Spectrum of weighted composition operators part VI: essential spectra of d-endomorphisms of Banach C(K)-modules

Abstract

We investigate properties of essential spectra of disjointness preserving operators acting on Banach C(K)-modules. In particular, we prove that under some very mild conditions the upper semi-Fredholm spectrum of such an operator is rotation invariant. In the last part of the paper we provide a full description of the spectrum and the essential spectra of operators acting on Kaplansky modules of the form $T = wU$, where $w \in C(K)$, U is a d-isomorphism, and the spectrum of U is a subset of the unit circle.

Appendix 1: Properties of disjointness preserving operators

In this section we will review the properties of disjointness preserving operators that we used throughout the paper. Initially we will assume that X, Y are complex Banach lattices and $T:X\rightarrow Y$ is a (bounded) disjointness preserving operator. That is $x\bot y \Rightarrow T(x)\bot T(y)$ for all $x,y\in X.$

The properties of disjointness preserving operators on real Banach lattices and more generally those of order bounded disjointness preserving operators on real Riesz spaces are well-known. In [1] these properties are developed in even more generality on complex Riesz spaces for order bounded disjointness preserving operators. However this is not entirely routine and some details in [1] are missing. We will present a complete discussion of the passage to the complex case for Banach lattices.

Let X be a complex Banach lattice. Then there exists a real Banach lattice $X_{r} = X_{+}-X_{+}\subseteq X$ such that $X_{r}\cap iX_{r}=\left\{ 0\right\}$ and $X=X_{r}\oplus iX_{r}$. Moreover, let $z=x+iy \in X$ with $x,y\in X_{r}$ then

$$\begin{aligned} |z|= & {} sup\left\{ xcos\theta +ysin\theta :\theta \in [0,2\pi ]\right\} \\= & {} sup\left\{ |x|cos\theta +|y|sin\theta :\theta \in [0,\pi /2]\right\} \end{aligned}$$

and $\left\| z\right\| =\left\| |z|\right\| .$ When X is represented as a Banach lattice of functions with respect to pointwise operations on some set (as it always can be), it is familiar that the above formula yields the usual

$$\begin{aligned} |z|=\sqrt{x^2+y^2}. \end{aligned}$$

For each operator $T:X \rightarrow Y$ there exist operators $T_r$ and $T_i$ from $X_r$ into $Y_r$ such that for each $z=x+iy\in X$ with $x,y\in X_r$, one has

$$\begin{aligned} T(z)=T_r (x)- T_i (y)+i(T_i (x)+T_r (y)). \end{aligned}$$

Suppose that T is a disjointness preserving operator. Then $T_r , T_i$ are disjointness preserving operators from $X_r$ into $Y_r$, since for each $x\in X_r,$ one has that both $|T_r (x)|, |T_i (x)|$ are less than or equal to |T(x)| in the partial order of $Y_r.$

In the case of real Banach lattices, Abramovich [2] proved that disjointness preserving operators are order bounded, and therefore by a result of Meyer [24], they are regular. Elementary proofs of these results were given by de Pagter [27] and Bernau [4], respectively (See also [25, Thorems 3.1.4 and 3.1.5]). An operator $T:X\rightarrow Y$ is by definition regular if $T_r, T_i$ are regular in the usual sense. So our final observation in the above paragraph shows that if T is disjointness preserving, then T is regular and order bounded. (In the sense that: Given $z\in X$ there exists $w\in Y_+$ such that, for any $u\in X$, $|u|\le |z|$ implies that $|T(u)|\le w$.)

When X is a Banach lattice we will denote by ${\widehat{X}}$, the Dedekind completion of X where ${\widehat{X}}={\widehat{X}}_r\oplus i{\widehat{X}}_r$. (Here ${\widehat{X}}_r$ is the Dedekind completion of $X_r$.) It is familiar that there exists a stonean compact Hausdorff space Q(X) such that ${\widehat{X}}$ is an ideal in $C_\infty (Q(X))$ and $Z({\widehat{X}})=C(Q(X))$. As usual $C_\infty (Q(X))$ denotes the algebra and the Dedekind complete Riesz space of all continuous functions $f:C(Q(X))\rightarrow \mathbb {C}\cup \left\{ \infty \right\}$ (where $\mathbb {C}\cup \left\{ \infty \right\}$ is the one point compactification of $\mathbb {C}$) such that $f^{-1}(\infty )$ is a nowhere dense subset of Q(X).

