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1, then LP
and fJP are complete complex linear metric spaces. Clearly A can be identified with a closed linear subspace of f i via the map which assigns to each f E A its mequivalence class. Let g be an analytic function in the open unit disc D\"oD such that for malmost all
z E "OD there exists the radial limit, limrt 19(rz). Then the measurable function z limrt Ig(rz) (defined malmost everywhere on "oD) is called the boundary value function of g. Recall (cf. [Du, Chapters 2 and 3) that every f E IfP (0 < p .;:;; 00) is a boundary value function of a unique analytic function in D\3D; which is called the analytic extension of f; we denote this function, unless otherwise stated, also by f. It satisfies the inequality (0.2)
sup
MpCr, f)
<
00
O
where Mp(r, f) = faD If(rz)I P m(dz) for 0 < p < 00 and Moo(r, f) = ess sUPzEaD If(rz)l. Conversely, if f is an analytic function in D\3D which satisfies (0.2) for some p with
o < p .;:;; 00, then An
f
E
there exists a boundary value function of f and it belongs to HP.
A iff f extends to a continuous function on the disc D which is analytic in
D\"oD. For fE LI we put
BANACH SPACES
fen)
=
If f E HI and if we identify
fan f(z)zn m(dz) f
(n
5
= 0,
± 1, ... ).
with its analytic extension, then
(0.3)
/(n)
for n
fn)(O)
=
n!
< o.
Clearly if f E HI , then fen) = 0 for n
~ O.
The converse implication is also true; it follows
from 0.1
THEOREM
Iff E LP (l ~ P
(THE FEJER THEOREM).
f
lim n an
fez)
f n k=n+
then
P
_.l
For the proof cf. [H, p. 23].
< 00),
(n  Ikl)/(k)zk
m(dz) =
o.
I
0
We introduce now the operator H called the Hilbert transform which plays the crucial role in the study of "classical" spaces of analytic functions. For u E L
1 we define the Hilbert transform of u to be the unique function v =
+ iv is a boundary = Janu dm and
such that u
f(O)
Re f(re i {})
= (21T)1
f
value function of an analytic function
27T lr2 0 u(e i {})        2 r cos {} r2
+
d{}
f
H(u)
in D\aD such that
for re i {} E D\aD.
For gEL I the Hilbert transform H(g) is defined by
M(g) = H(Re g) where Re g
+ i H(Im g)
= \,1(g + g). Img = (g  i)/2i.
The next result summarizes the most important properties of the Hilbert transform (for the proof cf. [Du, Chapter 4]).
0.2. (i) If 00 > p > 1, then H(f) E LP whenever f E LP; moreover there exists an absolute constant a independent of p such that THEOREM
IIH(!)llp ~ a
p2
p=r IIfllp
for every f E LP
and
IIH(f)llp
> aI
p2
p=IlIfllp
for some f= fp E LP.
(ii) (the Kolmogorov theorem) Iff ELI, then H(!) E LP for every p there exists a constant Cp such that
IIH(f)lI p
~ Cpllfll~
< 1.
Hence
for every f ELI.
(iii) The operator f  H(f) is of weak type (11), i.e., there is an absolute constant CI such that for every K > 0
ALEKSANDER PEt.CZyNSKI
6
O.III. Absolutely summing operators and their relatives. General references to this part are the papers by Grothendieck [Grl], [Gr3], Pietsch [Pi], Persson and Pietsch [PP] and Seminaire MaureySchwartz 19721973, 19731974, 19741975 [MS].
B(X. Y) stands for the space of all the bounded linear operators from a Banach space X into a Banach space Y. Let 1 ,;;;; p
<
00.
Let S be a compact Hausdorff space and let /1 E [C(S)lt. By i
Il.P
we denote the natural injection of C(S) into LP(/1), i.e., the operator which assigns to each
fE C(S) its /1equivalence class regarded as an element of LP(J1). If E is a closed linear subspace of C(S), then (unless otherwise stated) E Il •p denotes the subspace of LP(J1) being the closure in LP(/1)norm of ill.p(E) and i: p denotes the restriction of ill .p to E regarded as an operator to E Il.P' DEFINITION
0.1. Let I ,;;;; p
<
00.
A bounded linear operator T: X 
Y is pintegra!
(resp. strictly pintegral) if there are a compact Hausdorff space S, a /1 E [C(S)] bounded linear operators U: X 
C(S) and V: LP(J1) 
t
y** (resp. V: LP(J1) 
and
Y) such
that
Ky' T= Vill.pU
(0.4) where Ky: Y 
(resp. T= Vill.pU)
y** denotes the canonical embedding of Y into its second dual Y**.
A triple (U, V, ill,p) satisfying (0.4) is called a pintegral factorization of T
The p
integral nonn of T is the quantity
where the infimum is extended over all pintegral representations of T. DEFINITION
0.2. Let 1 ,;;;; p
<
00.
A bounded linear operator T: X 
summing if there are a compact Hausdorff space S, a /1 E [C(S)] E of C(S), and bounded linear operators U: X 
Y is pabsolutely
t, a closed linear subspace
E and V: E Il •p 
Y such that
(0.5)
T
) satisfying (0.5) is called a pabso!utely summing factorization of A triple (U, V. i~ r.P The pdbsolutely summing norm of T is the quantity
where the infimum is extended over all pabsolutely summing factorizations of T. We shall use the term "absolutely summing" instead of "Iabsolutely summing". The following result is a slightly improved version of the socalled GrothendieckPietsch theorem (cf. [Pi], [PP], [MtP], [P8]). 0.4. Let 1 ,;;;; p < 00 and let T E B(X, Y). (i) If T is pintegral and j: X  C(So) is a fIXed isometric isomorphic embedding of
THEOREM
X into C(So)' So arbitrary compact Hausdorff space, then there exists a /1 E [C(So)] t and a linear operator V: LP(J1) ipCT)
y** such that
= 11/111 I / p , IIVII = 1,
KyT= Vill,pj.
BANACH SP<\CES
(ii) If Tis pabsolutely summing and i: X
'>
7
C(So) is an isometric embedding of X
into C(So)' So arbitrary compact Hausdorff space, then there exists a !J. E [C(So)] linear operator V: E IJ.,P
'>
Y where E = i(X) such that
(0.6)
IWII (iii) If there is a constant C n
(0.7)
L
+and a
<
00
=
1, T
= v,/.E IJ..P " .
such that
n
IITxjll P :s;;; CP
j= 1
sup
L
II x * 11<;; 1 i= 1
Ix*(xj)iP for arbitrary
then T is pabsolutely summing with
'IT peT)
Xl' '"
= inf {C:
,xn in X(n = 1,2, ... ),
C satisfies (0.7)}. Conversely, if T is
pabsolutely summing, then T satisfies (0.7) with C = 'lTp(T).
0
Clearly every pintegral operator, say T, is pabsolutely summing and
'IT peT)
:s;;; ip(T).
Using the fact that every closed linear subspace of a Hilbert space is a range of the norm one projection from the whole space, it can be easily seen that every 2absolutely summing oper· ator is (strictly) 2integral. By Theorem O.S(ii), every p·absolutely summing operator from a C(S)space is (strictly) pintegral. Next observe that if Y is a dual Banach space (in particular Y is reflexive) or more generally if Y is complemented in its second dual, then every pintegral operator is strictly pintegral. A discrete analogue of strictly pintegral operators are the pnuclear operators which we define next. DEFINITION
0.3.
Let 1 :s;;; p
<
00.
A bounded linear operator T: X
if there are bounded linear operators U: X
(/\j) with :2:IAjIP
<
00
'>
l~, V:
IP
'>
'>
Y is pnuclear
Y, and a sequence of scalars
such that
T= VApU where Ap: l~
'>
IP is defined by Ap((~i)) = (A;~) for
(U E r.
The triple (U, V, Ap) is
called a pnuclear factorization of T. The pnuclear norm of T is the quantity
where the infimum is extended over all pnuclear factorizations of T.
It is easy to see that every pnuclear operator is compact. For p
=
1 we write neT) instead of n 1 (T) and we shall use the term "nuclear" instead
of "Inuclear". The set of nuclear operators from X into Y is denoted by N(X, Y). It is a Banach space under the norm n( . ). We shall need the following result due to Grothendieck [Gr3] , cf. also [DU] , and
[Per] for the reflexive case. THEOREM
0.5. If Y is a Banach space with a RadonNikodym property, then every
absolutely summing operator from a C(S)space into Y is nuclear and therefore compact.
<\LEKSANDER PEt.CZyNSKI
8
~cIKli
m{zEaD: I(Hf(z))I;;;'K}
aD
Ifldm
forfELI.
(iv) Iff satisfies the Lipschitz condition, then H(!) is continuous; if f is a COO smooth
function so is H(!). (v) If H(f) ELI, then
H(f)(n) =
I
> 0,
i/(n)
for n
/:0)
0, for n < o.
if(n)
for n =
0
The close relatives of the Hilbert transform are: the Riesz projection
R and the Boas
isomorphism B. They are defined via the Hilbert transform by
R(f) 2B(f)(z)
= rl(f+ ;H(!) + =
(I  i)/(O))
(fE LI),
+ zI [f iH(f)](z2) + (I  i)/(O)(I 
[f+ iH(!)](z2)
rl)
(f ELI,
Z
E aD).
Theorem 0.2 yields (for the proof of (jjjj) cf. Boas [80] and [KP].
0.3. (j) Let 1
Then R is a bounded projection from LP onto its subspace HP such that ker R = {f E LP: 1 E ff.Po}. Moreover, if II Rllp denotes the norm of this projection regarded as an operator on LP, then THEOREM
00.
2
2
b I  p  ~ IIRII ~ b  p pI P pI
where b is an absolute constant independent of p. (jj) (the Kolmogorov theorem) If f ELI, then Rf E H" for every p with 0 Hence there is a constant Cp such that
(0
< p < I.
< p < I).
(jjj) The operator f + R(f) is of weak type (II). (jjjj) Let I < p < 00. Then B maps isomorphically LP onto H"; we have 00
B(f)
= /(0) +
L
[/(n)z2n
+ j(n)z2nl]
n=1
(the series on the righthand side converges in LP). Moreover
4 l lIfllp ~ IIB(!)lIp
2
~ c p"
I IIfllp
(fE LP)
where c is an absolute constant independent of p. Finally observe that sometimes it is more convenient to deal with the models of LP and
H"
which consist of 21Tperiodic functions on the real line; in these models the Hilbert
transform is given by the formula
H(f)(t)
=  I
1T
for malmost all t.
lim 1: ..... 0
Sff I:
[get
+ s) 
get  s)] ds 2tg(s/2)
BANACH SPACES
9
Recall that a Bana;;h space Y has the RadonNikodym property if for every finite measure 11 on a sigma field of subsets of a set [2 every V E B(L t (11), Y) admits a repre~entation (0.8)
V(f) =
fn f(w)y(w)ll(dw),
fE Lt{J1),
where y( . ): [2 + Y is a IlBochner integrable Ilessentially bounded function. While the proof of Theorem 0.5 in full generality is slightly complicated, the fact that the operators in question are compact is easy. Indeed, if T: C(S) + Y is absolutely summing then, by Theorem O.4(ii), it admits a factorization
L 1 (11)
i,i \ C(S)
l
Y
T for some 11 E [C(S)]
t
and a V E B(L 1 (11), Y). Since V admits a representation (0.8), it
takes weakly convergent sequences in L t (11) into norm convergence sequences in Y (cL [DSI, Chapter VI, §8, Theorem 14]). On the other hand,
ill,l
is a weakly compact operator.
Hence Vill,t = T is compact. 0 Finally recall that if X is a finite dimensional Banach space and T E B(X, X), then the
trace of T is defined by dimX
tr(T)
=
L
el(Tej )
j=l
where
(ej)t<;;j<;;dim x
is any basis for X and
(ej)t<;;j<;;dim x
the sequence of coefficient func
tionals of this basis. The quantity tr(T) is independent of the choice of a basis. It satisfies the inequality Itr(T)1 ~ neT).
1. The F. and M. Riesz Theorem and Duals of the Disc Algebra This section has an introductory character. We outline a proof of the classical M. and F. RIesz theorem and use it to describe the duals of the disc algebra A and the quotient
qaD)/A. Let us regard the disc algebra A as a subspace of C(aD). The annihilator
Al
=
{p E [qaD)] *: faDfdP
=0
for fE A}
is completely described by the classical THEOREM 1.1 (F. AND M. RIEsz). Let m be Lebesgue measure on the unit circle aD
{z
E
=
C: Izl = I}. Then Al
=
{p E [qaD)]
* :p = h
. m for some h E H~}
where H ~ = (the closure in L 1 = L 1 (m) of all the polynomials
r.'J= 1 cjz j (n
= I, 2, ... )} .
Briefly Al = H~ if we identify (via the Radon·Nikodym theorem) the space L 1 with the subspace {p E [C(S)]
* : p «m}.
SKETCH OF THE PROOF. Step I. A closed set Fe aD with m(F)
A, i.e., there is an fF E A such that fF(z) We put fez)
=1
exp
=
I iff z E F, IfF(z)1
1 r21T z + e it 2),   . log wet) dt rr 0 z _ e,t
=0
is a peak set of
< 1 for z t{: F.
for
Izl
where w( . ) is a real 2rr·periodic continuous function such that 0 ~ w ~ 1, f~1Tlog wet) dt
>
00,
wet) = 0 iff e it E F, and w is infinitely many times differentiable at each point t t{: F. fF(z) = lim f(pz) for z E aD.
with e it
p>l
The
es~ential
technical difficulty is to show that the limit exists and that
f
is uniformly con·
Izl < I (cf. [H, p.80] , [St, p. 205] for details). Step II. If pEAl, then p «m. Pick any closed Fe aD with m(F) = 0 and define
tinuous for
fF as in Step I. Then
o = f f'j.. dp Hence p(F)
= limnff~ dp = O.
Thus p
«
for n
=
I, 2, . "
.
m, being a regular Borel measure which vanishes
on every closed subset of aD of Lebesgue measure zero. 10
11
BANACH SPACES
Step III. Ai = H~. By Step II and the RadonNikodym theorem, if p. EA 1 , then p. = h . m for some h E L I(m). Let f(t) = h(e it ). Then p. E Ai yields A
f(n)
I f2IT . t = 2; Jo f(t)e 1n
dt
=0
for n
= 0,
I, 2, ...
Thus, by the FeJer theorem, f is the limit in L I of polynomials of the form ~:= I cie iit . Equivalently h E H~. Thus H~ J Ai. Obviously zi E Ai for j = 1,2, . . . . Hence H~ C Ai. 0 The description of the dual of A. If Y is a subspace of a Banach space X, then the
+
map which assigns to the coset {x* phism from
X*/y1
onto Y*.
Y 1 } the restriction of x* to Y is an isometric isomor
Thus, by the M. and F. Riesz theorem, A* is isometrically iso
morphic to [C(aD») */H~. By the Lebesgue decomposition theorem
(1.1 )
[C(aD)] *
= L1
$1
Vsing .
(The equality in (1.1) and (1.2) below means "isometric isomorphism" and the symbol X $1 Z means the [I sum of X and Z.) Thus we have
(I.2)
A*
= LI/H~
$1 Vsing .
It follows from the definition of 11 that
(H~)l = {fE L 00 = LOO(m): f fgdm = 0 for g E H~} = 11. Thus a standard duality argument gives (L I /H ~)*
(1.3)
A ** = 11
= H oo •
Now (I .2) yields
$00 V:ing·
(The symbol X $00 Z means the 1 sum of X and Z.) Observe that Vsing is an L 1 (v)space. 00
Hence
V:
i ng is a C(K)space for an extremely disconnected compact Hausdorff space K. Our next aim is to state some consequences of the F. and M. Riesz theorem and (1.2)
in terms of isomorphic invariants of Banach spaces. A subspace £ of a C(S)space is said to have a small annihilator if there is a v E [C(S») such that £1 C L I (v). Obviously if £1 is a norm separable subspace of [C(S») *, then £ has a small anmhilator. Conversely if S is metrizable and £ C C(S) has a small annihilator then £1 is norm separable. We will distinguish two properties of a Banach space X. Property I. X is isomorphic to a space with a small annihilator. Property II. X* is a separable distortion of an L 1 (v), i.e., X* is isomorphic to a product M $ V where M is a separable space and V is an L I (v) space.
Observe the following: 1°. The disc algebra has both Properties I and II. 2°. If X is separable then Property I implies Property II. 3°. If n ;;:. 2 then ndimensional Hilbert space is not isometrically isomorphic to any subspace of a C(S)space with a small annihilator. However, every finite dimensional Banach space (being isomorphic to a C(S)space) has Property I.
+
ALEKSANDER PH.CZyNSKI
12
We end this section by stating a description of a predual and the dual of HI which is also a simple consequence of the F. and M. Riesz theorem and the following general fact. If Y is a subspace of a Banach space X then the dual of X/Y is naturally isometrically isomorphic to yl and (yl)* is naturally isometrically isomorphic to X**/Y**. We put Ao
= zA
and H~ = zH"" . THEOREM
1.2. (i) The map f ~ x; where x;( {g
+ Ao} )
=
f <wfg dm for g
E q3D)
is an isometric isomorphism between HI and the dual (q3D)/A o )*'
(ii) The map {g + H~} ~ IP; for gEL"" where isometric isomorphism from L "" /H~ onto (HI )*.
+ A o}
~ {f
+ H~}
1P;(n = f il Dfg dm for f E HI
for f E q3D) can be identified with the canonical embedding of q3D)/A o into its second dual identified with L "" /H~. 0 (iii) The map {f
is an
2. Absolutely Summing Operators from the Disc Algebra The main result of this section is Theorem 2.3 describing pabsolutely summing operators from the disc algebra. It is preceded by a description of pabsolutely summing operators from a subspace of a C(S)space with a small annihilator (Theorem 2.2). The technique is based upon Bishop's generalized RudinCarleson theorem [BI], [Gil] which with the improvement of [Gm2], [P2], [P3], [MP] (cf. also [Gm 1], [St]) can be stated as follows. THEOREM 2.1. Let X be a subspace of a C(S)space. Let F be a closed subset of S such that
(2.1)
p(F 1 ) = 0
for every closed Fl C F and every p E Xl.
Then for every u E C(F), every fuCs)
(2.2)
> 0 and every open = u(s) for s E F,
E
Ifu(s)1
<E
G :J F there is an fu E X such that
for s $. G and IIJlI
Moreover, if X is separable then the map u linear isometry.
+
= lIuli.
fu from C(F) into X can be chosen to be a
Let us observe that if X has a small annihilator in C(S), say Xl eLl (A) for some X E [C(S)]
t, then
a closed F C S satisfies (2.1) whenever X(F)
= O.
Now we are ready to prove THEOREM 2.2 (MITJAGINPELCZyNSKI). Let X be a subspace of a C(S)space with Xl eLl (X) for some probability measure X E [C(S)] *.
Let 1 .;;;; p space
<
00
and let T: X
+
E be a pabsolutely summing operator into a Banach
E.
Then there exists a strictly pintegral operator V: X such that
II(T  V)(nll .;;;; (fslf(S)IPh(S) dA) IIp
(2.3)
+
E and a nonnegative h ELI (A)
for f EX,
(2.4) PROOF. Smce Tis pabsolutely summing, there exists apE [C(S)]
1rp(T)
=
IIpllllP
and IITfll .;;;;
(fslf(s)I P
unique bounded linear operator B: X IJ.
t
such that
dp)l/p for f EX (cf. §O.III). Hence there is a +
E such that B(ilJ.,p(n)
13
= T(f)
for f E X.
For a
ALEKSANDER PEl.CZyNSKI
14
r E IC(S)]
+we denote by XT,p the closure of X (more precisely, of iT,p(X» = a + v with a «A, v 1 A and
Lebesgue decomposition theorem yields p
Ilpll
in LP(r). The
=
lIall
+ Ilvll.
We shall identify e(p) with the [Pdirect sum e(a) EBp LP(v). The crucial point of the proof is to establish
(2.5)
X/1,P
= Xa,p
EBp LP(v).
Assume that (2.5) has been proved. We then define V: X > E by BivQvi/1,P where Q v : e(a) EBp e(v) > e(v) is the natural projection,i v : e(v) > e(a) EBp LP(v) the natural embedding (i.e., iv(b)
= (0, b)
for bE e(v». By (2.5), ivQvi/1,p(X) C X/1,P; thus V is well
defined. Clearly Qvi/1,P = iv,p' and IIBiv ll .:;;; IIB1i1liv11 .:;;; 1. Thus V is strictly pintegral with ip(V)':;;; Ilvll l / P . Next for eachfEX
II(T  VXf)1I
= IIBi/1,p(f) 
BivQ)/1,p(f)1I
.:;;; lIi/1,pCf)  i vQv i /1,p(f)II.
e
If h E LP(p) = e(a) EBp (v) then h = i aQa(h) + ivQv(h) where Q a : L I (J1) > L I (a) and i a: L I (a) > L 1 (p) are the natural projection and the natural embedding respectively. Thus for h
= i/1,p(f), we
have (f)  ivQvi/1,p(f)
II(T V)(f) II .:;;; lIia,pCf)1I Hence we get (2.3) with h
= da/dA.
= i aQ ai/1,p(f) = ia,p(f).
= (fs lflP do rIP
It follows that
for IE X
Finally
(1Tp(T»)P  (ipCV»P
~
Ilpll  Ilvll
=
lIall
=
Ish dA
which proves (2.4).
(2.5). The inclusion X/1,P C Xa,p EBp LP(v) is trivial. For proving the reverse one, pick a z in Xa,p EBp LP(v), i.e., pick an a E Xa,p and bE LP(v) so that z = a + b PROOF OF
and
(2.6) Next fix finite M
> °and pick an x E X so that f sla  xl P do < EP. Since b ELP(v), there is a > 1 such that if Z = {s E S: Ib(s)IP .:;;; MP} , then
E
f
(2.7)
JS\Z
IblP dv
< EP
and
v(S\Z)' MP
< EP.
Combining (1.6), (1.7) with the Lusin theorem and the fact that v, a, A are regular Borel measures with v 1 A, and a A and using the fact that the measure B > f B Ib(s)I P dv is
«
absolutely continuous with respect to v, we construct a compact FeZ and an open G :J F such that
b restricted to F is continuous, A(F)
= 0,
(2.8) max{v(S\F), a(G), fZ\FlbiP dV}
< EP/MP.
Now we apply Theorem 2.1 for u E C(F) defined by u(s)
= b(s) 
xes) for s E F.
15
BANACH SPACES
Let fu E X satisfy
(2.2). Then IIfull
« sUPSEFlb(s)1 + IIxll « M + Ilxll because Fez.
Hence, we get
« (fsla 
(fslZ  (x + f)IP dp yIP
x  fu lP da yIP + (fs Ib  x  fui P dv YIP.
Estimating each of the terms separately we get
(fsla  x  fui P da yIP
« (fsla 
xl P da )I IP + (fslfu lP da rIP
«€+ (fS\GlfulPdaYIP + (fGlfuIPda)"P
« € + a(S)I/P€ + (a(G)llfull)l lp « €(1 + a(S)l/p + 1 + IIxli/M); = (fS\F Ib x  f.IP dV)I IP 5SIb x  f.IPdv u u « (fS\F IbI P)l IP + (JS\F Ix + f.u IP dV)I/P
«
(J
Z\F
Ibl P dv)l lP
+
(5
S\Z
Ibl P dv)I/P
+
[V(S\F)]'/Plix
+ f.u II
« (c + € + €(21Ixll + M))M I « €(2 + 211xll + M)MI. Letting
€
tend to zero we infer that
Z
E X p."
