# Stability and instability of weighted composition operators

**ABSTRACT** Let $\epsilon >0$. A continuous linear operator $T:C(X) \ra C(Y)$ is said to be {\em $\epsilon$-disjointness preserving} if $\vc (Tf)(Tg)\vd_{\infty} \le \epsilon$, whenever $f,g\in C(X)$ satisfy $\vc f\vd_{\infty} =\vc g\vd_{\infty} =1$ and $fg\equiv 0$. In this paper we address basically two main questions: 1.- How close there must be a weighted composition operator to a given $\epsilon$-disjointness preserving operator? 2.- How far can the set of weighted composition operators be from a given $\epsilon$-disjointness preserving operator? We address these two questions distinguishing among three cases: $X$ infinite, $X$ finite, and $Y$ a singleton ($\epsilon$-disjointness preserving functionals). We provide sharp stability and instability bounds for the three cases.

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**ABSTRACT:**If X and Y are realcompact and T: C(X) → C(Y) is a ring isomorphism then two things happen: (1) there is a homeomorphism h of X onto Y and (2) T is a composition map: for all ƒ in C(X), Tƒ = ƒ ∘ h. We strengthen that result in Propositions 2 and 3. We discard the multiplicative property of T and assume only that T is biseparating in the sense that ƒg = 0 ⇔ TƒTg = 0. It remains true that X is homeomorphic to Y but this time, if T is also linear, it is a weighted composition map: for some weight function a in C(Y), Tƒ = a(ƒ ∘ h).Journal of Mathematical Analysis and Applications 01/1995; 192(1):258-265. · 1.05 Impact Factor - [Show abstract] [Hide abstract]

**ABSTRACT:**Let T, S be non-void completely regular locally compact Hausdorff spaces and Let C o (T) (resp. Coo (T)) be the Banach space of all real-valued continuous functions on T which are zero at infinity (resp. the normed space of all continuous functions with compact support). A linear map H defined from a subalgebra A of Co(T ) into a subalgebra B of Co(S ) is said to be separating or disjointness preserving (also called d-homomorphisms) if x �9 y - 0 implies H x - H y - 0 for all x, y e A.Archiv der Mathematik 07/1994; 63(2):158-165. · 0.38 Impact Factor -
##### Article: On approximate isometries. (Reprint)

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**ABSTRACT:**See the review of the original in Bull. Am. Math. Soc. 51, 288-292 (1945; Zbl 0060.26404).

Page 1

arXiv:0801.2477v1 [math.FA] 16 Jan 2008

STABILITY AND INSTABILITY OF WEIGHTED

COMPOSITION OPERATORS

JES´US ARAUJO AND JUAN J. FONT

Abstract. Let ǫ > 0. A continuous linear operator T : C(X) −→

C(Y ) is said to be ǫ-disjointness preserving if ?(Tf)(Tg)?∞≤ ǫ,

whenever f,g ∈ C(X) satisfy ?f?∞= ?g?∞= 1 and fg ≡ 0. In

this paper we address basically two main questions:

1.- How close there must be a weighted composition operator to

a given ǫ-disjointness preserving operator?

2.- How far can the set of weighted composition operators be

from a given ǫ-disjointness preserving operator?

We address these two questions distinguishing among three cases:

X infinite, X finite, and Y a singleton (ǫ-disjointness preserving

functionals).

We provide sharp stability and instability bounds for the three

cases.

1. Introduction

Suppose that a mathematical object satisfies a certain property ap-

proximately. Is it then possible to approximate this object by objects

that satisfy the property exactly? This stability problem appears in

almost all branches of mathematical analysis and is of particular in-

terest in probability theory and in the realm of functional equations.

Within this context, considerable attention has been mainly given to

approximately multiplicative maps (see [11], [12], [8], and [15]) and to

approximate isometries (see [5], [6], [2], and [7]).

Recently, G. Dolinar ([3]) treated a more general problem of stability

concerning a kind of operators which ”almost” preserves the disjoint-

ness of cozero sets (see Definition 1.2).

2000 Mathematics Subject Classification.

47B33.

Research of the first author was partially supported by the Spanish Ministry of

Science and Education (Grants numbers MTM2004-02348 and MTM2006-14786).

Research of the second author was partially supported by European Union

(FEDER) and the Spanish Ministry of Science and Education (Grant number

MTM2004-07665-C02-01), and by Bancaixa (Projecte P1-1B2005-22 codi 05I343).

1

Primary 47B38; Secondary 46J10,

Page 2

2JES´US ARAUJO AND JUAN J. FONT

We need some notation. Let K denote the field of real or complex

numbers. Topological spaces X and Y are assumed to be compact and

Hausdorff. Also C(X) stands for the Banach space of all K-valued

continuous functions defined on X, equipped with its usual supremum

norm.

