Commutative Banach algebras

A Banach space B over the field of complex numbers C is a linear space over

C (thus, in B there are defined two operations — addition, and multiplication by complex scalars) which is provided with anorm, i.e a non-negative function : B −→R+= [ 0,∞) with the following properties:

(i) λa=|λ|afor eacha∈B and any complex scalarλ∈C.

(iii) 0 is the only element inB whose norm is zero.

(iv) B is a complete space with respect to the topology generated by the norm

By completeness we mean that every Cauchy sequence{an} ∞ n=1 of elements inB is convergent.

A Banach space B over Cis called aBanach algebra, ifB is provided with an associative operation (calledmultiplication) which is distributive with respect to addition, and if the inequality

In a Banach algebra, the inequality \( ab \leq ab \) is valid for all elements \( a, b \) in the set \( B \) A Banach algebra is classified as commutative if its multiplication operation is commutative Additionally, it is considered to have a unit if there exists a multiplicative identity element, typically represented by \( e \) or \( 1 \).

In a commutative Banach algebra B with a unit, an element f is considered invertible if there exists an element g in B such that their product equals the identity element e The inverse element, denoted as f −1, is uniquely defined for each invertible element f, satisfying the equation f −1 f = e.

B −1 of all invertible elements of B under multiplication is a subgroup ofB A simple example of a commutative Banach algebra with unit is the set of complex numbersC.

Proposition 1.1.1 Let B be a commutative Banach algebra with unit e Every element of the open unit ball centered ateis invertible, i.e.

{h∈B: h−ef Letg=f /s (1/s)f By the hypothesis g=f/|s|f Conse- quently,σ(f)⊂D(f), as claimed

Corollary 1.1.3 indicates that the spectrum of any element \( f \) in \( B \) is a bounded set in \( \mathbb{C} \), leading to the conclusion that \( \mathbb{C} \setminus \sigma(f) = \emptyset \) It is evident that \( B^{-1} \) forms an open subset of \( B \), with the mapping \( f \mapsto f^{-1} \) establishing a homeomorphism of \( B^{-1} \) onto itself Specifically, \( B^{-1} \) is an open group under multiplication within \( B \), and the mapping \( f \mapsto f^{-1} \) acts as a group automorphism of \( B^{-1} \) Additionally, since the spectrum \( \sigma(f) \) is a closed and bounded set, it qualifies as a compact subset of \( \mathbb{C} \) The value \( r_f = \max \) is also significant in this context.

|z|:z∈σ(f) is called thespectral radius of f ∈ B Since r f ≤ f, we have σ(f) ⊂D(r f )⊂ D(f) The spectral radiusr f can be expressed explicitly in terms off (e.g.[G1, S4, T2]) Namely, r f = lim n→∞ n f n ≤ lim n→∞ n f n =f (1.2)

Definition 1.1.4 Theperipheral spectrumof an elementfin a commutative Banach algebraB is the set σπ(f) z∈σ(f) :|z|=rf

A commutative Banach algebra B with a unit can be naturally represented by continuous functions on a compact topological space Central to this representation and the broader theory of commutative Banach algebras are complex-valued homomorphisms, which are linear multiplicative functionals of the algebra Specifically, a linear multiplicative functional of B is defined as any non-zero complex-valued function ϕ on B that satisfies certain properties.

(ii) ϕ(ab) =ϕ(a)ϕ(b) for every a, b∈ B, and all scalarsλ, à ∈ C The set M B of all non-zero linear multiplicative functionals ofBis called themaximal ideal space(or, thespectrum) ofB.

For a fixed a ∈ B with ϕ(a) = 0 we have ϕ(a) =ϕ(ea) = ϕ(e)ϕ(a), thus ϕ(a) ϕ(e)−1

= 0 Consequently,ϕ(e) = 1 for every linear multiplicative func- tionalϕofB Since aa −1 =e for everya∈B −1 , we have 1 =ϕ(e) =ϕ(aa −1 ) ϕ(a)ϕ(a −1 ), thusϕ(a)= 0 for every invertible elementa∈B.

Lemma 1.1.5 Every linear multiplicative functionalϕ∈ M B is continuous onB, andϕ= 1.

Proof Let f ∈ B, and let |z| > f for some z ∈ C Hence, ze−f ∈ B −1 by Corollary 1.1.3 According to the previous remark, ϕ(ze−f) = 0, and hence ϕ(f) = z ϕ(e) = z for every ϕ ∈ M B Consequently, the number ϕ(f) belongs to the disc z ∈C:|z| ≤ f

, i.e ϕ(f) ≤ f, and this holds for everyf ∈

B Therefore, the functional ϕ is bounded, thus continuous, and ϕ ≤ 1 By definition,ϕ is the least numberM withϕ(f)≤Mffor allf ∈B For any suchM we haveM ≥ 1, since 1 = ϕ(e) ≤ Me =M Hence, ϕ ≥ 1, and thereforeϕ= 1

In a compact Hausdorff set X, the space C(X) comprises all continuous functions on X, forming a commutative Banach algebra under pointwise operations and the uniform norm defined as f = max x∈X f(x) Among the linear multiplicative functionals of C(X), the point evaluation functional ϕx at a fixed point x ∈ X is notable, where ϕx(f) = f(x) for every f in C(X) It is evident that ϕx belongs to the set of multiplicative functionals MC(X), and it can be demonstrated that every element in C(X) can be related to such functionals.

MC(X) is of typeϕx for somex∈X Consequently,MC(X)and X are bijective spaces We usually identify them as sets without mention, and write them as

(b) LetD=D(1) z:|z| 0 Let F be a closed neighborhood of t1 in R+, which does not contain t2 There exists a functiong∈CR f(X) such that sup

Note thatg is a uniform limit of polynomials onf(X)⊂R Hence, the function g◦f belongs toA Since supp (à) =X and à

X g◦f dà=c 0, in contradiction with the previous equality Therefore, every real-valued function inAis constant, and consequently,Ais an antisymmetric algebra

The spaceC(X) for a compact Hausdorff setX is a uniform algebra LetK be a compact subset of the maximal ideal spaceM A of a uniform algebraA on

X Consider the algebra A K of restrictions of Gelfand transformsf , f ∈A on

In general, the algebra A K is not a closed subalgebra of C(K), meaning that A K does not always qualify as a uniform algebra However, the closure AK of A K within C(K) is indeed a uniform algebra, with the property that M A K is a subset of M A If the maximal ideal space M A K does not intersect the boundary ∂A, then the topological boundary ∂AK corresponds to b(M A K) in the context of the Gelfand topology, which follows as a direct consequence of the preceding statements.

Theorem 1.2.2 (Rossi’s Local Maximum Modulus Principle) If U is an open subset ofM A , then sup m∈U f(m)= max m∈bU ∪(∂A∩U) f(m) for every functionf ∈A.

Let A be a uniform algebra on X, and its maximal ideal space M_A is a compact set The point evaluation functional ϕ_x, defined as ϕ_x(f) = f(x), is a linear multiplicative functional and belongs to M_A for every x in X, allowing us to view X as a subspace of M_A The Gelfand transform of a function f in A is continuous on M_A, satisfying the relation f(ϕ_x) = ϕ_x(f) = f(x) for any point x in X Consequently, f can be regarded as a continuous extension of itself on the space.

M A Moreover, in a certain senseMAis the largest set for natural extension of all functions inA Recall that according to Lemma 1.1.5 the norm of anyϕ∈ M A is

The Gelfand transformation Λ establishes an isometric isomorphism between the algebra A and its Gelfand transform A, leading to the conclusion that A is closed in C(X) Additionally, the algebra of restrictions A|∂A on the Shilov boundary ∂A is isometrically isomorphic to A, resulting in the equivalence A ∼= A ∼= A ∂A Consequently, we treat the disc algebra A(D) and its restriction algebra A(T) on the Shilov boundary ∂A(D) = T as indistinguishable Furthermore, the relationship (m1−m2)(f) = m1(f)−m2(f) ≤ m1 + m2 f illustrates the properties of these algebras.

