Perturbations of norm-additive maps between continuous function spaces

Abstract

Let $X, Y$ be two locally compact Hausdorff spaces and $T:C_0(X)\rightarrow C_0(Y)$ be a standard surjective ɛ-norm-additive map, i.e.

\begin{equation*}\big|\|T(f)+T(g)\|-\|f+g\|\big|\leq \varepsilon,\;{\rm for\;all}\; f, g\in C_0(X).\end{equation*}

Then there exist a homeomorphism $\varphi:Y\rightarrow X$ and a continuous function $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ such that

\begin{equation*}|T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X).\end{equation*}

The estimate ‘$\frac{3}{2}\varepsilon$’ is optimal. And this result can be regarded as a new nonlinear extension of the Banach–Stone theorem.

1. Introduction

Let X be a locally compact Hausdorff space. The space $C_0(X)$ will stand for the Banach space of all continuous real-valued functions which vanish at infinity on X (i.e. $\{x\in X: |f(x)|\geq\varepsilon\}$ is compact in X for every $f\in C_0(X)$ and every ɛ > 0) equipped with the supremum norm. The following result is well-known as the Banach–Stone theorem (see [2, 3, 21]).

Theorem 1.1. Let $X, Y$ be two locally compact Hausdorff spaces and $T:C_0(X)\rightarrow C_0(Y)$ be a linear surjective isometry. Then there exists a homeomorphism $\varphi:Y\rightarrow X$ and a continuous function $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ such that

\begin{equation*} T(f)(y)=\lambda(y)f(\varphi(y)),\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation*}

The Banach–Stone theorem describes a deep fact that the linear metric structure of $C_0(X)$ determines the topology of X. And it has found a large number of generalizations and variants in many different contexts (see [15] for a survey of corresponding results). The classical Mazur–Ulam theorem [18] states that every standard surjective isometry between two real Banach spaces must be linear. Thus, the existence of a standard surjective isometry between $C_0(X)$ and $C_0(Y)$ can also guarantee that X and Y are homeomorphic. Instead of isometries, Amir and Cambern investigated the linear isomorphisms between $C_0(X)$ and $C_0(Y)$, where $X, Y$ are compact Hausdorff spaces or locally compact Hausdorff spaces ([1, 5–7]) . They showed that if the linear isomorphism $T:C_0(X)\rightarrow C_0(Y)$ satisfies that $\|T\|\cdot\|T^{-1}\| \lt 2$, then the underlying spaces X and Y are homeomorphic, and the universal constant ‘2’ is optimal (see [9]).

In another direction, the nonlinear extension of the Banach–Stone theorem has attracted a large number of mathematicians’ attention (see [11–14, 17, 23]). Recently, Galego and Porto da Silva [14] studied the bijective coarse quasi-isometries between $C_0(X)$ and $C_0(Y)$ and they obtained an optimal nonlinear extension of the Banach–Stone theorem.

Let $E, F$ be two Banach spaces. A map $T:E\rightarrow F$ is said to be a coarse quasi-isometry (or coarse (M,ɛ)-quasi-isometry) for some constants $M\geq1$ and $\varepsilon\geq0$ provided

\begin{equation*} \frac{1}{M}\|u-v\|-\varepsilon\leq\|T(u)-T(v)\|\leq M\|u-v\|+\varepsilon, \end{equation*}

for all $u,v\in E$. T is called an ɛ-isometry when M = 1 and an isometry when M = 1 and ɛ = 0. If $T(0)=0$, then T is called standard.

Theorem 1.2. (Galego-Porto da Silva)

Let $X, Y$ be two locally compact Hausdorff spaces and $T:C_0(X)\rightarrow C_0(Y)$ be a standard bijective map such that both T and T ⁻¹ are coarse $(M,\varepsilon)$-quasi-isometries with $M \lt \sqrt{2}$. Then there exists a homeomorphism $\varphi: Y\rightarrow X$ and a continuous function $\lambda: Y\rightarrow\{\pm1\}$ such that

\begin{equation*} |MT(f)(y)-\lambda(y)f(\varphi(y))|\leq (M^2-1)\|f\|+\Delta\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X), \end{equation*}

where Δ does not depend on f and y.

The upper bound $\sqrt{2}$ on M is optimal even in the linear case when T is a linear isomorphism [9]. When M = 1, i.e. $T:C_0(X)\rightarrow C_0(Y)$ is a bijective standard ɛ-isometry, Theorem 1.2 yields that

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq 2\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation*}

As it follows from a result of Omladič and Šemrl [20], T can be weaken as a standard surjective ɛ-isometry and the estimate ‘2ɛ’ is optimal (see, also, [4, p. 360]).

In view of geometry, the isometry T from Banach space E to another Banach space F preserves the length of one diagonal of the parallelogram generated by two vectors. But, one may ask what happens if T preserves the length of another diagonal of the parallelogram instead, that is,

\begin{align*} \|T(u)+T(v)\|=\|u+v\|,\;{\rm for\;all}\; u,v\in E. \end{align*}

By letting $g=-f$ in the above equation, it is clear that T is an isometry with $T(-f)=-T(f)$ and $T(0)=0$. Thus, the Banach–Stone theorem (Theorem 1.1) still holds when T is surjective. Such transformations are called norm-additive maps and have stronger properties than isometries when the domain is symmetric. And these maps have been studied recently in [8, 10, 16, 19, 22].

Let $E, F$ be two Banach spaces and $T:E\rightarrow F$ be a map, $\varepsilon\geq0$. T is called an ɛ-norm-additive map provided

\begin{equation*} \big|\|T(u)+T(v)\|-\|u+v\|\big|\leq \varepsilon,\;{\rm for\;all}\; u, v\in E. \end{equation*}

In this paper, we mainly study the properties of the ɛ-norm-additive map between $C_0(X)$ and $C_0(Y)$ which is a natural and interesting generalization of norm-additive map to the perturbed case. It is worth noting that although the proof of Theorem 1.2 for the sharp estimate ‘2ɛ’ of the surjective ɛ-isometries is very skilful [14], it cannot be applied to ɛ-norm-additive mappings for hunting the sharp estimate because the ɛ-norm-additive mapping between $C_0(X)$ and $C_0(Y)$ may be a strictly 2ɛ-isometry (see Example 2.3).

