PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 127, Number 6, Pages 1715–1721 S 0002-9939(99)04917-5 Article electronically published on February 11, 1999
INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS ´ ´ MANUEL GONZALEZ AND JOAQU´IN M. GUTIERREZ (Communicated by Dale Alspach) Abstract. We characterize the holomorphic mappings f between complex Banach spaces that may be written in the form f = g ◦ T , where g is another holomorphic mapping and T is an operator belonging to a closed injective operator ideal. Analogous results are previously obtained for multilinear mappings and polynomials.
In recent years, several authors [1], [6], [7] have studied conditions on a holomorphic mapping f between complex Banach spaces so that it may be written in the form f = g ◦ T , where g is another holomorphic mapping and T is 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 T belongs to the ideal Co of 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 ◦ T with T ∈ Co, then f is 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 f between 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 T in the ideal WCo of weakly compact operators, and F = C the complex field. In [6, Satz 2.1], using interpolation techniques, the factorization f = g ◦ T has been characterized in terms of the derivatives of f , for T in 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 T in U. As a consequence, we prove that a holomorphic map f : E → F admits such a factorization if and only if f is uniformly continuous on bounded sets with respect to a suitable topology τU . In the case U = Co, the topology τU is 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, multilinear 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 Mathematical Society
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´ ´ MANUEL GONZALEZ AND JOAQU´IN M. GUTIERREZ
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 operators, 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, F and G will denote Banach spaces, and BE will 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 E into F , endowed with the usual operator norm. If E1 , . . . , Ek are also Banach spaces, the notation Lk (E1 , . . . , Ek ; F ) stands for the space of all (continuous) k-linear mappings from E1 × · · · × Ek into F . . . to mean that the i th coordinate is We shall use the notational convention .[i] not involved. For instance, Lk−1 (E1 , .[i] . ., Ek ; F ) := Lk−1 (E1 , . . . , Ei−1 , Ei+1 , . . . , Ek ) . To each A ∈ Lk (E1 , . . . , Ek ; F ) and i ∈ {1, . . . , k} we associate an operator . ., Ek ; F ) Ai : Ei −→ Lk−1 (E1 , .[i] given by Ai (xi )(x1 , .[i] . ., xk ) := A(x1 , . . . , xk )
(xj ∈ Ej ; 1 ≤ j ≤ k) .
The space of all (continuous) k-homogeneous polynomials from E into F is denoted by P(kE, F ). To each P ∈ P(kE, F ) we associate an operator P¯ : E −→ P(k−1 E, F ) given by P¯ (x)(y) := Pˆ x, y, (k−1) . . . , y where Pˆ is the unique symmetric, k-linear mapping associated to P . For complex spaces E and F , H(E, F ) will denote the space of all holomorphic mappings from E into F , and Hb (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 ⊂ E is circled if for every x ∈ A and complex λ with |λ| = 1, we have λx ∈ 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 U is said to be injective [14, 4.6.9] if given an operator T ∈ L(E, F ) and an injective isomorphism i : F → G, we have that T ∈ U whenever iT ∈ U. We say that U is closed [14, 4.2.4] if, for all E and F , the space U(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 U is closed and injective if and only if for an operator T ∈ L(E, F ) to belong to U it is both necessary and sufficient that for each > 0 there exist a Banach space G and an operator S ∈ U(E, G ) so that (x ∈ E) . kT xk ≤ kS (x)k + kxk
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INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS
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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 τi be a vector topology on a Banach space Ei (1 ≤ i ≤ k). Given a k-linear mapping A ∈ Lk (E1 , . . . , Ek ; F ) with associated operators Ai : Ei → Lk−1 (E1 , .[i] . ., Ek ; F ), we have that A is uniformly τ1 × · · · × τk -continuous on bounded sets if and only if the operators Ai are (uniformly) τi -continuous on bounded sets. Proof. Suppose A is uniformly τ1 × · · · × τk -continuous on bounded sets. Given > 0, we can find τi zero neighbourhoods Ui ⊂ Ei so that kA(x1 , . . . , xk ) − A(y1 , . . . , yk )k < whenever xi , yi ∈ BEi satisfy xi − yi ∈ Ui (1 ≤ i ≤ k). Take j ∈ {1, . . . , k}, zi ∈ BEi for i = 1, .[j] . ., k, and xj , yj as above. Then
. ., zk = kA(z1 , . . . , xj , . . . , zk )−A(z1 , . . . , yj , . . . , zk )k < .
