MARILYN BREEN

A Q U A N T I T A T I V E K R A S N O S E L ' S K I I T H E O R E M IN R d

1. INTRODUCTION

This work will be concerned with a quantitative version of Krasnosel'skii's theorem in R d, and the following result will be proved: Let S be a d-dimension- al compact set whose set Q of lnc points is a union of n disjoint closed convex sets C i, each essential, 1 ~< i ~< n. If every 2n or fewer points of S see via S a common k-dimensional neighborhood of radius e >/0, then ker S contains such a k-dimensional e-neighborhood, 0 ~< k ~< d.

We begin with some preliminary definitions. Let S be a subset of R d. A point x in S is said to be a point of local convexity of S if and only if there is some neighborhood N of x such that S n N is convex. In case S fails to be locally convex at point q in S, then q is called a point of local nonconvexity (lnc point) of S. For points y and z in S, we say y sees z via S if and only if the corresponding segment [y, z] lies in S. Set S is called starshaped if and only if there is some point p in S such that, for every x in S, p sees x via S, and the set of all such points p is called the (convex) kernel of S, denoted ker S.

The well-known Krasnoserskii theorem [-4] states that if S ~ J~ is a compact set in R a, then ker S is nonempty if and only if every d + 1 points of S see a common point via S. Moreover, a quantitative analogue in 1-1] shows that for S compact in R d, ker S contains a d-dimensional neighbor- hood of radius e > 0 if and only if every d + 1 points of S see via S such a d-dimensional e-neighborhood. While other analogues of the theorem have been obtained for sets whose kernel is k-dimensional, 1 ~< k ~< d [1], little can be said about the size of this kernel when k < d. Therefore, an interesting problem is that of characterizing those sets S in R a whose kernels contain a k-dimensional neighborhood of radius e > 0, 1 ~< k < d. In [3], an example and theorem in the plane indicate that such a characterization should involve the number of lnc points of set S, and it is our purpose to provide an appropriate analogue of this result in R a.

The following terminology will be used: conv S, aft S, cl S, and int S will be used to denote the convex hull, affine hull, closure, and interior of set S, respectively. When S is convex, dim S will be the dimension of set S. For x 4= y, L(x, y) will represent the line determined by x and y, and R(x, y) will be the ray emanating from x through y.

2. T H E RESULTS

The following definition from I-2] will be needed.

DEFINITION. Let Q = U~'= a Ci denote the set of lnc points of the closed

GeometriaeDedicata 12 (1982) 219-226. 0046-5755/82/0122-0219501.20. Copyright (~ 1982 by D. Reidel Publishing Co., Dordrecht, Holland, and Boston, U.S.A.

220 M A R I L Y N BREEN

set S, with each C i set convex. We say C i is essential if and only if for every x in C i, there is some neighborhood N' ofx such that for N a convex neighbor- hood of x and N ~_ N', (S n N) ... C i is connected.

T H E O R E M 1. Let S be a compact subset of R d where d = dim aft S, and assume that the set Q of lnc points of S may be written as a union of n disjoint closed convex sets C~, each essential, 1 <. i <. n. If, for some k with 0 <. k <<. d, every 2n or fewer points of S see via S a common k-dimensional neighborhood of radius e >i 0, then ker S contains such a k-dimensional e-neighborhood. The number 2n is best possible.

Proof. Throughout the argument, we will assume that d/> 2 and that Q :/: q~, for otherwise S will be convex [5] and the result will be trivial. By [3, Lemma 2], it suffices to prove that for F any finite subset of S, F sees via S a k-dimensional neighborhood of radius e, so this will be our goal.

We begin the argument with a preliminary lemma.

LEMMA 1. For each set C i, there is a neighborhood N' of C i such that for N a convex neighborhood of C i and N ~_ N', S n N ".. C~ is connected.

