# On Markushevich bases {$x^\lambda-n$}_n = 1^\infty$ for their closed span in weighted $L^2(A)$ spaces over sets $A \subset [0, \infty)$ of positive Lebesgue measure, Hereditary completeness, and Moment problems

Elias Zikkos

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Elias Zikkos. On Markushevich bases {$x^\lambda-n$}_n = 1^\infty$ for their closed span in weighted $L^2(A)$ spaces over sets $A \subset [0, \infty)$ of positive Lebesgue measure, Hereditary completeness, and Moment problems. 2025. hal-05238739

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# On Markushevich bases {$x^{\lambda_n}$}$_{n=1}^\infty$ for their closed span in weighted $L^2(A)$ spaces over sets $A \subset [0, \infty)$ of positive Lebesgue measure, Hereditary completeness, and Moment problems

Elias Zikkos

Department of Mathematics, Khalifa University of Science and Technology,
Abu Dhabi, United Arab Emirates

email address: elias.zikkos@ku.ac.ae and eliaszikkos@yahoo.com

September 3, 2025

## Abstract

Inspired by the work of Borwein and Erdelyi [11] on generalizations of Müntz's theorem, we investigate the properties of the system {$x^{\lambda_n}$}$_{n=1}^\infty$ in weighted $L^p(A)$ spaces, for $p \ge 1$, denoted by $L_w^p(A)$, where

(I) A is a measurable subset of the real half-line $[0, \infty)$ having positive Lebesgue measure,

(II) w is a non-negative integrable function defined on A, and

(III) {$\lambda_n$}$_{n=1}^\infty$ is a strictly increasing sequence of positive real numbers such that $\inf\{\lambda_{n+1} - \lambda_n\} > 0$ and $\sum_{n=1}^\infty \lambda_n^{-1} < \infty$.

Based on the **Remez – type inequality** ([11, Theorem 5.6]), we find a sharp lower bound for the distance between the function $x^{\lambda_n}$ and the closed span of the system {$x^{\lambda_k}$}$_{k \ne n}$ in the $L_w^p(A)$ spaces. Then intrigued by the “Clarkson-Erdős-Schwartz Phenomenon” regarding the closed span of an incomplete system {$x^{\lambda_n}$}$_{n=1}^\infty$ in the $L_w^p(A)$ spaces ([11, Theorem 6.4]), we prove that a function $f$ in $\overline{\operatorname{span}}\{x^{\lambda_n}\}^\infty_{n=1}$ in the Hilbert space $L_w^2(A)$, admits the **Fourier – type series** representation $f(x) = \sum_{n=1}^\infty \langle f, r_n \rangle_{w,A} x^{\lambda_n}$ a.e on A, where {$r_n$}$_{n=1}^\infty$ is the unique biorthogonal family of {$x^{\lambda_n}$}$_{n=1}^\infty$ in $\overline{\operatorname{span}}\{x^{\lambda_n}\}^\infty_{n=1}$ in $L_w^2(A)$. As a result, we show that the system {$x^{\lambda_n}$}$_{n=1}^\infty$ is a **Markushevich basis** for $\overline{\operatorname{span}}\{x^{\lambda_n}\}^\infty_{n=1}$ in $L_w^2(A)$. Furthermore, we consider a **moment problem**: assuming certain growth conditions on a sequence {$d_n$}$_{n=1}^\infty$ of real numbers, we find a function $f \in \overline{\operatorname{span}}\{x^{\lambda_n}\}^\infty_{n=1}$ in $L_w^2(A)$, serving as a solution to

$$ \int_A f(x) \cdot x^{\lambda_n} \cdot w(x) dx = d_n, \quad n = 1, 2, \dots $$

Finally, if $m \le w(x) \le M$ on A for some positive numbers m and M and the set A contains an interval $[a, r_A]$, where $a \ge 0$ and $r_A$ is the essential supremum of A, we prove that the system {$x^{\lambda_n}$}$_{n=1}^\infty$ is hereditarily complete in $\overline{\operatorname{span}}\{x^{\lambda_n}\}^\infty_{n=1}$ in the space $L_w^2(A)$. As a result, a general class of compact operators on the closure is constructed that admit spectral synthesis.

Keywords: Müntz-Szász theorem, Distances, Closed Span, Biorthogonal Families, Markushevich bases, Hereditary completeness, Moment Problems.

AMS 2010 Mathematics Subject Classification. 30B60, 30B50, 46E15, 46E20, 46B20, 41A30, 34A35.

## 1 Introduction and the Main Results

This work has been motivated from the Borwein-Erdelyi results [11] on generalizations of the classical Müntz-Szász theorem which in turn answered a question posed by S. N. Bernstein regarding the completeness of systems {$x^{\lambda_n}$}$_{n=1}^\infty$ in the spaces $C_0[0, 1]$ and $L^p(0, 1)$ for $p \ge 1$, when $\Lambda = \{\lambda_n\}^\infty_{n=1}$ is a strictly increasing sequence of positive real numbers diverging to infinity. By the Müntz-Szász theorem the span of the system {$x^{\lambda_n}$}$_{n=1}^\infty$ is dense in the spaces $C_0[0, 1]$ and $L^p(0, 1)$ for $p \ge 1$, if and only $\sum_{n=1}^\infty \lambda_n^{-1} = \infty$.
Contributions to the Müntz-Szász problem were given later on by J. A. Clarkson, P. Erdős, L. Schwartz, W. A. J. Luxemburg, J. Korevaar, P. Borwein, T. Erdelyi, W. B. Johnson (see [15, 25, 10, 11]) where the interval $[0, 1]$ is replaced by an interval $[a, b]$ away from the origin for $a > 0$ or even by a compact set $K \subset [0, \infty)$ of positive Lebesgue measure. The reader may consult relevant survey articles and books such as [28, 4, 10, 21]. There is still an ongoing research on Müntz-Szász type problems (see [1, 2, 13, 27, 3, 24, 19, 20, 18, 22, 29, 33, 34, 35]).

When $\sum_{n=1}^{\infty} \lambda_n^{-1} < \infty$ the closed span of the system $\{x^{\lambda_n}\}_{n=1}^{\infty}$ is a proper subspace of $L^p(0, 1)$. In this case we have the “Clarkson-Erdős-Schwartz Phenomenon”: any function $f \in \text{span}\{x^{\lambda_n}\}_{n=1}^{\infty}$ in $L^p(0, 1)$, is extended to an analytic function throughout the interior of the slit disk $\mathbb{D}^* := \{z : |z| < 1\} \setminus (-1, 0)$ admitting a series representation of the form $f(z) = \sum_{n=0}^{\infty} a_n z^{\lambda_n}$ which converges uniformly on compacta (see [15], [21, Corollary 6.2.4] and [25, Theorem 8.2]). A converse result appears in [23], [21, Corollary 6.2.4] and [25, Theorem 8.2]. We also note that in [16, 33, 35] one finds results on the closed span in $L^2(-\infty, 0)$ and in $L^2(a, b)$ of exponential systems of the form

$$ \{x^k e^{\lambda_n x} : n \in \mathbb{N}, k = 0, 1, 2, \dots, \mu_n - 1\}. $$

A beautiful generalization of the “Clarkson-Erdős-Schwartz Phenomenon” was given by Borwein and Erdelyi in [11, Theorems 6.1 and 6.4] where the interval $(0, 1)$ is replaced by a measurable subset of the real half-line $[0, \infty)$ having positive Lebesgue measure. The crucial tool was a Remez-type inequality [11, Theorems 5.1 and 5.6] which is recalled below with several other results form [11] and right after we state ours. But first we need to introduce some notation and definitions following the work of [11].

## 1.1 Notations and definitions

Let $A \subset [0, \infty)$ be a measurable subset of the real half-line $[0, \infty)$ having positive Lebesgue measure and let $r_A$ be the essential supremum of $A$, that is

$$ r_A := \sup\{x \in [0, \infty) : m(A \cap [x, \infty) > 0\}, \quad \text{where } m \text{ is the Lebesgue measure.} \quad (1.1) $$

Let $w$ be a real-valued non-negative integrable function defined on $A$, with $\int_A w(x) dx < \infty$, and such that

$$ m(\{x \in A : w(x) = 0\}) = 0. \quad (1.2) $$

Let

$$ r_w := \sup \left\{ x \in [0, \infty) : \int_{A \cap (x, \infty)} w(t) dt > 0 \right\}. \quad (1.3) $$

**Remark 1.1.** Clearly, it always holds that $r_w \le r_A$. Also, it is easy to see that if there is a positive number $m$ so that $m \le w(x)$ for all $x \in A$, then $r_w = r_A$.

We also consider the slit disk

$$ D_{r_w} := \{z \in \mathbb{C} \setminus (-\infty, 0] : |z| < r_w\}. \quad (1.4) $$

For $p \ge 1$, we denote by $L_w^p(A)$ the space of **real-valued** measurable functions defined on $A$ such that $\int_A |f(x)|^p w(x) dx < \infty$, equipped with the norm

$$ \|f\|_{L_w^p(A)} := \left( \int_A |f(x)|^p w(x) dx \right)^{\frac{1}{p}}. $$

The space $L_w^2(A)$ is a real Hilbert space once endowed with the inner product

$$ \langle f, g \rangle_{w,A} := \int_A f(x)g(x) \cdot w(x) dx. $$
If $w \equiv 1$, then $L^p(A)$ is the space of functions such that $\int_A |f(x)|^p dx < \infty$ and we let

$$ ||f||_{L^p(A)} := \left( \int_A |f(x)|^p dx \right)^{\frac{1}{p}} \quad \text{and} \quad \langle f, g \rangle_A := \int_A f(x)g(x) dx. $$

Given a weight $w$, if $A$ is an interval $[a, b]$ we use the notations

$$ ||f||_{L^p_w([a,b])} := \left( \int_a^b |f(x)|^p \cdot w(x) dx \right)^{\frac{1}{p}} \quad \text{and} \quad \langle f, g \rangle_{w, [a,b]} := \int_a^b f(x)g(x) \cdot w(x) dx, $$

and in addition, if $w \equiv 1$, we use the notations

$$ ||f||_{L^p([a,b])} := \left( \int_a^b |f(x)|^p dx \right)^{\frac{1}{p}} \quad \text{and} \quad \langle f, g \rangle_{[a,b]} := \int_a^b f(x)g(x) dx. $$

**Remark 1.2.** We also use the notations $||f||_A := \sup_{x \in A} |f(x)|$ and $||f||_{[a,b]} := \sup_{x \in [a,b]} |f(x)|$.