By the support of a set $A\subset {\widehat{X}},$ with a slight abuse of language, we will denote $\chi _A,$ the band projection of ${\widehat{X}}$ onto $A^{dd}.$ We may also think of $\chi _A$ as the characteristic function of the clopen set in Q(X) that is the closure of all points $t\in Q(X)$ such that some element of A takes a finite non-zero value at t. Given an operator $T:X\rightarrow Y,$ we will denote by $\chi _T$ the support of the range of T in ${\widehat{Y}}.$

Now we can state and prove the main result for disjointness preserving operators on complex Banach lattices in [1].

Theorem 7.1

Let X, Y be Banach lattices and $T:X\rightarrow Y$ be an operator. Then the following are equivalent:

• (1) T is a disjointness preserving operator.

• (2) For all $x,y\in X$, $|y|\le |x|$ implies that $|T(y)|\le |T(x)|.$

• (3) T admits a unique factorization $T=VT_l$ such that $T_l:X\rightarrow Y$ is a lattice homomorphism and $V\in Z({\widehat{Y}})$ such that $|V|=\chi _T.$

Proof

Consider $(1)\Rightarrow (2).$ Given any $x\in X,$ we may assume that I(x) (the ideal generated by |x| in X) is lattice isomorphic to $C(K_x)$ for some compact Hausdoff space $K_x.$ Moreover, by our observations preceding the statement of the theorem, T is order bounded. Hence we may assume that T(I(x)) is contained in an ideal I(w) in Y for some $w\in Y_+.$ So I(w) is also lattice isomorphic to $C(K_w)$ for some compact Hausdorff space $K_w$. Now, since T is order bounded, when restricted to I(x), as a linear mapping on $C(K_x)$ into $C(K_w)$, T is bounded with respect to the sup-norms on these spaces. For any fixed $t\in K_w$, consider the bounded linear functional $l:C(K_x)\rightarrow \mathbf{C }$ defined by $l(y)=T(y)(t).$ To establish $(1)\Rightarrow (2)$, it is sufficient to show that $|l(y)|\le |l(x)|$ whenever $|y|\le |x|$ for some $y\in X.$ Let the closed set $F\subset K_x$ be the support of the regular Borel measure $l\in C(K_x)^{'}.$ It is clear that the result would be established if we show that F is a singleton.

Suppose that F is not a singleton. Suppose that l is a positive linear functional and $s_1, s_2$ are distinct points in F. We can find Urysohn functions $y_1,y_2\in C(K_x)$ such that $y_1(s_1)=1, y_2(s_2)=1$ and $y_1\bot y_2$. Then $l(y_1)l(y_2)>0$. But that $T(y_1)\bot T(y_2)$ implies that at least one of them should be zero. This yields a contradiction. Suppose l is real. Then we may suppose that $F_+,F_-$ are the non-empty supports of $l^+, l^-$ respectively such that $F=F_+\cup F_-$ and $F_+\cap F_-=\emptyset .$ If we have $s_1\in F_+$ and $s_2\in F_-$, again we may find Urysohn functions $y_1,y_2\in C(K_x)$ such that $y_1\bot y_2$ and $l(y_1)>0>l(y_2)$. Since $T(y_1)\bot T(y_2)$, this still leads to a contradiction. Now if l is a complex measure and, without loss of generality the support of its real part is not a singleton, we can repeat the above argument to find two functions $y_1,y_2\in C(K_x)$ such that $y_1\bot y_2$ and both $real(l(y_1))$ and $real(l(y_2))$ are non-zero. This means that $l(y_1)\ne 0$ and $l(y_2)\ne 0$ with $T(y_1)\bot T(y_2)$, and leads to a contradiction. Hence we are reduced to the case where the supports of the real part and the imaginary part of l are distinct singletons. Obviously that also will yield a contradiction. Therefore $F= \left\{ s_t \right\}$ for some $s_t\in K_x$. This proves $(1)\Rightarrow (2).$

The converse $(2)\Rightarrow (1)$ is almost immediate. It is well known for real Banach lattices that for any $x,y\in X$ we have

$$\begin{aligned} x\bot y \Leftrightarrow |y|\le |y+\lambda x| \end{aligned}$$

(42)

for all scalars $\lambda$. But the same is true for complex Banach lattices as well when $\lambda \in \mathbf{C }$. (Namely when x, y are complex numbers and $\lambda \in \mathbf{C }$ then it is easy to see that (42) is true. To see (42) for complex Banach lattices, simply think of the Banach lattice represented as a lattice of functions on some base space and use the preceding fact at each point of the base space.) So if $x\bot y$ in X, we have, by (42) and (2), $|T(y)|\le |T(y)+\lambda T(x)|$ for all $\lambda \in \mathbf{C }.$ Therefore $T(x)\bot T(y)$ in Y. Hence $(2)\Rightarrow (1).$

