simultaneous approximation by universal series

Math. Nachr. 283, No. 6, 909 – 920 (2010) / DOI 10.1002/mana.200710021

Simultaneous approximation by universal series

N. Tsirivas∗1

1 School of Mathematical Sciences, University College Dublin, Belfield, Dublin 4, Ireland

Received 16 February 2007, revised 31 December 2008, accepted 7 January 2009Published online 17 May 2010

Key words Universal approximation, universal series, Taylor series, Laurent series, generic property, algebraicgenericity

MSC (2000) 41A28, 30B30, 30E10, 47A16

We show that, if individual universal series exist, then we can choose a sequence of universal series performingsimultaneous universal approximation with the same sequence of indices. As an application we derive theexistence of universal Laurent Series on an annulus using only the existence of universal Taylor Series on discs.Our results are generic.

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0 Introduction

In the early 70’s W. Luh [12] and independently Chui and Parnes [6] proved the existence of universal Taylorseries in simply connected domains Ω with respect to a fixed center ζ in Ω. Namely there exists a holomorphicfunction f ∈ H(Ω) such that the partial sums of its Taylor expansion with center ζ uniformly approximate allpolynomials on any compact set K ⊂ (Ω)c with connected complement.

In 1986 W. Luh [13] proved the existence of a holomorphic function f ∈ H(Ω) which is a universal Taylorseries with respect to every center ζ ∈ Ω. Furthermore, if a polynomial h and a compact set K as previously aregiven, then the indices λn defining the subsequence of the partial sums which approximate h on K can be takenas the same for all centers ζ ∈ Ω. Thus we have a first example of simultaneous universal approximation.

In the previous example the universal approximation is not requested to be valid on the boundary of Ω. How-ever in 1996 V. Nestoridis [15] strengthened the notion of the universal Taylor series allowing the compact setK to meet ∂Ω; that is K ⊂ Ωc and the universal approximation is also valid on ∂Ω. Furthermore he obtainedsimultaneous universal approximation with the same sequence of indices and uniformly when the center ζ varieson compact subsets of the simply connected domain Ω [14, 16]. The last gives us an example of simultaneousuniversal approximation.

In [14] another generic property of holomorphic functions f on a simply connected domain Ω is found.Namely subsequences of the partial sums of the Taylor development of f approximate f uniformly on compactanot only of a disc but of the whole simply connected domain Ω.

The two generic approximations, one inside Ω towards f and the other outside Ω are performed simultane-ously by the same subsequence of partial sums [14].

In [8] Costakis and Vlachou have proven a theorem where 5 generic approximations are performed simulta-neously by the same sequence of indices.

Recently an abstract theory of universal series has been obtained [2, 18]. This abstract framework allows usto view in a unified way most of the known results and provides extremely simple and short proofs. Furtherthe simplicity of this point of view allows us to obtain several new results. The purpose of the present paperis to establish a generic result for simultaneous universal approximation for a finite or infinite denumerable setof universalities in the form of universal series according to [18] and [2]. We also obtain algebraic genericity(see [1]), that is, the existence of dense linear manifolds of series satisfying such universalities.

∗ e-mail: [email protected], Phone: +353 17 16 2589, Fax: +353 17 16 1172

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

910 Tsirivas: Simultaneous approximation by universal series

As an application we show the existence of universal Laurent Series [7] on an annulus, using only the existenceof universal Taylor series on discs [15].

Finally, we mention that simultaneous universal approximation appears naturally in several situations. For in-stance it appears in the proof of the existence of universal solutions of the wave equation, which under translationsapproximate all other solutions [5]. Recall that an operator is called hypercyclic provided that it admits a denseorbit. Concerning hypercyclic operators T , S the problem of simultaneous approximation is equivalent to thequestion whether T ⊕ S is hypercyclic (see [9], [10]) if different vectors x, y are allowed for T , S, respectively.If one impose x = y then the problem is equivalent to whether T, S are d-hypercyclic (see [4] and [3]).

For the importance of Baire’s theorem in various branches of Analysis we refer to [11] and [9].

1 Preliminaries

We recall initially the elements of the abstract theory we will use [2, 18].Let X be a metrizable topological vector space over the field K = R or C. We may assume that the topology

of X is induced by a translation-invariant metric ρ. Let x0, x1, x2, . . . be a fixed sequence of elements in X .In the sequel we shall write N = {1, 2, . . .} and N0 = N ∪ {0}. Fix a subspace A of KN and assume that it

carries a complete metrizable vector space topology, induced by a translation-invariant metric d.We make the following assumptions:

(A1) The projections pm : A → K, pm(a) = am are continuous for any m ∈ N0.

(A2) The set of polynomials C00 = {a = (an)n≥0 ∈ KN0 : {n ∈ N : an �= 0} is finite} is contained in A.

(A3) C00 is dense in A.

As usual, the symbol (en)n≥0 stands for the canonical basis of KN0 .Let μ = (μn)n∈N be a subsequence of N0.

Definition 1.1 A sequence a ∈ A belongs to UμA if, for every x ∈ X , there exists a subsequence λ = (λn)n∈N

of μ such that

(i)∑λn

j=0 ajxj → x as n → ∞, and

(ii)∑λn

j=0 ajej → a as n → ∞.If the sequence μ is the sequence of natural numbers we write Uμ

A = UA.