Throughout this article, in our results $\Lambda := \{\lambda_n\}_{n=1}^\infty$ is a strictly increasing sequence of positive real numbers so that

$$ \sum_{n=1}^{\infty} \frac{1}{\lambda_n} < \infty \qquad (1.5) $$

satisfying the gap condition

$$ \inf_{n \in \mathbb{N}} (\lambda_{n+1} - \lambda_n) > 0. \qquad (1.6) $$

We associate to $\Lambda$ the system

$$ M_{\Lambda} := \{e_n : n \in \mathbb{N}\} \quad \text{where} \quad e_n(x) := x^{\lambda_n}. \qquad (1.7) $$

**Remark 1.3.**

(1) We denote by $\text{span}(M_\Lambda)$ the set of all finite linear combinations of elements from $M_\Lambda$ with real coefficients.

(2) We say that a function $f: A \to \mathbb{R}$ belongs to the closed span of the system $M_\Lambda$ in $L^p_w(A)$, if for every $\epsilon > 0$ there is some $g_\epsilon \in \text{span}(M_\Lambda)$ so that $||f - g_\epsilon||_{L^p_w(A)} < \epsilon$.

## 1.2 Borwein-Erdelyi results

We state below several important results from [11] on Remez-type inequalities, extensions of the “Clarkson-Erdős-Schwartz Phenomenon”, and Müntz-Szász type theorems.

### 1.2.1 Remez – type inequalities

Inequality (1.8) is considered by Borwein and Erdelyi to be the central result of their work while inequality (1.9) is the $L^p$ version.

**Theorem A.** [11, Theorems 5.1 and 5.6] Suppose a sequence $\Lambda$ satisfies the condition (1.5). Let $s > 0$ and $p \in (0, \infty)$. Then the following are true.

(I) There exists a constant $c$ depending only on $\Lambda$ and $s$, (and not on $\rho$, $A$, or the “length” of $f$) so that

$$ ||f||_{[0,\rho]} \le c \cdot ||f||_A, \qquad (1.8) $$

for every $f \in \text{span}(M_\Lambda)$ and for every set $A \subset [\rho, 1]$ of Lebesgue measure at least $s$.

(II) There exists a constant $c$ depending only on $\Lambda$, $s$ and $p$, (and not on $\rho$, $A$, or the “length” of $f$) so that

$$ ||f||_{[0,\rho]} \le c \cdot ||f||_{L^p(A)}, \qquad (1.9) $$

for every $f \in \text{span}(M_\Lambda)$ and for every set $A \subset [\rho, 1]$ of Lebesgue measure at least $s$.
### 1.2.2 On the “Clarkson-Erdős-Schwartz Phenomenon”

Based on Theorem A, Borwein and Erdelyi obtained the following result for the closed span of an incomplete system $M_Λ$ in the $L_w^p(A)$ spaces. Apart from condition (1.5), the gap condition (1.6) is also imposed this time.

**Theorem B.** [11, Theorem 6.4]. Let A be a measurable subset of the real half-line $[0, \infty)$ of positive Lebesgue measure and let w be a non-negative integrable function defined on A, with $r_w$ as in (1.3). Let $Λ = \{λ_n\}_{n=1}^∞$ be a sequence of distinct positive real numbers that satisfies (1.5) – (1.6). Then the span of $\{x^{λ_n}\}_{n=1}^∞$ is not dense in $L_w^p(A)$, and every function $f \in L_w^p(A)$ belonging to the $L_w^p(A)$ closure of $\text{span}(M_Λ)$ extends analytically in the slit disk $D_{r_w}$ (1.4) and can be represented as

$$f(x) = \sum_{n=1}^{\infty} a_n x^{\lambda_n}, \quad x \in A \cap [0, r_w). \qquad (1.10)$$

### 1.2.3 Müntz-Szász results on $L_w^p(A)$

From Theorem B, Borwein and Erdelyi extended the Müntz-Szász theorem as follows.

**Theorem C.** [11, Theorem 6.5]. Let A be a measurable subset of the real half-line $[0, \infty)$ of positive Lebesgue measure. Let $Λ = \{λ_n\}_{n=1}^∞$ be a strictly increasing sequence of positive real numbers, diverging to infinity. Let w be a non-negative integrable function defined on A. Then span of $\{x^{λ_n}\}_{n=1}^∞$ is dense in $L_w^p(A)$ if and only if $\sum_{n=1}^∞ λ_n^{-1} = \infty$.

## 1.3 Our Results

Our first result, Theorem 1.1, is obtained from the Remez-type inequality (1.9) and we will refer to it as the *Distance result*. It plays a crucial role in the derivation of our other theorems.

Consider a sequence $Λ = \{λ_n\}_{n=1}^∞$ that satisfies conditions (1.5) – (1.6) and from the system $M_Λ$ (1.7) exclude an element $e_n(x) = x^{λ_n}$. The resulting system is denoted by $M_{Λ_n}$, that is

$$M_{Λ_n} := M_Λ \setminus \{e_n\}. \qquad (1.11)$$

Let $D_{A,w,p,n}$ stand for the **distance** between $e_n$ and the closed span of $M_{Λ_n}$ in $L_w^p(A)$, that is

$$D_{A,w,p,n} := \inf_{g \in \text{span}(M_{Λ_n})} ||e_n - g||_{L_w^p(A)}.$$

We will derive the sharp lower bound (1.12) for $D_{A,w,p,n}$.

**Theorem 1.1.** Let A be a measurable subset of the real half-line $[0, \infty)$ of positive Lebesgue measure and let w be a real-valued non-negative integrable function defined on A, with $r_w$ as in (1.3). Let $Λ = \{λ_n\}_{n=1}^∞$ be a sequence of distinct positive real numbers that satisfies (1.5) – (1.6). Then, for every $ϵ > 0$ there is a constant $u_ϵ > 0$, independent of $p \ge 1$ and $n \in N$, but depending on $Λ$, A and w, so that

$$D_{A,w,p,n} \ge u_ϵ(r_w - ϵ)^{λ_n}. \qquad (1.12)$$

We then revisit Theorem B and give some more details regarding the coefficients $a_n$ appearing in (1.10). If $P_j(x) = \sum_{n=1}^{p(j)} a_{j,n}x^{λ_n}$, for $j = 1, 2, \dots$, are the functions in the span of $M_Λ$ such that $||f - P_j||_{L_{w,A}^p} \to 0$ as $j \to \infty$, we show that for each fixed $n \in N$, the $a_{j,n}$ coefficients of the $P_j$'s tend to the $a_n$ coefficient of (1.10) as $j \to \infty$, see (1.14). In turn, this result is essential for deriving Lemma 3.1.

**Lemma 1.1.** Consider all the assumptions of Theorem B. Let $f \in L_w^p(A)$ belong to the $L_w^p(A)$ closure of $\text{span}(M_Λ)$, thus there exists a sequence $\{P_j\}_{j=1}^∞$ in $\text{span}(M_Λ)$ where $P_j(x) = \sum_{n=1}^{p(j)} a_{j,n}x^{λ_n}$, such
that $||f - P_j||_{L^p_{w,A}} \to 0$ as $j \to \infty$. Then $f$ extends analytically in the slit disk $D_{r_w}$ (1.4) and can be represented as

$$f(z) = \sum_{n=1}^{\infty} a_n z^{\lambda_n}, \quad z \in D_{r_w}, \qquad (1.13)$$

with the series converging uniformly on compact subsets of $D_{r_w}$ and such that

$$a_n = \lim_{j \to \infty} a_{j,n} \quad \text{for } n = 1, 2, \dots \qquad (1.14)$$

We will show below that more is true regarding these $a_n$ coefficients in the case of the Hilbert space $L^2_w(A)$. First we note that by (1.12) the distances $D_{A,w,2,n}$ are positive thus $M_\Lambda$ is what we call a minimal system in its closed span in $L^2_w(A)$. It is known that a family $\{f_n\}_{n \in \mathbb{N}}$ of functions in a separable Hilbert space $\mathcal{H}$ endowed with an inner product $\langle \cdot, \cdot \rangle_{\mathcal{H}}$ is minimal, if and only if it has a biorthogonal sequence in $\mathcal{H}$ (see [14, Lemma 3.3.1]): in other words, there exists a sequence $\{g_n\}_{n \in \mathbb{N}} \subset \mathcal{H}$ so that

$$\langle g_n, f_m \rangle_{\mathcal{H}} = \begin{cases} 1, & m = n, \\ 0, & m \neq n. \end{cases}$$

A family $\{f_n\}$ in $\mathcal{H}$ is said to be complete if its closed span in $\mathcal{H}$ is equal to $\mathcal{H}$. If a family is both complete and minimal then it has a unique biorthogonal family in $\mathcal{H}$, however this family is not necessarily complete in $\mathcal{H}$ (see [32]).

Hence, assuming conditions (1.5) – (1.6), the system $M_\Lambda$ has a unique biorthogonal family in its closed span in $L^2_w(A)$. We denote this family by $r_\Lambda$ and we will show that $r_\Lambda$ is complete in that closure, thus $M_\Lambda$ and $r_\Lambda$ are Markushevich bases for the closed span of $M_\Lambda$ in $L^2_w(A)$. To achieve this, we first obtain the sharp upper bound (1.15) for the norms of the elements in the family $r_\Lambda$, and then we derive the Fourier-type series representation (1.16) for functions in the closed span of $M_\Lambda$ in $L^2_w(A)$. Regarding (1.16), we show that for functions $f$ in the closed span of $M_\Lambda$ in $L^2_w(A)$, the coefficient $a_n$ in the series expansion $f(x) = \sum_{n=1}^\infty a_n x^{\lambda_n}$ as in (1.13), is equal to the inner product $\langle f, r_n \rangle_{w,A}$.

