# Disjointness Preserving and Functional Type Disjointness Preserving Operators

Authors: C. Ganesa Moorthy; C. T. Ramasamy
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

A new concept called functional type disjointness preserving operators is introduced and structure of disjointness preserving and functional type disjointness preserving operators on some function spaces are analysed.


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## Body

## 1. Introduction

There are many articles for construction of unbounded disjointness preserving operators. The paper [1] of Abramovich and Lipecki develops techniques to construct unbounded disjointness preserving linear functionals on any infinite dimensional F-lattice, and this is also mentioned in Remark  2.7 of the paper [2]. This paper [2] is devoted to construct unbounded disjointness preserving linear functionals on function spaces C
0
(
X
), when X is a locally compact Hausdorff space. The paper [3] of Jeang and Wong presents a simplified procedure to construct unbounded disjointness preserving functionals on function spaces C
0
(
X
). Jarosz [4] presents a construction of disjointness preserving operators on C
(
X
), where X is a compact Hausdorff space. Disjointness preserving operators are defined on many structures, namely, lattices, function spaces, algebras, and so forth. The present paper is restricted to study disjointness preserving operators on function spaces. A linear operator T between two function spaces is said to be disjointness preserving if T
f
·
T
g
=
0 whenever f
·
g
=
0. The second section of the paper exercises the standard techniques applicable to derive an essential structure of disjointness preserving operators. This is done on the function spaces C
c
(
X
)
,
C
0
(
X
) for a normal space X, and on 𝒟
(
ℝ
)
,
C
0
∞
(
ℝ
). Although results hold for both real and complex cases, it is assumed that the results on 𝒟
(
ℝ
)
,
C
0
∞
(
ℝ
) are over real fields.For a disjointness preserving linear functionalT we have T
f
=
0 or T
g
=
0, whenever f
·
g
=
0. Jarosz [4] constructed an unbounded disjointness preserving operator T from C
(
X
) onto E, for any given infinite compact Hausdorff spaces X and Y, and for any given linear subspace E of C
(
Y
) such that 1
≤
dim
⁡
E
≤
c (cardinality of continuum). These mappings have the following property: f
·
g
=
0 implies T
f
=
0 or T
g
=
0. So, let us give a new name to this type of mappings.Definition 1.
A linear operatorT from a function space into a function space is said to be a functional type disjointness preserving (FTDP) if f
·
g
=
0 implies T
f
=
0 or T
g
=
0.An essential structure of FTDP-mappings is derived on some function spaces in Section3.For a real or complex valued functionf on X, we define the cozero set of f by coz
(
f
)
=
{
x
∈
X
:
f
(
x
)
≠
0
}, and zero set of f by z
(
f
)
=
{
x
∈
X
:
f
(
x
)
=
0
} and if X is a topological space, then we define the support of f by supp
⁡
f
=
coz
(
f
)
¯, closure of cozero of f. For a nonempty set Y, F
(
Y
) will denote the linear space of all real or complex valued functions defined on Y. To each y
∈
Y, δ
y will denote the evaluation functional defined on F
(
Y
) by δ
y
(
f
)
=
f
(
y
), for every f
∈
F
(
Y
). All the topological spaces to be considered are Hausdorff spaces. For a topological space X, C
(
X
) will denote the linear space of all real or complex valued bounded continuous functions; C
c
(
X
) will denote the linear space of all real or complex valued continuous functions with compact support in X; C
0
(
X
) will denote the linear space C
0
(
X
)
=
{
f
:
X
→
ℝ
or
ℂ
:
f is continuous on X, and for every ϵ
>
0 there is a compact subset K of X such that |
f
(
x
)
|
<
ϵ, for every x
∈
X
∖
K
}. Let us use the usual notation C
∞
(
ℝ
) to denote the linear space of all real valued functions defined on the real line ℝ which are infinitely many times differentiable at every point in ℝ. 𝒟
(
ℝ
) will denote the linear space of all test functions on ℝ (see [5]). That is, 𝒟
(
ℝ
)
=
{
f
∈
C
∞
(
ℝ
)
:
f has compact support in ℝ
}. Let us use the notation C
0
∞
(
ℝ
) to denote the linear space: C
0
∞
(
ℝ
)
=
{
f
∈
C
∞
(
ℝ
)

