Internat. J. Math. & Math. Sci.
VOL. 13 NO. 4 (1990 817-820
817
EXAMPLE OF A
SEQUENTIALLYINCOMPLETE REGULAR INDUCTIVE LIMIT OF BANACH SPACES
JAN KUCERAandKELLY McKENNON
Department
of Mathematics WashingtonStateUniversity Pullman,Washington 99164 (Received August 22, 1989)ABSTRACT.
A
sequentially incomplete regularinductive limitofasequence of Banachspaces is constructed.KEY WORDS AND PHRASES. Regular locallyconvex inductive limit, sequentially complete locallyconvexspace.
1980MATHEMATICSSUBJECT CLASSIFICATION CODE
(1985 Revision):
Primary46A12,secondary46M40.
I. INTRODUCTION.
In
[I]
JorgeMujica asks: Iseveryregularinductive limitof Banachspacescomplete? Apartial answer is in[2]
with anexample ofa regular inductive limit which is quasi-incomplete. It is conjectured in[2]
that the regularinductivelhnitmight evenbe sequentially incomplete. Hereweprove the conjecture. On the otherhand,regularinductive limits ofBanachspaces always havesomecompleteness property,e.g. theyarefastcomplete,see
[3].
2. MAIN RESULT.
Let N
{I;2,...}
andRbe the space of real numbers. Foreach z{zj}
ERNxNand nENweput z
I[,.= max{, max{Izj[;] < n},sup{Izj[; ] >_ n}},
Forbrevity,wealsoput
LEMMA
I. The mapzII = II.: . - .
norm onE,
and each functional]’j:z-
zjE R
is continuous.818 JAN KUCERA AND KELLY McKENNON
LEMMA
2. EachE.
isBanach.PROOF.
Let{xC);k }
beCauchy. Thenforeach i,j EN,
thesequence{J’d,iCz(k));k N}
CauchyinR
andthus convergestosome x,i R. Put z{z,i}.
Foranyi,k,nN,
wehave lim
z(k) z(m) II,.=11 () I1,.,
whichimplieszE
d liraz(k)
zI1=
0.indlim
LEMMA
3. Theinductive limitE
n oo, is regular.PROOF. Let BCEbenotboundedin any
E,.
Withoutaloss of generalitywemay assume that foranyn 5N thereexistsx(n) B
suchthatx{n) I1,>
n. Thisimplies the existence of sequences{i(n)}, {j(n)}
CNsuchthat: eitherj(n) >_
n andx(n)q},i{,} I>
n orj(n)<
n and(),l. !> i() .,.
Foreach k
N,
chooser >
0 sothatr; max{i(n);
n_ k}
anddenotebyVtheconvexhull of
U{E(r);k N).
Takek,nN,x {x,.} E(r),
and distinguishthreecases:(a) jCn) >
k,whichimplies zq,la.{,}i_<ii
zII,< ,, < < ; z(n),.l,. l, (b) i(n) <
k&
k>_
n,whichimplies-" (-),(,)..,,)
(c) j(n) <
k&
k<
n,whichimplies xi{,}.j{,}I_ i(n). I1_< i(n)r _ i(n) <
(,’,,),.,,,, I.
Fo ,., I<
g(),c-a. I.
Since xE(r),k
5N,
was arbitrary, theelement
-xCn)
cannot be expressed as a convex combination of elements fromU{E(r);k e N},
i.e.x(n)
V.Hence
the 0-neighborhoodVinE
doesnotabsorbB
andB
isnotbounded inE.LEMMA
4.For
each i,j,nN,
putx(n),
d j-if<_ n,x(n),
d 0if i>
n, andx(n) {x(n)id}.
Then:Ca)
Foreh.e N, zCn) e El(1).
() ((,)} c
i.
(c) {zC)}
does notconverse m
E.PROOF.
Ca)
isevident.Cb)
Let V beaneighborhoodin E.For
ehN,
chooser. >
0so thatE.(r.)
C V.Fther,choose p, q Nothat W
>
2dqr>
2. Form,n>
q, definey, zE
by:yi..
xCrn)id xCn)id
for j_>
p, Yid 0otherwise,z. x(rn),
dx(n),
d for j<
p,z,
0otherwise.REGULAR INDUCTIVE LIMIT OF A SEQUENCE OF BANACH SPACES 819
Since
O for
I=(),, =(),, I<_ -
1 forfor ii>q,j<p>
q,j>
p Wehavev I1 sup{I
v,jI;i e
,i>
(c)
Assumez(n)
z inE. Sinceeach functional fj, defined in Lemma 1, is continuouso=
= .,
it Io =o=tin,o,on. Z=
we ,. f,.() limf,.(())
lim
zCn),# -I.
Hence zIll,n= max(7, )
and lim zI[,.=
0 i.e. zE
foranyn.N.
BycombiningLemmato1-4,weget:
THEOREM.Regularinductive limit ofBanachspacesmay be sequentially incomplete.
REFERENCES