0
COROLLARY 2.1. Let X satisfy the assumption of Theorem 2.1. If E has the Radon
Nikodym property and p = 1 then the operator V satisfying (2.3) and (2.4) is nuclear with n(V) = i l (V) and therefore compact. PROOF. Use the fact that every strictly integral operator to a space with the RadonNikodym property is nuclear (cf. §O.III). REMARK. or E
= HI,
The assumption of Corollary 2.1 is satisfied if E is either a Hilbert space
because HI being a separable dual (by Theorem 1.2) has the RadonNikodym
property. For the disc algebra Theorem
2.2 can be strengthened as follows:
THEOREM 2.3. Let E be a Banach space, I
«p <
and let TA + E be a pabsolutely summing operator. Then there is a strictly pintegral operator V: A + E and an outer function
(2.9)
(2.1 0)
I
< 1,
16
ALEKSANDER paCZYNSKI
(2.11) Let V and hELl be defined to satisfy (2.3) and (2.4) (for S Let
PROOF.
X
= A).
.p(z)
1
= exp
2
r27Teit+z . log [h(e 't ) 0 e'tz
J
11
I
•
+
1] dt
for Izl
= aD,
l\
= m,
< 1.
It is well known [Ou, Chapter 2] that
l.p(eit)1
=
lim l.p(pe it) 1 = h(e it )
+
1
almost everywhere
p+I
which implies (2.9) and (2.11).
< 1 we
Since h ;;;, 0, for z = re u with r 1 S27T I.p(z) 1 = exp 2 11
= exp 
0
Re«te,t + zz) .10g(I + h(e
1 J27T
211
0
have
1  r2 1
+ r2
 2rcos(t  argz)
it
» dt
10g(I
.
+ h(e 't » dt
;;;, 1, which proves (2.1 0). THEOREM 2.4 (MITJAGINPEt.CZyNSKI). Let 1 < p < 00. Then every pabsolutely summing operator from A is strictly pintegral. More precisely, there is a constant Cp with p2
(2.12)
Cp .;;;
p=I" b
(where b is an absolute positive constant independent ofp) such that ip(T)';;; Cp11p(T) for every pabsolutely summing operator from A. PROOF.
Let
restriction of the natural injection i lIP 1m ,p
:
C(aD) ~ LP (I
from A into HfIP'm where HfIPlm is the closure of iIIPlm,p(A) in LP(I
(2.13)
A ip(iIIPlm,p) .;;; IIRllp
(5avl
Here IIRllp denotes the norm of the orthogonal (Riesz) projection from LP onto HP regarded as an operator on LP. Let us consider the diagram C(aD)
(2.14)
rj A
ilIPlm,p
)
LP(I
where j denotes the natural inclusion,
MtJ;
~
uL
IfP
1
MtJ;l ) HP IIPI'm
VI is the outer function defined by
BANACH SPACES
I/I(z) = exp ( 
1
p
1
(21T Z
2 J, 1r
0
17
+ e.it log Re .,o(t) dt)
Z _
for Izl
e,t
and M!J; and M!J;I the operators of multiplications by 1/1 and 1/1
M!J;(f) = 1/1  I
and
I
< 1, respectively, i.e.,
M r 1 (f) = 1/I 1f.
It is known that II/II P = I'PI and 1/1 E HP (because 'P E HI). Hence M!J; is an isometric isomorphism from LP(I'Plm) onto LP. We have to check that
(2.15)
M!J; carries Hf
This will show that the diagram (2.14) is commutative and therefore i~lm,p is strictly pintegral. To prove (2.15) observe that obviously M!J; (Hf
(2.16)
if g E HP, then g/1/l E Hf
= I/I(rz) for Izl ~ 1 and 0 < r < 1. Clearly I/I r E A and, by (2.10), 1I/I,(z)1 1lp l.,o(rz)1 ;;;. 1 for Izl ~ 1. Observe that if lEA then IN EHf
=
lim
r= 1
J ~ 1I

'Y r
'Y
Ipl'Pl dm
~ IIfll 00
lim} II/I r  I/IIP dm 11/1 riP
r= 1
~ IIflloo r=lim1 SII/I r 
I/IIP dm = 0
(we use here that 1/1 E HP (cf. [Dll, Chapter 2] )). Next given g E HP, there is a sequence
(In) in A with 0 = limnfaD Ig  In IP dm = limnfaD IgN  InN IP I'PI dm. Since InN E Hf
i~lm,p = M!J;IRM!J;i l
= IIRllp(fI'P1 dm)l Ip _ This
= 11M!J;111 = lUll = 1, we have completes the proof of (2.13).
Now, let E be an arbitrary Banach space, 1 lutely summing operator with 1rp(T)
= 1.
00,
ip(i~lm,p) ~
and let T: A ~ E be a pabso
By Theorem 2.3, there is a strictly pintegral
operator V: A ~ E and an outer 'P E HI satisfying the conditions (2.9)(2.11). It follows from (2.9) that there is a unique bounded linear contraction B: Hf
= (T
VXI) for lEA, equivalently B· i~lm,p ip(T  V)
Since 1rpCT)
=
~ IIBllip(il!,lp) "
= T
V. Thus, by (2.13)
IIRllp(f I'PI dm riP.
1, (2.11) yields (f1'Pldm)l /p " 21/p
< 2.
Hence, again using (2.11), ip(T)
+ ip(V) ~ 211Rllp + 2 = Cpo This proves the last assertion of Theorem 2.4. Finally (2.12) follows from the estima
"ip(T  V)
tion for IIRllp (cf. §O.lI). REMARK.
0
Let us observe that the estimation for Cp given in (2.12) is the best possi
ble in the following sense infip(T) ~ ap2/(p  1)
for 1 < p
<
00
where a is an absolute constant independent of p and inf is extended over all operators
18
ALEKSANDER PEI.CZyNSKI
trom A with 1fpCT)
1f p(i!,p)
=I
=
1. In fact for the natural embedding i!,p: A
+
HP we have
and
(2.17)
i/i!.p)
= IIRll p'
(2.17). Since i!,p = R im,pj we have IIRllp ~ ip(i!,p) (here j: A + qaD) denotes the natural inclusion). To prove the reverse inequality we use the standard averaging PROOF OF
technique (cf. [PlO] for a more general treatment). Pick fJ. E [qaD)]
+with 1IfJ.1I = I so
that i!,p = BilJ.,pj for some operator B: LP(fJ.) + HP with IIBII = ip(i!,p)' (The existence of B follows from the definition of pintegral operators (cf. §O.III) and the fact that HP is reflexive for 1
00
and therefore every integral operator to HP is strictly integral.) Let
us set
(Saf)(z) =/(exz)
B(n =
for IELI and
f SaBSaI I m(dex)
iexl = Izl = 1,
for IE qaD).
Then, we have, for every IE qaD),
IIBfll~ ~ fIlSaBSaIfll~ m(dex) = fIlBSaIfll~ m(dex)
~ IIBIIJ (JISa1/IP dP.) m(dex) = IIBllfl/lP dm. (The last equality follows from the observation that because of the uniqueness of the Haar measure JSaIh dp.m(dex) = Jh dm for every probability measure fJ. E qaD)* and every nonnegative h ELI.) Thus extends to a bounded linear operator from LP into ffP which co
B
incides with R because BI = I for lEA.
TIlUS
IIRllp ~ IIBII
= ip(i!,p)'
0
3. Absolutely Summing Operators from the Disc Algebra into Hilbert Space The operator f  * (/(2 n )) is an example of an absolutely summing map from A onto [2. The existence of this phenomenon leads to a new proof of Grothendieck's theorem that
every bounded operator from [I into [2 is absolutely summing. It also emphasizes the difference between the disc algebra and C(S)spaces, because every absolutely summing operator from a C(S)space into [2 is nuclear, hence compact. The main result of this section Theorem 3.2gives a description of absolutely summing operators with a closed range from A into 12. At the end of this section we discuss the problem whether every bounded operator from A into [2 is 2absolutely summing. The following concept plays an important role throughout this section. DEFINITION 3.1. is a K
> 0 such
An orthogonal projection P: H2  * H2 is a Paley projection if there
that
IIPfl12 ~ Kllflll
(3.1)
for f E H2.
EXAMPLE 3.1. Let (n k ) be a sequence of nonnegative integers such that lim n k + 1 /n k
> 1.
k
Then the orthogonal projection Pf = 'Lk/(n k )zn k, f E H2 (where /(n)
= JaDf(z )zn m(dz) I, ... ) is a Paley projection. This fact is simply a restatement of a classical result of Paley (cf. [Z], [Du, Chapter 6], [Pal]). If P: H2  * H2 is an operator then PA denotes the restriction of P to A regarded as an operator from the disc algebra A into H2, formally PA = Pi~. 2' Our next result shows that every Paley projection induces an absolutely summing surjection from A onto a Hilbert for n
= 0,
space. Precisely we have PROPOSITION 3.1. If P is a Paley projection, then PAis an absolutely summing surjection from A onto P(H 2 ). PROOF. Clearly (3.1) implies (in fact is equivalent to) (3.2)
liPA (f)1I
Hence PAis absolutely summing and
~K
S
aD
IfI dm
1T 1 (P A) ~
for
f
E A.
K.
To prove that PAis onto P(H2) we show that PA: [PA (H2)]
*  * A * is
an isomorphic
embedding. Clearly we may identify the Hilbert space [PA (H 2 )] * with P(H2) and the adjoint PA with the operator which assigns to each f E P(H 2 ) the functional x; defined by 19
A.LEKSA.NDER PELCZYNSKI
20
xf
for g EA.
Here we used the fact that the projection P is orthogonal. A*
By (1.2), the functional
x; E
= L I /H~ x Vsing can be identified with the coset {f + H~} or, more precisely, with the + H ~}, 0). Therefore to prove that PA is an isomorphic embedding it suffices to
pair ( {f
show that there is K I
I~A (nil
(3.3)
=
> 0 such that inf
S
o aD
hEHI
If(z)
+ h(z)1 m(dz) ~ K IIIfll2
for f E p(H2).
By the Kolmogorov theorem the "orthogonal projection" annihilating H~ is a bounded linear operator from L I into L 1/2 (cf. §O.II). Hence there is a c
> 0 such
that
(3.4) Next, combining (3.1) with the Schwarz inequality applied to the product Ifl l / 4 IfI 3 / 4 = If I, we get
Hence
(3.5) Combining (3.4) with (3.5) we get (3.3) with K 1 = cK 4 .
0
COROLLARY 3.1. There exists a map from the disc algebra onto 12 which is absolutely summing. PROOF. Let (n k ) be a sequence of positive integers with limkn k + link> 1. Then the operator T: A  ) 0 12 defined by Tf = (/(n k » for f E A has the desired properties. This follows immediately upon combining Example 3.1 with Proposition 3.1. Corollary 3.1 has a surprising application. THEOREM 3.1 (GROTHENDIECK [Grl]). Every bounded linear operator from [I into /2 is absolutely summing. PROOF. Let S: [I
)0
[2
be any bounded linear operator and let T: A
)0
[2
be an
absolutely summing surjection. By the open mapping principle, there is a constant K 1 such that given x E [2 there is an f E A with IIfll ~ K Illxll and Tf
= x.
>0
Hence if ej is the
jth unit vector of [I ' then there is an /j E A such that T/j = Se. and 11[,·11 ~ K IIiSIi. Now 1 "" I J,....., ,. . . " we define the lifting S: [  ) 0 A by S((tj» = r.tj/j for (tj ) E [I. Then S = TS and liS II ~
KIIiSIi. Hence
1T 1 (S)
~
K 11T I (T)IISII.
0
BANACH SPACES
21
Our next result shows that up to a Banach space automorphism of a Hilbert space and of the disc algebra every absolutely summing surjection from A into a Hilbert space is a nuclear perturbation of a Paley projection. Instead of discussing "surjection onto Hilbert spaces" it will be more convenient for us to work with operators into 12 with a closed range.
*"
0 for Izl < 1 then the operator M.p defined by M/t) = If IP is a function in A with
12 be an absolutely summing operator with a closed range. Then there is a nuclear operator V: A  12 such that for every E > 0 there exist a multiplication automorphism M.p: A  A, a Paley projection P: H2  H2 and an invertible operator U: P(H 2 )  T(A) such that 1T 1(T(UP A M.p + V) < E. PROOF. By Corollary 2.1, the remark after Corollary 2.1 and Theorem 2.2, there exists a nuclear operator V: A 
(2.11) for p
=
1. Let Tl
[2 and an outer function
= T  V.
IP satisfying the conditions (2.9)
Then the inequality (2.9) can be rewritten
II Tl fll ,;;;; flfl IIPI dm
for f E A.
Hence there exists a unique linear operator Q: H~.plm 
Qi~lm,l. Next fix r with 0 (3.6) where IPrCz)
Qy(h)
= IP(rz)
for
=
12 with IIQII ,,;;; 1 such that Tl 1 and define Qy: H11.pylm + 12 by
Q(h . IPy/IP)
for h E Hr.pylm
Izl ,,;;; 1. To show that Qy is "well defined" we have to check that
if h E Hf.py,m then hIPy/IP E H ~.plm. To this end fix 8
~ > flh
 fllIPyl dm = flhIP y  fIPyl dm
In view of [Du, Chapter 2] there exists an R
< 1 so
>0
=
f
and pick f E A so that
Ih~  f~IIIPldm.
close to 1 that
Then, by (2.10),
I
I
IPy  f IPR IPy IIPI dm ,,;;; IlfIPyll~ f lIP  IPR I dm f f;
8 < 2"
Hence
Since fIPy/IPR E A, the last inequality yields that hIPy/IP E H~.plm. Hence Qy is well defined. Now put Ty
= Qyif!,ylm,l:
A _/2. Then, by (3.6), for fEA,
II(T.  Ty)(t) II = IIQ(f 
f·~ ) II ,;;;; f
,,;;; flfllIP  IPrl dm.
\t  ~ f
IIIPI dm
ALEKSANDER PEI:.CZyNSKI
22
Thus in view of [Du, Chapter 2] , lim
(3.7)
r= 1
1T 1 (T 1 
It follows from (2.1 0) that for 0
Tr)
<, < I
=
lim flIP  IPrl dm
r= 1
= O.
the operator M'Pr is a multiplication automorphism.
Hence, by the definition of Tr ' we have
IITrM
0~)
1
~
(f)11 = IIQ(M
1 (f)
~
. IPrllPll
= IIQ(fIP 1 )11 ..;; Jill dm
..;; (J1/12 dm)t/2.
Hence there exists a linear contraction, say Sr: H2  [2, such that Sr(f) = TrM 1 (f) for 'Pr lEA. Since A is dense in H2 in the norm II . Ib, Sr is uniquely determined and closure Sr(H2) = closure (TrM l(A». Let P = [l..r>: H2  H2 be the orthogonal projection with 'Pr ker P = ker Sr' Then there exists a onetoone bounded linear operator Ur = U: p(H2) = [2 such that Sr = UP. Now assume that U has a bounded inverse, i.e., there exists a bounded linear operator V: U(p(H2»  p(H2) such that VU is the identity on P(H 2). Then, by (3.8) for I E A, 11P(f)11";; 1IV11I1UP(f)1I = IIVllllTrM 1(f)1I 'Pr ..;; II VII fill dm. Thus 1IP(f)1I ..;; II VIIJl/I dm for IE H2 (because for every IE H2 there is a sequence (fn) in A such that limnilln  1112 = limnil/n  1111 = 0). Hence, if U has a bounded inverse, then P is a Paley projection and
,
Tr and given e
= TrM'Pr_lM.~Tr = UPAM'P r
> 0, 1T 1 (T
(V
+ UPAM'Pr» = 1T 1 (T 1

Tr)
<e
for, sufficiently close to 1 (by (3.7». We complete the proof by showing that there is an '0 with 0 < '0 < 1 such that if '0 < , < 1, then U = Ur does have a bounded inverse. To this end observe that Tl = T  V has a closed range because T does and V being nuclear is compact. Hence there is an e > 0 such that if T: A  [2 is an operator with IITl  Til < €, then T has a closed range. Thus, by (3.7), there is an '0 with 0 < '0 < I such that if '0 <, < 1, then Tr has a closed range. Therefore, TrM 1 also has a closed range and so does Sr' This implies that U has a closed 'Pr range because Sr = UP and P is a projection onto the domain of U. Thus U being onetoone has a bounded inverse. 0 The results of this section show that absolutely summing operators from the disc algebra into a Hilbert space are in general quite different in nature than absolutely summing operators from a C(S)space into a Hilbert space. The situation seems to be different in the case of 2absolutely summing operators. Prob[em 3.1. Is it true that (i) every bounded linear operator from A into [2 is 2absolutely summing?
BANACH SPACES
23
Let us observe that (i) is equivalent to each of the following properties:
(ii) every operator from
{2
into
{2
which factors through A is HilbertSchmidt;
(iii) every bounded linear operator from A into
{2
extends to a bounded linear opera
tor from C(aD) into /2; (iv) every operator from L I IH~ into (v) every operator from
{2
into
{2
{2
is 2absolutely summing;
which factors through L I IH~ is HilbertSchmidt;
(vi) for every linear operator T: L I IH~ +
= T where q:
{2
there exists an operator
T:
LI
+ {2
L I IH~ denotes the quotient map. The equivalences (i) <== (ii) and (iv) <== (v) follow immediately from the definition of
such that Tq
LI
+
a 2absolutely summing map. The equivalence of (i) and (iii) follows from the extension property of 2absolutely summing operators and the fact that every bounded linear operator from a C(S) space into
{2
is 2absolutely summing (cL [Grl] , [LPI]). The equivalences
(ii) <== (v) and (iii) <== (vi) follow by a standard duality argument (we apply formula (1.2) and use the fact that every bounded linear operator from L I (resp. C(aD)) into
{2
is 2·sum·
mingo
Notes and remarks to § §2 and 3. The results of §2 are due to Mitjagin and Pelczynski. Theorem 2.2 generalizes Proposition I of [MtP] where the case p = 1 is considered. The proofs of Theorems 23 and 2.4 have not been published previously. The following is open (the answer is "yes" if H~ has the bounded approximation property):
Problem 3.2. Is every p·absolutely summing operator from H~ into an arbitrary Banach space pintegral? The results of §3 are due to Peiczynski and Wojtaszczyk. The fact that if
~(nk+ 1 Ink) > I then the operator f + (/(n k )) maps the disc algebra onto (cL, e.g., [Fourl], [Four2] and [Vinl]).
{2
is not new
4. The Nonexistence of Local Unconditional Structure for the Disc Algebra and for its Duals In this section we shall show that the disc algebra A as a Banach space is in a certain sense very different from a C(S)space. A is not isomorphic to a quotient of any C(S)space, nor is it complemented in any Banach space with a local unconditional structure. Most of the argument which leads to the above results is based upon the observation that the natural embedding
i!
I: A ~ HI does not factor through any L Ispace. To illustrate our approach
we begin with the following simple PROPOSITION 4.1. The disc algebra and H oo are not isomorphic to quotients of C(S)
spaces. PROOF. Let us consider the following property of a Banach space X
(4.1)
there is a noncom pact absolutely summing operator from X into a separable duai Banach space. Note that 10. If X satisfies (4.1), then so does Y whenever there is an operator from
Y onto X. (Hint. Use the Banach open mapping principle.) 2°. A and H"" do have (4.1); the desired operator is i~,I: A ~ HI, resp., itf,':'t: H"" ~ HI (HI is a separable dual by Theorem 1.2). 3°. Every C(S)space fails to have (4.1). Every absolutely summing operator from a C(S)space to a separable dual is nuclear and therefore compact (cL §O.III).
0
COROLLARY 4.1. Neither the disc algebra nor H oo is isomorphic to a complemented
subspace of a C(S)space (Pe[czynski [P2] , Rosenthal [Rl]); in particular A is uncomplemented in C(aD) and H"" is uncomplemented in L"" (Rudin [Rul]). We shall strengthen Proposition 4.1 in various directions. To state our results we recall some concepts and facts. An operator T: X ~ Y is LIfactorable if there is an LI(v)space and operators U: X ~ L I (v) and V: L I (v) ~ Y with VU = T.
A Banach space X is said to have CL l.u.st
(= GordonLewis local unconditional structure) if there is a K
> 0 and
a function E ~
(UE • VE• FE) which assigns to every finite dimensional subspace E of X a finite dimensional space FE with a basis (I;) satisfying the condition II~t;1; II = II~ It; 11;11 for all scalars t I' t 2 , .. tn' and operators UE : E ~ FE' VE : FE ~ X such that VEUE(e) = e for e E E and IIVE"IIUEII < K. X is said to have l.u.st if moreover the above UE and VE are isomorphisms from E onto FE and from FE onto E respectively [DPR]. 24
BANACH SPACES
25
REMARKS. R.1. If Tis L Ifactorable so is T**. The converse is true assuming that the range of T is isomorphic to a complemented subspace of a dual Banach space [G·! ] .
[FJT]. R.Il. X has GL l.u.st iff X** is isomorphic to a complemented subspace of a (complex) Banach lattice. Hence if X has GL l.u.st so do X* and complemented subspaces of X
[FJ·T]. R.1l1. If a Banach lattice Z does not contain an isomorph of
Co
then Z is comple
mented in Z** (cf. [Sch, Theorem 2.10 and Proposition 5.15]). The following important result due to Gordon and Lewis [GL] links Banach lattices and spaces with l.u.st with absolutely summing operators. THEOREM 4.1. Let X have GL I.u.st.
Then every absolutely summing operator from
X into a dual Banach space is L Ifactorable. COROLLARY 4.2. If X is a domain of a nonL Ifactorable absolutely summing opera
tor whose range is a dual Banach space then X and all the duals of X do not have GL I.u.st. PROOF. The second adjoint of an absolutely summing operator is absolutely summing. Combine this fact with Theorem 4.1, R.l and R.Il.
0
Now we are ready to state the main result of this section. THEOREM 4.2. (i) Every L Ifactorable operator from A into HI is compact.
(ii) A and all the duals of A do not have GL l.u.st. (iii) If either A or H~ is a quotient of a Banach space Y with GL I.u.st, then Y contains a complemented isomorph of 11. (iv) If L 1 /H~ is a subspace of a Banach space Z with GL I.u.st, then Z contains I~
uniformly, i.e., given any positive integer nand
E
> 0 there is isomorphic embedding
Un:
I~ ~ Z with IIUnllllU;;111 < I + E. (v) If L 1 /H~ is isomorphic to a subspace of a Banach lattice Z then Z contains an
isomorph of co' PROOF. (i) Let T
= VU:
A ~ HI with V: L 1 (v) ~ HI and U: A ~ L 1 (v). Since
HI is a separable dual, V takes weakly compact sets in Ll(V) into a compact set in HI (cf. [DSI, Chapter VI, 8]). To complete the proof it is enough to show that U is weakly compact or equivalently that U*: [L 1 (v)]
* ~ L 1 /H~
x Vsing has the same property. Recall
that every operator from a C(S)space into a Banach space which does not contain isomorphs of Co is weakly compact [PI]. Thus every operator from [L 1 (v)]
*
into L 1 /Ho is weakly
compact. L 1 /H~ does not contain isomorphs of co' because L 1 /H~ is separable and complemented in a dual Banach space A *, isomorphs of
Co
are complemented in every separable
Banach space in which they are embedded (Sobczyk [S], cf. also [Ve]), and isomorphs of Co
are never complemented in dual Banach spaces [BP]. Since Vsing is an L 1 (p)·space ev
ery operator from a C(S)space [L 1 (v)] ator from [L 1 (v)]
*
* into
Vsing is weakly compact. Hence every oper
into the product L 1 /H~ x Vsing is weakly compact.
(ii) The natural embedding i~,I: A ~ HI is absolutely summing but not compact.
ALEKSANDER PELCZYNSKI
26
A I is not L 1 factorable. Now we use Corollary 4.2. Hence by (i), i m, (iii) Let T = iHe<>1 q: Y  4 HI where q: Y  4 He<> is the quotient map. Assume that m,
Tis L Ifactorable. (By Theorem 4.1, this is the case where Y has GL I.u.st.) Consider the diagram
By the open mapping principle, there is a C > 0 and Y n in Y with llYn II < c and Ty n = zn forn = 1,2, . . . . Clearly Ilz n zmIlHI): 1 forn,*m. Hencethesequence(Uyn)CLI(v) does not contain weak Cauchy subsequences, because V: L I (v)  4 HI takes weak Cauchy sequences into norm Cauchy sequences [DSI, Chapter VI.8]. (HI is a separable dual.) Therefore, by a result of [KPJ and [PRJ, there is an infinite sequence (UYnk) which is equivalent to the unit vectors of /1 such that there is a projection P from L1 (v) onto the closed linear span of the UYnk's. Hence PU maps Y onto /1, which yields the desired conclusion for HOC because [I has the lifting property. The proof for A is the same. (iv) If L 1 IH~ C Z and Z has GL I.u.st, then He<>
= (L I IH~)*
is a quotient of Z*.
