# Generalized Numerical Index of Function Algebras

Authors: Han Ju Lee
Journal: Journal of Function Spaces
 (2019)

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## Abstract

LetX be a complex Banach space and Cb(Ω:X) be the Banach space of all bounded continuous functions from a Hausdorff space Ω to X, equipped with sup norm. A closed subspace A of Cb(Ω:X) is said to be an X-valued function algebra if it satisfies the following three conditions: (i) A≔{x⁎∘f:f∈A,x⁎∈X⁎} is a closed subalgebra of Cb(Ω), the Banach space of all bounded complex-valued continuous functions; (ii) ϕ⊗x∈A for all ϕ∈A and x∈X; and (iii) ϕf∈A for every ϕ∈A and for every f∈A. It is shown that k-homogeneous polynomial and analytic numerical index of certain X-valued function algebras are the same as those of X.


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## Body

## 1. Introduction

In this paper, we consider only complex nontrivial Banach spaces. Given a Banach spaceX, we denote by BX and SX its closed unit ball and unit sphere, respectively. Let X∗ be the dual space of X. If X and Y are Banach spaces, a k-homogeneous polynomialP from X to Y is a mapping such that there is a k-linear continuous mapping L from X to Y such that P(x)=L(x,…,x) for every x in X. The Banach space of all k-homogeneous polynomials from X to Y is denoted by P(kX:Y) endowed with the polynomial norm P=supx∈BXP(x). We refer to [1] for background knowledge on polynomials.We are mainly interested in the following spaces. For two Banach spacesX, Y and a Hausdorff topological space Ω, (1)CbΩ:Y≔f:Ω→Y:fis  a  bounded  continuous  function  onΩ,AbBX:Y≔f∈CbBX:Y:fis  holomorphic  onBX∘AuBX:Y≔f∈AbBX:Y:fis  uniformly  continuous,where BX∘ is the interior of BX. Then Cb(Ω:Y) is a Banach space under the sup norm f≔sup⁡ftY:t∈Ω and both Ab(BX:Y) and Au(BX:Y) are closed subspaces of Cb(BX:Y). In case that Y is the complex scalar field C, we just write Cb(BX), Ab(BX), and Au(BX). Let (2)ΠX≔x,x∗:x=x∗=1=x∗x.Thespatial numerical range of f in Cb(BX:X) is defined by (3)Wf=x∗fx:x,x∗∈ΠX,and thenumerical radius of f is defined by (4)vf=sup⁡λ:λ∈Wf.LetX be a Banach space. The k-homogeneous polynomial numerical indexn(k)(X) is defined in [2] by (5)nkX≔inf⁡vP:P∈PXk:X,P=1.The b-analytic numerical indexnba(X) and u-analytic indexnua(X) are defined, respectively, by (6)nbaX≔inf⁡vf:f∈AbBX:X,f=1,nuaX≔inf⁡vf:f∈AuBX:X,f=1.It is clear from the definitions that 0⩽nba(X)⩽nua(X)⩽n(k)(X)⩽1 for all k≥1.Choi, García, Kim, and Maestre showed [3] that n(k)(A)=1 and nua(A)=1 for uniform algebras A. In general, it is not difficult to see that if A is a (unital) function algebra on a Hausdorff space, then, by the Gelfand transform, A is isometric to a (unital) uniform algebra on Δ where Δ is the maximal ideal space of A. We present this fact in Proposition 2 for the completeness of the paper. In this paper, we introduce a X-valued function algebra and the Gelfand transform does not work in this case. In the proof of [3], they used a very useful Urysohn type theorem, which was obtained by Cascales, Guiro, and Kadets [4]. Recently, Kim and the author found [5] that a Urysohn type theorem holds for some function algebras. It plays an important role in the main results of this paper. For