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Geometriae Dedicata 51: 1-13, 1994. 1 (~) 1994 Kluwer Academic Publishers. Printed in the Netherlands. Common Supports of Families of Sets ROBERT DAWSON Department of Mathematics and Computing Science, Saint Mary's University, Halifax, Nova Scotia, Canada B3H 3C3 (Received: 18 December 1991; revised version: 10 June 1993) Abstract. For a family .A of closed bounded convex subsets of a Banach space, sufficient conditions are given for the existence of a closed hyperplane which supports each member of A. Mathematics Subject Classifications (1991): 52A07. 0. Introduction While separation theorems usually refer to the positions of two or more sets in a linear space with respect to a hyperplane, most support theorems involve a single set. A notable exception in this regard is a theorem of Bishop and Phelps [1, p. 30] involving both support and separation at the same time. Later results combining both concepts may be found in [2], [5], [7], [11]. In this paper we introduce a relation between a flat and a family of bodies in a Banach space which generalizes separation, and which, in a suitable Banach space, implies the existence of an associated closed hyperplane which supports each member of the family. If a hyperplane H supports each member of a family A of bodies, and each member of the family is contained in the same closed halfspace determined by//, then we call H a definite common support; otherwise, a separating one. Clearly, for a hyperplane to be a definite common support for a family of bodies, there must be a continuous linear functional f with two properties: the 'analytic' property of attaining its supremum on each of the bodies; and the 'geometric' property of these suPrema being equal. The 'analytic' property holds trivially for closed bounded bodies in reflexive Banach spaces, although in co, even for a family of two symmetric bodies, there may be no such functional [8]. The 'geometric' property, on the other hand, may fail even in a finite-dimensional space ([6], [7]). I. Definite Common Supports In this section we present our main result on the existence of common supports. In order to do this, we introduce two related concepts: that of a fiat surrounded by a family of sets, and that of a strongly exposed face. Suppose A and B are nonempty subsets of a linear topological space. We denote

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Page 1: Common supports of families of sets

Geometriae Dedicata 51: 1-13, 1994. 1 (~) 1994 Kluwer Academic Publishers. Printed in the Netherlands.

Common Supports of Families of Sets

ROBERT DAWSON Department of Mathematics and Computing Science, Saint Mary's University, Halifax, Nova Scotia, Canada B3H 3C3

(Received: 18 December 1991; revised version: 10 June 1993)

Abstract. For a family .A of closed bounded convex subsets of a Banach space, sufficient conditions are given for the existence of a closed hyperplane which supports each member of A.

Mathematics Subject Classifications (1991): 52A07.

0. Introduction

While separation theorems usually refer to the positions of two or more sets in a linear space with respect to a hyperplane, most support theorems involve a single set. A notable exception in this regard is a theorem of Bishop and Phelps [1, p. 30] involving both support and separation at the same time. Later results combining both concepts may be found in [2], [5], [7], [11]. In this paper we introduce a relation between a flat and a family of bodies in a Banach space which generalizes separation, and which, in a suitable Banach space, implies the existence of an associated closed hyperplane which supports each member of the family.

If a hyperplane H supports each member of a family A of bodies, and each member of the family is contained in the same closed halfspace determined b y / / , then we call H a definite common support; otherwise, a separating one. Clearly, for a hyperplane to be a definite common support for a family of bodies, there must be a continuous linear functional f with two properties: the 'analytic' property of attaining its supremum on each of the bodies; and the 'geometric' property of these suPrema being equal.

The 'analytic' property holds trivially for closed bounded bodies in reflexive Banach spaces, although in co, even for a family of two symmetric bodies, there may be no such functional [8]. The 'geometric' property, on the other hand, may fail even in a finite-dimensional space ([6], [7]).

I. Definite Common Supports

In this section we present our main result on the existence of common supports. In order to do this, we introduce two related concepts: that of a fiat surrounded by a family of sets, and that of a strongly exposed face.

Suppose A and B are nonempty subsets of a linear topological space. We denote

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2 ROBERT DAWSON

the closure of A by el(A), the affine hull of A by aft(A), and the convex hull of A by co(A). By a body in a linear topological space X we will understand a closed, convex, and bounded subset. Af la t is a translate of a subspace.

