An extension of the universal power series of Seleznev

TL;DR

This paper proves the generic existence of power series with complex coefficients that can approximate any polynomial uniformly on certain compact sets, extending Seleznev's universal power series concept.

Contribution

It introduces a broad class of power series with coefficients depending on previous terms, generalizing previous universal approximation results.

Findings

Power series can approximate any polynomial uniformly on specified compact sets.

The constructed power series have coefficients that depend continuously on previous coefficients.

The results include the case where the coefficients are linear functions of previous coefficients.

Abstract

We show generic existence of power series a with complex coefficients a_n, such that the sequence of partial sums of a new power series where its coefficients b_n are functions of a_0, a_1, ..., a_n approximate every polynomial uniformly on every compact set K not containing the origin and with connected complement. The functions b_n are assumed to be continuous and such that for every complex numbers a_0, a_1, ... , a_{n - 1}, c there exists a complex number a_n such that b_n(a_0, a_1,..., a_{n-1}, a_n) = c. This clearly covers the case of linear functions b_n.

Equations

$$\mathbb{C}\ni z\to b_n(a_0,a_1,\dots ,a_{n-1},z)$$

$$E(m,j,s,N):=\Big{\{}a\in\mathbb{C}^{\aleph_{0}}:\sup_{z\in K_{m}}\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}<\frac{1}{s}\Big{\}}.$$

$$U^{\mu }=\bigcap _{m=1}^{\infty }\bigcap _{j=1}^{\infty }\bigcap _{s=1}^{\infty }\bigcup _{N\in \mu }E(m,j,s,N).$$

$$a\in \bigcap _{m=1}^{\infty }\bigcap _{j=1}^{\infty }\bigcap _{s=1}^{\infty }\bigcup _{N\in \mu }E(m,j,s,N).$$

$$\sup_{z\in K}\big{|}T_{N}(a)(z)-h(z)\big{|}<\varepsilon.$$

$$\sup_{z\in K}\big{|}h(z)-f_{j}(z)\big{|}<\frac{\varepsilon}{2}.$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}<\frac{1}{s}.$$

$$\sup_{z\in K}\big{|}T_{N}(a)(z)-h(z)\big{|}<\varepsilon.$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}<\frac{1}{s}.$$

$$\varepsilon=\dfrac{\frac{1}{s}-\sup_{w\in K_{m}}\big{|}T_{N}(a)(w)-f_{j}(w)\big{|}}{2(N+1)M}>0.$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}<\frac{1}{s}$$

$$\sqrt{\sum\limits_{k=0}^{n}|c_{k}-a_{k}|^{2}}<\sqrt{\sum\limits_{k=0}^{n}\Big{(}\frac{\delta}{\sqrt{N+1}}\Big{)}^{2}}\leq\sqrt{\sum\limits_{k=0}^{N}\frac{\delta^{2}}{N+1}}=\delta\leq\delta_{n}$$

$$\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}\leq\big{|}T_{N}(c)(z)-T_{N}(a)(z)\big{|}+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}=$$

$$=\big{|}\sum_{n=0}^{N}b_{n}(c_{0},c_{1},\dots,c_{n})z^{n}-\sum_{n=0}^{N}b_{n}(a_{0},a_{1},\dots,a_{n})z^{n}\big{|}+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}\leq$$

$$\leq\sum_{n=0}^{N}|b_{n}(c_{0},c_{1},\dots,c_{n})-b_{n}(a_{0},a_{1},\dots,a_{n})|\cdot|z|^{n}+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}<$$

$$<\sum_{n=0}^{N}\varepsilon M+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}=$$

$$=\sum_{n=0}^{N}\dfrac{\frac{1}{s}-\sup_{w\in K_{m}}\big{|}T_{N}(a)(w)-f_{j}(w)\big{|}}{2(N+1)}+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}=$$

