Separately polynomial functions

It is known that if f:R2→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f:{{\mathbb R}}^2\rightarrow {\mathbb R}$$\end{document} is a polynomial in each variable, then f is a polynomial. We present generalizations of this fact, when R2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb R}}^2$$\end{document} is replaced by G×H\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$G\times H$$\end{document}, where G and H are topological Abelian groups. We show, e.g., that the conclusion holds (with generalized polynomials in place of polynomials) if G is a connected Baire space and H has a dense subgroup of finite rank or, for continuous functions, if G and H are connected Baire spaces. The condition of continuity can be omitted if G and H are locally compact or one of them is metrizable. We present several examples showing that the results are not far from being optimal.

On topological Abelian groups we have to distinguish between the class of polynomials and the wider class of generalized polynomials (see the next section for the definitions). The two classes coincide if the group contains a dense subgroup of finite rank. Now, the scalar product on the square of a Hilbert space is an example of a continuous function which is a polynomial in each variable, is a generalized polynomial on the product, but not a polynomial (see Example 1 below). Therefore, the appropriate problem is to find conditions on the groups G and H ensuring that whenever a function on G × H is a generalized polynomial in each variable, then it is a generalized polynomial.
This problem was considered already by Mazur and Orlicz in [10] in the case when G and H are topological vector spaces. They proved that if X , Y , Z are Banach spaces and the map f : (X × Y ) → Z is a generalized polynomial in each variable, 1 then f is a generalized polynomial [10,Satz IV]. They also considered the case when continuity is not assumed, and X , Y , Z are linear spaces without topology [10, Satz III] (see also [2,Lemma 1]). The topic has an extensive literature; see [14,16] and the references therein.
In this note we consider the analogous problem when G and H are topological Abelian groups. We show that if G is a connected Baire space, H has a dense subgroup of finite rank, and if a function f : (G × H ) → C is a generalized polynomial in each variable, then f is a generalized polynomial on G × H (Theorem 1). The same conclusion holds if G and H are both connected Baire spaces, and one of them is metrizable or, if both are locally compact (Theorem 3).
If G and H are connected Baire spaces, f : (G × H ) → C is a generalized polynomial in each variable, and if f has at least one point of joint continuity, then f is a generalized polynomial ((iii) of Theorem 2).
It is not clear if the extra condition of the existence of points of joint continuity can be omitted from this statement (Question 1). The problem is that a generalized polynomial must be continuous by definition, and a separately continuous function on the product of Baire spaces can be discontinuous everywhere, as it was shown recently in [11]. In our case, however, there are some extra conditions: the spaces are also connected, and the function in question is a generalized polynomial. It is conceivable that continuity follows under these conditions. As for the biadditive case, see [4].
There are several topological conditions implying that separately continuous functions on a product must have points of joint continuity. In fact, the topic has a vast literature starting with the paper [12]. See, e.g., the papers [5][6][7]13].

Preliminaries
Let G be a topological Abelian group. We denote the group operation by addition, and denote the unit by 0. The translation operator T h and the difference operator Δ h are We say that a continuous function f : G → C is a generalized polynomial, if there is an n ≥ 0 such that Δ h 1 . . . Δ h n+1 f = 0 for every h 1 , . . . , h n+1 ∈ G. The smallest n with this property is the degree of f , denoted by deg f . The degree of the identically zero function is −1. We denote by GP = GP G the set of generalized polynomials defined on G.
A function f : G → C is said to be a polynomial, if there are continuous additive functions a 1 , . . . , a n : G → C and there is a P ∈ C[x 1 , . . . , x n ] such that f = P(a 1 , . . . , a n ). It is well-known that every polynomial is a generalized polynomial. It is also easy to see that the linear span of the translates of a polynomial is of finite dimension. More precisely, a function is a polynomial if and only if it is a generalized polynomial, and the linear span of its translates is of finite dimension (see [9,Proposition 5]). We denote by P = P G the set of polynomials defined on G.
Let f be a complex valued function defined on X × Y . The sections f x : Y → C and f y : Let G, H be topological Abelian groups. A function f : (G × H ) → C is a separately polynomial function if f x ∈ P H for every x ∈ G and f y ∈ P G for every y ∈ H . Similarly, we say that f : (G × H ) → C is a separately generalized polynomial function if f x ∈ G P H for every x ∈ G and f y ∈ G P G for every y ∈ H .
In general we cannot expect that every separately polynomial function on G × H is a polynomial; not even if G = H is a Hilbert space.