As usual, a closed hyperplane 11 separates A and B if A is contained in one closed halfspace determined by H and B is contained in the other. If v + H separates A and B for all v in some neighborhood of the origin, 11 is said to strongly separate A and B.

Let ,4 be a family of subsets of a linear topological space X. A is said to surround a closed fiat F in X if F intersects el(co(U {z: A E .A))) but fails, for each proper subfamily a of A, to intersect cl(co((.J{A: A C a})) . If, furthermore, there exists a neighborhood V of the origin such that, for every v E V, A surrounds v + F , then A is said to strongly surround F.

Clearly, in a finite-dimensional space, a family of n convex sets can surround a hyperplane exactly when n = 2 and the sets are nonempty, separated by the hyperplane, and disjoint from it. On the other hand, a fiat of codimension greater than 1 may be surrounded by two bodies whose interiors intersect.

1.1 EXAMPLE. In the plane, let A - {(x, y): (x - 1) 2 + y2 < 4} and B = {(x, y): (x + 1) 2 + y2 _< 4}. These surround the point (0, 2), but do not surround any line.

In a finite-dimensional space, bodies are compact. It thus follows that if a fiat in a finite-dimensional space is surrounded by a family of bodies, it is strongly surrounded by them. However, if the bodies are replaced by noncompact sets, or if the space is infinite-dimensional, this need not be true.

1.2 EXAMPLE. In l~ , let al = (½, 0, 0, 0 , . . . ) , a2 -- (0, ¼, 0, 0 , . . . ) , and, generally, let the nth coordinate of an be (1 - 2 -n ) and all others 0.

Let bl = (~, 0, 0, 0 , . . . ) , b2 = (0, ~, 0, 0 , . . . ) a n d , generally, let the nth coordinate o fbn be (1 + 2 -n ) and all others 0. Let A = cl(co({ai: i = 1, 2 , . . . } ) ) and B = cl(co({bi: i = 1, 2 , . . . } ) ) . It is evident that {A, B} surrounds {x: E :ci = 1}, but for small positive e, B intersects {x: ~ xi = 1 + e}, while A intersects { x : E z i = l - e }

(Note: This example, and all the examples hereafter, may be constructed in any infinite-dimensional Banach space whatsoever; they do not depend upon special properties of l~ . )

1.3 LEMMA. Let ,A = (A~: 1 < i < n) be a family o f sets in a locally convex space X with d im(X) = n - 1. I f .A surrounds a point p, then no hyperplane intersects all o f the ( Ai ).

Proof. Suppose, for a contradiction, that there exists a set of points (a/: 1 _< i _< n) with ai E Ai, and that these lie in a hyperplane: that is, in a flat of dimension n - 2. Every n - 1 of the convex sets co{aj: j # i}, i = 1 , . . . , n, intersect; therefore, by Helly's theorem [9], some point q is contained in all of

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COMMON SUPPORTS OF FAMILIES OF SETS 3

them. Hence, IJ~=l(co(I, Jj¢i Aj)) is starshaped about q. If A surrounds p, then p C co{bi: 1 _<-i _< n} with bi E Ai; clearly the boundary of this simplex is contained in U~=l(co(I..Jj¢i Aj)) . But every starshaped set is contractible; so LJi~l (co(I.Jj#i A j)) also contains p, a contradiction.

1.4 LEMMA. Let .,4 = (Ai: 1 <_ i <_ n) be a family of sets in a locally convex space X with dim(X) >_ n. l f .A strongly surrounds a flat F, then the following hold:

(a) No element o f A is empty. (b) The codimension o f F >_ n - 1. (c) I f the codimension o f _P = n - 1, then a convex set which intersects each

member of¢4 also intersects F.

Proof. (a) If some set Ai were empty, then we would have

but by hypothesis these two sets have different intersections with F. (b) Suppose that the codimension of F is less than n - 1. Let ~r be the projection

of X onto a subspace complementary to F. Then, i f F intersects cl(co((U~=l Ai))), we have

Thus, in every neighborhood V of the origin, there exists a v E V such that

e U co 7rAi ; j = l

whence F + v intersects U~=1 (co(Ui~j Ai)), and F is not strongly surrounded by A.

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4 ROBERT DAWSON

(c) If,A strongly surrounds F, then there is a neighborhood V of the origin such that for all v E V,

7r(F + v) E cl co 7rAi .

