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Internat. J. Math. & Math. Sci.

VOL. 13 NO. 4 (1990 817-820

817

EXAMPLE OF A

SEQUENTIALLY

INCOMPLETE 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. Here

weprove 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 nEN

weput z

I[,.= max{, max{Izj[;] < n},sup{Izj[; ] >_ n}},

Forbrevity,wealsoput

LEMMA

I. The mapz

II = II.: . - .

norm on

E,

and each functional]’j:z

-

zj

E R

is continuous.

(2)

818 JAN KUCERA AND KELLY McKENNON

LEMMA

2. Each

E.

isBanach.

PROOF.

Let

{xC);k }

beCauchy. Thenforeach i,j E

N,

thesequence

{J’d,iCz(k));k N}

Cauchyin

R

andthus convergestosome x,i R. Put z

{z,i}.

Foranyi,k,n

N,

we

have lim

z(k) z(m) II,.=11 () I1,.,

whichimpliesz

E

d lira

z(k)

z

I1=

0.

indlim

LEMMA

3. Theinductive limit

E

n oo, is regular.

PROOF. Let BCEbenotboundedin any

E,.

Withoutaloss of generalitywemay assume that foranyn 5N thereexists

x(n) B

suchthat

x{n) I1,>

n. Thisimplies the existence of sequences

{i(n)}, {j(n)}

CNsuchthat: either

j(n) >_

n and

x(n)q},i{,} I>

n orj(n)

<

n and

(),l. !> i() .,.

Foreach k

N,

choose

r >

0 sothat

r; max{i(n);

n

_ k}

anddenotebyVtheconvex

hull of

U{E(r);k N).

Takek,n

N,x {x,.} E(r),

and distinguishthreecases:

(a) jCn) >

k,whichimplies zq,la.{,}

i_<ii

z

II,< ,, < < ; 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 x

E(r),k

5

N,

was arbitrary, the

element

-xCn)

cannot be expressed as a convex combination of elements from

U{E(r);k e N},

i.e.

x(n)

V.

Hence

the 0-neighborhoodVin

E

doesnotabsorb

B

and

B

isnotbounded inE.

LEMMA

4.

For

each i,j,n

N,

put

x(n),

d j-if

<_ n,x(n),

d 0if i

>

n, and

x(n) {x(n)id}.

Then:

Ca)

Foreh.

e N, zCn) e El(1).

() ((,)} c

i

.

(c) {zC)}

does not

converse m

E.

PROOF.

Ca)

isevident.

Cb)

Let V beaneighborhoodin E.

For

eh

N,

choose

r. >

0so that

E.(r.)

C V.

Fther,choose p, q Nothat W

>

2dqr

>

2. Form,n

>

q, definey, z

E

by:

yi..

xCrn)id xCn)id

for j

_>

p, Yid 0otherwise,

z. x(rn),

d

x(n),

d for j

<

p,

z,

0otherwise.

(3)

REGULAR INDUCTIVE LIMIT OF A SEQUENCE OF BANACH SPACES 819

Since

O for

I=(),, =(),, I<_ -

1 forfor ii>q,j<p

>

q,j

>

p Wehave

v I1 sup{I

v,j

I;i e

,i

>

(c)

Assume

z(n)

z inE. Sinceeach functional fj, defined in Lemma 1, is continuous

o=

= .,

it Io =o=tin,o,on

. Z=

w

e ,. f,.() limf,.(())

lim

zCn),# -I.

Hence z

Ill,n= max(7, )

and lim z

I[,.=

0 i.e. z

E

forany

n.N.

BycombiningLemmato1-4,weget:

THEOREM.Regularinductive limit ofBanachspacesmay be sequentially incomplete.

REFERENCES

[1] MUJICA, J.,

Functional Analysis, Holomorphy andApproximationTheory

II,

North Hol- land1984.

[2] KUCERA, J., MCKENNON, K.,

Completeness of regular inductive limits,Int. J.Math.

&

MathSci.Vol. 12,No. 3

C1989),

425-428.

[3] DEWILDE, M.,

Closed Graph Theorems and WebbedSpaces,Pitman, London 1978.

参照

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