We want to show $(1)\Rightarrow (3).$ In the proof of $(1)\Rightarrow (2),$ we showed that the support of the linear functional $l:C(K_x)\rightarrow \mathbf{C }$ is a singleton. That is for a fixed $t\in K_w,$

$$\begin{aligned} l(y)=T(y)(t)=\lambda _t y(s_t) \end{aligned}$$

(43)

for some $s_t\in K_x$ and some $\lambda _t \in \mathbf{C }$, for all $y\in C(K_x)=I(x).$ Note that, by (2),  we may choose $w=|T(x)|\in Y.$ Given $x_1, x_2 \in X_+$, let $x=x_1 + x_2$. Then for any $t\in K_w$, by (43), we have $|T(x_1 + x_2)(t)|=|T(x_1)(t)|+|T(x_2)(t)|$. Since t is arbitrary, in fact, $|T(x_1 + x_2)|=|T(x_1)|+|T(x_2)|$ for all $x_1, x_2 \in X_+$. This yields a well defined positive operator $(T_l)_r :X_r\rightarrow Y_r$ where $(T_l)_r( x_1-x_2)=|T(x_1)|-|T(x_2)|$ for all $x_1, x_2 \in X_+$. In fact $(T_l)_r$ is a lattice homomorphism. This can be seen easily from (43), if we take $x_1\bot x_2$ in above and $x=x_1+x_2$. We can write the complexification of $(T_l)_r$, to obtain $T_l:X\rightarrow Y$ defined by $T_l(z)=(T_l)_r(u)+i(T_l)_r(v) =|T(u^+)|-|T(u^-)|+i(|T(v^+)|-|T(v^-)|)$ when $z=u+iv$. Let $x=|z|$ in (43). Then we get $|T_l(z)|=|T(z)|$. Now apply (2) to $|z|\le |z|$, to get $|T(z)|=|T(|z|)|.$ But $|T(|z|)|=T_l(|z|)$. Hence $T_l$ is a lattice homomorphism.

It remains to see that T factors through $T_l$ unimodularly on the support of the range of T. Fix a representation of ${\widehat{Y}}$ in $C_\infty (Q(Y))$. Consider the ranges of T and $T_l$ as subsets of ${\widehat{Y}}$ in $C_\infty (Q(Y))$. For each $x\in X_+,$ we have $T_l(x)=|T(x)|.$ Since ${\widehat{Y}}$ is Dedekind complete, there exists $V_x\in Z({\widehat{Y}})$ such that $T(x)=V_xT_l(x)$ and $|V_x|=\chi _{T(x)}.$ (Note that $\chi _{T(x)}$ is also the support of $T_l(x).$) We want to show

$$\begin{aligned} \chi _{T(x)}V_y=\chi _{T(y)}V_x. \end{aligned}$$

That is $V_x$ and $V_y$ are equal on the intersection of the supports of T(x) and T(y). Since $T_l$ is a positive operator, we have $\chi _{T(x+y)}=max\left\{ \chi _{T(x)} ,\chi _{T(y)}\right\} .$ Now

$$\begin{aligned} V_{x+y}T_l(x+y)=T(x+y)=T(x)+T(y)=V_xT_l(x)+V_yT_l(y). \end{aligned}$$

That is

$$\begin{aligned} (V_{x+y}-V_x)T_l(x)=(V_y-V_{x+y})T_l(y). \end{aligned}$$

Now multiplying both sides of the above equation by $\chi _{T(x)}\chi _{T(y)}$, we have

$$\begin{aligned} \chi _{T(y)}(V_{x+y}-V_x)T_l(x)=\chi _{T(x)}(V_y-V_{x+y})T_l(y). \end{aligned}$$

(44)

Suppose for some $t\in Q(Y)$ both $T_l(x)(t)$ and $T_l(y)(t)$ are positive. Let $k=\frac{T_l(x)(t)}{T_l(y)(t)} >0.$ Now evaluate (44) at t and simplify to obtain $(k+1)V_{x+y}(t)=kV_x(t)+V_y(t).$ Since $|V_{x+y}(t)|=|V_x(t)|=|V_y(t)|=1,$ we have $k+1=|kV_x(t)+V_y(t)|$ and therefore $V_x(t)=V_y(t).$ This proves that $\chi _{T(x)}V_y=\chi _{T(y)}V_x.$

