multifunction fromh−1,0itoRwhich had no continuous selection

CONTINUOUS SELECTIONS FOR LIPSCHITZ MULTIFUNCTIONS

I. KUPKA

Abstract. In [11] an example presented a Hausdorff continuous, u.s.c. and l.s.c. multifunction fromh−1,0itoRwhich had no continuous selection. The multifunction was not locally Lipschitz. In this paper we show that a locally Lipschitz multifunction fromRto a Banach space, which has ”locally finitely dimensional“ closed values does have a continuous selection.

1. Introduction

The research in the selection theory was started by Michael in 1956 (see for example [15], [16]) by proving several continuous selection theorems. Then, the problem of the existence of selections of various types – linear e. g.

[7], measurable [13], Carath´eodory [8], quasicontinuous [10], [14], Lipschitz [3], [6] etc. – was studied in many papers. A Lipschitz selection theorem for compact-valued multifunctions defined on a closed interval, with values in a metric space, was proved in [5]. Recent results concerning selections are listed in [18].

In general, there is no guarantee that a ”nice“ multifunction will have a continuous selection. Even closed- valued continuous multifunctions defined on compact interval and with values in Rneed not have a continuous selection (see[11]). In this paper, we show, in particular, that if such a multifunction is locally Lipschitz, it does have a continuous selection. This will be a consequence of a more general assertion, Theorem3.

Received January 15, 2004.

2000Mathematics Subject Classification. Primary 54C65; Secondary 54C30 . Key words and phrases. Continuous selection, Lipschitz multifunction.

2. Notation and terminology

For definiton of basic notions: multifunction, selection, l.s.c. u.s.c. and Hausdorff continuous multifunction, Hausdorff metric etc see e.g. [12] and [17].

In what follows we denote byNthe set of all positive integers, byRthe real line with its usual topology and by Ban arbitrary Banach space overR. IfX is a metric space,x∈X andris a positive real number, we denote the closed ball with the centerxand diameterrbyB(x, r). Throughout this paper we consider only multifinctions with nonvoid values.

IfK is a positive real number, and (X, d), (Y, %) are metric spaces, we say that a multifunction F fromX to Y is K-Lipschitz if for every x1, x2 from X the inequality H%(F(x1), F(x2))≤Kd(x1, x2) is true. (By H% we denote a Hausdorff metric on 2^Y − {∅}derived in a natural way from %).

Before proving our main results we need the following technical lemma:

Lemma 1. Let Y be a Banach space overR. Leta∈R, letm be a positive real number. Let I=ha, a+mi (I =ha−m, ai)⊂R. Let F :I →Y be a K-Lipschitz multifunction. Letr > 0, r < K. Letb ∈F(a). Then there exists anM-Lipschitz function f :I→Y such thatM = (K+r),f(a) =b and for each xin I

d(f(x), F(x)) = inf{d(f(x), t);t∈F(x)}< r.

Moreover f(I)⊆B(b,2Km)holds.

Proof. Let us consider the caseI=ha, a+mi. The caseI=ha−m, aiis symmetrical.

Letn∈N be such thatK^m_n < ^r₆ and ^m_n < ¹₃. Let us define x_i=a+^m_ni fori= 0,1,2, . . . n. Denote b=y₀. SinceF is K-Lipschitz, there exists a pointy₁∈F(x₁) such that

d(y0, y1)5H(F(x0), F(x1)) +rm 2n 5Kd(x₀, x₁) +rm

2n 5Km n +rm

2n 5 K+r

2 m

n.

By final induction we can find a set{y0, y1, . . . , yn}such that∀i= 0,1,2, . . . , n, yi∈F(xi) and d(y_i, y_i+1)5

K+r 2

m

n for i5n−1.

Let us define a continuous functionf :ha, a+mi →Y in this way: f(xi) =yi,i= 0,1,2, . . . , n f(x) = 1

m[n(x−x_i)y_i+1+n(x_i+1−x)y_i] if x∈(x_i, x_i+1).

We will prove thatf is (K+^r₂)-Lipschitz onha, a+mi.