(Aj xj − Aj yj ) z1 , .[j] Hence, Aj is uniformly τj -continuous on bounded sets. Conversely, let Ai be τi -continuous on bounded sets, for 1 ≤ i ≤ k. Given > 0, there is a τi zero neighbourhood Ui ⊂ Ei so that kAi x − Ai yk < whenever x, y ∈ BEi satisfy x − y ∈ Ui . Given xi , yi ∈ BEi with xi − yi ∈ Ui (1 ≤ i ≤ k), we have kA(x1 , . . . , xk ) − A(y1 , . . . , yk )k ≤ kA1 (x1 − y1 )(x2 , . . . , xk )k + kA2 (x2 − y2 )(y1 , x3 , . . . , xk )k + · · · +kAk (xk − yk )(y1 , . . . , yk−1 )k ≤ k , and so, A is uniformly τ1 × · · · × τk -continuous on bounded sets. Let U be an injective operator ideal. On every Banach space E we consider the topology τU generated by the seminorms pT (x) := kT xk ,
for T ∈ U(E, F ) , F any Banach space.
We then have (see [10, §3]): (1)
U(E, F ) = L ((E, τU ), F ) .
Proposition 3. Given a closed injective ideal U and S ∈ L(E, F ), we have that S ∈ U(E, F ) if and only if S is τU -continuous on bounded subsets. Proof. Note that, since U is closed, τU is the finest locally convex topology that agrees with τU on bounded subsets (see [10, Proposition 4.2]). Then apply (1). One of our key results is the next one. Theorem 4. Given A ∈ Lk (E1 , . . . , Ek ; F ), let U be a closed injective operator ideal. The following assertions are equivalent: . ., Ek ; F ) belongs to (a) for every 1 ≤ i ≤ k, the operator Ai : Ei → Lk−1 (E1 , .[i] U; (b) there are Banach spaces Yi and operators Ti ∈ U(Ei , Yi ) (1 ≤ i ≤ k) so that kA(x1 , . . . , xk )k ≤ kT1 x1 k · . . . · kTk xk k; (c) there are Banach spaces Yi , operators Ti ∈ U(Ei , Yi ) (1 ≤ i ≤ k) and a mapping D ∈ Lk (Y1 , . . . , Yk ; F ) so that A = D ◦ (T1 , . . . , Tk ); (d) A is uniformly τU -continuous on bounded subsets.
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´ ´ MANUEL GONZALEZ AND JOAQU´IN M. GUTIERREZ
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Proof. (a) ⇒ (b). Consider the seminorms qi (x) := inf nk−1 kAi xk + n−1 kxk n∈N
(x ∈ Ei )
for 1 ≤ i ≤ k. Letting Yi be the completion of (Ei /ker(qi ), qi ), define Ti : Ei → Yi by Ti x := x + ker(qi ), for x ∈ Ei . For every n ∈ N and 1 ≤ i ≤ k, we have kTi xk = qi (x) ≤ nk−1 kAi xk + n−1 kxk
(x ∈ Ei ).