some geometric properties on k-homogeneous polynomial (analytic) numerical index, refer to [6, 7].Let us briefly review some necessary notions. A nontrivial·∞-closed subalgebra of A of Cb(Ω) is called afunction algebra on a Hausdorff space Ω. For a Banach space X, a nontrivial subspace A of Cb(Ω:X) is said to be an X-valued function algebra if it satisfies three conditions: (i) A≔{x∗∘f:f∈A,x∗∈X∗} is a function algebra on Ω; (ii) A⊗X={ϕ⊗x:ϕ∈A,x∈X}, where (ϕ⊗x)(t)=ϕ(t)x for t∈Ω; and (iii) ϕf∈A for every ϕ∈A and f∈A, where (ϕf)(t)=ϕ(t)f(t) for t∈Ω. A subset T of Ω is said to benorming for A if f=sup⁡ft:t∈T holds for all f∈A. Byunital function algebra, we mean a function algebra containing all constant functions. A function algebra A on a compact Hausdorff space K is said to be auniform algebra if A separates the points of K (that is, for every x≠y in K, there is f∈A such that f(x)≠f(y). Note that the definition of function algebra in this paper is different from the usual one in [8].Letf be an element of an X-valued function algebra A. The f is said to be apeak function at t0 if there exists a unique t0∈Ω such that f=f(t0). A peak function f is said to be astrong peak function at t0∈Ω if f=f(t0) and for every open subset V containing t0 we get (7)supft:t∈Ω∖V<f.The corresponding point t0 is called astrong peak point for A. We denote by ρA the set of all strong peak points for A. It is easy to see that if Ω is compact, then every peak function is a strong peak function. It is worth remarking that if A is a nontrivial separating separable subalgebra of C(Ω) on a compact Hausdorff space Ω, then ρA is a norming subset for A [9]. There is a compact Hausdorff space K such that ρC(K) is an empty set [10]. For more information about peak functions and points, refer to [8, 10].For anX-valued function algebra A, let A={x∗∘f:x∗∈X∗,f∈A}. Then ρA=ρA. Indeed, if f∈A is a strong peak function at t0, then choose x∗∈SX∗ such that x∗f(t0)=f(t0)=f and it is clear that x∗∘f∈A is a strong peak function in A at t0. Therefore, ρA⊂ρA. Conversely, if g∈A is a strong peak function at t, then choose x∈SX. Therefore, g⊗x∈A is a strong peak function at t. Hence we have ρA⊂ρA. In addition, if ρA is norming for A, then it is also norming for A since g⊗x in A has the same norm as g for every g∈A and x∈SX.The following lemma will be useful to get main results. In proofs of the main results, the denseness of the strong peak functions in anX-valued function algebra A is an important part and equivalent to the fact that the set of strong peak points is norming for A. That means that the fact that every element in A can be approximated by the sequence of strong peak functions is equivalent to the fact that the norm of every element in A can be approximated on the set of strong peak points for A. The approximation by strong peak functions will prove to be useful to deal with the geometric properties of function algebras especially those related to generalized numerical indices of Banach spaces.Lemma 1 (see [5]).
LetA be a function algebra on Ω and fix ω0∈ρA. Then, given 0<ϵ<1 and for every open subset U containing ω0, there exists a strong peak function ϕ∈A such that ϕ=1=|ϕ(ω0)|, supω∈Ω∖U|ϕ(ω)|<ϵ, and for all ω∈Ω, (8)ϕω+1-ϵ1-ϕω⩽1.