: for every ϵ
>
0, there is a compact subset K of ℝ such that |
f
(
x
)
|
<
ϵ, for all x
∈
ℝ
∖
K
}. We consider the linear spaces C
(
X
)
,
C
c
(
X
)
,
C
0
(
X
)
,
𝒟
(
ℝ
), and C
0
∞
(
ℝ
) as normed spaces with the supremum norms on them. If X is normal, then C
c
(
X
) is dense in C
0
(
X
). For if f
∈
C
0
(
X
) and ϵ
>
0, then there is a g
∈
C
c
(
X
) such that ∥
g
∥
=
1, 0
≤
g
≤
1, g has value 1 on {
x
∈
X
:
|
f
(
x
)
|
≥
ϵ
}, g has value 0 on {
x
∈
X
:
|
f
(
x
)
|
≤
ϵ
/
2
}, f
·
g
∈
C
c
(
X
), and ∥
f
·
g
-
f
∥
≤
ϵ. Similarly, 𝒟
(
ℝ
) is also dense subspace of C
0
∞
(
ℝ
), which follows from Lemma 1 of Section 1.8 in [6].


## 2. Disjointness Preserving Mappings

The following theorem is a variation of the results obtained by Jarosz [4] and by Jeang and Wong [3].Theorem 2 (see [3, 4]).
LetT be a linear disjointness preserving mapping from C
(
X
) (or C
0
(
X
)) to F
(
Y
) where X is a compact Hausdorff space (or locally compact Hausdorff space X) and Y is a nonempty set. Let
(1)
Y
1
=
{
is a nonzero continuous linear functional
y
∈
Y
:
δ
y
∘
T
i
s
a
n
o
n
z
e
r
o
c
o
n
t
i
n
u
o
u
s
l
i
n
e
a
r
f
u
n
c
t
i
o
n
a
l
}
,
Y
2
=
{
y
∈
Y
:
δ
y
∘
T
i
s
d
i
s
c
o
n
t
i
n
u
o
u
s
}
,
Y
3
=
{
y
∈
Y
:
δ
y
∘
T
=
0
}
.
Then there is a function ϕ
:
Y
1
∪
Y
2
→
X (or ϕ
:
Y
1
∪
Y
2
→
X
∞
,
X
∞
=
X
∪
{
∞
} is the one point compactification of X) such that T
f
(
y
)
=
α
y
f
(
ϕ
(
y
)
) for all f
∈
C
(
X
) (or C
0
(
X
)), for some fixed scalar α
y, and for all y
∈
Y
1 such that δ
y
∘
T
=
0 on {
f
∈
C
(
X
) (or C
0
(
X
)) :

ϕ
(
y
)
∉
supp
⁡
f
}, for all y
∈
Y
1
∪
Y
2. Moreover, if every function in the range of T is a bounded function on Y, then ϕ
(
Y
2
) is a finite subset of X (or X
∞, resp., and ϕ
(
y
)
∈
X
=
X
∞
∖
{
∞
} for y
∈
Y
1).These results depend on the Urysohn lemma and existence of partitions of unity. Let us also use them in deriving the following theorem.Theorem 3.
LetX be a normal space and Y be a nonempty set. Let T be a linear disjointness preserving mapping from C
c
(
X
) into F
(
Y
). Let Y
1, Y
2, and Y
3 be defined as in Theorem 2. Then there is a function ϕ
:
Y
1
∪
Y
2
→
X such that δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on C
c
(
X
) for some fixed scalar α
y, and for all y
∈
Y
1 such that δ
y
∘
T
=
0 on {
f
∈
C
(
X
)
:
ϕ
(
y
)
∉
supp
⁡
f
}, for all y
∈
Y
1
∪
Y
2.Proof.
To eachy
∈
Y, define Φ
(
y
)
=
{
x
∈
X