Definition 1.1. An operator S : C(X) −→ C(Y ) is said to be a

weighted composition map if there exist a ∈ C(Y ) and a map h : Y −→

X, continuous on c(a) := {y ∈ Y : a(y) ?= 0}, such that

(Sf)(y) = a(y)f(h(y))

for every f ∈ C(X) and y ∈ Y .

Obviously every weighted composition map is linear and continuous.

We also include the case that S ≡ 0 as a weighted composition map

(being c(a) = ∅).

Recall that a linear operator T : C(X) −→ C(Y ) is said to be dis-

jointness preserving (or separating) if, given f,g ∈ C(X), fg ≡ 0 yields

(Tf)(Tg) ≡ 0. Clearly every weighted composition map is disjointness

preserving. Reciprocally, it is well known that if a disjointness preserv-

ing operator is continuous, then it is a weighted composition. On the

other hand, automatic continuity of disjointness preserving operators

can be obtained sometimes (see for instance [9], [1], [4], [10])).

Definition 1.2. Let ǫ > 0. A continuous linear operator T : C(X) −→

C(Y ) is said to be ǫ-disjointness preserving if ?(Tf)(Tg)?∞≤ ǫ, when-

ever f,g ∈ C(X) satisfy ?f?∞= ?g?∞= 1 and fg ≡ 0 (or, equiva-

lently, if ?(Tf)(Tg)?∞≤ ǫ?f?∞?g?∞whenever fg ≡ 0).

Obviously the study of ǫ-disjointness preserving operators can be

restricted to those of norm 1, because if T ?= 0 is ǫ-disjointness pre-

serving, then T/?T? is ǫ/?T?2-disjointness preserving. On the other

hand, every such T has the trivial weighted composition map S ≡ 0 at

distance 1. That is, giving any bound equal to or bigger than 1 does

not provide any information on the problem. Apart from this, it can

be easily checked that every continuous linear functional on C(X) of

norm 1 is 1/4-disjointness preserving and, consequently, every contin-

uous linear map T : C(X) −→ C(Y ) with ?T? = 1 is 1/4-disjointness

preserving. Thus, if we consider again the trivial weighted composition

map S ≡ 0, then ?T − S? = 1. We conclude that our study can be

restricted to ǫ belonging to the interval (0,1/4).

In [3] the author, following the above stability questions, studies

when an ǫ-disjointness preserving operator is close to a weighted com-

position map.The main result in [3] reads as follows: Let ǫ > 0

Page 3

STABILITY AND INSTABILITY3

and let T : C(X) −→ C(Y ) be an ǫ-disjointness preserving opera-

tor with ?T? = 1. Then there exists a weighted composition map

S : C(X) −→ C(Y ) such that

?T − S? ≤ 20√ǫ.

In view of the above comments we conclude that Dolinar’s result is

meaningful only for ǫ ∈ (0,1/400).

Apart from the general case, Dolinar also concentrates on the study

of linear and continuous functionals, where the bound given is 3√ǫ (see

[3, Theorem 1]).

On the other hand, notice that when X has just one point, we are

in a situation of ”extreme stability”, because every continuous lin-

ear operator is a weighted composition map. But in general, given

an ǫ-disjointness preserving operator, we do not necessarily have a

weighted composition map arbitrarily close. Instability questions deal

with bounds of how far apart an ǫ-disjointness preserving operator can

be from all weighted composition maps.

In the present paper we improve Dolinar’s result by showing, under

necessary restrictions on ǫ, that a weighted composition map is indeed

much closer. If fact we address the following two questions. Given any

ǫ-disjointness preserving operator,

(1) Stability. How close there must be a weighted composition

map? That is, find the shortest distance at which we can be

certain that there exists a weighted composition map.

(2) Instability. How far the set of all weighted composition maps

can be? That is, find the longest distance at which we cannot

be certain that there exists a weighted composition map.

How close. We prove that, for every ǫ < 2/17, the number?17ǫ/2

smallest in every case, as we give an example such that, for every

ǫ < 2/17, no number strictly less than

9.6).

The question appears to be very related to the following: Find the

biggest set I ⊂ (0,1/4) such that every ǫ ∈ I has the following property:

Given an ǫ-disjointness preserving operator T : C(X) −→ C(Y ) with

?T? = 1, there exists a weighted composition map S : C(X) −→ C(Y )

such that ?T − S? < 1. We prove that I = (0,2/17) (Theorem 2.1 and

Example 9.3).

is a bound valid for every X and Y (Theorem 2.1). It is indeed the

?17ǫ/2 satisfies it (Example

Page 4

4JES´US ARAUJO AND JUAN J. FONT

We will also study the particular case when X is finite. Here the

bound, which can be given for every ǫ < 1/4 and every Y , is the

number 2√ǫ, and is sharp (Theorem 4.3 and Example 12.1).

