# On exact systems $t \alpha \bullet e^{2\pi i n} \in \mathbb{Z}$ in $L_2(0, 1)$ which are not Schauder Bases and their generalizations

Elias Zikkos

► To cite this version:

Elias Zikkos. On exact systems $t \alpha \bullet e^{2\pi i n} \in \mathbb{Z}$ in $L_2(0, 1)$ which are not Schauder Bases and their generalizations. 2024. hal-04633820

HAL Id: hal-04633820

https://hal.science/hal-04633820v1

Preprint submitted on 3 Jul 2024

**HAL** is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
# On exact systems {$t^\alpha \cdot e^{2\pi int}$}$_{n \in \mathbb{Z} \setminus A}$ in $L^2(0, 1)$ which are not Schauder Bases and their generalizations

Elias Zikkos

Khalifa University of Science and Technology

Abu Dhabi, United Arab Emirates

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

June 20, 2024

## Abstract

Let {$e^{i\lambda_n t}$}$_{n \in \mathbb{Z}}$ be an exponential Schauder Basis for $L^2(0, 1)$, for $\lambda_n \in \mathbb{R}$, and let {$r_n(t)$}$_{n \in \mathbb{Z}}$ be its dual Schauder Basis. Let $A$ be a non-empty subset of the integers containing exactly $M$ elements. We prove that for $\alpha > 0$ the weighted system

$$ \{t^\alpha \cdot r_n(t)\}_{n \in \mathbb{Z} \setminus A} $$

is exact in the space $L^2(0, 1)$, that is, it is complete and minimal in $L^2(0, 1)$, if and only if

$$ M - \frac{1}{2} \le \alpha < M + \frac{1}{2}. $$

We also show that such a system is not a Riesz Basis for $L^2(0, 1)$. In particular, the weighted trigonometric system {$t^\alpha \cdot e^{2\pi int}$}$_{n \in \mathbb{Z} \setminus A}$ is exact in $L^2(0, 1)$, if and only if $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$, but it is not a Schauder Basis for $L^2(0, 1)$.

Keywords: Weighted Exact Systems, Completeness and Minimality, Schauder Bases, Riesz Bases, Vandermonde matrices.

AMS 2010 Mathematics Subject Classification. 42C30, 42A65

# 1 Introduction and our Result

In [2] Heil and Yoon investigated the properties of systems of weighted exponentials of the form {$g(t) \cdot e^{2\pi int}$}$_{n \in \mathbb{Z} \setminus A}$ where $g \in L^2(0, 1) \setminus \{0\}$ and $A$ is a non-empty finite subset of the integers containing exactly $M$ elements. Amongst their results, they prove in [2, Theorem 4] that the system

$$ \{t^J \cdot e^{2\pi int}\}_{n \in \mathbb{Z} \setminus A}, $$

where $J$ is a positive integer, is complete and minimal in $L^2(0, 1)$ if and only if $J = M$.

They also raise the interesting question [2, page 100] whether similar results hold for the system

$$ \{t^\alpha \cdot e^{2\pi int}\}_{n \in \mathbb{Z} \setminus A} \qquad (1.1) $$

where $\alpha$ is a real positive number, not necessarily an integer.

Motivated by the above and the work of Shukurov [6], we provide the following answer: we prove that when the cardinality $|A|$ of the set $A$ is equal to $M$, then the system (1.1) is complete and minimal in $L^2(0, 1)$ if and only if

$$ M - \frac{1}{2} \le \alpha < M + \frac{1}{2}. \qquad (1.2) $$

We also show that such a complete and minimal system is not a Schauder Basis for $L^2(0, 1)$.
**Example 1.1.** The system {$t^\alpha \cdot e^{2\pi i n t}$}$_{n \in \mathbb{Z} \setminus A}$ is complete and minimal in $L^2(0,1)$ if

$$|A| = 1, \quad \text{and} \quad \frac{1}{2} \le \alpha < \frac{3}{2},$$

$$|A| = 2, \quad \text{and} \quad \frac{3}{2} \le \alpha < \frac{5}{2},$$

$$|A| = 3, \quad \text{and} \quad \frac{5}{2} \le \alpha < \frac{7}{2},$$

and so on, but no such system is a Schauder Basis for $L^2(0,1)$.

