One-sided invertibility of Toeplitz operators on the space of all holomorphic functions on finitely connected domains

Abstract

We prove that if the symbol of a Toeplitz operator acting on the space of all holomorphic functions on a finitely connected domain is non-degenerate and vanishes then the range of this operator is not complemented. As a result, we obtain that a Toeplitz operator on the space of all holomorphic functions on finitely connected domains is left invertible if and only if it is an injective Fredholm operator. Also, such an operator is right invertible if and only if it is a surjective Fredholm operator.

1 Introduction

Let $\gamma _{0},\gamma _{1},\ldots ,\gamma _{n}$ be $C^{\infty }$ smooth Jordan curves in the Riemann sphere $\mathbb {C}_{\infty }$. We assume that the closures of the domains bounded by the curves $\gamma _{1},\ldots ,\gamma _{n}$ are disjoint and that these curves are contained in the domain bounded by the curve $\gamma _{0}$. These domains are denoted throughout the paper by

$$\begin{aligned} D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n}). \end{aligned}$$

In case there is no curve $\gamma _{0}$ we obtain the unbounded (in the complex plane) domain $D(\gamma _{1},\ldots ,\gamma _{n})$. If there are no curves $\gamma _{1},\ldots ,\gamma _{n}$ we consider the simply connected domain $D(\gamma _{0})$, i.e. the interior $I(\gamma _{0})$ of the curve $\gamma _{0}$. We denote by $H(D)=H(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n}))$ the Fréchet space of all functions holomorphic in D with the usual topology of uniform convergence on compact subsets of D. If the domain D is unbounded we consider the case of the spaces $H_{0}(D)$ of all functions holomorphic in D, which vanish at the infinity and also the spaces H(D) without any assumption on the value at $\infty$. In any case we denote the corresponding function space by X.

We study one-sided invertibility of the operators of the form

$$\begin{aligned} (T_{F}f)(z):=\sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(\zeta )\cdot f(\zeta )}{\zeta -z}d\zeta ,\, f\in H(D). \end{aligned}$$

(1.1)

Here $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}$ is a function holomorphic in some open neighborhood of bD in D. That is, in the set $U\cap D$, where U is an open neighborhood of bD in the Riemann sphere, the function $F_{i}$ is holomorphic in D close to the boundary curve $\gamma _{i}$. Strictly speaking, F is rather the equivalence class of such a function with respect to a relation of the sort which defines germs—we shall make it precise in the next section. Functions (germs) F are called symbols and the algebra of all symbols is denoted by $\mathfrak {S}(D)$. The symbol $(\gamma _{i})_{\varepsilon }$ for small $\varepsilon >0$ stands for an appropriately defined dilatation of the boundary curve $\gamma _{i}$. The number $\varepsilon >0$ is small enough to guarantee that $z\in D((\gamma _{0})_{\varepsilon };(\gamma _{1})_{\varepsilon },\ldots ,(\gamma _{n})_{\varepsilon })$. In [17] we showed that it is legitimate to call operators of form (1.1) Toeplitz operators on the space H(D). We remark here that operators of the form (1.1) appear in a natural way in various branches of mathematics to mention only approximation theory, number theory and even theoretical computer science. We refer the reader to [18, Introduction] for more information.

Our main result is the following theorem.

Theorem 1.1

(Main Theorem) Assume that the symbol $F\in \mathfrak {S}(D)$ is non-degenerate and vanishes. Then the range of the operator $T_{F}:X\rightarrow X$ is not complemented in X. As a result, the operator $T_{F}$ is left invertible if and only if it is an injective Fredholm operator.

Let us recall that a closed subspace $\mathcal {F}$ of a locally convex space $\mathcal {E}$ is complemented if there exists a continuous linear projection $P:\mathcal {E}\rightarrow \mathcal {E}, P^{2}=P$ onto $\mathcal {F}$. Equivalently, if there exists a closed linear space $\mathcal {G}$ such that $\mathcal {E}=\mathcal {F}\oplus \mathcal {G}$.

We explain the meaning of the assumptions in the theorem assuming that $X=H(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n}))$. The (germ) function $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}$ is non-degenerate if no function $F_{i}$ is identically equal to zero. It vanishes if there is a sequence $z_{k}\in D$ with at least one accumulation point in bD such that $F(z_{k})=0$. Equivalently, if for every representative F of the germ F there is $z\in D$ such that $F(z)=0$. We emphasize that the assumption that F vanishes implies not only that there is a zero of the function F in D. It means that the zeros of F accumulate on bD (and maybe somewhere else as well). In [17] we proved the following result:

Theorem 1.2

Consider the Toeplitz operator $T_{F}:X\rightarrow X$ with the symbol $F\in \mathfrak {S}(D)$. The operator $T_{F}$ is a Fredholm operator if and only if F is non-degenerate and does not vanish. If F is non-degenerate and does not vanish, then

$$\begin{aligned} \mathrm{index\,} T_{F}=-\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma )_{\varepsilon }}\frac{F^{'}(\zeta )}{F(\zeta )}d\zeta , \end{aligned}$$

where $(\gamma )_{\varepsilon }$ is the cycle $(\gamma _{0})_{\varepsilon }+\cdots +(\gamma _{n})_{\varepsilon }$ in the case $X=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$, $(\gamma _{1})_{\varepsilon }+\cdots +(\gamma _{n})_{\varepsilon }$ in the case $X=H_{0}(D)$ with $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and the sum of the circle $|z|=\frac{1}{\varepsilon }$ and the cycle $(\gamma _{1})_{\varepsilon }+\cdots +(\gamma _{n})_{\varepsilon }$ when $X=H(D)$ where $D=D(\gamma _{1},\ldots ,\gamma _{n})$.

Furthermore, we completed the picture with the following theorem:

Theorem 1.3

Consider the Toeplitz operator $T_{F}:X\rightarrow X$ with the symbol $F\in \mathfrak {S}(D)$, which does not vanish identically.

• (i) If F is non-degenerate and vanishes then the operator $T_{F}$ is injective. The range of the operator $T_{F}$ is a closed subspace of X of infinite codimension.

• (ii) If F is degenerate and does not vanish then the kernel of the operator $T_{F}$ is of infinite dimension. The range of the operator $T_{F}$ is a closed subspace of X of infinite codimension.

• (iii) If F is degenerate and vanishes then the operator $T_{F}$ is injective. The range of the operator is not closed in X and is of infinite codimension.

If an operator is right invertible then it is a surjection. Theorem 1.3 implies the next result.

Corollary 1.1

Consider the Toeplitz operator $T_{F}:X\rightarrow X$ with the symbol $F\in \mathfrak {S}(D)$. The $T_{F}$ is right invertible if and only if is a surjective Fredholm operator.

Operators of form (1.1) are natural object of study. Let us repeat that we described in [15] and [18] how they appear in approximation theory and number theory (however on rather different spaces than considered in this paper). We showed in [17, Theorem 4.2, Theorem 4.3 and Theorem 4.4] that certain matrices which are associated with the operators on H(D) are Toeplitz matrices for operators given by (1.1) and this property characterizes operators of form (1.1). Therefore we call the operators $T_{F}$ Toeplitz operators. We refer the reader to our previous papers, especially to [18] and [17], for the history, motivation and the literature concerning Toeplitz operators on some locally convex spaces of analytic functions.

The problem of characterizing one-sided invertible operators in various classes of concrete operators is old and it drew considerable attention. As far as Toeplitz operators on the Hardy space $H^{2}$ are concerned, it is known that a Toeplitz operator $T_{\varphi }$ with an unimodular symbol $\varphi \in L^{\infty }$, is left invertible on $H^{2}$ iff $\text {dist}_{L^{\infty }}(\varphi ,H^{\infty })<1$, it is right invertible iff $\text {dist}_{L^{\infty }}(\varphi ,\overline{H^{\infty }})<1$ ( [10, Theorem 2.20], see also [24, 25] and [12]). It seems a natural question to ask whether there is a characterization of this sort in the case which we investigate. Note that the operator $T_{\varphi }$ is left invertible on $H^{2}$ if and only if there is $\varepsilon >0$ such that $\Vert T_{\varphi }\Vert \ge \varepsilon \Vert f\Vert$ for $f\in H^{2}$. The reason why our Toeplitz operators fail to be left invertible is according to Main theorem much more involved—its range is not complemented.

The monograph [13] contains an in-depth study of linear singular integral operators. It follows from [13, Theorem 6.6.1] that a singular integral operator (so also a Toeplitz operator) is one-sided invertible (from the left or from the right) if and only if it is a Fredholm operator. In [17] we proved the following analog of the Coburn–Simonenko theorem.

Theorem 1.4

Consider the Toeplitz operator $T_{F}:X\rightarrow X$ with the symbol $F\in \mathfrak {S}(D)$.

• (i) If F is non-degenerate then either $\ker T_{F}$ is trivial or $\ker T_{F}^{'}$ is trivial.

• (ii) The operator $T_{F}$ is invertible if and only is F is a Fredholm operator of index zero.

It is a consequence of this result that a Fredholm operator $T_{F}:X\rightarrow X$ is either surjective or it is injective. Therefore in perfect analogy with [13, Theorem 6.6.1] (compare also [24, Corollary 1]) we have the following theorem.