By Lemma 1 we have that Ti ∈ U for 1 ≤ i ≤ k. It is enough to show that given xi ∈ Ei with kTi xi k < 1, for 1 ≤ i ≤ k, we have kA(x1 , . . . , xk )k < 1. Given i ∈ {1, . . . , k}, there is ni ∈ N so that kAi xi k + n−1 nk−1 i i kxi k < 1 . Assume nj = max{n1 , . . . , nk }. We have kxi k < nj for all 1 ≤ i ≤ k. Moreover, . Hence, kAj xj k < n1−k j
. ., xk ≤ kAj xj k · kx1 k· .[j] . . ·kxk k < 1 . kA(x1 , . . . , xk )k = Aj xj x1 , .[j] (b) ⇒ (c). Since U is injective, we can assume that Ti (Ei ) is dense in Yi , for 1 ≤ i ≤ k. Define D(T1 x1 , . . . , Tk xk ) := A(x1 , . . . , xk ) If Ti xi =
Ti x0i
(xi ∈ Ei , 1 ≤ i ≤ k).
for all 1 ≤ i ≤ k, we then have A(x1 , . . . , xk ) − A(x01 , . . . , x0k ) = A(x1 − x01 , x2 , . . . , xk ) + A(x01 , x2 − x02 , . . . , xk ) + · · · + A(x01 , . . . , x0k−1 , xk − x0k )
= 0, x0i )
= 0 for 1 ≤ i ≤ k. So, D is well defined and continuous, with since Ti (xi − kDk ≤ 1. Hence we can extend it to Y1 × · · · × Yk and, denoting the extension by D as well, we have A = D ◦ (T1 , . . . , Tk ). (c) ⇒ (a). Assume A = D ◦ (T1 , . . . , Tk ) with Ti ∈ U(Ei , Yi ), for 1 ≤ i ≤ k. For each i, define . ., Ek ; F ) Bi : Yi −→ Lk−1 (E1 , .[i] by (i)
. ., uk ) = D(T1 u1 , . . . , y , . . . , Tk uk ) (Bi y)(u1 , .[i]
(y ∈ Yi ; u1 ∈ E1 , .[i] . ., uk ∈ Ek ) .
Clearly, Bi is continuous, with . . ·kTk k · kDk , kBi k ≤ kT1 k· .[i] and Ai = Bi ◦ Ti . Hence Ai ∈ U. (a) ⇔ (d) by Propositions 2 and 3. If U is the ideal of compact operators, then the assertion (d) of the Theorem states that A is 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 U such that every k-linear mapping 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 = Co the assertion
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INJECTIVE FACTORIZATION OF HOLOMORPHIC MAPPINGS
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is true. It is obviously true if U is the ideal of all bounded operators. We proved in [7] that it fails for U = WCo, and also for U the completely continuous operators. Corollary 5. Given P ∈ P(kE, F ), let U be a closed injective operator ideal. The following assertions are equivalent: (a) the operator P¯ : E → P(k−1E, F ) belongs to U; (b) there are a Banach space Y and an operator T ∈ U(E, Y ) so that kP (x)k ≤ kT (x)kk for all x ∈ E; (c) there are a Banach space Y , an operator T ∈ U(E, Y ) and a polynomial Q ∈ P(k Y, F ) so that P = Q ◦ T ; (d) P is 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 , . . . , Tk ), where A 6= 0, D ∈ Lk (Y1 , . . . , Yk ; F ), and Ti ∈ L(Ei , Yi ) for 1 ≤ i ≤ k. Then the spaces Yi may be renormed so that kAk = kDk · kT1 k · . . . · kTk k. Proposition 7. Let U be a closed injective operator ideal. If f ∈ H(E, F ) is uniformly τU -continuous on bounded sets, a ∈ E and k ∈ N, then dkf (a) is uniformly τU -continuous on bounded sets. Proof. From the Cauchy integral formula [13, Corollary 7.3], we obtain
1 Z
1 k 1 f (a + λx) − f (a + λy)
d f (a)(x) − dkf (a)(y) = dλ
k!