**Theorem 1.2.** Let $A$ be a measurable subset of the real half-line $[0, \infty)$ of positive Lebesgue measure and let $w$ be a real-valued non-negative integrable function defined on $A$, with $r_w$ as in (1.3). Let $\Lambda = \{\lambda_n\}_{n=1}^\infty$ be a sequence of distinct positive real numbers that satisfies (1.5) – (1.6). Then there exists a family of functions

$$r_\Lambda = \{r_n : n \in \mathbb{N}\} \subset L^2_w(A)$$

so that it is the unique biorthogonal sequence to $M_\Lambda$ in $L^2_w(A)$ which belongs to the closed span of the system $M_\Lambda$ in $L^2_w(A)$, and the following are true:

(I) For every $\epsilon > 0$ there is a constant $m_\epsilon > 0$, independent of $n \in \mathbb{N}$, but depending on $\Lambda$, $A$ and $w$, so that

$$||r_n||_{L^2_w(A)} \leq m_\epsilon(r_w - \epsilon)^{-\lambda_n}, \quad \forall n \in \mathbb{N}. \qquad (1.15)$$

(II) Each function $f$ in the closed span of the system $M_\Lambda$ in $L^2_w(A)$, extends analytically in the slit disk $D_{r_w}$ (1.4), so that $f$ admits the Fourier-type series representation

$$f(z) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} z^{\lambda_n}, \qquad (1.16)$$

converging uniformly on compact subsets of the slit disk $D_{r_w}$.

(III) For each function $f \in L^2_w(A)$, its associated series

$$f^*(z) := \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} z^{\lambda_n}, \qquad (1.17)$$

is analytic in $D_{r_w}$ and $f^*$ belongs to the closed span of the system $M_\Lambda$ in $L^2_w(A)$.

(IV) The system $M_\Lambda$ is a Markushevich basis in its closed span in $L^2_w(A)$, that is,

$$\overline{span}(r_\Lambda) = \overline{span}(M_\Lambda) \quad \text{in} \quad L^2_w(A).$$
**Remark 1.4.** It follows from (III) above that for $f(x) = 1$ on A, the series

$$ \sum_{n=1}^{\infty} \langle 1, r_n \rangle_{w,A} z^{\lambda_n}, $$

is analytic in $D_{r_w}$ and belongs to the closed span of the system $M_\Lambda$ in $L_w^2(A)$.

Next, now that we know that the families $M_\Lambda$ and $r_\Lambda$ are Markushevich bases for the closed span of $M_\Lambda$ in $L_w^2(A)$, we ask if they are also

### Strong Markushevich Bases

for this closure. This means whether for any disjoint union of two sets $N_1$ and $N_2$, such that $N = N_1 \cup N_2$, the closed span of the mixed system

$$ \{x^{\lambda_n} : n \in N_1\} \cup \{r_n : n \in N_2\} $$

is equal to the closed span of $M_\Lambda$ in $L_w^2(A)$. Another term used instead of the phrase “strong Markushevich basis” is the phrase “hereditarily complete system”.

**Remark 1.5.** The answer to the above question is affirmative in the special case when $m \le w(x) \le M$ on A for some positive numbers m and M and the set A contains an interval $[a, r_A]$ for some $0 \le a < r_A$ where $r_A$ is the essential supremum of A (see Theorem 1.3).

**Remark 1.6.** It would be very interesting to investigate also the case when A does not contain such an interval.

Hereditary completeness of exponential systems $\{e^{i\lambda_n t} : n \in \mathbb{N}\}$ in $L^2(-a, a)$ has been studied in [7] where it was proved that a complete and minimal exponential system in $L^2(-a, a)$ is hereditarily complete up to one-dimensional effect ([7, Theorem 1.1]). In [34, Theorem 1.1] we proved that if $\Lambda$ satisfies the conditions (1.5) – (1.6), then the exponential system $\{e^{\lambda_n t}\}_{n=1}^\infty$ is hereditarily complete in its closed span in $L^2(a, b)$. The result below extends the work of [34].

**Theorem 1.3.** Consider all the assumptions of Theorem 1.2. Suppose however that $m \le w(x) \le M$ on A for some positive numbers m and M thus by Remark 1.1 we have $r_w = r_A$. Suppose also that the set A contains an interval $[a, r_A]$ for some $a \ge 0$. Then the system $M_\Lambda$ is hereditarily complete in its closed span in $L_w^2(A)$.

We remark that hereditary completeness is related to the Spectral Synthesis problem for linear operators [26, 7, 8, 9]. Let T be a bounded linear operator in a separable Hilbert space $\mathcal{H}$ such that T has a set of eigenvectors which is complete in $\mathcal{H}$. Then T admits Spectral Synthesis if for any invariant subspace A of T, the set of eigenvectors of T contained in A is complete in A. In [30] J. Wermer proved that if an operator T is both compact and normal, then T admits spectral synthesis. The conclusion might not be true if one of these two conditions does not hold (see [30, Theorem 2] and [26, Theorem 4.2]). To this end we point out that A. S. Markus [26, Theorem 4.1] proved that if a compact operator T has a trivial kernel and non-zero simple eigenvalues with corresponding eigenvectors $\{f_n\}$, then T admits spectral synthesis if and only if the family $\{f_n\}$ is hereditarily complete in $\mathcal{H}$.

Intrigued by the above, we present below a class of compact, non-normal operators on the closed span of $M_\Lambda$ in $L_w^2(A)$ that admit spectral synthesis.

**Theorem 1.4.** Consider all the assumptions of Theorem 1.3. Denote by $[\overline{span}(M_\Lambda)]_{w,A}$ the closed span of the system $M_\Lambda$ in $L_w^2(A)$. Fix a sequence $\{u_n\}_{n=1}^\infty$ of distinct non-zero real numbers such that

$$ |u_n| \le \rho^{\lambda_n} \quad \text{for some} \quad 0 < \rho < 1. \qquad (1.18) $$

Then $T : [\overline{span}(M_\Lambda)]_{w,A} \to [\overline{span}(M_\Lambda)]_{w,A}$ defined as

$$ T f(x) := \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot u_n \cdot x^{\lambda_n}, $$

is an operator which is compact, not normal, and admits Spectral Synthesis. In particular, this is true for the operator $T_\rho(f) := f(\rho x)$ for any fixed $0 < \rho < 1$.
Our final result in this article is on Moment problems (see Theorem 1.5). We note that when {$\lambda_n$}$_{n=1}^\infty$ is a sequence satisfying (1.5)-(1.6) or similar conditions, and {$r_n$}$_{n=1}^\infty$ is the unique biorthogonal family to the system {$e^{-\lambda_n x}$}$_{n=1}^\infty$ in the closed span of the latter in $L^2(0, T)$, then searching for solutions $f$ to moment problems such as

$$ \int_0^T f(x) \cdot e^{-\lambda_n x} dx = d_n \quad \forall n \in \mathbb{N}, $$

is very useful in Control Theory for Partial Differential Equations ([17, 5, 6]), starting with the pioneering work of Fattorini and Russell [17] and followed by a vast amount of work done after that. Inspired by this we derived the following result.

**Theorem 1.5.** Let A be a measurable subset of the real half-line $[0, \infty)$ of positive Lebesgue measure and let w be a real-valued non-negative integrable function defined on A, with $r_w$ as in (1.3). Let $\Lambda = \{\lambda_n\}_{n=1}^\infty$ be a sequence of distinct positive real numbers that satisfies (1.5) – (1.6). Consider a sequence {$d_n$}$_{n=1}^\infty$ of real numbers such that

$$ |d_n| = O(a^{\lambda_n}) \quad \text{for some} \quad a \in [0, r_w). \qquad (1.19) $$

Then there exists a unique function $f \in \overline{span}(M_\Lambda)$ in $L^2_w(A)$, serving as a solution to the moment problem

$$ \int_A f(x) \cdot x^{\lambda_n} \cdot w(x) dx = d_n, \quad n = 1, 2, \dots \qquad (1.20) $$

**Remark 1.7.** Certain concepts from Non-Harmonic Fourier series such as Bessel sequences and Riesz-Fischer sequences will be utilized to prove the above result.

The rest of this article is organized as follows: in Section 3 we revisit Theorem B and prove Lemma 1.1 as well as some auxiliary result. The proofs of Theorems 1.1, 1.2, 1.3, and 1.5 are given respectively in Sections 2, 4, 5, and 6. The proof of Theorem 1.4 is given in the Appendix section since it is almost identical to the one given for [34, Theorem 4.1].

## 2 Proof of Theorem 1.1 on Distances

Fix some small positive $\epsilon$ and let $\delta_\epsilon := r_w - \frac{\epsilon}{2}$. By the definition of $r_w$ in (1.3), we claim that there exists an $\alpha_\epsilon > 0$, which depends on $\epsilon$, so that the set

$$ B_\epsilon := \{x \in A \cap (\delta_\epsilon, \infty) : w(x) > \alpha_\epsilon\} \qquad (2.1) $$

has **positive** Lebesgue measure. Let us justify this argument by writing the set

$$ A \cap (\delta_\epsilon, \infty) $$

as a countable union of subsets

$$ \{x \in A \cap (\delta_{\epsilon}, \infty) : w(x) = 0\} \cup \{x \in A \cap (\delta_{\epsilon}, \infty) : w(x) > 1\} \cup \left( \bigcup_{n=1}^{\infty} \left\{ x \in A \cap (\delta_{\epsilon}, \infty) : \frac{1}{n+1} < w(x) \le \frac{1}{n} \right\} \right). $$

If for each positive $\alpha$ the measure of the set {$x \in A \cap (\delta_\epsilon, \infty) : w(x) > \alpha$} is equal to zero, then the measure of the set $A \cap (\delta_\epsilon, \infty)$ will be equal to the measure of the set {$x \in A \cap (\delta_\epsilon, \infty) : w(x) = 0$}. This means that $\int_{A \cap (\delta_\epsilon, \infty)} w(t) dt = 0$ which contradicts the definition of $r_w$ in (1.3). Hence for some $\alpha_\epsilon > 0$ there exists a set $B_\epsilon$ of positive Lebesgue measure so that $w(x) > \alpha_\epsilon$ on $B_\epsilon$.

