Operators on C_{0}(L,X) whose range does not contain c_{0}

0.13 MB·English

by  Jarno Talponen

OPERATORS ON C (L,X) WHOSE RANGE
0
DOES NOT CONTAIN c
0
8
0 JARNOTALPONEN
0
2
Abstract. Thispapercontains thefollowingresults:

n (a)SupposethatX6={0}isaBanachspaceand(L,τ)isanon-emptylocally
a compact Hausdorff space without isolated points. Then each linear operator
J T:C0(L,X)→C0(L,X)whoserangedoesnotcontainanisomorphiccopyof
5 c00 satisfiestheDaugavet equality||I+T||=1+||T||.
1 (b)LetΓbeanon-emptysetandX,YbeBanachspacessuchthatXisreflexive
andYdoesnotcontainc0isomorphically. Thenanycontinuouslinearoperator
] T:c0(Γ,X)→Y isweaklycompact.
A
F
.
h
t 1. Introduction
a
m
InwhatfollowsX,YandZarerealBanachspacesdifferentfrom{0},Lisanon-
[ empty locally compact Hausdorff space and C (L,X) denotes the ||·|| -normed
0 ∞
BanachspaceofX-valuedcontinuousfunctionsonLvanishingatinfinity. Thiswork
1
v is related to the result due to P. Cembranos [2] that if K is an infinite compact
4 and X is an infinite-dimensional Banach space, then the Banach space C(K,X) of
1 X-valued continuous functions on K contains a complemented copy of c . See also
0
3 the related paper [1] where the Dieudonn´e property of C(K,X) is studied.
2
Taking another direction, we will study continuous linear operators of the type
.
1 (∗) T: C (L,X)→Y, where Y does not contain c isomorphically.
0 0
0
AbovethereisastructuraldisparitybetweenspacesC (L,X)andY,sincetypically
8 0
the former space contains copies of c in abundance. This difference has a strong
0 0
: impact on the properties of T. Namely, it turns out that the range of T in Y is
v
small in some sense.
i
X If L does not contain isolated points, then an operator T: C (L,X)→C (L,X)
0 0
r of type (∗) satisfies the Daugavet type equality ||I+T|| = 1+||T|| (see Theorem
a
2.1). See [5] for recent discussion on matters related to the Daugavet property.
If L is discrete, X is reflexive and c is not contained in Y isomorphically, then
0
an operator T: C (L,X)→Y is weakly compact (see Theorem 2.3).
0
Preliminaries. Here X and Y denote real Banach spaces. The closed unit ball
and the unit sphere of X are denoted by B and S , respectively. An identity
X X
mapping is denoted by I. An operator T: X → Y is weakly compact if T(B ) is
X
weakly compact. If X 6= {0}, then we say that X is non-trivial. For given sets
A ⊂ B the mapping χ : B → {0,1} is determined by χ (t) = 1 if and only if
A A
t ∈ A. We refer to [3], [4] and [6] for suitable background information including
definitions and basic results.
Date:February2,2008.
1991 Mathematics Subject Classification. Primary46B20;Secondary 46B28.
1
2 JARNOTALPONEN
2. Results
Theorem 2.1. Let X be a non-trivial Banach space and (L,τ) a non-empty lo-
cally compact Hausdorff space without isolated points. Then each linear operator
T: C (L,X) → C (L,X) whose range does not contain an isomorphic copy of c
0 0 00
satisfies the Daugavet equality
||I+T||=1+||T||.
Letusfirstmakesomepreparationsbeforegivingtheproof. Itiseasytoseethat
the range of T contains c isomorphically if and only if the closure of the range
00
contains c .
0
The assumption that L does not contain isolated points cannot be removed.
Indeed, if L is not a singleton, t ∈ L is an isolated point and X contains no iso-
0
morphic copyof c , then the linear operatorT: C (L,X)→X; F 7→−χ (·)F(·)
0 0 {t0}
is of type (∗) and satisfies ||T||=||I+T||=1.