Corollary 1.2

A Toeplitz operator $T_{F}:X\rightarrow X$ is one-sided invertible (from the left or from the right) if and only if $T_{F}$ is a Fredholm operator.

Toeplitz operators on multiconnected domains on the Hardy spaces were studied by Abrahamse [1]. In our research we are guided by the classical Hardy space theory of Toeplitz operators. There are excellent monographs on this subject by Böttcher and Silbermann [10] and the recent ones by Nikolskii [21] and [22]. As for the functional analysis background we refer the reader to the in-depth monograph by Meise and Vogt [20]. There are also the classical monographs [19] by Köthe and [14] by Jarchow and the recent nice book [23]. Our method strongly relies on the Weierstrass factorization theory of entire functions. For this we refer the reader to Conway’s book [11]. There is now a growing interest on operator theory on spaces which could be named non-standard, in particular such spaces of holomorphic functions. We were influenced by the works of Albanese et al. [2,3,4,5], Bonet et al. [7] and Bonet and Taskinen [9]. We especially refer the reader to the very recent survey by Bonet on operators on weighted spaces of holomorphic functions [6]. We believe that our results are an interesting part of this line of research.

The paper is divided into four sections. First we briefly discuss the idea of the proof, then we provide the necessary background. The last section is entirely devoted to the proof of the main result.

2 The idea of the proof of main theorem

We sketch here the idea of the proof concentrating only on the key facts. The reader may prefer to read the next section first. In [18] we proved that if the symbol of a Toeplitz operator on the space of all real analytic functions on the real line $\mathcal {A}(\mathbb {R})$ has real zeros going to infinity but has no non-real zeros accumulating at a real point (see [15, p. 5] for the explanation) then the range of the Toeplitz operator is not a complemented subspace of $\mathcal {A}(\mathbb {R})$ [18, Theorem 3]. A dual result was also proved for the adjoint Toeplitz operator [18, Theorem 4]. Roughly speaking, if the symbol vanishes then the operator is not one-sided invertible. The method of the proof was rather involved and technical and it was simplified in [16]. We apply it also in the situation which we study in this paper. We strive to make the proof concise but on the other hand we do not avoid the (necessary) technical details. The aim of this section is to provide the idea of the argument. The statements which we make here are carefully proved in the last section.

The method consists of two steps. First we show that the range of a Toeplitz operator, the symbol of which is non-degenerate and vanishes, is equal to the intersection of the kernels of a certain sequence of continuous linear operators. Here it is important that the range of such an operator is closed (Theorem 1.3). Then we solve an interpolation problem induced by these functionals. The solution is based on Eidelheit’s theorem [20, Theorem 26.27], which we recall below. As we shall see, this suffices to complete the proof that $T_{F}$ is not left invertible when F is non-degenerate and vanishes. The fact that the domain is finitely connected makes the arguments more involved.

Assume that $F\in \mathfrak {S}(D)$ is non-degenerate and vanishes. That is, $F=F_{0}\oplus \cdots \oplus F_{n}$, no $F_{i}\equiv 0$ and there is a sequence $(z_{k})$ with at least one accumulation point on bD such that $F(z_{k})=0$. Using Weierstrass theory we factor out the zeros of F

$$\begin{aligned} F(z)=\prod _{k=0}^{\infty }E_{k}(z)\cdot F^{0}(z). \end{aligned}$$

(2.1)

Then the operator $T_{F^{0}}$ is a Fredholm operator and by the argument principle we may assume that

$$\begin{aligned} \text {index}T_{F_{0}}=-1. \end{aligned}$$

In view of Theorem 1.4, this implies that there is a continuous linear functional $\xi :H(D)\rightarrow \mathbb {C}$ such that

$$\begin{aligned} \text {range} T_{F_{0}}=\ker \xi . \end{aligned}$$

The functional $\xi$ is by the Köthe–Grothendieck–da Silva theorem (we recall this result on p. 8 below) on the dual space of the space H(D) represented by a (germ) function $\varphi =\varphi _{0}\oplus \varphi _{1}\oplus \ldots \oplus \varphi _{n}\in H(D^{c})$. This implies that

$$\begin{aligned} 0=\sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}^{0}(z)\cdot f(z)\cdot \varphi _{i}(z)\text {d}z, \end{aligned}$$

for every $f\in H(D)$ since, as we show in Lemma 3.1, the functionals do not distinguish between Toeplitz operators and multiplication operators, just as in the Hardy space case. But then

$$\begin{aligned} \langle T_{F}f,\frac{\varphi }{(z-z_{k})}\rangle =\sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{\prod _{j}E_{j}(z)}{z-z_{k}}\cdot F_{i}^{0}(z)\cdot f(z)\cdot \varphi _{i}(z)\text {d}z=0 \end{aligned}$$

(2.2)

by Cauchy’s theorem, since we prove that $F^{0}\cdot \varphi$ defines a function in H(D) and $z-z_{k}$ divides $\prod E_{j}(z)$. This is the key observation. The functionals $\xi _{m}$ which correspond to the functions

$$\begin{aligned} \frac{\varphi _{0}}{z-z_{k}}\oplus \frac{\varphi _{1}}{z-z_{k}}\oplus \cdots \oplus \frac{\varphi _{n}}{z-z_{k}}, k\in \mathbb {N} \end{aligned}$$

and the function $\varphi$ itself vanish on the range of the operator $T_{F}$. We use the Hahn–Banach theorem to prove that actually the equality holds

$$\begin{aligned} \text {range}\,T_{F}=\bigcap _{m=1}^{\infty }\ker \xi _{m}. \end{aligned}$$

In the second step we refer to Eidelheit’s theorem. Our goal is to solve the interpolation problem for the functionals $\xi _{m}$.

Theorem 2.1

([20], Theorem 26.27) Let E be a Fréchet space, $(U_{k})_{k\in \mathbb {N}}$ be a fundamental system of zero neighborhoods in E and let $(A_{j})_{j\in \mathbb {N}}$ be linearly independent, continuous linear forms on E. Then the infinite system of equations

$$\begin{aligned} A_{j}x=y_{j} \,\text {for all}\, j \,\in \mathbb {N} \end{aligned}$$

is solvable for each sequence $y\in \omega$ ($\omega$ the space of all sequences) if, and only if, the following holds:

$$\begin{aligned} \dim ((E^{'})_{U_{k}^{\circ }}\cap \textrm{span}\, \{A_{j}:j\in \mathbb {N}\})<\infty \,\text {for all}\, k\in \mathbb {N}. \end{aligned}$$

The key property which needs to be shown is

$$\begin{aligned} \dim ((H(D)^{'})_{U_{k}^{\circ }}\cap \mathrm{span\,}\{\xi _{j}:j\in \mathbb {N}\})<\infty \,\text {for all} \, k\in \mathbb {N}. \end{aligned}$$

(2.3)

Let us recall that if $U\subset E$, then its polar $U^{\circ }$ is the set

$$\begin{aligned} U^{\circ }=\{\xi \in E^{'}:|\xi (x)|\le 1, x\in U\}. \end{aligned}$$

Also, if $B\subset F$ is absolutely convex subset of a locally convex space F, then

$$\begin{aligned} F_{B}=\text {span}\, B=\bigcup _{t>0}tB. \end{aligned}$$

Thus, the symbol $(E^{'})_{U_{k}^{\circ }}$ stands for the span in $E^{'}$ of the set $U_{k}^{\circ }\subset E^{'}$, which is the polar of the zero neighborhood $U_{k}\subset E$.

Property (2.3) amounts to the following: if for $\xi =\sum \alpha _{l}\xi _{l}$ it holds that

$$\begin{aligned} |\xi (f)|\le Cp(f) \end{aligned}$$

for some continuous seminorm p, then starting some M, $\alpha _{M}=\alpha _{M+1}=\cdots =0$ with M depending only on the seminorm p. To prove this we construct functions $g_{\nu }$ such that $\xi _{i}(g_{\nu })=0$ for all but one i, say $i=M$. Furthermore, the seminorms $p(g_{\nu })$ are uniformly bounded but $\xi _{M}(g_{\nu })\rightarrow \infty$ as $\nu \rightarrow \infty$. This will imply that $\alpha _{M}=0$. A similar construction can be done for $M+1$. The construction uses again the Toeplitz operators and the Weierstrass elementary factors and again amounts to a formula of type (2.2). Specifically, the functions $g_{\nu }$ are defined as the value of the Toeplitz operator $T_{F^{0}}$, with $F^{0}$ given in (2.1), on carefully chosen holomorphic functions.

It follows from Eidelheit’s Theorem that the map

$$\begin{aligned} \Xi :H(D)\ni f\mapsto (\xi _{m}(f))\in \omega , \end{aligned}$$

where $\omega$ is the Fréchet space of all sequences, is surjective. This implies that $\ker \Xi$, which is $\text {range}\,T_{F}$ is not complemented and, as a result, the operator $T_{F}$ is not left invertible.