Evidently $\chi _T=sup\left\{ \chi _{T(x)}:x\in X_+\right\} .$ The set of $t\in Q(Y)$ such that T(x)(t) is finite and non-zero for some $x\in X_+$ forms a dense open subset of the support of the range of T in Q(Y). For any such $t\in Q(Y)$ and $x\in X_+$ define $V(t)=V_x(t)$ and whenever $\chi _T(t)=0$ for some $t\in Q(Y)$ define $V(t)=0$. Since Q(Y) is a stonean compact Hausdorff space and V as defined is continuous and bounded on a dense open subset of Q(Y), we can extend it to a unique $V\in C(Q(Y)).$ Therefore there exists a unique $V\in Z({\widehat{Y}})$ such that $T(x)=VT_l(x)$ for all $x\in X_+$ and $|V|=\chi _T.$ This means $T=VT_l$ as claimed and $(1)\Rightarrow (3).$

On the other hand, if $T=VT_l$ as in (3), since $T_l$ and V are both disjointness preserving, we have that T is disjointness preserving. That is $(3)\Rightarrow (1)$ trivially. $\square$

We note the following interesting consequence of the above result.

Corollary 7.2

Suppose X and Y are complex Banach lattices and $T_r:X_r\rightarrow Y_r$ is a lattice homomorphism. Then its complexification $T:X\rightarrow Y,$ defined by $T(x+iy)=T_r(x)+iT_r(y)$ for all $x,y\in X_r,$ is a lattice homomorphism.

Proof

T is disjointness preserving. Namely, suppose $|x+iy|\bot |u+iv|$ for some $x,y,u,v\in X_r$. Since $max\left\{ |x|,|y|\right\} \le |x+iy|$ and $max\left\{ |u|,|v|\right\} \le |u+iv|$, $T_r(x)$ and $T_r(y)$ are disjoint from each of $T_r(u)$ and $T_r(v)$. That is, $|T_r(x)|+|T_r(y)|$ is disjoint from $|T_r(u)|+|T_r(v)|$. Hence $|T(x+iy)|\bot |T(u+iv)|$.

Then by part (3) of Theorem 7.1, $T=VT_l$ for some lattice homomorphism $T_l$, where $|V|=\chi _T.$ But for each $x\in X_+$, we have $T_l(x)=|T(x)|=|T_r(x)|=T_r(x)$. This means that $T=T_l,$ and T is a lattice homomorphism. $\square$

Next result is another application of Theorem 7.1. In the case of Dedekind complete real Banach lattices it is due to Hart [7]. A result due to Luxemburg and Schep [22] and to Lipecki [20] shows that a lattice homomorphism on real Banach lattices may be extended to a lattice homomorphism on their Dedekind completions. It allows us to extend Hart’s result to arbitrary Banach lattices (see Theorem 7.4 below).

Remark 7.3

Indeed, an analog of Hart’s result is true even in the case of arbitrary Riesz spaces (see [1, Theorem 3.4 and Theorem 6.7]), but the corresponding proof in [1] is rather complicated, long, and some details are compressed. For the sake of completeness and the reader’s convenience we decided to provide an independent and much simpler proof in the case of Banach lattices, which is sufficient for our purposes in the current paper.

In the statement of the theorem below, $Y_T$ denotes the closed ideal generated by the range of the operator T in the (complex) Banach lattice Y.

Theorem 7.4

Suppose X, Y are Banach lattices and $T:X\rightarrow Y$ is a disjointness preserving operator. Then there exists an algebra homomorphism $\gamma :Z(X)\rightarrow Z(Y_T)$ such that