$$=\frac{1}{2s}-\frac{1}{2}\sup_{w\in K_{m}}\big{|}T_{N}(a)(w)-f_{j}(w)\big{|}+\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}.$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}\leq\frac{1}{2s}<\frac{1}{s}$$

$$|c_n-a_n|<\epsilon \text{ for }n\le N_0.$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}<\frac{1}{s}.$$

$$\sup\limits_{z\in K_{m}}\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}=\sup\limits_{z\in K_{m}}\big{|}\sum_{n=0}^{N}b_{n}(c_{0},c_{1},\dots,c_{n})z^{n}-f_{j}(z)\big{|}=$$

$$=\sup\limits_{z\in K_{m}}\big{|}\sum_{n=N_{0}+1}^{N}b_{n}(c_{0},c_{1},\dots,c_{n})z^{n}+\sum_{n=0}^{N_{0}}b_{n}(a_{0},a_{1},\dots,a_{n})z^{n}-f_{j}(z)\big{|}=$$

$$=\sup\limits_{z\in K_{m}}\big{|}z^{N_{0}+1}\sum_{n=N_{0}+1}^{N}b_{n}(c_{0},c_{1},\dots,c_{n})z^{n-N_{0}-1}+\sum_{n=0}^{N_{0}}b_{n}(a_{0},a_{1},\dots,a_{n})z^{n}-f_{j}(z)\big{|}=$$

$$=\sup\limits_{z\in K_{m}}|z^{N_{0}+1}|\cdot\big{|}\sum_{n=N_{0}+1}^{N}b_{n}(c_{0},\dots,c_{n})z^{n-N_{0}-1}-\dfrac{f_{j}(z)-\sum_{n=0}^{N_{0}}b_{n}(a_{0},\dots,a_{n})z^{n}}{z^{N_{0}+1}}\big{|}\leq$$

$$\leq M\sup\limits_{z\in K_{m}}\big{|}\sum_{n=N_{0}+1}^{N}b_{n}(c_{0},c_{1},\dots,c_{n})z^{n-N_{0}-1}-\dfrac{f_{j}(z)-\sum_{n=0}^{N_{0}}b_{n}(a_{0},a_{1},\dots,a_{n})z^{n}}{z^{N_{0}+1}}\big{|}.$$

$$\sup\limits_{z\in K_{m}}\big{|}p(z)-\dfrac{f_{j}(z)-\sum_{n=0}^{N_{0}}b_{n}(a_{0},a_{1},\dots,a_{n})z^{n}}{z^{N_{0}+1}}\big{|}<\frac{1}{2Ms}.$$

$$\mathbb{C}\ni z\to b_{N_0+1}(a_0,a_1,\dots ,a_{N_0},z)$$

$$\sup_{z\in K_{m}}\big{|}T_{N}(c)(z)-f_{j}(z)\big{|}\leq\frac{1}{2s}<\frac{1}{s}.$$

Full text

AMS classification number: 30K05.

Key words and phrases: Universal Taylor series, Baire’s theorem, generic property, Mergelyan’s theorem.

1 Introduction

A classical result of Seleznev states that there exists a formal power series $\sum\limits_{n=0}^{\infty}a_{n}z^{n},a_{n}\in\mathbb{C}$, such that its partial sums $S_{N}(z)=\sum\limits_{n=0}^{N}a_{n}z^{n}$ have the following universal approximation property:

For every compact set $K\subset\mathbb{C}\setminus\{0\}$ with connected complement and for every function $h:K\rightarrow\mathbb{C}$, which is continuous on $K$ and holomorphic in the interior of $K$, there exists a strictly increasing sequence of natural numbers $(\lambda_{m})_{m\in\mathbb{N}}$, such that $(S_{\lambda_{m}}(z))_{m\in\mathbb{N}}$ converges to $h$ uniformly on $K$, as $m\longrightarrow+\infty$ [1], [5].

