Dynamics, Operator Theory, and Infinite HolomorphyView this Special Issue
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Homomorphisms between Algebras of Holomorphic Functions
For two complex Banach spaces and , in this paper, we study the generalized spectrum of all nonzero algebra homomorphisms from , the algebra of all bounded type entire functions on , into . We endow with a structure of Riemann domain over whenever is symmetrically regular. The size of the fibers is also studied. Following the philosophy of (Aron et al., 1991), this is a step to study the set of all nonzero algebra homomorphisms from into of bounded holomorphic functions on the open unit ball of and of all nonzero algebra homomorphisms from into .
The study of homomorphisms between topological algebras is one of the basic issues in this theory. Two are the main topological algebras that we come across when we deal with holomorphic functions on infinite dimensional spaces (see Section 2 for precise definitions): , the holomorphic functions of bounded type (which is a Fréchet algebra), and , the bounded holomorphic functions on the open unit ball (which is a Banach algebra). Here, as a first step in the study of the set of homomorphisms between spaces, we mainly focus on algebras of holomorphic functions of bounded type and homomorphisms between them; (i.e., continuous, linear, and multiplicative mappings). These were already considered in . There the focus was to study the homomorphisms as “individuals,” seeking properties of single ones. We have here a different interest: we treat them as a whole, considering the set We will call this set the generalized spectrum or simply the spectrum. Our main aim is to study and to define on it a topological and a differential structure.
This problem has the same flavor as considering , the spectrum of the algebra (i.e., the set of nonzero continuous, linear, and multiplicative ). This was studied in [2, 3], where a structure of Riemannian manifold over the bidual was defined on it (see also [4, Section 3.6] for a very neat and nice presentation and [5–7] for similar results). Our approach is very much indebted to that in  and we get up to some point analogous results, defining on a Riemann structure over (note that ). We will also be interested in the fibers of elements of .
The outline of the paper is the following. In Section 3, for two complex Banach spaces and , we study the generalized spectrum of all nonzero algebra homomorphisms from to . We endow it with a structure of Riemann domain over whenever is symmetrically regular. In Section 4, we focus on the sets (fibers) of elements in that are projected on the same element of . The size of these fibers is studied and we prove that they are big by showing that they contain big sets. Following the philosophy of , all about is a step to study in Section 5 the spectrum of all nonzero algebra homomorphisms from to of bounded holomorphic functions on the open unit ball of . Finally, in Section 6, we deal with the generalized spectrum of all nonzero algebra homomorphisms from to .
2. Definitions and Preliminaries
Unless otherwise stated capital letters such as will denote complex Banach spaces. The dual will be denoted by and the open ball of center and radius by . If and we just write . The space of continuous, linear operators from to will be denoted by ; this is a Banach space with the norm . The adjoint operator of will be denoted by .
We will denote the canonical inclusion of a space into its bidual by .
Given two complex locally convex spaces and , a mapping is a continuous -homogeneous polynomial if there is a continuous -linear mapping such that Throughout the paper every polynomial and multilinear mapping will be assumed to be continuous.
An -linear mapping is called symmetric if for every permutation of . Each -homogeneous polynomial has a unique symmetric mapping (which we denote by ) satisfying (2). If is a continuous -homogeneous polynomial between Banach spaces, the following expressions define norms for -linear mappings and for -homogeneous polynomials, respectively: The polarization formula gives [4, Corollary 1.8] Given an -linear mapping , the notation will mean that is repeated times and is repeated times.
A mapping is holomorphic on the open set if for every in , there exist , each one of them an -homogeneous polynomial, such that the series converges uniformly in some neighborhood of contained in . This is called the “Taylor series expansion” of at .
If and are Fréchet spaces, then is Gâteaux holomorphic if for every the function , is holomorphic on some neighborhood of . It is known [4, page 152-153] that is holomorphic if and only if it is Gâteaux holomorphic and continuous.
In the case that , and being complex Banach spaces, is holomorphic on if and only if it is Fréchet differentiable on and in that case for every . Also, by the Cauchy inequalities [4, Proposition 3.2] we have, for every , A holomorphic function is of bounded type if it sends bounded subsets of to bounded sets of . We denote by the space of holomorphic functions of bounded type from to . If we simply write . Given we write With this notation for a Banach space is a Fréchet algebra endowed with the topology of uniform convergence on the bounded sets, whose seminorms are (for ) Let be a differential manifold on a complex Banach space and a locally convex space. A mapping is said to be holomorphic (of bounded type) if is holomorphic (of bounded type) for every chart of .
Given , we write for the evaluation mapping at ; that is, for all holomorphic .
Let be a continuous bilinear form. Fix and for let be a net in weak-star convergent to . Since , then there exists . Now, fix and for let be a net in weak-star convergent to . Since , then there exists Following this idea Aron and Berner showed in  that every function admits an extension to the bidual, called the Aron-Berner extension, . By [9, Theorem 3], for every -homogeneous polynomial , we have .