It follows that the singleton set {p} = 7rF is in co(U~= 1 a'Ai); but it is not even in the closed convex hull of any proper subfamily. Thus, p = r la l + . . . + r~a,~, with al C co(TrAi) and ri > 0. To show that F intersects any convex set C that intersects each member of,A, it suffices to show that the points {ai} may be picked arbitrarily from the sets {co(7rAi)}, while the barycentric coordinates {ri) remain positive.

Let P = {(Yl, Y2,. . . , Yn): Yi E co(TrAi)}. The elements of P may be consid- ered as elements t) = (Yi: 1 < i < n) of an (n 2 - n)-dimensional vector space with the Euclidean topology, and P is path-connected as a subset of that vector space. By Lemma 1.3, aff{yi} = 7rX. Thus, the functions 8j(t)), defined by the equation sly1 + s2Y2 + " " + SnYn = p, are well-defined and continuous functions of O. Each sj is positive at a = (ai: 1 < i < n). We shall show that this holds for all t) E P , proving (c).

Suppose that there exists b such that some sj(b) is negative. Let t)(t) be a path in P with r)(0) = ct, 0(1) = b. Then the function sj(t) = sj(~(t)) is positive for t = 0, and negative for t = 1; so it is zero for some t C (0, 1). Let to be the least t C [0, 1] such that any of the 8i(t) are zero. Then the remaining barycentric coordinates si(to) are positive, and the point p is in the convex hull of the corresponding proper subset of .,4, contrary to our assumptions.

Of course, in a finite-dimensional Banach space, we may vacuously strengthen Lemma 1.4 by merely requiring ¢4 to surround F. In an infinite-dimensional space, the first conclusion would still hold i f F was only surrounded; but the 'strengthened' forms of the second and third conclusions do not hold in any infinite-dimensional Banach space whatsoever. The following counterexample illustrates this in loo.

1.5 EXAMPLE. In l ~ , l e t a0 = (1, 1, 0, 0, 0 , . . . ) , al = (½, 0, 0, 1, 0, 0 , . . . ) , and, generally, let (an)0 = 2 -'~, (a,~)2n+l = 1 and let all other coordinates be 0. Let A = cl(co(ai: i = 0, 1 , . . . ) .

Define the bodies B and C analogously, except that (b~)0 = 2 -n and (bn)2n+2 = 1 for each n, while (cn)o = 2 -n and (cn)2n+1 = (c,~)2,~+2 = - 1 for each n. Then the following are easily shown:

(a) The origin 0 = limi(ai + bi + cD/3, and so 0 is in cl(co(A U B U C)). (b) No point x with x0 = 0 is in cl(co(A U B)), cl(co(A U C)), or cl(co(B U

C)). Suppose, for instance, that p C co(A tO B). Then we have

P = E (/~2i+lai + ,~2i+2bi), i=!

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COMMON SUPPORTS OF FAMILIES OF SETS 5

O O . where Ei=l/~Z --- 1 and )~i > 0 for all i. It follows that Ei=lP i = 1; and hence if O o x E cl(co(A U B)), ~i=12Ci - " 1 also. Therefore, for some n, x,z > 0. Suppose

some sequence of points (yj) in co(A U B) converges to x. Then (Yj)0 --+ 0, while (yj)n ---+ xn. But this is impossible, as (Yj)0 >_ 2(n-2)/2(yj)~.

It follows from these observations that {A, B, C} surrounds the hyperplane {x: x0 = 0}. Furthermore, co{a0, b0, c0} intersects all three bodies, but fails to intersect the surrounded flat {x: x0 = xl = 0} which has codimension 2.

1.6 DEFINITION. Let A be a closed subset of a Banach space X. An exposed face of A is a nonempty subset F which is the intersection of A and a supporting hyperplane, say f -1 (a), where f C X* and a = sup{f(x): x E A}. We call £1 a strongly exposed face of A if, in addition, whenever {x~} is a sequence in A and f{xn} --+ a, then {xn} has a cluster point in F. A point of a closed set A which is also a strongly exposed face of A is called a strongly exposed point of el.

1.7 Remark. It is easily shown that A contains all strongly exposed points of cl(co(A)). Similarly, if F is a stronlgy exposed face of A, then F ' = e l (co(F)) is a strongly exposed face of cl(co(A)); and, conversely, if F ~ is a strongly exposed face of cl(co(A)), then F = F ~ Cl A is a strongly exposed face of A.

