Functional Analysis: Surveys and Recent Results I II K.D. Bierstedt and B. Fuchssteiner (eds.) 0 Elsevier Science Publishers B.V. (North-Holland), 1984
349
SOME RESULTS ON CONTINUOUS LINEAR MAPS BETWEEN FRECHET SPACES Dietmar Vogt
Universitat-Gesamthochschule Wuppertal Fachbereich Mathematik
D-5600 Wuppertal 1, GauBstr.
20
1 The p r e s e n t a r t i c l e i s m a i n l y concerned w i t h t h e d e r i v e d f u n c t o r s E x t (E,.) t h e f u n c t o r s L(E,.),
of
E f i x e d , a c t i n g f r o m t h e c a t e g o r y o f F r e c h e t spaces t o
t h e c a t e g o r y o f l i n e a r spaces ( s e e Palamodov [17] ), w i t h c o n d i t i o n s f o r 1 1 E x t (E,F) = 0 (see [31]) and t h e i r a p p l i c a t i o n s . E x t (E,F) = 0 means t h a t a l l e x a c t sequences 0
--*
F
+
G
--*
E
+
0 s p l i t ( c f . Thm. 1 . 8 . ) .
I n a f i r s t s e c t i o n we g i v e a s h o r t i n t r o d u c t i o n t o t h e t h e o r y o f t h e f u n c t o r s 1 E x t (E,.) ( c f . , c31.) w i t h s p e c i a l emphasis on t h e c o n c r e t e r e p r e s e n t a t i o n s 1 f o r E x t (E,F) and t h e c o n n e c t i n g maps i n case E o r F i s n u c l e a r . The necessary 1 r e s p . s u f f i c i e n t c o n d i t i o n s f o r E x t (E,F) = 0 (Thm. 1 . 9 . ) a r e p r e s e n t e d w i t h o u t 1 p r o o f ( s . [31]). I n s t e a d we g i v e i n $2 a d i r e c t p r o o f f o r E x t ( s , s ) = 0 which l e a d s v i a t h e permanence p r o p e r t i e s d e r i v e d i n $ 1 d i r e c t l y t o t h e s p l i t t i n g theorem f o r e x a c t sequences 0
+
F
+
G
+
E
+
0 where F i s a q u o t i e n t and E a
subspace o f s ( c f . 1231, 1341, 11243 ) . The d e s c r i p t i o n o f t h e c l a s s e s 1 1 E x t (E,s) = 0 and E x t (s,F) = 0 g i v e n i n Thm. 2.5. shows t h e s t r o n g c o n n e c t i o n 1 between t h e t h e o r y o f E x t and t h e s t r u c t u r e t h e o r y o f n u c l e a r F r e c h e t ( - , a )
spaces as developed f o r t h e case o f s i n [23]
and [34].
I n $ 3 we p r o v e by use o f t h e p r o p e r t i e s o f Ext'(w,.)
and a lemma f r o m [293
(Lemma 2.1.
i n t h e present paper) t h a t a h y p o e l l i p t i c p a r t i a l d i f f e r e n t i a l
o p e r a t o r on
IR" w i t h c o n s t a n t c o e f f i c i e n t s has no r i g h t i n v e r s e i n Cm(Q),
Q c Din open ( s e e [34 w i t h a d i f f e r e n t p r o o f ) . T h i s extends a r e s u l t o f Grothendieck on e l l i p t i c o p e r a t o r s .
1 I n $ 4 we show how t h e knowledge o f c o n d i t i o n s f o r E x t (E,F) = 0 can be used f o r t h e i n v e s t i g a t i o n o f t o p o l o g i c a l p r o p e r t i e s ( b a r r e l l e d n e s s e t c . ) o f spaces
Ei
F '2 Lb(E,F),
E,F F r g c h e t spaces, one o f them n u c l e a r ( c f . Grothendieck [ 8 ] ,
11, $ 4 ) . I n p a r t i c u l a r we g i v e i n Thm. 4.9. a complete b a s i s f r e e d e s c r i p t i o n o f t h e c l a s s e s {E;Lb(E,F,) b a r r e l l e d } and { F : Lb(Eo,F) b a r r e l l e d } i f Eo o r Fo i s a power s e r i e s space s a t i s f y i n g a s t a b i l i t y c o n d i t i o n , so e x t e n d i n g t h e r e s u l t s o f Grothendieck l o c . c i t . . We a l s o g i v e a s o l u t i o n f o r t h e problem o f c l a s s i f i c a t i o n o f complex m a n i f o l d s V a c c o r d i n g t o t o p o l o g i c a l p r o p e r t i e s ( h e r e
350
D. Vogt
barrelledness of Lb(E,F))as proposed in Grothendieck [81, 11, p. 128, a t l e a s t i f V i s a Stein manifold. A more direct and systematic treatment i s contained in 1331. In 5 4 we make use of the r e s u l t s of [31], 5 4 and 5 5 on E x t 1( E , F ) = 0 which are * * based on (S1) and ( S 2 ) . The proofs are rather complicated so we do n o t present * * them here. Instead we use in the final sectionconditions (S1) and (S2) t o investi1 gate which spaces E or F can occur in a nontrivial way in the relation E x t ( E , F ) = O . These are essentially the countably normed spaces E (more precise: the spaces E satisfying a condition (DN ) ) and the quasinormable spaces F . A special role play cp the spaces F which are quojections or do n o t s a t i s f y the condition ( * ) of BellenotDubinsky [3]. Some r e s u l t s of !j 5 are strongly related t o the r e s u l t s of
"1.
0. We will use standard notation of the theory o f locally convex spaces a s in [12] , 1201. For nuclear spaces we refer t o [ 8 ] , [19], for sequence spaces t o [ 5 ] and f o r concepts of homological algebra t o 1151 and p 5 ] .
Throughout the paper E,F,G,H always denote locally convex Frechet spaces, over a fundamental system of seminorms. L ( E , F ) i s the IK = R o r E, 1) I l l c 11 112 5 linear space of continuous linear maps from E t o F. On the dual space E'=L(E,IK) of E we consider the dual (RtU{t-)-valued) norms IIyIlk = SUP \ y ( X ) ) : X E, Ilxl/k 5 1).
...
We p u t
*
El; := { y C El : lIyl/k 4
t
m)
.
A Frechet space E i s said t o have property ( D N ) (resp. ( Q ) ) i f there e x i s t s a 2 *2 fundamental system of seminorms such that I1 l l k - 5 II 1tk-l /I IIk+l (resp. II Ilk 5
II
II
for all k
(see [ 2 4
1341
POI ) .
A nuclear Frgchet space i s isomorphic t o a subspace of s i f f i t has property (DN)
( s . [23]), i t i s isomorphic t o a quotient space of s i f f i t has property (a), ( s . [ 3 4 ) . s denotes the nuclear Frschet space of a l l rapidly decreasing sequences: S = {X =
(x1,x2,
... )
:
[lx/lk =
4 J
lxjl j
k
<
+
m
for all k}
be an i n f i n i t e matrix such t h a t 0 More generally l e t A = ( a j . k ) j ,k > 0 for a l l j and k . Then we define sup a k
5
.
aj,k5aj,k+l,
j,k
E N these are Frechet spaces. They Equipped with t h e i r respective seminorms (11 are called Kothe sequence spaces. They are nuclear i f f f o r every k there i s p such t h a t (with 0 = 0 ) :
351
Continuous linear maps between Frechet spaces
2
j
a
j , k a.
(Grothendieck-Pietsch c r i t e r i o n ) .
< + m
J,k+P
I n t h i s case we have h ( A ) = h m ( A ) , and v i c e v e r s a . a.
...~r
...
(a.). : a 1 5 a 2 s f+-, J J t h e n h ( A ) ( r e s p . h m ( A ) ) i s c a l l e d power s e r i e s space and denoted by n r ( a ) ( r e s p .
o2
and
I f aj,k
= okJ w i t h sequences 0 c
!,;(a)).
I t depends o n l y on r and t h e sequence a. F o r f i x e d a a l l spaces Ar(a)
+
(resp. nF(a)) w i t h r < t o t h e cases r = 1, +
m.
-
p1
<
<
a =
a r e i s o m o r p h i c . T h e r e f o r e we can r e s t r i c t o u r a t t e n t i o n
n l ( a ) i s c a l l e d power s e r i e s space o f f i n i t e t y p e , !,_(a)
o f i n f i n i t e t y p e . Power s e r i e s spaces o f f i n i t e t y p e
and o f i n f i n i t e t y p e can
n e v e r be i s o m o r p h i c . By
w :=
by #(V)
lK’
we denote t h e p r o d u c t o f c o u n t a b l y many c o p i e s of t h e s c a l a r f i e l d ,
t h e n u c l e a r F r k c h e t space o f holomorphic f u n c t i o n s on a complex mani-
f o l d V. 1. I n t h i s f i r s t s e c t i o n we p r e s e n t some b a s i c f a c t s on t h e d e r i v e d f u n c t o r s k E x t (E,.),k = O , l , o f t h e f u n c t o r L(E,.). A l l these functors are considered
...,
t o a c t f r o m t h e c a t e g o r y o f F r k c h e t spaces t o t h e c a t e g o r y o f l i n e a r spaces over
lK
=
IR o r a. E denotes a f i x e d F r 6 c h e t space.
We do n o t g i v e a c o n s t r u c t i o n f o r these f u n c t o r s b u t t a k e t h e i r e x i s t e n c e as g r a n t e d by homological a l g e b r a ( s e e Palamodov [17]).
I n s t e a d we t a k e an
a x i o m a t i c approach, i . e . we s t a t e p r o p e r t i e s o f t h e s e f u n c t o r s and t h i s w i l l be t h e o n l y i n f o r m a t i o n w h i c h we w i l l use.
k The f u n c t o r s E x t (E,.)
have t h e f o l l o w i n g p r o p e r t i e s :
(0)
E x t o ( E , * ) = L(E;)
(I)
To e v e r y s h o r t e x a c t sequence 0 maps
k
tik : E x t (E,H)
such t h a t 0
+
L(E,F)
*
L L(E,G)-
+
q*
+
F
L
+
E x t k + l (E,F),
L(E,H)-
G
q --f
H
+
0 t h e r e a r e assigned
k = 0,1,
...