A Banach space is symmetrically regular if for all continuous symmetric bilinear form it follows that
We are going to work with the set of nonzero algebra homomorphisms between spaces of holomorphic functions of bounded type defined in (1). Observe that an idempotent element in satisfies that is a subset of and is a connected set. Hence, either or . So, for any , we should have .
3. The Differential Structure of
Our aim in this section is to endow with a structure of Riemann domain over . On a first step, for each linear operator , we define, inspired by [12, Lemma 1], It is plain that is an algebra homomorphism and . This defines a natural inclusion: On the other hand there is also a projection: These clearly satisfy for every .
Given we consider the mapping defined by . It is a well-known fact that fixed in the mapping given by is a holomorphic function of bounded type. See [4, Proposition 6.30] or more in general the proof of [3, Theorem 2.2].
With this, for each and each , we can define by for all and . Let us see that is well defined. To do that we need to check that the function belongs to . We see first that it is holomorphic. The following function of two variables is holomorphic on each variable. Then Hartogs’ theorem gives that it is holomorphic and hence it is so when restricted to the diagonal . This gives that (15) is holomorphic.
We see now that is of bounded type. Given there exists such that for all . Hence since for all and all .
On the other hand it is easy to see that is an algebra homomorphism, which finally gives .
As a second step we show that Indeed, for each and , we have Thus, Finally, given and , we consider the set Now we assume that is symmetrically regular. We proceed as in the definition of the Riemann domain structure of the spectrum of and we omit the details (see [4, Section 6.3] for a complete and detailed explanation of the procedure). First of all, if , then , with . Hence, for any , Therefore, for , we have and is a neighborhood basis at .
Also, for , we have that if , then , for all and if , then for . This gives that the topology generated by is Hausdorff.
Let us note that for each in the subset is the connected component containing .
Summing all this up we have proved the following result.
Proposition 1. If is a symmetrically regular Banach space and is any Banach space, is a Riemann domain over and each connected component of is homeomorphic to .
Our aim now is to show that each function can be extended, in some sense, to a function on of bounded type. We do it with the following sort of Gelfand transform: and showing that this, when restricted to each connected component, is holomorphic of bounded type. To do that we need the following lemma.
Lemma 2. If and are complex Banach spaces and is an element of , then the mapping defined by for belongs to . Conversely, given in the mapping belongs to .
Proof. Let be in , and let be the Taylor series expansion of at . We have
Note that if we take by the polarization formula (4) and Cauchy inequalities (6), we have
By the properties of convergent double series of nonnegative numbers we obtain that, for each fixed , the series converges absolutely and uniformly on any product of balls in and with finite radii. Hence is a continuous -homogeneous polynomial and actually is an entire function from to that, by above inequalities, is of bounded type.
Conversely, consider in and define by . By definition, for each , belongs to . If we fix now , we have that is a continuous linear form on and , implying that is the composition of holomorphic mappings. Thus is holomorphic for every . By Hartogs’ theorem, . Finally, for fixed we have that ) is a bounded subset of . Hence and is bounded on the bounded subsets of .
Proposition 3. Let be a symmetrically regular Banach space and let be any Banach space. Given a function consider its extension defined in (24). We have that is a holomorphic function of bounded type. That is, for every .
Proof. The point is to prove that the function
is holomorphic of bounded type. For that, we introduce an auxiliary mapping , defined by
As above, we only need to check that it is separately holomorphic to conclude that is holomorphic. First we fix and and denote . We have and this belongs to . Now we fix and take . This mapping is holomorphic (of bounded type) since it is the composition of the linear mapping defined by with the holomorphic mapping of bounded type:
Finally we fix and denote . Again, this is the composition of a linear mapping defined by with the same holomorphic mapping (30). We conclude that is holomorphic.
Let now . Given there exists such that for every . Hence and is of bounded type. Since the mapping defined by is obviously holomorphic of bounded type we have that defined by is holomorphic of bounded type. Now a direct application of Lemma 2 gives that the mapping belongs to .
The above proposition is related to the study of extension of functions of bounded type given in .
4. The Size of the Fibers of
We focus now on the sets of elements in that are projected on the same element of . This is called the “fiber” of and is defined by Our aim in this section is to find out how big these fibers can be. To begin with, each fixed defines a set On the other hand every defines a composition homomorphism given by . This gives an inclusion which maps the set into the fiber . Let us check that is injective. Given , , there exists such that and since separates points of we can find with . Thus and this gives .
There is also a projection The mapping belongs to (and hence is well defined). This follows from the fact that is holomorphic. Clearly, . Also, note that determines the values that takes when restricted to . This means that when finite type polynomials are dense in , determines . So, the only homomorphisms in are the ’s and we have the following result, which is closely related to [13, Lemmas 4.5 and 4.6] and in [1, Theorem 21].
Proposition 4. Let and be Banach spaces. If finite type polynomials are dense in then for each there exists such that . Also, .
Let us note that the mapping is actually injective and, by Proposition 4, if finite type polynomials are dense, surjective. This means that even in the case when finite type polynomials are dense in (i.e., the space is rather small), the fibers are thick. Let us see now that if this is not the case (i.e., when there is a polynomial in that is not weakly continuous on bounded sets), this mapping is no longer surjective and we can find even more homomorphisms inside each fiber.