The following result gives us sufficient and necessary conditions so that UA �= ∅.

Theorem 1.2 [2, Theorem 1] Under the previous assumptions, the following are equivalent:(1) UA �= ∅.(2) For every p ∈ N, x ∈ X and ε > 0, there exist n ≥ p and ap, ap+1, . . . , an ∈ K such that

ρ

⎛⎝ n∑j=p

ajxj , x

⎞⎠ < ε and d

⎛⎝ n∑j=p

ajej, 0

⎞⎠ < ε.

(3) For every x ∈ X and ε > 0, there exist n ∈ N0 and a0, a1, . . . , an ∈ K such that

ρ

⎛⎝ n∑j=0

ajxj , x

⎞⎠ < ε and d

⎛⎝ n∑j=0

ajej, 0

⎞⎠ < ε.

(4) For every increasing sequence μ of positive integers, UμA is a dense Gδ subset of A.

(5) For every subsequence μ of N, UμA contains, except 0, a dense vector subspace of A.

The previous result is generalized in the case where we do not only have a vector space X but a sequence ofspaces Xi, i = 1, 2, . . . , as follows:

Let (Xk)k≥1 be a sequence of metrizable, topological vector spaces over K whose topologies are induced bytranslation-invariant metrics ρk, k = 1, 2, . . . , and fix a sequence

(xk

i

)i∈N0

, in every space Xk, k = 1, 2, . . .

Let μ be a subsequence of N0.

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Math. Nachr. 283, No. 6 (2010) / www.mn-journal.com 911

Definition 1.3 A sequence a ∈ A belongs to UμA if for every k and x ∈ Xk there exists a subsequence

λ = (λn)n∈N of μ such that

(i)∑λn

j=0 ajxkj → x as n → ∞ and

(ii)∑λn

j=0 ajej → a as n → ∞.

If μ is the sequence of natural numbers we write UμA = UA.

Theorem 1.4 [2, Theorem 3] Under the above assumptions, the following are equivalent:

(1) UA �= ∅.

(2) For every p ∈ N0, x ∈ X and ε > 0, there exist n ≥ p and ap, ap+1, . . . , an ∈ K such that

ρ

⎛⎝ n∑j=p

ajxj , x

⎞⎠ < ε and d

⎛⎝ n∑j=p

ajej, 0

⎞⎠ < ε.

(3) For every k ∈ N, x ∈ Xk and ε > 0, there exist n ∈ N0 and a0, a1, . . . , an ∈ K such that

ρ

⎛⎝ n∑j=0

ajxkj , x

⎞⎠ < ε and d

⎛⎝ n∑j=0

ajej, 0

⎞⎠ < ε.

(4) UμA is a dense Gδ subset of A for every subsequence μ of N0.

(5) UμA ∪ {0} contains a dense vector subspace of A, for every subsequence μ of N0.

Remark 1.5 By the above Theorem 1.4 we have common universality with respect to a denumerably infinitefamily of spaces.

2 Main result

Suppose that two spaces A1 and A2 are given as A above with metrics d1 and d2. Let also X1 and X2 be twospaces with metrics ρ1 and ρ2 with corresponding fixed sequences x1

j ∈ X1, j = 0, 1, 2, . . . and x2j ∈ X2,

j = 0, 1, 2, . . .We ask the following question:Can we find some ai ∈ Ai for i = 1, 2 such that for every xi ∈ Xi, for i = 1, 2, there exists a subsequence

λ = (λn)n∈N of N so that

(i)∑λn

j=0 aijx

ij → xi as n → ∞, and

(ii)∑λn

j=0 aijej → ai as n → ∞, for i = 1, 2?

We emphasize that the sequences of indices performing the approximations in the spaces A1, X1 and A2, X2

can be chosen to be the same.This can be proven on the basis of the abstract theory [2, Theorem 1] or [2, Theorem 3].Furthermore the simultaneous approximation can be generalized to the case of any finite or denumerable

infinite set of spaces.As an application of this result we are able to show the existence of universal Laurent series in an annu-

lus, using only the existence of universal Taylor series on discs. Now we will show a result of simultaneousapproximation for any denumerable infinite set of spaces.

Let Ai, i ∈ N, be metrizable topological vector spaces that are subspaces of KN0 .Let di be corresponding translation-invariant metrics.Let X i

k, i, k ∈ N, be metrizable topological vector spaces. Let ρik be corresponding translation-invariant

metrics.Let xi

k,j ∈ X ik, i, k ∈ N, j ∈ N0.

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Following [2, Theorem 1] and the previous Definition 1.1, let U i,kAi

, i, k ∈ N, be the set of all ai ∈ Ai suchthat, for any x ∈ X i

k there exists some sequence (λn) with

ρik

⎛⎝ λn∑j=0

aijx

ik,j , x

⎞⎠ −→ 0 and di

⎛⎝ λn∑j=0

aijej , a

i

⎞⎠ −→ 0 as n −→ ∞.

