Separately polynomial functions

https://doi.org/10.1007/s00605-021-01614-9

Separately polynomial functions

Gergely Kiss¹·Miklós Laczkovich²

Received: 13 January 2021 / Accepted: 16 August 2021 / Published online: 6 September 2021

© The Author(s) 2021

Abstract

It is known that if f:R²→Ris a polynomial in each variable, then f is a polynomial.

We present generalizations of this fact, whenR²is replaced byG×H, whereGand H are topological Abelian groups. We show, e.g., that the conclusion holds (with generalized polynomials in place of polynomials) ifGis a connected Baire space and H has a dense subgroup of finite rank or, for continuous functions, ifGand H are connected Baire spaces. The condition of continuity can be omitted ifGandH are locally compact or one of them is metrizable. We present several examples showing that the results are not far from being optimal.

Keywords Polynomials·Generalized polynomials·Functions on product spaces Mathematics Subject Classification 22A20·54E45·54E52

1 Introduction

It was proved by Carroll in [3] that if f:R²→ Ris a polynomial in each variable, then f is a polynomial. Our aim is to find generalizations of this fact, when R²is replaced by the product of two topological Abelian groups.

Communicated by Karlheinz Gröchenig.

G. Kiss was supported by a Premium Postdoctoral Fellowship of the Hungarian Academy of Sciences.

Both authors were supported by the Hungarian National Foundation for Scientific Research, Grant No.

K124749.

B

miklos.laczkovich@gmail.com

1 Alfréd Rényi Institute of Mathematics, Reáltanoda utca 13-15, H-1053 Budapest, Hungary 2 Eötvös Loránd University, Pázmány Péter sétány 1/C, 1117 Budapest, Hungary

On topological Abelian groups we have to distinguish between the class of poly- nomials 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 con- tinuous function which is a polynomial in each variable, is a generalized polynomial on the product, but not a polynomial (see Example1below). Therefore, the appro- priate problem is to find conditions on the groupsGandHensuring that whenever a function onG×His 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 GandHare topological vector spaces. They proved that ifX,Y,Zare Banach spaces and the map f:(X×Y)→ Zis a generalized polynomial in each variable,¹then f is a generalized polynomial [10, Satz IV]. They also considered the case when continuity is not assumed, andX,Y,Zare 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 ifGis a connected Baire space,Hhas a dense subgroup of finite rank, and if a function f:(G×H) → Cis a generalized polynomial in each variable, then f is a generalized polynomial onG×H (Theorem1). 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 (Theorem3).

IfGandHare connected Baire spaces, f:(G×H)→Cis a generalized poly- nomial in each variable, and if f has at least one point of joint continuity, then f is a generalized polynomial ((iii) of Theorem2).

It is not clear if the extra condition of the existence of points of joint continuity can be omitted from this statement (Question1). 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 func- tions 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 [5–7,13].

2 Preliminaries

LetGbe a topological Abelian group. We denote the group operation by addition, and denote the unit by 0. The translation operatorTh and the difference operatorΔhare

1 Actually Mazur and Orlicz only assume thatX,Y,ZareF-spaces; that is, topological vector spaces whose topology is induced by a complete invariant metric.

defined byThf(x)= f(x+h)andΔhf(x)= f(x+h)−f(x)for every f: G→C andh,x∈G.

We say that a continuous function f: G→Cis ageneralized polynomial, if there is ann ≥0 such thatΔh1. . . Δhn+1f =0 for everyh1, . . . ,hn+1∈G. The smallestn with this property is thedegree of f, denoted by deg f. The degree of the identically zero function is−1. We denote byGP =GPG the set of generalized polynomials defined onG.

A function f: G → Cis said to be apolynomial, if there are continuous addi- tive functions a1, . . . ,an: G → C and there is a P ∈ C[x1, . . . ,xn] such that f = P(a1, . . . ,an). It is well-known that every polynomial is a generalized poly- nomial. 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 byP =PGthe set of polynomials defined onG.

Let f be a complex valued function defined onX×Y. Thesections fx:Y →C and f^y: X →Cof f are defined by fx(y)= f^y(x)= f(x,y) (x∈ X, y∈Y).