Proof. For convenience of notation, let C = C i. For each point c of C, select a convex neighborhood N' c of c satisfying the definition of essential, N ' disjoint from the remaining C l sets. Using the compactness of C, reduce to a finite subcollection Jff ' of these neighborhoods which cover C, and let N' = u {N' :N' in JV'}.

We assert that N' satisfies the lemma. Let N be a convex neighborhood of C with N _ N', and let JV = {N c :No = N n N'~, N ' in JV'}. To see that S n N ~ C is connected, select N 1 in X . Clearly for some N 2 in JV ~ N1, N 1 n N 2 :p ~ . Also, S n N 1 c~ N 2 .-. C :p ~ , for otherwise S c~ N 1 n N 2 would lie in C and hence be convex, impossible since C _ Q. Thus the sets S c~ N1 "" C and S n N 2 ~ C are not separated, and their union S n (N1 • N2) ~ C is con- nected. An easy induction finishes the proof.

Returning to the proof of the theorem, let C = C~ and use the argument in lemma 1 to select a closed convex neighborhood N of C disjoint from the remaining C i sets such that S n N ~ C is connected. This allows us to employ results from [2] to study S c~ N. By [-2, Corollary to Theorem 2], dim C- - = d - 2, and by [2, Theorem 3], S ~ N is a union of two convex sets. Using arguments from that theorem, select x and y in int ( (Sn N).-. affC) such that [x ,y] ~ S and yCaf f (Cu {x}). Let A' and B' denote the hyperplanes determined by C u {x} and C w {y}, respectively, with R the closed convex region determined by A' and B' which contains Ix, y]. Letting S x denote the subset of S seen by x via S, the sets Sx c~ R c~ N and Sy c~ R n N are convex sets whose union is S c~ R c~ N. Moreover, these sets contain C and intersect in a subset of affC.

Again by arguments in [2, Theorem 3], the disjoint convex sets

A QUANTITATIVE KRASNOSEL'SKII THEOREM IN R d 221

S x c~ R n N ~ aft C and Sy n R c~ N ,-, aft C may be separated by a hyperplane M with C _ M. Furthermore, A' c~ B' = A' c~ M = B' c~ M = aft C. We assert that M is disjoint from at least one of Sxc~Rc~N, , , a f fC and Sy c~ R c~ N ~ aft C: otherwise, for s in M c~ (Sxn R c~ N ~ aff C) and t in M n (Sy c~ R n N ~ aft C), conv(s u C) and cony (t w C) would intersect in a set of dimension d - 1, impossible since Sx r~ R c~ N and Sy r~ R n N meet only in aft C. For future reference, label the halfspaces determined by M so that S x r ~ R c ~ N ~ _ c l M 1 and S y c ~ R c ~ N ~ _ c l M 2.

We will need the following two lemmas.

LEMMA 2. l f J is any hyperplane containing aft C, then J c~ S c~ N is convex. Proof. We begin with the case in which J = M. Let a, b e M c ~ S n N to

show that In, b] _ S. Recall that by a previous argument, M c~ R c~ N -,, aft C is disjoint from at least one of S and Sy, and without loss of generality we assume that it is disjoint from Sy. Thus if both a and b are in R ~ aft C, then both are in the convex set Sx c~ R c~ N, and I-a, b] _ S. By an argument in [2, Theorem 3], the set S c~ M n N ~ R is a nonempty subset ofker (S c~ N). Hence if one of a or b is in S n M n N ~ R, then again [-a, b] __q S. Therefore, we need only consider the case in which a, b ~ R and at least one of the points, say a, is in aft C. Select z in the nonempty set S c ~ M c ~ N ~ R ~ ker(Sc~ N). (Note that z ~ aft C.) If b ¢ aft C, then b ~ M c~ R c~ N ,,- aff C so b e Sx. By results in [2], b~cl ( S n N n M 1 ) , and there is a sequence {bn} in S~c~Nc~M 1 converging to b. Then for each n, la, z-] u [z, b, ] ~_ S. Clearly there are no lnc points of S n N in conv {a, z, b,} ,-- [-a, b~], so by a variation of a lemma by Valentine [7, Corrollary 2], [a, bn] __- S. Since this is true for every n and S is closed, we have [-a, b] ___ S. Similarly, if b~aff C, there can be no lnc point in conv {a, z, b} ~ [-a, b], and I-a, b] _ S. We conclude that M c~ S c~ N is convex.