1 1 1 ql Ext ( E , F ) L Ext ( E , G ) - - + Ext
€io
i s e x a c t and depends f u n c t o r i a l l y on t h e s h o r t e x a c t sequence.
(11) F o r e v e r y i n j e c t i v e space
k I we have E x t ( E , I )
= 0 for k =
1,2,
... .
352
D. Vogt
We used (and w i l l use) t h e f o l l o w i n g n o t a t i o n s : c p O A = :Ay assigned t o A by t h e f u n c t o r s E x t
k
(E,.), k
(pkA
... . A
= 1,2,
i s t h e map
space I i s c a l l e d
i n j e c t i v e i f f o r each space E, each c l o s e d subspace Eo c E , and each cp c L ( E o , I )
t h e r e e x i s t s an e x t e n s i o n 9
c
L(E,I).
Examples o f i n j e c t i v e spaces a r e t h e
spaces l m ( M ) , M a n i n d e x s e t , and t h e i r p r o d u c t s .
I f F i s a F r e c h e t space t h e n an e x a c t sequence
I k i n j e c t i v e f o r a l l k, i s c a l l e d an i n j e c t i v e r e s o l u t i o n . k We can use any i n j e c t i v e r e s o l u t i o n o f F t o c a l c u l a t e t h e space E x t (E,F),
as
the f o l l o w i n g p r o p o s i t i o n shows. The p r o o f i s s t a n d a r d . I n f a c t u s u a l l y one uses k i n j e c t i v e r e s o l u t i o n s t o prove t h e existence o f f u n c t o r s Ext ( E , . ) w i t h the
-
p r o p e r t i e s d e s c r i b e d above.
1.1. P r o p o s i t i o n : If 0
k = 0,1,
F
+
Lo
I0
11-
L1
I2
-t
... is
an injective r e s o l u t i o n ,
... .
for k = 1,2,
proos:
+
We p u t Zo = F , Zk = k e r
... a 0
L~
= im
L
~
...
f o- r ~k = 1,2,
and o b t a i n f o r each
s h o r t e x a c t sequence
+
Zk
-
Ik
Lk
zk+l
+
0
which l e a d s by ( I ) and ( 1 1 1 ) t o a l o n g e x a c t sequence
*
0
+
L(E,Zk)
L(E,Ik)-
--t
S i n c e L(E,Zk+l)
= ker
Lk
L
*
L(E,Zk+l)
1 Ext (E,Zk)
f o r k = 1,2,
L;+~
/ im
ExtJ+'(E,Zk)
k (E,F) 2 E x t 1 (E,Zk-l)
... .
0
+
+
1
+
E x t (EyZk+l)
2
E x t (E,Zk)
+
... , j
0
+
... .
= 1,2,
...
:
*
L~
.
Combining these e q u a t i o n s and u s i n g Zo = Ext
+
t+h i s~ g i v e s us f o r a l l k = 0,1,2,
~
1 E x t (E,Zk) 2 '' ker Extj(E,Zk+l)
+
F we o b t a i n
ker
L;
/ im
LL-~
I t can e a s i l y be seen t h a t e v e r y Banach space F has an i n j e c t i v e r e s o l u t i o n o f Banach spaces. We need o n l y t o know t h a t e v e r y Banach space X can be n a t u r a l l y imbedded i n t o l " ( B o ) where Bo i s t h e u n i t b a l l i n X ' . We a p p l y t h i s t o F and p u t
353
Continuous linear maps between Frechet spaces
t h e n t o I o / F and o b t a i n 1 1 , e t c . .
I. = l " ( B o ) ,
1 . 2 . C o r o l l a r y : If E is nuclear and F then E x t k (E,F) = 0 for k = 1,2, ,..
=
J
F . is a product of Banaeh spaces F . , J J
Proof: We a p p l y t h e above d e s c r i b e d procedure s e p a r a t e l y t o t h e F . and o b t a i n an J i n j e c t i v e r e s o l u t i o n o f t h e form
~
o
+
n j
F . --t J
n
n ~j
j
l ajl
j
n
0
j
where t h e IJk a r e Banach spaces. From t h e p r o p e r t i e s o f a n u c l e a r space ( s . Grothendieck [83 , I I , § 3, nO1) and Prop. 1.1. we conclude t h e a s s e r t i o n .
The p r e c e d i n g remark i s v e r y u s e f u l as one sees f r o m t h e f o l l o w i n g lemma which i s contained i n
[3u
[ln Thm. 5 . 2 ( o r Cor. 5 . 1 . ) and a p r o o f o f which can be found
in
(Lemma 1 . 1 . ) . We need t h e f o l l o w i n g n o t a t i o n :
A p r o j e c t i v e spectrum P
~ : , F1 ~+
Fk (1 2 k ) o f Banach spaces i s c a l l e d a
fundamental system o f Banach spaces f o r t h e F r P c h e t space F i f ( i ) F = l i m p r o j Fk ( i i ) f o r e v e r y k t h e r e i s an 1 2 k such t h a t pk
: F
+
pk
F i s dense i n
p1
,k F1, where
Fk denotes t h e c a n o n i c a l map.
1 . 3 . Lemma: If P , , ~ : F1
+
Fk is a fundumenta2 system of Banaeh spuces f o r F
then the sequence ("eanonieal r e s o l u t i o n " )
is exact, where
L
: x
--+
( P ~ x and ) ~ q : ( x ~ )* ~( P k + l , k ' k + l
-
'k)k
.
T h i s y i e l d s as an i m e d i a t e consequence (see [ 3 q , Thm. 1 . 2 . ; c f . [16],
[17]
, p-
7.1.,
51) :
1.4. Theorem: If E o r F is nucZear then E x t k (E,F) = 0 f o r k 2 2.
Proof: I f
F i s nuclear, then F
morphic t o 1".
has a fundamental system o f Banach spaces i s o -
Hence t h e Fk i n Lemma 1 . 3 . can be chosen i n j e c t i v e , which means
t h a t t h e s h o r t e x a c t sequence i n Lemma 1 . 3 . i s an i n j e c t i v e r e s o l u t i o n o f l e n g t h 1 . The a s s e r t i o n f o l l o w s f r o m Prop. 1.1. I f E i s n u c l e a r t h e n we choose any fundamental system o f Banach spaces and
354
D. Vogt
a p p l y ( I ) t o t h e c a n o n i c a l r e s o l u t i o n (see 1.3.).
... + 0 ... .
f o r k = 2,3,
-+
0
k E x t (E,F)
-+
+
0
+
0
We o b t a i n
... .
The z e r o s come f r o m Cor. 1.2.
From Thm. 1.4. we o b t a i n a permanence p r o p e r t y :
If E o r F i s nuclear, E x t 1(E,F)
1.5. C o r o l l a r y :
then a l s o Extl(E,Fo)
M:L e t
q : F
+
=
= 0 and Fo
a quotient of F,
0.
Fo be t h e q u o t i e n t map, G := k e r q. Then
s h o r t e x a c t sequence
O + G + F + F
0
(I) a p p l i e d t o t h e
-+O
gives
...
-+
1 E x t (E,F)
1 E x t (E,Fo)
+
2 E x t (E,G)
7-
+
... .
The f i r s t term i s z e r o by assumption, t h e t h i r d by 1.4.,
hence a l s o t h e m i d d l e
term.
1
Our n e x t t a s k i s t o g e t more i n f o r m a t i o n on E x t (E,F)
i n case one o f t h e spaces
i s nuclear. 1.6. Theorem: F,
Let $ =
be a fundamental system of Banach spaces f o r
{Fk, pl,k}
l e t E o r F be nuclear. Then
where
1 E x t (E,F) 2r1 L(E,Fk) k B(E,F) = {(Ak)k
c
/ B(E,3)
n
L ( E y F k ) : ZI(Bk)k E rI L(E,Fk) such t h a t k
k
Ak = Pk+l,k B k + l
proos: I f
E i s n u c l e a r we a p p l y ( I
obtain
... + n k
L(E,Fk)
q"
n k
- Bk
f o r a l l k).
t o t h e c a n o n i c a l r e s o l u t i o n (see 1.3.).
L E,Fk)
’ 6
1 Ext (E,F)
+
0
+
We
...
where t h e z e r o comes f r o m Cor. 1 . 2 I f F i s n u c l e a r and p
: F1 -+ Fk i s any fundamental system o f Banach spaces t h e n 1Y k : FY + F i o f i n j e c t i v e Banach spaces and we can f i n d a fundamental system po 1 ,k maps qk : FL -+ Fk such t h a t
355
Continuous linear maps between Frechet spaces
i s commutative. Hence we o b t a i n a c o m n u t a t i v e diagram
where
cp
i s i n d u c e d by t h e q k . We a p p l y ( I ) t o t h i s diagram and o b t a i n
. .. + n
... + n
Q
1."
L(E,Fk)
n
q
0
FQ
0
6
L(E,Fk)
+E x t
1(E,F)
+
...
