology on the space L1 and derivation of convergent sequences. 189

9. Derivation of a convergent equi-absolutely continuous sequence of functions Consider the following situation: (9.1)

-00

< a < b < 00, n

= 1,2, ... ;

(9.2) a sequence of functions CPk : [a, bj absolutely continuous;

--t

]Rn,

k

1,2, ... , is equi-

(9.3) the sequence {
Our aim is to prove the fulfilment of the condition

(9.4)

t

E

cp~(t) E

n{ cc( {cp~(t): i = k, k + 1, ... }): k = 1,2, ... } for almost all

[a,bj.

Remark 9.1. The fulfilment of conditions (9.1-3) implies that the function CPo is also absolutely continuous, see Lemma 8.2. Remark 9.2. The fulfilment of the conditions (9.1-3) implies that:

(9.5) for every c > 0 there exists a 8 > 0 such that if a set M ~ [a, bj is measurable and J.t(M) < 8, then II J cp~(t)dtll < c for every k = 1,2, ... , M

see Remark 8.2. Remark 9.3. By Remark 9.2 and Theorem 8.1 we have proved: Lemma 9.1. Let conditions (9.1 - 3) hold. Then the sequence of /unctions {
L1 •

•

Remark 9.4. By Lemma 9.1 and Assertion 8.2 we can assume in addition that the sequence {cp~: k = 1,2, ... } converges weakly to a function '¢. Since integral is a continuous functional, for every t E [a, bj we have t

CPo(t) - cpo(a) =

i~~(cpk(t) -

CPk(a)) =

i~~J cp~(s)ds = a

t

J 'Ij;(s)ds. a

Therefore cp~(t) = 'Ij;(t) for almost all t E [a, bj. Remark 9.5. By Remark 4.2 the function 'Ij; is the limit in L1 of convex combinations ~1' ~2"" of elements of the sequence {cp~ : k = 1, 2, ... }. So ~i(t) E cc({cp~(t):

k = 1,2, ... })

for almost all t E [a, bj and for every i = 1,2, ....

190

CHAPTER 5

By the reasoning in the proof of Theorem 5.1 we can choose a sequence i = 1,2, ... } in a manner that ~i(t) -+ 1f;(t) for almost all t E [a, b]. Therefore {~i:

(9.6)

cp~(t)

E cc( {cp~(t): i = 1,2, ... }) for almost all t E [a, b].

We can now shorten the sequence {CPk : k = 1,2, ... } and reject its first (k - 1) members. For the new sequence the analog of (9.6) remains true, namely cp~(t) E

cc( {cp~(t): i = k, k

+ 1, k + 2, ... })

for almost all t E [a, b].

.

Since this holds for every k = 1,2, ... we obtain: Assertion 9.1. Let conditions (9.1- 3) hold. Then condition (9.4) holds ~.

CHAPTER 6

BASIC PROPERTIES OF SOLUTION SPACES

Our aim in this chapter and in part of the next one is to point out a quite short list of properties of solution sets of ordinary differential equations which may be considered as the axiomatics for an appreciable part of the theory. In fact, the framework of the new theory will contain not only solution spaces of ordinary differential equation, but other objects too, although they are close to solution spaces with respect to their properties. Such an enlargement of the domain of action is not now essential, but the fact that new tools suit well the investigation of equations with different singularities in the right hand side, of corresponding differential inclusions, and of equations with the control, etc., turns out to be important. As axioms of our theory we take the simplest properties of solution sets, which are in fact constantly used in the theory of ordinary differential equations, although sometimes their statements, which are applicable to our investigation, look unusual in comparison with the corresponding conditions of the classical theory. The presence of some of these properties in solution spaces follows directly from the definition of a solution. For other properties we point out sufficient conditions. We do not want to finish the investigation in this chapter. Here our aim is quite narrow: we show that the axioms of our theory correspond to known properties of solutions of equations. Here we do not restrict ourselves to the simplest statements of sufficient conditions. As far as can be done, and with the use, in essence of the material of the previous chapters we prepare the ground for the latter account. We try to avoid here any complications in proofs and the introduction of cumbersome conditions which can make the initial grasp of the problem difficult.

1. Ordinary differential equations An equality of the type (1.1)

cI>1 (t, y(t), y'(t), ... ,y(m)(t)) = cI>2 (t, y(t), y'(t), ... , y(m)(t)) ,

or, in shortened notation, cI>l(t,y,y', ... ,y(m») = cI>2(t,y,y', ... ,ym), is called an ordinary differential equation. It relates values of the argument t

192

CHAPTER 6

and the corresponding values y(t), y'(t), ... , y(m)(t) of the unknown function y and its derivatives. The argument t takes real values and the function y (and its derivatives) take values in the space ]Rn (respectively, in the space en). The functions <1>1 and <1>2 are defined on corresponding subsets of the product ]R x ]Rn x ... x]Rn (respectively, ]R x en x ... X en) and take values in the space ]Rk or k . In fact, in what follows we consider equations of a more special type, namely of the type y' = <1>(t, y); however, this is often sufficient. For instance, when <1> and y in (1.1) are scalar functions then equation (1.1) may often be solved with respect to highest derivative of y. Our equation turns into an equation of the type

e

y (m) -_

Under the change Y1 = y, Yk changes into the system

=

y(k-1) for k

= 2, ... , m the last equation

y~ = Y2, y~ = Y3,

y~-l = Ym, y~ =
The new system may be written as a single vector equality y'

= <1>(t, y) with

<1>(t,X1,""Xm) = (X2,X3, ... ,Xm,
Often such transforms are possible for the vector functions <1> and y. Of course, here we can meet refinements related to the equivalence in such passages, but we will not digress to discuss it. The complex plane e may be considered as a vector space over R In this sense it is isomorphic to the space ]R2 (we associate to a complex number a + ib the vector {a, b} ). The space em may be also considered in an analogous way as a vector space over R In this meaning it may be identified with the space ]R2m. Because of the existence of the isomorphisms mentioned the consideration of an equation with a complex unknown function reduces to the real case, although in some situations the use of complex numbers makes finding solutions easier. 2. Solutions of differential inclusion y' E F(t, y)

Apart from ordinary differential equations we consider differential inclusions. Differential inclusions (== equations with multi-valued right hand

193

Basic properties of solution spaces.

sides) generalize differential equations. In the previous section we restricted our considerations to equations of a particular type. In this section we consider inclusions of a particular type also, namely, of the type mentioned in the title. Such inclusions are interesting because they both appear in applications and they are constantly used as tools in the investigation of differential equations. As in the case of an equation the complex case here in general reduces to the real one. So in what follows we study the real case only. Let F be a multi-valued mapping of an open subset U of the space IR x IRn into the space IRn. (Here we do not impose on the mapping F any conditions as the continuity or the measurability and we do not assume the nonemptiness of its values.) A function Z E G.(U) is called a solution of the differential inclusion (= solves the inclusion) y' E F(t, y) (or in the expanded form y'(t) E F(t, y(t)) ) if either the domain of the function z consists of one point only or the domain of the function z is nontrivial, the function z is generalized absolutely continuous, the approximate derivative z'(t) exists and belongs to the set F(t, z(t)) at almost every point t E 7r(z). In the case when the mapping F is single valued the differential inclusion y' E F(t, y) is usually written in the form of the differential equation y' = F(t, y), moreover the above definition of a solution of an inclusion turns into the definition of a solution of an equation. The solution set of a differential inclusion (equation) y' E F(t, y) will be denoted by D(F). For a subset M of the set U we denote D(F, M) = {z: z E D(F), Gr(z) ~ M}. The set D(F) lies in the topological space G.(U). With respect to the induced topology the set D(F) is also called the solution space of the inclusion y' E F(t, y). Notice some properties of solution sets following immediately from our definitions. Let Z = D(F). Then: (1) if z E Z and a segment I lies in 7r(z), then

Z2

ZII

E Zj

(2) if the domains of functions Zl, Z2 E Z intersect and the functions coincide on the set 7r(zd n 7r(Z2), then the function for t E 7r(zd for t E 7r(Z2),

Zl,

194

CHAPTER 6

7f(z) = 7f(zd U 7f(Z2), belongs to Z (see Lemma 4.9.1). These conditions are satisfied not only by solution sets of equations and inclusions. They are satisfied, for instance, by the space Cs(U), which is not the solution space of an equation or an inclusion, because it contains functions that is not generalized absolutely continuous. In order to deal freely with such sets let us introduce the following notation. Let M be a subset of the product lR x X of the real line lR and a topological space X. Denote by R(M) (respectively, by Ri(M)) the set of all subset Z of the space Cs(M) satisfying conditions (1) and (2) (respectively, condition (1)). Evidently conditions (1) and (2) are insufficient to develop a very valuable theory. Below we introduce other conditions also, which are satisfied (or may be satisfied) by the solution sets of equations and inclusions. We denote them for the convenience of references by small letters of the latin alphabet, for instance, (c), (e), (k) and so on. If * is an arbitrary set of such letters (conditions), then R~i) (M) denotes the set of all elements of the set R( i) (M) satisfying all conditions from *. We meet two first such conditions in the following lemma and its corollary. Lemma 2.1. Let F : U --t lRn be a multi-valued mapping and Z = D(F). Then

(q) the set Z consists of generalized absolutely continuous functions and if a function z E Cs(U) is generalized absolutely continuous, M is a closed subset of 7f(z), J1.(M) = 0 and ZII E Z for every segment I ~ 7f(z) \ M, then z E Z. Proof. The first part of (q), i.e., the assertion about the generalized absolute continuity, follows immediately from the definitions. Check the fulfilment of the second part of (q). By virtue of the openness of the set 7f(z) \ M in 7f(z) (and the regularity of a segment, see Lemma 2.3.9) for every point t of the set 7f(z) \ M there exists a segment It ~ 7f(z) \ M, which contains the point t in its interior (with respect to the segment 7f( z)). The family "( = {(It): t E 7f( z) \ M} is an open cover of the set 7f(z) \ M. By Theorem 1.6.10 the family "( contains an (at most) countable cover "(1 = {(It;}: i = 1,2, ... } of the set 7f(z) \ M. By our assumptions ZIIt. E D(F) for every i = 1,2, .... By the definition of a solution this me~ns the existence of a set Mi ~ It. (may be, empty) of measure zero such that at every point s E It. \ Mi the approximate derivative z' (s) exists and belongs to F(s,z(s)). By Theorem 4.1.2 the measure of the set M* = M U (U{Mi: i = 1,2, ... }) is equal to zero. At every point s of the set 7f( z) \ M* the approximate derivative z' (s) exists and belongs to the set

Basic properties of solution spaces.

195

F(s, z(s)). This means that the function z solves the inclusion y' E F(t, y). The lemma is proved. • It is convenient to highlight a particular case of the property of solution spaces from Lemma 2.1. When we take as the set M of the lemma the set consisting of two endpoints of the segment 7r(z), we obtain: Corollary. Let F : U --t ~n be a multi-valued mapping, Z = D(F). Then (p) the set Z contains all functions from Cs(U) defined on one point sets and if z E Cs(U), 7r(z) = la, b], a < b, and ZII E Z for every segment I ~ (a, b), then z E Z.

Notice that the first part of (p) follows immediately from definitions. In order to prove the fulfilment of the second part of (p) with the help of Lemma 2.1, we need to show the generalized absolute continuity of the function z. By our assumptions and by the definition of a solution the function ZII is generalized absolutely continuous for every segment I ~ (a, b). This implies immediately what was required. •

3. The inequality lIy'(t)11 ~ cp(t) Let -00 ~ ao < bo ~ 00, U = (ao, bo) x ~n. Let a non-negative function cp be defined and be locally Lebesgue integrable on the interval (ao, bo). The inequality from the title of the section may be understood as the differential inclusion y' E G(t) with G(t) = 0,(0, cp(t)). By Lemma 4.7.6 all solutions of this inclusion are absolutely continuous. Let q, be an arbitrary primitive of the function cp on the interval (ao, bo). For Z ~ Cs(U) and M ~ U put ZM = {z: z E Z, Gr(z) ~ M} (Thus D(F, M) = (D(F))M). Lemma 3.1. Let Z = D(G). Then (s) for every compactum K ~ U the set ZK is equicontinuous.

Proof. The projection Kl ~ ~ of the compactum K in the first factor of the product ~ x ~n is compact (see Lemma 2.1.1 and Theorem 1.7.8). The points a = inf Kl > -00 and b = sup Kl < 00 belong to Kl (see Theorem 1.6.1) and [a, b] ~ (ao, bo). Take an arbitrary number c: > O. According to the condition in the definition of the absolute continuity (see §4.5) find a number 8 > 0 such that for every family 'Y = {(ai,l3i): i = 1, ... , k} satisfying condition (*,8, [a, b]) of §4.5

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If now z E D(G,K), a,f3 E 7r(z) and la - f31 < 8 (hence the family {( a, f3)} satisfies the condition (*,8, [a, b])) then {3

{3

J

Ilz(f3) - z(a)1I =

J

~

zl(t)dt

cp(t)dt =
~c

'"

'"

(see Lemma 4.7.6). The lemma is proved. • Lemma 3.2. For every segment I ~ (ao, bo) the set D(G) n 7r- 1 (I) is closed in the space C(I, IRn). Proof. Let a sequence {Zj : j = 1,2, ... } ~ D( G) n 7r- 1 (I) converge uniformly to a function z (E C (I, IR n ) ). By Lemma 5.8.2 the function z is absolutely continuous. If the segment I consists of one point, then our assertion is trivial. In the opposite case for every point s of a subset M ~ I of full measure the derivatives Zl (s) and
Ilz(s + 8) - z(s)1I =

~im

)->00

Ilzj(s + 8) - zj(s)11 = )->00 ~im

8+6

J

z;(t)dt

8

~

8+6

J

cp(t)dt = I
8

Therefore

Ilz(S+8l-z(s)1I ~ 1
When we pass to the limit as 8

Ilz'(S)11

~

---t

0, we obtain

I
•

So z E D(G). The lemma is proved. Lemma 3.3. Let Z = D( G). Then

(n) all functions of Z are generalized absolutely continuous and if functions Zj E Z, j = 1,2, ... , are defined on a segment I of the real line and the sequence {Zj: j = 1,2, ... } converges uniformly to a function z E Z, then for almost all tEl the derivatives Zl(t), zj(t), j = 1,2, ... , exist and (3.1)

Z' ( t) E

cc ( { zj (t) : j = 1, 2, ... }).

Proof. The proofreduces to referring to Assertion 5.9.1. Condition (9.2) is needed for this referring. Its fulfilment follows from the estimate

J

zj(t)dt

M

~

J

cp(t)dt

M

Basic properties of solution spaces.

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•

(see also condition (9.5) ). 4. Some properties of solution spaces

Properties established in lemmas of the previous section are helpful in the investigation of solution spaces. They hold place in situations that are very remote from the hypotheses of the lemmas. So it is convenient to make these properties a subject of an independent investigation without the assumption of any direct relation with restrictions on right hand sides of equations and inclusions. In this section we make some simplest remarks about this topic. First of all we enlarge the list of conditions under consideration and introduce the condition:

(c) for every compact urn K

~

U the set ZK is compact (with respect to the topology of the space C s (U) ). We will impose it on spaces Z E R(U), where U is a locally compact subset of the product IR x X of the real line IR and of a metric space X. We will also consider the condition of closedness in C s (U) of spaces under consideration. Lemma 4.1. A set Z ~ ~ Cs(U) is closed in the space if and only if for every compactum K ~ U the set Z K is closed in .

Proof. Necessity of the stated condition follows from the closedness of the set Cs(K) (see the definitions of the Vietoris topology and of the space Cs(M» and from the representation ZK = Z n Cs(K). Sufficiency. Let Z E [Z]. Lemma 2.4.2 implies the existence of a compactum K ~ U such that Gr( z) ~ (K). The set V = C s( (K)) is a neighborhood of the point z in the space Cs(U). The set ZK is closed. If it does not contain z then the set W = (V n without elements of Z. But the existence of such a neighborhood contradicts the condition z E [Z] <1>. Thus z E Z K ~ Z. The lemma is proved. • Lemma 4.2. Let ~ Cs(U), Z E Rp(U). Let for every compactum K ~ U and for every segment I ~ IR the set ZK n 7r- 1 (1) n be closed in (== in C(I, X) n , see Theorem 3.5.3). Then the set Z n is closed in . Proof. By Lemma 4.1 it is sufficient to show the closedness of the set ZK n in the space for every compactum K ~ U. By virtue of the IIletrizability of the space Cs(U) it is sufficient to show that the limit z E of every (convergent in belongs to ZK (see Theorem 1.5.3). Let 7r(z) = [a, b]. If a = b, then Z E Z by the first part of condition (p). Let a < b. Take arbitrarye E (0, b;a).

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Since the mapping 7r is continuous, 7r(Zj) 2 [a + c, b - c], beginning with some j. By Theorem 3.5.4 (4.1) E ZKn7r- 1 ([a+c,b-cD. The closedness of the set ZK n 7r-l([a + c, b - cD n and (4.1) imply that ZI[a+c,b-c] E ZK. By conditions (1) and (p) of §2 Z E ZK. The lemma is proved. • Theorem 3.7.1 and Lemma 4.2 (with = Cs(U)) imply: Theorem 4.1. Let Z E Rps(U). Let for every compactum K ~ U and every segment I ~ IR the set ZK n 7r- 1 (I) be closed. Then the space Z We have:

Zjl[a+c,b-c]

satisfies condition (c).

•

When we compare Theorem 4.1 with Lemmas 3.1 and 3.2, we obtain: Corollary. Under the assumptions of §3 D(G) E Rc(U). • Let us introduce some new notation. For a set M ~ IRx X and points t E R,yEXputMt={x: xEX, (t,x)EM}andMY={x: xEIR, (x,y)E M}. For a multi-valued mapping F : M ---t B (B is an arbitrary set) denote: Ft(x) = F(t,x) and FY(x) = F(x,y) (so we define the mappings Ft : M t ---t Band FY : MY ---t B). Return now to equations and inclusions. For X = IR n and an open subset U of the product IR x IR n denote by Q*(U) the set of all multi-valued mappings F : U ---t IR n with closed convex values satisfying at every point (t, y) E U the condition

(4.2) F(t,y) 2 n{cc(Ft(Oy)): Oy is a neighborhood of the point y in the space Ut (~ IRn)}. By Theorem 2.5.2 if the set F(t, y) is nonempty, convex and compact, then (4.2) is equipotent to the upper semicontinuity of the mapping F t at y.

Lemma 4.3. A space Z ~ Cs(U) satisfies condition (n) if and only if

(n') all functions of Z are generalized absolutely continuous and if functions Zj E Z, j = 1,2, ... , are defined on a segment I, the sequence {Zj: j = 1,2, ... } converges uniformly to a function Z E Z, then for almost all tEl the derivatives z'(t), zj(t), j = 1,2, ... , exist and (4.3)

Z'(t) En{cc({zj(t): j=k,k+1,k+2, ... }): k=1,2, ... }.

Proof. Condition (4.3) is equipotent to the fulfilment of condition (3.1) for all sequences

{Zj: j

= k, k + 1, k + 2, ... },

k = 1,2, ... ,

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199

at once. This gives what was required. The lemma is proved. • Theorem 4.2. Let a space n 7[-1(1). Let a sequence {Zj : j = 1,2, ... } ~ D(F) n E Rc(U). Then Z E Rc(U). Proof. Let K be a compact subset of the set U. By virtue of the obvious representation ZK = Z n of solutions of this inequality satisfies condition (n). By Corollary of Theorem 4.1 it satisfies condition (c). By Theorem 4.2 the space D(f, Ud is closed in
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200

we say 'initial values' in the plural we have in mind that Yo and values of the unknown function belong to the space ~n and the full coordinatewise record of the vector equation y( to) = Yo consists of n equalities). The Cauchy problem for an inclusion y' E F(t, y) with the initial values y(t o) = Yo (without the mentioning of the segment I) means the problem of the finding of a solution Z E D(F) such that to E (-7r(z)) and z(t o) = Yo. In this section we consider the existence of a solution (== the solvability) of the Cauchy problem in the particular case when the right hand side of the inclusion is single valued (Le., we deal with an equation y' = f(t, y) ) and satisfies the additional conditions:

a) for every t E ~ the function ft is continuous and for every y E ~n the function fY is measurable; b) -00 < ao < bo < 00, functions
for every (t, Yl), (t, Y2) E U. By virtue of the openness of the set U in ~ x ~n for every point (to,yo) E U there exists c > 0 such that (to - 2c, to + 2c) X G2 ,yo ~ U. Hence [to - c, to + cl x [G,yol ~ U. The function fl[to-"to+']x[O,yo] satisfies condition (4.13.1). By Theorem 4.13.3 it satisfies (4.13.2) too. By Corollary of Theorem 4.13.1 if z E Cs(U) and Gr(z) ~ [to - c, to + cl x [G,yol, then the function 7jJ(t) = f(t, z(t)) is measurable. By virtue of the compactness of the set Gr(z) and the continuity of the mapping a(t) = (t, z(t)) (see Theorem 2.1.1) the domain of every function z E Cs(U) may be covered by a finite number of segments II"'" h in a manner that for every j = 1, ... ,k the graph of the function ZII; lies in a set of the mentioned type. Our remark and Theorem 4.2.3 imply the measurability of the function 7jJ(t) = f(t, z(t)) for every z E Cs(U). Lemma 4.7.6 implies the Lebesgue integrability of the function 7jJ. If a function z solves the Cauchy problem y' = f(t, V),

(5.1)

y(t o) = Yo

on a segment 13 to, then for every point tEl (5.2)

z(t) =

(Z(t o) +

1

z'(s)ds

~)

t

Yo

+

J

f(s, z(s))ds.

to

Vice versa, by virtue of the above remark the function in the last integral of (5.2) is Lebesgue integrable. Hence every function z satisfying (5.2) is

201

Basic properties of solution spaces.

absolutely continuous. The derivation of the both sides in (5.2) implies that the function z solves the equation y' = f(t,y). The substitution t = to implies that z(t o) = Yo, i.e., every function z satisfying (5.2) solves the Cauchy problem (5.1). Theorem 5.1. Let a function f : U ~ lRn satisfy conditions a) and b). Then for every (to, Yo) E U there exists c > 0 such that for every segment 1 ~ (to - c, to + c), 1 3 to, there exists a unique solution of the problem (5.1) on I. Proof. For i = 0,1 and t E lao, bol put t


J

=

'Pi(s)ds.

to

Since U is open in lR x lRn , (to - 20, to + 20) X 026Yo The functions

~

U for some 0 > o.


space C(I, [06YO]) is complete. By virtue of the continuity of the mapping Q : C(I, [06YO]) ~ ]Rn, Q(z) = z(t o), the set X = Q-1(yO) is closed in C(1, [06YO]). Hence it is complete too. Associate to an arbitrary function z E Cs(U), 7I"(z) = 1, the function Az, t

Az(t) = Yo

+

J

f(s, z(s))ds.

to

By Lemma 4.7.6 t

IIAz(t) -

Yoll

=

J

~

f(s, z(s))ds

to

t

J

'Po(s)ds < 0

to

for tEl. So Az E X. We have the mapping A : Cs(U) n 71"-1(1) ~ X. By above remarks, a function z E Cs(U), 7r(z) = I, solves the problem (5.1) if and only if Az = z. Thus solutions of the Cauchy problem (5.1) on the segment I are fixed points of the mapping Alx : X ~ X. By Lemma 4.7.6 t

IIAz2(t) - Az1(t)1I

~

=

t

J

J

to

to

f(S,Z2(S))ds -

f(S,Z1(S))ds

t

J

IIf(s,Z2(S)) - !(s,z1(s))lIds

to

~

t

J

'P1(s)lIz2(S) - z1(s)lIds

to

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1

~ IIz2 - zlll·lcpl(t) - CPl(to)1 ~ 2"IIZ2 -

zd·

for Zl, Z2 E X and tEl. Since this is true for every tEl,

1 IIAz2 - AZll1 ~ 2"llz2 - zlll· To complete the proof it remains to refer to Theorem 1.7.4. The theorem is proved. • We will use the words 'Cauchy problem' not only in connection with differential equations or inclusions, but also in connection with an arbitrary space Z E R(U). Here U lies in the product ~ x X of the real line ~ and a metric space X. The Cauchy problem for a space Z with initial values y(t o) = Yo (on the segment I :1 to) means the problem of the finding of a function Z E Z, (7r(z)) :1 to (respectively, 7r(z) = I), z(t o) = Yo. When here the solution space of a corresponding differential equation or inclusion appears as the space Z, the new notion coincides with the notion of the above Cauchy problem. Lemma 5.1. Let Z E R(U). Then the following conditions are equipotent: a) for every point (to, Yo) of the set U there exists a number c > 0 such that for every segment I ~ (to - c, to + c), I :1 to, there exists no more than one solution of the Cauchy problem Z E Z, z(t o) = Yo on the segment I; b) for every point (to, Yo) of the set U there are its neighborhood V ~ U and a number c > 0 such that for every segment I ~ (to - c, to + c), 1:1 to, there exists no more than one solution of the Cauchy problem Z E Zv, z(t o) = Yo on the segment I; c) if the domains of functions Zl, Z2 E Z coincide and Zl(t) = Z2(t) at a point t of their common domain, then Zl = Z2; d) if the domains of functions Zl, Z2 E Z intersect and Zl(t) = Z2(t) at a point t E 7r(zd n 7r(Z2), then ZIi1l"(ztln1l"(Z2) = z2111"(ztln1l"(z2)' Proof. a=> b. It is obvious: we put V = U. b=>c. Show that the set M = {t: t E 7r(Zl), Zl(t) = Z2(t)} is open in the segment 7r(zd. Take an arbitrary to E M. Find a neighborhood V ~ U of the point (to, Zl (to)) and a number c > 0 according to condition b. By virtue ofthe continuity of the functions a(t) = (t, Zl (t)) and ,B(t) = (t, Z2(t)) (see Theorem 2.1.1) there exists a number 8 E (0, c) such that (t, Zl (t)) E V and (t, Z2(t)) E V for every point t E (to - 28, to + 28) n7r(zd. The functions Zl and Z2 coincide on the segment [to - 8, to +8] n7r(Zl)' Hence they coincide on the open subset (to - 8, to + 8) n 7r(Zl) of the segment 7r(zd. In view of the arbitrariness of the point to E M this proves the openness of the set M in 7r(zd. By Lemma 1.7.4 the set M is closed in 7r(Zl)' The set M is nonempty. The connectedness of the segment (see Example 2.6.1) implies the equality M = 7r(zd. .