(I) Letx, x⁰∈ hxi, xi+1i, for somei∈ {0,1, . . . , n}, x < x⁰. We obtain d(f(x), f(x⁰))

= 1

mkn(x⁰−xi)yi+1+n(xi+1−x⁰)yi−n(x−xi)yi+1−n(xi+1−x)yik

= n

mk(x⁰−x)y_i+1−(x⁰−x)y_ik5 n

m|x⁰−x| · k(yi+1−y_i)k 5 n

m|(x⁰−x)|

K+r 2

m n 5

K+r 2

|x⁰−x|.

(II) In general, ifx < x_i< x_i+1. . . , x_i+k < x⁰ for somei, k∈ {0,1, . . . , n}, i+k < nthen, because of (I) d(f(x), f(x⁰))

5d(f(x), f(xi)) +d(f(xi), f(xi+1)) +. . .+d(f(xi+k−1), f(xi+k)) +d(f(xi+k), f(x⁰))

5 K+r

2

|xi−x|+ K+r

2

|xi+1−xi|+. . .+ K+r

2

|x⁰−xi+k|

= K+r

2

|x⁰−x|.

Now, letx∈ ha, a+mi, thenx∈ hxi, xi+1ifor somei∈ {0,1, . . . , n}. So d(f(x), F(x)) = inf{d(f(x), t), t∈F(x)}

= infn

n

m(x−xi)yi+1+ n

m(xi+1−x)yi−t

;t∈F(x)o

SinceF is K-Lipschitz there exists a pointpfromF(x) such thatd(p, y_i+1)5(K+^r₂)(x_i+1−x) therefore d(f(x), p)5d(f(x), y_i) +d(y_i, y_i+1) +d(y_i+1, p)

5 K+r

2

(x−xi) + K+r

2 m

n + K+r

2

(xi+1−x) 5

K+r 2

(xi+1−xi) + K+r

2 m

n 52 K+r

2 m

n 52r 6 +rm

n < r.

sod(f(x), F(x))< rfor eachxfrom ha, a+mi.

Now, sincef(a) =bandf is a (K+r)-Lipschitz function, forrsuch thatr < Kand for eachxfromha, a+mi we have

d(b, f(x)) =d(f(a), f(x))5(K+r)|x−a|52K|a+m−a|52Km

sof(ha, a+mi)⊆B(b,2Km).

Theorem 1. Let B be a finitely dimensional Banach space. Let a∈ R, letl be a positive real number. Let I = ha, a+li (ha−l, ai). Let F : I → B be a K-Lipschitz multifunction with closed values. Then F has a K-Lipschitz selection onI.

Proof. We will prove the Theorem only for the caseI=ha, a+li. According to Lemma1there exists a sequence {fi}^∞_i=1of functionsf_i:ha, a+li →Bsuch that for each indexifromNand eachxfromha, a+li d(f_i(x), F(x))<¹_i is true. Moreover each functionfiis K+¹_i

-Lipschitz andfi(ha, a+li)⊂B(b,2Kl). This implies that for every xfrom X the set{fi(x);i= 1,2, . . .}is precompact.

SinceBis finitely dimensional, according to Arzela-Ascoli theorem the setM ={fi;i∈1,2, . . .}is precompact.

So there exists a continuous functionf :ha, a+li →B such thatf is a uniform limit of a sequence{fi_j}^∞_j=1 (a subsequence of{fi}^∞_i=1) of functions fromM.

Let us consider anε >0. As we have proved above there exists an indexk such thatfi_j is (K+ε)-Lipschitz for eachj=k. That means that the functionf is also (K+ε)-Lipschitz. f is proved to beK-Lipschitz.

Now it is simple to realize thatf is a selection ofF. For eachε >0 there exists an indexmsuch that for each xfrom X

d(f_im(x), F(x))< ε and sup

x∈ha,a+li

|fim(x)−f(x)|< ε.

So for everyx from X d(f(x), F(x))< 2ε. Since ε was an arbitrary positive real number, for eachx from X d(f(x), F(x)) = 0 is true. F has closed values sof is a selection ofF.

3. Main results

Theorem 2. Let B be a finitely dimensional Banach space over R. Let F :R→B be a K-Lipschitz multi- function with closed values. ThenF has a K-Lipschitz selection on R.

