NUMERICAL PEAK POINTS AND NUMERICAL Å ILOV BOUNDARY ...

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 12, December 2008, Pages 4339–4347 S 0002-9939(08)09402-1 Article electronically published on June 3, 2008

NUMERICAL PEAK POINTS ˇ AND NUMERICAL SILOV BOUNDARY FOR HOLOMORPHIC FUNCTIONS SUNG GUEN KIM (Communicated by N. Tomczak-Jaegermann)

Abstract. In this paper, we characterize the numerical and numerical strongpeak points for A∞ (BE : E) when E is the complex space l1 or C(K). We also prove that {(x, x∗ ) ∈ Π(l1 ) : |x∗ (en )| = 1 for all n ∈ N} is the numerical ˇ Silov boundary for A∞ (Bl1 : l1 ).

1. Introduction Throughout this paper we will just consider complex Banach spaces. For a Banach space E, SE and BE will be the unit sphere and the closed unit ball of E, respectively. If E and F are Banach spaces, Cb (BE : F ) denotes the Banach space of the bounded continuous functions f : BE → F , endowed with the supremum norm. In the case that F = C, we write Cb (BE ). An N -homogeneous polynomial P from E to F is a mapping such that there is an N -linear (and bounded) mapping L from E to F satisfying P (x) = L(x, . . . , x),

∀x ∈ E.

The set of all N -homogenous polynomials from E to F is denoted by P(N E : F ). We denote by A∞ (BE : F ) the Banach space of the bounded continuous function f : BE → F such that f is holomorphic on the open unit ball, endowed with the supremum norm. If F = C, we write A∞ (BE ). ˇ A result of Silov asserts that if A is a unital separating subalgebra of C(K) (K is a compact Hausdorff topological space), there is a smallest closed subset S ⊂ K such that every function of A attains its norm at some point of S ([7], Theorem I.4.2). Bishop [5] proved that if K is metrizable, in fact, there is a minimal subset of K satisfying the above condition for every separating subalgebra of C(K). That subset is the set of peak points for A. Globevnik [8] introduced the corresponding concepts of the boundary of a subalgebra A of Cb (Ω), the set of bounded and continuous functions on a topological space Ω not necessarily compact, and studied Received by the editors September 9, 2006, and, in revised form, July 16, 2007, October 18, 2007, and October 23, 2007. 2000 Mathematics Subject Classification. Primary 46A22; Secondary 46G25. Key words and phrases. Numerical peak points, numerical Silov boundaries. The author thanks the referee for invaluable suggestions and for help with an earlier version of this paper. c 2008 American Mathematical Society Reverts to public domain 28 years from publication

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SUNG GUEN KIM

the boundaries for Ω = Bc0 and A a certain space of holomorphic functions. A subset B ⊂ BE is a boundary for a subspace A ⊂ Cb (BE ) if f  = sup |f (z)|,

∀f ∈ A .

z∈B

An element x ∈ SE is called a peak point for A if there exists h ∈ A such that |h(x)| = hBE and |h(z)| < hBE for every z ∈ BE \{x}. In this case we say that h peaks at x. A peak point x is called a strong peak point for A if there exists h ∈ A such that |h(x)| = hBE and, given
> 0, there exists a δ > 0 such that for every z ∈ BE with z − x >
, we have |h(z)| < hBE − δ. In this case we say that h peaks strongly at x. In 1971, Harris [9] introduced the definition of a spatial numerical range for a bounded and holomorphic function defined on a Banach space and the corresponding concept of numerical radius. The spatial numerical range of a bounded function f from BE to E is given by W (f ) := {x∗ (f (x)) : (x, x∗ ) ∈ Π(E)} , where we denoted by Π(E) the following subset: Π(E) := {(x, x∗ ) ∈ SE × SE ∗ : x∗ (x) = 1}. The numerical radius v(f ) is just the number v(f ) := sup{|λ| : λ ∈ W (f )}. For more background and information about numerical ranges and radii, we refer the reader to [6]. For a linear space A ⊂ Cb (BE : E), M. Acosta and the author [3] gave the corresponding definition of numerical boundary for A. We say that B ⊂ Π(E) is called a numerical boundary for A if sup (x,x∗ )∈B

|x∗ (f (x))| = v(f ),

∀f ∈ A .