We mainly prove that if $X, Y$ are two locally compact Hausdorff spaces and $T:C_0(X)\rightarrow C_0(Y)$ is a standard surjective ɛ-norm-additive map, then there exists a homeomorphism $\varphi:Y\rightarrow X$ and a continuous function $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ such that

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation*}

The constant ‘$\frac{3}{2}$’ is optimal.

2. Main results

We start this section with the following observation which reveals the relationship between ɛ-isometries and ɛ-norm-additive maps on Banach spaces.

Proposition 2.1. Suppose that E and F are Banach spaces and $T:E\rightarrow F$ is an ɛ-norm-additive map. Then T is a 2ɛ-isometry.

Proof. For any $u\in E$, by the definition of T, we have

\begin{equation*} \big|\|T(u)+T(-u)\|-\|u-u\|\big|\leq\varepsilon, \end{equation*}

i.e.

\begin{equation*} \|T(u)+T(-u)\|\leq\varepsilon. \end{equation*}

For $u,v\in E$, we obtain that

\begin{align*} \|T(u)-T(v)\|&=\|(T(u)+T(-u))-(T(-u)+T(v))\|\\ &\leq \|T(-u)+T(v)\|+\|T(u)+T(-u)\|\\ &\leq\|u-v\|+2\varepsilon, \end{align*}

and

\begin{align*} \|T(u)-T(v)\|&=\|(T(u)+T(-u))-(T(-u)+T(v))\|\\ &\geq\|T(-u)+T(v)\|-\|T(u)+T(-v)\|\\ &\geq\|u-v\|-2\varepsilon. \end{align*}

Thus, T is a 2ɛ-isometry and the proof is complete.

Although every ɛ-norm-additive map is actually a 2ɛ-isometry, the converse is not true in general.

Example 2.2. Define $T:c_0\rightarrow c_0$ by $T(u)=(\|u\|,u_1,u_2,\ldots)$ for $u=(u_n)_{n=1}^\infty\in c_0$. Then T is a standard (0−)isometry, but it is not a δ-norm-additive map for any $\delta\geq0$.

The following example shows that the constant ‘2’ in Proposition 2.1 is sharp.

Example 2.3. Let $X=\lbrace a\rbrace$, then $C(X)=\mathbb{R}$. Fix ɛ > 0, define $T:\mathbb{R}\rightarrow\mathbb{R}$ by

\begin{equation*}T(u)=\begin{cases} 0&\text{if\; \textit{u}\,=\,0},\\ -\frac{\varepsilon}{2}&\text{if\;$u=\varepsilon$},\\ u+\frac{\varepsilon}{2}&\text{if\;$u\neq 0,\varepsilon$}. \end{cases}\end{equation*}

Then T is a standard ɛ-norm-additive map. Note that $\big||T(2\varepsilon)-T(\varepsilon)|-|2\varepsilon-\varepsilon|\big|=2\varepsilon$. This and Proposition 2.1 together show that T is a strictly 2ɛ-isometry.

Let $X, Y$ be two locally compact Hausdorff spaces and $T:C_0(X)\rightarrow C_0(Y)$ be a standard surjective ɛ-norm-additive map. Combining Proposition 2.1, Theorem 1.2 and the Omladič–Šemrl’s theorem [20], we have the following result.

Theorem 2.4. Let $T:C_0(X)\rightarrow C_0(Y)$ be a standard surjective ɛ-norm-additive map. Then there exists a homeomorphism $\varphi:Y\rightarrow X$ and a continuous function $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ such that

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq4\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation*}

However, the constant in the estimate above is not the best. Our main goal is to obtain a sharp version of the above theorem by reducing 4ɛ to $\frac{3}{2}\varepsilon$. And then we will show that $\frac{3}{2}\varepsilon$ is optimal. To begin with, we establish the following useful lemmas.

Lemma 2.5. Let $x\in X$ and $f_1, f_2, ..., f_n\in C_0(X)$ with $(f_1(x),f_2(x),...,f_n(x))\neq 0$. Then there exists a $g\in C_0(X)$ with $g(z)\leq 0$ for all $z\in X$ such that

\begin{equation*} \|g\|=-g(x)\;{\rm and}\;\|g+f_i\|=-g(x)-f_i(x),\;\forall\;i\in\{1,2,...,n\}. \end{equation*}

Proof. Let $I=[-\|f_1\|,\;\|f_1\|]\times[-\|f_2\|,\;\|f_2\|]\times...\times[-\|f_n\|,\;\|f_n\|]\subset\mathbb{R}^n$. Define $F:X\rightarrow \mathbb{R}^n$ by

\begin{equation*} F(z)= (f_1(z),f_2(z),...,f_n(z)),\;\forall\;z\in X. \end{equation*}

Then F is well defined and continuous. Let $\alpha=\max\limits_{1\leq i\leq n}\|f_i\|$. For every $(u_1,u_2,...,u_n)\in I$, let

\begin{equation*} h((u_1,u_2,...,u_n))=\max\limits_{1\leq i\leq n} \lbrace f_i(x)-u_i-3\alpha, -3\alpha\rbrace. \end{equation*}

Then $h:I\rightarrow \mathbb{R}$ is well defined. It is clear that h is continuous and $-3\alpha\leq h(F(z))\leq 0$ for every $z\in X$. By the Urysohn lemma, there exists a continuous function $P: I\rightarrow [0,1]$ such that $P((0,0,...,0))=0$ and $P(F(x))=1$. Define $g:X\rightarrow \mathbb{R}$ by

\begin{equation*} g(z)=(P\cdot h)(F(z))(=P(F(z))\cdot h(F(z))),\; \forall\; z\in X. \end{equation*}

Then g is well defined. Since $P, h, F$ are continuous, g is also continuous.