A Banach space X will be said to have the property (cr) if every body of X is the closed convex hull of its strongly exposed points. Any separable conjugate Banach space has this property, but it is also possessed by other Banach spaces, including 11 (F) for any index set I' (see [4]). For characterizations of the property (or), the reader is referred to Phelps [10].

The following lemma is the generalization of Lemma 1 of [5] to arbitrary finite families of closed sets. The proof is exactly analogous, and will not be given here.

1.8 LEMMA. Let B = cl(Bl + • • • + Bn), where each Bi is a closed subset of a Banach space X and ( B1 + . " + Bn ) is the Minkowski sum {xl + " . + xn: xi C Bi, 1 < i < n}. I fb E B is a point strongly exposed by f E X*, then b is the sum of points bi C Bi, i = 1 , . . . , n, strongly exposed by f .

1.9 LEMMA. Let ¢4 = {Ai: i = 1 , . . . , n} be a family of closed bounded sets in a Banach space X with dim(X) > n. If¢4 surrounds aflat F, then there exists a linear functional ¢ C X* such that sup{q~(x): x E Ai} = 1 for i = 1 , . . . , n; and there exists a closed halfspace K bounded by a hyperplane {x: ~b(x) = k} such that { Ai M K: i = 1 , . . . , n} strongly surrounds a flat G of codimension (n - 1).

Proof Choose p E F so that .A surrounds that 0-flat {p}. For each A~ there is a hyperplane Hi strongly separating {p) and U { Aj: j 7~ i}. Let W be the intersection of the n open halfspaces Ui which are determined by these hyperplanes and which contain p; note that W does not intersect the closed convex hull of any proper subfamily of ¢4.

Let / /~ be the translate ofl l i which contains p. Then we have {p} C [-1 Hi C_ W. As W is open, there exists a neighborhood V of the origin such that for every v E V,

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6 ROBERTDAWSON

v + p E W. Thus, for every v E V, v + ['1//[ C W; and so .,4 strongly surrounds f] H i. It is clear that codim(N HI) < n. If codim(f] Hi) = n - 1, then we may take that fiat to be G, and take K to be any halfspace containing .4.

Otherwise, by Lemma 1.4(b), codim(N Hi) = n; so H~ intersects, but does not contain, (]{H~: i ~ n) . But then H~ intersects N{Hi: i ~ n} so that ['1 Hi is nonempty. Without loss of generality, we may let 0 E N Hi. We note that p

Hi, and that ('1 Hi has codimension n. Let L be a subspace complementary to N Hi, and let r be the projection of X onto L with kernel r] Hi. Let R be the ray in L from 0 through 7rp. R is convex, unbounded, and intersects r (c l (co(U Ai))); so it contains a boundary point q of cl(co(Tr(U Ai))). Furthermore, R \{0} C 7rUj for each j ; so q is not in c l (co(U{rA/: i ~ j} ) ) for any j .

As L is finite-dimensional, there is a linear functional f E L* supporting cl(TrC) at q = limj~r~=lAijaij, where a/j E Ai and ~n=lA/j = 1. As q is not in any cl(co(U{TrAi: i ~ j} ) ) , we may assume the Aij to be bounded away from 0. Let fa" = ¢; then ¢(q) > sup{¢(x): x E di ) for i = 1 , . . . , n. But ¢(q) = limjE~=lAij(a/j); so ¢(q) = sup{¢(x): x E Ai) for each i. Without loss of generality, take ¢(q) = 1. Let g = {x: ¢(x) _> ½); then the bodies {Ai N K: 1 < i < n} are nonempty.

Let G = r - l ( R O - R ) , which is a flat with codimension n - 1. We claim that this flat is strongly surrounded by {Ai O K: 1 < i < n}. Certainly, G intersects cl(co(U{A5 M K })); it remains to show that it does not intersect cl(co(Ui~j {Aj n K}) ) for any i. Let E = K \ W ; then r E is finitely generated by the union and intersection of closed halfspaces in a finite-dimensional space. As r E does not intersect the line R U - R , it is bounded away from it. Consequently, E is also bounded away from G. As cl(co(U{Aj U K: j # i ) ) ) _C E , our claim is proved.