Basic properties of solution spaces.

203

c {:}d. It is obvious. c =>a. It is obvious. The lemma is proved. • For the convenience in the further account we will word the just established and discussed properties in the following way:

(e) for every point (t, y) of the set U the Cauchy problem Z E Z, z(t) = y has a solution (Le., there exists z E Z, for which t E (-rr(z)) and z(t) = y); (u) if the domains of the functions Zl, Z2 E Z coincide and at a point t of their common domain Zl(t) = Z2(t) then Zl = Z2 (Le., the functions coincide on the entire domain of definition). The following assertion is obvious: Lemma 5.2. Let Z ~ Cs(U). Assume that for every point x of the set U there exists its neighborhood Ox ~ U such that the space ZOx satisfies condition (e) (on the set Ox). Then the space Z satisfies condition (e) (on the entire set U). • Pay attention to two particularities offacts proved in Lemma 5.1. First, the condition of the uniqueness, which occurs in Theorem 5.1 and which is mentioned in the lemma as condition a), turns out to be equipotent to the simpler in the statement conditions c), (i.e., (u)), and d). Secondly, the fulfilment of the condition of uniqueness (in every version) locally (see b) ) implies it is fulfiled on the entire set U. Thus, in particular, as a direct consequence of Lemma 5.1 we obtain: Lemma 5.3. Let Z E R(U). Assume that for every point x of set U there exists its neighborhood Ox ~ U such that ZOx E Ru (Ox). Then • Z E Ru(U). In order to be rid of necessity of using the word 'set' too often in connection with the sets R.(U), where * denotes an arbitrary set of conditions under consideration (properties of solution spaces), sometimes we will also call them classes of solution spaces. Their elements need not necessarily be solution spaces of ordinary differential equations or inclusions but they possess many properties of real solution spaces. In considerations below the classes Rceu(U) ~ Rce(U) will be most important. We can expand to them many assertions describing properties of solution spaces of ordinary differential equations. The use of other conditions in our list will often be related to the proof of the belonging of a space to these classes. In particular, we mean the verification the fulfilment of condition (c). Example 5.1. Let a function f : U --+ IR n satisfy the hypotheses of Theorem 5.1. In this theorem we have shown that the space DU) satisfies condition (e). Theorem 5.1 and Lemma 5.1 imply that it satisfies condition (u) too. When we apply Lemma 3.4 and Theorem 4.2 we obtain the closedness of the space DU) in the solution space of the inequality

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204

Ily'(t)II :::; CPo(t). The Corollary of Theorem 4.1, Theorem 4.2, and Lemma 4.4 imply the fulfilment of condition (c). Thus D(J) E Rceu(U). In many situations the following assertion is helpful. Lemma 5.4. Let a single valued function h : V --t ~n be defined and be continuous on an open convex subset V of the space ~I and have continuous partial derivatives (:~ are vector functions with values in the space ~n). Let the derivatives be bounded in norm by a number M. Then for every two points p = (PI, ... ,PI) and q = (q1, ... , ql) of the set V we have Ilh(p) - h(q)11 :::; Milip - qll (i.e., the function h satisfies the Lipschitz condition with the constant MI). Proof. Let

cp(s)

=

d ds (h(q

+ s(p -

q»)

(

I 8h =~ 8Xi (q + s(p -

) q»(Pi - qi) .

We have

h(p) - h(q)

=

J1

cp(s)ds

o

I (1J::i (q + s(p - q»ds) (Pi - qi)·

=~ 0-1

0

Since II :~ II :::; M, Ilh(p) - h(q)11 :::; Milip - qll. The lemma is proved. • Example 5.2. Let a single valued continuous function f : U --t ~n have continuous partial derivatives in y. Show that D(J) E Rceu(U). Take an arbitrary point (to, Yo) E U. Find c > 0 such that [to -c, to +c] x [Goyo] ~ U. Let M = sup{llf(x)II, II -tt-(x) II, ... , II /t(x)11 : x E [to - c, to + c] x [GoYol} « 00, see Remark 2.3.4), V = (to - c, to + c) X Geyo. The function flv satisfies condition a) and the first part of condition b) (with CPo == M). By Lemma 5.4 it satisfies the second part of b) (with CP1 == Mn) too. Theorem 5.1 may be applied to the equation y' = f(t, y) in the region V. Therefore D(J) E Reu(U) (see also Lemmas 5.1,5.2 and 5.3). By reasoning of Example 4.1 D(J) E Rc(U). Thus D(J) E Rceu(U). It is not difficult to mention concrete equations satisfying the restriction of Examples 5.1 and 5.2. These examples give a sufficient number of situations where our near considerations are valuable.

6. Maximal extensions of solutions Let U be an open subset of the product ~ x ~n. For Z E R(U) denote by Z+ (respectively, by Z-) the set of all continuous functions z : [a, b) --t IRn (respectively, z : (a, b] --t IRn , 00 and -00 may be open endpoints) satisfying the conditions: zlI E Z for every segment I ~ 7r(z) and there is no element of the space Z extending the function z.

Basic properties of solution spaces.

205

Denote by Z-+ the set of all continuous function Z : (a, b) ---t ]Rn such that for some (then by virtue of the condition Z E R{U) for every) t E (a, b) we have zl(a,t] E Z- and Zl[t,b) E Z+. Lemma 6.1. Let Z E Rc(U). Let a function z : [a, b) ---t ]Rn belong to Z+. Then the graph of the function z goes out from every compact subset of the set U as t ---t b, i.e., for every compactum K ~ U there exists a number to E [a, b) such that (t, z(t» E U \ K for t E (to, b). Proof. I. Notice that there is no compact subset A of the set U containing Gr(z). In the opposite case the equality b = 00 is impossible by virtue of the compactness of the set A. Next, by virtue of the compactness of the set ZA the sequence {zl[a,b-2- k (b_a)]: k = 1,2, ... } has a subsequence converging to a function Zo E Z A. The function Zo extends the function z. This contradicts the condition z E Z+. II. Let us go on to the main part of the proof. Let K be an arbitrary compact subset of the set U. Assume that the compactum K does not satisfy the mentioned condition. By Lemma 2.4.2 there exists a compact subset A of the set U containing in its interior the compactum K. By I we have a possibility to choose points S1 < t1 < S2 < t2 < S3 < t3 < ... of the half interval [a, b) in a manner that (Si' Z(Si» E K, (ti' z(t i E vA and Gr{zi) ~ A, where Zi = ZI[Si,ti)" By virtue of the compactness of the set ZA ;> {Zi: i = 1,2, ... } the sequence {Zi: i = 1,2, ... } has a subsequence converging to a function Z· E Z A. By virtue of the continuity of the mapping 7r, 7r(z*) consists of one point limi-+oo Si = limi-+oo t i . On the other hand its graph must meet both the sets K and vA. The sets are disjoint. So our assumption is false. The lemma is proved. • Evidently analogous assertions are true for Z- and Z-+ too. Lemma 6.2. Let Z E Rce(U) and Zo E Z. Then there exists a function Z E Z-+ extending Zo. Proof. Let )..(t,y) = {7r{z): Z E Z, SUP7r(z) = t, z{t) = y} for (t,y) E U. The set )..(t,y) consists of segments with the right endpoints t. The union M = U)..(t, y) is connected (see Example 2.6.1 and Theorem 2.6.1). By Theorem 2.6.3 the set M is either a segment, either a half interval, or an interval. The set M contains the point t and does not meet the interval (t,oo). Thus M is either a segment or a half interval with the right endpoint t E M. The space Z satisfies condition (e). This implies immediately the openness of the set M in (-00, t]. Thus the set M is a half interval (a{t, y), t]. Construct now successive extensions Zi, i = 1,2, ... , of the function Zo in the left. Assume that we have a function Zi, i = 0,1,2, .... Let [ai, bo] = 7r{Zi)' The definition of a{t, y) implies the existence of a function z;+1' 7r{z;+1) = [ai+1,ai], taking the value zi{ai) at ai; moreover, the left endpoint ai+1 of 7r{z;+1) belongs to (a{ai' zi{ai», a(ai' zi(ai) + 2- i ) if

»

206

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a(ai' Zi(ai)) > -00, and satisfies the inequality ai+! < -i - 1 if we have a(ai' zi(ai)) = -00. Define the function for t E [ai+!, ail, for t E [ai, bo], 7r(Zi+!) = [ai+!, bolo Evidently under this construction

and if a = limi->oo a(ai' zi(ai)), then a = limi->oo ai' The functions Zi, i = 1,2, ... , sequentially extend each other. We have the possibility of defining a function z* on the half interval (a, bol = U~l[ai,bol by putting z*(t) = Zi(t) for t E [ai,bol. Decuase of our construction the restriction of the function z* to every segment I ~ (a, bol belongs to Z (because I ~ [ai, bol for some i = 1,2, ... ). Assume that there exists a function Z E Z extending the function z*. Since Z E Re(U), we can assume in addition that a E (7r(z)). So the points a(ai' zi(ai)) and ai lie in (7r(z)) beginning with some i:

This contradicts the definition of the number a(ai' zi(ai))' So z* E Z-. Likewise we extend the function Zo in the right to z** E Z+. The function: Z* (t) for t E 7r(z*), z(t) = { z**(t) for t E 7r(z**), is defined on the interval7r(z*)U7r(z**) and satisfies all imposed conditions. The lemma is proved. • Lemma 6.3. Let Z E Rc(U). Then for every compactum K ~ U there exists a number 8 > 0 such that for every function Z E ZK and every extension Zo E Z-+ of the function Z we have Oo7r(z) ~ 7r(zo) (i.e., if 7r(z) = [a, bl, then (a - 8, b + 8[~ 7r(zo)). Proof. By Lemma 2.4.2 there exists a compactum A ~ U containing the compactum K in its interior. Assume that the stated condition is not fulfilled. Lemma 6.1 implies that for every k = 1,2, ... there exists a function Zk E Z A with the graph meeting both the sets K and 8A and with l7r (Zk) < 2- k • By virtue of the compactness of the set Z A the sequence {Zk: k = 1,2, ... } contains a subsequence converging to a function z* E ZA' The continuity of the mapping 17r implies that 17r(z*) = O. Hence 7r(z*) consists of one point only. On the other hand, the graph of the function z* must meet both the sets K and 8A (by virtue of our definition

Basic properties of solution spaces.

207

of the topology of Cs(U) the sets {z: z E Cs(U), Gr(z) n K =F 0} and {z: z E Cs(U), Gr(z) n 8A =F 0} are closed). Our assumption has led to a contradiction. The lemma is proved. • 7. Continuity of the dependence of solutions on initial values As before let U denote an open subset of the product lR x lRn. Theorem 7.1. Let Z E Rc(U). Let K be a compact subset of the set U. Let G(x) = {z: z E ZK, Gr(z) 3 x} for x E K. Then the (multi-valued) mapping G : K --t ZK is upper semicontinuous. Proof. The product K x ZK is compact (Theorem 2.1.3). Theorem 3.3.2 implies the closedness of the set A = {(x,z): x E K, z E ZK, Gr(z) 3 x} in the product K x ZK (see also Theorem 1.5.3). Theorem 1.6.1 implies its compactness. Apply Corollary of Theorem 1.7.8 to the restriction to the compactum A of the projection of the product K x ZK into the first factor K. It implies that the multi-valued mapping g associating to a point x E K the set ({x} x Z K) n A is upper semicontinuous. The mapping G is the composition of the mapping g and of the projection of the product K x ZK onto the second factor Z K. This implies the upper semicontinuity of the mapping G. The theorem is proved. • Theorem 7.2. Let hypotheses of Theorem 7.1 hold. Let I be a segment of the real line. Let G 1 (x) = G(x) n 7["-1(1) for x E K. Then the (multivalued) mapping G 1 : K --t ZK is upper semicontinuous. Proof. The proof reduces to referring to Theorems 7.1 and 2.5.3. (We apply the last one to the mapping G and the constant mapping F == Z K n

7["-1(1)).

•

Corollary. Let with the notation of Theorem 7.2 for every point x of a set X ~ K the set G 1 (x) consist strictly of one function. Then the (single • valued) mapping G1 1x is continuous. In this corollary we obtain the continuity of the dependence of solutions of the Cauchy problem on initial values in the sense which is close to the most common understanding of these words. The closeness turns into full coincidence when in the definition of the mapping G 1 we replace the compactum K to the entire set U, for instance, if it is known in advance that the graphs of all solutions of the Cauchy problem with given and near initial values lie in a compact urn K ~ U j and so the mention of the compactum K in the definition of G 1 does not carry any real information. Theorem 7.3. Let Z E Rce(U) and (to, Yo) E U. Then there are numbers J.L > 1/ > 0 such that the set K = [to - 1/, to + 1/J x [OJLYoJ lies in U and for every function z E Z-+, the graph of which meets the set A = [to - 1/, to + 1/J x [OIlYO], the conditions [to - 1/, to + 1/J ~ 7["{z) and z([to - 1/, to + 1/]) ~ OJLYo hold.

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CHAPTER 6

Proof. Take a number 1£ > 0 such that the compactum

P = [to - 1£, to

+ 1£]

x [0I'Yo]

lies in U. Find a number 8> 0 according to Lemma 6.3 for the compactum P. By Theorem 3.6.1 the set Zp is equicontinuous. So there exists a number v E (O,minH, ~}) such that IIZ(8) - z(t)11 < ~ for every function z E Zp and every points 8, t E 7r(z) satisfying the condition 18 - tl < 2v. Show that the numbers 1£ and v satisfy the imposed conditions. Let the graph of a function Z E Z-+ meet the set A and for some t E [to - v, to + v] n 7r(z) the point z(t) belong to the set [O"Yo]. By our choice of the numbers 8 and v we have [to - v, to + v] ~ (t - 8, t + 8) ~ 7r(z). Let J(8) denote the segment with the endpoints 8 and t. By the choice of

v, if 8 E [to - v, to

+ v]

and Gr (zIJ(s))

Ilyo - z(8')11 ~ lIyo - z(t)11

~ P, then

+ Ilz(t) - z(8')11 < ~ + ~ = 1£

for every 8' E J(8). Therefore Z(J(8)) ~ 0I'Yo. Now the condition Z E Re(U) implies the openness of the set

M = {8: 8 E [to - v, to

+ v],

Gr(zIJ(s)) ~ K}

in the segment [to - v, to + v]. By virtue of the continuity of the function z the set M is closed in the segment [to - v, to + v]. The connectedness of the segment implies that M = [to - v, to + v]. Therefore z([to - v, to + v]) ~ 0I'Yo. The theorem is proved. • Let us return to the discussion of Theorem 7.2 and its Corollary. In Theorem 7.3 for an arbitrary point (to, Yo) of the set U we find its neighborhood (A), a compactum K and a segment I = [to - v, to + v] such that the graph of every solution of the Cauchy problem on the segment I with initial values from (A) lies in K. Therefore the mentioning of the compactum K in the definition of G 1 may be omitted. Thus in this case our assertions establish, respectively, the upper semicontinuity and the continuity of the dependence on initial values of solutions of the Cauchy problem on the segment I with initial values from (A). If in addition Z E Ru(U) then we can take the set A as the set X in Corollary of Theorem 7.2. This relates to, in particular, the solution spaces of equations in Examples 5.1 and 5.2.

CHAPTER 7

CONVERGENT SEQUENCES OF SOLUTION SPACES

In this chapter we discuss the continuity of the dependence of solutions of an equation on parameters of the right hand side. The notion mentioned in the title of the chapter is a convenient tool of investigation. Further we will see that we can apply it not only in the discussion of the continuity but in the investigation of other properties of solution spaces, for instance, in the proof of the existence theorems. In particular, remarks of §2 below, which are not used in practice in the classical theory of Ordinary Differential Equations, are strong tools in the investigation of properties of solutions. We can prove the convergence of sequences of solution spaces in framework of the classical theory with the help of the corresponding versions of theorems on the continuous dependence of solutions on parameters (see, for instance, Theorem 1.2.4 in [Hal). In this chapter we will show how we can prove the inessentiality of singularities of the right hand side and keep intact the continuity of the dependence of solutions on the parameters. We introduce here the topological space Rc(U). It is closely related to the concept of convergent sequences of solution spaces. Both the notion mentioned in the title of the chapter and the topological space Rc(U) are the main topological notions of our theory. 1. Continuity of the dependence of solutions of equations on parameters of right hand sides

In idea, the property which we will speak about, looks as follows. We have a differential equation y' = f(t, y, J-L). The right hand side of the equation depends on a parameter J-L. We also have a point (to, Yo) E U and a segment I 3 to. For every value of the parameter J-L let the Cauchy problem y' = f(t, y, J-L), y(to) = Yo on the segment I have an unique solution z/-" Then the mapping associating the function z,.. to the parameter J-L must be Continuous. However, here we immediately meet the question about the existence and the uniqueness of solutions of the mentioned Cauchy problems. First, such a solution need not exist even under the hypotheses of Theorem 6.5.1

CHAPTER 7

210

if the segment I is too large. Secondly, if a solution exists, in contrast to the situation of Theorem 6.5.1 it need not be unique. Therefore the in general case we can speak about any analog of the continuity of the corresponding multi-valued mapping. The consideration of simplest examples shows that the upper semicontinuity suits here. The classical theory suppose that the right hand side of the equation y' = f(t, y, /L) is continuous in the totality of arguments (Le., with respect to the Tychonoff topology of the product U x M, where U is a open subset of the space ~ x ~n, M is the set of values of the parameter /L). When we assume that the set of parameters M is a metric space the condition of the continuity of the function f(t, y, /L) in the totality of arguments turns into the condition: (1.1) for every compactum K ~ U and for every sequence of parameters /Lj ~ /Lo the sequence of functions CPj(t,y) = f(t,y,/Lj), (t,y) E K, converges uniformly to the function CPo (defined by the same formula). (By virtue of Theorems 3.2.1 and 3.2.4, Coroll<;try of Lemma 1.7.1 and Corollary of Theorem 3.5.4 these conditions are equipotent.) When we discuss here technical aspects of the continuity of the dependence of solutions on parameters in connection with (1.1), we need not consider the entire set M at once. We can consider a particular sequence of parameters /Lj ~ /Lo and in statement of results we need not mention its members explicitly, Le., we can consider a sequence of functions CPj converging to a function CPo in the sense of a corresponding version of condition (1.1), namely with respect to the compact open topology on the space C(U, ~n) (see Theorem 3.2.3). Further applications suggest that we should not restrict our investigation by the above situation. We will obtain a little stronger result, although with a complication in the proof. This complication is related not to the complication of constructions in this section. It is related to the use of the larger material amount in the previous chapters. See also [Ar]. Let F o, F : U ~ ~n be multi-valued mappings, K be a compact subset of the set U, z E Z ~ Cs(U). Put

oz(Fo, F) = inf

{rl1r(cp) cp(t)dt:

7r(cp) = 7r(z) and F(t, z(t))

cP is a positive measurable function,

~ Ocp(t)Fo(t, z(t)) for almost all t E 7r(cp) }

and oz,K(Fo,F) = sup{oAFo,F): Z E ZK}. We say that a sequence of multi-valued mappings Fi : U ~ IR n , i = 1,2, ... , converges to a mapping Fo : U ~ IRn with respect to a

Convergent sequences of solution spaces.

211

space Z ~ Cs(U) (in the set U) if oz,K(Fo, F i ) --+ 0 for every compact subset K of the set U. We say that a multi-valued mapping F : U --+ jRn is weakly continuous with respect to a space Z ~ Cs(U) if for every segment la, bj, -00 < a < b < 00, of the real line and for every sequence {Zk: k = 1,2, ... } ~ Z, 7r(Zk) = la, b], converging uniformly to a function Z E Z:

if Denjoy integrable functions o.k, k = 1,2, ... are defined on a segment la, bj, o.k(t) E F(t, Zk(t)) for almost all tEla, bj and the sequence of the functions

{A,(t) =

i

a,(s)ds: k=1,2, ... }

converges uniformly to a function A then the function A is generalized absolutely continuous and A'(t) E F(t, z(t)) for almo.st all tEla, bj.

Example 1.1. Let f : U --+ jRn be a single valued continuous function. Then the function f is weakly continuous with respect to the space Cs(U): by virtue of the single valuedness of the function f every function o.k, which occurs in the definition, coincides on a set of full measure with the continuous function rpk (t) = f(t, Zk (t)), and the fulfilment of the required condition follows from the continuity of the integral with respect to the metric the uniform convergence (see, for instance, Theorems 3.3.1 and 3.7.2). If a mapping F : U --+ jRn is weakly continuous with respect to a space Z and Zl ~ Z then the mapping F is weakly continuous with respect to the space Zl' This follows immediately from our definition. Denote by Q/(U) the set of all mappings F E Q*(U) with nonempty compact values satisfying the condition if K ~ jRn is a compact, I is a segment of the real line, and I x K then the mapping Fl 1xK satisfies condition (4.13.2).

~

U

The graph of every function Z E Cs(U) (by virtue of its compactness) may be covered by a finite number of sets of the type mentioned in this condition. Thus Corollary of Theorem 4.13.1 and Theorem 4.12.2 imply: Lemma 1.1. Let F E Q/(U) and Z E Cs(U). Then the mapping cp: 7r(z) --+ jRn, rp(t) = F(t,z(t)), is measurable. • Lemma 1.2. Let F : la, bj --+ jRn be a measurable multi-valued mapping of a segment la, bj, a < b, of the real line into the space jRn. Let values of the mapping F be closed. Let 'l/J : la, bj --+ jRn be a measurable single valued /unction, rp : la, bj --+ jR be a measurable single valued n~m-negative function and G(t) = F(t) n O/('l/J(t), rp(t)) for t E [a, b]. Then the mapping G is measurable.

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CHAPTER 7

Proof. The proof consists in referring to Lemmas 4.12.3 and 4.12.4 . • Lemma 1.3. Let F E Q/(U), Z E C.(U). Let

such that the sequence {Zi: i = 1,2, ... } converges to the function z. Fix positive integrable functions
J


~ O"'i(t)F(t, Zi(t)).

71"( z;)

By Lemma 1.3 there exist measurable functions f3i : 71'(Zi) - JRn satisfying the conditions f3i(t) E F(t, Zi(t)) and IIz;(t) - f3i(t) II ~
Convergent sequences of solution spaces.