Let G,H be topological Abelian groups. A function f: (G ×H) → C is a separately polynomial function if fx ∈ PH for every x ∈ G and f^y ∈ PG for everyy ∈ H. Similarly, we say that f: (G×H)→ Cis aseparately generalized polynomial functionif fx ∈G PHfor everyx ∈Gand f^y∈G PGfor everyy∈ H.

In general we cannot expect that every separately polynomial function onG×H is a polynomial; not even ifG=H is a Hilbert space.

Example 1 LetGbe the additive group of an infinite dimensional Hilbert space. Then the scalar product f(x,y)= x,yonG²is a separately polynomial function, since its sections are continuous additive functions. In fact, f^y is a linear functional and fx 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) onG², it is not a polynomial onG², because the dimension of the linear span of its translates is infinite. Indeed, let g(x) = x,x = x² for every x ∈ G. ThenΔhg(x) = 2h,x + h²for everyh ∈ G. It is easy to see that the functionsh,x(h ∈ G) generate a linear space of infinite dimension, and then the same is true for the translates ofgand 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 byr0(G)the torsion free rank of the groupG; that is, the cardinality of a maximal independent system of elements ofGof infinite order. Thusr0(G)=0 if and only ifGis torsion. In the sequel by the rank of a group we shall mean the torsion free rank. It is known that ifGhas a dense subgroup of finite rank, then the classes of polynomials and of generalized polynomials onGcoincide (see [9, Theorem 9]).

The set of roots of a function f:G → Cis denoted byZf. That is,Zf = {x ∈ G: f(x)=0}. We put

NP =NP(G)= {A⊂G: ∃p∈PG, p=0, A⊂Zp}

and

NG P =NG P(G)= {A⊂G: ∃p∈G PG, p=0, A⊂Zp}.

It is easy to see thatNPandNG Pare proper ideals of subsets ofG. LetN_P^σandN_{G P}^σ denote theσ-ideals generated byNPandNG P, respectively. Note thatNP ⊂NG P

andN_P^σ ⊂N_{G P}^σ .

IfG is discrete, thenN_P^σ andN_{G P}^σ are not proper σ-ideals (except whenG is torsion), according to the next observation.

Proposition 1 Let G be a discrete Abelian group. If G is not torsion, then G∈N_P^σ. Proof Leta ∈Gbe an element of infinite order. Thenφ(na)=n(n ∈Z)defines a homomorphism from the subgroup generated bya intoQ, the additive group of the rationals. SinceQis divisible,φcan be extended toGas a homomorphism fromG intoQ. Letψbe such an extension.

Thenpr =ψ+ris a nonzero polynomial onGfor everyr ∈Q. Ifx∈G, thenx is the root of pr, wherer = −ψ(x)∈Q. Therefore,G=

r∈QZpr ∈N_P^σ. A simple sufficient condition forG∈/N_{G P}^σ is given by the next result.

Lemma 1 If G is a connected Baire space, then theσ-idealsN_P^σandN_{G P}^σ are proper;

that is, G ∈/N_P^σ and G∈/N_{G P}^σ .

Proof It is enough to prove that every element ofNG P is nowhere dense. Suppose A∈NG Pis dense in a nonempty open setU. Letp∈G PGbe a nonzero generalized polynomial vanishing onA. SinceA⊂ZpandZpis closed, we haveU ⊂Zp. Since G is connected, every neighbourhood of the origin generatesG. 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 thatpis identically

zero, which is impossible.

3 Main results

Theorem 1 Let G,H be topological Abelian groups, and suppose that (i) N_{G P}^σ (G)is a properσ-ideal in G, and

(ii) H has a dense subgroup of finite rank.

If f:(G×H) → Cis a separately generalized polynomial function, then f is a generalized polynomial on G×H .

Theorem 2 Let G,H be topological Abelian groups, and suppose that N_{G P}^σ (G)is a properσ-ideal in G, andN_{G P}^σ (H)is a properσ-ideal in H . Then the following statements are true.