: for every open neighbourhood U of x, there is an f
∈
C
c
(
X
) such that coz
(
f
)
⊂
U and T
f
(
y
)
≠
0
}. First, we claim that to each y
∈
Y, Φ
(
y
) contains at most one point. On the contrary, suppose x
1
,
x
2
∈
Φ
(
y
) for some y
∈
Y, with x
1
≠
x
2. Let U
1 and U
2 be two disjoint open neighbourhoods of x
1 and x
2 respectively. Then there are f
1
,
f
2
∈
C
c
(
X
) such that T
f
i
(
y
)
≠
0 and coz
(
f
i
)
⊂
U
i, i
=
1,2. This is a contradiction to the assumption that T is disjointness linear preserving. So, Φ
(
y
) contains at most one point, for every y
∈
Y. We next claim that Φ
(
y
) is nonempty for every y
∈
Y
1
∪
Y
2. On the contrary we assume that Φ
(
y
) is empty for some y
∈
Y
1
∪
Y
2. Note that δ
y
∘
T
≠
0. Fix f
∈
C
c
(
X
). Let K
=
supp
⁡
f. To each x
∈
K, let U
x be an open neighbourhood of x such that T
g
(
y
)
=
0 whenever g
∈
C
c
(
X
) and coz
(
g
)
⊂
U
x. Then we can find a finite subfamily {
U
1
,
U
2
,
…
,
U
n
} of {
U
x
:
x
∈
X
} such that the subfamily covers K. Let {
g
1
,
g
2
,
…
,
g
n
}
⊂
C
(
K
) be a continuous decomposition of the identity subordinate to {
U
1
∩
K
,
U
2
∩
K
,
…
,
U
n
∩
K
}. Then, by the Tietze extension theorem, there are functions f
1
,
f
2
,
…
,
f
n in C
(
X
) such that f
i
∣
K
=
g
i and 0
≤
f
i
≤
1, for every i. Then f
·
f
i
∈
C
c
(
X
) and coz
(
f
·
f
i
)
⊂
U
i and hence T
(
f
·
f
i
)
(
y
)
=
0, for every i. So, T
f
(
y
)
=
T
(
∑
i
=
1
n
f
·
f
i
)
(
y
)
=
∑
i
=
1
n
T
(
f
·
f
i
)
(
y
)
=
0. Thus δ
y
∘
T
=
0 on C
c
(
X
), which is a contradiction. Hence Φ
(
y
) is a singleton set for each y
∈
Y
1
∪
Y
2. Define a function ϕ
:
Y
1
∪
Y
2
→
X by {
ϕ
(
y
)
}
=
Φ
(
y
). To each y
∈
Y
1
∪
Y
2, write J
y
=
{
f
∈
C
c
(
X
)
:
ϕ
(
y
)
∉
supp
⁡
f
}, and K
y
=
{
f
∈
C
c
(
X
)
:
f
(
ϕ
(
y
)
)
=
0
}. If y
∈
Y
1
∪
Y
2 and f
∈
J
y, then ϕ
(
y
)
∉
supp
⁡
f and there is a g
∈
C
c
(
X
) with coz
(
g
)
⊂
X
∖
supp
⁡
f such that T
g
(
y
)
≠
0. Since T is disjointness preserving,  T
f
(
y
)
=
0. Thus J
y
⊂
ker
⁡
(
δ
y
∘
T
). Since X is normal, if f
∈
K
y, y
∈
Y
1
∪
Y
2, and ϵ
>
0, then there is a g
∈
C
c
(