If we identify the formal power series $\sum\limits_{n=0}^{\infty}a_{n}z^{n}$ with the sequence $a=(a_{0},a_{1},\dots)$, then the previous fact holds on a $G_{\delta}$ and dense subset of $\mathbb{C}^{\aleph_{0}}$ endowed with the product topology [1].

It can easily be seen that the previous power series have zero radius of convergence. For universal Taylor series with strictly positive radius of convergence we refer to [2], [3] and [4]; see also [1].

In this paper we extend the result of Seleznev in the case where the universal approximation is not achieved by $S_{N}(z)=\sum\limits_{n=0}^{N}a_{n}z^{n}$, but it is achieved by $T_{N}(a)(z)=\sum\limits_{n=0}^{N}b_{n}(a_{0},\dots,a_{n})z^{n}$, where $b_{n}:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$ are given functions. Our assumptions are the continuity of such $b_{n}$ and that for every fixed $a_{0},a_{1},\dots,a_{n-1}\in\mathbb{C}$, the function $\mathbb{C}\ni z\rightarrow b_{n}(a_{0},a_{1},\dots,a_{n-1},z)$ is onto $\mathbb{C}$. In particular, our results are valid if the functions $b_{n}$ are linear: $b_{n}(a_{0},\dots,a_{n})=\sum\limits_{k=0}^{n}\lambda_{n,k}a_{k},\lambda_{n,k}\in\mathbb{C},\lambda_{n,n}\neq 0$.

In this case, we prove that the universal approximation property is generic topologically and algebraically. That is, the set $U$ of universal power series $a\in\mathbb{C}^{\aleph_{0}}$ is a $G_{\delta}$ and dense subset of $\mathbb{C}^{\aleph_{0}}$ (topological genericity) and it contains a dense vector subspace except 0 (algebraic genericity).

We also notice that our results easily imply the fact that for the generic power series $a=\sum\limits_{n=0}^{\infty}a_{n}z^{n}$, the power series $\sum\limits_{n=0}^{\infty}a_{n}z^{n}$, $\sum\limits_{n=0}^{\infty}\frac{a_{0}+\dots+a_{n}}{n+1}z^{n}$ have zero radius of convergence.

2 Main Result

Definition 1**.**

For every integer $n\geq 0$ let $b_{n}:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$ be a continuous function such that for every $a_{0},a_{1},\dots,a_{n-1}\in\mathbb{C}$, the function

[TABLE]

is onto $\mathbb{C}$. Let $a=(a_{0},a_{1},\dots)\in\mathbb{C}^{\aleph_{0}}$. For every integer $N\geq 0$ and $z\in\mathbb{C}$ we set $T_{N}(a)(z)=\sum\limits_{n=0}^{N}b_{n}(a_{0},\dots,a_{n})z^{n}$. Let $\mu$ be an infinite subset of $\mathbb{N}$. We define $U^{\mu}$ to be the set of $a\in\mathbb{C}^{\aleph_{0}}$, such that for every compact set $K\subset\mathbb{C}\setminus\{0\}$ with connected complement and for every function $h:K\rightarrow\mathbb{C}$, which is continuous on $K$ and holomorphic in the interior of $K$, there exists a strictly increasing sequence of integers $(\lambda_{m})_{m\in\mathbb{N}},\lambda_{m}\in\mu$ such that $(T_{\lambda_{m}}(a)(z))_{m\in\mathbb{N}}$ converges to $h(z)$ uniformly on $K$, as $m\longrightarrow+\infty$.

We notice that if we assume that there exists a sequence of integers $(\lambda_{m})_{m\in\mathbb{N}},\lambda_{m}\in\mu$, not necessarily strictly increasing, such that $(T_{\lambda_{m}}(a)(z))_{m\in\mathbb{N}}$ converges to $h(z)$ uniformly on $K$ then the two definitions are equivalent; see [6].