By R.Il, Z* has GL l.u.st. Hence, by (iii), Z* contains a complemented isomorph of [I which implies that Z contains
1';;
uniformly [GJ].
(v) Assume to the contrary that there is an isomorphism j: L I IH~
4
Z where Z is
a Banach lattice which does not contain isomorphs of co' By R.Ill, there is a projection
P: Z**
~
onto
Z_ Let us consider the diagram
CIAo
~ LIIH~ ~Z
~
(4.2)
.Ip C(S)~
Z**
where i*: CIAo
4
L I IH~ is defined by i*( {[ + Ao})
= {[ + H~}.
Clearly (i*)*
= ifh~.
Thus the assumption that Z has GL l.u.st combined with R.Il and Theorem 4.1 yields that
(ji*)*: Z*  4 HI factors through an L I (v) space. Let (ji*)* = VU be the factorization. Then (ji*)** = U*V*: HOC  4 Z** factors through a C(S}space (L I (v))*. Let V: denote the restriction of V*: L IH';  4 C(S) into a subspace CIAo identified with its canonical image in C" IH'; (cL Theorem 1.2(iii)). It is easy to check that the diagram (4.2) commutes, i.e.,ji* = PU*V:_ Since Z does not contain co' PU* is weakly compact [PI] and therefore PU* takes weak Cauchy sequences in C(S) into norm Cauchy sequences in Z [DSI, Chapter VI] _ Since the space CIAo has a separable dual (namely, HI), every bounded sequence in CIAo contains a weak Cauchy sUbsequence. This implies that PU*V: is compact. On the other hand, ji* is not compact because j is an isomorphic embedding and i* is not compact (we have Ili*( {zn + Ao}  {zm + Ao} )II L I /H~ ~ I for n m). Thus PU*V: ji*, a 00
'*
contradiction.
0
'*
BANACH SPACES
27
The argument of Theorem 4.2(i) shows, in fact, the following: Let X be a Banach space whose dual is a separable distortion of L 1 (v) (cf § I). Then REMARK.
(a) Every operator from X into an L 1 (v) is weakly compact. (b) Every L Ifactorable operator from X into a separable dual Banach space is compact.
5. Application to Unifonn Algebras In this section we extend results of §4 and some results of §3 to uniform algebras with nontrivial Gleason parts, and to uniform algebras with a separable annihilator as well. We close by discussing GJicksberg's conjecture on uniform algebras on S complemented in C(S) and related problems. Recall that a uniform algebra on a (compact Hausdorff) space S is a closed subalgebra which separates the points of S and contains the constant functions. We begin with the case of a uniform algebra whose annihilator is norm separable. Our first proposition combines a result of [P4] with a recent observation due to Wojtaszczyk
[WI] . For p E Xl = {v E [C(S)] *: Ix dv = 0 for x E X} we define Til: X + L 1 (Ipi) by Tix) = xg for x E X, where g = dp/dlpl is the RadonNikodym derivative of p with respect to Ipl.
5.1. Let X be a uniform algebra on S with a norm separable annihilator Then the following conditions are equivalent:
PROPOSITION
Xl
c
[C(S)]
*.
(i) X = C(S). (ii) If p E Xl, then fJ. is purely atomic. (iii) If p E Xl, then Til is compact.
(i) ~ (iii) obvious. (iii) ~ (ii) (Wojtaszczyk). Assume that there is a nonpurelyatomic p E Xl, so that PROOF.
Til is compact. Then TZ: L DO (1fJ.i) + X* is also compact. Since fJ. is not purely atomic, there is a sequence (rn) in L DO(IfJ.i) which has the same distribution as the sequence of Rademacher functions in L 1 [0, I]. Observe that (5.1)
fr
n
w*
0
as n +
00
for every fE LDO(IfJ.i)
(here "~" denotes convergence in the a(L" (luI), L 1 (Lui)topology). Since T! is weakstar continuous and compact, (5.1) yields
(5.2)
limIlT~(frn)lIp n
=0
for every fELDO(IfJ.I).
Using (5.2) we define inductively an increasing sequence of the indices n(I)
so that (5.3)
IIT.~(w,· " " )lI • ,1'2"·"$ x
< 1/3 k
28
for j.~
=k
(k
= 1,2, ...
)
< n(2) < ...
29
BANACH SPACES
where
s
Wilh.···.is
kQ rn(lk)
for any finite increasing sequence of positive integers
Uk)l<;;k<;;S'
Since X* can be identified
with [C(S)] */Xl. it follows from (5.3) that for every wil.i2 ..... i s there is a Vil.h ..... i s E Xl such that Ilw.
(5.4)
. glpl  v·
. IllC(S)I' 'I' .. ·.'s
'I····"s
(because T~( wi 1 '
. . . , is)
< 3 i s,
= {w'l' ... 'i sglp I + Xl}). Let F del ~te the closed linear subspace
of L 1 (Ipl) spanned by all the "likeWalsh" functions wiJ h .... ,is. It is easy to see that E is isometrically isomorphic to L 1 [0, I], by the isometry which assigns to the Walsh functions (in the Paley order) the functions (w'lJ .....n k)' Since g = dp/dlpl is a unimodular function, the sequence (gw n I .... ,n k) is also equivalent in L 1 (Ipl) to the WalshPaley sequence in L 1 [0, I] and the subspace gE is isometric to L 1 [0, I]. Now (5.4) yields that the perturbed sequence
(vi 1.i2 , .... is) is also equivalent to the WalshPaley sequence in L 1 [0, I], i.e., the map wi! ,.... i s + Vil ..... i s extends to an isomorphic embedding of E into Xl (because the subspaces Ek = span {wil ,i2 ,.... is with js = k} form a Schauder decomposition of 1:' and dim Ek = 2 k  I ). Hence Xl contains a subspace isomorphic to L 1 [0, I] and therefore, by a result of [PB] , is not isomorphic to a separable dual. On the other hand, Xl is isometrically isomorphic to the dual of C(S)/ X. Thus Xl is not separable, a contradiction. (ii) => (i) [P4]. Case I: S metrizable. If p E xl, then, by (ii), p =
L
as(p)1]s
sES
where 1]s denotes the unit mass at the point sand asCp) are the complex numbers with Iipli = ~sEslaip)l.
Let M(X) denote the Choquet boundary of X. Then, for every s E M(X), there is an
Xs E X such that xis) = I and Ixs(t) I < 1 for t
°= f lim n
Therefore if p E
xl,
S
x: dp
then Ipl(M(X))
= lim n
= 0.
"* s (cf. [Ph I]).
L
tES
aip)xn(t)
Hence
= aip)·
(Here we use the fact that if S is metrizable then
M(X) is a Go and therefore measurable.) To complete the proof in Case I, it is enough to show that M(X) = S. This combined with the previous remark will imply Xl = {o}, i.e., X = C(S). Suppose to the contrary that there is an s in S\M(X). By the ChoquetBishopde Leeuw representation theorem [PhI] there is a v E [C(S)] * concentrated on M(X) such that 1]/x)
= xes) =
f x dv = ( S
JM(X)
x dv
for x EX
.
1]s E Xl, and, by the preceding remark, v(M(X)) = p(M(X)) + 1]/M(X)) = f M(X) I dv = I, a contradiction. Case II: S arbitrary compact Hausdorff space. Then C(S)/X is separable because
Hence, p
=v
On the other hand v(M(X))
= 0.
30
ALEKSANDER PELCZYNSKI
(C(S)/X)*
Xl is separable. Hence a standard construction yields the existence of a compact metric space SI and a continuous map h: S~ SI such that if he: C(SI) ~ C(S) is defined by ho(n = f ° h and if XI = {fE C(SI): hO(n E X} then the map H: C(Sl)/XI ~ C(S)/X defined by H( {f + XI}) = {hOCn + X} is an isometric isomorphism onto C(S)/X. Next we check that the operator (ho)*: [C(S)] '" ~ C(SI) takes purely atomic measures on S onto purely atomic measures on S I and restricted to the annihilator of Xl coincides with the isometric isomorphism H*: Xl ~ xt. Thus XI is an algebra which satisfies the assumption of Case I. Hence X = {O} and therefore Xl = {O}. 0 REMARK. Chevalier [Chv] has extended the argument of Case 1. He showed that for every compact Hausdorff space S, C(S) is the only uniform algebra on S with a purely atomic =
t
annihilator. Proposition S.l allows us to extend Theorem 4.2 to the case of an arbitrary proper uniform algebra X with a separable annihilator. We replace A by X, H~ by Xl ( a separable dual!), L I (m)/H~ by L I (A)/Xl where A E C(S)t is any separable probability measure such that Xl eLI (A) and such that whenever /1 ELI (A) and /1 is singular to then /1
= 0; H""
Ivl for every v e Xl,
is replaced by (Xl)l C L "" (A). The natural embedding i~,1 is replaced by
any operator i:,I' where the latter is defined to be TJ1. regarded as an operator from X into
Xl whenever /1 E Xl is chosen so that TJ1. is not compact; this is possible, by Proposition S.1. In particular we have COROLLARY S.l. Let X be a uniform algebra on a compact Hausdorff space S with a norm separable annihilator. Then (a) There is a measure /1 E Xl such that the operator i:,1 : X ~ Xl is absolutely summing, noncompact and nonL Ifactorable. (b) X and all duals of X do not have GL l.u.st. (c) X is not isomorphic to a quotient of any C(K)space; in particular, X is not complemented in C(S).
Next we shall show that the properties of the disc algebra exhibited in §4 are shared by a large class of uniform algebras; every uniform algebra with a nontrivial Gleason part belongs to this class. Unfortunately the technical condition which characterizes the class is slightly complicated. DEFINITION
S.1.
Let X be a uniform algebra on S. Then p is called a Hardy measure
for X if there is a triplet (p, f n , F) such that (S.5)
p is a probability Borel measure on S,
(S.6) (S.7)
fn EX,
F(s)
= lim fn(s)
IIfnll <; 1 and
for n = 1,2, ... ,
IF(s) I = 1
for palmost all s,
n
(S.8) Our next proposition explains the role of Hardy measures. It is the main technical result of the present section.
BANACH SPACES
31
PROPOSITION 5.2. Let X be a uniform algebra on S. Assume that there exists a Hardy measure p for X. Then there are operators T: X ~ Xl and Q: Xl ~ H~ such that
(i) Q is a contractive projection, that is, IIQII ,,;; I and there is an isometric embedding
J: H~ ~ Xl such that QJ = idHI. (ii) The natural embeddini i~ ~I : Ao ~ H~ factors through (QD**, i.e., there are operators U: Ao ~ X** and 1T: (Hb)** ~ H~ such that i~?l = 1T(QD**U. (iii) T and QT are absolutely summing, noncompact and nonL Ifactorable operators. PROOF. Let (p, fn' F) be as in Definition 5.1. Put T(x) = xFp for x E X. T(x) E Xl. Clearly T: X ~ Xl is absolutely summing and IITII ,,;; 1T(T)
=
By (5.8),
1. The construc
tion of Q is more complicated. Given f E C(aD),! denotes the continuous function on the unit disc D which extends f and is harmonic for Iz I < I. Now fix n = I, 2, . . . and assign to every v E Xl the unique measure a E [C(aD)) * which represents the linear functional
faD fda Note that, by (5.6),1
0
=
(!Un(s))v(ds)
Js
for fE C(aD).
fn is well defined and if f E A [regarded as a subspace of C(aD)),
then! is the analytic extension of f onto the interior of the unit disc; hence it admits uniform approximation by polynomials in z. This implies that!
0
fn E X; hence f aDf da = 0
(because v E Xl). Thus by the M. and F. Riesz theorem, a = h . m for some h E H~. We put Qn(v)
= h.
Clearly Q n : Xl ~ H ci is a linear operator with IIQ n II ,,;; I, for n
=
1, 2, ....
Let us put Q(v)
= LimQn(v)
for v E Xl.
n
Here Limnh n denotes a fixed Banach limit of a bounded sequence (h n ) with respect to the a(H~, C(aD)/A)topology. Since H~ is the dual of C(aD)/A, the Banach limit exists and limnh n is a cluster point of the set U;=I {h n } in the a(H~, C(aD)/A)topology. Clearly Q: Xl ~ H~ is a linear operator with IIQII ,,;; 1. Moreover if v E Xl
n LI(p),
then Qn(v) ~ Q(v)
in the a(H~, C(aD)/A)topology;
(5.9)
and if g
= dv/dp,
f aD Q(v)· f dm =
then, for every f E C(aD),
fs f(F(s))g(s)p(ds).
Indeed, given f E C(aD), by (5.7), we have limnfUn(s))
= !(F(s)) = f(F(s))
for pal
most all s. Hence, by the Lebesgue theorem,
Ii:
faD Qn(V) . fdm
Thus f aDQ(v)f dm
=
Ii:
fs!(fn(s))g(S)P(dS)
= f sf(F(s))g(s)p(ds) because
= fsf(F(s)lg(s)p(ds).
a w*convergent sequence has a unique w*
cluster point. Next we construct an isometric embedding J: H~ ~ Xl with QJ = id I
first that q E Ho is a polynomial, say q chJck that
k'
= r,j=laj . z'.
We put J(q)
= q(F)
I. HO
Assume
. p. We shall
32
ALEKSANDER PH.CZyNSKI
J(q) E
(5.10)
xl,
(5.11)
(F k )k=O.±1.±2•... is an orthonormal sequence in L 2 (p).
(5.12)
IIJ(q)1I 1 ~ IIq1l1.
E X, f~Fk dp = limnfsxl~1Fdp = o. This proves (5.10). By (5.7), F(s)1 = F(s) for palmost all s. Thus, if k = ±I, ±2, ... , then fF k dp = fFk dp = O. This yields (5.11) because by (5.5) fFkF k dp = fl dp = 1. The verification of (5.12) is a little bit tricky. By (5.11), J extends to an isometric embedding, say], of H~ into Xl. By the factorization theorem (cf. [Du, Chapter 2]), given a By (5.8), Fp E Xl; if k
> 1, then, by (5.7), given x
polynomial q E H~ there are 11 and 12 in H~ with q
~ (fI11(F)12 dp
IIJ(q)lIl = flq(F)1 dp
= 1112
r
and IIqll1
= 111111 2111211 2.
/2 (II12(F)1 2 dp
r
Thus
/2
= 11]([1)11 211]([2)11 2 = 11111121112112 = Ilqll1· This proves (5.12). It follows from (5.12) that J extends to an operator (which will be denoted also by J) of norm ~ 1 from H~ into Xl.. To complete the proof of (i) it is enough to show that QJ(zk)
= zk
for k
=
1, 2, . . . . This will imply that QJ(h)
=h
for h E H~
and therefore IIJ(h)11 ;;;. IIhll, which together with (5.12) yields IIJ(h)1I = IIhll. Clearly J(Zk)
= Fkp for k
= 1,2, . . . .
Letl(z)
= "i:,j=_rajzj
and let r >k. By (5.9) and (5.11) we get
faD Q(Fkp)ldm =a k = f aD l(z)Zkm(dz). Hence QJ(zk) = zk for k = 1, 2, . . . . This proves (i). We identify X with its canonical image in X* * .
1*
Let 1* =
Z 1 I
and Un (t) =
In for IE Ao. Since the disc algebra operates on every uniform algebra, Un: Ao X is a welldefined linear operator with II Un II ~ 1, for n = 1, 2, . .. . Let us put U(t) = Limn Un (f) for I E Ao where "Limn" denotes here the Banach limit in the a(X**, X*)topology. Finally define "IT to be a contractive projection from (H~)** onto H~. (Since H~ is a dual Banach space, the "IT in question exists.) Since T is weakly compact, (QT)**(X**) C H~. Thus "IT(QT)**(X**) = (QT)**(X**) E X**. To complete the proof of (ii) it suffices to show that (QT)**U(zk) = zk for k = 1, 2, . .. . Fix k. By construction, U(zk) is a a(X**, X*)cluster point of the sequence (Ur(zk». Thus (QT)**(U(zk» is a a((H~)**, (H~)*)cluster point of the sequence (QT**(Ur(zk». Since (QT)**(X**) :::> H~, it is in fact a a(H~, qaD)/A)cluster point. We have (QT)**(Ur(Zk» = QT(f~1) = Q(fr1 Fp). Hence, by (5.9), for every IE qaD), 0
l
f
JaD
Q(f~1Fp)ldm =
J.S1~1FI(F)dp
for r
= 1,2, . . . .
Now, by (5.7), we infer that the sequence (Q(f~1 Fp» converges in the a(H~, qa)/A)topology to the functional zk (we have limrfsl~1 FI(F) dp = fsFkl(F) dp = faDj(z)zk m(dz); the last equality has been verified in the proof of (i». This proves (ii).
BANA.CH SPACES
33
Since T is absolutely summing and i~?1 is a noncompact and nonL Ifactorable operator, (iii) follows directly from (ii), Remark R.I in §4 and the fact that a second adjoint of a compact operator is compact.
0
COROLLARY 5.2. If a uniform algebra X admits a Hardy measure, then there is an absolutely summing surjection from X onto 12. PROOF. Let P: H2 ~ H2 be a Paley projection such that P(H 2 ) C H~. It follows easily from Definition 3.1 that P(H2) is a closed linear subspace isomorphic to 12 and P extends to a bounded linear projection, say P, from HI onto P(H 2 ). The desired operator is
PQT. It is clearly absolutely summing. By (ii), (PQT)** is a surjection. Since P and Tare weakly compact and the unit ball of X is dense in the unit ball of X** in the a(X**, X*)topology, it follows that PQT maps the unit ball of X onto a set whose norm closure contains a ball of P(H~). Hence PQT is surjective. 0 Combining Proposition 5.2 with an obvious adaptation of the argument used in the proof of Theorem 4.2 we obtain THEOREM 5.1. Let X be t1 uniform algebra on a compact Hausdorff space S. Assume that X admits a Hardy measure. Then G) X and all the duals of X do not have GL l.u.st (jj) if X is a quotient of a Banach space Y with GL l.u.st, then Y contains a complemented isomorph of 11 ; (jjj) if C(S)*/X1 is a subspace of a Banach space Z with GL l.u.st, then Z contains
I; uniformly; (jjjj) if C(S)*/X1 is isomorphic to a subspace of a Banach lattice Z, then Z contains an isomorph of co. 0 COROLLARY 5.3. If X is a uniform algebra on S which admits a Hardy measure, then X is not complemented in C(S). 0 Our last result gives a sufficient condition in order that a uniform algebra have a Hardy measure, in particular this is the case if a uniform algebra X has a nononepoint Gleason part, i.e., if there are two different homomorphisms of X into complex numbers, say ..p and "', such that sup IIxll=l,xEX
I..p(x)  1/J(x) I < 2.
PROPOSITION 5.3. Let X be a uniform algebra on a compact Hausdorff space S. Let us consider the following three properties: (l) X has a Gleason part which contains at least two different points. (2) There exists an ideal leX and a multiplicative linear functional ..p on X such
that if a = 11..pllll = SUPtEJ, IItll = 1 1..p(f)1, then 0 (3) There is a Hardy measure for x. Then (l) => (2) => (3).
< a < 1.
34
ALEKSANDER PEtCZYNSKI
(I) => (2). If
PROOF.
J
= ker t/I
t/I, then
(2) => (3). Let 11 be a Borel measure on S such that If dll =
a
= lim
lim I fn dll· n
To complete the proof we shall check that the triplet (p, fn' F) satisfies the conditions (5.5)(5.8) of Definition 5.1.
+ a 11l1aimost everywhere. Hence 1111 IF(s) I = 1 palmost everywhere. Since IIfn II
Clearly IF(s) I = 1 and I  a ~ 11  aF(s) I ~ 1 is absolutely continuous with respect to P and 1 and
~
0< a
= 111111 = lim Ifn dll = lim n
n
f ;!, dllll,
it follows that limnfn(s) = F(s) 11l1almost everywhere and therefore palmost everywhere. Finally given f E X we have
IIplli IfF dp
=
IfFll  aFl 2 dill I = I jF211  aFI2 dll
= I fF2(1
 aF){l  aF) dll
= lim I f(f~{l + a2 )  af~
 afn] dll
n
n
Thus Fp E Xl. This completes the proof.
0
Notes and remarks to § §4 and 5. The results of § §4 and 5 are closely related to the following problem due to Glicksberg [GI2].
Problem 5.1. Is C(S) the unique uniform algebra on a compact Hausdorff space S which is complemented in C(S)? Glicksberg (using Rudin's averaging technique [Rul]) proved that the answer is "yes" for S being a homogeneous space of a compact topological group G and for uniform algebras on S which are invariant under the action of G on C(S): (Tgf)(s) g E G,
= f(gs)
for f E C(S),
s E S. For details and generalization to locally compact groups cf. [G12] and [RI].
It was shown in [P4] that the answer is "yes" for an arbitrary compact Hausdorff space S and for uniform algebras on S with norm separable annihilators (cf. our Corollary 5.1). For uniform algebras with nontrivial Gleason parts the same fact (cf. our Corollary 5.3) was proved by Kisliakov [Kis 1] and with some additional assumptions by Etcheberry [Etch]. While Glicksberg's problem is still open, it seems to be more natural to ask
BAN I\CH SPACES
35
Problem 5.2. Conjecture. Let X be a proper uniform algebra on a compact Hausdorff space S (i.e., C(S)\X is nonempty). Then (a) X as a Banach space is not isomorphic to any C(K)space (b) X is not isomorphic to any complemented subspace of a C(K)space (c) X is not isomorphic to any quotient space of a C(K)space (d) X does not have GL l.u.st. It was observed in [P5] that if Sis metrizable and X is a uniform algebra on S which
is complemented in a C(K)space then X is isomorphic as a Banach space to C(S). Varopoulos [Va] proved recently that if a semisimple algebra X is isomorphic as a Banach space to a C(K)space, then X is an operator algebra. Bya result of Bishop (cf. [B2], [Gml, pp. 6062]), it follows that every uniform algebra has a quotient antisymmetric algebra; thus in (c) one may assume without loss of generality that X is antisymmetric. It is not clear whether the criterion of Proposition 5.2 can be applied for uniform algebras with norm separable annihilators. This suggests the following
Problem 5.3. Let X be a uniform algebra with a norm separable annihilator. Does X admit a Hardy measure? Does X have a nononepoint Gleason part? Another approach to attacking Problem 5.2 is to eliminate the assumption of the norm separability of the annihilator in Proposition 5.1 (cf. the Remark after the proof of Proposition 5.1). This suggests Problem 5.4. Let X be a proper uniform algebra on S. Does there exist a /1 E Xl such that the operator Til: X  L I (1/11) is not compact where T/x) == xdp/dipi for x EX? Theorem 4.2(i) was proved in [P7] while (ij) and (iii) were observed in [P6], (iv) was first proved using a different technique of [Pisl] by Pisier, (v) is due to W. B. Johnson (oral communication). The equivalence (iii) <=> (i) of Proposition 5.1 is due to Wojtaszczyk [W I] , while (i) <=> (ii) is taken from [P4]. The concept of a Hardy measure is implicitly contained in Kisliakov's work [Kis I]. Proposition 5.2 in the present form was stated and proved by Wojtaszczyk. A slightly weaker fact was established in [P7]. Proposition 5.3 goes back to Bishop and Wermer (cf. [Gm 1, Chapter VI, proof of Theorem 7.1]). ADDED IN PROOF.
[Kis3].
A similar result to Proposition 5.2 has been obtained by Kisliakov
6. Uniformly Peaking Families of Functions in A and
H~ .