+
0
/id
L ( E , F i ) L n L(E,Fi)k k
So
1
E x t (E,F)
The z e r o comes f r o m ( 1 1 ) . O b v i o u s l y t h e a s s e r t i o n f o l l o w s f r o m t h i s diagram. From Thm. 1.6. a g a i n we o b t a i n a permanence p r o p e r t y :
1.7. C o r o l l a r y : If E or F is nuclear, E x t 1(E,F) = 0 and Eo a closed subspace of E, then Extl(Eo,F)
proos: We
=
0.
choose a fundamental system o f Banach spaces f o r F which f o r n u c l e a r
F we assume t o c o n s i s t
o f i n j e c t i v e Banach spaces. Hence i n b o t h cases t h e
r e s t r i c t i o n maps Rk : L(E,Fk)
0 = E x t 1(E,F)
-+
L(Eo,Fk) a r e s u r j e c t i v e f o r a l l k
( f o r E nuclear
The Rk induce t h e map R i n t h e f o l l o w i n g diagram. R i s s u r j e c t i v e .
see [8],11,93).
k
L(E,Fk)
/ B(E,F)-n
R
k
L(Eo,Fk)
/ B(E,,,.3)
1 E x t (Eo,F)
T h i s shows t h e a s s e r t i o n . I n t h e f o l l o w i n g p a r t o f t h i s s e c t i o n we want t o e x p l a i n , what i n terms o f t h e c o n c r e t e r e p r e s e n t a t i o n o f 1.6. (and 1.4.) sequence ( I ) "means".
1
E x t (E,F)
resp. t h e long exact
L e t us remark t h a t t h e isomorphism which we used i n 1.6. i s
356
D. Vogt
i n a c a n o n i c a l way determined by t h e fundamental system o f Banach spaces. I t i s induced by t h e ’6 assigned t o t h e c a n o n i c a l r e s o l u t i o n (see 1 . 3 . ) .
I n the following
d i s c u s s i o n we w i l l always t a l k a b o u t t h i s isomorphism. Let 0
--f
F
L
G
q
H
-f
+
0 be an e x a c t sequence. We assume t h a t e i t h e r E o r F,G
3 ,b be fundamental
and H a r e n u c l e a r . L e t 5 ,
systems o f Banach spaces such t h a t
f o r e v e r y k we have an e x a c t sequence o f Banach spaces Lk
0 + Fk
Gk-
f
'k
Hk
0
+
and such t h a t f o r a l l 1 > k t h e diagram L
q
0-F-G-H
4
J
L1
0-F-G 1
0
1
4
1'O
-4
Lk qk F k - + Gk
--t
+O
.1
91 -ti
Hk
+
0
i s commutative. We can o b t a i n such fundamental systems e.g.
by t a k i n g for
Banach spaces generated by a fundamental system o f seminorms on G and f o r
fi?
the
5 and
@ t h e Banach spaces generated by t h e induced r e s p . coinduced fundamental systems
o f seminorms on F r e s o . H. We have t h e f o l l o w i n g s i t u a t i o n Sor
L(E,H)
/
\ 0
L
E x t 1(E ,F)
1 t
It?
n
I
L(E,Fk)/B(E,F)-
k
Ext'(E,G)
n k
q1 +
Il? L(E,Gk)/B(E,J)-
Q
1 E x t (E,H)
n k
It?
-
L(E,Hk)/B(E,*)
where D,I,Q denote t h e maps induced by means o f t h e isomorphisms f r o m F o , We want t o d e s c r i b e these maps. We a p p l y t h e r e f o r e ( I ) columnwise t o t h e f o l l o w i n g diagram:
0 J
F
.1
0
0 .
-
L
G
-
J
H
J
0
0
9
J
5.
0
0
+
0
L',
q
1
.
357
Continuous linear mam between Frechet spaces
where t h e columns a r e c a n o n i c a l r e s o l u t i o n s . We o b t a i n by use o f ( I ) and ( t h e p r o o f o f ) 1.6. t h e f o l l o w i n g commutative diagram w i t h e x a c t columns. The rows a r e e a s i l y
seen t o be e x a c t .
L
nL k
$
1
Ib ",;
------in
ll L(E,Fk) k
k
-n
L(E,Gk)
k 1
16O
E x t 1(E,G)
Extl(E,F)
L(E,Hk) J16O
q
Extl(E,H)
1
1
.1
0
0
0
We see i m m e d i a t e l y t h a t t h e maps I and Q a r e t h e n a t u r a l maps between t h e r e s p e c t i v e spaces, i . e . t h e maps induced by
n k
n
ck and
k
qk on t h e q u o t i e n t s .
Moreover we can by a s t a n d a r d procedure o f homological a l g e b r a ( s . [15]) 1 a map d : L(E,H) + E x t (E,F) as f o l l o w s : F o r 9 E L(E,H) we have, because o f t h e n u c l e a r i t y assumptions, acp E i m
*
b
define
* (n q k ) . b
We
choose j, E n L(E,Gk) such t h a t (Il q k ) j, = acp. Since (TI q k ) b 9 = c ( n q k ) j, = c acp = 0 we can f i n d x E n L(E,Fk) such t h a t (n L,;) x = b 9. We p u t d cp := 6 ’ x .
I t i s n o t d i f f i c u l t t o p r o v e t h a t d i s w e l l d e f i n e d , l i n e a r and makes t h e
f o l l o w i n g sequence e x a c t
... Hence ’6
+
L(E,G)
*
J--+ L(E,H)
d
+
1 E x t (E,F)
A o d where A i s an automorphism o f k e r 1 f r o m L(E,H) + E x t (E,F). =
w
We now p u t Dcp :=
[x]
where
that also
L~
1
E x t (E,H)
+
..
and ’ 6 i s t h e one a c t i n g
[ ] denotes t h e e q u i v a l e n c e c l a s s i n r~ L ( E , F k ) / B ( E , Y ) .
1 k L(E,Fk) + E x t (E.F) induces an isomorphism) k i s w e l l d e f i n e d and l i n e a r . S t r a i g h t f o r w a r d c a l c u l a t i o n shows t h a t
I t f o l l o w s e a s i l y (so a c t i n g f r o m
O
1
N
n
N
= B o 0 where B i s some automorphism o f k e r I , o r e q u i v a l e n t l y : t h a t i m 0 = i m 0, N
k e r D = k e r D. We have t o e x p l a i n
E.
I t d e s c r i b e s t h e f o l l o w i n g ( M i t t a g - L e f f l e r - ) approach t o t h e
s o l u t i o n o f t h e l i f t i n g problem
3 58
D. Vogt
F i r s t we s o l v e a17 t h e " l o c a l " l i f t i n g problems 0
Fk
+
Gk qk+ Hk
+
0
--t
E
which we can do b y o u r n u c l e a r i t y assumptions. The qk a r e t h e maps i n d u c e d by qY i . e . t h e components o f aq. The q k g i v e t h e components o f acp. Then we a p p l y b, i . e . o I$f we ~ + t h i n~ k o f L as an imbedding, t h e s e can be f o r m t h e maps ~ ~ + ~ , -~ qk. c o n s i d e r e d as maps i n L(E,Fk),
X
E
t h e y d e f i n e t h e components xk o f an element
i.e.
f L(E,Fk)*
Our o r i g i n a l l i f t i n g problem i s e a s i l y seen t o be s o l v a b l e i f and o n l y i f we can f i n d (Bk)k E
nk
-
L(E,Fk)
($k+l - Bk+l)
pk+l,k
=
such t h a t P k + l , k o $ k + l - $k = Pk+l,k 0 B k + l BkA hence - Bk f o r a l l k . T h i s means n o t h i n g e l s e t h a n EQ=[x] = 0
qk
and t h a t i s e q u i v a l e n t t o EQ = 0.
Hence t h e a x i o m a t i c approach d e s c r i b e s (under o u r n u c l e a r i t y assumptions) v i a t h e c o n c r e t e r e p r e s e n t a t i o n n o t h i n g e l s e t h a n t h e p o s s i b i 1 it y o f t h e " n a t u r a l 'I way o f 1 s o l u t i o n o f t h e l i f t i n g problem. k e r L d e s c r i b e s t h e o b s t r u c t i o n a g a i n s t t h i s i n 1 t h e c o n c r e t e s i t u a t i o n . E x t (E,F) d e s c r i b e s t h e " s t r u c t u r a l " o r "maximal" obs t r u c t i o n i n any s i t u a t i o n . I f i t i s z e r o t h e procledure always works. We a d m i t w i t h o u t f u r t h e r p r o o f t h e f o l l o w i n g theorem ( s .
pl],
Thm 1.8.):
1.8. Theorem: The following are equivaZent: 111
Ext'(E,F)
(2)
Every ezact sequence 0
= 0
0 splits.
F
+
G
+
E
131
For every exact sequence 0
+
F
+
G
+
H
+
0 and
cp E
L(E,H)
(4)
For every exact sequence
+
H
4G
+
E
+
o
cp E
L(H,F) there e x i s t s
+
E
L(E,G) m t h
cp =
$ E L(G,F) w i t h q =
-f
q o $.
Ji o
o
--f
9
and
there e x i s t s
L.
The r e s t o f t h i s paper w i l l be m a i n l y devoted t o a d i s c u s s i o n o f c o n d i t i o n s f o r 1 E x t (E,F) = 0 and t h e i r a p p l i c a t i o n s . We use t h e f o l l o w i n g c o n d i t i o n s on two
3 59
Continuous linear maps between Frkchet spaces
*
F r g c h e t spaces E and F ( s . [31]
and f o r (S1) A p i o l a [ 2
] ) . 11 (I1 5 11 Il2 5 ... llyll," = s u p 1 / y ( x ) I :
denotes a fundamental system of seminorms on E o r F r e s p . , Jlxllk 5 11 t h e dual norm o f
*
(S1)
/Ixllm llYllk
*
(S2)
Vp
I n [31],
Thm. 3.7.
then
s
*
(S1)
*
( /Ixlln llYllK +
3no,k VK,m
1.9. Theorem: I f
I n [13]
5
for y c
E’ o r F ' .
