### Article

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

**Published online** March 31, 2020

Copyright © Kyungpook Mathematical Journal.

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

Mohammad Ali Abolfathi∗ and Ali Ebadian

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

e-mail : m.abolfathi@urmia.ac.ir and a.ebadian@urmia.ac.ir

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

### Abstract

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

**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}_{x}_{∈}_{X}_{x}_{A}_{A}_{x}

Let (

The subalgebra of those functions

is denoted by _{α,K}_{X}_{α,K}

It is interesting to note that

Let

exists. We call ^{th}^{(}^{n}^{)}(

### Definition 1.1

Let

K is calledregular if for eachz _{0}∈K there exists a constantC such that for allz ∈K ,δ (z ,z _{0}) ≤C |z −z _{0}|.K is calleduniformly regular if there exists a constantC such that for allz ,w ∈K ,δ (z ,w ) ≤C |z −w |.

### Definition 1.2

Let ^{n}^{th}^{n}

Let

For each _{0} ∈ ^{1}(

This inequality implies that ^{1}(

### Definition 1.3

The algebra of complex-valued functions ^{(}^{j}^{)} ∈ ^{n}^{n}^{n}^{n}

The algebra of functions ^{(}^{j}^{)} ∈ ^{(}^{j}^{)} ∈ ^{∞}(^{∞}(^{∞}(^{∞}(

Let

Whenever we refer to

### Definition 1.4

Let

For

For convenience, we regard ^{n}^{n}_{j}

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

We show that, the completeness of ^{1}(^{n}^{n}

### Theorem 2.1

^{1}(_{1} = ||_{X}_{K}_{K}. Then Lip

**Proof**

Obviously _{n}_{X}_{α,K}_{n}^{n}_{j}^{1}(^{(}^{j}^{)} = _{j}^{(}^{j}^{)} ∈ _{n}_{Lip}_{(}_{X,K,M,α}_{)} → 0 as _{1} ∈ ℕ such that for every _{1} and

By letting _{1} and

Therefore for every _{1},

which implies (_{n}_{N}_{1})+_{N}_{1}) ∈ _{n}_{Lip}_{(}_{X,K,M,α}_{)} → 0 as

In the following let _{Lip}_{(}_{X,K,M,α}_{)} be the evaluation map, defined by _{z}_{z}

It is easy to see that _{Lip}_{(}_{K,M,α}_{)}_{Lip}_{(}_{X,K,M,α}_{)}.

### Theorem 2.2

_{Lip}_{(}_{X,K,M,α}_{)} = _{Lip}_{(}_{K,M,α}_{)} ∪

**Proof**

By the above, the inclusion _{Lip}_{(}_{K,M,α}_{)} ∪ _{Lip}_{(}_{X,K,M,α}_{)} is clear. Conversely, let _{Lip}_{(}_{X,K,M,α}_{)}. If _{K}_{0}(_{X}_{}_{K}

_{A}_{K}_{Lip}_{(}_{K,M,α}_{)}. We first show that _{1}|_{K}_{2}|_{K}_{1}, _{2} ∈ _{1} − _{2} ∈ _{1} − _{2}) = 0 hence _{1}) = _{2}). Therefore, _{Lip}_{(}_{K,M,α}_{)}. It follows that _{Lip}_{(}_{K,M,α}_{)}.

_{A}_{A}_{0}(_{X}_{}_{K}_{0} ∈ _{0}). We take an open set _{0} ∈

Obviously

So for every

Hence _{z}_{0}, for _{0} ∈ _{Lip}_{(}_{X,K,M,α}_{)} = _{Lip}_{(}_{K,M,α}_{)} ∪

### Corollary 2.3

**Proof**

By the naturality of

Hence

By similarly way the algebra

### Corollary 2.4

^{n}^{n}

### Lemma 2.5

D ^{n}^{+1}(X,K )⊆ Lip (^{n}X,K , 1)⊆ ℓip (^{n}X,K, α ),The standard norms of D ^{n}^{+1}(X,K )and Lip (^{n}X,K , 1)are equivalent on D ^{n}^{+1}(X,K ),D ^{n}^{+1}(X,K )is close subalgebra of Lip (^{n}X,K , 1).

**Proof**

(i) Obviously ^{n}^{n}^{n}^{+1}(^{(}^{j}^{)} ∈ ^{1}(

consequently,

Hence ^{(}^{j}^{)} ∈ ^{n}^{n}^{+1}(^{n}

(ii) Let ^{n}

Now for _{0} ∈ _{0}) and ^{n}^{+1}(

and hence, |^{(}^{n}^{+1)}(_{0})| ≤ _{1}_{,K}^{(}^{n}^{)}) and because _{0} is ordinary on

Hence for all ^{n}^{+1}(

Therefore standard norms of ^{n}^{+1}(^{n}^{n}^{+1}(

(iii) It follows immediate from (ii).

### Definition 2.6

Let

where _{0}(

It is easy to see that

### Theorem 2.7

_{A} is homoeomorphic to M_{Ā} if and only if_{X} for all f

**Proof**

See [5].

### Lemma 2.8

**Proof**

For

and it is also known that

therefore

and the proof is complete.

### Corollary 2.9

**Proof**

Straightforward calculations show that

for all ^{n}

hence

Therefore

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