The above Theorem2.1 holds analogouslyfor T: CB(L,X)→CB(L,X), essen-
tially with the same proof. Here CB(L,X) is the ||·|| -normed Banach space of
∞
X-valued bounded continuous functions on L.
For a linear operator T: C (L,X)→Y we denote
0
osc (A)=sup{||TF||: F ∈B , L\U ⊂F−1(0)} for A⊂L.
T C0(L,X)
Lemma 2.2. Let T: C (L,X)→Y be a linear operator, where Y does not contain
0
c0 isomorphically. Suppose that (Vn)n∈N is a sequence of pair-wise disjoint non-
empty open subsets of L. Then osc (V )→0 as n→0.
T n
Proof. By passing to a subsequence it suffices, without loss of generality, to show
that infn∈NoscT(Vn)=0. Indeed, assume to the contrary that there is some d>0
such that osc (V ) ≥ d for all n ∈ N. This means that one can find a sequence
T n
(Fn)n∈N ⊂(d1+1)BC0(L,X)suchthatFnissupportedinVnand||T(Fn)||=1forn∈
N. Note that for each finite subset I ⊂N it holds that F ∈(1 +1)B
i∈I i d C0(L,X)
as V are pair-wise disjoint. Since T is linear and continuous, we obtain that
n
P
1
sup T(ǫ F ) ≤ +1 ||T||,
i i
d
ǫ,I (cid:12)(cid:12)(cid:12)(cid:12)Xi∈I (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) (cid:19)
(cid:12)(cid:12) (cid:12)(cid:12)
wherethesupremumistake(cid:12)n(cid:12)overallsign(cid:12)(cid:12)sǫ: N→{−1,1}andfinitesubsetsI ⊂N.
(cid:12)(cid:12) (cid:12)(cid:12)
Recall the well-known result due to Bessaga and Pelczynski (see e.g. [4, p.202])
that in a Banach space Y a sequence (y )⊂S is equivalent to the standard unit
n Y
vectorbasisofc ifandonlyifsup || ǫ y ||<∞(supremumtakenasabove).
0 ǫ,I i∈I i i
By placing y = T(F ) we obtain that the range of T contains c isomorphically,
i i 00
which contradicts the assumptions. HePnce infn∈NoscT(Vn)=0. (cid:3)
Proof of Theorem 2.1. Recallthat as (L,τ) is a locallycompactHausdorffspace it
is completely regular, that is, for each closed set C ⊂ L and t ∈ L\C there is a
continuous map s: L→R such that s(C)={0} and s(t)=1.
Suppose thattherearenoisolatedpoints in(L,τ). LetT: C (L,X)→C (L,X)
0 0
be a linear operator. If the operator norm of T is 0 or ∞, then the Daugavet
equationholdstrivially,sothatwemayconcentrateonthe case||T||=C ∈(0,∞).
Let k ∈ N. Fix F ∈ S such that G = TF satisfies ||G|| > C − 1. Consider
C0(L,X) k
the open subspace U ={t∈L: ||G(t)||>C− 1} of L.
k
OPERATORS ON C0(L,X) WHOSE RANGE DOES NOT CONTAIN c0 3
Wecanpickasequence(Vn)n∈N ⊂U ofpair-wisedisjointopensubsetsasfollows.
Clearly U is also a locally compact (even compact) Hausdorff space which does
not contain isolated points. Hence U itself is not a singleton, and we may take
two points t ,t ∈ U, t 6= t . Since U is a Hausdorff space, there are disjoint
0 1 0 1
open neighbourhoods U ,U ⊂ U of t and t , respectively. By repeating the
0 1 0 1
same reasoning, pick t ,t ∈ U , t 6= t and disjoint open neigbourhoods
10 11 1 10 11
U ,U ⊂U oft andt ,respectively. Similarly,pickt ,t ∈U , t 6=t
10 11 1 10 11 110 111 11 110 111
and the corresponding disjoint open neighbourhoods U ,U ⊂U . Proceeding
110 111 11
in this manner yields a sequence of pair-wise disjoint open subsets by letting V =
n
U , s∈{1}n×{0} for n∈N.
