# Isomorphisms from Extremely Regular Subspaces ofC0K into C0S,X Spaces

Authors: Manuel Felipe Cerpa-Torres; Michael A. Rincón-Villamizar
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

For a locally compact Hausdorff spaceK and a Banach space X, let C0K,X be the Banach space of all X-valued continuous functions defined on K, which vanish at infinite provided with the sup norm. If X is ℝ, we denote C0K,X as C0K. If AK be an extremely regular subspace of C0K and T:AK⟶C0S,X is an into isomorphism, what can be said about the set-theoretical or topological properties of K and S? Answering the question, we will prove that if X contains no copy of c0, then the cardinality of K is less than that of S. Moreover, if TT−1<3 and AK is also a subalgebra of C0K, the cardinality of the αth derivative of K is less than that of the αth derivative of S, for each ordinal α. Finally, if λX>1 and TT−1<λX, then K is a continuous image of a subspace of S. Here, λX is the geometrical parameter introduced by Jarosz in 1989: λX=infmaxx+λy:λ=1:x=y=1. As a consequence, we improve classical results about into isomorphisms from extremely regular subspaces already obtained by Cengiz.


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## Body

## 1. Introduction

All Banach spaces here are assumed to be real. For a locally compact Hausdorff spaceK and Banach space X, let C0K,X be the Banach space of all X-valued continuous functions defined in K which vanish at infinite, endowed with the supremum norm. If K is compact, we write CK,X instead of C0K,X. When X is ℝ, these spaces are denoted by C0K and CK, respectively. For a set A, we denote its cardinality by A. We denote X∼Y if X and Y are isomorphic Banach spaces and X ↪ Y if Y contains a subspace isomorphic to X. If X∼Y and T:X⟶Y is an isomorphism, the number TT−1 is called distortion of T; if TT−1<λ, 1<λ<+∞, we write X∼<λY. Other notations and definitions used here can be found in [1].The classical Banach–Stone theorem established that ifC0K and C0S are isometric, then K and S are homeomorphic [2–5]. Amir and Cambern, independently, showed that the conclusion of Banach–Stone theorem holds if we assume that C0K∼<2C0S [6, 7]. Moreover, it is well-known that 2 is the best number in this result [8]. On the other hand, the number 2 can be replaced by 3 in the case of countable compact metric spaces [9].Many authors have obtained generalizations of the Banach–Stone theorem, both in the scalar and vector-valued case (see [10–14]). Among the many generalizations of this theorem, we would like to highlight the following one. First, we recall a definition introduced by Cengiz [15].Definition 1.
A closed subspaceAK of C0K is called extremely regular if for each k∈K, U⊂K open with k∈U and 0<ε<1, there is f∈AK such that 1=fk=f>ε>fw for all w∈K∖U. An extremely regular subspace AK of C0K is called extremely regular subalgebra if f⋅g∈AK whenever f,g∈AK.
For examples of extremely regular subspaces (subalgebras), see [15]. The next result was obtained by Cengiz [16].Theorem 1.
LetAK and ℬS be extremely regular subspaces of C0K and C0S, respectively. If AK∼<2ℬS, then K and S are homeomorphic.Despite this, the conclusion of the Banach–Stone is far from to be valid for arbitrary isomorphisms. Thus, it is natural asking.Problem 1.
Which set-theoretical and topological properties are preserved by isomorphisms of extremely regular subspaces?
So, the aim of this paper is to give answers to the problem posed above. In this direction, Cengiz showed that, under additional topological hypothesis onK and S, existence of an isomorphism between extremely regular subspaces of C0K and C0S, respectively, implies that K=S [17, 18]. Our first result shows that Cengiz’s result holds for arbitrary locally compact Hausdorff spaces. So, we prove the following.Theorem 2.
LetAK and ℬS be extremely regular subspaces of C0K and C0S, respectively. If AK∼ℬS, then K=S.In order to prove Theorem2, we establish the following vector-valued result, which generalizes ([19], Theorem 2).Theorem 3.
LetK be infinite. For an extremely regular subspace AK of C0K and a Banach space X, we have(1)AK↪C0S,X⟹c0↪XorK≤S.Recall that ifS is a topological space, the derivative of S, denoted by S1, is obtained by deleting from S its isolated points. If α is an ordinal, we define the αth derivative Sα of S as follows: S0=S, Sα+1=Sα1, and Sα=∩β<αSβ, in the case where α is a limit ordinal.The following result is an extension of ([19], Theorem 5) for extremely regular subspaces.Theorem 4.
LetX be a Banach space containing no copy of c0. Let AK be an extremely regular subalgebra of C0K and T:AK⟶C0S,X an into isomorphism. For each ordinal α such that Sα is infinite, we have(2)TT−1<3⟹Kα≤Sα.Moreover, ifSα is finite, then so is Kα.On the other hand, a well-known generalization of the Banach–Stone theorem was given by Holsztyński who showed that ifC0K is isometrically isomorphic to a subspace of C0S, then K is a continuous image of a subset of S [20]. Jarosz extended this result for extremely regular subspaces of C0K and into isomorphisms with distortion less than 2 [21]. Moreover, in the setting of metric spaces, Jarosz obtained a vector-valued extension of Holsztyński’s theorem for isomorphisms from extremely regular subspaces of C0K into C0S,X spaces [14]. Our last theorem is a vector-valued version of the previously mentioned Jarosz’s result which is valid for all locally compact Hausdorff spaces. Before stating it, we recall a parameter introduced by Jarosz [14] for all Banach spaces X:(3)λX=infmaxx+λy:λ=1:x=y=1.Theorem 5.
LetAK be an extremely regular subspace of C0K and X a Banach space satisfying λX>1. If T:AK⟶C0S,X is an into isomorphism satisfying TT−1<λX, then K is a continuous image of a subset of S.Paper is organized as follows: in the first section, we will establish some properties for extremely regular subspaces ofC0K. In the remaining sections, we will prove Theorems 2, 3, 4, and 5. The idea of the proof of these theorems is inducing a set-valued map from isomorphisms defined on extremely regular subspaces and then improving the arguments given in [19] and [22] in order to obtain the wanted conclusions.


