Additive jointly separating maps and ring homomorphisms

TL;DR

This paper characterizes additive jointly separating maps and ring homomorphisms between spaces of vector-valued continuous functions, generalizing recent results and providing new insights into their structure and properties.

Contribution

It offers a partial description of additive jointly separating maps and characterizes continuous ring homomorphisms between Banach algebras of vector-valued functions.

Findings

Partial description of additive jointly separating maps

Characterization of continuous ring homomorphisms

Generalizations of recent unital homomorphism results

Abstract

Let $X$ and $Y$ be compact Hausdorff spaces, $E$ and $F$ be real or complex normed spaces and $A(X,E)$ be a subspace of $C(X,E)$. For a function $f\in C(X,E)$, let $\coz(f)$ be the cozero set of $f$. A pair of additive maps $S,T: A(X,E) \lo C(Y,F)$ is said to be jointly separating if $\coz(Tf)\cap \coz(Sg)=\emptyset$ whenever $\coz(f)\cap \coz(g)= \emptyset$. In this paper, first we give a partial description of additive jointly separating maps between certain spaces of vector-valued continuous functions (including spaces of vector-valued Lipschitz functions, absolutely continuous functions and continuously differentiable functions). Then we apply the results to characterize continuous ring homomorphisms between certain Banach algebras of vector-valued continuous functions. In particular, the results provide some generalizations of the recent results on unital homomorphisms between vector-valued Lipschitz algebras, with a different approach.

Equations

$$Tf(y)=J_y(f(\phi (y)))(f\in A(X,E))$$

$$L(f)=\underset{x\ne y}{\sup }\frac{\Vert f(x)-f(y)\Vert }{d^{\alpha }(x,y)}<\infty .$$

$$\Vert f\Vert =\sum _{i=0}^n\frac{\Vert f^{(i)}{\Vert }_{\mathrm{Lip}}}{i!}(f\in {\mathrm{Lip}}^n([0,1],E).$$

$${\phi }_2(f(x_{1,2}))={\phi }_2(f)={\phi }_2(f(x_{2,3}))(f\in \mathcal{A}).$$

$${\phi }_2(c)={\phi }_2(cf(x_{1,2}))={\phi }_2(cf)={\phi }_2(cf(x_{2,3}))=0,$$

$${\phi }_j(f)={\phi }_j(f(x))(f\in \mathcal{A}).$$

$${\Phi }_1(f)=\mathrm{Re}({\phi }_1(f))-i\mathrm{Re}({\phi }_1(if))(f\in A(X)),$$

$${\Phi }_2(f)=\mathrm{Re}({\phi }_2(f))-i\mathrm{Re}({\phi }_2(if))(f\in A(X)).$$

$${\Phi }_1(f)={\int }_{X}fd\mu \mathrm{and}{\Phi }_2(f)={\int }_{X}fd\nu$$

$$|\mu |(K_1)|\nu |(K_2)=0.$$

$$\mathrm{Re}({\phi }_1)=\mathrm{Re}({\alpha }_1{\delta }_x)$$

$$\mathrm{Re}({\phi }_2)=\mathrm{Re}({\alpha }_2{\delta }_x).$$

$$\mathrm{Im}({\phi }_1)=\mathrm{Re}({\beta }_1{\delta }_x),\mathrm{Im}({\phi }_2)=\mathrm{Re}({\beta }_2{\delta }_x).$$

$$\mathrm{Re}(i{\phi }_1)=\mathrm{Re}({\lambda }_1{\delta }_{x^{\prime }})$$

$$\mathrm{Re}({\phi }_2)=\mathrm{Re}({\lambda }_2{\delta }_{x^{\prime }}).$$

$$0=\mathrm{Re}({\alpha }_{2}f(x))=\mathrm{Re}({\phi }_2(f))=\mathrm{Re}({\lambda }_{2}f(x^{\prime }))=|{\lambda }_2{|}^2\ne 0,$$

$$\mathrm{Im}({\phi }_1)=-\mathrm{Re}(i{\phi }_1)=\mathrm{Re}(-{\lambda }_1{\delta }_x),$$

$${\phi }_1(f)={\phi }_1(1)\mathrm{Re}(f(x))+{\phi }_1(i)\mathrm{Im}(f(x))(f\in \mathcal{A}).$$

$$Y_{0}=\big{(}\cup_{e\in E}{\rm coz}(T(c_{e}))\big{)}\cap\big{(}\cup_{e\in E}{\rm coz}(S(c_{e}))\big{)},$$

$$Y_c=\{y\in Y_0:{\delta }_y\circ T,{\delta }_y\circ S:A(X,E)⟶F\mathrm{are}\Vert \cdot {\Vert }_{\infty }-\mathrm{weak}\mathrm{continuous}\}.$$

$$Tf(y)={\Lambda }_y(f(\Phi (y)))(f\in A(X,E),y\in Y_c),$$

$$Sf(y)={\Lambda }_y^{\prime }(f(\Phi (y)))(f\in A(X,E),y\in Y_c).$$

$${\phi }_{y,v^{\ast }}=v^{\ast }\circ {\delta }_y\circ T\mathrm{and}{\psi }_{y,v^{\ast }}=v^{\ast }\circ {\delta }_y\circ S.$$

$${\phi }_{y,v^{\ast }}^e(f)={\phi }_{y,v^{\ast }}(fe)\mathrm{and}{\psi }_{y,v^{\ast }}^e(f)={\psi }_{y,v^{\ast }}(fe)(f\in A(X)).$$

$$H_y=\{(v_1^{\ast },v_2^{\ast },e_1,e_2)\in F^{{\ast }^2}\times E^2:{\phi }_{y,v_1^{\ast }}^{e_1}\mathrm{and}{\psi }_{y,v_2^{\ast }}^{e_2}\mathrm{are}\mathrm{nonzero}\}.$$

$${\phi }_{y,v_1^{\ast }}^{e_1}={\phi }_{y,v_1^{\ast }}^{e_1}(1)\mathrm{Re}({\delta }_x)+{\phi }_{y,v_1^{\ast }}^{e_1}(i)\mathrm{Im}({\delta }_x),$$

$${\psi }_{y,v_2^{\ast }}^{e_2}={\psi }_{y,v_2^{\ast }}^{e_2}(1)\mathrm{Re}({\delta }_x)+{\psi }_{y,v_2^{\ast }}^{e_2}(i)\mathrm{Im}({\delta }_x).$$

$${\phi }_{y,v_1^{\ast }}^{e_1}$$

$${\psi }_{y,v_2^{\ast }}^{e_2}$$

$${\phi }_{y,w_1^{\ast }}^{e_1^{\prime }}={\alpha }_1^{\prime }\mathrm{Re}({\delta }_{x^{\prime }})+{\beta }_1^{\prime }\mathrm{Im}({\delta }_{x^{\prime }}),{\psi }_{y,w_2^{\ast }}^{e_2^{\prime }}={\alpha }_2^{\prime }\mathrm{Re}({\delta }_{x^{\prime }})+{\beta }_2^{\prime }\mathrm{Im}({\delta }_{x^{\prime }}),$$

$${\phi }_{y,v_1^{\ast }}^{e_1}=\alpha \mathrm{Re}({\delta }_z)+\beta \mathrm{Im}({\delta }_z),\mathrm{and}{\psi }_{y,w_2^{\ast }}^{e_2^{\prime }}={\alpha }^{\prime }\mathrm{Re}({\delta }_z)+{\beta }^{\prime }\mathrm{Im}({\delta }_z),$$

Full text

1. Introduction

For a compact Hausdorff space $X$ and a real or complex Banach space $E$, let $C(X,E)$ be the Banach space of all continuous $E$-valued functions on $X$ under the supremum norm $\|f\|_{\infty}=\sup_{x\in X}\|f(x)\|$. In the scalar cases, that is the cases where $E=\mathbb{C}$, respectively $E=\mathbb{R}$, this space is denoted by $C(X)$, respectively $C_{\mathbb{R}}(X)$.

