On the existence of a pseuda-regular basis in some Köthe spaces

ON THE EXISTENCE OF A PSEUDO-REGULAR BASIS

IN SOME KOTHE SPACES

Z. Ertugrul, M. K ocatepe

Abstract

In this note, we study the relationship between the vanishing of the functor Ext (Ex F, E x F) for two Kothe spaces E and F , one of which having property (DN), and the existence of a pseudo-regular basis in the space Ex F and E®"F. Introduction

In [2], Kocatepe has shown that if Ext (>.(B), >.(A)) = 0, for two Schwartz regular Kothe spaces with >.(A) having propetry (DN) and >.(B) having property

(D)

,then

>.(A) x >.(B) and >.(A)@1r>.(B) have regular bases. In this note, we drop one of the

conditions: (D). We take E regular, F with (DN) and by using some characteristics of the vanishing of Ext functor given in [4], show that if Ext (Ex F, Ex F)

=

0, then Ex F and E@1rF have pseudo-regular bases, which is weaker than regularity (still unknown

whether strictly weaker), but is strong enough to obtain almost all of the results can be obtained using regularity, especially the quasi-equivalence property, [1]. We note that in its most generality, the vanishing of Ext 1_{(>.(A), >.(A))}

=

_{0, does not imply the existence} of a pseudo-regular basis in >.(A), since an example of such a space with no pseudo-regular basis has been given in [2].

Preliminaries

Let E

=

>.(A) be a Frechet Kothe space where A

=

(a}) is an infinite Kothe matrix such that for all i, k E N, we have O

<

a} ::; a7+1 .

LetE

=

>.(A), F

=

>.(B) be two Kothe Sl)aces. For the definition of the functor Ext 1 _{(E, F) and proof of the following fact we refer the reader to Vogt [5].}

Ext 1 ( E, F)

=

0 if and only if every short exact sequence O ----+ F ----+ G ----+ E ----+ 0 of Frechet spaces and continuous linear maps Sl)lits.

Also we use in this paper Ext(E,F) instead of Ext 1_(E,F).

The pair (E, F) is said to satisfy the condition (S*), in [4], briefly (E, F) E (S*), if the following holds

This condition has been rewritten in an apparently strengthened form in [4], Lemma 1.2. Namely, if (E, F) E (S*) then either E

=

l1 _{or (E, F) satisfy the following condition:}

(S*)0 'vµ3n0,k'vK,m,R,> 03n,S'vi,j:

:r �

max

{s:r 1 :r}.

J J J

So, (E, F) E (S*)0 is equivalent to

In the text, we shall also be using (F, E) E (S*)0, which we write as (S*)2 'vµ3n0,k'vK,m,R>0 3n,S'vi,j:

b� < max

{s

bf I_ bJ_o}.

af - af, Raµ

It was shown in [4] that (E,F) E (S*) if and only if Ext (E,F) (Ex F, Ex F)

=

0,then we have

Ext(E,E) =Ext(F,F) =Ext(E,F) =Ext(F,E)

=

0. A Kothe space E is said to have the property

0. If Ext

(DN) if 3n0 'vm3n,C > O'vi: (a7')2 � Ca�0af. If E has the property (DN)

and Ext(E,E)

=

0, then E has a regular basis, [3]. A Kothe space E is said to be regular if

'viand k, and it is called pseudo-regular, [1],if

Lemma Let E, F be two Kothe space. If Ext(E x F, E x F)

=

0, then there exist increasing sequences (m(k))k, (n(k))k, (Sk)k, (Ck)k such that\:/ k, i,j we have

and S k+2am(k+2) i < _{{ S} k+3ai m(k+3) S m(k+l)k+lai _} C bn(k+l) - max C bn(k+2) ' C bn(k) k+l j k+2 _j k j C bn(k+2) { C b n(k+3) C bn(k+l) } k+2 · _{1 < max} k+3 · ₁ k+l · ₁ S m(k+l) - S m(k+2)' S m(k) k+l ai k+2ai kai (1) (2) Moreover, if F has the property (DN) , then we can choose (n(k))k and(Ck)k in

such a way that they satisfy (1) and (2) and also

(3)