## 2. Main Results

The proof of [3, Theorem 2.1] shows that nba(A)=1 if A is a uniform algebra. Since a function algebra is isometric to a uniform algebra by the Gelfand transform, we have the following.Proposition 2.
LetA be a function algebra on a Hausdorff space Ω. Then it is isometric to a uniform algebra on a compact Hausdorff space and nba(A)=1.Proof.
LetA be a function algebra and Δ be the set of all nonzero algebra homomorphisms from A to C. The maximal ideal space Δ is a compact Hausdorff space with the Gelfand topology. The Gelfand transform f^ of f∈A is defined by f^(ϕ)=ϕ(f) for ϕ∈Δ. For t∈Ω, let δt be the dirac delta function by δt(f)=f(t) for f∈A. Fix a nonzero f∈A and let Ωf={t:f(t)≠0}; then δt∈Δ for all t∈Ωf and (9)f^=supf^ϕ:ϕ∈Δ⩽f=supft:t∈Ωf=sup⁡δtf:t∈Ωf⩽f^.Since the Gelfand transform f↦f^ is a homomorphism, A is isometrically isomorphic to the image A^, where A^ is the image of the Gelfand transform. Then A^ is a closed subalgebra of C(Δ) and it is separating the points of Δ. Thus, it is a uniform algebra on the compact Hausdorff space Δ.
For the second part, the proof used in [3, Theorem 2.1] to show nua(A)=1 can be applied to show that nba(A)=1 for uniform algebras A.Proposition2 gives a positive answer to the third question raised by Acosta and Kim [11].Theorem 3.
LetX be a Banach space and suppose that A is an X-valued function algebra on a Hausdorff space Ω such that ρA is a norming subset for A. Then we have (i)
n(k)(A)≥n(k)(X) for every k≥1,(ii)
nua(A)≥nua(X) and(iii)
nba(A)≥nba(X).Proof.
We provenba(A)≥nba(X) holds. The proofs for the other two cases are exactly the same. It is well-known that nba(X)>0 for all complex Banach spaces X [12].
LetA={x∗∘f:f∈A}. Then A is a function algebra. Let P∈Ab(BA:A) with P=1 and 0<ϵ<bba(X) be given. Choose f0∈SA so that P(f0)>1-ϵ/6. Since ρA is norming for A, find t0∈ρA such that P(f0)(t0)>1-ϵ/6. Since P is continuous, there is 0<δ<1 such that P(f0)-P(g)<ϵ/6 for every g∈BA with f0-g<δ.
LetW={t∈Ω:f0(t)-f0(t0)<δ/6,P(f0)(t)-P(f0)(t0)<ϵ/3} and W be an open subset of Ω containing t0. Then by Lemma 1, there is a strong peak function ϕ∈A such that ϕ=ϕ(t0)=1 and |ϕ(t)|<δ/6 for every t∈Ω∖W, and (10)ϕt+1-ϵ61-ϕt⩽1for every t∈Ω.
DefineΨ:X→A by Ψ(x)=1-δ/6(1-ϕ)f0+ϕ⊗x for all x∈X. It is easy to check that Ψ is well-defined and Ψ(x)⩽1 for all x∈BX. Then, let x0=f0(t0), (11)f0-Ψx0=supt∈Ωf0t-1-δ61-ϕtf0t-ϕtf0t0⩽supt∈Ωδ6f0t+ϕtf0t-f0t0+δ6ϕtf0t<δ6+δ3+δ6<δ.Then we have the following. (12)PΨx0t0≥Pf0t0-Pf0t0-PΨx0t0>1-ϵ6-ϵ6>1-ϵ.Choose x0∗∈SX∗ such that x0∗[P(Ψ(x0)(t0)]>1-ϵ and find a complex number z0 with |z0|⩽1 and a proper x~0∈SX satisfying x0=z0x~0. Then the function φ(z)=x0∗[P(Ψ(zx~0))(t0)] is an element of Au(BC). By the maximum modulus theorem, there exists z1 with |z1|=1 such that φ takes its maximum modulus on BC. Hence, (13)PΨz1x~0t0≥x0∗PΨz1x~0t0≥x0∗PΨz0x~0t0>1-ϵ.Let x1=z1x~0, choose x1∗∈SX∗ with x1∗(x1)=1, and define the function Φ:X→A by (14)Φx=x1∗x1-δ61-ϕf0+ϕ⊗xfor x∈X. Then Φ(x1)=Ψ(x1)=Ψ(z1x~0), and hence PΦx1t0>1-ϵ. Let Q(x)=P(Φ(x))(t0) for x∈X. Then Q∈Ab(BX:X). Then (15)1-ϵ<PΦx1t0=Qx1⩽Q⩽1.Since 0<ϵ<bba(X), there is (x2,x2∗)∈Π(X) so that (16)x2∗Qx2Q>vQQ-ϵ≥bbaX-ϵ>0.Note that (Φ(x2),x2∗∘δt0)∈Π(A) because Φ(x2)(t0)=x2. Hence we have (17)vP≥x2∗∘δt0PΦx2=x2∗PΦx2t0=x2∗Qx2≥QbbaX-ϵ≥1-ϵbbaX-ϵ.Since 0<ϵ<bba(X) is arbitrary, v(P)≥bba(X). This holds for all P∈Ab(BA;A) with P=1. Therefore, we get bba(A)≥bba(X).A version of the Bishop-Phelps-Bollobás type theorem for holomorphic functions has been shown [5, 13]. In the following theorem, we present a similar result. However the main focus is the denseness of the set of all strong peak functions, which is different from that of the results in [5].Theorem 4.
