Vol. 2, No. 3 (2009) 445–452 c
World Scientific Publishing Company
SEPARATING LINEAR MAPS OF CONTINUOUS FIELDS OF BANACH SPACES∗
Chi-Wai Leung Department of Mathematics
The Chinese University of Hong Kong, Hong Kong cwleung@math.cuhk.edu.hk
Chung-Wen Tsai†and Ngai-Ching Wong‡ Department of Applied Mathematics
National Sun Yat-sen University Kaohsiung, 80424, Taiwan
†tsaicw@math.nsysu.edu.tw,‡wong@math.nsysu.edu.tw
In this paper, we give a complete description of the structure of separating linear maps between continuous fields of Banach spaces. Some automatic continuity results are ob- tained.
Keywords: Separating linear maps; automatic continuity; continuous fields of Banach spaces; Banach bundles.
AMS Subject Classification: 46L40, 46H40, 46E40
1. Introduction
LetT be a locally compact Hausdorff space, called base space. Suppose for each t inT there is a (real or complex) Banach spaceEt. Avector field xis an element in the product spaceQ
t∈TEt, that is,x(t)∈Et, for allt∈T.
Definition 1.1 ([5,3]). A continuous fieldE= (T,{Et},A)of Banach spacesover a locally compact space T is a family {Et}t∈T of Banach spaces, with a set Aof vector fields, satisfying the following conditions.
(i) Ais a vector subspace ofQ
t∈TEt.
(ii) For everytinT, the set of allx(t) withxin Ais dense inEt.
(iii) For everyxin A, the function t7→ kx(t)kis continuous onT and vanishes at infinity.
∗This work is jointly supported by Hong Kong RGC Research Grant (2160255), and Taiwan NSC grant (NSC96-2115-M-110-004-MY3).
445
(iv) Let x be a vector field. Suppose for everyt in T and every >0, there is a neighborhoodU of tand ay in Asuch thatkx(s)−y(s)k< for allsin U. Thenx∈A.
Elements inAare called continuous vector fields.
When allEt equal to a fixed Banach spaceE, andAconsists of all continuous functions fromT intoEvanishing at infinity, we callEaconstant field. In this case, we writeA=C0(T, E), or A=C(T, E) whenT is compact, as usual.
It is not difficult to see thatAbecomes a Banach space under the normkxk= supt∈Tkx(t)k. Ifgis a bounded continuous scalar-valued function onT, andx∈A, thent7→g(t)x(t) defines a continuous vector fieldgxonT. The set of allx(t) with xinAcoincides withEt for everytinT. Moreover, for any distinct pointss, tinT and anyαinEsandβinEt, there is a continuous vector fieldxsuch thatx(s) =α andx(t) =β (see, e.g., [5,12]).
A map θ : A→B is called a homomorphism between two continuous fields of Banach spaces (X,{Ex}x,A) and (Y,{Fy},B) if there is a mapϕ:Y →X and a linear mapHy :Eϕ(y)→Fy for eachyin Y such that
θ(f)(y) =Hy(f(ϕ(y)), for allf ∈A, for ally∈Y. (1.1) A mapθ is said to beseparating (orstrictly separating as in [1]) if
kf(x)kkg(x)k= 0, for allx∈X, implies kθ(f)(y)kkθ(g)(y)k= 0, for ally∈Y.
The study of when a separating linear map is a homomorphism has been the focus of much research in the past. For example, in [10], Jarosz gives a complete description of an unbounded separating linear mapθ:C(X)→C(Y), where X, Y are compact Hausdorff spaces, and this is extended to locally compact spaces in [7,11]. On the other hand, Jamison and Rajagopalan [9] show that every bounded separating linear map θ : C(X, E) → C(Y, F) between continuous vector valued function spaces carries a standard form (1.1). Chan [2] extends this to bounded separating linear maps between two function modules.
In this paper, we present a complete description of separating linear maps θ : A→Bbetween continuous fields of Banach spaces (X,{Ex}x,A) and (Y,{Fy},B) on locally compact Hausdorff base spaces. Essentially, these maps carry the standard form (1.1). In caseθ is bijective, and both θ and θ−1 are separating, we shall see thatϕ:Y →X is a homeomorphism. Moreover,θ, as well as the fiber linear maps Hy, is automatically bounded in many situations. Our results unify and extend those shown in [9,10,2,7,11,1,8].