3n,S Vx E E, y c FL
3no Vw 3k VK,m
*
/I / I k
3,s
Vx E E,
*
llYllw
IIXII
1
y CFA
i n c o n n e c t i o n w i t h 3.9. i t i s shown:
E is countabZy normable o r F r e f l e x i v e and one of them nuclear, =
1 E x t (E,F)
i t i s shown t h a t
( S *2 )
=
0
*
(S2)
3
.
i s necessary and s u f f i c i e n t f o r E x t 1(E,F) = 0 i n t h e
case o f Kothe spaces. A necessary and s u f f i c i e n t c o n d i t i o n f o r E x t 1(E,F) = 0 i n t h e g e n e r a l case i s c o n t a i n e d i n [33]. F u r t h e r r e s u l t s on s p l i t t i n g r e l a t i o n s , g e n e r a l i z a t i o n s and i n v e s t i g a t i o n s o f 1 c o n d i t i o n s f o r an e x a c t sequence t o s p l i t i f E x t (E,F) # 0 a r e c o n t a i n e d i n [1
, ~ 1 8 1[XI, ~ [XI.
1,
2. I n t h i s s e c t i o n a l l spaces a r e assumed t o be n u c l e a r . I n s t e a d o f p r e s e n t i n g d e t a i l s o f t h e r a t h e r c o m p l i c a t e d p r o o f o f Thm. 1.9. we d i s c u s s w i t h f u l l p r o o f s
1
an example which i m m e d i a t e l y l e a d s t o t h e most i m p o r t a n t cases o f E x t (E,F) = 0 w i t h r e s p e c t t o a p p l i c a t i o n s . I t shows i n a v e r y n i c e way t h e i n t e r p l a y between t h e s t r u c t u r e t h e o r y o f ( s ) as developed i n [231,
[34]
( c f . a l s o [5],
[24],
[35])
and t h e p r e s e n t t h e o r y . The f o l l o w i n g lemma 2 . 1 . i s i n an e q u i v a l e n t f o r m (s. Thm. 1.8.) c o n t a i n e d i n [23],
1.5..
We g i v e a d i r e c t p r o o f u s i n g t h e m a t r i c e s o f t h e
r e s p e c t i v e maps. 2.1. Lemma:
E x t 1( s , s ) = 0
proof: We a p p l y 1.3. o r 1.5.
t o t h e fundamental system
of i n j e c t i v e Banach spaces. L e t Ak E L(s,Fk), i s r e p r e s e n t e d by a m a t r i x (ai!j)v,j
~
with
k = 1,2,
...
be given. Each Ak
360
D. Vogt
f o r some n = n ( k ) , d = d ( k ) . We have t o determine m a t r i c e s (b(k!) ;J - b(k! f o r a l l k,v,j. V,J
s a t i s f y i n g analogous e s t i m a t e s such t h a t
ack! = b ( k + f ) v,J V,J
We p u t D ( l ) = d ( l ) , D ( k + l ) = max ( d ( k + l ) , 2 D ( k ) + k
+
3 ) and N ( l ) = n ( l ) ,
N ( k + l ) = max ( n ( k + l ) , 2 N ( k ) ) f o r a l l k. We w i l l determine i n d u c t i v e l y m a t r i c e s ( u ( ~ ! ) f o r k = 2 , 3 , k = 1,2, (1)
...
lUiyI v;pj
V
with
,J
...
and (v!;)
z ( k - 2 ) j 5 2-k 2kj
p(k)+N(k)v
1 k,v,j. We s t a r t w i t h v ( ' ) = 0 f o r a l l v , j . V ,J
1, =
t(v,j)
I1 = t(v,j)
: D(k)
+
L e t vi:j
N(k)v
E I.
we o b t a i n
k
: D(k) + N(k)v + k
Hence ( 3 ) i s f u l f i l l e d by d e f i n i t i o n . For (v,j)
+
be determined. We d e f i n e
+ 2
+
< j}
2 2 j }
for
361
Continuous linear maps between Frhchet spaces
~
-
21+D( k)+N( k ) v + j 22D(k)+2N(k)v+k+3
5 2D ( k + l ) + N ( k + l ) v which p r o v e s ( 1 ) and (2) Now we d e f i n e
The second e q u a l i t y comes f r o m ( 3 ) . The s e r i e s converges because o f ( 1 ) . We o b t a i n t h e f o l l o w i n g e s t i m a t e :
f o r a p p r o p r i a t e C = C ( k ) and a l l k , v , j .
E N d e f i n e s a map Bk E L(s,Fk).
Hence (bi!j)v,j
We f u r t h e r have because o f ( 3 )
which means t h a t
pk+l,k
0
Bk+l - Bk = Ak
.
From 1 . 5 . and 1.7. we g e t i m m e d i a t e l y
2.2. Theorem:
1 E x t (E,F) = 0.
I f E i s a subspace o f
s and F a q u o t i e n t space of s t h e n
F o r t h e f o l l o w i n g r e s u l t we use t h e method o f [23], t h e remark a t t h e end o f p r o o f o f S a t z 1.7.
2.3. Lemma:
rf
Extl(E,s)
= 0 then E
u: There e x i s t s a n e x a c t O - s - s ~ s - ~0
Satz 1.7.,
i t i s indicated i n
.
i s isomorphic t o a subspace of
sequence ([23]
, 1.6.)
S.
362
D. Vogt
set E
According t o T. and Y . Komura's theorem ( [ l l ] ) =
-1 E and o b t a i n an e x a c t sequence q
we can assume E imbedded i n sN. We
N
O+s+E+E+O which s p l i t s by assumption (and Thm. 1.8.). The E x t ' ( s , - )
Hence we have an imbedding E
case i s a l i t t l e b i t more d i f f i c u l t .
N
+
E
+
s.
The p r o o f uses t h e method f i r s t
used i n [3q i n t h e s l i g h t l y changed f o r m o f [24]. 2.4.
Lemma:
?
F is isomorphic to a quotient space of s.
= 0 then
We c o n s i d e r F as imbedded i n s',
Proof: as
If E x t 1(s,F)
N
c a l l Q t h e q u o t i e n t space and o b t a i n Q
i n t h e p r e v i o u s p r o o f . kle g e t t h e f o l l o w i n g diagram
0
0
Since
5
1
c s we g e t f r o m E x t (s,E)
H
second row s p l i t s which g i v e s
= 0 and =
E
.:+a
1.7. t h a t E x t 1(Q,E)
= 0. T h e r e f o r e t h e
The f i r s t column o f t h e diagram above g i v e s t h e f i r s t row o f t h e diagram below. Our s t a n d a r d e x a c t sequence g i v e s t h e r i g h t column. The r e s t i s c o n s t r u c t e d a l o n g t h e same l i n e as above
.
0
0
t
9
O + s + H + s N + O
t
t
t
T
O + S + G + S S
S
t
t
0
0
+
O
363
Continuous linear maps between Frkhet spaces
E
Because o f 2.1. t h e second row s p l i t s . Hence H
s
G
61s
@
Q i s a quotient o f
2’ s and t h e r e f o r e a l s o E.
Now we have a l l i n g r e d i e n t s f o r t h e f o l l o w i n g theorem which d e s c r i b e s t h e e x a c t
1
s o l u t i o n c l a s s e s o f E x t (E,s)
1
= 0, E x t (s,E)
= 0. Remember t h a t a l l spaces i n t h i s
s e c t i o n a r e assumed t o be n u c l e a r .
2.5. Theorem: Ial E x t 1(E,s) = 0 if and o n l y if E is isomorphic to a subspace of s. Ibi Extl(s,E) = 0 if and onZy if E is isomorphic to a quotient space of 5. The c l a s s e s of subspaces and q u o t i e n t spaces o f s have been d e s c r i b e d i n [23J,
[34]
by t o p o l o g i c a l l i n e a r i n v a r i a n t s ( D N ) and (9). I t i s i n t e r e s t i n g w i t h r e s p e c t t o t h i s t o compare Thm. 2.5. w i t h [31], 5 4 where we d e s c r i b e e.g. t h e c l a s s e s 1 1 { E : E x t (E,s) = 0 1 and { F : E x t (s,F) = 01 by t o p o l o g i c a l i n v a r i a n t s .
3. I n t h i s s e c t i o n we use t h e t h e o r y o f
5
1 t o show t h a t h y p o e l l i p t i c p a r t i a l
d i f f e r e n t i a l o p e r a t o r s w i t h c o n s t a n t c o e f f i c i e n t s have no r i g h t i n v e r s e s i n C"(Q), where o i s an open s e t i n R n . T h i s extends a r e s u l t o f Grothendieck on e l l i p t i c o p e r a t o r s ( s . [24, [ 2 2 ] ) . I t i s shown by a d i f f e r e n t p r o o f and i n g r e a t e r g e n e r a l i t y i n [32]. L e t P ( D ) be a h y p o e l l i p t i c l i n e a r p a r t i a l d i f f e r e n t i a l o p e r a t o r w i t h c o n s t a n t c o e f f i c i e n t s , Q c lRn open and P(D)-convex. We p u t
f(Q)
= { f E C"(s)
: P(D)f = O j
From t h e e x a c t sequence
we o b t a i n by ( I ) f o r any n u c l e a r F r 6 c h e t space E an e x a c t sequence 0
--f
L(E.
&a))
8
L(E,Cm(B))-
P(W* L(E,C"(a))
0
6
E x t 1 (E,fl(a))
L1
E x t 1(E,C"(B))+..
From [ Z q , 2 . 1 . we t a k e t h e f o l l o w i n g lemma. I t i s shown t h e r e under t h e assumption o f e l l i p t i c i t y b u t t h e p r o o f needs o n l y h y p o e l l i p t i c i t y . F o r t h e sake o f completeness we g i v e a p r o o f . 3 . 1 . Lemma.:
Proof:
L'
=
0.
We choose a sequence Q1 cc
a2
and denote by Hk t h e c l o s u r e o f J ( n )
c c . . . c c Q o f open s e t s such t h a t B =
1 Ek
i n C(3,)
c o n s i d e r e d as a space o f
u slk k
f u n c t i o n s on Q ~ W . e o b t a i n i n a n a t u r a l way a commutative diagram w i t h e x a c t l i n e s :
.
364
D. Vogt
0
+
Crn(Q)
n
--t
k
t L
0 L
--f
d(a)
q
cm(")
tj
C"(Qk)
k
9
Hk
+
n
-+
nHk
--f
+
fj
k
0
0 .
