SYMMETRIC POLYNOMIALS AND HOLOMORPHIC FUNCTIONS ON INFINITE DIMENSIONAL SPACES

. A survey of general results about spectra of uniform algebras of symmetric holomorphic functions and algebras of symmetric analytic functions of bounded type on Banach spaces is given.


SYMMETRIC POLYNOMIALS ON REARRANGEMENT-INVARIANT FUNCTION SPACES
Let X, Y be Banach spaces over the field K of real or complex numbers. A mapping P : X → Y is called an n-homogeneous polynomial if there exists a symmetric n-linear mapping A : X n → Y such that for all x ∈ X P(x) = A(x, . . . , x).
A polynomial of degree n on X is a finite sum of k-homogeneous polynomials, k = 0, . . . , n. Let us denote by P ( n X, Y) the space of all n-homogeneous continuous polynomials P : X → Y and by P (X, Y) the space of all continuous polynomials.
It is well known ( [13], XI §52) that for n < ∞ any symmetric polynomial on C n is uniquely representable as a polynomial in the elementary symmetric polynomials (G i ) n i=1 , G i (x) = ∑ k 1 <···<k i x k 1 . . . x k i .
Symmetric polynomials on p and L p [0, 1] for 1 ≤ p < ∞ were first studied by Nemirovski and Semenov in [16]. In [11] González, Gonzalo and Jaramillo investigated algebraic bases of various algebras of symmetric polynomials on so called rearrangement-invariant function spaces, that is spaces with some symmetric structure. Up to some inessential normalisation, the study of rearrangement-invariant function spaces is reduced to the study of the following three cases: 1. I = N and the mass of every point is one; 2. I = [0, 1] with the usual Lebesgue measure; These are well-defined symmetric polynomials on X(I) and we will call them the elementary symmetric polynomials on X(I).

SYMMETRIC POLYNOMIALS ON SPACES WITH A SYMMETRIC BASIS
Let X = X(N) be a Banach space with a symmetric basis {e n }. A polynomial P on X is symmetric if for every permutation σ ∈ G(N) a i e σ(i) .
We consider the finite group G n (N) of permutations of {1, . . . , n} and the σ-finite group G 0 (N) = ∪ n G n (N) as subgroups of G(N). By continuity, a polynomial is symmetric if and only if it is G 0 (N)-invariant. Indeed, if P is G 0 (N)-invariant and σ ∈ G(N), Recall that a sequence {x n } is said to have a lower p-estimate for some p ≥ 1, if there is a constant C > 0 such that a i x i for all a 1 , . . . , a n ∈ R.
Note that X ⊂ r if and only if the basis has a lower r-estimate, and therefore we have in this case J (X) = {r ∈ N : {e n } has a lower r-estimate}. Now we define n 0 (X) = inf J (X), where we understand that the infimum of the empty set is ∞. The elementary symmetric polynomials are then where r ≥ n 0 (X).
Theorem 1.1. [11] Let X be a Banach space with a symmetric basis e n , let P be a symmetric polynomial on X and consider k = deg P and N = n 0 (X).
2. If k ≥ N, then there exists a real polynomial q of several real variables such that f m for all f ∈ X, where m = min{n ∞ (X), k}. Theorem 1.3. [11] Let X[0, ∞) be a separable rearrangement-invariant function space, let P be a G 0invariant polynomial on X[0, ∞) and consider k = deg P. Let n 0 and n ∞ be defined as above.
2. If n 0 ≤ n ∞ and n 0 ≤ k, then there is a real polynomial q in several real variables such that where m = min{n ∞ , k}.