In the case that B is a (  × w∗ )-closed numerical boundary for A that is minˇ imal under the previous conditions, B is said to be the numerical Silov boundary. In [3], the authors studied numerical boundaries for holomorphic functions on some classical Banach spaces. Parallel to the concepts of peak and strong peak-points, we introduce the corresponding definition of numerical and numerical strong-peak points. An element (x, x∗ ) ∈ Π(E) is called a numerical peak point for A if there exists h ∈ A such that |x∗ (h(x))| = v(h) and |z ∗ (h(z))| < v(h) for every (z, z ∗ ) ∈ Π(E)\{(x, x∗ )}. In this case we say that h peaks numerically at (x, x∗ ). Also, (x, x∗ ) ∈ Π(E) is called a numerical strong-peak point for A if there exists a function h ∈ A such that |x∗ (h(x))| = v(h), and for any (xn , x∗n ) ∈ Π(E) with limn→∞ |x∗n (h(xn ))| = v(h) we have that xn → x in norm and x∗n → x∗ in w∗ -topology. In this case we say that h peaks strong-numerically at (x, x∗ ). It is clear that every numerical strong-peak point is a numerical peak point. Note that if E is a finite dimensional space, then every numerical peak point is also a numerical strong-peak point. It is immediate from definition that every (  × w∗ )-closed numerical boundary for A∞ (BE : E) contains all numerical strong-peak points. In Section 2, we characterize the numerical peak points for A∞ (Bl1 : l1 ) and prove that every numerical peak point for A∞ (Bl1 : l1 ) is also a numerical strongpeak point. We prove that {(x, x∗ ) ∈ Π(l1 ) : |x∗ (en )| = 1 for all n ∈ N} is the ˇ numerical Silov boundary for A∞ (Bl1 : l1 ).

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ˇ NUMERICAL PEAK POINTS AND NUMERICAL SILOV BOUNDARY

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In Section 3, we characterize the numerical peak points for A∞ (BC(K) : C(K)) when K is a compact metrizable space. If K is an infinite compact Hausdorff topological space, we prove that there are no numerical strong-peak points for A∞ (BC(K) : C(K)). ˇ 2. Numerical peak points and numerical Silov boundary for holomorphic functions on l1 Lemma 2.1. If (x, x∗ ) ∈ Π(l1 ) is a numerical peak point for A∞ (Bl1 : l1 ), then |x∗ (en )| = 1 for all n ∈ N. Proof. Let x := (vn )n ∈ Sl1 , x∗ := (wn )n ∈ Sl∞ . Assume that there exists a positive integer n0 such that |wn0 | < 1. Since the subset of peak points in SE for A∞ (Bl1 ) is invariant under surjective linear isometries on l1 , we may assume that n0 = 1, so ∗ |w1 | < 1. Let h ∈ A∞ (Bl1 : l1 ) peak
numerically at (x, x ) with h := (hn )n for some ∗ hn ∈ A∞ (Bl1 ). Let xλ := λe1 + n>1 wn en for |λ| ≤ 1. We claim that v1 = 0. If v1 = 0, then (x, x∗λ ) ∈ Π(l1 ) for |λ| ≤ 1. We define the polynomial ψ : D(0, 1) → C by  ψ(λ) := λh1 (x) + wn hn (x) (|λ| ≤ 1). n>1

Since max|λ|≤1 |ψ(λ)| = v(h) = |x∗ (h(x))| = |ψ(w1 )|, by the Maximum Modulus Theorem, ψ(λ) = v(h) for all |λ| ≤ 1. Choose any complex number β with |β| = 1. Then (x, x∗β ) = (x, x∗ ) ∈ Π(l1 ) and |ψ(β)| = |x∗β (h(x))| = v(h). Since h peaks numerically at (x, x∗ ), we have a contradiction. Thus v1 = 0. We define the 1degree polynomial u : D(0, 1) → C by  u(λ) := λv1 + wn vn (|λ| ≤ 1). n>1

