on the approximation properties for the space h∞
TRANSCRIPT
N~~tth. N ~ w . 1% (1085) 19-27
On the Approximation Properties for the Space H“
By JEAN BOURGAIN of Brussels and OLEG REINOV~ of Leningrad
(Received July 12, 1983)
It is unknown whether the HARDY space H - has the approximation property. However, it will be shown that for each p f 1 Ha has the approximation property AP,, defined below (see also [6]), and, moreover, Hn has the approximation property ‘‘up to log n” (see Theorem 9).
Definition. -4 BANACH space ,Y has the property -4P, if the canonical mapping Y * g,,X +N,( Y , X ) is one-to-one for m y Y . If p z 1 X has the property C-?WAPp if for any Y the canonical mapping Y* G P X -Ip( Y , X**) is a C-isometric im- bedding.
Here. for p z 1 , l i p and I , are the ideals of p-nuclear and strictly p-integral operators respectively ; Y * G,X is the completion of the tensor product Y* @X in the “finite p-nuclear” norm (see [4]). If p < 1 then
Y* H,(Y ,X)=(TEL(Y, X ) : T = z y ; @ z j , z(lly;ll llzjll)p--) ;
,X is the subspace of the projective tensor product Y * 6 IX, associated with
Evidently, AP, implies AP, if 1 zs z r > O . Also, i f X** (respectively, X*) has the property -4Pp, then S has the property AP,, for every p (respectively, for p s 1 ) . On the other hand every RANACR space has the properties ,4P2 and APSI3. For each p=-2 /3 , p * 2 , there is a BANACH space without the property AP,. For more (recent) results on the ,4P’s the reader is refered to [5], [6].
Theorem 1. The space €I- has the property AP, for any p r Q , p*1. Xoreover, if p z - 1 then H’ and all its even duals have the property l -XAP,; if p c 1 then all the duals of H‘ have the property AP,,.
Theorem 1 is an immediate consequence of the following theorems (2, 3 , 5 and 6).
Theorem 2. Let p w 1 be given and let X be a BANACH space suck that 17,.(X, Z ) = =I,,(X, 2) as sets for any (reflexive) BANACH space Z. Then the space X and it8 even duals have the property l-MAP,.
Theorem 3. (cf. (61). If pc[l, 21 and X has the property AP, then X has the
N],( Y , X).
1) Supported in part by the Swedish Institute Bilateral Scholarship Program 4.
20 Math. Wach. 12% (1985)
property AP,, where l /p + l/s = 2. Thus if X has the property A P , for any p =- I then X has the property AP,, for any p=-O, p + l .
It follows from Theorems 2 and 3
Corollary 4. Let p>l and X be a BANACH space such that IT,,(X, 2) = I p r ( X , 8) as sets for any (reflexive) BANACH space 2. If l/s + 1/p = 2 then the space X and all its rJuuls have the property AP, .
In what follows the equality l I p f ( X , 2) = I p r ( X , 2) means that i,(T) s K n p ( T ) for each TEL?,,(X, 2).
K
6: Theorem 5 . (cf. [3]). If q r l and l I , (X , Z)=I,(X, 2) for uny2, then 17,(X**, 2)
It follows from Theorem 5
Theorem 6. (cf. [2], [3]). IT,@", Z)=Iq(Hm, 2) for any q > l , where C,=
To be selcontained we will give here a simple proof of Theorem 8 and also a proof of Theorem 3.
Proof of Theorem 3. Suppose that X does not have the property AP,. Then we cnn find zEY*G,,,Y such that trace z = 1 and trace Rz=O for each finite rank operator R in 5. Let
K =I , (X**. 2)
CP
=pZ/ ( rp -1 ) .
z= zajx;@Xzi where ci, 2 0 , (aj) EZ', llx~ll11xJ i 1. Define an operator t : X -1' by
t = zu;'p'x; 8 ei
(here ( e j ) is the standard basis in c ~ ) , and a tensor element
w = ~ a ~ - " f e i ~ x j . dt is easy to see that tEIT,,.(X, P ) and tuEN,(P, X) (since 1 -s/p'= 1 - 8 + s / p and J' (1 -s) =sp' (l/s - 1) = 9).