In case J :~ M, then for z in S n M n N ~ R ~_ ker (SnN) , clearly zCJ. Thus for a, b in J n S c~ N, there can be no point of C (and hence no lnc point of ScaN) in conv {a, z, b} ~ [a, b]. Again by Valentine's lemma, I-a, b] _ J c~ S n N and J n S n N is convex. This finishes the proof of Lemma 2.

LEMMA 3. Let rI be any plane in R d, q an lnc point for S n FI. l f qeC, then q is the only lnc point o fS n FI belonging to C. Moreover, i fA and B are distinct hyperplanes containing aft C, then A c~ 17 and B n II are distinct lines at q.

Proof. Clearly if q is an lnc point for S n l-I, then q is an lnc point for S, and hence q belongs to some C~ set, say C. By Lemma 2, i f J is any hyperplane containing aft C, then J n S will be locally convex at each point of C. Since q is an lnc point for S ~ I-I, it follows that S c~ I1 ~ J c~ S c~ I1, and 11 ~ J. Thus J ~ H must be a line at q. Furthermore, H contains only one point of aft C: Otherwise aft C ~ H would be a line, and for d e I - I ~ a f f C ~ ~5, a f f (d~ C) would be a hyperplane containing both 17 and affC, impossible

222 M A R I L Y N BREEN

by the argument above. We conclude that C n H = {q}. For future reference, notice that each C i set contributes at most one lnc point of S n 11, 1 ~< i ~< n, and S n 1-I has at most n lnc points.

I fA and B are distinct hyperplanes containing aft C, then A c~ H and B n H are lines at q by the a rgument above. Moreover , since A n B = affC, A c~ B n H = aft C n FI = {q}, and A n H and B c~ H are distinct lines at q. This completes the p roof of Lemma 3.

Returning to the p roof of Theorem 1, let W' denote the set of points w' in F n i n t R for which w' sees some point of C via S. For each w' in W', w' sees via S some ceC, and we may select a point w in [w', c)c~N ~_ int R. Let W (possibly empty) denote the set of points w selected, one for each w' in W'. Clearly each w point belongs either to S x n R c~ N or to S~ n R c~ N, not both.

We choose hyperplanes A and B as follows: If W n Sx = ~ , let A = A'. If W n S x ~ ~ , rotate M about aft C toward A' until the first point of W n S~ is met, and let A denote the resulting hyperplane. (Of course A will be M in case W c~ S x n M ~ ~;.) A parallel a rgument applied to W n Sy and B' yields the hyperplane B. Since at least one of S~ or Sy is disjoint f rom M c~ R c~ N ,~ aft C, at least one of A or B is distinct f rom M, and A ~ B.

Label the halfspaces determined by A, A', B, B' so that x~cl B 1 u cl B' 1 and y~cl A~ u c l A' r No te that cl A~ c~S and cl B~ n S are locally convex at each point of C. Also, recall that M ~ B' and M ~ A'.