1.10 THEOREM. Let .4 = {Ai: i = 1, 2, . . . , n) be a family of closed bounded sets in a Banach space X with dim(X) >_ n and with the property (a). I f ,4 surrounds a flat, then ,4 has a common support which strongly exposes a face of each member of,4. Furthermore, all these faces lie in a common flat of dimension (n- 1).

Proof. It follows from 1.7 that it is sufficient to consider the case in which each member of .,4 is convex. By Lemma 1.9, there exists a halfspace K which may be represented as {x: ¢(x) > ½}, such that the truncated bodies {Ai fl K: 1 < i < n} strongly surround a flat F of codimension (n - 1). We may, and shall, assume F to be a subspace. Let L be a subspace complementary to F , and r the projection of X onto L with kernel F.

There exists a neighborhood V of the origin in X such that if v E V, then v + G is surrounded by {Ai n K}. Let { v l , . . . , vn} be the vertices of an (n - 1)- dimensional simplex in L n V, with vl + . - . + vn = 0. Let C = cl(co(U{Ai n K})) , B = F M C, and Bi = (vi + F ) n C for i = 1 , . . . , n. We claim that B = 1/n(B1 + . . . + B,~).

C l e a r l y B D 1/n(Bl + ' " + B n ) , a s 1 / n ( v l + . . . + Vn) = 0. T o s e e t h e

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COMMON SUPPORTS OF FAMILIES OF SETS 7

opposite inclusion, suppose that x E F is a convex combination of points a/ E Ai N K, i = 1 , . . . , n. Part (c) of Lemma 1.4 implies that co{a1 , . . . , a~ } intersects vi + F in a point wi E Bi. But 1 /n (wl + . . . + wn) E F N c o { a l , . . . , an) = {x). Thus B C 1/n(B1 + . . . + Bn).

Since B is the closed convex hull of its strongly exposed points, there is a point b E B which is strongly exposed by some f E X*. Furthermore, as sup{~b(x): x E B) = 1, we may choose b to make ~b(b) as close to 1 as we like. Let us take ¢(b) > 1 - d/(2d + 2D), where

d = rain inf{llxi[: x E co{vj: j 56 i}}

D = max sup{lly - xll: y E B, x E n K)} .

Lemma 1.8, applied to B and the sets Bi, implies that b = 1 /n(b l + . . . + b,~), where each bi is a point of Bi strongly exposed by f . As F has already been assumed to be a subspace, we may assume that b -- 0. Let S be the simplex with vertices {bl , b 2 , . . . , bn}, and let M = aft(S); this is a subspace complementary to F. The subspace spanned by f - l ( 0 ) N F and M is a closed hyperplane H, and hence the kernel of a nonzero g E X* with sup{g(x): x E B} = 0. It follows also that sup{g(x): x E Ai N K ) = 0 for i = 1 , . . . , n.

To show that {x: g(x) = 0} is a common support for the sets {A,. N K: 1 < i < n ) , it suffices to show that if (Xm) is a sequence in any Ai and i fg(xm) ~ 0, then (Xm) has a cluster point in M. (Since sup{g(x): x E Ai} = 0, such sequences exist.) This is certainly the case when (Xm) has a subsequence in M. Thus, we may assume (Xm) C_ X \ M so that, for each m, xm and S define an n-simplex Sm. Now, Sm fq F is a line segment with endpoints 0 and x~m. Clearly, g(xm) ~ 0

! because g(Xm) _< g(X~m) < 0. Since {x~m} C_ F N C, x m ~ 0. For each Xm, let

t A m x m + ( 1 Am)ymwi th0_< Am < 1. Ym E S be the point satisfying x m = Clearly (Ym) has a subsequence (Ym(j)) converging to a point y E S, and we may also assume that (Am(j)) ~ A. Since y # 0, A # 0. It follows easily that (Xm(j)) converges to the point (1 - 1/) ,)y E M.

Therefore, the hyperplane H supports each truncated set Ai N K at some point ai. It remains to show that H also supports the original sets at these points; for this, it will suffice to show that ¢(a/) > ½. Consider the sequence (Ym) defined above.

X t X t If q~(ym) _< ¢ ( m ) , then ¢(Xm) _> ~b( m); and if this occurs for infinitely many values of m, ¢(ai) >_ ¢(b) > ½.