Considering the set $U^{\mu}$ as a subset of the space $\mathbb{C}^{\aleph_{0}}$ endowed with the product topology, we shall prove that $U^{\mu}$ is a countable intersection of open dense sets. Since $\mathbb{C}^{\aleph_{0}}$ is a metrizable complete space, Baire’s theorem is at our disposal and so $U^{\mu}$ is a dense $G_{\delta}$ set.

The following lemma is well known [1], [3]:

Lemma 2**.**

There exists a sequence of infinite compact sets $K_{m}\subset\mathbb{C}\setminus\{0\},\\ m=1,2,\dots$ with connected complements, such that the following holds: every non-empty compact set $K\subset\mathbb{C}\setminus\{0\}$ having connected complement is contained in some $K_{m}$.

We fix now a sequence $K_{m},m=1,2,\dots$ as in Lemma 2. Let $f_{j},j=1,2,\dots$ be an enumeration of all polynomials having coefficients with rational coordinates. For any integers $m,j,s,N$ with $m\geq 1,j\geq 1,s\geq 1,N\geq 0$, we denote by $E(m,j,s,N)$ the set

[TABLE]

Lemma 3**.**

$U^{\mu}$* can be written as follows:*

[TABLE]

Proof.

The inclusion $U^{\mu}\subseteq\bigcap\limits_{m=1}^{\infty}\bigcap\limits_{j=1}^{\infty}\bigcap\limits_{s=1}^{\infty}\bigcup\limits_{N\in\mu}E(m,j,s,N)$ follows obviously from the definitions of $U^{\mu}$ and $E(m,j,s,N).$ Let

[TABLE]

We shall show that $a\in U^{\mu}$. Let $K\subset\mathbb{C}\setminus\{0\}$ be a non-empty compact set having connected complement and $h:K\rightarrow\mathbb{C}$ a function, which is continuous on $K$ and holomorphic in the interior of $K$. Let $\varepsilon>0$. We have to determine an integer $N\in\mu$, such that

[TABLE]

By Mergelyan’s theorem there exists a polynomial $f_{j},j=1,2,\dots$ having coefficients whose coordinates are both rational, such that

[TABLE]

There exists a compact set with connected complement $K_{m},m=1,2,\dots$ given by Lemma 2, such that $K\subseteq K_{m}$. We can determine an $s$, such that $\frac{1}{s}<\frac{\varepsilon}{2}$. Then we have $a\in\bigcup\limits_{N\in\mu}E(m,j,s,N)$. Thus, there exists an integer $N\in\mu$, such that

[TABLE]

As we have \sup\limits_{z\in K}\big{|}h(z)-f_{j}(z)\big{|}<\frac{\varepsilon}{2}, \sup\limits_{z\in K_{m}}\big{|}T_{N}(a)(z)-f_{j}(z)\big{|}<\frac{1}{s}<\frac{\varepsilon}{2} and $K\subseteq K_{m}$, the triangular inequality implies

[TABLE]

This proves that $a\in U^{\mu}$ and completes the proof. ∎

Lemma 4**.**

For every integer $m\geq 1,j\geq 1,s\geq 1$ and $N\in\mu$, the set $E(m,j,s,N)$ is open in the space $\mathbb{C}^{\aleph_{0}}$.

Proof.

Let $a=(a_{0},a_{1},\dots)\in E(m,j,s,N)$. Then we have

[TABLE]

Let M:=\max\big{\{}1,\sup_{z\in K_{m}}|z|^{N}\big{\}}. We set now:

[TABLE]

For $n=0,1,\dots,N$ the function $b_{n}$ is continuous at $(a_{0},a_{1},\dots,a_{n})$, so there exists $\delta_{n}>0$ such that $|b_{n}(c_{0},c_{1},\dots,c_{n})-b_{n}(a_{0},a_{1},\dots,a_{n})|<\varepsilon$ for $(c_{0},c_{1},\dots,c_{n})\in\mathbb{C}^{n+1}$ with $\sqrt{\sum\limits_{k=0}^{n}|c_{k}-a_{k}|^{2}}<\delta_{n}$. We set $\delta=\min\{\delta_{0},\delta_{1},\dots,\delta_{N}\}$. Suppose that $c=(c_{0},c_{1},\dots)\in\mathbb{C}^{\aleph_{0}}$ satisfies $|c_{k}-a_{k}|<\frac{\delta}{\sqrt{N+1}}$ for $k=0,1,\dots,N$. We shall show that