Another example of continuous fields of Banach spaces comes fromBanach bun- dles. (The readers are referred to [4,6] for the definitions.) For a Banach bundle ξ = (p, E, T), define Γ0(ξ) to be the set of all continuous cross sections of ξ which vanishes at infinity. In this case, we writeE= (T,{Et},Γ0(ξ)). It is not difficult to see that Γ0(ξ) satisfies the conditions (i), (iii), (iv) in Definition 1.1. We refer to Appendix C in [6] where it is shown that ifT is locally compact, then for any point
xin Ethere is a continuous cross sectionf such thatf(p(x)) =x. Thus, condition (ii) follows. Therefore, all results in this paper apply to Banach bundles. For further development in this line, readers are referred to [13].
2. The results
For a locally compact Hausdorff spaceX, we write X∞=X∪ {∞},
for its one-point compactification. If X is already compact, then the point ∞ at infinity is an isolated point in X∞. Moreover, we identify
C0(X) ={f ∈C(X∞) :f(∞) = 0},
and other similar spaces for those of continuous functions onX vanishing at infinity.
For a continuous field (X,{Ex}x,A) of Banach spaces, set for eachx inX the sets Ix={f ∈A:f vanishes in a neighborhood inX∞ ofx},
Mx={f ∈A:f(x) = 0}.
Theorem 2.1. Letθ:A→Bbe a separating linear map between continuous fields of Banach spaces (X,{Ex},A), (Y,{Fy},B) over locally compact Hausdorff spaces X, Y, respectively. Set
Y0=\
{kerθ(f) :f ∈A}.
Then, ∞ ∈ Y0 is compact and there is a continuous map ϕ : Y \Y0 → X∞ such that
θ(Iϕ(y))⊆Iy, for ally∈Y \Y0. Set
Y1={y∈Y \Y0:θ(Mϕ(y))⊆My}, Y2={y∈Y \Y0:θ(Mϕ(y))*My}.
Then there is a linear mapHy:Eϕ(y)→Fy for each y inY1 such that θ(f)(y) =Hy(f(ϕ(y)), for ally∈Y1.
The exceptional set Y2 is open in Y∞, and ϕ(Y2) consists of finitely many non- isolated points inX∞.
Moreover, θ is bounded if and only if Y2 =∅ and all Hy are bounded. In this case,
kθk= sup
y∈Y
kHyk.
We divide the proof into several lemmas as in [10,11]. Clearly, Y0 is compact and contains∞. For eachy in Y \Y0, let
Zy ={x∈X∞:θ(Ix)⊆Iy}.
Lemma 2.1. Zy is a singleton, for all y inY \Y0.
Proof. Suppose on the contrary thatZy=∅for someyinY \Y0. Then for eachx inX∞there is anfxin Ix vanishing in a compact neighborhoodUxofxsuch that θ(fx)∈/Iy. By compactness,
X∞=Ux0∪Ux1∪ · · · ∪Uxn
for some pointsx0=∞, x1, . . . , xn inX∞. Let 1 =h0+h1+· · ·+hn
be a continuous partition of unity such that hi vanishes outside Uxi for i = 0,1, . . . , n. For anyginA, the separating property ofθ implies that the product of the norm functions ofθ(hig) andθ(fxi) is always zero, and then
θ(fxi)∈/Iy implies θ(hig)(y) = 0, i= 0,1, . . . , n.
Hence,θ(g)(y) = 0 for allg∈A. This gives a contradiction that y∈Y0.
Next letx1,x2be distinct points inZy. In other words,θ(Ixi)⊆Iy fori= 1,2.
Choose compact neighborhoodsV, U of x1 in X∞ such that V is contained in the interior ofU, andx2∈/U. Letg∈C(X∞) such thatg= 1 onV andg= 0 outside U. Then for allf inA, the facts (1−g)f ∈Ix1 andgf∈Ix2 ensure thatθ(f)∈Iy. In particular,y∈Y0, a contradiction again.