For every pair of elements \( m_1, m_2 \in M_A \) and a function \( f \in A \), the norm of the linear functional \( m_1 - m_2 \in A^* \) is bounded by 2, indicating that the diameter of the set \( M_A \subset A^* \) does not exceed this value The condition \( m_1 - m_2 < 2 \) establishes a transitive relation within \( M_A \), which can be verified as an equivalence relation The equivalence classes formed under this relation are known as the Gleason parts of \( A \) Notably, points at the extremes of the diameter, where \( m_1 - m_2 = 2 \), belong to separate Gleason parts.

A homomorphism Φ:A−→B between two uniform algebras naturally gen- erates an adjoint continuous mapΦ ∗ : M B −→ M A between their maximal ideal spaces, defined by Φ ∗ (ϕ)

IfΦ:A−→B preserves the norm, i.e if Φ(f)

B =f A for everyf ∈A, thenΦis called anembeddingofAintoB Clearly,Φ ∗ (∂B)⊂ M A

Proposition 1.2.3 Let A and B be uniform algebras, and let Φ: A −→ B be a homomorphism that does not increase the norm, i.e for whichΦ(f)

B ≤ f A , f ∈A Then Φis an embedding of A into B if and only if the range Φ ∗ (∂B) of Φ ∗ contains the Shilov boundary∂A.

Proof For everyf ∈Awe have m∈Φmax ∗ (∂B) m(f) = max ϕ∈∂B Φ ∗ (ϕ)

If∂A⊂Φ ∗ (∂B), then f A = max ϕ∈∂A f(ϕ)= max ϕ∈∂A ϕ(f)≤ max ϕ∈Φ ∗ (∂B) ϕ(f)=Φ(f)

Conversely, if Φ:A−→B is an isometry, thenΦ ∗ (∂B) is a boundary for A, since by (1.8) ϕ∈Φmax ∗ (∂B) f(ϕ)= max ϕ∈Φ ∗ (∂B) ϕ(f)=Φ(f)

Corollary 1.2.4 A homomorphism Φ of A onto B which does not increase the norm is an embedding if and only ifΦ ∗ (∂B) =∂A.

The proof of Proposition 1.2.3 demonstrates that it suffices to establish that Φ ∗ (∂B) is a subset of ∂A Assuming the opposite, if Φ ∗ (∂B) is not contained in ∂A, we can find a point ϕ0 in Φ ∗ (∂B) that is outside of ∂A By Corollary 1.1.8, there exists a neighborhood U around ϕ0 in M A \∂A, such that for any function f in A, the maximum value of f within U is less than or equal to the maximum value of f in the complement of U within M A.

By the assumed Φ(A) =B, we see that max ϕ∈(Φ ∗ ) −1 (U) g(ϕ)≤ max ϕ∈(Φ ∗ ) −1 (M A \U) g(ϕ) for everyg ∈B Consequently, (Φ ∗ ) −1 (M A \U) is a closed boundary ofB, and (Φ ∗ ) −1 (ϕ0)⊂(Φ ∗ ) −1 (U)⊂ M B \(Φ ∗ ) −1 (M A \U)⊂ M B \∂B, in contradiction with the initially assumed propertyϕ0∈Φ ∗ (∂B) HenceΦ ∗ (∂B)⊂∂A

Every embedding Φ: A(T) −→ A(T) of the disc algebra onto itself is an isometric isomorphism between A(T) and ΦA(T)

Consequently, the adjoint map Φ ∗ : M Φ(A(T)) −→ M A(T) generates a homeomorphism of D onto D, and Φ ∗ ∂(Φ(A(T))

=∂A(T) =T, i.e Φ ∗ (T) =T,Φ ∗ (D) =D, and hence the function Φ ∗ is a finite Blaschke product(cf [G2]) onD, i.e.

Therefore, for any embeddingΦ: A(T)−→A(T) ofA(T) onto itself there exists a finite Blaschke productB(z) onDwithΦ◦f =f ◦B, i.e such that Φ f(z)

Let A ⊂ C(X) be a uniform algebra on a compact set X One can easily identify certain points as elements of the Shilov boundary∂Aof a uniform algebra

A point \( x_0 \in X \) is defined as a peak point of a uniform algebra \( A \) if there exists a function \( f \) in \( A \) such that \( f(x_0) = 1 \) and \( f(x) < 1 \) for every \( x \in MA \setminus \{x_0\} \) Every peak point is included in the Shilov boundary \( \partial A \) Although the set of peak points does not always form a boundary for \( A \), it does serve as the minimal boundary for algebras with metrizable maximal ideal spaces.

A function \( f \in A \) is defined as a peaking function if it satisfies \( f = 1 \) and either \( f(x) = 1 \) or \( |f(x)| < 1 \) for all \( x \in M_A \) The corresponding peak set, denoted as \( P(f) = \{ x \in M_A : f(x) = 1 \} \), consists of all points where the function reaches its peak A function \( f \) is considered a peaking function if and only if \( \sigma_\pi(f) = \{ 1 \} \) If a subset \( K \subset M_A \) equals \( P(f) \) for some peaking function \( f \), we say that \( f \) peaks on \( K \) Additionally, \( K \) qualifies as a peak set if there exists a function \( f \in A \) such that \( f|_K \equiv 1 \) and \( |f(m)| < 1 \) for all \( m \in M_A \setminus K \).

A generalized peak point, or p-point, of a set A in a space M is defined as a point x that corresponds to the intersection of various peak sets of A Specifically, x is a p-point if, for every neighborhood V of x, there exists a peaking function f such that x belongs to the peak set P(f) contained within V The Choquet boundary, denoted as δA, encompasses all generalized peak points of A and serves as a boundary for A, with its closure aligning with the Shilov boundary ∂A, thus establishing the relationship δA = ∂A It is important to note that the peak points of A are not generally dense within ∂A, except in cases where MA is metrizable.

In this section, we will consider A as a uniform algebra on its maximal ideal space M A = X We define F(A) as the collection of all peaking functions within A For a specific point x in X, we denote F x (A) as the set of peaking functions of A that satisfy P(f) ∋ x, meaning that f(x) = 1.

Lemma 1.2.5 Let A ⊂ C(X) be a uniform algebra If f, g ∈ A are such that f h ≤ gh for all peaking functionsh∈ F(A), then|f(x)| ≤ |g(x)|on ∂A.

Assuming that \( f h \leq g h \) for every \( h \in F(A) \), if \( |f(x_0)| > |g(x_0)| \) at some \( x_0 \in \partial A \), we can choose a \( \gamma > 0 \) such that \( |g(x_0)| < \gamma < |f(x_0)| \) and find an open neighborhood \( V \) around \( x_0 \) where \( |g(x)| < \gamma \) Selecting a peaking function \( h \in F_{x_0}(A) \) with \( P(h) \subset V \) allows us to raise the power of \( h \) so that \( |g(x)h(x)| < \gamma \) for every \( x \in \partial A \setminus V \) This inequality also holds in \( V \), leading us to conclude that \( g h < \gamma \).

Therefore,|f(x0)|< γin contradiction with the choice ofγ Consequently,|f(x)| ≤

Corollary 1.2.6 If the functions f, g∈Asatisfy the equality f h=gh for all peaking functionsh∈ F(A), then|f(x)|=|g(x)| on∂A.

Lemma 1.2.7 If the functions f, g∈A satisfy the inequality ξ∈∂Amax

|g(ξ)|+|k(ξ)| for allk∈A, then |f(x)| ≤ |g(x)|for every x∈∂A.

Proof The proof follows the line of proof of Lemma 1.2.5.

In the context of the boundary ∂A, we have |f(x0)| > |g(x0)| for some point x0 Without loss of generality, we can assume x0 is part of δA By selecting a γ > 0 such that |g(x0)| < γ < |f(x0)|, we can find an open neighborhood V around x0 where |g(x)| remains less than γ Additionally, we define R > 1 so that f is bounded by R and the maximum value of |g(ξ)| on the boundary ∂A does not exceed R We can utilize a peaking function k ∈ F x0 (A) that is contained within V By adjusting k to a sufficiently high power, we ensure that the inequality |g(x)| + |Rk(x)| < R + γ holds for every x in ∂A excluding V Since this condition also applies within V, we conclude that |g(x)| + |Rk(x)| < R + γ for all x in ∂A.

Therefore,|f(x0)|< γin contradiction with the choice ofγ Consequently,|f(x)| ≤

Corollary 1.2.8 If the functions f, g∈A satisfy the equality ξ∈∂Amax

|g(ξ)|+|k(ξ)| for allk∈A, then |f(x)|=|g(x)|for every x∈∂A.