+
- fk)k.
and j a r e t h e i d e n t i c a l imbeddings, q i s d e f i n e d by q ( ( f k ) k ) = (fk+llek
The lower l i n e i s a c a n o n i c a l r e s o l u t i o n (see 1 . 3 . ) .
To show exactness i n t h e
upper l i n e we have t o p r o v e s u r j e c t i v i t y o f q. We even g i v e a r i g h t i n v e r s e . T h e r e f o r e we choose q k c D ( Q k ) , ((fk)k) = (
qlk
k v=l
-
qVfv
= 1 on 52 k - l fk)k
( Q :~= pl
) and p u t
*
n
T h i s d e f i n e s a c o n t i n u o u s , l i n e a r map R :
k
C"(Q,)
-+nCrn(Qk). k
Since
on Q~ we have q o R = i d . A p p l i c a t i o n o f ( I ) g i v e s t h e f o l l o w i n g commutative diagram w i t h e x a c t l i n e s
... + n
*
L n
L(E,Crn(Gk))
k
... + n k
L(E,Cm(Qk))
6O E x t 1(E,Crn(Q))
--+
-
tj*
’ 6
L(E,Hk)
...
-+
?L1
Extl(E,
. d(~)) 0 +
q * i n t h e upper l i n e i s s u r j e c t i v e s i n c e q has a r i g h t i n v e r s e t h e r e . Hence ’ 6 = 0 i n t h e upper l i n e and t h e r e f o r e
L~
o ’6
= 0 where 6'
(lower l i n e ) i s s u r j e c t i v e .
T h i s proves t h e r e s u l t .
I f P f D ) has a r i g h t i n v e r s e , t h e n c l e a r l y P(D)* i s s u r j e c t i v e f o r e v e r y F. Hence 1 E x t ( F , ~ ( Q ) ) = 0 f o r e v e r y F, i n p a r t i c u l a r f o r F = w : = (I;' . 3.2. Lemma:
If H is a Fr6chet space such that E x t 1 (w,H)
Banach space f o r every continuous seminorm
=:
=
0 then H/ker
1)
11 i s
a
I1 II on H.
1 Since t h e assumption i m p l i e s E x t (w,H/L) = 0 f o r any c l o s e d subspace L c H i t s u f f i c e s t o show: If H i s a F r i k h e t space w i t h c o n t i n u o u s norm such t h a t 1
E x t (w,H) = 0 t h e n H i s a Banach space. We may assume t h a t
11
5
11 112
5
...
i s a fundamental system o f c o n t i n u o u s norms
on H. We c a l l Hk t h e c o m p l e t i o n o f t h e normed space (H,ll (for k
3
1),
and P ~ :, Hk ~
-+
H,
1 ) t h e c a n o n i c a l e x t e n s i o n o f t h e i d e n t i t y . Then we have t h e c a n o n i c a l
resolution
365
Continuous linear maps between Frkhet spaces
where q ( ( x k ) k )
=
(Pk+l,k
-
'k+l
.
'k)k
If H i s n o t a Banach space t h e n f o r each k pk,l
ak f H. Otherwise we would have
>
1 we can f i n d ak
C
Hk such t h a t
Hk = H f o r some k which, by t h e c l o s e d
pk,l
graph theorem, would i m p l y t h a t H i s i s o m o r p h i c t o t h e Banach space Hk/ker okk,l
(c1, c2,
.
Then by E~ az, ... ) c n H k k' assumption f o r A E L(w, n Hk) t h e r e e x i s t s B E L ( w , n Hk) such t h a t A = q o B. We k p u t B e = (bl,j, bz,j, ... ) where e j = ( 6k , j )k and ko b t a i n : j For 5 =
...) E
w
we p u t A(5) =
bk + l , j
'k+l,k
al,
f o r k > j and k
bk,j
=
(cl
=.
j
=
0 for j
- b. .
(2)
a j= o j + l , j
(3)
F o r each k we have j k such t h a t b
bj+l,j
j.j
k ,j
2 jk
( 1 ) i m p l i e s t h a t t h e r e i s b . F H such t h a t b = b . f o r a l l k 2 j . I f we choose J k,j J 3 j,, we have because o f ( Z ) , t h e second p a r t o f (1) and ( 3 )
j
P
~
a ,j =~ b.
J
-
'j,l
b
. = b . - b = b . E H J 1,j J
j,J
which c o n t r a d i c t s t h e c h o i c e of
a
j'
We r e t u r n t o o u r h y p o e l l i p t i c o p e r a t o r P ( 0 ) . I f i t has a r i g h t i n v e r s e i n C"(Q) 1 we know t h a t E x t ( w . ~ ( s ? ) ) = 0 . We choose a compact K c M w i t h non empty i n t e r i o r
Ilfll = su I f ( t ) 1 . Then ~ ( Q ) / k e rI1 112 d ( Q ) IK . i s a n u c l e a r Banach space, hence f i n i t e i i m e n s i o n a l . T h i s i s p o s s i b l e o n l y f o r
and a p p l y Lemma 2 t o t h e seminorm
c-R
n = 1. T h e r e f o r e we p r o v e d t h e main r e s u l t o f t h i s s e c t i o n :
3.3.
Theorem:
If n
in c - ( Q ) .
2
2 and P(D) is hypoeZZiptic then P(D) has no r i g h t inverse
F o r t h e e l l i p t i c case t h i s r e s u l t was f i r s t p r o v e d by Grothendieck ( s . [21J, Appendix C o r [22],
p . 40 f ) . Theorem 3 i m p l i e s t h a t a l s o p a r a b o l i c o p e r a t o r s
( t h e h e a t e q u a t i o n e.g.)
have n o r i g h t i n v e r s e s . F o r h y p e r b o l i c o p e r a t o r s i t i s
known t h a t t h e y have r i g h t i n v e r s e s a t l e a s t i n C m ( W n ) ) .
4 . I n Grothendieck [83, I1 § 4 and a l s o i n [27]
r e l a t i o n s between Kothe m a t r i c e s
r e s p . F r c c h e t spaces a r e i n v e s t i g a t e d which l e a d i n c o n c r e t e examples t o r e s u l t s very s i m i l a r t o t h e r e l a t i o n
1
E x t (E,F) = 0.
366
D. Vogt
We explain,up t o some extent,the connection. In p a r t i c u l a r we show how the res u l t s on E x t 1(E,F) = 0 can be used t o solve a problem o f Grothendieck i n [ 8 ] , 11,
5
4, p . 118 ( s . Thm. 4.6. below) and t o complete h i s i n v e s t i g a t i o n of the r e l a t i o n E b Gfl F i n the case t h a t one of t h e spaces i s a power s e r i e s space s a t i s f y i n g a c e r t a i n s t a b i l i t y condition. I n f a c t we get a r a t h e r complete answer. A more systematic approach can be found in 1331. The crucial lemma 4 . 4 . i s a s p e c i a l ized version of a r e s u l t proved t h e r e . For spaces with basis see [13].
We assume E
= h"(A),
F
= h(B), E
a Schwartz space and p u t P = E b
Gn
F. P can be
considered as a space of double indexed sequences o r of matrices. The dual P ' can be i d e n t i f i e d with the space of a l l matrices belonging t o bounded l i n e a r maps from F t o E. By P*we denote t h e Kothe dual o f P , i . e . t h e space o f a l l matrices v = (v. .). such t h a t 1,J 1 , j E N v. . j
i;j
1
,J
<
+
"
f o r a l l u = ( uI. , J. ) i , j c l N E P . E, means t h e s e t of nonnegative real sequences i n E = k(A). 4.1. Theorem (Grothendieck): The following are equivalent: ( a ) P is bornological
ibi icj
(di
P is barrelled P ' = P* Vno 3no Vm,
h c E+ 3 R
0 Vi,j
:
I t i s easy t o see t h a t we can w r i t e down (Si) in t h e following way:
*
(S2)
Vno 3mo Vm 3 n , R >
0 Vi,j :
Both conditions on the matrices a r e obviously r e l a t e d . We have even (see [13)):
*
4.2. Proposition: Conditions 4.1. ( d i and (S2) are equivalent.
A t l e a s t f o r one p a r t of 4.2. we shall now give a proof not involving any special assumptions on E o r F, in p a r t i c u l a r they a r e not assumed t o be Kothe spaces.
W e recall t h a t f o r any complete l o c a l l y convex space X t h e following implications
367
Continuous linear maps between Frkchet spaces
hold:
X bornological
3
X b a r r e l l e d = X b s e q u e n t i a l l y complete.
We s h a l l show t h a t s e q u e n t i a l completeness o f
*
hV
P i , P = Lb(E,F) o r P = E '
implies (S2). F o r t h e p r o o f which has some s i m i l a r i t y w i t h t h e p r o o f o f Theorem 7 i n
[ 41
F
we
use t h e f o l l o w i n g n o t a t i o n : A p r o j e c t i o n P i n F i s c a l l e d 12 - a d m i s s i b l e i f t h e range o f P i s f i n i t e dimensional and
z x J.
Px =
j
y.(x) J
with y. t F'
.
4 . 3 . Lemma:
( S 2 ) i s equivalent t o the following condition: For euery
J
P
*
there
e x i s t no.k such t h a t f o r a22 K,m there e x i s t n,S and a F-adniissible p r o j e c t i o n P with
f o r a l l x E E and y E ( k e r P ) '
F.
Proof:
We p u t
f o r j = cl,k,K
Q
* 1 lw
e t c . denote the dual norms i n ( k e r P ) ' .
K. With C such t h a t 11Qx11. * J C I y IJ. f o r t h e s e j .
= i d - P and assume IJ. < k <
we have
*
I y l j 5 lly
*
o Qllj
5
5
Cllxllj
We o b t a i n
for a l l x
c
E, y E ( k e r P ) '
IJ.'
hence f o r a l l y E F '
IJ.'