Example 1
Let G be the additive group of an infinite dimensional Hilbert space. Then the scalar product f (x, y) = x, y on G 2 is a separately polynomial function, since its sections are continuous additive functions. In fact, f y is a linear functional and f x is a conjugate linear functional for every x, y ∈ G. Thus the sections of f are polynomials. Now, while the scalar product is a generalized polynomial (of degree 2) on G 2 , it is not a polynomial on G 2 , because the dimension of the linear span of its translates is infinite. Indeed, let g(x) = x, x = x 2 for every x ∈ G. Then Δ h g(x) = 2 h, x + h 2 for every h ∈ G. It is easy to see that the functions h, x (h ∈ G) generate a linear space of infinite dimension, and then the same is true for the translates of g and then for those of f as well.
Therefore, the best we can expect is that, under suitable conditions on G and H , every separately generalized polynomial function on G × H is a generalized polynomial.
We denote by r 0 (G) the torsion free rank of the group G; that is, the cardinality of a maximal independent system of elements of G of infinite order. Thus r 0 (G) = 0 if and only if G is torsion. In the sequel by the rank of a group we shall mean the torsion free rank. It is known that if G has a dense subgroup of finite rank, then the classes of polynomials and of generalized polynomials on G coincide (see [9,Theorem 9]).
The set of roots of a function f : It is easy to see that N P and N G P are proper ideals of subsets of G. Let N σ P and N σ G P denote the σ -ideals generated by N P and N G P , respectively. Note that N P ⊂ N G P and N σ P ⊂ N σ G P . If G is discrete, then N σ P and N σ G P are not proper σ -ideals (except when G is torsion), according to the next observation.
Proof Let a ∈ G be an element of infinite order. Then φ(na) = n (n ∈ Z) defines a homomorphism from the subgroup generated by a into Q, the additive group of the rationals. Since Q is divisible, φ can be extended to G as a homomorphism from G into Q. Let ψ be such an extension.
A simple sufficient condition for G / ∈ N σ G P is given by the next result.
Proof It is enough to prove that every element of N G P is nowhere dense. Suppose A ∈ N G P is dense in a nonempty open set U . Let p ∈ G P G be a nonzero generalized polynomial vanishing on A. Since A ⊂ Z p and Z p is closed, we have U ⊂ Z p . Since G is connected, every neighbourhood of the origin generates G. It is known that in such a group, if a generalized polynomial vanishes on a nonempty open set, then it vanishes everywhere (see [15,Theorem 3.2,p. 33]). This implies that p is identically zero, which is impossible.  By Lemma 1, (i) of Theorem 1 can be replaced by the condition that G is a connected Baire space. Similarly, the condition of Theorem 2 can be replaced by the condition that G and H are connected Baire spaces.

Theorem 1 Let G, H be topological Abelian groups, and suppose that
As for (iii) of Theorem 2 note the following facts. Comparing these with (iii) of Theorem 2 we obtain the following.
Proof Let GP <d denote the set of generalized polynomials f ∈ GP H of degree < d. Clearly, GP <d is a linear space over C.
Let K be a dense subgroup of H with r 0 (K ) = N < ∞. Let {h 1 , . . . , h N } be a maximal set of independent elements of K of infinite order, and let L denote the subgroup of K generated by the elements h 1 , . . . , h N . If k = (k 1 , . . . , k N ) ∈ Z N , then we put k = max 1≤i≤N |k i |. We abbreviate the sum N i=1 k i · h i by k, h . Then we have L = { k, h : k ∈ Z N }. We put First we prove that if p ∈ G P <d vanishes on A, then p = 0.
Suppose p = 0. Since p is continuous and K is dense in H , there is an x 0 ∈ K such that p(x 0 ) = 0. The maximality of the system {h 1 , . . . , h N } implies that nx 0 ∈ L with a suitable nonzero integer n. It is easy to see that there is a polynomial P ∈ C[x] such that p(mx 0 ) = P(m) for every integer m. Since P(1) = p(x 0 ) = 0, it follows that P = 0, hence P only has a finite number of roots. Thus p(mnx 0 ) = P(mn) = 0 for all but a finite number of integers m. Fix such an m. Then mnx 0 ∈ L, and thus mnx 0 = k, h with a suitable k ∈ Z N . We find that p( k, h ) = 0 for some k ∈ Z N .
Let k = (k 1 , . . . , k N ) ∈ Z N be such that p( k, h ) = 0 and k is minimal. If k ≤ [d/2], then k, h ∈ A, and we have p( k, h ) = 0 by assumption. Thus we Then we have k − j < k for every j = 1, . . . , d. By the minimality of k we have p( k − j , h ) = 0 for every j = 1, . . . , d. Put which is impossible. This proves p = 0. The set of functions V = {p| A : p ∈ GP <d } is a finite dimensional linear space over C. The map p → p| A is linear from GP <d onto V and, as we proved above, it is injective. Therefore GP <d is of finite dimension.
Let b 1 , . . . , b s be a basis of GP <d . Since the functions b 1 , . . . , b s are linearly independent, there are elements x 1 , . . . , x s such that the determinant det |b i (x j )| is nonzero (see [1, Lemma 1, p. 229]). Put X = {x 1 , . . . , x s }. Then b 1 | X , . . . , b s | X are linearly independent, and thus the map f → f | X is bijective and linear from G P <d onto the set of functions f : X → C.
Then there are functions q 1 , . . . , q s ∈ G P <d such that q i (x i ) = 1 and q i (x j ) = 0 for every i, j = 1, . . . , s, i = j.
Let p ∈ G P <d be given. Then p − s i=1 p(x j )q j is a generalized polynomial of degree < d vanishing on X , hence on H . That is, we have p = s i=1 p(x j )q j . Proof of Theorem 1 Let f : (G × H ) → C be a separately generalized polynomial function. Put G n = {x ∈ G : deg f x < n} (n = 1, 2, . . .). Since N σ G P (G) is a proper σ -ideal in G, there is an n such that G n / ∈ N G P (G). Fix such an n. By Lemma 2, there are points y 1 , . . . , y s ∈ H and generalized polynomials q 1 , . . . , q s ∈ G P H such that p = s i=1 p(y i ) · q i for every p ∈ G P H with deg p < n. Therefore, we have for every x ∈ G n and y ∈ H . If y ∈ H is fixed,