For compact Hausdorff spaces $X$ and $Y$ and real or complex Banach spaces $E$ and $F$, an additive map $T$ from a subspace $A(X,E)$ to $C(Y,F)$ is said to be separating or disjointness preserving if for any pair $f,g$ of elements of $A(X,E)$ with disjoint cozeros, their images $Tf$ and $Tg$ also have disjoint cozeros. Clearly in the scalar case, $T$ is separating if and only if it is zero product preserving, in the sense that $f\cdot g=0$ implies that $Tf\cdot Tg=0$. In particular, in this case, all (ring) homomorphisms between subalgebras of $C(X)$ (as well as $C_{\mathbb{R}}(X)$) are separating.

Weighted composition operators are standard examples of separating maps between spaces of functions. More generally, if $X,Y$ are compact Hausdorff spaces, $E,F$ are Banach spaces and $A(X,E)$ and $A(Y,F)$ are subspaces of $C(X,E)$ and $C(Y,F)$, respectively, then any additive map $T:A(X,E)\longrightarrow A(Y,F)$ of the form

[TABLE]

where $\varphi:Y\longrightarrow X$ is a continuous map and $\{J_{y}\}_{y\in Y}$ is a family of additive maps from $E$ to $F$ is a separating map.

Linear separating maps between various spaces of continuous functions (in either of scalar or vector valued case) have been studied for many years, see for instance [4, 7, 8, 14, 15]. In most cases, it is shown that for certain subspaces of continuous functions any continuous linear separating map is of the above form, which is called a generalized weighted composition operator. Continuous bilinear maps $\phi$ from $C^{1}[0,1]\times C^{1}[0,1]$ to a Banach space $E$ such that $f\cdot g=0$ implies $\phi(f,g)=0$ have been studied in [1]. A similar problem has been considered in [2] for the Banach algebra of little Lipschitz functions instead of $C^{1}([0,1])$. Clearly any separating map $T:\mathcal{A}{\longrightarrow}\mathcal{B}$ between spaces of scalar-valued functions $\mathcal{A}$ and $\mathcal{B}$, satisfies this implication for $\phi(f,g)=Tf\cdot Tg$.

Linear separating maps between vector-valued function spaces have been considered, for instance, in [3, 9, 11, 16]. Linear separating bijections between spaces of vector-valued continuous functions whose inverses are also separating , were studied in [9] and it was shown that such a map induces a homeomorphism between the underlying topological spaces. Similar results were given in [16] for such maps between vector-valued little Lipschitz function spaces. In [6], Dubarbie studied separating linear bijections on spaces of vector-valued absolutely continuous functions defined on compact subsets of the real line.

On the other hand, in [5], Botelho and Jamison characterized unital homomorphisms between vector-valued Banach algebra of Lipschitz functions with values in the Banach algebras $c$ and $\l_{\infty}$, as generalized weighted composition operators. Then in [17], Oi extended the result for $C(K)$-valued Lipschitz algebras, where $K$ is a compact Hausdorff space. More general cases of unital homomorphisms between certain Banach algebra-valued continuous functions have been studied recently in [10].

In this paper we first study a pair $T,S$ of additive maps between certain spaces of vector-valued continuous functions (on a compact Hausdorff space) which jointly preserve disjointness of cozero sets of functions. The results can be applied, for instance, whenever $T$ and $S$ are defined between vector-valued Lipschitz functions, absolutely continuous functions and ($n$-times) continuously differentiable functions on the unit interval. Then considering the case that the target spaces are Banach algebras, we give some results concerning continuous unital ring homomorphisms between some vector-valued algebras of functions. Hence, the results provide some generalizations of the results of [5] and [17] with a different approach.

2. Main results

We use the notation $\mathbb{K}$ for the field of real or complex numbers. Let $X$ be a compact Hausdorff space and $E$ be a Banach space over $\mathbb{K}$. For $f\in C(X,E)$, we denote the cozero set of $f$ by ${\rm coz}(f)$, that is ${\rm coz}(f)=\{x\in X:f(x)\neq 0\}$. By a constant function in $C(X,E)$ we mean a function on $X$ sending all points of $X$ to a fixed element $e\in E$. For $e\in E$ we denote its corresponding constant function by $c_{e}$. For a subspace $A(X,E)$ of $C(X,E)$ and $x\in X$, the map $\delta_{x}:A(X,E)\longrightarrow E$ is defined by $\delta_{x}(f)=f(x)$. For $f\in C(X)$ and $h\in C(X,E)$, $fh\in C(X,E)$ is defined by $fh(x)=f(x)h(x)$ for all $x\in X$.

Definition 2.1**.**

Let $X$ be a compact Hausdorff space and $E$ be a normed space over $\mathbb{K}$.

(i) A $\mathbb{K}$-subspace $A(X)$ of $C(X,\mathbb{K})$ is called boundedly normal if there exists a constant $M>0$ such that for any pair $K,K^{\prime}$ of disjoint closed subsets of $X$ there exists $f\in A(X)$ with $\|f\|_{\infty}\leq M$, $f=0$ on $K$ and $f=1$ on $K^{\prime}$.

(ii) We say that a subspace $A(X,E)$ of $C(X,E)$ is nice if there exists a boundedly normal subspace $A(X)$ of $C(X,{\mathbb{K}})$ containing constants such that $A(X)\cdot A(X,E)\subseteq A(X,E)$, where $A(X)\cdot A(X,E)=\{fh:f\in A(X),h\in A(X,E)\}$. **

Here are some examples of nice subspaces of $C(X,E)$.

Example 2.2**.**

Let $E$ be a $\mathbb{K}$-normed space.