We show that there are points u and v in bdry S c~ N c~ cl A 2 ~ cl B 2 such that u and v see via S no c o m m o n point in S ~ (el A1 c~cl B~): Since at least one of S~ c~ R c~ N ,-~ aft C and Sr c~ R c~ N ~ aft C is disjoint f rom M, assume that (S~ c~ R n N --~ aft C) n M = ~ (and thus A ~ M). We assert that there exists some u in bdry S nNc~cl A 2 ~ c l B E such that u~Sx and u sees via S no point of M c~ A 2 c~ B 2 : If no such u existed, then every point of S~ c~ bdry S c~ N n cl A 2 n cl B 2 would see a point of M n A' 2 N B 2 and hence see a point of M n A 2 n B 2 ~ N. However, using s tandard arguments, since S is compac t and S~ n N is convex, this would force S~ c~ N n R ~ aft C to meet M, impossible. In case S r n N n R ~ a f fC is disjoint f rom M, then by a similar a rgument we may select point v in bdry S n N n cl A 2 n cl B 2 such that v ~ S r and v sees via S no point of M n A 2 n B 2. In case S r c~ N n R ,-~ aft C meets M, we may choose any point v in their intersection. Then clearly v e b d r y S n N n c l A 2 n c l B2, and by previous comments , v see via S no point of M r

It remains to show that points u and v have the appropria te property. Suppose on the cont rary that u and v see some c o m m o n point s in S ,-, (cl A ~ n n cl B~). Then one of the segments I-u, s], rv, s] would meet M at a point t, and clearly t~ cl A 1 n cl B r Since M _ (cl A 1 ~ cl B1)u (A 2 n B2), this forces t to belong to M n A 2 n B 2. Since u sees no point o f M n A 2 c~ B~, [-u, s] cannot meet M so Iv, s] must meet M at t. Also, u is in M 1 so s must lie in M~. Since v

A Q U A N T I T A T I V E K R A S N O S E L ' S K I I THEOREM IN R a 223

sees poin t t in M n A' 2 n B'z, v must have been chosen in (S r n N c~ R ~ aft C) n c~ M. But then v cannot see poin t s in M t , and we have a contradict ion. O u r a s sumpt ion is false, and points u and v see via S no point o f S ,-~ (cl A 1 n cl B~), the desired result.

N o w for each C i set, use the cons t ruc t ion above to select hyperplanes A(i), B(i) and points ui, v i so that u~ and v i see no c o m m o n point of S ,~ ~ (cl A(i) 1 n c l B(i)~), 1 <~ i <~ n. Clearly any point which sees u t , v 1 . . . . ,u , , v

via S lies in n {cl A(i) 1 n el B(i)l : 1 <~ i <~ n} =- K e , so by our hypothesis, K F contains a k-dimensional ne ighborhood u of radius e.

We want to p rove that for t in S n K r and s in F, Is, t] _~ S. The following l emma serves as a pre l iminary step.

L E M M A 4. The set S n K F is convex. Proof. Select points s and t in S n K F to show that Is, t] ~ S n K F. Since

every 2n points of S see a c o m m o n point via S, we m a y choose a point p which sees s, t, u 2, v 2 . . . . , u,, v via S. Select r in conv {s, p, t} so that [s, r] u u It , t] ~ S and so that the dis tance f rom r to line L((s, t) is minimal. If points r, s, t are collinear, then Is, t] ~ S and the a rgumen t is finished. Hence we assume that this is not the case to reach a contradict ion.

Let H denote the plane de termined by s, r, t. By remarks in the p roof of L e m m a 3, each C i set contr ibutes at mos t one lnc point of S n H, 1 ~< i ~< n, and so S n H has at mos t n lnc points. We have [s, r] u [r, t] _~ S n conv {s, r, t}, Is, t] ~ S n conv {s, r, t}, so by a var ia t ion of Valentine 's l emma, [7, Coro l la ry 2], there is some lnc point q of S n cony {s, r, t} in conv {s, r, t} ~ Is, t]. Clearly q is an lnc point for S n II. Fur the rmore , since S n H has only finitely m a n y lnc points and s sees no point on It, r), it is easy to show that q m a y be chosen on (s, r]. Similarly, (t, r] contains some lnc point q' of S c~ conv {s, r, t}.

F o r convenience of nota t ion, let q belong to C = Cl and let A = A(l) and B = B(1) denote the cor responding hyperplanes selected in our previous construct ion. By L e m m a 3, A c~ H and B c~ H are distinct lines at H. Since s, t~cl A 1 n cl B 1 n lq, we have s, t~c l (A 1 n H ) ~ cl(B 1 n H). Let R~ denote A n F I n B z and let R B denote B n H n Az , each an open ray at q. Fur ther , let L 1 denote the open halfplane in H determined by L = L(s, r) and contain- ing t, ~ the opposi te open halfplane.

