DISJOINTNESS PRESERVING

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49(2003), 61–75

DISJOINTNESS PRESERVING

FREDHOLM LINEAR OPERATORS OF C

0

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JYH-SHYANG JEANG and NGAI-CHING WONG

Communicated by S¸erban Str˘atil˘a

Abstract. Let X andY be locally compact Hausdorff spaces. We give a full description of disjointness preserving Fredholm linear operators T from C0(X) intoC0(Y), and show that T is continuous if either Y contains no isolated point or T has closed range. Our task is achieved by writing T as a weighted composition operatorT f =h·f◦ϕ. Through the relative homeomorphismϕ, the structure of the range space ofT can be completely analyzed, andX andY are homeomorphic after removing finite subsets.

Keywords:Auto-continuity, Fredholm operators, disjointness preserving operators, weighted composition operators.

MSC (2000):46J10, 47B33, 47A53.

1. INTRODUCTION

A (not necessarily bounded) linear operator S from a Banach space E into a Banach space F is said to be Fredholm if it has finite nullity and finite corank;

i.e. nullity(S) = dim kerS <∞and corank(S) = dimF/ran(S)<∞. Fredholm composition operators betweenL²(µ) spaces (see e.g. [23], [14], [24]) and Hilbert spaces of analytic functions (see e.g. [15], [25]) have been well studied and proven to have many applications.

Let X and Y be locally compact Hausdorff spaces. Let C0(X) and C0(Y) be Banach spaces of continuous (real- or complex-valued) functions defined onX and Y vanishing at infinity, respectively. A linear operator T :C0(X)→ C0(Y) is disjointness preserving or separatingif T f·T g = 0 whenever f ·g = 0. If, in addition,T is bounded thenT is a (weighted) composition operatorsT f=h·f◦ϕ (see Theorem 2.4).

In this paper, we shall give a full description of the structure of disjointness preserving Fredholm operatorsT :C0(X)→C0(Y) (Theorem 3.14). When such an operator exists, X and Y are homeomorphic after removing finite subsets.

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Moreover, T is bounded if either Y contains no isolated point or T has closed range. These extend the well-known fact that if T is bijective then X and Y are homeomorphic and T is automatically bounded (see [19], [11], [20]). As an application, the information about the range space of T, a finite co-dimensional subspace of C₀(Y), given in Theorem 3.14 is utilized to give a Gleason-Kahane- Zelazko type result (Corollary 4.2). Finally, we remark that our results are very useful in the investigation of shift operators on continuous function spaces [7], [21], which is a subject attracts increasing interests from researchers recently (see e.g.

[8], [17], [13], [9], [16], [26], [4]).

2. PRELIMINARIES

LetX_∞(respectivelyY_∞) be the one-point compactificationX∪{∞}(respectively Y ∪ {∞}) of a locally compact Hausdorff spaceX (respectivelyY). We note that

∞ is an isolated point in X∞ if and only ifX is compact. For each y in Y, let δ_y denote the point evaluation aty, that is, δ_y is the linear functional of C₀(Y) defined byδ_y(g) =g(y).

We begin with the following two elementary observations. The first of them enables us to assume freely that the underlying fieldKis the complex scalars C, while the second suggests us a way to look into the problem of automatic continuity of a linear operator betweenC0(X) spaces.

Lemma 2.1. Let Tr be a real linear operator from the real Banach space C0(X,R) intoC0(Y,R). Let Tc :C0(X,C)→C0(Y,C) be the complexification of T_r defined by

Tc(f1+ if2) =Trf1+ iTrf2, f1, f2∈C0(X,R).

Then, we have:

(i)Tris bounded if and only if Tc is bounded;

(ii) T_r has closed range if and only if T_c has closed range;

(iii) nullity(T_r) = nullity(T_c)andcorank(T_r) = corank(T_c);

(iv)Tr is disjointness preserving if and only if Tc is disjointness preserving.

Proof. Most of the arguments are straightforward. We just mention that if f = f₁+ if₂, g = g₁+ ig₂ with f₁, f₂, g₁ and g₂ in C₀(X,R) then f ·g = 0 is equivalent tof_j·g_k= 0 forj, k= 1,2.

Lemma 2.2. Let T be a linear operator fromC₀(X)intoC₀(Y). Then T is bounded if and only ifδy◦T is bounded for ally inY.

Proof. It is an easy consequence of the Uniform Boundedness Principle. Al- ternatively, one can make use of the Closed Graph Theorem.

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Definition 2.3. In view of Lemma 2.2, we divide Y into three disjoint parts, the nullity part Y0, thecontinuous part Yc and the discontinuous partYd, where

Y0=

y∈Y :δy◦T ≡0 , Yc=

y∈Y :δy◦T is nonzero and continuous and

Y_d=

y∈Y :δ_y◦T is discontinuous . Accordingly,T is bounded if and only ifYd=∅.

Theorem 2.4. ([19], [20]; see also [1], [10]) Let T be a disjointness preserving linear operator from C₀(X)intoC₀(Y). Then:

(i)Y0 is closed andYd is open;

(ii) a unique continuous mapϕfromY_c∪Y_d intoX_∞ exists such that ϕ(y)6∈supp(f)⇒T(f)(y) = 0, ∀f ∈C₀(X);

(iii)ϕ(Y_c)⊆X andϕ(Y_d)is a finite set of non-isolated points;

(iv)a unique continuous non-vanishing scalar function h on Yc exists such that

T f_|Y_c=h·f◦ϕ, T f_|Y₀ ≡0.

Theorem 2.4 can be improved ifTis also Fredholm. Notations in Theorem 2.4 will be used throughout this paper. Recall that a bounded linear operatorT from a Banach spaceEinto a Banach spaceF is called aninjectionif there is anr >0 such that kT xk >rkxk. It follows from the Open Mapping Theorem that T is an injection if and only if T is injective and has closed range. See [2] for more information.