(i) For a compact metric space $(X,d)$ and $\alpha\in(0,1]$, let ${\rm Lip}_{\alpha}(X,E)$ be the space of all functions $f:X\longrightarrow E$ such that

[TABLE]

Then ${\rm Lip}_{\alpha}(X,E)$ is a normed space with respect to the norm $\|f\|_{\rm Lip}=\|f\|_{\infty}+L(f)$, $f\in{\rm Lip}_{\alpha}(X,E)$, which is complete whenever $E$ is a Banach space. For $\alpha\in(0,1)$, the closed subspace ${\rm lip}_{\alpha}(X,E)$ of ${\rm Lip}_{\alpha}(X,E)$ consists of all functions $f\in{\rm Lip}_{\alpha}(X,E)$ satisfying $\lim_{d(x,y)\to 0}\frac{\|f(x)-f(y)\|}{d^{\alpha}(x,y)}=0$. It is easy to see that ${\rm Lip}_{\alpha}(X,\mathbb{K})$, for $\alpha\in(0,1]$, and ${\rm lip}_{\alpha}(X,\mathbb{K})$, for $\alpha\in(0,1)$, are boundedly normal subspaces of $C(X,\mathbb{K})$. Meanwhile, ${\rm Lip}_{\alpha}(X,E)$ and ${\rm lip}_{\alpha}(X,E)$ are both nice subspaces of $C(X,E)$. Moreover, they are Banach algebras if so is $E$. For the case that $\alpha=1$, we use the notation ${\rm Lip}(X,E)$ for ${\rm Lip}_{\alpha}(X,E)$.

(ii) For $n\in{\mathbb{N}}$, let $C^{n}([0,1],E)$ be the space of all continuously $n$-times differentiable functions $f:[0,1]\longrightarrow E$. Then $C^{n}([0,1],E)$ is a normed space with the norm $\|f\|=\sum_{i=0}^{n}\frac{\|f^{(i)}\|_{\infty}}{i!}$, $f\in C^{n}([0,1],E)$ which is complete if $E$ is a Banach space. Also $C^{n}([0,1],E)$ is a nice subspace of $C([0,1],E)$. The same is true for the subspace ${\rm Lip}^{n}([0,1],E)$ of $C^{n}([0,1],E)$ consisting of all Lipschitz functions $f:[0,1]\longrightarrow E$ such that for all $i=1,...,n$, $f^{(i)}\in{\rm Lip}([0,1],E)$. We should note that if $E$ is a Banach algebra, then $C^{n}([0,1]),E)$ is a Banach algebra with respect to the defined norm. Similarly, ${\rm Lip}^{n}([0,1],E)$ is a Banach algebra with respect to the following norm

[TABLE]

(iii) For a compact subset $X$ of the real line, let ${\rm AC}(X,E)$ be the space of all absolutely continuous $E$-valued functions on $X$. Then ${\rm AC}(X,E)$ is a nice subspace of $C(X,E)$. Moreover, $\|f\|=\|f\|_{\infty}+{\rm var}(f)$, $f\in{\rm AC}(X,E)$, defines a norm on ${\rm AC}(X,E)$, where ${\rm var}(f)$ is the total variation of $f\in{\rm AC}(X,E)$ and ${\rm AC}(X,E)$ is complete whenever $E$ is a Banach space. If $E$ is a Banach algebra, then so is $({\rm AC}(X,E),\|\cdot\|)$. **

Definition 2.3**.**

Let $X,Y$ be compact Hausdorff spaces, $E,F$ be $\mathbb{K}$-normed spaces and $A(X,E)$ be a subspace of $C(X,E)$. A pair $T,S:A(X,E)\longrightarrow C(Y,F)$ of additive maps is said to be jointly separating if ${\rm coz}(f)\cap{\rm coz}(g)=\emptyset$ implies that ${\rm coz}(Tf)\cap{\rm coz}(Sg)=\emptyset$ for all $f,g\in A(X,E)$.**

We should note that for any pair of functions $h,k\in C(Y,F)$ we have ${\rm coz}(h)\cap{\rm coz}(k)=\emptyset$ if and only if for each $v^{*}\in F^{*}$ and $y\in Y$, $\nu^{*}(h(y))\cdot\nu^{*}(k(y))=0$ if and only if for each $v^{*},w^{*}\in F^{*}$ and $y\in Y$, $\nu^{*}(h(y))\cdot w^{*}(k(y))=0$. Hence a pair of additive maps $T,S:A(X,E)\longrightarrow C(Y,F)$ is jointly separating if and only if $(v^{*}\circ Tf)\cdot(v^{*}\circ Sg)=0$ holds for all $v^{*}\in F^{*}$ whenever $f,g\in A(X,E)$ with ${\rm coz}(f)\cap{\rm coz}(g)=\emptyset$. This is also equivalent to say that $(v^{*}\circ Tf)\cdot(w^{*}\circ Sg)=0$ holds for all $v^{*},w^{*}\in F$ for such functions $f,g\in A(X,E)$.

Next lemma gives, in particular, the general form of continuous additive jointly separating functionals on boundedly normal subspaces of $C(X)$.

Lemma 2.4**.**

Let $X$ be a compact Hausdorff space, and ${\mathcal{A}}$ be a boundedly normal $\mathbb{K}$-subspace of $C(X,\mathbb{K})$ containing constants. Let $n\in\mathbb{N}$ and $\varphi_{1},...,\varphi_{n}:{\mathcal{A}}\longrightarrow\mathbb{K}$ be continuous nonzero additive functionals on ${\mathcal{A}}$ such that $\Pi_{j=1}^{n}f_{j}=0$ implies that $\Pi_{j=1}^{n}\,\varphi_{j}(f_{j})=0$ for all $f_{1},...f_{n}\in{\mathcal{A}}$. Then there exists $x\in X$ such that $\varphi_{j}(f)=\varphi_{j}(f(x))$ for all $f\in{\mathcal{A}}$ and $j=1,...,n$. In particular, in the case that $\mathbb{K}=\mathbb{C}$, $\varphi_{j}=\varphi_{j}(1)\,{\rm Re}(\delta_{x})+\varphi_{j}(i)\,{\rm Im}(\delta_{x})$ for all $j=1,...,n$ and in the case that $\mathbb{K}=\mathbb{R}$, $\varphi_{j}=\varphi_{j}(1)\,\delta_{x}$ for all $j=1,...,n$.

Proof.

We note that, by continuity assumption, $\varphi_{1},...,\varphi_{n}$ are real-linear. We first show that the proof can be reduced to the case $n=2$. Suppose that the lemma has been proven in this case and choose $u_{j}\in A$ with $\varphi_{j}(u_{j})\neq 0$, $j=1,...,n$. Then for any pair of functions $f_{1},f_{2}\in{\mathcal{A}}$ with $f_{1}f_{2}$=0 we have $f_{1}f_{2}u_{3}\cdots u_{n}=0$. Then by hypothesis, $\varphi_{1}(f_{1})\varphi_{2}(f_{2})=0$. Hence, by the case where $n=2$, there exists $x_{1,2}\in X$ such that $\varphi_{1}(f)=\varphi_{1}(f(x_{1,2}))$ and $\varphi_{2}(f)=\varphi_{2}(f(x_{1,2}))$, for all $f\in{\mathcal{A}}$. A similar argument shows that there exists a point $x_{2,3}\in X$ such that $\varphi_{2}(f)=\varphi_{2}(f(x_{2,3}))$ and $\varphi_{3}(f)=\varphi_{3}(f(x_{2,3}))$ for all $f\in{\mathcal{A}}$. Thus

[TABLE]

We now show that $x_{1,2}=x_{2,3}$. Since $\varphi_{2}$ is nonzero, the above equalities show that there exists a scalar $c\in{\mathbb{K}}$ such that $\varphi_{2}(c)\neq 0$. If $x_{1,2}\neq x_{2,3}$, then we can find a function $f\in{\mathcal{A}}$ such that $f(x_{1,2})=1$ and $f(x_{2,3})=0$. Then $cf\in{\mathcal{A}}$ and, using the above equalities, we have

[TABLE]

a contradiction. Hence $x_{1,2}=x_{2,3}$. This argument can be applied for all pairs $\varphi_{j},\varphi_{k}$, where $j,k\in\{1,...,n\}$ are distinct, to conclude that there exists a point $x\in X$ such that for all $j=1,...,n$

[TABLE]

By the above argument we can assume without loss of generality that $n=2$.