Proof. This is a simple consequence of Theorem1 so we will only give an outline of the proof. Let b be an element of the setF(0). Using Theorem1, we can define by inductionK-Lipschitz selectionsf1, f2, . . . f2i, f2i+1, . . . of F such that for each nonnegative integer i the function f2i (f2i+1) is defined on h2i,2i+ 2i (h−2i−2,

−2ii) and f_2i(2i+ 2) =f_2(i+1)(2i+ 2) (f_2i+1(−2i−2) =f_2(i+1)+1(−2i−2)) and such that f₁(0) =f₂(0) =b.

It is easy to see that a function f : R→ B defined by f(x) =f_2i(x) if x∈ h2i,2i+ 2iand f(x) = f_2i+1(x) if x∈ h−2i−2,−2iiis correctly defined and it is aK-Lipschitz selection ofF. Theorem2 is true for certain multifunctions with non-convex and non-compact values. It is a generalization of a result, obtained for multifunctions with convex compact values:

Corollary 1. [6, Corollary 2] Let n be a positive integer, let B = Rⁿ. Let F : R → B be a K-Lipschitz multifunction with convex compact (and nonvoid) values. ThenF has a K-Lipschitz selection onR.

In the following lemma we shall use the following assumption concerning a multifunctionF fromRto a Banach spaceB:

Assumption LFD.For everyxfromRthere exists an open neighborhoodOx⊂Rand a finitely dimensional setBx⊂Bsuch thatF(Ox)⊂Bx.

We say that a multifunctionF :R→Bislocally Lipschitz if for every realxthere exists an open intervalU_x and a positive real constantKxsuch thatx∈Ux andF is Kx-Lipschitz onUx.

Lemma 2. Let B be a Banach space. Let F : R→B be a locally Lipschitz mutifunction with closed values.

Let F satisfy the assumption LFD. Let a∈Randb∈F(a). Then for every real c,d,c < d satifyingc≤a≤d there exists a Lipschitz selectionf :hc, di →Bof F such, thatf(a) =b.

Proof. It suffices to show thatF is Lipschitz onhc, diand that there exists a finitely dimensional subsetZ of Bsuch thatF(hc, di)⊂Z. After that we can apply Theorem1.

We proceed by a usual ”locally on compact implies globally on compact“ procedure. Obviously for every x fromhc, dithere exists an open intervalU_x, a positive real numberK_x and a finitely dimensional subsetB_xofB such thatx∈U_x, F(U_x)⊂B_xandF isK_x-Lipschitz onU_x.

Consider the following open coverC ofhc, di: C={U_x;x∈ hc, di}. There exists a finite subcoverS ofCand a positive integernsuch thatS={Ux₁, Ux₂, . . . , Ux_n}. Let us denoteM = max{Kx₁, Kx₂, . . . , Kx_n}. ThenF is M-Lipschitz on each intervalUx_i fori∈ {1,2, . . . , n}. The facthc, di ⊂U :=Sn

i=1Ux_i impliesF isM-Lipschitz onhc, di.

Moreover,F(hc, di)⊂F(U)⊂Z :=Sn

i=1Bx_i, and we can see thatZ is finitely dimensional.

If c < a < dTheorem 1 impliesF has an M-Lipschitz selection h (g ) onhc, ai (ha, di) such that g(a) = h(a) =b. So ifc < a < dthe functionf :hc, di →Bdefined byf(x) =g(x) onhc, aiandf(x) =h(x) onha, diis a Lipschitz selection ofF onhc, di. The proof for the casesa=c,a=dis even easier.

To realize that the assumptions of our final result, Theorem 3, can hardly be weakened let us compare the following three assertions:

(1) There exists a finitely valued Lipschitz multifunction from a unit circle into R² that has no continuous selection. (See Example 1. Of course, each multifunction with values in R² or R automatically satisfies the assumption LFD.)

(2) There exists a Hausdorff continuous multifunction from the compact intervalh−1,0itoRwith closed values, which is locally Lipschitz in every point ofh−1,0) and has no continuous selection (See Example2).