213

The right hand side of the inequality

IIBi(t) - (Z(t) - z(a + c:))11 ~

~

IIBi(t) - (Zi(t) - Zi(a

+ c:))11 + II(Zi(t)

- Zi(a + c:)) - (Z(t) - z(a + c:))11

t

J

I"Si(S) - z:(s)lIds

+ IIZi(t)

- Z(t) II

+ IIZi(a + c:) -

z(a + c:))11

tends to zero as i - t 00 (for a + c: ~ t ~ b - c:). This estimate does not depend on t. We have IIBi - (z - z(a + c:))11 - t 0, i.e., Bi - t Z - z(a + c:). By virtue of the weak continuity of the mapping F with respect to the space Z the inclusion (z(t) - z(a + c:))' E F(t, z(t)) is true for almost all t E [a + c: < b - c:l, i.e., z'(t) E F(t, z(t)). Hence ZI[a+e,b-e] E D(F). Now the arbitrariness in the choice of c: implies that Z E D(F). The theorem is proved. • The result obtained does not give a possibility to analyze completely the situation described in the beginning of the section. In order to make a last step introduce the set s(U) of all sequences {Zk: k = 1,2, ... } ~ Ri(U) satisfying the condition: for every compact subset K of the set U and for every infinite set A ~ N every sequence Zk E (Zk)K' k E A, is equicontinuous. We say that a sequence {Zk: k = 1,2, ... } of subspaces of the space C8 (U) converges to a space Z ~ C.(U) in U if for every compact subset K of the set U and for every sequence of functions Zk E (Zk)K' k E A, there are a function Z E Z and a subsequence {Zk: k E B} of the sequence {Zk: k E A} converging to the function z: Zk - t Z as k E B, k - t 00 (notice that when we introduce this definition we prepare the using of Theorem 2.5.1). In what follows we will allow the use of these notions and notation in more general situations, when U is a locally compact subset of the product R X X of the real line lR and of a metric space X. In particular, Remark 1.1 and Lemma 1.4 below concern just this general case. Theorem 3.6.1 implies easily: Remark 1.1. Every convergent sequence {Zk: k = 1,2, ... } ~ Ri(U)

belongs to s(U). On the other hand: Lemma 1.4. Let a

= {Zk: k = 1,2, ... } E s(U) and lim top sup Zk lie in Z ~ C 8 (U). Then the sequence a converges in U to Z. k-+oo

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CHAPTER 7

Proof. Let K be an arbitrary compact subset of the set U. Consider a sequence of functions Zk E (Zk)K' k E A. By virtue of the condition {Zk : k = 1,2, ... } E s(U) and Theorem 3.7.1 we can pass to a subsequence {Zk: k E Ad converging to a function Z E lim topsUPk--+oo Zk. This gives what was required. The lemma is proved. • Lemma 1.4 and Theorem 1.1 imply immediately: Theorem 1.2. Let hypotheses of Theorem 1.1 hold and {Z n D(Fk ) : k = 1,2, ... } E s(U). Then the sequence {Z n D(Fd: k = 1,2, ... } converges in U to the space Z n D(F). • When we compare the last result with Theorem 2.5.1 we obtain the possibility of proving the upper semicontinuity of the mapping which associates to a mapping F : U ~ IR n the set D(F, K)j see §6.2. The restriction that we consider solutions with graphs lying in a compactum K is not very essential. By this we mean that if we restrict the consideration to solutions with small domains of definition then, as rule, we can point a compact subset of the set U which contains the graphs of all solutions of the Cauchy problem with given and near initial values (see Theorem 6.7.3 and Lemma 8.1.1). Let us return to the situation discussed in the beginning of this section. Theorem 1.3. For every compact subset K of the set U the multi-valued mapping q,K: C(U,lRn) ~ Cs(U), q,K(f) = D(f,K), is upper semicontinuous. Proof. By virtue of the metrizability of the spaces C(U, IRn) (Theorem 3.2.2) and Cs(U) we can use Theorem 2.5.1. Consider an arbitrary sequence of functions {h: j = 1, 2, ... } ~ C (U, IRn) converging in the space C (U, IRn) to a function fo. Let Zj E q,K(fj) for j = 1,2, .... By Lemma 2.4.2 there exists a neighborhood V of the compactum K, the closure of which is compact and lies in U. By Theorem 3.2.3 we have the uniform convergence fjl[V] ~ fol[V]. By Theorems 3.5.3 and 1.7.8 (see the proof of Theorem 3.6.1) and by the definition of the topology of the space of partial mappings we obtain the compactness of the set U{Gr(fjl[V]) : j = 0,1,2, ... }. Lemma 2.3.6 implies the existence of a number m ~ 0 such that IIfj(x)II ~ m for every j = 0,1,2, ... and x E [V]. Thus the set U{D(fj, V): j = 0,1,2 ... } lies in the solution space of the inequality Ily'(t)II ~ m. By Lemma 6.3.1 {D(fj, V): j = 0,1,2, ... } E s(V). The uniform convergence fjl[V] ~ fol[V] implies the convergence with respect to the space Cs(V). In view of the remarks of Example 1.1 referring to Theorem 1.2 completes the proof. The theorem is proved. • In Theorem 1.3 we have proved the continuity of the dependence of solutions of equations on the right hand sides (with respect to the compact open topology of the space C (U, IRn)). Although here we do not mention any parameters of the right hand side, we immediately obtain a corresponding

215

Convergent sequences of solution spaces.

wording 'with parameters' when we compare Theorem 1.3 with Theorems 3.2.3 and 3.2.4. Theorem 1.3 may be added by an assertion about the continuity of the dependence of solutions on initial values. To do it we prove first the following auxiliary assertion. Lemma 1.5. Let F : X -+ Cs(U) be a multi-valued mapping of a metric space X into the space C s (U). Let for every compact subset K of the set U the mapping FK : X -+ Cs(U), FK(X) = F(x) n Cs(K), have compact values and be upper semicontinuous. Then the multi-valued mapping

G : X x expc U x expc U -+ Cs(U), G(x, K 1 ,K2) = {z: z E F(x), Gr(z) n Kl -=1= 0, Gr(z)

~

K 2},

is upper semicontinuous. Proof. By virtue of the metrizability of the domain of the mapping G and of the space Cs(U) we can use Theorem 2.5.1. Consider an arbitrary sequence {(Xj, Kf, K4): j = 1,2, ... } of points of the domain of the mapping G converging to a point (x, K 1 , K2). Let Zj E G(xh Kf, K4) for j = 1,2, .... By Lemma 2.1.2 Xj -+ x, Kf -+ Kl and K4 -+ K 2. By Lemma 2.4.2 there exists a neighborhood V of the compactum K 2 , the closure of which is compact and lies in U. By virtue of the convergence K4 -+ K2 we have K4 ~ V, beginning with some j = jo. Hence Gr(zJ ~ V. Apply Theorem 2.5.1 to the mapping F rv ). Since Zj E Frv)(xj), the sequence {Zj : j = 1,2, ... } contains a subsequence converging to a function Z E F(V)(x). Theorem 3.3.2 implies that Gr(z) ~ K 2 . Lemma 2.5.3 implies the nonemptiness of the intersection Gr(z) n K 1 . By Theorem 2.5.1 the lemma is proved. • Theorem 1.3 and Lemma 1.5 imply immediately: Theorem 1.4. The multi-valued mapping W : C(U, IRn) x expc U x expc U

w(f, K 1 , K 2 )

= {z: z E D(f, K 2 ),

Cs(U), Gr(z) n Kl -=1= 0} -+

•

is Upper semicontinuous.

2. Properties of convergent space sequences Lemma 2.1. Let a sequence {Zk: k a space Z E R(U). Let Zk E z;+ for beginning with some k = ko, and the COnverge uniformly to the function z. oj the function Z and a subsequence

= 1,2, ... } ~ Rc(U) k = 1,2, .... Let Z E

converge in U to Z, 7r(z) ~ 7r(Zk),

sequence {zkl1r(z): k = ko, ko + 1, ... } Then there are an extension z* E Z-+ {Zk: k E A} such that I ~ 7r(Zk) for

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CHAPTER 7

every segment I ~ 7r(z*), beginning with some k = kl E A, and the sequence {zkII: k E A, k ~ kd converges uniformly to the function Z*II. Proof. I. Show first that for some 8 > 0 there exists a subsequence {Zk : k E B} of the sequence {Zk: k = 1,2, ... } such that 7r(Zk) ;2 [a - 8, b + 8], where k E B, [a, b] = 7r(z), and the subsequence converges uniformly on [a - 8, b + 8] to an extension Zo E Z of the function z. Fix a compactum K ~ U which contains the set Gr(z) in its interior. By virtue of the convergence Zk I... (z) ---t z, by Theorem 3.5.3 and by the definition of the topology of the space Cs(U) we have Gr(zkl ... (z») ~ (K), beginning with some k = ko. Lemma 6.6.1 implies the existence of points ak < bk of the interval7r(zk) such that [a,b] ~ [ak,b k], Gr(zkl[ak.bkl) ~ K and (ak' zk(ak)), (b k , zk(bk)) E 8K. Because of the convergence of the sequence {Zk: k = 1,2, ... } to Z in U the sequence {Zk = Zkl[ak,bkl: k = ko, ko + 1, ... } has a subsequence {zk: k E A *} converging to a function Zo E Z K. By Theorem 3.5.1 and its Corollary 7r(Zk) ---t 7r(zo) or ak ---t ao and bk ---t bo, where lao, bo] = 7r(Zo), k E A*. By Theorem 3.5.4 (ak' zk(ak)) ---t (ao, zo(ao)) and (bk, zk(b k )) ---t (b o, zo(b o)) (k E A*). Hence (ao, zo(ao)), (b o, zo(b o)) E 8K. The function Zo is the limit of a sequence converging on the segment 7r(z) to the function z. Thus zo extends z. Therefore the graph of Zo on the segment 7r(z) ~ n{[ak' bk] : k E A *} lies in (K). Hence the points ao and bo do not belong to the segment 7r(z). So 7r(z) ~ (ao, bo). Take as 8 an arbitrary positive number which is less than the distance from the points ao and bo to the segment 7r(z). For this choice of 8 the segment [a - 8, b+ 8] lies in the segment [ak' bk], k E A*, beginning with some k = kl E A* (use the continuity of the mapping 7r). The convergence of the sequence {Zk I[a-6,bHl: k E A*, k ~ kd to the function ZOI[a-6,bHl follows now from Theorem 3.5.4. II. Now we repeat with some modifications and complications the proof of Lemma 6.6.2. Choose subsequences:

1m

= {Zk:

k E Am}, m

= 1,2, ... ,

({I, 2, ... }

= Ao

;2

Al ;2 A2 ;2 ... )

and fix numbers 'T/o( = a) ~ 'T/l ~ 'T/2 ~ ... in such a manner that on the segment ['T/m, bJ the sequence 1m converges uniformly to a function z;:: E Z. Let Am be the set of all segments [c,'T/mJ, on which we can choose a subsequence of the sequence 1m converging to an element of the space Z. Theorem 2.6.1 implies the connectedness of the set M = UA m. Because of! the set is open in the half interval (-OO,'T/mJ. By Theorem 2.6.3 the set M is a half interval with the open left endpoint: M = (am,'T/mJ, where am < 'T/m· Take as'T/m+l an arbitrary number from (am,min{am + 2- m ,'T/m}) if am > -00, and from (-00, min{ -m, 'T/m}) if am = -00. By our definition

Convergent sequences of solution spaces.

217

of am there exists a subsequence I'm+! of the sequence I'm converging on the segment [1]m+l,1]mj to an element of the space Z. The sequence I'm converges on the segment [1]m, bj. Thus the sequence I'm+! converges on the segment [1]m+!, bj (to an extension z;';;'+! E Z of the function z;;). Under this construction

For 1] = limm-+oo am we have 1] = limm-+oo 1]m' Define the function z** on the half interval (1], bj = U{[1]m, bj m = 1,2, ... } by putting z**(t) = z;;(t) for t E [1]m, b], m = 1,2, .... Now fix an arbitrary increasing sequence of indices A' = {k m m = 1,2, ... }, where km E Am. If a segment I lies in (1], b], then I ~ [1]m, b] for some m = 1,2, .... On the segment I the subsequence {zkll: k E A', k ~ k m} of the sequence {zkll : k E Am} converges uniformly to the function Z"'II = Z;';;'II' Show that z ... E Z-. Assume the opposite. This holds only if there exists an extension z E Z of the function z**. By Lemma 2.4.2 there exists a compactum K ~ U containing in its interior the graph of the function z. Repeat the construction as in I, but for the sequence {Zk: k E A'} instead of the initial sequence 1'0' The function Zo appearing in this construction extends the function z**. The endpoints ao < bo of 7r(z**) satisfy the condition ao < 'f] < b < boo If 8* is an arbitrary positive number less than 'f] - ao and bo - b, then (ao + 8*, bo - 8*) ;2 ['f], b]. We have lao + 8*, bo - 8*] ~ [ab bk] beginning with some k = kl E A' (to prove the latter fact use the continuity of the mapping 7r as in I). Theorem 3.5.4 implies the convergence of the sequence {zkl[ao+6*,b o-O*j: k E A'} to the function ZOI[ao+6*,bo-O*j. We have 'f] = limm-+oo am E (ao + 8*, bo - 8*). So Q m E (ao + 8*, bo - 8"), beginning with some m = mo. This contradicts the definition of am, m = mo, mo + 1, mo + 2, .... Thus our assumption is false and z** E Z-. III. By analogy with II construct an extension in the right z*** E Z+ of the function z and choose in the sequence {Zk: k E A'} a subsequence {Zk: k E A} converging uniformly on every segment I ~ 7r(z***) to a function z"'*. Define on the interval 7r(z"') U 7r(z"'*) the function: z*

z**(t)

{ (t ) -- z ..... (t)

for t E 7r(z**), for t E 7r(z·"').

The verification of the fulfilment of all imposed conditions is not difficult. The lemma is proved. • The following assertion is convenient for the proof of the existence the; orems (Le., for the verification the fulfilment of condition (e) ) for solution

218

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spaces of equations with various types of right hand sides. In the next chapter we meet examples of such situations. But the reader now has Theorem 6.5.1 (see also Examples 6.5.1 and 6.5.2) and he may use this result for independent exercises. Theorem 2.1. Let a sequence {Zk: k = 1,2, ... } ~ Rce(U) converge in U to a space Z E R(U). Then Z E Re(U). Proof. Take an arbitrary point (t, y) of the set U. Denote by Z the function defined on the one point set {t} and taking the value y at t. The membership Zk E Re(U) implies that Z E Zk for every k = 1,2, .... By virtue of the convergence of the sequence {Zk: k = 1,2, ... } in U to the space Z the latter fact implies the belonging of the function Z to the space Z. By Lemma 6.6.2 for every k = 1,2, ... there exists an extension Zk E Z;;+ of the function z. When we apply Lemma 2.1 we obtain the existence of a function z* E Z-+, the domain of which contains the point t and which takes the value y at t. This implies immediately what was • required. The theorem is proved. Let us now state an assertion closely related to the continuity of the dependence of solutions on (parameters of) the right hand side and initial values. Theorem 2.2. Let a sequence {Zk : k = 1,2, ... } ~ Rce(U) converge in U to the space Z E R(U), Zk E Z;;+, tk E 7r(Zk). Let the sequence {(tk,Zk(t k )): k = 1,2, ... } converge to a point (t,y) E U. Then there are a function z* E Z-+, 7r(z*) 3 t, z(t) = y, and a subsequence {Zk: k E A} such that for every segment I ~ 7r(z*) we have I ~ 7r(zd beginning with some k = ko E A, and the sequence {Zk II: k E A, k ~ k o} converges uniformly to the function z*II. Proof. Fix a compact subset K of the set U containing the point (t, y) in its interior. Then (tk' Zk(t k )) E (K) beginning with some k = k 1 . Lemma 6.6.1 implies that there are numbers ak < bk such that tk E [ak' bk ] ~ 7r(Zk), Gr (zkl[ak,h]) ~ K and (ak, zk(ad), (b k , zk(bd) E oK. Show that t E [ak' bk] beginning with some k = k2 > k 1 . Assume the opposite. The convergence of the sequence {Zk: k = 1,2, ... } to the space Z in U and our additional assumption imply that we can choose a subsequence {zkl[akobk] : k E Ao} converging to a function Zo E ZK. We have t tt [ak' bk] for every k E Ao. Since It - tkl ---+ 0, we can require in addition the fulfilment of one of the conditions It - ak I ---+ or It - bkI ---+ 0, k E Ao· Theorem 3.5.4 implies that (t, zo(t)) = (t, y) E oK. This contradicts the choice of the compactum K. The obtained contradiction gives what was required. The sequence {zk = zkl[ak,bk]: k = k2' k2 + I, ... } contains a convergent subsequence {Zk: k E Ad. Theorem 3.5.4 implies that Zk(t) = zZ(t) ---+ y, k E A l . Now the application of Lemma 2.1 to the sequence {Zk: k E Ad

°

Convergent sequences of solution spaces.

219

and to the function z = zol{t} gives what was required. The theorem is proved. • Theorem 6.7.3 is convenient for the comparison of local and global properties of solution spaces of ordinary differential equations. In this context it is helpful to have the following assertion. Theorem 2.3. Let Z E R(U). Let V be an open subset of the set U, Zv E Rce (V). Let K be a compact subset of the space IRn. Let OK be its neighborhood (in the space IRn), I be a segment of the real line, and I x OK ~ V. Assume that if the graph of a function z E (Zv)-+ meets the set I x K, then I ~ 7r(z) and z(I) ~ OK. Then if the graph of a function Zo E Z-+ meets the set I x K, then I ~ 7r(zo), zolr E Zv and zo(I) ~ OK. Proof. Let Zo E Z-+, tl E In 7r(zo), zo(it) E K. Denote by M the set of all s E I such that the segment J(s) with the endpoints it and s lies in 7r(zo) and z(J(s» ~ OK. Evidently the set M is either a segment either a half interval or an interval. The point tl lies in (M) [" Let a < b be the endpoints of M. Fix numbers

in a manner that a = limk--+oo ak and b = limk--+oo bk . By Lemma 6.6.2 for every k = 1,2, ... there exists an extention Zk E (Zv)-+ of the function ZOhak,bkj' Apply Lemma 2.1 with Zk == Zv for k = 1,2, ... , z = ZOI{td' Lemma 2.1 implies the existence of an extention z* E (Zv)-+ of the function ZOIM' Then I ~ 7r(z*), z*(I) ~ OK and the compactum Gr(z*lr) (~ V) contains Gr (ZOIM)' By virtue of the membership Zv E Rc(V) this implies the closedness of the set M (in the segment I). Since

zo(M) = z*(M)

~

z*(I)

~

OK

the set M is open in the segment I. By virtue of the connectedness of the segment (see Example 2.6.1) we have M = I. This gives what was required. • The theorem is proved.

3. Equicontinuity condition Let U denote (as before) an open subset of the product IR x IRn. We saw

in §1 that the proof of the continuity of the dependence of solutions of an equation on parameters reduces to the proof of the convergence of seqUences of solution spaces. Remarks in §1 do not exhaust our possibilities in this connection. Lemma 1.4 splits the question about the convergence of a sequence a = {Zk: k = 1,2, ... } ~ R(U) to a space Z ~ Cs(U) into two: the question about the membership a E s(U) and the question about the inclusion lim topsUPk--+oo Zk ~ Z. In the next sections we point out new

CHAPTER 7

220

methods of the investigation of these questions. They differ very much from the methods of §1. Here we consider the first question. Our general plan consists in estimating the geometry of the set of singularities: on a part W of the region U the corresponding convergence (or any other property of solution spaces) it is assumed proved (for instance, with the help of versions of methods of the classical theory) and we seek in what case, and for what geometry of the set U \ W, the corresponding conclusion remains true for the entire set U. Let a multi-valued mapping (J : U --t lRn be fixed. For (t, y) E U put r~ = {{y} U H : H is a open subset of the set Ut and for some E > 0 (G~y n (J(t,y)) \ {y} ~ H} (Figure 7.1). For t E lR denote by T((J, t) the topology on the set Ut generated by the directed sets r~, y E Ut (see details in §1.2 and see the notation in II· )1 §6.4). Theorem 3.1. Let H a = {Zk: k = 1,2, ... } ~ Ri(U). For every point Figure 7.1 let (t,y) E U lim topsup{Gr (ZI[Sl,S2]) : Z E Zk, o,£-+O,k---+oo

[SI' S2] ~ (t - E, t

+ E) n 7r(Z),

Z([SI' S2]) n Gey

i= 0}

~ (J(t, y).

Let I be a family of open subsets of U. Let {(Zk)V : k = 1,2, ... } E s(V) for every V E I' For every point let t E lR the set (U \ (U'))t contain no nontrivial connected subsets on which the topology T( (J, t) and the Euclidean topology of lRn coincide (i.e., induce the same topology). Then a E s(U). Proof. Assume the opposite, i.e., that the condition a E s(U) is not fulfilled. Then there exist a compactum K ~ U and a sequence Zi E (Zi)K' i E A, which is not equicontinuous. The latter fact implies the existence of a number c > 0, of a subsequence {Zi: i E AI} and of points Si < ti from 7r(Zi), i E Al such that ti - Si --t 0 and Ilzi(S;) - Zi(t i ) II ;;:: c. The set expK is compact. Therefore the sequence {Gr(zil[Si,ti]) : i E Ad has a subsequence converging to a set F E exp K. Since

Convergent sequences of solution spaces. ti - Si

- t 0, the projection of the set F in consists of one point

t

= i-+oo,iEA lim

ti 2

~

= i-+oo,iEA lim

(in the product

221 ~

x

~n)

S i. 2

For every point y E F we have F ~ e(t,y). Therefore on the set F t the topology T(e, t) and the Euclidean topology of ~n coincide. Theorem 3.3.3 implies the connectedness of the set F. The condition IIZi(Si) - Zi(ti)11 ~ c implies the estimate diamF ~ c. Hence the set F contains more than one point. Therefore F n (U,,) =1= 0 and F n V =1= 0 for some V E 'YTake an arbitrary point y E F n V and put

8=

~min{p((t,y),(~ x ~n) \ V),c}

(see Lemma 2.3.8). Because of the convergence of the sequence {Gr(zi![Si,ti]) : i E A 2 } to the set F and by Theorem 3.4.2 we can choose a subsequence {Gr(zil[Si,t;j): i E A 3 } and points s~ E lSi, til, i E A 3 , in such a manner that (s~, Zi(S~)) E 06(t, Yl) and Zi(S~) - t y. Here Gr(zil[Si,ti]) \ 06(t, y) =1= 0, because in the case where Zi([Si, til) ~ 06Y we have:

which contradicts our choice of the functions Zi, i E A 1 • We can select points t~ E lSi, ti], i E A 3 , in such a manner that on the segment li with the endpoints s~ and t~ the graph of the function Zi lies in the compactum Kl = [06(t, y)] and (t~, Zi(tD) E {)06(t, y). The sequence {Zi!L. : i E A 3 } has no convergent subsequences, because in the opposite case the domain of the limit function would consist of one point t only and its graph would contain the point (t, y) and meet {)06(t, y) simultaneously, which is impossible. Theorem 3.7.1 implies that the sequence {zilli : i E A 3 } is not equicontinUOUs. The graphs of all its members lie in the compactum Kl ~ V. This contradicts the condition

{(Zk)V: k = 1,2, ... } E s(V). Thus our assumption is false and the theorem is proved. • Notice a particular case. The mapping e == ~n satisfies all conditions imposed above on the mapping e. The proved assertion implies: Corollary 1. Let O!

= {Zk:

k

= 1,2, ... } ~ R(U),

CHAPTER 7

222

, be a family of open subsets of the set U, for every element V of which {(Zk)V: k = 1,2, ... } E s(V). Let for every point t E IR the set (U \ (U'))t do not contain connected • nontrivial subsets. Then a E s(U). By Assertion 2.4.1 we obtain as a particular case of Corollary 1: Corollary 2. Let a

= {Zk:

k

= 1,2, ... } ~ R(U),

, be a family of open subset of the set U, for every element V of which {(Zk)V: k = 1,2, ... } E s(V). Let the set (U \ (U,)) be at most countable. Then a E s(U). • So an one point singularity does not mar the fulfilment of the condition a E s(U). In the general case the idea of Theorem 3.1 may also be realized in the following different way. Assertion 3.1. Let X and M be metric spaces, K be a compact subset of the product IR x X, cp : K - t M be a continuous mapping, and {Zi: i = 1,2, ... } ~ Ri(K), a

= {{cp(t,z(t)): z

E Zd: i

= 1,2, ... } E s(1R x cp(K)).

Then for every point (to, xo) E K the set P

z E Zj, j = k, k + 1, ... , 7r(z) ~ (to - c, to Im(z) n Ooxo # 0}]: k = 1,2, ... , c > 0, 8> O}

= n{[U{Im(z):

+ c),

lies in ({to} x X}) n cp-lcp(to, xo). Proof. Assume the opposite. Then there exists a point

By virtue of the definition of the set P this means the existence of functions Zk E Zjk and of points Sk, tk E 7r(zd, jl < h < iJ < .. , such that 7r(Zk) ~ (to - 2- k , to + 2- k ),

Z(Sk)

-t

Yo,

z(tk) - t Yl'

By virtue of the continuity of the mapping cp we have

Convergent sequences of solution spaces.

223

as k ---+ 00. Since cp(to,xo) =1= cp(to,xd, the condition a E s(lRx cp(K» is not fulfilled. The contradiction obtained gives what was required. The assertion is proved. • Assertion 3.2. Let

a = {Zk: k = 1,2, ... } ~ Ri(U),

X be a metric space, cp : U ---+ X be a continuous mapping. For every sequence Zi E Zkn kl < k2 < k3 < ... , let the sequence cp(t, Zi(t)), t E 7r(Zi), i = 1,2, ... , be equicontinuous. Let, be a family of open subsets of the set U, for every element V of which {(Zdv: k = 1,2, ... } E s(V). For every t E IR and x E X let the set (U \ (UY»t n cp-l (x) contain no nontrivial connected subsets. Then a E s(U). Proof. Assume the opposite. There are a compactum K ~ U and a sequence Zi E (Zk.)K, kl < k2 < k3 < ... , which is not equicontinuous. By

virtue of the definition of the equicontinuity we can assume in addition that 11r(zi) ---+ 0 and diamlm(zi) ~ c for some c > O. By Theorem 3.4.3 we can assume in addition the convergence of the sequence {Gr(zi): i = 1,2, ... } with respect to the Hausdorff metric to a set H E exp K. Since 11l'(zi) ---+ 0 Theorem 3.5.1 implies that the projection of the set H in the time axis (Le., in the first factor of the product IR x IR n ;2 H) consists of one point only; see Figure 7.5 below. Denote it by s. By Theorem 3.3.3 the limit H of the sequence of continua Gr(zi)' i == 1,2, ... , is connected. The estimate diamlm(zi) ~ c implies that diamH ~ c. Thus the continuum H contains more than one point. By Assertion 3.1 H ~ cp-l(X) for a point x E X. By virtue of the last hypothesis of our assertion H n V =1= 0 for some V E ,. Now we can repeat the choice of the sequence {Zi: i = 1,2, ... } and of the number c > 0 which is made in the beginning of our reasoning (see the proof of Theorem 3.1) in order to obtain the fulfilment of the following condition: there exists a compactum Kl

~

V, such that Gr(zi)

~

Kl for of all

i == 1,2, ....

This contradicts the condition {(Zk)v: k = 1,2, ... } E s(V). Thus our assumption has led to a contradiction. The assertion is proved. Let us now see how results of this kind work. Example 3.1. Consider the system (3.1)

== -x + y + ).2 x - l e - l,\y/x: 1 y' == -x - y - ).2 x - l e -l.\y/x I.