First assume that $\mathbb{K}=\mathbb{C}$. Then ${\mathcal{A}}$ is a complex subspace of $C(X)$ and $\varphi_{1},\varphi_{2}:{\mathcal{A}}\longrightarrow{\mathbb{C}}$ are nonzero real-linear jointly separating functionals. Hence for $f,g\in{\mathcal{A}}$ with disjoint cozero sets, at least one of the equalities $\varphi_{1}(f)=0$ and $\varphi_{2}(f)=0$ holds. This easily implies that for distinct $j,k\in\{1,2\}$, the pair ${\rm Re}(\varphi_{j})$, ${\rm Re}(\varphi_{k}):{\mathcal{A}}\longrightarrow{\mathbb{R}}$ and also the pair ${\rm Re}(\varphi_{j})$, ${\rm Im}(\varphi_{k}):{\mathcal{A}}\longrightarrow{\mathbb{R}}$ of real-linear functionals on the complex space ${\mathcal{A}}$ are jointly separating. Replacing $\varphi_{1}$ and $\varphi_{2}$ by $i\varphi_{1}$ and $i\varphi_{2}$ if it is necessary, we may assume that ${\rm Re}(\varphi_{1})$ and ${\rm Re}(\varphi_{2})$ are both nonzero. Now we consider the complex-linear functionals $\Phi_{1},\Phi_{2}:{\mathcal{A}}\longrightarrow{\mathbb{C}}$ defined by

[TABLE]

and

[TABLE]

If $f,g\in{\mathcal{A}}$ such that $f\cdot g=0$, then $f\cdot ig=if\cdot g=0=if\cdot ig$ and since ${\rm Re}(\varphi_{1})$ and ${\rm Re}(\varphi_{2})$ are nonzero jointly separating we deduce that $\Phi_{1},\Phi_{2}$ are also nonzero jointly separating. We note that $\Phi_{1},\Phi_{2}$, as continuous nonzero complex linear functionals on ${\mathcal{A}}$, can be extended to continuous complex-linear functionals on $C(X)$. Hence there exist (nonzero) complex regular Borel measures $\mu$ and $\nu$ on $X$ satisfying

[TABLE]

for all $f\in{\mathcal{A}}$. Using the fact that ${\mathcal{A}}$ is boundedly normal and $\Phi_{1},\Phi_{2}$ are jointly separating, we can easily deduce that for any pair $K_{1},K_{2}$ of disjoint closed subsets of $X$ we have

[TABLE]

We note that ${\rm supp}(\mu)={\rm supp}(\nu)=\{x\}$ for some $x\in X$. Indeed, if there are distinct points $x_{1}\in{\rm supp}(\mu)$ and $x_{2}\in{\rm supp}(\nu)$, then choosing neighborhoods $U_{1}$ and $U_{2}$ of $x_{1}$ and $x_{2}$, respectively with disjoint closures, it follows from the above argument that $|\mu|(\overline{U_{1}})\,|\nu|(\overline{U_{2}})=0$, which is impossible. This clearly implies that ${\rm supp}(\mu)={\rm supp}(\nu)=\{x\}$ for some $x\in X$ and consequently we have $\Phi_{1}=\alpha_{1}\delta_{x}$ and $\Phi_{2}=\alpha_{2}\delta_{x}$ for some $\alpha_{1},\alpha_{2}\in\mathbb{C}\backslash\{0\}$. Thus

[TABLE]

and

[TABLE]

We claim that there are $\beta_{1},\beta_{2}\in{\mathbb{C}}$ such that

[TABLE]

We prove the first equality, since the second one is proven in a similar manner. If ${\rm Im}(\varphi_{1})=0$ then clearly the desired equality holds for $\beta_{1}=0$. Hence assume that ${\rm Im}(\varphi_{1})\neq 0$. Then ${\rm Re}(i\varphi_{1})\neq 0$ and since ${\rm Re}(\varphi_{2})\neq 0$, using the above argument for jointly separating functionals $i\varphi_{1}$ and $\varphi_{2}$ we conclude that there exist $x^{\prime}\in X$ and scalars $\lambda_{1},\lambda_{2}\in{\mathbb{C}}\backslash\{0\}$ such that

[TABLE]

and

[TABLE]

Hence ${\rm Re}(\alpha_{2}\delta_{x})={\rm Re}(\varphi_{2})={\rm Re}(\lambda_{2}\delta_{x^{\prime}})$. This implies that $x=x^{\prime}$. Indeed, if $x\neq x^{\prime}$ then we can choose $f\in{\mathcal{A}}$ satisfying $f(x)=0$ and $f(x^{\prime})=\overline{\lambda_{2}}$. Then

[TABLE]

which is a contradiction. Hence $x=x^{\prime}$ and consequently

[TABLE]

as desired. Thus (2.1) hold for some $\beta_{1},\beta_{2}\in{\mathbb{C}}$. This implies that $\varphi_{1}(f)={\rm Re}(\alpha_{1}f(x))+i{\rm Re}(\beta_{1}f(x))$ for all $f\in{\mathcal{A}}$. In fact, an easy calculation shows that

[TABLE]

In particular, $\varphi_{1}(f)=\varphi_{1}(f(x))$ holds for all $f\in{\mathcal{A}}$. Similarly $\varphi_{2}=\varphi_{2}(1)\,{\rm Re}(\delta_{x})+\varphi_{2}(i)\,{\rm Im}(\delta_{x})$, as desired.

