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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(1): 117-125

Published online March 31, 2020

### The Maximal Ideal Space of Extended Diﬀerentiable Lipschitz Algebras

Department of Mathematics, Urmia University, P. O. Box 165, Urmia, Iran

Received: May 14, 2018; Revised: December 25, 2018; Accepted: December 26, 2018

In this paper, we first introduce new classes of Lipschitz algebras of infinitely differentiable functions which are extensions of the standard Lipschitz algebras of infinitely differentiable functions. Then we determine the maximal ideal space of these extended algebras. Finally, we show that if X and K are uniformly regular subsets in the complex plane, then R(X,K) is natural.

Keywords: Banach function algebra, diﬀ,erentiable Lipschitz algebras, extended Lipschitz algebra, maximal ideal space, rational functions. This work was supported by Urmia University.

### 1. Introduction

Let X be a compact Hausdorff space, and let C(X) be the Banach algebra of all continuous complex-valued functions on X under the uniform norm, ||f||X = supxX |f(x)| for f in C(X). A subalgebra A of C(X) which separates the points of X, contains the constants, and which is a Banach algebra with respect to some norm ||.||, is a Banach function algebra on X. If the norm of a Banach function algebra is equivalent to the uniform norm, then it is a uniform algebra. If A is a function algebra on X, then Ā, the uniform closure of A, is a uniform function algebra on X. Let A be a Banach function algebra on X. Then for every xX the map ex : ff(x), A → ℂ, is a complex homomorphism on A and it is called the point evaluation homomorphism at x. A Banach function algebra A on X is natural if MA, the maximal ideal space of A, is X; equivalently, if every homomorphism on A is given by evaluation at a point of X. Define the map J : XMA by J(x) = ex.

Let (X, d) be a compact metric space, K be a compact subset of X and 0 < α ≤ 1. We define

$pα,K(f)=sup{∣f(x)-f(y)∣dα(x,y):x,y∈K,x≠y},Lip(X,K,α):={f∈C(X):pα,K(f)<∞}.$

The subalgebra of those functions f in Lip(X,K, α) such that

$limd(x,y)→0∣f(x)-f(y)∣dα(x,y)=0, (x,y∈K,x≠y),$

is denoted by ℓip(X,K, α). It is easy to see that these extended Lipschitz algebras are both Banach algebras under the norm ||f||α,K = ||f||X + pα,K(f). In fact, Lip(X,K, α) is a natural Banach function algebra on X for α ≤ 1 and ℓip(X,K, α) is a natural Banach function algebra on X whenever α < 1, .

It is interesting to note that Lip(X,K, β) is dense in ℓip(X,K, α) for each β where α < β ≤ 1. This result is similar to the density of Lip(X, β) in ℓip(X, α) , using the measure theory and duality. For further general facts about Lipschitz algebras the reader is referred to [1, 9, 10].

Let X be a perfect, compact plane set. A complex-valued function f : X → ℂ is complex-differentiable at aX if

$f′(a)=lim{f(z)-f(a)z-a:z∈X,z→a}$

exists. We call f′(a) the complex derivative of f at a. Also, we denote the nth derivative of f at aX by f(n)(a). Now we introduce the type of compact sets which we shall consider next, .

### Definition 1.1

Let K be a compact plane set which is connected by rectifiable arcs, and suppose δ(z,w) is the geodesic metric on K, the infimum of the lengths of the arcs joining z and w.

• K is called regular if for each z0K there exists a constant C such that for all zK, δ(z, z0) ≤ C|zz0|.

• K is called uniformly regular if there exists a constant C such that for all z,wK, δ(z,w) ≤ C|zw|.

### Definition 1.2

Let K and X be compact plane sets such that KX. Let Dn(X,K) be the algebra of continuous functions on X with continuous nth derivatives on K. For fDn(X,K), we define the norm by

$‖f‖n=‖f‖X+∑j=1n‖f(j)‖Kj!.$

Let K and X be compact plane sets such that KX. If K is a finite union of regular sets then the following property is holds:

For each z0K there exists a constant C such that for every zK and fD1(X,K),

$∣f(z)-f(z0)∣≤C∣z-z0∣(‖f‖X+‖(f∣K)′‖K).$

This inequality implies that D1(X,K) is complete . Throughout this paper we always assume that X and K are nonempty compact plane sets, KX and 0 < α ≤ 1 for Lip and 0 < α < 1 for lip.