X
) such that ∥
g
∥
=
1, 0
≤
g
≤
1, g has value 1 on {
x
∈
X
:
|
f
(
x
)
|
≥
ϵ
}, and g has value 0 on {
x
∈
X
:
|
f
(
x
)
|
≤
ϵ
/
2
}
,
f
·
g
∈
C
c
(
X
)
,
f
·
g
∈
J
y, and ∥
f
·
g
-
f
∥
≤
ϵ. Thus J
y is dense in K
y. If y
∈
Y
1, then ker
⁡
(
δ
y
∘
T
) is closed and hence K
y
⊂
ker
⁡
(
δ
y
∘
T
). So K
y
=
ker
⁡
(
δ
y
∘
T
) and δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on C
c
(
X
), for some α
y. This proves the theorem.Corollary 4.
LetX be a normal space and Y a nonempty set. Let T be a linear disjointness preserving mapping from C
0
(
X
) to F
(
Y
). Suppose δ
y
∘
T is continuous on C
0
(
X
), for every y
∈
Y. Let
(2)
Y
1
=
{
δ
y
∘
T
is
a
nonzero
continuous
linear
functional
y
∈
Y
:
δ
y
∘
T
i
s
a
n
o
n
z
e
r
o
c
o
n
t
i
n
u
o
u
s
l
i
n
e
a
r
f
u
n
c
t
i
o
n
a
l
}
,
Y
2
=
{
y
∈
Y
:
δ
y
∘
T
=
0
}
.
Then there is a function ϕ
:
Y
1
→
X such that δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on C
0
(
X
) for some fixed scalar α
y and for all y
∈
Y
1.Proof.
Consider the restriction ofT to C
c
(
X
). Then we conclude that δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on C
c
(
X
), for some fixed scalar α
y, because δ
y
∘
T
=
0 on C
c
(
X
) if and only if δ
y
∘
T
=
0 on C
0
(
X
). Continuity assumption on δ
y
∘
T also implies that the relation δ
y
∘
T
=
α
y
δ
ϕ
(
y
) is true on C
0
(
X
) also because C
c
(
X
) is dense in C
0
(
X
).
The classical Urysohn lemma and the result on existence of partitions of unity have their version in𝒟
(
ℝ
) (see: [6, Section  1.8 Lemma  1] and [5, Theorem  6.20]). So we have the following version of the previous theorem to the space 𝒟
(
ℝ
).Theorem 5.
LetY be a nonempty set. Let T be a linear disjointness preserving mapping from 𝒟
(
ℝ
) to F
(
Y
). Let Y
1, Y
2, and Y
3 be defined as in Theorem 2. Then there is a function ϕ
:
Y
1
∪
Y
2
→
ℝ such that δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on 𝒟
(
ℝ
) for some fixed scalar α
y and for all y
∈
Y
1 and δ
y
∘
T
=
0 on {
f
∈
𝒟
(
ℝ
)
:
ϕ
(
y
)
∉
supp
⁡
f
}, for all y
∈
Y
1
∪
Y
2.Proof.
To eachy
∈
Y, define Φ
(
y
)
=
{
x
∈
ℝ