Assume now that $\mathbb{K}=\mathbb{R}$. Then ${\mathcal{A}}$ is a real subspace of $C_{\mathbb{R}}(X)$ and $\varphi_{1},\varphi_{2}:{\mathcal{A}}\longrightarrow{\mathbb{R}}$ are jointly separating real-linear functionals on ${\mathcal{A}}$. Extending $\varphi_{1},\varphi_{2}$ to continuous real-linear functionals on $C_{\mathbb{R}}(X)$, the same argument can be applied to show that $\varphi_{j}=\varphi_{j}(1)\delta_{x}$, $j=1,2$, for some point $x\in X$. ∎

Let $X,Y$ be compact Hausdorff spaces and $E,F$ be normed spaces over $\mathbb{K}$. For a pair $T,S:A(X,E)\longrightarrow C(Y,F)$ of additive maps on a subspace $A(X,E)$ of $C(X,E)$ containing constants, we put

[TABLE]

and

[TABLE]

It is obvious that if there exists at least one constant function whose images under $T$ and $S$ are constant, then $Y_{0}=Y$. We denote the space of all bounded real-linear operators from $E$ to $F$ by $B_{\mathbb{R}}(E,F)$.

Theorem 2.5**.**

Let $X,Y$ be compact Hausdorff spaces, $E,F$ be normed spaces over $\mathbb{K}$ and $A(X,E)$ be a nice subspace of $C(X,E)$ containing constants. Let $T,S:A(X,E)\longrightarrow C(Y,F)$ be additive jointly separating maps. Then there exist a continuous map $\Phi:Y_{c}\longrightarrow X$ and two families $\{\Lambda_{y}\}_{y\in Y_{c}}$ and $\{\Lambda^{\prime}_{y}\}_{y\in Y_{c}}$ of real-linear operators from $E$ to $F$ such that

[TABLE]

and

[TABLE]

In particular, if $T,S$ are continuous, then $Y_{c}=Y_{0}$ and, moreover, for each $y\in Y_{c}$, $\Lambda_{y}\in B_{\mathbb{R}}(E,F)$ and, furthermore, $y\mapsto\Lambda_{y}$ is a continuous map from $Y_{c}$ to $B_{\mathbb{R}}(E,F)$ with respect to the strong operator topology on $B_{\mathbb{R}}(E,F)$.

We prove the theorem through the subsequent lemmas.

In the sequel we assume that $X,Y$ and $E,F$ and also $A(X,E)\subseteq C(X,E)$ and $T,S:A(X,E)\longrightarrow C(Y,F)$ are as in the theorem.

By hypotheses there exists a boundedly normal subspace $A(X)$ of $C(X,\mathbb{K})$ such that $A(X)\cdot A(X,E)\subseteq A(X,E)$. For each $(y,v^{*})\in Y\times F^{*}$ , we put

[TABLE]

Then clearly for each $y\in Y_{c}$ and each $v^{*},w^{*}\in F^{*}$, the maps $\varphi_{y,v^{*}}$ and $\psi_{y,w^{*}}$ are continuous real-linear functionals on $A(X,E)$ which are jointly separating.

Also for each $(y,v^{*},e)\in Y\times F^{*}\times E$ we define the jointly separating additive functionals $\varphi_{y,v^{*}}^{e},\psi_{y,v^{*}}^{e}:A(X)\longrightarrow\mathbb{C}$ by

[TABLE]

Let $y\in Y_{c}$. Then $y\in Y_{0}$ and so there are $e_{j}\in E$ and $v_{j}^{*}\in F^{*}$, $j=1,2$, such that $\varphi_{y,v_{1}^{*}}^{e_{1}}\neq 0$ and $\psi_{y,v_{2}^{*}}^{e_{2}}\neq 0$. Motivated by this, for each $y\in Y_{c}$ we consider the following set

[TABLE]

In the next lemmas we assume that $\mathbb{K}=\mathbb{C}$. Similar results hold for the case $\mathbb{K}=\mathbb{R}$.

Lemma 2.6**.**

Let $y\in Y_{c}$. Then there exists a unique point $x\in X$ (depending only on $y$) such that for all $(v_{1}^{*},v_{2}^{*},e_{1},e_{2})\in H_{y}$,

[TABLE]

and

[TABLE]

Proof.

For any $(v_{1}^{*},v_{2}^{*},e_{1},e_{2})\in H_{y}$, $\varphi_{y,v_{1}^{*}}^{e_{1}}$ and $\psi_{y,v_{2}^{*}}^{e_{2}}$ are nonzero jointly separating continuous real-linear functionals on $A(X)$. Thus by Lemma 2.4 there exists a point $x\in X$ (depending on $v_{1}^{*},v_{2}^{*},y,e_{1},e_{2}$) such that

[TABLE]

where $\alpha_{1}=\varphi_{y,v_{1}^{*}}^{e_{1}}(1)$, $\beta_{1}=\varphi_{y,v_{1}^{*}}^{e_{1}}(i)$ and similarly $\alpha_{2}=\psi_{y,v_{2}^{*}}^{e_{2}}(1)$ and $\beta_{2}=\psi_{y,v_{2}^{*}}^{e_{2}}(i)$.

We now show that the point $x$ depends only on $y$. For this, let $(w_{1}^{*},w_{2}^{*},e^{\prime}_{1},e^{\prime}_{2})$ be another element of $H_{y}$. Then by the above argument there exists $x^{\prime}\in X$ such that

[TABLE]

where $\alpha^{\prime}_{1}=\varphi_{y,w_{1}^{*}}^{e^{\prime}_{1}}(1)$, $\beta^{\prime}_{1}=\varphi_{y,w_{1}^{*}}^{e^{\prime}_{1}}(i)$ and also $\alpha^{\prime}_{2}=\psi_{y,w_{2}^{*}}^{e^{\prime}_{2}}(1)$ and $\beta^{\prime}_{2}=\psi_{y,w_{2}^{*}}^{e^{\prime}_{2}}(i)$. Since $\varphi_{y,v_{1}^{*}}^{e_{1}}$ and $\psi_{y,w_{2}^{*}}^{e^{\prime}_{2}}$ are also nonzero continuous real-linear jointly separating functionals on ${\mathcal{A}}(X)$ we also have

[TABLE]

for some $z\in X$ and scalars $\alpha,\alpha^{\prime},\beta,\beta^{\prime}\in\mathbb{C}$. Hence,

[TABLE]

and, as before, we can conclude that $x=z$. Similarly $x^{\prime}=z$, that is $x=x^{\prime}$.

It is easy to see that the point $x\in X$ with the desired property is unique. ∎

By the above lemma, the point $x\in X$ is the same for all quadruples $(v_{1}^{*},v_{2}^{*},e_{1},e_{2})\in H_{y}$. Hence we can define a map $\Phi:Y_{c}\longrightarrow X$ which associates to each point $y\in Y_{c}$ the unique point $x\in X$ such that (2.2) and (2.3) hold for all $(v_{1}^{*},v_{2}^{*},e_{1},e_{2})\in H_{y}$.

Lemma 2.7**.**

Let $y\in Y_{c}$. Then for all $e\in E$ and $v^{*}\in F^{*}$, we have

[TABLE]

and

[TABLE]

Proof.

We prove the first equality, since the other one is proven similarly.