3 Background

We refer the reader to [17, Background] for the technical details and definitions. Let us here only repeat that $D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ is the domain in the Riemann sphere bounded by the $C^{\infty }$ smooth Jordan curves $\gamma _{0},\gamma _{1},\ldots ,\gamma _{n}$. The curves $\gamma _{1},\ldots ,\gamma _{n}$ are contained in the interior of the curve $\gamma _{0}$ and the closures of the domains bounded by the curves $\gamma _{1},\ldots ,\gamma _{n}$ are disjoint. If there is no curve $\gamma _{0}$, the unbounded domain is denoted by $D(\gamma _{1},\ldots ,\gamma _{n})$. If there are no curves $\gamma _{1},\ldots ,\gamma _{n}$, the simply connected domain bounded by the curve $\gamma _{0}$ is denoted by $D(\gamma _{0})$—this is just the interior $I(\gamma _{0})$ of the curve $\gamma _{0}$. We assume the standard orientation of the curves $\gamma _{0},\gamma _{1},\ldots ,\gamma _{n}$, i.e. the one induced by the standard orientation of the plane. We now define the function spaces which we work in.

Definition 3.1

Let $\gamma _{0},\gamma _{1},\ldots ,\gamma _{n}\subset \mathbb {C}$ be $C^{\infty }$ smooth Jordan curves such that $\overline{I(\gamma _{i})}\cap \overline{I(\gamma _{j})}=\emptyset$ for $i,j=1,\ldots ,n$ with $i\ne j$ and

$$\begin{aligned} \gamma _{1},\ldots ,\gamma _{n}\subset I(\gamma _{0}). \end{aligned}$$

Then X denotes the Fréchet space H(D) of all functions holomorphic in $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$.

Let $\gamma _{1},\ldots \gamma _{n}\subset \mathbb {C}$ be $C^{\infty }$ smooth Jordan curves such that $\overline{I(\gamma _{i})}\cap \overline{I(\gamma _{j})}=\emptyset$, $i\ne j$. Then

$$\begin{aligned} D:=D(\gamma _{1},\ldots ,\gamma _{n}):=E(\gamma _{1})\cap \ldots \cap E(\gamma _{n}), \end{aligned}$$

where $E(\gamma )$ stands for the exterior of the Jordan curve $\gamma$ and X denotes either the Fréchet space $H_{0}(D)$ of all functions holomorphic in D which vanish at $\infty$ or the space H(D) of all functions holomorphic in D (in general with no value at $\infty$).

If $K\subset D$ run through all compact subsets of D then the seminorms

$$\begin{aligned} p_{K}(f):=\sup _{z\in K}|f(z)| \end{aligned}$$

form a fundamental system of seminorms of the space X.

Consider the inductive family

$$\begin{aligned} \{H(U\cap D),r_{U,V}\}_{U,V\supset \gamma _{i}} \end{aligned}$$

where the sets $U\supset V$ run through open (in $\mathbb {C}$) neighborhoods of $\gamma _{i}$ and for $U\supset V$ the map $r_{U,V}$ is the restriction $r_{U,V}(f):=f\vert _{V\cap D}$.

Definition 3.2

If $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ then the symbol space

$$\begin{aligned} \mathfrak {S}(D)=\mathfrak {S}(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})) \end{aligned}$$

is

$$\begin{aligned} \mathfrak {S}(D):=\bigoplus _{i=0}^{n}\mathrm{lim ind\,} \{H(U\cap D),r_{U,V}\}_{U,V\supset \gamma _{i}}. \end{aligned}$$

If $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and the space X is equal to $H_{0}(D)$ then the symbol space $\mathfrak {S}(D)=\mathfrak {S}(D(\gamma _{1},\ldots ,\gamma _{n}))$

$$\begin{aligned} \mathfrak {S}(D):=\bigoplus _{i=1}^{n} \mathrm{lim ind\,} \{H(U\cap D),r_{U,V}\}_{U,V\supset \gamma _{i}}. \end{aligned}$$

If $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and the space X is equal to H(D) then the symbol space $\mathfrak {S}(D)=\mathfrak {S}(D(\gamma _{1},\ldots ,\gamma _{n}))$

$$\begin{aligned} \mathfrak {S}(D):=\bigoplus _{i=1}^{n} \mathrm{lim ind\,} \{H(U\cap D),r_{U,V}\}_{U,V\supset \gamma _{i}}\oplus \mathfrak {S}_{\infty }, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {S}_{\infty }:=\{ \mathrm{lim ind\,}H(U\setminus \{\infty \}),r_{U,V}\}_{U,V\ni \infty }, \end{aligned}$$

the open sets U run through open in $\mathbb {C}_{\infty }$ neighborhoods of $\infty$ and $r_{U,V}$ are as usual the restriction maps when $U\supset V$, i.e. for $F\in H(U{\setminus } \{\infty \}), r_{U,V}(F)=F\vert _{V{\setminus } \{\infty \}}$.

In Definition 3.2 we abuse the notation and use the same symbol $\mathfrak {S}(D)$ to denote different objects. However it always follows from the context which one we mean.

Let $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ and $F\in \mathfrak {S}(D)$. Then F is the equivalence class of functions defined in some sets of the form $\mathcal {U}\cap D$, with $bD\subset \mathcal {U}$ and $\mathcal {U}$ open, with respect to the relation

$$\begin{aligned} F_{1}\sim F_{2} \Leftrightarrow F_{1}\vert _{U\cap D}=F_{2}\vert _{U\cap D}. \end{aligned}$$

Here $F_{i}\in H(U_{i}\cap D), i=1,2$, $U_{i}$ are open neighborhoods of bD and also $U\supset bD$ is open and satisfies $U\subset U_{1}\cap U_{2}$. The relation closely resembles the one which defines germs. Observe however that the intersection of the domains of definition of the functions F is empty unlike in the germ case. Also, $F_{\infty }\in \mathfrak {S}_{\infty }$ is the equivalence class with respect to a similar equivalence relation of the functions defined in punctured neighborhoods of $\infty$ in $\mathbb {C}_{\infty }$.

We assign now to symbols in $\mathfrak {S}(D)$ continuous linear operators on the spaces H(D). We will give all the details in the case of the bounded domains

$D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ and then rather briefly comment on the other two cases.

Assume first that $X=H(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n}))$ and $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}\in \mathfrak {S}(D)$ is represented by a function defined in some neighborhood of bD in D denoted by still the same symbol $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}$. We set for $f\in X$ and $z\in D$

$$\begin{aligned} (T_{F}f)(z):=\sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta , \end{aligned}$$

where $(\gamma _{i})_{\varepsilon }$ is an appropriately defined dilatation of the curve $\gamma _{i}$—we refer again the reader to [17, p. 47] for the technical details. We only remark here that the number $\varepsilon >0$ above is small enough to guarantee that

• (i) $(\gamma _{i})_{\delta }$ is contained in the domain of definition of $F_{i}$ for every $0<\delta \le \varepsilon$ (that is, there are no unnecessary holes);

• (ii) the point z belongs to

  $$\begin{aligned} I((\gamma _{0})_{\varepsilon })\cap E((\gamma _{1})_{\varepsilon })\cap \cdots \cap E((\gamma _{n})_{\varepsilon }). \end{aligned}$$

It follows from Cauchy’s theorem that the definition is correct. It does not depend on the representative F chosen nor on $\varepsilon >0$.

If $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and $X=H_{0}(D)$ then for $F\in \mathfrak {S}(D)$, $F=F_{1}\oplus \cdots \oplus F_{n}$ we put

$$\begin{aligned} (T_{F}f)(z):=\sum _{i=1}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta \end{aligned}$$

for $f\in X$.

If $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and $X=H(D)$ then for $F\in \mathfrak {S}(D)$, $F=F_{1}\oplus \cdots \oplus F_{n}\oplus F_{\infty }$ we put

$$\begin{aligned} (T_{F}f)(z):=\sum _{i=1}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta +\frac{1}{2\pi \mathrm i}\int \limits _{|\zeta |=R}\frac{F_{\infty }(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta \end{aligned}$$

for $f\in X$. This time we need also to guarantee that R is large enough. The following fact is immediate.

Proposition 3.1

The operator $T_{F}$ with $F\in \mathfrak {S}(D)$ maps continuously the space X into itself.

Sometimes we find it convenient to use the notion of cycles. Thus, for instance $\gamma =\gamma _{0}+\gamma _{1}+\cdots +\gamma _{n}$ and with this notation the expression

$$\begin{aligned} \sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta \end{aligned}$$

can simply be written as

$$\begin{aligned} \frac{1}{2\pi \mathrm i}\int \limits _{(\gamma )_{\varepsilon }}\frac{F(\zeta )\cdot f(\zeta )}{\zeta -z}\text {d}\zeta . \end{aligned}$$

In [17] we showed that it is legitimate to call the operators $T_{F}:X\rightarrow X$ Toeplitz operators. Namely we proved that one can assign to every operator $T:X\rightarrow X$ $n+1$ infinite matrices. Such an operator is an operator of the form $T_{F}$ for some $F\in \mathfrak {S}(D)$ if and only if these matrices are Toeplitz matrices and the operator satisfies a certain compatibility condition. These are [17, Theorem 4.2, Theorem 4.3, Theorem 4.4].

We assume that the symbol F in each case is non-degenerate. This means that no component of F is identically equal to zero.