Now, the corresponding series with simultaneous approximation are the following:

Definition 2.1 Let μ be a fixed subsequence of the positive integers.Then Uμ

∞ is the set of all (ai)i≥1 ∈ ∏i≥1 Ai such that, for any sequence (ki) of N and any xi ∈ ∏

i≥1 X iki

there exists a subsequence (λn) of μ such that, for all i ≥ 1,

ρiki

⎛⎝ λn∑j=0

aijx

iki,j , xi

⎞⎠ −→ 0 and di

⎛⎝ λn∑j=0

aijej, a

i

⎞⎠ −→ 0 as n −→ ∞.

And we set U∞ = U (n)∞ .

Remark 2.2 If U∞ �= ∅ then U i,kAi

�= ∅ for any i, k ∈ N. This is obvious by the definitions of sets U∞ and

U i,kAi

(Definition 1.1).

Later in this section we will see that the reverse holds, under some assumptions.

Remark 2.3 If U∞ �= ∅ then the spaces X ik, i, k ∈ N, are separable. This is obvious by the definitions. We

remark that the definition of Uμ∞ is complicated. However we can express the set Uμ∞ in relation to some othersimpler sets.

Let μ be a subsequence of N. Let N, n, s ∈ N. We define the sets

Fμ(N, n, s) :=

⎧⎨⎩(aj) ∈∏i≥1

Ai

/∀ 1 ≤ i ≤ N, di

⎛⎝ μn∑j=0

aijej , a

i

⎞⎠ <1s

⎫⎬⎭ .

We suppose that the spaces X ik, i, k ∈ N, are separable. Let (yi

k,�)�≥1 be dense sequences in the spaces X ik,

i, k ∈ N. Let i, k, n, �, s ∈ N. We define the sets

Eμ(i, k, n, �, s) :=

⎧⎨⎩(aj) ∈∏j≥1

Aj

/ρi

k

⎛⎝ μn∑j=0

aijx

ik,j , y

ik,�

⎞⎠ <1s

⎫⎬⎭ .

By the previous terminology we have the following lemma.

Lemma 2.4 Suppose that μ is a subsequence of N and that each X ik, i, k ∈ N, is separable where the

(yik,�)�≥1’s are dense sequences in X i

k. Then

Uμ∞ =

⋂N,s≥1

⋂(k1,k2,...,kN )∈NN

(�1,�2,...,�N )∈NN

⋃n≥1

(N⋂

i=1

Eμ(i, ki, n, �i, s) ∩ Fμ(N, n, s)

).

The previous is obvious; see the proof of Theorem 1 in [2].

Remark 2.5 The previous lemma holds without the assumptions (A1)–(A3) and the fact that the spaces Ai,i ≥ 1, can be complete. Furthermore, we note that we have not imposed any topology in the space

∏i≥1 Ai.

From now on we suppose that the space∏

i≥1 Ai carries the product topology. We also suppose that thecondition (A1) holds in every space Ai, i = 1, 2, . . . The sets Fμ(N, n, s) and Eμ(i, k, n, �, s) are open forevery i, �, k, N, n, s ∈ N and a subsequence μ of N. This is obvious. By this and Lemma 2.4 we obtain that theset Uμ

∞ is a Gδ subset of∏

i≥1 Ai.Now in addition to the previous assumptions we suppose that the spaces Ai, i ∈ N are complete and the

conditions (A2)–(A3) are satisfied.



We have a similar theorem as in [2, Theorem 1].

Theorem 2.6 The following assertions are equivalent

(i) U∞ �= ∅;(ii) for any N ∈ N any (k1, k2, . . . , kN ) ∈ N

N , any xi ∈ X iki

(1 ≤ i ≤ N), p ∈ N0 and any ε > 0 there issome n ≥ p and ai

p, aip+1, . . . , a

in ∈ K (1 ≤ i ≤ N) such that, for 1 ≤ i ≤ N ,

ρiki

⎛⎝ n∑j=p

aijx

iki,j, xi

⎞⎠ < ε and di

⎛⎝ n∑j=p

aijej , 0

⎞⎠ < ε;

(iii) the condition ii) for p = 0, that is, for any n ∈ N any (k1, k2, . . . , kN ) ∈ NN , any xi ∈ X iki

(1 ≤ i ≤ N)and any ε > 0 there is some n ∈ N and ai

0, ai1, . . . , a

in ∈ K (1 ≤ i ≤ N) such that, for 1 ≤ i ≤ N ,

ρiki

⎛⎝ n∑j=0

aijx

iki,j, xi

⎞⎠ < ε and di

⎛⎝ n∑j=0

aijej , 0

⎞⎠ < ε;

(iv) for any subsequence μ of the positive integers, Uμ∞ is a dense Gδ-subset of

∏i≥1 Ai;

(v) for any subsequence μ of the positive integers, Uμ∞ ∪ {0} contains a dense vector subspace of

∏i≥1 Ai.

P r o o f. The proof of the equivalence of (i)–(iv) is almost exactly the same as that of Theorem 1 of [2](Theorem 1.2 in Section 1 in the preliminaries of the present paper).

We remind that the metric d in the space∏

1≥1 Ai is the usual one. That is if (ai), (bi) ∈ ∏i≥1 Ai then

d((ai), (bi)) =∞∑

i=1

12i

di(ai, bi)1 + di(ai, bi)

.