Since 1 = x∗ (x) = |u(w1 )|, by the Maximum Modulus Theorem, u(λ) = 1 for all |λ| ≤ 1. By the same reason as in the above argument, (x, x∗β ) ∈ Π(l1 ) and |ψ(β)| = |x∗β (h(x))| = v(h) for any β ∈ C with |β| = 1. Thus we have a contradiction. Therefore |w1 | = 1.
Theorem 2.2. S := {(x, x∗ ) ∈ Π(l1 ) : |x∗ (en )| = 1 for all n ∈ N} is the set of all numerical strong-peak points for the space of 2-degree polynomials in A∞ (Bl1 : l1 ) and every numerical peak point for the space of 2-degree polynomials in A∞ (Bl1 : l1 ) is a numerical strong-peak point. Proof. By Lemma 2.1, it is enough to show the first statement of the theorem. Let (y0 , y0∗ ) ∈ S. We will prove that (y0 , y0∗ ) is a numerical strong-peak point for the space of 2-degree polynomials in A∞ (Bl1 : l1 ). Let J = supp(y0 ). Since the subset of peak points in SE for A∞ (Bl1 ) is invariant under surjective linear isometries on l1 , we can assume that y0 (k) > 0 for all k ∈ J. Let an := y0∗ (en ) for all n ∈ N. Then ak = 1 for all k ∈ J. By Theorem 2.6 in [4] there exists a 2-degree polynomial f ∈ A∞ l1 ) which peaks strongly at y0 with f  = 1. Let
(B ∞ λn := sign(an ) and z0 := n=1 λn bn en ∈ Sl1 with bn > 0 for all n ∈ N. We define a 2-degree polynomial h ∈ A∞ (Bl1 : l1 ) by h(x) := f (x)z0 (x ∈ l1 ). Clearly y0∗ (h(y0 )) = 1 = h = v(h). We claim that h peaks strong-numerically at (y0 , y0∗ ). Consider a sequence {(xn , x∗n )} in Π(l1 ) such that limn→∞ |x∗n (h(xn ))| = v(h). We

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SUNG GUEN KIM

will prove that xn → y0 in norm and x∗n → y0∗ in w∗ -topology. Since 1 = lim |x∗n (h(xn ))| ≤ lim |f (xn )| ≤ 1 n→∞

n→∞

and f peaks strongly at y0 , xn → y0 in norm. Write x∗n := (aj )j ∈ Sl∞ and (n)

(n)

xn := (cj )j ∈ Sl1 for all n, j ∈ N. Let x := (dj ) ∈ l1 . First we claim that (n)

limn→∞ ak

(n )

= 1 for all k ∈ J. Let i0 ∈ J be fixed. Let {ai0 i } be any subsequence

(n)

(n )

of {ai0 }. As the set {ai0 i } is a bounded subset of C, there exist a subsequence (ni )

(nil )

{ai0 l } and a complex number r with |r| ≤ 1 such that liml→∞ ai0 1 = x∗n (xn ) for all n ∈ N, we have 1=

∞ 

(nil ) (nil ) cj

= r. Since

for all l ∈ N.

aj

j=1 (n)

Since xn → y0 in norm, we have limn→∞ ci0 = y0 (i0 ) > 0. For a sufficiently large (nil )

l, ci0

> 0. It follows that, for a sufficiently large l, (nil ) (nil ) c i0

1 = a i0

(nil )

≤ |ai0

+

(nil )

| c i0

∞ 

(nil ) (nil ) cj

aj

j=i0 ∞ 

+

(nil )

|aj

(nil )

| |cj

|

j=i0

≤

∞ 

(nil )

|cj

| = 1.

j=1 (nil )

Thus 1 = ai0

(nil )

= |ai0

(n)

| for a sufficiently large l, so r = 1. Therefore limn→∞ ai0 = (n)

1. We claim that limn→∞ aj (n ) {aj0 i }

= aj for all j ∈ N\J. Let j0 ∈ N\J be fixed. Let (n)