Proposition 5. If is symmetrically regular and there exists a polynomial on not weakly continuous on bounded sets at a point , then, for each , there is such that , for all .
Proof. It is enough to prove the result for because we can change fibers through the mapping
Also, if , then , with .
Let be a polynomial that is not weakly continuous on bounded sets at and a bounded net weakly convergent to such that , for all . For an ultrafilter containing the sets , let be given by Then is a homomorphism in (actually in ), that is, not of composition type. Indeed, since is a constant function on it follows that for every in and so . If for certain , then we have This says that , for all . Hence, which is a contradiction.
Something more can be said when there is a polynomial that is not weakly continuous on bounded sets. In this case, we can insert in each fiber a big set of homomorphisms that are not of composition type. We do it in detail in the fiber of 0.
First, note that if there is a polynomial not weakly continuous on bounded sets, then there is a homogeneous polynomial which is not weakly continuous on bounded sets at 0. So, the Aron-Berner extension of is not -continuous at any point [14, Corollary 2, Proposition 3] and [15, Proposition 1] (see also [16, Proof of Proposition 2.6]). Thus, we fix, for each a bounded net in which -converges to , but , for all . For each we fix also an ultrafilter containing the sets . Consider the set and define the following mapping where This mapping is well defined because ; then . The mapping is also injective. Indeed, if , then , for all and . Then, Therefore, , for all and, evaluating at 0, we obtain that and, hence, .
Note, also, that the homomorphisms are not of composition type. Indeed, if , for certain , , and , then, for all , If this were the case we would have , for all . Now, Evaluating at 0 leads to a contradiction.
Let us finish this analysis by studying : the composition of the inclusion defined in (35) with the inclusion of into . Then the mapping turns out to be holomorphic (endowing with an appropriate topology). We prepare the proof of this fact with a lemma that is a variant of a classical Dunford result; see also [17, Theorem 3].
Lemma 6. let be a Gâteaux holomorphic mapping from a complex Fréchet space to a Banach space such that is holomorphic for every , where is a norming subset of the closed unit ball of (i.e., for every ). Then is holomorphic on .
Proof. Let be a finite dimensional compact subset of . By hypothesis, is a compact subset of and hence it is bounded. Thus, there exists such that Hence, the family of scalar valued holomorphic functions is bounded on the finite dimensional compact subsets of . But as is a Fréchet space, then it is also Baire, and by  this family is locally bounded: given there exists an open neighborhood of such that We have obtained that the Gâteaux holomorphic function is locally bounded. Then it is holomorphic by [4, Proposition 3.7].
The following proposition gives that is holomorphic. The proof may seem at some points similar to that of Proposition 3, but the fact that in the target space we have now a nonmetrizable locally convex topology makes the whole situation much more delicate. We are going to consider now in the topology defined by the following fundamental system of seminorms: for , where and is a bounded subset of .
Proposition 7. The mapping is injective and holomorphic if we consider in the -topology.
Proof. Clearly is well defined. Our first step is to prove that is Gâteaux holomorphic. Let , , , and . Here we have . If is the Taylor series expansion of at on , then
for every and the convergence is absolute and uniform on the bounded subsets of [8, 9]. Thus
where our last step is simply formal. We are going to concentrate our effort now to show that this formal last equality holds in our setting. If we denote by the closure of , then is a bounded subset of for . Thus there exists such that , for every in and . We fix and we take such that . We have
Hence, by the properties of summability of double series of nonnegative numbers, the double series below is convergent in :
and its sum is again . On the other hand, by using the polarization formula (4), [9, Theorem 3], and Cauchy’s inequalities (6) we get
By applying now (55) we obtain
Since the function is the composition of a continuous multilinear mapping and two holomorphic mappings of bounded type, it is holomorphic of bounded type on . By using (57), we obtain that the series
-converges in ; hence belongs to for every in . Actually, if we consider , this is a linear operator. By (57), it is also continuous, since given there exists such that
for every . Then
We have, for with , again by (57),
As a consequence the series defined on with values in , -converges uniformly on the compacts of . Hence it is an entire function. Since we have proved that all series involved converge absolutely, we can apply the reordering of absolutely convergent double series to conclude that the last formal equality of (53) actually holds and then
is an entire function on for every . This gives that is Gâteaux holomorphic.
We fix now , a continuous seminorm of the fundamental system defined in (50) and denote by the completion of the normed space . Given and we define the continuous linear functional by . Clearly the quotient mapping , defined by , belongs to . On the other hand the set is a norming subset of since We can consider that remains Gâteaux holomorphic. Thus is Gâteaux holomorphic for every and every . If we show that it is continuous, then it will be holomorphic and, by Lemma 6, we will get that is holomorphic for every seminorm . Since is a complete space, we can conclude that is holomorphic.
Let . Now for fixed and , we consider and we choose such that As is uniformly continuous on bounded subsets of , for a given , there exists such that if with , then Let with . Clearly . Thus This gives that is continuous for every and and completes the proof.
We recall that the topology on is the topology of the pointwise convergence on the points of .
Corollary 8. The mapping