## 2. On Extremely Regular Subspaces

In this section, we prove some properties about extremely regular spaces. We start by stating the notation used throughout the paper.The unit ball of a Banach spaceX is denoted by BX. We identify dual space C0K∗ with the space of regular countably additive bounded measures and denote it by MK ([23], p. 222). The space MK will be equipped with the weak∗ topology inherited from C0K∗. The total variation of a measure μ∈MK on a Borel set E is denoted by μE and its norm by μ=μK. As usual, δk denotes the Dirac measure at k∈K.Letk∈K be given and Vk is a fundamental system of open neighborhoods of k. Consider the set Ck=Vk×0,∞ and define a partial order in Ck as follows: U,t≺V,s if and only if V⊂U and s<t. Note that Ck is a directed set. Let AK be an extremely regular subspace of C0K. It is easy to see that there exists a net fU,tU,t∈Ck in AK satisfying(1)
fU,t=fU,tk=1(2)
fU,tw<t for all w∈K\UWe writeU,t,fU,tU,t∈Ck⟷k to indicate that above conditions are satisfied.The next result is proved in ([24], Lemma 1) for compact spaces. The proof also works in the locally compact case, so we restate it as follows.Lemma 1.
LetAK be an extremely regular subspace of C0K and k∈K given. Suppose that U,t,fU,tU,t∈Ck⟷k. If μ∈MK, then(4)limU,t∈Ck∫KfU,tdμ=μk.It is well known that ifK is infinite, then C0K contains a subspace isometric to c0. We will prove an analogous result for extremely regular subspaces of C0K.Proposition 1.
LetK be infinite and AK an extremely regular subspace of C0K. Then AK contains a subspace isomorphic to c0.Proof.
Letkn be a sequence of distinct elements of K and Un:n∈ℕ a sequence of open subsets of K satisfying Un∩Um=∅ if m≠n and kn∈Un for all n∈ℕ. For each n∈ℕ, let fn∈AK be such that(1)
fnkn=fn=1;(2)
fnw<1/2n, if w∈K\Un.
We claim thatfn is a weakly unconditionally Cauchy sequence. Let an be a bounded sequence and m∈ℕ fixed. Then(5)α0≔sup∑j=1majfjk:k∈K\∪j=1mUj≤∑j=1maj2j≤supn∈ℕan.
On the other hand, ifp∈1,…,m is given, we have(6)αp≔sup∑j=1majfjk:k∈Up≤ap+∑j=1,j≠pmaj2j≤2supn∈ℕan.
From inequalities (5) and (6), it follows that(7)∑j=1majfj=sup∑j=1majfjk:k∈K=maxα0,α1,…,αm≤2supn∈ℕan.
Sincem∈ℕ is arbitrary, we have(8)supn∈ℕ∑j=1najfj≤2supn∈ℕan.
So, we prove the claim. Now, observe thatinffn:n∈ℕ>0. By the well-known Bessaga–Pełczyński theorem ([1], Theorem 2.4.11), we conclude that c0 embeds in AK.