Then one has

$$ \|f\|_{L^p_w(A)} \ge \|f\|_{L^p_w(B_\epsilon)} \ge \alpha_\epsilon^{1/p} \cdot \|f\|_{L^p(B_\epsilon)}. $$

Since $B_\epsilon \subset [\delta_\epsilon, r_w]$ it then follows from Theorem A, relation (1.9), that for all functions $f$ in the span of $M_\Lambda$, there exists a positive constant $c_\epsilon$ which depends on $\epsilon$ and on the measure of the set $B_\epsilon$, such that

$$ \|f\|_{L^p(B_\epsilon)} \ge c_\epsilon \cdot \|f\|_{[0, \delta_\epsilon]}. $$
Choose any $a \in (0, \delta_\epsilon)$. Clearly one has $||f||_{[0, \delta_\epsilon]} \ge ||f||_{[a, \delta_\epsilon]}$.

Then, combining all of the above shows that

$$ ||f||_{L_w^p(A)} \ge \alpha_\epsilon^{1/p} \cdot c_\epsilon \cdot ||f||_{[a, \delta_\epsilon]}. \quad (2.2) $$

Now, let $g$ belong to the span of the system $M_{\Lambda_n}$ and let $e_n(x) = x^{\lambda_n}$, thus $e_n - g$ belongs to the span of $M_\Lambda$. Then from above we get

$$ ||e_n - g||_{L_w^p(A)} \ge \alpha_\epsilon^{1/p} \cdot c_\epsilon \cdot ||e_n - g||_{[a, \delta_\epsilon]}. \quad (2.3) $$

Next, we recall a lower bound for $||e_n - g||_{[a, \delta_\epsilon]}$ due to Luxemburg and Korevaar. They proved in [25, Theorem 7.1 and relation (1.9)] that for every positive $\eta$ there exists a positive constant $M_{\epsilon,\eta}$, which depends on $\eta$ and on the interval $[a, \delta_\epsilon]$, hence on $\epsilon$, such that

$$ \inf_{g \in \text{span}(M_{\Lambda_n})} ||e_n - g||_{[a, \delta_\epsilon]} \ge M_{\epsilon, \eta} (\delta_\epsilon - \eta)^{\lambda_n}. $$

For the fixed $\epsilon > 0$, take $\eta = \epsilon/2$ and recall that $\delta_\epsilon = r_w - \epsilon/2$ hence $\delta_\epsilon - \epsilon/2 = r_w - \epsilon$. Thus there exists a constant $M_\epsilon$, such that

$$ \inf_{g \in \text{span}(M_{\Lambda_n})} ||e_n - g||_{[a, \delta_\epsilon]} \ge M_\epsilon \cdot \left(\delta_\epsilon - \frac{\epsilon}{2}\right)^{\lambda_n} = M_\epsilon \cdot (r_w - \epsilon)^{\lambda_n}. $$

Together with (2.3) shows that for all $g \in \text{span}(M_{\Lambda_n})$, we have

$$ ||e_n - g||_{L_w^p(A)} \ge \alpha_\epsilon^{1/p} \cdot c_\epsilon \cdot M_\epsilon \cdot (r_w - \epsilon)^{\lambda_n}. $$

Finally, letting $m_\epsilon = \alpha_\epsilon^{1/p} \cdot c_\epsilon \cdot M_\epsilon$ gives us the distance result (1.12), that is

$$ \inf_{g \in \text{span}(M_{\Lambda_n})} ||e_n - g||_{L_w^p(A)} \ge m_\epsilon \cdot (r_w - \epsilon)^{\lambda_n}. $$

# 3 Proof of Lemma 1.1 and some auxiliary result

In this section we first revisit Theorem B by proving Lemma 1.1.

## 3.1 Proof of Lemma 1.1

As in the proof of Theorem 1.1, fix some small positive $\epsilon$ and let $\delta_\epsilon := r_w - \epsilon/2$. Then there exists an $\alpha_\epsilon > 0$, which depends on $\epsilon$, so that the set $B_\epsilon$ as in (2.1) has positive Lebesgue measure. Since $||f - P_j||_{L_{w,A}^p} \to 0$ as $j \to \infty$ then $||f - P_j||_{L^p(B_\epsilon)} \to 0$ as $j \to \infty$ as well. Thus $\{P_j\}_{j=1}^\infty$ is a Cauchy sequence in $L^p(B_\epsilon)$. It follows by Theorem A relation (1.9), that $\{P_j\}_{j=1}^\infty$ is uniformly Cauchy on $[0, \delta_\epsilon]$. Hence there exists a function $g_{\delta_\epsilon} \in C[0, \delta_\epsilon]$, the space of continuous functions on $[0, \delta_\epsilon]$, such that $|g_{\delta_\epsilon}(x) - P_j(x)| \to 0$ uniformly on $[0, \delta_\epsilon]$ as $j \to \infty$. Thus we also have

$$ ||g_{\delta_\epsilon} - P_j||_{L^p(0, \delta_\epsilon)} \to 0 \quad \text{as} \quad j \to \infty, \quad \text{and} \quad ||P_j||_{L^p(0, \delta_\epsilon)} \to ||g_{\delta_\epsilon}||_{L^p(0, \delta_\epsilon)} \quad \text{as} \quad j \to \infty. \quad (3.1) $$

From the pre-mentioned result by Luxemburg and Korevaar [25, Theorem 7.1 and relation (1.9)] it follows that for every positive $\eta$ there exists a positive constant $M_\eta$, such that

$$ \inf_{h \in \text{span}(M_{\Lambda_n})} ||e_n - h||_{L^p([0, \delta_\epsilon])} \ge M_\eta (\delta_\epsilon - \eta)^{\lambda_n}. \quad (3.2) $$

Now, $P_j(x) = \sum_{n=1}^{p(j)} a_{j,n} x^{\lambda_n}$: fix $k \in \{1, 2, \dots, p(j)\}$ and write

$$ P_j(x) = a_{j,k} \cdot \left( x^{\lambda_k} + \sum_{n=1, n \neq k}^{p(j)} \frac{a_{j,n}}{a_{j,k}} x^{\lambda_n} \right). $$
Then let $H_{j,k}(x) := \sum_{n=1, n \neq k}^{p(j)} \frac{a_{j,n}}{a_{j,k}} x^{\lambda_n}$ thus $H_{j,k}$ belongs to the space $M_{\Lambda_k}$ (1.11). Combined with (3.2), we get

$$ \|P_j\|_{L^p(0,\delta_\epsilon)} = |a_{j,k}| \cdot \|e_k + H_{j,k}\|_{L^p(0,\delta_\epsilon)} \ge |a_{j,k}| \cdot M_\eta \cdot (\delta_\epsilon - \eta)^{\lambda_k}. $$
$$ |a_{j,k}| \le m_\eta \cdot \|P_j\|_{L^p(0,\delta_\epsilon)} \cdot (\delta_\epsilon - \eta)^{-\lambda_k}. \tag{3.3} $$

Letting $m_\eta = 1/M_\eta$ gives

Similarly one gets

$$ |a_{i,k} - a_{j,k}| \le m_{\eta} \cdot \|P_i - P_j\|_{L^p(0,\delta_{\epsilon})} \cdot (\delta_{\epsilon} - \eta)^{-\lambda_k}. \tag{3.4} $$

Now, it follows from (3.1) that $\{P_j\}_{j=1}^\infty$ is Cauchy in $L^p(0, \delta_\epsilon)$. Thus, from (3.4) we see that for every fixed $k \in \mathbb{N}$, $\{a_{j,k}\}_{j=1}^\infty$ is a Cauchy sequence of real numbers, hence

$$ a_{j,k} \to a_k \quad \text{for some} \quad a_k \in \mathbb{R}. \tag{3.5} $$

Moreover, from (3.1) and (3.3) we get

$$ |a_k| \le m_\eta \cdot \|g_{\delta_\epsilon}\|_{L^p(0,\delta_\epsilon)} \cdot (\delta_\epsilon - \eta)^{-\lambda_k}. \tag{3.6} $$

If we fix the index $i$ in (3.4) and let the index $j \to \infty$, we then get from (3.1) and (3.4)

$$ |a_{i,k} - a_k| \le m_\eta \cdot \|P_i - g_{\delta_\epsilon}\|_{L^p(0,\delta_\epsilon)} \cdot (\delta_\epsilon - \eta)^{-\lambda_k}. \tag{3.7} $$

We will need the estimates (3.6) – (3.7) below. First of all, it follows from (3.6) that

$$ F_{\delta_{\epsilon}}(z) := \sum_{n=1}^{\infty} a_n z^{\lambda_n} \tag{3.8} $$

defines an analytic function in the slit disk

$$ D_{\delta_{\epsilon}} := \{z \in \mathbb{C} \setminus (-\infty, 0] : |z| < \delta_{\epsilon}\}. $$

converging uniformly on compact subsets of $D_{\delta_\epsilon}$.