We point out that the condition (1.2) is necessary and sufficient in order for more general systems of the form {$t^\alpha \cdot r_n(t)$}$_{n \in \mathbb{Z} \setminus A}$ to be complete and minimal in $L^2(0,1)$, where {$r_n$}$_{n \in \mathbb{Z}}$ is the dual Schauder basis for an exponential Schauder basis {$e^{i\lambda_n t}$}$_{n \in \mathbb{Z}}$ for $L^2(0,1)$. That is the content of our result which reads as follows.

**Theorem 1.1.** Let {$e^{i\lambda_n t}$}$_{n \in \mathbb{Z}}$ be an exponential Schauder Basis for $L^2(0,1)$, for $\lambda_n \in \mathbb{R}$, and let {$r_n$}$_{n \in \mathbb{Z}}$ be its Dual Schauder Basis for $L^2(0,1)$. Let $A \subset \mathbb{Z}$ so that $|A| = M$ for some positive integer $M$. Then the system

$$\{t^\alpha \cdot r_n(t)\}_{n \in \mathbb{Z} \setminus A} \qquad (1.3)$$

is complete and minimal in the space $L^2(0,1)$, if and only if $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$.

However, for such values of $\alpha$ the system (1.3) is not a Riesz Basis for $L^2(0,1)$ and neither a frame. In particular, the weighted trigonometric system {$t^\alpha \cdot e^{2\pi i n t}$}$_{n \in \mathbb{Z} \setminus A}$ is complete and minimal in the space $L^2(0,1)$, if and only if $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$, but it is not even a Schauder Basis for $L^2(0,1)$.

We point out that under the assumption that the elements of the set $A$ are consecutive integers, the non-basicity of the system (1.1) follows from the work of Kazarian [4, Corollary 1]. In such a case, a complete and minimal system (1.1) is a Markushevich Basis [5, Theorem 1].

**Remark 1.1.** The proof of Theorem 1.1 depends on the invertibility of Vandermonde matrices.

## 2 Preliminaries

For the terminology presented below, one may consult the books of Young [7], Heil [3], and Christensen [1].

Let $\mathcal{H}$ be a separable Hilbert space endowed with an inner product $\langle \cdot, \cdot \rangle$ and a norm $||\cdot||$. Let $F := \{f_n\}_{n \in J}$ be a countable family of vectors in $\mathcal{H}$ with $J \subset \mathbb{Z}$. We say that $F$ is complete if the closed span of $F$ in $\mathcal{H}$ is equal to $\mathcal{H}$ and we say that $F$ is minimal if each element $f_n$ does not belong to the closed span of the remaining vectors of $F$ in $\mathcal{H}$. If $F$ is both complete and minimal then it is called **exact**. It is known that the minimality of $F$ is equivalent to the existence of a biorthogonal sequence $\{g_n\}_{n \in J}$ in $\mathcal{H}$, that is

$$\langle f_n, g_m \rangle = \begin{cases} 1, & n = m \\ 0, & n \neq m \end{cases}.$$

This biorthogonal family $\{g_n\}_{n \in J}$ is unique if $F$ is exact, but $\{g_n\}_{n \in J}$ itself does not have to be exact. If both $F$ and $\{g_n\}_{n \in J}$ are exact, we say that they are Markushevich bases for $\mathcal{H}$.

An exact system $F$ is a Schauder Basis for $\mathcal{H}$, if for any $f \in \mathcal{H}$, there exist unique numbers $d_{f,n}$ so that

$$f = \sum_{n \in J} d_{f,n} f_n,$$

with convergence in the $L^2(0,1)$ norm. In this case, the unique biorthogonal family $\{g_n\}_{n \in J}$ is also a Schauder Basis for $\mathcal{H}$, called the dual Schauder basis of $F$, and for any $f \in \mathcal{H}$ we have

$$f = \sum_{n \in J} \langle f, g_n \rangle f_n = \sum_{n \in J} \langle f, f_n \rangle g_n,$$
with convergence in the $L^2(0, 1)$ norm ([3, Theorem 5.12 and Corollary 5.22] and [2, relation (1)]). A Schauder basis may converge conditionally, whereas a Riesz Basis for $\mathcal{H}$ is a bounded Schauder basis that converges unconditionally ([3, Theorem 7.11]). We recall that $\{f_n\}_{n \in J}$ is a Riesz basis for $\mathcal{H}$ if $f_n = V(e_n)$ for all $n \in J$ where $\{e_n\}_{n \in J}$ is an orthonormal basis for $\mathcal{H}$ and $V$ is a bounded bijective operator from $\mathcal{H}$ onto $\mathcal{H}$.