Proof. First we observe that, if for a given µ, n0 and k satisfy (S*)0, then for the same µ, (S*)0 is satisfied for larger values of n0 and k. We determine (m(k))k, (n(k))k, (Sk)k,

(Ck)k inductively. Put m(l) = n(l) = 1 and Ci= Si= 1 where i = 1, 2, 3. Forµ= n(l)

in (S*)i andµ= m(l) in (S*)2 there exists n0, k, n₀, k. We put n(2) = max{k, n0, n(l)+ 1} and m(2) = max{n0,k,m(l) + 1}. Forµ= n(2), andµ= m(2), there exist again

n0, k, n₀, k. We put n(3) = max{k, n₀, n(2) + 1} and m(3) = max{n₀, k, m(2) + 1}.

Now in(S*)i, putµ= n(l). By the above observation n0 = m(2), k = n(2) satisfy

(S*)i. Taking K = n(3), m

=

m(3) and R

=

1 there exist ii = ii4 and S = S4. In

(S*)2, put µ = m(l) then we have n0 = n(2), k = m(2) taking K = m(3), m = n(3)

and R = 1 there exist n = n4 and S = S4. For µ = n(3) and µ = m(3) there exist

n0, k, n0, k. We put n( 4) = max{ k, n0, n(3) + 1, n₄}, m( 4) = max{ n0, k, m(3) + 1, ii4} and S4 = S4S3C3, C4 = S4S3C3 ·

Let m(l), ... , m(k + 2); n(l), ... , n(k + 2); S1, ... , Sk+2; Ci, ... , Ck+2 be determined in

this way.

Now, for µ = n(k) we have n0 = m(k +

1),

k = n(k +

1).

We apply (S*)i to

K = n(k + 2), m = m(k + 2), R = 05k_±�ck _{and obtain ii = ii}

k+3, S = Sk+3. For

k+l k+l

µ = m(k), we have n0 = n(k + 1), k = m(k + 1), we apply (S*)2 to K = m(k + 2), m =

n(k + 2), R = _cCk_+�Sk _{and obtain n = n}

k+3 .and Sk+3· To find m(k + 3) and n(k + 3),

k+l k+l

we again consider µ

=

n(k + 2), and µ = m(k + 2), then there exist n0, k, n0, k. We put n(k + 3)

=

max{k, n0, n(k + 2) + 1, nk+3} and m(k + 3)

=

max{n0, k, m(k + 2) + 1, iik+3}

such that

and bn(k+2) { 5- bn( k+3) C S bn(k+I) } j k+3 j k+l k+I j m(k+l) :S max m(k+2) ' m(k) _· ai ai SkCk+2ai

We put S = 5k+35k+20k±2 and C = 5k±30k±25k±2 _{. Hence we have(l)and (2).}

k+3 ck+_, k+3 sk+i

Finally, when F has (DN), we may choose

) d C Sk+3Ck+2Sk+2

n(k +3 2'.max{k,n0,n(k+2)+1,nk+3} an k+32'. _S k+l

such that both (2) and

(3)

hold.

Now we can prove the main result of this note : D

Theorem Let E and F be Schwartz Kothe spaces where E is regular and F has property

(DN). If Ext(E

x

F, Ex F)

=

0 then Ex F and E®1rF have pseudo-regular bases.

Proof. Let A = (a}) and B

=

(b]) be the Kothe matrices for E, F, respectively. We choose sequences according to the above Lemma and use (Ska";(k)), (CkbJ(k)) which are

equivalent to A and B respectively. Then we have (1) and (2). By using Proposition 1.5. in [4],(1) gives S _k+2ai m(k+2) S m(k+3)

<

k+3_{ai for some k} C bn(k+l) _{C bn(}k+2) k+l j k+2 j S m(l+2) _S m(l+3)

=} 1+2ai

<

z+3ai for all l 2'. k .