UNIFORM ALGEBRAS OF SYMMETRIC HOLOMORPHIC FUNCTIONS
Let X be a Banach sequence space with a symmetric norm, that is, for all permutations σ : N → N, and x = (x n ) ∈ B also (x σ (1) , . . . , x σ(n) , . . .) ∈ B, where B is an open unit ball.
A holomorphic function f : B → C is called symmetric if for all x ∈ B and all permutations σ : N → N, the following holds: Our interest throughout this section will be in the set A us (B) = { f : B → C| f is holomorphic, uniformly continuous, and symmetric on B}.
The following result is straightforward.  [5] Let P : c 0 → C be an n-homogeneous polynomial and ε > 0. Then there is N ∈ N and an n-homogeneous polynomial Q : Corollary 2.4. [4] For all n ∈ N, n ≥ 1, the only n-homogeneous symmetric polynomial P : c 0 → C is P = 0.
Since any function f ∈ A us (B) can be uniformly approximated on B by finite sums of symmetric homogeneous polynomials, it follows that A us (B) consists of just the constant functions when B is the open unit ball of c 0 .
So, for all N, x j x k , Symmetric Polynomials and Holomorphic Functions... 27 where a = P(e 1 ) and b = A(e j , e k ). From this, we can conclude, that for X = 1 , the space of symmetric 2-homogeneous polynomials on 1 , P s ( 2 1 ), is 2-dimensional with basis {∑ j x 2 j , ∑ j =k x j x k }. On the other hand, the corresponding space P s ( 2 2 ) of symmetric 2-homogeneous polynomials on 2 , is 1-dimensional with basis {∑ j x 2 j }. For 1 < p < 2, P s ( 2 p ) is also the one-dimensional space generated by This argument can be extended to all n and all p, and we can conclude that for all n, p, the space of symmetric n-homogeneous polynomials on p , P s ( n p ), is finite dimensional. Consequently, since for all f ∈ A us (B), f is a uniform limit of symmetric n-homogeneous polynomials, we have reasonably good knowledge about the functions in A us (B). So we can say that A us (B), for B the open unit ball of an p space, is a "small" algebra.

THE SPECTRUM OF A us (B)
Recall that the spectrum (or maximal ideal space) of a Banach algebra A with identity e is the set M(A) = {ϕ : A → C |ϕ is a homomorphism and ϕ(e) = 1}. We recall that if ϕ ∈ M(A), then ϕ is automatically continuous with ϕ = 1. Moreover, when we consider it as a subset of A * with the weak-star topology, M(A) is compact.
We will examine M(A us (B)) when B = B p . The most obvious element in M(A us (B)) is the evaluation homomorphism δ x at a point x of B (recalling that since the functions in A us (B) are uniformly continuous, they have unique continuous extensions to B). Of course, if x, y ∈ B are such that y can be obtained from x by a permutation of its coordinates, then δ x = δ y . It is natural to wonder whether M(A us (B)) consists of only the set of equivalence classes {δ x |x ∈ B}, where x ∼ y means that x and y differ by a permutation. Example 2.6. [1,4] For every n ∈ N define F n : B → C by F n (x) = ∑ ∞ j=1 x n j . To simplify, we take B = B 2 (so that F n will be defined only for n ≥ 2). It is known that the algebra generated by {F n |n ≥ 2} is dense in A us (B). For each k ∈ N, let It is routine that each v k has norm 1, that δ v k (F 2 ) = 1 for all k ∈ N, and that for all n ≥ 3, Since M(A us (B)) is compact, the set {δ v k |k ∈ N} has an accumulation point ϕ ∈ M(A us (B)). It is clear that ϕ(F 2 ) = 1 and ϕ(F n ) = 0 for all n ≥ 3. It is not difficult to verify that ϕ = δ x for every x ∈ B. This construction could be altered slightly, by letting v k = 1 √ k (α 1 e 1 + ... + α k e k ), where each |α j | ≤ 1. Thus, with this method we give a small number of additional homomorphisms in M(A us (B)) that do not correspond to point evaluations.
It should be mentioned that it is not known whether M(A us (B p )) contains other points. However, in [1] was given a different characterization of M(A us (B p )). In order to do this, we first simplify our notation by considering only B 1 . For each n ∈ N, define F n : B 1 → C n as 28 I.V. Chernega follows: Let D n = F n (B 1 ), and let [D n ] be the polynomially convex hull of D n (see, e.g., [12]). Let , for all n ∈ N.} In other words, Σ 1 is the inverse limit of the sets [D n ], endowed with the natural inverse limit topology.
The analogous results, and the analogous definitions, are valid for Σ p and M(A us (B p )).
The basic steps in the proof of Theorem 2.7 are as follows: First, since the algebra generated by {F n |n ≥ 1} is dense in A us (B 1 ), each homomorphism ϕ ∈ M(A us (B 1 )) is determined by its behavior on {F n }. Next, every symmetric polynomial P on 1 can be written as P = Q • F n for some n ∈ N and some polynomial Q : C n → C. Finally, to each (b i ) ∈ Σ 1 , one associates ϕ = ϕ (b i ) : A us (B 1 ) → C by ϕ(P) = Q(b 1 , . . . , b n ). This turns out to be a well-defined homomorphism, and the mapping ) is a homeomorphism. Unlike the infinite dimensional case, the following result holds.