X' {

•

224

CHAPTER 7

We put the last terms equal zero for x = o. The right hand sides of the equations contain the real parameter A E [0, (0). Let Ai --t 00 and Zi denote the solution space of the system (3.1) for A = Ai. Let a = {Zi: i = 1,2, ... }. Show that a E s(JR x JR2). I. Let V be a region in the space JR x JR2 of variables t, x, y, the closure [V] be compact and y do not meet the plane y = o. On the set [V] the last terms of the equations (3.1) converge uniformly to zero. Therefore there exists a number m > 0, such that J. Ix'(t)1 ~ m and ly'(t)1 ~ m for evx ery solution u = (x, y) E U{(Zi)V: i = 1,2, ... }. This implies that {(Zdv: i = 1,2, ... } E s(V). II. Let V be a region in the space JR x JR2 of variables t, x, y. t =const Let the closure [V] be compact and lie either in the half space Figure 7.2 JR x ((0, (0) x JR), or in the half space JR x (( -00,0) x JR). Consider the mapping cp : V --t JR, cp(t, x, y) = x + y (Figure 7.2). Apply Assertion 3.2 to the set E = {(t, x, y): t, x E JR, y = o}. By I and the estimate ••••• _ _ _ _ _ _ _ _ _ _ _ _ o

(3.2)

I(cp(t, u(t)))~1 ~ m

for some m > 0 we have {(Zi)V: i = 1,2, ... } E s(V). III. Regions mentioned in I and II cover the plane t = const except the point x = y = O. Now the membership a E s(U) follows from Corollary of Theorem 3.1. In results proved we deal not with the right hand side of an equation (inclusion), but we deal directly with a solution space. Therefore when we investigate an particular equation, we need preliminary considerations before we obtain any possibility of refering to these results. In order to obtain a corresponding result dealing with right hand sides of equations (inclusions), introduce the class of multi-valued mappings F : U --t ~n satisfying the conditions: (3.3) values of the mapping Fare nonempty convex one sided cones with the top at the origin of coordinates ({O'} may be the value too). Assume that:

225

Convergent sequences of solution spaces.

(3.4) every multi-valued mapping F, Fi : U condition (3.3);

--t

lRn ,

i = 1,2, ... , satisfies

(3.5) F(x) 2 n{ CC(U{Fi(OX) : i = k, k+1, ... }UF(Ox» : k = 1,2, ... , Ox runs over the set of all neighborhoods of x E U} for every point x E U; (3.6) the primitives of Lebesgue integrable functions

(}:i

i = 1,2, ... , constitute the equicontinuous family, Si(t)

Ilull

:

lR

--t

[0,(0), E lR,

= {u: u

~ (}:i(t)};

(3.7) Gi(t, y) = Fi(t, y) + Si(t) (t, y) E U and i = 1,2, ....

{u

+v

Multivalued mappings of these conditions will play the role of majorants of right hand sides in the investigation of particular equations and inclusions. Its geometry allows us to estimate the structure of sets of singularities. Their role is close to the role of the mapping () in Theorem 3.1. Its unique advantage with respect to the mapping () is its simple and close relation (majorization) with right hand sides of equations (inclusions) under consideration. For a cone V with the top at the origin of coordinates and c > 0 denote by Po the c-neighborhood of the subset p = {u: u E V, Ilull = I} of the unit sphere S and denote by Ve the cone

{O'} U

{u: uE lR ui= 0, n,

11:11

E Pc}

(see Figure 7.3).

/ -.... ,

Figure 7.3

Figure 7.4

Denote by r(F, t) the topology on the set Ut generated by the system

of neighborhoods {y ± (F(t, Y»c n 0 6 0: c,8 > O}, y E Ut (see Figure 7.4).

CHAPTER 7

226

Theorem 3.2. Let conditions (3.3 - 7) hold,

{Zi: i = 1,2, ... }

~

R(U),

Zi ~ D(G i ) for all i = 1,2, .... Let, be a family of open subsets of the set U. Let {(Zi)V: i = 1,2, ... } E s(V) for every V E ,. Let E = U\ (U,). For every t E ~ the set E t let contain no nontrivial continua on which the topology r(F, t) and the Euclidean topology coincide. Then {Zi: i = 1,2, ... } E s(U). ~

Proof. I. Assume the opposite. For a compactum K a sequence Zj

E (Zij)K,

il

U there exists

< i2 < ...

which is not equicontinuous. This means that for some c > 0 for every m = 1,2, ... the set Am of indices kEN satisfying the condition (3.8) there are points IIZk(t) - zk(s)11 ~ c,

S

< t of 7r(zm) such that t -

S

< 2- m and

is infinite. Since the formula

jl

= minAI,

ji

= min(Ai \ {il, ... ,ji-d) for i

= 2,3, ... ,

defines an infinite sequence of indices A = {i1,i2, ... }. For jm E A fix a pair of points Sjm < tjm according to (3.8): t·3Tn - s·J'fn

< 2- m and liz'3Tn (t·3m ) - z·]Tn (s·3Tn )11

2r c.

Thus (3.9)

tj -

Sj --+

0 as j E A and j

--+ 00,

(3.10)

The projection P of the compactum K in the first factor of the product IR x IRn is compact. Therefore the sequence {Sj : j E A} has a subsequence {Sj: j E B} converging to a point S E P. Due to (3.9) S = ~im tj. 1EB.

j-+oo

Convergent sequences of solution spaces.

The space exp K is compact, see Theorem 3.4.3. Therefore the sequence {Gr{zj): j E B} contains a subsequence {Gr{zj): j E Bd converging in exp K to a limit H. Since S

=

.lim

t--+ 00

Sj

=

227

!

/: ,, .

! i

.lim tj

1--+00

the projection of the compactum H into the first factor of the product IR x IRn coincides with the point s. Condition (3.10) implies that diam H ;;:: c. Therefore the set H contains more than one point. The compactum H is connected, see Theorem 3.3.3 and Figure 7.5. II. Show that

:

~

:

:

-s;------~--s;

--72

S3i3

S

Figure 7.5

H ~ U\ (U'Y).

Assume the opposite, i.e., the existence of a point (t, Yo) E H For some 'TJ > 0

(3.11)

H \ 0T/Yo

n (U,).

=1= 0,

(3.12) the set [OT/{t, Yo)] lies in an element V of the family 'Y.

By virtue of the condition (t, Yo) E H and Theorem 3.4.2 we can fix points Pj E [Sj, tj] in order that Zj{pJ -+ Yo as j E BI and j -+ 00. The last convergence implies the infiniteness of the set B2 = {j : j E B I , (Pj, Zj(pj)) E 0T/(t, Yo)}. For j E B2 denote by I j the maximal segment satisfying the conditions Pj E I j ~ [Sj,tj], Gr(zjlIJ ~ [OT/{t, Yo)]. Because of (3.11) the sequence {Gr(ZjIIJ:j E B2} has limit points outsides 0T/Yo, Thus the sequence {ZjII' : 'j E B 2 } is not equicontinuous. J This contradicts conditions (3.12) and {{Zi)V: i = 1,2, ... } E s{V). III. Assume that the continuum H t lies in a plane A ~ IRn. Let A denote the space of vectors of the plane A. {In the connection with the notation H t and with the analogous notation below recall that H ~ {t} x IRn and

228

CHAPTER 7

H t ~ ]Rn). If at all points of the set H t or of a nonempty open subset of it their above neighborhoods in Ut with respect to the topology reF, t) contain Euclidean 8-balls of the plane A, then the induced topologies coincide. For this case the theorem is proved. Analyze now the case when for a point Xl = (t, YI) E H the cone F(t, YI) does not contain a half-space of the space A. The set CC(U{Fi(Oxd: i = k, k + 1, ... } U F(Oxd) in (3.5) is a closed cone. We have two possibilities: A. The zero vector 0 of the space A does not belong to the interior of the set CC(U{Fi(Oxd: i = k, k + 1, ... } U F(Oxd) n A. B. The zero vector 0 of the space A belongs to the interior of the set

CC(U{Fi(Oxd: i = k,k

+ I, ... } UF(Oxd) nA.

Then CC(U{Fi(Oxd: i = k, k + I, ... } U F(Oxd) = A. If B is true for every neighborhood OXI of the point Xl and for every k = 1,2, ... then by virtue of (3.5) F(xd ;2 A, which leads out the framework of the case under consideration. Thus for a neighborhood OXI of the point Xl condition A holds. For some 1-£ > 0 we have [t - 1-£, t + I-£J x [OJ.lyd ~ OXI· Let el be an arbitrary point of the unit sphere of the space A, which does not belong to the cone

By Theorem 2.4.2 there exist a linear functional h : ]Rn ~ ]R and numbers p < q such that heel) ~ q and h(S) ~ (-oo,p). Since S is a cone, we can take p = o. There exists a vector eo E ]Rn, such that h( u) = (eo, u) for all u E ]Rn. For the vector e = -eo/lleoll we have: (e, u) ~ 0 for every vector u E CC(U{Fi([t - 1-£, t i = k, k + 1, ... } U F([t - 1-£, t + I-£J x [OJ.lyd)). (3.13)

+ I-£J

x [OJ.lYIJ):

IV. Return to the notation of I. Let a linear functional 9 : satisfy the condition

(3.13') g(u) ~ 0 for every vector u E CC(U{Fi([t - 1-£, t + 1, ... } U F([t - 1-£, t + I-£J x [OJ.lyd))·

+ I-£J

]Rn ~

lR

x [OJ.lYIJ)

i = k, k

In particular, by (3.13) this holds for the functional g(u) = (e, u). By Theorem 3.4.2 we can fix points rj E [Sj, tjl such that zj(rj) ~ YI as j E B 1 , j ~ 00. So the set B2 = {j: j E B 1 , r j E (t - 1-£, t + 1-£), zj(rj) E OJ.lYl} is infinite.

229

Convergent sequences of solution spaces.

Now let a set B3 ~ B2 be infinite. Let for j E B3 points Pj ~ qj of the segment [t - f.£, t + f.£] are such that Zj([Pj, qj]) ~ [OILYI]. Let CPj(t) = g(Zj(t)) for t E [pj, qilThe function CPj is generalized absolutely continuous and by (3.13') and

(3.7)

Prove that (3.15) if cPj(Pj)

-+

G, CPj(qj)

-+

D as j E B3 and j

-+ 00,

then G ~ D.

By (3.14) the function t

'ljJj(t) = CPj(t)

+

J

aj(s)ds

Pi

cPj(qj) - CPj(Pj)

~

qi

-

J

aj(s)ds

-+ 0

Pi by virtue of the equicontinuity of the primitives of the functions aj. Thus

This gives the fulfilment of (3.15). V. Keep the notation of III and IV. Assume that:

Gr(zj![Pi,qi]) -+HI ~ Hn({t} x (OILYI n>.))

as

j E B,

J -+ 00,

and that the mapping f(u) = (e, u) is injective on the compactum (Hdt. Show that the topology r(F, t) and the Euclidean topology induce on (HI)t the same topology. The continuity of the mapping f, the connectedness and the compactness of the set HI with respect to the Euclidean topology imply that the set f((Hdt) ~ IR is a segment (or an one point set). Denote it by I. By virtue of the compactness of the set (Hdt, of the continuity and of the bijectivity of the mapping f : (Hdt -+ I the inverse mapping

CHAPTER 7

230

9 : I -+ (Hdt is continuous with respect to the Euclidean topology on the set HI (Theorem 2.2.1). So it is sufficient to check the continuity of the mapping 9 with respect to the topology r(F, t). Assume the opposite. For a point (t, a) of the set HI and for a number c > 0 for every j = 1,2,... there exists a point aj E ). satisfying the condition

By virtue of the compactness of the unit ball we can pass to a subsequence

of the sequence

{ II:; =:11 : j

= 1,2, ... }

converging to a vector Uo E A. We have lIuoll = 1 and Uo (j. ±(F(t,a)hc. Conditions (3.3) and (3.5) and Lemma 2.5.2 imply that

(F(t, a»)e: 2 CC(U{Fi(O(t, a»: i = k, k + I, ... }) for some k = 1,2, ... and for some neighborhood O(t, a) of the point (t, a) in U. By virtue of the continuity of the mapping 9 with respect to the Euclidean topology g([f(a) - 8, f(a) + 8]) ~ O(t, a) for some 8 > O. The vector Uo does not belong to the set

Lo

= CC(U{Fi(O(t, a)):

i

= k, k + 1, ... })

nor to the set

Ll

=-

CC(U{Fi(O(t, a»: i

= k, k + 1, ... }).

Theorem 2.4.2 and the membership 0 E Lo n Ll imply the existence of linear functionals fi : JRn -+ JR, i = 1,2, satisfying the condition

By virtue of the convergence aj -+ a with respect to the Euclidean topology and the continuity of the functional f we have f(aj) -+ f(a). Thus f(aj) E [f(a) - 8, f(a) + 81 beginning with some j = k.

Convergent sequences of solution spaces.

231

The segment Mj with endpoints f(aj) and f(a) lies in the segment + 8]. Therefore

[f(a) - 8, f(a) (3.16)

We obtain: If(aj) - f(a)1 < 8, aj E O(t, a) and g(Mj ) ~ O(t, a). By virtue of the continuity of the functional Ii, i = 1,2, we have fi ((a - ai)/ilai - all) < 0 beginning with some j = kl ~ k. Therefore for j ~ kl (3.17) Nw take an arbitrary index j ~ k l . Since the mapping f is injective, f(ai) =I f(a). Consider the case when f(ai) > f(a). By Theorem 3.4.2 for j E B3 we can fix points ej, dj E [pj, qj] such that

Zj(Cj)

-t

a,

zj(dj ) - t ai.

The continuum HI is homeomorphic to a segment. By (3.16) between the points a and ak it does leave the set O(t, a). Then the graph of the function Zj between Cj and dj lies in set O(t, a), beginning with some j = jo E B 3 • By (3.15) Cj < dj , beginning with some j E B 3 • The latter fact, the inclusions Gr (Zjl[Cj,d j]) C;;; O(t,a), fo(Lo) C;;; [0,00), and the considerations of IV with 9 = fo imply that fo(a) ~ fO(ai), which contradicts (3.17). The case f(ai) < f(a) may be considered in an analogous way with the change fo to fl. Thus our assumption is false. Hence the mapping 9 is continuous with respect to the topology r(F, t). In the beginning of our reasoning we noticed that it implies the coincidence on the set HI of the topology r(F, t) and of the Euclidean topology. VI. Return to the notation of III and put for j E B2

Pj = inf{p: p E [Sj, Tj], p ~ t - J.L, Zj([P, TjD

C;;; [O~yd},

qj = sup{q: q E [Tj, tj], q ~ t + J.L, Zj([Tj, qD

C;;; [O~yd}·

The set HI from V contains more than one point. If the corresponding mapping f is injective we obtain what was required from V. In the opposite case for some mE f«Hdt) the set H2 = ({y} xf-l(m))nHl contains more than one point. Let a and b be its arbitrary two different points. According to Lemma 3.4.2 fix points Pj, qj E [pj, qj] in order to obtain the convergence Zj(Pj) - t a, Zj(qj) - t b.

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Assume for definiteness (with a passage to a subsequence, if it is necessary) that pj < qj. Let L = ]-1(0) n A. If for every (t, y) E H2 the set F(t, y) contains entirely the space L then the topology r(F, t) coincides on (H2 )t with the Euclidean topology. In the opposite case we return to the beginning of the reasoning of VI with the change A to An ]-l(m), Pj, qj to Pj, qj and we do it till the fulfilment of assumptions of V. Since in every step the dimension of the corresponding plane A decrease by 1, not late than in n-th step our process finishes. We cannot come down to the dimension zero, because in this case a nontrivial continuum would be a subset of the plane of the dimension zero. The theorem is proved. • Example 3.2. Return to the system (3.1). Our aim here is the same as in Example 3.1. We repeat the steps I and III of the reasoning of Example 3.1 word by word. In the step II we have a possibility to use Theorem 3.2 with Si(t) == m (where m is from (3.2) ) and Fi(t, y) = {(A, -A): A ~ O}, that gives what was required: with the notation of Example 3.2 the sequence of solution spaces {Zi: i = 1,2, ... } belongs to s(JR. x JR.2).

4. Estimates of set location Let us go on to the second of the questions stated in the beginning of the previous section. Here we make some preliminary considerations. As before U denotes an open subset of the product JR. x JR.n. Let M ~ U, Mo ~ JR. and 0 =1= 4> ~ Cs(U). The upper bound j.£} is called a measure of the set M with respect to the space 4>. For rp E 4> denote by l(Mo, rp) the set of all connected open subsets ofthe segment 7r( rp) meeting the set Mo. For e > 0 denote by 1* (Mo, rp, e) the set of all finite pairwise disjoint families 'Y of elements of l(Mo, rp) with db) ~ e. If the set l*(Mo, rp, e) is empty we put w;,o(Mo) = o. In the opposite case we put w;,o(Mo) = sup{E{lIrp(sup J) - rp(inf 1)11 : J E A}: A E l*(Mo, rp, e)}. Notice that w* e > o. Therefore the limit w;(Mo) = limo--+o w;,o(Mo) exists. Let w}. The set M is called rp-w-thin if the set {t: t E 7r(rp) , (t, rp(t)) E M} may be represented as a union of a countable number of its subsets M k , k = 1,2, ... , with w;(Mk ) = o. The set M is called 4>-w-thin if it is rp-w-thin for every function cp E 4>. The set M is called cp-thin (respectively, 4>-thin)

Convergent sequences of solution spaces.

233

if /-,
h(t, z{t» - h{s, z{s»

~

c{t - s) for every two points s < t of 7r{z).

Lemma 4.2. Under the imposed conditions the set Ml(h, c) is closed in the space Cs{U). Proof. Let a sequence {Zk: k = 1,2, ... } ~ Ml{h,c) converge to a function Z E Cs(U) and points s < t belong to 7r{z). By Theorem 3.4.2 we can choose points Sk, tk E 7r{Zk) such that (Sk' Zk{Sk» -+ (s, z(s» and (tk' Zk(t k ) -+ (t, z(t)). By virtue of the convergences Sk -+ sand tk -+ t we have Sk < t k, beginning with some k (see Lemmas 2.1.1 and 1.7.2). Since Zk E Ml(h,c),

By virtue of the continuity of the function h and Lemma 1.7.2 the limit passage in the last inequality implies that

h{t, z{t» - h(s, z(s»

~

c(t - s).

E Ml (h, c). The lemma is proved. • Lemma 4.3. Let a mapping h : U -+ 1R be continuous, c> 0, M ~ U and /-,(h{M» = O. Then /-'M1(h,c){M) = O. Proof. Take an arbitrary function Z E Ml (h, c) and put Mo = {t : t E 7r(z), (t, z{t» EM}. The function cp{t) = h(t, z(t» is defined on the segment 7r(z). It is continuous and increasing. Therefore it maps bijectively the segment 7r(z) onto a segment I (see Theorems 1.7.8, 2.6.3 and Corollary 1 of Theorem 2.6.4). By Theorem 2.2.1 the inverse mapping ¢: 1-+ 7r(z) is continuous. The membership Z E Ml (h, c) implies that the mapping'I/J satisfies the Lipschitz condition with the constant c- l . Therefore it is absolutely continuous (see Example 4.5.1). By Corollary of Theorem 4.5.1 the set Mo ~ 'I/J(I n h(M» has the measure zero. The lemma is proved. • Theorem 4.1. Let a mapping h : U -+ 1R be continuous, c> 0, M ~ U, . U(h(M» = 0 and Z ~ M 1 (h, c). Then JL[zj(M) = O.

So

Z

CHAPTER 7

234

Proof. Lemma 4.2 implies the inclusion [Z] ~ Ml (h, c). Referring to Lemma 4.3 completes the proof. • Let as in the previous reasoning a function h : U -+ IR be continuous. Let c ~ O. Denote by M 2(h, c) the set of all functions Z E Cs(U) satisfying the condition

Ilz(t) - z(s)11 7r(z).

~

c(h(t, z(t)) - h(s, z(s))) for every two points s < t of

Analogously Lemma 4.2 we prove: Lemma 4.4. Under the imposed conditions the set M 2(h, c) is closed in the space Cs(U). • Lemma 4.5. Let a mapping h : U -+ IR be continuous, c > 0, M be a closed subset of the set U and {L(h(M)) = O. Then WM2 (h,c)(M) = O. Proof. Take an arbitrary function z E M2 (h, c) and put Mo = {t : t E 7r(z), (t, z(t)) EM}, cp(t) = h(t, z(t)). We have cp(Mo) ~ h(M) and {L(h(M)) = O. By Lemma 4.1.1 {L(cp(Mo)) = O. Therefore for every c > 0 there exists an at most countable family al of interval such that Ual 2 Mo and d(al) ~ c-1c. The set a2 of connected components of the (open) set G = Ual covers Mo. By Theorem 4.1.2 {L(G) ~ c-1c. Elements of the set a2 are pairwise disjoint. By Theorem 2.6.7, Lemma 4.1.3, and Theorem 4.1.1 d(a2) ~ c-1c. The family {HA = Mo n cp-l(A): A E ad consists of pairwise disjoint sets. The closedness of the set M implies the closedness of the set Mo ~ 7r(z) (see Theorems 2.1.1 and 1.7.5). By Theorem 1.6.1 the set Mo ~ 7r(z) is compact. By the definition of the induced topology the set H A , A E a2, is open in the compactum Mo. The representation HA = Mo \ (U{HB: B E a2 \ {A}}) implies now its closedness. Elements of the family {HA : A E a2, HA =1= 0} are pairwise disjoint and open (in Mo). The compactness of the set Mo implies the finiteness of this family. Therefore the number 1/ = min{p(HA' 7r(z) \ cp-l(A)): A E a2, HA =1= 0} is positive. The membership z E M 2 (h, c) implies that the function cp is non-decreasing. For, E l*(Mo, z, 1/) we have

l:{llz(b)-z(a)1I : (a, b) E ,} ~ cl:{lcp(b)-cp(a)1 : (a, b) E ,} ~ c(c-1c) = c. So wz(M) = O. The last estimate gives immediately what was required. The lemma is proved. • Theorem 4.2. Let a mapping h : U -+ IR be continuous, c > 0, M be a closed subset of the set U, {L(h(M)) = 0 and Z ~ M 2(h, c). Then w[zj(M) = O. Proof. Lemma 4.4 implies the inclusion [Z] ~ M 2 (h, c). By Lemma 4.5 this gives what was required. The theorem is proved. •

Convergent sequences of solution spaces.

235

We say that a set M ~ U is at most countable (respectively, finite) with respect to a space Z ~ Cs(U) if for every function z E Z the set {t: t E 7l"(z), (t, z(t») EM} is at most countable (respectively, finite). The remarks of Example 4.1.1 imply: Theorem 4.3. Let Z ~ Cs(U). Let a set M ~ U be at most countable • with respect to Z. Then I-Lz(M) = o. The following assertion is obvious: Lemma 4.6. If z E Cs(U) and a set M ~ U is represented as a union of (an at most countable number of) its z-w-thin subsets M k , k = 1,2, ... , then the set M is z-w-thin. • Evidently an one point set M ~ U is z-w-thin. This fact and Lemma 4.6 imply: Theorem 4.4. Let Z ~ Cs(U). Let a set M ~ U be at most countable • with respect to Z. Then the set M is Z -w-thin.

5. Estimates of the upper limit of a space sequence As before U denotes an open subset of the product IR x IRn. Lemma 5.1. Let a family "I of open connected subsets of a segment [a, b], a < b, of the real line satisfy the conditions a) if VI, V2 E"I and VI n V2 =F 0, then VI U V2 E "I; b) for every nonempty closed subset F of the segment [a, b] there exists a (nonempty) set V E "I, which meets the set F. Then [a, b] E "f. Proof. I. The set G = U"I is open (in the segment [a, b]). Therefore the set F = [a, b] \ G is closed. Condition b) implies the emptity of the set F, i.e., the equality G = U"f. II. The set H = U"Io, where "10 = {V: V E "I, V 3 a}, is open in the segment [a, b]. Theorem 2.6.1 implies the connectedness of the set H. By Theorem 2.6.3 the set H either coincides with the segment [a, b], or is a half interval [a, c) for some c E (a, b]. In the second case I implies the existence of a set VI E "I containing the point c. The equality [H] = [a, c] implies the existence of a point t E H n VI. The definition of the set H implies the existence of a set Vo E "10 containing the point t. By condition a) we have Va U VI E "I. Since a E Vo ~ Va U VI, Va U Vi E "10. Hence c E VI ~ Va U VI ~ H. This contradicts the definition of the point c. In the first case there exists an element Vi of the family "11 containing the point b. So Vi = [a, b]. The lemma is proved. • Theorem 5.1. Let ~ E Ri(U) (see §6.2), Z E Rq(U), "I be a family of open subsets of the set U. Let ~v ~ Zv for every V E "I. Let the set U \ (U"I) be ~-thin. Then ~ ~ Z.

