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PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 127, Number 6, Pages 1715–1721
S 0002-9939(99)04917-5
Article electronically p ublished on February 11, 1999
INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS
MANUEL GONZ ´
ALEZ AND JOAQU´
IN M. GUTI ´
ERREZ
(Communicated by Dale Alspach)
Abstract. We characterize the holomorphic mappings fbetween complex
Banach spaces that may be written in the form f=g◦T,wheregis another
holomorphic mapping and Tis an operator belonging to a closed injective oper-
ator ideal. Analogous results are previously obtained for multilinear mappings
and polynomials.
In recent years, several authors [1], [6], [7] have studied conditions on a holo-
morphic mapping fbetween complex Banach spaces so that it may be written in
the form f=g◦T,wheregis another holomorphic mapping and Tis a (linear
bounded) operator belonging to certain classes of operators.
In this paper, we look at this problem in the setting of operator ideals, thus
finding more general conditions so that these factorizations occur. Our results
include all the previous ones, with simpler proofs, and apply to many new cases.
A linear mapping Tbelongs to the ideal Coof compact operators if and only if it
is weakly (uniformly) continuous on bounded subsets. So, if a mapping f:E→F
may be written as f=g◦Twith T∈Co,thenfis weakly uniformly continuous
on bounded subsets of E(the mappings with this property have been studied by
numerous authors, for instance [2], [3], [5]). The authors have shown in [7] that the
converse also holds: if a holomorphic mapping fbetween Banach spaces is weakly
uniformly continuous on bounded subsets, then it can be factorized in the form
f=g◦T, with T∈Co. A similar result was proved in [1] for Tin the ideal WCo
of weakly compact operators, and F=Cthe complex field.
In [6, Satz 2.1], using interpolation techniques, the factorization f=g◦Thas
been characterized in terms of the derivatives of f,forTin any closed, injective
and surjective operator ideal, and F=C.
In this paper, for any closed injective operator ideal U, we find several conditions
on a polynomial P:E→F(or a multilinear map) so that it may be written as
P=Q◦T, with Tin U. As a consequence, we prove that a holomorphic map
f:E→Fadmits such a factorization if and only if fis uniformly continuous
on bounded sets with respect to a suitable topology τU.InthecaseU=Co,the
topology τUis the finest l.c. topology that coincides with the weak topology on
bounded sets. Consequently, using a simpler proof, we recover the result in [7].
Received by the editors September 10, 1997.
1991 Mathematics Subject Classification. Primary 46G20; Secondary 47D50.
Key words and phrases. Factorization, holomorphic mapping between Banach spaces, multi-
linear mapping, polynomial, operator ideal.
The first author was supported in part by DGICYT Grant PB 97–0349 (Spain).
The second author was supported in part by DGICYT Grant PB 96–0607 (Spain).
c
1999 American Mathemati cal Society
1715
1716 MANUEL GONZ ´
ALEZ AND JOAQU´
IN M. GUTI´
ERREZ
Many usual operators ideals are closed and injective: for instance (see [14]),
the (weakly) compact operators, the (weakly) completely continuous operators,
the Rosenthal operators, the unconditionally converging operators, the (weakly)
Banach-Saks operators [11, §3], the strictly singular operators, the operators with
separable range, the decomposing (Asplund) operators, the Radon-Nikodym oper-
ators, and the absolutely continuous operators [12, §3]. Our results apply to all of
them.
The conditions that we find so that the factorization occurs are quite natural
and have been used by many authors in various contexts (see, e.g., [1, 2, 6, 8]).
Throughout, E,Fand Gwill denote Banach spaces, and BEwill denote the
closed unit ball of E. When we deal with holomorphic mappings, these spaces will
be complex. Otherwise, they can be either real or complex.
We denote by L(E,F) the space of all operators from Einto F, endowed with
the usual operator norm. If E1,... ,E
kare also Banach spaces, the notation
Lk(E1,... ,E
k;F) stands for the space of all (continuous) k-linear mappings from
E1×···×E
kinto F.