## 3. On Isomorphisms from Extremely Regular Spaces intoC0S,X Spaces

The aim of this section is proving Theorem3, and as a consequence, we give a proof of Theorem 2 which was proved by Cengiz under additional topological hypothesis. First, we state two lemmas. A proof of the following one can be found in ([25], Proposition 5.1).Lemma 2.
LetX be a Banach space. If x,y∈X satisfy minx,y≥η for some η>0, then there are a1,a2∈K with maxa1,a2≤1 such that a1x+a2y≥ηλX.The following result was proved for compact spaces in ([24], Theorem 2.4). By modifying slightly the proof showed in [24], we see that result also holds in the locally compact case.Theorem 6.
LetAK be an extremely regular subspace of C0K and T:AK⟶C0S be an into isomorphism. For each k∈K, we have(9)supT∗δsk:s∈S≥1TT−1.To go on, we introduce a set-valued map which plays a fundamental role in our proofs.Notation 1.
LetAK be an extremely regular space and T:AK⟶C0S,X an into isomorphism. For any r>0, we define a set-valued map from S to K as follows:(10)Ωk,r=s∈S:T∗x∗⋅δsk≥rfor somex∗∈BX∗,where x∗⋅δs∈C0S,X∗ is defined by x∗⋅δsg=x∗gs for all g∈C0S,X.
Given a set-valued mapΛ from K to S, for any F⊂S, we define(11)Λ−1F=k∈K:Λk∩F≠∅.
In the case whereF=s is a singleton, we write Λ−1s instead of Λ−1s.Lemma 3.
LetX be a Banach space containing no copy of c0 and AK an extremely regular space of C0K. Let T:AK⟶C0S,X be an into isomorphism. For any r>0 and s∈S, Ωr−1s is finite.Proof.
Suppose that there iss0∈S such that Ωr−1s0 is infinite and take kn:n∈ℕ a sequence of distinct elements of Ωr−1s0. For each n∈ℕ, there is xn∗∈BX∗ such that(12)T∗xn∗⋅δs0kn≥r.
Fixa∈ℝ with 0<a<r. By Lemma 1, if U,t,fU,tU,t∈Ck1⟷k1 we have(13)limU,t∈Ck1∫KfU,tdT∗x1∗⋅δs0=T∗x1∗⋅δs0k1≥r>a.
So, there existsU1,t1∈Ck1 such that if U,t∈Ck1 and U1,t1≺U,t then(14)∫KfU,tdT∗x1∗⋅δs0>a.
SetU1≔U1 and choose n1∈ℕ satisfying 1/2n1<t1. Thus, U1,t1≺U1,1/2n1 and(15)TfU1,1/2n1s0≥x1∗TfU1,1/2n1s0=∫KfU1,1/2n1dT∗x1∗⋅δs0>a.
By settingf1≔fU1,1/2n1, we have(1)
f1=f1k1=1(2)
f1w<1/2n1 for all w∈K\U1(3)
Tf1s0>a
Now by induction, assume that there are disjoint open subsetsU1,…,Um of K, f1,…,fm∈AK, and n1,n2,…,nm∈ℕ with n1<n2<⋯<nm satisfying(1)
fj=f1kj=1(2)
fjw<1/2nj for all w∈K\Uj(3)
Tfjs0>a
for allj∈1,…,m. Hence, by Lemma 1, if U,t,fU,tU,t∈Ckm+1⟷km+1, we have(16)limU,t∈Ckm+1∫KfU,tdT∗xm+1∗⋅δs0=T∗xm+1∗⋅δs0km+1≥r>a.
Thus, there existsUm+1,tm+1∈Ckm+1 such that if U,t∈Ckm+1 and Um+1,tm+1≺U,t, then(17)∫KfU,tdT∗xm+1∗⋅δs0>a.