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CHAPTER 7

Proof. I. Take an arbitrary function cP E 0 we have 'IjJ(Oot) ~ V. By virtue of the conditions

O. The segment I = [t - 8, t + 8] n 11"( cp) may be represented as the union of a countable number of segments 10 = {t}, Ik = In (-00, t - 2k] for k = -1,-2, ... and h = In[t+2- k ,(0) for k = 1,2, .... For every k = ±1, ±2, ... the segment h does not meet the set Mo. Thus it is covered by the family v* = {h n G: G E vo}. This family and the segment h satisfy the condition b) of Lemma 5.1. By Lemma 4.9.1 condition a) holds too. Lemma 5.1 implies now the generalized absolute continuity of the function cplh•. The function cplIo is also generalized absolutely continuous. This implies the generalized absolute continuity of the function cplI. So I E v. The membership (t E) (I) E Vo contradicts the choice of the point t: t E Mo. This gives what was required. III. Show that UVo = 1I"(cp), i.e., that Mo = 0. By virtue of the cp-w-thinness of the set U \ (U,) and Lemma 4.1 the closed subset Mo of the segment 11"( cp) may be represented as the union of an at most countable family {Mi: i = 1,2, ... } with w;(Mi ) = O. By virtue of the definition of the induced topology Baire theorem 1.7.6 implies the existence of points c < d of the segment 1I"(cp) and of a number i = 1,2, ... such that 0 =1= Mo n (c, d) ~ [Mi] n (c, d). Denote M; = Mo n (c, d). For arbitrary 'f} > 0 there exists 8 > 0 such that w;,o(Mi ) < 'f}. Let a family r satisfy condition (*,8, M;) of §4.5. Enumerate elements of the family r according to its order on the line: r = {(ai, bi ): i = 1, ... , k}, where a1 < b1 :::; a2 < ... :::; ak < bk · Put 101

= min( {bi -ai : i = 1, ... ,k }U{ ai+l -bi

:

i

= 1, ... ,k-1, aH1 -bi

=1=

O}).

Let £ E (0,101/2) and 8 E U{{ai,b i }: i = 1, ... ,k}. The point 8 E M; is not isolated in the set Mo ~ [Mil. Thus the set (8 - £,8 + c) n Mi

Convergent sequences of solution spaces.

237

is infinite. Therefore we can fix a point SE of this set such that the sets (8 - c, SE) n Mi and (SE, S + c) n Mi are nonempty. The sets (s - c, s + c), where s E U{ {ai, b;}: i = 1, ... ,k}, are pairwise disjoint. Thus the family fe = {((ait, (bit): i = 1, ... ,k} consists of pairwise disjoint (nonempty) intervals. For c < 1/2k(o - E{b i - ai: i = 1, ... ,k}) we have d(f':) < o. Because of the choice of the points SE every element of the family re meets the set Mi' Thus

E{lIcp((a;)E) - cp((bit)11 : i = 1, ... , k} ~ 'f/. By virtue of the continuity of the function cp the passage to the limit in this inequality as c ---+ 0 gives the estimate

This proves the absolute continuity of the function cp on the set M;. By Theorem 1.6.10 the family Vl = {I: [E vo, [ ~ ((c, d) n1f(cp)) \M;} contains an at most countable subfamily {Ik: k = 1,2, ... }, the union of which coincides with UV1' Lemmas 5.1 and 4.9.1 imply the generalized absolute continuity of the restrictions of the function cp on every segment [Ie, k = 1,2, .... Since (c, d) n 1f(cp) = M; U U{[k: k = 1,2, ... } we obtain the generalized absolute continuity of the restriction of the function cp to the set (c, d) n 1f(cp). Next we obtain the emptity of the set Mo. IV. Lemmas 5.1 and 4.9.1 imply now the generalized absolute continuity of the function cpo V. Let a segment [ lie in 1f(cp) \ M. Denote by e the set of all connected open subsets G of the segment [satisfying the condition: cplJ E Z for every segment J ~ G. The condition Z E R(U) implies, that the set e satisfies condition a) of Lemma 5.1. Our choice of [ and the definition of M imply that e satisfies condition b). By Lemma 5.1 CPII E Z. Since J.L(M) = J.L",(U \ (U,)) = 0 and Z E Rq(U), cp E Z. In view of the arbitrariness of cp E we have ~ Z. The theorem is proved. • Lemma 5.2. Let {Zk: k = 1,2, ... } ~ Ri(U) and Z = lim tOPSUPZk' Then Z E Ri(U). k->oo Proof. Let z E Z. Let a segment [a, bJ lie in the domain [s, tJ of a function z. Select a sequence Zk E Zk, k E A, converging to the function z (see Lemma 1.5.1). Let 1f(Zk) = [Sk' tkJ. By Corollary of Theorem 3.5.1 Sle --.. sand tk ---+ t. Let Sk (respectively, t k ) be a point of the segment [Sk' tkJ nearest to the point a (respectively, to the point b). We have {a, b} ~ [s, tJ, Sic --.. sand tk ---+ t. Therefore Sk ---+ a and tk ---+ b. In addition st, ~ tt, and [sk' tkJ ~ 1f(Zk)' By Theorem 3.5.4 zlcl[s;,t;) ---+ Zl[a,b)' The lemma is proved. •

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CHAPTER 7

When we apply the just proved lemma to a stationary sequence Z k == Z, we obtain: Corollary. Let Z E Ri(U). Then [Z] E Ri(U). • Theorem 5.2. Let Z E Rq(U). Let 'Y be a family of open subsets of the set U. Let the set Zv be closed in Cs(V) for every V E 'Y. Let the set U \ (U'Y) be [Z]-thin. Then the set Z is closed in the space Cs(U). Proof. By Corollary of Lemma 5.2 the space cI> = [Z] belongs to Ri(U). By our hypotheses cI>v = Zv for every element V of the family 'Y. Theorem 5.1 implies the inclusion cI> ~ Z. The inverse inclusion Z ~ cI> is obvious. The theorem is proved. • Notice that the application of this theorem may remain valuable and convenient when the family 'Y covers the entire set U. In this case we have: Corollary. Assume that Z E Rq(U), 'Y is an open cover of the set U for every element V of which the set Zv is closed in the space Cs(V). Then the set Z is closed in the space Cs(U), • We will reinforce this corollary in §6 and we will change the condition Z E Rq(U) to the condition Z E R(U). In respect of other applications of Theorem 5.2 pay attention to Theorems 4.2-6 wich help to obtain the estimates occuring in Theorem 5.2. Theorem 5.3. Let Z E Rq(U), 0 = {Zk: k = 1,2, ... } ~ Ri(U), cI> = lim top sup 0 ~ w. Let 'Y be a family of open subsets of the set U. Let lim top SUPk-> 00 (Zk)V ~ Zv (in Cs(V») for every V E 'Y. Let the set U \ (U,) be w-thin. Then cI> ~ Z. Proof. By Lemma 4.1 the set U \ (U'Y) is cI>-thin. Referring to Theorem 5.2 completes the proof. • By Lemma 1.4 Theorem 5.3 implies: Theorem 5.4. Let under hypotheses of Theorem 5.3 a E s(U). Then the sequence a converges to the space Z in U. • Theorems of this section in their full generality are directed to the investigation of singularities (ofright hand sides with parameters), including rather complicated ones. Now, in order not to cause the reader to digress from the scheme of application of results of such a kind by technical difficulties, consider examples of the investigation of the simplest singularities. Example 5.1. The right hand side of the scalar equation (5.1 )

,

y =

0 0 2

+ t2 + y2

is defined for all real values of 0, t and y, except (the one point of the three-dimensional space ]R3) 0 = t = Y = O. Define the right hand side there by putting its equal to zero for this values of 0, t and y. By remarks of Example 6.5.2 for every value of the parameter 0 the solution space of our equation satisfies conditions (c), (e) and (110). By remarks of §1 solutions of

Convergent sequences of solution spaces.

239

the equation depend continuously on the parameter a for a #- o. Consider an arbitrary sequence ak ~ 0, ak i- 0, and denote by Zk the solution space of our equation with a = ak. Denote by Z the solution space of the equation with a = O. For every point (t, y) i- (0,0) of the plane ~2 denote by V(t, y) the neighborhood of the point (t, y) of the radius tv't2 + y2. Put 'Y == {V(t, y): t, y E ~2, (t, y) i- (0, By remarks of §1 for every V E , the sequence {(Zk)V: k = 1,2, ... } converges to the space Zv in V. The set E = ~2 \ (U,) consists of one point (0,0). By Remark 1.1 and Corollary 2 of Theorem 3.1 the sequence 'fJ = {Zk: k = 1,2, ... } belongs to S(~2). By Theorems 4.3 and 4.4 the set E is 'IT-thin for 'IT = C8(~2). By Theorem 5.4 the sequence 'fJ converges to the space Z in ~2. The latter fact, in view of the arbitrariness of the sequence ak ~ 0 and Theorem 2.5.1, means the continuity of the dependence of solutions of our equation on the parameter a for a = O. Thus solutions of our equation depend continuously on the parameter a (for every its value). In a completely analoguous way we can consider the equation

On.

(5.2) where the function f : IR x IR ~ ~ is continuous. Example 5.2. Let a function f : ~ x ~ ~ [2, 00) be continuous, a E R Show that solutions of the equation (5.3)

y' = f(t, y)

+

a2

lal

+y2

depend continuously on the parameter a (as in the previous example for a = 0 we put the fraction equal to zero independently on the value of V). For a i- 0 the right hand side of equation (5.3) is continuous in the totality of the variables a, t, y. Here we obtain what was required from Theorem 1.3. Thus it remains to prove that for every sequence ai ~ 0, ai i- 0, the sequence of the solution spaces of the equations

(5.4) converges to the space Z of the solutions of the equation y' = f(t, V). Outside the line y = 0 the right hand side of the equation (5.3) depends continuously on the totality of the variables a, t, y. By Theorem 1.3 (Zi)V ~ Zv as i ~ 00, where V = ~x (~\ {O}) and Zi denotes the solution space of the equation (5.4). By Remark 1.1 and Corollary 1 of Theorem 3.1 the sequence of the spaces {Zi: i = 1,2, ... } belongs to s(~ x ~). To complete the reasoning

CHAPTER 7

240

we need to obtain estimates of the set (~ x ~) \ V = ~ x {O} according to considerations of §4. By putting h(t, y) = y and c = 1 we obtain the estimates JlM,(h,c)(E) = WM2(h,c)(E) = O. The solution spaces of all equations under consideration lie in the closed subspace W = Ml (h, c) n M 2 (h, c) of the space Cs (~ x ~) Now Theorem 5.4 implies the convergence Zi ~ Z. We obtain what was required. Results of the previous section and last examples prepare us for the consideration of the question about the convergence Zi ~ Z in steps: we can prove first the convergence (Zi)V ~ Zv for an arbitrary element V of a family of open subsets of the set U, and then pass to estimates of the geometry of the rest U \ U,. The convergence (Zi)V ~ Zv may be proved, for instance, with the use of classical theorems or with the use of any new versions of the corresponding classical theorems. The possibility of proving the convergence (Zi)V ~ Zv with the use of our scheme itself following our estimates of the geometry of the singularity set is not very obvious. But we can repeat the estimating of the geometry of singularity set several times, even we can use here a transfinite induction. Example 5.3. Keep the assumptions of Example 5.2. Consider the equation

y' = f(t, y)

(5.5)

+

lal

a 2 + y2

+

lal

a 2 + (y - sint)2

+

lal

a 2 + (y - cost)2

As in Examples 5.1-2 solutions of equation (5.5) depend continuously on the parameter a (as in two previous examples for a = 0 we put the fraction equal to zero independently on the values of t and y). For a i- 0 the right hand side of the equation (5.3) is continuous in the totality of the variables a, t, y. We obtain what was required from Theorem 1.3. In order to analyze the situation for a = 0 consider an arbitrary sequence ai ~ 0, ai i- O. Let Zi denote the solution space of the equation

y

, = f(t

,y

)+

lail

ai 2 + y2

+

lail

ai 2 + (y - sin t)2

+

lail

ai 2 + (y - cos t)2

Let W denote the complement in the plane ~ x ~ to the union of the x-axis L 1 , of the graph L2 of the function y = sin t and of the graph L3 of the function y = cos t. The convergence (Zi)W ~ Zw as i ~ 00 follows froIll Theorem 1.3. Every point x of the set E = (~ x ~) \ W belonging to one only of the sets L 1 , L2 or L3 possesses a neighborhood Ox without points of the other two of these three sets. The convergence (Zi)Oo: ~ Zoo: as i ~ 00 may be proved as was done in Example 5.2.

Convergent sequences of solution spaces.

241

Figure 7.6

The set (L1 n L 2) U (L2 n L 3 ) u (L1 n L 3 ) (Figure 7.6) consists of isolated points. We have the result of the previous paragraph. Now we repeat the reasoning of Example 5.1 which gives the convergence Zi -+ Z. We obtain what was required. Example 5.4. Return to system (3.1). Denote by Z the solution space of the system (5.6)

{

Xi

= -x

yl =

+y

-x - y.

Let U = JR X (JR 2 \ {O}). Show that with the notation of Example 3.1 the sequence ao = {(Z;)u : i = 1,2, ... } converges to Zu. I. With the notation of step I of Example 3.1 (5.7)

(Zi)V

-+

Zv·

This follows from classical results (see §1). II. Let e > 0, V = JR x ((2e,3e) x (-e,e)), h(t,x,y) = -y and W= M1(h,e). The spaces (Zl)V, (Z2)V, (Z3)V,'" lie in w. By I it remains to investigate the set E = V n (JR x (JR x {o})). It is finite with respect to w. By Theorem 5.7 we have (5.7). III. If e > 0 and V = JR x ((-3.::, -2.::) x (-.::, .::)), then we have (5.7) This is analogous to II. IV. The regions V studied in I-III cover the set U entirely. By Theorem 5.7 we have what was required. The origin of coordinates 0 is an asymptotically stable stationary point ofthe system (5.6). The relation between the systems (3.1) and (5.6), which is lllentioned in Example 5.4, is sufficient to state that the point 0 is also an asymptotically stable stationary point of the system

242

CijAPTER 7

Here the usual effects of 'first approximation' keep their place, although in our case in the x-axis (Le., for y = 0) perturbation terms can be written as ~ and they tend to the infinity as x ---+ O. This lead us far from common conditions of the 'first approximation', see Chapter XV. 6. Additional remarks

As before U denotes an open subset of the product ~ x ~n. Theorem 6.1. Let cP E Ri(U), Z E Rp(U)" be a family of open subsets of the set U, for every element V of which CPv ~ Zv. Let the set U \ (U,) be at most countable with respect to CP. Then cP ~ Z. Proof. Take an arbitrary function cp E CPo Denote by /J the set of all segments I ~ 7r(cp) such that cpll E Z. Put /Jo = {(I): IE /J}, M = {t : t E 7r(cp), (t, cp(t)) E U \ (U,)}. The set M is closed, at most countable and 7r(cp) \ (U/Jo) ~ M. The family /J satisfies obviously condition a) of Lemma 5.1. Show that it satisfies condition b). It is sufficient to consider the case when F ~ M (in the opposite case the corresponding condition is satisfied obviously). Then the set F is at most countable. By Assertion 2.4.1 there exists an isolated point to of this set. Take 8 > 0 such that 02cton(F\ {to}) = 0. The family /Jo covers the segments II = [to - 8, to] n 7r(cp) and 12 = [to, to + 8] n 7r(cp) except its endpoint to. The condition Z E Rp(U) implies that cplll' cplh E Z and cplltul2 E Z. So we have shown the fulfilment of condition b). By Lemma 4.6 cp E Z. The theorem is proved. • Theorem 6.1 implies easily (see the proof of Theorem 5.4) Theorem 6.2. Let a = {Zk: k = 1,2, ... } ~ Ri(U), a E s(U), lim top SUPk-+oo Zk ~ \If ~ Cs(U), Z E Rp(U). Let, be a family of open subsets of the set U. Let the sequence {(Zk)V: k = 1,2, ... } converge in V to the space Zv for every V E ,. Let the set U \ (U,) be at most countable with respect to \If. Then the sequence a converges in U to the space Z. • This assertion remains valuable in the case when the family, covers the entire set U although here is better to use the following simplified analog of Theorem 6.1. It is a direct consequence of Lemma 5.1 (see the proof of Theorem 6.1): Theorem 6.3. Let cP E Ri(U), Z E R(U), , be an open cover of the • set U, for every element V of which CPv ~ Zv. Then cP ~ Z. Corollary. Let Z E R(U), , be an open cover of the set U, for every element V of which the set Zv is closed in the space C s(V). Then the space Z closed in C 8 (U). • As an other consequence of this theorem we obtain: Theorem 6.4. Let a = {Zk: k = 1,2, ... } ~ Ri(U), Z E R(U). Let , be an open cover of the set U. Let the sequence {(Zk)V: k = 1,2, ... }

Convergent sequences of solution spaces.

243

converge in V to Zv. for every V E 'Y. Then the sequence a converges in

U to Z. Proof. The proof repeats the proof of Theorems 5.4 and 6.2 with the use of referring to Theorem 6.3. The fulfilment of the condition a E s(U) follows here from Corollary of Theorem 3.1 and from the emptity of the set U \ (U'Y). • Remark 6.1. The convergence of a stationary sequence Zk == Z as k -+ 00 in U to the space Z E R(U) is equipotent to the condition

Z E Rc(U). Therefore Theorem 6.4 implies: Corollary. Let Z E R(U). Let'Y be an open cover of the set U for every element V of which Zv E Rc(V). Then Z E Rc(U), • Likewise from Theorems 5.4 and 6.2 we obtain the following assertions. Theorem 6.5. Let Z E Rqs(U), lJI ;2 Z be a closed subspace of the space Cs(U), Let'Y be a family of open subsets of the set U. Let Zv E Rc(V) for every V E 'Y. Let the set U \ (U'Y) be lJI-thin. Then Z E Rc(U). • Theorem 6.6. Let Z E Rps(U), lJI ;2 Z be a closed subspace of the space Cs(U). Let'Y be a family of open subsets of the set U. Let Zv E Rc(V) for every V E 'Y. Let the set U \ (U'Y) be at most countable with respect to lJI. Then Z E Rc(U). • The following particular case of Theorem 3.1 may be helpful to verify the fulfilment of the condition Z E Rs(U) when we use Theorems 6.5 and 6.6. Theorem 6.7. Let Z E R(U), () be a multi-valued mapping of U in ~n. For every point (t, y) E U let lim topsup{Gr(zl[Sl,S2]): z E Z, [SI' S2] ~ (t - c, t

+ c n 7r(Z)),

6,0->0

Z([SI' S2])

n G6y i= 0}

~ (}(t,

y).

Let 'Y be a family of open subsets of the set U. Let Zv E Rs (V) for every V E 'Y. For every point t E ~ let the set (U \ (U'Y))t contain no nontrivial connected subsets on which the topology r((), t) and the Euclidean topology coincide. Then Z E Rs(U). • As in the case of Theorem 3.1, we obtain: Corollary. Let Z E R(U). Let'Y be a family of open subsets of the set U, for every element V of which Zv E Rs(V). For every point t E ~ let the 8et (U \ (U'Y))t contain no nontrivial connected subsets. Then Z E Rs(U). •

7. R(U) as a topological space As before U denotes an open subset of the product ~ x ~n.

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For M ~ U and V ~ C.(U) put O{M, V} = {Z: Z E R(U), ZM ~ V}. Introduce the topology on the set R(U) generated by the sub-base consisting of the sets O{M, V}, where M runs over the set of all compact subsets of U and V runs over the set of all open subsets of the metric space C.(U). Notice that the topology introduced is not Hausdorff, but this fact is not an obstacle to its using. Lemma 7.1. Let Z E Rc(U). Then the point Z of the space R(U) possesses a countable base. Proof. I. Fix a countable base (3 of the set U. Denote by (30 the set of elements of (3 with compact closures lying in U. The set (30 is countable and constitutes a base. Theorem 1.1. 7 implies that the set (31 of all finite subsets of the set (30 is countable. Thus the family x = {[U,]: , E (3d is countable. It consists of compacta lying in U. By Theorem 1.1. 7 this implies the countability of the family 8 = {O{K, O(ZK' 2-i)}: K E x, i=1,2, ... }. II. Take an arbitrary compactum K ~ U and an arbitrary open subset V :2 ZK of the space C.(U). Show that for some c > 0 we have O,(K, c) ~ U and ZO,(K,E) ~ V. Assume the opposite. For i = 1,2, ... and Ci = 2-ip(K, (~ x ~n) \ U), there exists a function Zi E ZO,(K,Ei) \ V. The set ZO,(K,Ed is compact. It contains all elements of the sequence a = {Zi: i = 1,2, ... }. Therefore the sequence a has a subsequence converging to some Z E Z. Then Z E n{ZO,(K,E;}: i = 1,2, ... } = ZK and Z (j. V. This is impossible by the choice of K and V. III. With the notation of II for every point x E K fix an element Bx of the base (30 such that x E Bx ~ O( K, c). The cover {Bx: x E K} of the compactum K contains a finite sub cover {B X1 ' ... ' BXj }. Put K1 = [U{Bxw .. ,Bxj }]. Evidently K1 E x and K ~ K1 ~ O,(K, c). By virtue of the last inclusion ZK ~ ZK 1 ~ V. By the Corollary of Lemma 2.3.8 O(ZKl' 2- i ) ~ V for some i = 1,2, .... Therefore

The set O{K1' O(ZKIl2-i)} belongs to 8. By virtue of the arbitrariness in the choice of K and V the inclusion Z E O{K1' O(ZKIl2-i)} ~ O{K, V} means that the family 8 constitutes a sub-base at the point Z. The family of intersections of finite subfamilies of 8 is countable and constitutes a base at the point Z. The lemma is proved. • Lemma 7.2. A sequence a = {Zi: i = 1,2, ... } ~ R(U) converges to Z E Rc(U) in the space R(U) if and only if it converges to Z in U in the sense of the definition of §1.

Convergent sequences of solution spaces.

245

Proof. By virtue of Lemma 1.7.3 the convergence of the sequence a to Z in the space R(U) is equipotent to the condition (7.1) for every K E expc U the sequence {(Zd K : i = 1,2, ... } converges in the space Cs{U) to the set ZK in the sense of condition (1.7.1).

By Lemma 2.5.1 condition (7.1) is equipotent to the convergence of the sequence a in U to the space Z in the sense of §1. The lemma is proved .• Now Lemmas 7.1 and 7.2 and Theorem 1.5.3 imply: Theorem 7.1. A point Z E Rc{U) belongs to the closure of a set M ~ R{U) if and only if there exists a sequence of elements of the set M converging in U to the space Z. • So tools developed in previous sections of this chapter are applicable to . the description of the topology of the subspace Rc{U) of the space R{U). On the other hand, the same relation of the convergence of sequences with the topology of the space Rc{U) simplifies the using of results about the preservation of properties of solution spaces under limit passages. For instance, Theorem 2.1 states the closedness of the set Rce{U) in the space Rc{U). When they establish properties of solution spaces of equations (inclusions) often it is convenient to construct corresponding approximating sequences of spaces. In order to use Theorem 2.1 to prove the fulfilment of condition (e), we may construct a space sequence converging to the space in question (and satisfying the other hypotheses of Theorem 2.1). We will meet such situations later on. In the proof of the convergence of such sequences sometimes the following arguments turn out to be helpful. Let PI be an arbitrary metric on the set U (corresponding to the topology of U). The metric PI generates a Hausdorff metric on the set expc U. The last one induces a metric P2 on the space Cs{U) (by the embedding Gr). The metric P2 may be used to define a deflection a on the set of closed subsets of the space Cs(U): a{FI' F2 ) = inf{ {oo }U{e: e > 0, F2 ~ OeFd) (see §3.4, where we have considered the deflection on the set of compact subsets of a metric space). With this notation for Zo E Rc{U), K E expc U and c > 0 we have O{K, O«ZO)K, = {Z: Z E R(U), a{{ZO)K' ZK) < c}. By virtue of facts established in the step II of the proof of Lemma 7.1 we have: Assertion 7.1. A sequence {Zi: i = 1,2, ... } ~ R(U) converges to a space Zo E Rc(U) (in U or, that is equipotent, with respect to described topology of the space R(U)) if and only if a«ZO)K' (Zi)K) ---* 0 for every COmpactum K ~ U. • Now let

en

(7.2) Uo be an open subset of the set Uj

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(7.3) Ml ~ M2 ~ M3 ~ ... ~ U; (7.4) for every compactum K ~ Uo there exists an index i = 1,2, ... such that K ~ Mi. As examples of such sequences we may take Mi == Uo, or Mi == [Uo]. See also the proof of Theorem 3.2.2 (there X = Uo = U). Theorem 7.2. Let conditions (7.2 - 4) hold. Let Zo E Rc{U), {Zi : i = 1,2, ... } ~ R{U) and a{{ZO)Mi' (Zi)M) ~ O. Then the sequence {(Zd uo : i = 1,2, ... } converges to {Zo)uo in Uo. Proof. Take an arbitrary K ~ Uo and a sequence f3 = {Zi : i E B} ~ Cs(U) such that Zi E (Zi)K for every i E B. By (7.4) K ~ M i , beginning with some i = i o. By Lemma 2.3.4 for every i E B, i ~ i o, there exists a function z; E (Zo) Mi such that (7.5)

By the Corollary of Lemma 2.3.8 O,{K, co) ~ U for some co E (O, (0). We have a((ZO)Mi' (Zi)MJ + 2- i < co, beginning with some i = i 1 ~ i o· The last estimate and (7.5) imply that Gr(z;) ~ O(Gr(zi), co). Hence Gr(z;) ~ O,(K, co) E expc U. Since Z E Rc{U), the sequence {z;: i E B, i ~ id has a subsequence {z;: i E Bd converging to a function z* E Z. For i E Bl

Therefore the sequence {Zi: i E Bd converges to the function z*. This • gives what was required. The theorem is proved.

CHAPTER 8

PEANO, CARATHEODORY AND DAVY CONDITIONS

In Chapter 6 we have shown that in some quite simple cases the solution spaces of an equation satisfy the conditions which are suitable as axioms for the construction of a general theory. Although the theory is applicable to equations of very various types, for this time we have the possibility to use it only in quite narrow framework of the examples of Chapter 6. In this chapter we expand the sphere of applications of our theory. We expand it, in particular, to the cases which are covered by the classical theory of ordinary differential equations. In the using of the terms 'Caratheodory conditions' and 'Davy conditions' I follow a tradition of Russian mathematical texts. Many authors out of respect for the pioneering role of Caratheodory, reserve his name for denoting the most general situation here.