Definition 3.3

Assume that $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ and $X=H(D)$ or $D=D(\gamma _{1},\ldots ,\gamma _{n})$ and $X=H_{0}(D)$. We say that $F\in \mathfrak {S}(D)$ vanishes if for every F which represents F there is $z\in D$ such that $F(z)=0$. Equivalently, if there exists a sequence $(z_{k})$ with at least one accumulation point on bD such that $F(z_{k})=0$ for every representative F, provided $F(z_{k})$ is defined.

If $D=D(\gamma _{1},\ldots ,\gamma _{n})$, $X=H(D)$ and $F=F_{1}\oplus \cdots \oplus F_{n}\oplus F_{\infty }$. Then we say that F vanishes if $F_{1}\oplus \cdots \oplus F_{n}$ vanishes in the above sense or there is a sequence $(z_{k})$ which converges to $\infty$ such that $F_{\infty }(z_{k})=0$ for every $F_{\infty }$ which represents $F_{\infty }$, provided $F(z_{k})$ is defined.

Our result is based on the explicit description of the dual space of the space H(D) known as the Köthe–Grothendieck–da Silva duality (see [19, pp. 372–378]). Namely,

$$\begin{aligned} H(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n}))^{'}_{b}\cong H_{0}(D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})^{c}) \end{aligned}$$

and the duality is given by

$$\begin{aligned} \langle f,\phi \rangle :=\int \limits _{(\gamma )_{\varepsilon }}f\cdot \phi \text {d}z=\sum _{i=0}^{n}\int \limits _{(\gamma _{i})_{\varepsilon }}f\cdot \phi _{i}(z)\text {d}z, \end{aligned}$$

(3.1)

where $\varepsilon >0$ is sufficiently small. Here, $\phi =\phi _{0}\oplus \phi _{1}\oplus \cdots \oplus \phi _{n}$ is a germ of a holomorphic function which vanishes at $\infty$ on the closed set

$$\begin{aligned} D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})^{c}=\overline{E(\gamma _{0})}\cup \overline{I(\gamma _{1})}\cup \cdots \cup \overline{I(\gamma _{n})}. \end{aligned}$$

Similarly,

$$\begin{aligned} H_{0}(D(\gamma _{1},\ldots ,\gamma _{n}))^{'}_{b}\cong H(D(\gamma _{1},\ldots ,\gamma _{n})^{c}) \end{aligned}$$

with the duality analogous to (3.1) with no $i=0$ term. Furthermore,

$$\begin{aligned} H(D(\gamma _{1},\ldots ,\gamma _{n}))^{'}_{b}\cong H(D(\gamma _{1},\ldots ,\gamma _{n})^{c})\oplus H_{0}(\infty ), \end{aligned}$$

where $H_{0}(\infty )$ stands for the space of all germs of holomorphic functions which vanish at $\infty$ on the point $\infty$. This time for $\phi =\phi _{1}\oplus \cdots \oplus \phi _{n}\oplus \phi _{\infty }$ it holds that

$$\begin{aligned} \langle f,\phi \rangle =\sum _{i=1}^{n}\int \limits _{(\gamma _{i})_{\varepsilon }}f\cdot \phi _{i}\text {d}z+\int \limits _{|z|=R}f\cdot \phi _{\infty }\text {d}z \end{aligned}$$

for $R>0$ large enough.

Lemma 3.1

Assume that $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ and F is a symbol. Let $\phi \in H(D^{c})$. Then for every $f\in H(D)$,

$$\begin{aligned} \langle T_{F}f,\phi \rangle =\int \limits _{(\gamma )_{\varepsilon }}F(z)\cdot f(z)\cdot \phi (z)dz=\sum _{i=0}^{n}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}(z)\cdot f(z)\cdot \phi _{i}(z)dz \end{aligned}$$

for sufficiently small $\varepsilon >0$. The same formula holds for the space

$H_{0}(D(\gamma _{1},\ldots ,\gamma _{n}))$. If $X=H(D(\gamma _{1},\ldots ,\gamma _{n}))$ then

$$\begin{aligned} \langle T_{F}f,\phi \rangle =\sum _{i=1}^{n}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}(z)\cdot f(z)\cdot \phi _{i}(z)dz+\int \limits _{|z|=R}F_{\infty }(z)\cdot f(z)\cdot \phi _{\infty }(z)dz \end{aligned}$$

Proof

The proof follows easily from Cauchy’s theorem and Fubini’s theorem. We show only the first formula. We have

$$\begin{aligned} \langle T_{F}f,\phi \rangle&=\sum _{j=0}^{n}\int \limits _{(\gamma _{j})_{\delta }}(T_{F}f)(\zeta )\cdot \phi _{j}(\zeta )\text {d}\zeta \\ {}&=\sum _{j=0}^{n}\int \limits _{(\gamma _{j})_{\delta }}\left( \sum _{i=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}(z)\cdot f(z)}{z-\zeta }\text {d}z\right) \cdot \phi _{j}(\zeta )\text {d}\zeta \end{aligned}$$

for small $0<\varepsilon <\delta$. By Fubini’s theorem

$$\begin{aligned} \langle T_{F}f,\phi \rangle&=\sum _{i=0}^{n}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}(z)\cdot f(z)\cdot \left( \sum _{j=0}^{n}\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma _{j})_{\delta }}\frac{\phi _{j}(\zeta )}{z-\zeta }\text {d}\zeta \right) \text {d}z \end{aligned}$$

Now it suffices to apply Cauchy’s integral formula, Cauchy’s theorem and the fact that $\phi _{0}$ vanishes at $\infty$ to obtain

$$\begin{aligned} \langle T_{F}f,\phi \rangle =\int \limits _{(\gamma )_{\varepsilon }}F(z)\cdot f(z)\cdot \phi (z)\text {d}z. \end{aligned}$$

$\square$

4 One-sided invertibility

We show the following representation of the range of the operator $T_{F}$.

Proposition 4.1

Assume that F is non-degenerate and vanishes. Then there exists a sequence of functionals $\xi _{n}\in X^{'}$ such that

$$\begin{aligned} \mathrm{range\,}T_{F}=\bigcap _{k=1}^{\infty }\ker \xi _{k}. \end{aligned}$$

Proof

We consider first the case of the space H(D) for $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$. Assume that $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}$ is non-degenerate. That is, no $F_{i}$ vanishes identically. Furthermore, assume that there exists a sequence $(z_{k})\subset D$ such that $F(z_{k})=0, k\in \mathbb {N}$. The accumulation points of the sequence $(z_{k})$ are on $\gamma _{0}\cup \gamma _{1}\cup \cdots \cup \gamma _{n}$. The sequence $(z_{k})$ is the union of the sequences

$$\begin{aligned} (z_{k})=(z_{k}^{0})\cup (z_{k}^{1})\cup \cdots \cup (z_{k}^{n}) \end{aligned}$$

such that $(z_{k}^{i})$ is the set of zeros of $F_{i}$. Naturally, not every subsequence $(z_{k}^{i})$ must appear in this representation but, according to the assumption, at least one does. $\square$

We factorize the symbol F using Weierstrass theory

$$\begin{aligned} F_{j}(z)=\prod _{i=0}^{n}\prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{j}(z), \,j=0,1,\ldots ,n, \end{aligned}$$

(4.1)

where $F^{0}_{j}$ does not vanish (in the sense of Definition 3.3) and $m_{k}^{i}$ is multiplicity of $z_{k}^{i}$ as zero of $F_{i}$. We shall also write

$$\begin{aligned} F(z)=\prod _{k=0}^{\infty }E_{k}(z)\cdot F^{0}(z). \end{aligned}$$

(4.2)

Then close to $\gamma _{j}$

$$\begin{aligned} F(z)=F_{j}(z)=\prod _{k=0}^{\infty }E_{k}(z)F_{j}^{0}(z) \end{aligned}$$

with

$$\begin{aligned} \prod _{k=0}^{\infty }E_{k}(z)=\prod _{i=0}^{n}\prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z). \end{aligned}$$

All this requires some comment. We refer to and rely on [11, pp. 164–173]. We don’t claim that the functions $E_{k,i}$ are the Weierstrass elementary factors. We only claim that there exist functions $E_{k,i}$ such that:

• (i) the function $E_{k,0}$ is holomorphic in $I(\gamma _{0})$, $E_{k,i}$ is holomorphic in $E(\gamma _{i})\cup \{\infty \}\subset \mathbb {C}_{\infty }$ when $i=1,\ldots ,n$;

• (ii) $E_{k,i}(z_{k}^{i})=0$ and $z_{k}^{i}$ is simple and the only zero of the function $E_{k,i}$;

• (iii) the product

  $$\begin{aligned} \prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z) \end{aligned}$$

  converges in $H(I(\gamma _{0}))$ if $i=0$ and in $H((E(\gamma _{i})\cup \{\infty \}))$ if $i=1,\ldots ,n$.

These assumptions imply that the product

$$\begin{aligned} \prod _{i=0}^{n}\prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z) \end{aligned}$$

converges in $D=I(\gamma _{0})\cap E(\gamma _{1})\cap \cdots \cap E(\gamma _{n})$.

As we shall see in a moment there may be points z in D such that $F^{0}(z)=0$. Let us emphasize this: we only claim that $F^{0}$ defines a symbol which does not vanish in the sense of Definition 3.3. This is equivalent to saying that the zeros of $F^{0}$ do not accumulate on bD or that there is only a finite number of zeros of $F^{0}$ in D. Observe that (4.2) is a joint factorization, the products converge in D, however the open set of existence of F and $F^{0}$ is in general not connected, it is, as we know, the union of neighborhoods of the boundary curves $\gamma _{i}$.