Suppose that $y\in Y_{c}$ and let $e\in E$ and $v^{*}\in F^{*}$ be arbitrary. If $\varphi_{y,v^{*}}^{e}=0$, then $\varphi_{y,v^{*}}^{e}(1)=0=\varphi_{y,v^{*}}^{e}(i)$ and so the equality is obvious. Hence we assume that $\varphi_{y,v^{*}}^{e}\neq 0$. Since $y\in Y_{c}\subseteq Y_{0}$ it follows easily that there exist $w^{*}\in F^{*}$ and $e^{\prime}\in E$ such that $\psi_{y,w^{*}}^{e^{\prime}}\neq 0$. Therefore, $(v^{*},w^{*},e,e^{\prime})\in H_{y}$ and consequently, by lemma above, we have again the desired equality. ∎

Now for each $y\in Y_{c}$ we define real-linear operators $\Lambda_{y},\Lambda^{\prime}_{y}:E\longrightarrow F$, by

[TABLE]

Let $y\in Y_{c}$, and $v^{*}\in E^{*}$. Put $x=\Phi(y)$. Using the above lemma and the real-linearity of $\varphi_{y,v^{*}}$, for each $e\in E$ and $f\in A(X)$ we have

[TABLE]

Hence for each $y\in Y_{c}$, $e\in E$ and each $v^{*}\in F^{*}$ we have $\varphi_{y,v^{*}}(fe)=\varphi_{y,v^{*}}^{e}(f)=\varphi_{y,v^{*}}^{e}(f(\Phi(y)))$, that is

[TABLE]

Similarly

[TABLE]

Lemma 2.8**.**

Let $y\in Y_{c}$. Then for each $h\in A(X,E)$ with $h(\Phi(y))=0$ there exists a sequence $\{h_{n}\}_{n=1}^{\infty}$ in $A(X,E)$ such that each $h_{n}$ vanishes on a neighborhood of $\Phi(y)$ and $\|h_{n}-h\|_{\infty}\to 0$ as $n\to\infty$.

Proof.

Assume that $x=\Phi(y)$. For each $n\in{\mathbb{N}}$ we put

[TABLE]

Let $V_{n}$ be a neighborhood of $x$ with $\overline{V_{n}}\subseteq U_{n}$. Then $x\in U_{n}$ for all $n\in\mathbb{N}$ and, using bounded normality of $A(X)$, there exist $M>0$ and a sequence $\{g_{n}\}$ in $A(X)$ such that $\|g_{n}\|_{\infty}<M$, $g_{n}=1$ on $X\backslash U_{n}$ and $g_{n}=0$ on $V_{n}$, for all $n\in{\mathbb{N}}$. Put $h_{n}=g_{n}h$, $n\in{\mathbb{N}}$. Then $h_{n}\in A(X,E)$ and, furthermore, $h_{n}=0$ on $V_{n}$ and $\|h-h_{n}\|_{\infty}<\frac{1}{n}(1+M)$, that is $h_{n}\to h$ in $A(X,E)$. ∎

Lemma 2.9**.**

For each $y\in Y_{c}$ and $v^{*}\in F^{*}$

[TABLE]

and

[TABLE]

hold for all $h\in A(X,E)$.

Proof.

Let $y\in Y_{c}$ and $v^{*}\in F^{*}$. Put $x=\Phi(y)$. To prove the first equality, it suffices to show that for each $h\in A(X,E)$ with $h(x)=0$ we have $\varphi_{y,v^{*}}(h)=0$. Indeed, if we prove this implication, then for each $h\in A(X,E)$, the element $e=h(x)$ in $E$ satisfies $(h-c_{e})(x)=0$ and consequently $\varphi_{y,v^{*}}(h-c_{e})=0$ which implies, by (2.4), that

[TABLE]

as desired.

First consider the case where $h\in A(X,E)$ vanishes on a neighborhood $U$ of $x$. Using the normality of $A(X)$, there exists a function $g\in A(X)$ with $g(x)=1$ and $g=0$ on $X\backslash U$. We note that since $y\in Y_{c}\subseteq Y_{0}$ there exists $e\in E$ and $w^{*}\in F^{*}$ such that $w^{*}(S(c_{e})(y))\neq 0$, that is $w^{*}(\Lambda^{\prime}_{y}(e))=\psi_{y,w^{*}}^{e}(1)\neq 0$. Hence for $G=ge\in A(X,E)$, using (2.5), we have

[TABLE]

Since ${\rm coz}(h)\cap{\rm coz}(G)=\emptyset$ we have $\varphi_{y,v^{*}}(h)\cdot\psi_{y,v^{*}}(G)=0$ and being $\psi_{y,v^{*}}(G)\neq 0$ we get $\varphi_{y,v^{*}}(h)=0$, as desired.

Now, the general case that $h\in A(X,E)$ and $h(x)=0$ follows immediately from the lemma above and the continuity of $\varphi_{y,v^{*}}$. ∎

Let $y\in Y_{c}$ and $v^{*}\in F^{*}$. By the above lemmas for each $h\in A(X,E)$ we have

[TABLE]

and

[TABLE]

Since these equalities holds for all $v^{*}\in F^{*}$ we get

[TABLE]

for all $y\in Y_{c}$ and $h\in A(X,E)$. Hence next lemma completes the proof of the first part of the theorem.

Lemma 2.10**.**

The map $\Phi:Y_{c}\longrightarrow X$ is continuous.

Proof.

The proof is straightforward. ∎

The second part of the theorem is easily verified.

Corollary 2.11**.**

Let $X$ and $Y$ be compact Hausdorff spaces, $E$ be a complex commutative unital normed algebra and $A(X,E)$ be a nice subalgebra of $C(X,E)$ containing constants. If $T:A(X,E)\longrightarrow C(Y)$ is a unital ring homomorphism, then there exists a subset $Y_{c}$ of $Y$, a continuous map $\varphi:Y_{c}\longrightarrow X$ and a family $(\Lambda_{y})_{y\in Y_{c}}$ of real-linear ring homomorphisms such that

[TABLE]

Moreover, if $T$ is continuous, then $Y_{c}=Y$, each $\Lambda_{y}$ is continuous and $y\to\Lambda_{y}$ is continuous with respect to the strong operator topology on $B_{\mathbb{R}}(E,{\mathbb{C}})$.

Proof.

The result follows immediately, since $T$ is easily verified to be a separating map. ∎

For a unital Banach algebra $E$, let $E^{-1}$ denote the open subset of invertible elements of $E$ and for $x\in E$ let $\rho(x)$ denote its spectral radius.

Definition 2.12**.**

Let $X$ be a compact Hausdorff space and $E$ be a unital Banach algebra. We say that a subalgebra ${\mathcal{A}}$ of $C(X,E)$ is inverse closed if each element $f\in{\mathcal{A}}$ with $f(x)\in E^{-1}$ for all $x\in X$, is invertible in ${\mathcal{A}}$, that is the function $f^{-1}\in C(X,E)$ defined by $f^{-1}(x)=f(x)^{-1}$, $x\in X$, is an element of ${\mathcal{A}}$.**

Clearly, if ${\mathcal{A}}$ is an inverse closed subalgebra of $C(X,E)$ which is a Banach algebra with respect to some norm, then for each $f\in{\mathcal{A}}$,

[TABLE]

Assume that $X$ is a compact Hausdorff space and $E$ is a (complex) unital Banach algebra. For each invertible element $f\in C(X,E)$, and distinct points $s,t\in X$ we have f^{-1}(s)-f^{-1}(t)=f^{-1}(s)\big{(}f(t)-f(s)\big{)}f^{-1}(t). Using this equality we can easily see that all Banach algebras introduced in Example 2.2 are inverse closed.