$$\begin{aligned} T(fx)=\gamma (f)T(x) \end{aligned}$$

for all $f\in Z(X)$ and $x\in X.$

Proof

Suppose that $T:X\rightarrow Y$ is a lattice homomorphism and X, Y are Dedekind complete. Let $e\in Z(X)$ be an idempotent and $x\in X_+.$ Then $0\le T(ex)\le T(x)$ implies that there exisxts a minimal ${\widehat{e}} _x\in Z(Y)$ (i.e. $\chi _{T(x)} {\widehat{e}} _x={\widehat{e}} _x$) such that ${\widehat{e}} _xT(x)=T(ex).$ Similarly there exists a minimal $\widehat{(1-e)} _x\in Z(Y)$ such that $\widehat{(1-e)} _xT(x)=T((1-e)x).$ But, since $T(ex)\bot T((1-e)x)$, we have ${\widehat{e}} _x+\widehat{(1-e)} _x=\chi _{T(x)}$ and ${\widehat{e}} _x\bot \widehat{(1-e)} _x.$ That is ${\widehat{e}} _x$ is a minimal idempotent. If $0\le y \le x$, then that $0\le T(y)\le T(x),$ implies that ${\widehat{e}} _y\le {\widehat{e}} _x.$ Now define an idempotent $\gamma (e)\in Z(Y)$ by $\gamma (e)=sup\left\{ {\widehat{e}} _x: x\in X_+\right\}$. Then $\chi _{T(x)}\gamma (e)={\widehat{e}} _x$ and $T(ex)=\gamma (e)T(x)$ for all $x\in X_+.$ By linearity, we have $T(ex)=\gamma (e)T(x)$ for all $x\in X.$ As defined $\gamma (e)\in Z(Y)$ is unique. So, given idempotents $e_1, e_2\in Z(X)$, such that $T(e_1(e_2x))=\gamma (e_1)\gamma (e_2)T(x)$ for all $x\in X$, implies that $\gamma (e_1e_2)=\gamma (e_1)\gamma (e_2).$ Let $sp(\mathbf{B }_X)$ denote the linear span of the band projections in Z(X). Then it is immediate that $\gamma :sp(\mathbf{B }_X)\rightarrow sp(\mathbf{B }_Y)$ is a well-defined algebra homomorphism with $\left\| \gamma \right\| =1$ where $\gamma (1)=\chi _T$ and $\gamma (\sum \lambda _i e_i)=\sum \lambda _i \gamma (e_i)$ for any finite set of $e_i\in \mathbf{B }_X$ and $\lambda _i \in \mathbf{C }$ with $i=1,2,...n$. Since Q(X) is extremally disconnected, $sp(\mathbf{B }_X)$ is dense in $Z(X)=C(Q(X))$ in the sup-norm. Using density and continuity, we extend to an algebra homomorphism $\gamma :Z(X)\rightarrow Z(Y)$ such that $\gamma (1)=\chi _T$ and

$$\begin{aligned} T(fx)=\gamma (f)T(x) \end{aligned}$$

for all $f\in Z(X)$ and $x\in X.$

Returning to the general case, suppose $T:X\rightarrow Y$ is a disjointness preserving operator. By Theorem 7.1 part (3), we have $T=VT_l$ where $T_l:X\rightarrow Y$ is a lattice homomorphism and $V\in Z({\widehat{Y}})$ such that $|V|=\chi _T.$ When restricted to $X_r$, $T_l$ is a (real) lattice homomorphism into $Y_r.$ By the above quoted result of Luxemburg and Schep [22] and of Lipecki [20], there is an extension of $(T_l)_{|X_r}$ to $\widehat{T_l}:\widehat{X_r}\rightarrow \widehat{Y_r}$ that is a lattice homomorphism. By Corollary 7.2, its complexification $\widehat{T_l}: {\widehat{X}}\rightarrow {\widehat{Y}}$ is also a lattice homomorphism. Therefore, by the first part of the proof, there exists an algebra homomorphism $\gamma :Z({\widehat{X}})\rightarrow Z({\widehat{Y}})$ such that

$$\begin{aligned} \widehat{T_l}(fx)={\widehat{\gamma }}(f)\widehat{T_l}(x) \end{aligned}$$

for all $f\in Z({\widehat{X}})$ and $x\in {\widehat{X}}.$ Clearly ${\widehat{T}}:{\widehat{X}}\rightarrow {\widehat{Y}}$ defined by ${\widehat{T}}=V\widehat{T_l}$ is a disjointness preserving operator that extends T. Also it satisfies

$$\begin{aligned} {\widehat{T}}(fx)={\widehat{\gamma }}(f){\widehat{T}}(x) \end{aligned}$$

for all $f\in Z({\widehat{X}})$ and $x\in {\widehat{X}}.$ (Note that V commutes with ${\widehat{\gamma }}$) Hence if we let $\gamma ={\widehat{\gamma }}_{|Z(X)}$ and recall $T={\widehat{T}}_{|X}$, we almost obtain the statement of the theorem. The exception is that, for each $f\in Z(X)$, we have $\gamma (f)\in Z({\widehat{Y}})$ instead of $\gamma (f)\in Z(Y_T).$ But for each $f\in Z(X),$ $\gamma (f)$ acts only on the range of the operator T and leaves it invariant. Evidently, if we show that $\gamma (f)$ also leaves invariant the closed ideal generated by the range of T, then we would have $\gamma (f)\in Z(Y_T)$ and the theorem would be proved.

