VOL. 18 NO. 4 (1995) 659-664
SOMEREMARKSABOUT MACKEYCONVERGENCE
JOZEFBURZYK
Institute of Mathematics Polish Academy of Science
Wieczorka 8 Katowice, 40-01 3, Poland
THOMASE.GILSDORF
Department
of Mathematics University of North Dakota Grand Forks, ND 58202-8376, USA(Received September 16, 1993 and in revised form March 5, 1994)
ABSTRACT. In this paper, we examine Mackey convergence with respect to K-convergenceand bornological (Hausdorff locally convex) spaces. In particular, we prove that: Mackeyconvergenceand local completeness imply property K; there are spaces having K- convergent sequences that are not Mackey convergent; thereexistsaspacesatisfying the Mackey convergencecondition, is barrelled, but is not bornological; and if a space satisfies the biackey
convergence condition and every sequentially continuous seminorm is continuous,thenthespaceisbornological.
KEY WORDS ANDPHRASES. Mackeyconvergence, property K, barrelled space,
bornological space.
1991 MATHEMATICS
SUBJECT
CLASSIFICATION CODE. Secondary 46A08.Primary 46A17"
I. INIIIIIBIIIION.
This notecontainsa somewhatmiscellaneouscollection of results that do or do nothold for aspace ("space" here means: Hausdorfflocallyconvex space) that satisfies the Mackey convergence condition; specifically, we consider the followingpropertiesofaspace (X,):
(a) (X, x) satisfies the
Macke2
conver_ence
condition [1; 5.1.29, page 158]" For every null sequence (x(n)) there is there is a closed, bounded absolutely convexset (i.e.,a disk B,such that x(n) 0 inXB where XB is the linearspanofB equippedwith thetopologyofthegaugeofB. Equivalently,
[1"
5.1.3,660 J. BURZYK AND T.
,
(X, v,) satisfies the Katowice property (also property_K
or is a _K- space.)[1;1.2.15, page 20], [2; page 19]" For every null sequence (x(n)) there is a
subsequence (y(n)) of(x(n)) suchthat
y(n)
convergesinX.n =1
.
(X, v) isbornological.Spaces that satisfy (a) have been studied recently with regard to topics such as inductive limits. For example, the reader may peruse [3] and the referencestherein. Property(b) was originallydefinedbyMazurandOrl|cz (see
[1; 1.4, page 30]) and was used in [4] tofind some sequential conditions under whichaspaceisbornological.
We first show that Mackey convergence and local completeness imply propertyK, wherewe recall thataspace (X, ) is locally complete [1;5.1.6, page 152] if for each closed bounded disk B, the space XB is a Banach space.
However,we also prove that there are spaces having K- convergent sequences
thatarenot Mackeyconvergent and thataweak Mackey convergentsequenceis a Mackeyconvergent sequence for the original topology. Hence, a space that satisfies property (a) with respect to its weak topology also satisfies Shur’. property- every weakly convergent sequence is convergent for the original
topology.
Antosikand thefirstauthor askin [4] whether abarrelledspace forwhich
every sequentially continuous seminorm is continuous must be bornological.
Bonet
[5]
proves the answer is "no". We prove here that if the barrelled assumption is replaced by (a), then the answer is "yes". Thisslightly improves the last result of[4].Finally, we exhibit a space that satisfies (a), is barrelled, but is not bornological. Thisexampleisrelatedtoan old problem ofGrothendieck(see [6]
and [7])" Is there a quasibarrelled (DF) -space satisfying (a)
tout
not bornologJcal?2. MAIN RESULTS.
THEOREM I.
.
{,}c.
(i) If (X, )islocally complete and satisfiestheMackeyconvergencecondition, thenitsatisfiesproperty K.
(ii) Anyweak Mackey convergent sequenceisMackey convergentwith respect tothe originaltopology.