(i) If f:(G×H)→Cis a separately generalized polynomial function, then f is a generalized polynomial on G×H with respect to the discrete topology.

(ii) Every joint continuous separately generalized polynomial function f:(G×H)→ Cis a generalized polynomial on G×H .

(iii) If G and H are connected and a separately generalized polynomial function f:(G× H) → Chas at least one point of joint continuity, then f is a gen- eralized polynomial on G×H .

By Lemma1, (i) of Theorem1can be replaced by the condition thatGis a connected Baire space. Similarly, the condition of Theorem2can be replaced by the condition thatGandH are connected Baire spaces.

As for (iii) of Theorem2note the following facts.

• If X,Y are nonempty topological spaces, X is Baire, Y is first countable and f:(X×Y)→Cis separately continuous, then f has at least one point of joint continuity. (See, e.g. [17, p. 441].)

• A topological group is first countable if and only if it is metrizable.

• IfX,Yare nonempty locally compact andσ-compact topological spaces, f:(X× Y)→Cis separately continuous, then f has at least one point of joint continuity.

(See [12, Theorem 1.2]).

• Every connected and locally compact topological group isσ-compact.

Comparing these with (iii) of Theorem2we obtain the following.

Theorem 3 Suppose that the topological Abelian groups G,H are connected and Baire, and either

(i) at least one of G and H is metrizable, or (ii) G and H are locally compact.

If f:(G×H) → Cis a separately generalized polynomial function, then f is a generalized polynomial on G×H .

Question 1 Are the conditions (i) and (ii) necessary in the statement of Theorem3?

(See the introduction.)

We prove Theorems1 and2in the next section. In Sect.5 we present examples showing that some of the conditions appearing in Theorems1and2cannot be omitted.

4 Proof of Theorems1and2

Lemma 2 Let H be a topological Abelian group, and suppose that H has a dense subgroup of finite rank. Then, for every positive integer d, there are finitely many points x1, . . . ,xs ∈ H and there are generalized polynomials q1, . . . ,qs ∈GPH of degree<d such that p=_s

i=1p(xi)·qi for every p∈G PH withdegp<d.

Proof LetGP^<ddenote the set of generalized polynomials f ∈GPHof degree<d.

Clearly,GP^<dis a linear space overC.

Let K be a dense subgroup of H withr0(K) = N < ∞. Let{h1, . . . ,hN}be a maximal set of independent elements of K of infinite order, and let L denote the

subgroup of K generated by the elementsh1, . . . ,hN. Ifk = (k1, . . . ,kN)∈ Z^N, then we putk =max1≤i≤N|ki|. We abbreviate the sum_N

i=1ki·hibyk,h. Then we haveL= {k,h: k∈Z^N}. We put

A= {k,h: k∈Z^N, k ≤ [d/2]}.

First we prove that ifp ∈G P^<dvanishes onA, then p=0.

Suppose p = 0. Since p is continuous andK is dense inH, there is anx0 ∈ K such thatp(x0)=0. The maximality of the system{h1, . . . ,hN}implies thatnx0∈ L with a suitable nonzero integern. It is easy to see that there is a polynomialP∈C[x]

such that p(mx0)=P(m)for every integerm. Since P(1)= p(x0)=0, it follows thatP =0, hencePonly has a finite number of roots. Thusp(mnx0)=P(mn)=0 for all but a finite number of integersm. Fix such anm. Thenmnx0 ∈ L, and thus mnx0= k,hwith a suitablek∈Z^N. We find that p(k,h)=0 for somek∈Z^N. Letk = (k1, . . . ,kN) ∈ Z^N be such that p(k,h) = 0 andk is minimal. If k ≤ [d/2], thenk,h ∈ A, and we have p(k,h)=0 by assumption. Thus we havek>[d/2]. Put=(1, . . . , N), where

i =

⎧⎪

⎨

⎪⎩

1 ifki >[d/2], 0 if|ki| ≤ [d/2],

−1 ifki <−[d/2]

(i =1, . . . ,N).

Then we havek−j<kfor every j=1, . . . ,d. By the minimality ofkwe havep(k− j,h)=0 for every j =1, . . . ,d.

