ON THE STRUCTURE OF SUPPORT POINT SETS

Let X be a metrizable compact convex subset of a locally convex space. Using Choquet's Theorem, wc determine the structure of the support point set of X when X has countably many extreme points. We also


INTRODUCTION.
Let X be a subset of a locally convex space E A continuous linear functional J on X is said to be associated with fE X if ReJ(f)=max{ReJ(g):g E X} and ReJ is non constant on X.In this case we call asuormrtpoint of X The set of support points of X will be denoted by Supp X The set of extreme points of a convcx subset F of E will be denoted by Ext F Let D {z: [z[ < C} and equip the space A of functions analytic in D with the topology of uniform convergence on compact subsets of D This topology is metrizable [1, p.1].Every continuous linear functional on A is induced by a sequence {bn}n__ 0 which satisfies lim sup ]bnl 1/n oo o0 <1 and J(0 E anbn for f(z)= E an zn e n=0 n=0 A [1, p.36].Recently, the support points of many subclasses of A have been studied.For more details see [1] and [2].
Ill Section 2, we consider a metrizable compact convex set X in a locally convex space.Using Choquet's theorem we determine the structure of Supp X when Ext X is countable (Theorem 2.1).
In Section 3, we consider the classes: P(p) {f(z) an zn e A ]an] p <_ 1}, _< p < .In Theorem n-1 n-I 3.4, we determine Supp P(p) Indeed, it is shown that Supp X is in 1-1 correspondence with a proper subset of Supp Ball(p) 2.
SUPPORT POINTS OF SETS WITH COUNTABLY MANY EXTREME POINTS.
Let E be a locally convex space, and suppose that X is a metrizable compact convex subset of E A theorem by Choquct [3, p.19] says that if x E X then there exists a probability measure Px on X supported by Ext X such that L(x) I L dpx for every L in E* In case Ext X is countable (possibly finite), we have the following: Ext X CIIOQUET'S THEOREM (Countable Case).Suppose Ext X {fn} is countable.Then X {' Anfn: A n _> 0 !i for each n and A n 1}.n PROOF.Let (/ X By Choquet's Theorem, there exists a probability measure pf on X supported by {fn} such that L(f) I L dpf.Thus L(f) [ #f(fn) L(fn) Hence L(f-npf(fn)fn) O. n {fn} Since'this is true for every L in E*, we get f Pf(fn) fn as required.n We proceed to the main result of this section.THEOREM 2.1.Let X be a metrizable compact convex subset of a locally compact space E such that Ext X {fn is countable.For each positive integer n, set Kn equal to the closed convex hull of {fi: n}.Then (1) Supp X is contained in the union of those Kn which are proper subsets of X.
(2) Kn C_ Supp X if and only if fn closed affine hull of (fi:i :/: n}.PROOF.To prove (1), let f Supp X.By Choquet's Theorem, we can write .A ir with each A > 0 and .A Let b be a continuous linear functional associated with f.Then Re b(f) .A Re t(fi) < A Re (f) Re b(f) Hence we must have Re b(fi) Re (f) whenever A > 0. On the other hand, since Re is non-constant on X, we must have Re b(fi) Re b(t') for some i.We conclude that A 0 for some i, as required.
To prove (2), suppose that fn does not belong to the closed affine hull H of {fi: :/: n} and fix g ( Ku Then H g is a closed real subspace of E not continuing fn g A version of the Hahn-Banach theorem [4, page 59] gives a functional J in E* whose real part vanishes on H-g while b(fn -g)=-1.Set b(fn+l)= b.Then b(fn) =bwhile b(fi) b(fn+l) =b for every n.Thus, b(g) =b for all g (/ K n For any h in X.by Choquet's Theorem, we have h .f/i fi with fi -> 0 and .i 1.Thus b(h) /n(b-1) + /i b in =b-/n _ b.This shows that g SuppX.
Conversely, assume that Kn C_ Supp X For ease of notation we take n and assume Ext X {fn}n= is infinite.By. assumption, fi is a support point of X.Let be an associated linear functional in E* and of {fi'i = 1} C_ S. On the other hand, in view of Choquet's Theorem, if fl E S then Re would l,c constant o,, X.Thus fl S and consequently, fl closed affine hull of {fi: = 1} EXAMPLES.(1) Let X be a triangle in R 2 with vertices fl' f2 and f3" Tl,,,sc vertices are the extreme points of X and the affine hull of any two of them is a line, not containing the third.The theorem guarantees that 3 Supp X J Kn which is indeed the boundary of X n=l Let X be a square in R 2 with vertices fl' f2' f3 and f4 The affine hull of any three of the fi's is all of (2) R 2 In particular, each fi / am,le hull of -{f:: :/: i}.The theoren guarantees that no Kn is contained in S,,pp X.
fact, Supp X the boundary of X has no interior.
(3) Let T be the fmnily of all functions which are analytic and univalent in D and take the for,u f(z) x> zn zl zan ,an >0.By [5],Ext.T={fn}=l where fl(z)=z and fn(z)=z-for n>l.For ,>l,itis rim2 clear that fn does not belong to the closed affine hull of the remaining {f i} so K n C_ Supp X by the second part n=2 of the Theorem.Since fl is a limit point of the remaining fi's K X and Supp X [.J Kn by the first part of n=2 the Theorem.COROLLARY 2.2.Let X be as in Theorem 2.1.Then Supp X U -6 (Ea) where each Ea is a subset of Ext X. PROOF.Suppose (/ Supp X and is an associated linear functional with f.Writing .A ifi, we see that Re(f)=Re(fi) whenever A :/:0.Take Ea={fi ]A :/:0}.Then ffi(Ea) C_ SuppX.
The theorem says these Ea are proper subsets of Ext X i.e., they cannot be "too big".The next proposition implies that they can't all be singletons, i.e., "too small" PROPOSITION 2.3.Let X be a compact convex subset of a locally convex space.If X ha.s more tl,an two extreme points, then Supp X is uncountable.
PROOF.Without loss of generality we may assume that 0 X Let fl and f2 be two independent clenents of X, and let 1 and 2 be continuous and linear functionals such that l(fl) 2(f2) and l(f2) d.2(fl) 0.
Define : X R 2 by ,(f) (bl(f), 2(f)).Then (X) is a compact convex subset of R 2 with non empty interior.Since (X) has uncountably many boundary points, Supp((X)) is uncountable.Since -1 takes support points to support points, we see that Supp X is uncountable too.2ri c EXAMPLE.Take fn e-'n'for n 1, 2 and X -6 {fn} in R 2 Then Supp X U co {fn,fn+l} n:l llere all the Ea's have cardinality two even though Ext X is infinite.COROLLARY 2.4.Let X be as in Theorem 2.1.Then Ext X Supp X if and only if X has two extreme points.