THE SPECTRUM OF
Theorem 2.8. [1,4] Every homomorphism ϕ : A s (B) → C is an evaluation at some point of B.
We describe below the main ideas in the proof of this result.
Proposition 2.9. [1,4] Let C ⊂ C n be a compact set. Then C is symmetric and polynomially convex if and only if C is polynomially convex with respect to only the symmetric polynomials.
In other words, C is symmetric and polynomially convex if and only if Proposition 2.10. [1,4] Let B be the open unit ball of a symmetric norm on C n . Then the algebra generated by the symmetric polynomials R 1 , . . . , R n is dense in A s (B).
Symmetric Polynomials and Holomorphic Functions... 29 To prove Theorem 2.8, let us consider the symmetric polynomials P 1 = R 1 − ϕ(R 1 ), . . . , P m = R m − ϕ(R m ). If ker P 1 ∩ · · · ∩ ker P m = ∅, then Lemma 2.11 implies that there are symmetric polynomials Q 1 , . . . , Q m on C n such that ∑ m j=1 P j Q j ≡ 1. This is impossible, since ϕ(P j Q j ) = 0. Therefore, there exists some x ∈ C n such that P j (x) = 0 for all j, which means ϕ(R j ) = R j (x) for all j. By Proposition 2.10, ϕ(P) = P(x), for all symmetric polynomials P : C n → C.
So, for all such P, |ϕ(P)| = |P(x)| ≤ P . This means that x belongs to the symmetrical polynomial convex hull of B. Since B is symmetric and convex, it is symmetrically polynomially convex (by Proposition 2.9). Thus x ∈ B.

THE ALGEBRA OF SYMMETRIC ANALYTIC FUNCTIONS ON p
Let us denote by H bs ( p ) the algebra of all symmetric analytic functions on p that are bounded on bounded sets endowed with the topology of the uniform convergence on bounded sets and by M bs ( p ) the spectrum of H bs ( p ), that is, the set of all non-zero continuous complex-valued homomorphisms.

THE RADIUS FUNCTION ON M bs ( p )
Following [3] we define the radius function R on M bs ( p ) by assigning to any complex homomorphism φ ∈ M bs ( p ) the infimum R(φ) of all r such that φ is continuous with respect to the norm of uniform convergence on the ball rB p , that is |φ As in the non symmetric case, we obtain the following formula for the radius function where φ n is the restriction of φ to P s ( n p ) and φ n is its corresponding norm.
Proof. To prove (3.1) we use arguments from [3, 2.3. Theorem]. Recall that Then there is a sequence of homogeneous symmetric polynomials P j of degree n j → ∞ such that P j = 1 and |φ(P j )| > t n j . If 0 < r < t, then by homogeneity, and φ is not continuous for the r norm. It follows that R(φ) ≥ r, and on account of the arbitrary choice of r we obtain R(φ) ≥ lim sup n→∞ φ n 1/n .