The following lemma, due to Bishop, helps to localize elements of uniform algebras.

Lemma 1.2.9 (Bishop’s Lemma) If E⊂X is a peak set for A, andf ≡0 onE for somef ∈A, then there exists a peaking function h∈ F(A)which peaks on E and such that

Proof Iff ∈Aand max ξ∈E|f(ξ)|=M >0 For any naturaln∈Ndefine the set

Clearly,E⊂Un⊂Un−1for everyn >1 Choose a functionk∈ F(A) which peaks on E, and let kn be a big enough power of k so that |kn(x)|< 1

=M We claim that|f(x)h(x)|< M for everyx /∈E In what follows,xis a fixed element inX\E.

(i) Letx /∈U1 Then x /∈Un for alln∈N, and hence|kn(x)|< 1

(ii) Letx∈Un−1\Unfor somen >1 Thenx∈Uifor every 1≤i≤n−1, andx /∈Ui for alli≥n Hence|f(x)|< M

2 i for alli≥n Sincex∈Un−1, we see that|f(x)|< M

If also σπ(f h) = σπ(gh) for all h ∈ F(A), then we have a much stronger result than in Corollary 1.2.6 Namely,

Lemma 1.2.10 If f, g∈A satisfy the equality σπ(f h) =σπ(gh) (1.11) for every peaking functionh∈A, thenf(x) =g(x)on∂A.

In the proof, we establish that \( f = g h \) holds true, as indicated by the condition \( |f(x)| = |g(x)| \) on the boundary \( \partial A \) For any point \( y \) in the boundary \( \delta A \), if \( f(y) = 0 \), it follows that \( g(y) = 0 \) as well, allowing us to assume \( f(y) = 0 \) without loss of generality We can select an open neighborhood \( V \) around \( y \) in \( X \) and utilize a peaking function \( k \) from \( F_y(A) \) such that \( P(k) \subset V \) By applying Bishop’s Lemma, we find another peaking function \( h \) in \( F_y(A) \) with \( P(h) = P(k) \), ensuring that the functions \( f \) and \( g h \) reach their maximum modulus exclusively within \( P(h) \) Consequently, we conclude that \( f(x_V) = g(z_V) \) for some \( z_V \in P(h) \), solidifying the relationship between \( f \) and \( g h \).

Since in every neighborhoodV ∋ythere are pointsxV andzV inV withf(xV) g(zV), then f(y) =g(y) by the continuity of f and g Consequently, f =g on

The next lemma is an additive version of Bishop’s Lemma (Lemma 1.2.9).

Lemma 1.2.11 (Additive analogue of Bishop’s Lemma) If E ⊂X is a peak set for A, and f ≡0 on E for some f ∈ A, then there exists a function h∈ F(A) which peaks onE and such that

|f(x)|+N|h(x)| 1 Choose a function k ∈ F(A) which peaks on E, and let kn be a big enough power of k so that R|kn(x)|< M

2 n knbelongs toF(A) Moreover,P(h) =h −1 {R}E, |h(x)| < 1 on X\E, and max ξ∈E

|f(x)|+R|h(x)|< M+R for everyx /∈E In what follows,xis a fixed element inX\E.

(i) Letx /∈U1 Thenx /∈Un for alln∈N, and henceR|kn(x)|< M

(ii) Let x∈Un−1\Un for some n >1 Thenx∈Ui for all 1≤i≤n−1, andx /∈Uifor eachi≥n Hence|f(x)|< M

2 i for eachi≥n Sincex∈Un−1, we see that|f(x)|< M

Un, then |f(x)| ≤ M, whence |f(x)|+R|h(x)| < M +R since|h(x)| Rfor the functionhconstructed above. Indeed,

Corollary 1.2.12 Let E be a peak set of A, x0∈E,f ∈A, N ≥ f, andα∈T be such that|f(x0)| = max ξ∈E |f(ξ)| >0 and f(x0) =α|f(x0)| If h is the peaking function ofA withP(h) =E, constructed in Lemma1.2.11, then

(a) ff(x) +(x0)+αN h(x)N for all≤ |xf∈(x)X|+\E, andN|h(x)| 0, there exists a compact set K within D such that the essential supremum of f over K is less than ε Notably, the boundary value mapping from H ∞ (D) to H ∞ (T) is an isometry; however, this mapping does not extend to the corresponding L ∞ algebras or the Bourgain algebras of H ∞ Nonetheless, it is important to note that the boundary value mapping can extend isometrically from H ∞ (D) to a specific algebra.

U(D) = [H ∞ (D), H ∞ (D)] generated by H ∞ (D) and H ∞ (D) Indeed, it extends to the generators ofU(D) and a closure argument provides a further extension to

Note that U(D) H ∞ (D), H ∞ (D) ∼= C(MH ∞ (D)), and isometries on

H ∞ (D) induced by automorphisms ofDextend naturally to isometries ofU(D).

Consider the algebras H ∞ (T) and H ∞ (D) Note that the maximal ideal spaces of the corresponding algebrasL ∞ are the corresponding sets∂H ∞ Since the mappingΛ:H ∞ −→C(∂H ∞ ) :f −→f ∂H ∞ is an isometry, we have

This follows immediately from (1.21), (1.22) and Corollary 1.4.6 Note that Corollary 1.4.7(ii) implies, in particular, thatH ∞ (D)

∂H ∞ (D)is a closed subalgebra ofL ∞ (D) ∂H ∞ (D) since Bourgain algebras are automatically closed.

The Bourgain algebra A B b contains important information aboutA IfA is an algebra of continuous functions on a setΩ, thenA B b contains also information aboutΩ.

Proposition 1.4.8 If U1 andU2 are biholomorphically equivalent domains in C n , then the corresponding Bourgain algebrasH ∞ (U1) U(U b 1 ) andH ∞ (U2) U(U b 2 ) are iso- metrically isomorphic.

Proof Let U1 and U2 be biholomorphically equivalent and τ: U2 −→ U1 be a biholomorphic mapping Define the mapT:Cb(U1)−→Cb(U2) by

T(f) (z1, z2, , zn) =f τ(z1, z2, , zn) for all f ∈ Cb(U1) and (z1, z2, , zn) ∈ U2 The mapping T is an isometric algebra isomorphism with respect to the sup-norms on U1 and U2 Moreover,

In what follows we apply completely continuous Hankel type operators related with the Bourgain algebras of the corresponding spacesH ∞ relative to the algebras

U = [H ∞ , H ∞ ] ⊂ L ∞ on the unit ball B n and the unit polydisc D n in C n , respectively, to the problem of biholomorphic equivalence of domains inC n Recall that every functionf ∈H ∞ (D n ) hasradial limits, rր1limf r(z1, z2, , zn)

=f ∗ (z1, z2, , zn), at almost every point (z1, z2, , zn)∈T n , and theradial boundary value function f ∗ (z1, z2, , zn) of any f ∈ H ∞ (D n ) belongs to H ∞ (T n ) (e.g [R6], Theorem 2.3.2]).

Lemma 1.4.9 Let g∈ U(D n ) and let the corresponding Hankel type operatorSg :

H ∞ (D n ) −→ U(D n )/H ∞ (D n ) be completely continuous If g ∗ is the boundary value function ofgonT n , then the operatorSg ∗ :H ∞ (T n )−→ U(T n )/H ∞ (T n )is also completely continuous.

This follows directly from the fact that every weakly null sequence {fn} in

H ∞ (T n ) is of the formfn=ϕ ∗ n where{ϕn}is a weakly null sequence inH ∞ (D n ).

Below we apply the Bourgain algebra technique to provide an alternative proof of the Poincar´e Theorem for analytic functions in several complex variables.

Theorem 1.4.10 (Poincar´e Theorem) The setsB n andD n are not biholomorphi- cally equivalent if n≥2.