(n )

be any subsequence of {aj0 }. As the set {aj0 i } is a bounded subset of C, (nil )

there exist a subsequence {aj0 (ni ) liml→∞ aj0 l

} and a complex number ρ with |ρ| ≤ 1 such that

= ρ. Since 1 = lim |x∗n (h(xn ))| ≤ lim |f (xn )| |x∗n (z0 )| ≤ 1, n→∞

n→∞

we have 1 = lim |x∗n (z0 )| = lim | n→∞

n→∞

∞ 

(n)

aj λj bj |.

j=1

It follows that for fixed i0 ∈ J, 1 = ≤ ≤

lim |

l→∞

∞ 

(nil )

aj

λj bj |

j=1 (nil )

lim |aj0

l→∞

bj0 + bi0 +

(nil )

λj0 bj0 + ai0 

bi0 | +

 j∈N\{i0 ,j0 }

bj = 1.

j∈N\{i0 ,j0 }

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bj

ˇ NUMERICAL PEAK POINTS AND NUMERICAL SILOV BOUNDARY

Thus

(nil )

lim |aj0

l→∞

(nil )

λj0 bj0 + ai0

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bi0 | = |ρλj0 bj0 + bi0 | = bj0 + bi0 ,

showing ρλj0 = 1. We have ρ = aj0 . Therefore limn→∞ aj0 = aj0 . Since the w∗ topology in l∞ is the convergence in each coordinate, we complete the proof.
(n)

ˇ boundary for A∞ (Blp : lp ) for By Theorem 2.3 in [3], Π(lp ) is the numerical Silov each 1 < p < ∞. The following is an application of Theorem 2.2 to the numerical ˇ Silov boundary. Theorem 2.3. S := {(x, x∗ ) ∈ Π(l1 ) : |x∗ (en )| = 1 for all n ∈ N} is the numerical ˇ Silov boundary for A∞ (Bl1 : l1 ). Proof. It is easy to show that S is (  × w∗ )-closed. By Theorem 2.2 and the fact that every (  × w∗ )-closed numerical boundary for A∞ (Bl1 : l1 ) contains all numerical strong-peak points, it suffices to prove that S is a numerical boundary for A∞ (Bl1 : l1 ). Let h ∈ A∞ (Bl1 : l1 ) and let
> 0. Choose (x0 , x∗0 ) ∈ Π(l1 ) such that v(h) −
< |x∗0 (h(x0 ))|. Let x0 := (vn )n , x∗0 := (wn )n and h := (hn )n for some hn ∈ A∞ (Bl1 ). We claim that there exists a complex sequence {λn } in the unit disk such that if   zn∗ := λj ej + wj ej j>n

1≤j≤n

∗ for all n ∈ N, then (x0 , zn∗ ) ∈ Π(l1 ) and |zn+1 (h(x0 ))| ≥ |zn∗ (h(x0 ))| > v(h) −
for all n ∈ N. If |w1 | = 1, let λ1 := w1 . Otherwise w1 = 0; hence  (x0 , λe1 + wn en ) ∈ Π(l1 ) for all |λ| ≤ 1. n>1

If h1 (x0 ) = 0, let λ1 := 1. Assume that h1 (x0 ) = 0. We define a nonconstant 1-degree polynomial ψ1 : D(0, 1) → C by  wn hn (x0 ) (|λ| ≤ 1). ψ1 (λ) := λh1 (x0 ) + n>1

Since max|λ|=1 |ψ1 (λ)| ≥ |ψ1 (w1 )| = |x∗0 (h(x0 ))|, by the Maximum Modulus Theorem, there exists a complex number λ1 with |λ1 | = 1 such that |ψ1 (λ1 )| ≥ |x∗0 (h(x0 ))| > v(h) −
.