C bn(l+l) _l+l j C bn(l+2) l+2 _j Similarly from (2) we get

C bn(k+2) _{C bn(}k+3) k+2 j

<

k+3 j _{for some k} S k+lai m(k+l) - S k+2ai m(k+2) C bn(l+2) C bn(l+3) -'- z+2 j

_<

z+3 j f 11 l

>

k ---,- S _z+1ai m(l+l) - S z+2am(l+2) or a i - . (4)

(5)

Since E and F are Schwartz space, can find increasing sequences of indices (mi) and ( ni) such that

If ( ei) and (fj ) denote the canonical bases for E and F respectively, we have that

is a 2seudo-regular basis for E x F. To see this, let m < n, m, n E N. The following four cases are possible

(1) Xm

=

er, Xn

=

ls withmi+l Sr S mi+1,nj+l S s S nj+1, i S j. Thenr <

s and we have and so by (5), we get S m(k+l) C bn₍k+2) k+lar _{< k+2} s _{'v k > 2_} S m(k) - C bn(k+I) kar k+l s

(2) Xm

=

er, Xn

=

es with mi + 1 Sr S mi+1,mj + 1 S s S mj+1, i S j. This

case follows from the regularity of the matrix (Ska'('(k)).

(3) Xm

=

lr,Xn

=

es withnj-1 + 1 Sr S nj,mi + 1 S s S mi+l,j Si. In this

case we have

Then using ( 4), we get

C bn(k+I) S m(k+2) k+l r _< k+2as

\;/ k 2 1. C_kb;(k_{) - S}

k+la;'(k+I)

(4) Xm = lr,Xn = ls v · '1ni + l Sr S ni,nj + 1 S s S nj,i S j. This case also

follows from the regularity of the matrix (Ckb�(k)).

Hence these four cases show that Ex F has a pseudo-regular basis. Next we consider E®1rF. We set

The matrices (Af) and (BJ) are regular and

52

A; = l,A7 2 l'vi,k,B}

=

l,BJ 2 l'vj,k and(BJ)2 _{S (B})(Bj+}1

Then E@1rF is isomorphic to the Kothe space >.(C) where (C� n)

=

(A�B�). We define n0

=

0 and and 00 I = LJ{(m,n): mi+ 1

:=;

m

:=;

_mi+l,n

:=;

n;}. i=l 00 J

=

LJ{(m,n): m

:=;

mi+l,ni

+ 1

:=;

n

:=;

ni+i}. i=O

Then I U J = N x N, In J =

0.

Now we define a matrix _(D�,n) by

Dk _{ A�

m,n

=

_B�

if (m,n) E J

if (m,n) E J

First we show that the matrices (C�,n) and (D�,n) are equivalent. If (m,n) E I, then there is a unique i such that mi

+ 1

:=;

m ::; _mi+ 1, n

:=;

ni. Then

C bn(k+l) ( ) S m(k+2) Ak+l k+l ni _� k+2a_{mi+l _ mi+l}

Ckbnn,(,k) - S k+lami+l m(k+l) � mi+l where ( *) follows from (6) and ( 4). So

from which it follows that Ck = _Bk _Ak

<

_Ak+l _{= D}k+1. If (m, _{n) E J, then there}

m,n n m - m m,n

is a unique i such that m ::; _{mi+I, ni}

+

l ::; n

:=;

ni+l · We have

S m(3) (6) C bn(2) (DN) C bn(4) C bn(3) C bn(4) 3a_mi+1

<

_{2 ni+l}

<

_{4 n} i+l =} 3 ni+l

<

4 n,+l S m(2) - C bn(l) C bn(3) S m(2) - S m(3) · 2a_{m,+1 1 ni+l 3 n,+1 2}a_{m;+1 3}a_mi+1 So by (5), we have Then C bk+l n_n(_i+l k+l)