Let P = t ( X ) and v=image of 20 in Y*g,,X. Evidently, trace vt= 1 and v = O as an operator. Therefore X does not have the property AP,.
Proof of Theorem 5. We may assume 8 reflexive. Let TEIT,,(X**, 2) and U cZ*@X**. Using standard argument on the selection of (1 +e)-norming finite dimensional subspace in Y* for a given finite dimensional subspace E'c P (and applying it to the space Y = ZziGk(2)) we obtain easily that U admits a factoriza- tion
U : Z % Z o Z X * *
such that 11811=1 andnpe(V)s( l+e)np4U), dimZ,<- Thus
trace UT =trace VQT si,(QT) zp*( V ) .
Bourgain/Re,inov, On the Approximation Properties 21
Since n,,(X, 2,) fIp(X, 2,) and Z0 is finite dimensional, by duality we have li np(X**, 2,) =Ip(X**, 20). Thus
Itrace UTI s n p t ( U ) (1 + E ) Knp(T) . Therefore T € ( Z * @ X * * , n,,,)*=lp(X+*,2), and i,(T) sKnP(T) . Theorem 2 is a consequence of Theorem 5 and the following lemmas
Lemma ‘7. For p w 1 the following are equivalent; 1) X has the property AP, (rwpectively, the property K - NAPp) ; 2) for any reflezive space Y the canonical mapping
Y*&J +Ip( Y , X**) is one-to-one (respectively, a K-isometric imbedczing) ;
3) given ewe, n reflexive space Y , tin operator T<I?,>(X, Y ) and a weakty p’- swmmable sequence (zk)EX. there ezists a finite rank operator R from X to Y such that
(respectively. and ,z,,(R) s K n J T ) ) . 21 I T X k - Rz,l y’ -= E
Remark. Lemma 7 answers a question of SAPHAR ([7], p. 388).
Lemma S. Let p 1 and .X be a BANACII space szirh that there i s a projection P from X** onto X. Then S ltav the property AP, if and only if i t has the property
Proof of Theorem 2. We shall show that X satisfies the third condition in Lemma 7. Let E > O be given and let T€IT,,,(X, Y ) , where Y is a reflexive RANACH space. By hypothesis the operator !Z’ can be factored as follows:
IlPll -HAP,.
T : X ~ C ( K ) -LLJJ’(K, m1-E: Y where U, S are linear bounded; m is a RADON probability on a compact space K , j is the identity map. For a weakly p’-summable sequence (zk) CX consider a representation of (jS(z,)) (that isnorm p-summable) as a product (ak( jSCz,) a;‘)),, where (a,) is a sequence of reals such that ak-0 and A = ZlljS(z,) a;’ll’‘ceo. Let 6 z 0 be such that A6 < E.
We can find a finite rank operator Q: LP(K, m ) - Y that approximates U on the compact set { ( j S ( z k ) ak)/II j 8 (~ , )11]~ :
22 Math. Pu’achr. 122 (1985)
Now, it follows from Lemma 7 that X has the property AP,, and from Lemma
Thus, it remains to prove Lemmas 7 and 8.
Proof of Lemma 7. We have to prove only implications 2 ) = - 3 ) 3 1 ) . 2)*3). Suppose that there exists E z O , a reflexive space Y , a weakly p’-summahle se- quence (xk)cS and an operator TEITpt(X, Y ) . np8(T) 5 1 , such that:
8 and Theorem 5 that X and all its even duals have the property 1 -lMAP,.
a) if REX* @ Y then 2 llTxk - RxkllP’ r e , or, respectively b) if R€X*@ Y and n,,(R) S K then ~~~Txk-RxkI[”’~~. We may consider T and R as elements of the space P’( Y ) : let T(n) = Tx, and
a) 1 1 2’- Rll,,,, =-e“p’, or, respectively. b) if npp(R) sK then [I!?’ -Rlllp’(y, =-e‘lP’ . It follows that there exists a functional cp € (P”’( Y ) ) * = lp ( Ir’) such that a) (cp, R ) = o for any REX*@ Y , and (cp, 5?’)+0, or. respectively, b) (cp, !?‘)=-sup ((cp, R ) : R E X @ Y ; npe(R) S K I . Put a= zcp(n)@xnc Y * G p X . Then a) trace RG = (cp, 3) = 0 for any REX * @ Y, and trace TG = (rp, 5?’> 8 0, or 1)) Il@llz-.,+;‘trace T@=(rp, !?‘)>sup ((9, R ) : R E X @ I’, n J B ) s K )
R(n) = R X ~ . Then we have :
=sup {trace R@: R E X @ Y , z,.(R)sK) = ~ l l ~ l l I * ~ ~ , ~ * * ) *
3 ) s l ) . Let e>O be given and let Z be a BANACR space and t ~ Z + 6 J . Iltll=l. Then there is an operator U E R ~ . ( X , Z**) such that nP, (U)=l and trace U t = 1.