Consider the Toeplitz operator $T_{F^{0}}$.

Claim

We may assume that

$$\begin{aligned} \mathrm{index\,}T_{F^{0}}=-1. \end{aligned}$$

(4.3)

Proof

To prove the claim we use the argument principle and Theorem 1.2. If

$$\begin{aligned} \nu :=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{(F^{0}_{i}(z))^{'}}{F^{0}_{i}(z)}\text {d}z>1, \end{aligned}$$

then we take instead of the function $F^{0}$ the function

$$\begin{aligned} \tilde{F}^{0}(z):=\frac{F^{0}(z)}{\prod _{k=1}^{\nu -1}E_{k,j}(z)} \end{aligned}$$

(4.4)

for some $0\le j\le n$. Such a choice is possible since the sequence $(z_{k})$ is infinite. The function

$$\begin{aligned} \prod _{k=1}^{\nu -1}E_{k,j}(z) \end{aligned}$$

has precisely $\nu -1$ zeros. Then by Theorem 1.2 and the argument principle

$$\begin{aligned} \mathrm{index\,}T_{F_{0}}&=-\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma )_{\varepsilon }}\frac{(\tilde{F}^{0})^{'}(z)}{\tilde{F}^{0}(z)}\text {d}z\\ {}&=-\frac{1}{2\pi \mathrm i}\int \limits _{(\gamma )_{\varepsilon }}\frac{(F^{0})^{'}(z)}{F^{0}(z)}\text {d}z+(\nu -1)=-1. \end{aligned}$$

Then

$$\begin{aligned} F(z)=F^{0}(z)\cdot \prod _{i=0}^{n}\prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z)=\tilde{F}^{0}(z)\cdot \prod _{k=1}^{\nu -1}E_{k,j}(z)\cdot \prod _{i=0}^{n}\prod _{k=1}^{\infty }E_{k,i}^{m_{k}^{i}}(z). \end{aligned}$$

(4.5)

If

$$\begin{aligned} \nu :=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{(F^{0}_{i}(z))^{'}}{F^{0}_{i}(z)}\text {d}z<1. \end{aligned}$$

(4.6)

then we take

$$\begin{aligned} \tilde{F}^{0}(z):=F^{0}(z)\cdot \prod _{k=1}^{\nu +1}E_{k,j}(z). \end{aligned}$$

(4.7)

Again it follows from Theorem 1.2 and the argument principle that

$$\begin{aligned} {index\,}T_{\tilde{F}^{0}}=-1 \end{aligned}$$

and we have a decomposition similar to (4.5).

Observe that the functions (4.4) and (4.7) define symbols. Indeed, there is only a finite number of poles in (4.4), so the corresponding function is holomorphic on a smaller neighborhood of bD in D. In view of the above comments both choices of the function $\tilde{F}^{0}$ define symbols which do not vanish.$\square$

Since $\text {index}\,T_{F_{0}}=-1$, there exists a functional $\xi \in H(D)^{'}$ such that

$$\begin{aligned} \text {range}\,T_{F^{0}}=\ker \xi . \end{aligned}$$

Indeed, it follows from Theorem 1.4 in view of (4.3) that the operator $T_{F^{0}}$ is injective and the range of $T_{F^{0}}$ is a closed subspace of H(D) of codimension 1.

Thus there exists

$$\begin{aligned} \varphi =\varphi _{0}\oplus \varphi _{1}\oplus \cdots \oplus \varphi _{n}\in H_{0}(\overline{E(\gamma _{0})})\oplus H(\overline{I(\gamma _{1})})\oplus \cdots \oplus H(\overline{I(\gamma _{n})}) \end{aligned}$$

such that for every $f\in H(D)$,

$$\begin{aligned} 0=\int \limits _{(\gamma )_{\varepsilon }}F^{0}(z)\cdot f(z)\cdot \varphi (z)\text {d}z=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}F^{0}_{i}(z)\cdot f(z)\cdot \varphi _{i}(z)\text {d}z. \end{aligned}$$

We used Lemma 3.1.

For every $f\in H(D)$, by Cauchy’s integral formula for $\varepsilon>\delta >0$,

$$\begin{aligned} \begin{aligned} 0&=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}^{0}(z)\cdot \left( \sum _{j=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{j})_{\delta }}\frac{f(\zeta )}{\zeta -z}\text {d}\zeta \right) \cdot \varphi _{i}(z)\text {d}z\\ {}&=\sum _{j=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{j})_{\delta }}\left( \sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}^{0}(z)\varphi _{i}(z)}{\zeta -z}\text {d}z\right) \cdot f(\zeta )\text {d}\zeta . \end{aligned} \end{aligned}$$

(4.8)

The integral

$$\begin{aligned} \sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}^{0}(z)\varphi _{i}(z)}{\zeta -z}\text {d}z \end{aligned}$$

defines a functional in $H(D)^{'}\cong H_{0}(\overline{E(\gamma _{0})})\oplus H(\overline{I(\gamma _{1})})\oplus \cdots \oplus H(\overline{I(\gamma _{n})})$. It follows from (4.8) that

$$\begin{aligned} \sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}^{0}(z)\varphi _{i}(z)}{\zeta -z}\text {d}z\equiv 0. \end{aligned}$$

(4.9)

From Cauchy’s integral formula for z close to $\gamma _{j}$,

$$\begin{aligned} F^{0}(z)\cdot \varphi (z)&=F_{j}^{0}(z)\cdot \varphi _{j}(z)=\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{j})_{\delta }}\frac{F_{j}^{0}(\zeta )\cdot \varphi _{j}(\zeta )}{\zeta -z}\text {d}\zeta -\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{j})_{\varepsilon }}\frac{F_{j}^{0}(\zeta )\cdot \varphi _{j}(\zeta )}{\zeta -z}\text {d}\zeta \\ {}&=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\delta }}\frac{F_{i}^{0}(\zeta )\cdot \varphi _{i}(\zeta )}{\zeta -z}\text {d}\zeta -\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}^{0}(\zeta )\cdot \varphi _{i}(\zeta )}{\zeta -z}\text {d}\zeta \end{aligned}$$

with $0<\delta <\varepsilon$ such that $z\in I((\gamma _{j})_{\delta }){\setminus } \overline{I((\gamma _{j})_{\varepsilon })}$ if $j=0$ or $z\in E((\gamma _{j})_{\delta }){\setminus } \overline{E(\gamma _{j})_{\varepsilon })}$ if $j=1,\ldots ,n$.

It follows from (4.9) that for every $j=0,1,\ldots ,n$,

$$\begin{aligned} F_{j}^{0}(z)\cdot \varphi _{j}(z)=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\delta }}\frac{F_{i}^{0}(\zeta )\cdot \varphi _{i}(\zeta )}{\zeta -z}\text {d}\zeta . \end{aligned}$$

That is, $F^{0}\cdot \varphi$, a priori defined only close to bD in D, extends to a holomorphic function in D. In other words, $F^{0}\cdot \varphi$ defines an element of the space H(D).

Let $1\le m_{j}$ be the multiplicity of $z_{j}$ as a zero of the function $F_{\nu }$ for some $\nu =0,1,\ldots ,n$. We consider the functionals defined by the functions

$$\begin{aligned} \psi :=\frac{\varphi _{0}}{(z-z_{j})^{l_{j}}}\oplus \frac{\varphi _{1}}{(z-z_{j})^{l_{j}}} \oplus \cdots \oplus \frac{\varphi _{\nu }}{(z-z_{j})^{l_{j}}}\oplus \cdots \oplus \frac{\varphi _{n}}{(z-z_{j})^{l_{j}}} \end{aligned}$$

(4.10)

for $1\le l_{j} \le m_{j}$. For such a function $\psi$ we have for $f\in H(D)$,

by Cauchy’s theorem since $(z-z_{j})^{l_{j}}$ divides $\prod _{k=1}^{\infty }E_{k}(z)$ and $F^{0}\cdot \varphi$ defines, as we have shown, a function in H(D). This shows that

$$\begin{aligned} \text {range}\,T_{F}\subset \bigcap _{k=1}^{\infty }\ker \xi _{k}, \end{aligned}$$

(4.11)

where the functionals $\xi _{k},k\in \mathbb {N}$ correspond to the functions of the form (4.10) for the zeros $z_{k}$ of F and the function $\varphi$ itself. We claim that the equality holds. We justify this statement now.