S i n c e P has f i n i t e dimensional
range an analogous i n e q u a l i t y w i t h Q r e p l a c e d by P i s t r i v i a l . We add t h e s e i n e q u a l i t i e s and observe t h a t Ily o it
llyllj f o r
j = p,k,K.
*
QIIj +
*
1Iy o P l l j i s e q u i v a l e n t t o
T h i s proves t h e a s s e r t i o n .
4.4. P r o p o s i t i o n : If P b i s sequentially complete f o r P=Lb(E,F)
or P = E i
&,,
F,
then E and F s a t i s f y condition ( S 2 ) .
Proof:
*
Under t h e assumption t h a t ( S 2 ) does n o t h o l d we c o n s t r u c t d o u b l e indexed
sequences x
k ,n
i n E and y
k ,n
i n F ' such t h a t t h e s e r i e s
k$n < Y k , n ' ' P X k , n ) converges u n i f o r m l y on e v e r y e q u i c o n t i n u o u s s e t o f maps t h e map A E L(F,E)
d e f i n e d by
cp
c.L(E,F)
and such t h a t
368
D. Vogt
i s n o t bounded. S i n c e t h e e q u i c o n t i n u o u s subsets o f L(F,E) a r e t h e bounded s e t s i n Lb(E,F),
since
f u r t h e r m o r e t h e c a n o n i c a l map E ' F + Lb(E,F) i s c o n t i n u o u s and under t h e I -b n, c a n o n i c a l i d e n t i f i c a t i o n s (Eb @n F ) corresponds t o t h e bounded maps i n L(F,E) ( i f t h e i r image i s c o n t a i n e d i n E c E " ) t h i s proves t h e a s s e r t i o n .
W e a p p l y lemma 4 . 3 .
Then by assumption we have p (we can assume IJ. = 1) such t h a t
f o r e v e r y k ( w i t h no = k-1) t h e r e e x i s t K (we can assume K = k + l ) and m (we can assume m = k ) such t h a t f o r a l l n and C and f o r a l l 1 - a d m i s s i b l e p r o j e c t i o n s P we have x E E, y E ( k e r P ) ; w i t h
We choose a b i j e c t i o n
on
V.
For
v
=
v
+
( k ( v ) , n f v ) ) f r o m D\1 o n t o h’
a c c o r d i n g t o ( * ) . We choose a
k ,n
E F w i t h llak,nlJk
L e t x ~ , ~Y ,~ , ~ ak,n , be chosen f o r k = t h a t (*) i s s a t i s f i e d and
W e p u t ~ ( x =) put k =
x
IN
and s e t up an i n d u c t i o n
1 and k = k ( l ) , n = n ( 1 ) we p u t C = 8, P = 0 and choose x ~ , ~yk,n ,
,dl
k,l
1, Yk,n(ak,n)
n = n ( p ) and p
k(p),
= 6
5
6
n,m
=
1,
=
1.
...
,V
such
'
x ( x ) . P i s a 1 - a d m i s s i b l e p r o j e c t i o n . We k ( P ) ,niIJ.) Y k b ) ,n(lL) k ( v + l ) , n = n ( v + l ) , Q = i d - P and choose Co such t h a t IlQxll. 5 Co llxllj,
f o r j = l,k,k+l.
Now we s e t C = Co 2k+n+1 and f i n d x
a c c o r d i n g t o (*), i . e . such t h a t w i t h y
k ,n =
yk,n o
k ,n
Q
E E,
yk,n c
J
(ker P)
l
,.W e choose ak,n E k e r P = i m Q such t h a t /lak,nllk = 1 and yk,n (ak,n) = Yk,n ( a k y n ) = 1.
369
Continuous linear maps between Frechet spaces
Every e q u i c o n t i n u o u s subset o f L(E,F) i s c o n t a i n e d i n one o f t h e f o r m
{v
6 L(E,F)
and C ( k )
B
=
:Ilqxilk 5 C ( k ) ~ ~ x l ~ n f( ok r) a l l k } w i t h i n c r e a s i n g sequences n ( k ) i n
z 0.
Given B we p u t 1 = n ( l ) , rn
=
n ( l + l ) and o b t a i n f o r
ip
N
E B
S i n c e t h e e s t i m a t e i s termwise, t h e s e r i e s on t h e l e f t hand converges u n i f o r m l y on B. F o r A E L(F,E) as d e f i n e d above we have A ak,n
=
x
~ hence , ~ IIA ak,nllk
= n while
Jlak,nllk = 1 f o r a l l k and n. T h e r e f o r e f o r e v e r y k t h e s e t { A x : llxllk 5 11 i s
n o t bounded, i . e . A i s n o t bounded.
1 S i n c e a l l t h e necessary c o n d i t i o n s f o r E x t (E,F)
it
=
0 i n 1311 a r e based on ( S p )
t h e y a l l a r e necessary c o n d i t i o n s f o r t h e b o r n o l o g i c i t y o r b a r r e l l e d n e s s o f Lb(E,F).
I n s t e a d o f a converse o f Prop. 4 . 4 .
wecomplement i t by a s u f f i c i e n t
condition.
4.5. P r o p o s i t i o n :
(B), then Lb(E,F)
If E and F are nuczear, E has property ( D N ) and F has property I an F is bornoZogica2.
= Eb
d .,
prooS: From G r o t h e n d i e c k ' s r e s u l t [81,11,§4,n03, we know t h a t Lb(s,s) i s b o r n o l o g i c a l . From [24], sequences
and 4 . 5 . ,
Prop. 3.3.
Cor. 2
( c f . 4.1. and 4.2.)
by a d d i n g complements we o b t a i n e x a c t
L
O + E + s + G - 0 q
O+H-s-F+O where G and H a r e complemented subspaces o f s . Since E x t 1(E,H) = 0 and E x t 1( G , s )
= 0 we o b t a i n t h a t t h e map
370
D. Vogt
i s s u r j e c t i v e . By [ 8 ] , I , Hence E ' b
6n
5
1, no 2, Prop. 3
L*
8 q i s a t o p o l o g i c a l homomorphism.
F = Lb(E,F) i s b o r n o l o g i c a l .
L e t V be a complex a n a l y t i c m a n i f o l d . I n [ 8 ] , 11, that Lb(X(V), %(V))
i s b o r n o l o g i c a l if V =
0
4, p . 128 Grothendieck s t a t e s
Ic, whereas f o r V
=
D (unit disc) i t
i s n o t . He asks f o r a c l a s s i f i c a t i o n o f complex m a n i f o l d s w i t h r e s p e c t t o l i n e a r t o p o l o g i c a l p r o p e r t i e s o f X ( V ) . Now f o r t h e p r e s e n t case we can g i v e a complete answer.
4.6. Theorem: The following are equivaZent:
i s bornoZogicaZ i s barrelzed
(1)
L ~%(V), (
(2)
Lb(X(V), & ( V ) )
(3)
g ( V ) has property (DN)
(4)
every pZurisubhamonic f u n c t i o n on V,
&(V))
which i s bounded from above i s constant
(strong LiouvilZe p r o p e r t y ) .
Proof:
We have (1) = 1 2 )
=
Pt; s e q u e n t i a l l y complete f o r P = L b ( & ( V ) , & ( V ) ) .
t h i s implies (S2) f o r E = F
4.4.
=a(V). Since
on (S;) we know f r o m t h a t theorem t h a t ( 3 ) = ( 1 ) f o l l o w s f r o m 4.5.
x(V)has
[3l],
Thm. 7.2.
By
i s c o m p l e t e l y based
(DN).
s i n c e x ( V ) i s n u c l e a r and has p r o p e r t y
( a ) ( c f . [31]
§ 7,B). The e q u i v a l e n c e o f ( 3 ) and ( 4 ) i s shown i n L371. B e f o r e we t r e a t t h e case o f power s e r i e s spaces we need some i n f o r m a t i o n about t h e c o n n e c t i o n between t h e r e s u l t s o f [27]
and t h e p r e s e n t paper. By LB(F,E) we
denote t h e space o f a l l bounded l i n e a r maps f r o m F t o
E, i . e . t h o s e maps which
send some neighbourhood o f z e r o i n t o a bounded s e t . We o b t a i n f r o m [27],
1.3.
and 1.4.
4.7. Theorem: If E = x"(A) o r F = X(B) then the foZZowing are e q u i v a l e n t : (1)
L(F,E) = LB(F,E)
(21
f o r every sequence K(N) t h e r e e x i s t s K such t h a t f o r each n we have
and C w i t h
f o r alZ x f E, y
F'.
No
371
Continuous linear maps between Frtichet spaces
One p a r t o f
[Zq,
7.3. says t h a t a t l e a s t f o r r e f l e x i v e E t h e e q u a l i t y L(F,E)
i m p l i e s t h a t E b @n
LB(F,E)
=
F, hence a l s o E L &n F i s b a r r e l l e d .
We r e c a l l t h a t a F r g c h e t space F has p r o p e r t y
( 5 )i f
t h e f o l l o w i n g h o l d s ( s . [36]
,
c2q 1: (5)
*2 * * VY E F ' : I l ~ l l K5 C IIYIIL IIyllk.
Vk 3K VL 3C > 0
One c o n n e c t i o n between
[21
and t h e p r e s e n t paper i s g i v e n by t h e f o l l o w i n g lemma
w h i c h e x p l a i n s i n a v e r y s a t i s f a c t o r y way t h e c o i n c i d e n c e o f t h e c o n d i t i o n s i n [31]
, 4.2. and [ 2 n , 4.3. as w e l l as i n [31],
4.3.
( r = l ) and [27],
2.1. a t l e a s t
f o r nuclear iil(u).
(L)then
4.8. Lemma: If E = X m ( A ) or F = X(B) and F has property L(F,E)
=
Proof:
*(ST) y i e l d s n o .
f r o m (S1)
LB(F,E).
