Polynomial approximation avoiding values in countable sets

TL;DR

This paper extends classical polynomial approximation theorems to ensure the approximating polynomials avoid specified countable sets of values, broadening the scope of uniform approximation on complex domains.

Contribution

It generalizes Lavrenté's and Mergelyan's theorems to include polynomial approximations that avoid given countable sets of values on complex domains.

Findings

Polynomial approximations can avoid countable sets on compact sets with connected complements.

Extension of Lavrenté's theorem to avoid countable sets.

Extension of Mergelyan's theorem for finite unions of Jordan domains.

Abstract

We generalize a version of Lavrent\'ev's theorem which says that a function that is continuous on a compact set K with connected complement and without interior points can be uniformly approximated as closely as desired by a polynomial without zeros on the set K, so that the polynomial can avoid values from any given countable set. We also prove a corresponding version of Mergelyan's theorem when the interior of K is a finite union of Jordan domains, pairwise separated by a positive distance.

Equations

$$\underset{z\in K}{\max }\left|f(z)-p(z)\right|<\epsilon .$$

$$\underset{z\in K}{\max }\left|f(z)-p(z)\right|<\epsilon ,$$

$$\underset{z\in K}{\max }\left|f(z)-p(z)\right|<\epsilon .$$

$$\underset{z\in K}{\max }\left|f(z)-p(z)\right|<\epsilon ,$$

$$\left|f(z)-p(z)\right|<\epsilon ,z\in K,$$

$$\underset{z\in K}{\max }\left|p(z)-q(z)\right|<\epsilon ,$$

$$g(z)=c_0\prod _{k=1}^m(z-z_k)$$

$$g_n(z)=c_0\prod _{k=1}^m(z-z_{k,n}).$$

$$|g_n(z)-g(z)|<\epsilon ,z\in K.$$

$$|p(z)-q(z)|<\epsilon ,z\in K.$$

$$\underset{z\in K}{\max }\left|q(z)-p(z)\right|<\epsilon ,$$

$${\delta }_j:=\underset{z\in \partial K}{\min }\left|p_j(z)-a_j\right|>0,$$

$$|p_j(z)-p_{j-1}(z)|<\frac{{\epsilon }_{j-1}}{2},z\in K,$$

$${\epsilon }_j<\min \left({\delta }_j,\frac{{\epsilon }_{j-1}}{2}\right).$$

$$\begin{array}{rl}\left|p_l(z)-p_k(z)\right|& \le |p_l(z)-p_{l-1}(z)|+\cdots +|p_{k+1}(z)-p_k(z)|\\ & <\frac{{\epsilon }_{l-1}}{2}+\cdots +\frac{{\epsilon }_k}{2}<\sum _{j=1}^{l-k}\frac{{\epsilon }_k}{2^j}<\sum _{j=1}^{\infty }\frac{{\epsilon }_k}{2^j}={\epsilon }_k,\end{array}z\in K.$$

$$p(z)=\underset{j\to \infty }{\lim }{p}_j(z)$$

$$|p(z)-p_j(z)|\le {\epsilon }_j,j\ge 0,z\in K.$$

$$|p(z)-a_j|\ge |p_j(z)-a_j|-|p(z)-p_j(z)|\ge {\delta }_j-{\epsilon }_j>0,z\in \partial K,$$

$$f(z)=f({\varphi }_j^{-1}({\varphi }_j(z))),z\in \overline{O_j}.$$

$$g(z)=f({\varphi }_j^{-1}((1-{\xi }_j){\varphi }_j(z))),$$

$$|g(z)-f(z)|\le \frac{\epsilon }{3},z\in \overline{O_j}.$$

$$\left|g(z)-f(z)\right|\le \frac{\epsilon }{3},z\in K.$$

$$\left|g(z)-a\right|\ge \delta ,z\in \overline{K^{\circ }},a\in \overline{A}.$$

$$|q(z)-g(z)|<\min (\epsilon /3,\delta /2),z\in K.$$

$$\left|p(z)-q(z)\right|<\min \left(\epsilon /3,\delta /2\right),z\in K,$$

$$\underset{z\in K}{\max }\left|f(z)-p(z)\right|<\epsilon .$$

Full text

1 Introduction

Motivated by the Voronin universality theorem for the Riemann zeta-function, the first author [1], [2] generalized the Lavrentév theorem [8] and the Mergelyan theorem [9] so that in certain cases the approximating polynomial may be assumed to be zero-free on the compact set $K$. In particular the following conjecture111Gauthier (unpublished) also considered this problem in the seventies, see [2, Remark 1]. [2, Conjecture 2] was stated

Conjecture 1**.**

Assume that $K$ is a compact set with connected complement and that $f$ is a continuous function on $K$ which is analytic in the interior of $K$, such that $f$ is zero-free on $K^{\circ}$. Then given any $\varepsilon>0$ there exists a polynomial $p$ which is zero-free on $K$ such that

