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For a holomorphic function f of bounded type on a complex Banach space E, we show that its derivative df:E→E∗ takes bounded sets into certain families of sets if and only if f may be factored in the form f=g○S, where S is in some associated operator ideal, and g is a holomorphic function of bounded type. We also prove that the multilinear and polynomial mappings factor in an analogous way if and only if they are “K-bounded.”
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J. Math. Anal. Appl. 348 (2008) 444–453
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Journal of Mathematical Analysis and Applications
www.elsevier.com/locate/jmaa
Derivative and factorization of holomorphic functions
Fernando Bombal a, Joaquín M. Gutiérrez b,, Ignacio Villanueva a
aDepartamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense de Madrid, 28040 Madrid, Spain
bDepartamento de Matemática Aplicada, ETS de Ingenieros Industriales, Universidad Politécnica de Madrid, C. José Gutiérrez Abascal 2, 28006 Madrid,Spain
article info abstract
Article history:
Received 10 May 20 08
Avail able online 18 July 200 8
Submitted by Richard M. Aron
Keywords:
Factorization
Polynomial
Holomorphic functions on Banach spaces
Operator ideal
Derivative
For a holomorphic function fof bounded type on a complex Banach space E,weshow
that its derivative df :EEtakes bounded sets into certain families of sets if and only
if fmay be factored in the form f=gS,whereSis in some associated operator ideal,
and gis a holomorphic function of bounded type. We also prove that the multilinear and
polynomial mappings factor in an analogous way if and only if they are “K-bounded.”
©2008 Elsevier Inc. All rights reserved.
1. Introduction and preliminaries
It is known that a holomorphic function of bounded type fHb(E)admits a factorization of the form f=gS,with
Sa compact (linear) operator and ga holomorphic function of bounded type if and only if the derivative df :EEtakes
boundedsetsof Einto relatively compact sets of E, and this is also equivalent to fbeing weakly uniformly continuous on
bounded subsets of E: see [1, Theorem 1.7], [15, Theorem 8], [13, Satz 2.1].
A similar result, in the weakly compact case, was obtained in [2, Theorem 5].
In the present paper, we extend these results to many other operator ideals and the associated families of bounded sets.
Throughout, E,F,G,X, and Ywill denote complex Banach spaces, Nwill be the set of natural numbers, Cwill represent
the complex field, and wwill stand for the weak-star topology on a dual Banach space. We use BEfor the closed unit ball
of E, and Efor the dual of E.If DEis a subset, then aco(D)represents the absolutely convex hull of D, and aco w(D)
is the wclosure of aco(D).Givenaset Vin the dual pair E,E,weuse Vfor its polar.
The completed projective tensor product of Eand Fis represented by E
πF.
We denote by L(E,F)the space of all (bounded linear) operators from Einto F, endowed with the supremum norm.
Given an operator TL(E,F),weuseTL(F,E)for its adjoint operator.
If E1,...,Ekare Banach spaces, the notation Lk(E1,...,Ek;F)stands for the space of all k-linear (continuous) mappings
from E1×··· × Ekinto F. A mapping P:EFis a k-homogeneous (continuous) polynomial if there is a k-linear mapping
A:E×(k)
··· ×EFsuch that P(x)=A(x,...,x)for all xE. The space of all such polynomials is denoted by P(kE,F).
Recall that with each PP(kE,F)we can associate a unique symmetric k-linear mapping
P:E×(k)
···×EFso that
P(x)=
Px,(k)
...,x(xE).
The first and the third named authors were partially supported by Dirección General de Investigación, MTM 2005–00082, Spain and by Universidad
Complutense de Madrid, UCM–910346. The second named author was supported in part by Dirección General de Investigación, MTM 2006–03531, Spain.
*Corresponding author.
E-mail addresses: bombal@mat.ucm.es (F. Bombal), jgutierrez@etsii.upm.es (J.M. Gutiérrez), ignaciov@mat.ucm.es (I. Villanueva).