Assume that $\eta \in H(D)^{'}$ vanishes on the range of the operator $T_{F}$. Assume that $\eta$ is represented by a function $\phi =\phi _{0}\oplus \phi _{1}\oplus \cdots \oplus \phi _{n}$. By Lemma 3.1 we have

$$\begin{aligned} 0=\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}F_{i}(z)\cdot f(z)\cdot \phi _{i}(z)\text {d}z \end{aligned}$$

for every $f\in H(D)$. The same arguments as above give that $F\cdot \phi$ defines an element of H(D). We conclude that for some function $h\in H(D)$ it holds that

$$\begin{aligned} F_{i}(z)\cdot \phi _{i}(z)=h(z) \end{aligned}$$

(4.12)

for $i=0,1,\ldots ,n$ and z sufficiently close to $\gamma _{i}$, say for $z\in U_{i}\cap D$, $U_{i}$ an open neighborhood of $\gamma _{i}$. Observe that this implies also that no $\phi _{i}$ is identically equal to 0. Also, only a finite number of the points $z_{1}^{i},\ldots ,z_{N_{i}}^{i}$ does not belong to $U_{i}$. The rest can be divided out. Thus

$$\begin{aligned} \prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F_{i}^{0}(z)\cdot \phi _{i}(z)=h(z) \end{aligned}$$

for a different function $h\in H(D)$. In other words,

$$\begin{aligned} \phi _{i}(z)=\frac{h(z)}{\prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{i}(z)} \end{aligned}$$

(4.13)

for some numbers $N_{i} \in \mathbb {N}$ and z close to $\gamma _{i}$. Recall: the function $E_{k,i}$ is a factor which vanishes only at $z_{k}^{i}$. The zero is simple. The number $m_{k}^{i}$ is the multiplicity of $z_{k}^{i}$.

If S is defined in the following way

$$\begin{aligned} \left( \prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{0}(z)\right) \oplus \left( \prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{1}(z)\right) \oplus \cdots \oplus \left( \prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{n}(z)\right) , \end{aligned}$$

then

$$\begin{aligned} \langle T_{S}f,\phi \rangle =\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }} \frac{\prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)}{\prod _{i=0}^{n}\prod _{k=1}^{N_{i}}E_{k,i}^{m_{k}^{i}}(z)}\cdot f(z)\cdot h(z)\text {d}z=0 \end{aligned}$$

by Cauchy’s theorem since $f\cdot h$ is holomorphic in D. We conclude

$$\begin{aligned} \text {range}\,T_{S}\subset \ker \eta . \end{aligned}$$

Recall that the functional $\eta$ is defined by the function $\phi =\phi _{0}\oplus \phi _{1}\oplus \cdots \oplus \phi _{n}$ with $\phi _{i}$ satisfying (4.13).

The symbol S does not vanish (in the sense of Definition 3.3) and, as a result, the operator $T_{S}$ is a Fredholm operator. Also, it follows from Theorem 1.2 and the argument principle that

$$\begin{aligned} \text {index}\,T_{S}=-1-\sum _{i=0}^{n}\sum _{k=1}^{N_{i}}m_{k}^{i}. \end{aligned}$$

We claim that

$$\begin{aligned} \text {range}\,T_{S}=\bigcap _{k=1}^{M}\ker \xi _{k}, \end{aligned}$$

(4.14)

where the functionals $\xi _{k}$ correspond to the function $\varphi$ and the functions

$$\begin{aligned} \begin{aligned} \frac{\varphi _{0}}{(z-z_{k_{i}}^{i})^{l_{k}^{i}}}\oplus \cdots \oplus \frac{\varphi _{n}}{(z-z_{k_{i}}^{i})^{l_{k}^{i}}}, \end{aligned} \end{aligned}$$

(4.15)

where $1\le l_{k}^{i}\le m_{k}^{i}$, $1\le k_{i}\le N_{i}$ and $i=0,1,\ldots ,n$.

First of all, we notice that

$$\begin{aligned} \text {range}\,T_{S}\subset \bigcap _{k=1}^{N}\ker \xi _{k}. \end{aligned}$$

Indeed,

$$\begin{aligned} \langle T_{S}f,\xi _{k}\rangle&=\sum _{j=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{\prod _{i=0}^{n}\prod _{k=1}^{M_{i}}E_{k,i}^{m_{k}^{i}}(z)\cdot F^{0}_{j}(\zeta )}{(z-z_{\nu })^{l_{\nu }}}\cdot f(\zeta )\cdot \varphi _{j}(\zeta )\text {d}\zeta \\ {}&=\sum _{j=0}^{n}\int \limits _{(\varphi _{j})_{\varepsilon }}F_{j}^{0}(\zeta )\cdot \left( \frac{\prod _{i=0}^{n}\prod _{k=1}^{M_{i}}E_{k,i}^{m_{k}^{i}}(z)}{(z-z_{\nu })^{l_{\nu }}}\cdot f(\zeta )\right) \cdot \varphi _{j}(\zeta )\text {d}\zeta \\ {}&=\langle T_{F_{0}}g,\varphi \rangle . \end{aligned}$$

for some function $g\in H(D)$. But $\text {range}\,T_{F^{0}}=\ker \xi$ and $\xi$ is represented by $\varphi$. We infer that

$$\begin{aligned} \langle T_{S}f,\xi _{k}\rangle =0 \end{aligned}$$

for every $f\in H(D)$ and $k=1,\ldots ,N$.

The functions defined in (4.15) are linearly independent. The cardinality of the set of these functions is

$$\begin{aligned} 1+\sum _{i=0}^{n}\sum _{k=1}^{N_{i}}m_{k}^{i}. \end{aligned}$$

Hence the codimension of the space

$$\begin{aligned} \bigcap _{k=1}^{N}\ker \xi _{k} \end{aligned}$$

is equal, in view of Theorem 1.4, to the codimension of the range of the operator $T_{S}$. This proves the equality (4.14). That is, if $\eta$ vanishes on the range of $T_{F}$ then

$$\begin{aligned} \bigcap _{k=1}^{N}\ker \xi _{k}\subset \ker \eta . \end{aligned}$$

This means that if a functional $\eta \in H(D)^{'}$, represented by a function $\phi$, which vanishes on the range of the operator $T_{F}$ then it is a linear combination of the functions given in (4.15).

Since the range of the operator $T_{F}$ is closed by Theorem 1.3 it follows from the Hahn–Banach theorem that

$$\begin{aligned} \text {range}\,T_{F}=\bigcap _{k=1}^{\infty }\ker \xi _{k}. \end{aligned}$$

Indeed, assume that

$$\begin{aligned} f\in \Bigl (\bigcap _{k=1}^{\infty }\ker \xi _{k}\Bigr ) \setminus \text {range}\,T_{F}. \end{aligned}$$

Since the range of $T_{F}$ is closed there exists a functional $\xi \vert _{\text {range}\,T_{F}}\equiv 0$ such that $\xi (f)\ne 0$. This is impossible since, as we have shown, if $\xi$ vanishes on the range of $T_{F}$ then $\xi$ is a linear combination of functionals defined by functions (4.15). Hence $\xi (f)=0$, which is a contradiction.

We now provide comments concerning the case of the domain $D=D(\gamma _{1},\ldots ,\gamma _{n})$. First consider the spaces $H_{0}(D)$. If G is meromorphic in

$$\begin{aligned} E(\gamma _{1})\cap \cdots \cap E(\gamma _{n}) \end{aligned}$$

and holomorphic at $\{\infty \}$, then the argument principle can be generalize to say that the integral

$$\begin{aligned} \sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{G^{'}(\zeta )}{G(\zeta )}\text {d}\zeta \end{aligned}$$

is equal to the number of zeros of G in $E((\gamma _{1})_{\varepsilon })\cap \cdots \cap E((\gamma _{n})_{\varepsilon })$ minus the number of poles of G in the same region. This can be applied to the functions $E_{k}$ in factorization (4.1), since in this case they can be assumed to be of the form

$$\begin{aligned} E_{p_{k}}\left( \frac{z_{k}-w_{k}}{z-w_{k}}\right) \end{aligned}$$

(4.16)

for some points $w_{k}\in D^{c}$ (see [11, Theorem 5.15]). Here, the symbol $E_{p_{k}}$ stands for the standard Weierstrass elementary factor

$$\begin{aligned} E_{p_{k}}(z)=(1-z)\exp \left( z+\frac{z^{2}}{2}+\cdots +\frac{z^{p_{k}}}{p_{k}}\right) . \end{aligned}$$

(4.17)

We conclude that as in the case of the domains $D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ we can assume that $\text {index}\,T_{F^{0}}=-1$ for an appropriately defined symbol $F^{0}$, which does not vanish (i.e. the zeros of $F^{0}$ do not accumulate on bD). Also,

for $\varphi$ which represents the functional $\xi$ such that $\text {range}\,T_{F^{0}}=\ker \xi$ and $1\le l\le m_{j}$. This follows from the fact that

$$\begin{aligned} \frac{\prod _{k=1}^{N}E_{k}(z)}{(z-z_{j})^{l}} \end{aligned}$$

vanishes at $\infty$, which is a consequence of (4.16). Furthermore we know that if $\phi$ vanishes on the range of $T_{F}$ then

$$\begin{aligned} \phi (z)=\frac{h(z)}{\prod _{k=1}^{N}E_{k}(z)\cdot F^{0}(z)} \end{aligned}$$

for some function h in $H_{0}(D)$. The fact that h vanishes at $\infty$ allows us to repeat the rest of the arguments in this case.

In the case of the space H(D) for $D=D(\gamma _{1},\ldots ,\gamma _{n})$ the zeros of F may accumulate on $bD=\gamma _{1}\cup \cdots \cup \gamma _{n}$ or/and go to infinity. We can write

$$\begin{aligned} (z_{k})=(z_{k}^{0})\cup (z_{k}^{\infty }), \end{aligned}$$

where $(z_{k}^{0})$ is a sequence which accumulates on bD and $(z_{k}^{\infty })$ goes to $\infty$.