Lemma 2.5. Let T f(y) = h(y)f(ϕ(y)) be a weighted composition operator from C0(X) into C0(Y). Here, h is a continuous non-vanishing scalar-valued function onY andϕis a continuous map fromY intoX. ThenT is continuous.

If, in addition,T has closed range then there exist positive constantsrandRsuch that

(2.1) 0< r6 sup

y∈ϕ⁻¹({x})

h(y)

6R for allxinϕ(Y).

Proof. Sinceδy◦T =h(y)·δ_ϕ(y) for ally in Y, it follows from Lemma 2.2 thatT is continuous. Moreover, ifT is injective and has closed range thenT is an injection. So there are constants r, R >0 such that rkfk 6kT fk6Rkfk. It is obvious that sup

y∈ϕ⁻¹({x})

h(y)

6Rfor allxinϕ(Y). On the other hand, for eachx₀ inϕ(Y) letU andV be open neighborhoods ofx0withU ⊆V. Let 06fU V 61 in C0(X) satisfy the conditions thatfU V|U = 1 =kfU Vk, and fU V(x) = 0 ifx6∈V. Then

r=r fU V

6

T fU V

= sup

y∈Y

T fU V(y) = sup

y∈Y

h(y)fU V(ϕ(y))

6 sup

y∈ϕ⁻¹(V)

h(y) .

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Therefore, we are able to choose a net{yλ}from Y andε >0 such that ϕ(y_λ)→x₀ and

h(y_λ)

> r−ε >0.

By passing to a subnet if necessary, we can assume that {yλ} converges in Y_∞. Since

T f_{U V}(y_λ) =

h(y_λ)

> r−ε >0 eventually for all neighborhoodsU andV ofx₀ withU ⊆V, we have {y_λ} converges to some y₀ 6=∞. Clearly,ϕ(y₀) =x₀ and we have

r−ε6 h(y0)

6 sup

y∈ϕ⁻¹({x0})

h(y) .

Sinceεcan be arbitrary small, the desired inequality follows.

Finally, suppose that T has closed range but not necessarily injective. Let ϕ(Y) be the closure of ϕ(Y) in X. Consider the disjointness preserving linear operator Te : C0(ϕ(Y)) → C0(Y) defined by Tefe(y) = T f(y) = h(y)·f(ϕ(y)), wheref is any Tietze extension inC0(X) offe. It is plain thatTe is injective and ran(Te) = ran(T) is closed. The desired assertion thus follows from the first part of the proof.

Remark 2.6. Note that “sup” cannot be dropped in (2.1). We would like to thank Martin Stanev for providing us a counterexample for this. Lemma 2.5 fixes a related bug in Proposition 4 of [20].

The following statement is a consequence of the Open Mapping Theorem.

Its proof can be found in, for example, 28A of [5].

Proposition 2.7. Let T be a bounded linear operator from a Banach space E into a Banach spaceF with finite corank. Then T has closed range.

3. MAIN RESULTS

In the following lemmas, we always assume that T is a disjointness preserving linear operator fromC₀(X) intoC₀(Y).

Lemma 3.1. Let T have finite corank. Then there exist positive scalars r andR such that

0< r6 sup

y∈ϕ⁻¹({x})∩Yc

|h(y)|6R for allxin ϕ(Yc).

Proof. Since T has finite corank, there exist g1, . . . , gn in C0(Y) such that C0(Y) is the linear span ofg1, . . . , gn and ran(T). So, for everyg inC0(Y), there are scalarsλ₁, . . . , λ_n and anf inC₀(X) such thatg=λ₁g₁+· · ·+λ_ng_n+T f.

Let T⁰ : C0(X) → C0(Yc ∪Y0) be the linear operator defined by T⁰f = T f_|Y_c_∪Y₀. Then T⁰ is continuous by Lemma 2.2. Since Y_d is an open set in Y, Y_c∪Y₀∪ {∞} = Y_∞ \Y_d is closed in Y_∞. By Tietze’s extension theorem, every continuous function g⁰ in C0(Yc ∪Y0) can be extended to a g in C0(Y).

Consequently,

g⁰=g_|Y_c_∪Y₀ =λ₁g_1|Y

c∪Y₀+· · ·+λ_ng_n|Y

c∪Y₀+T f_|Y_c_∪Y₀

=λ₁g_1|Y

c∪Y₀+· · ·+λ_ng_n|Y

c∪Y₀+T⁰f,

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for some scalars λ1, . . . , λn and an f in C0(X). This shows that T⁰ has finite corank. By Proposition 2.7, ran(T⁰) is closed. SinceT⁰f_|Y₀ ≡0, the induced map Te : C₀(X) → C₀(Y_c) defined by T fe = T⁰f_|Y_c has closed range as well. Then, Lemma 2.5 applies.

Lemma 3.2. LetT have finite nullitym. Thenϕ(Y_c)∩X =ϕ(Y_c∪Y_d)∩X, and

X\ϕ(Yc) ={x1, . . . , xm}

consisting of exactly m isolated points, where the closure is taken in X∞. More- over,

kerT = span{χ_{x₁_}, χ_{x₂_}, . . . , χ_{x_m_}},

whereχ_{x_i_} is the characteristic functions of {xi}fori= 1,2, . . . , m.