### Definition 1.3

The algebra of complex-valued functions f on X whose derivatives up to order n exists on K and for each j(1 ≤ jn), f(j)Lip(K, α) is denoted by Lipn(X,K, α). The algebra ℓipn(X,K, α) is defined in a similar way. For f in Lipn(X,K, α) or in ℓipn(X,K, α), let

$‖f‖α,K,n=‖f‖X+pα,K(f)+∑j=1n‖f(j)‖α,Kj!=‖f‖X+pα,K(f)+∑j=1n‖f(j)‖K+pα,K(f(j))j!.$

The algebra of functions f in Lip(X,K, α)(ℓip(X,K, α)) with derivatives of all orders for which f(j)Lip(K, α)(f(j)ℓip(K, α)) for all j(j = 1, 2, 3 . . .) is denoted by Lip(X,K, α)(ℓip(X,K, α)). Now, we introduce certain subalgebra of Lip(X,K, α) and ℓip(X,K, α) as follows:

Let $M={Mj}j=0∞$ be a sequence of positive numbers such that

$M0=1andMjMr.Mj-r≥(jr)(r=0,1,…,j).$

Whenever we refer to $M={Mj}j=0∞$ we mean this sequence satisfies the above conditions.

### Definition 1.4

Let

$Lip(X,K,M,α)={f∈Lip∞(X,K,α):(‖f‖X+pα,K(f)+∑j=1∞‖f(j)‖α,KMj)<∞},ℓip(X,K,M,α)={f∈ℓip∞(X,K,α):(‖f‖X+pα,K(f)+∑j=1∞‖f(j)‖α,KMj)<∞}.$

For f in Lip(X,K,M, α) or in ℓip(X,K,M, α), set

$‖f‖=‖f‖X+pα,K(f)+∑j=1∞‖f(j)‖α,KMj.$

For convenience, we regard Lipn(X,K, α) and ℓipn(X,K, α) as being algebras of the type Lip(X, , K,M, α) and ℓip(X,K,M, α), respectively, by setting Mj = j! (j = 0, 1, 2 . . ., n) and $1Mj=0 (j=n+1,…)$. In this paper, we determine the maximal ideal of Lip(X,K,M, α) and ℓip(X,K,M, α). Finally by using a interesting method we obtain the maximal ideal of new classes of uniform algebras generated by rational functions, which defined by T. G. Honary and S. Moradi in .

### 2. The Maximal Ideal Space of Extended Differentiable Lipschitz Algebras

We show that, the completeness of D1(X,K) implies that Lipn(X,K, α) and ℓipn(X,K, α) are Banach function algebras.

### Theorem 2.1

Let X be a compact plane set which is connected by rectifiable arcs and sequence$M={Mk}k=0∞$satisfies the condition above and D1(X,K) complete under norm ||f||1 = ||f||X + ||(f|K)′||K. Then Lip(X,K,M, α) or ℓip(X,K,M, α) is Banach functions algebra under norm$‖f‖=‖f‖X+pα,K(f)+∑j=1∞‖f(j)‖α,KMj$on X.

Proof

Obviously Lip(X,K,M, α), is normed algebra and contains the constant functions and separates the points of X. Now, we prove that the algebra Lip(X,K,M, α) is complete. Let ${fn}n=1∞$ be a Cauchy sequence in Lip(X,K,M, α), so ${fn}n=1∞$ is a Cauchy sequence in Lip(X,K, α). By completeness of Lip(X,K, α), there exists fLip(X,K, α) such that ||fnf||X + pα,K(fnf) → 0 as n → ∞. On the other hand ${fn(j)}n=1∞$ is a Cauchy sequence in Lip(K, α) for all j(j = 1, 2, 3 . . .). Since Lipn(K, α) is complete , there exists gjLip(K, α) such that $‖fn(j)-gj‖α,K→0$ as n→ ∞ for all j(j = 1, 2, 3 . . .). Consequently by the completeness of D1(X,K) f(j) = gj and f(j)Lip(K, α) for all j. We show that fLip(X,K,M, α) and ||fnf||Lip(X,K,M,α) → 0 as n→∞. For ɛ = 1 there exists N1 ∈ ℕ such that for every m, nN1 and i ∈ ℕ,

$(‖fn-fm‖X+pα,K(fn-fm)+∑j=1i‖fn(j)-fm(j)‖α,KMj)≤(‖fn-fm‖X+pα,K(fn-fm)+∑j=1∞‖fn(j)-fm(j)‖α,KMj)<1.$

By letting m → ∞ for every nN1 and i ∈ ℕ we have

$(‖fn-f‖X+pα,K(fn-f)+∑j=1i‖fn(j)-f(j)‖α,KMj) ≤1.$

Therefore for every nN1,

$(‖fn-f‖X+pα,K(fn-f)+∑j=1∞‖fn(j)-f(j)‖α,KMj) ≤1,$

which implies (fnf) ∈ Lip(X,K,M, α) and consequently f = ((ffN1)+fN1) ∈ Lip(X,K,M, α). It is clearly ||fnf||Lip(X,K,M,α) → 0 as n → ∞. By similarly way the algebra ℓip(X,K,M, α) is Banach function algebra.