We assert that $g\in C_0(X)$. For any convergent net $\lbrace x_\lambda\rbrace_{\lambda\in\Lambda}$ with x_λ converges to infinity, we have $ F(x_\lambda)\rightarrow (0,0,...,0)\in\mathbb{R}^n$. Hence $ P(F(x_\lambda))\rightarrow 0$. Note that $|h(F(x_\lambda))|\leq 3\alpha$ for any $\lambda\in\Lambda$, we have $g(x_\lambda)\rightarrow 0$. Thus $g\in C_0(X)$.

For every $z\in X$, $0\leq P(F(z))\leq 1$ and $-3\alpha\leq h(F(z))\leq 0$. Thus $-3\alpha\leq g(z)\leq 0$ for every $z\in X$. Note that $g(x)=-3\alpha$, then $\|g\|=-g(x)$. For every $i\in\lbrace 1,2,...,n\rbrace$ and every $z\in X$, we have

\begin{align*} \alpha&\geq g(z)+f_i(z)=P(F(z))\cdot h(F(z))+f_i(z)\\ &\geq h(F(z))+f_i(z)\geq f_i(x)-f_i(z)-3\alpha+f_i(z)\\ &=f_i(x)-3\alpha=g(x)+f_i(x). \end{align*}

Then

\begin{equation*} |g(z)+f_i(z)|\leq 3\alpha-f_i(x)=-g(x)-f_i(x),\;\forall\;z\in X. \end{equation*}

This implies that $\|g+f_i\|=-g(x)-f_i(x)$ and the proof is complete.

Similar to Lemma 2.5, we have the following lemma.

Lemma 2.6. Let $x\in X$ and $f_1, f_2, ..., f_n\in C_0(X)$ with $(f_1(x),f_2(x),...,f_n(x))\neq 0$. Then there exists a $g\in C_0(X)$ with $g(z)\geq 0$ for all $z\in X$ such that

\begin{equation*} \|g\|=g(x)\;{\rm and}\;\|g+f_i\|=g(x)+f_i(x),\;\forall\;i\in\{1,2,...,n\}. \end{equation*}

For every $x\in X$ and $f\in C_0(X)$, let

\begin{align*} P(f,x)&=\lbrace (y,\lambda)\in Y\times\lbrace\pm1\rbrace:\lambda T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon\rbrace,\\ \bar{P}(f,x)&=\lbrace (y,\lambda)\in Y\times\lbrace\pm1\rbrace:\lambda T(f)(y)\leq f(x)+\frac{3}{2}\varepsilon\rbrace. \end{align*}

Put

\begin{equation*} P_+(x)=\cap_{f\in C_0(X),\;f(x)\geq 0}P(f,x),\;P_-(x)=\cap_{f\in C_0(X),\;f(x)\leq 0}\bar{P}(f,x). \end{equation*}

Lemma 2.7. For every $x\in X$, $P_+(x)$ is non-empty.

Proof. Fix $x\in X$, the proof is divided into three steps.

Step I. We prove that for $f\in C_0(X)$ with $f(x) \gt \frac{3}{2}\varepsilon$, $P(f,x)$ is a non-empty compact set. Let

\begin{align*} P_1=\lbrace y\in Y:T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon\rbrace,\;\; P_2=\lbrace y\in Y:-T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon\rbrace. \end{align*}

Since T is an ɛ-norm-additive map,

\begin{equation*} 2\big|\|T(f)\|-\|f\|\big|=\big|\|T(f)+T(f)\|-\|f+f\|\big|\leq \varepsilon, \end{equation*}

i.e.

\begin{equation*}\|f\|-\frac{1}{2}\varepsilon\leq\|T(f)\|\leq \|f\|+\frac{1}{2}\varepsilon.\end{equation*}

This yields that at least one of the $P_1,\;P_2$ is non-empty. Note that $f(x) \gt \frac{3}{2}\varepsilon$, P ₁ and P ₂ are two compact sets. Since $P(f,x)=P_1\times\lbrace 1\rbrace\cup P_2\times\lbrace -1\rbrace$, $P(f,x)$ is a non-empty compact subset of $Y\times \lbrace\pm1\rbrace$.

Step II. Fix a $f_0\in C_0(X)$ with $f_0(x) \gt \frac{3}{2}\varepsilon$, we prove that $P(f_0,x)\cap P(f,x)$ is a non-empty compact set for every $f\in C_0(X)$ with $f(x)\geq 0$. It is clear that $P(f,x)$ is closed in $Y\times \lbrace\pm1\rbrace$ and hence $(P(f_0,x)\cap P(f,x))\subset P(f_0,x)$ is compact. It remains to prove that $P(f_0,x)\cap P(f,x)$ is non-empty. By Lemma 2.5, there exists a $g\in C_0(X)$ such that

\begin{equation*} \|g\|=-g(x),\;\|g+f\|=-g(x)-f(x),\;\|g+f_0\|=-g(x)-f_0(x). \end{equation*}

Since T is an ɛ-norm-additive map, one has

\begin{equation*} \|T(g)\|\geq \|g\|-\frac{1}{2}\varepsilon,\;\|f+g\|+\varepsilon\geq \|T(f)+T(g)\|,\;\|f_0+g\|+\varepsilon\geq \|T(f_0)+T(g)\|. \end{equation*}

Pick $(y,\lambda)\in Y\times\lbrace\pm1\rbrace$ such that $\|T(g)\|=\lambda T(g)(y)$. Then

\begin{align*} -g(x)-f_0(x)+\varepsilon&=\|g+f_0\|+\varepsilon\geq\|T(g)+T(f_0)\|\geq \lambda T(g)(y)+\lambda T(f_0)(y)\\ &\geq\|g\|-\frac{1}{2}\varepsilon+\lambda T(f_0)(y)=-g(x)-\frac{1}{2}\varepsilon+\lambda T(f_0)(y). \end{align*}

This implies $-\lambda T(f_0)(y)\geq f_0(x)-\frac{3}{2}\varepsilon$. Similarly, we can get $-\lambda T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon$. Thus $(y,-\lambda)\in P(f_0,x)\cap P(f,x)$ and $P(f_0,x)\cap P(f,x)$ is a non-empty compact subset of $Y\times \lbrace\pm1\rbrace$.