By [12], for a compact metric space $X$ and a commutative unital (complex) Banach algebra $E$, the maximal ideal space $M_{{\rm Lip}(X,E)}$ of the Banach algebra ${\rm Lip}(X,E)$ is homeomorphic to $X\times M_{E}:=\{\delta_{x}\times\psi:x\in X,\psi\in M_{E}\}$, where for each $x\in X$ and $\psi\in M_{E}$, $(\delta_{x}\times\psi)(f)=\psi(f(x))$ for all $f\in{\rm Lip}(X,E)$. Next corollary, in particular, gives a similar result for $\|\cdot\|$-continuous complex ring homomorphisms on ${\rm Lip}(X,E)$.

Corollary 2.13**.**

Let $X$ be a compact Hausdorff space, $E$ be a complex commutative unital Banach algebra and ${\mathcal{A}}$ be an inverse closed subalgebra of $C(X,E)$ which is a Banach algebra with respect to a norm $\|\cdot\|$ with $\|\cdot\|\geq\|\cdot\|_{\infty}$. Assume that $\varphi:{\mathcal{A}}\longrightarrow{\mathbb{C}}$ is a nonzero ring homomorphism. Then

(i)* $\varphi$ is $\|\cdot\|_{\infty}$-continuous if and only if it is continuous with respect to $\|\cdot\|$.*

(ii)* If $\varphi$ is $\|\cdot\|$-continuous, then there exists a point $x\in X$ and a continuous ring homomorphism $\Lambda:E\longrightarrow\mathbb{C}$ such that*

[TABLE]

Proof.

(i) For the if part, assume that $\varphi$ is continuous with respect to the norm $\|\cdot\|$. Then there exists $K>0$ such that $|\varphi(f)|\leq K\|f\|$ for all $f\in{\mathcal{A}}$. Hence for all $n\in{\mathbb{N}}$

[TABLE]

which implies that $|\varphi(f)|\leq K^{\frac{1}{n}}\|f^{n}\|^{\frac{1}{n}}$ for all $f\in{\mathcal{A}}$. Tending $n$ to infinity, we get $|\varphi(f)|\leq\rho(f)\leq\|f\|_{\infty}$, that is $\varphi$ is $\|\cdot\|_{\infty}$-continuous.

The only if part is trivial.

(ii) We note that the ring homomorphism $\varphi$ is unital, since it is nonzero. Hence, the result is immediate from Corollary 2.11. ∎

Corollary 2.14**.**

Let $X$ be a compact metric space, $K$ be a compact Hausdorff space and $\varphi:{\rm Lip}(X,C(K)))\longrightarrow\mathbb{C}$ be a nonzero continuous ring homomorphism. Then there exist points $x\in X$ and $t\in K$ such that either

[TABLE]

or

[TABLE]

Proof.

It is obvious that $\varphi(1)=1$. Hence, using Corollary 2.13, we conclude that there exist a point $x\in X$ and a continuous ring homomorphism $\Lambda:C(K)\longrightarrow\mathbb{C}$ such that

[TABLE]

Using Corollary 2.13 once again, we can find a point $t\in K$ and a continuous ring homomorphism $\tau:\mathbb{C}\longrightarrow\mathbb{C}$ such that

[TABLE]

Since $z\mapsto z$ and $z\mapsto\overline{z}$ are the only continuous ring homomorphisms on $\mathbb{C}$ we get the desired description for $\varphi$. ∎

Remark 2.15**.**

Let $E$ be a commutative unital Banach algebra. As we noted before all Banach algebras ${\rm lip}_{\alpha}(X,E)$, for a compact metric space $X$ and $\alpha\in(0,1)$, $C^{n}([0,1],E)$, ${\rm Lip}^{n}([0,1],E)$ and $AC(X,E)$, for a compact subset $X$ of the real line, are inverse closed. Since their norms also satisfies $\|\cdot\|\geq\|\cdot\|_{\infty}$, the above corollary holds true whenever ${\rm Lip}(X,C(K))$ is replaced by any of these Banach algebras for $E=C(K)$. **

In [5], Botelho and Jamison gave the description of unital homomorphisms $T:{\rm Lip}(X,E)\longrightarrow{\rm Lip}(Y,E)$ for compact metric spaces $X$ and $Y$ where $Y$ is connected and the target (Banach) algebra $E$ is either $c_{0}=C_{0}(\mathbb{N})$ or $\l_{\infty}=C(\beta\mathbb{N})$. We note that semisimplicity of $E$ implies that ${\rm Lip}(X,E)$ is semisimple, as well. Hence such a homomorphism $T$ is continuous with respect to the Lipschitz norms on both sides. In the next theorem (see also Remark 2.18) we give a generalization of [5] for continuous ring homomorphisms between some subalgebras of $C(X,E)$, where $X$ is compact and $E$ is a certain subalgebra of $C(K)$ for an arbitrary compact Hausorff space $K$.

For a (complex) Banach algebra $A$, let $\mathcal{R}(A)$ denote the set of all nonzero continuous ring homomorphisms $\varphi:A\longrightarrow{\mathbb{C}}$.

Theorem 2.16**.**

Let $X$ and $Y$ be compact metric spaces such that $Y$ is connected. Let $K$ and $K^{\prime}$ be compact Hausdorff spaces and $T:{\rm Lip}(X,C(K))\longrightarrow{\rm Lip}(Y,C(K^{\prime}))$ be a unital ring homomorphism. If $T$ is continuous with respect to the Lipschitz norms on both sides, then there are a continuous function $\tau:K^{\prime}\longrightarrow K$, a family $\{\psi_{s}\}_{s\in K^{\prime}}$ of continuous functions from $Y$ to $X$ and a clopen subset $\mathcal{C}$ of $K^{\prime}$ such that

[TABLE]

for all $f\in{\rm Lip}(X,C(K))$ and $y\in Y$.

Proof.

Since $T$ is a continuous unital ring homomorphism, it follows that for each $\Psi\in\mathcal{R}({\rm Lip}(Y,C(K^{\prime})))$, the map $\Psi\circ T:{\rm Lip}(X,C(K))\longrightarrow\mathbb{C}$ is an element of $\mathcal{R}({\rm Lip}(X,C(K)))$. Hence, by Corollary 2.14 there exist points $x\in X$ and $t\in K$ such that either

[TABLE]

or

[TABLE]

Now for each $y\in Y$ and $s\in K^{\prime}$, the map $\Psi:{\rm Lip}(Y,C(K^{\prime}))\longrightarrow\mathbb{C}$ defined by $\Psi(g)=g(y)(s)$, $g\in{\rm Lip}(Y,C(K^{\prime}))$, is an element of $\mathcal{R}({\rm Lip}(Y,C(K^{\prime})))$, and so there are points $x\in X$ and $t\in K$ satisfying one of the above equalities. Thus we can define a map $\sigma:Y\times K^{\prime}\longrightarrow X\times K$ by $\sigma(y,s)=(x,t)$, such that either

[TABLE]

or

[TABLE]

Clearly $\sigma$ is continuous with respect to the product topology. Now we claim that the point $t\in K$ in the above equalities is independent of the choice of $y\in Y$, that is, it depends only on the point $s\in K^{\prime}$.