The functions $E_{k}$ in (4.1) are either of the form (4.16), when they correspond to a zero in $(z_{k}^{0})$ or just the Weierstrass elementary factors, if they correspond to points in $(z_{k}^{\infty })$. The arguments in this case can be repeated. One needs however to remember that according to Theorem 1.2,

$$\begin{aligned} \text {index}\,T_{F}=-\sum _{i=1}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}\frac{F_{i}^{'}(\zeta )}{F_{i}(\zeta )}\text {d}\zeta -\frac{1}{2\pi \textrm{i}}\int \limits _{|\zeta |=R}\frac{F_{\infty }^{'}(\zeta )}{F_{\infty }(\zeta )}\text {d}\zeta . \end{aligned}$$

This completes the proof. We define now a map

$$\begin{aligned} \Xi :H(D)\rightarrow \omega , \end{aligned}$$

where $\omega$ stands for the Fréchet space of all sequences, by the formula

$$\begin{aligned} \Xi (f)=(\xi _{k}(f))_{k\in \mathbb {N}}. \end{aligned}$$

The functionals $\xi _{k},k\in \mathbb {N}$ were constructed in Proposition 4.1.

Theorem 4.1

Assume that F is non-degenerate and vanishes. Then the operator $\Xi$ is a surjection.

Proof

To prove the theorem we use Eidelheit’s theorem [20, Theorem 26.27] and Weierstrass factorization theory as presented in [11, pp. 164–173].

Assume that $D=D(\gamma _{0};\gamma _{1},\ldots ,\gamma _{n})$ for some $C^{\infty }$ smooth Jordan curves $\gamma _{0},\gamma _{1},\ldots ,\gamma _{n}$ as in Definition 3.1.

Choose $\alpha \in D$ and $r>0$ such that $z_{k}\notin B(\alpha ,r)$ for $k\in \mathbb {N}$. Let $T(z)=(z-\alpha )^{-1}$. To factorize the symbol we consider the domain $G:=T(D)\subset \mathbb {C}_{\infty }$, for which it holds that

$$\begin{aligned} \{z:|z|>R\}\subset G \end{aligned}$$

for some $R>0$. Denote $t_{k}:=T(z_{k})$. We have $|t_{k}|\le R$ for $k\in \mathbb {N}$. Furthermore, we assume that

$$\begin{aligned} \text {dist}\,(t_{1},\mathbb {C}\setminus G)\ge \text {dist}\,(t_{2},\mathbb {C}\setminus G)\ge \cdots \end{aligned}$$

For $\varepsilon >0$ denote

$$\begin{aligned} L_{\varepsilon }:=\{z\in G :\text {dist}\,(z,\mathbb {C}\setminus G)\ge \varepsilon \}. \end{aligned}$$

Choose a monotone sequence $\varepsilon _{k},k\in \mathbb {N}$, $\varepsilon _{k}\rightarrow 0$ such that no point $t_{k}$ lies on the level set

$$\begin{aligned} \{t \in G :\text {dist}\,(t,\mathbb {C}\setminus G)=\varepsilon _{j}\} \end{aligned}$$

for $j\in \mathbb {N}$. The sets $L_{\varepsilon _{k}}$ are compact subsets of $G\subset \mathbb {C}_{\infty }$. Let $K_{j}:=T^{-1}(L_{\varepsilon _{j}})$. Then $K_{j}$ are compact subsets of the domain D and the seminorms

$$\begin{aligned} p_{j}(f):=\sup _{z\in K_{j}}|f(z)| \end{aligned}$$

form a fundamental system of seminorms in H(D) (note that $0\notin G$, since $\infty \notin D$). Let

be a fundamental system of zero neighborhoods in H(D). Note that the functionals $\xi _{k}, k\in \mathbb {N}$ are linearly independent. This follows from the construction, since the functions $\varphi ,\frac{\varphi }{z-z_{1}},\ldots$ are linearly independent. In order to use Eidelheit’s theorem (Theorem 2.1), we need therefore to show that

$$\begin{aligned} \dim \bigl ((H(D)^{'})_{U_{j}^{\circ }}\cap \text {span}\, \{\xi _{l}:l\in \mathbb {N}\}\bigr )<\infty \end{aligned}$$

for every $j\in \mathbb {N}$ – the symbol $U_{j}^{\circ }$ stands for the polar of $U_{j}$.

Fix $j\in \mathbb {N}$ and assume that

$$\begin{aligned} z_{1},\ldots ,z_{N}\in \text {int}\, K_{j} \end{aligned}$$

and $z_{N+1},\ldots \notin K_{j}$. Note that this implies that

$$\begin{aligned} \text {dist}\,(t_{N+1},\mathbb {C}\setminus G)<\text {dist}\,(t_{N},\mathbb {C}\setminus G). \end{aligned}$$

As before $m_{j}$ is multiplicity of $z_{j}$ as zero of F. Let also $\xi _{j}$ for $1\le j\le M$ be those of the functionals $\xi _{k},k\in \mathbb {N}$ which correspond to the points $z_{1},\ldots ,z_{N}$. That is every $\xi _{j}, 1\le j\le M$ is represented by some function

$$\begin{aligned} \frac{\varphi }{(z-z_{l})^{m}}=\frac{\varphi _{0}}{(z-z_{l})^{m}}\oplus \cdots \oplus \frac{\varphi _{n}}{(z-z_{l})^{m}} \end{aligned}$$

for $1\le l\le N$ and $1\le m\le m_{l}$.

To prove the theorem it suffices to show that if $\xi =\sum \alpha _{l}\xi _{l}$ satisfies

$$\begin{aligned} |\xi (f)|\le Cp_{j}(f) \end{aligned}$$

for every $f\in H(D)$, then $\alpha _{M+1}=\alpha _{M+2}=\cdots =0$.

We consider the factorization of the symbol

$$\begin{aligned} F(z)=\prod _{k=1}^{\infty } E_{p_{k}}^{m_{k}}\left( \frac{t_{k}-w_{k}}{T(z)-w_{k}}\right) \cdot F^{0}(z), \end{aligned}$$

(4.18)

where $w_{n}\in \mathbb {C}\setminus G$ is chosen in such a way that

$$\begin{aligned} |w_{n}-t_{n}|=\text {dist}\,(t_{n},\mathbb {C}\setminus G). \end{aligned}$$

Naturally, $E_{p_{k}}$ stands for the Weierstrass elementary factor

$$\begin{aligned} E_{p_{k}}(z)=(1-z)\exp \left( z+\frac{z}{2}+\cdots +\frac{z^{p_{k}}}{p_{k}}\right) \end{aligned}$$

with the natural numbers $p_{k}$ chosen in such a way to guarantee that (4.18) converges.

This notation requires an explanation, since as we know $F=F_{0}\oplus F_{1}\oplus \cdots \oplus F_{n}$. Formula (4.18) means that

$$\begin{aligned} F_{j}(z)=\prod _{k=1}^{\infty } E_{p_{k}}^{m_{k}}\left( \frac{t_{k}-w_{k}}{T(z)-w_{k}}\right) \cdot F^{0}_{j}(z) \end{aligned}$$

for z close to $\gamma _{j}$.

Observe that although T has a singularity at $\alpha \in D$, the factorization (4.18) is well-defined in D. This is justified in [11, p. 171].

Consider the functions

$$\begin{aligned} g[f](z):=\prod _{k\ne N+1}E_{p_{k}}^{m_{k}}\left( \frac{t_{k}-w_{k}}{T(z)-w_{k}}\right) \cdot f(z), \end{aligned}$$

where f at this moment is only assumed to belong to H(D).

Then for every $f\in H(D)$,

$$\begin{aligned} \langle T_{F^{0}}(g[f]),\frac{\varphi }{(z-z_{l})^{m}}\rangle =\sum _{i=0}^{n}\frac{1}{2\pi \textrm{i}}\int \limits _{(\gamma _{i})_{\varepsilon }}F^{0}_{i}(z)\cdot g[f](z)\cdot \frac{h(z)}{F^{0}_{i}(z)}\frac{\text {d}z}{(z-z_{l})^{m}}=0 \end{aligned}$$

by Cauchy’s theorem if $l\ne N+1$ and $1\le m\le m_{l}$. We used representation (4.13) and for simplicity, without loss of generality, assumed that

$$\begin{aligned} \varphi (z)=\frac{h(z)}{F^{0}(z)} \end{aligned}$$

for z close to bD. For the general representation (4.13) the argument given below works under the assumption that N is large enough.