Lemma 3 Let G, H be discrete Abelian groups. A function f : (G × H ) → C is a generalized polynomial if and only if the sections f x (x ∈ G) and f y (y ∈ H ) are generalized polynomials of bounded degree.
Then, for every y ∈ H , we have Δ x 1 . . . Δ x d f y = 0 for every x 1 , . . . , x d ∈ G, and thus f y is a generalized polynomial of degree < d for every y ∈ H . A similar argument shows that f x is a generalized polynomial of degree < d for every x ∈ G, proving the "only if" statement.
Now suppose that f : (G × H ) → C is such that f x (x ∈ G) and f y (y ∈ H ) are generalized polynomials of degree < d. Then we have for every h 1 , . . . , h d ∈ G, and for every k 1 , . . . , k d ∈ H . In order to prove that f is a generalized polynomial of degree < 2d, it is enough to show that Proof of Theorem 2 (i) Suppose f satisfies the conditions. By Lemma 3, it is enough to show that the degrees deg f x and f y are bounded.
is a proper σ -ideal, there is an n such that A n / ∈ N G P (G). We fix such an n, and prove that for every h 1 , . . . , h n ∈ H . Let g denote the left hand side of (4). Then g(x, y) = s i=1 a i f (x, y + b i ), where s = 2 n , a i = ±1 and b i ∈ H for every i. Let y ∈ H be fixed. Then g y = s i=1 a i f y+b i , and thus g y is a generalized polynomial on G. If x ∈ A n , then deg f x < n, and thus g x = 0. Therefore g y (x) = 0 for every x ∈ A n . Since g y is a generalized polynomial and A n / ∈ N G P (G), it follows that g y = 0. Since y was arbitrary, this proves (4). Thus deg f x < n for every x ∈ G.
A similar argument shows that, for a suitable m, deg f y < m for every y ∈ H . Statement (ii) of the theorem is clear from (i).
Suppose that G and H are connected. Now we use the fact that if f is a discrete generalized polynomial on an Abelian group which is generated by every neighbourhood of the origin, and if f has a point of continuity, then f is continuous everywhere. (See [15, Theorem 3.6]) or, for topological vector spaces, [2, Theorem 1].) In our case the group G × H is connected, so the condition is satisfied, and we conclude that f is continuous everywhere on G × H . Thus (iii) follows from (ii).

Examples
In Theorem 1 none of the conditions on G and H can be omitted. First we show that without condition (i) the conclusion of Theorem 1 may fail. We shall need the easy direction of Lemma 3.

Example 2
Let G, H be discrete Abelian groups. We show that if none of G and H is torsion, then there is a separately polynomial function f : is not a proper σ -ideal; that is, G = ∞ n=1 A n , where A n = ∅ and A n ∈ N P (G) for every n. Let p n ∈ P G be such that p n = 0 and A n ⊂ Z p n . Then p n is not constant; that is, deg p n ≥ 1.
Let P n = p 1 · · · p n ; then P n (x) = 0 for every x ∈ n i=1 A i , and we have 0 < deg P 1 < deg P 2 < . . .. (Here we use the fact that deg pq = deg p + deg q for every p, q ∈ G P G , p, q = 0.) Note that for every x ∈ G we have P n (x) = 0 for all but a finite number of indices n.
Similarly, we find polynomials Q n ∈ P H such that 0 < deg Q 1 < deg Q 2 < . . ., and for every y ∈ H we have Q n (y) = 0 for all but a finite number of indices n.
We put f (x, y) = ∞ n=1 P n (x)Q n (y) for every x ∈ G and y ∈ H . If y ∈ H is fixed, then the sum defining f is finite, and thus f y ∈ P G . Similarly, we have f x ∈ P H for every x ∈ G.
The degrees deg f y (y ∈ H ) are not bounded. Indeed, for every N , there is an y ∈ H such that Q N (y) = 0. Then f y = M n=1 Q n (y) · P n with an M ≥ N , where the coefficients Q n (y) are nonzero if n ≤ N . Therefore, deg f y ≥ deg P N ≥ N , proving that the set {deg f y : y ∈ H } is not bounded. By Lemma 3, it follows that f is a not a generalized polynomial.
By the example above, if G and H are discrete Abelian groups of positive and finite rank, then the conclusion of Theorem 1 fails. That is, G / ∈ N σ G P (G) cannot be omitted from the conditions of Theorem 1.
Next we show that the condition on H cannot be omitted either.