Step III. We prove that the set family $\lbrace P(f,x)\cap P(f_0,x)\rbrace_{f\in C_0(X),\;f(x)\geq 0}$ has the finite intersection property. Fix $f_1,f_2,...,f_n\in C_0(X)$ with $f_i(x)\geq 0$ for every $i\in\{1,2,...,n\}$, by Lemma 2.5, there exists a $g\in C_0(X)$ such that

\begin{equation*} \|g\|=-g(x)\;\;{\rm and}\;\; \|g+f_i\|=-g(x)-f_i(x),\; \forall i\in\{0,1,2,...,n\}. \end{equation*}

Pick $(y,\lambda)\in Y\times\lbrace\pm1\rbrace$ such that $\|T(g)\|=\lambda T(g)(y)$. For every $i\in\{1,2,...,n\}$, by the same argument as in Step II, we obtain

\begin{equation*} (y,-\lambda)\in P(f_i,x)\cap P(f_0,x). \end{equation*}

Thus

\begin{equation*} \cap_{f\in C_0(X),\;f(x)\geq 0}(P(f,x)\cap P(f_0,x))\neq \emptyset. \end{equation*}

Note that

\begin{equation*} P_+(x)=\cap_{f\in C_0(X),\;f(x)\geq 0}P(f,x)=\cap_{f\in C_0(X),\;f(x)\geq 0}(P(f,x)\cap P(f_0,x)). \end{equation*}

The proof is complete.

The following Lemma 2.8 is analogous to Lemma 2.7 and we omit its proof.

Lemma 2.8. For every $x\in X$, $P_-(x)$ is non-empty.

Remark 2.9. Fix $x\in X$, it is not difficult to verify that if $(y,\lambda)\in P_+(x)$, then $(y,-\lambda)\notin P_+(x)$. Suppose on the contrary that $(y,\lambda), (y,-\lambda)\in P_+(x)$, pick $f\in C_0(X)$ with $f(x) \gt \frac{3}{2}\varepsilon$, then

\begin{align*} \lambda T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon \gt 0,\;\; -\lambda T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon \gt 0. \end{align*}

This leads to a contradiction. By a similar argument as above, we can see that if $(y,\lambda)\in P_-(x)$, then $(y,-\lambda)\notin P_-(x)$.

From now on, we assume that $T: C_0(X)\rightarrow C_0(Y)$ is a standard surjective ɛ-norm-additive map. For every $y\in Y$ and $g\in C_0(Y)$, define

\begin{align*} Q(g,y)&=\lbrace (x,\lambda)\in X\times\lbrace\pm1\rbrace:\lambda f(x)\geq g(y)-\frac{3}{2}\varepsilon,\;f\in T^{-1}(g)\rbrace,\\ \bar{Q}(g,y)&=\lbrace (x,\lambda)\in X\times\lbrace\pm1\rbrace:\lambda f(x)\leq g(y)+\frac{3}{2}\varepsilon,\;f\in T^{-1}(g)\rbrace. \end{align*}

Put

\begin{equation*} Q_+(y)=\cap_{g\in C_0(Y),\;g(y)\geq 0}Q(g,y), \;Q_-(y)=\cap_{g\in C_0(Y),\;g(y)\leq 0}\bar{Q}(g,y). \end{equation*}

Lemma 2.10. For every $y\in Y$, $Q_+(y),\;Q_-(y)$ both are non-empty.

Proof. Let $y\in Y$, we just prove $Q_+(y)$ is non-empty, the case for $ Q_-(y)$ is similar. The proof is divided into three steps.

Step I. Assume that $g\in C_0(Y)$ with $g(y) \gt \frac{3}{2}\varepsilon$. We first prove $Q(g,y)$ is a non-empty compact set. By Lemma 2.5, there exists $\bar{g}\in C_0(Y)$ such that

\begin{equation*} \|\bar{g}\|=-\bar{g}(y)\;\;{\rm and}\;\;\|\bar{g}+g\|=-\bar{g}(y)-g(y). \end{equation*}

Since T is surjective, there exists $h\in C_0(X)$ such that $T(h)=\bar{g}$. Find $(x,\lambda)\in X\times\{\pm 1\}$ such that $\|h\|=\lambda h(x)$. For every $f\in T^{-1}(g)$,

\begin{align*} \|\bar{g}\|-\frac{1}{2}\varepsilon+\lambda f(x)&\leq\|h\|+\lambda f(x)=\lambda (h(x)+f(x))\\ &\leq\|h+f\|\leq\|T(h)+T(f)\|+\varepsilon\\ &=\|\bar{g}+g\|+\varepsilon=-\bar{g}(y)-g(y)+\varepsilon\\ &=\|\bar{g}\|-g(y)+\varepsilon. \end{align*}

Thus $-\lambda f(x)\geq g(y)-\frac{3}{2}\varepsilon$. This implies that $(x,-\lambda)\in Q(g,y)$ and $Q(g,y)$ is non-empty. For every $f\in T^{-1}(g)$, let

\begin{align*} R(f)_1=\{x\in X:f(x)\geq g(y)-\frac{3}{2}\varepsilon\},\;\; R(f)_2=\{x\in X:-f(x)\geq g(y)-\frac{3}{2}\varepsilon\}, \end{align*}

\begin{equation*} R(f)=\{(x,\lambda)\in X\times\lbrace\pm1\rbrace:\lambda f(x)\geq g(y)-\frac{3}{2}\varepsilon\}. \end{equation*}

By the above argument, $R(f)\neq\emptyset$ and $R(f)=R(f)_1\times \{1\}\cup R(f)_2\times\{-1\}$. Since $g(y) \gt \frac{3}{2}\varepsilon$, $R(f)_1,\;R(f)_2$ are compact. By the Tychonoff theorem, R(f) is a non-empty compact subset of $X\times \lbrace\pm1\rbrace$. Note that $Q(g,y)=\cap_{f\in T^{-1}(g)}R(f)$, this implies that $Q(g,y)$ is a non-empty compact set.