To show this, let us fix $f\in Z(X)$ and $x\in X_+$. Consider the ideal I(T(x)) that is generated by |T(x)| in Y. Since $|T(x)|=|T_l(x)|=T_l(x)$, we have $I(T(x))=\left\{ aT_l(x): a\in Z(Y(T(x)))\right\}$. Let us also fix a representation of ${\widehat{Y}}$ in $C_{\infty }(Q(Y))$. For any $a\in Z(Y(T(x)))$, we define a unique function ${\widehat{a}}\in C(Q(Y))=Z({\widehat{Y}})$ as follows. If $\widehat{T_l (x)}(t)$ is finite and non-zero for some $t\in Q(Y)$ with $\chi _{T(x)}=1$, let ${\widehat{a}}(t)=\widehat{aT_l (x)}(t)/\widehat{T_l (x)}(t)$. If $\chi _{T(x)}(t)=0$ for some $t\in Q(Y)$, let ${\widehat{a}}(t)=0$. Hence ${\widehat{a}}$ is defined as a bounded continuous function on a dense open subset of the stonean compact set Q(Y). So there is a unique extension ${\widehat{a}}\in C(Q(Y))=Z({\widehat{Y}})$. Since $\widehat{aT(x)}={\widehat{a}}\widehat{T(x)}$, we have

$$\begin{aligned} \widehat{\gamma (f)(aT(x))} = \widehat{\gamma (f)}{\widehat{a}}\widehat{T(x)}={\widehat{a}}\widehat{\gamma (f)}\widehat{T(x)}=\widehat{aT(fx)}. \end{aligned}$$

That is $\gamma (f)(aT(x))=aT(fx) \in I(T(x))$, since $T(fx) \in I(T(x))$ by part 2 of Theorem 7.1. Hence, the right hand side of the last equality is in $Y_T$. This means that $\gamma (f)(I(T(x))\subset Y_T$. But since $T_l$ is positive and additive on $X_+$, it is immediate that $Y_T=cl(\cup \left\{ I(T(x)):x\in X_+\right\} )$. Therefore $\gamma (f)(Y_T)\subset Y_T$. This completes the proof. $\square$

In the rest of this section we will be reviewing the concept of disjointness in Banach C(K)-modules and the basic properties of disjointness preserving operators on Banach C(K)-modules. The general conditions we assume on Banach C(K)-modules are stated in Sect. 2.2 of this article. The only exception is that we will not assume that m(C(K)) is weak-operator closed in L(X), since it is not necessary for the concepts we will discuss here.

First we will review the notion of disjointness of elements in a Banach C(K)-module. We will define a pair of elements $x,y\in X$ to be disjoint, if there exists some $u\in X$ such that $x,y\in X(u)$ and $x\bot y$ in X(u) when the cyclic space X(u) is represented as a Banach lattice as explained in Sect. 2.2 ( see also [16, Lemma 2.4]). This is equivalent to the definition given in [1, Definition 4.4, p.21]. But to show this, we need to introduce the following object defined in [1, Definition 4.1, p.20]. Namely, for any $x\in X$, we have

$$\begin{aligned} \Delta (x)=cl\left\{ fx: f\in C(K), \ \left\| f\right\| \le 1\right\} \end{aligned}$$

where ’cl’ is norm-closure in X. (As remarked in [1], the above defined set $\Delta (x)$ represents an analog for Banach C(K)-modules of the notion of an order interval in a Banach lattice.) According to Definition 4,4 in [1], $x,y\in X$ are called disjoint (xdy) if $\Delta (x+y)=\Delta (x)+\Delta (y)$ and $\Delta (x)\cap \Delta (y)=\left\{ 0\right\} .$

First we want to show that if $x,u\in X$ and $x\in X(u)$, then indeed $\Delta (x)$ is the closed order interval generated by |x| in the complex Banach lattice X(u). That is, $y\in X(u)$ and $|y|\le |x|$ if and only if there exists $\left\{ f_n\right\}$ in C(K) with $\left\| f_n\right\| \le 1$ such that $f_nx\rightarrow y$ in norm. To see this note that $|y|\le |x|$ implies y is in the ideal generated by |x| in X(u). But the closed ideal generated by |x| in X(u) is X(x). Assume without loss of generality that $\left\| x\right\| =1.$ Then parts (4) and (5) of Lemma 2.4 in [16] applied to X(x) imply the existence of the required sequence. The other direction is self-evident.