Let z1∗ := λ1 e1 + n>1 wn en . Then (x0 , z1∗ ) ∈ Π(l1 ) and |z1∗ (h(x0 ))| ≥ |x∗0 (h(x0 ))| > v(h) −
. If |w2 | = 1, let λ2 := w2 . Otherwise w2 = 0; hence  (x0 , λ1 e1 + λe2 + wn en ) ∈ Π(l1 ) for all |λ| ≤ 1. n>2

If h2 (x0 ) = 0, let λ2 := 1. Assume that h2 (x0 ) = 0. We define a nonconstant 1-degree polynomial ψ2 : D(0, 1) → C by  ψ2 (λ) := λ1 h1 (x0 ) + λh2 (x0 ) + wn hn (x0 ) (|λ| ≤ 1). n>2

Since max|λ|=1 |ψ2 (λ)| ≥ |ψ2 (b2 )| = |ψ1 (λ1 )| > v(h0 )−
, by the Maximum Modulus Theorem, there exists a complex number λ2 with |λ2 | = 1 such that |ψ2 (λ2 )| ≥ |ψ1 (λ1 )| > v(h) −
.
∗ ∗ Let := 1≤j≤2 λj ej + n>2 wn en . Then (x0 , z2 ) ∈ Π(l1 ) and |z2 (h(x0 ))| ≥ ∗ |z1 (h(x0 ))| > v(h) −
. Continuing this process, we can get a complex sequence z2∗




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SUNG GUEN KIM

{λn } in the unit disk satisfying the claim. Let z ∗ := (λn )∞ n=1 ∈ l∞ . We will show that (x0 , z ∗ ) ∈ S. Indeed, it follows that for each n ∈ N,   |λj − wj | |vj | ≤ 2 |vj | → 0, |z ∗ (x0 ) − 1| = |z ∗ (x0 ) − zn∗ (x0 )| ≤ j>n

j>n

as n → ∞. Thus z ∗ (x0 ) = 1. We will show that lim |zn∗ (h(x0 ))| = |z ∗ (h(x0 ))|.

n→∞

Let h(x0 ) := (βn )∞ n=1 ∈ l1 . It follows that   |λj − wj | |βj | ≤ 2 |βj | → 0, |z ∗ (h(x0 )) − zn∗ (h(x0 ))| ≤ j>n

j>n

as n → ∞. Thus we have v(h) −
< lim sup |zn∗ (h(x0 ))| = |z ∗ (h(x0 ))| ≤ n→∞

sup (x,x∗ )∈S

|x∗ (h(x))| ≤ v(h),

which shows that sup(x,x∗ )∈S |x∗ (h(x))| = v(h). Since h ∈ A∞ (Bl1 : l1 ) is arbitrary, we complete the proof.
3. Numerical peak points and numerical strong-peak points on C(K) Theorem 3.1. Let K be a compact Hausdorff topological space with at least two points. If (x0 , x∗0 ) ∈ Π(C(K)) is a numerical peak point for A∞ (BC(K) : C(K)), then: (a) There exists a unique t0 ∈ K such that |x0 (t)| = 1, ∀t ∈ K and x∗0 = sign(x0 (t0 ))δt0 . Hence x0 is an extreme point of BC(K) . (b) x0 is a peak point for A∞ (BC(K) ). Proof. Let h ∈ SA∞ (BC(K) :C(K)) peak numerically at (x0 , x∗0 ). Then v(h) = |x∗0 (h(x0 ))|. By Theorem 2.7 in [2], 1 = v(h) = h. Choose an element t0 ∈ K such that 1 = h(x0 ) = |δt0 ◦ h(x0 )| = δt0 ◦ h. We claim that |x0 (t)| = 1, ∀t ∈ K. For every complex number λ in the unit disk, the function x0 + λ(1 − |x0 |) ∈ C(K), and for every t ∈ K, it is satisfied that |x0 (t) + λ(1 − |x0 (t)|) | ≤ 1. Define the continuous function φ : D(0, 1) → C by φ(λ) := δt0 (h(x0 + λ(1 − |x0 |))) (|λ| ≤ 1). Note that φ is holomorphic on D(0, 1) and |φ(λ)| ≤ 1 for every λ in the unit disk. Also |φ(0)| = |δt0 ◦ h(x0 )| = 1. Since φ attains its maximum modulus at 0, φ is constant. We choose a complex number λ0 satisfying the facts that |λ0 | = 1 and |x0 (t0 )+λ0 (1−|x0 (t0 )|) | = 1. So φ(0) = φ(λ0 ). The element z0 := x0 +λ0 (1−|x0 |) is in the unit ball of C(K) and |z0 (t0 )| = 1. Then (z0 , z0 (t0 )δt0 ) ∈ Π(C(K)). Since |φ(λ0 )| = |z0 (t0 )δt0 (h(z0 ))| = v(h) = 1, we have z0 = x0 and x∗0 = x0 (t0 )δt0 . Thus λ0 (1−|x0 (t)|) = 0 for all t ∈ K, so |x0 (t)| = 1 for all t ∈ K. So x∗0 = sign(x0 (t0 ))δt0 . The uniqueness of t0 follows from Urysohn’s Lemma. Therefore we have proved assertion (a).