Proof. Suppose that there were distinct points x1, x2, . . . , xm+1 in X \ ϕ(Yc∪Yd). Let V1, V2, . . . , Vm+1 be disjoint compact neighborhoods ofx1, x2, . . . . . . , x_m+1inX\ϕ(Yc∪Y_d), respectively. For eachi= 1,2, . . . , m+1, let 06f_i61 in C0(X) satisfy that fi(xi) = 1 andfi = 0 outsideVi. Then ϕ(y) 6∈ supp(fi), and thus T fi(y) = 0, for all y in Yc∪Yd by Theorem 2.4. Note that T f vanishes on Y₀ for all f in C₀(X). Hence f_i ∈ kerT. Since

f₁, f₂, . . . , f_m+1 is linearly independent, we have dim(kerT)>m+ 1, a contradiction. So the open setX\ϕ(Yc∪Yd) ={x1, x2, . . . , xk} consists of isolated points, andk6m.

As a finite set,ϕ(Y_d) is closed (Theorem 2.4). Therefore, (3.1) X\ϕ(Yc)⊆(X\ϕ(Yc∪Yd))∪ϕ(Yd).

Since bothX\ϕ(Y_c∪Y_d) and ϕ(Y_d) are finite, X\ϕ(Y_c) is a finite open subset ofX. This implies thatX\ϕ(Yc) consists of isolated points. Sinceϕ(Yd) contains only non-isolated points inX_∞ (Theorem 2.4),ϕ(Y_d)∩X ⊆ϕ(Y_c), and thus (3.2) X\ϕ(Yc) =X\ϕ(Yc∪Yd) ={x1, x2, . . . , xk}

by (3.1). Consequently,ϕ(Y_c)∩X=ϕ(Y_c∪Y_d)∩X.

Finally, we prove thatX\ϕ(Yc) consists of exactlymisolated points whose characteristic functions span kerT. Since T f = h·f ◦ϕ on Y_c and h is non- vanishing, we havef_|ϕ(Y

c)= 0 iff ∈kerT. Conversely, iff vanishes onϕ(Yc) then f =

k

P

i=1

λiχ_{x_i_}. Therefore, supp(f)∩ϕ(Yd) ={x1, . . . , xk} ∩ϕ(Yd) =∅ by (3.2).

By Theorem 2.4 again,T f_|Y_d = 0. AsT f_|Y_c_∪Y₀ = 0, we havef ∈kerT. It follows that kerT = span{χ_{x₁_}, . . . , χ_{x_k_}}. Since{χ_{x_i_}}^k_i=1is linearly independent and the dimension of kerT ism, we havek=m.

Remark 3.3. In the proof of Lemma 3.2, we have shown that T f_|Y_c= 0 impliesT f = 0,

providedT has finite nullity.

The following result ensures that T is bounded whenever Y contains no isolated point.

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Lemma 3.4. Let T be Fredholm. Then:

(i) Y_d consists of finitely many isolated points. In fact, the cardinality of Ydis less than or equal to n, the corank ofT;

(ii)ϕ(Y_c) is closed inX, and

ϕ(Yc) =ϕ(Yc∪Yd)∩X.

Proof. (i) We first claim that if supp(g)⊆Y_d theng 6∈ran(T), unless g = 0. In fact, it is a direct consequence of Remark 3.3. Now, suppose there were distincty1, y2, . . . , yn+1 in Yd. SinceYdis open (Theorem 2.4), there are disjoint neighborhoods V1, V2, . . . , Vn+1 of y1, y2, . . . , yn+1 in Yd, respectively, which are open in Y. Let Ui be a compact neighborhood of yi contained in Vi and let g_i 6= 0 be in C₀(Y) with supp(g_i) ⊆ U_i ⊆ V_i for i = 1,2, . . . , n + 1. Since dim (C0(Y)/ran(T)) = n, there are some not all zero scalars λ1, . . . , λn+1 such that 06=

n+1

P

i=1

λ_ig_i ∈ran(T). But suppⁿ⁺¹ P

i=1

λ_ig_i

⊆

n+1

S

i=1

V_i ⊆ Y_d. Consequently,

n+1

P

i=1

λigi6∈ran(T), a contradiction. Hence the cardinality ofYdis at mostn. Being finite and open,Yd consists of isolated points.

(ii) By Lemma 3.2, it suffices to show thatϕ(Yc) is closed inX. To this end, letx0 ∈ ϕ(Yc)∩X. First, we note that x0 6=∞. If x0 is isolated in ϕ(Yc)∩X then x0 ∈ ϕ(Yc). So we assume there exist yλ in Yc such that ϕ(yλ)6= x0 and ϕ(yλ)→x0. It follows from Lemma 3.1 that we can assume h(yλ) is away from zero. By passing to a subnet, if necessary, we can also assume yλ →y0 in Y_∞. Since all points in Yd are isolated by (i), we havey06∈Yd. Suppose that y0∈Y0

ory₀=∞. We have 0 = lim

λ→∞T f(yλ) = lim

λ→∞h(yλ)f(ϕ(yλ)) ∀f ∈C0(X).

Sinceh(yλ) is away from zero, f(x₀) = lim

λ→∞f(ϕ(y_λ)) = 0 ∀f ∈C₀(X).

This implies x₀ =∞, a contradiction. Soy₀∈Y_c, and thenx₀ =ϕ(y₀)∈ϕ(Y_c).

Henceϕ(Yc) =ϕ(Yc)∩X is closed inX.

Let #(S) denote the cardinality of a setS.

Definition 3.5. We define an equivalence relation∼onYc such thaty∼y⁰ if and only ifϕ(y⁰) =ϕ(y); or equivalently, kerδy◦T = kerδy⁰◦T (Theorem 2.4).

Lety be a point inYc. Denote by [y] the equivalence class inYc represented byy.

We call y amerging point ofT if [y] contains more than one points. In this case, we callϕ(y) themerged point ofT in X for the class [y]. LetM be the set of all merging points ofT and

m(T) =X

{#([y])−1 : [y]∈Y_c/∼}= #(M)−#(ϕ(M)).