Step II. Fix $g_0\in C_0(Y)$ with $g_0(y) \gt \frac{3}{2}\varepsilon$, we will show that for every $g\in C_0(Y)$ with $g(y)\geq 0$, $Q(g_0,y)\cap Q(g,y)$ is a non-empty compact set. Note that $Q(g,y)$ is closed in $X\times \lbrace\pm1\rbrace$. Hence $(Q(g_0,y)\cap Q(g,y))\subset Q(g_0,y)$ is compact. By Lemma 2.5, there exists a $\bar{g}\in C_0(Y)$ such that

\begin{equation*} \|\bar{g}\|=-\bar{g}(y),\;\|\bar{g}+g\|=-\bar{g}(y)-g(y),\;\|\bar{g}+g_0\|=-\bar{g}(y)-g_0(y). \end{equation*}

Pick $h\in C_0(X)$ such that $T(h)=\bar{g}$. Find $\{x,\lambda\}\in X\times \{\pm1\}$ such that $\|h\|=\lambda h(x)$. For every $f\in T^{-1}(g)$,

\begin{align*} \|\bar{g}\|-\frac{1}{2}\varepsilon+\lambda f(x)&\leq\|h\|+\lambda f(x)=\lambda (h(x)+f(x))\\ &\leq\|h+f\|\leq\|T(h)+T(f)\|+\varepsilon\\ &=\|\bar{g}+g\|+\varepsilon=-\bar{g}(y)-g(y)+\varepsilon\\ &=\|\bar{g}\|-g(y)+\varepsilon. \end{align*}

This implies that $-\lambda f(x)\geq g(y)-\frac{3}{2}\varepsilon$ and $(x,-\lambda)\in Q(g,y)$. Similarly, we can show $(x,-\lambda)\in Q(g_0,y)$. Therefore, $Q(g_0,y)\cap Q(g,y)$ is a non-empty compact set.

Step III. We check that the set family $\lbrace Q(g,y)\cap Q(g_0,y)\rbrace_{g\in C_0(Y),\;g(y)\geq 0}$ has the finite intersection property. Fix $g_1,g_2,...,g_n\in C_0(Y)$ with $g_i(y)\geq 0$ for every $i\in\{1,2,...,n\}$, by Lemma 2.5, there exists a $\bar{g}\in C_0(Y)$ such that

\begin{equation*} \|\bar{g}\|=-\bar{g}(y),\; \|\bar{g}+g_i\|=-\bar{g}(y)-g_i(y),\; \forall i\in\{0,1,2,...,n\}. \end{equation*}

Pick $h\in C_0(X)$ such that $T(h)=\bar{g}$. Find $\{x,\lambda\}\in X\times \{\pm1\}$ such that $\|h\|=\lambda h(x)$. By the same argument as in Step II, we have

\begin{equation*} (x,-\lambda)\in Q(g_0,y)\;\;{\rm and}\;\;(x,-\lambda)\in Q(g_i,y),\;\forall i\in\{1,2,...,n\}. \end{equation*}

Thus

\begin{equation*} \cap_{g\in C_0(Y),\;g(y)\geq 0}(Q(g,y)\cap Q(g_0,y))\neq \emptyset. \end{equation*}

Note that

\begin{equation*} Q_+(y)=\cap_{g\in C_0(Y),\;g(y)\geq 0}Q(g,y)=\cap_{g\in C_0(Y),\;g(y)\geq 0}(Q(g,y)\cap Q(g_0,y)). \end{equation*}

Then $Q_+(y)$ is non-empty and the proof is complete.

Lemma 2.11. For every $x\in X$, $P_+(x)$ is a singleton.

Proof. Let $x\in X$, by Lemma 2.7, $P_+(x)$ is non-empty. Let $(y,\lambda)\in P_+(x)$, by Lemma 2.10, $Q_+(y)$ and $Q_-(y)$ are non-empty. For every $f\in C_0(X)$ with $f(x) \gt 3\varepsilon$, we have

\begin{equation*} \lambda T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon. \end{equation*}

If λ = 1, then $T(f)(y)\geq f(x)-\frac{3}{2}\varepsilon \gt 0$. For every $(x',\mu)\in Q_+(y)$ and every $f\in C_0(X)$ with $f(x) \gt 3\varepsilon$, one has

(2.1)\begin{equation} \mu f(x')\geq T(f)(y)-\frac{3}{2}\varepsilon\geq f(x)-3\varepsilon \gt 0. \end{equation}

We assert that $x=x'$. Suppose on the contrary that $x\neq x'$, by the Urysohn lemma, there exists a $f\in C_0(X)$ with $f(x) \gt 3\varepsilon$ and $f(x')=0$. This contradicts to (2.1). And hence we have $x=x'$ and µ = 1. Thus $Q_+(y)=\lbrace (x,\lambda)\rbrace$.

If $\lambda=-1$, then $T(f)(y)\leq -(f(x)-\frac{3}{2}\varepsilon) \lt 0$. For every $(x',\mu)\in Q_-(y)$ and every $f\in C_0(X)$ with $f(x) \gt 3\varepsilon$, one has

\begin{equation*} \mu f(x')\leq T(f)(y)+\frac{3}{2}\varepsilon\leq -(f(x)-\frac{3}{2}\varepsilon)+\frac{3}{2}\varepsilon=-f(x)+3\varepsilon \lt 0.\end{equation*}

By the same argument as above, we have $x=x'$ and $\mu=-1$. Thus $Q_-(y)=\lbrace (x,\lambda)\rbrace$.