We will now show below that

$$ F_{\delta_{\epsilon}}(x) = f(x) \quad \text{almost everywhere on} \quad A \cap [0, \delta_{\epsilon}). \tag{3.9} $$

**Fix** any $\rho \in (0, \delta_\epsilon)$ and choose $\eta > 0$ such that $\delta_\epsilon - \eta > \rho$, thus $\rho/(\delta_\epsilon - \eta) < 1$. Then the series

$$ \sum_{n=1}^{\infty} (\delta_{\epsilon} - \eta)^{-\lambda_n} \cdot x^{\lambda_n} = \sum_{n=1}^{\infty} \left( \frac{x}{\delta_{\epsilon} - \eta} \right)^{\lambda_n} $$

converges uniformly on the interval $[0, \rho]$. Therefore,

$$ \sum_{n=p(j)+1}^{\infty} (\delta_{\epsilon} - \eta)^{-\lambda_n} \cdot x^{\lambda_n} \to 0 \quad \text{uniformly on } [0, \rho] \text{ as } j \to \infty \tag{3.10} $$

and there exists some $M > 0$ so that

$$ \sum_{n=1}^{p(j)} (\delta_{\epsilon} - \eta)^{-\lambda_n} \cdot x^{\lambda_n} < M \quad \text{for all} \quad x \in [0, \rho]. \tag{3.11} $$

Then for all $x \in [0, \rho]$ write

$$ \begin{aligned} |F_{\delta_{\epsilon}}(x) - P_j(x)| &= \left| \sum_{n=1}^{\infty} a_n x^{\lambda_n} - \sum_{n=1}^{p(j)} a_{j,n} x^{\lambda_n} \right| \\ &= \left| \sum_{n=p(j)+1}^{\infty} a_n x^{\lambda_n} - \sum_{n=1}^{p(j)} (a_{j,n} - a_n) x^{\lambda_n} \right| \\ &\le \sum_{n=p(j)+1}^{\infty} |a_n| x^{\lambda_n} + \sum_{n=1}^{p(j)} |a_{j,n} - a_n| x^{\lambda_n}. \end{aligned} $$
If we replace the upper bounds from (3.6) – (3.7) we get for all $x \in [0, \rho]$

$$|F_{\delta_{\epsilon}}(x) - P_{j}(x)| \leq m_{\eta} \cdot \|g_{\delta_{\epsilon}}\|_{L^{p}(0,\delta_{\epsilon})} \cdot \sum_{n=p(j)+1}^{\infty} (\delta_{\epsilon} - \eta)^{-\lambda_{n}} \cdot x^{\lambda_{n}} + m_{\eta} \cdot \|P_{j} - g_{\delta_{\epsilon}}\|_{L^{p}(0,\delta_{\epsilon})} \cdot \sum_{n=1}^{p(j)} (\delta_{\epsilon} - \eta)^{-\lambda_{n}} \cdot x^{\lambda_{n}}.$$

It then follows from (3.1), (3.10) and (3.11) that

$$|F_{\delta_{\epsilon}}(x) - P_{j}(x)| \to 0 \quad \text{uniformly on } [0, \rho] \quad \text{as } j \to \infty.$$

Thus for every positive $\tau > 0$, there exists some $j_{\tau} \in \mathbb{N}$ such that for all $j \ge j_{\tau}$ one has $|F_{\delta_{\epsilon}}(x) - P_{j}(x)| < \tau$ for all $x \in [0, \rho]$. Hence

$$\int_{A \cap [0, \rho]} |F_{\delta_{\epsilon}}(x) - P_{j}(x)|^{p} \cdot w(x) \, dx \le \tau^{p} \cdot \int_{A} w(x) \, dx, \quad \text{for all } j \ge j_{\tau}.$$

Since $\int_A w(x) dx < \infty$ we get

$$\|F_{\delta_{\epsilon}} - P_{j}\|_{L_{w,A}^{p}[0,\rho]} \to 0 \quad \text{as} \quad j \to \infty. \qquad (3.12)$$

But $\|f - P_j\|_{L_{w,A}^p[0,\rho]} \to 0$ as $j \to \infty$ since $\|f - P_j\|_{L_{w,A}^p} \to 0$ as $j \to \infty$. Combined together shows that

$$\|f - F_{\delta_{\epsilon}}\|_{L_{w,A}^{p}[0,\rho]} = 0.$$

Due to (1.2) we see that

$$F_{\delta_{\epsilon}}(x) = f(x) \quad \text{almost everywhere on } A \cap [0, \rho].$$

However, this is true for any fixed $\rho \in (0, \delta_{\epsilon})$. Hence we conclude that

$$f(x) = F_{\delta_{\epsilon}}(x) \quad \text{almost everywhere on } A \cap [0, \delta_{\epsilon}).$$

In other words,

$$f(x) = \sum_{n=1}^{\infty} a_n x^{\lambda_n} \quad \text{almost everywhere on } A \cap [0, \delta_{\epsilon}). \qquad (3.13)$$

Using the same rational, and repeating the above arguments, if we choose $\delta_1$ such that $r_w > \delta_1 > \delta_{\epsilon}$, then there is a function $F_{\delta_1}(x)$,

$$F_{\delta_1}(x) = \sum_{n=1}^{\infty} b_n x^{\lambda_n}, \quad b_n \in \mathbb{R}, \qquad (3.14)$$

defining an analytic function in the slit disk

$$D_{\delta_1} := \{z \in \mathbb{C} \setminus (-\infty, 0] : |z| < \delta_1\},$$

converging uniformly on compact subsets of $D_{\delta_1}$, and such that

$$f(x) = F_{\delta_1}(x) \quad \text{almost everywhere on } A \cap [0, \delta_1).$$

Observe however, that due to the **Uniqueness** of limits, the limit in (3.5) is unique. Thus the coefficients $b_n$ in (3.14) are equal to the respective coefficients $a_n$ in (3.8). Hence, $F_{\delta_1}(x)$ is identical to $F_{\delta_{\epsilon}}(x)$. This implies that (1.13) is valid for all $z \in D_{r,w}$ (in the almost everywhere sense on A).
## 3.2 An auxiliary result

**Lemma 3.1.** Suppose that $f \in \overline{span}(M_\Lambda)$ in $L_w^p(A)$, $p \ge 1$, thus by Theorem B and Lemma 1.1 we have $f(x) = \sum_{n=1}^\infty a_n x^{\lambda_n}$ almost everywhere on $A \cap [0, r_w)$. Then, for any fixed element $x^{\lambda_n}$, the function

$$f(x) - a_n x^{\lambda_n}$$

belongs to the closed span of the system $M_{\Lambda_n}$ (1.11) in $L_w^p(A)$.

*Proof.* Since $f \in \overline{span}(M_\Lambda)$ in $L_w^p(A)$, then a sequence $\{P_j(x)\}_{j=1}^\infty$ in $\overline{span}(M_\Lambda)$ exists, where $P_j(x) = \sum_{n=1}^p a_{j,n} x^{\lambda_n}$, such that $||f - P_j||_{L_w^p(A)} \to 0$ as $j \to \infty$. By Theorem B and Lemma 1.1, $f$ is extended analytically in the slit disk $D_{r_w}$ (1.4) such that $f(x) = \sum_{n=1}^\infty a_n x^{\lambda_n}$ for almost all $x \in A \cap [0, r_w]$ and $a_n = \lim_{j \to \infty} a_{j,n}$ (see relation (1.14)). Hence, for every fixed $n \in \mathbb{N}$ we have

$$(a_{j,n} - a_n)x^{\lambda_n} \to 0 \quad \text{as} \quad j \to \infty, \quad \text{uniformly on } [0, r_w].$$

That is, for any $\epsilon > 0$ there exists some $j_\epsilon \in \mathbb{N}$ so that for all $j > j_\epsilon$ and for all $x \in [0, r_w]$ one has $|(a_{j,n} - a_n)x^{\lambda_n}|^p < \epsilon$. Therefore for all $j > j_\epsilon$ we have

$$\int_{A \cap [0, r_w]} |(a_{j,n} - a_n)x^{\lambda_n}|^p \cdot w(x) dx < \epsilon \int_{A \cap [0, r_w]} w(x) dx.$$

Thus

$$\lim_{j \to \infty} \int_{A \cap [0, r_w]} |(a_{j,n} - a_n)x^{\lambda_n}|^p \cdot w(x) dx = 0.$$

Clearly one also has

$$\lim_{j \to \infty} \int_A |(a_{j,n} - a_n)x^{\lambda_n}|^p \cdot w(x) dx = 0.$$

But $||f - P_j||_{L_w^p(A)} \to 0$ as $j \to \infty$. Combined together and applying the Minkowski inequality shows that

$$\lim_{j \to \infty} \int_A |[f(x) - P_j(x)] - [a_n x^{\lambda_n} - a_{j,n} x^{\lambda_n}]|^p \cdot w(x) dx = 0.$$

We rewrite this as

$$\lim_{j \to \infty} \int_A |[f(x) - a_n x^{\lambda_n}] - [P_j(x) - a_{j,n} x^{\lambda_n}]|^p \cdot w(x) dx = 0. \quad (3.15)$$

Clearly the function $P_j(x) - a_{j,n}x^{\lambda_n}$ belongs to the span of the system $M_{\Lambda_n}$ (1.11). We then conclude from (3.15) that the function $f(x) - a_n x^{\lambda_n}$ belongs to the closed span of the system $M_{\Lambda_n}$ in $L_w^p(A)$.

□

# 4 Proof of Theorem 1.2

In this section we prove Theorem 1.2 on the existence of a biorthogonal family $r_\Lambda$ to the system $M_\Lambda$ and the Fourier-type series representations (1.16) for functions in the closed span of $M_\Lambda$ in $L_w^2(A)$. We also show that $M_\Lambda$ is a Markushevich basis for its closed span in $L_w^2(A)$.