In the following let J : XMLip(X,K,M,α) be the evaluation map, defined by J(z) = ez, where ez is the evaluation homomorphism on Lip(X,K,M, α). Let also

$M˜Lip(K,M,α)={φ˜:φ˜(f)=φ(f∣K),φ∈MLip(K,M,α),f∈Lip(X,K,M,α)}.$

It is easy to see that Lip(K,M,α)⊆ MLip(X,K,M,α).

### Theorem 2.2

Let K and X be perfect compact subsets of the complex plane such that KX and Lip(K,M, α) is complete. Then MLip(X,K,M,α) = Lip(K,M,α)J(XK).

Proof

By the above, the inclusion Lip(K,M,α)J(XK) ⊆ MLip(X,K,M,α) is clear. Conversely, let ψMLip(X,K,M,α). If A = {fC(X) : f|K = 0} then C0(XK) = A|XK  and M(A) ≅ XK. Moreover, ALip(X,K,M, α). There are two cases for ψ :

Case 1: ψ|A = 0. In this case we prove that if φ(f|K) = ψ(f) for fLip(X,K,M, α) then φMLip(K,M,α). We first show that φ is well-defined. Let f1|K = f2|K, where f1, f2Lip(X,K,M, α). Obviously f1f2A so ψ(f1f2) = 0 hence ψ(f1) = ψ(f2). Therefore, φ is well-defined. Clearly, φ is a homomorphism and hence φMLip(K,M,α). It follows that ψ = φ̃Lip(K,M,α).

Case 2: Let ψ|A ≠ 0. So ψ|A is a non-zero homomorphism on A. By C0(XK) = A|XK, there exists an z0XK such that for every fA, ψ(f) = f(z0). We take an open set U in X such that KU and z0XU. By Urysohn’s Lemma there exists a continuous function h on X such that 0 ≤ h ≤ 1 and

$h(z)={1z∈K0z∈XU.$

Obviously hLip(X,K,M, α) and moreover,

$ψ(h)=ψ(1-(1-h))=ψ(1)-ψ(1-h)=1-(1-h) (z0)=1-1=0.$

So for every fLip(X,K,M, α),

$ψ(f)=ψ(fh-f(1-h))=ψ(fh)+ψ(f(1-h))=ψ(f)ψ(h)+f(1-h) (z0)=ψ(f).0+f(z0)=f(z0).$

Hence ψ = ez0, for z0XK. Therefore, MLip(X,K,M,α) = Lip(K,M,α)J(XK).

### Corollary 2.3

With the same assumptions in the theorem, if moreover, Lip(K,M, α) is natural then Lip(X,K,M, α) is also natural.

Proof

By the naturality of Lip(K,M, α)

$M˜Lip(K,M,α)={φ˜:φ˜(f)=ez(f∣K)=f(z),f∈Lip(X,K,M,α)}=J(K).$

Hence

$MLip(X,K,M)=J(K)∪J(XK)=J(X)≅X.$

By similarly way the algebra ℓip(X,K,M, α) is natural.

### Corollary 2.4

The algebras Lipn(X,K, α) and ℓipn(X,K, α) both are natural.

### Lemma 2.5

Let X and K be uniformly regular subsets of the complex plane such that KX. For n ≥ 0,

• Dn+1(X,K) ⊆ Lipn(X,K, 1) ⊆ ℓipn(X,K, α),

• The standard norms of Dn+1(X,K) and Lipn(X,K, 1) are equivalent on Dn+1(X,K),

• Dn+1(X,K) is close subalgebra of Lipn(X,K, 1).

Proof

(i) Obviously Lipn(X,K, 1) ⊆ ℓipn(X,K, α). If fDn+1(X,K) then fC(X) and f(j)D1(K) for all j = 0, 1, 2, . . . , n. Since K is uniformly regular, there exists C > 0 such that for all j = 0, 1, 2, . . . , n and z, w in K we have

$∣f(j)(z)-f(j)(w)∣≤C‖f(j+1)‖K∣z-w∣$

consequently,

$p1,K(f(j))=sup{∣f(j)(z)-f(j)(w)∣∣z-w∣:z,w∈K,z≠w}≤C‖f(j+1)‖K.$

Hence fLip(X,K, 1) and f(j)Lip(K, 1) for all j = 1, 2, . . . , n. It follows fLipn(X,K, 1) therefore Dn+1(X,K) ⊆ Lipn(X,K, 1).