<

C bk+2 nni+l (k+2) S m(k) _S m(k+l) for all k 2:: 2, kam,+1 k+l a_mi+1

Hence

from which it follows that Ck

<

_B2k-l = _D2k-l.

m,n - n m,n

nt,n

::;;

c�,n

for all k, m, n is obvious

Finally we show that the matrix (D;:,_, n) is pseudo-regular when the elements (m, n)

are ordered as follows :

(1, n0 + 1), · · ·, (m1, n0 + 1), (1, n0 + 2), · · ·, (m1, n0 + 2), · · · · · ·, (1, n1), · · ·, (m1, n1), (m1 + 1, 1), · · ·, (m1 + 1, n1), (m1 + 2, 1), · · ·, (m1 + 2, ni), · · ·

· · ·, (m2,

1), · · ·,

(m2, ni), · · ·

'

(1, ni-1 + 1), ···,(mi, ni-l + 1), (1, ni-1 + 2), ···,(mi, ni-l + 2),

· · ·, (1,

ni), ···,(mi, ni), (mi+ 1, 1), ···,(mi+ 1, ni), (mi+ 2, 1), ···,(mi+ 2, ni), · · ·

· · ·, (mi+l,

1), · · ·,

(mi+l, ni), (1, ni + 1), · · ·, (mi+l, ni + 1), (1, ni + 2), · · ·, (mi+l, ni + 2), · · ·

· · ·, (1,

ni+i), · · ·, (mi+l, ni+i), · · ·

Let the ordered pair ( r, u) appear before the ordered pair ( s, v) in the above ordering. We have four cases :

(l)(r, u) _EI, (2)(r, u) _EI, (3)(r,u) _{E J,} (4)(r, u) _{E J,} (s,v) E J (s,v) E J (s,v) E J (s,v) E J

Now we show the pseudo-regularity of (D;:,_, n) in each case:

(1) mi+ 1::;; r::;; mi+l,u::;; ni,mj

+

1::;; s::;; mj�1,v::;; nj,i::;; j. Then r::;; s. Then by regularity of (Af),

nk+1 nk+1 which is eguivalent to

;t

::;;

;t

.

r,u s,v

(2) mi+ 1::;; r::;; mi+1,u::;; ni,s::;; mj+1,nj + 1::;; v::;; nj+1,i::;; j, so r

<

v. As we have shown before, we have

(3) r S mi, ni-l

+

1 Su S ni, v S nj, mj

+

1 S s S mj+l, i S j. Then u S s. So we have

(4) mi +1 Sr S mi+i, u S n;, s S mH1, nj + 1 S v S nj+l, i S j. This case can be shown as case ( 1), it follows from the regularity of (Bf) .

D

References

[1] Crone, L., Dubinsk y, E., Robinson, W.B.: Regular bases in products of power series spaces.

J. Funct. Anal. 24, 211-222 (1977)

[2] Kocatepe, M.: A note on vanishing of the functor Ext for Kothe spaces. manus. math. 71, 113-119 (1991)

(3] Krone, J.: Zur topologischen Charakterisierung von Unter-und Quotientenrii umen spezieller

nuklearer Kotheraume mit der Splittingmethode. Diplomarbeit, Wuppertal, 1984

(4] Krone, J., Vogt, D.: The splitting relation for Kothe spaces. Math. Z. 180, 387-400 (1985) (5] Vogt, D.: On the functors Ext1_{(E, F) for Prechet spaces. Studia Math. 85, 163-197 (1987)}

BAZI KOTHE UZAYLARINDA YAKLA�IK-DUZGUN TABANLARIN VARLIGI HAKKINDA

Ozet

Bu makalede biri (DN) ozelligine sahip iki Kothe uzaymm c,arp1m uzaylannm Ext funktorunun s1fir olmas1 ile c,arp1m uzaylarmm yakla§ik -diizgiin (pseudo-regular) tabanlarmm olmas1 arasmdaki ili§ki i;ah§1ld1.

Z. Ertugrul & M. Kocatepe University of Bilkent