: for every open neighbourhood U of x, there is an f
∈
𝒟
(
ℝ
) such that coz
(
f
)
⊂
U and T
f
(
y
)
≠
0
}. It is easy to verify that Φ
(
y
) contains at most one point, for every y
∈
Y. We next claim that Φ
(
y
) is nonempty for every y
∈
Y
1
∪
Y
2. On the contrary we assume that Φ
(
y
) is empty for some y
∈
Y
1
∪
Y
2. Note that δ
y
∘
T
≠
0. Fix f
∈
𝒟
(
ℝ
). Let K
=
supp
⁡
f. To each x
∈
K, let U
x be an open neighbourhood of x such that T
g
(
y
)
=
0 whenever g
∈
𝒟
(
ℝ
) and coz
(
g
)
⊂
U
x. Then we can find a finite subfamily {
U
1
,
U
2
,
…
,
U
n
} of {
U
x
:
x
∈
X
} such that the subfamily covers K. By [5, Theorem  6.20] there are functions f
1
,
f
2
,
…
,
f
n in 𝒟
(
ℝ
) such that 0
≤
f
i
≤
1, for every i, f
1
+
f
2
+
⋯
+
f
n
=
1 on K and supp
⁡
f
i
⊂
U
i, for all i. Then f
·
f
i
∈
𝒟
(
ℝ
) and coz
(
f
·
f
i
)
⊂
U
i and hence T
(
f
·
f
i
)
(
y
)
=
0, for every i. So, T
f
(
y
)
=
T
(
∑
i
=
1
n
f
·
f
i
)
(
y
)
=
∑
i
=
1
n
T
(
f
·
f
i
)
(
y
)
=
0. Thus δ
y
∘
T
=
0 on 𝒟
(
ℝ
), which is a contradiction. Hence Φ
(
y
) is a singleton set for each y
∈
Y
1
∪
Y
2. Define a function ϕ
:
Y
1
∪
Y
2
→
ℝ by {
ϕ
(
y
)
}
=
Φ
(
y
). To each y
∈
Y
1
∪
Y
2, write J
y
=
{
f
∈
𝒟
(
ℝ
)
:
ϕ
(
y
)
∉
supp
⁡
f
}, and K
y
=
{
f
∈
𝒟
(
ℝ
)
:
f
(
ϕ
(
y
)
)
=
0
}. If y
∈
Y
1
∪
Y
2 and f
∈
J
y, then ϕ
(
y
)
∉
supp
⁡
f and there is a g
∈
𝒟
(
ℝ
) with coz
(
g
)
⊂
X
∖
supp
⁡
f such that T
g
(
y
)
≠
0. Since T is disjointness preserving,T
f
(
y
)
=
0. Thus J
y
⊂
ker
⁡
(
δ
y
∘
T
). If f
∈
K
y, y
∈
Y
1
∪
Y
2, and ϵ
>
0, then by [6, Lemma 1 of Section  1.8] there is a g
∈
𝒟
(
ℝ
) such that ∥
g
∥
=
1, 0
≤
g
≤
1, g has value 1 on {
x
∈
ℝ
:
|
f
(
x
)
|
≥
ϵ
}, and g has value 0 on {
x
∈
ℝ
:
|
f
(
x
)
|
≤
ϵ
/
2
}
,
f
·
g
∈
J
y, and ∥
f
·
g
-
f
∥
≤
ϵ. Thus J
y is dense in K
y. If y
∈
Y
1, then ker
⁡
(
δ
y
∘
T
) is closed and hence K
y
⊂
ker
⁡
(
δ
y
∘
T
). So K
y
=
ker
⁡
(
δ
y
∘
T
) and δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on 𝒟
(
ℝ
), for some α
y. This proves the theorem.Observe that𝒟
(
ℝ
) is dense in C
0
∞
(
ℝ
) in view of [6, Lemma  1 in Section  1.8]. So, we have the following corollary.Corollary 6.
LetY be a nonempty set. Let T be a linear disjointness preserving mapping from C
0
∞
(
ℝ
) to F
(
Y
). Suppose δ
y
∘
T is continuous on C
0
∞
(
ℝ
), for every y
∈
Y. Let Y
1, Y
2 be defined as in Corollary 4. Then there is a function ϕ
:
Y
1
→
X such that δ
y
∘
T
=
α
y
δ
ϕ
(
y
) on C
0
∞
(
ℝ
) for some fixed scalar α
y, and for all y
∈
Y
1.Remark 7.
One may changeℝ into any open region in ℝ
n in Theorem 5 and Corollary 6. C
0
∞
(
ℝ
) may be replaced in Corollary 6 by any of its linear subspace containing 𝒟
(
ℝ
). C
0
(
X
) may be replaced in Corollary 4 by any of its linear subspace containing C
c
(
X
).