## 4.1 Constructing the Biorthogonal family and deriving the upper bound (1.15)

In (1.12) we derived a positive lower bound for $D_{A,w,2,n}$ which is the distance between a function $x^{\lambda_n}$ and the closed span of the system $M_{\Lambda_n}$ (1.11) in $L_w^2(A)$. Since $L_w^2(A)$ is a separable Hilbert space, it then follows from the *Closest Point Theorem* that there exists a unique element in $\overline{span}(M_{\Lambda_n})$ in $L_w^2(A)$, that we denote by $\phi_n$, so that

$$||e_n - \phi_n||_{L_w^2(A)} = \inf_{g \in \overline{span}(M_{\Lambda_n})} ||e_{n,k} - g||_{L_w^2(A)} = D_{A,w,2,n}.$$
The function $e_n - \phi_n$ is orthogonal to all the elements of the closed span of $M_{\Lambda_n}$ in $L^2_w(A)$, hence to $\phi_n$ itself. Therefore

$$ \langle e_n - \phi_n, e_n - \phi_n \rangle_{w,A} = \langle e_n - \phi_n, e_n \rangle_{w,A}. $$

Hence

$$ (D_{A,w,2,n})^2 = \langle e_n - \phi_n, e_n \rangle_{w,A}. $$

Next, we define

$$ r_n(x) := \frac{e_n(x) - \phi_n(x)}{(D_{A,w,2,n})^2}. $$

It then follows that $\langle r_n, e_n \rangle_{w,A} = 1$ and $r_n$ is orthogonal to all the elements of the system $M_{\Lambda_n}$. Thus $\{r_n : n \in \mathbb{N}\}$ is biorthogonal to the system $M_{\Lambda}$ in $L^2_w(A)$. Since $\phi_n \in \overline{\operatorname{span}}(M_{\Lambda_n})$ in $L^2_w(A)$ then $r_n \in \overline{\operatorname{span}}(M_{\Lambda})$ in $L^2_w(A)$.

**Remark 4.1.** Clearly we have $||r_n||_{L^2_w(A)} = \frac{1}{D_{A,w,2,n}}$ thus (1.15) follows from (1.12).

Next we show that $\{r_n\}$ is the unique biorthogonal sequence to the system $M_{\Lambda}$, which belongs to its closed span in $L^2_w(A)$. Indeed, if there is another such biorthogonal sequence, call it $\{q_n\}$, then for all $n \in \mathbb{N}$ we have

$$ \langle r_n - q_n, e_m \rangle_{w,A} = 0, \quad \forall m \in \mathbb{N}. $$

But this in turn implies that $r_n - q_n = 0$ almost everywhere on $A$ since the system $M_{\Lambda}$ is complete in its closed span in $L^2_w(A)$.

## 4.2 The Fourier-type series representations (1.16)

By Theorem B, a function $f \in \overline{\operatorname{span}}(M_{\Lambda})$ in $L^2_w(A)$ extends analytically in the slit disk $D_{r_w}$ (1.4) and $f(x) = \sum_{n=1}^{\infty} a_n x^{\lambda_n}$ almost everywhere on $A \cap [0, r_w)$. To obtain (1.16) we will show that

$$ \langle f, r_n \rangle_{w,A} = a_n. \quad (4.1) $$

Due to biorthogonality we get

$$ \begin{aligned} \langle f, r_n \rangle_{w,A} &= \int_A f(x) \cdot r_n(x) \cdot w(x) dx \\ &= \int_A r_n(x) \cdot a_n x^{\lambda_n} \cdot w(x) dx + \int_A r_n(x) \cdot [f(x) - a_n x^{\lambda_n}] \cdot w(x) dx \\ &= a_n + \int_A r_n(x) \cdot [f(x) - a_n x^{\lambda_n}] \cdot w(x) dx. \end{aligned} \quad (4.2) $$

Now, it follows from Lemma 3.1 that the function $f_n(x) := f(x) - a_n x^{\lambda_n}$ belongs to the closed span of the system $M_{\Lambda_n}$ (1.11) in $L^2_w(A)$. Hence, for every $\epsilon > 0$, there is a function $g_{\epsilon}$ in the span of $M_{\Lambda_n}$ so that $||f_n - g_{\epsilon}||_{L^2_w(A)} < \epsilon$. Due to the biorthogonality we have

$$ \int_A r_n(x) \cdot g_{\epsilon}(x) \cdot w(x) dx = 0. $$

Combining with the Cauchy-Schwarz inequality we get

$$ \begin{aligned} \left| \int_A r_n(x) \cdot f_n(x) \cdot w(x) dx \right| &= \left| \int_A r_n(x) \cdot [f_n(x) - g_{\epsilon}(x)] \cdot w(x) dx \right| \\ &= \left| \int_A \left( r_n(x) \cdot \sqrt{w(x)} \right) \cdot \left( [f_n(x) - g_{\epsilon}(x)] \cdot \sqrt{w(x)} \right) dx \right| \\ &\le \epsilon \cdot \|r_n\|_{L^2_w(A)}. \end{aligned} $$

The arbitrary choice of $\epsilon$ implies that $\int_A r_n(x) \cdot f_n(x) \cdot w(x) dx = 0$, that is

$$ \int_A r_n(x) \cdot [f(x) - a_n x^{\lambda_n}] \cdot w(x) dx = 0. $$

Replacing in (4.2) shows that (4.1) holds.
## 4.3 The associated function $f^*$ of $f \in L_w^2(A)$

Clearly $f(x)$ can be written uniquely as

$$f(x) = g(x) + h(x)$$

where

(I) $g$ belongs to the closed span of the system $M_\Lambda$ in $L_w^2(A)$, and

(II) $h$ belongs to the orthogonal complement of the closed span of the system $M_\Lambda$ in $L_w^2(A)$, thus $\langle h, r_n \rangle_{w,A} = 0$ for all $r_n \in r_\Lambda$.

Since $h = f - g$, we then have

$$\langle g, r_n \rangle_{w,A} = \langle f, r_n \rangle_{w,A} \quad \text{for all } n \in \mathbb{N}.$$

It follows from (1.16) that

$$g(x) = \sum_{n=1}^{\infty} \langle g, r_n \rangle_{w,A} \cdot x^{\lambda_n} \quad \text{almost everywhere on } A.$$

Combining the above shows that

$$f^*(z) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot z^{\lambda_n}$$

belongs to the closed span of $M_\Lambda$ in $L_w^2(A)$.

## 4.4 The system $M_\Lambda$ is a Markushevich basis for its closed span in $L_w^2(A)$

We have to show that

$$\overline{\operatorname{span}}(r_\Lambda) = \overline{\operatorname{span}}(M_\Lambda) \text{ in } L_w^2(A).$$

Let us denote $\overline{\operatorname{span}}(M_\Lambda)$ in $L_w^2(A)$ by $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ and let $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}$ be the closed span of $r_\Lambda$ in $L_w^2(A)$. Obviously $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}$ is a subspace of $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$. Let $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}^{\perp}$ be the orthogonal complement of $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}$ in $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$, that is

$$[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}^{\perp} = \{f \in [\overline{\operatorname{span}}(M_\Lambda)]_{w,A} : \langle f, g \rangle_{w,A} = 0 \text{ for all } g \in [\overline{\operatorname{span}}(r_\Lambda)]_{w,A}\}.$$

Now, if $f \in [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ then as shown earlier

$$f(x) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} x^{\lambda_n}, \quad \text{almost everywhere on } A \cap [0, r_w].$$

But if $f \in [\overline{\operatorname{span}}(r_\Lambda)]_{w,A}^{\perp} \subset [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ then $\langle f, r_n \rangle_{w,A} = 0$ for all $r_n \in r_\Lambda$. We conclude that $f = 0$ almost everywhere on $A \cap [0, r_w]$, hence $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A}^{\perp}$ contains just the zero function. Therefore $[\overline{\operatorname{span}}(r_\Lambda)]_{w,A} = [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$.

This completes the proof of Theorem 1.2.

# 5 Proof of Theorem 1.3 on Hereditary completeness

Let us write the set $\mathbb{N}$ as an arbitrary disjoint union of two sets $N_1$ and $N_2$. In order to obtain the hereditary completeness of $M_\Lambda$ in its closed span in $L^2(A)$, we must show that the closed span of the mixed system

$$M_{1,2} := \{e_n : n \in N_1\} \cup \{r_n : n \in N_2\},$$
in $L_w^2(A)$ is equal to $\overline{\mathrm{span}}(M_\Lambda)$ in $L_w^2(A)$, which it was denoted earlier by $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$. Denote by $W_{\Lambda_{1,2}}$ the closed span of $M_{1,2}$ in $L_w^2(A)$. Obviously $W_{\Lambda_{1,2}}$ is a subspace of $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$. Let $W_{\Lambda_{1,2}}^{\perp}$ be the orthogonal complement of $W_{\Lambda_{1,2}}$ in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$, that is

$$W_{\Lambda_{1,2}}^{\perp} = \{f \in [\overline{\mathrm{span}}(M_\Lambda)]_{w,A} : \langle f, g \rangle_{w,A} = 0 \text{ for all } g \in W_{\Lambda_{1,2}}\}.$$

Now, if $f \in W_{\Lambda_{1,2}}^{\perp} \subset [\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$, then by (1.16) we have

$$f(z) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot z^{\lambda_n}, \quad z \in D_{r_w}$$

with the series converging uniformly on the slit disk $D_{r_w}$ (1.4), and of course $\langle f, r_n \rangle_{w,A} = 0$ for all $n \in N_2$. Thus,

$$f(x) = \sum_{n \in N_1} \langle f, r_n \rangle_{w,A} \cdot x^{\lambda_n}, \quad \text{almost everywhere on } [0, r_w]. \qquad (5.1)$$

**Remark 5.1.** Since $w$ satisfies $m \le w(x) \le M$ on $A$ for some positive numbers $m, M$, then by Remark 1.1 we have $r_A = r_w$.

Since $f \in L_w^2(A)$ and by assumption for some $a \in [0, r_A)$ the interval $[a, r_A]$ is a subset of $A$, then $f \in L_w^2([a, r_A])$. Then the inequality $m \le w(x)$ implies that $f \in L^2(a, r_A)$. Moreover, one has $f \in L^2(0, r_A)$ as well since $f$ is continuous on $[0, r_A)$ due to the analytic extension of $f$ to the disk $D_{r_A}$.

To this end we point out that there is a **converse** result (see [21, Corollary 6.2.4], [25, Theorem 8.2], [23]) to the “Clarkson-Erdős-Schwartz Phenomenon” which reads as follows.

> “If $g \in L^2(0, 1)$ and $g(x) = \sum_{n=1}^{\infty} a_n \cdot x^{\lambda_n}$ on $(0, 1)$, then $g$ belongs to the closed span of $\{x^{\lambda_n}\}_{n=1}^{\infty}$ in $L^2(0, 1)$”.