We call #(Y_d) the discontinuity index, #(Y₀) the vanishing index andm(T) the merging indexofT, respectively.

Remark 3.6. It is easy to see that ifg∈ran(T) then

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(i)g(y) = 0 for ally in Y0;

(ii)g(y) = 0 for someyin Yc if and only ifg(y⁰) = 0 for ally⁰ in [y].

Lemma 3.7. Let T be continuous and Fredholm and have finite corank n.

Then the sum of the merging and vanishing indices ofT is equal to n, i.e., m(T) + #(Y0) =n.

Proof. Suppose first that the inequality

(3.3) m(T) + #(Y0)6n

does not hold, i.e., there exist y⁽⁰⁾₁ , . . . , y_l⁽⁰⁾

0 in Y0 and merged points x1, . . . , xk

in ϕ(Y_c) with corresponding merging pointsy⁽ⁱ⁾₁ , . . . , y_l⁽ⁱ⁾

i in ϕ⁻¹(x_i)∩Y_c fori = 1, . . . , k, such that

k

X

i=1

(li−1) +l0>n+ 1.

Fori= 0,1,2, . . . , k, letg_j⁽ⁱ⁾be inC0(Y) such thatg_j⁽ⁱ⁾(y_j⁽ⁱ⁾) = 1 andg_j⁽ⁱ⁾(y⁽ⁱ

0) j⁰ ) = 0 wheneveri⁰ 6=i or j⁰ 6=j for 16 j 6li−1 (16j 6l0 when i = 0). Without loss of generality, we can assume all g_j⁽ⁱ⁾’s have disjoint supports. Then we have at leastn+ 1 suchg⁽ⁱ⁾_j in C₀(Y). By Remark 3.6, allg₁⁽⁰⁾, . . . , g_l⁽⁰⁾

0 ,g⁽¹⁾₁ , . . . , g⁽¹⁾_l

1−1, g₁⁽²⁾, . . . , g_l⁽²⁾

2−1, . . . , g₁^(k), . . . , g^(k)_l

k−1 are not in ran(T). Moreover, they are linear independent inC0(Y) modulo ran(T). In fact, ifg=P

λ⁽ⁱ⁾_j g_j⁽ⁱ⁾∈ran(T), we shall show λ⁽ⁱ⁾_j = 0 for all i, j. Note that g(y⁽ⁱ⁾_l

i ) = 0 for 16i6k. Then, by Remark 3.6 again,λ⁽⁰⁾_j =g(y_j⁽⁰⁾) = 0 forj= 1, . . . , l₀, andλ⁽ⁱ⁾_j =g(y_j⁽ⁱ⁾) = 0 for all indices (i, j) with 16i6k and 16j 6l_i−1. So dim (C₀(Y)/ran(T))>n+ 1. This contradiction says that inequality (3.3) holds.

Since T is continuous, Yd = ∅. Let Y0 = {y⁽⁰⁾₁ , . . . , y_l⁽⁰⁾

0 }, and ϕ(M) = {x1, . . . , xk} with ϕ⁻¹(xi)∩Yc = {y⁽ⁱ⁾₁ , . . . , y_l⁽ⁱ⁾

i } for i = 1, . . . , k. Let g_j⁽ⁱ⁾ be defined as above. Let A be the span of g⁽ⁱ⁾_j ’s. As shown in the first paragraph, ran(T)∩ A ={0}. We will show that C0(Y) = ran(T)⊕ A, and thus m(T) +

#(Y₀) = dim(A) =n.

For eachginC0(Y), chooseλ⁽ⁱ⁾_j ’s inCsuch thatg⁰=g−P

i,j

λ⁽ⁱ⁾_j g_j⁽ⁱ⁾satisfying g⁰(y) = 0 for all y in Y₀, and ^g

0(y⁽ⁱ⁾₁ )

h(y⁽ⁱ⁾₁ ) = · · · = ^g

0(y⁽ⁱ⁾

li ) h(y⁽ⁱ⁾

li) for i = 1, . . . , k. Define a scalar-valued function f⁰ by setting f⁰(ϕ(y)) = ^g_h(y)⁰^(y) for all y in Y_c. Then f⁰ is continuous and well-defined onϕ(Yc) which is closed inX by Lemma 3.4 (ii). By Tietze’s Extension Theorem, there is a continuous functionf in C₀(X) such that f(x) =f⁰(x) for allxin ϕ(Yc). Note thatT f_|_Y

0 = 0 andT f(y) =h(y)f(ϕ(y)) = g⁰(y) fory in Yc. SinceYd=∅, we haveg⁰=T f. That is,

g=T f +X

i,j

λ⁽ⁱ⁾_j g⁽ⁱ⁾_j ∈ran(T)⊕ A.

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Lemma 3.8. Let T be Fredholm (but not necessarily continuous) and have finite corankn. Then

(3.4) m(T) + #(Y0) + #(Yd) =n.

Proof. By Lemma 3.4 (i),Yd={y1, . . . , yk}consists ofk6nisolated points.

LetTe:C₀(X)→C₀(Y₀∪Y_c) be a disjointness preserving linear operator defined by

T fe =T f_|Y₀_∪Y_c.

