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EXTREME INTEGRAL POLYNOMIALS ON A COMPLEX BANACH SPACE

SEÁN DINEEN

Abstract

We obtain upper and lower set-theoretic inclusion estimates for the set of extreme points of the unit balls ofPI(nE)andPN(nE), the spaces ofn-homogeneous integral and nuclear polynomials, respectively, on a complex Banach spaceE. For certain collections of Banach spaces we fully characterise these extreme points. Our results show a difference between the real and complex space cases.

1. Introduction

Geometric properties of spaces of polynomials, e.g. smoothness, extreme points, exposed points, norm attaining polynomials, etc. have been invest- igated by a number of authors in recent years. We refer to [1], [3], [7], [8], [13], [12], [18], [20]. In particular, Ryan-Turett [18] and Boyd-Ryan [3] in their investigations examined the extreme points of the unit ball of the space of integral polynomials defined on arealBanach space. In this paper we study the extreme points of the unit ball of the space of integral polynomials defined on acomplex Banach space. Throughout this paperEwill, unless otherwise stated, denote a Banach space over the complex numbersC. We letBE orB denote the open unit ball ofE. We refer to [11] for basic facts on polynomials on Banach spaces and to [9], [10] for the geometry of Banach spaces.

2. Extreme Points

IfEis a Banach space overCwe let

ExtR(E)= {xE:x =1,x+λy ≤1 for−1≤λ≤1 impliesy=0} and

ExtC(E)= {x∈E:x =1,x+λy ≤1 for|λ| ≤1 impliesy=0}.

The author wishes to thank Bogdan Grecu and Manolo Maestre for helpful remarks.

Received September 20, 2000.

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Thus, ExtR(E)is the set of all (real) extreme points of the unit ball ofE and ExtC(E) is the set of complex extreme points of the same set. Clearly ExtR(E) ⊂ ExtC(E). The inclusion may be proper since ExtR(L1(0,1))is empty while ExtC(L1(0,1))consists of all unit vectors.

A Banach space is strictly convex (or rotund) (respectively strictlyc-convex) if

ExtR(E)= {xE:x =1}

(respectively ExtC(E) = {xE : x = 1}). As noted above the Banach spaceL1(0,1)is strictlyc-convex but is not strictly convex.

The role of extreme points in functional analysis, convexity theory, linear programming and optimisation theory is well documented. Complex extreme points were introduced by Thorpe and Whitley [21] in order to prove a strong maximum modulus principle for Banach-valued holomorphic functions. Since their introduction they have proved useful in the study of the Shilov boundary, complex geodesics, invariant metrics, bounded symmetric domains and JB- triple systems. We refer to [2], [4], [5], [14], [15], [22], [23] for details.

3. Integral and Nuclear Polynomials

IfEis a complex Banach spaceP(nE) denotes the space of continuousn- homogeneous polynomials onEendowed with the norm·:= ·B of uni- form convergence over the unit ball B of E. A polynomial PP(nE) is said to beintegralif there exists a regular Borel measureµon the unit ball of E, BE endowed with the weaktopology, such that

(1) P (x)=

BE

φ(x)ndµ(φ)

for allxinE. We letPI(nE)denote the space of alln-homogeneous integral polynomials onEand we endow this space with the norm ·I := infµ where the infimum is taken over allµsatisfying (1).

Ann-homogeneous polynomialP onEisnuclearif there exists a bounded sequencej)jEandj)jl1such that

(2) P (x)=

j=1

λjφj(x)n

for allx inE. The space of all nuclear n-homogeneous polynomials on E, PN(nE), is a Banach space when PN is defined to be the infimum of

j=1j| φjntaken over all representations ofP satisfying (2).

Proposition3.1. LetEdenote a complex Banach space.

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(a) The spacePI(nE)is isometrically isomorphic to

n,s,E . (b) PN(nE)PI(nE)P(nE)and· ≤ ·I ≤ ·N. (c) IfφEthenφnPN(nE)and

φn = φnI = φnN

(d) IfEhas the approximation property thenPN(nE)is isometrically iso- morphic to

n,s,πEandPN(nE)=P(nE)isometrically.