1. Kneser condition Let U be an open subset of the product ~ x ~n. The name of the following properties of a space Z in the title of the section:

~

C s (U) is mentioned

(k) for every point (t, y) of the set U there exists a number 6 > 0 such that for every point s E (t - 6, t + 6) the set {z(s): Z E Z, s, t E 7f(z), z(t) = y} is connected. The fulfilment of condition (u) implies the fulfilment of condition (k), because a one point set is connected. Lemma 1.1. Let Z E Rce(U). Let t E ~ and K be compact subsets of the set Un « -00, t] X ~n) (respectively, of the set Un ([t, 00) x ~n)). For every function Z E Z+ (respectively, Z E Z-) let t E 7f{z) ifGr(z)nK f:. 0. Then the set q,

= {z:

Z

E Z, SUP7l"(z)

= t,

(inf7l"(z),z(inf7f(z))) E K}

(respectively, q,

= {z:

Z

E Z, inf7l"(z)

= t,

(SUP7l"(z),z(SUP7l"(z))) E K})

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is compact. Proof. The lemma will be proved in main case. Assume the opposite. Then by Theorem 1.6.5 and remarks of §1.6 we can choose a sequence {CPk: k = 1,2, ... } ~ q, without convergent subsequences. By virtue of the compactness of the set K we can assume in addition that the sequence {(tk' CPk(t k )): k = 1,2, ... } ~ K, tk = inf7r(cpd, converges to a point (to, y). For every k = 1,2, ... fix an extension Zk E Z-+ of the function 'Pk (such an extension exists by Lemma 6.6.2). Apply Theorem 7.2.2 to the sequence of the functions {Zk: k = 1,2, ... } and to the sequence of the points {t k : k = 1,2, ... } with Zk = Z for every k = 1,2, .... Let z* E Z-+ be a function existing by Theorem 7.2.2. It takes the value y at the point to. Let {Zk: k E Ad be the corresponding subsequence. Since to E 7r(z*) and (to, z*(to» E K, t E 7r(z*). Take a number 8 > 0 such that [to - 8, t + 8] ~ 7r(z*). By virtue of the condition imposed in Theorem 7.2.2 on the subsequence {Zk: k E Ad we have [to - 8, t + 8] ~ 7r(Zk), beginning with some k = kl E AI, and on the segment [to - 8, t + 8] the sequence of the functions {Zk: k E AI, k ~ kl } converges uniformly to the function z*. We have tk E (to - 8, t + 8), beginning with some k = k2 ~ kl . Hence 7r( CPk) = [tk' t] ~ [to - 8, t + 8]. Theorem 3.5.4 implies the convergence of the sequence {CPk: k E AI, k ~ k2} to the function z*l[to,tj' This contradicts the choice of the sequence {CPk: k = 1,2, ... }. Thus our assumption is false and the lemma is proved. • Lemma 1.2. Under the hypotheses of Lemma 1.1 let Z E Rk(U) and the compactum K be connected. Then the set

S = {z(t):

Z

E Z-+, Gr(z) nK

i=

0}

is compact and connected (Le., it is a continuum). Proof. The lemma (as in the case of Lemma 1.1) will be proved in the main case. Theorem 3.5.4 implies the continuity of the mapping a : q, ---+ IRn , a(cp) = cp(t). By Lemma 1.1 the set q, is compact. Theorem 1.7.8 implies the compactness of the set S = a(q,). Assume that the set S is disconnected. Then S = Xl UX2, where the sets Xl and X 2 are nonempty, closed and disjoint. By virtue of the continuity of the mapping a and Theorem 1.7.5c the sets q,i = a-I(Xd, i = 1,2, are closed in q,. By Theorem 1.6.2 they are compact. Theorem 1. 7.8 implies the compactness of the sets Ki = U Gr(q,i), i = 1,2. By Theorem 1.6.1 these sets are ·closed. Theorem 1.3.4 implies the closedness of the sets K n Kl and KnK2. The nonemptiness of the sets KnK l and KnK2 is obvious in view of the nonemptiness of the sets q,l and q,2. If Z E Z-+ and Gr(z) n K i= 0, then 7r(z) 3 t. Thus the condition Z E Rce(U) and Lemma 6.6.2 imply that the sets K n Kl and K n K2 cover K. The connectedness of K implies the nonemptiness of the intersection (K n Kd n (K n K2)' So Kl n K2 i= 0.

Peano, Caratheodory and Davy conditions.

249

The projection of the nonempty set Kl n K2 ~ ~ x ~n in the first factor is compact. Its upper bound s « 00, see Remark 2.3.4) belongs to it. Take a point y E ~n such that (s, y) E Kl n K 2. We have s < t (the case s = t is impossible, because ({t} x ~n) n (Kl n K 2) = {t} x (Xl n X 2) = 0). Let for i = 1, 2 <1>;

= {z: Z E Z, 71'(z) = [s, t], z(s) = y, z(t) E Xd, K;* = UGr(<1>;).

The set <1>;, i = 1,2, is closed in the compactum {z: z E Z, sup 71'(z) = t, (inf7l'(z), (inf7l'(z» = (s,y)} (see Lemma 1.1). Therefore it is compact. By Theorem 1. 7.8 the set K; is compact. By the choice of s and by the definition of the sets K~ and K; we have K~ n K; = {( s, y)}. Therefore

for s' E (s, t). The sets ({ s'} So the set

({s'}

X ~n)

X ~n)

n (K: UK;)

n K~ and ({ s'} X ~n )nK; are nonempty.

= {z(s'): z E Z, s,s' E 71'(z), z{s) = y}

is disconnected, which contradicts condition (k). Thus our assumption is false. The set S is connected. The lemma is proved. Theorem 1.1. Let a sequence

•

{Zk: k = 1,2, ... }

~

Rcek(U)

converge in U to a space Z E Rc(U), Then Z E Rcek(U), Proof. I. By Theorem 7.2.1 Z E Re(U). It remains to prove that

Z E Rk(U), II. Let (to, Yo) be an arbitrary point ofthe set U. Fix numbers f.L > v > 0 according to Theorem 6.7.3 (for Z). Assume that for a number tl E [to - v, to + v] the set

S = {z(tl ) : z E Z, 71'(z) = [to - v, to

+ v],

z(t o) = Yo}

is disconnected: S = Xl U X 2 , where the sets Xl and X 2 are disjoint, nonempty and closed. Put for definiteness that to < t l . For i = 1,2 fix a function Zi E Z such that 71'(Zi) = [to, tIl, Zi(tO) = Yo and zi(td E Xi' Let H == Gr(zd U Gr(z2)' III. Let us show that, beginning with some k = kb we have: if z E Zk, 1r(z) == [a, b] ~ [to, to + v] and (a, z(a» E H, then Gr(z) ~ Op.Yo.

CHAPTER 8

250

Assume the opposite. Then there exist functions z;. E Zk, k E A, such that 7r(z;') = [ak, bk] ~ [to, to + v], Im(z;') ~ [OIL Yo] , (ak' z;'(ak)) E Hand Ilz;'(b k ) - Yoll = /-to By virtue of the convergence of the sequence {Zk : k = 1, 2, ... } to the space Z the sequence {z;': k E A} contains a subsequence converging to a function z E Z. Denote: 7r(z) = [a, b]. We have [a, b] ~ [to, to + v], z(a) = zi(a), where i = 1 or 2, and Ilz(b) - Yoll = /-to Consider the function z* E Z defined on the segment [to, b] by the formula *( ) = {Zi(t)

z t

z(t)

for t E [to, a], for t E [a, b].

The function z* cannot be extended to an element of Z-+ by virtue of our choice of /-to The obtained contradiction gives what was required. IV. Let WI and W 2 be disjoint neighborhoods of the sets Xl and X 2 , respectively, in the space ]Rn. Let

Bk

= {z: z

E

Zk, (inf7r(z),z(inf7r(z))) E H, SUP7r(z)

= td

for k = 1,2, .... By III

u Gr(Bd ~ (to - v, to + v) X GlLyo for k ~ k l . For k = kl' kl + 1, ... put X; = {z(tr): z E Bd. By Lemma 1.2 the set X; is connected. This observation and the nonemptiness of the sets Xl and X 2 imply the nonemptiness ofthe set X; \ (WI UW2 ). Fix a function z;. E Bk such that

By III

Gr(zZ) ~ [to - v, to

+ v]

x [OIlYO].

By virtue of the convergence of the sequence {Zk: k = 1,2, ... } to the space Z in U the sequence {z;': k = kl' kl + 1, ... } has a subsequence converging to a function z E Z. We have (a, z(a)) E H, where a = inf 7r(z), and z(tr) E ]Rn \ (WI U W 2 ), i.e., z(tr) rt. Xl U X 2 = S. Let z(a) = zi(a), where i = 1 or 2. Consider the function z* E Z such that 7r(z*) = [to, td and for t E [to, a] *( ) = {Zi(t) z t z(t) for t E [a, tl]. Let z** E Z-+ be an arbitrary extension of the function z*. By the choice of v we have [to-v, to+v] ~ 7r(z**). Next, z**(to) = Yo and z**(tr) = z(t 1 ) rt. S.

Peano, Caratheodory and Davy conditions.

251

This contradicts the definition of the set S. The obtained contradiction gives what was required. The theorem is proved. • Later on we will use Theorem 1.1 in proofs of the fulfilment of the Kneser condition. In order to follow this way we need to construct a space sequence (with properties which are mentioned in the statement of the theorem) converging to the space under consideration. In the scalar case the situation is simpler: Theorem 1.2. Let Uo be an open subset of the plane IR x IR and

Z E Rce(Uo). Then Z E Rcek(UO)' Proof. Take an arbitrary point (to, Yo) of the set Uo and find numbers J.I. > 11 > 0 according to Theorem 6.7.3. Our aim will be achieved when we show that for every point tl E (to - 11, to + 11) and for every points al < a2 of the set S = {z(td: Z E Z, ?fez) :3 to, t l , z(t o) = Yo} the segment [aI, a2] lies in S. Let for definiteness to < t l . Assume the opposite. Then there exists a point a E [aI, a2] \ S. Take an arbitrary function z E Z-+,

I

~a2 ~a

I I ,

to

S

Figure 8.1

tl

al

)

Na I

I

l )0

tl

Figure 8.2

the domain of which contains the point tl and which takes the value a at the point t l . For i = 1,2 fix functions Zi E Z such that ?f(Zi) = [to, tIl, Zi(t O) = Yo and zi(td = ai' Let I = [to, tIl n ?fez). The set I is either a segment or a half interval with an open left endpoint. If at a point s E I the graph of the function Z meets the graph of the function Zi, i = 1 or 2, then the function Z*(t) = {Zi(t) for t E [to, s] z(t) for t E Is, tIl

(1I'(Z*) = [to, tIl) belongs to Z. This contradicts the condition a ~ S (Figure 8.1). Thus Zl(t) < z(t) < Z2(t) for every point tEl (Figure 8.2). Hence Gr(ziI ~ [to - 11, to + 11] x [O~Yo]. By Lemma 6.6.1 the last fact may be true only when [to, tl] ~ ?fez). Then to E I and ZI(tO) < z(to) < Z2(t O) = Zl(t O ), that is impossible. The obtained contradiction gives what was required. The theorem is proved. •

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Theorem 1.3. Let Z E Rcek{U), (t, Yo) E U, t E [a, b]. Assume that if Z E Z-+, t E 7r{z) and z{t) = Yo then [a, b] ~ 7r{z). Then the set «P = {z : z E Z, 7r{z) = [a, b], z{t) = Yo} is connected. Proof. Let «Pa = {z/[a,tj: Z E «p} and «Pb = {Z/[t,bj: Z E «p}. The condition Z E R{U) implies, that «P = «Pa x «Pb (moreover this formula relates the topologies of these spaces too, see Theorems 3.5.4 and 3.5.5). By Lemma 1.1 the spaces «P, «Pa and «Pb are compact. The Corollary of Theorem 1.7.8 and Lemma 2.1.1 imply the upper semicontinuity of the mapping which is inverse to the projection of the product «Pa x «Pb into every factor. By Theorem 2.6.4 our assertion reduces to the cases t = a and t = b. Both these cases are analogous. For definiteness put t = a. Assume the opposite. Let «P = «P1 U «P2, where «P1 and «P2 are nonempty disjoint closed subsets of the compactum «P. Fix a countable dense subset T = {t k : k = 1,2, ... } of the segment [a, b] and for every k = 1,2, ... define the mapping h : «P -7 (1~n)k by the formula fk{Z) = (z{td,.·· ,z{tk )). Show that h{«P 1) n fk{«P 2) = 0 for some k = 1,2, .... If we assume the opposite, then for every k = 1,2, ... there are functions z~ E «Pi, i = 1,2, such that fk{Zl) = fk(Zn. By Theorem 1.7.8 the set U Gr{«p) is compact. It contains the graphs of all functions zL k = 1,2, ... , i = 1,2. Condition (c) implies a possibility to pass to subsequences {zl: k E A} and {z~ : k E A} converging, respectively, to functions Zl E «P1 and Z2 E «P2' For every k = 1,2, ... we have Zl{t k ) = Z2{t k ). The density of the set T in the segment [a, b] and Lemma 1.7.4 imply the coincidence of the functions Zl and Z2. Therefore «P1 n «P2 =J 0, which contradicts the choice of the sets «P1 and «P2' Thus for a number k the set fk{«P) is disconnected. To continue the reasoning it is convenient to word the obtained result in an other way: there exist such points Sl < ... < Sk of the segment [a, b] that for the mapping f : «P -7 (1~n)k, f{z) = (z{sd, ... , z{sd), the set f{«p) is disconnected. Fix such a mapping f, for which the number k is the smallest of all possible ones. Lemma 1.2 immediately implies, that k ~ 2. Let the mapping 9 : «P -7 (Rn)k-1 be defined by the formula g{z) = (z{sd, ... , z{sk-d)· Let h : f (<
•

Peano, Caratheodory and Davy conditions.

253

Consider the vectors el = {1,0}, e2 = {O, -I}, e3 = -el, e4 = -e2 and the subsets Ml = {(x,y): x ~ 0, -x:::;; y:::;; x}, M2 = ((x,y): y ~ 0, -y:::;;x:::;;y},M3 ={(x,y): x:::;; 0, x:::;;y:::;;-X},M4={(X,y): y:::;;O, y:::;; x:::;; -V} of the plane. Let U = jR X jR2. Let Zi be the solution space of the equation y' = ei, i = 1,2,3,4. Define the space Z by the condition: ZIRxMi = ZilRXMi for -' t i = 1,2,3,4 (see Figure 8.3 and §9.5) below. Evidently such a space Z E R(U) exists and is defined uniquely by this condition. The reader can show that Z E Rce(U) and that at the point (t, 0, 0) the Kneser condition is not fulfilled. Notice now that Theorem Figure 8.3 7.2.3 implies Lemma 1.3. Let Z E Rce(U) and'Y be an open cover of the set U, for every element V of which Zv E Rk(V), Then Z E Rk(U), •

,,

..,

, ", -,

-,

"

2. Caratheodory and Davy conditions In this section we investigate a class of equations and inclusions studied by Caratheodory (case of equations, see [CLD and by Davy (a larger case of inclusions, see [DaD. Start with preliminary remarks. See also [LR] in the connection with the approximation of right hand sides. Lemma 2.1. Let multi-valued mappings Fk, k = 0,1,2, ... , be defined on a topological space X, take their values in the Euclidean space Rn and be upper semicontinuous. Assume that for every x E X the set F(x) = CC(U{Fk(X): k = 0,1,2, ... }) is compact and

(2.1) for every neighborhood OFo(x) of the set Fo(x) in ~n there are a neighborhood Ox of the point x in X and a number ko = 1,2, ... such that U{Fk(OX): k = ko, ko + 1, ko + 2, ... } ~ OFo(x). Then the mapping F : X ~ jRn is upper semicontinuous. Proof. Take an arbitrary point x E X and an arbitrary neighborhood W ofthe set F(x). By Corollary of Lemma 2.3.8 026F(X) ~ W for a number 6> O. For the neighborhood 06FO(X) of the set Fo{x) find a neighborhood Ox and a number ko according to (2.1). For k = 0,1, ... , ko -1 by virtue of the upper semicontinuity of the mapping Fk there exists a neighborhood Vk of the point x in X such that Fk{Vk ) ~ 06Fk{X). For t E oxnv1n·· ·nVko - l

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we have

By Lemma 2.4.3 the set OcF(x) is convex. Lemma 2.4.5 implies the convexity of the set [OcF(x)]. The definition of the closed convex hull implies the inclusion F(t) ~ [OcF(x)] ~ W. The lemma is proved. • Let as before U denote an open subset of the product ~ x ~n. Lemma 2.2. Let Fk : U --t ~n, k = 1,2, ... , be multi-valued mappings. Let Fl(X) 2 F2(x) 2 F3(X) 2 ... and F(x) = n{Fk(X): k = 1,2, ... } for every point x E U. Then D(F) = n{D(Fk): k = 1,2, ... }. Proof. Evidently D(F) ~ D(Fk ) for every k = 1,2,3, .... Therefore D(F) ~ n{D(Fk): k = 1,2, ... }. Prove the inclusion n{D(Fk): k = 1,2, ... } ~ D(F). Let Z E n{D(Fk): k = 1,2, ... }. If 1l'(z) consists of one point only then Z E D(F) by definitions. Let 1l'(z) contain more than one point. For every k = 1,2, ... we can point in the segment 1l'(z) a subset Mk of measure zero such that for every point t E 1l'(z) \ Mk the derivative z'(t) exists and belongs to the set Fk(t, z(t». By Theorem 4.1.2 the measure of the set M = U{Mk: k = 1,2, ... } is equal to zero. For every point t E 1l'(z) \ M we have z'(t) E n{Fk(t, z(t»: k = 1,2, ... } = F(t, z(t». Thus z E D(F). The lemma is proved. • Lemma 2.3. Let a = {Zk: k = 1,2, ... } ~ Rc(U), Zl 2 Z2 2 ... and Z = na. Then the sequence a converges in U to the space Z. Proof. The fulfilment of the condition a E s(U) follows from Theorem 3.6.1, because all elements of a lie in Zl E Rc(U). By Lemma 6.4.1 and Theorem 1.6.1 elements of the sequence a are closed in Cs(U). Referring to remarks of Example 1.5.3 and to Lemma 7.1.4 completes the proof. • The following assertion is obvious. Lemma 2.4. If a sequence {Zk: k = 1,2, ... } of subspaces of the space Cs(U) converges in U to a space Z ~ Cs(U) and Z; ~ Zk for every k = 1,2, ... then the sequence {Z;: k = 1,2, ... } converges to the space Z in U. • Let us now turn to the main topic of this section. Let (beginning here and to the end of the section) cp denote a non-negative function which is defined and locally Lebesgue integrable on an interval (a, b), -00 :::;; a < b :::;; 00, of the real line and U ~ (a, b) x ~n. Denote by Qd('P)(U) the set of all multi-valued mappings F : U --t ~n with compact convex values satisfying the conditions: a) there exists a single valued mapping f : U --t ~n such that for every y E ~n the mapping fY is measurable and f(t, y) E F(t, y) ~ 0,(0, cp(t» for every (t, y) E Uj

Peano, Caratheodory and Davy conditions.

255

b) for every t E (a, b) the mapping Ft is upper semicontinuous. These are the Davy conditions. For F(t,y) == {j(t,y)} they go into Caratheodory conditions. The function cp is called a (Davy, respectively, Caratheodory) majorant of the right hand side F. Notice that Qd('P)(U) ~ Q*(U) (see §6.4 and Theorem 2.5.2). By results of §§6.4 and 6.5 for F E Qd('P)(U) the solution space D(F) satisfies conditions (c) and (n). Now our goal is to prove the fulfilment of conditions (e) and (k). For t E (a, b) and Y E ~n put if (t, y) E U, in the opposite case. Evidently for every y E ~n the function all t E (a, b) we have IIf~(t)1I :::;; cp(t). Define on the space ~n the function:

Jr

.

If

is measurable. For almost

n

Ilx - yll : :; k'

in the opposite case, where k = 1,2, ... , Y E ~n. By Theorem 1.7.6 it is continuous. A simple direct calculation with an analogous using of Theorem 1.7.6 allows to check the existence and the continuity of its derivatives. Let ~(x,k) = (~, ... ,~) for x= (k1, ... ,kn ) E Nn and k = 1,2, .... ~or every k = 1,2,... and y = (Yl, ... , Yn) E ~n there exists x = (k 1, ... , k n ) E Nn such that IkYi - kil < 1 for i = 1, ... ,n. Then

IIY -

~(x, k)1I

1/

2

2

...;n < k' n

= k Y (kYl - kd + ... + (kYn - kn) :::;; k G:k,{(x,k) (y) > o.

On the other hand, if O(y, n/k) n O(~(x, k), n/k) 1= 0, where x = (k 1, ... , kn ) E Nn, k = 1,2, ... , then p(y,~(x, k)) < 2n/k,

IkYl - kl l2 + ... + IkYn -

knl2 < 4n 2 , IkYl -

kll < 2n, ... , IkYn

- knl < 2n.

Only a finite number of x E Nn satisfy the last condition. Therefore the function 13Z(x) = E{G:k,{(x,k) (x) : x EN} is continuous and has continuous derivatives. For k = 1,2, ... , x E Nn and x E ~n put (.l

fJk,x

()

x

=

G:k,{(x,k) (x)

13Z(x)'

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256

These functions are continuous. They have continuous derivatives. For every point x E IRn:

(2.2)

!3k,,.,(X)

~

0;

(2.3) there exists a neighborhood Ox of the point x, on which only finite number of functions of the family {!3k,,.,: x E Nn} do not equal identically zero;

For every k = 1,2, ... define on the set (a, b) x IRn the function:

Let now (to, Yo) E U. Take an arbitrary number "I > 0 satisfying the condition [to - "I, to + "II x O,(Yo, "I) ~ u. By virtue of the compactness of the set O,(Yo, "I) and (2.3) the set A = {x: x E Nn, SUP!3k,,.,(O,(yo, "I)) > O} contains only a finite number lk of elements. Put mk = sup{ 8:~,>< (x) : x E A, j = 1, ... ,n, x E O,(Yo, "I)} J « 00, see Remark 2.3.4). For every t E [to - "I, to + "II the norms of all derivatives of the functions ik(t, y) in yare not greater than lkmkcp(t). By Lemma 6.5.4 (2.5) for every point p, q E OT/Yo, By virtue of our assumptions, of Theorems 4.2.6 and of (2.5) the function h satisfies on the set Uo = (to -"I, to +1J[ xOT/Yo conditions a) and b) of §6.5. Remarks in Example 6.5.1 imply the membership D(fk, Uo) E Rceu(Uo). By Lemmas 6.5.2 and 6.5.3 and the arbitrariness in the choice of the point (to, Yo) we have D(h, U) E Reu(U). This fact and remarks in the beginning of the reasoning imply the membership D(A, U) E Rceu(U), Show that for every t E (a, b) at every point x E Ut the sequence of the mappings Fo = Ft , Fk = (fklu)t for k = 1,2, ... satisfies (2.1). Let o Fo (x) be an arbitrary neighborhood of the set Fo (x) in the space IRn. By Corollary of Lemma 2.3.8 OeFO(x) ~ OFo(x) for some c > O. By virtue of the upper semicontinuity of the mapping Fo there exists a neighborhood Ox of the point x in Ut such that Fo(Ox) ~ OeFO(x). There are numbers 8 > 0 such that 025X ~ Ox and ko = 1,2, ... such that ::a < 8. If Y E Oe x, k = k o, ko + 1, ko + 2, ... , x E A and !3k,x(y) =f= 0, then O(~(x, k), V :3 y. By virtue of the triangle inequality ~(x, k) E 025X ~ Ox ~ UtI Hence (2.6)

J.(t, ~(x, k» = J(t, ~(x, k))

E

OeFo(x).

Peano, Caratheodory and Davy conditions.

257

By (2.2) and (2.4) the point Ik{t, y) belongs to the convex hull of the set {!(t,~(x, k)): x E Nn, f3k,x(Y) -=I O}. By Lemma 2.4.3 the set OeFO(x) is convex. Now (2.6) implies that Ik(t, y) E OeFO{x) ~ OFo(x). In view of the arbitrariness of k = ko, ko + 1, ko + 2, ... this means the fulfilment of (2.1). Lemma 2.1 implies the upper semicontinuity in y of the mappings

Gk : U

-+ ~n,

Gk(t,y)

= cc(F(t,Y)U{!i{t,y): i = k,k+1,k+2, ... }), k =

1,2, .... Thus the mapping G k satisfies condition b). Evidently the mapping satisfies condition a) too. Therefore G k E Qd(rp)(U), D{G k) E Rc{U). Consider an arbitrary point (t, y) of the set U. We noticed above that the sequence of the mappings Fo = Ft , Fk = (fklu)t, k = 1,2, ... , satisfies (2.1). By Lemmas 2.4.2 and 2.4.3 we have: F(t,y) = n{OeF{t,y) : c > O} 2 n{ Gk(t, y): k = 1,2, ... }. Since the inverse inclusion is obvious, F(t, y) = n{Gk(t, y): k = 1,2, ... }. By Lemma 2.2 D(F) = n{D{G k ) : k = 1,2, ... }. By Lemmas 2.3 and 2.4 we obtain the convergence in U of the sequence {D(fk, U): k = 1,2, ... } to the space D(F). The proved convergence D(fk, U) -+ D(F) (in U), our remarks, and Theorem 1.1 imply: Theorem 2.1. Let FE Qd(rp)U. Then D(F) E Rcekn(U), Now give several simple examples in order to familiarize ourselves with the scheme of the using of Theorem 2.1. Example 2.1. The right hand side of the scalar equation I

y=

sin(t + y)

vm

is defined on the entire plane except the line t = O. The manner in which

we define it on this line does not affect on the notion of a solution of this equation. All changes are possible only for the unique value of the argument t = O. A solution need satisfy our equation almost everywhere only (see §6.2). Under such a change a solution remains a solution and new solutions do not appear. For instance, we can extend the right hand side of the equation to a function I(t, y) by putting 1(0, y) = 0 for every y E R. The function 1 satisfies the Caratheodory conditions. As a majorant 'P here we can take the function for t -=I 0 for t = O. The function 'PI is locally Lebesgue integrable, because almost everywhere (more precisely except the point t = 0) it is the derivative of the non-

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decreasing function w(t) = 21tl t sgn t, see Lemma 4.9.5. Recall that -I

sgnt

=

{

~

if t < 0, if t = 0, if t > 0.