The Havin Lemma. This section has a preparatory character for the next one. The main result is a very technical Proposition 6.1 due to Havin [Hvl]. This is a generalization of a quantitative character of the fact that every closed subset of the unit circle of Lebesgue measure zero is a peak set for the disc algebra (cf. § 1, Step I in the proof of the F. and M. Riesz theorem). Roughly speaking it is shown that if a closed subset e of the unit circle has a small Lebesgue measure then there is a function in the disc algebra of norm one whose values are close to zero except for a set of small measure and are close to one at each point of e. Moreover all the estimations on the measure of the exceptional set and how the values of the function are close either to zero or to one depend on the measure of e only. A similar construction leads to a result of Amar and Lederer [AL] on peak sets for Hoc (Proposition 6.2) and to a characterization of exposed points of Hoc (Theorem 6.1). Recall that m denotes the normalized Lebesgue measure on the unit circle aD = {z E C: Izl = I}. Given a measurable function u: aD '> C, we denote (if it exists) by H(u) the Hilbert transform of u (cf. §O.II) and by Ii the harmonic extension of u onto D, i.e.,
PROPOSITION 6.1 (THE HAVIN LEMMA). There exists a positive function 8 '> €(8) for 0<8 < 1 with lim.s=o€(8) = 0 such that given a measurable subset e of aD with 0 < m(e) < 8 there exist fe and ge in HOC with the following properties: (6.1 )
(6.2)
(6.3) (6.4)
Ife(z)1
sup Ife(z) 
+ Ig/z)1 = 1
for z E aD,
11 = 0(8 1 / 2 ),
zEe
sup Ige(z)1
zEe
= 0(8 1 / 2 ),
f
aD
Ige(z) 
11 dz < €(8),
if e is closed, then fe and ge are in A.
PROOF. Step I. Construction. Fix e C aD measurable with m(e) < 8 B = 8 1 / 2 , b = 8 1 / 2 . If e is not closed we define we = w by 36
~
1 and put
BANACH SPACES
w(z)
B
={
b
37
if z E e, if z f e.
If e is closed we first pick finite families of closed intervals (.6.;)1 ~i~N and (.6./')1 ~i~N such that if e' = UN .6.~ and e" = UN .6.~' then J= 1 J J= 1 J
e C e',
(6.5)
.6.~
m(e")
< 0,
.6.; C int.6./'
n .6.; = .6.~ n .6./, = ¢
and
= 1, 2,
for k =1= j (k, j
... , N),
and we define w to be a C~ function on aD such that B ,,;;; w(z) ,,;;; b for all z E aD and
w(z)
=
{
B
for z E e'
b
for z
' tf. e".
Next we put
v
= H(w);
h
= w + iv;
fe
= exp h I .
We shall check
h(z)
=1=
for zED,
0
If/z)1 ,,;;; 1
for z E aD,
(6.6)
faD 10g(1
 Ifel) dm
>
00.
Then we define ge to be the outer function with Ige(z)1
=
1  Ife(z)1 for z E aD almost
everywhere. Precisely we put 
(6.7)
1 rr
ge(z) = exp  2 ge(z)
=
S21T 0
lim ge(Pz)
tt. + z 10g(I £It _ z
.
 Ife(e't)1 dt
for Izl
< 1,
for z E aD almost everywhere.
p=1
Step II. Verification of (6.6). By the maximum principle for harmonic functions, B";;; Re h(z) = w(z) ,,;;; b
for zED.
Hence h(z) =1= 0 for zED and Re
h 1 =
_,w'_ ";;;b <0.
W2+lJ2
Since wE L ~, v = H(w) exists and v E LP for all p (cf. §O.II). Thus
< 1.
11 and h I
exist and
I; hence fe E H~ and Ife(z)I";;; 1 for z E aD almost everywhere. It remains to show that f aD(I  Ife I) dm > 00; this will imply that ge defined by (6.7) does exist and belongs to H~ (cf. [Du, Chapter 2]). Let Zo = {z E aD: V(z)2 ,,;;; 2} and Zk = {z E aD: 2k ,,;;; V2(z) < 2k+1} for k = 1, 2, . . . . Observe that there is a C> 0 such that m(Zk) < CT k for k = 0, I, ... because vEL 2 • If z E Z k' then are analytic for Izl
Thus
Furthermore lehII ,,;;; e Reh 
1 ,,;;;
ALEKSANDER PEl.CZyNSKI
38
flog(l  If I) dm
=
£. f
k=O
log(l  lexp hl(z)l) dz
Zk
Step Ill. Verification of (6.4). If e is closed, then wand therefore v are COO functions < Ife(eit)1 < 1 is a 21Tperiodic C oo _ function, hence ge E A (cf. [Du, Chapter 5]). Step IV. Verification of (6.1) and (6.2). The formula (6.7) yields that Ige(z)1 = 1 Ife(z)1 for z E aD almost everywhere (cf. [Du, Chapter 2]). This yields (6.1). If z E e, then
(cf. §O.II). Hence h, h I and fe are in A. Moreover 0
_1__ Ih(z)1 Hence remembering that B 11  fiz)1
~
1
v'w
2 (z)
+
v2 (z)
_1_ _  I "'" lw(z)1  B
> 1, we get = 11
 exp hl(z)1 ~ explhl(z)1  1
~ Ihl(z)lexplhl(z)1 ~ BIexp(I)
= 0(0 1 / 2 ).
The proof of the second inequality of (6.2) is slightly harder. To treat simultaneously the case where e is a closed set and e is not closed we put e* = e" if e is closed, and e* = e otherwise. Let X
=
{z E aD: Iv(t)1 ~ 77} for some 77
> O.
Since the Hifbert transform is of
the weak type (1.1) there is an absolute constant C I such that
m(X)~Cl171f
aD
because faDlwldm
Iwldm~ClBo171 +C1 b77 1 ,
= faD\e.lwldm + fe.lwldm
~Bo
+ b.
Thus, taking into account that
if z ($ X U e* then
Ife(z)1
= exp RehI(z) = exp b
~exp
b2
we get
faDlfel dm
w 2(z)
w(z)
+ V2(z)
~b+172bl,
+ 17 2
= f)fel dm + fx\e.lfel dm + faD\Xue.lfel dm ~ m(e*)
+ CI(B77IO + b77 I ) + b + 17 2 b 1
~ 0(0 1 / 6 ), provided we put 77 = This completes Step IV. Step V. Verification of (6.3). Let ue = 1l0g(J  Ifel)l. Then ge = exp(u e  iH(u e
01/ 3 .
»·
The crucial point is to show that there is a constant M such that
BANACH SPACES
JaD U; dm < M r
(6.8)
39
for all measurable e
< 3D.
Having done this we complete the proof as follows: By (6.1) and (6.2), ue whenever m(e) + O. Thus, by (6.10), faDlueldm + 0 whenever m(e) Hilbert transform is of weak type (1.1), the last relation yields
m{z E 3D: IH(ue)(z) I
+
+
0 in measure
O. Since the
~ (faDIUel dm y/2} .; ; C1 (IaD IUel dm y/2
where C 1 is an absolute constant (cf. §O.lI). Thus if m(e)
+
0 then H(u e )
+
0 in measure
1 in measure. Since Ilgell"" .;;;; 1 for all measurable e < 3D, we conclude that flg e  11 dm + 0 whenever m(e) + 0 which proves the second inequality of (6.3). The
and ge
+
first formula of (6.3) follows directly from (6.1) and the first formula of (6.2). To establish (6.8) put
=
y
Clearly 0
00
.ll
IRe
h
and
ue
= log
Iwl
+ v2
.
(1)1/6
+ yl)1/3
JyI/3 dm
w2
1 1  exp(y) .;;;; 6 1  exp(y)
=6(1+ Therefore u; .;;;; 36 (I
=
1
exp(y)  1
.;;;; 36 (I
)1/6';;;;6(l+yl)I/6.
+ yl/3).
Finally
= J(lv2 1 + IWI)I/3 dm .;;;;J~dm + Jlwll/3 dm. Iwll/3
w
We estimate each term separately. By Kolmogorov's theorem (cf. §O.II), there is an absolute constant C2/3 such that flvl 2/3 dm .;;;; C2 / 3(flwl dm)2/3. Hence
2/3 dm';;;; b1/3C2/3 (JIwl dm )2 / 3 Jl/3 w V
~ C ~1/6(2d/2)2/3 "'" 2/3 u U
while flwl 1/3 dm ';;;;B 1/ 3S
+ bl /3
.;;;;
2.
~
22/3 C
2/3'
"'"
0
€(S) which follows from the argument in Step V seems to be unsatisfactory. We get only €(S) .;;;; Cilogil  eoal 12a REMARK.
for a
The estimation for the order of the function S
+
< 1/2. The next result, due to Amar and Lederer [AL], uses a similar technique to that
of Havin's lemma. It is, however, formulated in a different language, which we now recall. Let A be the maximal ideal space of L"" and let
f
+
f be the algebraic isometric iso
morphism from L"" onto C(A). Recall that A is a compact Hausdorff space. If Xe is the characteristic function of a measurable subset e of 3D, then e = (t E A: Xe(t) = I} is a
40
\LEKSANDER PELCZYNSKI
e
m
clopen set; all the sets form a basis for the topology of Ll. By we denote the unique regular Borel measure on Ll which corresponds via the RIesz representation theorem to the functional
f ~ II dm, i.e., mis defined by
f j dm = f I dm
for
IE L
00 •
H oo can be regarded as the uniform algebra on Ll, in fact Ll is the Shilov boundary for ~ (cf. [H, Chapter 10)). Now we are ready to state PROPOSITION 6.2. Let F be a closed Go in Ll with m(F)
= O.
Then F is contained in a nontrivial peak set lor H , i.e., there exists a nonconstant I E ~ such that 11/11 = 1, F C jI(1) = {t E Ll: f(t) = I} and jI(1) = {t E Ll: I/(t) I = I}. oo
PROOF. The assumption on F yields the existence of a decreasing sequence (en) of
n;=
measurable subsets of aD such that F = len and m(en ) = m(en ) ~ 0 as n ~ 00. Passing to a subsequence, if necessary, we may assume without loss of generality that e1
m(en ) <
n 3
for n
= 1,2, ....
= Ll,
Now we put 00
L  n Xe n\en+1 '
w
=
h
= w + iv,
v = H(w),
n=1
1= exp(1/h).
Since wE LP for p < 2, h E 1fP. Thus h is a boundary value of an analytic function, say h in the unit disc. Furthermore, Re h . ;;; 1 , because Re h ..;;; 1. Thus h
"* 0 and I is a nonconstant
~ function with 11/1100 ..;;; 1. Furthermore if t E F then f(t) = 1 because for z I/(t)  11 = lexp(1/h)  11";;; lexp{l/Ihl)  11";;; e lln  1. Hence
00
[1(1)
=
n
E en'
00
{t E Ll: If(t)  11 ..;;; e lln  I} ::)
n=1
n
en
= F.
n=1
Finally replacing if necessary I by Yz(f 6.2. 0
+ 1), we
obtain the last assertion of Proposition
It is interesting to compare Proposition 6.2 with the next one. PROPOSITION 6.3. There is a closed nonempty G/jset Fell with m(F) not a peak set lor H oo •
= 0 which
is
PROOF. Suppose that the assertion of Proposition 6.3 is false. Then given a closed G/j set Fell with m(F)
= 0 there exists an IF E H
oo
such that
iF
peaks exactly on F.
Hence
(6.9) Indeed, by regularity of J.l, it is enough to observe that J.l(F) C Ll with m(F)
= 0".
=0
for every closed G/j set F
The last follows from
J.l(F)
f
= lim j'j. dJ.l = 0 n
(because fr;. converges pointwise to the characteristic function of F). The map I ~ f for induces an isometric isomorphism from L l(m) onto L I (m) which carries H~ onto
IE L
00
41
BANACH SPACES
LI(m) n (H~)l. Now (6.9) implies that the annihilator (H~l coincides with LI(m) n (~)1; hence (H~)l, being isometrically isomorphic to Hb, is norm separable. Since (L ~ /~)* is isometrically isomorphic to (H~)l, we infer that (L ~ /H~) is separable. This leads to a contradiction because (Hb)* is isometrically isomorphic to L ~ /H~ and Hb contains a subspace isomorphic to II. 0 Problem 6.1. Characterize the subsets of the Shilov boundary of H~ which are peak sets for H~. We end this section with an application of Proposition 6.2 of a geometric character. We shall characterize the exposed points of the unit ball of H~. Recall that if X is a Banach space then an x E X with IIxll = I is an exposed point of the unit ball B X = {y EX: Ilyll <: I} if there is an x* E X* with Ilx*11 = 1 which strictly supports Bx at x, i.e., x*(x) = 1 and Rex*(y) < 1 for every y E Bx\{x}. BH~
THEOREM 6.1. A function f E iff
(6.10)
m(£)
>0
H~
with Ilfll
= 1 is
an exposed point of the unit ball
where E == {z E aD : If(z)1 == l}.
PROOF. Let f be an exposed point for B H~ and let x* E (H~)* be a linear functional which strictly supports B H"" at f. Let <1>* E (L ~)* be a norm preserving extension of x*. Given measurable e c aD we denote by <1>: the linear functional on L ~ defined by :(g) = *(Xeg) for gEL ~. Clearly <1>* = <1>: + ~D\e and 11<1>*11 = 11<1>:11 + 1I~D\ell. Furthermore if e satisfies the condition If(z) I <: a < 1 for z E aD\e almost everywhere, then <1>: = <1>*. Indeed, otherwise 1I~D\ell > 0 and
a contradiction. Now let en = {z E aD: 1  If(z) I < n I }. By the previous remark :n = <1>* for n = 1,2, . . . . Clearly e l :J e 2 :J ... . Let e~ = new To show thatfsatisfies (6.10) it is enough to establish that m(e~) > O. Suppose that m(e~) = 0; then F = n;=len is a closed Gosubset of ~ with m(F) = O. Thus, by Proposition 6.2, there is a nonconstant fF E H"" with IlfF11 = 1 such that iFI (I) :J F. For n = 1, 2, ... , we have 1*(n  *(fFnl == <1>;
n
«(1  fF)n
<: Ilill sup 11  fF(z)l. zEe n
n;
Taking into account thatel:J e2 :J ... :J F = = I en and iFI (I) :J F, we infer that limnsuPzEen 11  fF(z) I = O. Hence *(n = *(f. f F ) which contradicts the definition of the exposed point, becausefF being nonconstant implies thatf· fF EBH",,\{f}. Hence m(e",,) > o. Now assume that anfEH"" with Ilfll == 1 satisfies (6.10). Let <1>* E (L"")* be defined by
*(g)
=
m(E)l fEg(z)signf(z)m(dz) for gEL"".
ALEKSANDER PElCZYNSKI
42
Let x* be the restriction of <1>* onto H~. Clearly IIx*1I = Ix*(f)1 = 11<1>*11 = 1. Hence Re x*(g) :(; 1 for g E B H~' Finally if, for some g E B~, *(g) = x*(g) = I, then Ig(z)1 = 1 and g(z)sign/(z) = 1 for z E E almost everywhere. Hence g(z) = I(z) for z E E almost everywhere. Since m(£)
> 0, we
infer that
I
= g.
0
REMARK. Note that if I is an exposed point for B H~ then a linear functional which strictly supports B H~ at I can be chosen from L 1/H ~, a predual of H~.
Notes and Remarks to §6. The proof of Havin's lemma is taken with some modification from [Hvl]. The condition (6.4) of the lemma was observed by Kisliakov [Kis2] and independently by Pe1czynski (unpublished). Wojtaszczyk [W2] observed recently that Havin's lemma can be generalized to uniform algebras with unique representing measures for linear multiplicative functionals. The proof of Proposition 6.2 is taken from [AL]. The construction is classical (cf. [Z, Vol. I, p. 105]). Proposition 6.3 shows that the "Remarque" in
[AL] is false. Theorem 6.1 is due to Amar and Lederer [AL] and Fisher [Fi]. An analogous result for the disc algebra is due to Phelps [Ph2]. Let us recall that an IE BH~ is an extreme point of B H~ iff I aDlog(l  III) dm = 00 (cf. [H, p. 138]). Finally observe that no point of BH~ is strongly exposed, i.e., there is no IE B H~ and x* E (H~)* with x*(f) = 11/11 = IIx*1I = 1 and such that for every sequence
(fn) in B H"" if x*(fn) + x*(f) then II/n  III + 0. Indeed, regard H~ as a subspace of C(Ll) and note that for every f.1 E C(Ll)* there exists a peak set, say F, for H~ with f.1(F) = 0. Suppose IF E H~ peaks exactly at F, i.e., {s E Ll: iF(s) = I} = {s Ell: liF(s) I = I} = F. E BH~ put
Given
I
for n
= 1, 2,
In
= 1(1  f~·)/ill

I~II for n = 1,2, . . . . Then
limnilln 
III = 11/11
... while limnI/~ df.1 = Iidf.1. Various interesting properties of H~ and L 1 /H~ related to the material of this section
are discussed in the expository paper by Havin [Hv2]. ADDED IN PROOF.
Delbaen [De3] recently generalized Havin's Lemma for a represent
ing measure of a linear multiplicative functional on a uniform algebra such that the representing measures of the functional form a weakly compact set.
7. Characterizations of Weakly Compac(Sets in L 1 /Hb and in A * In this and in the next section we shall show that, despite the results of §4, the spaces
A. A * and L 1 /Hb share various properties of C(aD), C(aD)*, and L 1. In particular a specific characterization of weakly compact sets in L 1 spaces [Gr2] , [P9] as those sets on which weakly unconditional series of functionals converge uniformly can be carried out to L 1 /Hb and A *. Another peculiar property of a weakly compact set in L 1 /Hb says that the set is an image under the quotient map of a weakly compact set in L 1. The main result of this section, Theorem 7.1, was proved recently by Delbaen [D I] and independently by Kisliakov
[Kis2] . Recall that a sequence (xn) of elements of a Banach space X is w.u.s. (weakly uncon
ditionally summable) if :E:=llx*(xn)1 < 00 for every x* E X*. Clearly if (x n ) is w.u.s. then there is a K > 0 such that :Elx*(xn)1 ~ Kllx*1I for x* E X*. A sequence (yn) in X is weakly Cauchy if limnx*(y n) exists for every x* E X*. The following classical result goes back to Orlicz. LEMMA
7.1. Let W be a subset of X such that every sequence of elements of W con
tains a weak Cauchy subsequence. Then (7.1)
lim n
PROOF.
sup Ix~(x)1 xEW
=0
for every w.u.s. sequence (x~) in X*.
Define T: X + 11 by Tx
= (x~(x».
Then T is a bounded linear operator
which carries W into a totally bounded subset of 11 , because in 11 every weakly Cauchy sequence is norm convergent. Hence limN:E:=Nlx~(x)1 (7.1). 0
=0
uniformly for x E W. This yields
In view of formulas (1.1) and (1.2) of §1, in this and in the next section we identify L 1 /Hb with a subspace of A * and A with a subspace of (L 1 /Hb)*. The duality is given by the bilinear form (BE denotes the unit ball of a Banach space E)
(w, x)
+
faDxwdm
Moreover, SUPXEBAJaDxwdm
for x EA and w
= Ilw Il L 1/H A and
=
{w
+ H~}
E
Ll/H~.
SUPWEBL1/HAJaDxwdm
= IIxII A .
Let Y be a closed linear subspace of a Banach space X. A map (in general nonlinear) r: X/Y + X is called the nearest point crosssection if r(w) E wand
IIwllx/y
=
inf lIy yEY
+ r(w)lI x = IIr(w)lIx
for w E X/Yo
The nearest point crosssection need not exist. It does in the case where Y is 43
Hb
and X is
ALEKSANDER PELCZYNSKI
44
either L I or X is qaD)*, because by the F. and M. Riesz theorem, H ~ is a closed subspace of qaD)* in the a(qaD)*, qaD»topology. In fact in these cases it is unique (cf. [Kh 1] , [Hv2]). Now we are ready to state THEOREM 7.1. tet W be a subset of L I /H~. Then the' following conditions are equivalent:
(1) There is a weakly compact subset V of L I such that q(V)::::l W where q: LI
>
L I /H~ is the quotient map. More precisely, for V one can take the weak closure of the image of W under the nearest point crosssection.
(2) The weak closure of W in L 1 /H~ is weakly compact. (3) Every sequence of elements of W contains a weak Cauchy subsequence.
o for every w.u.s.
(4) Iimn(sup{w+HA}EwIJaD
PROOF. Clearly for arbitrary Banach space (I)
~
(2)
=\>
(3)
sequence (
(4). (Use Lemma 7.1
and the "easier" implication of the EberleinSmulian theorem.) To prove that (4) ~ (1), we observe first that without loss of generality one may assume W is an infinite countable set,
(7.2)
+ H ~}
every {w (7.3)
E W regarded as a linear functional on
A attains its norm on the unit ball of A.
For (7.2) note that L 1/H ~ is separable and observe that if a subset of L I / H ~ satisfies (1) so does its closure. By the open mapping principle, if ({w~
1
For (7.3) note:
0
+ H~}) and ({w~ + H~}) + H~} 
are enumeration of countable sets W' and W" respectively and if II {w~
{w~
+
<
H~} IlL I/HI
the BishopPhelps
Tn for n
t~eorem
=
1, 2, . . . , then W' satisfies (I) iff W" does.
20
[BPh], the subset A ZP of A * consisting of the linear functionals
which attain their norms on the unit ball of A is dense in A *. Since A * is the IIsum of
L I /H~ and Vsing (cf. § 1), the natural projection of A * onto L 1 /H~ which annihilates Vsing maps A~p into itself. Hence A~p
+ H~})
Now let ({w n Yn E A so that (7.4)
faD WnYn dm
=
n L 1 /H~
is dense in L 1 /H~.
be an enumeration of W. It follows from (7.3) that there is a
inf
l hEHO
faD IWn
+ hi dm
and IIYnl1
=
1 for n
=
1,2, ... 
By the F. and M. Riesz theorem, H~ is a closed subspace of qaD)* in the a(qaD)*, qaD»topology; hence there are h n E H~ such that
(7_5)
inf
1
f
aD
Iw n
+ hi dm
=
f
aD
Iw n
+ h n I dm
for n = 1, 2, . .. .
hEHO
Let Vn
= wn + h n
for n
=
1,2, ... and let V
=
U~=l {vn}' Our goal is to show
By
45
BANACH SPACES
(7.6)
the weak closure of V in L 1 is weakly compact.
Since q(V) = Wand T( {w n + H~}) = vn for n = 1,2, ... , this will complete the proof. We left to the reader a simple checking that if W satisfies (4) then it is norm bounded. This implies that sUPnllvnll = M < 00. To apply the wellknown criterion of weak compact· ness in L 1 it remains to show that the vn's are uniformly integrable or equivalently that the un's are uniformly integrable, where un = Y nVn (or n = I, 2, . .. . It follows from (7.4) and (7.5) that
[
)aD
= )aD [ lu n I dm
un dm
for n
= 1,2, . . . .
Thus un ~ 0 for n = 1,2, . . . . Suppose that the functions un's are not uniformly integrable. Then using a standard gliding hump procedure (as in [KP, Lemma 2]) we construct a subsequence (uic) of (un) and a sequence (e k ) of mutually disjoint closed subsets of 3D such that for some l) > 0 (7.7)
for k
(7.8)
the functions XaD\e
Since sUPnllunlll
= sUPnllvnlll <
00,
k
=
1, 2, ... ,
. Uic are uniformly integrable.
we can choose the sequence (uic) so that
there exists a tJ E C(3D)* which is the limit (7.9)
Let tJ
of the sequence (uic) in the a(C(3D) *, C(3D))topology, i.e., limkf aDxuic dm x E C(3D).