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ˇ NUMERICAL PEAK POINTS AND NUMERICAL SILOV BOUNDARY

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We will show that δt0 ◦ h peaks at x0 . Let y ∈ SC(K) such that 1 = |δt0 ◦ h(y)| = δt0 ◦ h. By the same argument as in the proof of assertion (a), we have |y(t0 )| = 1. Note that (y, sign(y(t0 ))δt0 ) ∈ Π(C(K)). Since 1 = |sign(y(t0 ))δt0 (h(y))| = δt0 ◦ h = h = v(h), we have y = x0 and x∗0 = sign(y(t0 ))δt0 , which show assertion (b).




Theorem 3.2. (1) Let K be a compact metrizable space. Then M := {(x, sign(x(t))δt) : x ∈ extBC(K) , t ∈ K} is the set of all numerical peak points for the space of 1-degree polynomials in A∞ (BC(K) : C(K)). (2) If K is any infinite compact topological space, then there are no numerical strong-peak points for A∞ (BC(K) : C(K)). Proof. (1): By (a) of Theorem 3.1, it suffices to show that if (x0 , sign(x0 (t0 ))δt0 ) ∈ M, then it is a numerical peak point for the space of 1-degree polynomials in A∞ (BC(K) : C(K)). Since the subset of peak points in SC(K) for A∞ (BC(K) ) is invariant under surjective linear isometries on C(K), we can assume that x0 (t) = 1 for all t ∈ K. Since K is a metrizable space, there exists a dense subset {ln } in K. Choose (αn ) ∈ Sl1 with αn > 0 for all n ∈ N. We claim that there exists a function y0 ∈ SC(K) such that y0 (t0 ) = 1 and 0 ≤ y0 (t) < 1 for all t ∈ K\{t0 }. Indeed, let d be a metric in K. Let 1 An := {t ∈ K : d(t, t0 ) ≥ } (n ∈ N). n Clearly An is a closed subset of K with t0 ∈ / An for all n ∈ N. By the Tietze Extension Theorem, for each n ∈ N, there exists a sequence {zn } in C(K) such that 0 ≤ zn ≤ 1, zn (t0 ) = 1 and zn (An ) = {0}. We define the function y0 (t) :=

∞  1 zn (t) (t ∈ K). 2n n=1

We define a 1-degree polynomial h ∈ A∞ (BC(K) : C(K)) by h(x) :=

∞ 

αn (1 + x(ln ))y0 (x ∈ C(K)).

n=1

We claim that h peaks numerically at (x0 , δt0 ). Let (z0 , z0∗ ) ∈ Π(C(K)) be such that v(h) = |z0∗ (h(z0 ))|. Since 2 ≥ h ≥ v(h) ≥ δt0 (h(x0 )) = 2, we have 2 = v(h) = h. Since 2 = |z0∗ (h(z0 ))| = h(z0 ), we have z0 (ln ) = 1 for all n ∈ N. Since {ln } is a dense subset of K, z0 (t) = 1 = x0 (t) for all t ∈ K. Thus z0 = x0 . By the Riesz Representation Theorem on C(K)∗ , there exists a unique regular complex Baire measure µ = v + iw on K (v and w are positive measures) satisfying    z0∗ (x) = x(t) dµ = x(t) dv + i x(t) dw (x ∈ C(K)) K