I.V. Chernega
Let now be s > lim sup n→∞ φ n 1/n so that s m ≥ φ m for m large. Then there is c ≥ 1 such that φ m ≤ cs m for every m. If r > s is arbitrary and f ∈ H bs ( p ) has Taylor series expansion Hence Thus φ is continuous with respect to the uniform norm on rB, and R(φ) ≤ r. Since r and s are arbitrary,

AN ALGEBRA OF SYMMETRIC FUNCTIONS ON THE POLYDISK OF 1
Let us denote It is easy to see that D is an open unbounded set. We shall call D the polydisk in 1 .
and and notice that Symmetric Polynomials and Holomorphic Functions... 31 Note that F is an analytic mapping from D into 1 since F (x) can be represented as a convergent series F (x) = ∞ ∑ k=1 F k (x)e k for every x ∈ D and F is bounded in a neighborhood of zero (see [9], p. 58).
For any fixed x ∈ D and t ∈ C such that tx ∈ D, let g(F (tx)) = ∑ ∞ j=1 t j r j (x) be the Taylor series at the origin. Then Let us compute r m (x). We have It is easy to see that the sum on the right hand side of (3.2) is finite. Since g(F (x)) = 0 for every x ∈ D, then r m (x) = 0 for every m. Further being F 1 , . . . , F n algebraically independent q k,i 1 ...i n = 0 in (3.2) for an arbitrary k < m, k 1 i 1 + . . . + k n i n = m. As this is true for every m then Q n ≡ 0 for n ∈ N. So g(x) ≡ 0 on 1 .
Let us denote by H 1 s (D) the algebra of all symmetric analytic functions which can be repre- with the topology that turns the bijection Ψ an homeomorphism. This topology is the weakest topology on H 1 s (D) in which the following seminorms are continuous: Note that Ψ is a homomorphism of algebras. So we have proved the following proposition: There is an onto isometric homomorphism between the algebras H 1 s (D) and H b ( 1 ).
The following example shows that there exists a character on H 1 s (D), which is not an evaluation at any point of D.
Example 3.6. [6] Let us consider a sequence of real numbers (a n ), 0 ≤ |a n | < 1 such that (a n ) ∈ 2 \ 1 and that the series ∞ ∑ n=1 a n conditionally converges to some number C. Despite (a n ) / ∈ 1 , evaluations on (a n ) are determined for every symmetric polynomial on 1 . In particular, F 1 ((a n )) = C, F k ((a n )) = ∑ a k n < ∞ and {F k ((a n ))} ∞ k=1 ∈ 1 . So (a n ) "generates" a character on H 1 s (D) by the formula ϕ( f ) = Ψ( f )(F ((a n ))). Since (a n ) ∈ 2 , then F k ((a π(n) )) = F k ((a n )) , k > 1. Notice that there exists a permutation on the set of positive integers, π, such that ∞ ∑ n=1 a π(n) = C = C. For such a permutation π we may do the same construction as above and obtain a homomorphism ϕ π "generated by evaluation at (a π(n) )", ϕ( f ) = Ψ( f )(F ((a π(n) ))). Let us suppose that there exist x, y ∈ D such that ϕ( f ) = f (x) and ϕ π ( f ) = f (y) for every function f ∈ H 1 s (D). Since ϕ(F k ) = ϕ π (F k ), k ≥ 2, then by [1] Corollary 1.4, it follows that there is a permutation of the indices that transforms the sequence x into the sequence y. But this cannot be true, because F 1 (x) = ϕ(F 1 ) = ϕ π (F 1 ) = F 1 (y). Thus, at least one of the homomorphisms ϕ or ϕ π is not an evaluation at some point of D.
Note that the the homomorphism "generated by evaluation at (a n )" is a character on P s ( 1 ) too, but we do not know whether this character is continuous in the topology of uniform convergence on bounded sets.