Otherwise, there must be infinitely many values of m for which ¢(Ym) > ¢(x~ ). r is on the boundary of the simplex Sin; so it follows that each Now, each x m

Ym is on the boundary of S. Hence, (Ym) has a convergent subsequence (Yn(j)) t which is contained in one face of S, and for which (Yn(j)) > ¢(xn(j)). But

IIx'(j)- y,~(j)ll >- d, while I Ix ' ( j ) - xn(j)ll <_ D. Thus

X t 1 - A n ( j ) tl n(j) - x tj)ll D = X I < A-(i) If .(j) Y . ( j ) I I - d "

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8 ROBERT DAWSON

Now,

¢(xn( j ) )

Hence,

¢(ai)

_>

, A ~ ( j ) - 1 1 ¢(xn(j)) + ¢(Yn(j))

X ! ¢ ( ~ ( j ) ) + 1 - A,~(j) (¢(x~( j ) ) - ¢(Yn(j))) An(j)

, D ¢(x (j)) + ~- (¢(x~(j)) - ¢(Yn(/))) (as ¢(Yn(j)) > ¢(x~(j))

d + D D d ¢(X~n(J)) - d ¢(Y~(J))

d + D D > ----d-- ¢(x~(j)) - ~- (as ¢(y.(j)) _< 1).

= lim ¢(x,~(j)) j--*oo

J - ~ - - ¢ ( x . ( j ) ) -

d + D D lim ¢(x~(j)) - - -

d j~oo d

d + D D - ~ ¢ ( b ) - ~

d + D ( d ) > ~ 1 2 (d+ D)

1

2"

D d

We conclude that the hyperplane H supports the sets {Ai} at the points {ai}. Furthermore, the intersections {H N Ai} lie in the (n - 1)-flat M, and are strongly exposed faces.

2. Infinite Families of Sets

While the results of the previous section apply to infinite-dimensional Banach spaces, a key 'geometric' part of the construction takes place in a finite-dimensional subspace, relying on Carath6odory's theorem. In this section, we will exhibit some intrinsically infinite-dimensional results applicable to infinite families of sets, in which the geometry and analysis interact closely.

2.1 LEMMA. Let .4 be a family of closed bounded sets in a Banach space X. Sup-

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COMMON SUPPORTS OF FAMILIES OF SETS 9

pose that a line ! exists which is surrounded by A, and which is not wholly contained in el(co(U.4)). Then an f E X* exists such that f # 0 and {sup f[Ad: a i C .4} is a singleton.

Proof Let C be the closed convex hull of U = [.J.4. Let z be a point of intersection of I with the boundary of C. Let f E X* be such that f (z ) = 1, and f (x ) _ 1 for all x E C; we claim that such an f satisfies the conclusion of the lemma. If not, there is a member of .4, say Aa, such that sup f[A1] = 1 - ~ for some o~ > 0. Let (xn) be a sequence in co{A/: Ai E `4} which converges to z. We may assume that, for each n, x,~ is a linear combination of points in the first re(n) elements of`4; then xn = Ekm_-l)~nka~k, where 0 _< )~k, ~ = 1 ) ~ = 1 and an\ C Ak.

Now,

x n = M l a n l + (1 - ; ~ l ) E 1 - - /~nl a n k . k=2

It follows that lim inf{Anl: n = 1, 2 , . . . } > 0. For otherwise, there exists a subsequence { An~ 1 } converging to 0. But then the sequence

~(~') Ani k b~ = ~ 1 - - A n i 1 anlk

k=2

converges to z. This however is impossible, as {bn~ } is in the convex hull of the union of a proper subset of `4, and by hypothesis the closure of such a convex hull does not intersect i.

Thus we can find n such that )~,~l > 0. For any such n,

f ( z ) _< Anlf(anl) + (1 - Anl) _< Anl(1 - (~) + (1 - Anl) = 1 - c~Anl < 1,

a contradiction.

2.2 Remark. The hypotheses of the preceding lemma imply that every element of .,4 must contain a point from the (countable) set {a,~k} which is contained in no other element of .4 (otherwise z would be contained in the closed convex hull of a proper subfamily). Thus the family .4 cannot be uncountable. However, Example 2.5 below exhibits a countable family of bodies that satisfy the hypotheses of the lemma, so the lemma and the following theorem are not vacuous.

2.3 THEOREM. Let .4 be a family of weakly compact bounded sets in a Banach space X, and suppose that 1, C, z, and f are as in the preceding lemma. Then {x: f (x ) = f (z)} is a (definite) common support of members of.4.