How far. Of course, an answer valid for every case would be trivial,

because if we take X with just one point, then every continuous linear

operator is a weighted composition map, so the best bound is just 0.

If we avoid this trivial case and require X to have at least two points,

then we can see that again the problem turns out to be trivial since

the best bound is now attained for sets with two points. The same

happens if we require the set X to have at least k points.

In general, it can be seen that the answer does not depend on the

topological features of the spaces but on their cardinalities. If we as-

sume that Y has at least two points, then the number 2√ǫ is a valid

bound if X is infinite (Theorem 3.1), and a different value plays the

same rˆ ole for each finite set X (Theorem 4.1).

We also prove that these estimates are sharp in every case (Theo-

rems 3.2 and 4.2). But here, instead of providing a concrete counterex-

ample, we can show that the bounds are best for a general family of

spaces Y , namely, whenever Y consists of the Stone-ˇCech compactifi-

cation of any discrete space.

On the other hand, unlike the previous question, the answer can be

given for every ǫ < 1/4.

The case of continuous linear functionals. The context

when Y has just one point, that is, the case of continuous linear func-

tionals, deserves to be studied separately. We do this in Sections 6

and 7. In fact some results given in this case will be tools for a more

general study. Various situations appear in this context, depending on

ǫ. Namely, if ǫ < 1/4, then the results depend on an suitable splitting

of the interval (0,1/4) (based on the sequence (ωn) defined below), as

well as on the cardinality of X (Theorem 5.1).

Also, as we mentioned above, when X has just one point, every

element of C(X)′is a weighted composition map, that is, a scalar

multiple of the evaluation functional δx. We will see that a related

phenomenon sometimes arises when X is finite (see Remark 5.1).

In every case our results are sharp.

Notation. Throughout K = R or C. X and Y will be (nonempty)

compact Hausdorff spaces. To avoid the trivial case, we will always

assume that X has at least two points. In a Banach space E, for e ∈ E

and r > 0, B(e,r) and B(e,r) denote the open and the closed balls of

center e and radius r, respectively.

Page 5

STABILITY AND INSTABILITY5

Spaces and functions. Given any compact Hausdorff space Z, we

denote by cardZ its cardinal. C(Z) will be the Banach space of all K-

valued continuous functions on Z, endowed with the sup norm ?·?∞.

C(Z)′will denote the space of linear and continuous functionals defined

on C(Z). If a ∈ K, we denote by ? a the constant function equal to a on

by 1. For f ∈ C(Z), 0 ≤ f ≤ 1 means that f(x) ∈ [0,1] for every x ∈

Z. Given f ∈ C(Z), we will consider that c(f) = {x ∈ Z : f(x) ?= 0}

is its cozero set, and supp(f) its support. Finally, if A ⊂ Z, we denote

by clA the closure of A in Z, and by ξAthe characteristic function of

A.

Continuous linear functionals and measures: λϕ, |λ|, δx. For ϕ ∈

C(X)′, we will write λϕto denote the measure which represents it. For

a regular measure λ, we will denote by |λ| its total variation. Finally, for

x ∈ X, δxwill be the evaluation functional at x, that is, δx(f) := f(x)

for every f ∈ C(X).

The linear functionals Tyand the sets Yr. Suppose that T : C(X) −→

C(Y ) is linear and continuous. Then, for each y ∈ Y , we define a con-

tinuous linear functional Tyas Ty(f) := (Tf)(y) for every f ∈ C(X).

Also, for each r ∈ R we define Yr:= {y ∈ Y : ?Ty? > r}, which is an

open set. It is clear that, if ?T? = 1, then Yr is nonempty for each

r < 1.

The sets of operators.We denote by ǫ − DP(X,Y ) the set of

all ǫ-disjointness preserving operators from C(X) to C(Y ), and by

WCM(X,Y ) the set of all weighted composition maps from C(X) to

C(Y ). When Y has just one point, then ǫ−DP(X,Y ) and WCM(X,Y )

may be viewed as subspaces of C(X)′. In this case, we will use the no-

tation ǫ−DP(X,K) and WCM(X,K) instead of ǫ−DP(X,Y ) and

WCM(X,Y ), respectively. That is, ǫ − DP(X,K) is the space of all

ϕ ∈ C(X)′which satisfy |ϕ(f)||ϕ(g)| ≤ ǫ whenever f,g ∈ C(X) sat-

isfy ?f?∞= 1 = ?g?∞and fg ≡ 0, and WCM(X,K) is the subset of

C(X)′of elements of the form αδx, where α ∈ K and x ∈ X.