IVe have to show that there exiets REX*@Z** such that a) trace Rt * 0 ,
b) trace R T z ( 1 -&) and n,,(R) sk’ . We may assume that the space Z is reflexive since t can he considered as an
element of Z***G,X with I(tl(=l in this space. and the operator U admits the following factorization
or, respectively,
u : X T Y 2 Z * * where Y is a reflexive space, llij’ll s 1 and n,,(T) s 1.
By hypothesis we can find an operator REX*@Z such that
(respectively, and n,@) s K ) .
Now, let t=X<@x, , , where zll<lIPS1 and sup {z[(x’. %,)Ip’: 11z’[1 ~ 1 ) - +-.
2 IIRx,, - Ux,,IIp’-=Ep’
It follows tha t trace Rt = trace Ut -trace ( U T - Rt) = 1 - 2 (2:. Ux, - Rx,,)
Zl-(ZIIZnll P ) !I, ( ~ ~ ~ U x ~ - ~ ~ , , ~ ~ ~ ’ ) ~ ’ ~ ’ ~ l - & . Proof of Lemma 8. According to Lemma 7 we have to show that if the space
Y has the property AP, then the canonical mapping Y * g p X - I p ( Y , X**) is a
Boorgain/Reinov, On the Approximation Properties 23
/\PI/-isometric imbedding for any reflexive BANACH space Y . Eut it is an easy consequence of the well-known fact that N,( Y, 2) =Ip( Y, 2) for any reflexive Y .
One can show that a BANACH space X has the property AP,, p s 1, iff for each (zn-)EP(X) with l / q = l / p - l andevery e - 0 thereis a finiterankoperator R: X- -X such tlint sup llRxk-xn-l1<~ (for q = +- F ( X ) means co(X) ) . Now, it follows
from this and from the above results that for each p w 0 the identity map id: H" -lZ" can be approximated by finite rank operators uniformly on every com- pact subset K such that K = ~ ( ( X ~ ) ~ ) , the closed absolutely convex hull in Ha of a, set ( x ~ ) ~ with ~ I / X ~ ~ ~ ~ - C +a. We can prove, however, a stronger result. Namely:
Theorem 9. There is ct junction B(E) . E w O , such that if ( x ~ ) ~ is cb sequence in H", satisfying
k
--
then there exists ctn opercitor T H"-H" such that 1) s11p l J T X j -zjll <&.
3) l lTll<-w .
i 2) rank T -= B(E) ,
For the proof we need several lemmas. Throughout C denotes some (different) constants.
Lemma 12. There is C=-0 mch that if fEL:. I w O them there exists ( p e r such that
a) 11d1-~31 b) I d f -=A (on TI, c) 111 -f?4L-=~~-~llflll - Lemma 13. There is C>O such that if ai=-O (i= 1, 2, ..., N ) and zaisl then
1 log - sc . ai
'log(i+l) ai
24 Math. Nachr. 122 (1985)
The proofs of Lemmas 10 and 12 can be found explicitely in [l]. The proof of Lemma I 1 is standard. Lemma 13 is a rather simple exercise, so we omit its proof too.
Proof of Theorem 9. A separation argument shows that it is enough t o prove that if (f i)cLl, Zll{illl sl, then there is T:.H'-H-satisfying 2) and 3) such that 2 I(zi - Tzj, fj)l S E .