0022-247X/$ – see front matter ©2008 Elsevier Inc. All rights reserved.
doi:10.1016/j.jmaa.2008.07.032
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A mapping f:EFis holomorphic if, for each xE, there are r>0 and a sequence (Pk)of polynomials, with
PkP(kE,F), such that fmay be given by its Taylor series expansion around x:
f(y)=
k=0
Pk(yx),
uniformly for yx<r.Weusethenotation
Pk=1
k!dkf(x),
while H(E,F)stands for the space of all holomorphic mappings from Einto F. The symbol Hb(E,F)represents the space
of all mappings fH(E,F)of bounded type, that is, bounded on bounded sets.
If the range space is omitted in the above notations, it is understood to be the complex field C, for instance, H(E):=
H(E,C),P(kE):= P(kE,C).
Several authors [2,3,8,24] have studied the K-bounded polynomials. Given a bounded subset KE,wedenote
xK:= supφ(x):φK.
A polynomial PP(kE,F)is K -bounded if there is a constant C>0 such that
P(x)
Cxk
K(xE).
An analogous definition may be given for k-linear mappings. The K-bounded polynomials, for Kcompact, have been con-
sidered in [3,24]. The weakly compact case is contained in [2] and the authors of [8] study the sets Kspanning a finite
dimensional subspace, and other particular cases. It is proved in the mentioned articles that, if Kis (weakly) compact
(resp., Kspans a finite dimensional subspace), then the K-bounded polynomials Pare exactly those factoring in the form
P=QSwith Sa (weakly) compact operator (resp., an operator of finite rank). We also extend these results to many
otheroperatorideals.
Following [2], we write [i]
... to mean that the ith element is left out, for instance,
E1×[i]
···×Ek:= E1×··· × Ei1×Ei+1×··· × Ek.
For a general introduction to polynomials and holomorphic mappings, the reader is referred to [10,20]. The definition and
general properties of operator ideals may be seen in [22]. For the definitions and facts from Banach space theory, see [9].
An operator ideal Uis said to be injective [22, 4.6.9] if, given an operator TL(E,F)and an into isomorphism i:FG,
we have that TUwhenever iT U.TheidealUis surjective [22, 4.7.9] if, given TL(E,F)and a surjective operator
q:GE, we have that TUwhenever TqU.WesaythatUis closed [22, 4.2.4] if for all Eand F, the space
U(E,F):= TL(E,F):TU
is closed in L(E,F). Examples of injective and surjective operator ideals may be seen in [16].
If Uis an operator ideal, the dual ideal Udis the ideal of all operators Tsuch that the adjoint Tbelongs to U.Itiseasy
to see that
Uis injective ⇒ Udis surjective,
Uis surjective ⇒ Udis injective,
Uis closed ⇒ Udis closed.
The paper is organized as follows: Section 2 contains the factorization results and, in Section 3, we give examples of
factorization through operators belonging to particular ideals.
We consider complex Banach spaces, but all the results are true in the real case, excepted those concerning holomorphic
mappings.
2. Factorization results
If DEis a bounded subset, then xDis a continuous seminorm on E,so
D:= xE:xD=0
is a closed subspace of E. As in [8, p. 161], we consider the quotient space E/Dand the canonical quotient map
π:EE/D.
Then, it is easy to show that the function
π(x)
=
x+D
:= xD
is a norm on E/D.Wedenoteby EDthe completion of E/D, which is a Banach space.
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446 F. Bombal et al. / J. Math. Anal. Appl. 348 (2008) 444–453
Lemma 1. Given a bounded subset D E,letT :EF be a linear map such that
T(x)
CxD(xE)
for some constant C >0. Then there is an operator TL(ED,F)such that T =Tπ(see the commutative diagram below).In
particular, T is continuous.