The sequences (ωn) and (An). We define, for each n ∈ N,

ωn:=n2− 1

Z. In the special case of the constant function equal to 1, we denote it

4n2

and

An:= [ω2n−1,ω2n+1),

It is clear that (An) forms a partition of the interval [0,1/4).

The sequences (ωn) and (An) will determine bounds in Sections 4

and 5.

Page 6

6JES´US ARAUJO AND JUAN J. FONT

2. Main results I: How close. The general case

In this section we give the best stability bound in the general case.

This result is valid for every X in general, assuming no restrictions on

cardinality (see Section 8 for the proof).

Theorem 2.1. Let 0 < ǫ < 2/17, and let T ∈ ǫ − DP(X,Y ) with

?T? = 1. Then

?

2

BT,

?17ǫ

?

∩ WCM(X,Y ) ?= ∅.

Remark 2.1. Theorem 2.1 is accurate in two ways. On the one hand,

for every ǫ ∈ (0,2/17), the above bound is sharp, as it can be seen in

Example 9.6. On the other hand, we have that (0,2/17) is the maximal

interval we can get a meaningful answer in. Namely, if ǫ ≥ 2/17, then

it may be the case that ?T − S? ≥ 1 for every weighted composition

map S (Example 9.3). But, as it is explained in the comments after

Definition 1.2, this is not a proper answer for the stability question.

3. Main results II: How far. The case when X is infinite

We study instability first when X is infinite. Our results depend on

whether or not the space X admits an appropriate measure.

Theorem 3.1. Let 0 < ǫ < 1/4. Suppose that Y has at least two

points, and that X is infinite. Then for each t < 1, there exists T ∈

ǫ − DP(X,Y ) with ?T? = 1 such that

B?T,2t√ǫ?∩ WCM(X,Y ) = ∅.

Furthermore, if X admits an atomless regular Borel probability mea-

sure, then T can be taken such that

B?T,2√ǫ?∩ WCM(X,Y ) = ∅.

We also see that the above bounds are sharp when considering some

families of spaces Y .

Theorem 3.2. Let 0 < ǫ < 1/4. Suppose that Y is the Stone-ˇCech

compactification of a discrete space with at least two points, and that

X is infinite. Let T ∈ ǫ − DP(X,Y ) with ?T? = 1. Then

B?T,2√ǫ?∩ WCM(X,Y ) ?= ∅.

Furthermore, if X does not admit an atomless regular Borel proba-

bility measure and Y is finite (with cardY ≥ 2), then

B?T,2√ǫ?∩ WCM(X,Y ) ?= ∅.

Page 7

STABILITY AND INSTABILITY7

The proofs of both results are given in Section 10.

Remark 3.1. The property of admitting an atomless regular Borel prob-

ability (or, equivalently, complex and nontrivial) measure can be char-

acterized in purely topological terms. A compact Hausdorff space ad-

mits such a measure if and only if is scattered (see [14, Theorem 19.7.6]).

4. Main results III: How far and how close. The case

when X is finite

Next we study the case when X is finite. Here, the best instability

bounds depend on the sequence (ωn), and the cardinality of Y does not

play any rˆ ole as long as it is at least 2. We define o′

for every finite set X (recall that we are assuming cardX ≥ 2). We

put

Theorem 4.1. Let 0 < ǫ < 1/4. Assume that Y has at least two

points, and that X is finite. Then there exists T ∈ ǫ−DP(X,Y ) with

?T? = 1 such that

B (T,o′

X: (0,1/4) −→ R,

o′

X(ǫ) :=

2

?

2(n−1)√ǫ

n

(n−1)ǫ

n+1

n−1

if n := cardX is odd and ǫ ≤ ωn

if n := cardX is odd and ǫ > ωn

if n := cardX is even

n

X(ǫ)) ∩ WCM(X,Y ) = ∅.

The next result says that Theorem 4.1 provides a sharp bound, and

gives a whole family of spaces Y for which the same one is a bound for

stability as well. As we can see in Example 12.1, our requirement on

these Y is not superfluous.

Theorem 4.2. Let 0 < ǫ < 1/4. Suppose that Y is the Stone-ˇCech

compactification of a discrete space with at least two points, and that

X is finite. Let T ∈ ǫ − DP(X,Y ) with ?T? = 1. Then

B (T,o′

X(ǫ)) ∩ WCM(X,Y ) ?= ∅.

The instability bounds are special when the space X is finite. In the

following theorem we study the stability bounds in this particular case.

Example 12.1 shows that the result is sharp.

Theorem 4.3. Let 0 < ǫ < 1/4. Suppose that that X is finite, and let

T ∈ ǫ − DP(X,Y ) with ?T? = 1. Then

B?T,2√ǫ?∩ WCM(X,Y ) ?= ∅.