We assert that $P_+(x)$ is a singleton. Suppose on the contrary that there exist $(y_1,\lambda_1)\neq(y_2,\lambda_2)\in P_+(x)$. By Remark 2.9, $y_1\neq y_2$. Without loss of generality, we assume that $\lambda_1=1$. Then $Q_+(y_1)=\lbrace (x,1)\rbrace$. By the Urysohn lemma, there exists $g\in C_0(Y)$ such that $g(y_1) \gt 3\varepsilon$ and $g(y_2)=0$. Then for every $f\in C_0(X)$ with $T(f)=g$, we have

\begin{equation*} f(x)\geq T(f)(y_1)-\frac{3}{2}\varepsilon=g(y_1)-\frac{3}{2}\varepsilon \gt 0. \end{equation*}

Thus

\begin{equation*} 0=\lambda_2g(y_2)=\lambda_2T(f)(y_2)\geq f(x)-\frac{3}{2}\varepsilon\geq g(y_1)-3\varepsilon \gt 0. \end{equation*}

This leads to a contradiction. Hence $P_+(x)$ is a singleton and the proof is complete.

Similar to Lemma 2.11, we have the following result.

Lemma 2.12. For every $x\in X$, $P_-(x)$ is a singleton.

Lemma 2.13. For every $x\in X$, $P_+(x)=P_-(x)$.

Proof. Fix $x\in X$, by Lemmas 2.11 and 2.12, there exist $y_1,\;y_2\in Y$ and $\lambda_1,\;\lambda_2\in \{\pm1\}$ such that $P_+(x)=(y_1,\lambda_1)$ and $P_-(x)=(y_2,\lambda_2)$. According to the proofs of Lemma 2.11 and 2.12, we have

\begin{equation*} (x,\lambda_1)= \begin{cases} &Q_+(y_1),\;\text{if}\;\lambda_1=1,\\ &Q_-(y_1),\;\text{if}\;\lambda_1=-1, \end{cases}. \;\;{\rm and}\;\; (x,\lambda_2)= \begin{cases} &Q_-(y_2),\;\text{if}\;\lambda_2=1,\\ &Q_+(y_2),\;\text{if}\;\lambda_2=-1. \end{cases} \end{equation*}

We assert that $y_1=y_2$. Suppose on the contrary that $y_1\neq y_2$, by the Urysohn lemma, there exists a $g\in C_0(Y)$ such that

(2.2)\begin{equation} \lambda_1 g(y_1) \gt \lambda_2 g(y_2)+3\varepsilon \gt 6\varepsilon. \end{equation}

For every $f\in C_0(X)$ with $T(f)=g$,

\begin{equation*}\begin{cases} f(x)\geq g(y_1)-\frac{3}{2}\varepsilon&\text{if $\lambda_1=1$},\\ -f(x)\leq g(y_1)+\frac{3}{2}\varepsilon &\text{if $\lambda_1=-1$}. \end{cases}\end{equation*}

Thus

(2.3)\begin{equation} f(x)\geq \lambda_1 g(y_1)-\frac{3}{2}\varepsilon. \end{equation}

On the other hand, for every $f\in C_0(X)$ with $T(f)=g$,

\begin{equation*}\begin{cases} f(x)\leq -g(y_2)+\frac{3}{2}\varepsilon&\text{if $\lambda_2=1$},\\ -f(x)\geq -g(y_2)-\frac{3}{2}\varepsilon &\text{if $\lambda_2=-1$}. \end{cases}\end{equation*}

Thus

(2.4)\begin{equation} f(x)\leq-\lambda_2g(y_2)+\frac{3}{2}\varepsilon. \end{equation}

Combining (2.2), (2.3) and (2.4), we have

\begin{align*} -\lambda_2g(y_2)+\frac{3}{2}\varepsilon&\geq f(x)\geq\lambda_1 g(y_1)-\frac{3}{2}\varepsilon \gt \lambda_2g(y_2)+\frac{3}{2}\varepsilon. \end{align*}

This implies that $\lambda_2g(y_2) \lt 0$. By (2.2), $\lambda_2g(y_2) \gt 0$. This leads to a contradiction and hence $y_1=y_2$.

Next we prove that $\lambda_1=\lambda_2$. Choose $f\in C_0(X)$ such that $f(x) \gt 3\varepsilon$, then

\begin{equation*} \lambda_1T(f)(y_1)\geq f(x)-\frac{3}{2}\varepsilon \gt \frac{3}{2}\varepsilon,\; \lambda_2T(-f)(y_1)\leq-f(x)+\frac{3}{2}\varepsilon \lt -\frac{3}{2}\varepsilon. \end{equation*}

If $\lambda_1\neq\lambda_2$, then

\begin{equation*} \varepsilon\geq\|T(f)+T(-f)\|\geq|T(f)(y_1)+T(-f)(y_1)| \gt 3\varepsilon. \end{equation*}

This leads to a contradiction. Hence $\lambda_1=\lambda_2$ and $P_+(x)=P_-(x)$. The proof is complete.

By a similar argument as Lemmas 2.11, 2.12 and 2.13, we can show the following result. To simplify this article, we omit its proof.

Lemma 2.14. For every $y\in Y$, $Q_+(y)=Q_-(y)$ is a singleton.

Remark 2.15. For every $x\in X$, $y\in Y$ and $\lambda\in\{\pm1\}$, by Lemmas 2.11, 2.12, 2.13 and 2.14, one has

\begin{equation*} \{(x,\lambda)\}=Q_+(y)\;\;\Longleftrightarrow\;\;\{(y,\lambda)\}=P_+(x). \end{equation*}

Now we are ready to show the main result of this paper.