The space∏

1≥1 Ai is complete. Thus, Baire’s Theorem is at our disposal. For the proof that (iv) implies (v)one needs to use the following modification of the proof of Theorem 3 of [2], more precisely, the index k ≥ 1 inthat proof has to be replaced by the multi-index (N, k1, . . . , kN ), where N ≥ 1, (k1, k2, . . . , kN ) ∈ NN .

We now have the following simple but significant

Observation 2.7 Condition (iii) of the Theorem 2.6 is equivalent to the condition

(vi) for any i, k ∈ N, U i,kAi

�= ∅.

P r o o f. We need only show that (vi) implies (iii). Thus, suppose that (vi) holds. We fix N ≥ 1, (k1, . . . , kN ) ∈N

N , xi ∈ X iki

(1 ≤ i ≤ N) and ε > 0. For i = 1, . . . , N , we have that U i,ki

Ai�= ∅.

By Theorem 1 of [2] (i.e., Theorem 1.2 in the present paper), we obtain ni ∈ N and ai0, . . . , a

ini

∈ K

(1 ≤ i ≤ N) such that, for 1 ≤ i ≤ N ,

ρiki

⎛⎝ ni∑j=0

aijx

iki,j, xi

⎞⎠ < ε and di

⎛⎝ ni∑j=0

aijej , 0

⎞⎠ < ε;

Now we set n = max(n1, . . . , nN ) and aij = 0 for i = 1, . . . , N , ni < j ≤ n. Then condition (iii) clearly

holds.

Remark 2.8 The case of finitely many spaces Ai, i = 1, 2, . . . , n, n ≥ 2, follows from Theorem 2.6 becauseany finite such system may be embedded in an infinite one.



Remark 2.9 We note that we can obtain Theorem 2.6 as an application of the abstract theory we formulatedin Section 1. The proof uses the simple idea of identifying a countable collection of sequence spaces with a singlesequence space but it is complicated. The proof we present here is much shorter and easier to follow.

Now we will present another proof of a similar result. This proof is due to V. Nestoridis. This proof is simpleas the previous proof of Theorem 2.6 and Observation 2.7 but more general. Especially we do not suppose thatthe conditions (A2) and (A3) hold. We only suppose that the condition (A1) holds. Firstly we state the necessaryterminology.

Let (Ai, di) be the spaces as previously for i = 1, 2, . . . , without the assumptions (A2) and (A3), but withthe assumption (A1). For every i = 1, 2, . . . let also the spaces (X i

k)k=1,2,... with the sequences (xik,j), j ≥ 0,

respectively where the spaces (X ik, ρi

k), (i, k) ∈ N2, are defined as previously.

For every subsequence μ of N let Uμi , i ≥ 1, be the set of universal sequences in Ai as in Definition 1.3 with

respect to the spaces (X ik)k≥1, the fixed sequences

(xi

k,j

)j≥0

(in X i

k

)and the sequence μ.

By the previous terminology we have the following Theorem 2.10.

Theorem 2.10 We suppose that for every i ∈ N and for every subsequence μ of N the sets Uμi are dense in the

spaces (Ai, di). Then for every subsequence μ of N the sets Uμ∞ are dense, Gδ subsets of

∏i≥1 Ai and contain a

dense vector subspace of∏

i≥1 Ai except 0 where the space∏

i≥1 Ai is endowed with the product topology.

P r o o f. The metric d of the space∏

i≥1 Ai is the usual. Without loss of generality we can suppose thatthe metrics di, i ≥ 1, are uniformly bounded, especially that di(x, y) ≤ 1 for every (x, y) ∈ A2

i , for every

i = 1, 2, . . . . The space(∏

i≥1 Ai, d)

is a complete metric space. For every (i, k) ∈ N2 the space X i

k is

separable. Let (yik,�)�≥1 be a dense sequence in the space X i

k for every (i, k) ∈ N2.The proof follows three steps:

First step: We show that Uμ∞ �= ∅ for every subsequence μ of N.

Second step: We show that for every subsequence μ of N the set Uμ∞ is a Gδ , dense subset of

∏i≥1 Ai.

Third step: We show that for every subsequence μ of N the set Uμ∞ ∪ {0} contains a dense vector subspace of∏

i≥1 Ai.

First step: Let μ be a subsequence of N. Let a1 ∈ Uμ1 . Then for every (k1, �1) ∈ N2 there exists a

subsequence μ1,k1,�1 of μ so that

ρ1k1

⎛⎝μ1,k1 ,�1n∑j=0

a1jx

1k1,j , y

1k1,�1

⎞⎠ −→ 0 and d1

⎛⎝μ1,k1 ,�1n∑j=0

a1jej , a

1

⎞⎠ −→ 0 as n −→ +∞.

By the assumptions and Baire’s Theorem we conclude that the set⋂∞

k1,�1=1 Uμ1,k1 ,�1

2 is Gδ and dense in (A2, d2).