Theorems 4.1 and 4.2 are proved in Section 11, and Theorem 4.3 in

Section 12.

Page 8

8JES´US ARAUJO AND JUAN J. FONT

5. Main results IV: The case of continuous linear

functionals

In some of the previous results, we assume that the space Y has at

least two points. Of course the case when Y has just one point can

be viewed as the study of continuous linear functionals. In this section

we give the best stability and instability bounds in this case, and see

that both bounds coincide. Here we do not require X to be finite, and

Theorem 5.1 is valid both for X finite and infinite. Anyway, the result

depends on the sequence (ωn) and its relation to the cardinal of X.

We first introduce the map oX: (0,1/4) −→ R as follows: For n ∈ N

and ǫ ∈ An,

We use this map to give a bound both for stability and instability

(see Section 7 for the proof).

oX(ǫ) :=

2n−1−√1−4ǫ

2n

k−1−√1−4ǫ

if 2n ≤ cardX

if k := cardX < 2n and k is even

if k := cardX < 2n and k is odd

k

k−1

k

Theorem 5.1. Let 0 < ǫ < 1/4. If ϕ ∈ ǫ − DP(X,K) and ?ϕ? = 1,

then

B (ϕ,oX(ǫ)) ∩ WCM(X,K) ?= ∅.

On the other hand, there exists ϕ ∈ ǫ − DP(X,K) with ?ϕ? = 1

such that

B (ϕ,oX(ǫ)) ∩ WCM(X,K) = ∅.

Remark 5.1. Sometimes the information given by the number ǫ is re-

dundant, in that ǫ is too ”big” with respect to the cardinal of X. This

happens for instance when X is a set of k points, where k ∈ N is odd.

This is the reason why the definition of oX(and that of o′

necessarily depend on ǫ.

X) does not

6. The bounds 1/4 and 2/9 for continuous linear

functionals

We start with a lemma that will be broadly used.

Lemma 6.1. Let 0 < ǫ < 1/4. Let ϕ ∈ ǫ−DP(X,K) be positive with

?ϕ? = 1. If C is a Borel subset of X, then

?1 −√1 − 4ǫ

λϕ(C) / ∈

2

,1 +√1 − 4ǫ

2

?

.

Page 9

STABILITY AND INSTABILITY9

Proof. Suppose, contrary to what we claim, that there is a Borel subset

C such that?1 −√1 − 4ǫ?/2 < λϕ(C) <?1 +√1 − 4ǫ?/2. This im-

with (λϕ(C) − δ)(1 − λϕ(C) − δ) > ǫ.

By the regularity of the measure, there exist two compact subsets,

K1 and K2, such that K1 ⊂ C and K2 ⊂ X \ C and, furthermore,

λϕ(K1) > λϕ(C) − δ and λϕ(K2) > 1 − λϕ(C) − δ.

On the other hand, let us choose two disjoint open subsets U and V

of X such that K1⊂ U and K2⊂ V . By Urysohn’s lemma, we can

find two functions f1and f2in C(X) such that 0 ≤ f1≤ 1, 0 ≤ f2≤ 1,

f1≡ 1 on K1, f2≡ 1 on K2, supp(f1) ⊂ U and supp(f2) ⊂ V . Clearly,

f1f2≡ 0 and

ϕ(fi) =

X

for i = 1,2. Besides, ?f1?∞= ?f2?∞= 1. However,

|ϕ(f1)||ϕ(f2)| ≥ (λϕ(C) − δ)((1 − λϕ(C) − δ) > ǫ,

which contradicts the ǫ-disjointness preserving property of ϕ, and we

are done.

plies that λϕ(C)(1 − λϕ(C)) > ǫ and, consequently, we can find δ > 0

?

fidλϕ≥ λϕ(Ki)

?

If ϕ ∈ C(X)′, then let us define

|ϕ|(f) :=

?

X

fd|λϕ| =

?

X

fdλϕ

d|λϕ|dλϕ

for every f ∈ C(X).

Lemma 6.2. Given ϕ ∈ C(X)′, |ϕ| is a positive linear functional on

C(X) with ?|ϕ|? = ?ϕ?. Moreover, if ǫ > 0 and ϕ ∈ ǫ − DP(X,K),

then |ϕ| ∈ ǫ − DP(X,K) and λ|ϕ|= |λϕ|.

Proof. The first part is apparent. As for the second part, using Lusin’s

Theorem (see [13, p. 55]), we can find a sequence (kn) in C(X) such

that

?

and ?kn?∞≤ 1 for every n ∈ N. This implies that, for all f ∈ C(X),

|ϕ|(f) = limn−→∞ϕ(fkn), and we can easily deduce that |ϕ| is ǫ-

disjointness preserving. It is also clear that λ|ϕ|= |λϕ|.