(ii) Let fLipn(X,K, 1). Then we have

$‖f‖Lipn(X,K,1)=‖f‖X+p1,K(f)+∑j=1n‖f(j)‖K+p1,K(f(j))j!=‖f‖X+p1,K(f)+∑j=1n‖f(j)‖Kj!+∑j=1np1,K(f(j))j!≤‖f‖X+∑j=1n‖f(j)‖Kj!+∑j=0n-1C‖f(j+1)‖Kj!+C‖f(n+1)‖Kn!=‖f‖X+∑j=1n‖f(j)‖Kj!+∑j=1nC‖f(j)‖K(j-1)!+C‖f(n+1)‖Kn!=‖f‖X+∑j=1n(1+Cj)‖f(j)‖Kj!+C‖f(n+1)‖Kn!≤(1+(n+1)C) (‖f‖X+∑j=1n+1‖f(j)‖Kj!)=(1+(n+1)C) ‖f‖Dn+1.$

Now for z0K and for all zK which (zz0) and fDn+1(X,K) we have,

$∣f(n)(z)-f(n)(z0)∣∣z-z0∣≤p(f(n)),$

and hence, |f(n+1)(z0)| ≤ p1,K(f(n)) and because z0 is ordinary on K, thus

$‖f(n+1)‖K≤p1,K(f(n)) (f∈Dn+1(X,K)).$

Hence for all f in Dn+1(X,K),

$‖f‖n+1=‖f‖X+∑j=1n+1‖f(j)‖Kj!=‖f‖X+∑j=1n‖f(j)‖Kj!+‖f(n+1)‖K(n+1)!≤‖f‖X+∑j=1n-1‖f(j)‖Kj!+‖f(n)‖Kn!+p1,K(f)(n+1)!=‖f‖X+∑j=1n-1‖f(j)‖Kj!+(n+1)‖f(n)‖K+p1,K(f)(n+1)!≤‖f‖X+p1,K(f)+∑j=1n-1‖f(j)‖Kj!+(n+1)‖f(n)‖K+p1,K(f)n!≤(n+1) (‖f‖X+p1,K(f)+∑j=1n‖f(j)‖α,Kj!)=(n+1)‖f‖Lipn(X,K,1).$

Therefore standard norms of Dn+1(X,K) and Lipn(X,K, 1) are equivalent on Dn+1(X,K).

(iii) It follows immediate from (ii).

### Definition 2.6

Let K and X be compact subset of ℂ such that KX. We define

$R(X,K)={f∈C(X):f∣K∈R(K)}$

where R(K) is the uniform closure of R0(K), the algebra of all rational functions with poles off K, [4, 8].

It is easy to see that R(X,K) is uniform algebra and hence it is Banach function algebra.

### Theorem 2.7

Let X be a compact Hausdorff space, and let A be a Banach algebra on X. Then MA is homoeomorphic to MĀ if and only if |||| = ||f||X for all fA.

Proof

See .

### Lemma 2.8

Let n ≥ 0 and K and X be uniformly regular subsets of the complex plane such that KX. Then$Dn+1(X,K)¯=ℓipn(X,K,α)¯=Lipn(X,K,α)¯=R(X,K)$.

Proof

For n ≥ 0, we have

$Dn+1(X,K)⊆ℓipn(X,K,α)⊆D1(X,K)⊆R(X,K),$

and it is also known that

$R0(X,K)⊆D1(X,K)⊆R(X,K),$

therefore$Dn+1(X,K)¯=ℓipn(X,K,α)¯=Lipn(X,K,α)¯=R(X,K),$

and the proof is complete.

### Corollary 2.9

Let K and X be uniformly regular subsets of the complex plane such that KX. Then R(X,K) is natural.

Proof

Straightforward calculations show that

$‖(fm)(j)‖K≤2n2δnmn‖fm‖Km-n, (0≤j≤n)pα,K((fm)(j))≤2n2δn-1mn+1λ‖fm‖Km-n, (0≤j≤n)$

for all m > n, where δ and λ are constants independent of m and j. Therefore for fLipn(X,K, α) we have,

$‖fm‖α,K,n=‖fm‖X+pα,K(fm)+∑j=1n‖(fm)(j)‖α,Kj!,=‖fm‖X+pα,K(fm)+∑j=1n‖(fm)(j)‖K+pα,K((fm)(j))j!,≤‖fm‖X+∑j=1n1j!2n2δnmn‖fm‖Km-n+∑j=0n1j!2n2δn-1mn+1λ‖fm‖Km-n,≤‖f‖Xm+‖fm‖Km-n((n+2)2n2δn-1mn) (δ+λm),≤‖f‖Xm+‖fm‖Xm-n((n+2)2n2δn-1mn) (δ+λm),≤‖f‖Xm(1+(n+2)2n2δn-1mn) (δ+λm),$

hence

$‖fm‖α,K,n1m≤(‖f‖Xm)1m(1+(n+2)2n2δn-1mn)1m(δ+λm)1m.$

Therefore $‖f^‖=limm→∞‖fm‖α,K,n1m≤‖f‖$. By Theorem 2.7 and Lemma 2.8 completes the proof of corollary.

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