ThenTeis continuous. By Remark 3.3, kerT = kerTe. We claim that Te has finite corank. Sincey1, . . . , yk are isolated points inY, allχ_{y₁_}, . . . , χ_{y_k_} are functions in C0(Y). By Remark 3.3, χ_{y_i_} 6∈ran(T) for all i= 1, . . . , kand they are linear independent. Without lose of generality, we can assume that g1, . . . , gn form a basis ofC₀(Y) modulo ran(T), whereg_i=χ_{y_i_}for alli= 1, . . . , k. In particular, for all g in C₀(Y), g=λ₁g₁+· · ·+λ_ng_n+T f for somef in C₀(X) and scalars λi. Now, for alleg inC0(Y0∪Yc),

eg=g_|Y₀_∪Y_c=λ₁g_1|Y

0∪Y_c+· · ·+λ_ng_n_|Y

0∪Y_c+T f_|Y₀_∪Y_c

=λk+1gk+1|Y0∪Yc+· · ·+λngn|Y0∪Yc+T f,e

where g is any Tietze extension ofeg. This implies corank(Te) 6n−#(Y_d). By Lemma 3.7, we have corank(Te) =m(Te) + #(Y₀) =m(T) + #(Y₀). Hencem(T) +

#(Y0) + #(Yd)6n. The equality (3.4) then follows in the same manner as in the proof of Lemma 3.7.

Lemma 3.9. Let T be Fredholm. Then h is bounded and away from zero, that is, there exist positive constants randR such that

0< r6|h(y)|6R for ally in Yc.

Proof. It follows from Lemma 3.8 that the merging indexm(T) ofT is finite.

By Lemma 3.1, we see that the non-vanishing scalar function h is bounded and away from zero.

Recall that a map ϕ from Y into X is said to be proper if preimages of compact subsets ofX underϕare compact inY. It is obvious that ϕis proper if and only if lim

y→∞ϕ(y) =∞. As a consequence, a proper continuous mapϕfromY ontoX is a quotient map, i.e. ϕ⁻¹(O) is open inY if and only ifO is open inX.

Lemma 3.10. Let T be Fredholm. Thenϕis proper. More precisely, ϕhas a continuous extension from Y_∞ toX_∞ by settingϕ_|Y₀ ≡ ∞ andϕ(∞) =∞. If, in addition,X is compact then the finite set Y0 consists of isolated points.

Proof. It is enough to show that ify_λ ∈ Y_c such that y_λ → p∈Y₀∪ {∞}

thenϕ(yλ)→ ∞inX_∞. Forf inC0(X), we have 0 =T f(p) = limT f(yλ) = lim

λ→∞h(yλ)f(ϕ(yλ)).

Sincehis away from zero by Lemma 3.9, we have lim

λ→∞f(ϕ(y_λ)) = 0, ∀f ∈C₀(X).

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This implies ϕ(yλ) → ∞ in X∞. Finally, we note that ∞ is an isolated point in X_∞ when X is compact. Thus, ϕ⁻¹(∞) is open in Y_∞ in this case. Since Y₀ ⊆ϕ⁻¹{∞} ⊆ Y_d∪Y₀∪ {∞} are all finite (the last inclusion is provided by Theorem 2.4), the open setϕ⁻¹{∞}consists of isolated points.

In the following example, we shall see thatY0may contain non-isolated points whenX is not compact.

Example 3.11. Letc0(respectivelyc) be the Banach space of null (respectively convergent) sequences. In other words,c₀=C₀(N) andc=C(N∞). LetT be the canonical embedding fromc0intoc. In this case,X =N,Y =N∞,Yc=N, Y0 ={∞} and ϕ: N → N is the identity map. We note that ∞ is the unique cluster point inN∞.

With a little more efforts, we have a similar example in whichX =Y andϕ is a homeomorphism.

Example 3.12. LetX be the disjoint union in R²ofI_n⁺={(n, t) : 0< t6 1} and I_n⁻ = {(n, t) : −1 < t < 0} for n = 1,2, . . .. Let p be the point (1,1) and letX1=X\ {p}. Letϕbe the homeomorphism fromX1 ontoX by sending the intervals I₁⁺\ {p} onto I₁⁻, I_n+1⁺ onto I_n⁺, and I_n⁻ ontoI_n+1⁻ in a canonical way for n= 1,2, . . .. Then the corank one disjointness preserving linear isometry T f =f◦ϕ from C0(X) into C0(X) has exactly one vanishing point, i.e., p. We note that p is not an isolated point in X. In a similar manner, one can even construct an example in whichX is connected (by adjoining each I_n^± a common base point, for example).

In view of Theorem 2.4 and Lemmas 3.2, 3.4 and 3.8, the following result implies thatX andY are homeomorphic after removing finite subsets. This ex- tends the well-known fact that if there is a disjointness preserving bijective linear operator between C0(X) and C0(Y) thenX and Y are homeomorphic (see [19], [11], [20]).

Lemma 3.13. Let T be Fredholm. Then ϕ : (Y_c, M) → (ϕ(Y_c), ϕ(M)) is a relative proper homeomorphism. More precisely, ϕ : Yc\M → ϕ(Yc)\ϕ(M) is a proper homeomorphism, and the induced map ϕe: Yc/∼ → ϕ(Yc) is also an homeomorphism, where “∼” is the equivalence relation such that y₁ ∼y₂ if and only ifϕ(y1) =ϕ(y2).

Proof. It follows from Lemma 3.10 that ϕ : Y_c \M → ϕ(Y_c)\ ϕ(M) is bijective, proper and continuous. We claim that ϕ⁻¹ : ϕ(Yc)\ϕ(M) →Yc\M is continuous, i.e., y_λ →y₀ inY_c\M wheneverϕ(y_λ)→ϕ(y₀) inϕ(Y_c)\ϕ(M).

Without loss of generality, we can assume thatyλconverge to a non-isolated point y⁰ in Y∞. Ify⁰ ∈Y0or y⁰ =∞, then

0 = lim

λ→∞T f(y_λ) = lim

λ→∞h(y_λ)f(ϕ(y_λ)), f ∈C₀(X).