Suppose that xdy for some $x,y\in X$ as in Definition 4.4 in [1]. Then $x,y\in \Delta (x+y)$ implies that $x,y\in X(u)$ with $u=x+y.$ Also that $\Delta (x)\cap \Delta (y)=\left\{ 0\right\}$ implies that the order intervals generated $|x|, \ |y|$ in X(u) are disjoint, that is $|x|\wedge |y|=0.$ Hence $x\bot y$ in X(u). Conversely, suppose for some $x,y\in X(u)$ we have $x\bot y$ in the Banach lattice X(u). Then $\Delta (x)\cap \Delta (y)=\left\{ 0\right\}$ and $|x+y|=|x|+|y|.$ Since $f,g\in C(K)$ are in the unit ball of C(K) implies that $|fx+gy|\le |x|+|y|=|x+y|$, we obtain $\Delta (x) +\Delta (y)=\Delta (x+y).$

Now we want to show that the definition of a disjointness preserving operator that we give here (Definition 2.6) is equivalent to the one given in [1, Definition 4.5, p.21].

Let X be a Banach $C(K_X)$-module and Y be a Banach $C(K_Y)$-module. Let $T:X\rightarrow Y$ be an operator. We say (Definition 2.6) T is a disjointness preserving operator if, for each $x\in X,$ (1) $T(X(x))\subset Y(T(x))$ and (2) T restricted to X(x) is a disjointness preserving operator when X(x) and Y(T(x)) are considered as Banach lattices. On the other hand the same operator T is called disjointness preserving in [1, Definition 4.5, p.21], if for all $x,y\in X$ with xdy, we have T(x)dT(y) in Y. It is easy to see if the operator T satisfies Definition 2.6 above, then it also satisfies Definition 4.5 in [1]. To show this assume that xdy in X. Then $x\bot y$ in $X(x+y).$ Therefore, by Definition 2.6, $T(x)\bot T(y)$ in $Y(T(x+y))$. But $T(x+y)=T(x)+T(y)$, so T(x)dT(y) in Y. Hence T satisfies Definition 4.5 in [1]. But proving the converse implication, requires some work.

We need two lemmas from [1]. Initially, given $x\in X$ and $x^*\in X'$ define $x^*\Box x\in C(K_X)'$ by

$$\begin{aligned} x^*\Box x(f)=x^*(fx) \end{aligned}$$

for all $f\in C(K_X)$ (see [1, Definition 4.7, p.23]). The lemma below is from [1, Lemma 4.8, part(3), p.23].

Lemma 7.5

Let $x,y\in X.$ Then the following are equivalent.

• (1) $y\in X(x)$ and $|y|\le x$ in the Banach lattice X(x).

• (2) For each $x^*\in X'$, $\left\| x^*\Box y\right\| \le \left\| x^*\Box x\right\|$.

• (3) For each $x^*\in X'$, $|x^*\Box y|\le |x^*\Box x|$.

Proof

$(1)\Rightarrow (3)$ is almost immediate. Assume (1). Consider the complex Banach lattice $C(K_X)'$ as a module over its center $Z(C(K_X)')=C(K_X)'^{\prime }$. Given any $f\in C(K_X),$ it follows from definitions that $x^*\Box fx=f\cdot (x^*\Box x)$ where, on the right hand side, we consider f acting as a central operator on $x^*\Box x$. Then $|x^*\Box fx|\le \left\| f\right\| |x^*\Box x|$. If $|y|\le x$ in X(x), there exists a sequence $\left\{ f_n\right\}$ in the unit ball of $C(K_X)$ such that $f_nx\rightarrow y$ in norm. (See the discussion on $\Delta (x).$)Then $|x^*\Box y|\le |x^*\Box x|$ for all $x^*\in X'.$

That $(3)\Rightarrow (2)$ is evident.