[TABLE]

and therefore that $c\in E(m,j,s,N)$. This will prove that $E(m,j,s,N)$ is indeed open. For $n=0,1,\dots,N$ we have

[TABLE]

and so $|b_{n}(c_{0},c_{1},\dots,c_{n})-b_{n}(a_{0},a_{1},\dots,a_{n})|<\varepsilon$. For $z\in K_{m}$, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence,

[TABLE]

and the proof is completed. ∎

Lemma 5**.**

For every integer $m\geq 1,j\geq 1$ and $s\geq 1$, the set $\bigcup\limits_{N\in\mu}E(m,j,s,N)$ is open and dense in the space $\mathbb{C}^{\aleph_{0}}$.

Proof.

By Lemma 4 the sets $E(m,j,s,N),N\in\mu$ are open. Therefore the same is true for the union $\bigcup\limits_{N\in\mu}E(m,j,s,N)$. We shall prove that this set is also dense. Let $a=(a_{0},a_{1},\dots)\in\mathbb{C}^{\aleph_{0}},N_{0}$ be an integer such that $N_{0}\geq 0$ and $\varepsilon>0$. It suffices to find $N\in\mu$ and $c=(c_{0},c_{1},\dots)\in E(m,j,s,N)$, such that

[TABLE]

Let $M:=\sup\limits_{z\in K_{m}}|z|^{N_{0}+1}$. We set $c_{n}=a_{n}$ for $n\leq N_{0}$ and so $b_{n}(c_{0},c_{1},\dots,c_{n})=$

$=b_{n}(a_{0},a_{1},\dots,a_{n})\text{ for }n\leq N_{0}.$ We need to find $N\in\mu$ such that

[TABLE]

We have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $0\notin K$ and $K^{c}$ is connected, by Mergelyan’s theorem there exists a polynomial $p(z)=p_{0}+p_{1}z+\dots+p_{m}z^{m}$ such that

[TABLE]

The function

[TABLE]

is onto $\mathbb{C}$ so the equation $b_{N_{0}+1}(a_{0},a_{1},\dots,a_{N_{0}},z)=p_{0}$ has a solution $c_{N_{0}+1}\in\mathbb{C}$. Similarly, we can find $c_{N_{0}+2},\dots,c_{N_{0}+m+1}$ such that $b_{N_{0}+n+1}(c_{0},c_{1},\dots,c_{N_{0}+n+1})=p_{n}$ for $n=1,2,\dots,m$ and $c_{N_{0}+m+2},c_{N_{0}+m+3},\dots$ such that $b_{N_{0}+n+1}(c_{0},c_{1},\dots,c_{N_{0}+n+1})=0$ for $n>m$. By choosing $N\in\mu$ such that $N\geq m+N_{0}+1$ we have

[TABLE]

This proves that the set $\bigcup\limits_{N\in\mu}E(m,j,s,N)$ is indeed dense. ∎

Theorem 6**.**

Under the above assumptions and notation, the set $U^{\mu}$ is a $G_{\delta}$ and dense subset of the space $\mathbb{C}^{\aleph_{0}}$.

Proof.

The result is obvious by combining the previous lemmas with Baire’s Theorem. ∎

Theorem 7**.**

Under the above assumptions and notation, assuming in addition that the functions $b_{n}$ are linear, then the set $U^{\mu}\cup\{0\}$ contains a vector space, dense in $\mathbb{C}^{\aleph_{0}}$.