Proof Suppose thatD n andB n are biholomorphically equivalent and letτ:D n −→

B n be a biholomorphic mapping between them DefineT:Cb(B n )−→Cb(D n ), as before, by

T(f) (z1, z2, , zn) =f τ(z1, z2, , zn) for allf ∈Cb(B n ) and (z1, z2, , zn)∈D n Proposition 1.4.8 implies that

Letf be a fixed non-constant function in the algebraA(B n ) Observe that

We claim that the mapping S T (f) : H ∞ (D n ) −→ U(D n )/H ∞ (D n ) is com- pletely continuous Note that f ∈ C(B n )| B n and alsof ∈ U(B n ) An argument from [I] implies that

C(B n )| B n ⊂H ∞ (B n ) L b ∞ (B n ) Consequently, by the remark immediately following Definition 1.4.1 we have f ∈H ∞ (B n ) L b ∞ (B n ) ∩ U(B n ) =H ∞ (B n ) U(B b n ) , and hence

=H ∞ (D n ) U(D b n ) , by Proposition 1.4.4, i.e S T (f) is a completely continuous operator, as claimed. From Lemma 1.4.9 it follows that forn ≥2 the radial boundary value function

T(f)∗ of the non-constant anti-analytic function T(f) belongs to H ∞ (T n ) H ∞ (T n ) L b ∞ (T n ) =H ∞ (T n ) U(T b n ) (see [I, Y]), which is impossible

Note that the boundary value technique avoids the need of a direct reference to the more complicated Bourgain algebras ofH ∞ (D n ) andH ∞ (B n ).

Bourgain algebras can be extended to commutative topological algebras, where a commutative algebra A over C is defined as a topological algebra if it is equipped with a topology that ensures the continuity of its basic operations In this context, let B represent a commutative topological algebra with A as its subalgebra, and let c bw 0 (A) denote the space of bounded weakly null sequences within A.

Definition 1.4.11 TheBourgain algebraA B b of a commutative topological algebra

Arelative toB is the set of all elementsf ∈B for whichSf c bw 0 (A)

⊂c0(B/A), i.e A B b consists of all f ∈ B such that for every {ϕn} ∈ c bw 0 (A) there exists a sequence{gn} inAfor which n→∞lim(ϕnf −gn) = 0 (1.23)

It is straightforward to see that ifA is an algebra, thenA⊂A B b

Proposition 1.4.12 Let A ⊂ B be commutative topological algebras Every com- pletely continuous Hankel type operatorSf:A−→πA(f A) maps bounded weakly Cauchy sequences inA onto Cauchy sequences inB/A.

In this proof, we consider a bounded weakly Cauchy sequence {gn} in A, where the sequence {πA(f gn)} is not Cauchy in B/A We identify a neighborhood U of 0 in B/A such that for every natural number M > 0, we can find integers nM and mM ≥ M, ensuring that πA(f gnM) - πA(f gmM) is not in U Consequently, the sequence πA(f(gnM - gmM)) does not converge to 0 in B/A Due to the complete continuity of Sf, the bounded sequence {gnM - gmM} for M=1 is not weakly null in A, leading to the conclusion that F(gnM - gmM) does not approach 0 for some F in A* Thus, the sequence {F(gn)} is not Cauchy, indicating that {gn} cannot be a weakly Cauchy sequence.

Note that the dual space B ∗ does not always separate the points of B for every commutative topological algebra B Local convexity of B is a sufficient condition for this.

Theorem 1.4.13 Let B be a commutative topological algebra andA be a subalge- bra of B The Bourgain algebra A B b of A relative to B is a closed commutative topological subalgebra ofB.

In the context of functional analysis, let \( f \) be an element of the set \( A \) and \( B \) Given a bounded weakly null sequence \( \{ \phi_n \} \) in \( c_w^0(A) \) where \( \phi_n \in A \), it follows that there exist elements \( h_n \in A \) such that \( \phi_n f - h_n \) converges weakly to 0 Importantly, the sequence \( \{ h_n \} \) is also a bounded weakly null sequence in \( A \), as \( \phi_n f \) is bounded and tends weakly to 0.

Let \( f_1, f_2 \in A B_b \) and consider a bounded weakly null sequence \( \{ \phi_n \} \) in \( A \) According to the established results, there exist \( h_n \in A \) such that \( \phi_n f_1 - h_n \to 0 \) As noted, \( \{ h_n \} \) is also a bounded weakly null sequence in \( A \) Consequently, there are \( k_n \in A \) such that \( h_n f_2 - k_n \to 0 \) This leads to the conclusion that \( f_1 f_2 \phi_n - k_n = f_2 (f_1 \phi_n - h_n) + (f_2 h_n - k_n) \to 0 \) Therefore, \( f_1 f_2 \in A B_b \), confirming that \( A B_b \) forms an algebra.

Polynomial extensions of Banach algebras

In this section we derive a method for expanding commutative Banach algebras, in which polynomials play a crucial role.

Let A and B be commutative Banach algebras with units, where B is considered an extension of A if there exists a homomorphism from A into B that preserves the unit Additionally, A[x] represents the algebra of polynomials in the variable x over A Similar to the scalar case, the degree of a polynomial p(x) = an x^n + an−1 x^(n−1) + + a0, with coefficients in A, is defined by the highest power of x present in the polynomial.

A is said to be again the greatest integern with a0 = 0 In this case a0 ∈ A is called again theleading coefficient ofp(x) Ifa0 = 1 thenp(x) is called amonic polynomial.

Let \( p(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_0 \) be a polynomial in \( A[x] \) We define an associated extension of \( A \) by considering the ideal \( I = p(x)A[x] \subset A[x] \) The quotient algebra \( B = A[x]/I \) is equipped with the norm given by \( n-1 \sum_{i=0}^n c_ix^i + p(x) \in A[x] \), where the parameter \( t > 0 \) satisfies the condition \( t^n \geq c_1t^{n-1} + c_2t^{n-2} + \ldots + c_n \).

The Arens-Hoffman extension of a commutative Banach algebra A, denoted as B, is characterized by the natural homomorphism from A into B, which serves as an embedding This extension is associated with a polynomial p(x) and exhibits several notable properties.

Any element \( b \in B = A[x]/I \) can be uniquely expressed in the form \( a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \ldots + a_0 \), where \( a_i \in A \) This indicates that the algebra \( B \) functions as a free module over \( A \), with the functions \( x_1, x_2, \ldots, x_{n-1} \) forming an \( A \)-basis for \( B \) The norm of \( B \) corresponds to the component-wise norm, defined as \( \max_{0 \leq i \leq n-1} a_i \) Additionally, the set of linear multiplicative functionals \( M_B \) is homeomorphic to the set of pairs \( (m, z) \in M_A \) such that the polynomial \( z^n + m(a_{n-1})z^{n-1} + m(a_{n-2})z^{n-2} + \ldots + m(a_0) = 0 \).

.(c) Every b ∈ B is an integral element over A, i.e there exists a polynomial q(x) =x n +c1x n−1 +ã ã ã+cn withci∈A, such that q(b) = 0.

We observe that according to the property (b) the map π:M B −→ M A : (m, z)−→mis surjective By the Fundamental Theorem of Algebra, cardπ −1 (m)

In the case of polynomials with multiple variables, the degree of a polynomial \( p(x_1, x_2, \ldots, x_n) = a_{i_1, i_2, \ldots, i_n} x_1^{i_1} x_2^{i_2} \cdots x_n^{i_n} \) is determined by identifying the highest sum of the indices \( i_1 + i_2 + \ldots + i_n \) The notation \( A[x_1, x_2, \ldots, x_n] \) represents the algebra of polynomials in \( n \) variables, where the coefficients are drawn from the set \( A \).

Let A and B be commutative Banach algebras with units, where B is defined as a polynomial extension of A This means there exists an isomorphism Φ from A to a Banach subalgebra C within B that shares the same unit as B Additionally, there is a homomorphism Ψ that maps A[x1, x2, , xn] onto B, establishing a structured relationship between these algebras.

A[x1, x2, , xn] is commutative, where id:A−→Aã1 is the natural embedding ofAintoA[x1, x2, , xn].

Example 1.5.1 (a) If A ⊂ B and A is isometrically isomorphic to B, then B is a trivial polynomial extension of A Indeed, one can take the set of constant polynomialsAã1 forA[x1, x2, , xn] in (1.27).

Let B represent the algebra of all complex-valued functions f(z, t) defined on the domain D × [0, 1], where the function z → f(z, t) is in the disc algebra for each fixed t in the interval [1/4, 1] Additionally, let A be a subset of B, which includes functions f for which z → f(z, t) is also in the disc algebra for every fixed t in the narrower range of [1/2, 1] It is evident that A is a proper subset of B.