(e) If1

n,s,E (in particular ifE has the Radon-Nikodým Property) thenPI(nE)andPN(nE)are isometrically isomorphic.

(f) ExtR(PI(nE))⊂ {φn :φE,φ =1}. (g) ExtR(PI(nE))⊂ExtR(PN(nE)).

Proof. Parts (a), (b), (c) and (d) are well known. Parts (e) and (f) are due independently to C. Boyd-R. A. Ryan [3] and D. Carando-V. Dimant [6] while (g) can be deduced from the proof of [3, Theorem 2].

Remarks. (i) By (b) and (g) we have

PN(nE)PI(nE) and ExtR(PI(nE))⊂ExtR(PN(nE)).

This contrasting information is quite useful since ExtR(PI(nE))has good ab- stract properties while calculations are easier with elements of ExtR(PN(nE)). (ii) In general we do not know ifPN(nE)is a dual Banach space but, by (a) and (g), ExtR(PN(nE))is non-empty.

ForrealBanach space we also have the following result which is in contrast to the results we obtain in the next section forcomplexBanach spaces.

Proposition3.2 ([3], [18]). IfEis a reflexive Banach space overRand n >1then

ExtR(PI(nE))= { ±φn :φE,φ =1}.

4. Extreme Polynomials

Proposition4.1. IfEis a complex Banach space andn≥1then {φE :φn∈ExtR(PN(nE))} ⊂ExtC(E).

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Proof. Supposeφ = 1 andφ ∈ExtC(E). Then there existsψE, ψ =0, such thatφ+λψ ≤1 for allλ∈C,|λ| ≤1. Ifω=en+12πi then

n j=0

+ωjψ)n=n

k=0

n

k ψkφn−k n

j=0

ωkj . Ifk=0 thenn

j=0ωkj =0. Hence 1

n+1 n j=0

+ωjψ)n =φn.

Sinceφ = 1 and φ +ωjψ ≤ 1 for all j, Proposition 3.1(c) implies φnN =1 and,(φ+ωjψ)nN ≤1 for allj. Henceφn/ExtR(PN(nE)). This completes the proof.

IfEis a complex Banach space we say thatxandyinEare equivalent if there existsα∈C,|α| =1, such thatx =αy. IfAEwe denote by[A] the cardinality of the set of equivalence classes inA. For example it is well known that[ExtR(n1)]= n. For complex extreme points we obtain a different type of result.

Corollary 4.2. IfE is a finite dimensional complex Banach space of dimension≥2then

[ExtC(E)]= ∞.

Proof. By Proposition 4.1 and the Krein-Milman Theorem dim(PI(nE))≤ [ExtC(E)]

for any positive integer n. Since the monomials zjwn−j, j = 1,2, . . . , n are linearly independent, dim(PI(nE))= dim(P(nE))is at leastn. Hence [ExtC(E)]≥nfor alln. This completes the proof.

IfEis a Banach space we say thatAE isE-transitiveif for allθ,φ inAthere exists an isometry ofEonto itself,T, such thatφT =θ, that is

tT (φ)=θwheretT denotes the transpose ofT.

Corollary4.3. If the set of complex extreme points of the unit ball ofE isE-transitive then

ExtR(PI(nE))= {φn:φE, φ∈ExtC(E)} and, in particular,

ExtR(E)=ExtC(E).

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Proof. By Proposition 3.1(a), (f) and (g) and Proposition 4.1, ExtR(PI(nE)) contains an element φn where φ ∈ ExtC(E). Let ψ ∈ ExtC(E). By E- transitivity of the set of complex extreme points there exists an isometryT of E such that ψT = φ. If PPI(nE) and ψn ± PI ≤ 1 then ψn◦T±P◦TI ≤1. Sinceψn◦T =◦T )n=φnandφn∈ExtR(PI(nE)) this impliesP◦T =0. HenceP =0 andψn ∈ExtR(PI(nE)). An application of Proposition 4.1 completes the proof.

Example 4.4. LetA denote a W-algebra , i.e. aC-algebra which is also a dual Banach space. We can suppose, without loss of generality, thatA is aC-subalgebra ofB(H), the space of bounded linear operators on the Hilbert spaceH. By [15] the real and complex extreme points of the unit ball of aC-algebra coincide.

