Acta Math. Univ. Comenianae Vol. LXXXI, 1 (2012), pp. 141–142
141
A NOTE ON MUTIPLICATION OPERATORS
ON K ¨OTHE-BOCHNER SPACES
S. S. KHURANA
Abstract. Let (Ω,A, µ) is a finite measure space, E an order continuous Ba-nach function space over µ, X a Banach space and E(X) the K¨othe-Bochner space. A new simple proof is given of the result that a continuous linear op-erator T:E(X) → E(X) is a multiplication operator (by a function in L∞
) iff
T(ghf, x∗
ix) =ghT(f), x∗
ixfor everyg∈L∞
, f∈E(X), x∈X, x∗ ∈X∗
.
1. Introduction and Notation
In this paper all vector spaces are taken over the real fieldR. (Ω,A, µ) is a finite measure space andL∞
(µ) =L∞
, L1(µ) =L1 have their usual meanings. Eis an ideal in the vector latticeL1, E ⊃L∞
, and has the norm k.kE so that (E,k.kE) is a Banach lattice and is called K¨othe function space relative to the measureµ
([3]). The order inE is the natural order of functions inL1. Also the inclusions L∞
⊂ E ⊂ L1 are continuous. (X,k.kX) is another Banach space such that the Banach space (E(X),k.kE(X)) is the associated K¨othe-Bochner function space relative toE. ThusE(X) consists of all strongly measurable functionsf:Ω→X
for which the real functionsω→ kf(ω)kbelongs toE andkfkE(X)=kkf(.)kXkE ([3]). For measure theory we refer to [1]. IfY is a Banach space,Y∗
will denote its dual and for ay∈Y, y∗
∈Y∗
, hy, y∗
iwill also be used fory∗ (y).
In ([2]) a result is proved about the mutiplication operators in K¨othe-Bochner
spaces. The proof is quite sophisticated and, besides several lemmas, makes use of Markushevich bases. In this note we give a simple elementary proof.
2. Main Theorem
Now we come to the main theorem
Theorem 1. SupposeE an order continuous K¨othe function space overµ, X
a Banach space and E(X) the associated K¨othe-Bochner space. Let T:E(X)→
E(X)be a continuous linear operator. The following statements are equivalent:
Received Septembr 29, 2011.
2010Mathematics Subject Classification. Primary 47B38, 46B42; Secondary 28A25.
142 S. S. KHURANA
(i) There is ag0∈L∞ such that T(f) =g0f for all f ∈E(X). (ii) T(ghf, x∗
ix) = ghT(f), x∗
ix for every g ∈ L∞
, f ∈ E(X), x ∈ X, and
x∗ ∈X∗
.
Proof. (i) =⇒(ii): Obvious.
(ii) =⇒(i): For anh∈E,x∈X,g∈L∞
, we haveT((ghhx, x∗
i)x) =ghhT(x), x∗ ix; take anyx∗
∈X∗
withhx, x∗
i= 1. We getT(ghx) =ghpx for somep∈E (note since|hT(x)(.), x∗
i| ≤ kT(x)(.)k, we havehT(x)(.), x∗
i ∈E) and soghpx∈E(X).
pmay depend onx. Suppose T(x1) =p1x1 andT(x2) =p2x2. We claimp1=p2. Ifx1, x2are linearly dependent, there is nothing to prove; otherwisex1, x1−x2are linearly independent. Take anx∗
∈X∗
such thathx1, x∗i= 1,:hx1−x2, x∗i= 0. This means 0 = T(hx1−x2, x∗iz) = hT(x1−x2), x∗iz = hp1x1−p2x2, x∗iz = (p1−p2)z, for allz∈E. From this it follows thatp1=p2.
Now we want to prove that p is bounded. Suppose this is not true. Select a strictly increasing sequence{cn} of positive real numbers such that (i) cnin3, (ii)µ(Qn)i0 whereQn=|p|−1(cn, cn+1). For eachn, choose positiveαnso that, for the functionsfn=αnχQn,kfnkE= 1. Fix ay∈X withkykX= 1. The function
f =P∞
n=1n12fn is inE andf ≥
1
n2fn. This givesf|p| ≥
1
n2fn|p| ≥
1 n2fnn
3 and sokf|p|kE≥nfor all n. NowkT(f y)kE(X)=kf pykE(X)=kf|p|kE≥nfor alln, which is a contradiction. Sop∈L∞
. We putg0=p. ThusT(gx) =gg0x, for all
x∈X,g ∈L∞
and soT(h) =g0hfor all simple functionsh∈E(X). Since E is order continuous, simple functions are dense, and so the result follows.
References
1. Diestel J. and Uhl J. J.,Vector Measures, Amer. Math. Soc. Surveys vol. 15 Amer. Math. Soc., 1977.
2. Calabuig J. M., Rodriguez J. and Sanchez-Perez, E. A., Multiplication operators in K¨ othe-Bochner spaces. J. Math. Anal. Appls.373(2011), 316–321.
3. Lin, P. K.,K¨othe-Bochner function spaces. Birkhauser Boston Inc., MA, 2004.
S. S. Khurana, Departemt of Mathematics, Univerisity of Iowa, Iowa City, Iowa 52242, U. S. A.,