with z0∗  = |µ| = 1. Since 1 = z0∗ (x0 ) =

K



K

 x0 dv + i

K

x0 dw = v(K) + iw(K), K

v(K) = 1, w(K) = 0. Thus w = 0 and  ∗ z0 (x) = x(t) dv (x ∈ C(K)). K

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4346

SUNG GUEN KIM

It follows that 2 = |z0∗ (h(x0 ))| = |





K

 y0 (t)dv| ≤ 2

h(x0 )dv| = 2| K

|y0 (t)|dv ≤ 2. K

 Thus 1 = K |y0 (t)|dv. We claim that v({t0 }) = 1. Otherwise, by the regularity of v and the choice of y0 , there exists an open subset θ0 of K containing t0 such that v(K\θ0 ) > 0. Let δ0 := maxt∈K\θ0 |y0 (t)| < 1. It follows that  |y0 (t)|dv ≤ v(θ0 ) + δ0 v(K\θ0 ) < v(K) = 1, 1= K

which is impossible. Thus v({t0 }) = 1 and v(K\{t0 }) = 0. Therefore, we have  z0∗ (x) = x(t) dv = x(t0 ) v({t0 }) = x(t0 ) = δt0 (x) K

for all x ∈ C(K), showing z0∗ = δt0 . (2): By (a) of Theorem 3.1, it suffices to show that if (x0 , x∗0 ) ∈ Π(C(K)) with x0 ∈ extBC(K) , then (x0 , x∗0 ) is not a numerical strong-peak point. Let {tn } in K be the sequence such that there is a sequence {xn } in BC(K) such that 0 ≤ xn ≤ 1, xn (tn ) = 1, for all n and supp(xn ) ∩ supp(xm ) = ∅ (n = m). Let h ∈ A∞ (BC(K) : C(K)) such that 1 = v(h) = |x∗0 (h(x0 ))|. We will show that h cannot peak strongnumerically at (x0 , x∗0 ). By Theorem 2.7 in [2], we have v(h) = h = 1. There exists a t0 ∈ K such that sign(x0 (t0 ))δt0 ◦ h ∈ A∞ (BC(K) ) and 1 = h(x0 ) = |δt0 (h(x0 ))| = |sign(x0 (t0 ))δt0 (h(x0 ))| = δt0 ◦ h. Let zn := x0 (1 − xn ) for all n ∈ N. Since the support of xn are pairwise disjoint, there is a positive integer N such that xn (t0 ) = 0 for all n > N . Let λn := sign(zn (t0 )) for all n > N . Thus (zn , λn δt0 ) ∈ Π(C(K)) for all n > N. Since {xn } is equivalent to a c0 -basis, then it converges weakly to 0. By the Rainwater theorem, the sequence {zn } is in the unit ball of C(K) and converges weakly to x0 . Since C(K) has the Dunford-Pettis property, then it has also the polynomial DunfordPettis property [10], and so, if we follow the argument in the proof Proposition 4.1 in [1], then |λn δt0 ◦ h(zn )| → 1. It follows that zn − x0  = x0 xn  ≥ |x0 (tn )xn (tn )| = 1 for all n ∈ N. Therefore, we have proved that (x0 , x∗0 ) is not a numerical strong-peak point.
ˇ It is known in [3], Theorem 5.2, that there is no numerical Silov boundary for A∞ (BC(K) : C(K)) if K is an infinite compact Hausdorff topological space. n ) : |x(k)| = Corollary 3.3. Let n ∈ N. Then M := {(x, sign(x(t))δt) ∈ Π(l∞ ˇ 1 for all k = 1, 2, . . . , n, for some t = 1, 2, . . . , n} is the numerical Silov boundary n n : l for A∞ (Bl∞ ∞ ).

Proof. (⊆): By Proposition 5.1 in [3], it follows. (⊇): Clearly M is (  × w∗ )-closed. By Theorem 3.2 (1), M is the set of all n n n : l numerical peak points for A∞ (Bl∞ ∞ ). Since l∞ is finite dimensional, M is the set of all numerical strong-peak points.


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ˇ NUMERICAL PEAK POINTS AND NUMERICAL SILOV BOUNDARY

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