(3) Each locally Lipschitz multifunction with closed values fromRto a Banach space, satisfying the assumption LFD has a continuous selection. (See Theorem 3).

The examples presented below are based on ideas, used in examples published in [4] and [11].

Example 1. LetK= cos(t) + i·sin(t); t∈ h0,2π) be the unit circle in the complex plane.

For eacht fromh0,2π) let us denote

a_t= cos(t) + i·sin(t), b_t= cos t

2

+ i·sin t

2

ct= cos

π+ t 2

+ i·sin

π+t

2

Let us define a two-valued multifunctionF :K→KbyF(at) ={bt, ct} for every tfrom h0,2π).

This multifunction has compact (even finite) values and is Lipschitz.This can be seen by two ways.

An intuitive way is the easier one. If we draw a picture of our circle, we realize, that witht ”moving“ from 0 towards 2πthe pointat is moving from the point [1,0] to [0,1], then [−1,0] and finally to [1,0] again. In this time the two-tuple [bt, ct] travels around the circle too, but its speed is the half of the speed of at.

Now we show in an exact way thatF is 1-Lipschitz. Lett1, t2be fromh0,2π),t1> t2. We have

|a_t1−a_t2|=p

(cos(t₁)−cos(t₂))²+ (sin(t₁)−sin(t₂))²

=p

2−2 cos(t1) cos(t2)−2 sin(t1) sin(t2) =p

2(1−cos(t1−t2))

=√ 2p

1−cos(t1−t2)).

Similarly

|b_t1−b_t2|=√ 2

s 1−cos

t₁−t₂ 2

. And, of course,

|ct₁−ct₂|=|bt₁−bt₂|.

Moreover

|bt1−c_t2|=|ct1−b_t2|=√ 2

s 1−cos

t₁−t₂

2 −π

=√ 2

s 1 + cos

t₁−t₂ 2

. Therefore

H(F(a_t1), F(a_t2)) =H({bt1, c_t1},{bt2, c_t2})≤min{|bt1−b_t2|,|bt1−c_t2|}

= min (√

2 s

1−cos

t1−t2

2

,√ 2

s 1 + cos

t1−t2

2 )

Now it is sufficient to show that min

(s 1−cos

t1−t2

2

, s

1 + cos

t1−t2

2 )

≤p

1−cos(t1−t2) = 1

√2|at₁−at₂| for allt1, t2, 2π > t1> t2≥0.

So the last thing we need to verify is that for alll∈ h0,2π) min

1−cos

l 2

,1 + cos l

2

≤1−cos(l) or equivalently∀l∈ h0,2π):

cos l

2

−cos(l)≥0 or cos l

2

+ cos(l)≤0.

(∗) Since

cos l

2

−cos(l) = 2 sin 3

4l

sin l

4

cos l

2

+ cos(l) = 2 cos 3

4l

cos l

4

it is easy to verify that

cos l

2

−cos(l)≥0 ∀l∈

0,4 3π

cos l

2

+ cos(l)≤0 ∀l∈ 2

3π,2π

Therefore (∗) is verified and for allt1, t2 fromh0,2π),t1> t2,

H(F(at₁), F(at₂))≤ |at₁−at₂|.

F is proved to be 1-Lipschitz.

Nevertheless, F has no continuous selection onK. It has two natural continuous selections on each Kε⊂K where the setKεis defined byKε={at;t∈ h0,2π−ε)}for every positiveε <2π. These selections are: f(at) =bt

andg(at) =ctfor eachat fromKε.

However, no of these selections can be prolonged toK, For examplef(a0) =b0 = [1,0] , but lim

t→2π⁻f(at) = lim

t→2π⁻bt= [−1,0].

Example 2. [11] LetF :h−1,0i →Rbe defined as follows:

F(0) =R F(x) =

n(n+ 1)

2 x+ k

2ⁿ;k∈Z

∪

n(n+ 1)2ⁿ+ 1

2ⁿ⁺¹ x+n+ 1 2ⁿ⁺¹ + k

2ⁿ;k∈Z

for every positive integernand everyx∈D

−¹_n,−_n+1¹ E .