To show $(2)\Rightarrow (1)$ : suppose $y\notin \Delta (x)$. Since $\Delta (x)$ is an absolutely convex closed set in X, by standard separation theorems on real Banach spaces there exists a bounded real linear functional $x^*_r$ on X such that

$$\begin{aligned} sup\left\{ |x^*_r(z)|:z\in \Delta (x)\right\}<1<x^*(y). \end{aligned}$$

Then, as standard, define $x^*(z)=x^*_r(z)-ix^*_r(iz)$ for each $z\in X.$ We know that $x^*\in X'$ and $\left\| x^*\right\| =\left\| x^*_r\right\|$. For each f in the unit ball of $C(K_X)$ and for some real $\theta$, we have

$$\begin{aligned} |x^*(fx)|=e^{i\theta }x^*(fx)=x^*(e^{i\theta }fx)=x^*_r(e^{i\theta }fx)<1<x^*_r(y)\le |x^*(y)|. \end{aligned}$$

(Note that $e^{i\theta }fx\in \Delta (x).$) Hence $\left\| x^*\Box x\right\| <\left\| x^*\Box y\right\|$ and $(2)\Rightarrow (1)$. $\square$

The next lemma is also from [1, Lemma 4.8, part(4), p.23].

Lemma 7.6

Let $x,y\in X$. Then the following are equivalent.

• (1) $x,y\in X(x+y)$ and $x\bot y$ in $X(x+y).$

• (2) For each $x^*\in X'$, $x^*\Box x\bot x^*\Box y$ in $C(K_X)'.$

Proof

To see $(1)\Rightarrow (2)$: Note $x\bot y$ implies $|x|\le |x+\lambda y|$ for all $\lambda \in \mathbf{C }$ in $X(x+y)$. Then, by Lemma 7.5 part(3), for each $x^*\in X'$, we have $|x^*\Box x|\le |x^*\Box (x+\lambda y)|$ for all $\lambda \in \mathbf{C }$. Hence $x^*\Box x\bot x^*\Box y$ in $C(K_X)'.$

To see $(2)\Rightarrow (1)$: Note that (2) implies that for each $x^*\in X'$, $|x^*\Box x|\le |x^*\Box (x+\lambda y)|$ for all $\lambda \in \mathbf{C }$. When $\lambda =1$, by Lemma 7.5 part(1), we have that x is, and therefore also y is in $X(x+y)$. Then, the full inequality, again by Lemma 7.5 part(1), yields $|x|\le |x+\lambda y|$ for all $\lambda \in \mathbf{C }.$ Hence $x\bot y$ in $X(x+y).$ This completes the proof. $\square$

Our next result is the claimed equivalence between Definition 2.6 of this paper and Definition 4.5 in [1]. The interested reader should see [1, Theorem 6.6, p.37] for the idea behind the proof. Given an operator $T:X\rightarrow Y$ between Banach modules X, Y by the notation $T_{|X(x)}$, we denote the restriction of T to the cyclic subspace X(x) of X.

Theorem 7.7

Suppose X, Y are Banach $C(K_X)$ and Banach $C(K_Y)$ modules respectively. Let $T:X\rightarrow Y$ be an operator. Then the following are equivalent.

• (1) For any $x,y\in X,$ xdy in X implies T(x)dT(y) in Y.

• (2) For each $x\in X,$ $T(X(x))\subset Y(T(x))$ and $T_{|X(x)}:X(x)\rightarrow Y(T(x))$ is a disjointness preserving operator.

Proof

We already showed $(2)\Rightarrow (1)$ when we introduced the two definitions. To see $(1)\Rightarrow (2)$: Fix an $x\in X$ and take any $y^*\in Y'.$ Define an operator $U:X(x)\rightarrow C(K_Y)'$ by

$$\begin{aligned} U(z)=y^*\Box T(z), \end{aligned}$$

for all $z\in X(x).$ If $y\bot z$ in X(x), then $y\bot z$ in $X(y+z)$($\subset X(x)$). By (1), we have $T(y)\bot T(z)$ in $Y(T(y)+T(z))$. Then Lemma 7.6 part(2) gives $y^*\Box T(y) \bot y^*\Box T(z)$ in $C(K_Y)'.$ So U is a disjointness preserving operator between the given Banach lattices. By Theorem 7.1 part(2), we have $|y|\le |z|$ in X(x) implies $|y^*\Box T(y)|\le |y^*\Box T(z)|$ in $C(K_Y)'.$ Since $y^*\in Y'$ is arbitrary, by Lemma 7.5 part(1) we have

$$\begin{aligned} T(y)\in Y(T(z)) \; \text {and} \, |T(y)|\le |T(z)|. \end{aligned}$$

When we take $z=x$ in the above inequality, we get $T(X(x))\subset Y(T(x)).$ On the other hand, the displayed inequality shows that the operator $T_{|X(x)}:X(x)\rightarrow Y(T(x))$ satisfies Theorem 7.1 part(2). Therefore $T_{|X(x)}$ is a disjointness preserving operator between the given Banach lattices. So $(1)\Rightarrow (2)$ $\square$