# 3 Proof of Theorem 1.1

First we will show that the system (1.3) is complete in $L^2(0, 1)$ if $\alpha \ge M - \frac{1}{2}$ and then show that it is minimal if $\alpha < M + \frac{1}{2}$. We then prove that for $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$ the exact system (1.3) is not a Riesz basis for $L^2(0, 1)$, while the exact system (1.1) is not even a Schauder Basis for $L^2(0, 1)$.

## 3.1 If $\alpha \ge M - \frac{1}{2}$, the system (1.3) is Complete

Since $\{r_n(t)\}_{n \in \mathbb{Z}}$ is the dual Schauder Basis of $\{e^{i\lambda_n t}\}_{n \in \mathbb{Z}}$ for $L^2(0, 1)$, then for any $f \in L^2(0, 1)$ we have

$$f(t) = \sum_{n \in \mathbb{Z}} \langle f, r_n \rangle e^{i\lambda_n t}$$

with the series converging in the $L^2(0, 1)$ norm.

For notation purposes, let $t_\alpha(t) := t^\alpha$. Now, if the system (1.3) is not complete in $L^2(0, 1)$, then there exists a non-zero function $h(t) \in L^2(0, 1)$ so that

$$\langle h, t_\alpha r_n \rangle = 0, \quad \text{for all } n \in \mathbb{Z} \setminus A.$$

Since $|t^\alpha| \le 1$ for $t \in [0, 1]$, then $ht_\alpha$ belongs to $L^2(0, 1)$. Combining the above shows that

$$h(t) \cdot t^\alpha = \sum_{n \in A} \langle h t_\alpha, r_n \rangle e^{i \lambda_n t},$$

almost everywhere on $[0, 1]$, and such that at least one of the coefficients $\langle h t_\alpha, r_n \rangle$ for $n \in A$ is not equal to zero.

Rewrite now the set $\{\lambda_i : i \in A\}$ as $\{\Lambda_1, \Lambda_2, \dots, \Lambda_M\}$ such that $\Lambda_1 < \Lambda_2 < \dots < \Lambda_M$. Then

$$h(t) \cdot t^\alpha = \sum_{j=1}^{M} a_j e^{i \Lambda_j t}, \quad a_j \in \mathbb{C}.$$

The analytic extension to the complex plane of the above right hand side is the entire function

$$H_{\alpha}(z) := \sum_{j=1}^{M} a_j e^{i \Lambda_j z}, \qquad (3.1)$$

so $H_\alpha(t) = h(t) \cdot t^\alpha$ for $t \in [0, 1]$. We will prove below that $H_\alpha$ cannot vanish $M$ times or more at the point $z = 0$. Suppose it does: then

$$0 = H_{\alpha}(0) = H'_{\alpha}(0) = H''_{\alpha}(0) = \dots H^{(M-1)}_{\alpha}(0) \qquad (3.2)$$

where $H_\alpha^{(k)}(z)$ is the $k$th derivative of $H_\alpha$ with respect to the variable $z$. From (3.1) and (3.2), we get

$$\begin{aligned} 0 &= a_1 + a_2 + a_3 + a_4 + \dots + a_M \\ &= a_1 \cdot \Lambda_1 + a_2 \cdot \Lambda_2 + a_3 \cdot \Lambda_3 + a_4 \cdot \Lambda_4 + \dots + a_M \cdot \Lambda_M \\ &= a_1 \cdot \Lambda_1^2 + a_2 \cdot \Lambda_2^2 + a_3 \cdot \Lambda_3^2 + a_4 \cdot \Lambda_4^2 + \dots + a_M \cdot \Lambda_M^2 \end{aligned}$$

and so on, until the last equation $(0 = H_\alpha^{(M-1)}(0))$

$$0 = a_1 \cdot \Lambda_1^{M-1} + a_2 \cdot \Lambda_2^{M-1} + a_3 \cdot \Lambda_3^{M-1} + a_4 \cdot \Lambda_4^{M-1} + \dots + a_M \cdot \Lambda_M^{M-1}.$$
We can write this system of equations as