E
π
TF
ED
T
Proof. Assume π(xy)=0. Then xyD=0, and so T(xy)=0. Hence, the formula T(π(x)) := T(x)defines a linear
map
T:E/DF
such that
Tπ(x)
=
T(x)
CxD=C
π(x)
,
so Tis continuous and has a continuous extension, still denoted by T,toED.2
Given an operator ideal Uand a Banach space E, we consider the family of sets
CU(E):= AE:AT(BF)for some TU(E,F).
Proposition 2. Given an injective operator ideal U,letD CU(E).Then
πU(E,ED).
Proof. There are a Banach space Gand an operator SU(E,G)such that DS(BG).SinceUis injective, we can assume
that S(E)is dense in G.ForxE,deneU(S(x)) := π(x).IfS(x)=0, we have π(x)=xD=0. Therefore, U:S(E)ED
is a well-defined linear map. Moreover, we have
US(x)
=
π(x)
=
xD
=supx:φD
supx,S ):ψBG
=supS(x), ψ :ψBG
=
S(x)
,
so Uis continuous and has a continuous extension to an operator GED,stilldenotedbyU, and the following diagram
is commutative
E
S
πED
G
U
Thus, π=USU.2
Proposition 3. Let T L(E,F)be an operator, and let Ube an injective operator ideal. Then the following assertions are equivalent:
(a) TU(E,F);
(b) T(BF)CU(E);
(c) there are a subset D CU(E)and a constant C >0such that
T(x)
CxD(xE).
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Proof. (a) (b) by the definition of CU(E).
(b) (c). Take D:= T(BF). Then,
T(x)
=supT(x), ψ :ψBF
=supx,T ):ψBF
=
x
D.
(c) (a). By Lemma 1, there is an operator TL(ED,F)such that T=Tπ.ByProposition2,πU(E,ED),so
TU(E,F).2
We now give some properties of the family CU(E). To this end we need the following:
Lemma 4. (See [18, Theorem 20.7.3].) An operator ideal Uis closed and injective if and only if for an operator T L(E,F)to belong
to Uit is both necessary and sufficient that for each >0there exist a Banach space Gand an operator SU(E,G)so that
T(x)
S(x)
+x(xE).
Proposition 5. Let Ube an operator ideal, and let E be a Banach space. Then the following properties hold:
(a) if A CU(E)and B A, then B CU(E);
(b) if A CU(E), then the absolutely convex wclosed hull of A belongs to CU(E),andλACU(E)for every scalar λ;
(c) if A1,..., AnCU(E),thenn
i=1AiCU(E)and n
i=1AiCU(E).
Moreover, if Uis closed and injective, we also have:
(d) given a subset A E, if for every >0there is a set ACU(E)such that A A+BE,thenACU(E).
Proof. (a) and (b) are obvious, and (c) is easy.
(d)GivenasubsetAE, suppose that, for every >0, there is a set ACU(E)such that AA+BE.Let
π:EEAand π:EEAbe the canonical quotient maps onto E/Aand E/A, respectively. An easy application of
Lemma 1 yields:
Aπ(B(EA)). (1)
Therefore, it is enough to show that πU(E,EA).ByProposition2,πU(E,EA). Since, for every xE,
π(x)
=
xA
xA+BE
xA+x
=
π(x)
+x,
an application of Lemma 4 yields the result. 2
The following result is, in some sense, a converse to Proposition 5.
Proposition 6. Given a Banach space E , let C(E){{0}} be a class of bounded subsets of Ewith the following properties:
(a) for all A ,BC(E)and λC,wehave A+BC(E)and λAC(E);
(b) if A C(E)and B A, then B C(E);
(c) for every operator S L(X,E),wehaveS
(C(E)) C(X);
(d) if A C(E),thenaco w(A)C(E).
Then, letting Ube the class of operators such that T L(E,F)belongs to Uif and only if T (BF)C(E),wehavethatUis an
operator ideal, and
C(E)=CU(E).