2πi |λ|=1
k! λk+1 ≤
sup kf (a + λx) − f (a + λy)k .
|λ|=1
Lef B be a circled, bounded subset of E. Given > 0, there is a circled τU zero neighbourhood U in E so that kf (x0 ) − f (y 0 )k < whenever x0 , y 0 ∈ a + B satisfy x0 − y 0 ∈ U . Therefore, given x, y ∈ B with x − y ∈ U , we easily get from the above inequality:
1 k
d f (a)(x) − 1 dkf (a)(y) < .
k!
k! If (Yk ) is a sequence of Banach spaces, we denote by c0 (Yk ) the Banach space of all sequences (yk ) so that yk ∈ Yk and kyk k → 0, endowed with the supremum norm. In the proof of the following Theorem we shall use the well-known fact that if B ⊂ E is bounded and f : B → F is a uniformly continuous mapping, then f (B) is bounded in F . Theorem 8. Given f ∈ H(E, F ), let U be a closed injective operator ideal. Then f is uniformly τU -continuous on bounded subsets if and only if there are a Banach space G, an operator T ∈ U(E, G) and a mapping g ∈ Hb (G, F ) so that f = g ◦ T . Proof. Suppose f = g ◦ T , with g and T as in the statement. Let A ⊂ E be bounded. Then T is uniformly continuous from (A, τU ) into (T A, k · k). Since g is bounded on bounded sets, it is norm-to-norm uniformly continuous on bounded sets. Hence, f is uniformly continuous from (A, τU ) into (F, k · k).
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´ ´ MANUEL GONZALEZ AND JOAQU´IN M. GUTIERREZ
Conversely, suppose f is uniformly τU -continuous on bounded sets. We write the Taylor series expansion of f at the origin as f (x) =
∞ X
Pk (x)
(x ∈ E).
k=0
By Proposition 7, Pk is uniformly τU -continuous on bounded sets, for each k. By Theorem 4, there are a Banach space Yk , an operator Tk ∈ U(E, Yk ) and a polyˆ k k · kTk kk nomial Qk ∈ P(k Yk , F ) such that Pk = Qk ◦ Tk . We can get kPˆk k = kQ (Lemma 6). Since τU is coarser than the norm topology, it follows that f ∈ Hb (E, F ), and so lim kPk k1/k = 0. From the inequalities kk kPk k kPk k ≤ kPˆk k ≤ k! (see [13, Theorem 2.2]), and using the Stirling formula, we get lim kPˆk k1/k = 0. ˆ k k1/k → 0 and kTk k → 0. Define Therefore, we can assume that kQ T : E −→ Y := c0 (Yk ) by T x := (Tk x)k . Clearly, T ∈ U. Denoting πk : (yi ) ∈ Y 7−→ yk ∈ Yk , P∞ 1/k = we define g : Y → F by g(y) := k=1 Qk ◦ πk (y). Since lim kQk ◦ πk k 1/k = 0, we get that g is a holomorphic mapping of bounded type that lim kQk k 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 : `∞ → `2 such that q(B`∞ ) ⊇ B`2 . Letting q(x)i be the i th coordinate of q(x) ∈ `2 , we define P (x) :=
∞ X
xi q(x)2i ,
for x = (xi ) ∈ `∞ .
i=1
The associated operator P¯ : `∞ → P(2 `∞ ) is given by ∞
∞
2X 1X xi q(y)2i + yi q(x)i q(y)i . P¯ (x)(y) := 3 i=1 3 i=1 We only have to show that P¯ ∈ / WCo (see Corollary 5) or, equivalently, that P¯ is not completely continuous. Denoting by en the sequence (0, . . . , 0, 1, 0, . . . ) with 1 in the nth position, we select a sequence (xn ) ⊂ B`∞ so that q(xn ) = en . Then, 3P¯ (en )(xn ) = 1 + 2xnn q(en )n . Since q(en ) → 0, we have 3P¯ (en )(xn ) → 1 and, therefore, kP¯ (en )k does not converge to 0.
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[email protected] ´ ticas, ETS de Ingenieros Industriales, Universidad Polit´ Departamento de Matema ecnica de Madrid, C. Jos´ e Guti´ errez Abascal 2, 28006 Madrid, Spain E-mail address:
[email protected]
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