THE SYMMETRIC CONVOLUTION
Recall that in [3] the convolution operation " * " for elements ϕ, θ in the spectrum, M b (X), of H b (X), is defined by In [6] we have introduced the analogous convolution in our symmetric setting. It is easy to see that if f is a symmetric function on p , then, in general, f (· + y) is not symmetric for a fixed y. However, it is possible to introduce an analogue of the translation operator which preserves the space of symmetric functions on p .
Let us indicate some elementary properties of the intertwining. Proposition 3.8. [6] Given x, y ∈ p the following assertions hold.
(3) F n (x • y) = F n (x) + F n (y) for every n ≥ p. Proof. Note that x • y = x • 0 + 0 • y and that the map x → x • 0 is linear. Thus the map x → x • y is analytic and maps bounded sets into bounded sets, and so is its composition with f . Moreover, f (x • y) is obviously symmetric. Hence it belongs to H bs ( p ).
The mapping f → T s y ( f ) where T s y ( f )(x) = f (x • y) will be referred as to the intertwining operator. Observe that T s Proposition 3.10. [6] For every y ∈ p , the intertwining operator T s y is a continuous endomorphism of H bs ( p ).
Proof. Evidently, T s y is linear and multiplicative. Let x belong to p and x ≤ r. Then x • y ≤ p r p + y p and |T s y f (x)| ≤ sup So T s y is continuous. Using the intertwining operator we can introduce a symmetric convolution on H bs ( p ) . For any θ in H bs ( p ) , according to (3.4) the radius function R(θ • T s y ) ≤ p R(θ) p + y p . Then arguing as in [3, 6.1. Theorem] , it turns out that for fixed f ∈ H bs ( p ) the function y → θ • T s y ( f ) also belongs to H bs ( p ). Definition 3.11. For any φ and θ in H bs ( p ) , their symmetric convolution is defined according to (φ θ)( f ) = φ(y → θ(T s y f )). Corollary 3.12. [6] If φ, θ ∈ M bs ( p ), then φ θ ∈ M bs ( p ).
[7] a) For every ϕ, θ ∈ M bs ( p ) the following holds: b) The semigroup (M bs ( p ), ) is commutative, the evaluation at 0, δ 0 , is its identity and the cancelation law holds.
Proof. Observe that for each element F k in the algebraic basis of polynomials, {F k }, we have To check that the convolution is commutative, that is, φ θ = θ φ, it suffices to prove it for symmetric polynomials, hence for the basis {F k }. Bearing in mind (3.5) and also by exchanging parameters (θ ϕ)(F k ) = θ(F k ) + ϕ(F k ) = (ϕ θ)(F k ) as we wanted.
It also follows from (3.5) that the cancelation rule is valid for this convolution: If ϕ θ = ψ θ, then ϕ(F k ) + θ(F k ) = ψ(F k ) + θ(F k ), hence ϕ(F k ) = ψ(F k ), and thus, ϕ = ψ. Example 3.14. [7] There exist nontrivial elements in the semigroup (M bs ( p ), ) that are invertible: In [1, Example 3.1] it was constructed a continuous homomorphism ϕ = Ψ 1 on the uniform algebra A us (B p ) such that ϕ(F p ) = 1 and ϕ(F i ) = 0 for all i > p. In a similar way, given λ ∈ C we can construct a continuous homomorphism Ψ λ on the uniform algebra A us (|λ|B p ) such that Ψ λ (F p ) = λ and Ψ λ (F i ) = 0 for all i > p : It suffices to consider for each n ∈ N, the element v n = λ n 1/p (e 1 + · · · + e n ) for which F p (v n ) = λ, and lim n F j (v n ) = 0. Now, the sequence {δ v n } has an accumulation point Ψ λ in the spectrum of A us (|λ|B p ). We use the notation ψ λ for the restriction of Ψ λ to the subalgebra H bs ( p ) of A us (|λ|B p ). It turns out that ψ λ ψ −λ = δ 0 since for all elements F j in the algebraic basis, Therefore, we obtain a complex line of invertible elements {ψ λ : λ ∈ C}.
As in the non-symmetric case [3] Theorem 5.5, the following holds: Proof. For every z ∈ C, consider the composition operator L z : H bs ( p ) → H bs ( p ) defined according to L z ( f )((x n )) := f ((zx n )), and then, the restriction L * z to M bs ( p ) of its transpose map. Now put ϕ z : For each f ∈ H bs ( p ) the self-map of C defined according to z ϕ z ( f ) is entire by [3] Lemma 5.4.(i). Therefore, the mapping z ∈ C ϕ z ∈ M bs ( p ) is analytic. Since ϕ = δ 0 , the set Σ := {k ∈ N : ϕ(F k ) = 0} is non-empty. Let m be the first element of Σ, so that ϕ(F m ) = 0. Then if ϕ z = ϕ w , one has z m ϕ(F m ) = w m ϕ(F m ), hence z m = w m . Taking the principal branch of the m th root, the map ξ ϕ m √ ξ is one-to-one.
Recall that a linear operator T : H bs ( p ) → H bs ( p ) is said to be a convolution operator if there is θ ∈ M bs ( p ) such that T f = θ f . Let us denote H conv ( p ) := {T ∈ L(H bs ( p )) : T is a convolution operator}.