LetVm+1⊂K be open with km+1∈Vm+1 and Vm+1∩Uj=∅ for all j∈1,…,m and set Um+1=Vm+1∩Um+1. If nm+1∈ℕ satisfies nm<nm+1 and 1/2nm+1<tm+1, then Um+1,tm+1≺Um+1,1/2nm+1 and(18)TfUm+1,1/2nm+1s0≥xm+1∗TfUm+1,1/2nm+1s0=∫KfUm+1,1/2nm+1dT∗xm+1∗⋅δs0>a.
Finally, we setfm+1=fUm+1,1/2nm+1. Thus(1)
fm+1=f1km+1=1(2)
fm+1w<1/2nm+1 for all w∈K\Um+1(3)
Tfm+1s0>a
Hence, there exists a sequence of mutually disjoint open subsetsUm:m∈ℕ of K, a sequence fm:m∈ℕ in AK, and a strictly increasing sequence of natural numbers nm:m∈ℕ such that(1)
fm=f1km=1(2)
fmw<1/2nm for all w∈K\Um(3)
Tfms0>a
for allm∈ℕ. By arguing as in Proposition 1, it is easy to check that Tfns0 is a weakly unconditional Cauchy sequence. Since(19)infTfns0:n∈ℕ≥a>0,we conclude that c0 embeds in X by ([1], Theorem 2.4.11), contradicting the hypothesis. Thus, Ωr−1s must be finite for each s∈S.Lemma 4.
If0<r<1/TT−1, then for all k∈K, we have Ωk,r≠∅.Proof.
Notice thatC0S,X embeds isometrically in C0S×BX∗ via the isomorphism f⟼f^, where f^s,x∗=x∗fs for all s,x∗∈S×BX∗. Let T^:AK⟶C0S×BX∗ be the into isomorphism given by T^f=Tf^ for f∈AK. By Theorem 6, we have(20)supT^∗δs,x∗k:s,x∗∈S×BX∗≥1T^T^−1.
SinceT^−1≤T−1 and T=T^, it follows that(21)supT^∗δs,x∗k:s,x∗∈S×BX∗≥1TT−1.
Thus, there iss,x∗∈S×BX∗, satisfying T^∗δs,x∗k≥1/TT−1>r.
Now, ifU,t,fU,tU,t∈Ck⟷k, from Lemma 1 we obtain(22)T^∗δs,x∗k=limU,t∈CkT^∗δs,x∗fU,t=limU,t∈Ckx∗TfU,ts=limU,t∈CkT∗x∗⋅δsfU,t=T∗x∗⋅δsk.
Whence, there ares∈S and x∗∈BX∗ such that T∗x∗⋅δsk≥r; that is, Ωk,r≠∅.Proof of Theorem 1.
LetT:AK⟶C0S,X be an into isomorphism. We claim that S is infinite. Indeed, if S were finite, say S=n, then C0S,X≅Xn. Since K is infinite, c0↪AK by Proposition 1. Whence, c0 embeds in Xn and by ([26], Theorem 1) we conclude that c0↪X which is impossible. So, S must be infinite. Now, let 0<r<1/TT−1 be fixed. If k∈K is given, by Lemma 4, we have Ωk,r≠∅. By taking s0∈Ωk,r, we obtain that k∈Ωr−1s0. Thus,(23)K=∪s∈SΩr−1s.
On the other hand, by Lemma3, Ωr−1s is finite for all s∈S. Since S is infinite, the equation above implies that(24)K=∪s∈SΩr−1s≤sups∈SΩr−1sS≤ℵ0S=S.
Thus,K≤S.Proof of Theorem 2.
Suppose thatAK∼ℬS, then AK↪C0S and ℬS↪C0K. If K were finite, then S also must be finite by Theorem 3. By ([15], Main theorem), we have AK=C0K≅ℝn and ℬS=C0S≅ℝm, where K=n and S=m. So, m=n. If K and S are infinite, then by Theorem 2, we have K≤S and S≤K. Thus, by the Cantor–Schröder–Bernstein theorem, we conclude that K=S.