We shall use the notational convention [i]
... to mean that the ith coordinate is
not involved. For instance,
Lk−1(E1,[i]
...,E
k;F):=L
k−1
(E
1
,... ,E
i−1,E
i+1 ,... ,E
k).
To each A∈L
k
(E
1
,... ,E
k;F)andi∈{1,... ,k}we associate an operator
Ai:Ei−→ L k−1( E 1,[i]
...,E
k;F)
given by
Ai(xi)(x1,[i]
...,x
k):=A(x
1
,... ,x
k)(x
j
∈E
j
;1≤j≤k).
The space of all (continuous) k-homogeneous polynomials from Einto Fis de-
noted by P(k
E,F). To each P∈P(
k
E,F) we associate an operator
¯
P:E−→ P ( k−1E,F )
given by ¯
P(x)(y):= ˆ
P
x, y, (k−1)
... ,y
where ˆ
Pis the unique symmetric, k-linear
mapping associated to P.
For complex spaces Eand F,H(E,F) will denote the space of all holomorphic
mappings from Einto F,andH
b
(E,F) will denote the subspace of all f∈H(E,F)
which are bounded on bounded sets. In the complex case too, we say that a subset
A⊂Eis circled if for every x∈Aand complex λwith |λ|=1,wehaveλx ∈A.
For a general introduction to polynomials and holomorphic mappings, the reader
is referred to [4, 13]. The definition and general properties of operator ideals may
be seen in [14].
An operator ideal Uis said to be injective [14, 4.6.9] if given an operator T∈
L(E,F) and an injective isomorphism i:F→G,wehavethatT∈Uwhenever
iT ∈U.WesaythatUis closed [14, 4.2.4] if, for all Eand F,thespaceU(E, F )
is closed in L(E,F).
The following basic result will be used:
Lemma 1 ([9, 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 S∈U(E, G)
so that
kTxk≤kS
(x)k+kxk(x∈E).
INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS 1717
Considering now multilinear mappings, we first wish to relate some of their
topological properties to those of the associated operators. To this end, we need
the following result, whose proof, which is standard, is given for completeness.
Proposition 2. Let τibe a vector topology on a Banach space Ei(1 ≤i≤k).
Given a k-linear mapping A∈L
k
(E
1
,... ,E
k;F)with associated operators Ai:
Ei→L
k−1
(E
1
,[i]
...,E
k;F), we have that Ais uniformly τ1× ···× τ
k-continuous
on bounded sets if and only if the operators Aiare (uniformly) τi-continuous on
bounded sets.
Proof. Suppose Ais uniformly τ1× ··· × τ
k-continuous on bounded sets. Given
>0, we can find τizero neighbourhoods Ui⊂Eiso that kA(x1,... ,x
k)−
A(y
1,... ,y
k)k<whenever xi,y
i∈B
E
isatisfy xi−yi∈Ui(1 ≤i≤k). Take
j∈{1,... ,k},z
i∈B
E
ifor i=1,[j]
...,k,andx
j
,y
jas above. Then
(Ajxj−Ajyj)z1,[j]
...,z
k
=kA(z
1,... ,x
j,... ,z
k)−A(z
1,... ,y
j,... ,z
k)k<.
Hence, Ajis uniformly τj-continuous on bounded sets.
Conversely, let Aibe τi-continuous on bounded sets, for 1 ≤i≤k.Given
>0, there is a τizero neighbourhood Ui⊂Eiso that kAix−Aiyk<whenever
x, y ∈BEisatisfy x−y∈Ui.Givenx
i
,y
i∈B
E
iwith xi−yi∈Ui(1 ≤i≤k), we
have
kA(x1,... ,x
k)−A(y
1,... ,y
k)k
≤kA
1
(x
1−y
1
)(x2,... ,x
k)k+kA
2(x
2−y
2)(y1,x
3,... ,x
k)k+···
+kA
k(x
k−y
k)(y1,... ,y
k−1)k
≤k ,
and so, Ais uniformly τ1×···×τ
k-continuous on bounded sets.