Let a2 ∈ ⋂∞k1,�1=1 Uμ1,k1 ,�1

2 . Then for every k1, �1, k2, �2 ∈ N there exists a subsequence μ2,k1,�1,k2,�2 ofμ1,k1,�1 such that

ρ2k2

⎛⎝μ2,k1 ,�1,k2,�2n ∑

j=0

a2jx

2k2,j , y

2k2,�2

⎞⎠ −→ 0 and d2

⎛⎝μ2,k1 ,�1,k2,�2n ∑

j=0

a2jej , a

2

⎞⎠ −→ 0 as n −→ +∞.

Inductively we construct a sequence (ai)i≥1 ∈ ∏i≥1 Ai and for every m ∈ N and (k1, �1, . . . , km, �m) ∈

N2m we construct a sequence μm,k1,�1,...,km,�m so that for every m ∈ N and (k1, �1, . . . , km+1, �m+1) ∈N2(m+1) the sequence μm+1,k1,�1,...,km+1,�m+1 is a subsequence of μm,k1,�1,...,km,�m and for every m ∈ N and(k1, �1, . . . , km, �m) ∈ N2m

ρmkm

⎛⎝μm,k1 ,�1,...,km,�mn ∑

j=0

amj xm

km,j , ymkm,�m

⎞⎠ −→ 0 and dm

⎛⎝μm,k1,�1,...,km,�mn ∑

j=0

amj ej , a

m

⎞⎠ −→ 0 as n −→ +∞.



Using the denseness of (yik,�)�≥1 in X i

k for every (i, k) ∈ N2 and the triangle inequality it is easy to see thatfor every m ∈ N, (k1, k2, . . . , km) ∈ N

m and (x1, x2, . . . , xm) ∈ ∏mi=1 X i

kithere exists a subsequence λ =

(λn)n∈N of μ so that

ρiki

⎛⎝ λn∑j=0

aijx

iki,j, xi

⎞⎠ −→ 0 (2.1)

and

di

⎛⎝ λn∑j=0

aijej, a

i

⎞⎠ −→ 0 (2.2)

as n → +∞ for every i = 1, 2, . . . , m.Let any (k1, k2, . . . ) ∈ NN and any (xi)i≥1 ∈ ∏

i≥1 X iki

.

By the convergence of (2.1) and (2.2) we can construct a subsequence λ = (λn)n∈N of μ so that for everym = 1, 2, . . .

ρiki

⎛⎝ λm∑j=0

aijx

iki,j , xi

⎞⎠ <1m

and di

⎛⎝ λm∑j=0

aijej, a

i

⎞⎠ <1m

for every i = 1, 2, . . . , m. This implies that for every i = 1, 2, . . . .

ρiki

⎛⎝ λn∑j=0

aijx

iki,j , xi

⎞⎠ −→ 0 and di

⎛⎝ λn∑j=0

aijej, a

i

⎞⎠ −→ 0 as n −→ +∞.

Thus Uμ∞ �= ∅ and the first step was completed.

Second step: We remind that for every subsequence μ of N the set Uμ∞ is a Gδ subset of∏

i≥1 Ai. Let anyB = (b1, b2, . . . ) ∈ ∏

i≥1 Ai and ε0 > 0. We will find some a ∈ Uμ∞ with d(B, a) < ε0.

There exists n0 ∈ N such that∑∞

j=n0+112j < ε0

2 .Following the former procedure in first step we can choose some a = (a1, a2, . . . ) ∈ Uμ∞ such that di(ai, bi) <

ε02 for every i = 1, 2, . . . , n0 because every ai, i = 1, 2, . . . , belongs to a dense subset of Ai.

Thus a ∈ Uμ∞ and d(B, a) < ε0. So the set Uμ∞ is dense in(∏

i≥1 Ai, d)

for every subsequence μ of N.

Third step: The space∏

i≥1 Ai is separable. Now for every subsequence μ of N we consider the set

Uμ,1∞ =

{a = (a1, a2, . . . ) ∈

∏i≥1

Ai

/∀m ∈ N, ∀ (k1, k2, . . . , km) ∈ N

m and

∀ (x1, x2, . . . , xm) ∈m∏

i=1

Xiki

∃λ = (λn)n∈N a subsequence of μ so that

ρiki

(λn∑j=0

aijx

iki,j , xi

)→ 0 and di

(λn∑j=0

aijej , a

i

)→ 0 as n → +∞ ∀ i = 1, 2, . . . , m

}.

Making a diagonal argument we easily see that Uμ∞ = Uμ,1

∞ for every subsequence μ of N. Now we follow theproof as in Theorem 3 of [2] for the above set Uμ,1

∞ instead of Uμ∞.

We use the modification as in Theorem 2.6 of the proof of Theorem 3 of [2], more precisely the index k ≥ 1in that proof has to be replaced by the multi-index (N, k1, k2, . . . , kN ), where N ≥ 1, (k1, k2, . . . , kN ) ∈ NN .The proof of Theorem 2.10 is now complete.

Remark 2.11 The previous result concerns classes of type UA. However the same proof works more generallyfor classes of the form U(L, M) of [2] under the assumptions of Theorem 29 of [2].



3 A more general setting

In this section we consider a more general setting according to [2]. Let Ei, i = 1, 2, . . . , be a sequence ofcomplete and metrizable topological vector spaces over the field K = R or C whose topologies are induced bytranslation-invariant metrics di, i = 1, 2, . . .