We assert that at least one of the rays Ra and R n must intersect int cony {s, r, t} :Le t H denote the hyperp lane aff (C u L) and label the open half- spaces determined by H so that L1 _c H1" Using the no ta t ion in L e m m a 2, select a point z in S c~ M n N ,,~ R ___ ker (S n N). Not ice that s, r, and t cannot lie in the same closed half space de termined by M, for otherwise S n conv {s, r, t} would be locally convex at q, a contradict ion. I f neither R A nor R B met int conv {s, r, t}, this would force z to belong to H I , and by a rguments similar to those in L e m m a 2, S n cl L~ would be locally convex at q. Clearly this is

224 MARILYN BREEN

impossible, so one o fR A and R B must meet int conv {s, r, t}, and the assertion is established.

By the paragraph above, R a and R B cannot both lie in cl L 2 . Furthermore, since tecl (A 1 n Fl)n cl(B 1 n II), R a and R B cannot both lie in cl L r Thus we may assume that R A _ LI and R n c L2, with R a meeting int cony {s, r, t}. Notice that if q = r, then s and t would lie in opposite open halfplanes deter- mined by R A, impossible since s, t e c l ( A 1 n i l ) . It follows that r < q < s. Hence, exactly one of r and s must lie in the open convex region determined by R A and R B. That is, exactly one of r and s lies in A 2 r iB 2 n i l . Since secl A~ n cl B~, it follows that t e A 2 n B2n H. Moreover, since tecl A l n c l B 1 , this forces both r and p to lie in A 2 n B 2. Recall that p sees u 2, v 2, . . . , u,, v via S and hence pec l A(i)x n c l B(i)l for 2 ~< i ~< n. We conclude that {A, B} ~ {A(i), B(i)} for 2 ~< i~< n, so {A, B} = {A(1), B(1)} and C = C l. We have proved that if q is an lnc point of S n c o n v {s,r, t} and qe(s, r], then q ~ r and q e C r Since S n 11 has at most one lnc point in C~, q is the only lnc point of S n conv {s, r, t} on (s, r].

Repeating the argument for q', we conclude that q' ~ r and q 'e C 1. Clearly q ~ q', so we have contradicted the fact that 11 n C~ = {q}. Our assumption that s, r, t are not collinear is false, the points must be collinear, and Is, t] ~_ S. Therefore the set S n K F is convex, and the proof of Lemma 4 is complete.

Finally, we must show that for s in F and t in S n KF, [s, t] _~ S. In case s belongs to K r , the result is an immediate consequence of Lemma 4. Hence we assume that for at least one i, s¢cl A(i)~ n cl B(i)r Without loss of general- ity, suppose that s¢c lA(1 )~nc lB(1 )~ , and select some point p such that s, t, u 2, v 2 . . . . , u , v, see p via S. Let r denote the point of [t, p] Closest to t for which Is, r] ~ S. In case r = t, the argument is finished, so assume t < r ~< p to reach a contradiction. Using the argument from Lemma 4, there is some lnc point q of S n cony {s, r, t} on (s, r]. As in that lemma, let q~Ct = C, with A = A(1) and B = B(l) the corresponding hyperplanes at C, and let R a = A c~ 11n B2, R e = B n 11n A 2. Furthermore, by arguments from Lemma 4, one of R A and R n meets int conv {s, r, t}, and R A and R e neces- sarily lie in opposite open halfplanes determined by L(s, r).

We assert that s must lie in the open convex region determined by R A and RB: If {A, B} = {A(i), B(i)} for any 2 ~< i ~< n, then [p, t] ~_ cl A(i)~ n n cl B(i) I , forcing s to lie in A 2 n B 2 n 11, the appropriate region. If {A, B} = = {A(1), B(1) ~, then s¢cl A 1 n cl B 1 and again s e A 2 n B 2 n H.