Proposition 3.16. [7]
A continuous homomorphism T : H bs ( p ) → H bs ( p ) is a convolution operator if, and only if, it commutes with all intertwining operators T s y , y ∈ p .
It is, clearly, bijective. Moreover we obtain a representation of the convolution semigroup Proof.-First, notice that using the above proposition, . Thus the statement follows.
We say that a function f ∈ H bs ( p ) is finitely generated if there are a finite number of the basis functions {F k } and an entire function q such that f = q(F 1 , . . . , F j ).

Theorem 3.21. A function f ∈ H bs ( p ) is w p -continuous if and only if it is finitely generated.
Proof. Clearly, every finitely generated function is w p -continuous. Let us denote by V n the finite dimensional subspace in p spanned by the basis vectors {e 1 , . . . , e n }. First we observe that if there is a positive integer m such that the restriction f | V n of f to V n is generated by the restrictions of F 1 , . . . , F m to V n for every n ≥ m, then f is finitely generated. Indeed, for given n ≥ k ≥ m we can write for some entire functions q 1 and q 2 on C n . Since (see e. g. [1]) and f | V n is an extension of f | V k we have q 1 (t 1 , . . . , t n ) = q 2 (t 1 , . . . , t n ). Hence be the representation of f | V n (x) for some entire function q on C n . Since { (F 1 (x), . . . , F m (x)) : x ∈ V n } = C m , q(t 1 , . . . , t n ) must be bounded on the set {|t 1 | < ε, . . . , |t m | < ε}. The Liouville Theorem implies q(t 1 , . . . , t n ) = q(t 1 , . . . , t m , 0 . . . , 0), that is, f | V n is generated by F 1 , . . . , F m . Since it is true for every n, f is finitely generated.
n! is not w p -continuous. Proposition 3.22. w p is a Hausdorff topology.

Proposition 3.23.
On bounded sets of M bs ( p ) the topology w p is finer than the weak-star topology w(M bs ( p ), H bs ( p )).
Proof. Since (M bs ( p ), w p ) is a first-countable space, it suffices to verify that for a bounded sequence (ϕ i ) i which is w p convergent to some ψ, we have lim i ϕ i ( f ) = ψ( f ) for each f ∈ H bs ( p ) : Indeed, by the Banach-Steinhaus theorem, it is enough to see that lim i ϕ i (P) = ψ(P) for each symmetric polynomial P. Being {F k } an algebraic basis for the symmetric polynomials, this will follow once we check that lim i ϕ i (F k ) = ψ(F k ) for each F k . To see this, notice that given ε > 0, ϕ i ∈ U ε,k for i large enough, that is, there is θ i such that ϕ i = ψ θ i with |θ i (F k )| < ε.
Symmetric Polynomials and Holomorphic Functions... 37 Proposition 3.24. If (M bs ( p ), ) is a group, then w p coincides with the weakest topology on M bs ( p ) such that for every polynomial P ∈ H bs ( p ) the Gelfand extension P is continuous on M bs ( p ).