[TABLE]

While the conjecture in general seems quite difficult, it was proved in [1] that the conjecture is true if the interior of $K$ is empty, and more generally in [2, Theorem 6] that the conjecture is true if the interior of $K$ is a finite union of Jordan domains, pairwise separated by a positive distance. The conjecture has since then been proved in increasing generality. Gauthier-Knese [6] proved that the conjecture is true for “chains of Jordan domains”. Khruschev [7] proved the conjecture to be true if $K$ is locally connected. Andersson-Gauthier [3] gave an independent proof222Actually a somewhat more general statement was proved, see e.g. [3, Theorem 4]. of the “trees of Jordan domains” case. Furthermore it was proved by Danielyan [5, Theorem 1] that the conjecture is true for some $f$ with any given zero-set $Z\subset\partial K$. However he did not prove it for all functions with such a zero-set $Z$ and the conjecture is thus still open for more complicated sets like the Cornucopia set333Example suggested by A. G. O-Farrell. See discussion in [2, Section 4]..

2 Main results

This paper deals with the generalization444Parts of the results in this paper are included in the undergraduate thesis [11] written by the second author and supervised by the first author of the present paper. of the zero-free approximation problem to approximation by polynomials which avoids any countable set $A$ on the set $K$. Our main result is the following theorem.

Theorem 1**.**

Let $A\subset{\mathbb{C}}$ be any countable set and let $K\subset{\mathbb{C}}$ be a compact set with connected complement, such that its interior $K^{\circ}$ is the union of a finite number of Jordan domains, pairwise separated by a positive distance. Let $f$ be a continuous function on $K$ which is analytic in its interior $K^{\circ}$ such that $f(z)\not\in A$ if $z\in K^{\circ}$. Then given any $\varepsilon>0$ there exists some polynomial $p$ such that $p(z)\not\in A$ if $z\in K$ and such that

[TABLE]

In the proof of Theorem 1 (see section 4), we use similar arguments as in the proof of the corresponding result for the zero-free case [2, Theorem 6]. In particular we use rescaling of Riemann mappings and the Carathéodory theorem. However, to treat the parts of the set $K$ that are not in $\overline{K^{\circ}}$ we need some more complicated argument, see Lemma 2 in section 3. With Conjecture 1 in mind we may ask if the corresponding result holds in the finite or countable case.

Question 1**.**

Does Conjecture 1 hold

if we replace “zero-free” with “avoid any set $A$ with two elements”? 2. 2.

if we replace “zero-free” with “avoid any finite set $A$”? 3. 3.

if we replace “zero-free” with “avoid any countable set $A$”?

It is clear that the truth of Conjecture 1 does not change if we replace “zero-free” with ”avoid any set $A$ with one element”, since we may always consider a shifted function (see proof of Lemma 1 in section 3). While we have managed to answer Question 1 in the affirmative in the case when the interior of $K$ is the union of finitely many separated Jordan domains, by using the proof method of [2], it is not clear to us how the methods of Gauthier-Knese [6], Andersson-Gauthier [3] or Khruschev [7] would generalize to this problem. In fact we do not even know how to treat the case when $A$ is a set with two elements and $K$ is the union of two closed discs that intersects at one point. We suggest this as an open problem.

Problem 1**.**

Let $K=\{z:|z+1|\leq 1\}\cup\{z:|z-1|\leq 1\}$. Prove or give a counterexample: For any $\varepsilon>0$ and continuous function $f$ on $K$ that is analytic on $K^{\circ}$ such that $f(z)\not\in\{0,1\}$ if $z\in K^{\circ}$ there exists some polynomial $p$ such that

[TABLE]

and such that $p(z)\not\in\{0,1\}$ if $z\in K$.

In the same way that the Lavrentév theorem [8] is a direct consequence of the Mergelyan theorem [9], for compact sets $K$ with empty interior, Theorem 1 gives us the following version555See [11, Sats 3.2.2]. This result also follows easily by using Lavrentév’s theorem to approximate $f$ by a polynomial $q$ and then using Lemma 2 to approximate the polynomial $q$ by the polynomial $p$. of Lavrentév’s theorem which generalizes the zero-free version [1, Theorem 1.1].