(S;)
impZies
F o r a g i v e n sequence K(N) we p u t p = K(no) and o b t a i n
a number k such t h a t f o r a l l K and m we have n and S w i t h
f o r a l l x E E, y E F ' . F o r k we choose K (S;),
a C
3
k according t o
0 such t h a t
(L)and
o b t a i n f o r L = K ( n ) , n = n(K,m)
as i n
for all y E F'. We o b t a i n f o r x E E, y E F ' :
*
I I X I I ~IIYIIK
5
(C
*
+
S ) max (lIxlln I I Y I I K ( ~ ,lIx/I )
*
/ I Y I I K ( ~ ~1 )
which p r o v e s t h e a s s e r t i o n .
*
*
A s i m i l a r p r o o f shows t h a t (S1) and ( S 2 ) a r e e q u i v a l e n t i f E has p r o p e r t y (DN).
Remark:
I n t h e f o l l o w i n g theorem we assume f o r ( 3 ) , a.and
p. t h a t
SLPI
an+,/
an < +
m
372
D. Vogt
and f o r ( 3 ) , y. t h a t l i m / an = 1 ( s t a b i l i t y assumptions). n A F r 6 c h e t space F has p r o p e r t y (E) i f t h e f o l l o w i n g h o l d s ( s . [27] ; [31] ) :
(E)
Vk,
E
> O
3K VL
Obviously property
3C > O
(n) i m p l i e s
*ltE
Vy E F ' : IIYIIK
C-
*
+€
C IIYIIL IIYIIk
.
('5').
The theorem extends ( i n t h e n u c l e a r case) Prop. 15 i n 181, 11, 54, n03 which g i v e s t h e case a. f o r E a Kothe space. 4.9.
Theorem: If E and F are nuclear and one o f them a power s e r i e s space
satis-
f y i n g our s t a b i l i t y asswlrptions, t h e n t h e following a r e e q u i v a l e n t : (1)
Lb(E,F) i s bornological
(2)
Lb(E,F) i s b a r r e l l e d
( 31
a . i f F = h r ( a ) : E has property (DN) p . i f E = AJu)
: F has property ( Q )
y . i f E = A l ( a ) : F has property
As i n 4.6. we have (1)
proos: 4.1.,
4.2.,
*
3
(2) = (S2).
(1) f o l l o w s f r o m 4.5.,
3
S i n c e t h e n e c e s s i t y p a r t s o f thms.
a r e based on ( S z ) , t h e s e theorems y i e l d ( 2 )
4.3. i n [31]
t h e cases and p. ( 3 )
-
(5).
5
(3). I n
i n t h e case y . c o n d i t i o n (S;)
is
s a t i s f i e d as we know from [3], 4.2. t h e s u f f i c i e n c y p a r t o f w h i c h i s based on (S1).
4.8. y i e l d s L(F,E) = LB(F,E), which i m p l i e s b a r r e l l e d n e s s o f E L 6~~F Lb(E,F) ( s . [2g, 7.3.). F i n a l l y Theorem 14.3. b i n [ 8 ] , 11, 5 4 no 2 t e l l s t h a t i n t h i s case ( 2 )
3
(1).
5 . We want t o d e t e r m i n e t h o s e F r 6 c h e t spaces E and F which can o c c u r i n a non-
1
t r i v i a l way i n t h e r e l a t i o n E x t (E,F)
= 0. We s h a l l use t h e t h e o r y o f 131).
t h e main t o o l i s an i n v e s t i g a t i o n o f t h e r e l a t i o n s (S1)
*
Hence
and ( S 2 ) .
F o r a g i v e n fundamental system o f seminorms i n F we denote by Fk t h e c a n o n i c a l Banach spaces. The dual o f Fk can be i d e n t i f i e d w i t h t h e space F;C, i t s norm with
[3]
/I . ;1
We use t h e f o l l o w i n g c o n d i t i o n i n t r o d u c e d by B e l l e n o t and Dubinsky
t o c h a r a c t e r i z e t h e Fr'echet spaces which have n u c l e a r Kothe q u o t i e n t s w i t h
c o n t i n u o u s norm
(*I
3w Vk 3K :
SUP
*
{llyIJk : y E F;L
*
, I J y / J5K 1)
=
+
-.
T y p i c a l examples f o r spaces n o t s a t i s f y i n g (*) a r e p r o d u c t s o f Banach spaces o r , more genera1,quojections s a t i s f y (*) i s
W.
(s. [3]).
The o n l y F r 6 c h e t Monte1 space which does n o t
373
Continuous linear maps between Frichet spaces
5.1. P r o p o s i t i o n :
If F does n o t s a t i s f y ( * ) then (S1) i s s a t i s f i e d f o r every
FrQchet space E. The
proof
i s obvious.
W i t h r e s p e c t t o t h e f o l l o w i n g a n a l y s i s (Lemma 5.3. and Thm. 5 . 5 . )
the p r o p o s i t i o n
*
means t h a t f o r F n o t s a t i s f y i n g ( * ) we cannot g e t any i n f o r m a t i o n on E f r o m (S1)
*
.I).
o r f r o m ( S 2 ) . However i t i s o n l y known under a d d i t i o n a l assumptions t h a t (S1)
1
p l i e s E x t (E,F)
= 0 ( s e e e.g.
Thm. 1.9. o r
139,
5
im-
3).
I f we r e s t r i c t o u r a t t e n t i o n m a i n l y t o t h e cases where e i t h e r E o r F i s c o n t a i n e d i n t h e c l a s s o f n u c l e a r spaces t h e n t h e case where F i s n u c l e a r causes no d i f f i -
*
1
c u l t y s i n c e i n t h i s case (S1)
i m p l i e s E x t (E,F)
= 0.
I n t h e o t h e r case we can g i v e a p r e c i s e c h a r a c t e r i z a t i o n o f t h e spaces F f o r which 1 E x t (E,F) = 0 f o r e v e r y E. F o l l o w i n g B e l l e n o t and Oubinsky we c a l l q u o j e c t i o n s t h e F r g c h e t spaces F where F / k e r norm
11 11.
131
11 11
i s a Banach space f o r any c o n t i n u o u s semi-
These a r e t h e spaces o c c u r i n g i n Lemma 3 . 2 . .
From Lemma 3 . 2 . and an
argument i n [ 3 g , we o b t a i n : 5.2. Theorem: 111
The following are e q u i v a l e n t f o r a FrQchet space F:
(31
F i s a quojection 1 E x t (w,F) = 0 E x t 1(E,F) = 0 f o r every nucZear FrQchet space E.
From
131
(2)
we know t h e f o l l o w i n g . I f F i s a q u o j e c t i o n t h e n i t does n o t s a t i s f y
( * ) . The converse i s t r u e f o r r e f l e x i v e F. I t i s unknown i f i t i s t r u e always, i.e.
if
F i s a quojection
i f f i t does n o t s a t i s f y ( * ) . Hence we do n o t know i f
( f o r E r e s t r i c t e d t o t h e c l a s s o f n u c l e a r spaces) 5.1. B u t 5.1.
says t h a t 5.5.
and 5.5. c o v e r e v e r y t h i n g .
(1) i s o p t i m a l as l o n g as we use c o n d i t i o n s (S1)
and (S;)
f o r our analysis. W h i l e i n t h e p r e c e d i n g we d i s c u s s e d cases where we cannot g e t any i n f o r m a t i o n on Y .I). 1 E from (S1) r e s p . ( S 2 ) o r f r o m E x t (E,F) = 0, o u r main o b j e c t i v e i s now t o i n -
v e s t i g a t e t h e case o f spaces w i t h (*). We need t h e f o l l o w i n g g e n e r a l i z a t i o n o f (ON) ( c f . [Z],
(O,+-)
--f
[34]). L e t
(O,+-),
cp
always denote a s t r i c t l y i n c r e a s i n g c o n t i n u o u s f u n c t i o n
l i m cp(r) = r=-
(DNcp ) 3no ’dm
+
m.
3, C
b'x
c
E, r > 0 :
374
5.3.
D. Vogt
( ONcp) f o r some
cp
.
We choose
Proof:
*
( S E ) then E has property
I f F s a t i s f i e s (*) and E and F s a t i s f y
Lemma:
)I
*
a c c o r d i n g t o ( * ) and o b t a i n f r o m (S1) no and k . ( * ) a g a i n
y i e l d s K and p u t t i n g t h i s i n t o (S;)
we f i n a l l y o b t a i n : F o r e v e r y m t h e r e e x i s t s
n and S such t h a t f o r a l l x E E and y f 0 E F ' we have:
*
*
E F ' such t h a t
We choose a sequence y,
and a f u n c t i o n
for all v
cp
c IN.
I-L
P
such t h a t f o r a l l S
3.
0 we have C
7
0 with
For
we o b t a i n lIxllm c
C'P ( r ) Ilxll
+
1
.
llxlln
0 which proves
By i n c r e a s i n g C i f necessary we have t h i s i n e q u a l i t y f o r a l l r (ONcp)
5.4.
*
Lemma:
I f E has property (DN ) f o r some cp
then there e x i s t s a nuclear
cp
it
Kiithe space k ( B ) with continuous nomi such t h a t (S1) i s s a t i s f i e d .
Proof: $(l)
=
-
We choose a s t r i c t l y i n c r e a s i n g f u n c t i o n $ : (0,tm)
1, l i m $(r) =
for all v
r-te-
c N.
+
+ (O,t-),
$(r)
5 r,
such t h a t
Moreover we choose a sequence p .
>
J -
1 such t h a t 2
J
-1
p. J
< t
-.
Cr
r
r
375
Continuous linear maps between Frkhet spaces
We p u t b. J9
1
= 1
for all j
+
bj,*
= $
-1
o
-1 ..o+ ( p j ) f o r a l l j
( j -1)ti mes
and d e f i n e b
j,k
f o r k > 2 i n d u c t i v e l y by
We o b t a i n f o r j 2 k b. J'k+l b j,k
$ o
=
...
(bj,2)
Q $
=
$
-1
... 0 ~r-1
(pj)
( j - k ) t imes
( k - 1) t i mes
Hence x(B) i s n u c l e a r . I t has c o n t i n u o u s norm.