Proof. (1) Denote by i:CCthe identity map on C.Thereisasubset{0} = AC(C). Choose 0 = λ0A. Then,
aco(A)C(C), by (d). The set
αλ0:|α|1aco(A)
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belongs to C(C), by (b). Therefore, the closed unit disk
D:= αC:|α|1=1
λ0αλ0:|α|1
belongs to C(C),by(a).Moreover,i(D)=D,soiU(C,C).
(2) Using (a), we obtain that U(E,F)is a vector subspace of L(E,F).
(3) Let
SL(X,E), TU(E,F), VL(F,Y).
Then,
V(BY)⊆VBF
TV(BY)⊆VT(BF)
TV(BY)C(E), by (a), (b), and the definition of U
(VTS)(BY)=STV(BY)C(X), by (c)
VTSU(X,Y).
Therefore, Uis an operator ideal [22, 1.1.1].
Suppose now that ACU(E). Then, there is an operator TU(E,F)such that AT(BF). By (b) and the definition
of U, we have that AC(E),so
CU(E)C(E).
Conversely, let AC(E).SinceAis a bounded subset, we get the inclusion (1) of the proof of (d) in Proposition 5, so
acow(A)π(B(EA)).
Let φπ(B(EA)), and choose ψB(EA)so that π) =φ.GivenxA,wehave
x=x,π(ψ)
=π(x), ψ
π(x)
=
xA1,
so φA◦◦ =acow(A)[17, Theorem 3.3.1], and we conclude that
acow(A)=π(B(EA)),
and this set belongs to C(E),by(d).BythedenitionofU,wehaveπU(E,EA),soACU(E), and
C(E)CU(E).
Therefore,
C(E)=CU(E). 2
With each k-linear mapping ALk(E1,..., Ek;X)we associate operators
Ai:EiLk1E1,[i]
..., Ek;X(1ik),
given by
Ai(xi)x1,[i]
...,xk:= A(x1,...,xk)(xjEj,1jk).
With ALk(E1,...,Ek)we also associate (k1)-linear mappings
Bi
A:E1×[i]
···×EkE
i(1ik),
so that
Bi
Ax1,[i]
...,xk(xi):= A(x1,...,xk)(xjEj,1jk).
The following result is well known. A proof may be seen in [2, pp. 184–185].
Proposition 7. Let A Lk(E1,...,Ek;X)be a k-linear mapping, and let V iEibe subsets (1ik). Then A is bounded by a
constant M >0on V1×···×Vkif and only if it is bounded by M on V ◦◦
1×··· ×V◦◦
k.
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Theorem 8. Let A Lk(E1,...,Ek;X)be a k-linear mapping, and let Ube a closed injective operator ideal. The following assertions
are equivalent:
(a) for every 1ik, the operator Ai:EiLk1(E1,[i]
..., Ek;X)belongs to U;
(b) there are sets D iCU(E
i)(1ik)and a constant C >0so that
A(x1,...,xk)
Cx1D1...xkDk(xiEi,1ik);
(c) there are Banach spaces Yi,operatorsS
iU(Ei,Yi),for1ik, and a k-linear mapping Q Lk(Y1,...,Yk;X)so that
A=Q(S1,...,Sk).
Moreover, if X =C, the preceding assertions are also equivalent to:
(d) for every 1ik, the mapping B i
A:E1×[i]
···×EkE
itakes bounded sets into sets in CU(E
i).
Proof. (a) (b). Let KiE
ibe the image by A
iof the unit ball of the dual space Lk1(E1,[i]
..., Ek;X). Then, KiCU(E
i)
(1ik). As in the proof of [2, Theorem 1], let
Di:=
j=1jk1Ki+j1BE
i(1ik).
By Proposition 5(b) and (d), DiCU(E
i). It is enough to show that
A(x1,...,xk)
1
whenever xiDi1(1ik).Now,xiDi1 if and only if
xiD
i=
j=1jk1Ki+j1BE
i◦◦
(see the proof of [17, Proposition 3.3.2]), since the set jk1Ki+j1BE
iis absolutely convex and wclosed, for each jN.