By Theorem 2.1 the solution space belongs to Rcekn(lR x 1R). Notice that all solutions of this equation are absolutely continuous. This is a direct consequence of Lemma 4.7.6. Example 2.2. Consider now the differential inclusion with the right hand side F(t ) = [sin2 (t + y) sin2 (t + y)].

JItT

2J1tT'

,y

As in the previous example the right hand side is not defined for t = 0. Define it for t = by putting F(t, y) = {o}. We obtain the differential inclusion with the same solution set, see the analogous reasoning in Example 2.1. The function
°

for t

# 0,

for t = 0, as the function f of Davy conditions. Thus the solution space of the differential inclusion y' E F(t, y), or (with the other notation) the solution space of the inequality

belongs to Rcekn(IR x 1R). As in the previous example all solutions of our inclusion (respectively, inequality) are absolutely continuous. Example 2.3. Give an example of a simplest situation which cannot be investigated completely by the application of Theorem 2.1. The space of Example 1.1 is the solution space of the differential equation with the right hand side

{ {l,O} f(t,x,y)=

{-1,0} {O,-I}

{O,I}

for for for for

x ~ 0, x < 0, y > 0, y < 0,

-x ~ y ~ x, x ~ y ~ -x, -y < x < -y, y < x < -yo

Peano, Caratheodory and Davy conditions.

259

The equation u' = f(t,u) (here u = (x,y» satisfies all Caratheodory conditions, except the condition of the continuity of the right hand side in u. This difference with hypotheses of the theorem leads to the non-fulfilment of the Kneser condition. Pass now to the continuity of the dependence of solutions on parameters of the right hand side. Since the question about the uniqueness of solutions we leave apart, in view of Theorem 2.5.1 it is sufficient to study the question about the convergence of sequences of solution spaces. In fact, the largest part of this investigation is done in §7.1 and it remains to compare the obtained results with the Caratheodory and Davy conditions. Lemma 2.5. Let F : U - t IR n be a multi-valued mapping with compact convex values. Let F(t, y) ~ 0,(0, cp(t» for every point (t, y) E U and the mapping Ft be upper semicontinuous. Then the mapping F is weakly continuous with respect to the space Cs(U). Proof. Let [a, bJ be a segment of the real line. Let functions z, Zk E C.(U), k = 1,2, ... , be defined on the segment [a, bJ and Zk - t z. Let elements of a sequence {ak: k = 1,2, ... } be Denjoy integrable functions defined on the segment [a, bJ and satisfy for almost all t E [a, bJ the condition D:k(t) E F(t, Zk, (t». Let the sequence of the functions

{ A.(t)

~

j

a.(s)ds: k

~ 1,2 ... }

converge uniformly to a function A. All functions A k, k = 1,2, are solutions of the inequality Ily'(t)1I ~ cp(t). By Lemma 6.3.2 the function A solves this inequality too. By our definitions and Lemma 6.3.3 for every point t from a subset M of full measure of the segment [a, bJ we have ak(t) = A~(t) E F(t, Zk(t» for of all k = 1,2, ... , the derivative A'(t) exists and (2.7)

A'(t)En{cc{A~(t):

k=j,j+1, ... }: j=1,2, ... }= =n{Cc{ak(t): k=j,j+1, ... }: j=1,2}.

For every neighborhood V of the point z{t) we have Zk{t) E V, beginning with some k. Theorem 2.5.2 and (2.7) imply the belonging A'(t) E cc(Ft(V». By virtue of the upper semicontinuity of the mapping F t and Theorem 2.5.1 the conditions ak(t) E F{t, Zk{t» and Zk{t) - t z{t) imply the nonemptiness ofthe set F{t, z{t». Remarks of §6.4 imply now the inclusion F(t,z{t» ;2 n{Ft{V): V is a neighborhood of the point z{t) in Ud 3 A'{t). The lemma is proved. •

CHAPTER 8

260

Lemma 2.6. Let hypotheses of Lemma 2.5 hold. Let a sequence of multivalued mappings Fi : U --t IRn, i = 1,2, ... , converge in U with respect to a space Z ~ Gs(U) to a mapping F. Then {D(Fi) n Z: i = 1,2, ... } E s(U). Proof. Let K be an arbitrary compact subset of the set U. Let Zi E D(Fi' K) n Z for i = 1,2, ... and {'Pi: i = 1,2, ... } be a sequence of Lebesgue integrable positive functions such that

Fi(t, Zk(t)) ~ O(F(t, Zi(t)), 'Pi(t)) and

J

< bZ,K(F,Fi ) +

'Pi(s)ds

t·

7r(Zi)

If now 0:, {3 E 1f(Zi), i = 1,2, ... , and 0: ~ {3, then (3

II Z i({3) - zi(o:)11

JIlz~(s)llds ~J +J ~J + ~

a

(3

(2.8)

(3

'P(s)ds

'PI (s)ds

(3

bZ,K(F,Fi ) + Iii.

'P(s)ds

a

For arbitrary c > 0 take an indexj = 1,2, ... such that bZ,K(F, Fi )+l/i < ~ for i = i + 1, i + 2, ... (it may be done by virtue of the convergence of the sequence {Fi: i = 1,2, ... } to F in U with respect to Z) and a number b > 0 such that

J

'P(s)ds <

M

~,

J

'Pi(s)ds

<

~

M

for i = 1, ... ,j and for every segment M ~ 1f(Zi), i = 1, ... ,j, of the length < b (it may be done by virtue of the absolute continuity of the Lebesgue integral). If now 0:, {3 E 1f(z;), i = 1,2 ... , and 10: - {31 < b, then Ilz({3) - z(o:)11 ~ c by (2.8). It means the equicontinuity of the sequence {Zi : i = 1,2, ... }. Hence {D(Fi) n Z : i = 1,2, ... } E s(U). The lemma is proved. • Lemmas 2.5 and 2.6 and Theorem 7.1.2 imply immediately: Theorem 2.2. Let F E Qd(cp)(U) n Q/(U). Let a sequence of multivalued mappings Fk : U --t IRn, k = 1,2, ... , converge with respect to the space Gs(U) to the mapping F. Then the sequence {D(Fk): k = 1,2, ... } converges in U to the space D(F). •

Peano, Caratheodory and Davy conditions.

261

By Theorem 4.13.3 every single valued mapping cP E Qd(cp)(U) belongs to the set Q/(U). So: Theorem 2.3. Let a single valued function f : U ~ ]Rn belong to Qd(CP)(U). Let a sequence of multi-valued mappings Fk : U ~ IRn, k = 1,2, ... , converge in U with respect to the space Cs(U) to the function f. Then the sequence {D(Fk): k = 1,2, ... } converges in U to the space D(f). • Remark 2.1. To prove the convergence of a sequence of multi-valued mappings Fk : U ~ IRn, k = 1,2, ... , to a mapping F : U ~ Rn we have a simple sufficient condition. Let positive Lebesgue integrable functions CPk, k = 1,2, ... , be defined on a segment [a, b]' U ~ (a, b) x Rn, Fk(t, y) ~ Of(F(t, y), IPk(t») for every (t, y) E U and k = 1,2, .... Let IPk(s)ds ~ 0 as k ~ 00. Then the sequence {Fk: k = 1,2 ... } converges in U with respect to C s (U) to the mapping F. Proof of this fact is obvious, see the definitions in §7.1. Many assertions about the continuity of the dependence of solutions on parameters may be obtained as partiCular case of these theorems. Give an example of their specific use. Example 2.4. The equation

J:

, y=

sin(t + y)

Jltf

is considered in Example 2.1. In an analogous way we may investigate the equation , _ sin(t + y) I la yJltf +t, where -1 < Q < 1. Theorem 2.3 implies the continuity of the dependence of solutions on the parameter Q. Use Theorem 2.3 and consider the function 1/JaIJ(t) = Iiti a - ItlIJI to estimate the deflection of the right hand side (see the function CPk in Remark 2.1). We obtain the needed estimate from the Convergence 1

f

-1

f Isa 1

1/JaIJ(s)ds

=2

0

sIJlds

=

21_1_ -_1_1 ~ Q+1 .8+1

0

as .8 ~ Q. The same calculations allow to obtain an analogous result for the equation ,_ sin(t+y) Ilah( ) y Jltf + t t, Y , where h( t, y) is an arbitrary continuous bounded function.

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Se also Theorem XII.2.3 below.

3. Localization principle The investigation of properties of solution spaces may be essentially simplified by a possibility to do it locally with a further carryover of obtained results to the entire domain of the right hand side. A possibility of such a carryover is based on Lemmas 6.6.2 (for (e», 6.6.2 (for (u», 1.3 (for (k) under the fulfilment of (c) and (e)), on Theorem 7.6.4 (for the convergence of space sequences) and on its Corollary (for (c». For instance, this scheme of reasoning was used in the proof of Theorem 2.1. Point a corresponding result for condition (n). We have in mind its other applications too. The following its a little more general statement will be convenient. Let U be an open subset of the product IR x IRn. Lemma 3.1. Let a space Z E R(U) consist of generalized absolutely continuous functions. Let 'Y be a family of open subsets of the set U, for every element V of which Zv E Rn(V). Let J.Lz(U \ (U'Y» = O. Then Z E Rn(U). Proof. Let Zk E Z, 7r(Zk) = [a,b], k = 1,2 .... Let the sequence {Zk: k = 1, 2 ... } converge uniformly to a function Z E Z. Denote by v the family of all segments lying in [a, b] and satisfying the condition Gr(zlr) ~ V for some V E 'Y.

The hypotheses of the lemma imply that the measure of the set Mo = [a, b] \ (U{ (I) : I E v}) is equal to zero. By Theorem 1.6.10 the family v contains an (at most) countable subfamily Vo = {II, 12, ... } with U{(I): I E v} = U{(I): IE vol. For every j = 1,2, ... we can associate a subset Mi of measure zero of the segment I j such that for every point t E Ii \ M j all derivatives z'(t), z~(t), k = 1,2, exist and satisfy condition (6.3.1). By Theorem 4.1.2 the measure of the set M = U{ Mi : j = 0, 1, 2, ... } is equal to zero. At every point t E [a, b] \ M condition (6.3.1). holds. The lemma is proved. • In framework of the purpose of this section notice: Corollary. Let Z E R(U). Let 'Y be an open cover of the set U, for every element V of which Zv E Rn(V). Then Z E Rn(U). (The estimate J.L(U \ (U'Y)) = 0 is obvious. The generalized absolute continuity of elements of Z follows easily from Lemmas 4.9.1 and 7.5.1: in the last one for Z E Z we need take as 'Y the family of all connected open subsets of the segment 7r(z), on which the function z is generalized • absolutely continuous.) Look now what the idea of the localization of the investigation gives in particular cases.

Peano, Caratheodory and Davy conditions.

263

Example 3.1. Let U ~ (a, b) x Rn, where -00 ~ a < b ~ 00. Let functions aj, j = 1, ... , d, be locally Lebesgue integrable on the interval (a, b). Let functions h : U --t ~n, j = 1, ... ,d, be continuous. Let

Consider the equation

By remarks of §2.1 for every t E ~ the function It is continuous. By Theorem 4.2.6 for every Y E ~ the function IY is measurable. We can use Theorem 2.1 if we present an integrable majorant of the right hand side. Under our assumptions this turns out to be impossible. However Theorem 2.1 is applicable locally. Our general remarks about the localization principle allow to complete the investigation. In fact every point x of the set U possesses a neighborhood V with the compact closure lying in U (see Lemma 2.4.2). By Remark 2.3.4 for every j = 1, ... , d and for mj = sup{lI/j(t, Y)II: (t, y) E [V]} we have mj < 00. By Theorem 4.7.1 and remarks of Example 4.7.7 the function cp(t) = mI!al(t)1 + ... + mdlad(t)1 (defined on (a,b)) is locally Lebesgue integrable. It majorizes the function I on the compactum [V] and all the more on the set V: 11/(t, y)1I ~ cp(t) for (t, y) E V. Naturally, remarks about the function I remain true for the function 9 = I Iv: for every t E ~ the function gt is continuous and for every y E ~n the function gY is measurable. The function 9 has a locally Lebesgue integrable majorant. Thus the function 9 satisfies the Caratheodory conditions. By Theorem 2.1 the space D(J, V) = D(g) belongs to Rcekn(V), Such a neighborhood V may be pointed for every x E U. Therefore D(J) E Rcekn(U), Example 3.2. Let under assumptions of Example 3.1 the functions Ii, j = 1, ... , d, have continuous partial derivatives in y. For arbitrary (to, Yo) E U find c > 0 such that [to - c, to + c] x [OEYO] ~ U. Let K

=

[to - c, to

+ c]

x [OEYO] and M

= SUP{II~(t,Y)II:

(t,y) E K,

n}.

By Remark 2.3.4 M < 00. The function CPl(t) = Mn(lal(t)1 + ... + lad(t)l) is locally Lebesgue integrable (see Theorem 4.7.1 and Example 4.7.7). By Lemma 6.5.4

j == 1, ... , d, k = 1, ... ,

II/(t,p) - f(t,q)1I ~ Mn(l a l(t)1

+ ... + lad(t)I)lIp - qll =

cpdt) lip -

qll

for t E (to - c, to + c) and p, q E OeYo. Thus on the set V = (to - c, to + c) X OEYO the right hand side of the equation y' = I(t, y) satisfies condition b) of §6.5. Our new set V satisfies

264

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all conditions imposed on the set V in Example 3.1. Thus on the set V the right hand side of the equation under consideration satisfies all hypotheses of Theorem 6.5.1. So D(f, V) E Rceun(V). We can point such a neighborhood V for every point of the set U. Remarks and lemmas of this section imply that D(f) E Rceun(U). Example 3.3. Let real functions a and (3 be defined and be Lebesgue integrable on the interval (a, b), -00 ~ a < b ~ 00. Evidently the right hand side of the equation y' = a(t)y + (3(t) satisfies all condition imposed in Example 3.2 (with d = 2, f1(t,y) == y and f2(t,y) == 1). Thus the solution space of this equation belongs to Rceun(V), where V = (a, b) x ]Rn. Let to E (a, b) and A(t) = It: a(s )ds. A direct verification can show that the formula (3.1)

y(t) =

1

(YO +

e- A(')/3(S)dS) eA('1

defines a solution of our equation with the initial value y(t o) = Yo. By virtue of condition (u) this is the unique solution with this initial value. Notice a particularity of the formula (3.1): it defines the function y(t) on the entire interval (a, b). Let now z be an arbitrary solution of our equation and t1 E 71'(z). The function Zl (t)

= ( Z (t) 1 +

f.' a(u)du j e-.r a(u)du(3( s)d) se t

'I

'I

tt

is defined on the interval (a, b) and is also a solution of the equation under . consideration with the initial value zl(td = z(td. Therefore by virtue of condition (u) it coincides with z on 71'( z). Since to E 71'( Zl) (and Zl is a solution), for some value Yo the function Zl is defined by (3.1). Thus the formula (3.1) represents all solutions of our equation. Expressions of the type
remark.

Peano, Caratheodory and Davy conditions.

265

Let a non-negative function 0, e E IRn and G(t) = {u: u E IRn , (u -
= 0/(0,
for t E (a, b). Show that every solution z of the differential inclusion z' (t) E H (t) is absolutely continuous. Let 7r(z) = [e, dj and M be a set of all points t E [e, dj, at which the derivative z'(t) exists and Ilz'(t) -
Lebesgue integrable. By Lemma 4.9.2 the function t

z*(t) = z(t) -

J

'l/J1(s)ds

c

is generalized absolutely continuous on [e, dj and

(z*)'(t) = z'(t) - z'{t)

+

(almost always) for t E M and (z*)'(t) = z'(t) E H(t) for t rt M. Since (z*)'(t) rt 0/(
~

.!.'l/J2(t) a

Lemma 4.7.6 implies the Lebesgue integrability of the function (z*)'(t) -
J t

z.... (t)

= z"(t) -


c

CHAPTER 8

266

is absolutely continuous. By Theorem 4.7.1 this implies the absolute continuity of the function z* (t) and the absolute continuity of the function t

z(t) = z*(t)

+

J

'1Pt (s)ds,

c

which gives what was required. Evidently Remark 4.1. If U is an open subset of the product ~ x ~n, Z E R(U) and, is an open cover of the set U, for every element V of which the set Zv consists of absolutely continuous functions, then the set Z consists of absolutely continuous functions too. 5. Continuously differentiable solutions. Peano and Picard conditions In this section we speak mainly about solutions of a differential equation y' = f(t, y) (with the right hand side f defined on an open subset U of the product ~ x ~n). Lemma 5.1. Let z E D(f), 7l'(z) = [a, b], where a < b. Let a function

fIGr(z) be continuous. Then for every t E [a, b] the derivative z'(t) exists and is equal to f(t, z(t)). We speak here not about the approximate derivative. We speak here about the derivative in the common sense and as a consequence we obtain the continuity of the derivative. Proof. Let M be a subset of full measure of the segment [a, b]. Let the approximate derivative z'(s) be defined and satisfy the equality z'(s) == f(s, z(s)) for every s E M. Take an arbitrary point to of the segment [a, b]. By Theorem 4.10.1

z(q) - z(P) E cc(cp(M n [p, q])) q-p

for every p < q, a ~ p ~ to ~ q ~ b. Pass here to the limit as p, q - t. Lemma 2.5.2 and the continuity of the function fIGr(z) imply the existence of the limit of the quotient in the left-hand side (i.e., the existence of the usual derivative z'(t)) and its equality to f(t, z(t)). The lemma is proved .• The proved lemma implies immediately that under the assumption of the continuity of the right hand side (this is just the Peano condition) every solution of our differential equation is continuously differentiable. By remarks of Example 3.1 in this case the solution space satisfies conditions

Peano, Caratheodory and Davy conditions.

267

(c), (e), (k) and (n). Under these assumptions the last condition is fulfilled in the following reinforced version: (n*) all functions from (the solution space of the equation under consideration) Z are continuously differentiable and if Zk E Z, 7r(Zk) = I, k = 1,2, ... , and the sequence {Zk: k = 1,2, ... } converges uniformly to a function Z E Z, then the sequence {z~: k = 1,2, ... } converges uniformly to the function z'.

(As in (n*) we speak about the convergence ofthe sequence offunctions I(t, Zk(t)), then we can deduce (n*) from Theorems 3.3.1 and 3.5.3.) The Picard condition consists in the requirement of the continuity of the right hand side I of the equation under consideration and of the existence of a number M ;;:: 0 such that II/(t, Yl) - I(t, Y2)11 ~ MllYl - Y211 for every two points (t, Yl) and (t, Y2) (with coinciding first coordinates) of the set U. Properties of equations satisfying the Peano condition, naturally, remain true for equations satisfying the Picard condition. In addition the second part of condition b of §5.5 is fulfilled too. Therefore the solution space of the equation with the Picard condition satisfies condition (u). This remain true for equations with right hand sides locally satisfying the Picard condition (see §3). In fact the requirement of the continuity of the right hand side for the using of Lemma 5.1 is too strong. We may point out other cases of its application. Let x = (t, y) E U and Z ~ Cs(U). The set G ~ U is called the left Z -neighborhood (respectively, right Z -neighborhood, Z -neighborhood) of the point x, if it contains the point x, the set G \ {x} is open with respect to the Euclidean topology and for every function Z E Z, Gr(z) 3 x, there exists a number {j > 0 such that Gr(zl ... (z)n[t_6,t)) ~ G (respectively, Gr(zl ... (z)n[t,t+6)) ~ G, Gr(zl7r(z)n[t-6,t+6)) ~ G)). The set GEx (where c > 0) is a Z-neighborhood of the point x for every Z ~ Cs(U). Every Z-neighborhood of the point x is its left and right Z-neighborhood. If G 1 is a left and G 2 is a right Z-neighborhoods of the point x, then the set G 1 U G 2 is a Z-neighborhood of the point x. The set of all Z-neighborhoods of the point x is directed to the point :t (see §1.2). The topology rZ generated by the system of neighborhoods x E U} coincides on the graph of every function from Z with the Euclidean topology. Therefore Lemma 5.1 implies Theorem 5.1. Let a (single valued) function f : U --+ ]Rn be continuous with respect to the topology rD(f). Then for every function Z E Z (with the domain containing more than one point) at every point t E 7r(z) the (Usual) derivative z'(t) exists, is equal to f(t, z(t» and the function z'(t) is COntinuous. • z~(t) =

r;

{r;:

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CHAPTER 8

The direct using of Theorem 5.1 needs a full description of the topology In order to obtain the description in most cases we need know well properties of solutions. Already the situation may be easily corrected by the following remark. TD(f).

Let a topology T be a part of the topology TD(f) (i.e., T ~ TD(f)). If a function is continuous with respect to the topology T then it is continuous with respect to the topology TD(f) too. Therefore to apply Theorem 5.1 it is sufficient to point a topology or a system of neighborhoods of points of U, which estimates properties of the right hand side of the equation in a needed way. A left contingent cont,(Z,x) (respectively, right contingent contr(Z,x), contingent cont(Z, x)) of the space Z ~ Cs(U) at a point x = (t, y) E U is the set

.

hm top sup 0--+0

{z(s) - z(t) S -

t

: z E Z, s, t E 7l'(z),

S

E (t - e, t), z(t) = Y

}

(respectively, . {z(s) - z(t) : zEZ, s,tE7l'(Z), SE(t,t+c), z(t)=y } , 11m top sup 0--+0 s- t

.

hm top sup 0--+0

{z(s) - z(t) S -

t

s E (t - c, t) U (t, t

: z E Z, s,t E 7l'(z),

+ c),

z(t) = y} (= cont,(Z, x) U contr(Z, x))).

Let F : U -+ ]R.n be an arbitrary multi-valued mapping, G be a left D(F)-neighborhood of a point x E U. Theorem 4.10.1 implies that cont,(D(F), x) ~ cc(F(G)). By Theorems 2.4.1 and 1.7.2 we obtain: Theorem 5.2. Let F : U -+ ]R.n be a multi-valued mapping. Let for

a left D(F)-neighborhood G of a point x = (t,y) E U the set F(G) be bounded. Then for every neighborhood V of the set cont,(D(F), x) and for every c > 0 the set G-(V, c) = {x} U {(t*,y*): t* E (t - e,t), y* E Uto, ~::::r E V} is a left D(F)-neighborhood of the point x. • Analogous remarks and theorems are true for right D(f)-neighborhoods, right contingents and the set G+(V, c) = {x} U {(t*, y*): t* E (t, t + c), y* E Uto, ~::::r E V} and for D(F)-neighborhoods, contingents and the set

G(V, c) = G-(V, c) U G+(V, c). Example 5.1. Consider the scalar differential equation I

Y

=

y2 t

2

+y 2

+ t.

Peano, Caratheodory and Davy conditions.

269

Its right hand side is defined on the entire plane except the point (0,0). The solution space Z of the equation does not depend on the definition of the right hand side at the point (0,0). Define the right hand side at the point by putting 1(0,0) = O. The function 1 is continuous on the entire plane except the point (0,0). In order to use Theorem 5.1 in investigation of our equation we will establish the continuity of the function 1 with respect to the topology rZ. A crude estimate of the right hand side gives the inequality ly'(t)l:::;; 1 + Itl. Therefore cont(Z,(O,O)) ~ [-l,lJ. Let a = inf{a: a;?: 0, cont(Z,(O,O)) ~ [-a, a]}. Evidently cont(Z,(O,O)) ~ [-a,aJ. Show that a = 0. Assume the opposite. Take an arbitrary 6 > 0 and apply Theorem 5.2 to the set V = (-a - 6, a + 6). For every point (t, y) (f (0,0)) of the set G(V, c) we have

Ity I :::;;a+6,

2

y2

t ;?: (a+6)2' ly'(t)l:::;;x(6,c),

1)) (a+<5)2 h were x ( u, c = 1+(a+6)2 + c. The last estimate implies that cont(Z, (0, 0)) this is true for every 6 > 0 and c > 0,

~

a2

a2

[-x(6, c), x(6, c)J. Since

cont(Z,(O,O))~ [ -1+a 2 '1+a 2

] •

Since 0 < a :::;; 1, then 1~:2 < a2 :::;; a. The obtained inclusion contradicts the definition of the number a. Thus our assumption is false and hence a = O. By Theorem 5.2 for 6 > 0 and c > the set G(( -6,6), c), is a Z-neighborhood of the point (0,0). For every point (t, y) of this set we have II(t,y)1 :::;; 1!~2 + c. This implies the continuity of the function 1 with respect to the topology rZ at the point (0,0). By remarks of the first paragraph of our reasoning we have the continuity on the entire plane. By Theorem 5.1 all solutions of our equation are continuously differentiable.