First we note that $-1=T(-1)=T(i^{2})=(Ti)^{2}$. Hence, for each $y\in Y$ and $s\in K^{\prime}$, $Ti(y)(s)\in\{i,-i\}$. For each $s\in K^{\prime}$ we put

[TABLE]

Obviously, $A_{s}$ is a clopen subset of $Y$ and since $Y$ is connected we conclude that for each $s\in K^{\prime}$ we have either $A_{s}=Y$ or $A_{s}=\emptyset$. Thus for each $s\in K^{\prime}$ we have either

Case I. $Ti(y)(s)=i$ for all $y\in Y$, or

Case II. $Ti(y)(s)=-i$ for all $y\in Y$.

Let $\pi_{2}:X\times K\longrightarrow K$ be defined by $\pi_{2}(z,u)=u$ for $(z,u)\in X\times K$. To prove the claim, fix an element $s\in K^{\prime}$. Assume, furthermore, that Case I holds for $s$. Then for each $y\in Y$, by the above description of $T$, we have

[TABLE]

where $(x,t)=\sigma(y,s)$. Choose an arbitrary $y\in Y$ and put $t=\pi_{2}(\sigma(y,s))$. Since $Y$ is connected, it suffices to show that the open subset $U_{s}=\{z\in Y:\pi_{2}(\sigma(z,s))\neq t\}$ of $Y$ is closed. For this, assume that $\{y_{n}\}$ is a sequence in $U_{s}$ converging to a point $y_{0}\in Y\backslash U_{s}$. Put $(x_{0},t_{0})=\sigma(y_{0},s)$ and for each $n\in{\mathbb{N}}$, put $(x_{n},t_{n})=\sigma(y_{n},s)$. Then by hypothesis we have $t_{0}=t$ and $t_{n}\neq t$ for all $n\in{\mathbb{N}}$. For each $\lambda\in C(K)$, let $c_{\lambda}:X\longrightarrow C(K)$ be the constant function defined by $c_{\lambda}(x)=\lambda$, $x\in X$. Then we have

[TABLE]

and

[TABLE]

for all $\lambda\in C(K)$ and $n\in{\mathbb{N}}$. Therefore,

[TABLE]

Since $T$ is continuous we have

[TABLE]

for all $\lambda\in C(K)$. This shows that

[TABLE]

where $\delta_{u}:C(K)\longrightarrow\mathbb{C}$ is the evaluation functional at a point $u\in K$. We note that for any pair of distinct points $u,v\in K$, we have $\|\delta_{u}-\delta_{v}\|=2$. Then, by the above inequality we have $d(y_{n},y_{0})\geq\frac{2}{\|T\|}$ for all $n\in{\mathbb{N}}$, which is impossible. This argument shows that $U_{s}$ is a clopen subset of $Y$. Being $Y$ connected we get $U_{s}=\emptyset$, which proves our claim in Case I.

Assume now that Case II holds for the given $s\in K^{\prime}$. Then for each $y\in Y$ we have

[TABLE]

where $(x,t)=\sigma(y,s)$. Using the same argument, as in the previous case, we can show that the open subset $U_{s}=\{z\in Y:\pi_{2}(\sigma(z,s))\neq t\}$ of $Y$ is closed. Hence $U_{s}=\emptyset$, which proves our claim in Case II.

By the above argument we can define a function $\tau:K^{\prime}\longrightarrow K$ which associates to each point $s\in k^{\prime}$, the unique point $t\in K$ satisfying either

[TABLE]

or

[TABLE]

where, $\psi_{s}(y)=\pi_{1}(\sigma(y,s))$. It is obvious that $\psi_{s}:Y\longrightarrow X$ is continuous for each $s\in K^{\prime}$.

To conclude the theorem, we put

[TABLE]

Then, by the Cases I,II we have

[TABLE]

that is $\mathcal{C}$ is a clopen subset of $K^{\prime}$. Moreover, the above description of $T$ shows that if $s\in\mathcal{C}$, then $Tf(y)(s)=f(\psi_{s}(y))(\tau(s))$ for all $f\in{\rm Lip}(X,C(K))$ and $y\in Y$, and if $s\notin\mathcal{C}$, then $Tf(y)(s)=\overline{f(\psi_{s}(y))(\tau(s))}$ for all $f\in{\rm Lip}(X,C(K))$ and $y\in Y$, as desired. ∎

Theorem 2.17**.**

Let $K$ and $K^{\prime}$ be compact Hausdorff spaces, and $T:{\mathcal{A}}\longrightarrow{\mathcal{B}}$ be a continuous unital ring homomorphism where ${\mathcal{A}}$ is either $C^{n}([0,1],C(K))$ or ${\rm Lip}^{n}([0,1],C(K))$ and, similarly ${\mathcal{B}}$ is either $C^{n}([0,1],C(K^{\prime}))$ or ${\rm Lip}^{n}([0,1],C(K^{\prime}))$ for some $n\in\mathbb{N}$. Then there are a continuous function $\tau:K^{\prime}\longrightarrow K$, a family $\{\psi_{s}\}_{s\in K^{\prime}}$ of continuous functions on $[0,1]$ and a clopen subset $\mathcal{C}$ of $K^{\prime}$ such that

[TABLE]

for all $f\in A$ and $y\in[0,1]$.

Proof.

The proof is basically the same proof as in Theorem 2.16. It suffices to note that for the inequality (2.6) we may use the vector-valued mean value theorem to conclude that

[TABLE]

Hence we have again

[TABLE]

for all $\lambda\in C(K)$. ∎

Remark 2.18**.**

The proof of the above theorems work if $C(K)$ and $C(K^{\prime})$ are replaced by natural uniform algebras $A$ and $B$ on $K$ and $K^{\prime}$, respectively, such that for each $t\in K$, $\delta_{t}$ is an isolated point of $K$ with respect to the operator norm, that is for each $t\in K$ there exists $c_{t}>0$ such that $\|\delta_{t}-\delta_{s}\|>c_{t}$ holds for all points $s\in K$ distinct from $t$. Here $\|\delta_{t}-\delta_{s}\|$ denotes the operator norm of $\delta_{t}-\delta_{s}$ on $A$. In particular, if $A$ is a natural uniform algebra on a compact Hausdorff space $K$ and $M>0$ such that for all distinct points $s,t\in K$ there exists a function $f\in A$ with $f(s)=0$, $f(t)=1$ and $\|f\|_{\infty}\leq M$, then $A$ has the desired property. **