Putv = ,h. Since degp <d, it follows thatΔ^d_−vp(x)=0 for everyx ∈ H.

Now we have

0=Δ^d_−vp(k,h)= d

j=0

(−1)^d−j d j

p(k,h − jv)=

=(−1)^dp(k,h)+ d

j=1

(−1)^d−j d j

p(k−j,h)=

=(−1)^dp(k,h), which is impossible. This provesp =0.

The set of functionsV = {p|A: p ∈GP^<d}is a finite dimensional linear space overC. The map p→ p|Ais linear fromGP^<dontoV and, as we proved above, it is injective. ThereforeGP^<dis of finite dimension.

Let b1, . . . ,bs be a basis ofGP^<d. Since the functionsb1, . . . ,bs are linearly independent, there are elementsx1, . . . ,xs such that the determinant det|bi(xj)|is nonzero (see [1, Lemma 1, p. 229]). PutX = {x1, . . . ,xs}. Thenb1|X, . . . ,bs|X are linearly independent, and thus the map f → f|Xis bijective and linear fromG P^<d onto the set of functions f: X →C.

Then there are functionsq1, . . . ,qs ∈G P^<dsuch thatqi(xi)=1 andqi(xj)=0 for everyi,j =1, . . . ,s, i = j.

Letp∈G P^<dbe given. Thenp−_s

i=1p(xj)qj is a generalized polynomial of degree<d vanishing onX, hence onH. That is, we havep=_s

i=1p(xj)qj. Proof of Theorem1 Let f:(G×H) → Cbe a separately generalized polynomial function. PutGn = {x∈G:deg fx <n}(n=1,2, . . .). SinceN_{G P}^σ (G)is a proper σ-ideal inG, there is annsuch thatGn∈/NG P(G). Fix such ann.

By Lemma 2, there are points y1, . . . ,ys ∈ H and generalized polynomials q1, . . . ,qs ∈ G PH such that p = s

i=1p(yi)·qi for every p ∈ G PH with degp<n. Therefore, we have

f(x,y)= s i=1

f(x,yi)qi(y)

for everyx∈ Gnandy∈ H. Ify∈ H is fixed, then f(x,y)−_s

i=1 f(x,yi)qi(y) is a generalized polynomial onGvanishing onGn. SinceGn∈/NG P(G), it follows that f(x,y)=_s

i=1 f(x,yi)qi(y)for every(x,y)∈G×H. By f^yi ∈ G PG and

qi ∈G PH, we obtain f ∈G PG×H.

Lemma 3 Let G,H be discrete Abelian groups. A function f:(G×H)→ Cis a generalized polynomial if and only if the sections fx (x ∈ G)and f^y (y ∈ H)are generalized polynomials of bounded degree.

Proof Suppose f:(G×H)→Cis a generalized polynomial of degree<d. Then Δ₍x₁,0). . . Δ₍x_d,0)f =0 for everyx1, . . . ,xd ∈ G. Then, for everyy∈ H, we have Δx₁. . . Δx_d f^y =0 for everyx1, . . . ,xd ∈G, and thus f^yis a generalized polynomial of degree<d for everyy ∈ H. A similar argument shows that fx is a generalized polynomial of degree<d for everyx ∈G, proving the “only if” statement.

Now suppose that f:(G×H)→Cis such that fx(x ∈G)and f^y(y∈ H)are generalized polynomials of degree<d. Then we have

Δ₍h₁,0). . . Δ₍h_d,0)f =0 (1)

for everyh1, . . . ,hd ∈G, and

Δ₍0,k1). . . Δ₍0,kd)f =0 (2)

for everyk1, . . . ,kd ∈ H. In order to prove that f is a generalized polynomial of degree<2d, it is enough to show that

Δ₍a₁,b₁). . . Δ₍a_2d,b_2d)f =0 (3) for every(ai,bi)∈G×H (i =1, . . . ,2d). The identityΔu+v=TuΔ_v+Δugives

Δ₍ai,bi)=T₍ai,0)Δ₍0,bi)+Δ₍ai,0)

for every i. Therefore, the left hand side of (3) is the sum of terms of the form TcΔc₁. . . Δc_2d f, wherec∈G× {0}, andci ∈(G× {0})∪({0} ×H)for everyi. If

there are at leastd indicesi withci ∈ (G× {0}), then (1) givesΔc₁. . . Δc_2d f =0.