Proof The result follows from the lemma, as f attains, by weak compactness, its supremum on each member of .4.

It might at first appear that the requirement that 1 should intersect X \ C is redundant. It seems plausible that, ifl intersects C but not any c l (co(U\Ai)) , weak

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10 ROBERT DAWSON

compactness of the bodies Ai might force ! fq C to be compact. However, this is not the case; in any infinite-dimensional Banach space, we can find a countable family of line segments such that the closed convex hull of their union contains an entire line while the closed convex hull of any proper subfamily misses this line completely. To construct this, and other instructive counterexamples, we will require a countable family of points {ai} and a point b which is in their closed convex hull but not in the closed convex hull of any proper subfamily.

This is less straightforward than consideration of finite families of points might suggest. Firstly, if we replace 'closed convex hull' by 'convex hull', no such con- figuration exists: recall that every point in the convex hull of an infinite family of points is in the convex hull of a finite (hence proper) subcollection. Secondly, in many Banach spaces the closed convex hull of a set of points can be surpris- ingly large: for instance, in 12(Z+), the origin is in the closed convex hull of an orthonormal basis. However, it is possible to construct such a configuration in any infinite-dimensional Banach space.

2.4LEMMA. lnanyinfinite-dimensionalBanachspaceX, letao = (1, 0, 0, 0 , . . . ), al = (--2, 1, 0, 0 , . . . ), and, generally, let (ai)i-1 = - 2 i, (ai)i = 1 and all other coordinates of ai be O. Then the origin is in cl(co({ai})) but not in the closed convex hull of any proper subcollection.

Proof. The nth term of

n

E 2(i-i2)/2ai i=0

n

E 2(i-i2)/2 i=0

is 2 ('~-n=)/2, and all other terms are O; thus

n

E 2(i-i2)/2ai

lim i=0 = (0, O, O, ) n " ' "

n--+c~ E 2(i-i2)/2 i=0

whence it is clear that 0 C C. On the other hand, if we omit any of the points, the origin is not in the closed

convex hull of the union of the remaining points. Specifically, if for any fixed n

b = ~ . N i a i , ~ Ni = 1 i#n i#n

then for some j > O, I bj I > 2-(n2+n+2)/2; thus no sequence of such convex combinations can converge in the norm topology to O. It remains to prove this claim.

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COMMON SUPPORTS OF FAMILIES OF SETS 11

Suppose that [ bj [ _< 2 -(n2+n+2)/2 for all j > 0. As bn-1 = Ar~-l,

I An-1 1<_ 2 -(n2+n+2)/2.

But bn_2 = A n - 2 - 2n - lA~- l , so

I An-2 [ < 2-(nZ+n+2)/2 + 2-(n2-n+4)/2 < 2-((n-1)z+(r~-l)+2)/Z"

By induction, we find that for i < n,

I Ai 1< 2 -(i2+i+z)/z ~_ 2 -(i+2).

Also, an induction shows that I Ai [ _< 2 -(i+2) for i > n; we conclude that E Ai < ½, contradicting our assumption above that E Ai = 1.

2.5 EXAMPLE. In loo, let A0 be the set {(z0, 1, 0, 0, 0 , . . . ): z0 C [ -1 , 1]}, and for i > 0 let Ai be the set of points ai with

(ai)0 = zi E [--2 (i2+i+2)/2, 2(i2+i+2)/2], (ai)i = --2 i, (ai)i+l = 1,

and all other terms zero. Then the line I = {(t, 0, 0, 0 , . . . ): t E R} is contained in cl(co([,J{Ai})), but for any proper subfamily a of {Ai}, l does not intersect cl(co(a)) . Furthermore, although these bodies are weakly compact, and every proper subfamily has a common support, the whole family does not. If we restrict the parameters z~ to the range [0, 2(iz+i+2)/2], the resulting family of bodies has exactly one definite common support.

2.6 Remark. In Lemma 2.1 and Theorem 2.3, we do not use both ends of the line i. It would be sufficient to require the existence of a ray r intersecting both C and X\C, but not intersecting the closed convex hull of any proper subfamily of {Ai}. Such a surrounded ray may exist even when no surrounded line does (see the following example).