(iii) If (X, :) satJsfJas theMackeyconvergencecondition with respectto its weak
PR(X)t.: it. Suppose (x(n)) is any null sequence in (X, T). Local completenessand theMackey convergence condition togetherimply that (x(n)) isa null sequence inXB for some closed bounded diskB,and XB is a Banach
space. As aBanach space satisfies property K, (x(n)) has a subsequence that is seriesconvergent in XB to some x XB. Because B is closed and the identity map fromXB imoX is continuous, thissubsequenceis seriesconvergentinX.
(ii). Now suppose (x(n)) is a weak Mackey convergent sequence in X. Then there isasequence (a(n)) ofpositivenumbers tendingto infinity, and such that
a(n)x(n) 0 weakly. The set la(n)x(n) n
Nt
is even-
bounded in X. Therefore, it suffices to find asequence (.(n)) of positive numbers such that (n)- 0 but a(n) (n).
(iii). Obviousconsequenceof(ii).
The authors do not know ifproperty Kimplies the Mackey convergence
conditioningeneral, but the nextexample shows that one can have a sequence
that satisfiesproperty K but isnotMackeyconvergent.
EXAMPLE I. Let X equipped with the following seminorms: For
each x- (x(n)) X,define:
a
For each k N,pk(x)-Ix(k)l.
/ Foreach y_l
,
py(X)=
y(k)x(k).
k-1We let be the locally convextopology generated by these seminorms. Note thatby ourdefinitionofT, asequence (x(n)) is T-convergent to x X ifand onlyif:
i) x(n) x coordinatewise; namely, x(n,k) x(k) as n
,
and(ii) Thereisan M0such that forevery k, n N, Ix(n,k)l
_
M.Define (z(n)) c Xby z(n) e(n), where e(n)- (0,0,..., 0, ,0,...) (1 in the nth place). Clearly, z(n) 0 and is even summable, hence (z(n)) is a
K-sequence. However,because of (i) and (ii), therecannot be a sequence ((n)) of positivenumbers tendingtoinfinity such that a(n)z(n) O.
We note here that the abo,’espace does not satisfy property K;consider thesequence z(n)-- (0,0,..., 0,1,,...) (0in the first n-1 places, then l’s). (z(n)) isa nullsequenceinX, buthasno summablesubsequence.
IIIF.[III[ 2. If (X, ) satisfies the Mackeyconvergence condition and
every sequentially continuous seminorm on X is continuous, then (X, T) is bornological.
662 J. BURZYK AND
(p(x(n)))isbounded. Then there isasequenceo1 positive scalars (n) oo such
that (n)x(n) 0, which in turnmeans that forsome M>O,p(c(n)x(n)) <_M for each positive integer n. tlence, p(x(n))- O. This shows that p is sequentially continuous, and by assumption, p is continuous. It follows that (X, T) is bornological.
We now present an example of a space that satisfies the Mackey
convergencecondition, is barrelled, but is not bornological. However, we will also show that thespacedoesnot have a fundamental sequenceofbounded sets soisnota solution tothe Grothendieckproblem.
EXAMPLE 2. Let X be the set of all real-valued functions on R that are constantexcept on a countable set. EquipXwith the locally convex topology generated by the family ofseminorms P
Pt
t R}, wherePt(X)
Ix(t)l, x X.Claim (X, ) satisfies the Mackey convergence condition. To see this,
suppose (x(nj) isany nullsequencein(X,). For each n N, let Tn denote the
setonwhich (x(n)) isnotconstant. We count Tn by Tn t(1,n), t(2,n)
1.