Theorem 2**.**

Let $A\subset{\mathbb{C}}$ be any countable set, let $K\subset{\mathbb{C}}$ be a compact set with connected complement and without interior points, and let $f$ be a continuous function on $K$. Then given any $\varepsilon>0$ there exists some polynomial $p$ such that $p(z)\not\in A$ if $z\in K$ and such that

[TABLE]

This result is stronger when $K$ is a larger set. Examples of large sets (in terms of area measure) that satisfies the conditions of the theorem are $K_{1}=S+iS$, $K_{2}=[0,1]+iS$ and $K_{3}=[0,1]e^{2\pi iS}$, where $S\subset[0,1]$ is a fat Cantor set. In fact we can choose $S$ to have one-dimensional measure arbitrarily close to $1$ so that $K_{j}\subset[0,1]+i[0,1]$ for $j=1,2$ have area measure arbitrarily close to $1$. The result is also stronger if $A$ is a larger set, such as a dense set in ${\mathbb{C}}$. An example of a dense set in ${\mathbb{C}}$ is when $A={\mathbb{Q}}+{\mathbb{Q}}i$ is the set of rational complex numbers. Even stronger, if we use $A$ as the set of (complex) algebraic numbers in Theorem 2 we obtain the following result on approximation by a polynomial that only takes transcendental values on the set $K$.

Corollary 1**.**

Let $K\subset{\mathbb{C}}$ be a compact set with connected complement and without interior points, and let $f$ be a continuous function on $K$. Then given any $\varepsilon>0$ there exists some polynomial $p$ such that

[TABLE]

and such that $p(z)$ is transcendental if $z\in K$.

While our results hold for a countable set $A$ it would be interesting to know whether they hold for some uncountable set $A$. We do not consider this problem here, but the following result gives some restriction on how we can choose the sets $K$ and $A$ if we want our approximation results to hold.

Proposition 1**.**

If we remove the condition that $A$ is countable in Theorem 2 then the conclusion of Theorem 2 is false if $K$ has some non trivial path-connected component and $A$ has some non trivial connected component.

Proof.

We give a proof by contradiction, by assuming that the conclusion of Theorem 2 holds for this choice of $A$ and $K$. Since $K$ has some non trivial path-connected component and by using the fact that a path-connected set in the complex plane is arc-connected666http://mathworld.wolfram.com/Arcwise-Connected.html we can find two points $z_{1},z_{2}\in K$ such that $z_{1}\neq z_{2}$, and some simple curve (Jordan arc) $B\subseteq K$ with parametrization $\beta:[0,1]\to{\mathbb{C}}$ such that $\beta(0)=z_{1}$, $\beta(1)=z_{2}$ and $\beta([0,1])=B$. Now let $a_{1},a_{2}\in A$ be such that $a_{1}\neq a_{2}$ and such that $a_{1}$ and $a_{2}$ are contained in the same connected component of $A$. Let us consider the smooth curve $\Gamma$ with parametrization $\gamma:[0,1]\to{\mathbb{C}}$ that is given in the figure below,

a_{1}$$a_{2}Intersection pointf(B)=\Gamma$$p(B)$$\varepsilon-neighborhood of $\Gamma$

and let the continuous function $f:K\to{\mathbb{C}}$ be defined by $f(z)=\gamma(\beta^{-1}(z))$ if $z\in B$, such that $f(B)=\Gamma$ and by the Tietze extension theorem [12, Theorem 20.4] as any continuous extension to $K$. Let us now use our assumption that the conclusion of Theorem 2 holds and construct a polynomial $p$ such that $p(z)\not\in A$ if $z\in K$ and such that

[TABLE]

holds. If $\varepsilon>0$ is sufficiently small, then $p(B)$ contains a Jordan curve $J\subseteq p(B)$ surrounding $a_{1}$ but not $a_{2}$. This is a consequence of the fact that the curve $p(B)$ follows the $\varepsilon$-neighborhood of $\Gamma$ (grey in figure) and by the crossed arcs lemma777http://www.cut-the-knot.org/blue/JCT/JCT_Part4.shtml must intersect in the marked square with side $2\varepsilon$ (see figure) centered at the intersection point of the curve $\Gamma$. It follows by the Jordan curve theorem that $J^{\complement}$ is not connected and that $a_{1}$ and $a_{2}$ do not lie in the same connected component of $J^{\complement}$. Since $A\subseteq p(K)^{\complement}\subseteq p(B)^{\complement}\subseteq J^{\complement}$ this contradicts our assumption that $a_{1}$ and $a_{2}$ lie in the same connected component of $A$. ∎

3 Some lemmas on polynomials approximating polynomials

In order to prove Theorem 1 we need some useful lemmas on polynomial approximation. In fact we only need Lemma 2, but in order to prove Lemma 2 we need the following Lemma.

Lemma 1**.**

Let $p$ be a polynomial and let $K$ be a compact subset of ${\mathbb{C}}$. Then given any $\varepsilon>0$ and any complex number $a$ there exists some polynomial $q$ of the same degree as $p$ such that

[TABLE]

and such that $q(z)\neq a$ for $z\in\partial K$.

The following proof is a shifted variant of the proof of [1, Theorem 1.1].

Proof.