LetX be a Banach space and A an X-valued function algebra on a Hausdorff space Ω. Then, given ϵ>0, whenever a norm-one element f in A and a point ω0 in ρA satisfy f(ω0)>1-ϵ/5, there is a norm-one strong peak function g∈A at ω0∈ρA such that f-g<ϵ.Proof.
Suppose thatf satisfies the prescribed conditions. Then (18)U1=ω∈Ω:fω0fω0-fω<ϵ5is an open set containing ω0. There exists ω1∈ρA∩U2. Using Lemma 1, take a strong peak function ϕ∈A such that ϕ(ω0)=1=ϕ, sup⁡ϕω:ω∈Ω∖U1<ϵ/5, and (19)ϕω+1-ϵ51-ϕω⩽1for all ω∈Ω. Set (20)gω=ϕωfω0fω0+1-ϵ51-ϕωfω.It is easy to check that g∈A and g(ω1)=1=g. Moreover, from the inequality (21)gω-fω⩽ϕωfω0fω0-fω+ϵ51-ϕωfω,we have that g(ω)-f(ω)⩽ϵ/5+2ϵ/5 for all ω∈Ω if we consider two cases ω∈U1 and ω∈Ω∖U1. Hence, we get f-g<ϵ and complete the proof since we know g is a strongly norm attaining function from the fact that ϕ is a strong peak function.From Theorem4, we have the following consequence.Corollary 5.
LetX be a Banach space and A be an X-valued function algebra on a Hausdorff space Ω. Then the set ρA is norming if and only if the set of strong peak functions in A is dense.Proof.
The necessity is proved by Theorem4. For the converse, assume that the set of strongly norm attaining functions in A is dense in A. Given f∈A, there is a sequence {fn} of strong peak functions in A such that limn→∞fn-f=0. For each n, let tn be the strong peak point corresponding to fn. Then (22)0=limn→∞fn-f≥limsupn→∞fntn-ftn≥limsupn→∞fntn-ftn.Thus, (23)0=limn→∞fntn-ftn=limn→∞fn-ftn.This means that limn→∞fn(tn)=limn→∞fn=f. This shows that ρA is a norming subset of A.Theorem 6.
LetX be a Banach space and A be an X-valued function algebra on a Hausdorff space Ω such that ρA is a norming subset for A. Fix P∈Au(BX:X) and define the map QP:BA→Cb(Ω:X) by QP(f)(t)=P(f(t)) for f∈BA and t∈Ω. Suppose that QP(f) is an element of A for every f∈BA and for every P∈Au(BX:X). Then we have nua(A)=nua(X).Proof.
By Theorem3, we have only to show that nuaA⩽nua(X). Consider the set (24)L=f,x∗∘δt:f∈SA,t∈Ω,x∗∈SX∗andx∗ft=1.Let π1:A×(A)∗→A be the natural projection. Then since ρA is norming for A, Corollary 5 shows that π1(L) is dense in SA. Then, it is shown [14] that for every Q∈Au(BA;A), we have (25)vQ=sup⁡x∗Qft:f,x∗∘δt∈L.Given P∈Au(BX:X) with P=1, we have QP∈Au(BA;A). Indeed, QP is a map from BA to BA. Since P is uniformly continuous on BX, given ϵ>0, there is δ>0 such that if x,y∈BX and x-y⩽δ, then P(x)-P(y)<ϵ. If f,g∈BA and f-g<δ, then P(f(t))-Pgt<ϵ for all t∈Ω. Hence Q(f)-Q(g)⩽ϵ. This shows that QP is uniformly continuous on BA. Now it is enough to show that QP is G-holomorphic on BA∘ [15]. Fix f∈BA∘ and g∈A, and let U(f,g)={z∈C:f+zg∈BA∘} be an open subset in the complex plane. Let φ(z)=QP(f+zg) for z∈U(f,g). Then φ(z) is a A-valued continuous function on U(f,g). For each t∈Ω, φ(z)(t)=QP(f+zg)(t)=P(f(t)+zg(t)) is holomorphic. Fix z0∈U(f,g) and choose δ>0 such that z0+δBC⊂U(f,g). The Cauchy integral formula shows that, for each t∈Ω, (26)φz0t=12πi∫z-z0=δφztz-z0dz.As a result, we have (27)φz0=12πi∫z-z0=δφzz-z0dz,since the continuity of φ(z) implies the Bochner integrability of the integral. This means that φ is holomorphic on U(f,g) and QP is holomorphic on BA∘ [15]. We also have QP=1 since there is a strong peak function g∈A at t0∈Ω such that g(t0)=1=g and g⊗x is in BA for each x∈BX. It is clear that v(QP)≥nua(A). For every ϵ>0, there is (f,x∗∘δt)∈L such that (28)nuaA-ϵ⩽vQP-ϵ<x∗∘δtQPf=x∗Pft⩽vP.Therefore, we get nua(A)⩽nua(X).The same proof shows the following.Theorem 7.
LetX be a Banach space and A be an X-valued function algebra on a Hausdorff space Ω such that ρA is a norming subset for A. Fix P∈P(kX:X) and define the map QP:A→Cb(Ω:X) by QP(f)(t)=P(f(t)) for f∈A and t∈Ω. Suppose that QP(f) is an element of A for every f in A and for every P∈P(kX:X)). Then we have n(k)(A)=n(k)(X).Proof.