By Proposition 7, it is enough to show that, for
xiVi:=
j=1jk1Ki+j1BE
i,
we have A(x1,...,xk)1.
Let xiVi(1ik). Then, there is jiNsuch that
xijk1
iKi+j1
iBE
i(1ik).
Clearly, xiji(1ik). We can assume that j1=max{ji:1ik}. Then,
A(x1,...,xk)
=
A1(x1)(x2,...,xk)
A1(x1)
x2···xk
x1K1jk1
1
1.
(b) (c). Denoting by EDithe completion of the normed space Ei/Di,letπi:EiEDibe the natural quotient map
onto Ei/Di.ByProposition2,πiU(Ei,EDi).Dene
Qπ1(x1),...,πk(xk):= A(x1,...,xk)(xiEi,1ik).
To see that Qis well defined, assume πi(xi)=πi(yi)(1ik).ThenxiyiDi=0for1ik. Therefore
A(x1,...,xk)A(y1,...,yk)
A(x1y1,x2,...,xk)
+
A(y1,x2y2,x3,...,xk)
+··· +
A(y1,...,yk1,xkyk)
=0.
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On the other hand,
Q(π1(x1),...,πk(xk))
=
A(x1,...,xk)
Cx1D1···xkDk
=C
π1(x1)
···
πk(xk)
,
so Qis continuous on E1/D1×···×Ek/Dk.Hence,Qhas a continuous extension to ED1×···×EDk,stilldenotedby Q.
Letting Si:= πi(1ik), the assertion (c) is proved.
(c) (a). Let Di:= S
i(BY
i)CU(E
i).ForxiEifixed, we have
Ai(xi)
=sup
Ai(xi)x1,[i]
...,xk
:xjBEj,j= i
=sup
A(x1,...,xk)
:xjBEj,j= i
=sup
QS1(x1),..., Sk(xk)
:xjBEj,j= i
supQ
S1(x1)
···
Sk(xk)
:xjBEj,j= i
=supQx1D1···xkDk:xjBEj,j= i
=CxiDi.
Now, Proposition 3 implies AiU.
(a) (d). Recall that (E1
π···
πEk)=Lk(E1,...,Ek).Since AiU, the adjoint A
i:Lk1(E1,[i]
..., Ek)E
itakes
bounded sets into sets in CU(E
i). A fortiori so does its restriction Ci
A:E1
π
[i]
···
πEkE
i. Therefore, the mapping Bi
A
takes bounded sets of E1×[i]
···×Ekinto sets in CU(E
i).
(d) (a). Since the unit ball of E1
π
[i]
···
πEkis the closed convex hull of BE1[i]
···BEk[23, Proposition 2.2], Ci
A
takes the unit ball of E1
π
[i]
···
πEkinto a set KiCU(E
i), by Proposition 5. The operator A
i:Lk1(E1,[i]
..., Ek)E
i
is w-to-wcontinuous, so it takes the unit ball of the domain space into the weak-star closure of Ki, which belongs to
CU(E
i), again by Proposition 5. By Proposition 3, AiU.2
The equivalence (a) (c) was obtained in [15] and, by interpolation techniques, in [7, Theorem 3.4]. The theorem was
obtained in [2] when Uis the ideal of weakly compact operators and X=C.
With each polynomial PP(kE,X)we associate an operator
P:EPk1E,X
given by P(x)( y):=
P(x,y,...,y)for all x,yE. Recall that the derivative of Pis the polynomial
dP Pk1E,L(E,X)
defined by
dP(x)( y)=k
P(x,...,x,y)(x,yE).
Corollary 9. Let Ube a closed injective operator ideal, and let P P(kE,X)be a polynomial. The following assertions are equivalent:
(a) the operator P:EP(k1E,X)belongs to U;
(b) there are a set D CU(E)and a constant C >0so that
P(x1,...,xk)
Cx1D···xkD(x1,...,xkE);
(c) there are a set D CU(E)and a constant C >0so that
P(x)
Cxk
D(xE);
(d) there are a Banach space Y , an operator S U(E,Y)and a polynomial Q P(kY,X)so that P =QS.