= U . m .+ v where U ~ L 1
= f aDx dtJ for
and v is singular with respect to m. By (7.7), U ~ 0, v ~ 0
and IItJlI ~ l) > O. We shall denote by (yic) and (vic) subsequences of (yn) and (v n ), respectively corresponding to the subsequence (uic) of (un); we have uic = Yicvic for k = 1,2, . . . . The crucial part of the proof is, exploiting the fact that tJ> 0, to construct in A a w.u.s. sequence (x,,) and a subsequence (Uk) so that the integrals fX"Uk s+ 1 dm = f(XsYk s+ 1 )vks+ 1 dm stay uniformly away from zero. This will contradict (4), because (x s ) w.U.S. yields (X sYk S +1) w.U.S. We shall consider two cases separately. The first one heavily depends on Havin's lemma (Proposition 6.1). In the second one we use only the fact that the annihilator of A is norm separable. Case I: v = O. Then a = f aDu dm = IltJlI > l). Let fk = fek and gk = gek for k = 1,2, ... ,where the functionsfek and gek are constructed in Havin's lemma (Proposition 6.1). Since ek are closed,fk and gk belong to A. Oearly (6.1)(6.3) yield (7.10) (7.11)
limJ k
aD
hgkdm=f
aD
hdm
lim sup Ifk(z) k
zEek
forhELl,
11 = o.
46
\LEKSANDER PH.CZyNSKI
Let us fix '7 > 0 so that 0  a1/ > 0/2. Using (7.9)(7.11) we construct an increasing sequence of the indices k j < k2 < ... so that if Go = 1 and Gs = nj== Igkj for s ;;;. 1, then for s = 1, 2, ... (7.12)
(7.13)
(7.14) Now we define a w.u.s. sequence (x s ) in A by
xs = Gsfk
for s = 1,2, ...
s+1
The sequence (x s ) is w.u.s. because, by (6.1), s s+ 1 Ix,iz) I ~ (lfkJz)1 + IgkJz)1) = 1
n
L
j== 1
j== 1
J
for z E
aD and for s = 1,2, . . . .
J
Combining (7.12) with (7.13) we get
IfaD GSUk
s+1
dm
I;;;. IJaD Gsu dm 1 2s 2a1/ ; ;. f aD udm  t
(7.15)
j==1
s
;;;. a 
L
If
aD
Gjudm  (
JaD
T~2a1/  2s 2a1/
> a(1
Gj_1udm
1
2S 2a1/
 1/).
j== 1
Clearly IIGslI"" ~ 1 (by (6.1», SUPzEeks+llfks+l(z)  11 0 as s sUPkllu';1I1 ~ sUPnllunli
=M < lim 8==""
00.
lim sup 8==00
f.eks + 1 11  fk
IIGslu k
s+1 _
= faDudm = a, we
f 3D 11  fk
8+ 1
(by 7.11), and
Thus
and (by (6.2» lim8==",,!aD\eks+lI1 fk8+1ldm
lims==""!oDuks+1 dm
00
IIGslu k
8+ 1
(7.16)
dm
dm
=0
Hence, by (7.8) and the relation
get
f aD\ek 11  fk IIGslu ks+ s+ ~ lim sup f Uk dm 8==00 aD\ek s+ ~ a  lim inf f. uk dm s==oo e ks + 1 8+1
= lim sup 8==00
0.
8+ 1
1
1
~a
Combining (7.15) with (7.16) we get
= 1.
s+1
8+ 1
1
dm
47
BANACH SPACES
lim inf I JaD ( xu' dm I 11=00 II k.l'+1
IJaD (G.I'u/c
?o lim inf .1'=00
dm Ilim sup (
.1'+1
JaD
.1'=00
11  Ik
.1'+1
IIG.I'luic
.1'+1
dm ?o Q.2·
Finally we put
4fJ.I' Since
(x.l')
='Yk .1'+ I . x.l'
for s
=
1, 2, ....
is w.u.s. and IlYic II = 1 for k = 1, 2, ... , the sequence ( 4fJ.I') is w.u.s, Clearly we have
(
4fJ
laD .I'
Uk'
dm = (
laD
.1'+1
x
II
Uk .1'+1 dm
for s = 1,2, ....
Thus lim sup .1'=00
{
sup I}
I(
laD
w+Ho EW
4fJ.I'w dm I ?o lim inf I (xllu k
laD
.1'=00
11+1
dm I ?o Q.2·
This contradicts (4) and completes the proof in Case I.
Case II: v> O. Since the measure v is regular and singular with respect to m, there is a closed set e C aD with m(e) = 0 such that vee) ?o 711vll/8 = 7a/8. Using Bishop's general RudinCarleson theorem (cf. §2, Theorem 2.1) we construct inductively a sequence (fn)n;;.O in A and a sequence (e n )n;;'1 of closed subsets of aD such that
10
(7.17)
=0,
11
In(z)
= Il/nll =
en + I
= en n
{
e 2 = {zEaD: Izwl~TI for wEe},
e l = aD,
= 1,
"itn(z) I < Tn
1 for z E e, z E aD: sup Iz 
wi
wEe
for z
tt en
~ Tn and j~ Ifj(z)  fj_1 (z)1 ..;; Tn f (n
= 2, 3, ... ).
Observe that
S(z)
=L
Ifj(z)  fj_1 (z)1 ~ 3
for z E aD.
;=1
Indeed, if z E e, then S(z)
= 1.
If z
tt e,
then z E e k  1 \e k for some k
= 2, 3, ...
2, then 00
S(z) ~ 2
L
00
Ifj(z)1 ~ 2
+2
j=1
if k
L
Tn ~ 3;
n=2
> 2, then 00
k2
S(z)
=
L
If;(z)  fj_l(z)1
+ I/k_l(z)1 + 2
;=1
j=k 00
~ 2 k+ 2
L
+ 1+2
L j=k
T;
< 3.
Ifj(z) I
; if k
=
48
\LEKSANDER PELCZYNSKI
Clearly n;:lej = e. Hence using (7.9) and (7.17) we construct inductively increasing sequences of the indices (n s ) and (k s ), so that for s = 1,2, ... ,
If
aD
fn Uk. dm s
/
I~
2sa,
for i
~ s,
r f.ns dIJ. I "'",.; : 2 s2 a. IJ'"aD f.nsu'kS+l dm  JaD Let us put no
= 0 and Xs = fns  fns+ 1 (s = 1,2, ... ). 00
L
IxsCz)I ~
s=1
L
Ifn/ z). f nS_ 1 (z)1 ~ ~ Ifj(z)  fjl (z)1 ~ 3.
s=1
/=1
Furthermore, for s = 1, 2, ... ,
I
Then (x s) is w.u.s. because, for z E aD,
00
faD X,U kS+ 1 dm
!~ IfaD f n, UkS+
1
dm
II
~ IlaD r f.n, dlJ.l 29 2a 
faD fns+ 1U,k s+ 1 dm
I
2s 1 a
~
IfaDfn,dvllfaDfn9udmlrS2arsla
~
If.e f.n, dv
~
II f
aD\e
vee)  v(aD\e)  T'a
f.n, dv
1
2s 2a  29 2a  2,l a
~ ~a,
Now we put I{J, = XsYk,+ 1 for s = 1, 2, . . . and we complete the proof as in Case I.
0
COROLLARY 7.1. The assertion of Theorem 7.1 remains valid if L 1 /H~ is rep/aced by A*, Ll by qaD)* and the condition (4) by
(4a) limn (sup {/J+H6}EW IfaDl{Jn dlJ.l)
= 0 for every w.u.s.
sequence (I{Jn) in A.
PRobF. To derive the implication (4a) ~ (1) from Theorem 7.1, denote by PI and P2
the natural projections from A* = qaD)*/H~ = Ll/H~ Ea1 Vsing onto LI/H~ and Vsing respectively, and let T: qaD)*/H~ ~ qaD)* and TL : L I /H~ ~ L I to be the nearest point crosssections. Now if W C A * satisfies (4a), then both of the sets PI (W) and P2 (W) have the same property. For subsets of Ll/H~, (4a) is equivalent to (4). Hence the weak closure of T(Pl(W» = TL(Pt(W» is weakly compact. Furthermore, for subsets of Vsing ' (4a) combined with the general RudinCarleson theorem and the identity TP2 = P2 yields lim n(sup/JEP2(W)lfaDl{Jn dlJ.l) = 0 for every w.u.s. sequence (I{Jn) in C(aD). Thus, by a result of [P9], the weak closure of the set T(P2(W» = P2(W) is weakly compact. Since
49
BANACH SPACES
we conclude that the weak closure of r(W) is weakly compact.
0
7.2. A weak closure of a uniformly bounded set We L 1 /H~ is weakly compact iff there exists a function e + K(e) for e > 0 such that for every {w + H~} E W there exists agE 11 with IlglI"" "" K(e) such that Ilq(w  g)II L 1 /HA < e(q: L 1 + L 1 /H~ denotes the quotient map). COROLLARY
Hint. Use Theorem 7.1 and the uniform integrability criterion for weak compactness of subsets of L 1. 0
8. Weakly Compact Operators from A, L I /H~ and A* and Complemented Subspaces of These Spaces In this section we reap the benefits of Theorem 7.1. We establish further similarities between A and C(aD), and between L I and L I /H~. We show that A, A * and L I /H~ are weakly complete and have the DunfordPettis property. We also show that weakly compact operators from A into an arbitrary Banach space are characterized by their behavior on subspaces isomorphic to co' exactly in the same way as the weakly compact operators from a C(S)space. In the second part of the section, we deal with complemented subspaces of A I and LI/H~. We show, in particular, that LI/H~ does not contain any complemented subspace isomorphic to L I. At the end of the section we discuss v~rious related open problems. Recall that a Banach space X is weakly complete if every weak Cauchy sequence in X converges weakly to some element of X. We say that X has the DunfordPettis property if every weakly compact operator from X into arbitrary Banach space takes weak Cauchy sequences into norm convergent sequences. The proofs of the Corollaries 8.18.5 below are, "modulo Theorem 7.1", easy modifications of analogous results for C(S)spaces and L Ispaces. COROLLARY 8.1. (a) The spaces L I /H~ and A * are weakly complete. (b) The space L I /H~, A * and A have the DunfordPettis property. PROOF. (a) follows immediately from the equivalence of the conditions (2) and (3) of Theorem 7.1 (resp. Corollary 7.1). To prove (b) for L I /H ~, resp. for A *, observe that the equivalence of conditions (1) and (3) of Theorem 7.1 (resp. Corollary 7.1) implies that every weak Cauchy sequence in L I /H~ (resp. in C(aD)*) is the image under a quotient map of a weak Cauchy sequence in L I (resp. in C(aD)*). Now we use the fact that L I (resp.
[C(aD)) *) has the DunfordPettis property (cf. [DSI, Chapter VI)). The assertion (b) for A follows from Grothendieck's observation that if X* has the DunfordPettis property then X does (cf. [Gr2), [P9]). 0 COROLLARY 8.2. If E is one of the spaces A, A*, LI /H~ and T: E + E a weakly compact operator, then T2 is compact. 0 COROLLARY 8.3. Let Y be a Banach space and let T: A + Y be a bounded linear operator. Then the following conditions are equivalent: (i) T is weakly compact. (ii) T takes weak Cauchy sequences into convergent sequences.
50
BA.NACH SPACES
51
(iii) T restricted to any isomorph of Co is not invertible. (iv) T takes every W.U.s. sequence in A into a sequence which converges to zero. PROOF. (i)
'* (ii).
Apply Corollary 8.1 (b) for A.
(ii) '* (iii). Obvious, because there are weak Cauchy sequences in Co which do not converge in norm. (iii) '* (iv). Note that a bounded linear operator takes W.U.s. sequences into w.u.s. ones and use the result of [B·P] which says that if (xn) is a w.u.s. sequence in a Banach space X which does not converge to zero in norm, then there is an isomorphic embedding from Co into X which takes the unit vectors of Co onto a subsequence of the sequence (xn)' (iv) '* (i). Observe that if T satisfies (iv), then W = T*(B A) satisfies the condition (4a) of Corollary 7.1. Hence, by this corollary, the weak closure of W in C(3D)* is weakly compact. Hence T* is weakly compact and therefore T is weakly compact (cf. [DSI, Chapter
VI]).
0
COROLLARY 8.4. Let W be a bounded subset of A * (resp. L 1 /H~) whose weak closure is not weakly compact. Then there is a sequence (x~) of elements of W which is equiv
alent to the unit vector basis of {1 and spans a complemented subspace of A * (resp. L 1 /H~). More precisely, there are operators T: {1 + [C(3D)] * (resp. T: 11 + L 1) and S: A * + {1 (resp. S: L1 /H~ + {1) such that SqT = id/ 1 and qTe n = x~ for n = 1,2, ... where q: [C(3D)] * + A * (resp. q: L 1 + L 1 /H~) denotes the restriction (the quotient) map and en denotes the nth unit vector of Z1 . PROOF. The assumption on W combined with Corollary 7.1 yields the existence of a W.U.s. sequence (IPs) in A and a sequence (y:) in W such that ly:(IPs) 1 ~ 1 for s
= 1,2, ....
Let M
~ max (sup IlY:II,' sup L s lIy·II=1 s
IY*('Ps)I).
< +00. Passing to a subsequence, if necessary, we may also assume that ~~=s+ 1 IY:(IPr) 1< 2 s 3 for s = 1, 2, .. , . Next we define inductively a sequence of the
Clearly M
indices (k(n));=o and a sequence of infinite subsets of the integers (Nn);=o such that No  the set of all positive integers, ko = 0, (8.1)
(n
(8.2)
if j E N n , then j > k(n) and 1xt(IPk(n») 1 < 2 n
= 1,2, ... ), 3
(n=1,2, ... ).
Suppose that for some q ~ 0 and for 0 <: r <: q the indices k(r) and the infinite sets N(r) have been constructed to satisfy (8.1) and (8.2). Fix an integer m > M2 q +4 and let Aq be a set of distinct m indices in N q which are> k(q). Then for each j
= 1, 2,
. .. there is an
index tV) E Aq such that Ixt(IPrU»)1 < 2 q4, because otherwise ~~= 1Ixt(IPt)1 ~ ~tEAq Ixt(IPr)1 > M which leads to a contradiction. Since N q is an infinite set and Aq is a finite one, we infer that there is an infinite subset Nq+ 1 of the set N q n {k(q) + 1, k(q) + 2, ... } and an index k(q + 1) E Aq such that if j E Nq+ 1> then Ixt(IPk(q + 1»)1 < q 4 .
r
This completes the induction.
ALEKSANDER PH.CZyNSKI
52
Now put 1/I n =
n»
every finite sequence of scalars c l , c 2 ,
. •• ,
cm (m = 1,2, ... )
(1+ *)k~1 Icjl ~ Ilj~1 cjS(xf) II ~ (1  *)j~ ICj I. Applying a standard stability argument (cf. [BP)) we construct an isomorphism V: II ~ such that VS(x!) ,,; en for n = 1,2, . . . . Finally define T: [I ~ [C(aD)] * by T(e n ) = T(X~) for n = 1, 2, ... where T: A * ~ [C(aD)] * is the nearest point crosssection. The proof for L I /H~ is the same. 0 Next we shall discuss properties of complemented subspaces of A, LI /H~ and H*. Our first corollary generalizes results for C(S) and L I (v) spaces (cf. [PI] and [R2]); it shows in particular that A, L I /H~ and A * do not have complemented infinite dimensional reflexive subspaces.
II
COROLLARY 8.5. Let X be an infinite dimensional Banach space. (a) If X is complemented in A, then X contains an isomorph of co. (b) If X is complemented in A and X* is not separable, then X contains an isomorph ofC(aD). (c) If X is complemented either in A * or in L I /H~, then X contains an isomorph of
(d) If X is a separable dual and is complemented either in A * or in L I /H~ then weak and norm convergence of sequences in X coincide. (e) If X has a GL l.u.st and X is complemented in LI /H~ then weak and norm convergence of sequences in X coincide. PROOF. (a) Apply Corollaries 8.l(b) and 8.2. (c) Combine Corollary 8.l(b) with Corollaries 8.2 and 8.4. (d) Use Corollary 8.I(b) combined with a simple observation that a complemented subspace of a Banach space with the DunfordPettis property has the DunfordPettis property, and apply a result of Grothendieck (cf. [Gr2] and [P4]) that if a separable dual space has the DunfordPettis property, then weak and norm convergence of sequences coincide in the space. PROOF OF (b). Let Y be a complementary subspace to X in A. Since A/Y is isomorphic to X, the hypothesis implies that (A/Y)* is not separable. Thus (C(aD)/Y)* is not separable and, by a result of Rosenthal [R2] , C(aD)/Y contains a subspace isomorphic to C(aD). By a result of [LP2] (cf. also [R2]), this implies that A/Y contains a subspace isomorphic to C(aD) because, by the F. and M. Riesz theorem the space (C(aD)/y)/(A/Y) has a separable dual. This completes the proof because A/Y is isomorphic to X. REMARK. In fact (b) is true for every Banach space which is isomorphic to a subspace of a C(S)space with separable annihilator, provided S is an uncountable compact metric space. PROOF OF (e). The argument is similar to that of Theorem 4.2(v). We consider the diagram
BA.NACH SPACES
53
where P: LI /H~ ~ X is a projection, q: LI + LI /H~ is the quotient map, and i*:
L 1 /H~ is defined by i*( {[ + Ao})
= {[ + H~}
for [E C(aD). To define oo operators Sand T we first note that (i*)* = iM~ : H + HI is an absolutely summing operator. Hence (Pi*)* is absolutely summing with values in the separable dual space HI. Since X has GL l.u.st, Theorem 4.1 combined with Remark R.IIl in §4 yields that itp* is L Ifactorable. Hence (Pi*)* = VU for some L 1 (v)space and operators U: X* + L 1 (v) and V: Ll(v) + HI. We now define C(K) to be the dual of LI(v), S to be the restriction of V* to C(aD)/A o [identified with its canonical image in (C(aD)/A o )**] and we put T = QU*, where Q is a projection from X** onto X. Note that L I /H~ is complemented in the dual Banach space A * and therefore every complemented subspace of L I /H~ is complemented in the second dual of the subspace. Next we observe that Pi* = TS is a compact operator. Indeed note that T: C(S) + X takes weakly Cauchy sequences into convergent sequences because C(S) has the DunfordPettis property (cL [DSI, Chapter VI)) and, by a result of [PI], T is weakly compact. (The range of T, being contained in the weakly complete space L 1 /H~, does not contain subspaces isomorphic to co.) Furthermore every sequence in ~e unit ball BC(aD)/Ao has a weak Cauchy subsequence because (C(aD)/A o )* = HI is separable. Hence TS(BC(aD)/Ao) is a totally bounded subset of X. Now suppose that (x n ) is a sequence which is weakly convergent to zero in X. By Theorem 7.1, there is a sequence (fn) in LI which consists of uniformly bounded and uniC(aD)/A o +
formly integrable functions, and q(fn) = xn for n = 1, 2, . " . The properties of (fn) combined with the density in L 1 of all the continuous functions yield that for every € > 0 there is an M = M(€) < +00 and a sequence (gn) in C(aD) such that Iign lioo ~ M and Iign  [nlll < € for n = 1,2, . . . . Therefore the elements of the sequence (Pi*( {gn form a totally bounded subset of X because Pi* is compact and sUPnll{gn + Ao}ll ~ sUPnlignll ~ M. On the other hand we have, for n = 1,2, ... ,
11Pi*( {gn + Ao})  xn II
= lIP( {gn + H~ }) 
+ Ao}))
Pq(fn)1I
Since € > 0 has been chosen arbitrarily, we infer that the set Un {x n} is totally bounded in X. Hence the weakly convergent sequence (xn) converges in norm. 0
ALEKSANDER PEl.CZyNSKI
54
Our next corollary which is an immediate consequence of Corollary 8.5 shows that L 1 does not have "the strong primary property" analogous to the property of C( [0, I]) discovered in [LP2]. Note that H~, being separable dual, does not contain isomorphs of Ll. COROLLARY
8.6. The space L 1 /H~ does not contain a complemented isomorph 01 L 1.
Problem 8.1. Does there exist a subspace of L 1 /H~ which is isomorphic to L I? Notes and remarks to §§7 and 8. In 1967 Piranian, Shields and Wells [PSW] raised the question whether L 1 /H~ is weakly complete in the following form. Given a bounded sequence (gn) in L 1 such that the limit limnfaDlgn dm = (f) exists for every 1 E H~, does there exist agE L 1 such that (f) = f oDlg dm for 1 E H~? Kahane [Kh2] observed that there is a g in question such that <1>(1) = f oDlg dm for 1 EA. The weak completeness of L 1 /H~ and therefore of A * (our Corollary 8.1(a)) was proved by Mooney [Moo] and independently by Havin [Hvl]. Applying the result of [AL] on peak sets in H~, Amar [A] adopted the method of [Kh2] to get an elegant proof of the MooneyHavin result. Chaumat [Cbm] proved that L 1 /H~ and therefore A * and A have the DunfordPettis property. A \different proof of this result and the "liftability" of weakly compact sets in L 1 /H~ to weakly compact sets in L 1 was obtained by Cnop and Delbaen [CO]. Chaumat's approach generalizes to the case of a predual of a uniform algebra in which an abstract analogue of the AmarLederer lemma (our Proposition 6.2) holds, while the Delbaen and Cnop method generalizes to uniform algebras satisfying some conditions imposed on sets of representing measures (cf. [CO] and [02]). Theorem 7.1 in the present form and most of the corollaries in §8 have been obtained independently by Delbaen [01] and Kisliakov [IGs2]. Wojtaszczyk [W2] recently generalized Theorem 7.1 to certain planar uniform algebras. The proof of Theorem 7.1 presented in the text follows with some modifications the argument in [Kis2]. The construction in Case II goes back to Chaumat [Chm] ; our argument is different. Corollary 8.5(e) and Corollary 8.6 are due to W. B. Johnson and are published here with his permission. Several questions remain open. Very little is known about the nature of the dual of H~. Kisliakov in [IGs2] claims that the assertion of Theorem 7.1 is also true in the case ot H~. Unfortunately his argument contains a gap which he has recognized and communicated to me. Let us consider the following properties of a Banach space X: (a) the DunfordPettis property of X, (b) the DunfordPettis property of the dual of X, (c) the weak completeness of the dual of X, (d) a set We X* is weakly compact if it is weakly closed and sUPW*EW1w*(fn)1 = 0 for every w.u.s. sequence (in) in X, (e) every sequence in X* which converges in the a(X*, X)topology converges weakly, (f) every complemented subspace of X is either finite dimensional or contains l~. Problem 8.2. Does H~ have the properties (a)(f)? Problem 8.3. Let S be a compact metric space and let X be a subspace of C(S) with
BANACH SP\CES
55
a norm separable annihilator. Does X have the properties (a)(d)? How about the special case in which it is assumed in addition that X is a uniform algebra on S? Problem 8.4. Does the ndisc algebra A(Dn) (resp. the nball A(Bn)) have the properties (a)(d) for n ~ 2? (For the definition of A(Dn) and A(Bn) see § 11.) Problem 8.S. What translation invariant subspaces of C(3Dn) have some of the properties (a)(d)? Our next problem concerns complemented subspaces of A and L 1 /H~. Problem 8.6. (a) Let E be an infinite dimensional complemented subspace of A. Assume that E has an unconditional basis (resp. E has a GL l.u.st). Is E isomorphic to Co (resp. is E isomorphic to a complemented subspace of C(3D))? (b) Let E be an infinite dimensional complemented subspace of L1/H~. Assume that E has one of the following properties: (1) E has an unconditional basis, (2) E has GL l.u.st, (3) E is a separable dual, (4) E has the RadonNikodym property. Is E isomorphic to 11? Our last problem is closely related to the fact that every weakly compact set in L 1 /H~ can be lifted to a weakly compact set in L 1 (cf. Theorem 7.1) .. It is also related to Problem 3.1. Problem 8.7. Let E be a reflexive subspace of L 1 /H~. Does there exist a reflexive subspace E 1 of L 1 such that the quotient map q: L 1 ~ L 1 / H ~ restricted to E 1 is an isomorphism from E 1 onto E? How about the special case when E is isomorphic to a Hilbert space? ADDED IN PROOF. 10 Delbaen [De3] generalized Theorem 7.1 for uniform algebras such that every linear multiplicative functional has a weakly cqmpact set of representing measures and there is no nontrivial measure which is Singular with respect to all the representing measures. 2 0 Kisliakov announces in [Kis4] that if E is a reflexive subspace of A * then there exists a bounded linear operator T: A * ~ A * such that TIE is an isomorphism and there exists an E1 C [C(3D)] * such that the quotient map maps isomorphically E1 onto E. 3 0 The answers on Problems 8.6(b)(3) and 8.6(b)(4) are negative (Delbaen private communication).