Let

A= {wN (H): trace(vw)=0 for allvA }.

whereN (H )is the space of trace class operators onH. By [17, Theorem 4.2.9],A:=N (H)/Ais the unique isometric predual ofA. The(A,A) duality is given by

(u, w+A)=trace(uw)=

x∈E

uw(x), x

whereuA, w+AAandEis an orthonormal basis forH. LetU denote the set of unitary elements inA, i.e.

U = {uA :uu=uu=1H }.

ClearlyuwAforuunitary inA andwinA. Hence the mapping ut :AA, ut(v+A):=uv+A

is well defined and easily seen to be an isometry ofA. Ifv1andv2are unitary elements inA thenv2v1is also unitary and for allw+AAwe have

(v2(v2v1)t, w+A)=(v2, v2v1w+A)=

x∈E

v2v2v1w(x), x

=

x∈E

v1w(x), x =(v1, w+A).

Hence

v1=v2(v2v1)t

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andU isA-transitive. Results in [16] (see also Example 4.6) imply that ExtC(A)=ExtR(A)=U

i.e. the real (and complex) extreme points coincide with the set of unitary elements whenA is one of the following :

(a) a commutativeW-algebra, (b)B(H),H finite dimensional, (c) a type II1factor.

Hence, by Corollary 4.3, ifA is any of the aboveW-algebras then ExtR(PI(nA))= {φn :φ∈ExtC(A)}.

If dim(A)≥2 then ExtC(A)is a proper subset of{x :xA,x =1}. (See for instance [5, Theorem 1.9]). For example ifA =mthen its predual ism1 and ExtR(PI(nm1))consists of allφn whereφlies in the distinguished boundary of the unit polydisc inCm.

The above method does not extend to allW-algebras. By [16, Corollary 2]

ExtC(B(H))= {uB(H):uu =IH oruu=IH }.

IfH is infinite dimensional then the forward shift,S, is a non-unitary extreme point of the unit ball ofB(H). By [16] isometries ofB(H )map extreme points to extreme points and unitaries to unitaries. Hence no isometry ofB(H) mapsStoIH and ExtC(B(H))=ExtR(B(H))is notB(H)-transitive.

Proposition4.5. IfEis a finite dimensional complex Banach space then {φn:φ∈ExtR(E)} ⊂ExtR(PI(nE))

for alln.

Proof. Letφ∈ExtR(E). Supposeφn/ExtR(PI(nE)). SinceEis finite dimensional Proposition 3.1(e) implies thatφn/ExtR(PN(nE)).

Hence there existP andQinPN(nE),P =Q, andλ∈ R, 0< λ < 1, such that

φn=λP +(1−λ)Q

and P = Q =1.

SinceE is finite dimensional so also is PN(nE). Hence the unit ball of PN(nE)is the convex hull of its extreme points and thus, by Proposition 3.1(f), there exists i)k+li=1E, φi = 0, such that λP = k

i=1φni, λPN = k

i=1φni,(1−λ)Q=k+l

i=k+1φni and(1−λ)QN =k+l

i=k+1φin.

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Hence

φnN = φn =1=k+l

i=1

φni.

Choosex0E,x0 = 1, such thatφ(x0) = 1. On differentiatingφn = k+l

i=1φni atx0we obtain

φ =φn−1(x0=k+l

i=1

φin−1(x0i. Hence

1= φ ≤k+l

i=1

in−1(x0)|φik+l

i=1

φin =1.

and we can choose for eachi,βi ∈C, with|βi| =1 such that φ=

k+l i=1

in−1(x0)| · φi · βiφi

φi.

Sinceφ ∈ExtR(E)this implies that for eachi,φi =αiφfor someαi ∈C.

HenceP = φn = Q. This contradicts our hypothesis and shows thatφn ∈ ExtR(PN(nE))=ExtR(PI(nE))and completes the proof.

Example 4.6. IfJ is a JB-algebra then, by [4, Lemma 4.1] and [15, Theorem 11],

ExtR(J)=ExtC(J).

Hence Propositions 4.1 and 4.5 imply

n :φ∈ExtR(J)} =ExtR(PI(nJ))

for any finite dimensional JB-algebra J. This result gives an independent proof of the result in Example 4.4 for finite dimensionalC-algebras.