Define a mapϕ:Y \Y0→X∞by
Zy={ϕ(y)}.
In other words,θ(Iϕ(y))⊆Iy, or
f ∈Iϕ(y) implies θ(f)∈Iy, for ally∈Y \Y0. (2.1) Lemma 2.2. ϕ:Y \Y0→X∞ is continuous.
Proof. Supposeyλ →y in Y \Y0, but xλ =ϕ(yλ)→x 6=ϕ(y). By Lemma 2.1, θ(Ix)*Iy. LetUx,Uϕ(y)be disjoint compact neighborhoods ofx, ϕ(y), respectively.
Letg∈C(X∞) such thatg= 1 onUxandg= 0 onUϕ(y). Sincexλ→x, for allf in A, (1−g)fis eventually inIxλ. Thus,θ((1−g)f)∈Iyλ eventually. By the continuity of the norm function, θ((1−g)f)(y) = 0. On the other hand, gf ∈ Iϕ(y) implies θ(gf)∈Iy. Hence,θ(f)(y) = 0 for all f ∈A. This givesy∈Y0, a contradiction.
Denote byδy the evaluation map aty inY, i.e., δy(g) =g(y)∈Fy, for allg∈B.
Lemma 2.3. Let{yn}be a sequence in Y \Y0 such that ϕ(yn)are distinct points inX∞. Then
lim supkδyn◦θk<+∞.
Proof. Suppose not, by passing to a subsequence if necessary, we can assume the normkδyn◦θk> n4, and there is anfn inAsuch thatkfnk ≤1 andkθ(fn)(yn)k>
n3, forn= 1,2, . . .. Letxn =ϕ(yn) andVn, Unbe compact neighborhoods ofxn in X∞such that Vn is contained in the interior ofUn, and Un∩Um=∅, for distinct n, m= 1,2, . . .. Letgn∈C(X∞) such thatgn= 1 onVn andgn = 0 outsideUn for n= 1,2, . . .. Observe
θ(fn)(yn) =θ(gnfn)(yn) +θ((1−gn)fn)(yn)
=θ(gnfn)(yn), as (1−gn)fn∈Ixn. So we can assumefn is supported inUn, for n= 1,2, . . .. Let
f = X∞
n=1
1
n2fn∈A.
Since n2f −fn ∈ Ixn, we have n2θ(f)(yn) = θ(fn)(yn) by (2.1), and thus kθ(f)(yn)k > n, for n = 1,2, . . .. As θ(f) in B has a bounded norm, we arrive at a contradiction.
Set
Y1={y∈Y \Y0:θ(Mϕ(y))⊆My}, Y2={y∈Y \Y0:θ(Mϕ(y))*My}.
Lemma 2.4. ϕ(Y2)is a finite set consisting of non-isolated points inX∞. Proof. Letx=ϕ(y) withy inY2. Then by (2.1) we have
θ(Ix)⊆Iy but θ(Mx)*My.
Since, by Uryshons Lemma,Ixis dense inMx, this implies the linear operatorδy◦θ is unbounded. By Lemma 2.3, we can only have finitely many of suchx’s. Soϕ(Y2) is a finite set. Moreover, if x is an isolated point in X∞ then Ix =Mx, and thus x /∈ϕ(Y2).
Proof. [Proof of Theorem 2.1] Lety ∈Y1, we haveθ(Mϕ(y))⊆My. Hence, there is a linear operatorHy:Eϕ(y)→Fy such that
θ(f)(y) =Hy(f(ϕ(y))), for allf ∈A. (2.2) Next we want to see thatY2is open, or equivalently,Y0∪Y1is closed inY∞. Let yλ→y withyλ inY0∪Y1. We want to show thaty∈Y0∪Y1. SinceY0 is compact, we might assumeyλ∈Y1for allλ.
In case there is any subnet of {ϕ(yλ)}consisting of only finitely many points, we can assume ϕ(yλ) = ϕ(y) for all λ. Then for all f in A, f(ϕ(y)) = 0 implies
f(ϕ(yλ)) = 0, and thusθ(f)(yλ) = 0 for allλby (2.2). By continuity,θ(f)(y) = 0.