Lemma 6.3. Let 0 < ǫ < 1/4. Let ϕ ∈ ǫ−DP(X,K), ?ϕ? = 1. Then

there exists x ∈ X with

|λϕ({x})| ≥√1 − 4ǫ.

lim

n−→∞

X

????kn−

dλϕ

d|λϕ|

????d|λϕ| = 0,

?

Page 10

10JES´US ARAUJO AND JUAN J. FONT

Furthermore, if 0 < ǫ < 2/9, then there exists a unique x ∈ X with

|λϕ({x})| ≥1 +√1 − 4ǫ

Proof. Let 0 < ǫ < 1/4. We prove the result first for positive func-

tionals.Suppose that for every x ∈ X, λϕ({x}) <

each x ∈ X, take an open neighborhood U(x) of x with λϕ(U(x)) <

√1 − 4ǫ. Since X is compact, we can find x1,x2,...,xnin X such that

X = U(x1)∪U(x2)∪···∪U(xn). Let r1:= λϕ(U(x1)), r2:= λϕ(U(x1)∪

U(x2)), ..., rn:= λϕ(U(x1)∪U(x2)∪···∪U(xn)), and suppose without

loss of generality that r1< r2< ··· < rn= 1. By Lemma 6.1, r1≤

?1 −√1 − 4ǫ?/2, and we can take i0:= max?i : ri≤?1 −√1 − 4ǫ?/2?.

against Lemma 6.1. This proves the first part of the lemma for positive

functionals.

If ϕ is not positive, then we use Lemma 6.2, and from the above

paragraph we have that there exists x ∈ X such that

|λϕ({x})| = |λϕ|({x}) ≥√1 − 4ǫ.

As for the second part, if follows immediately from Lemma 6.1

and the fact that

nally, if there exist two different points x1,x2such that |λϕ({xi})| ≥

?1 +√1 − 4ǫ?/2 (i = 1,2), then |λϕ|({x1,x2}) ≥ 1 +√1 − 4ǫ > 1,

Lemma 6.4. Let 0 < ǫ < 1/4. Let ϕ ∈ ǫ−DP(X,K), ?ϕ? = 1. Then

there exists x ∈ X with

?ϕ − λϕ({x})δx? ≤ 1 −√1 − 4ǫ.

Furthermore, if 0 < ǫ < 2/9, then there exists a unique x ∈ X with

?ϕ − λϕ({x})δx? ≤1 −√1 − 4ǫ

Proof. It is easy to see that

2

.

√1 − 4ǫ. For

We then see that ri0+1belongs to??1 −√1 − 4ǫ?/2,?1 +√1 − 4ǫ?/2?,

?1 −√1 − 4ǫ?/2 <√1 − 4ǫ for 0 < ǫ < 2/9. Fi-

against our assumptions. This completes the proof.

?

2

.

1 = ?ϕ?

= ?λϕ({x})δx? + ?ϕ − λϕ({x})δx?

= |λϕ({x})| + ?ϕ − λϕ({x})δx?,

and the conclusion follows from Lemma 6.3.

?

Corollary 6.5. Let 0 < ǫ < 1/4. Suppose that ϕ ∈ ǫ−DP(X,K) and

2√ǫ < ?ϕ? ≤ 1.

Page 11

STABILITY AND INSTABILITY 11

Then there exists x ∈ X such that

?ϕ − λϕ({x})δx? ≤ ?ϕ? −

?

?ϕ?2− 4ǫ.

Furthermore, if 0 < ǫ < 2/9 and

a unique x ∈ X such that

?9ǫ/2 < ?ϕ? ≤ 1, then there exists

?

2

?ϕ − λϕ({x})δx? ≤

?ϕ? −

?ϕ?2− 4ǫ

.

Proof. Let 0 < ǫ < 1/4. It is apparent that ϕ/?ϕ? has norm 1 and is

ǫ/?ϕ?2-disjointness preserving. Besides ǫ/?ϕ?2< ǫ/(2√ǫ)2= 1/4.

Hence, by Lemma 6.4, there exists x ∈ X with

????

and we are done. The proof of the second part is similar.

ϕ

?ϕ?− λ ϕ

?ϕ?({x})δx

????≤ 1 −

?

1 −

4ǫ

?ϕ?2

?

Corollary 6.6. Let 0 < ǫ < 2/9. Suppose that ϕ ∈ ǫ−DP(X,K) and

?9ǫ/2 < ?ϕ? ≤ 1, and that x ∈ X is the point given in Corollary 6.5.

?

whenever f ∈ C(X) satisfies |f(x)| = 1 = ?f?∞.