Therefore, since $f$ as in (5.1) belongs to $L^2(0, r_A)$, it follows that $f$ belongs to the closed span of $\{e_n\}_{n \in N_1}$ in the space $L^2(0, r_A)$. Clearly now $f$ belongs to the closed span of $\{e_n\}_{n \in N_1}$ in $L^2(A)$ as well. Finally, from the inequality $w(x) \le M$ on $A$ we deduce that $f$ belongs to the closed span of $\{e_n\}_{n \in N_1}$ in $L_w^2(A)$ also.

Hence, for every $\epsilon > 0$ there is a function $g_\epsilon$ in span($\{e_n\}_{n \in N_1}$ so that $||f - g_\epsilon||_{L_w^2(A)} < \epsilon$. Write

$$\langle f, f \rangle_{w,A} = \langle f, f - g_\epsilon \rangle_{w,A} + \langle f, g_\epsilon \rangle_{w,A}.$$

Since $f \in W_{\Lambda_{1,2}}^{\perp}$ then $\langle f, e_n \rangle_{w,A} = 0$ for all $n \in N_1$, thus $\langle f, g_\epsilon \rangle_{w,A} = 0$. Therefore,

$$||f||_{L_w^2(A)}^2 = \langle f, f \rangle_{w,A} = \langle f, f - g_\epsilon \rangle_{w,A} \le ||f||_{L_w^2(A)} \cdot ||f - g_\epsilon||_{L_w^2(A)} \le ||f||_{L_w^2(A)} \cdot \epsilon.$$

Hence

$$||f||_{L_w^2(A)} \le \epsilon.$$

This holds for every $\epsilon > 0$ thus $f(x) = 0$ almost everywhere on $A$, hence $W_{\Lambda_{1,2}}^{\perp} = \{0\}$. Thus $W_{\Lambda_{1,2}} = [\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$, meaning that the system $M_\Lambda$ is hereditarily complete in the space $L_w^2(A)$.

The proof of Theorem 1.3 is now complete.

# 6 Proof of Theorem 1.5 on the Moment problem

In order to prove Theorem 1.5, we first introduce some concepts from Non-Harmonic Fourier Series such as Bessel sequences and Riesz-Fischer sequences.
## 6.1 Bessel sequences and Riesz-Fischer sequences

Let $\mathcal{H}$ be a separable Hilbert space endowed with an inner product $\langle \cdot, \cdot \rangle$, and consider a sequence $\{f_n\}_{n=1}^\infty \subset \mathcal{H}$. We say that (see [31, p. 128 Definition]):

(i) $\{f_n\}_{n=1}^\infty$ is a **Bessel** sequence if there exists a constant $B > 0$ such that $\sum_{n=1}^\infty |\langle f, f_n \rangle|^2 < B||f||^2$ for all $f \in \mathcal{H}$.

(ii) $\{f_n\}_{n=1}^\infty$ is a **Riesz-Fischer** sequence if the moment problem $\langle f, f_n \rangle = c_n$ has at least one solution in $\mathcal{H}$ for every sequence $\{c_n\}$ in the space $l^2(\mathbb{N})$.

The following result stated by Casazza et al. is an interesting connection between Bessel and Riesz-Fischer sequences.

**Proposition A.** [12, Proposition 2.3, (ii)]

The Riesz-Fischer sequences in $\mathcal{H}$ are precisely the families for which a biorthogonal Bessel sequence exists. In other words

(a) Suppose that two sequences $\{f_n\}_{n=1}^\infty$ and $\{g_n\}_{n=1}^\infty$ in $\mathcal{H}$ are biorthogonal. Suppose also that $\{f_n\}_{n=1}^\infty$ is a Bessel sequence. Then $\{g_n\}_{n=1}^\infty$ is a Riesz-Fischer sequence.

(b) If $\{f_n\}_{n=1}^\infty$ in $\mathcal{H}$ is a Riesz-Fischer sequence, then there exists a biorthogonal Bessel sequence $\{g_n\}_{n=1}^\infty$.

And now a sufficient condition so that two biorthogonal families in $\mathcal{H}$ are Bessel and Riesz-Fischer sequences. The result follows from [14, Proposition 3.5.4] and Proposition A.

**Lemma 6.1.** Consider two biorthogonal sequences $\{u_n\}_{n=1}^\infty$ and $\{v_n\}_{n=1}^\infty$ in $\mathcal{H}$ and suppose there is some $M > 0$ so that

$$ \sum_{m=1}^{\infty} |\langle v_n, v_m \rangle| < M \quad \text{for all } n = 1, 2, 3, \dots $$

Then $\{v_n\}_{n=1}^\infty$ is a Bessel sequence in $\mathcal{H}$ and $\{u_n\}_{n=1}^\infty$ is a Riesz-Fischer sequence in $\mathcal{H}$.

## 6.2 Proof of Theorem 1.5

As before, let $[\overline{\text{span}}(M_\Lambda)]_{w,A}$ be the closed span of $M_\Lambda$ in $L^2_w(A)$. Let $\{d_n : n \in \mathbb{N}\}$ be the sequence of non-zero real numbers that satisfies (1.19). Then, for every $n \in \mathbb{N}$ define

$$ U_n(t) := \lambda_n d_n r_n(t) \quad \text{and} \quad V_n(t) := \frac{x^{\lambda_n}}{\lambda_n d_n}. $$

Now, it easily follows that the sets

$$ \{U_n : n \in \mathbb{N}\} \quad \text{and} \quad \{V_n : n \in \mathbb{N}\} $$

are biorthogonal in $[\overline{\text{span}}(M_\Lambda)]_{w,A}$.

We will show below that $\{U_n\}_{n=1}^\infty$ and $\{V_n\}_{n=1}^\infty$ are Bessel and Riesz-Fischer sequences respectively in $[\overline{\text{span}}(M_\Lambda)]_{w,A}$. First, recall that for every $\epsilon > 0$ there is $m_\epsilon > 0$ so that $||r_n||_{L^2_w(A)} \le m_\epsilon(r_w - \epsilon)^{-\lambda_n}$. Since $|d_n| = O(a^{\lambda_n})$ for some $a \in [0, r_w)$, then there is some $N > 0$ so that $|d_n| \le Na^{\lambda_n}$ for all $n \in \mathbb{N}$. Choose then

$$ \epsilon = \frac{r_w - a}{2} \quad \text{hence} \quad r_w - \epsilon = \frac{r_w + a}{2} \quad \text{thus} \quad ||r_n||_{L^2_w(A)} \le m_\epsilon \left( \frac{2}{r_w + a} \right)^{\lambda_n}. $$

Since $r_w > a$, we can choose some positive $\gamma$ so that

$$ 1 < \gamma < \frac{(r_w + a)}{2a}, \quad \text{hence} \quad \frac{2\gamma a}{r_w + a} < 1. $$
Since $\gamma > 1$ there exists some $M > 0$ so that $\lambda_n \le M\gamma^{\lambda_n}$ for all $n \in \mathbb{N}$.

Combining all of the above, shows that there is some positive number $\tau$ so that

$$ ||U_n||_{L^2_w(A)} \le \lambda_n \cdot |d_n| \cdot ||r_n||_{L^2_w(A)} \le \tau \cdot \gamma^{\lambda_n} \cdot a^{\lambda_n} \cdot \left( \frac{2}{r_w + a} \right)^{\lambda_n} $$

thus

$$ ||U_n||_{L^2_w(A)} \le \tau \cdot \left( \frac{2a\gamma}{r_w + a} \right)^{\lambda_n}. $$

By the Cauchy-Schwarz inequality we get

$$ |\langle U_n, U_m \rangle_{w,A}| \le \tau^2 \cdot \left( \frac{2a\gamma}{r_w + a} \right)^{\lambda_n} \cdot \left( \frac{2a\gamma}{r_w + a} \right)^{\lambda_m}. \qquad (6.1) $$

Thus

$$ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} |\langle U_n, U_m \rangle_{w,A}| < \tau^2 \cdot \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \left( \frac{2a\gamma}{r_w + a} \right)^{\lambda_n} \cdot \left( \frac{2a\gamma}{r_w + a} \right)^{\lambda_m} < \infty $$

with convergence justified by the fact that the fraction $\frac{2a\gamma}{r_w + a} < 1$.

It then follows from Lemma 6.1 that $\{U_n\}_{n=1}^\infty$ is a Bessel sequence in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ and its biorthogonal sequence $\{V_n\}_{n=1}^\infty$ is a Riesz-Fischer sequence in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$.

Therefore, the moment problem

$$ \int_A f(x) \cdot V_n(x) \cdot w(x) dx = c_n \quad \forall n \in \mathbb{N}, $$

has a solution $f$ in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ whenever $\sum_{n=1}^\infty |c_n|^2 < \infty$. Since $\sum_{n=1}^\infty 1/\lambda_n < \infty$, we can take $c_n = 1/\lambda_n$ for all $n \in \mathbb{N}$. Hence, recalling the definition of $V_n$, there is some function $f \in [\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ so that

$$ \int_A f(x) \cdot \left( \frac{x^{\lambda_n}}{d_n \lambda_n} \right) \cdot w(x) dx = \frac{1}{\lambda_n} \quad \forall n \in \mathbb{N}. $$

Thus

$$ \int_A f(x) \cdot x^{\lambda_n} \cdot w(x) dx = d_n \quad \forall n \in \mathbb{N}, $$

hence obtaining a solution $f \in [\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ to the moment problem. The proof is now complete.

# A Proof of Theorem 1.4 on a class of operators that admit Spectral Synthesis

In order to prove Theorem 1.4, we need the following result obtained by Markus.

## Theorem C. [26, Theorem 4.1]

Let $\mathcal{H}$ be a separable Hilbert space and let $T: \mathcal{H} \to \mathcal{H}$ be a compact operator such that

(i) its kernel is trivial and

(ii) its non-zero eigenvalues are simple.