Example4.7. IfE is a strictly convex finite dimensional Banach space then ExtR(PI(nE))= {φn :φE,φ =1}.

This is similar to the result for real Banach spaces quoted above (Proposi- tion 3.2).

To extend Proposition 4.5 to infinite dimensional spaces we requireweak- exposedpoints. For convenience we introduce these in complex form.

Definition4.8. LetEdenote a Banach overC. A linear functionalφE is aweak-exposed point of the unit ball ofE ifφ = 1 and there exists

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xE,x = 1, such that(φ(x)) = 1 and(ψ(x)) < 1 for allψE, ψ ≤ 1 and ψ = φ. We say thatx weak-exposesφ. We let Expω(E) denote the set of all weak-exposed points of the unit ball ofE.

Ifφ:E−→Cis complex linear thenφ˜ := (φ)is a real linear mapping.

Moreover,φ = ˜φandφ(x)= ˜φ(x)iφ(ix)˜ for allxinE. Hence ifφ1

andφ2are complex linear thenφ1=φ2if and only if(φ1)= (φ2). Ifxweak-exposesφandφ = φ12 2withφ1 = φ2 =1 then,|φ1(x)| ≤ 1 and |φ2(x)| ≤ 1 imply φ1(x) = φ2(x) = 1. Hence φ1 = φ2 = φ and Expω(E)⊂ExtR(E).

For our next result, which was motivated by [3, Proposition 5], we require the following Lemma [3, Lemma 4].

Lemma4.9. LetEbe a normed space and letφbe a unit vector inE. Sup- pose that for each finite dimensional subspaceF ofEthere exists a subspace GofE, FG, such thatφ|Gis an extreme point of the unit ball ofG. Then φis an extreme point of the unit ball ofE.

Proposition4.10. IfEis a complex Banach space then {φn:φ ∈Expω(E)} ⊂ExtR(PI(nE)).

Proof. Letφ∈Expω(E)and supposexweak-exposesφ. Given a finite dimensional subspace X of

n,s,E choose a finite dimensional subspace F of E such that xF and X

n,s,F. Since xF, φ|F = 1. Let ψF,ψ =1 andψ =φ|F. By the Hahn-Banach Theorem there exists ψ˜ ∈ E such thatψ|˜ F = ψ and ˜ψ = ψ = 1. Sinceψ˜ = φ we have (ψ(x))˜ = (ψ(x)) <1 andφ|F ∈Expω(F)⊂ExtR(F).

By Proposition 4.5

(φ|F)n =φn|F ∈ExtR(PI(nF ))=ExtR

n,s,

F .

By Proposition 3.1(a) and Lemma 4.9, φn∈ExtR

n,s,

E

=ExtR(PI(nE)).

This completes the proof.

Example4.11. IfEis a complex Banach space let

D(E)={xE:x =1 andf (·):= ·is real Gâteaux differentiable atx}.

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Then, see [19] and [10],

Expω(E)= {φE:φ=f(x)for somexD(E)}.

For many classical Banach spaces, e.g.p, 1< p <∞, this result can be used to show that

{φ :φE,φ =1} =Expω(E) and hence, by Proposition 4.10, that

ExtR(PI(nE))= {φn :φE,φ =1}.

IfE = 1thenx =(xn)nD(1)if and only ifxn = 0 for alln. Hence φ := (yn)n ∈ Expω()if and only if|yn| =1 for allnand Expω() = ExtR(). Since ExtR()=ExtC()this implies

{φn:φ∈ExtR()} =ExtR(PI(n1))

and we recover a special case of Example 4.4 since is aW-algebra.

Example4.12. LetmdenoteCmendowed with the supremum norm. By Proposition 3.1(f)

ExtR(PI(nm))⊂ {φn:φm1,φ =1}

forn ≥ 2. Ifφ := (w1, . . . , wm)m1,θ := 1, . . . , θm) ∈ Rm andσ is a permutation of{1, . . . , m}let

φθ,σ :=(e1wσ (1), . . . , emwσ(m)).