Proof. Let f ∈ C(X) be such that |f(x)| = 1 = ?f?∞. By Corollary

6.5, we have that

Then

?ϕ?2− 4ǫ ≤ |ϕ(f)|

||ϕ(f)| − |λϕ({x})|| ≤ |(ϕ − λϕ({x})δx)(f)| ≤

?ϕ? −

?

?ϕ?2− 4ǫ

2

.

Hence, by applying Lemma 6.3

|ϕ(f)| ≥ |λϕ({x})| −

?ϕ? −

?

?ϕ?2− 4ǫ

2

≥

?ϕ? +

?

?

?ϕ?2− 4ǫ

2

−

?ϕ? −

?

?ϕ?2− 4ǫ

2

=

?ϕ?2− 4ǫ.

?

Page 12

12JES´US ARAUJO AND JUAN J. FONT

7. The sequence (ωn) for continuous linear functionals

Recall that we have defined, for each n ∈ N,

ωn:=n2− 1

4n2

and

An:= [ω2n−1,ω2n+1).

The precise statement of the results in this section depends heavily

on the number n such that ǫ ∈ An, and on the cardinality of X.

Suppose that X is a finite set of k elements, and that ϕ ∈ C(X)′

has norm 1. Then it is immediate that there exists a point x ∈ X with

|λϕ({x})| ≥ 1/k. We next see that this result can be sharpened when k

is even and ϕ ∈ ǫ−DP(X,K), and also when X has ”many” elements

(being finite or infinite).

Proposition 7.1. Let 0 < ǫ < 1/4. Suppose that X is a finite set of

cardinal k ∈ 2N. If ϕ ∈ ǫ − DP(X,K) and ?ϕ? = 1, then there exists

x ∈ X such that

|λϕ({x})| ≥1 +√1 − 4ǫ

Proof. By Lemma 6.2, we can assume without loss of generality that

ϕ is positive. Suppose that k = 2m, m ∈ N. Notice that there cannot

be m different points x1,...,xm∈ X with

?1 −√1 − 4ǫ

for every i ∈ {1,...,m}, because otherwise

?1 −√1 − 4ǫ

against Lemma 6.1. This implies that there exist at least m+1 points

whose measure belongs to

?

Suppose that at least m different points x1,...,xm∈ X satisfy λϕ({xi}) ≤

?1 −√1 − 4ǫ?/k.

Since X \ {x1,...,xm} has m points, this obviously implies that there

exists x ∈ X \ {x1,...,xm} with λϕ({x}) ≥?1 +√1 − 4ǫ?/k, and we

k

.

λϕ({xi}) ∈

k

,1 +√1 − 4ǫ

k

?

λϕ({x1,...,xm}) ∈

2

,1 +√1 − 4ǫ

2

?

,

0,1 −√1 − 4ǫ

k

?

∪

?1 +√1 − 4ǫ

k

,1

?

.

Then λϕ({x1,...,xm}) ≤

?1 −√1 − 4ǫ?/2, and

consequently we have that λϕ(X \ {x1,...,xm}) ≥

?1 +√1 − 4ǫ?/2.

are done.

?

Page 13

STABILITY AND INSTABILITY13

Proposition 7.2. Let 0 < ǫ < 1/4, and let n ∈ N be such that ǫ ∈ An.

Suppose that cardX ≥ 2n. If ϕ ∈ ǫ − DP(X,K) and ?ϕ? = 1, then

there exists x ∈ X such that

|λϕ({x})| ≥1 +√1 − 4ǫ

2n

.

Proof. Let D := {x ∈ X : |λϕ({x})| > 0}. It is clear that D is a

countable set, and by Lemma 6.3 it is nonempty. Let M := {1,...,m}

if the cardinal of D is m ∈ N, and let M := N otherwise. It is obvious

that we may assume that D = {xi: i ∈ M} and that |λϕ({xi+1})| ≤

|λϕ({xi})| for every i.

Next let

?

i=1

J :=j ∈ M :

j

?

|λϕ({xi})| <1

2

?

and

R :=

?

i∈J

|λϕ({xi})|.

We have that R ≤ 1/2, and by Lemma 6.1 applied to the functional

associated to |λϕ|, we get R < 1/2. Take any open subset U of X

containing all xi, i ∈ J, such that |λϕ|(U) < 1/2, that is, |λϕ|(U) ≤

?1 −√1 − 4ǫ?/2, and suppose that |λϕ({x})| <√1 − 4ǫ for every x / ∈

U ∪U1∪···∪Uland |λϕ|(Ui) <√1 − 4ǫ for every i. If we consider, for

i ∈ {1,...,l}, bi:= |λϕ|

be an index i0with

?1 −√1 − 4ǫ

which goes against Lemma 6.1.