Let Ube an injective operator ideal. On every Banach space Ewe consider the
topology τUgenerated by the seminorms
pT(x):=kTxk,for T∈U(E,F),Fany Banach space.
We then have (see [10, §3]):
U(E,F)=L((E, τU),F).(1)
Proposition 3. Given a closed injective ideal Uand S∈L(E,F), we have that
S∈U(E, F )if and only if Sis τU-continuous on bounded subsets.
Proof. Note that, since Uis closed, τUis the finest locally convex topology that
agrees with τUon bounded subsets (see [10, Proposition 4.2]). Then apply (1).
One of our key results is the next one.
Theorem 4. Given A∈L
k
(E
1
,... ,E
k;F), let Ube a closed injective operator
ideal. The following assertions are equivalent:
(a) for every 1≤i≤k, the operator Ai:Ei→L
k−1
(E
1
,[i]
...,E
k;F)belongs to
U;
(b) there are Banach spaces Yiand operators Ti∈U(E
i
,Y
i)(1≤i≤k)so that
kA(x1,... ,x
k)k≤kT
1
x
1
k·...·kT
k
x
k
k;
(c) there are Banach spaces Yi,operatorsT
i∈U(E
i
,Y
i)(1≤i≤k)and a
mapping D∈L
k
(Y
1
,... ,Y
k;F)so that A=D◦(T1,... ,T
k);
(d) Ais uniformly τU-continuous on bounded subsets.
1718 MANUEL GONZ ´
ALEZ AND JOAQU´
IN M. GUTI´
ERREZ
Proof. (a) ⇒(b). Consider the seminorms
qi(x):= inf
n∈N
n
k−1
kA
i
xk+n
−1
kxk
(x∈E
i
)
for 1 ≤i≤k. Letting Yibe the completion of (Ei/ker(qi),q
i), define Ti:Ei→Yi
by Tix:= x+ker(q
i
), for x∈Ei. For every n∈Nand 1 ≤i≤k,wehave
kT
i
xk=q
i
(x)≤n
k−1
kA
i
xk+n
−1
kxk(x∈E
i
).
ByLemma1wehavethatT
i∈U for 1 ≤i≤k.
It is enough to show that given xi∈Eiwith kTixik<1, for 1 ≤i≤k,wehave
kA(x
1
,... ,x
k)k<1.
Given i∈{1,... ,k},thereisn
i∈Nso that
nk−1
ikAixik+n−1
ikxik<1.
Assume nj=max{n
1
,... ,n
k}.Wehavekx
i
k<n
jfor all 1 ≤i≤k. Moreover,
kAjxjk<n
1−k
j. Hence,
kA(x1,... ,x
k)k=
A
jx
jx
1,[j]
...,x
k
≤kA
j
x
j
k·kx
1
k· [j]
...·kxkk<1.
(b) ⇒(c). Since Uis injective, we can assume that Ti(Ei)isdenseinY
i
,for
1≤i≤k. Define
D(T1x1,... ,T
kx
k):=A(x
1
,... ,x
k)(x
i
∈E
i
,1≤i≤k).
If Tixi=Tix0
ifor all 1 ≤i≤k,wethenhave
A(x
1
,... ,x
k)−A(x
0
1,... ,x
0
k)
=A(x
1−x
0
1,x
2,... ,x
k)+A(x
0
1
,x
2−x
0
2,... ,x
k)
+···+A(x
0
1,... ,x
0
k−1,x
k−x
0
k)
=0,
since Ti(xi−x0
i)=0for1≤i≤k.So,Dis well defined and continuous, with
kDk≤1. Hence we can extend it to Y1×···×Y
kand, denoting the extension by
Das well, we have A=D◦(T1,... ,T
k).