(i). Fix E w 0 and positive integers J , K with J<K. We shall shorn that there exists T: H"-Ba such that
J
j - 1 a) C l(zj-Tzjt f j ) l < ~ ,
I b) z - - - T I I T * f i l l l e E -" d I l f i l l l 7
J-=jrli log 1 J + + K
c) ran]< T-= J ~ , d) 11T11< J log JE-' . To prove this concider a projection P: L"-Lm such that IlPll s.2, rank Pc J J ,
1 ' 1 +- with S small enough, 8-E . I IF11 I1 Define
TQ? = Zxi??i<8L(tiP(p))
where xiEN" will be chosen later in order to satisfy some conditions. Clearly, rank T< J ~ .
We have :
.I J
d + C C a i j l l - x ~ l l , . i
Thus to prove a) it will be enough t o get 111 -xilll I;Sclltilll for each i. Furthermore, since ' T*f = P*(i,S1(Ri6if?if)), it follows that IIT*flll s SZ and i
i
Now, apIJ1)' Lemma, 11 with ~ = ~ ~ F l ~ ~ ~ ' a ~ ' 2 i 8 i i ~ j and w = t i t o estimate the sum on the right side hy
Bourgnin/Reinov, On the Approximation Properties 25
and b) is obtained. Let e,€L", ef€(La)* be such that l[ek.lsl, Ile$llc? and
26 Math. Nadir. 122 1985
For each pair J,-=J,+I let T , he the operator constructed in (i). Then for each r
2 l(~j-T,.zj , fj)l 2 2 + C + C i i S J r J , < i S J p + l I - J , + i
J ,+I
j = J y - = E + ~ E - ~ C l i f j l l l + E .
Let T = 22-1 (T I + T2 + ... + TR). Then
2 [(zi-Txi. f j ) [ -=2e+2R-Je -" .
Now we have only to take 3 large enough. I
Using the same method one can prove also
Theorem 14. For each p-= + 00 there is cc firnction B,,(E) with the following proper- t y . If (z j ) is ct. seqwnce i n H" szrch that
~ ~ T J I ~ - o cmrl Z I X j ( 2 ) I P s 1 then there eri,pts (671 operator T : Ha -. H" szrch t h t
(i) ~ ~ T T , - z , ~ ~ < E ; j= 3 , 2, ...; (ii) rank T-=BJE);
(iii) llT]l-= BJE) . It is possil)lc also to ])rove a part of the above result in a more general setting.
Lemma 15. For p z 1 and c1 I~ANACH space S consider the following assertions; 1)* X has the property AP, ; 2), For each space Lp=Lp(p), for every 0perator.s U EL(L", X ) and T E N I ( X , Lp)
3), If (9) is cc seqwence i n X such thcd llzkl/ -0 c r d sup z[(zk, z')(''-== then
for each E ~ O fhere ezists ce f in i te 4rcbnk opercctor R : X -X mtch that sup llRxk -xkl1-=& W e have : l), *2), *3), and 1)102)1*3)1.
Proof . It is clear that (3)[*1),. 1)p=-2 )p . Let UEL(LP, X), !l"EN1(X, Lp) and
For this we need
i t follows from trace TU+O that UT=!=O;
IIlfP -1
k
trace T U + O . The operator UT admits the following factorization.
s 4 1,' -r: 1 1 5 LP z x - where A€UP,(X, P') and B is diagonal. Let Z = A ( X ) and V=CBj with j being the injection 24P". Since B*<I?, we have VEN,(Z, Lp), so U V € N , ( Z , X ) = =Z*G,,X. Now if UT=O then UV=O and trace ( U V ) (A)=O (trace U V A is well-defined since X has the property AP,). But i t is easy to see that trace (U V ) (A) =trace TU. Contradiction.
Bo urgnin/Reinov, On the Approximation Properties 27
The implication 2),=-3), can be proved by the same method as in the proof of
Now, it follows from Lemma 15 and Theorem 2
Corollary 16. Let yw 1 and let S be a BANACH space s w h that I7],.(X, 2) = I J X ,
Lemma, 7, 2) 3 3 ) .
2) for m y 2. The$% X h a s properties 21, nnd 3),, from L m t n a 15.
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Vrije Uniuersiteit BrwseE Depn rlwr ent of illnthe7nnlics F 7 - 1050 Bt-us.wl.9 Belgium
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