## 3. FTDP-Mappings

Theorem 8.
LetX be a compact Hausdorff space and Y a nonempty set. Let T
:
C
(
X
)
→
F
(
Y
) be a nonzero disjointness preserving mappings. Let ϕ
,
Y
1
,
Y
2, and Y
3 be defined as in Theorem 2. Consider the following statements. (i)
T
is a FTDP-mapping.
(ii)
There is a unique
x
0
∈
X
such that, for every open neighbourhood
U
of
x
0
, there is a
f
∈
C
(
X
)
with
c
o
z
(
f
)
⊂
U
such that
T
f
≠
0
in
F
(
Y
)
.
(iii)
ϕ
(
Y
1
∪
Y
2
)
is a singleton subset of
X
.
(iv)
∩
{
supp
⁡
f
:
f
∈
C
(
X
), T
f
≠
0
in
F
(
Y
)
}
is a singleton set. Then the implications (i)⇒(ii), (i)⇒(iii), (i)⇒(iv), and (iii)⇒(ii) are true. If (i) is true, then the singleton sets in (iii) and (iv) are {
x
0
} for x
0 given in (ii). If (ii) is true, then x
0
∈
ϕ
(
Y
1
∪
Y
2
)
¯.Proof.
(i)⇒(ii): Suppose T is a FTDP-mapping. Suppose there are two distinct points x
1
,
x
2
∈
X with the property given in (ii). We can find two disjoint open neighbourhoods U
1, U
2 of x
1, x
2, respectively. Then there are f
1
,
f
2
∈
C
(
X
) with coz
(
f
1
)
⊂
U
1
,
coz
(
f
2
)
⊂
U
2 such that T
f
1
≠
0 and T
f
2
≠
0. This is a contradiction. So, uniqueness of x
0 in (ii) is established. If there is no x
0 in X with the property mentioned in (ii), then for each x
∈
X, there is a neighbourhood U
x of x such that T
f
=
0 in F
(
Y
) whenever coz
(
f
)
⊂
U
x. Then we find a subcover {
U
1
,
U
2
,
…
,
U
n
} of {
U
x
:
x
∈
X
} for X. Then we find a partition g
1
,
g
2
,
…
,
g
n for unity in C
(
X
) such that coz
(
g
i
)
⊂
U
i, i
=
1,2
,
…
,
n. Then for every f
∈
C
(
X
), we have T
f
=
T
(
∑
i
=
1
n
f
g
i
)
=
∑
i
=
1
n
T
(
f
g
i
)
=
0. Thus T
=
0 on C
(
X
), a contradiction. So (ii) is true.
Note thatϕ
(
Y
1
∪
Y
2
) is nonempty, because T
≠
0 on C
(
X
). Moreover y
∈
Y
1
∪
Y
2 if and only if δ
y
∘
T
≠
0 on C
(
X
). We claim that x
0
∈
ϕ
(
Y
1
∪
Y
2
)
¯. On the contrary, we assume that x
0
∉
ϕ
(
Y
1
∪
Y
2
)
¯. Find an open neighbourhood U
0 of x
0 and a function f
0
∈
C
(
X
) such that coz
(
f
0
)
⊂
U
0, T
f
0
≠
0 in F
(
Y
), and U
¯
0
∩
ϕ
(
Y
1
∪
Y
2
)
¯
=
∅. Fix any y
∈
Y
1
∪
Y
2. Let x
1
=
ϕ
(
y
). Then there is an open neighbourhood U
1 of x
1 such that U
¯
0
∩
U
1
=
∅. Find a function f
1
∈
C
(
X
) such that coz
(
f
1
)
⊂
U
1 and T
f
1
(
y
)
≠
0 in C
(
X
). So T
f
0
(
y
)
=
0, because T is disjointness preserving. Thus T
f
0
(
y
)
=
0, for any y
∈
Y
1
∪
Y
2. So T
f
0
=
0, where T
f
0
≠
0 in C
(
X
). Therefore, we conclude that x
0
∈
ϕ
(
Y
1
∪
Y
2
)
¯, if (ii) is true.
(i)⇒(iii): Suppose T is a FTDP-mapping. Since T
≠
0, Y
1
∪
Y
2 is nonempty. Suppose ϕ
(
y
1
)
≠
ϕ
(
y
2
) for some y
1
,
y
2
∈
Y
1
∪
Y
2. Find two disjoint open sets U and V such that ϕ
(
y
1
)
∈
U and ϕ
(
y
2
)
∈
V. Find f
,
g
∈
C
(
X
) such that coz
(
f
)
⊂
U, coz