We deduce that there exists j ∈ M, j / ∈ J, such that |λϕ({xj})| ≥

√1 − 4ǫ. By the way we have taken D, this implies that |λϕ({xi})| ≥

√1 − 4ǫ for every i ∈ J, and obviously J must be finite, say J =

{1,...,m0}.

Let us see now that m0≤ n−1. We have that, since ǫ < ω2n+1, then

√1 − 4ǫ > 1/(2n + 1), which implies that

n√1 − 4ǫ >1 −√1 − 4ǫ

U. Then there exist open sets U1,...,Ulin X, l ∈ N, such that X =

?

U ∪?i

j=1Uj

?

, then we see that there must

bi0∈

2

,1 +√1 − 4ǫ

2

?

,

2

.

Page 14

14JES´US ARAUJO AND JUAN J. FONT

Consequently, if m0≥ n, then we get

R =

m0

?

1 −√1 − 4ǫ

2

i=1

|λϕ({xi})|

≥ n√1 − 4ǫ

>

,

which is impossible, as we said above. We conclude that m0≤ n − 1.

On the other hand, taking into account that

|λϕ({xi})| ≥1 +√1 − 4ǫ

m0+1

?

i=1

2

,

we have that

(m0+ 1)|λϕ({x1})| ≥1 +√1 − 4ǫ

2

,

which implies that

n|λϕ({x1})| ≥1 +√1 − 4ǫ

2

.

As a consequence we get

|λϕ({x1})| ≥1 +√1 − 4ǫ

2n

,

and we are done.

?

Proof of Theorem 5.1. Let us show the first part. By Propositions 7.1

(see also comment before it) and 7.2, there exists x ∈ X with |λϕ({x})| ≥

1 − oX(ǫ). If we define ψ := λϕ({x})δx, then we are done.

Let us now prove the second part. Suppose that ǫ belongs to An,

n ∈ N. It is clear that this fact implies that (2n − 1)√1 − 4ǫ ≤ 1.

If cardX ≥ 2n, then we can pick 2n distinct points x1,x2,...,x2nin

X, and define the map ϕ ∈ C(X)′as

ϕ :=1 +√1 − 4ǫ

2n

i=1

?2n−1

?

δxi

?

+1 − (2n − 1)√1 − 4ǫ

2n

δx2n.

It is easy to see that ϕ satisfies all the requirements.

To study the cases when cardX < 2n, put X := {x1,...,xk}.

Suppose first that k is even. Since (2n − 1)√1 − 4ǫ ≤ 1, we have

Page 15

STABILITY AND INSTABILITY15

(k − 1)√1 − 4ǫ < 1. We can easily see that if we define the map ϕ as

ϕ :=1 +√1 − 4ǫ

k

i=1

?k−1

?

δxi

?

+1 − (k − 1)√1 − 4ǫ

k

δxk,

then we are done.

Suppose finally that k is odd. It is clear that if we define

ϕ :=1

k

?

k

?

i=1

δxi

?

,

then ϕ is a norm one element of C(X)′, and is ωk-disjointness preserv-

ing, which implies that it is ǫ-disjointness preserving. It is also easy to

see that ?ϕ − ψ? ≥ 1−1/k for every weighted evaluation functional ψ

on C(X).

?

8. How close. The general case: Proofs

Let 0 < ǫ < 2/9, and let T : C(X) −→ C(Y ) be a norm one

ǫ-disjointness preserving operator. If we take any y ∈ Y√

Ty/?Ty? is a norm one ǫ/?Ty?2-disjointness preserving operator with

ǫ

?Ty?2<ǫ

By Lemma 6.3, there exists a unique xy∈ X such that??λTy({xy})??>

that

These fact can be summarized in the following lemma.

9ǫ/2, then

9ǫ

2

=2

9.

?Ty?/2. Thus, we can define a map hT: Y√

??λTy({hT(y)})??> ?Ty?/2 for each y ∈ Y√

Lemma 8.1. Let 0 < ǫ < 2/9, and let T ∈ ǫ−DP(X,Y ) with ?T? = 1.

If y ∈ Y√

9ǫ/2−→ X, in such a way

9ǫ/2.

9ǫ/2, then

??λTy

??({hT(y)}) ≥

?Ty? +

?

?Ty?2− 4ǫ

2

.

Proposition 8.2. Let 0 < ǫ < 2/9, and let T ∈ ǫ − DP(X,Y ) with

?T? = 1. Then the map hT is continuous.

Proof. We will check the continuity of this map at every point. To this

end, fix y0 ∈ Y√

hT(y0). We have to find an open neighborhood V (y0) of y0such that

hT(V (y0)) ⊂ U(hT(y0)).

9ǫ/2and let U(hT(y0)) be an open neighborhood of

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