Denotingby 1 thecharacteristicfunction, wemayrepresent x(n) Xby
where c(n) and
Mn,
k) areconstants, for n, k N. It is clear from this that(c(n))isa nullsequenceof real numbers. Next,denote by T the countableset
LJ{T
n n N},and labelitby T- t(1),t(2) }. Becausex(n, t(k))thereisasequence(a(n)) oo suchthatboth a(n)x(n, t(k)I 0and a(nc(n)
0, as n oo. Thus, (x(n))isa Mackey convergentsequence.
la|m 2 (X, ) is barrelled. If fX’ X’denotesthecontinuousdual of X), thenf isof the form
f(x) =c(1)x((1)) +... + (k)x(t(k)),
pointwise boundedmembers of X’. With this in mind, let (f(n)) (f(n,x)) be any sequence such that for every
x
X and every n N, If(n,x)l is bounded.Wewill finda c>0anda
pP
suchthat If(n,x)l<- Cp(x),n N,x X.Foreachn N,let Sn 1,n), (2,n) (k
n.n)
,
tbr which f(n,x).,(
n)x(ki,n
),-1
c(i,n) 0 constants. Put
S= S
n’nN].
S isfinite;forifnot,thentherewouldexistforeach nN,
\ [cJ{ Sk "k =1 ..,/3n_
1]].
For brevity, relablefin
by n, and thent(n)
Sn
denote by y an element of X that takes the values given by
y(t(kn,l))
lct(kn,n)l-ln.
Then f(n,y) nk n o as n oo, and this is bad because then (f(n)) isnotpointwise boundedaty.Hence, weareallowedtowrite S t(1), t(2) t(K for some K N,
and
K
f(n, x)
.c
(i,n)x(ti).
=1
Put (i) max[k(i,n)l: n
NI,
for 1 K. Each =(i) < because of the pointwise boundedness of thesequence if(n)). Itfollows thatwithc
rnax{a(i,) 1 K},we haveK
f(rt,x)l
5.,
(i)lx(ti)l.
t--I
_
cp(x),where pis the seminormcorresponding to theset S. This means that (f(nl) is equicontinuous.
Ela|m 5 (X, ) isnotbornological. To seethis, consider the linearfunctionalf on X given by f(x) c, where x(O c, except on a countable set. f maps
bounded sets in X to bounded sets in R, but is not continuous; hence by [1;
664 J. BURZYK AND T.
1a|m 4 (X,r) doesnot haveafundamental sequenceof bounded
sets(f.s.b.). Ioprove this, suppose for a contradictionthat (X, r) has a f.s.b.,
(Bn).
Then in particular,foreach n =1, 2 wecanfindan>0 such that for each x Bn,Ix(n) l_
an.
Let us defineAcX byA-lYrn’rnNt,
where we define eachYm
asfollows: put3rn(t)
0 if rn, andYm(m)
1 +am From this we see that forany 1 R,we have that if t rn0 for some m0 N,thensupllYrn(t)
I" mNt
is 1 +arn
0 otherwise sup{I
Ym(t)
mNI
O.Therefore, A is x- bounded. On the otherhand, it is easy to see that A is not containedinany Bn
ACKNOWLEDGEMENT. Research for the second namedauthor was supported by the University of North Dakota. The authors wouldalso like to thank Professor
P.Antosikand the referee for several usefulcomments concerning thispaper.
REFERENCES
1]
PIREZ-CARRERAS,
P.,BONET,J.
Barrelled LocallyConvex Spaces, NorthHolland Math.Studiesno. 131, 1987.
[2] SWARTZ, C. AnIntroduction to FunctionalAnalx/sis,Dekker, 1992.
[3] BIERSTEDT, K. D. Anintroduction tolocally convexinductivelimits,in FunctionalAnalysis anditsApplications, WorldSci.Pub., 1987.
[4] ANTOSIK, P., BURZYK,
J.
Sequentialconditionsfor barrelledness andbornology,Bull. Pol. Acad.Sci.,i5 no. 7-8, (1987), 457-459.
[5] BONEF,
J.
AquestionofAntosikand Burzyk on non-bornological barrelledspaces,Bull. Pol.Acad. Sci., 5 9,no. 3-4, (1991 ), 175-176.
[6]