The proof uses the result of Theorem 6, follows the lines of the implication $(3)\implies(4)$ of the proof of Theorem 3 in [1] and is omitted.

3 Remarks and Comments

The assumptions of the previous section are valid in particular when $b_{n}(a_{0},\dots,a_{n})=a_{n}$ which gives the classical result of Seleznev. Also, it covers the interesting case $b_{n}(a_{0},\dots,a_{n})=\frac{a_{0}+\dots+a_{n}}{n+1}$.

More generally, if $\psi_{n}:\mathbb{C}\rightarrow\mathbb{C},n=0,1,\dots$ are homeomorphisms and $\lambda_{n,k}\in\mathbb{C}$, $0\leq k\leq n$, $n\in\mathbb{N}$, $\lambda_{n,n}\neq 0$, we can set b_{n}(a_{0},\dots,a_{n})=\psi_{n}\big{(}\sum_{k=0}^{n}\lambda_{n,k}a_{k}\big{)} and the results of the previous section are valid.

Another remark is that in order to prove that $U^{\mu}$ is a $G_{\delta}$ set we only need the continuity of the functions $b_{n}:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$ (1). We do not need the assumption that for every $a_{0},a_{1},\dots,a_{n-1}\in\mathbb{C}$, the function $\mathbb{C}\ni z\rightarrow b_{n}(a_{0},a_{1},\dots,a_{n-1},z)$ is onto $\mathbb{C}$ (2). It is also true that using only assumption (2) we can prove that $U^{\mu}$ is dense in $\mathbb{C}^{\aleph_{0}}$.

Indeed, from the classical result of Seleznev, there exist formal power series $c_{0}+c_{1}z+c_{2}z^{2}+\dots$ such that for every compact set $K\subset\mathbb{C}\setminus\{0\}$ with connected complement and for every function $h:K\rightarrow\mathbb{C}$, which is continuous on $K$ and holomorphic in the interior of $K$, there exists a sequence of integers $(\lambda_{m})_{m\in\mathbb{N}},\lambda_{m}\in\mu$ such that $c_{0}+c_{1}z+\dots+c_{\lambda_{m}}z^{\lambda_{m}}\longrightarrow h$ uniformly on K, as $m\longrightarrow+\infty$. Also, we can modify a finite set of coefficients $c_{k}$ and still have the same universal approximation.

Let $a_{0},a_{1},\dots,a_{N_{0}}\in\mathbb{C}$ be fixed. It suffices to show that we can find $a_{N_{0}+1},a_{N_{0}+2},\dots\in\mathbb{C}$ such that $a=(a_{0},a_{1},\dots,a_{N_{0}},a_{N_{0}+1},a_{N_{0}+2},\dots)\in U^{\mu}$. We set $\delta_{k}=b_{k}(a_{0},a_{1},\dots,a_{k}),0\leq k\leq N_{0}$. As we have already mentioned, we can find a formal power series of Seleznev $c=(c_{0},c_{1},\dots)$ satisfying $c_{k}=\delta_{k},0\leq k\leq N_{0}$. Then, because the function $\mathbb{C}\ni z\rightarrow b_{N_{0}+1}(a_{0},a_{1},\dots,a_{N_{0}},z)$ is onto $\mathbb{C}$, we can find $a_{N_{0}+1}$ such that $b_{N_{0}+1}(a_{0},a_{1},\dots,a_{N_{0}},a_{N_{0}+1})=c_{N_{0}+1}$. Continuing in this way we can find $a=(a_{0},a_{1},\dots,a_{N_{0}},a_{N_{0}+1},a_{N_{0}+2},\dots)\in\mathbb{C}^{\aleph_{0}}$ such that $b_{n}(a_{0},a_{1},\dots,a_{n})=c_{n}$ for every $n\in\mathbb{N}$. Therefore $a\in U^{\mu}$. This proves that $U^{\mu}$ is dense.

Acknowledgment. We would like to thank G. Costakis for helpful communications.