s
Since c 6⊂ T(C (L,X)) we obtain by using Lemma 2.2 that osc (V ) → 0 as
0 0 T n
n → ∞. Fix n ∈ N such that osc (V ) < 1. Let u ∈ V . By using the complete
T n k 0 n
regularityofLonecanfindacontinuousmaps: L→[0,1]suchthats(L\V )={0}
n
and s(u ) = 1. Observe that the mappings s(·)F(·) and s(·) G(·) are
elements0of B . Hence max(1,||G(·)||X)
C0(L,X)
1 s(·) 1
(2.1) ||T(s(·)F(·))|| ≤ and T G(·) ≤
C0(L,X) k max(1,||G(·)|| ) k
(cid:12)(cid:12) (cid:18) X (cid:19)(cid:12)(cid:12)C0(L,X)
(cid:12)(cid:12) (cid:12)(cid:12)
by the definition of osc (V ). Note tha(cid:12)(cid:12)t (cid:12)(cid:12)
T n (cid:12)(cid:12) (cid:12)(cid:12)
s(t)
(1−s(t))F(t)− G(t) ≤(1−s(t))||F(t)|| +s(t)≤1
max(1,||G(t)|| ) X
(cid:12)(cid:12) X (cid:12)(cid:12)X
(cid:12)(cid:12) (cid:12)(cid:12)
for a(cid:12)l(cid:12)l t ∈ L. Hence E(·) =· (1−s(·))F(·)+(cid:12)(cid:12) s(·) G(·) defines an element
of B(cid:12)(cid:12) . Note that E is a kind of interp(cid:12)o(cid:12)lmaatxio(1n,||oGf(·F)||Xa)nd G.
C0(L,X)
Observe that
2
||G−T(E)||≤
k
according to (2.1), and that
||E+G|| ≥ ||(E+G)u || = s(u0) G(u )+G(u )
C0(L,X) 0 X ||G(u0)||X 0 0 X
= 1+||G(u )|| >1(cid:12)(cid:12)+C− 1. (cid:12)(cid:12)
0 X (cid:12)(cid:12) k (cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
Thus
3
||I+T||≥||E+T(E)||≥||E+G||−||G−T(E)||>1+C−
k
and by letting k →∞ we obtain that ||I+T||≥1+C =1+||T||. By the triangle
inequality ||I+T||≤1+||T|| and we have the claim. (cid:3)
Let us recall a few classical results due to James which are applied here fre-
quently: A closed convex subset C ⊂ X is weakly compact if and only if each
f ∈X∗ attains its supremum over C, and X is reflexive if and only if B is weakly
X
compact (see e.g. [4, Ch.3]).
Theorem 2.3. Let Γ be a non-empty set and X,Y be Banach spaces such that X
is reflexive and Y does not contain c isomorphically. Then any continuous linear
0
operator T: c (Γ,X)→Y is weakly compact.
0
Theaboveresultholdssimilarlyforℓ∞(Γ,X)inplaceofc (Γ,X),essentiallywith
0
thesameproof. NotethattheoperatorsI: c (N,X)→c (N,X)andT: c (N,Y)→
0 0 0
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Y; (y ) 7→ (y ,0,0,...) are not weakly compact for any non-trivial X and non-
n 1
reflexive Y according to the James characterizationof reflexivity. Hence neither of
theassumptionsaboutthereflexivityorthenon-containmentofc canberemoved.
0
Proof of Theorem 2.3. Let T: c (Γ,X) → Y be a continuous linear operator such
0
that T(Y) does not contain c . One may write c (Γ,X) = C (Γ,X) isometrically
0 0 0
where Γ on the right hand side is interpreted as a discrete topological space.