In other words: the intersection of the graph ofF with the setD

−_n¹,−_n+1¹ E

×Ris a system of segments joining the following couples of points: the point ₋₁

n,₂^mn

with the pointh

−_n+1¹ ,₂^mn+¹₂i and

−_n¹,₂^mn

with the point h−_n+1¹ ,₂^mn+¹₂+₂n+1¹

i

wheremis an arbitrary integer.

To show that F is locally Lipschitz on h−1,0) it is sufficient to show that it is n(n+ 1)-Lipschitz on I_n = D−1

n ,_n+1⁻¹ E

for everyn∈N,n >0.

Letx₁, x₂∈I_n. Lety₁∈F(x₁). Then there exists an integer ksuch that

y₁= n(n+ 1) 2 x₁+ k

2ⁿ or y₁=n(n+ 1)2ⁿ+ 1

2ⁿ⁺¹ x₁+n+ 1 2ⁿ⁺¹ + k

2ⁿ.

There exists alsoy2from F(x2) such that y₂= n(n+ 1)

2 x₂+ k

2ⁿ or y₂=n(n+ 1)2ⁿ+ 1

2ⁿ⁺¹ x₂+n+ 1 2ⁿ⁺¹ + k

2ⁿ so|y1−y2| equals

n(n+ 1)

2 |x1−x₂| or n(n+ 1)(2ⁿ+ 1)

2ⁿ⁺¹ |x1−x₂|.

In both cases we have

|y₁−y₂| ≤K_n|x₁−x₂|, where K_n=n(n+ 1).

(∗∗)

In the same way we can pick any2from F(x2) first and find a y1 fromF(x1) such that the inequality (∗∗) is true.

This means that for eachx₁, x₂ fromI_n H(F(x₁), F(x₂))≤K_n|x1−x₂|is true.

We have just proved thatF is locally Lipschitz onh−1,0). The Hausdorff continuity ofF onh−1,0iis proved in [11].

F has no continuous selection onh−1,0i: every continuous selectionf ofF defined on the set h−1,0) has the property lim

t→0⁻f(t) = +∞.

Next we will prove our main theorem:

Theorem 3. Let Bbe a Banach space overR. Let F :R→Bbe a locally Lipschitz mutifunction with closed values. LetF satisfy the assumption LFD. Leta∈R andb∈F(a). ThenF has a continuous selectionf onR such thatf(a) =b.

Proof. For n = 1,2,3. . . denote In = h−n, ni. In what follows we procced by induction. Let us suppose, without loss of generality, thata= 0.

(1) According to Lemma 2 there exists a Lipschitz selection f1 : T1 → B of F on the interval I1 such that f(a) =b. Let us denotef1(−1) =b1andf1(1) =c1.

(2) Let us suppose that for n in N, n= 1,2, . . . k there exist Lipschitz selections fn of F on In such that if l, m∈ {1,2, . . . k}, l > mthenfl(x) =fm(x) for eachxfromIm.

For each of thenconsidered let us denotefn(−n) =bn andfn(n) =cn.

Since bk ∈ F(−k) there exists a Lipschitz selection gk of F on h−k−1,−kisuch that gk(−k) = bk. Since ck ∈F(k) there exists a Lipschitz selectionhk ofF onhk, k+ 1isuch thathk(k) =ck.

Let us define a functionfk onIk by

f_k(x) = g_k(x) forxfromh−k−1,−ki fk(x) = f_k−1(x) forxfromh−k, ki f_k(x) = h_k(x) forxfromhk, k+ 1i.

We have just constructed by induction a sequence of Lipschitz selectionsfk ofF on the intervalsIksuch that if k1< k2thenfk₂(x) =fk₁(x) for allxfromIk₁. All functionsfk are continuous selections ofF on their domains.

Let us define a functionf :R→Bby

f(x) = f₁(x) forx∈ h−1,1i,

f(x) =fk(x) forx∈ h−k−1,−ki ∪ hk, k+ 1i, k = 1,2, . . .

The functionf is a selection ofF onR. It is continuous because all functions fk are continuous.

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I. Kupka, Faculty of Mathematics, Physics and Informatics of Comenius University, Mlynsk´a dolina, 842 48 Bratislava, Slovakia, e-mail:[email protected]