## 4. On Isomorphisms from Extremely Regular Spaces intoC0S,X Spaces with Distortion Less than 3

In this section, we prove Theorem4. We need the following lemma for isomorphisms from extremely regular subalgebras of C0K into C0S,X spaces. The proof follows by adapting the argument used in ([19], Proposition 3.1) for extremely regular subalgebras. First, we state a notation.Notation 2.
Forf∈C0K,X and r>0, we denote(25)Kf,r=k∈K:fk≥r.Lemma 5.
LetAK be an extremely regular subalgebra of C0K. Let T:AK⟶C0S,X be an into isomorphism with TT−1<3 and fix δ∈0,1 such that TT−1<3δ. If h,g∈AK satisfy 0≤h≤g≤1 and hKα>δ, then(26)∩h≤f≤gKTf,η/T−1∩Sα≠∅,where(27)η=3δ−TT−12.Proof of Theorem 3.
Letα be an ordinal and 0<δ<1 satisfying TT−1<3δ. Also let η be as in Lemma 5. We shall use the fact that closed subalgebras of C0K are lattices ([23], Lemma 4.48). Consider the collection(28)ℱk=f∈AK:0≤f≤1,fk>δ,for each k∈K and define(29)Λk=∩f∈ℱkKTf,ηT−1.
Observe that(30)KTf,ηT−1∩Sα:f∈ℱk,fork∈Kα,is family of compact subsets of S. So, if we prove that this family has the finite intersection property, we will immediately conclude that Λk∩Sα≠∅ for k∈Kα.
Letf1,…,fn∈ℱk then h=min1≤j≤nfj and g=max1≤j≤nfj are in ℱk. So, by Lemma 5, we have(31)∩j=1nKTfj,ηT−1∩Sα⊃∩h≤f≤gKTf,ηT−1∩Sα≠∅.
Thus by compactness, the familyKTf,η/T−1∩Sα:f∈ℱk has nonempty intersection. It follows that(32)Λk∩Sα=∩f∈ℱkKTf,ηT−1∩Sα≠∅.
Now, consider the set-valued map fromK to S given by k⟼Λk. We will prove that Λ−1s is finite for all s∈S. If not, there is s0∈S such that Λ−1s0 is infinite. Since K is regular, there is a sequence Un:n∈ℕ of open disjoint subsets of K such that Un∩Λ−1s0≠∅. For each n∈ℕ, pick kn∈Un∩Λ−1s0. Also, for each n∈ℕ, let fn∈AK be such that 0≤fn≤1, fnkn=1, and fnw<1/2n for all w∈K\Un. Note that fn∈ℱkn and in view of kn∈Λ−1s0, we have s0∈Λkn for each n∈ℕ. Whence,(33)infTfns0:n∈ℕ≥ηT−1>0.
It is not difficult to show thatTfns0 is weakly unconditionally convergent. By (33) and ([1], Theorem 2.4.11), we conclude that c0 embeds in X which is impossible. Henceforth, Λ−1s is finite for all s∈S.
Now, ifKα≠∅ and k∈Kα, we have Λk∩Sα≠∅ which implies that k∈Λ−1s where s∈Λk∩Sα Thus,(34)Kα⊂∪s∈SαΛ−1s.
Consequently, ifSα is finite, then so is Kα. Now if Sα is infinite, from the equation above, we have(35)Kα≤∪s∈SαΛ−1s≤sups∈SαΛ−1sSα≤ℵ0Sα=Sα,that is Kα≤Sα, and we are done.