We have proved that s ~ A 2 n B 2 ~ int R. However, since s sees q via S and s eF , this is impossible according to our selection of the hyperplanes A and B. Our assumption that r ~ t must be false, and Is, t] ~ S. Therefore, every point of F sees each point of S n K v via S. Since S n K F contains a k-dimensional e-neighborhood U, every point of F sees U via S, and by our opening remarks, ker S must contain a k-dimensional e-neighborhood. This completes the proof of the theorem.

A Q U A N T I T A T I V E K R A S N O S E L ' S K I I T H E O R E M IN R d 225

The following corollary is an immediate consequence of the arguments in 1-3, Lemma 2] and in.Theorem 1 above.

COROLLARY. Let d denote a family of compact sets which has the Bolzano- Weierstrass property relative to the Hausdorff metric. (See 1-6, pp. 37-38] for the necessary definitions. Let S be a d-dimensional compact set whose set Q of Inc points is a union of n disjoint closed convex sets Ca , . . . , C , each essential. I f every 2n or fewer points of S see via S a member of d , then ker S contains a member of d .

To see that the bound 2n is best possible, consider the following example.

EXAMPLE 1. Clearly the result is best for n = 1. When d = 2 and k = 1, Example 2 in [3] illustrates that the number 2n is best for every n ~> 2. (The reader is referred to [3] for the lengthy construction.) Furthermore, for d arbitrary and k = d - 1, this example may be extended to R d by replacing the unit circle with the unit d-sphere U and by replacing diameters of U with ( d - 1)-dimensional neighborhoods in U of radius 1. For 1 ~< k < d - 1 , appropriate adaptations may be made by intersecting U with suitable closed half spaces.

In conclusion, it is interesting I to notice that when k < d, the result in Theorem 1 fails without the requirement that the C sets be essential.

EXAMPLE 2. Let 2 ~< d, 1 ~< k < d, and let S 1 and S 2 be d-dimensional convex polytopes which share a (k -1 ) - face C - S 1 n S 2 so that for an appropriate e > 0 and for each x in S i ,,~ C, (w {R(x, c):c in C} ) n Sj contains a k-dimensional e-neighborhood i(=j, 1 ~ i,j<~ 2. Clearly C is the set of lnc points for S = S 1 w S 2 and C is not essential. Furthermore, every 2 (and in fact every 3) points orS see via S a common k-dimensional e-neighborhood. However, ker S = C is a set of dimension k - 1.

B I B L I O G R A P H Y

1. Breen, Marilyn: 'K-dimensional Intersections of Convex Sets and Convex Kernels' Discrete Math. 36 (1981), 233-237.

2. Breen, Marilyn : 'Points of Local Nonconvexity and Finite Unions of Convex sets,. Canad. J. Math. 27 (1975), 376-383.

3. Breen, Marilyn: 'A Quantitative Version of Krasnosel'skii 's Theorem in R 2'. Pacific J. Math. 91 (1980), 31-37.

4. Krasnosel'skii, M. A. : 'Sur un crit6re pour qu'un domain soit 6toi16'. Math. Sb. (61) 19 (1946), 309-310.

5. Tietze, H. : '~rber Konvexheit im kleinen und im grossen und fiber gewisse den Punkten einer Menge zugeordnete Dimensionzahlen'. Math. Z. 29 (1928), 697-707.

6. Valentine, F. A. : Convex Sets. McGraw-Hill, New York, 1964. 7. Valentine, F. A. 'Local Convexity and L, Sets'. Proc. Amer. Math. Soc. 16 (1965), 1305-13 I0.

226 MARILYN BREEN

(Received July 19,1980)

Author's address:

Marilyn Breen, Department of Mathematics, University of Oklahoma, Norman, Oklahoma 73019, U.S.A.