Since h is away from zero (Lemma 3.9), we have f(ϕ(y_λ)) → 0 as λ → ∞ for all f in C0(X). This implies that ϕ(y0) = ∞. It is impossible as ϕ(Yc) ⊆ X (Theorem 2.4). By Lemma 3.4 (i), we thus havey⁰∈Yc andϕ(y0) =ϕ(y⁰). Since y0 is not a merging point ofT in Yc, we havey0 =y⁰. Henceϕ⁻¹ is continuous, as asserted.

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Next, we claim thatϕe: Yc/∼ →ϕ(Yc) is an open map. Let Ue be an open set inY_c/∼, which lifts to an open setU inY_c. Sinceϕ:Y_c\M →ϕ(Y_c)\ϕ(M) is a homeomorphism, it suffices to show that ifc∈ϕ(e Ue) for some merged pointc ofT inX, thenc is an interior point ofϕ(eUe). Suppose not, and there werezλ in ϕ(Yc)\ϕ(eUe) such thatzλconverge to the non-isolated pointcinϕ(Yc). Without lose of generality, we can assume that all z_λ’s are not in the finite set ϕ(M). Let yλ=ϕ⁻¹(zλ) inYc. As the equivalence classes [yλ]6∈Ue implyyλ∈/U, there exists a convergent subnetyλα ofyλ in Y∞ such that y⁰ = limyλα ∈/ U. Ify⁰ ∈Yc then y⁰∈/ U implies [y⁰]∈/ U. But,e ϕ([ye ⁰]) =ϕ(y⁰) = lim

λ→∞ϕ(y_λ) = lim

λ→∞z_λ=c∈ϕ(e Ue), a contradiction. It is also plain thaty⁰∈/ Y_dsinceY_dcontains only isolated points by Lemma 3.4 (i). Soy⁰ ∈Y0or y⁰ =∞. Therefore,

0 = limT f(yλ_α) = limh(yλ_α)f(ϕ(yλ_α)).

Sincehis away from zero by Lemma 3.9, we havef(zλ_α) =f(ϕ(yλ_α))→0. Thus, f(c) = 0 for allf inC₀(X). This is a contradiction again. Socis an interior point ofϕ(eUe). This shows thatϕ(eUe) is an open set. Consequently,ϕe:Yc/∼ →ϕ(Yc) is an homeomorphism as asserted.

Now we are ready to state the main result in this paper.

Theorem 3.14. LetT be a disjointness preserving Fredholm linear operator fromC0(X)intoC0(Y) with nullitymand corank n. ThenY is a disjoint union

Y =Yc∪Yd∪Y0,

where the continuous partYc is open, the discontinuous partYdis finite and consists of isolated points, and the nullity part Y₀ is finite (and consists of isolated points when X is compact). There is a unique bounded and away from zero continuous scalar functionhonY_cand a unique continuous mapϕfromY_∞intoX_∞ withϕ(∞) =∞such that:

(i) T f =h·f ◦ϕ onYc andT f vanishes onY0,∀f ∈C0(X).

(ii) ϕ(Y0) ={∞},ϕ(Yc)⊆X andX\ϕ(Yc) ={x1, x2, . . . , xm} consists of misolated points.

(iii) ker(T) = span{χ_{x₁_}, χ_{x₂_}, . . . , χ_{x_m_}} is a closed ideal of C0(X) of dimension m.

(iv)LetM be the finite set of all merging points ofT in Y_c. Then ϕ: (Y_c, M)→(ϕ(Y_c), ϕ(M))

is a relative proper homeomorphism. The induced map

ϕe:Yc/∼ →ϕ(Yc)

is an homeomorphism, wherey∼y⁰ if and only if ϕ(y) =ϕ(y⁰).

(v)n=m(T) + #(Y0) + #(Yd), where the merging index m(T) = #(M)−

#(ϕ(M)).

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(vi)If, in addition,Y contains no isolated point orT has closed range then T is bounded. Moreover, ifT is bounded thenT has closed range

ran(T) =

g∈C₀(Y) :g(p₁) =g(p₂) =· · ·=g(p_k) = 0 and g(a⁽ⁱ⁾₁ )

h(a⁽ⁱ⁾₁ )

= g(a⁽ⁱ⁾₂ ) h(a⁽ⁱ⁾₂ )

=· · ·= g(a⁽ⁱ⁾_l

i ) h(a⁽ⁱ⁾_l

i )

,∀i= 1,2, . . . , j

.

HereY₀={p1, p₂, . . . , p_k}, andϕ(M) ={c1, c₂, . . . , c_j}is the set of merged points ofT inX whenϕ⁻¹(ci) ={a⁽ⁱ⁾₁ , a⁽ⁱ⁾₂ , . . . , a⁽ⁱ⁾_l

i }consists of the corresponding merging points of T inYc fori= 1,2, . . . , j.

Proof. We first note that assertions (i), (ii), (iii), (iv) and (v) are included in previous lemmas. Moreover, the description of the range space ofT in (vi) follows from the continuity of T and Lemma 3.7 (and its proof). When Y contains no isolated points,T is automatically bounded by Lemma 3.4 (i). By Lemma 2.2, it is then enough to verifyY_d=∅ whenT has closed range.

In view of Lemma 3.4 (i), we might suppose, on the contrary, that the open set

Yd={y1, . . . , yl} 6=∅

and l 6n. By Lemma 3.4 (ii), either there exists a y⁰_i in Yc such that ϕ(yi) = ϕ(y⁰_i)6=∞, orϕ(yi) =∞in which case we sety_i⁰=∞, for eachi= 1,2, . . . , l. For those y_i⁰ =∞, we let V_i be any open set in Y_c such that Y_c\V_i is compact and Vi is disjoint from the finite set M. For the others, we let Vi be any open set in Y_c containingϕ⁻¹(ϕ(y_i))∩Y_c. It follows from Lemma 3.13 thatϕ(V_i) is open in ϕ(Yc) =X\ {x1, x2, . . . , xm}, and thus open inX sincex1, x2, . . . , xmare isolated points inX, fori= 1,2, . . . , l.