Consider the function

$$\begin{aligned} f_{q,l}(z):=(z-z_{N+1})^{l}\frac{E_{q}\left( \frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\right) }{z-z_{N+1}} \end{aligned}$$

where $0\le l\le m_{N+1}-1$ and q to be chosen later. Note that $f_{q,l}$ belongs to H(D). Then, as we have just observed,

if $l\ne N+1$ and $1\le m\le m_{l}$. Furthermore,

by Cauchy’s integral formula. We also have

$$\begin{aligned} \lim _{z\rightarrow z_{N+1}}&\frac{E_{q}\left( \frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\right) }{z-z_{N+1}}=-\frac{t_{N+1}-w_{N+1}}{(T(z_{N+1})-w_{N+1})^{2}}\cdot T^{'}(z_{N+1})\cdot E_{q}^{'}(1)\\ {}&=-\frac{t_{N+1}-w_{N+1}}{(T(z_{N+1})-w_{N+1})^{2}}\cdot T^{'}(z_{N+1})\cdot \exp \left( 1+\frac{1}{2}+\cdots +\frac{1}{q}\right) \end{aligned}$$

and

for $1\le l<m_{N+1}$. Thus

$$\begin{aligned} \begin{aligned} \sum \alpha _{i}&\xi _{i}\left( T_{F^{0}}(g[f_{m,l}])\right) \\ {}&=\alpha _{\nu }\prod _{k\ne N+1}E_{p_{k}}^{m_{k}}\left( \frac{t_{k}-w_{k}}{T(z_{N+1})-w_{k}}\right) \cdot h(z_{N+1})\\ {}&\cdot \frac{t_{N+1}-w_{N+1}}{(T(z_{N+1})-w_{N+1})^{2}}\cdot T^{'}(z_{N+1})\cdot \exp \left( 1+\frac{1}{2}+\cdots +\frac{1}{q}\right) , \end{aligned} \end{aligned}$$

(4.19)

where $\xi _{\nu }$ corresponds to the functional defined by

$$\begin{aligned} \frac{\varphi }{(z-z_{N+1})^{m_{N+1}}}. \end{aligned}$$

Observe that as $q\rightarrow \infty$ this expression tends to $\infty$ if $h(z_{N+1})\ne 0$. We know that no $\varphi _{i}$ is identically equal to 0—this follows from (4.12). It follows therefore from (4.13) that $h(z_{N})$ can be equal to 0 only for a finite number of indices N.

Let now $\Gamma =\Gamma _{0}+\cdots +\Gamma _{n}$ by a cycle in D homologous to 0 in D such that

$$\begin{aligned}&K_{j}\subset I(\Gamma _{0})\cup E(\Gamma _{1})\cup \cdots \cup E(\Gamma _{n}),\\&z_{N+1},\ldots \in E(\Gamma _{0})\cup I(\Gamma _{1})\cup \cdots \cup I(\Gamma _{n}). \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} p_{j}\Bigl (T_{F^{0}}(g[f_{q,m_{N+1}-1}])\Bigr )&\le \frac{1}{\delta }\sup _{z\in \Gamma }\Bigl |F^{0}(z)\cdot (z-z_{N+1})^{m_{N+1}-1}\cdot \\ {}&\cdot \prod _{k\ne N+1}E_{p_{k}}^{m_{k}}\Bigl (\frac{t_{k}-w_{k}}{T(z)-w_{k}}\Bigr )\Bigr |\cdot \sup _{z\in \Gamma }\Bigl |\frac{E_{q}\Bigl (\frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\Bigr )}{z-z_{N+1}}\Bigr | \end{aligned} \end{aligned}$$

(4.20)

for some $\delta >0$. We shall use this estimate in a moment.

Let now $\beta _{0},\beta _{1},\ldots ,\beta _{n}$ be the boundary curves of T(G). Consider a dilatation of this cycle $(\beta )_{\varepsilon }$ defined in the following way

$$\begin{aligned} (\beta _{j})_{\varepsilon }=\{\zeta \in T(G) :\text {dist}\,(\zeta ,\beta _{j})=\varepsilon \}. \end{aligned}$$

This is, up to some constant, equivalent to our previous definition of the dilatations. We use this one, since we need to have precise control on the distance to bG. One easily checks that if $\varepsilon >0$ is small $(\beta _{j})_{\varepsilon }$ are $C^{\infty }$ smooth Jordan curves.

Choose now $\varepsilon >0$ in such a way that

$$\begin{aligned}&T(K_{j})\subset E((\beta _{0})_{\varepsilon })\cap E((\beta _{1})_{\varepsilon })\cap \cdots \cap E((\beta _{n})_{\varepsilon })\\&t_{N+1},\ldots \in I((\beta _{0})_{\varepsilon })\cup \cdots \cup I((\beta _{n})_{\varepsilon }). \end{aligned}$$

That is,

$$\begin{aligned} |t_{N+1}-w_{N+1}|=\text {dist}\,(t_{N+1},\mathbb {C}\setminus G)<\varepsilon . \end{aligned}$$

Please note that such a choice is possible by the definition of the compact sets $K_{j}$ and the definition of the number N.

For $\zeta \in (\beta )_{\varepsilon }$ we have

$$\begin{aligned} \Bigl |\frac{t_{N+1}-w_{N+1}}{\zeta -w_{N+1}}\Bigr |&\le |t_{N+1}-w_{N+1}|\text {dist}\,(w_{N+1},(\beta )_{\varepsilon })^{-1}\\ {}&\le |t_{N+1}-w_{N+1}|\text {dist}\,(\mathbb {C}\setminus G,(\beta )_{\varepsilon })^{-1}\\ {}&=\frac{\text {dist}\,(t_{N+1},\mathbb {C}\setminus G)}{\text {dist}\,(\mathbb {C}\setminus G,(\beta )_{\varepsilon })}<1. \end{aligned}$$

Define now $\Gamma$ as $T^{-1}((\beta )_{\varepsilon })$. With this choice of the curves we have

$$\begin{aligned} \sup _{z\in \Gamma } \Bigl |\frac{E_{q}\left( \frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\right) }{z-z_{N+1}}\Bigr |&\le c \left( 1+\sup _{z\in \Gamma }\Bigl |E_{q}\left( \frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\right) -1\Bigr |\right) \\ {}&\le c\left( 1+\sup _{z\in \Gamma }\Bigl |\frac{t_{N+1}-w_{N+1}}{T(z)-w_{N+1}}\Bigr |^{q+1}\right) \le c \end{aligned}$$

by Lemma 5.11 in [11].

In view of (4.20) we conclude that

$$\begin{aligned} p_{j}(T_{F^{0}}(g[f_{q,m_{N+1}-1}]))\le c \end{aligned}$$

uniformly for $q\in \mathbb {N}$.

It follows from (4.19) that the coefficient $\alpha _{\nu }$ which corresponds to the functional related to the function $\frac{\varphi }{(z-z_{N+1})^{m_{N+1}}}$ is equal to 0. In a similar way we show that the other coefficients which correspond to the zeros $z_{N+1},\ldots$ are equal to 0. We can use Eidelheit’s theorem to conclude that $\Xi$ is surjective.

Consider now the space $H_{0}(D)$ for $D=D(\gamma _{1},\ldots ,\gamma _{n})$. The proof is similar in this case. We just highlight the differences. First of all, there is no need to use the transformation T. That is,

$$\begin{aligned} g[f](z):=\prod _{k\ne N+1}E_{p_{k}}^{m_{k}}\left( \frac{z_{k}-w_{k}}{z-w_{k}}\right) \cdot f(z) \end{aligned}$$

with appropriately chosen points $w_{k}$ and $f\in H_{0}(D)$. Observe that the function g[f] belongs to $H_{0}(D)$, since

$$\begin{aligned} \lim _{z\rightarrow \infty }\prod _{k\ne N+1}E_{p_{k}}^{m_{k}}\Bigl (\frac{z_{k}-w_{k}}{z-w_{k}}\Bigr )=1. \end{aligned}$$

This is shown in [11, p 172]. We need to force the functions $f_{q,l}$ to belong to $H_{0}(D)$, that is, to vanish at $\infty$. This is done be taking

$$\begin{aligned} f_{q,l}(z):=(z-z_{N+1})^{l}\frac{E_{q}\Bigl (\frac{z_{N+1}-w_{N+1}}{z-w_{N+1}}\Bigr )}{z-z_{N+1}}\cdot \frac{1}{(z-\alpha )^{l}} \end{aligned}$$

for $\alpha \notin D$. The rest of the proof can be repeated.

The proof in the case of the space H(D) for $D=D(\gamma _{1},\ldots ,\gamma _{n})$ follows the first case since in $\mathbb {C}_{\infty }$ we have $bD=\{\infty \}\cup \gamma _{1}\cup \cdots \cup \gamma _{n}$. $\square$

Proof of Main Theorem and Corollary 1.2

Let us first prove that if F is non-degenerate and vanishes then $T_{F}$ is not left invertible. Since the map $\Xi :H(D)\rightarrow \omega$ is a surjection, the kernel $\ker \Xi$ is not complemented. Indeed, otherwise H(D) would contain a closed subspace without a continuous norm, which is impossible since H(D) has such a norm. Thus, $R(T_{F})=\ker \omega$ is not complemented and, as a result, $T_{F}$ is not left invertible.

We sketch the argument that if $T_{F}$ is an injective Fredholm operator, then $T_{F}$ is left invertible. It follows from the assumption that $T_{F}$ is an isomorphism from X onto $Y=T_{F}(X)$ and Y is a finite codimensional subspace of X. Thus there exists a continuous projection P from X onto Y. A left inverse is defined by $T_{F}^{-1}P$. The details can be found in [18, Theorem 1, p. 37].

Similarly, it is easy to show that if $T_{F}$ is a surjective Fredholm operator, then $T_{F}$ is right invertible. To prove this one follows the proof of [18, Theorem 2, p. 48]. $\square$