9. Complementation of Finite Dimensional Subspaces in A, L 1/H ~ and I I The technique of Theorem 7.1 provides no information on the degree of complementation of finite dimensional subspaces of A (resp. L 1 /H~). For instance Corollary 8.5(a) obviously implies that infinite dimensional hilbertian subspaces are un complemented in A but it yields no estimate of the growth of the norm of the best projection onto ndimensional Hilbert subspaces of A. Our approach to handle the above problem and similar questions concerning the "local" structure of A and L 1 /H~ is based upon Theorem 2.4. This theorem and the duality relation between A, L 1 /H~, and H"" are the only facts on spaces of analytic I functions which are used in this section, the rest is a good example of "purely formal" technique of the local theory of Banach spaces and the theory of absolutely summing operators and related Banach ideals. We shall show (cf. Theorems 9.1 and 9.2) that A, L 1 /H~, H"" do not contain nicely complemented l~'s for I < p < 00; A and H"" do not contain nicely complemented l~'s, equivalently L 1 /H~ does not contain uniformly (cf. Theorem 4.2(iv) for definition). We begin with introducing two concepts; the first one plays an important role in the local theory of Banach spaces, the second is a specific invariant suggested by Theorem 2.4. DEFINITION 9.1. Let X and Y be Banach spaces; a pair (S, T) of linear operators S: X * Y, T: Y * X with TS = id x is said to be a factorization of X through Y. The factorization constant of X through Y is the quantity
r;:
ry(X)
=
inf IISIlIITIl
1+00
if there are factorizations of X through Y, otherwise.
Here "inf" is extended over all factorization of X through Y. DEFINITION
9.2.
Let X be a Banach space and let I
< p < 00.
The ip 
7rP
ratio of
X is the quantity
where the supremum extends over all operators T from X into arbitrary Banach spaces with = 1. The concepts defined above are related between themselves and with projections. We have
7rp (T)
PROPOSITION
(a) kp(X)
9.1. Let
X. Yand Xl be Banach spaces and let I < p < 00. Then
< ry(X)kp(Y); 56
BANACH SP<\CES
(b) if X I is a subspace of Y and P: Y
~
57
X I a projection then
IIPII ;;. r y(X)h x I (X) provided min(ry(X), rXI(X)) < +00; (c) if X is finite dimensional and Xl is isomorphic to X then rX I (X) = d(X, Xl) where d(X, Xl) = inf {IISIlIiSIIlIS: X ~ Xl isomorphism} is the BanachMazur distance between X and X I ; (d) if X is finite dimensional then ry(X)
= ry •• (X).
PROOF. (a) Let us fix a linear operator U: X with 1rp(U)
=
E (E an arbitrary Banach space) I and let (S, T) be a factorization of X through Y. Then ipCU) ;::: ip(UTS)
< IISlIip(UT) < IISlIkpCy)1rpCUT)
< liS II II TllkpC Y)1rp (U). Thus
which yields the desired conclusion. We left to the reader a routine proof of (b) and (c). (d) Clearly ry(X) ;;. ry .. (X). To prove the reverse inequality fix e
> 0 and pick a
factorization (S, T) of X through y** so that IISIlIiTIl < ry .. (XXl + e). Let F = SeX). Next note that for a finite dimensional X the space [B(Y, X)] ** can be identified with the space B(Y**, X). Moreover the canonical embedding of Y into y** induces the canonical embedding of B(Y, X) into S(Y**, X)
= (B(Y,
X))**. A simple restatement of the Gold
stine theorem that canonical embedding of the unit ball of the space is weakstar dense in the unit ball of the second dual implies that there exists aTE B(Y, X) such that IITII
r,i=
<
(1 + e)IITil and T**Y** = Ty** for y** E F. Let T(y) = lyt(Y)X; for y E Y. Now we apply the LindenstraussRosenthallocal reflexivity principle [LR] in the improved form due to JohnsonRosenthalZippin [JRZ]. Hence there exists an isomorphic embedding U:
n
Y such that lIy**1I < II UY**11 < (1 + e)lIy**1I for y** E F and y**(y = y t(Uy**) for y** E F and for j = I, 2, . .. . Then (US, T) is a factorization of X through Y with
F
II USIlIiTIl
< (l + e)2I1SIIIITil < ry •• (XXI + e)3.
Letting e 
0 we get ry(X) <
ry • • (X). 0 We shall also need the following result on absolutely summing operators. PROPOSITION 9.2 (B. MAUREY [Maul]). Let X be a finite dimensional Banach space,
let p with 1 < p < 00 be fixed, and let k ;;. I. Then the following conditions are equivalent: (i) ip(T) < k p1rp(T) for every bounded linear operator T from X into arbitrary Banach space. (ii) For every space LP.·(v) with p* = p/(P  I) and every subspace Y of LP·(v) every operator u: Y  X extends to a linear operator U: LP·(v)  X with IIUII < kpllViI.
ALEKSANDER PELCZYNSKI
58
PROOF. (i) ~ (ii). Recall that if X is a finite dimensional Banach space and Z a reflexive one, then the space N(X, Z) of all nuclear operators from X into Z is reflexive and its dual is isometrically isomorphic to the space B(Z, X) of all bounded operators from Z into X; the duality is given by the trace
(U, V)
~
for U E N(X, Z), V E B(Z, X).
tr(UV)
The condition (ii) says that the restriction operator B(LP * (v), X) ~ B(Y, X) takes the ball in B(LP * (v), X) of radius kp and with the center at the origin onto the unit ball of B(Y, X). Hence a standard duality argument yields that (ii) is equivalent to the following condition: (ii)* n(V):S;;; kpn(JV) for every operator
v:
X
~
Y and every closed linear subspace
Y of every LP* (v)space. Here n( . ) denotes the nuclear norm of an operator and J: Y ~ LP * (v) denotes the inclusion map. We shall show that (i) ~ (ii)*. Pick V: X ~ Y with n(JV) = 1 and consider the best nuclear factorization diagram for JV, i.e., we write JV = Ti\S where T: X ~ [00 and S: [I ~ LP*(v) are operators of norm one and A: 1 ~ [I is the operator of ml.\ltipli00
cation by a sequence (A;) with A; ;;;;. 0 for j = 1, 2, . . . and 'i\ = 1. (The best nuclear factorization for JV does exist because X is finite dimensional and LP * (v) is reflexive.) Clearly A = Ap * Ap where Ap: [00 ~ [p and Ap *: [p ~ [I are the operators of multiplication by the sequences (A} Ip) and (A} Ip·) respectively. Let us consider the diagram
r
A p)
IP
Ap •
)
[I
U
S
Here Ep is the closure of ApS(X) in [p, and Wp
= ApS is regarded
as an operator from X
into Ep. Clearly TAp.(Ep) C Y. Let Wp' denote the restriction of TAp' to Ep regarded as an operator to Y. Obviously Wp is a pabsolutely summing operator with 1TpCWp) = ('i(A.I 1P f)l lp = 1. Hence, by (i), Wp is pintegral and ip(Wp) :s;;; k p1Tp(Wp ) = k p . By the Schwartz duality theorem (cL [Kw, Theorems 1 and 2]) TAp. is p*absolutely summing because (TA p .)* is p*. integral. Thus Wp.' being restriction of a p*absolutely summing operator, is p*absolutely summing and 1T p *(W p .) = 1T p .(TAp ') = ip ' [(TA p ')*] = ('iA;)I IP * = 1. Now we use a result of Persson and Pietsch [poP] or a result of Persson [Per] to conclude that V = Wp' Wp is integral and therefore nuclear (because X is finite dimensional), and
n(V)
= i(V) :s;;; 1T p *(Wp .)ip(W p) :s;;; k p .
(ii) ~ (i). By the Persson·Pietsch duality theory or [PP] , condition (i) is equivalent to the following
BANACH SPACES
59
(i)* npo(V) ~ kpnpo(jV) for every Banach space E and every linear operator V:E
x. Here npo( . ) denotes the p*nuclear norm of the operator and j: X ear isometric embedding. To show that (ii) '* (i)* fix e > 0 and an operator V: E consider the p*nuclear factorization diagram for jV
["'" a fixed lin
X with npo(jV)
= 1 and
Apo
E    .....V.,+l X ,;J;+l t°
where Ap 0: 1
00

IP ° is the multiplication operator by the sequence
Ol} /po) with Aj ~ 0
and 1 ~ "i:.Aj ~ 1 + e, and Sand T are operators of norms one. Let Epo be the closure of ~oS(X) in IPo, Vpo = ApoS regarded as an operator from E into Epo, and Tpo the restriction of T to E pO regarded as an operator to X. By (ii), there is an operator T po: IP 0  X which extends Tpo with IITpoll ~ kpllTpoll = k p . Thus V = TpoApoS and npo(V) ~ IISIlIITp.llnpo(Apo) ~kp(l + e)l/pO. Letting e  0, we get np.(V) ~ kp = kpn(V). Now we are ready to prove
0
THEOREM 9.1. If Yis one of the spaces A, A *, L I /H~, H oo then
nl / 2 'Yy(l~) ~ C 10g(n + 1)
for n
=
1,2, ...
where C is an absolute const~nt. Consequently if Xl is an ndimensional subspace of Y and P: tion, then IIPlI ~ C(n l / 2 /log(n + 1»· d(X I , I~)l.
Y~
Xl is a projec
PROOF. The second part of the theorem is an obvious consequence of the first one and Proposition 9.1(b) and (c). Next observe that by Proposition 9.1(a) and the standard duality relations between A, A *, L I /H~ and H"" (cf. §2) it is enough to consider the case Y = A only. Let CI • C2 , ••• denote positive constants independent of p and n. By Theorem 2.4, kp(A) ~ Cip for p ~ 2. Next we shall show (9.1)
kp(l~) ~ C2nl/2I/P
for p
> 2 and for n = 2, 3, ...
Assuming (9.1) and combining it with Proposition 9.2(a), we get k (12) "V (12) ~ ~ ~ C n l /2 1 / p • pl for p ~ 2. 'A n P kp(A) 3 Specifying p = (Iog(n + 1))1, we get 'YA (l~) ~ C4 n 1 / 2 /log(n + 1). To establish (9.1) it is enough, in view of Proposition 9.2, to show that there exists a
.\LEKSANDER PEl.CZyNSKI
60
constant Cs such that the condition (ii) of Proposition 9.1 is violated for X = l~ and for every constant kp < Csnl/21IP. By a recent result of Figiel, Undenstrauss and Milman [FLM] , there exists a constant C6 such that there exists an ndimensional subspace Y of and an isomorphism U: Y ~ I~ such that IIUII = 1 and IIU 1 11 0:;;;; 2 (note that p* = I) < 2). Let U: Ig: n ~ I~ be any extension of U. Then (U 1 , U) is a factorization
z'f;* n
p/& 
of I~ through l~:n. Let I: Ib 6 n ~ Ig: n be the natural embedding. It is easy to see that 11111 = 1 and IIrlll 0:;;;; (C6 n)I IP = C7 n 11P . Clearly (I 1U 1, UI) is a factorization of I~.
It is well known that for every Ll(lI) space, rLl(V)(I~);;;;' Can 1/2 . (This is for instance an easy consequence of Grothendieck's result that every bounded linear operator in 12 which is L Ifactorable is HilbertSchmidt (cf. [GRI] , [LPI] ).) Thus
Cs n 1/2 Hence IIUII;;;;' C9nI/21IP.
IIr 1U 111I1U· III
0:;;;;
0:;;;;
2n 11P IIUIi.
0
r A (l~) ;;;;. Cn 1/2? Observe that the positive answer to Problem 3.1 implies the positive answer to Problem 9.1. On the other hand Theorem 9.1 yields that operators on 12 which factor through A are "almost" of HilbertSchmidt type. We have Problem 9.1. Does there exist a C> 0 such that
COROLLARY 9.1. If an operator T: 12 ~ 12 factors through A, then T is compact
and if (Sj) is an enumeration in a non increasing sequence of nonzero eigenvalues of ITI, each eigenvalue repeated according to its multiplicity, then
L
s~
[log(j
~
1)) C
< +00
for every c> 3.
PROOF. Let (U, V) be a factorization of T through A. Since P is reflexive and A has the DunfordPettis property (cf. §8), T is compact. Next observe that the polar formula implies that an operator T factors through a Banach space iff its absolute value ITI factors through the same Banach space. Hence without loss of generality one may assume that T = ITI and that the sequence (Sj) is infinite (otherwise there is nothing to prove). Let (ej ) be the orthonormal sequence of eigenvectors of T corresponding to the sequence (Sj) of the eigenValues. Let us fix a positive integer n and denote by Pn the orthogonal projection from 12 onto the span (e j )l<;;;j<;;;n. Next define Sn: Pn{l2) ~Pn{l2) by Sn(e j ) = Sjl ej for
= 1,2, ... ,n and let Un be the restriction of U to Pn(l2). Clearly SnPn VUn Hence Theorem 9.1 yields
j
IISnllllPnllllVlIlIUnll ;;;;. C log(n
+ 1)
for n
=
= idPn(l2).
1,2, ... ;
since IIPnll = 1 and IlUnll 0:;;;; IIVlI, for C1 = C/IIUIIIIVIl we have IISnll ;;;;. C1 n 1/2 /10g(n + I) for n = 1,2, . . . . Using the relation Sl ;;;;. S2 ;;;;. ... ;;;;. sn' we easily obtain IISnl1 = S;;l. Thus
sn
0:;;;;
C~110g(n
+ 1)/n I/2 .
Hence
..
L n=l for every c > 3.
0
s2 [log(n
n
+
..
I)]C
0:;;;;
L
C 1 1 n=l
n[log(n
1
+ 1)]C
2
< +00
BANACH SPA..CES
61
Another consequence of Theorem 9.1 is COROLLAR Y 9.2. Let 1 < p < 00 and let p* = p/(P  I). Then there exist an absolute constant C > 0 and a number ex(p) with 0 < ex(p) <; \6 such that
I'A(l~)
= I'H",,(/~) = I'LI/H~{l~·);;' CnG:(p)
for n
= 1,2, ...
PROOF. Combine Theorem 9.1 with a result of Guran! Kadec and Macaev [GKM] that d(l~, I~) <; nII/2I/pl. 0 Next observe that Davis and Johnson [DJ] showed that if X is an infinite dimensional Banach space which does not contain I ~'s uniformly, then there is an ex with 0 <; ex <; \6 such that l' x(l~) <; CnG: for some absolute constant C. Combining this fact with Theorem 9.1 one can prove COROLLARY 9.3. If (Xn) is an increasing sequence of finite dimensional subspaces of
a space X with the above property, then lim I'A (Xn ) = lim l' ",,(Xn ) n
n
H
= +00.
Our next result extends Corollary 9.2 on the case p THEOREM 9.2. There is an ex with 0 that, for n = 1, 2, ... ,
CnG: <; I'A (l~)
=
1.
< ex <; \6 and an absolute constant C> 0 such
= 'YH""(/~) = inf{d(/;;',
£)1 E C LI /H~; dim E
= n}.
PROOF. It is enough to prove the lefthand inequality. The equalities follow from Proposition 9.1 (d) and the duality relations between A, H"" and L 1 /H ~ established in § 1. (Note that H"" is norm one complemented in A **.) By a result of Maurey and Pisier [MauPis] , to prove the inequality it is enough to show that limn l' A (l~)
= 00; hence
Theorem 2.4 and Proposition 9.1(a), it suffices to show that, for some p +00. Assume to the contrary that sUPnkp(l~)
= M < 00.
in view of
> 2, limnkpC/~) =
(Note that the sequence
(kp(l ~)n= 1,2,... is increasing.) Then, Proposition 9.2 would imply that for every subspace Y of LP*, every operator U: Y * I~ admits an extension V: LP· * 11 with /lV/I <; MlIU/I. Since L 1 contains an increasing sequence of subspaces isometrically isomorphic to I~'s whose union is dense in L 1 and since L 1 is complemented in its second dual, a standard compactness argument (for instance involving Banach limits; cf. § 5, proof of Proposition 5.2) would yield that every bounded operator from yeLP· into L 1 extends to a bounded operator from LP· into L 1. Hence to complete the proof it is enough to establish the following LEMMA 9.1. For some p
> 2 (in fact for every p > 2) there exists a bounded linear
operator from a subspace of LP· into L 1 which has no extension to a bounded linear operator from LP * into L 1 . PROOF. Let Y be a noncomplemented subspace of LP* which is isomorphic to 12 (for instance for every p
> 4 one can take
as Y a closed linear subspace of LP * spanned by
the characters {zn: n E Rud} where Rud is a A4 set constructed by Rudin, cf. [Ru2] and
ALEKSANDER PELCZYNSKI
62
[Rl] ; for 4 ;;. p > 2 the existence of Y in question follows from the recent result of five authors [BDGJN]). Let Eq (I ~ q
< 00) denote
the closed linear space in L q of the char
acters {z2 n : n = 1,2, ... }. It is well known (cf. [Z, Vol. I, Chapter V, §6]) that for every 1 ~ q 1
< q2 <
00
the natural embedding iq2Q 1: L q2 + L q 1 maps isomorphically Eq 2
onto Eq 1 (in other words all sets of functions E q coincide with E 2 and the norms
II IIq 2 and
II IIq 1 are equivalent on E 2 ). This implies easily For every q ;;. 1 and
€
> 0 there exists aD> 0
such that if e is a measurable subset of aD with
(9.2)
m(e) < 0 and if f E E2 with IIfll2 felfIPdm<€for 1 ~p~q.
= 1 then
Since both Y and El are isomorphic to 12 , there exists an isomorphism U: Y ~ E 1 . We shall show that U is the desired operator. Suppose to the contrary that there exists a bounded linear operator
U:
LP *
+
L 1 which extends U. Then, by a result of Maurey
[Mau2] , for every fixed r with p* > r > 1, there exists an operator V: LP * + L r and a multiplication operator Mg : Lr + L 1 defined, for some g E Lr*, by Mil)
= gf for f
E Lr
such that U = Mg V. Next fix q with r > q > 1 and using (9.2) pick 0 > 0 so small that if e is a measurable subset of aD with m(aD\e) < 0, then the operator Me: L S + L S of multiplication by the characteristic function Xe of e restricted to Es is an isomorphism for 1 ~ s ~ q*. Now we pick K > 0 so large that if e = {z E aD: Ig(z) I ~ K} then m(aD\e) < D. By a result of [KP] , the orthogonal projection Q: L q * + MAEq *) is bounded as an operator on Lq*; hence Q being orthogonal is also bounded as an operator from Lq into Me(Eq). Let ge = g . Xe' Clearly the operator Mge of multiplication by ge is bounded as an operator from L r into Lq. Since the restrictions to Eq of Me and of iq,l are isomorphisms, there is an isomorphism W: Me(Eq) ~ El such that WMe(f) = iq,1 (I) for f E Eq . Clearly WMge V(LP*) eEl and P: U 1 WMge V is a projection from LP· onto Y, a contradiction. 0 Before we pass to a discussion of possible strengthenings of Theorem 9.1 we would like to state a corollary of "global" character which is related to Problem 8.2. COROLLARY
9.4. Let X be a subspace oJ H oo satisfying one of the following proper
ties: (a) X is separable and contains an isomorph of co. (b) X does not contain I ~ uniformly (in particular X is superrejlexive), dim X = 00.
(c) X contains I~ uniformly complemented. Then X is uncomplemented in H oo • PROOF.
(a) Use Sobczyk's theorem [S] and a result of [BP] that isomorphs of Co are
uncomplemented in dual Banach spaces. (b) Use Theorem 9.1 and the result of [DJ] mentioned before Corollary 9.3. (c) follows from Theorem 9.2. REMARK.
Using slightly more carefully results of [BDGJN] and [BGN] one can
show that the exponents a in Theorem 9.2 can be made any number feel that the "real value" is a
=
< 1/8.
However, we
1/2.
Problem 9.2. (a) Does there exist a C> 0 such that 'YA(l~);;' Cn 1 / 2 for every n?
BANACH SPACES
63
(b) Does every operator from A into [I factor through [2? (c) Does L I /H~ have cotype 2, i.e., there is a C> 0 such thatJ~II~7=1 w;,/t)IIL I/HI dt ;;;. C(~f=lllwjllil/HA)I/2 for every wI' w 2 , . . . , wn in LI /H~ and for n = 1,2, ... ? 0 (Here (rj ) denotes the Rademacher functions.) It follows from a theorem of Maurey (cf. Tomczak Jaegermann [TJ]) that the positive answer on (c) implies the positive answer on (b); obviously the positive answer to (b) implies the positive answer on (a).
Recall that a scalar sequence (c n ) is an (A, [1) multiplier if there exists a (unique) bounded linear operator T(cnf A + [1 such that T(zn) = cno n for n = 0, 1, ... where on denotes the nth unit vector of [1. It is worthwhile to mention that the answer on Prob
lem 9 .2(b) is positive for (A, [1) multipliers. It is in fact a restatement of an old result of Paley [Pa2]. Precisely we have PROPOSITION 9.3 (PALEY). If (c n ) is an (A, [1)mu[tiplier then ~lcnl2
< +00; hence
T(c n ) factors through [2.
PROOF. The second part of the proposition is an obvious consequence of the first. We define U: A + [2 by U(f) = (f(n» for f E A and V: [2 + [1 by V«d n = (dnc n )
»
for (d n ) E [2. Then clearly T = Vu. To prove the first assertion denote by tional on
[1
(n
= 0,
the coset {cnz n
°
~
the nth unit vector of Co regarded as a func
°
1, ... ) and note that the adjoint operator T*: ["" + A * takes ~ into
+ Ll /H~}
for n
= 0,1,
... (remember that A*
= Ll /H~
ffi~ Vsing ).
Fix a positive integer N. By the Kolmogorov theorem (cf. §O.lI) there is a constant C> 0 such that for an arbitrary sequence (f) with
IIT*II ;;;.IIT*(.f 1=0
€j
= ± 1 (j =
€/jj) II = hEHl inf SaD o
;;;. C
(faD \jt
1, 2, ... , N) we have
I.i ClfZ j
+ h(z)
10
\m (dz)
j
c/jz \1/2 m(dZ») 2.
Averaging over all ± 1 sequences, i.e., integrating against the Rademacher functions and applying the Khinchine inequality for L 1/2 we get
C 1 1IT*II;;;'
;;;.
m(dz) dt SI(rJan \j~N ClP)z1.\1/2)2 0
(SIII ~ 0
aD
\N j~
= ( r JliN L JaD 0 j=O ;;;'C1 / 2 (
I
aD
(
LN
j=O.
Icjl
Cllt)zI.\1/2 m(dz}:1t)2
)2
Cllt)zI.\1/2 m(dz) dt
2)1/2 m(dz))2 =C (.LN 1/2
1=0
Icjl 2)1/2
ALEKSANDER PEt.CZyNSKI
64
where C1 / 2 is the constant in the Khinchine inequality. Letting N 
<
00.
00
we get ~;:olc/
0
A positive answer on our last problem will provide a "local" analogue of Corollary 8. Problem 9.3. Does there exist for k = 1, 2, ... and for C ~ 1 an integer n = n(k. C) such that if X is a subspace of A with dim X = n which is a range of a projection from A of norm";; C, then X contains a kdimensional subspace, say E with deE, 1';) < 2? Notes and remarks to §9. Most of the material of this section was obtained jointly by w. B. Johnson, G. Pisier and the author and is published here for the first time. Proposition 9.2 is due to Maurey [Maul]. The proof in the text is due to W. B. Johnson and G. Pisier and is published here with their permission. Proposition 9.3 is due to Paley [Pa2]. For other proofs see Helson [Hel] and Rudin [Ru3] _ The proof in the text is similar to Rudin's proof. ADDED IN PROOF. Every multiplier from L 1 /H~ into P is absolutely summing (cf. [KwP2]). This easily yields Proposition 7.3. It is unknown whether every bounded linear operator from L 1 /H~ into [2 is absolutely summing.