2.7 EXAMPLE. In lo~ let a = ( - 1 , 0, 0, 0 , . . . ) ; b0 = (1, 1, 0, 0, 0 , . . . ) ; and for i > 0, let bi be the point with (bi)0 = 1, (bi)i = - 2 i, (bi)i+l = 1, and all other terms zero. For i = 0, 1, 2 , . . . , let Ai be the line segment [a, bi]. Then the ray {(t, 0, 0, 0 , . . . ) : t > 0} intersects C and X\C, but not (Lemma 2.4) the closed convex hull of any proper subfamily of {Ai}. However, it is evident that for any line 1, each of the endpoints of the segment 1 U C must be in the closed convex hull of a proper subset of {a, b0, bl , . . . }, and at least one of these proper subsets must omit one of the {bi }. Thus, the corresponding endpoint is in the closed convex hull of a proper subset of the { Ai}.

3. Separating Common Supports

The problem of the existence of a separating common support for a family of sets can often be reduced to that of the existence of a definite common support for a

Page 12: Common supports of families of sets

12 ROBERTDAWSON

suitably defined new family. Suppose A and B are nonempty bounded subsets of a locally convex space which are strongly separated by some closed hyperplane H. Let I be the intersection of all closed halfspaces which contain B and whose boundary separates B and A. We shall mean by a shadow of A with respect to B any nonempty set A t which is strongly separated from A U B and which is the intersection of I with a translate of H. It is easily verified that any such A t is a closed bounded convex set.

3.1 PROPOSITION. Let X be an at least two-dimensional Banach space and suppose A and B are bodies, one o f which is weakly compact. Then {A, B} has a definite common support i f A U B fails to be convex, and a separating common support i f and only i f A and B are separated.

Proof. Suppose A U B fails to be convex; then {A, B) surrounds a flat F. Let HA and 1-IB be hyperplanes separating F from A and B respectively, let W be the intersection of the two closed halfspaces determined by these hyperplanes and containing F, and let C = co(A U B).

C is clearly convex and bounded; furthermore, if one of the bodies (without loss of generality, B) is weakly compact, C' is also closed. Thus, the Bishop-Phelps theorem [1] implies that W N C contains a support point p of C. Clearly any hyperplane which supports C' at p must support A and B and fail to separate them.

Suppose A and B are separated. If A and B have nonempty intersection, then any separating hyperplane is also a common support. If A and B are disjoint, then they are strongly separated. In this case, A has a shadow A t with respect to B. Since we may assume B is weakly compact, the first part of the proposition implies that {A', B ) has a nonseparating common suppor t / / . Suppose, for a contradiction, t h a t / / d o e s not support A and let a ~ E A t N / / . S ince / / separa tes A and B, co(A t_J {at}) must be disjoint from B and hence strongly separated from B. It follows that a t is strongly separated from A ~, which is impossible.

3.2 Remark. If neither A nor B is weakly compact, the conclusion need not hold: a counterexample in e0 is given in [8].

3.3 PROPOSITION. Let A and B be closed and bounded subsets of a Banach space X with dim(X) >_ 2 and with the property (~). I f A and B are strongly separated then {A, B} has a separating common support which strongly exposes a face o f each set; and we may assume that these faces lie in a common line.

Proof. Let A' be a shadow of A with respect to B. Theorem 1.10 implies that {A ~, B} has a definite common support f E X* which strongly exposes a face of each set, and that there exists a line L which contains both of these faces. We may assume that sup{f(x): x E 13} = sup{f(x): x E A'} = 0. Choose some a' E A' fq L and b E B fq L. Since inf{f(x): x E A} = 0, there is a sequence xn C_ A such that f(Xn) ~ 0. Let x~n be the point of intersection of A ~ and the line

t a t" through xn and b. Clearly f(xtn) ~ 0; hence x n ~ It follows that {xn} has a cluster point in L.

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COMMON SUPPORTS OF FAMILIES OF SETS ] 3

Acknowledgements

The results presented here were obtained in collaboration with Michael Edelstein. In particular, he outlined and developed much of the proof of Theorem 1.10. Theorem 2.3 and the results of the third section are entirely his. I am communicating these results with his permission.

I would also like to acknowledge the contributions of J. B. Collier, with whom Michael Edelstein initially discussed some of this material; of A. Thompson, who read the final drafts; and of the referee, who has suggested several significant changes from the original version.

This work was partially funded by a grant from NSERC.

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