(c) ⇒(a). Assume A=D◦(T1,... ,T
k) with Ti∈U(E
i
,Y
i), for 1 ≤i≤k.For
each i, define
Bi:Yi−→ L k−1( E 1,[i]
...,E
k;F)
by
(Biy)(u1,[i]
...,u
k)=D(T
1
u
1
,... ,(i)
y,... ,T
ku
k)(y∈Y
i
;u
1
∈E
1
,
[i]
...,u
k∈E
k).
Clearly, Biis continuous, with
kBik≤kT
1
k· [i]
...·kTkk·kDk,
and Ai=Bi◦Ti. Hence Ai∈U.
(a) ⇔(d) by Propositions 2 and 3.
If Uis the ideal of compact operators, then the assertion (d) of the Theorem
states that Ais weakly uniformly continuous on bounded sets. The result in the
compact case was proved in [7].
It is an open problem to characterize the ideals Usuch that every k-linear map-
ping which is τU-continuous on bounded sets is also uniformly τU-continuous on
bounded sets. It is a well-known result, proved in [2], that for U=Cothe assertion
INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS 1719
is true. It is obviously true if Uis the ideal of all bounded operators. We proved in
[7] that it fails for U=WCo,andalsoforUthe completely continuous operators.
Corollary 5. Given P∈P(
k
E,F), let Ube a closed injective operator ideal. The
following assertions are equivalent:
(a) the operator ¯
P:E→P(
k−1
E,F)belongs to U;
(b) there are a Banach space Yand an operator T∈U(E, Y )so that kP(x)k≤
kT(x)k
kfor all x∈E;
(c) there are a Banach space Y,anoperatorT∈U(E, Y )and a polynomial
Q∈P(
k
Y,F)so that P=Q◦T;
(d) Pis uniformly τU-continuous on bounded sets.
The argument in the proof of [4, Lemma 1.16] can be used to show that (b) ⇒
(c). The other parts are obtained by adapting the proof of Theorem 4.
Next, we shall extend the factorization theorem to holomorphic mappings. The
following result will be useful:
Lemma 6 ([6, Lemma 1]).Suppose A=D◦(T1,... ,T
k),whereA6=0,D∈
L
k
(Y
1
,... ,Y
k;F),andT
i∈L(E
i
,Y
i)for 1≤i≤k. Then the spaces Yimay be
renormed so that kAk=kDk·kT
1
k·...·kT
k
k.
Proposition 7. Let Ube a closed injective operator ideal. If f∈H(E,F)is uni-
formly τU-continuous on bounded sets, a∈Eand k∈N,thend
k
f(a)is uniformly
τU-continuous on bounded sets.
Proof. From the Cauchy integral formula [13, Corollary 7.3], we obtain
1
k!dk
f(a)(x)−1
k!dkf(a)(y)
=
1
2πi Z|λ|=1
f(a+λx)−f(a+λy)
λk+1 dλ
≤sup
|λ|=1
kf(a+λx)−f(a+λy)k.
Lef Bbe a circled, bounded subset of E.Given>0, there is a circled τUzero
neighbourhood Uin Eso that kf(x0)−f(y0)k<whenever x0,y
0∈a+Bsatisfy
x0−y0∈U. Therefore, given x, y ∈Bwith x−y∈U, we easily get from the above
inequality:
1
k!dk
f(a)(x)−1
k!dk
f(a)(y)
<.
If (Yk) is a sequence of Banach spaces, we denote by c0(Yk) the Banach space
of all sequences (yk)sothaty
k∈Y
kand kykk→0, endowed with the supremum
norm. In the proof of the following Theorem we shall use the well-known fact that
if B⊂Eis bounded and f:B→Fis a uniformly continuous mapping, then f(B)
is bounded in F.