For every i = 1, 2, . . . let (ein)n∈N be a sequence of elements in Ei and let Φi

n : Ei → K be a sequence(Φi

n)n∈N of linear and continuous functions.For every i = 1, 2, . . . we consider a sequence (X i

k)k≥1 of metrizable topological vector spaces over K whosetopologies are induced from translation-invariant metrics (ρi

k)k≥1.For every i = 1, 2, . . . and k = 1, 2, . . . let xi

k,0, xik,1, x

ik,2, . . . be a fixed sequence of elements in X i

k.For every a ∈ C00 we consider the polynomial gi

a =∑∞

j=0 ajeij ∈ Ei for every i = 1, 2, . . .

Further we make the following assumptions:

(E1) The set {gia : a ∈ C00} is dense in Ei for every i = 1, 2, . . .

(E2) For every a ∈ C00 the sets {n ∈ N : Φin(gi

a) �= 0} are finite for every i = 1, 2, . . .

(E3) For every a ∈ C00 and for every i = 1, 2, . . . it holds:

∞∑j=0

Φij(g

ia)ei

j =∞∑

j=0

ajeij .

(E4) For every a ∈ C00, i = 1, 2, . . . and k = 1, 2, . . . it holds:

∞∑j=0

Φij(g

ia)xi

k,j =∞∑

j=0

ajxik,j .

Definition 3.1 Let i, k ∈ N and let μ be a subsequence of N. An element f ∈ Ei belongs to U i,k,μ if forevery x ∈ X i

k there exists a subsequence λ = (λn)n∈N of μ such that

(i) ρik

(∑λn

j=0 Φij(f)xi

k,j , x)→ 0 as n → ∞, and

(ii) di

(∑λn

j=0 Φij(f)ei

j , f)→ 0 as n → ∞.

If μ = N we denote U i,k,μ = U i,k. Now we define the set of universal series in the product space∏∞

i=1 Ei

that perform simultaneous approximations with respect to all the spaces.

Definition 3.2 An element f = (f1, f2, . . . ) ∈ ∏∞i=1 Ei belongs to Uμ,∞, if for every (k1, k2, . . . ) ∈ NN

and every x = (x1, x2, . . . ) ∈ ∏∞i=1 X i

kithere exists a subsequence λ = (λn)n∈N of μ such that for every

i = 1, 2, . . .

ρiki

⎛⎝ λn∑j=0

Φij(f

i)xiki,j, xi

⎞⎠ −→ 0 as n −→ ∞ and di

⎛⎝ λn∑j=0

Φij(f

i)eij , f

i

⎞⎠ −→ 0 as n −→ ∞.

Theorem 3.3 Suppose, according to the previous terminology, that U i,k �= ∅ for every i, k ∈ N. Then forevery subsequence μ of N the set Uμ,∞ is a dense Gδ subset of

∏∞i=1 Ei and contains a dense vector subspace

of∏∞

i=1 Ei except 0.

We omit the proof of Theorem 3.3, because it repeats the arguments of the proof of Theorem 2 of the Section1 of [2].

Now we will generalize the previous result. For every i = 1, 2, . . . and k = 1, 2, . . . we consider a sequenceof continuous functions (T i

k,n)n∈N where T ik,n : Ei → X i

k. We suppose that the following condition is valid:



For every n ∈ N every a = (a1, a2, . . . , an) ∈ Cn00 (Cn

00 = C00 × C00 × · · · × C00, n times) and every(k1, k2, . . . , kn) ∈ Nn, there exists a subsequence (nv)v∈N of N such that

ρjkj

(T j

kj ,nv(gj

a),∞∑

λ=0

ajλxj

kj ,λ

)−→ 0 as v −→ ∞

for every j = 1, 2, . . . , n.

Let B(T i

k,n, i, k ∈ N, n ∈ N0

)=

{(nν)ν∈N ∈ NN / (nν)ν∈N is a subsequence of N such that there ex-

ists n ∈ N, a = (a1, a2, . . . , an) ∈ Cn00 and (k1, k2, . . . , kn) ∈ Nn so that for every j = 1, 2, . . . , n

ρjkj

(T j

kj ,nν(gj

a),∑∞

λ=0 ajλxj

kj ,λ

)→ 0 as ν → ∞

}.

A sequence γ ∈ KN belongs to Γ (T ik,n, i, k ∈ N, n ∈ N0), if for every b ∈ B(T i

k,n, i, k ∈ N, n ∈ N0) thesequences b and γ have at least a common subsequence. For example the sequence of natural numbers belongsto Γ (T i

k,n, i, k ∈ N, n ∈ N0). Therefore Γ (T ik,n, i, k ∈ N, n ∈ N0) �= ∅.

Now we define the set Uμ,∞ of universal series that perform approximations with respect to any denumerableinfinite set of spaces and the functions T i

k,n, (i, k) ∈ N2, n = 0, 1, 2, . . .