Moreover, if X =C, then the preceding assertions are also equivalent to:
(e) the derivative dP P(k1E,E)takes bounded sets of E into sets in CU(E).
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Proof. (a) (b) (d) (a) and (a) (e) are as in Theorem 8, using the symmetric k-linear mapping
Passociated to
the polynomial P.
(b) (c) is obvious.
(c) (b). Assume (b) does not hold. Then, we can find sequences (xn
i)
n=1E(1ik)such that xn
iD1
(1ik,nN)and
Pxn
1,...,xn
k
n
k!(nN). (2)
Using the polarization formula [20, Theorem 1.10], we obtain
Pxn
1,...,xn
k
1
k!2k
j1
1ik
P1xn
1+··· +kxn
k
C
k!2k
j1
1ik
1xn
1+···+kxn
k
k
D
Ckk
k!sup
xn
i
k
D:1ik,nN
Ckk
k!.(3)
Combining (2) and (3), we get nCkkfor every nN, which is impossible. 2
The proof of [16, Proposition 5] yields:
Proposition 10. Let Cbe a family of subsets of a Banach space F satisfying the assertions (a) through (d) of Proposition 5,replacing
in (b) the wclosure by the norm closure, and let f H(E,F). The following conditions are equivalent:
(a) each x E has a neighbourhood V xso that f (Vx)C;
(b) there is a zero neighbourhood V Esothat f(V)C;
(c) for every k Nand each x E, wehave dkf(x)(BE)C;
(d) for every k N,wehaved
kf(0)(BE)C.
We note that assertion (d) of Proposition 5 is only needed in the proof of (d) (a).
Theorem 11. Let Ube a closed injective operator ideal, and let f Hb(E). The following assertions are equivalent:
(a) the derivative df :EEtakes bounded sets of E into sets in CU(E);
(b) there are a Banach space Y , an operator S U(E,Y)and a function g Hb(Y)such that f =gS.
Proof. (a) (b). Let f=Pkbe the Taylor series expansion of faround 0. Then df =dPk.Ifdf takes bounded sets
of Einto sets in CU(E),thensodoesdPkfor all k(Proposition 10). By Corollary 9, there are Banach spaces Yk,operators
SkU(E,Yk)and polynomials QkP(kYk)so that Pk=QkSk. Proceeding as in the proof of [15, Theorem 8], we get the
desired result.
(b) (a). Suppose f=gS,withSU. By the chain rule,
df (x)=dgS(x)S=SdgS(x)(xE),
so the following diagram is commutative:
E
S
df E
Ydg Y
S
Since gHb(Y),dg is also of bounded type, by Cauchy’s inequality (see the proof of [2, Theorem 5]). Hence, df takes
bounded sets of Einto sets in CU(E).2
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Table 1
Properties of operator ideals
Operator ideals Properties
(Weakly) compact Closed, injective, and surjective
Completely continuous Closed and injective
Unconditionally converging Closed and injective
Rosenthal Closed, injective, and surjective
Banach–Saks Closed, injective, and surjective
Limited Closed and surjective
Grothendieck Closed and surjective
3. Examples
In order to give examples to which Theorem 11 may be applied, we recall that an operator TL(E,F)is (weakly)
compact if T(BE)is a relatively (weakly) compact subset of F;Tis completely continuous if it takes weakly convergent
sequences of Einto convergent sequences of F;Tis Rosenthal if every sequence in T(BE)has a weak Cauchy subsequence,
in other words, T(BE)is a Rosenthal set;Tis unconditionally converging if it takes weakly unconditionally Cauchy series of E
into unconditionally convergent series in F.
Asubset DEis an (L)-set [12] if, for each weakly null sequence (xn)E,wehave
lim
nsup
φDxn=0.(4)
An operator TL(E,F)is completely continuous if and only if T(BF)is an (L)-set in E(see [5, Lemma 4] and
[14, Proposition 3.2]).