Theorem 2.16. Let $T:C_0(X)\rightarrow C_0(Y)$ be a standard surjective ɛ-norm-additive map. Then there exists a homeomorphism $\varphi:Y\rightarrow X$ and a continuous function $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ such that

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation*}

Proof. For $y\in Y$, by Lemma 2.14, there exist $x\in X$, $\lambda\in\{\pm1\}$ such that

(2.5)\begin{equation} Q_+(y)=\{(x,\lambda)\}. \end{equation}

Define

\begin{equation*} \varphi(y)=x,\;\;\lambda(y)=\lambda, \end{equation*}

where $x, y, \lambda$ satisfy (2.5). Then $\varphi:Y\rightarrow X$ and $\lambda:Y\rightarrow\lbrace\pm1\rbrace$ are well defined. It follows from Remark 2.15 that φ is bijective. In what follows, we show that

(2.6)\begin{equation} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\;y\in Y,\;f\in C_0(X). \end{equation}

Given $y\in Y$ and $f\in C_0(X)$. The proof is divided into two cases.

Case I. $f(\varphi(y))\geq0$. By Remark 2.15, we have $\{(y,\lambda(y)\}=P_+(\varphi(y))$. Then

(2.7)\begin{equation} \lambda(y)T(f)(y)\geq f(\varphi(y))-\frac{3}{2}\varepsilon. \end{equation}

We assert that

(2.8)\begin{equation} \lambda(y)T(f)(y)\leq f(\varphi(y))+\frac{3}{2}\varepsilon. \end{equation}

Suppose on the contrary that $\lambda(y)T(f)(y) \gt f(\varphi(y))+\frac{3}{2}\varepsilon\geq0$. If $\lambda(y)=1$, we have $\{(\varphi(y),1\}=Q_+(y)$ and

\begin{equation*} f(\varphi(y))\geq T(f)(y)-\frac{3}{2}\varepsilon \gt f(\varphi(y)). \end{equation*}

This leads to a contradiction. If $\lambda(y)=-1$, we have $T(f)(y) \lt 0$ and $\{(\varphi(y),-1\}=Q_+(y)=Q_-(y)$. Then

\begin{equation*} -f(\varphi(y))\leq T(f)(y)+\frac{3}{2}\varepsilon \lt -f(\varphi(y)). \end{equation*}

This is a contradiction. Combining (2.7) and (2.8), we have

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon. \end{equation*}

Case II. $f(\varphi(y))\leq0$. By Remark 2.15 again, we have $\{(y,\lambda(y)\}=P_+(\varphi(y))=P_-(\varphi(y))$. Then

(2.9)\begin{equation} \lambda(y)T(f)(y)\leq f(\varphi(y))+\frac{3}{2}\varepsilon. \end{equation}

We assert that

(2.10)\begin{equation} \lambda(y)T(f)(y)\geq f(\varphi(y))-\frac{3}{2}\varepsilon. \end{equation}

Suppose on the contrary that $\lambda(y)T(f)(y) \lt f(\varphi(y))-\frac{3}{2}\varepsilon\leq0$. If $\lambda(y)=1$, we have $\{(\varphi(y),1\}=Q_+(y)=Q_-(y)$ and

\begin{equation*} f(\varphi(y))\leq T(f)(y)+\frac{3}{2}\varepsilon \lt f(\varphi(y)). \end{equation*}

This leads to a contradiction. If $\lambda(y)=-1$, we have $Tf(y) \gt 0$ and $\{(\varphi(y),-1\}=Q_+(y)=Q_-(y)$. Then

\begin{equation*} -f(\varphi(y))\geq T(f)(y)-\frac{3}{2}\varepsilon \gt -f(\varphi(y)). \end{equation*}

This is a contradiction. Combining (2.9) and (2.10), we have

\begin{equation*} |T(f)(y)-\lambda(y)f(\varphi(y))|\leq\frac{3}{2}\varepsilon. \end{equation*}

Next we prove that φ is a homeomorphism and λ is continuous. Suppose that $\{y_\alpha\}_{\alpha\in\Lambda}\subset Y$ is a convergent net with $y_\alpha\rightarrow y$. Choose a compact neighbourhood U of y, without loss of generality, we can assume that $\{y_\alpha\}_{\alpha\in\Lambda}\subset U$. By the Urysohn lemma, there exists a $g\in C_0(Y)$ such that $g|_U\equiv 3\varepsilon+1$. By (2.6), for any $f\in C_0(X)$ with Tf = g, we obtain that

\begin{align*} \frac{3}{2}\varepsilon&\geq|T(f)(y_\alpha)-\lambda(y_\alpha)f(\varphi(y_\alpha))|\\ &\geq|g(y_\alpha)|-|f(\varphi(y_\alpha))|\\ &=3\varepsilon+1-|f(\varphi(y_\alpha))|. \end{align*}

This implies that $|f(\varphi(y_\alpha))|\geq\frac{3}{2}\varepsilon+1$ and $\{\varphi(y_\alpha)\}_{\alpha\in\Lambda}$ is contained in the compact set $\lbrace x\in X: |f(x)|\geq\frac{3}{2}\varepsilon+1\rbrace$.