3.
SUPPORT POINTS OF CERTAIN CLASSES OF ANALYTIC FUNCTIONS.
For _< p < Co define P(p) {E an zn A: E lanl p -< 1] It is easy to see that the clscs P(p) are n=l n=l compact convex subsets of A These classes are closely related to Ball(gp) and we will find that Supp P(p) is in one- to-onc correspondence with a proper subset of Supp Ball(p) As a corollary, we determine the support l)oints of ccrtain fanilies of univalent functions.We use the notation a for the sequence {an}x)=l Ve begin with a simple observation.
PROPOSITION 3.1.Let X be the unit ball of a Bausch space E. Then Supp X {x X llxll l}.If is a.sociated with x, then (x) I111 PROOF.That every vector of norm one belongs to Supp X is a consequence of the tlahn-Banach theorem.Suppose conversely that the real part of X* achieves its maximum over X at x Since X is closed under multiplication by scalars of absolute value at most one, we have Re (x) sup Re (y) I111 Thus I111 Re (x) _< IIll Ilxll and so IIxll-1.Moreover (x)-I111 implies Re (x)_> I(x)l, so (x)is in fact real EXAMPLE.The family P(p) "looks like" the unit ball of p but we cannot immediately apply Proposition 3.1 support points.For example, the sequence n}n= l}n=l belongs to the unit sphere of 2 to filld its bu CO ] an zn is not a support point of P( 2).The problem is that any non-constant linear functional (bn)-I 2 which I1 its maximum at {an}nco=l must be a scalar multiple of {an}n=l.So lim sup n bn which does not a.sulne correspond to a continuous linear functional on A.
We find the support points of P(p) by making the remarks in the preceeding example more precise.PROPOSITION 3.2.Suppose T E F is a linear, injective, and continuous map between topological vector spaces E and F and let X be a subset of E Then Tx Supp TX if and only if x Supp X and some linear functional associated with x belongs to range T* PROOF.Recall that T* F* E* is defined by T* oT.Suppose Tx SuppTX and choose F* with Re(Tx)=max Re(Ty).Set b= ,oT; then range T* Re b(x) max Re(y),andinjectivityof T yX yX implies that Re is not constant on X Conversely, let range T* such that Re b(x) max Re (y) Write T F* Then Re (Tx) max (y), and Re cannot be constant on TX since Re b is not constant on X. yTX PROPOSITION 3.3.Let a X-Ball(p),(1 <p<oo),with Ilallp= 1,and b gq.Then: (1) If b is associated with a, then there exists f 0 with /lbnl q -lanl p for all n.
an lanl p-1 if an :/:0 PROOF.One way to see this, is to replace {p by ep(H) where it(n) n n 2,3 in the proof of Theorem 3.4.
R.EMARK.One can define P(o) {f (z) an zn suplanl < One can show.usi,,s a,, arg,,n,c,,t si,,,ilar !i--" ,:X:3 to tl,e proof of Theorem 3.4, that.Supp P(oo) {f(z) : anz n lanl for some n > 1}. n=l i=2 set S {g E E: Re b(g)= Re b(f)} Note that S is a closed affine subspace of E. Since Re b(f)= 2 i l --_ 1 Re b(fi) < 2 Re b(f) Re b(O, we have Re b(fi)= Re b(f) for alia >2.Thus the closed affine hull i=2 i=

( 2 )
If bn= { otherwise then b is associated with a. COltOLI, ARY 3.5.A function f(z) + al, n is a support point of Q(p) if and only if and I.1 , for om I,ositive integer N 2, if p n=2 n=2