Definition 3.4 Let μ ∈ Γ (T ik,n, i, k ∈ N, n ∈ N0). A f = (f1, f2, . . . ) ∈ ∏∞

i=1 Ei belongs to Uμ,∞ if forevery (k1, k2, . . . ) ∈ NN and every x = (x1, x2, . . . ) ∈

∏∞i=1 X i

kithere exists a subsequence λ = (λn)n∈N of μ

such that for every i = 1, 2, . . .

ρiki

⎛⎝ λn∑j=0

Φij(f

i)xiki,j

, xi

⎞⎠ −→ 0, di

⎛⎝ λn∑j=0

Φij(f

i)eij , f

i

⎞⎠ −→ 0 and ρiki

⎛⎝ λn∑j=0

T iki,j

(f i), xi

⎞⎠ −→ 0

as n → ∞.

We say that a sequence (nν)ν∈N ∈ B(T i

k,n, i, k ∈ N, n ∈ N0

)can be chosen independently from a ∈ Cn

00,(k1, k2, . . . , kn) ∈ Nn and n ∈ N, if for every n ∈ N, every a ∈ Cn

00 and every (k1, k2, . . . , kn) ∈ Nn it holds:

ρjkj

(T j

kj ,nν(gj

a),∞∑

λ=0

ajλxj

kj ,λ

)−→ 0 for every j = 1, 2, . . . , n.

Theorem 3.5 In addition to the previous assumptions we suppose that U i,k �= ∅ for every i, k ∈ N. Thenfor every μ ∈ Γ

(T i

k,n, i, k ∈ N, n ∈ N0

)the set Uμ,∞ is a dense Gδ subset of

∏∞i=1 Ei. If, in addition, the

continuous functions T ik,n are linear, for every (i, k) ∈ N2 and n ∈ N0 and the sequence (nv)v∈N can be chosen

independently from a ∈ Cn00, (k1, k2, . . . , kn) ∈ Nn and n ∈ N, then the set Uμ,∞ ∪ {0} contains a dense vector

subspace of∏∞

i=1 Ei.

The proof is in the same spirit as the proof of Theorem 3.3 and is omitted.

4 An application

Now we will prove the existence of universal Laurent series in an annulus using only the existence of universalTaylor series in a disc and the result of Section 2 in the case of two spaces A1 and A2.

We note that here we have a similar presentation as in Section 2.Let D(ζ0, R) be the open disc with center ζ0 ∈ C and radius R ∈ (0, +∞). We consider the space

H(D(ζ0, R)) of holomorphic functions in D(ζ0, R) that is supplied with the topology of uniform convergenceon compact. Now we define the set of universal functions in H(D(ζ0, R)) as in [15].

A function f ∈ H(D(ζ0, R)) belongs toU(H(D(ζ0, R))) if for every non void compact set K ⊂ C�D(ζ0, R)

with connected complement and every function h : K → C, continuous in K and holomorphic in◦K there exists

a subsequence of natural numbers (λn)n∈N in N0 such that

supz∈K

|Sλn(f, ζ0)(z) − h(z)| −→ 0 as n −→ +∞,



where

SN (f, ζ0)(z) =N∑

j=0

f (j)(ζ0)j!

(z − ζ0)j .

It is known (see [15]) that

U(H(D(ζ0, R))) �= ∅. (4.1)

It is well-known that the function space H(D(ζ0, R)) is isometric with the vector subspace of CN0 , AD(ζ0,R) =

T (H(D(ζ0, R))) where T is the linear isometry T (f) =(

f(n)(ζ0)n!

)∞

n=0, f ∈ H(D(ζ0, R)), where the metric in

the space AD(ζ0,R) is the natural metric that is induced by the linear function T .For the following lemma we refer to [15].

Lemma 4.1 There exists a sequence of compact sets Km ⊆ C � D(ζ0, R), m = 1, 2, . . . , with connectedcomplements, such that the following holds: every non-empty compact set K ⊆ C � D(ζ0, R) having connectedcomplement is contained in some Km.

Let (Km)m≥1 be a sequence as in the above Lemma 4.1.Now we consider the set of universal series Un

AD(ζ0,R)in the space AD(ζ0,R) with respect to space Xn :=

A(Kn) :={f : Kn → C, f continuous in Kn and holomorphic in

◦Kn

}endowed with the sup-norm and the

sequence xnv = (z − z0)v|Kn, in Xn = A(Kn), n = 1, 2, . . . , v = 0, 1, 2 . . . .

Now it is easy to see that

T (U(H(D(ζ0, R)))) =⋂n≥1

UnAD(ζ0,R)

. (4.2)

By (4.1) and (4.2) we have that

UnAD(ζ0,R)

�= ∅ for all n ≥ 1. (4.3)

We will use this result to show the following Theorem 4.2.

Theorem 4.2 Let D(ζ0, r, R) = {z ∈ C : r < |z − ζ0| < R} be an annulus, 0 < r < R < +∞. Thereexists a Laurent series

∑+∞n=−∞ cn(z − ζ0)n convergent in D(ζ0, r, R) such that, for every non void compact set

K ⊆ C � {ζ0} with K ∩D(ζ0, r, R) = ∅ and connected complement and every function h : K → C continuous

in K and holomorphic in◦K, there exists a subsequence λ = (λn)n∈N of natural numbers such that

supz∈K

∣∣∣∣∣λn∑

k=−λn

ck(z − ζ0)k − h(z)

∣∣∣∣∣ −→ 0 as n −→ ∞.