(
g
)
⊂
V, and T
f
(
y
1
)
≠
0, T
g
(
y
2
)
≠
0. Then f
·
g
=
0 in C
(
X
), but T
f
≠
0 and T
g
≠
0 in F
(
Y
). This is a contradiction. So, ϕ
(
Y
1
∪
Y
2
) is a singleton set, and ϕ
(
Y
1
∪
Y
2
)
=
ϕ
(
Y
1
∪
Y
2
)
¯.
(i)⇒(iv): Suppose (i) is true. Consider x
0 given in (ii). Suppose f
∈
C
(
X
)
,
T
f
≠
0 in F
(
Y
), and x
0
∉
supp
⁡
f. Then find g
∈
C
(
X
) such that coz
(
g
)
⊂
X
∖
supp
⁡
f and T
g
≠
0 in F
(
Y
). Thus f
·
g
=
0, T
f
≠
0 and T
g
≠
0, a contradiction. Thus x
0
∈
supp
⁡
f if T
f
≠
0. If x
1
∈
X and x
1
≠
x
0, find two disjoint open neighbourhoods U
0 and U
1 of x
0 and x
1, respectively. Find f
∈
C
(
X
) such that T
f
≠
0 and coz
(
f
)
⊂
U
0. Then x
1
∉
supp
⁡
f. So, {
x
0
}
=
∩
{
supp
⁡
f
:
f
∈
C
(
X
), T
f
≠
0 in F
(
Y
)
}, where x
0 the member mentioned in (ii). This proves (iv).
(iii)⇒(ii): Suppose (iii) is true, and suppose ϕ
(
Y
1
∪
Y
2
)
=
{
x
0
}. Then for every y
∈
Y
1
∪
Y
2 and for every open neighbourhood U of x
0 there is a function f
∈
C
(
X
) such that coz
(
f
)
⊂
U and T
f
(
y
)
≠
0, and hence T
f
≠
0 in F
(
Y
). Let x
1
∈
X be such that x
1
≠
x
0. Find two disjoint neighbourhoods of U
0, U
1 of x
0, x
1, respectively. Fix y
∈
Y
1
∪
Y
2 arbitrarily. Find a function f
∈
C
(
X
) such that coz
(
f
)
⊂
U
0 and T
f
(
y
)
≠
0. Then for every function g
∈
C
(
X
) with coz
(
g
)
⊂
U
1, we have f
·
g
=
0, and hence T
g
(
y
)
=
0, since T is disjointness preserving. Thus for every function g
∈
C
(
X
) with coz
(
g
)
⊂
U
1, we have T
g
(
y
)
=
0, for every y
∈
Y
1
∪
Y
2, and hence we have T
g
=
0 in F
(
Y
). This of course proves (ii), and it is proved that x
0 is the member mentioned in (ii).Corollary 9.
LetX be a compact Hausdorff space and Y be a nonempty set. Let T
:
C
(
X
)
→
F
(
Y
) a nonzero FTDP- mapping. Then the following are equivalent.(i)
δ
y
∘
T is continuous for every y
∈
Y.
(ii)
T
f
=
δ
x
0
(
f
)
g on C
(
X
) for some fixed function g
∈
F
(
Y
), for some fixed x
0
∈
X, and for all f
∈
C
(
X
).Proof.
Suppose (i) is true. Supposeϕ
(
Y
1
∪
Y
2
)
=
{
x
0
}. Then for every y
∈
Y
1
∪
Y
2, we have {
f
∈
C
(
X
)
:
x
0
∉
supp
⁡
f
}
⊂
ker
⁡
(
δ
y
∘
T
), and hence {
f
∈
C
(
X
)
:
f
(
x
0
)
=
0
}
=
ker
⁡
(
δ
y
∘
T
). Thus, ker
⁡
T
=
{
f
∈
C
(
X
)
:
f
(
x
0
)
=
0
}. This proves (ii). Another implication is obvious.Remark 10.
Theorem8 is extendable to C
c
(
X
) when X is normal and to 𝒟
(
ℝ
). Corollary 9 is extendable to C
c
(
X
), C
0
(
X
), 𝒟
(
ℝ
), and C
0
∞
(
ℝ
) when X is normal. Theorem 8 is extendable to C
0
(
X
) when X is locally compact with an additional assumption that the singleton set may be {
∞
} in X
∞. Corollary 9 is extendable to C
0
(
X
) when X is locally compact.



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*Source: 838692-2014-02-25.xml*