°

6. General approach to existence theorems

~

This section does not contain new essential information in comparison with the accounted material and its contents has a character of a comment. But this comment cost to give it. In textbooks on the theory of Ordinary Differential Equations authors pay more attention to theorems on the existence of solutions of the Cauchy problem than to theorems on the continuity of the dependence of solutions on parameters. Psychologically this is easy to understand: when an existence theorem is not proved then what is the continuity we must discuss?

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270

Continuity theorems play the role of a comment to existence theorems. The existence theorems take a central place. However, contrary to a continuing belief in an absolutely greater importance of the existence in comparison with the continuity, in reality the existence is a secondary property. We establish it easily each time when a corresponding result on the continuity in the form of the convergence of corresponding sequences of solution spaces is obtained. When the convergence result is proved it remains to use Theorem 7.2.1. We had such situations above. As a simple exercise the reader can deduce the Peano theorem from the Picard and Weierstrass theorems. Let U be an open subset of the product ~ x ~n. For an equation with the continuous right hand side the Peano theorem and remarks of §2 and of Example 6.4.1 give the following result. Theorem 6.1. Let a mapping f : U --+ ~n be continuous. Then D(J) E Rcek(U), Following completely just mentioned scheme remarks of §2 expand the class of equations and inclusion considered in Theorem 6.1 to the class of differential inclusions y' E F(t, y) with right hand sides F E Qd(
Pk(X)=

o k(P(x,E)-~) 1

for p(x, E) < for

1

k~

1

k'

p(x,E) ~

for p(x, E)

2

k'

2

> k'

The function gk(t, y) = Pk(t, y)f(t, y) is continuous. By Theorem 6.1 D(gk) E Rcek(V), The convergence D(gk, U) --+ D(J, U) is obvious: for ev~ ery point x E U there exist its neighborhood Ox and an indeX ko = 1,2, ... , such that gk lox = fl ox for k = ko, ko + 1, ko + 2, .... NoW

Peano, Caratheodory and Davy conditions. Theorem 6.6.2 implies the convergence D(9k) D(f) E Rcek(V).

---+

271

D(f). By Theorem 2.1

7. Majorants of right hand sides The proof of Theorem 2.1 used properties of the solution space of the inequality Ily'(t)11 ~ cp(t) investigated in §6.3. In this section we will try to expand the list of differential inclusions, which may be used as majorants in results as Theorem 2.1 (now this list consists only of the mentioned inequality). We will try to understand how we may prove that the solution space of a differential inclusion possesses the corresponding properties. As before U denotes an open subset of the product ]R x ]Rn. Denote by Q**(U) the set of all multi-valued mappings F E Q*(U) (see §6.4) satisfying the condition: for every y E ]Rn the mapping FY is measurable. The last condition is near to condition a) of §2 (see Theorem 4.13.3). The condition FE Q*(U) is near to condition b) of §2 (see remarks of §5.4). Denote by Q;*(U) the set of all mappings F E Q**(U) with compact values. For (arbitrary) multi-valued mapping G : U ---+ ]Rn denote by Q*c (U) the set of all multi-valued mappings F E Q*(U) satisfying the conditions:

(7.1) F(x)

~

G(x) for every x E Uj

(7.2) if x E U and G(x)

1= 0,

then F(x)

1= 0.

Put Q**C(U) = Q*C(U) n Q**(U), Q;*C(U) = Q;*(U) n Q**C(U). When we estimate the relation between F and G in the last definitions we can word condition (7.1) in the following way: the mapping G majorizes (is a majorant of) F. Following the way to our aim, select a set of all suitable majorants and introduce the set Qme(U) of all multi-valued mappings G : U ---+ ]Rn satisfying the conditions

(7.4) D(F) E Re(U) for every mapping FE Q;*C(U). With this notations the problem consists in the proof of the membership G E Qme(U). We will need the following auxiliary assertion. Lemma 7.1. Let -00 ~ a < b ~ 00, U ~ (a,b) x ]Rn, G E Q*(U). Let Ip : (a,b) ---+ (0,00). Let H(t,y) = O,(G(t,y),cp(t» for (t,y) E U. Then II E Q*(U).

CHAPTER 8

272

Proof. Let (t, y) E U. For every neighborhood V of the point y in Ut and for every c > 0 we have

Ht(V) = U{O,(G(t,y*),cp(t)): y* E V} ~ U{O(G(t,y*),cp(t) +c): y* E V} = O(Gt(V), cp(t) + c), c(Ht(V)) ~ c(O(Gt(V), cp(t) + c)) ~ O(c(Gt(V)), cp(t) + c) ~ O(cc(Gt(V)), cp(t) + c) (see Lemma 2.4.3),

If a point p belongs to the set n{ cc(Ht(V)): V is a neighborhood of the point y} then for every neighborhood Oy of the point y the set

O(p, cp(t)

+ 2c) n cc(Gt(Oy))

~

O,(p, cp(t)

+ 2c) n cc(Gt(Oy))

is nonempty. Therefore, the family {O,(p, cp(t) + 2c) n cc(Gt(V)): V is a neighborhood of the point y} of compact subsets of the set 0, (p, cp( t) + 2E) has finite intersection property. By Theorem 1.6.3 its intersection Me is nonempty. By the same arguments the set

M

= n{Me:

c> O} ~ n{O,(p, cp(t)

+ 2c) : c >

O}

= O,(p, cp(t))

is nonempty too (it lies in the compactum n{cc(Gt(V)): V is a neighborhood of the point y} ~ Gt(y) ). Hence p E Ht(y). In view of the arbitrariness in the choice of the point p this means, that n{cc(Ht(V)): V is a neighborhood of the point y} ~ Ht(y). In view of the arbitrariness in the choice • of the point (t, y) E U this implies our assertion. Lemma 7.2. Let under hypotheses of Lemma 7.1 G E Q**(U) and the function cp be measurable. Then H E Q**(U). Proof. The proof reduces to referring to Lemmas 7.1 and 4.12.3. • Lemma 7.3. Let F1,F2 E Q*(U). Let H(t,y) = F1(t,y) n F2 (t,y) for (t, y) E U. Then H E Q*(U). Proof. Let (t, y) E U. For i = 1,2 we have n{ cc{Ht(V)): V is a neighborhood of the point y} ~ n{CC{{Fi)t{V)): V is a neighborhood of the point y} ~ Fi(t, y). Therefore n{ cc({H)t{V)): V is a neighborhood of the point y} ~ Fl (t, y) n F2 {t, y) = H{t, y). In view of the arbitrariness in the choice of the point (t, y) E U this • implies our assertion.

273

Peano, Caratheodory and Davy conditions.

Lemma 7.4. Let under hypotheses of Lemma 7.3 F I , F2 Then H E Q**(U).

E

Q**(U).

Proof. The proof reduces to referring to Lemmas 7.3 and 4.12.4. • Lemma 7.S. Let G E Q**(U), D(G) E Rcn(U), Go,F E Q**G(U), Q;*Go(U) i= 0. Let for every point x E U either F(x) = G(x), or the set F(x) be nonempty and compact. For r > 0 define the mapping G r : U ---7 ~n, (7.5)

Then D(F) E [U{D(A): A E Q;*Gr(U)}: r ~ O}]Rc(U) n Rcn(U). Proof. I. Since D(F,K) = D(F) nD(G,K) for every compactum K ~ U, by Theorem 6.4.2 and Lemma 6.4.4 D(F) E Rc(U). The membership D(F) E Rn(U) is obvious. Thus the problem consists in the proof of the membership D(F) E [U{D(A): A E Q;*Gr(U)} : r ~ O}]Rc(U). Do it.

II. Fix a function A E Q;*Go(U). For x E U and k = 1,2, ... put n Of(A(x), k). Remark 2.3.3 and Theorem 2.4.1 imply the compactness of values of the mapping Ak : U ---7 ~n. By Lemmas 7.2 and 7.4 Ak E Q;*Gk(U). III. Put Mk = {x: x E U, F(x) U A(x) ~ 0(0, k), G(x) i= 0} for k = 1,2, .... By Theorem 4.12.1 for every y E ~n the set (Mk)Y is measurable. Theorems 2.5.2 and 1.7.5 imply that for every t E ~ the set (Mk)t is open (in the space ~n). For x E Mk the triangle inequality implies that F(x) ~ A2k(X). Therefore the mapping Bk : U ---7 ~n,

Ak(X) = G(x)

for x E M k , for x E U\ M k , belongs to Q~*G2k (U). IV. Show that the sequence a = {D(Bk): k = 1,2, ... } converges in U to the space D(F). Since all elements of the sequence a lie in D(G), the condition D(G) E Rc(U) implies the membership a E s(U). By Lemma 7.1.4 our aim will be achieved when we prove the inclusion llin top SUPk-+oo D(Bk ) ~ D(F). Let Z E lim topsUPk-+oo D(Bk). If 7r(z) consists of one point only, then Z E D(F) by definitions. Let the domain [a, b] of Z contain more than one point (i.e., a < b). By Lemma 1.5.1 we can choose functions Zj E D(BkJ for j = 1,2 ... (where kl < k2 < k3 < ... ) such that the sequence {Zj : i = 1,2, ... } converges to z. Let I be an arbitrary segment lying in (a, b). By virtue of the continuity of the mapping 7r we have I ~ 7r(Zj), beginning with some j = jo. By the Condition D(G) E Rn(U) for every point t of a subset P of full measure of

274

CHAPTER 8

the I the derivatives z'(t) and zj(t), j = jo,jo + l,jo + 2, ... , exist, belong to the sets G(t, z(t)) and G(t, Zj(t)), respectively, and z'(t) E n{ cc{ zj(t): j = j* ,j* + 1, ... } : j* = jo,jo + 1, ... }. For t E P we have two possibilities. 1. F(t, z(t)) = G(t, z(t)). Then z'(t) E F(t, z(t)) by the choice of the set P. 2. The set F(t, z(t)) is nonempty and compact. In this case (see Lemma 2.3.6) for some ko = 1,2, ... the point z(t) belongs to the open subset V = (Mko)t of the space ~n and V ~ (Mdt for k = ko + 1, ko + 2, .... So Bk; (t, y) = F(t, y) for y E V and kj ~ ko. The convergence Zj(t) --t z(t) and the condition F E Q*(U) imply that

z'{t) En {cC{{U{Bk;(t,Zj(t)) : j =j*,j* + 1,j* + 2, ... , kj ~ F{t, z(t)). We have proved that Z

ZII

~

ko} : j* = 1,2, ... }

E D(F). By Corollary of Lemma 6.2.1

E D(F).

V. Referring to Theorem 7.2.1 completes the proof. • A mapping Go E Q**G(U) is called a sufficient me-approximation of a mapping G : U --t ~n, if for every r > 0 the mapping G T : U --t ~n defined by (7.5) belongs to Qme(U). Theorem 7.1. Let a mapping G E Q**{U) possess a sufficient me-approximation Go and D(G) E Rcn(U). Let F E Q**G(U). Let for x E U either F(x) = G(x), or the set F(x) be nonempty and compact. Then D{F) E Rcen(U). Proof. The proof reduces to referring to Lemmas 7.5 and 7.7.2 and Theorem 7.7.1 and 7.2.1. • Corollary. Under hypotheses of Theorem 7.1 G E Qme(U) and D(G) E Rcen(U). • Denote by Qmek(U) the set of all multi-valued mappings G E Qme(U) satisfying the condition

(7.6) D(F) E Rk(U) for every mapping F E Q;*G(U). A mapping Go E Q**G(U) is called a sufficient mek-approximation of a mapping G : U --t ~n, if for every r > 0 the mapping G T : U --t ~n defined by (7.5) belongs to Qmek(U), Theorem 7.2. Let under hypotheses of Theorem 7.1 the mapping Go be a sufficient mek-approximation of the mapping G. Then D(F) E Rcekn(U). Proof. The proof is analogous to the proof of Theorem 7.1. • Corollary. Under hypotheses of Theorem 7.2 G E Qmek(U) and D(G) E Rcenk(U), •

Peano, Caratheodory and Davy conditions.

275

Lemma 7.6. Let -00 :::;; a < b :::;; 00, U ~ (a, b) x IRn, G E Q**(U). Let a function cp : (a, b) --+ (0, (0) be locally Lebesgue integrable. Let the set Go(t, y) = G(t, y) n 0,((5, cp(t)) be nonempty for every point (t, y) E U. Then the mapping Go : U --+ IRn is a sufficient mek-approximation of the mapping G. Proof. By Lemma 7.4 Go E Q**(U). By Theorem 4.12.3 the mapping Go satisfies the Davy conditions (with cp as majorant). Lemmas 7.2 and 7.4 imply that for every r > 0 the mapping G r : U --+ IRn defined by (7.5) satisfies the Davy conditions. This gives what was required. The lemma is proved. • Theorem 7.3. Let under the hypotheses of Lemma 7.6 D(G) E Rcn(U). Then G E Qmek(U). Proof. The proof reduces to referring to Lemma 7.6 and Corollary of • Theorem 7.2. Example 7.1. Let U = IR x R For (t, y) E U denote by G(t, y) the segment [1,1 + Iyl-l], if y =F 0, and the half interval [1,(0), if y = 0. For (t, y) E U put h( t, y) = y. The line y = 0 is finite with respect to the space M1(h, 1) (see §7.4), Ml(h, l) ;2 D(G). By Theorem 2.1 outside this line the space D(G) satisfies conditions (c) and (n). Theorem 7.6.6 and Lemma 7.4.2 implies that D(F) E Rc(U). By Theorem 7.4.3 and Lemma 3.1 D(F) E Rn(U). By Theorem 7.3 G E Qmek(U). Theorem 7.4. Let {G k : k = 1,2, ... } ~ Qme(U) (respectively, ~ Qmek(U)), D(Gd E Rcn(U). Let G1(x) ;2 G 2 (x) ;2 G 3 (x) ;2 ... and k = 1,2, ... } for every point x E U. Then G(x) = n{Gk(x): D(F) E Rcen(U) (respectively, D(F) E Rcekn(U)) for every mapping FE Q;*G(U). Proof. For k = 1,2, . .. and x E U put if F(x) = 0 if F(x) =F 0. Evidently Fk E Q**(U). By Theorem 7.1 D(Fk ) E Rce(U) (respectively, by Theorem 7.2 D(Fk ) E Rcek(U)). Since for every point x E U we have F(x) = n{Fk(X): k = 1,2, ... }, by Lemmas 2.2 and 2.3 the sequence {D(Fk): k = 1,2, ... } converges in U to D(F). By Theorem 7.2.1 (respectively, by Theorem 2.1) D(F) E Rce(U) (respectively, D(F) E Rcek(U)). The fulfilment of condition (n) is obvious. The theorem is proved. • Corollary. Let under hypotheses of Theorem 7.4 Q;*G(U) =F 0. Then G E Qme(U) (respectively, G E Qmek(U)). • Example 7.2. Let U = IR x JR. For (t, y) E U denote by G(t, y) the segment [1 + Iyl-l, 1 + 2Iyl-l] if y =F 0, and the empty set if y = o.

276

CHAPTER 8

For k = 1,2, ... and (t, y) E U denote by Gk(t, y) the segment G(t, y), if Iyl ~ k- 1 , the segment [1 + k, 1 + 2Iyl-l], if 0 < Iyl ~ k- 1 and the half interval [1 + k, 00), if y = O. When we repeat reasonings of Example 7.1 we show that G k E Qmek(U) for every k = 1,2, .... We have G1(x) 2 G 2 (x) 2 G 3 (x) 2 ... and G(x) = n{ Gk(x): k = 1,2, ... } for every point x E U. By Corollary of Theorem 7.4 G E Qmek(U),

CHAPTER 9

COMPARISON THEOREM

Explicit aim of this chapter is to expand results related to differential inequalities. They allow us to compare the behavior of solutions of two ordinary differential equations. The position of such assertions is well known. In comparison with the existing results on this subject we weaken essentially restrictions on the right hand side of the majorizing equation. In particular, we will consider equations with the multi-valued right hand sides (differential inclusions). Here in restrictions on the right hand side we move a little outside the framework of Davy (Caratheodory) conditions. The implicit, but main, aim of this chapter is to show how we can investigate discontinuous right hand sides of equations and inclusions in complicated situations (related here to differential inequalities). The author does not insist that the multidimensional version of the comparison theorem proposed here is the best of possible ones. On the contrary, in comparison with the corresponding theorem for equations with continuous right hand sides the proposed assertion looks cumbersome. It may be that this will stimulate the appearance of a new more 'understandable' assertion on this subject. On the other hand, in the statement of this result we do not go outside the framework of the main general ideas of our approach, and when the reader has any initial experience of insight within this circle of notions the mentioned assertion does not look so terrible. The scheme for the passage from the scalar case to the multidimensional one in the proof of the corresponding classical result (see, for instance, [RHL]) keeps its value for equation with a discontinuous right hand side too. We shall not discuss this question here. On the other hand, this simple method does not work in some situations, because even the addition of a constant term in the right hand side may lead to the necessity of proving anew the presence of basic properties of solution spaces (as (c), (e), etc.). Our method (more cumbersome in the statement and proof of the main result) may turn out to be simpler in the investigation of particular equations. Our case of the comparison theorem for the scalar case also does not represent its final version. Although here we make an essential step in weakening restrictions imposed on the right hand side, we have possibilities of

278

CHAPTER 9

moving further in this direction. But the employment of these possibilities will need further effort (and complications of statement and proofs). 1. Existence of upper solutions in the scalar case

Let U be an open subset of the plane ~ x R Theorem 1.1. Let Z E Rce(U) and (to, Yo) E U. Then there exists a function Zo E Z+ such that to = inf7r(zo), zo(to) = Yo and:

(1.1) if Z E Z, tl = inf 7r(z) E 7r(zo) and z(t 1 ) for every t E 7r(z) n 7r(zo) (Figure 9.1).

~

zo(td, then z(t)

~

zo(t)

Proof. I. There exists a number v > 0 such that

[to - v, to

+ v]

~

7r(z)

for every solution of the Cauchy problem y E Z-+, y(t o) = Yo, see Lemma 6.6.3. II. For t E [to, to + v] put a(t) = sup{z(t): Z E Z,

u i

+-----ro---

+ v], z(t o) = Yo}. tI Show that a E Z. Let {t k : k = 1,2, ... } be a Figure 9.1 dense subset of the segment [to, to + v]. For every k = 1,2,... fix a function Zk E Z such that 7r(Zk) = [to, to + v], Zk(t O) = Yo and zk(td = a(tk). Such a function exists because condition (c) implies that:

7r(z) = [to, to

a(tk) E {z(t k ): Z E Z, 7r(z) = [to, to Construct a sequence of functions {ak: k = 1,2, ... } ~ Z. Put al = Zl· Let functions al, ... , ak-l be pointed. Fix a function ak. By the definition of a( td we have: ak-l(t k ) ~ a(t k). If ak-l (t k) a(t k ), we put ak = ak-l· If ak-l(tk) < a(tk), then the open set M = {t :

t

E

[to, to

+ v],

z(t o) = Yo}.

tk "" -- .... _-_._-+.--=----..

J

+ v], ak-l (t) < Zk(t)}

is a neighborhood of the point t k , see Figure 9.2.

Figure 9.2

279

Comparison theorem.

Let J :3 tk be a connected component of the set M. The function: for for is defined on the segment [to, to

+ v).

t E J,

t E [to, to

+ v) \

J

It belongs to Z and

(1.2) Because of the membership Z E Rc (U) the sequence {ak : k = 1, 2, ... } contains a subsequence converging to a function (3 E Z, 7r({3) = [to, to + v). By (1.2) (3(t k ) = a(tk) for every k = 1,2, .... Show that (3(t) = a(t) for every t E [to, to + v). The definition of a and the membership (3 E Z imply the estimate a(t) ~ (3(t). Thus it remains to obtain the estimate a{t) ~ (3(t). If this is false then there exists a function Z E Z such that z{t) > (3(t). By virtue of the density of the set {tk: k = 1,2, ... } in the segment [to, to + v) we have z(t k ) > (3(t k) for some k = 1,2, .... As we have established above, this is false. Thus (3 = a, which gives what was required. III. The facts proved in I-II may be summed up in the following way: there exist a number v > 0 and a function a E Z such that: (1.3) (1.4)

7r(a)

= [to, to + v) and a(t o) = Yo;

if Z E Z, 7r(z) ~ [to, to + v) and z(inf 7r(z)) ~ a(inf 7r(z)), then z(t) ~ a(t) for every t E 7r(z).

In fact, if the function a E Z from II does not satisfy the condition (1.4) then there is a function Zl E Z such that 7r(zd ~ [to, to some tl E 7r(zd.

+ v),

zl(inf7r(zd) ~ a(inf7r(zd), and zl(td > a(td for

The continuous realfunction 6(t) = a(t)-zl (t) is defined on the segment 7r(zd· It must vanish at a point s on the segment [inf 7r(zd, td. The function

Z2 (t ) -_ {a(t) zdt)

for for

E

[to, s)

t E

t

Is, t l )

belongs to Z, to E 7r(Z2) and Z2(t O ) = a(t o) = Yo. By the definition of v in I the function Z has an extension Z3 E Z defined on the segment [to, to + v]. We have z3(td > a(td, which contradicts the definition of a(td in II and gives what was required.

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CHAPTER 9

IV. If for i = 1, 2 the pair consisting of the function ai (as a) and of [t ]. the number 1/i (as 1/) satisfies (1.3 - 4) and 1/1 ~ 1/2, then al = 0,111 Therefore the condition:

a21

to = inf1r'(z) and for every

1/

< SUP7r(z) the function

zl[to,v]

satisfies

(1.3-4) defines an unique function z E Z+. It satisfies all imposed conditions. The • theorem is proved.

2. Scorza-Dragoni property We will consider a differential inclusion, the right hand side F of which is defined on an open set U ~ ~ X ~n, moreover: (2.1) values of the mapping Fare nonempty compact and convex; (2.2) for every fixed t E

~

the mapping F(t, y) is upper semicontinuous in

y;

(2.3) if a compactum K ~ ~n and a segment I ~ ~ are such that IxK then for every open subset V of the space ~n the set

~

U,

M*(K, V) = {t: tEl, F({t} x K) ~ V}

is measurable. Compare conditions (2.1-3) with remarks of §§4.12-13 and 8.2. We say that a differential inclusion y' E F( t, y) satisfies the ScorzaDragoni condition if for every point x E U it has a neighborhood Ox ~ U and a subset M of the measure zero of the real line such that: if z E D(F, Ox) and t E (7r(z») \ M, then all limit points of the quotient z(t

(2.4) as h

+ h) -

z(t)

h --t

0 belong to F(t, z(t»).

(It seems that the first person who paid attention to properties of this kind was Scorza-Dragoni, [Sc2].) Lemma 2.1. Let conditions (2.1 - 3) hold. Assume that (2.5) there exists an integrable function cp : lR for (t, y) E U and u E F(t, y).

--t

[0,00) such that

lIull

~ cp(t)

281

Comparison theorem.

Then the differential inclusion y' E F(t, y) satisfies the Scorza-Dragoni condition. Proof. 1. Take an arbitrary x E U. Find a segment / ~ R and a bounded open set G ~ Rn such that

x E (/) x G

~

/ x [G] ~ U.

II. Let H be a closed subset of the segment / and X be the characteristic function of the set / \ H. For almost all t E H t+h

J

lim -hI

(2.6)

h--.O

X(s)
t

Let Q be the set of all density points t of the set H satisfying (2.6). Let the mapping FIHX[G] be upper semicontinuous. III. Let with the notation of I, II z E D(F), t E (7r(z») n Q and (t, z{t)) E (/) x G. Take arbitrary 8 > O. By virtue of the upper semicontinuity of the mapping F on the set H x [G] for some neighborhood W ~ (/) x G of the point (t, z(t») we have

F(W S

n (H

x [G])) ~ O/jF(t, z(t».

Let a > 0 be such that (t - a, t E (t - a, t + a). IV. Let hE (0, a). Then

J

=

h1

~

7r(z) and (s, z(s» E W for all

J t+h

Hh

z(t+h)-z(t) h

+ a)

z'(s)ds

h1

=

t

J

'

Hh

1 x(s)z'(s)ds+ h

(I-X(s»z (s)ds.

t

t

By (2.6) t+h

Hh

lJ

X(s)z'(s)ds ::;;

lJ

t

X(s)
-t

0

as

h

-t

O.

t

Take an arbitrary

U

E

f(s) =

F(t, z(t» and consider the function

{~(S)

for for

7r(z) n H, s E 7r(z) \ H. S

E

All its values belong to O/jF(t, z(t». It is Lebesgue integrable. Therefore its mean value Hh

lJ t

f(s)ds

CHAPTER 9

282

belongs to [06F(t, z(t))], see Theorem 4.10.1. In addition,

~

t+h

J t

f(s)ds -

~

Hh

*J

Hh

J

(1 - X(s))z'(s)ds =

t

X(s)ds

-7

0

as

h

-7

0,

t

because t is a density point of the set H. Thus limit values of the expression

~

Hh

J

(1 - X(s))z'(s)ds,

t

and limit values of the quotient (2.4) as h -7 0 and h > 0 belong to [06F(t, z(t))]. Since this is true for every 8 > 0, limit values of the quotient (2.4) as h -7 0 and h> 0 belong to F(t, z(t)). V. In complete analogy with IV we establish that limit values of the quotient (2.4) as h -7 0 and h < 0 belong to F(t, z(t)). VI. Theorem 4.13.2, VI and V imply what was required. The lemma is proved. • Lemma 2.2. Let conditions (2.1- 3) hold. Let H ~ lR be a closed set of measure zero. Let