10. Bases and the Approximation Property in Some Spaces of Analytic Functions In this section we review results on the approximation property and bases in HP spaces (I ~ p ~ 00) and in the disc algebra. Most of the results are stated without proofs. We discuss several open problems. Recall that a Banach space X has the approximation property (shortly A.P.) if for every compact set K in X and € > 0 there exists a finite rank operator T: X  X with IIT(x)  xII < € for x E K; moreover, if there exists a A ~ 1 (which depends on X only but not on K) such that the operator T in question can be chosen with II Til ~ A, then X is said to have the Abounded approximation property (shortly the AB.A.P.). X has the bounded approximation property (shortly the B.A.P.) if X has the AB.A.P. for some A ~ 1. A sequence (x n ) of elements of X is called a basis for X if for every x E X there is a unique sequence of scalars (en) such that x = ~cnxn. The basis (x n ) is unconditional if the series ~cnxn converges unconditionally to X for every x EX. Each of the above properties is stronger than the preceding one, i.e., unconditional basis
=>
basis
=>
B.A.P.
=>
A.P.
The table below shows what is known about the approximation property and the existence of bases for "classical" spaces of analytic functions. A.P. IB.A.P. basis unconditional basis
If'''
?
?


A
+ + +
+ + +
+ + +
+ ?
HP (l HI
< p < 00)
L oo/H";
?
?

CIAo = qaD)IA o
+ + +
+ + +
+ + +
LP /H~ (I LI/HA
< p < 00)
?
+ 
We pass to a discussion of the results contained in the table. The nonexistence of bases in H oo and L IH'; is trivial because these spaces are unseparable. The nonexistence of unconditional bases in A and L I /H~ follows for instance from the results of §4 because those spaces do not have any GL I.u.st and the existence of an unconditional basis in a Banach space implies the existence of a GL I.u.st. All the separable spaces in the table have IB.A.P. To this end first recall that a separable Banach space X has the AB.A.P. iff there is a sequence Tn: X  X of finite rank operators such that II Tn (x) xll 0 for every x E X and sUPnllTnll ~ A. For fELl we put 00
65
.\LEKSANDER PH.CZyNSKI
66
1
Tn (f)
= n
nl
"
(n
L..
= 1,2, ... ).
k=n+l
It is easy to see that the Tn's take HP into HP for I ~ p ~ 00, A into A, and induce finite
rank operators on LP IHg (I ~ p ~ 00) and CIAo. The Tn's regarded as operators in each of the above spaces are of norm one. By the Fejer theorem (cf. [8, p. 23]) IITn(f)  flip 
0
for fE HP;
IlqTn(f)  q(f)II LP /Hg 
0
for f E LP;
Here q: LP 
LP IHg (resp. q: qaD) 
IITn(f)  flloo IIqTn(f)  q(f)IICfA 0
for fE A,
0
for fE qaD).
CIAo) denotes the quotient map. Thus
10.1. Each of the spaces 11', LP /Hg (l ~ p
PROPOSITION

0
< 00), A, CIAo
has IB.A.P.
By a result of Boas [Bo] (cf. also [KwP] and §O.Il), the spaces HP and LP /Hg are isomorphic to LP for I < p < 00. By a result of Marcinkiewicz and Paley (cf., e.g., Burkholder and Gundi [BG)), the Haar system is an unconditional basis for LP (l < p < 00). Thus
10.2. If I
PROPOSITION
< p < 00,
then the spaces HP and LP IHg have unconditional
bases. The situation in the "limit cases" of HI, A, L I IH~ and CIAo is much more delicate. The existence of bases in these cases is essentially due to Billard [Bi] and Bockarev [Bt]. We shall briefly describe their approach. Let (gj) be an orthonormal system in the real space L~ [0, 27T] with go == 1/..[I1r. We define a new orthonormal system consisting of 27Tperiodic real even functions in (""11',7T) by
L1
g~(t) 1
= ~ gj(2t) g,.(2t)
for t E [0, 7T] ,
(j
for t E [7T, 0] ,
= 0,
I, ... )_
Next we put g# 
1
0
+i
g,~ =
vfS'
_1_
V2
(g~ + iH(g~)) 1
/
(j
= 1,2, ... )
where H(g~)(t) ,
=
lim eHO
1  7T
f.7r e
gj(t
+ s)  gj(t  s) 2 tg(sI2)
ds
is the Hilbert transform of gj. It is easy to see that (gt) is an orthonormal sequence in the complex space L 2 (7T, 7T), in fact in H2. (We identify H2 with the subspace of L 2 (7T, 7T) spanned by the characters (eint)n;;.o). Moreover, (gf) is an orthonormal basis for H2 if (gj) is a complete orthonormal
1.
system in L Starting with "nice" orthonormal systems in L ~ one can construct in that way orthonormal systems which form bases for HP (00 > P ~ l) and for A. In particular we have
BANACH SPACES PROPOSITION lO.3.(BILLARD c[Bi]).
67
1
Let (hn):=o be the Haar system in L [0, 21T] ,
i.e.,
~ ~ t < (2r + 1) ~,
(21T)1/2,2 k / 2 for 2r
2k + 1
(21T)1/22 k / 2 for (2r
2k + 1
+ 1) ~ ~ t < 2(r + 1) ~, 2k +1
o
2k + 1
otherwise, (r
= 0,
1, ... , 2k  1; k
= 0,
1, .. , ).
Then (h!):=o is a basis for HI . Let (fn) be the Franklin system, i.e., the orthonormal system in
L1 [0, 2'lT]
which is
obtained by the GramSchmidt orthogonalization of the sequence 1, J~ho(s) ds, J~hl (s) ds, where (hn):=o is the Haar system. Then we have PROPOSITION
10.4 (BOCKAREV [Bt]).
The system (f!) is a basis for A.
Here we identify A with a subspace of C [1T, 1T] spanned by the characters (eint)n?>o. A few words on the proof of Proposition 10.4. Since each fn is a piecewise linear function, H(f~) is continuous and therefore ff: E A for n = 0, 1, . .. . Next recall that the Franklin system is a basis for CR 10, 21T] (for an elegant proof of this fact cf. [Cl]) and therefore the ~equence (j~) is a basis for the subspace of (10.1 ) CR [1T, 1T] consisting of all the even functions. The crucial point of the proof is to establish the inequality (10.2)
IIHCto f~}f~ )II~ ~K(llfll~ + IIH(f)II~)
for every even fE CR [1T, 1T] and for N
= 1,2,
...
where K is an absolute constant. The proof of (10.2) is difficult. It uses a delicate real variable technique involving an analysis of the behavior of the Hilbert transform of piecewise linear functions with nodes in the "first 2k
+ r"
dyadic points of the interval [1T, 1T] and
some results of Ciesielski [C2]. Next it is convenient to introduce the real space AR
=
{(t, g) E CR [1T, 1T] x CR [IT, 1T] : g
= H(f)}.
Using (10.1) and (10.2) we easily check that the sequence (f~, H(f~)) is a basis for the closed linear subspace of A R which it spans. This subspace consists of all the pairs (f, g) E
AR with f even. (The last fact is not obvious but not hard to check.) Thus for each fE A with Re(f) even there is a real sequence (c/f))j?> 0 such that
ALEKSANDER PEl,CZYNSKI
68
f=
{I 0.3)
L clf)ff
;==0
(the series converges uniformly on [1T, 1T], i.e., in the norm topology of A). Hence if Ref is an odd function, then 00
f
=
L
ic;(if)fr
;==0
because if Re(f) is odd then Im(f) is even. Finally using the identity f(t)
= ~ [(f(t) + f(t)) + (f(t) 
fef»~]
we infer that for every f E A there is a sequence of complex numbers (Cj(f»r~o such that (10.3) holds. The sequence (cj(f)j<;.o is unique because the sequence (fr) is orthonormal. The proof of Proposition 10.3, being similar to that of Proposition 10.4, is even technically simpler. There are several consequences of Propositions 10.3 and 10.4. First we have COROLLARY 10.1. The spaces CIAo and L I IH~ have bases. PROOF. The space CIAo is a predual of HI (cf. § 1). Therefore the existence of a basis in CIAo follows from Proposition 10.3 and the result of Johnson, Rosenthal and Zippin [JRZ] that if X* has a basis then X does. A strengthening of this general result can be also used to show that L I /H~ has a basis. There exists however a less formal argument. The coefficient functionals of the Bockarev basis (fr)j;' 0 form a basis for the subspace of A * which they span. Since (fr)j;'o is an orthonormal system, the jth coefficient functionals can be identified with the coset {1l + H~} for j = 0,1, ... (under the natural identification of L I /H~ with a subspace of A *; cf. § 1). To complete the proof it is enough to show that the closed linear span E of the cosets ({1r + H~} )j~O coincides with LIIH~. To this end, fix n ~ 0 and let e int = r:;oc/fl(f) uniformly on [1T, 1T]. Then e int uniformly on [1T,1T] and therefore also in LInorm. Hence {e int + H~} =
= r:;oCjl!(f)
r:;oc/ if! + H~} in the norm topology of L I IH~.
This shows that each coset {e int + H~} belongs to E for n = 0, 1, .... Combining the last fact with the Fejer theorem we get the desired conclusion. 0 The space A(Dn) of all continuous functions on the ndisc analytic in its interior can be regarded as the n weak tensor product power of the disc algebra. By a result of Gelbaum and Gil de Lam ad rid [GGL] , the weak tensor product of Banach spaces with bases has a basis. Hence CoROLLARY 102. The space A(Dn) has a basis n = 1,2, .... Very little is known on the existence of bases in the spaces A(U) for other bounded closed domains of holomorphy U in (cf. § 11 for the. definition). In particular we have Problem 10.1. Does the spaceA(Bn ) have a basis? Here Bn = {A = (Zj) E C": 1:IZj 12 :e;;; I}. Let us observe that if U is a bounded closed
en
BANACH SP!\CES
69
circled domain in en then A(U) has the IB.A.P. For, define Tn: A(U) ~ A(U) to be the nth Fejer sum operator, i.e.,
Tn(f) =
f
n n+ ~ ; j [In]
for [EA(U) (n = 0,1, ... )
;=0
where [I n] denotes the homogeneous part of degree n of the Taylor expansion of [at zero. Next we turn to a discussion of the interrogation marks in the table. First we state Problem 10.2. (a) Does the space CIAo have an unconditional basis?
(b) Does HI have an unconditional basis? (c) Is H I isomorphic to a subspace of a space with an unconditional basis? (d) Does HI have any local unconditional structure? Obviously the positive answer to (a) implies the positive answer to (b). Furthermore, it has been observed in [KwP] and independently by Burkholder (unpublished) that the Billard basis (ht) for HI is not unconditional. We do not know whether the space HI is isomorphic to the space "martingale HI"  Mart HI where the last one is defined to be the space of all the functions [in L I whose FourierHaar expansion "'if=oc;(f)f; converges unconditionally in L I ; we admit sup 1e;1=1
If
;=0
E;C;(f)!j
III· L
Finally in connection with (b) and (d) it is interesting to mention that no example of a subspace of L I without any local unconditional structure is known. Next we pass to the Problem 10.3. Does H"" have the approximation property (resp. the bounded approxi
mation property)? The same problem is open for L"" IH';. The spaces H"", L "" IH'; and the space of all the bounded linear operators on an infinite dimensional Hilbert space are the most common Banach spaces naturally appearing in analysis for which the approximation property is unknown. All those spaces are nonseparable. However Problem 10.3 can be reduced to some questions on separable function algebras. It is known (cf., e.g., [J]) that a Banach space X has A.P. (resp. AB.A.P. for some A ~ 1) iff every separable subspace of X is contained in another separable subspace of X with A.P. (resp. with AB.A.P. for the same A). On the other hand, by a recent result of Marshall [M] the Blaschke products are linearly dense in H"". Thus the positive answer on Problem 10.3 would follow from the positive answer on
the following Problem 10.4. Does every closed sub algebra of H"" generated by a countable set of
Blaschke products have A.P. (resp. AB.A.P. for some absolute constant A)? We do not know the answer to Problem 10.4 even in the case of algebras generated by finitely many Blaschke products. Observe that, by a result of Milne [Mi] , there is a separable uniform algebra which fails to have the A.P. Another way of reducing Problem 10.3 to a question on the disc algebra involves the concept of the uniform approximation property introduced in [PR]. Recall that a Banach
\LEKSANDER PELCZYNSKI
70
space X has the uniform approximation property (V.A.P. for short) if there exists a A < 00 and a function k + N(k) (k = 1, 2, ... ) such that for every subset F C X of cardinality k there is an operator T: X
+
X such that T(f)
=f
for
f
E
F, II Til
~
A, dim T(X) ~ N(k).
Obviously the V.A.P. with a constant A implies AB.A.P. Furthermore, a complemented subspace of a Banach space with V.A.P. has the V.A.P. Lindenstrauss and Tzafriri [L TI] proved that a Banach space X has the V.A.P. iff X** does. From the results of § 1 it follows that 11 is a complemented subspace of A **. Thus Problem 10.5. Does the disc algebra A have the uniform approximation property? ADDED IN PROOF. Every finite dimensional in L 1 /H~ of norm one is one dimensional;
there is no sequence in A and L 1 /H~ finite dimensional projections of norm one which tends strongly to the identity operator; in fact the range of every finite dimensional projection in A of norm one is isometrically isomorphic to
t;; (cf. [W3].
11. The Polydisc Algebra and the nBaU Algebra, and Their Duals Let V be a bounded closed domain of holomorphy in the ndimensional complex vector space en, let av denote the boundary of U. By A(U) we denote the Banach space of all continuous complex valued functions on V which are holomorphic in V\aVthe interior of
U; the norm in A(V) is defined by IIfll = suPAEulf(A)1 = sUPAEa ulf(A)1 for f E A(V). In this section we shall primarily deal with the following domains: the ndisc, D n = {A = (Zj) E en: maxI <;;j<;;n IZjl ~ I}; the nball, B n = {A = (Zj) E en: ~j=llzl ~ I}. Clearly DI = B I = D is the unit disc. The spaces A(D n ) and A(Bn) are called the ndisc algebra and the nball algebra respectively. They are the most natural analogue of the disc algebra among the spaces of analytic functions of several complex variables. Very little is known about the linear topological classification of the spaces A(V). Clearly if there is a biholomorphic map from the interior of a domain VI onto the interior of a domain U2 which extends to a homeomorphism from VI onto V 2 , then the spaces A(VI ) and A(V2 ) are isomorphic (= linearly homeomorphic). This condition is not necessary as the example shows: VI = Dthe unit disc in e l ; V 2 = D U {z E e l : Iz  31 ~ I} the union of two disjoint discs. A reasonable conjecture says that the topological dimension of V, i.e., the number of variables, is a linear topological invariant of A(U). More precisely, if VI and V 2 are bounded closed domains of holomorphy in en and em respectively and if n =1= m, then the Banach spaces A(V I ) and A(V2 ) are not isomorphic. On the other hand a beautiful result due to Henkin [He2], [He3] shows that the spaces A(V) of the same number of variables may not be isomorphic; in particular, the ndisc algebra is not isomorphic as a Banach space to the nball algebra whenever n ;;;. 2 (cf. Theorem 11.1 below). A concrete open question related to the "dimension conjecture" mentioned above is the following Problem 11.1. Are the spaces A(Dn) and A(Dm ) (resp. A(Bn) and A(Bm)) not isomorphic whenever n =1= m? The next theorem summarizes what is known in the linear topological classification of the Banach spaces A(Dn) and A(Bm). THEOREM 11.1. (a) (Henkin [HeI]' [He2l). If n then A(Bn) is not isomorphic to A(Dm).
71
=
1,2, ... and m
= 2,
3, ...
ALEKSANDER PEI:.CZyNSKI
72
(b) (MitjaginPelczynski [MtP]). The disc algebra A phic to A(Bn) for n ~ 2.
= A(Dl) = A(B l ) is not isomor
The isomorphic invariants which enable us to obtain Theorem 11.1 are expressed in terms of the duals of the spaces A(Dn ) and A(Bm). and they have been introduced in § I. We recall those invariants. Let X be a separable Banach space. We say that X has Property I if it is isomorphic to a subspace of a C(S)space whose annihilator in [C(S)] * is norm separable; X has Property II if X* is a separable distortion of L 1 (v), i.e., X* is isomorphic to the Cartesian product M ffi V where M is a separable Banach space and V is an L 1 (v)space. Clearly the possessing by a Banach space X of Property I (resp. Property II) is an isomorphic invariant of X. Hence our Theorem 11.1 is an immediate consequence of the following. THEOREM 11.2. We have
A
= A(Dl) = A(B l )
Property I
Property II
Yes
Yes
A(Bn) (n
~
2)
No
Yes
A(Dn) (n
~
2)
No
No
Most of this section is devoted to the proof of Theorem 11.2. We begin with the "yes" entries of the table. The first row follows easily from the F. and M. Riesz theorem (cL § 1). The last "yes" in the second row is a consequence of Henkin's generalization of the F. and M. Riesz theorem which we present next. We begin with introducing the concept of an analytic measure (due to Henkin) which plays now an important role in the theory of analytic functions of several complex variables. Let U be a closed bounded domain of holomorphy in Cn . A sequence ~) in A(U) is a Montel sequence provided sup;lIfjll < 00 and lim;fj(A) = 0 for every A in the interior of U. DEFINITION 11.1. An analytic measure on aD is a complex Borel measure p. on au such that lim;foufjdp. = 0 for every Montel sequence (fj) in A(U). We shall denote by AM(U) the set of all the analytic measures on au. Clearly we have PROPOSITION 11.1. (i) AM(U) is a norm closed subspace of [C(au)] *. (ii) [A(U)] 1 C AM(U). (iii) If J.1. E AM(U) and f E A(U) then fp. E AM(U). (iv) If v is a representing measure for a point A E U\au, i.e., fa uf dv = f(A) for f E A(U), then v E AM(U). 0 The evaluation at a point AE U\au regarded as an element of [A(U)]
'Pr.
noted by Now we are ready to state Henkin's result.
* will be de
BANACH SPACES
73
THEOREM 11.3. (a) If II E [C(3Bn)] * and II « iJ.Li with J.L E AM(3Bn ), then II E AM(3Bn )· (b) The dual [A(Bn)] * is isometrically isomorphic to the product AM(Bn)/[A(Bn)] 1 $1 V where V is the L Ispace consisting of all measures in [C(3Bn)] * which are singular with respect to all positive analytic measures on 3Bn . (c) The space AM(Bn)/[A(B n)] 1 is separable and is isometrically isomorphic to the closed linear span of the IP~ for A E Bn \3Bn" PROOF. The difficult part of the Theorem is (a). Assuming that (a) has been established we deduce (b) and (c) as follows. Let V = {v E [C(3Bn)] *. v 1 iJ.Li for every J.L E AM(3B n)}. Then V is a band (= a complex sub lattice ) in [C(3B n)] *. Hence V is an L 1_ space. Furthermore by (a) and the Lebesgue decomposition theorem
Factoring through the annihilator [A(Bn)] 1 and taking into account that the dual of [A(Bn)] * is naturally isometrically isomorphic (via the restriction map) with the quotient
[C(3Bn)] */[A(Bn)] 1 and that [A(Bn)] 1 C AM(Bn ), we get [A(Bn)] *
= [C(3Bn )] */[A(bn )] 1 = (AM(Bn)/[A(B n)] 1) $1
This completes the proof of (b). To prove (c) we assume (b) and we shall identify the dual [A(Bn)]
V.
* with the
product
AM(Bn)/[A(Bn)] 1 $1 V and the quotient AM(Bn)/[A(Bn)] 1 with the subspace AM(Bn)/A(Bn)l $1 {OJ of that product. Let E be the norm closed subspace of [A(Bn)] * spanned by the IPt's for A E Bn \3Bn . Under the above identification each IPt corresponds to the coset {J.L + [A(Bn)]l} where J.L is any representing measure for the point A EBn\3Bn. Since J.L E AM(Bn ), IP: E AM(Bn)/[A(Bn)] 1. Hence E C AM(Bn)/[A(Bn )] 1. Now let x* E [A(Bn)] *\E. Then, by the HahnBanach theorem, there exists an x** in [A(Bn)] ** such that x**(x*) = 1 and x**(IP1) = 0 for AE Bn \3Bn . Thus, by Helly's theorem, for every sequence (Aj) in Bn \3Bn there is a sequence (fj) in A(Bn) such that IP,/fk ) = fk(Aj) = 0 for j = I, 2, ... , k; x*(fk) = 1; IIfkii :e;;; 2I1x**11 (k = I, 2, ... ). If we have chosen the sequence (Aj) to be dense in Bn \3Bn , the Montel theorem then yields that (fj) is a Montel sequence. Thus, if J.L is any extension of x* to a linear functional on [C(3Bn)] *, then J.L is not an analytic measure because lil!l I
f
aBn
fj dJ.L
= li~ I
x*(fj)
= 1.
This shows that E = AM(Bn)J [A (B n) ]1. lt remains to prove that E is separable. To this end fix r with 0
< r < 1 and let Qr :
C(rB n ) be the restriction map, where rBn = {A E Bn: r I A E Bn}. By the Montel theorem, Qr is compact. Hence Q:: [C(rBn)] *  [A(Bn)] * is also.compact. Thus the set {IPt: A E rB n} is totally bounded being contained in the image under of the unit ball of [C(rBn)] *. Therefore, the set
A(Bn) 
Q:
I\LEKSANDER PELCZYNSKI
74
is separable and so is E. This completes the proof of (c). PROOF OF (a). It is enough to show if p. E AM(Bn ), then ~p. E AM(Bn)
(11.1 )
for k
= 1,2, ...
, n.
Indeed (11.1) together with Proposition 11.1(iii) yields Pp. E AM(Bn) for every polynomial P in the 2n variables zl' z2' ... ,zn' zl' z2' ... ,zn By the StoneWeierstrass theorem the polynomials are dense in L1(1p.1). Thus Ll(Ip.1) C AM(Bn), because AM(Bn) is a closed subspace of C{aB n)*. This proves (a). Now let m denote the Lebesgue measure on aB n' Before proving (11.1) we need the observation that the assertion of Theorem 11.3(a) holds for p. = m. To this end recall the Cauchy formula for Bn (11.2) where cn is a numeric factor independent of f. From (11.2) we derive
f(O)
= cn
ak1 +k2+"'+kn ~kf(O) aZ kl .•. aZn n
= kl!
l
(
JaBn
f(w)m(dw),
k2! ... k n ! cn (
JaB n
few)
n k. n wi" .
m(dw)
,= 1
for arbitrary multiindex (k 1 • k 2 • ... ,kn ) and for fE A(Bn)' (Here w
= (wI'
w 2 ' ... ,
wn ).) Hence the Lebesgue measure m is analytic being a representing measure for the origin. By the Montel theorem, lim;
= 0 for every
Montel sequence
ni=
every multiindex (k 1 • k2' ... ,kn )· Therefore, the measures 1Zfi . m are analytic and the same application of the StoneWeierstrass theorem as in the beginning of the proof of (a) yields
Uj) in A(Bn} and
if (11.3)
1/
is a complex Borel measure on aBn which is absolutely
continuous with respect to the Lebesgue measure m, then
EAM(Bn}' Next we shall show how to derive (11.1) from (11.2), (11.3) and the following 1/
MAIN LEMMA.
Let
Then Tf extends to a continuous function on C{Bn} for each f E A(Bn }, and moreover the operator T: A(Bn)  C(Bn) is compact. PROOF OF (11.1). Without loss of generality one may assume that k
Sf= zlf Tf for ffiC.A(B n }.
= 1.