P r o o f. Let AΩ1,ζ0 be the vector subspace of CN that is defined for Ω1 = D(ζ0, R) and let BΩ2,ζ0 bethe vector subspace of CN that is defined for Ω2 = D

(0, 1

r

). From the above relation (4.3) there exists a =

(a0, a1, . . . ) ∈ AΩ1,ζ0 and b = (0, b1, b2, . . . ) ∈ BΩ2,ζ0 that perform simultaneous approximations. This is theonly point where we use the previous results of the present paper. Let K be a compact subset of C as in thetheorem. Then there exist K1, K2 disjoint compact sets, with connected complements such that K = K1 ∪ K2,K1 ⊂ D(ζ0, r), K2 ∩ D(ζ0, R) = ∅, ζ0 /∈ K1. We set

K1 ={

1z − ζ0

: z ∈ K1

}.

The function h(z) − ∑∞n=1 bn(z − ζ0)−n is continuous in K2 and holomorphic in

◦K2 and the function

h(

1z−ζ0

)−∑∞

n=0 an(z − ζ0)−n is continuous in K1 and holomorphic in◦K1. The series

∑∞n=1 bn(z − ζ0)−n

and∑∞

n=0 an(z − ζ0)−n converge uniformly in K2 and K1 respectively.



Let ε0 > 0. Because of the above facts and the fact that a and b perform simultaneous approximations we canfind m0 ∈ N such that

supz∈K2

∣∣∣∣∣m0∑k=1

bk(z − ζ0)−n −∞∑

k=1

bk(z − ζ0)−n

∣∣∣∣∣ <ε0

2, (4.4)

supz∈K1

∣∣∣∣∣m0∑k=0

ak(z − ζ0)−n −∞∑

k=0

ak(z − ζ0)−n

∣∣∣∣∣ <ε0

2, (4.5)

supz∈K2

∣∣∣∣∣m0∑k=0

ak(z − ζ0)k −(

h(z) −∞∑

k=1

bk(z − ζ0)−k

)∣∣∣∣∣ <ε0

2, (4.6)

supz∈K1

∣∣∣∣∣m0∑k=1

bk(z − ζ0)k −(

h

(1

z − ζ0

)−

∞∑k=0

ak(z − ζ0)−k

)∣∣∣∣∣ <ε0

2. (4.7)

By (4.4) and (4.6) we get

supz∈K2

∣∣∣∣∣m0∑k=0

ak(z − ζ0)k +m0∑k=1

bk(z − ζ0)−k − h(z)

∣∣∣∣∣ < ε0. (4.8)

By (4.5) and (4.7) we get

supz∈K1

∣∣∣∣∣m0∑k=1

bk(z − ζ0)k +m0∑k=0

ak(z − ζ0)−k − h

(1

z − ζ0

)∣∣∣∣∣ < ε0. (4.9)

Thus by (4.8) and (4.9) we obtain

supz∈K

∣∣∣∣∣m0∑k=0

ak(z − ζ0)k +m0∑k=1

bk(z − ζ0)−k − h(z)

∣∣∣∣∣ < ε0. (4.10)

For k = 0, 1, . . . , m0 we put ck = ak and for k = −1,−2, . . . ,−m0 we put ck = b−k. So by (4.10) we have

supz∈K

∣∣∣∣∣m0∑

k=−m0

ck(z − ζ0)k − h(z)

∣∣∣∣∣ < ε0.

So we conclude that there exists a sequence (λn)n∈N so that∑λn

k=−λnck(z − ζ0)k tends to h uniformly on K

and the theorem has been proved.

Let UD(ζ0,r,R) be the set of universal Laurent series that are described in Theorem 4.2. From the proof ofTheorem 4.2 we conclude also that the set UD(ζ0,r,R) is a dense Gδ subset of H(D(ζ0, r, R)) and contains adense vector subspace of H(D(ζ0, r, R)) except 0.

Remark 4.3 Theorem 4.2 can be generalized in the case of any finitely connected domain instead of anannulus (see [7], [19], [20]). The proof is similar but a little bit more complicated and is omitted. In additionto this case we can have universal Laurent series where the approximation is valid at the level of all derivativesprovided the functions to be approximated are polynomials only [17].

Moreover a similar result holds in the case of Laurent-Faber series [7].

Remark 4.4 We note that in Theorem 4.2 we have the symmetric sums∑λn

k=−λnck(z − ζ0)k, n = 1, 2, . . .

If the lower and upper limit were allowed to differ, the simpler theory of [2] would have sufficed to get theresult.



Acknowledgements During this research the author was supported by the State Scholarships Foundation of Greece (I.K.Y).I would like to express my gratitude to Professor V. Nestoridis for helpful discussions and suggestions and his interest in

this work. Many thanks to Professors F. Bayart and G. Costakis for their valuable remarks. I would also like to thank theanonymous referee for his suggestions, which improved the proofs and the presentation of the paper considerably.

I would like to express my gratitude to the State Scholarships Foundation of Greece (I.K.Y) for its support.Many thanks to Professor Stephen Gardiner for his interest in this work.

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