B are isomorphic, andB is a polynomial extension ofA by the property (a).

Let X = D × [0, 1] as defined previously, and let B represent the algebra of functions f(z, t) in C(X) For each fixed t, the function f(z, t) is a member of the disc algebra, indicating that B serves as a polynomial extension of the algebra A.

, since B = A+zA, M A = M B However, B is not an Arens-Hoffman extension ofA.

As part (c) of Example 1.5.1 shows, not all polynomial extensions are neces- sarily Arens-Hoffman extensions.

Theorem 1.5.2 If B is a polynomial extension ofA, then there is a nested family of Banach algebras,

A=A0⊂A1⊂A2⊂ ã ã ã ⊂An,such that there is a homomorphismΦfromAnontoBfor whichΦ A is the isomor- phismΦ B A from(1.27), andAi is an Arens-Hoffman extension ofAi−1, 1≤i≤n.

A[x1, x2, , xn] and Ψ(A) is isomorphic to

Assuming that set A is a subset of set B, we can identify elements b1, b2, , bn within B such that B can be expressed as A combined with these elements Let Bk represent the space of polynomials formed by b1, b2, , bn in A, where the degrees of these polynomials do not exceed k.

According to the Baire Category Theorem, there exists a value \( k_0 \geq 0 \) such that the space \( B_{k_0} \) is classified as a second category within \( B \) The space \( D_{k_0} \) consists of polynomials in variables \( x_1, x_2, \ldots, x_n \) over \( A \), with degrees not exceeding \( k_0 \) The elements \( x_1^{i_1} x_2^{i_2} \cdots x_n^{i_n} \) where \( i_1 + i_2 + \cdots + i_n \leq k \) form a basis for \( D_{k_0} \) Additionally, \( D_{k_0} \) operates as a free module over \( A \) and is recognized as a Banach space under the norm defined by \( \| a_{i_1, i_2, \ldots, i_n} x_1^{i_1} x_2^{i_2} \cdots x_n^{i_n} \| = a_{i_1, i_2, \ldots, i_n} \).

The operator T: ai 1 ,i 2 , ,i nx i 1 1 x i 2 2 ã ã ãx i n n −→ ai 1 ,i 2 , ,i nb i 1 1 b i 2 2 ã ã ãb i n n , is a continuous mapping fromDk 0 onto the space of second categoryBk 0 ofB Hence

B = Bk 0 by the Open Mapping Theorem For a fixed b ∈ B define a linear operatorS:Dk 0 −→ Dk 0 as follows For any basis element x i 1 1 x i 2 2 ã ã ãx i n n define S(x i 1 1 x i 2 2 ã ã ãx i n n ) to be an element inT −1 (bãT x i 1 1 x i 2 2 ã ã ãx i n n )

S ai 1 ,i 2 , ,i nx i 1 1 x i 2 2 ã ã ãx i n n ai 1 ,i 2 , ,i nS(x i 1 1 x i 2 2 ã ã ãx i n n ).

By using induction, we establish that for all natural numbers n, T(S^n) equals b^n T According to the Cayley–Hamilton Theorem, there exists a polynomial q(x) defined as x^n + a1x^(n-1) + + an over the ring A, such that q(S) equals zero Consequently, we find that T(q(S(d))) equals zero, which implies q(b)T equals zero Given that the unit of A corresponds to the unit of B and T preserves these units, it follows that q(b) must also equal zero Therefore, it can be concluded that every element b in B is an integral element over A.

Let the polynomialsqj(xj) =x n j +a1jx n−1 j +ã ã ã+anj be such thatqj(bj) = 0 for 1≤j≤n DefineA0=A, and letAj be the Arens-Hoffman extension ofAj−1 associated withqj(xj), j = 1,2, , n We obtain a nested sequence

In the context of Banach algebras, we consider a nested sequence of algebras A = A0 ⊂ A1 ⊂ A2 ⊂ ⊂ An, where each Aj is an Arens-Hoffman extension of the previous algebra Aj−1 for 1 ≤ j ≤ n Each element d in An can be uniquely expressed as d = ai1,i2, ,in x1^i1 x2^i2 xn^in, with 0 ≤ ik ≤ n Consequently, the norm on An is equivalent to the previously established norm (1.27) This leads to the conclusion that the homomorphism Φ: An → B, defined by Φ(ai1,i2, ,in x1^i1 x2^i2 xn^in) = ΦB(A)(ai1,i2, ,in bi1^i1 bi2^i2 bin^in), is continuous, and it holds that Φ|A = ΦB(A), as demonstrated.

Corollary 1.5.3 Any b∈B is an integral element overΦ B A (A).

Proof Since, as we have seen above, every a ∈ An is an integral element of A, thenais also an integral element ofΦ B A (A)

Corollary 1.5.4 If B is a polynomial extension of an algebraA, thenB is a finite A-module.

Proof Since An is a finite A-module and the surjective homomorphismΦ B A pre- serves the algebraA, thenB is also a finiteA-module

Note that the algebraB from Example 1.5.1(b) is a polynomial extension of

A, but there are nob1, b2, , bn∈B, n= 1,2, ,such thatB=A[b1, b2, , bn].

If A ⊂ B are Banach algebras with the same unit, we say that B is a strong polynomial extension of B, if there are b1, b2, , bn ∈ B such that B A[b1, b2, , bn].

In the context of Banach algebras, if A is a closed subalgebra of a Banach algebra B and b is an element of B such that B is generated by A and b, then there exists a closed subalgebra D of B that contains A Furthermore, the dimension of the quotient space B/D is equal to r(b) - 1, where r(b) represents the minimal degree of all monic polynomials q(x) with coefficients in A for which q(b) equals zero.

In this proof, we define r(b) as k and consider A1, the Arens-Hoffman extension of A linked to the polynomial q(x) = x^k + ak-1x^(k-1) + ak-2x^(k-2) + + a0, where q(b) = 0 According to Theorem 1.5.2, the homomorphism Φ: A1 → B = A[b] results in B being equal to A1, expressed as A + Ab + Ab^2 + + Ab^(k-1) This indicates that the algebra A1 functions as a free (k-1)-dimensional A-module Consequently, the null space of Φ, denoted as Null(Φ), consists of polynomials p(x) = ck-1x^(k-1) + ck-2x^(k-2) + + c0 in A1, satisfying the condition p(b) = 0.

For a fixed p(x)∈Null (Φ) let Kp m∈

, where c (p) 1 ∈ A is the leading coefficient of p(x) We claim that the family of sets{Kp} p has a nonempty intersection, i.e.

Indeed, if suppose, on the contrary, that K = ỉ, then for every p ∈ Null (Φ) one can find anm ∈ M S with m c (p) 1

In the context of polynomial equations, there exist polynomials p1(x), p2(x), , pl(x) within the null space of Φ, along with elements d1, d2, , dl from set A, such that the sum of the products of these elements and their corresponding polynomials equals one Consequently, the polynomial p(x) defined as the linear combination of these elements and polynomials results in p(b) equaling zero, indicating that its degree is k−1, with the leading coefficient at x raised to the power of k−1.

1 This contradicts the minimality property of r(b) = k Therefore, K = ỉ, as claimed.

Letp(x) =ck−1x k−1 +ck−2x k−2 +ã ã ã+c0 be a polynomial in Null (Φ) We claim thatm(ci) = 0 for every m∈K, and everyci,i= 1,2, , k According to (1.29) we have thatm(ck−1) = 0 Sinceq(b) = 0, whereqis the polynomial (1.28).

It follows that b k =−ak−1b k−1 −ak−2b k−2 − ã ã ã −a0, and therefore,

= (ck−2−ck−1ak−1)b k−1 + (ck−3−ck−1ak−2)b k−2 +ã ã ã+ (c0−ck−1a0)b

Consequently, the polynomial x p(x) = (ck−2−ck−1ak−1)x k−1 + (ck−3−ck−1ak−2)x k−2 +ã ã ã

Isomorphisms between uniform algebras

In this section we assume thatAandBare uniform algebras on their maximal ideal spacesXandY respectively We find conditions for peripherally multiplicative and peripherally additive operators to be algebraic isomorphisms.