## 5. On Isomorphisms from Extremely Regular Spaces intoC0S,X Spaces with Distortion Less than λX

Finally, we prove Theorem5. We remark that Jarosz obtained the statement of theorem in the metric setting ([14], Theorem 1). By following a new approach, we show that conclusion of theorem holds in the general topological case.Proof of Theorem 4.
LetT:AK⟶C0S,X be a into isomorphism with TT−1<λX. By considering the isomorphism A=T/T, we have A=1 and A−1=TT−1<λX. So, we may assume that T=1 and T−1<λX. Let r∈ℝ such that(36)1λX<r<1T−1.
The following claim is a consequence of Lemma4.Claim 1.
For eachk∈K, we have Ωk,r≠∅.Claim 2.
Fork1,k2∈K with k1≠k2, we have Ωk1,r∩Ωk2,r=∅.
Suppose thatΩk1,r∩Ωk2,r≠∅ and let s∈Ωk1,r∩Ωk2,r. There are x1∗,x2∗∈BX∗ satisfying(37)T∗x1∗⋅δsk1≥r,T∗x2∗⋅δsk2≥r.
Let0<ε<1 be given, then(38)T∗x1∗⋅δsk1>r−ε,T∗x2∗⋅δsk2>r−ε.
IfU1,t,fU1,tU1,t∈Ck1⟷k1 and U2,t,fU2,tU2,t∈Ck2⟷k2, it follows from Lemma 1 that(39)limU1,t∈Ck1T∗x1∗⋅δsfU1,t>r−ε,limU2,t∈Ck2T∗x2∗⋅δsfU2,t>r−ε.
Thus, there isU01,t1∈Ck1 such that if U1,t∈Ck1 and U01,t1≺U1,t, then(40)T∗x1∗⋅δsfU1,t>r−ε.
In a similar way, there existsU02,t2∈Ck2 such that if U2,t∈Ck2 and U02,t2≺U2,t, we have(41)T∗x2∗⋅δsfU2,t>r−ε.
ChooseU1 and U2 open subsets of K with k1∈U1⊂U01, k2∈U2⊂U02, and U1∩U2=∅, and let t∈0,∞ be such that t<mint1,t2,ε. Then, U01,t1≺U1,t and U02,t2≺U2,t. In view of inequalities (40) and (41), we obtain(42)T∗x1∗⋅δsfU1,t>r−ε,T∗x2∗⋅δsfU2,t>r−ε.
Hence,(43)TfU1,ts>r−ε,TfU2,ts>r−ε.
By Lemma2, there are a1,a2∈K with maxa1,a2≤1 satisfying(44)a1TfU1,ts+a2TfU2,ts>r−ελX.
Sincea1fU1,t+a2fU2,t≤1+ε and T=1, we infer that(45)1+ε≥Ta1fU1,t+a2fU2,t≥a1TfU1,ts+a2TfU2,ts>r−ελX.
Arbitrariness ofε>0 implies that 1≥rλX which contradicts (36). So, we are done.
Now, letS0=∪k∈KΩk,r and define ϕ:S0⟶K by ϕs=k if and only if s∈Ωk,r. Claims 1 and 2 show that ϕ is a surjective function.Claim 3.
The functionϕ is continuous.
Letsγγ∈Γ be a net in S0 converging to s∈S0 and assume that ϕsγ=kγ↛ϕs=k. Let 0<ε<1 be given. Since ϕs=k, there is x∗∈BX∗ such that(46)T∗x∗⋅δsk≥r>r−ε.
Now, ifU,t,fU,tU,t∈Ck⟷k, we have by Lemma 1(47)limU,t∈CkT∗x∗⋅δsfU,t>r−ε.
Sincekγ↛k, there is a compact neighborhood V⊂K with k∈V such that for any γ∈Γ, there exists γ′≥γ satisfying kγ′∉V. In view of (47), there is U1⊂K open with k∈U1⊂V and 0<t1<ε such that(48)TfU1,t1s≥T∗x∗⋅δsfU1,t1>r−ε,whereas sγ⟶s, there is γ1∈Γ such that kγ1∉V and(49)TfU1,t1sγ1>r−ε.
Because ofϕsγ1=kγ1, there is x1∗∈BX∗ such that(50)T∗x1∗⋅δsγ1kγ1≥r>r−ε.
So, ifU1,t,fU1,tU1,t∈Ckγ1⟷kγ1, by Lemma 1 we get(51)limU1,t∈Ckγ1T∗x1∗⋅δsγ1fU1,t>r−ε.
From (51), there is U2⊂K open with kγ1∈U2, U2∩V=∅, and 0<t2<ε satisfying(52)TfU2,t2sγ1≥T∗x2∗⋅δsγ1fU2,t2>r−ε.
It follows from (49), (52), and Lemma 2 that there exist a1,a2∈K satisfying maxa1,a2≤1 and(53)a1TfU1,t1sγ1+a2TfU2,t2sγ1>r−ελX.
Notice thata1fU1,t1+a2fU2,t2≤1+ε, so, from the inequality above, we obtain(54)1+ε≥Ta1fU1,t1+a2fU2,t2≥a1TfU1,t1sγ1+a2TfU2,t2sγ1>r−ελX.
Since0<ε<1 was chosen arbitrarily, we have 1≥rλX which is impossible by (36). This completes the proof of Theorem 4.



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