It is easily seen thatφθ,σn ∈ExtR(PI(nm))if and only ifφn ∈ExtR(PI(nm)). Hence to show

(3) ExtR(PI(nm))= {φn:φm1,φ =1}.

it suffices to show thatαn ∈ ExtR(PI(nm))whereα := 1, . . . , αm)sat- isfies 0≤ α1α2 ≤ · · · ≤ αmandm

i=1αi = 1. Letl denote the smallest positive integer such thatαl > 0. Ifl = mthenα ∈ ExtR(m1)and Proposi- tion 4.5 implies thatαn ∈ExtR(PI(nm)). We may thus suppose, from now on, thatl < m.

By Proposition 3.1(d), (PI(nm)) = P(nm1) with duality φn, P = P (φ)forPP(nm1)andφ(m)= m1. By the Krein-Milman Theorem

|P|,PP(nm1), achieves its maximum over the unit ball ofm1 at a pointφ whereφn∈ExtR(PI(nm)).

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Now consider the 2-homogeneous polynomial onm1

Pα(z1, . . . , zm):=

l≤i<j≤m

zizj+

k≥l

αkαl

2αk zk2. Since

|Pα(z1, . . . , zm)| = |Pα(0, . . . ,0, zl, . . . , zm)|

l≤i<j≤m

|zi||zj| +

k≥l

αkαl

2αk |zk|2

=Pα(0, . . . ,0,|zl|, . . . ,|zm|)

≤ Pα ·

k≥l

|zk|

2

it follows that|Pα|achieves its maximum,Pα:= PαBm1, at some point which has the forme(0, . . . ,0, xl, . . . , xm)wherexi ≥0 foril,m

i=lxi = 1 andθ ∈R.

Applying the method of Lagrange multipliers to the problem of maximizing

|Pα|on the setT:=

(0, . . . ,0, xl, . . . , xm):xi >0 and m

i=lxi =1 yields the equations

i>l

xi =λ=1−xl

and m

i=l

xixk+ αkαl

2αk 2xk =λ fork > l. Henceαlxk = αkxl forkl. Sincem

i=lxi = m

i=lαi = 1 this impliesαi =xi foril.

Hence

sup{ |Pα(x)|:xT} =

l≤i<j≤m

αiαj+

k≥l

αkαl

2αk αk2

= 1 2

m i=l

αi 2

− 1 2αl·

m k=l

αk

= 1

2(1−αl).

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Ifll,{l} ⊂S⊂ {l, . . . , m}let S=

x∈Rm:x =(0, . . . ,0, xl, . . . , xm),

xi >0 ifiS, xi =0 ifi /Sand m

i=l

xi =1

.

An analysis similar to the above shows that sup{ |Pα(x)|:xS} =

i∈Sαi

αl

2

i∈Sαi = 1 2

1− αl i∈Sαi

Sinceαi>0 wheni≥lthis shows thatPα = |Pα(z1, . . . ,zm)|,m

i=1|zi|=1, if and only if(z1, . . . , zm)=eαfor someθ ∈R. Henceα2∈ExtR(PI(2m)) and we have established (3) whenn=2.

Ifβ:=(1,1, . . . ,1)m thenβ =1 andβ, (z1, . . . , zm) =m

i=1zi

for(z1, . . . , zm)m1. Hence|β|achieves its maximum over the unit ball ofm1

preciselyatallpoints of the forme(x1, . . . , xm)wherexi ≥0,m

i=1xi =1 andθ ∈R.

Ifn > 2 thenQ := Pα ·βn−2 is an n-homogeneous polynomial on m1

and |Q| achieves its maximum over the unit ball of m1 precisely at those points which have the form eα where θ ∈ R is arbitrary. Hence αn ∈ ExtR(PI(nm))and we have established (3) for alln≥2 and allm.

Note that the function

f ((yn)n):=1+Pα

enyσ (n)m

n=1

satisfies|f (y)|<|f (x)|for ally =xin the closed unit ball of1. This may be used to show that all finitely supported unit vectors are peak points of different algebras of holomorphic functions (we refer to [2] for details).

REFERENCES

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DEPARTMENT OF MATHEMATICS UNIVERSITY COLLEGE DUBLIN BELFIELD, DUBLIN 4 IRELAND

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