Definition 1.6.1 An operatorT:A→B is said to be

(a) preserving the peripheral spectraof algebra elements if σπ(T f) =σπ(f) for everyf ∈A (1.31)

First we will consider the case of σπ-additive operators.

Lemma 1.6.2.If an operatorT:A→B isσπ-additive, then the following equalities hold for allf, g∈A.

If, in addition T is surjective, then it isR-linear.

Proof The equality (a) is obvious, since|z|or everyz∈σπ(f) (b) follows from (a) by letting f = g = 0 (c) follows from the σπ-additivity of T Indeed, σπ

T(−f) = 0, and therefore,T(−f) =−T f Equalities (d) and (f) follow from the σπ-additivity ofT and (a) correspondingly, by letting g= 0 (e) follows from (d) and theσπ-additivity ofT because ofσπ

=σπ(f+g) =σπ(T f+T g) (g) follows form (a) and (c) The last statement follows from the theorem of Mazur- Ulam [MU] (see also [V]), since, by (b) and (g),T(0) = 0, andT f−T g=f−g for allf, g∈A

Note that according to equality (f), the operator T preserves the norms of algebra elements, and therefore maps the unit ball of A in the unit ball of B.

By (d) T preserves the peripheral spectra, and by (g) it preserves the distances between algebra elements It is straightforward to see that theσπ-additivity ofT is equivalent to both (d) and (e).

LetC ∗ =C\ {0} In the sequel we will use the following notation: sF(A) ={s f:f ∈ F(A)}, wheres∈C ∗ , sF x (A) ={s f:f ∈ F x (A)}, wheres∈C ∗ ,

C∗-peaking functions of A are defined as elements of the family {s: F(A), s ∈ C∗} A function f is classified as a C∗-peaking function if and only if its spectrum σπ(f) is a singleton Similarly, a function g belongs to sF(A) if σπ(g) equals {s} For any function f in sF(A) with s ∈ C∗, the peak set P(f) is defined as f −1 {s} Notably, the peak sets satisfy the relation P(sf) = P(f) for any function f in F(A) and any s in C∗ Consequently, the collection of peak sets for the classes of peaking functions aligns with that of C∗-peaking functions of A.

Lemma 1.6.3 If T :A→B is a surjective operator which preserves the peripheral spectra of algebra elements, then

Proof Indeed, if s∈C ∗ then T sF(A)

⊂sF(B) follows by the preservation of the peripheral spectra byT Given a k ∈ sF(B), s∈ C ∗ , let k =T h for some h∈A Then h∈sF(A) sinceσπ(h) = σπ(T h) =σπ(k) ={s} Hencek =T h∈

, and thereforesF(B)⊂T sF(A) Consequently,T sF(A)

=sF(B) for anys∈C ∗ , as claimed

Definition 1.6.4 An operatorT:A→B is calledmonotone increasing in modulus if the inequality |f(x)| ≤ |g(x)| on ∂A implies |(T f)(y)| ≤ |(T g)(y)| on ∂B for everyf, g∈A.

Lemma 1.6.5 If a monotone increasing in modulus surjective operator T:A→B preserves the peripheral spectra of algebra elements, then for any generalized peak pointx∈δA the set

=sF(B) (1.36) for any s ∈ C ∗ Indeed, if s ∈ C ∗ then T sF(A)

⊂ sF(B) follows by the preservation of the peripheral spectra by T Given a k ∈ sF(B), s ∈ C ∗ , let k=T hfor some h∈A Thenh∈sF(A) since σπ(h) =σπ(T h) = σπ(k) = {s}. Hence k = T h ∈ T sF(A) , and therefore sF(B) ⊂ T sF(A)

=sF(B) for anys∈C ∗ , as claimed.

Let x be a generalized peak point of A First we show that the family

P(T h) :h ∈ C ∗ ã F x (A) has the finite intersection property If h1, h2, , hn belong to C ∗ ã F x (A), and hj ∈ sjF x (A), sj ∈ C ∗ , then, clearly, the function g = h1ãh2ã ã ãhn belongs to the space (s1ãs2ã ã ãsn)ã F x (A) Since |hj(x)| ≤

|sj|, j= 1, , n, we have|g(ξ)|=|h1(ξ)| ã |h2(ξ)| ã ã ã |hn(ξ)| ≤ j=k|sj|

|hk(ξ)| j=k|sj| hk(ξ) for every ξ ∈ ∂A and any fixed k = 1, , n The preserva- tion of peripheral spectra by T implies that T g ∈ (s1 ãs2ã ã ãsn)ã F(B) and

T hk ∈ skF(B) Hence, |(T hk)(y)| ≤ |sk| for every y ∈ Y By Lemma 1.6.2(h),

T isR-linear, and since it is also monotone increasing in modulus, it follows that

For every point \( y \) on the boundary of \( B \), the inequality \( |(T hk)(y)| \leq |(s1ãs2ã ã ãsn)| \) holds This leads to the conclusion that for any \( y \in Y \) where \( |(T g)(y)| = |s1ãs2ã ã ãsn| \), it follows that \( |(T hk)(y)| = |sk| \) Consequently, we can deduce that \( (T hk)(y) = sk \), which implies that the image of \( T g \) is a subset of the image of \( T hk \) This relationship is true for all \( k = 1, \ldots, n \), resulting in \( P(T g) \subset \bigcup_{k=1}^{n} P(T hk) \).

P(T hj) Conse- quently, the family

P(T h) :h∈C ∗ ã F x (A) has the finite intersection property, as claimed Hence it has a non-empty intersection, since all of its elements are closed subsets of the compact setY

The following lemma provides sufficient conditions for an operatorT:A→B to be monotone increasing in modulus.

Lemma 1.6.6 If a surjective operatorT:A→B satisfies the equality

|f(ξ)|+|g(ξ)| for everyf, g∈A, then it is monotone increasing in modulus.

Proof If|f(x)| ≤ |g(x)| on∂A, then, clearly, ξ∈∂Amax

|g(ξ)|+|k(ξ)| for anyk∈A By equality (ii) we have η∈∂Bmax

. Now from Lemma 1.2.7 and the surjectivity of T if follows that |(T f)(y)| ≤

Proposition 1.6.7 If a surjective operatorT:A→B satisfies the equalities (i) σπ(T f+T g) =σπ(f+g), and

|f(ξ)|+|g(ξ)| for allf, g∈A, thenT is bijective.

Proof We will show thatT is injective IfT f =T gfor somef, g∈A, then for any h∈Awe haveT f+T h=T g+T h Consequently,σπ(T f+T h) =σπ(T g+T h). Theσπ-additivity ofT implies σπ(f+h) =σπ(T f+T h) =σπ(T g+T h) =σπ(g+h).

By equality (ii) we have ξ∈∂Amax

|g(ξ)|+|h(ξ)| for everyh∈A Lemma 1.2.13 now implies that f =g HenceT is injective, and therefore bijective

Lemma 1.6.8 If the operatorT: A→B satisfies the assumptions of Proposition 1.6.7, then for generalized peak pointx∈δA the setEx is a singleton and belongs toδB.

In this article, we demonstrate that the operator T is monotone increasing in modulus, as established by equality (ii) and Lemma 1.6.6 Additionally, according to Lemma 1.6.2(d), T preserves the peripheral spectra of algebra elements Consequently, T meets the criteria outlined in Lemma 1.6.5, which is further validated through the proof process.