Otherwise there are at least d indices i with ci ∈ ({0} ×H), and then (2) gives

Δc1. . . Δc_2d f =0. This proves (3).

Proof of Theorem2 (i) Suppose f satisfies the conditions. By Lemma3, it is enough to show that the degrees deg fxand f^yare bounded.

PutAn= {x ∈G: deg fx <n}. ThenG =_∞

n=1An. SinceN_{G P}^σ (G)is a proper σ-ideal, there is annsuch thatAn∈/NG P(G). We fix such ann, and prove that

Δ₍0,h₁). . . Δ₍0,h_n)f =0 (4)

for everyh1, . . . ,hn ∈ H. Letg denote the left hand side of (4). Then g(x,y) = s

i=1aif(x,y+bi), wheres=2ⁿ,ai = ±1 andbi ∈ Hfor everyi. Lety∈ Hbe fixed. Theng^y=s

i=1aif^y+bi, and thusg^yis a generalized polynomial onG.

If x ∈ An, then deg fx < n, and thus gx = 0. Therefore g^y(x) = 0 for every x ∈ An. Sinceg^y is a generalized polynomial and An ∈/ NG P(G), it follows that g^y =0. Sinceywas arbitrary, this proves (4). Thus deg fx<nfor everyx∈G.

A similar argument shows that, for a suitablem, degf^y <mfor everyy∈ H.

Statement (ii) of the theorem is clear from (i).

Suppose thatGandH are connected. Now we use the fact that if f is a discrete generalized polynomial on an Abelian group which is generated by every neighbour- hood 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 groupG×His connected, so the condition is satisfied, and we conclude that f is continuous everywhere onG×H. Thus (iii) follows from (ii).

5 Examples

In Theorem1none of the conditions onGandHcan be omitted. First we show that without condition (i) the conclusion of Theorem1may fail. We shall need the easy direction of Lemma3.

Example 2 LetG,Hbe discrete Abelian groups. We show that if none ofGandHis torsion, then there is a separately polynomial function f:(G×H)→Csuch that f is not a generalized polynomial onG×H.

By Proposition1,N_P^σ(G)is not a properσ-ideal; that is,G =_∞

n=1An, where An = ∅and An ∈ NP(G)for every n. Let pn ∈ PG be such that pn = 0 and An⊂Zpn. Thenpnis not constant; that is, degpn≥1.

Let Pn = p1· · ·pn; then Pn(x)= 0 for every x ∈ n

i=1Ai, and we have 0<

degP1<degP2< . . .. (Here we use the fact that degpq=degp+degqfor every p,q ∈G PG,p,q =0.) Note that for everyx∈Gwe havePn(x)=0 for all but a finite number of indicesn.

Similarly, we find polynomials Qn∈PH such that 0<degQ1<degQ2< . . ., and for everyy∈ Hwe haveQn(y)=0 for all but a finite number of indicesn.

We put f(x,y) = _∞

n=1Pn(x)Qn(y)for every x ∈ G and y ∈ H. If y ∈ H is fixed, then the sum defining f is finite, and thus f^y ∈ PG. Similarly, we have

fx ∈PH for everyx∈G.

The degrees deg f^y (y ∈ H)are not bounded. Indeed, for every N, there is an y∈ H such thatQN(y)=0. Then f^y =_M

n=1Qn(y)·Pnwith anM ≥ N, where the coefficients Qn(y)are nonzero ifn ≤ N. Therefore, deg f^y ≥ degPN ≥ N, proving that the set{deg f^y: y∈ H}is not bounded. By Lemma3, it follows that f is a not a generalized polynomial.

By the example above, ifGandHare discrete Abelian groups of positive and finite rank, then the conclusion of Theorem1fails. That is,G∈/N_{G P}^σ (G)cannot be omitted from the conditions of Theorem1.