By (2.6), for $\alpha\in\Lambda$, we have

(2.11)\begin{equation} |Tf(y_\alpha)-\lambda(y_\alpha)f(\varphi(y_\alpha))|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\; f\in C_0(X). \end{equation}

For any convergent subnet $\{\varphi(y_{\alpha'})\}_{\alpha'\in \Lambda'}$ of $\{\varphi(y_\alpha)\}_{\alpha\in\Lambda}$ with $\varphi(y_{\alpha'})\rightarrow x$, let $\{\lambda(y_{\alpha''})\}_{\alpha''\in \Lambda''}$ be a convergent subnet of $\{\lambda(y_{\alpha'})\}_{\alpha'\in\Lambda'}$ with $\lambda(y_{\alpha''})\rightarrow \lambda$. By (2.11), we have

\begin{equation*} |T(f)(y)-\lambda f(x)|\leq\frac{3}{2}\varepsilon,\;{\rm for\;all}\; f\in C_0(X). \end{equation*}

This implies that $\{(x,\lambda)\}=Q_+(y)$ and hence $\varphi(y)=x$. Thus φ is continuous. By the same argument, we can get φ ⁻¹ is also continuous. Hence φ is a homeomorphism. For any convergent subset $\{\lambda(y_{\alpha'})\}_{\alpha'\in \Lambda'}$ of $\{\lambda(y_\alpha)\}_{\alpha\in\Lambda}$ with $\lambda(y_{\alpha'})\rightarrow \lambda$, by (2.11) again, we have

\begin{equation*} |T(f)(y)-\lambda f(\varphi(y))|\leq\frac{3}{2}\varepsilon,\; {\rm for\;all}\;f\in C_0(X). \end{equation*}

This implies that $\{(\varphi(y),\lambda)\}=Q_+(y)$ and $\lambda(y)=\lambda$. Hence $\lambda: Y\rightarrow \{\pm1\}$ is continuous. The proof is complete.

The following example shows that the estimate ‘$\frac{3}{2}\varepsilon$’ in Theorem 2.16 is optimal.

Example 2.17. Let $X=\{x_1, x_2\}$ with the discrete topology, then $C(X)=\ell_\infty^2$. Let

\begin{equation*} U_1=\{(a,b)\in C(X):b\leq-\frac{1}{2}\varepsilon,\;a+b\geq-\frac{1}{2}\varepsilon,\;(a,b)\neq(\varepsilon,-\frac{1}{2}\varepsilon)\} \end{equation*}

and

\begin{equation*} U_2=C(X)\setminus(U_1\cup\{(0,0),(\varepsilon,-\frac{1}{2}\varepsilon)\}). \end{equation*}

Define $T:C(X)\to C(X)$ by

\begin{equation*} T((a,b))= \left\{ \begin{array}{ll} (0,0) & (a,b)=(0,0),\\ (\varepsilon,\varepsilon) & (a,b)=(\varepsilon,-\frac{1}{2}\varepsilon),\\ (a-\frac{1}{2}\varepsilon,b) & (a,b)\in U_1,\\ (a,b-\frac{1}{2}\varepsilon) &(a,b)\in U_2.\\ \end{array}\right. \end{equation*}

It is not difficult to verify that T is a standard surjective map. In what follows, we show that T is an ɛ-norm-additive map. By the definition of T,

\begin{align*} \|T(f)-f\|\leq\frac{1}{2}\varepsilon,\;\forall\;f\in C(X),\; f\neq(\varepsilon,-\frac{1}{2}\varepsilon). \end{align*}

Then for any $f,g\in C(X)$ with $f,g\neq(\varepsilon,-\frac{1}{2}\varepsilon)$, we have

\begin{align*} \big|\|T(f)+T(g)\|-\|f+g\|\big|&\leq\|(T(f)+T(g))-(f+g)\|\\ &\leq\|T(f)-f\|+\|T(g)-g\|\leq\varepsilon. \end{align*}

The left case is $f=(\varepsilon,-\frac{1}{2}\varepsilon)$ and $g\in C(X)$. When $g=(0,0)$ or $(\varepsilon,-\frac{1}{2}\varepsilon)$, it is clear that

\begin{equation*} \|T(f)+T(g)\|=\|f+g\|. \end{equation*}

Let $g=(a,b)\in U_1$, we have

\begin{align*} |b+\varepsilon|\leq|b|\leq a+\frac{1}{2}\varepsilon. \end{align*}

Then

\begin{align*} \|T(f)+T(g)\|=&\max\{|a+\frac{1}{2}\varepsilon|,|b+\varepsilon|\}=a+\frac{1}{2}\varepsilon,\\ \|f+g\|=&\max\{|a+\varepsilon|,|b-\frac{1}{2}\varepsilon|\}=a+\varepsilon. \end{align*}

Thus

\begin{align*} \big|\|T(f)+T(g)\|-\|f+g\|\big|=\frac{1}{2}\varepsilon. \end{align*}

Let $g=(a,b)\in U_2$, we have

\begin{align*} \big|\|T(f)+T(g)\|-\|f+g\|\big|&=\big|\|(a+\varepsilon, b+\frac{1}{2}\varepsilon)\|-\|(a+\varepsilon, b-\frac{1}{2}\varepsilon)\|\big|\\ &\leq\|(a+\varepsilon, b+\frac{1}{2}\varepsilon)-(a+\varepsilon, b-\frac{1}{2}\varepsilon)\|\\ &=\|(0,\varepsilon)\|=\varepsilon. \end{align*}

Therefore, T is an ɛ-norm-additive map. The homeomorphism $\varphi:X\rightarrow X$ and $\lambda:X\rightarrow\{\pm1\}$ which satisfy Theorem 2.16 are

\begin{equation*} \varphi=Id_X,\;\;\lambda(x_i)=1,\; for\; i=1,2. \end{equation*}

When $f=(\varepsilon,-\frac{1}{2}\varepsilon)$, then

\begin{equation*} |T(f)(x_2)-\lambda(x_2)f(\varphi(x_2))|=|\varepsilon-(-\frac{1}{2}\varepsilon)|=\frac{3}{2}\varepsilon. \end{equation*}

This implies that the estimate $\frac{3}{2}\varepsilon$ in Theorem 2.16 is optimal.

Remark 2.18. The assumption of surjectivity of T in Theorem 2.16 is essential in general. For instance, let $X=\{a\}$, $Y=\{b,c\}$ be two discrete topological spaces, then $C(X)=\mathbb{R}$ and $C(Y)=\ell_\infty^2$. Define $T:\mathbb{R}\rightarrow \ell_\infty^2$ by $T(x)=(x,\sin x)$ for all $x\in\mathbb{R}$. Then T is a norm-additive map, but X and Y are not homeomorphic.