Consequently,θ(Mϕ(y))⊆My, and thusy∈Y0∪Y1.
In the other case, every subnet of {ϕ(yλ)} contains infinitely many points.
Lemma 2.3 asserts thatM= lim supkHyλk<+∞. This gives
kθ(f)(y)k= limkθ(f)(yλ)k= limkHyλ(f(ϕ(yλ)))k ≤Mkf(ϕ(y))k.
Thus, iff(ϕ(y)) = 0 we haveθ(f)(y) = 0. Consequently,y∈Y0∪Y1. Observe that the boundedness of θimpliesY2=∅. Moreover,
kθk= sup{kθ(f)k:f ∈Awithkfk= 1}
= sup{kHy(f(ϕ(y)))k:f ∈Awithkfk= 1, y∈Y1}
≤sup{kHyk:y∈Y1}.
The reverse inequality is plain.
Finally, we supposeY2=∅and allHy are bounded. We claim that supkHyk<
+∞. Otherwise, there is a sequence {yn} in Y1 such that limn→∞kHynk= +∞.
By Lemma 2.3, we can assume allϕ(yn) =xinX. Lete∈Exandf ∈Asuch that f(x) =e. Then
kHyn(e)k=kθ(f)(yn)k ≤ kθ(f)k, n= 1,2, . . . .
It follows from the uniform boundedness principle that supkHynk<+∞, a contra- diction. It is now obvious that θis bounded.
The following extends the results for constant fields shown in [1,8].
Theorem 2.2. Let(X,{Ex},A),(Y,{Fy},B)be continuous fields of Banach spaces over locally compact Hausdorff spacesX, Y, respectively. Letθ:A→Bbe a bijective linear map such that both θ and its inverse θ−1 are separating. Then there is a homeomorphismϕfromY ontoX, and a bijective linear operator Hy :Eϕ(y)→Fy
for eachy inY such that
θ(f)(y) =Hy(f(ϕ(y))), for allf ∈A, for ally∈Y.
Moreover, at most finitely manyHy are unbounded, and this can happen only when y is an isolated point in Y. In particular, if X (or Y) contains no isolated point thenθ is automatically bounded.
Proof. Since θ is onto, we have Y0 = {∞}. Because both θ, θ−1 are separating, there are continuous mapsϕ:Y →X∞ andψ:X →Y∞such that
θ(Iϕ(y))⊆Iy and θ−1(Iψ(x))⊆Ix, for allx∈X, y∈Y.
In caseψ(x)6=∞, this gives
θ(Iϕ(ψ(x)))⊆Iψ(x)⊆θ(Ix), or
Iϕ(ψ(x))⊆Ix.
It follows ϕ(ψ(x)) = x for all x in X with ψ(x) 6= ∞. Similarly, we will have ψ(ϕ(y)) = y for all y in Y with ϕ(y) 6= ∞. Set X3 = X \ψ−1(∞) and Y3 = Y \ϕ−1(∞). It is then easy to see thatϕ=ψ−1 induces a homeomorphism from Y3 onto X3. By the bijectivity of θ, the open sets X3 and Y3 containX1 and Y1, respectively.
Next, we want to see that Y2 =∅and Y1 =Y3 =Y. Indeed, by Theorem 2.1, Y2∩Y3is open, and a finite set (asϕ(Y2) is). HenceY2∩Y3consists of isolated points inY, and so doesϕ(Y2∩Y3). It then follows from Lemma 2.4 thatY2∩Y3is empty.
Consequently,Y1=Y3andϕ(Y2)⊆ {∞}. Similarly,X1=X3andψ(X2)⊆ {∞}. It follows from (2.1) and the injectivity ofθthatϕ(Y), and thusϕ(Y1) =X1, is dense inX. AsX1is closed inX, we see thatX =X1and thusX2=∅. Correspondingly, Y =Y1 andY2=∅. It turns out thatϕis a homeomorphism fromY ontoX with inverseψ.