Next we show that the condition onHcannot be omitted either.

Example 3 Let H be a discrete Abelian group of infinite rank. We show that ifGis a topological Abelian group such thatPG contains nonconstant polynomials, then there is a continuous separately polynomial function f onG×H such that f is not a generalized polynomial.

Leth_α(α < κ)be a maximal set of independent elements of Hof infinite order, whereκ≥ω. LetKdenote the subgroup ofHgenerated by the elementsh_α(α < κ).

Every element ofKis of the form

α<κk_αh_α, wherek_α ∈Zfor everyα, and all but a finite number of the coefficientsk_αequal zero.

Let p ∈PG be a nonconstant polynomial. We define f(x,y)=_∞

i=1ki· pⁱ(x) for everyx ∈ G andy ∈ K,y =

α<κk_αh_α. (Note that the sum only contains a finite number of nonzero terms for everyxandy.) In this way we defined f onG×K such that fxis additive onK for everyx ∈G.

If y ∈ H, then there is a nonzero integern such thatny ∈ K. Then we define f(x,y)= ¹_n· f(x,ny)for everyx ∈G. It is easy to see that f(x,y)is well-defined onG×H, and fx is additive on Hfor everyx ∈ G. Therefore, fx is a polynomial onGfor everyx ∈G.

Ify∈ Handny∈ Kfor a nonzero integern, then f^yis of the form_n¹·_N

i=1ki·pⁱ, and thus f^y ∈ PG. Since f^y is continuous for every y ∈ H and H is discrete, it follows that f is continuous onG×H.

Still, f is not a generalized polynomial onG×H, as the set of degrees deg f^y (y ∈ H)is not bounded: if y =hi, then f^y = pⁱ, and degpⁱ =i·degp ≥i for every(i =1,2, . . .).

In the example above we may chooseGin such a way thatG ∈/ N_{G P}^σ (G)holds.

(Take, e.g.,G=R.) In our next example this condition holds for bothGandH.

Example 4 LetE be a Banach space of infinite dimension, and letGbe the additive group ofEequipped with the weak topologyτ ofE. It is well-known that every ball inEis nowhere dense w.r.t.τ, and thusGis of first category in itself.

Still, we show that G ∈/ N_{G P}^σ (G). Indeed, the original norm topology of E is stronger thanτ, 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 onE, and thusNP(G)⊂NP(E)andN_P^σ(G)⊂

N_P^σ(E). SinceN_P^σ(E)is proper by Lemma1, it follows thatN_P^σ(G)is proper. The same is true forN_{G P}^σ (G).

Now letHbe an infinite dimensional Hilbert space, and letGbe the additive group ofHequipped with the weak topology ofH. Let f be the scalar product onH². 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 onG²(see Example 1).

However, f is not a generalized polynomial onG², since f is not continuous. In order to prove this, it is enough to show that f(x,x)= x²is not continuous onH w.r.t. the weak topology. Suppose it is. Then there is a neighbourhoodU of 0 such that x < 1 for every x ∈ U. By the definition of the weak topology, there are linear functionals L1, . . . ,Ln and there is aδ > 0 such that whenever|Li(x)| < δ (i=1, . . . ,n), thenx<1.

Since His of infinite dimension, there is anx=0 such thatLi(x)=0 for every i = 1, . . . ,n. (Otherwise every linear functional would be a linear combination of L1, . . . ,Ln, and thenH =H^∗would be finite dimensional.) Thenλx∈Ufor every λ∈Candλx<1 for everyλ∈C, which is impossible.

The example above shows that in (ii) of Theorem2the condition of joint continuity cannot be omitted. Note also that the groupGdefined in Example4is a topological vector space, hence connected. This shows that in (iii) of Theorem2the condition of the existence of points of joint continuity cannot be omitted either.

Acknowledgements We are indebted to the referee for calling our attention to important pieces of literature and for several suggestions that improved the paper considerably.

Funding Open access funding provided by ELKH Alfréd Rényi Institute of Mathematics.

Declarations

Conflict of interest The authors declare that they have no conflict of interest.

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