Claim1. LetginC0(Y) satisfy thatgvanishes on

l

S

i=1

ViandY0, and ^g(z_h(z¹⁾

1) =

g(z₂)

h(z2) wheneverz₁∼z₂in Y_c. Theng∈ran(T) if and only if gvanishes on Y_d. Letg=T f ∈ran(T). Note thatgvanishes on

l

S

i=1

Vi, andhis away from zero by Lemma 3.9. Sof vanishes on the open set

l

S

i=1

ϕ(Vi), and thus,ϕ(yi)6∈supp(f) fori= 1,2, . . . , l. By Theorem 2.4, we haveg(y_i) =T f(y_i) = 0.

On the other hand, ifgvanishes onYdwe definef(ϕ(y)) = _h(y)^g(y),∀y∈Yc, and f(x) = 0, ∀x∈X\ϕ(Y_c). By Lemmas 3.2 and 3.4,f ∈C₀(X) andT f(y) =g(y),

∀y∈Yc∪Y0. Note that T f vanishes on

l

S

i=1

Vi. By an argument similar to above, T f(yi) = 0 for alli= 1,2, . . . , l. Hence g=T f ∈ran(T).

Claim2. Letgbe inC0(Y) such thatgvanishes onYd∪Y0, and _h(z^g(z¹⁾

1)= ^g(z_h(z²⁾

2)

wheneverz1∼z2inYc. Suppose thatg(y) = 0 whenevery∈Yc andϕ(y) =ϕ(yi) for somei= 1,2, . . . , l. Theng∈ran(T).

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For allε > 0, we note that Uε ={y ∈ Y∞ : |g(y)| < ε} is an open subset of Y_∞. We choose some open sets V_i, described as in the paragraph just before Claim 1, such that Vi ⊆U^ε

2 for i= 1,2, . . . , l. Let 06 hε 6 1 be a continuous function inY such thath_ε≡1 outsideU_εandh_ε(y) = 0 foryinU^ε

2. Letg_ε=g·h_ε inC0(Y). Thengεvanishes onU^ε

2 andY0. Since the merging indexm(T) is finite, we can even assume that ^g_h(z^ε^(z¹⁾

1) = ^g_h(z^ε^(z²⁾

2), wheneverz₁ ∼ z₂ in Y_c, if ε is small enough. By Claim 1, we havegε∈ran(T). Clearly, kgε−gk62ε. Since ran(T) is closed, we haveg∈ran(T).

Claim3. Letf be inC0(X) such thatT f(y⁰_i) = 0 for alli= 1,2, . . . , l. Then T f(y_i) = 0 for alli= 1,2, . . . , l.

Note that the assumptionT fvanishes aty⁰_iimplies thatT f(y) =h(y)f(ϕ(y)) also vanishes at all othery inY_c withϕ(y) =ϕ(y_i⁰) =ϕ(y_i) fori= 1,2, . . . , l. De- fine a scalar-valued function g on Y byg(y) =T f(y) fory 6=y1, y2, . . . , yl, and g(y) = 0 fory=y1, y2, . . . , yl. Note thaty1, y2, . . . , ylare isolated points (Lemma 3.4 (i)). So g ∈ C0(Y). By Claim 2, we have g ∈ ran(T), and

l

P

i=1

λiχ_{y_i_} = T f−g ∈ran(T) for some scalarsλi. But

l

P

i=1

λiχ_{y_i_}6∈ran(T) unless all λi’s are zero by Remark 3.3. Therefore,T f(y_i) = 0 for alli= 1,2, . . . , l.

We now define linear operatorsS1, S2:C0(X)→F^l by

S1(f) =



 T f(y1)

... T f(yl)



 and S2(f) =



 T f(y₁⁰)

... T f(y_l⁰)



,

where F is the underlying scalar field. Let A : F^l → F^l be a linear operator satisfying that

A



 T f(y₁⁰)

... T f(y_l⁰)



=



 T f(y1)

... T f(yl)



.

SinceTker(δ_y⁰

i◦T)⊆Tker(δ_y_i◦T) by Claim 3, such a linear operator Aexists.

Moreover,A can be represented as anl×l matrix (a_ij)_l×l, andS₁ =AS₂. As a result,δ_y_i◦T =P

j

a_ij·δ_y⁰

j ◦T for alli= 1,2, . . . , l. Note thaty_j⁰ ∈Y_c∪ {∞} for allj= 1,2, . . . , l. Thus,δy_i◦T is continuous, butyi∈Ydfori= 1,2, . . . , l. This contradiction says thatYd=∅.

As a consequence of Proposition 2.7 and Theorem 3.14, we have

Corollary 3.15. Let T be a disjointness preserving Fredholm linear operator from C0(X) into C0(Y). Then T is bounded if and only if T has closed range.

Remark 3.16. One can find an example of an unbounded disjointness preserving linear operator in [19]. Consequently, there is an unbounded disjointness preserving linear functional%on someC0(X) (Lemma 2.2). LetY =X∪ {y}be a disjoint union. DefineT :C0(X)→C0(Y) by settingT f_|X=f andT f(y) =%(f)

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for eachf inC0(X). Then,T is an injective disjointness preserving linear map of corank 1. But,T is unbounded. It is easy to see thatT does not have closed range.

For example, letfnbe a null sequence inC0(X) such that%(fn) approaches 1. The Cauchy sequence {T fn} does not converge in ran(T). Note also thatY contains an isolated point in this case (cf. Lemma 3.4).

In [3], Araujo showed that an injective disjointness preserving linear operator on a class of continuous vector-valued function spaces is bounded if it has closed range. Compare this with the following result.