AsubsetDEis a (V)-set [21] if, for each weakly unconditionally Cauchy series xnin E, the relation (4) holds. An
operator TL(E,F)is unconditionally converging if and only if T(BF)is a (V)-set in E[5, Lemma 4].
Asubset AEis limited [6] if, for every weak-star null sequence n)in E,wehave
lim
nsup
xAx
n=0.
An operator TL(E,F)is limited if T(BE)is a limited subset of F.
Asubset AEis a Grothendieck set if, for every operator TL(E,c0),T(A)is relatively weakly compact. An operator
TL(E,F)is a Grothendieck operator if T(BE)is a Grothendieck set [11].
Asubset AEis a Banach–Saks set [19, Section 3] if every sequence (xn)Ahas a subsequence (xnk)such that the
sequence of arithmetic means
1
m
m
k=1
xnk
m=1
is norm convergent. An operator TL(E,F)is a Banach–Saks operator if T(BE)is a Banach–Saks set.
For a thorough study of different classes of sets in Banach spaces and their relation to operator ideals, the reader is
referred to [4].
The above mentioned classes of operators are, in fact, ideals which have the properties listed in Table 1 (see [16] and
references therein).
Corollary 12. Let f Hb(E)be a holomorphic function of bounded type. Then the assertions on each row of Table 2are equivalent.
Proof. It is enough to apply Theorem 11 to the injective ideal Uof (weakly) compact, completely continuous, or uncon-
ditionally converging operators, or to the dual ideal Ud(also injective), when Uis the ideal of Rosenthal, Banach–Saks,
limited, or Grothendieck operators.
We give the details in some particular cases. The others are dealt with similarly.
(a) Let Ube the ideal of operators with Rosenthal adjoints. Assume that, for every bounded set BE, we have that
acow(df (B)) is a Rosenthal set in E.Let
C(E):= AE:acow(A)is a Rosenthal set.
By Proposition 6, we have
C(E)=CU(E),
so df (B)CU(E). By Theorem 11, there are a Banach space Y,anoperatorSU(E,Y)(that is, Sis Rosenthal), and
afunction gHb(Y), such that f=gS.
Author's personal copy
F. Bombal et al. / J. Math. Anal. Appl. 348 (2008) 444–453 453
Table 2
Factorization of holomorphic functions
For each bounded set BE,wehave: f=gSwith:
df (B)is relatively compact Scompact
df (B)is relatively weakly compact Sweakly compact
df (B)is an (L)-set Scompletely continuous
df (B)is a (V)-set Sunconditionally converging
acow(df(B)) is a Rosenthal set SRosenthal
acow(df(B)) is a Banach–Saks set SBanach–Saks
acow(df(B)) is a limited set Slimited
acow(df(B)) is a Grothendieck set SGrothendieck
Conversely, if f=gSwith gand Sas above, by Theorem 11, df (B)CU(E)for every bounded set BE.By
Proposition 5(b),
acowdf (B)CU(E)
and, by the definition of CU(E),thesetaco w(df (B)) is Rosenthal.
(b) Let Ube the ideal of (weakly) compact operators. Assume that, for every bounded set BE,wehavethatdf (B)is
a relatively (weakly) compact set in E. Then there exist a Banach space Gand a (weakly) compact operator TL(G,E)
such that
df (B)T(BG)T∗∗(BG∗∗ )E.
Since Tis (weakly) compact, we have, by the definition of CU(E), that
df (B)CU(E),
and the proof proceeds as in (a).
(c) Let Ube the ideal of completely continuous operators. Assume that, for every bounded set BE,wehavethatdf(B)
is an (L)-set in E. By [14, Proposition 3.5], there are a Banach space Fand a completely continuous operator TL(E,F)
such that df (B)T(BF). By the definition of CU(E),wehavethat
df (B)CU(E),
and the proof proceeds as in (a). 2
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