Let $\{f_n\}_{n \in \mathbb{N}}$ be the corresponding sequence of eigenvectors. Then $T$ admits Spectral Synthesis if and only if $\{f_n\}_{n \in \mathbb{N}}$ is Hereditarily complete in $\mathcal{H}$.

We will show in Lemma A.1 that if $\{u_n\}_{n=1}^\infty$ is a sequence of real numbers satisfying (1.18), then

$$ T f(x) := \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot u_n \cdot x^{\lambda_n}, \qquad (A.2) $$

is an operator well defined on $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ to $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$, such that it is compact, with a trivial kernel, having non-zero simple eigenvalues $\{u_k\}_{k=1}^\infty$ and with $\{x^{\lambda_k}\}_{k=1}^\infty$ being the corresponding eigenvectors. But since the system $M_\Lambda$ is hereditarily complete in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$, it will then follow from Theorem C that $T$ admits Spectral Synthesis and the proof of Theorem 1.4 will finish.
**Lemma A.1.** The following are true about T.

1. T is a well defined bounded operator from $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ to $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ and moreover T is compact.

2. $\{u_k\}_{k=1}^\infty$ are eigenvalues of T and $\{e_k\}_{k=1}^\infty$ are the corresponding eigenvectors.

3. $\{u_k\}$ are eigenvalues of $T^*$ (the adjoint of T) and $\{r_k\}_{k=1}^\infty$ are the corresponding eigenvectors.

4. The kernel of T is trivial.

5. The spectrum of T is $\{0\} \cup \{u_k\}_{k=1}^\infty$.

6. Each eigenvalue of T is simple.

7. The operator T is not normal.

*Proof.*

1. Let $f$ be a function in $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$. Then $f$ extends analytically in the slit disk $D_{r_w}$

$$f(z) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot z^{\lambda_n},$$

with the series converging uniformly on compact subsets of $D_{r_w}$. From relation (1.15), for every $\epsilon > 0$ there exists some $m_\epsilon > 0$, independent of $n \in \mathbb{N}$, so that

$$|\langle f, r_n \rangle_{w,A}| \le ||f||_{L_w^2(A)} \cdot ||r_n||_{L_w^2(A)} \le ||f||_{L_w^2(A)} \cdot m_\epsilon(r_w - \epsilon)^{-\lambda_n}. \quad (A.3)$$

Since $\{u_n\}_{n=1}^\infty$ satisfies (1.18), then for some $\rho \in (0,1)$ one has $|u_n| \le \rho^{\lambda_n}$, hence $u_n \to 0$ as $n \to \infty$. Choose

$$\epsilon = \frac{r_w(1 - \rho)}{2}, \quad \text{thus} \quad r_w - \epsilon = \frac{r_w(1 + \rho)}{2}. \quad (A.4)$$

Combining (A.3) – (A.4) yields that there exists some $m_\epsilon > 0$, independent of $n \in \mathbb{N}$ and $f \in L_w^2(A)$, so that

$$|\langle f, r_n \rangle_{w,A} \cdot u_n| \le ||f||_{L_w^2(A)} \cdot m_\epsilon \cdot \left( \frac{2\rho}{r_w(1+\rho)} \right)^{\lambda_n}. \quad (A.5)$$

One deduces from (A.5) that $Tf(z)$ is a function analytic on the slit disk

$$\left\{ z \in \mathbb{C} \setminus (-\infty, 0] : |z| < r_w \cdot \left( \frac{1+\rho}{2\rho} \right) \right\}$$

converging uniformly on its compact subsets, and in particular on the interval $[0, r_w]$. The uniform convergence on $[0, r_w]$ and the inequality $m \le w(x) \le M$ on the set A, imply that

(i) $Tf(z)$ belongs to the space $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$.

(ii) The series $T(f)$ converges in the $L_w^2(A)$ norm.

It also follows from (A.5) and (A.2) that there exists some $N > 0$ so that

$$\|T(f)\|_{L_w^2(A)} \le N \|f\|_{L_w^2(A)} \quad \text{for all} \quad f \in [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}.$$

Therefore, $T : [\overline{\operatorname{span}}(M_\Lambda)]_{w,A} \to [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ defines a Bounded Linear Operator. We will use $\|T\|$ to denote the operator norm of T which is the supremum of the set

$$\{\|T(f)\|_{L_w^2(A)} : f \in [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}, \|f\|_{L_w^2(A)} = 1\}.$$

We also denote by $T^*$ the Adjoint operator of T.
Next we show that $T$ is compact. Let $e_n(x) = x^{\lambda_n}$. Define $T_m$ on $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ by

$$T_m(g)(x) = \sum_{n=1}^{m} \langle g, r_n \rangle_{w,A} u_n e_n(x).$$

Let $f$ be a unit vector in $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$. It is easy to see that

$$\|(T - T_m)(f)\|_{L_w^2(A)} \le \sum_{n=m+1}^{\infty} \|\langle f, r_n \rangle_{w,A} u_n e_n\|_{L_w^2(A)}.$$

It is easy to see that there is some $q > 0$ so that one has $\|e_n\|_{L_w^2(A)} < qr_w^{\lambda_n}$. Combined with (A.5) and with $\epsilon$ as in (A.4) gives

$$\|\langle f, r_n \rangle_{w,A} u_n e_n\|_{L_w^2(A)} \le m_\epsilon \|f\|_{L_w^2(A)} \left( \frac{2\rho}{1+\rho} \right)^{\lambda_n}.$$

Therefore,

$$\|(T - T_m)(f)\|_{L_w^2(A)} \le m_\epsilon \|f\|_{L_w^2(A)} \sum_{n=m+1}^{\infty} \left( \frac{2\rho}{1+\rho} \right)^{\lambda_n}.$$

Since $2\rho < 1 + \rho$ the series above converges. Hence, $\|T - T_m\|_{L_w^2(A)}$ tends to zero as $m$ tends to infinity. Thus, the finite rank operators $\{T_m\}$ converge to $T$ in the uniform operator topology. Therefore, $T$ is compact.

2. The families $M_\Lambda$ and $r_\Lambda$ are biorthogonal thus from (A.2) we get

$$T(e_k) = u_k e_k.$$

3. For fixed $k \in \mathbb{N}$ and any $n \in \mathbb{N}$ we have

$$\begin{aligned}
\langle T^* r_k - u_k r_k, e_n \rangle_{w,A} &= -u_k \langle r_k, e_n \rangle_{w,A} + \langle r_k, T e_n \rangle_{w,A} \\
&= -u_k \langle r_k, e_n \rangle_{w,A} + \langle r_k, u_n e_n \rangle_{w,A} \\
&= -u_k \langle r_k, e_n \rangle_{w,A} + u_n \langle r_k, e_n \rangle_{w,A}.
\end{aligned}
\qquad (A.6)$$

For $n = k$ (A.6) equals zero, and the same holds for all $n \ne k$ due biorthogonality. Therefore $\langle T^* r_k - u_k r_k, e_n \rangle_{w,A} = 0$ for all $n \in \mathbb{N}$, hence

$$T^* r_k - u_k \cdot r_k = 0.$$

4. For any $n \in \mathbb{N}$ we have

$$\langle T f, r_n \rangle_{w,A} = \langle f, T^* r_n \rangle_{w,A} = \langle f, u_n r_n \rangle_{w,A}.$$

If $Tf = 0$ then $\langle f, r_n \rangle_{w,A} = 0$ for all $n \in \mathbb{N}$. The completeness of the family $\{r_n\}_{n=1}^\infty$ in the space $[\overline{\mathrm{span}}(M_\Lambda)]_{w,A}$ means that $f = 0$. Hence the kernel of $T$ is the zero function.

5. Suppose now that $Tf = \lambda f$ for some $\lambda \notin \{u_n\}_{n=1}^\infty$ and $f \ne 0$. Then,

$$\begin{aligned}
\lambda \langle f, r_n \rangle_{w,A} &= \langle Tf, r_n \rangle_{w,A}, \\
&= \langle f, T^* r_n \rangle_{w,A}, \\
&= \langle f, u_n r_n \rangle_{w,A}, \\
&= u_n \langle f, r_n \rangle_{w,A}.
\end{aligned}
\qquad (A.7)$$

Thus,

$$(\lambda - u_n) \cdot \langle f, r_n \rangle_{w,A} = 0 \quad \text{for all } n \in \mathbb{N}.$$
Since $\lambda \notin \{u_n\}_{n=1}^{\infty}$ then $\langle f, r_n \rangle_{w,A} = 0$ for all $n \in \mathbb{N}$, and the completeness of the family $\{r_n\}_{n=1}^{\infty}$ in $[\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$ means that $f = 0$. We conclude that the $u_k$'s are the only non-zero eigenvalues of $T$ and form its compactness it follows that the spectrum of $T$ is

$$ \{0\} \cup \{u_k\}_{k=1}^{\infty}. $$

6. Suppose now that $Tf = u_k f$ for some $u_k \in \{u_n\}_{n=1}^{\infty}$ and some $f \in [\overline{\operatorname{span}}(M_\Lambda)]_{w,A}$. Then, using the same computation as in (A.7) above, we get

$$ (u_k - u_n) \langle f, r_n \rangle_{w,A} = 0 \quad \text{for all} \quad n \in \mathbb{N}. $$

But $u_n \neq u_k$ if $n \neq k$, thus $\langle f, r_n \rangle_{w,A} = 0$ for all $n \neq k$. Since

$$ f(x) = \sum_{n=1}^{\infty} \langle f, r_n \rangle_{w,A} \cdot e_n(x), $$

we get $f(t) = \langle f, r_k \rangle_{w,A} e_k$, meaning that every $u_k$ is simple.

7. Finally, suppose that $T$ is a normal operator, thus

$$ TT^*(e_k) = T^*T(e_k) \quad \text{for all} \quad k \in \mathbb{N}. $$

Any two eigenvectors that correspond to different eigenvalues of a normal operator are orthogonal. Clearly the set of eigenvectors $e_n = x^{\lambda_n}$ of $T$ is not orthogonal, therefore $T$ is not normal.

This concludes the proof of Lemma A.1. $\square$

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