Theorem 8. Given f∈H(E,F ), let Ube a closed injective operator ideal. Then
fis uniformly τU-continuous on bounded subsets if and only if there are a Banach
space G,anoperatorT∈U(E, G)and a mapping g∈H
b
(G, F )so that f=g◦T.
Proof. Suppose f=g◦T, with gand Tas in the statement. Let A⊂Ebe
bounded. Then Tis uniformly continuous from (A, τU)into(TA,k·k). Since g
is bounded on bounded sets, it is norm-to-norm uniformly continuous on bounded
sets. Hence, fis uniformly continuous from (A, τU)into(F, k·k).
1720 MANUEL GONZ ´
ALEZ AND JOAQU´
IN M. GUTI´
ERREZ
Conversely, suppose fis uniformly τU-continuous on bounded sets. We write the
Taylor series expansion of fat the origin as
f(x)=
∞
X
k=0
Pk(x)(x∈E).
By Proposition 7, Pkis uniformly τU-continuous on bounded sets, for each k.By
Theorem 4, there are a Banach space Yk, an operator Tk∈U(E,Yk)andapoly-
nomial Qk∈P(
k
Y
k
,F) such that Pk=Qk◦Tk. We can get kˆ
Pkk=kˆ
Qkk·kT
k
k
k
(Lemma 6). Since τUis coarser than the norm topology, it follows that f∈
Hb(E,F), and so lim kPkk1/k = 0. From the inequalities
kPkk≤kˆ
P
k
k≤ k
k
k!kP
k
k
(see [13, Theorem 2.2]), and using the Stirling formula, we get lim kˆ
Pkk1/k =0.
Therefore, we can assume that kˆ
Qkk1/k →0andkT
k
k→0. Define
T:E−→ Y:= c0(Yk)
by Tx:= (Tkx)k. Clearly, T∈U. Denoting
πk:(y
i
)∈Y7−→ yk∈Yk,
we define g:Y→Fby g(y):=P
∞
k=1 Qk◦πk(y). Since lim kQk◦πkk1/k =
lim kQkk1/k =0,wegetthatgis a holomorphic mapping of bounded type that
satisfies the requirement.
To finish up, we give a polynomial P∈P(
3
`
∞
) that cannot be written in the
form P=Q◦S, with S∈WCo. Note that many classes of operators on `∞
coincide, e.g., the weakly compact operators, the completely continuous operators,
the unconditionally converging operators, etc.
Consider a surjective operator q:`∞→`2such that q(B`∞)⊇B`2. Letting
q(x)ibe the ith coordinate of q(x)∈`2, we define
P(x):=
∞
X
i=1
xiq(x)2
i,for x=(x
i
)∈`
∞.
The associated operator ¯
P:`∞→P(
2
`
∞
)isgivenby
¯
P(x)(y):=1
3
∞
X
i=1
xiq(y)2
i+2
3
∞
X
i=1
yiq(x)iq(y)i.
We only have to show that ¯
P/∈WCo(see Corollary 5) or, equivalently, that ¯
Pis
not completely continuous. Denoting by enthe sequence (0,... ,0,1,0,...)with1
in the nth position, we select a sequence (xn)⊂B`∞so that q(xn)=e
n
. Then,
3¯
P(en)(xn)=1+2x
n
n
q(e
n
)
n.
Since q(en)→0, we have 3 ¯
P(en)(xn)→1 and, therefore, k¯
P(en)kdoes not con-
verge to 0.
INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS 1721
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De parta m en to de M atem ´
aticas, Facultad de Ciencias, Universidad de Cantabria,
39071 Santander, Spain
E-mail address:gonzalem@ccaix3.unican.es
De parta m en to de M atem ´
aticas, ETS de Ingenieros Industriales, Universidad Polit´
ec-
nica de Madrid, C. Jos´
eGuti
´
errez Abascal 2, 28006 Madrid, Spain
E-mail address:jgutierrez@math.etsii.upm.es