Example 3
Let H be a discrete Abelian group of infinite rank. We show that if G is a topological Abelian group such that P G contains nonconstant polynomials, then there is a continuous separately polynomial function f on G × H such that f is not a generalized polynomial.
Let h α (α < κ) be a maximal set of independent elements of H of infinite order, where κ ≥ ω. Let K denote the subgroup of H generated by the elements h α (α < κ). Every element of K is of the form α<κ k α h α , where k α ∈ Z for every α, and all but a finite number of the coefficients k α equal zero.
Let p ∈ P G be a nonconstant polynomial. We define f (x, y) = ∞ i=1 k i · p i (x) for every x ∈ G and y ∈ K , y = α<κ k α h α . (Note that the sum only contains a finite number of nonzero terms for every x and y.) In this way we defined f on G × K such that f x is additive on K for every x ∈ G.
If y ∈ H , then there is a nonzero integer n such that ny ∈ K . Then we define f (x, y) = 1 n · f (x, ny) for every x ∈ G. It is easy to see that f (x, y) is well-defined on G × H , and f x is additive on H for every x ∈ G. Therefore, f x is a polynomial on G for every x ∈ G.
If y ∈ H and ny ∈ K for a nonzero integer n, then f y is of the form 1 n · N i=1 k i · p i , and thus f y ∈ P G . Since f y is continuous for every y ∈ H and H is discrete, it follows that f is continuous on G × H .
Still, f is not a generalized polynomial on G × H , as the set of degrees deg f y (y ∈ H ) is not bounded: if y = h i , then f y = p i , and deg p i = i · deg p ≥ i for every (i = 1, 2, . . .).
In the example above we may choose G in such a way that G / ∈ N σ G P (G) holds. (Take, e.g., G = R.) In our next example this condition holds for both G and H .

Example 4
Let E be a Banach space of infinite dimension, and let G be the additive group of E equipped with the weak topology τ of E. It is well-known that every ball in E is nowhere dense w.r.t. τ , and thus G is of first category in itself.
Still, we show that G / ∈ N σ G P (G). Indeed, the original norm topology of E is stronger than τ , and makes E a connected Baire space. If a function is continuous w.r.t. τ , then it is also continuous w.r.t. the norm topology. Therefore, every polynomial p ∈ P(G) is also a polynomial on E, and thus N P (G) ⊂ N P (E) and N σ P (G) ⊂ N σ P (E). Since N σ P (E) is proper by Lemma 1, it follows that N σ P (G) is proper. The same is true for N σ G P (G). Now let H be an infinite dimensional Hilbert space, and let G be the additive group of H equipped with the weak topology of H . Let f be the scalar product on H 2 . Since the linear functionals and conjugate linear functionals are continuous w.r.t. the weak topology, it follows that f is a separately polynomial function on G 2 (see Example 1).
However, f is not a generalized polynomial on G 2 , since f is not continuous. In order to prove this, it is enough to show that f (x, x) = x 2 is not continuous on H w.r.t. the weak topology. Suppose it is. Then there is a neighbourhood U of 0 such that x < 1 for every x ∈ U . By the definition of the weak topology, there are linear functionals L 1 , . . . , L n and there is a δ > 0 such that whenever |L i (x)| < δ (i = 1, . . . , n), then x < 1.
Since H is of infinite dimension, there is an x = 0 such that L i (x) = 0 for every i = 1, . . . , n. (Otherwise every linear functional would be a linear combination of L 1 , . . . , L n , and then H = H * would be finite dimensional.) Then λx ∈ U for every λ ∈ C and λx < 1 for every λ ∈ C, which is impossible.
The example above shows that in (ii) of Theorem 2 the condition of joint continuity cannot be omitted. Note also that the group G defined in Example 4 is a topological vector space, hence connected. This shows that in (iii) of Theorem 2 the condition of the existence of points of joint continuity cannot be omitted either.