REPRESENTATIONS OF THE CONVOLUTION SEMIGROUP
and we refer to it as the basis of elementary symmetric polynomials. Lemma 3.25. We have that G n = 1/n! Proof. To calculate the norm, it is enough to deal with vectors in the unit ball of 1 whose components are non-negative. And we may reduce ourselves to calculate it on L m the linear span of {e 1 , . . . , e m } for m ≥ n. We do the calculation in an inductive way over m.
Since G n | L m is homogeneous, its norm is achieved at points of norm 1. If m = n, then G n is the product x 1 · · · x n . By using the Lagrange multipliers rule, we deduce that the maximum is attained at points with equal coordinates, that is at 1 n (e 1 + · · · + e n ). Thus |G n ( 1 n , n . . ., 1 n , 0, . . . )| = 1/n n ≤ 1 n! . Now for m > n, and x ∈ L m , we have G n (x) = ∑ ∞ k 1 <···<k n ≤m x k 1 · · · x k n . Again the Lagrange multipliers rule leads to either some of the coordinates vanish or they are all equal, hence they have the same value 1 m . In the first case, we are led back to some the previous inductive steps, with L k with k < m, so the aimed inequality holds. While in the second one, we have Let C{t} be the space of all power series over C. We denote by F and G the following maps from M bs ( 1 ) into C{t} Let us recall that every element ϕ ∈ M bs ( 1 ) has a radius-function where ϕ n is the restriction of ϕ to the subspace of n-homogeneous polynomials [6]. Proposition 3.26. The mapping ϕ ∈ M bs ( 1 ) G → G(ϕ) ∈ H(C) is one-to-one and ranges into the subspace of entire functions on C of exponential type. The type of G(ϕ) is less than or equal to R(ϕ).
Proof. Using Lemma 3.25, lim sup n→∞ n n!|ϕ n (G n )| ≤ lim sup n→∞ n n! ϕ n G n = lim sup n→∞ n ϕ n = R(ϕ) < ∞, hence G(ϕ) is entire and of exponential type less than or equal to R(ϕ). That G is one-to-one follows from the fact {G n } is a basis.
Theorem 3.27. The following identities hold: Proof. The first statement is a trivial conclusion of the properties of the convolution. To prove the second we observe that Therefore, Hence, being the series absolutely convergent, Example 3.28. Let ψ λ be as defined in Example 3.14. We know that F (ψ λ ) = λ. To find G(ψ λ ) note that According to well-known Newton's formula we can write for x ∈ 1 , nG n (x) = F 1 (x)G n−1 (x) − F 2 (x)G n−2 (x) + · · · + (−1) n+1 F n (x). Moreover, if ξ is a complex homomorphism (not necessarily continuous) on the space of symmetric polynomials P s ( 1 ), then Next we point out the limitations of the construction's technique described in 3.14.
Remark 3.29. Let ξ be a complex homomorphism on P s ( 1 ) such that ξ(F m ) = c = 0 for some m ≥ 2 and ξ(F n ) = 0 for n = m. Then ξ is not continuous.
Proof. Using formula (3.7) we can see that is not of exponential type. So if ξ were continuous, it could be extended to an element in M bs ( 1 ), leading to a contradiction with Proposition 3.26.
According to the Hadamard Factorization Theorem (see [14, p. 27]) the function of the exponential type G(ϕ)(t) is of the form where {a k } are the zeros of G(ϕ)(t). If ∑ |a k | −1 < ∞, then this representation can be reduced to Recall how ψ λ was defined in Example 3.14.