Let us set
75
BANACH SPACES
Then, by the first part of the Main Lemma, Sf E C(Bn). Furthermore, by (11.2), (SfX~)
r
wlf(w)
= en Ji
aBn [1  (a, w)}n
m(dw)
for
a E Bn \aBn ·
Therefore, (Sf Xa) depends analytically on & in the interior of the ball Bn; hence Sf is holomorphic in Bn \aBn , equivalently Sf E A(Bn). Thus S: A(Bn) + A(Bn) is a bounded linear operator. Next observe that for every fixed a E Bn \aBn the measure vI E [C(aBn)] * defined by VI (A) = JA (wd[1  (A, w)]n)m(dw) for measurable A c aBn is absolutely continuous with respect to m. Hence, by (11.3), vA E AM(Bn ). Therefore, for every Montel sequence
a E Bn \aBn
(fj) in A(Bn) and for every (11.4) and (11.5)
By the second part of the Main Lemma, (Tfj) is a sequence of equicontinuous functions (because sup;llfjll < 00) which, by (11.5), converges to zero on a dense subset of Bn. Hence
Tfj( a)
(11.6)
+
as i + 00 uniformly for
0
= o.
Thus, for every J.I. E [C(aB n )]*, lim;JaBnTfjdJ.l. sure J.I.,
r
lim
; JaBn
(Sfj) dJ.l.
aE Bn·
Furthermore, for every analytic mea
=0
because, by (11.4), if (fj) is a Montel sequence so is (Sfj). Therefore, for every Montel sequence (fj) in A(Bn) lim
;
f aBn fj(w)w
t
J.I.(dw)
= lim C· ; JaBn
Tfj dJ.l.
+ lim ;
JaBn Sfj dJ.l. = 0
which completes the proof of (11.1). PROOF OF THE
w
= (Zt  l )/[1 (A' w)]n for .4 = E aBn and A =1= w; K(a, w) = 0 for a = wE
MAIN LEMMA. Let us set K(a, w)
= (WI' ...
(Zt' ... ,zn) E Bn and w
,
W n)
aBn. First observe that it is enough to show lim cx; = 0
where cx; = sup
;
IEBn
f
IK( A' w )Im(dw)
A;(I)
(11.7) and A/A)
= {w E
aBn: Re(A, w)
Assuming (11.7) we complete the proof as follows. Let tinuous function defined by 1/I(t)
h
~
t
< 1.
for 0 ~ t
< ~,
l/i}.
1/1: [0,00]
+
= K(a' w)1/I(j{1
[0, 1] be a con
1/I(t) ~ 1 for t ~ 1, 1/I(t) linear for
Let
K;Ca' w) and let
=0
~ 1
 Re(A' w)))
for
i = 1,2, ...
76
ALEKSANDER PELCZYNSKI
Since the kernels K j are continuous on Bn x aBn • T/ E C(Bn) for f E C(Bn) and the Opt tors Tj are compact. We have also
(11.8) Hence, by (11.7), the sequence (Tjt) tends uniformly to Tf on Bn. Thus Tf is continuous for every fE C(Bn) and T is a bounded linear operator on C(Bn). Moreover, limjllTj  TIl = 0; hence T is compact.
The proof of (11.7) is easy for n
= 1 because
= 1 for A"* w.
then IK( A' w)1
If n ~ 2
we observe first that IK(A, w)1 ~ II~  wll/ll  (A, w)l n and Aj(A) C Aj(o/IIAII) for 0"* 0 and j = l, 2, .... Clearly, by the rotation invariantness of the Lebesgue measure m on aB n' the integral JAj(AIIAIII)ll o willI!  (A, w)l n m(dw) is a function of 110 11 only. Therefore, we may assume without loss of generality that 0 = ae 1 where e 1 = (1, 0, 0, ... , 0) and 0
=0
where (3j
=
j=oo
Ilae 1 sup
(
O":a":l
11 
JA j (el)
wll _ m(dw).

aWl
In
This will follow if we show (11.9)
wll _ ~ 2 n + 1/ 2 11  w11(12n)/2 11 aw 1 ln
lIae l

(0
1 and w E aNn)
and (11.10) To establish (11.9) we need elementary inequalities 1  az \ 1aZ\~I;  az 1 ~1~2
The second one yields (for z
(11.11)
(0
1, zED).
= WI) 2n 
1
1
=~=
11  aWl Inl
11wllnl
Applying both inequalities we get
la  w112 + 1  Iw I I2 11aw112 11  aw112
+ 1  Iw112
4(1  IW112) ~1+
(11.12)
11  aw 1 12 5  2 Re WI

11  wl12
31wl12
11  w112 ~
8(1  Re
WI)
IIw 1 12
~
1
811  wll
BANACH SPACES
77
Combining (11.11) with (11.12) we get (11.9). The proof of (11.10) is routine. After expressing the integral in terms of spherical coordinates (in the unit sphere in R2n) the problem reduces to the finiteness of the integral 1T
sin 2n  2a sin 2n  3 /3 da d/3
1T
So So . [(I  cosa)2 + sin2a cos2i3]
(2nl)/4 '
whose integrand is ~ C/(a 2 + (13  Tr/2)2)1/2 for some absolute constant C. 0 Our next goal is to show that if n ;;;;. 2 then the spaces A(Dn) and A(Bn) are not isomorphic to complemented subspaces of a C(S)space with a separable annihilator. This will establish the "no" entries in the first row of the table. We shall need the following concept. DEFINITION 11.2. A subset F of a closed bounded domain U C Cn is called a support for the disc algebra if there exists a map
'PF(D)
(11.14)
gF
0
= F,
=Z
<1
for every wE U\(F
(11.15)
~F(w)1
(11.16)
I F (!) =fo
for every zED,
n
aU),
for every fEA(U).
Our next result provides a useful criterion for nonisomorphism of some A(U)spaces to a disc algebra. THEOREM 11.4. Let U C Cn be a closed bounded domain of holomorphy. Suppose that
(*) there exists in U an uncountable family (F'Y)'YEr of supports of the disc algebra such that
Fa
n F~ n
au =
¢ whenever a =1= 13
(a, 13 E r).
Then the Banach space A(U) is not isomorphic to any subspace ofC(S)space with a separable annihilator.
PROOF. Let us set I",I
= IF' Y ,g", = gF I 'Y
for 'Y E
r, where the operators IF 'Y and the = F'Y' Let PA : A ~ 12 be the
functions gF'Y are the IF and gF of Definition 11.2 for F Paley operator defined by
and let P'Y = PAI'Y for 'Y E r. Since the operator PA is absolutely summing (cf. §3, Example 3.1), so are P'Y and Tr1(P'Y) ~ Tr1(PA ) = C for every 'Y E r. Now assume to the contrary that there is a closed linear subspace X of a C(S)space (S compact Hausdorff) such that the annihilator Xl eLl ()I.) for some Borel probability measure A on S, and there are bounded linear operators R: A(U) ~ X and Q: X ~ A(U) such
78
ALEKSANDER PElCZYNSKI
that QR = idA(u) is the identity on A(U). Then, by Corollary 2.1, for every 'Y E r, there exist a nonnegative function hy and a compact (even nuclear) operator Vy: X _[2 such that (11.17) Next fix € with 0 < € < (Vi  l)/(IIRII tion by E L (X) so that
+ 1) and, for every
'Y E
r, pick a nonnegative
func
00
(11.18) Let (m
= 1,2, ... ).
U;;;
Since =1 r m = r and since r is uncountable, at least one of the sets r m' say r m 0' is infmite. Now fix a positive integer M > 411Q IIm~ and distinct indices 'Y 1 ' 'Y 2' ... , 'YM in r mo. For simplicity we shall write in the sequel the index i instead of 'Yj; for instance Pj instead PYj' bj instead of bYj' etc. Let us set Yj,r =R«(gyrl)
fori
= 1,2, ... ,M;r = 1,2, . . . .
Clearly IIYJ• rll ~ 1IRIIIIg~rllloo ~ IIRII because by (11.13), (11.14) and (11.15), IIg~rlll ,
J
/
=
IIgjll = 1. Thus, using the fact that the operators Vj are compact, we extract an infinite increasing subsequence (r(k)) of the indices such that IlVj(Xj,k)11
<€
for i
= 1,' 2, ...
,M; k
= 1, 2, ...
where Xj,k = Yj,r(2kl)  Yj,r(2k)' Thus, by (1Ll7) and (11.18),
(11.19)
because IIxj,kll ~ IIYj,r(2k1)11 + IIYj,r(2k)1I ~ 211RII (j = 1,2, ... ,M; k = 1,2, ... ). On the other hand, using the definition of PA and remembering that QR = id x , we get PjQ(Yj,r)
= Pj(glrl) = PA ( z2rl) = er
where er denotes the rth unit vector of [2 (r
(j
= 1,2, ...
,M)
= 1, 2, ... ).
Hence IIPjQ(xj,k)1I = Vi because PjQ(Xj,k) is a difference of two orthogonal vectors each of norm one. Thus (11.20)
Is
IXj,k1bjdA
~Vi €(lIRIl + 1) ~ 1·
(j
= 1,2, ... ,M;k =
1,2, ... ).
79
BANACH SPACES
It follows from the assumption (*) that the sets FI n au, F2 n au, ... , FM n au are mutually disjoint. Combining this with (I 1.13), (I 1.14) and (11.15), we infer that there
exists an index ro such that, for r > ro' ~~llgrl(w)1 ~ 2 for w E U. Thus there exists an index k such that for arbitrary complex numbers c I' c 2 , . . . , cM , we have
IIjtl CjXj,k
II
~ IIj~
cj Yj,r<2kl)
II +
Ilj~ c
j Yj,r(2k)
II
qQII(t~ c;g(2HH I + t~ c;g(2kH II) ~
411QII
max
l';;;'j';;;'M
ICjl.
Hence ~~ 1 Ix/s) I ~ 411QII for every s E U. Therefore, M
L
;=1
Ix/s)lb/s) ~ 411Q11mo
because "I; E rmo (equivalently IIb;lIoo ~
which contradicts with the choice of M
for s E U
mol. Thus, by (11.14),
0
Now we are ready for COROLLARY 11.1. If n ;;;. 2 then each of the spaces (i) A(Bn), (ii) A(Dn) is not iso
morphic to any complemented subspace of a C(S)space with a separable annihilator. PROOF. It is enough to show that both domains Bn and D n (n ;;;. 2) satisfy the assumption of Theorem 11.4. (i) Let r
= (wI' w 2 , ... , wn ) E aBn: WI = Re WI > OJ. For w E r define = zw for Z ED; next defineg w EA(Bn) by gw(~) = (~, w) = = (zl' z2"'" zk) EBn, and put Fw = I{Jw(D)· Clearly each triple
=
{w
I{Jw: D Bn by I{Jw(z) ~~=lZkWk for ~
(F w' I{Jw' g w) satisfies the conditions (11.13)(11.16) of Definition 11.2, and Bn together with the family (Fw)wEr satisfies (*). (ii) Let r
=
{w
= (WI'
W 2 ' ... , W n  1 )
E Cn
1:
IWll
= IW21 = ... =
Iwn_11
=
I}.
For w E r define I{Jw: D  D n by I{Jw(z) = (wI' w 2 ' ... , w n 1 ' z) for zED, and gw E A(Dn) by gw(~) = znn~:!l rl(Zk + w k ) for ~ = (Zl' Z2' ... , zm) E Dn, and put Fw = I{JjD). Clearly each triple (Fw, I{JW' gw) satisfies the conditions (11.13)(11.16) of Definition 11.2, and D n together with the family (FW)WEr satisfies (*). 0 We turn to the last "no" entry in the table of Theorem 11.2. Again we shall prove more than is stated there; namely, we have THEOREM 11.5. If m ;;;. 2 then the dual [A(Dm)]
* is not isomorphic to any
subspace
<\LEKSANDER PELCZYNSKI
80
of the Cartesian product M EBI V where M is an arbitrary separable Banach space and V is an L I (v}space.
This stronger result immediately yields COROLLARY 11.2. If n
=
1, 2, ... and m
= 2,
3, ... then A(Dm) is not isomorphic
to any quotient space of A(Bn).
Theorem 11.5 itself is an immediate consequence of the following three lemmas stated below. If E is a Banach space and r a set of indices then by 1~(E) we denote the Banach space of all functions (X'Y)'YEr: r + E such that II(x'Y)'YEr ll = sUProcr~'YErollx'Yll where the supremum is extended over all finite subsets roof r.
<
00
LEMMA
11.1. The space L I /H~ is not isomorphic to any subspace of an L I (v}space.
LEMMA
11.2. If m ;;;;. 2 then the space [A(Dm)]
isomorphic to I~(L I /H~) for some uncountable set LEMMA
* contains a subspace isometrically
r.
11.3. Let E be a Banach space. Assume that for some uncountable set
r
the
space I ~(E) embeds isomorphically into a Cartesian product M EBI V where M is a separable Banach space and V is an L 1 (v}space. Then E is isomorphic to a subspace of some other L I (p. }space.
PROOF OF LEMMA 11.1. Use Corollary 4.1, and the fact that the dual of L I /H~ is isometrically isomorphic to H oo (cf. § 1). 0 PROOF OF LEMMA 11.2. Let r = aD. To each w E aD we assign the subspace E w of [A(Dn)] * consisting of the functionals
(11.21)
x*(f) = J( aD
x*
for which there is a gx. ELI such that
f(z, w, 0,0, ... , O)gx.(z)m(dz)
for fEA(D n ).
It is clear that the map x* + {H~ + gx.} ts, for every fixed w, an isometric isomorphism from Ew onto L I /H~. We shall show that the family of these isometric isomorphisms "extends" to an isometric isomorphism from the smallest closed linear subspace of [A(Dm)] * which contains all the subspaces E w ' for w E r = aD, onto I~(L I /H~). To this end it is clearly enough to show that if {WI' w2 ' ..• ,wM } is an arbitrary finite subset of r then
(I 1.22)
for every
Let gj = g x! ELI satisfy (I 1.21) with
x*
=
xt
xt
E Ew. (j /
= 1,2, ...
,M).
for j = 1, 2, . . . ,M. For j = 1, 2, . . . ,M,
we define an h j E A (D) so that hj(w k ) = 0 for k =1= j, hj(wj) = 1, Ihiw) I < 1 for w =1= Wj. The existence of the h/s with the above properties follows immediately from the RudinCarleson theorem (cf. Theorem 2.1). By the definition of the xl's, for f E A(D n ), for j = 1, 2, ... ,M, and for r = 1, 2, . .. we have (11.23)
xt(f) = xtUh'j);
xt(fh'k) = 0
whenever k =1= j;
IIfhjll
~
Ilfli.
81
BANACH SPA.CES
Here we identify hj with functions in A(B m ) whose value at a point (z\, Z2' ... , zn) is
[h j (Z2)r. Now fix E > 0 and pick fj E A(Dm) with Ilfjll = 1 so that xt(fj)  ElM for j = I, 2, ... , M. Next pick an integer r so large that
= Re xt(fj) ~ Ilx!11
(11.24)
Then, by (11.23), putting
x'(.~,
x*
=
hVk) ~
~f!= 1 xt, we have
J, x'(hUk) ~ k~' it. xWkh~) M
= L xt(fjhj) = j=
1
M
L j=
1
xt(fj) ~
M
L j=
1
Ilxtll  E,
while, by (11.23) and (11.24),
L~1 h~fk II ~ :~b 1;1 Ih~(z)lllfkll ~ 1 + E. Thus,
j~ Letting
IIx!11
E
~ IIx*1I ~ (1 + E)1 Ix*C~1 fkh~) I ~ C~
tend to 0 we get (11.22).
Ilxtll E)/(1 + E).
0
PROOF OF LEMMA 11.3. By [LPI, Proposition 7.1] , it is enough to show that there is a C;:' 1 such that for every finite dimensional subspace F of E there is a subspace Fl in
V and an isomorphism Q: F + F 1 with IIQIIIIQ 111 ~ C Let qM and q v denote the natural projection of M 611 V onto M and V respectively. Let, for a E r,
Clearly we may identify each EOI with E via the map (x')') + x Oi ' Let T be an isomorphic embedding of 1~(E) into M 611 V, let T 1 : Tl~(E) + 1~(E) denote the inverse of T and let TOI denote the restriction of T to E OI . Let us fix a finite dimensional subspace F of E and consider the space B(F, M) of all bounded linear operators from F into M. Since M is separable and F finite dimensional, B(F, M) is separable. Thus there is a pair of distinct indices 'Yl and h such that IlqMT')'1  qMT')'211 < (211 T 111)1 because the set {qMT')': 'Y E r} after the identification of each E')' with E is an uncountable subset of B(F, M). Next define W: F + 1~(E) by W(F)
for'Y
*
let Q
= q v TW:
'Y 1 and'Y
*
= (X~)')'Er
where
X?1 = X?2 = Yzf and x~ = 0
'Y2 (fE F). Clearly W is an isometric isomorphism embedding. Now,
F + V. Then for every
f
E
F we have
82
ALEKSANDER PEl:.CZyNSKI
IIQfII = IITWflllIqMTWfli
while II Qfll .;;; IIq villi Til IIWIIIIfli = II Tllllfli. Hence Q: F + V is an isomorphic embedding with II Q 1111 QIII .;;; 211 Til II T11I where Ql: Q(f) + F denotes the inverse of Q. This completes the proof.
0
Notes and remarks to § 11. Theorem 11.3 is due to Henkin [He2]. The proof presented in the text is a simplified version of the proof due to Cole and Range [CR]. The proof of the Main Lemma is due to T. Figiel. Similar operators and integrals have been considered in [Ru4]. Theorem 11.3 is a special case of a more general result for strictly pseudoconvex domains. Recall that a bounded closed domain U in Cn is strictly pseudoconvex with C 2 _ smooth boundary if there is a twice continuously differentiable real function p defined on an open neighborhood G of U such that U = {A E G: p( A) .;;; O}, au = {A E G: peA) = O}, grad p 0 at each point A E au, and, for every AE au, n a2 p(A) _ WjWk > 0
*
L
j.k=l
whenever 0
*
(WI'
w2 ,
... ,
azjazk
w n ) E Cn satisfies the condition
£ j= I
apO)
Regarding p as a function of 2n real variables admit
:: ~ +(:~ , ;;)
Wj
= O.
aZj XI' YI' . . . ,
x n • Y n where
:~ ~ ~'(::, +, :;')
Zj
= Xj + iYj we
(j=1,2, ... ,n).
We have
l1.3a (CF. HENKIN [He3]). The assertions (a), (b), (c) of Theorem 11.3 remain valid if Bn is replaced by arbitrary bounded closed strictly pseudoconvex domain with C 2 smooth boundary. THEOREM
Theorem 11.3a can be proved along the same line as Theorem 11.3. The main technical difficulty is a construction of an analogue of the Cauchy formula (11.2), i.e., an integral representation
83
BANACH SPA.CES
f(~)
fa
=
r/(w)K(w, a)a(dw)
(~
E u\au,[ EA(w))
where a is the (2n  I)dimensional surface Lebesgue measure and K: au x (U\au)
+
Cis
a kernel which is analytic in the first variable, and such that for the operator induced by this kernel the analogue of the Main Lemma is true (at this point the "geometry" of the boundary of the strictly pseudoconvex domain is exploited). The desired integral formula has been first constructed by Henkin [He3] and by Ramirez de Arellano [Rrn] for strictly pseudoconvex domains with C 3 smooth boundaries, and next, due to improvement by qlverlid [~], for strictly pseudo convex domains with C 2 smooth boundaries. The reader is referred to the excellent survey [CHe] for further information concerning the integral representations and analytic measures. Also other results of this section can be extended to the case of domains of holomorphy with C 2 smooth boundaries. In particular Corollary 11.1 to Theorem 1104 admits the following generalization (cL MitjaginPelczynski [MtP, Theorem I]). 11.6. If n ~ 2 and if U
THEOREM
C
Cn is a bounded closed domain of holomorphy
with the C 2 smooth boundary, then the Banach space A(U) is not complemented in any subspace of a C(S)space with a separable annihilator. Before stating a generalization of Theorem 11.5 let us observe that the Shilov boundary of the uniform algebra A(D m ) is the mtorus (aDr
={~ =(Zj) E em: Iz I =IZ21 =... =IZm I =I}, J
while the topological boundary of D m is aD m
={~= (Zj)
E Cm:
max IZjl l";;',";;'m
=I}.
Hence, if m ~ 2, then the Shilov boundary of A(Dm) is a proper subset of the topological boundary of D m . In general we have THEOREM
11.7. Let U
C
Cm (m ~ 2) be a bounded closed domain of holomorphy
C 2 smooth
with boundary. Assume that the Shilov boundary of the algebra A(U) is a proper subset of the topological boundary of U. Then [A(U)] * is not a separable distortion of an L 1 (v)space. Hence A(U) is not isomorphic to A(U1 ) for any strictly pseudo convex domain U 1 in Cn (n = 1,2, ... ) with C 2 smooth boundary. Theorem 11.7 is also due to Henkin (unpublished); the second part of Theorem 11.7 has been announced in a slightly weaker form in [He3] (cf. Theorem 1.6 of [He3]). Theorem 11.5 and Corollary 11.2 are essentially due to Henkin [He 1]. The proof presented in the text seems to be new. Let us outline briefly another proof that A (D) is not isomorphic to A(Dn) for n ~ 2. Let mn denote the normalized Haar measure on [aD] n and let f1P([aD] n) denote the closure of A(D n ) in If([aD]n), i.e., the completion of A(Dn) in the norm U[aD]lIIfI Pdm n )l/P. Let i~n): A(Dn) + HP( [aD] n) be the natural embedding. Clearly i~n) is a pabsolutely
84
ALEKSANDER PRCZYNSKI
summing operator with 'ITpu~n») = 1. Next, for p > 1, we evaluate the pintegral norm of i~n) using the in variance of i~n) with respect to action of the group (aDt on A(Dn) and
HP([aD] n) and the averaging technique exactly in the same way as in the proof of formula (2.17) of §2. We get ipu~n») = IIR(n)lI p where R(n) denotes the orthogonal projection from L2([aD]n) onto H2([aD]n) regarded as an operator from LP([aD]n) onto HP([aD]n). On the other hand a straightforward argument gives IIR(n)lI p = IIRII; where R is the Riesz projection. Thus ip(i~n»);;;. C 1 [p2/(p  1)]n where C 1 is an absolute constant (cf. §O.II). The above arguement combined with Theorem 2.4 shows that for n ;;;. 2, the Up  'lTp)ratio of A (D n) (cf. Definition 9.2) has a different behavior at 00 than the Up  'ITP )ratio of A (D n). In fact we have
lim kpCA(Dn))/k (A (D))
p=oc
p
=
00
which, in view of Proposition 9.l(a), yields that the Banach space A(D) is not isomorphic to
A(D n ) for n ;;;. 2.
0
REMARK. (1) A similar technique enables us to show HOC (aD) is not isomorphic to H""([aD]n) for n ;;;. 2. (2) However this approach does not seem to work to distinguish
between A(Bn) and A(D). The reason is that the norm of the analogue of the Riesz projectionthe rotation invariant projection from LP(aB n ) onto HP(aBn)is of order Cp2/(p  1). This can be deduced from the proof due to Koranyi and Vagi [KV] of the boundedness of this projection [B. S. Mitjagin, private communication]. (3) We do not know whether for n ;;;. 2 and for 1 < p * 2 < 00 the Up  'lTp)ratio of A(Dn) (resp. A(Bn)) is finite. If it is finite, we shall be able at least for the polydisc algebras to use this fact to solve Problem 11.1. Thus Problem 11.2. Let 1 < p < 00. (a) Is every pabsolutely summing operator from n A(D ) (resp. A(Bn)) pintegral? (b) Is it true that for n = 2, 3, ... , there exists an absolute constant Cn such that, for every finite rank operator T: A(D n ) ') E (E arbitrary Banach space), ip(T) ~ Cn(p2/(p  l)t'IT peT)? (c) Is it true that there is an absolute constant C such that for every finite rank operator T: A(Bn) ~ E (E arbitrary Banach space), ipCT) ~ Cp2/(p  I)?
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