*
To e s t a b l i s h (S1) we choose no such t h a t f o r a l l m we have n,C w i t h
(1)
llxllm
5
Cq (r)
llxll
+ "0
1
Ilxll,
f o r a l l r > 0 and x C E. F o r g i v e n p we p u t k = p + 1 and o b t a i n f o r a l l K > k and l a r g e j
I n (1) we p u t r =
hence
bj'Kand o b t a i n f o r l a r g e j
5
376
D. Vogt
f o r a l l j with appropriate S Lemma 5.3. and 5.4.
>
0, which proves t h e a s s e r t i o n ( c f . 4 . 2 . ) .
t o g e t h e r g i v e t h e f o l l o w i n g theorems. P r o p o s i t i o n 5.1.
t e l l s us t h a t t h e r e s t r i c t i o n on
F i s natural.
5 . 5 . Theorem: The f o l l o w i n g are e q u i v a l e n t : (1)
t h e r e e x i s t s F s a t i s f y i n g ( * ) such t h a t
1 E x t (E,F)
(2)
t h e r e exists anucZear Fnot isomorphic t o
~il such
131
E has property (DN ) for some ‘p
Remark:
= 0
1
t h a t E x t (E,F)
= 0
9.
A t r i v i a l consequence i s t h a t (1) o r ( 2 ) i m p l i e s t h a t E admits a
c o n t i n u o u s norm.
To g i v e a more c a r e f u l a n a l y s i s o f t h e meaning o f ( 3 ) i n 5.5. we use t h e
following condition (s.
1281):
(**) There e x i s t s po such t h a t f o r e v e r y p 2 p,
we have a q 3 p w i t h t h e f o l l o
E which i s Cauchy w . r . t . even converges t o 0 w . r . t . II II
wing p r o p e r t y : e v e r y sequence i n
11 1)
0 w.r.t.
Po
11 I1
9
-
and converges t o
P ’
T h i s c o n d i t i o n i s e q u i v a l e n t t o E b e i n g c o u n t a b l y normable ( s .
[ 7]),
i.e. to
t h e e x i s t e n c e o f a fundamental system o f norms on E whichmakes i t i n t o a c o u n t a b l y normed space i n t h e sense o f G e l f a n d 5.6. Lemma: (1) (2)
E
-
”
Silov.
The folzowing are equivaZent :
has property (DN ) f o r some cp
cp
t h e analogue t o ( * * ) w i t h “Cauchy” repZaced by ’%bounded” hoZds.
proos:
I t i s easy t o p r o v e t h a t (1) i m p l i e s ( 2 ) . We assume ( 2 ) and show t h a t t h e
property there implies:
for a l l x
6
E,
r > 0 and an a p p r o p r i a t e f u n c t i o n cp(r) = cp
P.Po39
F o r t h i s i t s u f f i c e s t o show t h a t f o r e v e r y r > 0 t h e r e e x i s t s all x E F llxllp
5
M llxll
PO
+
r1
IlXllq
E with
M such t h a t f o r
*
We assume t h a t t h i s were n o t t r u e . Then t h e r e e x i s t s a r M we have xM F
(r).
7
0 such t h a t f o r e v e r y
377
Continuous linear maps between F r k h e t spaces
+
11 > 0. Hence we can assume 11x 11 M P M P r f o r a l l M which c o n t r a d i c t s ( 2 ) .
= 1. But t h e n Jlx
I n p a r t i c u l a r IIx /Ix (1 mq
5
11
M Po
-+
0 and
We choose cp such t h a t
v PYP0.q ( r ) rf o r a l l p,po,q
= 0
Cp(t-1 which o c c u r . I t f o l l o w s t h a t E has p r o p e r t y (DN ) . cp
I f E i s a Schwartz space t h e n f o r e v e r y p we have which i s bounded w . r . t .
'i; such
t h a t e v e r y sequence
has a subsequence which i s Cauchy w . r . t .
p. T h i s proves
t h e second p a r t o f t h e f o l l o w i n g lemma, t h e f i r s t one i s obvious f r o m 5.6. ( 1 ) I f E has property (ON ) f o r some
5.7. Lemma:
cp
then it i s countably normatile.
(2) I f E i s a countabZynomable Schwnrtz space then i t hasproperty some
cp
.
Hence theorem 5.5. 5.8. Theorem:
(DN ) cp
For a Schwartz space
E t h e fofolowing a r e e q u i v a l e n t :
t h e r e e x i s t s a Schwarta spuce (nuclear space) F not isomorphic t o 1 t h a t E x t (E,F) = 0
121
E is countably normable.
[51,
for
g i v e s f o r t h e case o f Schwartz spaces:
(1)
I n Dubinsky
.
1 6 1 and i n
[7 1 ,
w
such
[28] t h e r e a r e g i v e n examples o f n u c l e a r
(F)-spaces w i t h c o n t i n u o u s norm b u t n o t c o u n t a b l y normable which t h e r e f o r e do n o t have t h e bounded a p p r o x i m a t i o n p r o p e r t y . From t h e p r e c e d i n g we conclude t h a t t h e r e a r e a l s o examples o f n u c l e a r spaces E w i t h c o n t i n u o u s norm f o r which t h e r e 1 i s no F s a t i s f y i n g ( * ) such t h a t E x t (E,F) = 0. F o r F s a t i s f y i n g ( * ) and n u c l e a r 1 E 5.1. i m p l i e s E x t (E,F) = 0 .
1 To determine c o n d i t i o n s on F f o r t h e e x i s t e n c e o f E such t h a t E x t (E,F) = 0 we r e c a l l t h e f o l l o w i n g concept which was i n t r o d u c e d by Grothendieck [ 9 ]
: A locally
convex space F i s c a l l e d quasi-normable i f f o r e v e r y e q u i c o n t i n u o u s s e t A c F ' t h e r e i s a neighbourhood o f z e r o V i n F such t h a t on A t h e t o p o l o g y o f FA c o i n c i d e s w i t h t h e t o p o l o g y o f u n i f o r m convergence on V . Equivalently (s.
[9],
p. 107): F i s quasi-normable i f and o n l y i f f o r e v e r y
neighbourhood of z e r o U t h e r e e x i s t s V such t h a t f o r each
a >
0 there exists a
378
D. Vogt
bounded s e t B c F such t h a t V C B In
[14]
t a u .
i t i s shown t h a t a F r g c h e t space F i s quasinormable i f and o n l y i f
t h e r e e x i s t s 9 such t h a t F has t h e p r o p e r t y
(QY) Vp 3k VK 3 C V r
>
0, y
c F'
:
* " 1 * l l ~ l 5l ~C cp ( r ) l l ~ l +l ~ IMlV -
5.9.
I f E and F s a t i s f y
Lemma:
*
( S p ) and E i s a proper inon normablel Frdchet
space, then F has property (9 ) f o r some 9 9
.
If F does n o t s a t i s f y c o n d i t i o n (*) t h e n i t o b v i o u s l y has p r o p e r t y (9 )
proos:
for every
cp
cp
and hence i t i s quasi-normable.
I f i t s a t i s f i e s c o n d i t i o n (*) t h e n we know f r o m t h e remark a f t e r 5.5. admits a c o n t i n u o u s norm. We can assume t h a t a l l
that E
11 Ilk on E a r e norms. I n an
analogous way as i n t h e p r o o f o f 5.3. we show t h a t F has p r o p e r t y (Q ) f o r some
Given
9
.
cp
*
IJ-
we f i n d no and k a c c o r d i n g t o ( S p ) . We choose m
>
no such t h a t t h e r e
e x i s t s a sequence xv E E such t h a t
T h i s i s p o s s i b l e s i n c e F i s n o t normable. F o r g i v e n K and m we o b t a i n n and S such t h a t f o r a l l y E F ' and
\I
:
We c o n t i n u e as i n t h e p r o o f o f 5.3, The e x i s t e n c e o f a n u c l e a r h(A), even w i t h some a d d i t i o n a l p r o p e r t i e s , f o r g i v e n
1
F w i t h p r o p e r t y ( a ),such t h a t E x t ( X ( A ) , F ) = 0 i s i m p l i c i t l y shown i n [14]. 9 B u t s i n c e we do n o t need here t h e a d d i t i o n a l p r o p e r t i e s we g i v e a d i r e c t p r o o f . 5.10. Lemma:
I f F has property (Q ) f o r some cp then there e x i s t s a nuclear *'p
Kothe space E = h ( A ) such that (S1)
Proof:
We can assume cp(r)
2
is s a t i s f i e d .
r, 911) = 1. We choose a > 1 such t h a t Z a j j j
< t
m
379
Continuous linear maps between Frkhet spaces
and p u t a
j ,1
= 1 for all j
F o r k > 2 we d e f i n e a
j ,k
i n d u c t i v e l y by
aj,k+l = aj,k
cP(aj,k).
An i n d u c t i o n argument shows t h a t @ ( a . ) 2 Jlk
U.
J
f o r k 3 2 hence
which i m p l i e s t h a t h(A) i s n u c l e a r . The p r o o f c o n t i n u e s as t h e p r o o f o f 5 . 4 . The e q u i v a l e n c e o f ( 2 ) and ( 3 ) i n t h e f o l l o w i n g theorem i s c o n t a i n e d i n theorem 6. I t s h o u l d
c o n s t r u c t i o n o f x(A) g i v e n i n (141. The e q u i v a l e n c e o f i n Lemmas 5.9. and 5.10.
5.11. Theorem:
[lq,
be n o t i c e d t h a t h e r e we do n o t need t h e more s o p h i s t i c a t e d
(I) and
(2) i s c o n t a i n e d
For a FrBchet space F t h e f o l l m i n g are equivazent: 1 E x t (E,F) = 0
there e x i s t s a proper Frdchet space E such t h a t
(1)
(2) F has property (D ) f o r some ‘9
F i s quasi-normable
131
.
cp
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