We claim that the sum osc ({γ}) is defined and finite. Indeed, other-
γ∈Γ T
wise one can extract pair-wise disjoint (open) subsets Γ ⊂ Γ, n ∈ N, such that
n
P
osc ({γ}) ≥ 1 for n ∈ N. However, since c 6⊂ T(Y), Lemma 2.2 yields
γ∈Γn T 0
that this case does not occur. Thus there exists a sequence (γn)n∈N ⊂Γ such that
P
osc (Γ\{γ } )→0 as k →∞.
T n n≤k
In order to verify the statement of the theorem we must show that T(B )
c0(Γ,X)
is weakly compact. In doing this we will apply the James characterization of
weakly compact sets. Fix f ∈ Y∗. It suffices to show that f attains its supre-
mum over T(B ). Observe that f ◦T defines an element of C (Γ,X)∗. For
c0(Γ,X) 0
each k ∈ N define a contractive linear projection P : c (Γ,X) → c (Γ,X) by
k 0 0
P f(·) = χ (·)f(·). Put g = f ◦T ◦P and Z = P (c (Γ,X)) for k ∈ N.
k {γn}n≤k k k k k 0
Note that g restricted to Z satisfies g | ∈ Z∗, where Z∗ = ℓ1({γ } ,X∗)
isometricallykfor k ∈ N. Clekarly g | k=Zkg | kfor k,l ∈ kN. Hencenthne≤rke is a
k+l Zk k Zk
sequence (x∗n)n∈N ⊂X∗ such that
k k
(2.2) g χ (·)y = x∗(y )
k {γn} n n n
!
n=1 n=1
X X
for k χ y ∈Z , k ∈N.
n=1 {γn} n k
Observe that since X is reflexive, there exists according to the James character-
izatPionof reflexivity a sequence (xn)n∈N ⊂SX such that x∗n(xn)=||x∗n|| for n∈N.
It follows that
k
(2.3) x∗(x )=||g || for k ∈N.
n n k
n=1
X
Now, since osc (Γ\{γ } )→0 as k →∞, we obtain that
T n n≤k
(2.4) lim ||T −T ◦P ||=0 and lim ||g −f ◦T||=0.
k k
k→∞ k→∞
By putting these observations together and using the continuity of T we obtain
that the sequence T k χ (·)x ⊂Y is Cauchy. On the other hand,
n=1 {γn} n k∈N
(cid:16) (cid:16) (cid:17)(cid:17)
k P
||f ◦T||≥f ◦T χ (·)x ≥||g ||−||g −f ◦T||→||f ◦T|| as k →∞
{γn} n k k
!
n=1
X
by using (2.2), (2.3) and (2.4). We conclude that
k
y =· lim T χ (·)x ∈T(B )
k→∞ {γn} n! c0(Γ,X)
n=1
X
satisfies f(y)=||f ◦T||=sup f(z), which completes the proof. (cid:3)
z∈T(Bc0(Γ,X))
Problem 2.4. We do not know if a linear operator T: C (L,X) → Y is weakly
0
compact if c 6⊂Y and L is 0-dimensional.
0
OPERATORS ON C0(L,X) WHOSE RANGE DOES NOT CONTAIN c0 5
References
[1] F. Bombal, P. Cembranos, The Dieudonn´e Property of C(K,E), Trans. Amer. Math. Soc.
Vol285,(1984), p.649-656.
[2] P. Cembranos, C(K,E) contains a complemented copy of c0, Proc. Amer. Math. Soc., Vol
91,(1984) p.556-558.
[3] J.Diestel,J.UhlJr.,Vectormeasures,MathematicalSurveys,No.15.AmericanMathemat-
icalSociety, Providence,R.I.,1977.
[4] P.Habala,P.Hajek,V.Zizler,Introduction to Banach Spaces I-II,matfyzpress,1996.
[5] D.Werner,RecentprogressontheDaugavetproperty,IrishMath.Soc.Bull.No.46(2001),
p.77-97.
[6] S.Willard,General Topology, DoverPublications,2004.
UniversityofHelsinki,DepartmentofMathematicsandStatistics,Box68,,(Gustaf
Ha¨llstro¨minkatu2b) FI-00014University of Helsinki, Finland
E-mail address: [email protected]