Corollary 3.17. Let T be a disjointness preserving linear operator from C₀(X)intoC₀(Y)with finite corank. ThenT is bounded if and only ifT has closed range and the kernel ofT is a closed ideal ofC₀(X). In this case, the conclusions in Theorem3.14, except for possibly(ii)and(iii), are valid forT. Instead, we have ϕ(Yc) is closed inX andkerT ={f ∈C0(X) :f vanishes onϕ(Yc)}.

Proof. By Lemma 2.1, we can assume that the underlying field is the complex scalars. The necessity follows from Theorem 2.4 and Proposition 2.7. For the sufficiency, we note that if kerT is a closed ideal then it must be in the form of {f ∈C₀(X) :f vanishes onXe}for some closed subsetXe ofX. Consequently, the quotient algebra ofC0(X) by kerT is isomorphic toC0(Xe). The induced injective linear operatorTefromC0(X) intoe C0(Y) also has closed range and finite corank.

We shall show thatTeis disjointness preserving. To this end we note that iffeand eg are non-negative functions in C0(X) withe fe·eg = 0, we can extend them to f and g in C0(X) with f ·g = 0, too. In fact, we can set eh =fe−eg and extend it to an h in C0(X) by Tietze’s Extension Theorem. Then the desired disjoint extensions are f = max{h,0} and g = max{−h,0}, respectively. Consequently, Tefe·Teeg=T f·T g= 0 for all non-negativefe,eginC0(X) withe fe·eg= 0. By writing each function in C0(Xe) as a linear sum of at most four non-negative functions, we can conclude that Te is disjointness preserving (cf. the proof of Lemma 2.1).

The other assertions follows from Theorem 3.14. In particular,Xe =ϕ(Y_c) in this case.

4. AN APPLICATION

The results of this paper are important tools in the study of shift operators ([7], [21]). Besides, we provide a supplement to the following interesting Gleason- Kahane-Zelazko type result of Chang-Pao Chen ([6]; see also Jarosz [18]) as an application. Recall that a subspace A of C₀(Y) is said to be an Z_n-subspace if everyg inAhas at leastndistinct zeroes inY. If every (closed) n-codimensional Z_n-subspace of C₀(Y) is of the form T

{kerδ_y_i : i = 1,2, . . . , n} for n distinct pointsy₁, y₂, . . . , y_ninY thenC₀(Y) is said to have the (closed)I_n-property. IfY is compact then the classical Gleason-Kahane-Zelazko Theorem ([12], [22]) states thatC(Y) hasI1-property.

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Theorem 4.1. ([6], Corollary 4.4) Let Y be a locally compact Hausdorff space and letnbe a positive integer greater than1. Then every closedZn-subspace of C0(Y) is the intersection of n maximal ideals of C0(Y) if and only if Y is σ-compact and every point in Y is aGδ set.

Theσ-compactness and Gδ conditions onY ensure that for every pointy in Y_∞ there is anf in C0(Y) vanishing exactly aty (and∞). As a consequence of this fact and Theorem 3.14 (without assuming Theorem 4.1, though), we have

Corollary 4.2. Let Abe an n-codimensional subspace of C0(Y), which is the range of a bounded disjointness preserving linear operatorT. If A is an ideal then A is a closed Zn-subspace. In this case, A is the intersection of nmaximal ideals ofC0(Y). The converse holds if Y is σ-compact and every point inY is a Gδ set.

Proof. In view of the proof of Corollary 3.17, we may assume that T is injective. By Theorem 3.14 (vi), we see thatA= ran(T) is a closed Z_n-subspace of C₀(Y) if ran(T) is an ideal of C₀(Y). In fact, ran(T) is the intersection of n maximal ideals ofC₀(Y) in this case. Conversely, suppose that ran(T) is an Z_n- subspace ofC0(Y). We letgbe a continuous scalar function onY vanishing exactly atp1, p2, . . . , pk and∞. In particular,gdoes not vanish in a neighborhood of each a⁽ⁱ⁾_j in the notation of Theorem 3.14 (vi). Note that there exists an everywhere positive continuous function f in C₀(Y). For each a⁽ⁱ⁾_j , one by one, let f_j⁽ⁱ⁾ be a non-negative continuous function on Y such that f_j⁽ⁱ⁾(a⁽ⁱ⁾_j ) = ¹₂^h(a

(i) j )

g(a⁽ⁱ⁾_j ) and f_j⁽ⁱ⁾ vanishes outside a neighborhood ofa⁽ⁱ⁾_j , which does not contain any ofp1, . . . , pkor othera⁽ⁱ_j0⁰⁾. Replacegbyg·

f_j⁽ⁱ⁾+¹₂^h(a

(i) j ) g(a⁽ⁱ⁾_j )

f f(a⁽ⁱ⁾_j )

, recursively. Theng(p₁) =· · ·= g(pk) = 0 and ^g(a

(i) j )

h(a⁽ⁱ⁾_j ) = 1. In this way, we can redefine g locally at eacha⁽ⁱ⁾_j such that g satisfies the remainingn−k linear equations stated in Theorem 3.14 (vi) without introducing additional zeroes to g. Then g is in the range of T having exactly kzeroes. This implies thatn=kand thus ran(T) is an ideal.

Acknowledgements. Partially supported by Taiwan National Science Council:

NSC 86-2115-M110-002, 37128F.

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JYH-SHYANG JEANG NGAI-CHING WONG

Institute of Mathematics Department of Applied Mathematics Academia Sinica National Sun Yat-sen University Taipei, Taiwan 11529 Kaohsiung, Taiwan 80424

REPUBLIC OF CHINA REPUBLIC OF CHINA

E-mail:[email protected] E-mail:[email protected] Received August 1, 2000.