The main difficulty in the proof of Theorem7 is to check that QP is in P(kA;A). Let L be the corresponding continuous k-linear map defining P. Let L~:Ak→A by L~(f1,…,fk)(t)=L(f1(t),f2(t),…,fk(t)) for fi∈A and t∈Ω. Then it is easy to check that L~ is a continuous k-linear map and QP(f)=L~(f,…,f) for f∈A. The other part of the proof is the same as the proof of Theorem 6.LetX,Y be Banach spaces and let A(BX:Y) be either Au(BX:Y) or Ab(BX:Y). Notice that A(BX:Y) are Y-valued function algebras over BX. If a Banach space X is finite dimensional, ρA(BX) is the set of all complex extreme points of BX as observed in [16, 17]. A strongly exposed point of BX is a strong peak point for A(BX), so if a strongly exposed point of BX is dense in SX, then ρA(B)¯=SX and it is norming for A(BX). It is also proved in [17] that if X is locally c-uniformly convex space and it is an order continuous sequence space, then ρAu(BX) is norming. The typical example of uniformly complex convex sequence space is l1. For the definitions related to various complex convexities and more examples, we refer to [9, 18–21].LetC be a closed convex and bounded set in a Banach space X. The set C has theRadon-Nikodým property if, for every probability space (Ω,B,μ) and every X-valued countably additive measure τ on B such that τ(A)/μ(A)∈C for every A∈B with μ(A)>0, there is a Bochner measurable f:Ω→X so that (29)τA=∫Afωdμω,A∈B.The spaceX is said to have theRadon-Nikodým property if its unit ball BX has the Radon-Nikodým property [22]. For the basic properties and useful information on the Radon-Nikodým property, see also [22–25]. It has been shown [9] that if X has the Radon-Nikodým property, then ρA(BX) is norming for A(BX).Corollary 8.
Suppose thatX satisfies one of the following conditions: (i) X has the Radon-Nikodým property; (ii) X is locally uniformly convex space; (iii)  X is a locally c-uniformly convex order continuous sequence space. Then we have (i)
n(k)(A(BX:Y))=n(k)(Y) for every k≥1,(ii)
nua(A(BX:Y))=nua(Y).Proof.
IfX satisfies one of the three conditions, ρA(BX:Y)=ρA(BX) and it is norming for A(BX:Y). Therefore, Theorem 3 implies that n(k)(A(BX:Y))≥n(k)(Y), nua(A(BX:Y))≥nua(Y). For the case (ii), fix P∈Au(BY:Y) and define the map QP:A(BX:Y)→Cb(BX:Y) by QP(f)(x)=P(f(x)) for f∈A(BX:Y) and x∈BX. Then QP(f)∈A(BX:Y). Consequently, Theorem 6 shows that (30)nuaABX:Y=nuaYand the proof of (ii) is complete. The remaining proof (i) can be finished in the same way by Theorem 7.By Theorem3, we get the following.Corollary 9.
LetΩ be a Hausdorff topological space and suppose that ρCb(Ω) is norming for Cb(Ω). If Y is a Banach space with nba(Y)=1, we have v(f)=f for all f∈Ab(BCb(Ω:Y):Cb(Ω:Y)).As we show in the next proposition, closed bounded convex sets with the Radon-Nikodým property satisfy the condition of Corollary9.Proposition 10.
Suppose thatΩ is a nonempty closed bounded convex subset of a Banach space and Ω has the Radon-Nikodým property. Then ρCb(Ω) is norming for Cb(Ω) and the set of strong peak functions of Cb(Ω) is dense.Proof.
It is enough to show that the set of strong peak functions ofCb(Ω) is dense by Corollary 5. Given f∈Cb(Ω) and ϵ>0, the Stegall perturbed optimization theorem [25] shows that there is x∗∈X∗ such that the function φ(x)=|f(x)|+Re⁡x∗x strongly attains its norm at x0∈Ω and x∗<ϵ. Choose a complex number z0∈SC such that (31)fx0+Re⁡x∗x0=fx0+z0Re⁡x∗x0.Then it is easy to check that g(x)=f(x)+z0Re(x∗(x)) is a strong peak function at x0 and f-g<ϵ. This shows the denseness of the set of strong peak functions on Cb(Ω).



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