P(T f) :f ∈ C ∗ ã F x (A) is a family of peak sets with non-empty intersection,

SinceT preserves peripheral spectra of algebra elements, equality (1.34) im- plies T −1

Ex∩ δB Let y ∈ Ex∩ δB, k ∈ F y (B), and let h = T −1 (k) To show that h∈ F x (A) it is enough to verify that h(x) = 1 Take an open neighborhoodV of x and a peaking function g ∈ F x (A) with P(g) ⊂ V Equality (1.34) yields

T g ∈ F(B) Since y ∈ Ex ⊂ P(T g) we have that (T g)(y) = 1, and therefore,

= 2, and there must be axV ∈∂A with h(xV) = 1 and g(xV) = 1 Therefore, xV ∈ P(g)⊂ V We deduce that any neighborhood

V of x contains a point xV with h(xV) = 1 The continuity of h implies that h(x) = 1, thush∈ F x (A) Consequently,T −1

Let y ∈ Ex∩δB If there were a z ∈ Ex\ {y}, there would be a peaking functionk ∈ F y (B) with |k(z)| < 1 For anyh ∈ T −1 (k)∩ F x (A) we have h∈

F x (A),k=T h∈ F(B), and P(k) =P(T h)⊃Ex Hence the functionk=T his identically equal to 1 onEx, contradicting|k(z)| r \), the open set \( W = \{ \xi \in \delta A : |h(\xi)| > r \} \) contains \( x \) By Lemma 1.6.10, for every \( \xi \in W \), we have \( |k(\tau(\xi))| = (T h)(\tau(\xi)) \geq |h(\xi)| > r \), implying \( \tau(\xi) \in V \) because \( |k(\eta)| < r \) on \( \delta B \setminus V \) Thus, \( \tau(W) \subset V \), demonstrating the continuity of \( \tau \) Similarly, considering the operator \( T^{-1}: B \to A \) and the mapping \( \tau^{-1}: \delta B \to \delta A \), we conclude that \( \tau^{-1} \) is also continuous, completing the proof.

When applied to the operatorT −1 :B→Aand the mappingτ −1 :δB→δA, Lemma 1.6.10 implies|g(y)| ≤(T −1 g)(τ −1 (y))for anyy∈δB and everyg∈B, or, ifg =T f, f ∈A, andy=τ(x), x∈δA, equivalently,(T f)(τ(x))≤ |f(x)|. Hence we have the following

Corollary 1.6.12 If the operator T: A→B satisfies the assumptions of Proposi- tion1.6.7, then|(T f)(τ(x))| ≤ |f(x)| for anyx∈δA and everyf ∈A.

Proposition 1.6.13 If T:A →B satisfies the assumptions of Proposition 1.6.7, then the equality

(T f)(τ(x)) =f(x) (1.40) holds for everyf ∈A and every generalized peak point x∈δA.

Proof If T satisfies the assumptions of Proposition 1.6.7, then, by Lemma 1.6.8 the mappingτ from (1.37) is well-defined, and, given anx∈δA, anyC ∗ -peaking functionh∈sF x (A), s∈C ∗ , satisfies the equality (1.38), i.e (1.40).

Let x0 be a generalized peak point of A and let f be in A with f T f = R Without loss of generality we can assume that f(x0) = 0, since in f(x0) = 0, then also (T f)(τ(x0)) = 0 by Lemma 1.6.9, applied to the operator

In the context of the mapping T −1 :B → A and the function T f ∈ B, we consider an open neighborhood V of x0 in X According to the additive version of Bishop’s Lemma, we can select an R-peaking function h ∈ Rã F x 0 (A) such that x0 is included in P(h) and lies within V This function |f(x)| + |h(x)| reaches its maximum solely within the set P(h) contained in V Furthermore, we identify ξV ∈ P(h) and αV ∈ T such that f(ξV) equals αV multiplied by the absolute value of f(ξV).

(f+αVh)(ξ)=f(ξV) +αVR Hence, by Corollary 1.2.12,

|f(ξV)|+R=f(ξV) +αVR=f+αVh, (1.41) whilef +γh ≤ f+αVh for anyγ ∈ T Therefore,f(ξV) +αVR ∈σπ(f+ αVh) =σπ

Hence there is a pointzV ∈Y with f(ξV) +αVR(T f+T(αVh)

We may assume thatzV ∈δB Indeed,|f(ξV)|+R=f+αVhis the maximum modulus of the functionf+αVh, and, according to Lemma 1.6.2(d), of the function

T f+T(αVh) as well Therefore, the function T f +T(αVh) attains the value

At a specific location on the Choquet boundary δB, we can select the point zV, which satisfies the equation |f(ξV)| + R The surjectivity of τ indicates that zV can be expressed as zV = τ(xV) for some xV in δA Furthermore, the relationships established by equality (1.42), Corollary 1.6.12 (ii), and equation (1.38) lead to the conclusion that f(ξV) + αVR equals (T f + T(αVh).

≤(T f)(τ(xV))+|(T(αVh))(τ(xV))| ≤ |f(xV)|+|αh(xV)|=|f(xV)|+|h(xV)|

Since this maximum is attained only within P(h), xV ∈ P(h), thus h(xV) = 1, and according to (1.41), (T(αVh))(zV) = (T(αVh))(τ(xV)) =αVh(xV) =αVR. Now equality (1.42) becomes f(ξV) +αVR (T f+T(αVh)

= (T f)(τ(xV)) +αVh(xV) = (T f)(τ(xV)) +αVR, thusf(ξV) = (T f)(τ(xV)) Therefore, any neighborhoodV ofx0 contains points ξV andxV such thatf(ξV) = (T f)(τ(xV)) The continuity off, T f, andτimplies thatf(x0) = (T f)(τ(x0))

Theorem 1.6.14 Let A ⊂ C(X) and B ⊂ C(Y) be uniform algebras on their maximal ideal spacesX andY correspondingly If a surjective operatorT:A→B satisfies the equalities

|f(ξ)|+|g(ξ)| for everyf andg in A, thenT is an isometric algebra isomorphism fromAonto B.

Proof Proof Since the operatorT satisfies the hypotheses of Proposition 1.6.7, then Proposition 1.6.13 implies that the equality (1.40), i.e (T f)(τ(x)) = f(x), holds for everyx∈δAand allf ∈A Therefore, the restricted operatorT ′ : A| δA →

B| δB defined by T ′ (f| δA ) = T f| δB , f ∈ A, is an algebra isomorphism between

A| δA andδB Since the Choquet boundary of an algebra is its boundary,A| δA ∼=A andB| δB ∼=B, and also T is uniquely determined byT ′ It follows thatT is an algebra isomorphism betweenAandB

Recall that an operatorT:A→Bis said to beT-homogeneous, if the equality(T f)(sf) =s(T f) holds for everyf ∈A and anys∈T={z∈C:|z|= 1}.

Lemma 1.6.15 Every additive operator T which preserves the peripheral spectra of algebra elements is σπ-additive, i.e satisfies the peripheral additivity property (i)of Theorem1.6.14.

=σπ(f+g). Lemma 1.6.16 If an operator T:A→B satisfies the equality

T f+αT g=f+αg (1.43) for everyf, g∈Aand any α∈T, thenT satisfies equality(ii)of Theorem1.6.14. Proof Letf, g∈A Ifα∈Tis such that η∈∂Bmax

The argument is reversible, and therefore, η∈∂Bmax

|f(ξ)|+|g(ξ)| i.e equality (ii) holds, as claimed

Clearly, Lemma 1.6.16 holds for any operator that satisfies the equality σπ(T f+αT g) =σπ(f +αg) for every f, g in Aand any α∈T In particular, it holds for anyT-homogeneous operator which isσπ-additive.

Lemma 1.6.17 Any C-linear operator T: A→B with T(1) = 1 which preserves the norms of algebra elements, preserves also their peripheral spectra.

Proof Let f ∈ A and z0 ∈ σπ(f) Then z0 = f(x0) for some x0 ∈ X, and

|z0|=|f(x0)|=f Clearly,f+z0= max ξ∈∂A f(x) +z0= 2|z0| The linearity of

T yieldsT(f+z0) =T f+T(z0) =T f+z0T(1) =T f+z0 The norm-preservation ofT implies|(T f)(y)| ≤ T f=f=|z0|, and hence

T f+z0=T(f+z0)=f+z0= 2|z0|,since |f(x)| ≤ |z0| for all x ∈ X Thus (T f)(y0) = z0 for some y0 ∈ Y, and therefore,z0∈σπ(T f), since|z0|=f=T f.

Let \( u_0 \in \sigma_\pi(T f) \) for some \( f \in A \), implying \( u_0 = (T f)(y_0) \) for a specific \( y_0 \in Y \) and \( |u_0| = |(T f)(y_0)| = T f \) The linearity of \( T \) leads to \( T(f + u_0) = T f + T(u_0) = T f + u_0 T(1) = T f + u_0 \) By the norm-preservation property of \( T \), it follows that \( |f(x)| \leq f = T f = |u_0| \) and \( f + u_0 = T f + T(u_0) = T(f + u_0) = T f + u_0 = 2|u_0| \), with \( |(T f)(y)| \leq |u_0| \) for all \( y \in Y \) Consequently, \( f(x_0) = u_0 \) for some \( x_0 \in X \), indicating that \( u_0 \in \sigma_\pi(f) \) since \( |u_0| = T f = f \) Thus, we conclude that \( \sigma_\pi(T f) = \sigma_\pi(f) \), leading to the subsequent corollary derived from Theorem 1.6.14 and Lemma 1.6.16.