Now Y = Y1 and X = X1 implies that both θ and θ−1 can be written as homomorphisms of continuous fields of Banach spaces:
θ(f)(y) =Hy(f(ϕ(y))), for allf ∈A, for ally∈Y, θ−1(g)(x) =Tx(g(ψ(x))), for allg∈B, for allx∈X.
It is easy to see that the linear mapHy:Eϕ(y)7→Fy has an inverseTϕ(y)for every y inY, and thus it is bijective.
By Lemma 2.3, at most finitely manyHyare unbounded. Letybe a non-isolated point in Y. We will show that the linear mapHy is bounded. Suppose not, then for each n = 1,2, . . . there is an fn in A of norm one such that kθ(fn)(y)k = kHy(fn(ϕ(y)))k> n4. By the continuity of the norm ofθ(fn), there are all distinct pointsyn ofY in a neighborhood ofy such thatkθ(fn)(yn)k> n3. Letxn=ϕ(yn) in X for n = 1,2, . . .. Since ϕ is a homeomorphism, we can also assume that all xn are distinct with disjoint compact neighbourhoods Un. By multiplying with a norm one continuous scalar function, we can assume each fn is supported in Un. Letf =P
n 1
n2fn inA. Sincen2f −fn∈Ixn, we haven2θ(f)(yn) =θ(fn)(yn) and thuskθ(f)(yn)k> nforn= 1,2, . . .. This contradiction tells us thatHyis bounded for all non-isolatedyin Y1.
The last assertion follows from Theorem 2.1, and we have kθk = supkHyk
<+∞.
Remark 2.1.
(1) Unlike the scalar case, if any fiberEx of the continuous field of Banach spaces (X,{Ex},A) is of infinite dimension, someHy can be unbounded in Theorem 2.2. This happens even for the constant fields based on compact spaces. See Example 2.4 in [8].
(2) There is a counterexample in ([8], Example 3.1) of a continuous bijective sepa- rating linear map between constant fields based on nonhomeomorphic compact
spaces, whose inverse is not separating. So the biseparating assumption in The- orem 2.2 cannot be dropped.
Acknowledgment
We would like to express our gratitude to the referee for many helpful comments in improving the presentation of this paper.
References
1. J. Araujo and K. Jarosz, Automatic continuity of biseparating maps,Studia Math., 155(2003) 231–239.
2. J. T. Chan, Operators with the disjoint support property, J. Operator Theory, 24 (1990) 383–391.
3. J. Diximier,C*-algebras, North-Holland publishing company, Amsterdam–New York–
Oxford, 1977.
4. M. J. Dupr´e and R. M. Gillette,Banach bundles, Banach modules and automorphisms of C*-algebras, Pitman Research Notes in Mathematics Series 92, 1983.
5. J. M. G. Fell, The structure of algebras of operator fields, Acta Math.,106 (1961) 233–280.
6. J. M. G. Fell and R. S. Doran,Representations of *-Algebras, Locally Compact Groups, and Banach *-Algebraic Bundles, Volume 1, Academic, New York (1988).
7. J. J. Font and S. Hern´andez, On separating maps between locally compact spaces, Arch. Math. (Basel),63(1994) 158–165.
8. H.-L. Gau, J.-S. Jeang and N.-C. Wong, Biseparating linear maps between continuous vector-valued function spaces, J. Australian Math. Soc., Series A,74 no. 1 (2003) 101–111.
9. J. E. Jamison and M. Rajagopalan, Weighted composition operator onC(X, E), J.
Operator Theory,19(1988) 307–317.
10. K. Jarosz, Automatic continuity of separating linear isomorphisms, Canad. Math.
Bull.,33(1990) 139–144.
11. J.-S. Jeang and N.-C. Wong, Weighted composition operators of C0(X)’s, J. Math.
Anal. Appl.201(1996) 981–993.
12. R.-Y. Lee, On the C*-algebras of operator fields, Indiana Univ. Math. J. 25 no. 4 (1978) 303–314.
13. C.-W. Leung, C.-K. Ng and N.-C. Wong, Automatic Continuity andC0(Ω)-Linearity of Linear Maps BetweenC0(Ω)-Modules, preprint.