Let $g(z)=p(z)-a$ be such that

[TABLE]

where $z_{k}$ denotes the zeros of $g(z)$. Since $\partial K$ has no interior points there exist sequences $z_{k,n}\in(\partial K)^{\complement}$ such that $z_{k,n}\rightarrow z_{k}$. Let

[TABLE]

Since $z_{k,n}\in(\partial K)^{\complement}$ we obtain $g_{n}(z)\neq 0$ for $z\in\partial K.$ Since the coefficients converge it is clear that $g_{n}(z)$ converges uniformly to $g(z)$ on $K$. Hence, there is some $n$ such that

[TABLE]

Let $q(z)=g_{n}(z)+a.$ Since $g_{n}(z)\neq 0$ for $z\in\partial K$ it follows that $q(z)\neq a$ if $z\in\partial K$, and since $p(z)-q(z)=g(z)-g_{n}(z)$ it follows from (3) that

[TABLE]

∎

Lemma 2**.**

Let $q$ be a polynomial and let $K$ be a compact subset of ${\mathbb{C}}$. Then given any $\varepsilon>0$ and any countable set $A\subset{\mathbb{C}}$ there exists some polynomial $p$ such that

[TABLE]

and such that $p(z)\not\in A$ for $z\in\partial K$.

Proof.

Let $A=\{a_{1},a_{2},\dots\}$ and let $m$ be the degree of $q$. Let $0<\varepsilon_{0}<\varepsilon$ and $p_{0}(z):=q(z)$. For $j=1,2,\ldots$ there is, according to Lemma 1, some polynomial $p_{j}$ of degree $m$ such that

[TABLE]

and such that

[TABLE]

where $\varepsilon_{j}>0$ for $j\geq 1$ is definied recursively so that

[TABLE]

By the inequalities (5), (6), and the triangle inequality we find for $0\leq k<l$ that

[TABLE]

By (6) and (7) it follows that $\{p_{j}\}_{j=1}^{\infty}$ is a Cauchy-sequence in the vector space of polynomials of degree at most $m$ equipped with the sup-norm on $K$. Since this space is complete then

[TABLE]

is a polynomial of degree at most888It is clear by the construction that the polynomial will in fact have degree exactly $m$. $m$. The inequality (7) yields

[TABLE]

For the polynomial $p$ definied by (8), the inequalites (4), (6), (9) and the triangle inequality gives us

[TABLE]

for all $a_{j}\in A.$ The conclusion of our lemma follows by (9) and (10), by recalling that $p_{0}(z)=q(z)$ and $0<\varepsilon_{0}<\varepsilon$. ∎

4 Proof of Theorem 1

Since $A\subseteq(K^{\circ})^{\complement}$ we also have $\overline{A}\subseteq\overline{(K^{\circ})^{\complement}}=(K^{\circ})^{\complement}.$ Thus we have that $f(z)\not\in\overline{A}$ if $z\in K^{\circ}$. Now let $K^{\circ}=\cup_{j=1}^{n}O_{j}$ where $O_{j}$ are Jordan domains such that $\overline{O_{j}}\cap\overline{O_{i}}=\varnothing$ if $i\neq j$. Let $D=\{z\in{\mathbb{C}}:|z|<1\}$ denote the open unit disc. By the Carathéodory theorem999Also called the Carathéodory-Osgood-Taylor theorem since it was proved independently by Carathéodory [4] and Osgood-Taylor [10]. [12, Theorem 14.19], the Riemann mappings $\phi_{j}:O_{j}\to D$ extend to homeomorphisms $\phi_{j}:{\overline{O_{j}}}\to\overline{D}.$ It is clear that

[TABLE]

Let us now define

[TABLE]

whenever $z\in\overline{O_{j}}$ for sufficiently small $0<\xi_{j}<1$ such that

[TABLE]

The Tietze extension theorem [12, Theorem 20.4] allows $g$ to be extended to a continuous function on $K$ such that

[TABLE]

By the construction it is clear that $g$ is continuous on $K$, that $g$ is analytic on $K^{\circ}$ and that $g(z)\not\in\overline{A}$ if $z\in\overline{K^{\circ}}$. Thus the compact set $g(\overline{K^{\circ}})$ and the closed set $\overline{A}$ are disjoint and must hence be separated by a positive distance $\delta$ so that

[TABLE]

By Mergelyan’s theorem (see [9] or [12, Theorem 20.5]) we can choose a polynomial $q$ such that

[TABLE]

By Lemma 2 there exists some polynomial $p$ such that

[TABLE]

and such that $p(z)\not\in A$ if $z\in\partial K$. By the inequalities (12), (13), (14) and the triangle inequality it follows that also $p(z)\not\in A$ if $z\in K^{\circ}$. Thus $p(z)\not\in A$ if $z\in K$. Finally it follows from the inequalities (11), (13), (14) and the triangle inequality that

[TABLE]

∎