Corollary 1.6.18 If a surjective additive operatorT:A→B preserves the periph- eral spectra of algebra elements and satisfies the equality

|f(ξ)|+|g(ξ)| for allf, g∈A, thenT is an isometric algebra isomorphism.

Theorem 1.6.14 and Lemma 1.6.16 imply the following

Proposition 1.6.19 If a surjective operator T:A→B satisfies the equality (i ′ ) σπ(T f+αT g) =σπ(f +αg) for everyf, g inA and anyα∈T, thenT is an isometric algebra isomorphism.

Proof Indeed, (i ′ ) implies (1.43) and also (i) of Theorem 1.6.14, and thereforeT satisfies both equalities (i) and (ii) of Theorem 1.6.14

As mentioned before, the equality (i ′ ) is satisfied automatically by any T- homogeneous operator which is σπ-additive The next proposition follows from Proposition 1.6.19.

Proposition 1.6.20 If T:A→B is a surjective,T-homogeneous, andσπ-additive operator withT(1) = 1, thenT is an isometric algebra isomorphism.

Theorem 1.6.21 Any surjectiveC-linear operatorT between two uniform algebras, which preserves the peripheral spectra of algebra elements, is an isometric algebra isomorphism.

Corollary 1.6.22 Any surjective C-linear operator T: A → B with T(1) = 1, which preserves the norms of invertible algebra elements, is an isometric algebra isomorphism.

Proof Clearly, if T preserves the norms of invertible elements and T(c) = c for anyc∈C, thenT preserves the norms of all algebra elements The result follows from Theorem 1.6.21 and Lemma 1.6.17

As a consequence from Corollary 1.6.22 we obtain Nagasawa’s theorem [N] (also [R3]) for uniform algebras.

In the context of uniform algebras A and B, consider their respective open unit balls, B A and B B If there exists a biholomorphic mapping F from B A to B B that satisfies the conditions F(0) = 0 and F(1) = 1, then F can be extended to A as an isometric algebra isomorphism.

According to T Ransford [R3], the operator F maintains the norms and extends to a C-linear isomorphism from space A to space B, while also preserving these norms Furthermore, Corollary 1.6.22 indicates that this extension is multiplicative We will now examine the scenario involving σπ-multiplicative operators.

Lemma 1.6.24 If an operator T:A→B with T(1) = 1is σπ-multiplicative, then the equalities

Proof The equality (a) follows from theσπ-multiplicativity ofT by lettingg= 1. (a) and theσπ-multiplicativity ofT imply (c), because ofσπ

Equalities (b) and (d) follow from (a) and (c) correspondingly, since|z|or everyz∈σπ(f)

The operator T maintains the peripheral ranges of algebra elements, as indicated by equality (a) Additionally, the σπ-multiplicativity property (1.33) is directly equivalent to both properties (a) and (c) Furthermore, property (b) is also relevant in this context.

T preserves the norms of algebra elements, and according to (d) it is norm- multiplicative.

Lemma 1.6.25 Any surjective and norm-multiplicative operator T: A → B is monotone increasing in modulus.

Proof If |f(x)| ≤ |g(x)| on ∂A, then clearlyf h ≤ gh for anyh ∈ A The surjectivity of T implies that for any k ∈ F(B) there is an h ∈ A such that k=T(h) By the norm-multiplicativity ofT we have

(T f)ãk=(T f)(T h)=f h ≤ gh=(T g)(T h)=(T g)ãk) for every k ∈ F(B) Now Lemma 1.2.5 implies that |(T f)(y)| ≤ |(T g)(y)| on

Proposition 1.6.26 Any σπ-multiplicative surjective operator T: A→B is bijec- tive.

Proof IfT f =T gfor somef, g∈A, then for anyh∈ F(A) we have (T f)(T h) (T g)(T h) Consequently,σπ

Lemma 1.2.10 now implies thatf =g Therefore,T is injective, and consequently it is bijective

Lemma 1.6.27 Let T: A → B be a surjective σπ-multiplicative operator with

T(1) = 1 Then for any generalized peak point x ∈ δA the set Ex defined in (1.35)is a singleton and belongs toδB.

According to Lemma 1.6.5, the set Ex is confirmed to be non-empty Let x be a generalized peak point of A within δA The σπ-multiplicativity property of the operator T indicates that T is norm-multiplicative Furthermore, as stated in Lemma 1.6.25, the operator T is monotone increasing in modulus Given that T(1) equals 1, it follows from Lemma 1.6.24(a) that specific properties of T are established.

T preserves the peripheral spectra of algebra elements HenceT satisfies the hy- potheses of Lemma 1.6.5 From its proof we know that

P(T f) :f ∈ F x (A) is a family of peak sets with non-empty intersection,Ex, hence it meetsδB(e.g [L1]). Consequently,Ex∩δB= ỉ.

SinceT preserves the peripheral ranges of algebra elements, then, by (1.33), we have thatT −1

For any \( y \in E_x \cap \delta B \), let \( k \in F_y(B) \) and define \( h = T^{-1}(k) \) The uniqueness of \( h \) follows from the injectivity of \( T \) To demonstrate that \( h \in F_x(A) \), it suffices to show \( h(x) = 1 \) Consider an open neighborhood \( V \) of \( x \) and a peaking function \( g \in F_x(A) \) with \( P(g) \subset V \) According to (1.33), \( T g \in F(B) \) Since \( y \in E_x \subset P(T g) \), we have \( (T g)(y) = 1 \), indicating \( T g \in F_y(B) \) By Lemma 1.6.24(c), it follows that \( k(y)(T g)(y) = 1 \geq h(g) = (T h)(T g) = k^\sim(T g) = 1 \).

Hence hg= 1, and there must be axV ∈∂A withh(xV) = 1 andg(xV) = 1. Therefore,xV ∈P(g)⊂V We deduce that any neighborhoodV ofxcontains a pointxV withh(xV) = 1 The continuity ofhimplies thath(x) = 1, soh∈ F x (A). Consequently,T −1

Let \( y \) be an element of the intersection of \( Ex \) and \( \delta B \), and assume there is another element \( z \) in \( Ex \) that is not equal to \( y \) It follows that a peaking function \( k \) exists within \( F_y(B) \) such that the absolute value of \( k(z) \) is less than 1 Given that \( h = T^{-1}(k) \), we find \( h(x) = 1 \) Consequently, \( Ex \) is a subset of \( P(T h) \), which is equivalent to \( P(k) \), leading to the conclusion that \( k(z) \) must equal 1, contradicting the earlier condition that \( |k(z)| < 1 \) Therefore, we can conclude that the set \( Ex \) contains exactly one point.

We see that under the assumptions of Lemma 1.6.27, for any x ∈ δA the elementτ(x) from (1.37) for which

P(T f) is well-defined, and so is the mappingτ:x−→τ(x) fromδA into δB Moreover, the equality (1.38), i.e.

(T h)(τ(x)) =h(x) holds for every C ∗ -peaking function h ∈ C ∗ ã F x (A) Under the assumptions of Lemma 1.6.27, the operator T is bijective, by Proposition 1.6.26 Let k ∈ sã

Fτ(x)(B), for some x ∈ δA, and let T −1 k = h ∈ Cã F(A) By (1.38) we have k(τ(x)) = (T h)(τ(x)) =h(x) = (T −1 k)(x) Therefore, the equality (1.39), i.e.

(T −1 k)(x) =k(τ(x)) also holds for everyx∈δAand any C ∗ -peaking functionk∈C ∗ ã Fτ(x)(B) ofB.

Lemma 1.6.28 Let T:A→B be a surjective andσπ-multiplicative operator with

T(1) = 1, and let f ∈ A If (T f)(τ(x0)) = 0 for some x0 ∈ δA, then also f(x0) = 0.