Another consequence of our analysis is the following remark. Corollary 3.31. Let Φ be a homomorphism of P s ( 1 ) to itself such that Φ(F k ) = −F k for every k. Then Φ is discontinuous.
We close this section by analyzing further the relationship established by the mapping G. It is known from Combinatorics (see e.g. [15, p. 3, 4]) that for every x ∈ c 00 . Formula (3.10) for G(δ x ) is true for every x ∈ 1 : Indeed, for fixed t, both the infinite product and G(δ x )(t) are analytic functions on 1 .
Taking into account formula (3.10) we can see that the zeros of G(δ x )(t) are a k = −1/x k for x k = 0. Conversely, if f (t) is an entire function of exponential type which is equal to the right hand side of (3.9) with ∑ |a k | −1 < ∞, then for ϕ ∈ M bs ( 1 ) given by ϕ = ψ λ δ x , where x ∈ 1 , x k = −1/a k and ψ λ is defined in Example 3.14, it turns out that G(ϕ)(t) = f (t). So we have just to examine entire functions of exponential type with Hadamard canonical product and λ ∈ C. Let us denote by δ (x,λ) a homomorphism on the algebra of symmetric polynomials P s ( 1 ) of the form x k n , k > 1.
Note that for further consideration the order of numbering does not matter. Proposition 3.36. For arbitrary x, y ∈ p we have (1) x y ∈ p and x y = x y ; (2) F k (x y) = F k (x)F k (y) ∀k ≥ p .
(3) If P is an n-homogeneous symmetric polynomial on p and y is fixed, then the function x → P(x y) is n-homogeneous.
Proof. It is clear that x y p = ∑ i,j |x i y j | p = ∑ i |x i | p ∑ i |y j | p = x p y p . Also F k (x y) = ∑ i,j (x i y j ) k = ∑ i x k i ∑ j y k j = F k (x)F k (y). Statement (3) follows from the equality λ(x y) = (λx) y.
Given y ∈ p , the mapping x ∈ p π y → (x y) ∈ p is linear and continuous because of Proposition 3.36. Therefore if f ∈ H bs ( p ), then f • π y ∈ H bs ( p ) because f • π y is analytic and bounded on bounded sets and clearly f (σ(x) y) = f (x y) for every permutation σ ∈ G. Thus if we denote M y ( f ) = f • π y , M y is a composition operator on H bs ( p ), that we will call the multiplicative convolution operator. Notice as well that M y = M σ(y) for every permutation σ ∈ G and that M y (F k ) = F k (y)F k ∀k ≥ p . Proposition 3.37. For every y ∈ p the multiplicative convolution operator M y is a continuous homomorphism on H bs ( p ).
Note that in particular, if f n is an n-homogeneous continuous polynomial, then M y ( f n ) ≤ f n y n . And also that for λ ∈ C, M λy ( f n ) = λ n M y ( f n ), because π λy (x) = λπ(x). Analogously, M y+z ( f n ) = f n • (π y + π z ), because π y+z = π y + π z . Therefore the mapping y ∈ p → M y ( f n ) is an n-homogeneous continuous polynomial.
Recall that the radius function R(φ) of a complex homomorphism φ ∈ M bs ( p ) is the infimum of all r such that φ is continuous with respect to the norm of uniform convergence on the ball rB p , that is |φ( f )| ≤ C r f r . It is known that where φ n is the restriction of φ to P s ( n p ) and φ n is its corresponding norm (see [6]).
Proposition 3.38. For every θ ∈ H bs ( p ) and every y ∈ p the radius-function of the continuous homomorphism θ • M y satisfies R(θ • M y ) ≤ R(θ) y and for fixed f ∈ H bs ( p ) the function y → θ • M y ( f ) also belongs to H bs ( p ).