$$ \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ \Lambda_1 & \Lambda_2 & \Lambda_3 & \Lambda_4 & \cdots & \Lambda_M \\ \Lambda_1^2 & \Lambda_2^2 & \Lambda_3^2 & \Lambda_4^2 & \cdots & \Lambda_M^2 \\ \Lambda_1^3 & \Lambda_2^3 & \Lambda_3^3 & \Lambda_4^3 & \cdots & \Lambda_M^3 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \Lambda_1^{M-1} & \Lambda_2^{M-1} & \Lambda_3^{M-1} & \Lambda_4^{M-1} & \cdots & \Lambda_M^{M-1} \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ a_4 \\ \vdots \\ a_M \end{pmatrix}. $$

The above square matrix is an invertible Vandermonde matrix since the $\Lambda_i$'s are all different. Clearly now we have $a_j = 0$ for all $j = 1, 2, 3, \dots, M$, thus $H_\alpha(z)$ is the Zero function. But $H_\alpha(t) = h(t) \cdot t^\alpha$ for $t \in [0, 1]$, thus $h(t)$ is identically equal to zero, a contradiction. Therefore $H_\alpha(z)$ vanishes at the point $z = 0$ at most $M-1$ times, hence $H_\alpha(z) = z^k \cdot g(z)$ for some integer $k$ in the set $\{0, 1, \dots, M-1\}$ and some entire function $g$ such that $g(0) \neq 0$. Since $H_\alpha(t) = h(t) \cdot t^\alpha$ for $t \in [0, 1]$, we have

$$ h(t) \cdot t^{\alpha} = t^{k} \cdot g(t), \quad \text{for some } k \in \{0, 1, 2, \dots, M-1\}, \quad g \in C[0, 1], \quad g(0) \neq 0. $$

Thus

$$ |h(t)|^2 = |g(t)|^2 \cdot t^{2k-2\alpha}. $$

Since $g(0) \neq 0$, then there are $\epsilon > 0$ and $\delta > 0$, so that $|g(t)| > \epsilon$ for all $t \in [0, \delta]$. So, for any $0 < \rho < \delta$, one has $|h(t)|^2 > \epsilon^2 \cdot t^{2k-2\alpha}$ for all $t \in [\rho, \delta]$. Therefore,

$$ \int_{\rho}^{\delta} |h(t)|^2 dt > \epsilon^2 \cdot \int_{\rho}^{\delta} t^{2k-2\alpha} dt. $$

Since $k \le M-1$ and $\alpha \ge M - \frac{1}{2}$, then $2k - 2\alpha \le -1$ and it follows that

$$ \int_{\rho}^{\delta} t^{2k-2\alpha} dt \to \infty \quad \text{as} \quad \rho \to 0^{+}. $$

Hence $\int_0^1 |h(t)|^2 dt = \infty$ which contradicts the assumption that $h \in L^2(0, 1)$. Therefore, if $\alpha \ge M - \frac{1}{2}$ then the system (1.3) is complete in $L^2(0, 1)$.

## 3.2 If $\alpha < M + \frac{1}{2}$, the system (1.3) is Minimal

Like before, the set $\{\lambda_i : i \in A\}$ is rewritten as $\{\Lambda_1, \Lambda_2, \dots, \Lambda_M\}$ such that $\Lambda_1 < \Lambda_2 < \dots < \Lambda_M$.

We will show that for each $n \in \mathbb{Z} \setminus A$, there exist real numbers $a_{n,j}$ for $j = 1, 2, \dots, M$, so that the entire function

$$ f_n(z) := e^{i\lambda_n z} + a_{n,1}e^{i\Lambda_1 z} + a_{n,2}e^{i\Lambda_2 z} + \dots + a_{n,M}e^{i\Lambda_M z} \quad (3.3) $$

vanishes at the point $z = 0$ at least $M$ times and

$$ f_n(t) = t^M \cdot g_n(t) \quad \text{for some continuous function } g_n \text{ on } [0, 1]. \quad (3.4) $$

The minimality of (1.3) follows from (3.3) and (3.4). Indeed, from (3.4) we have

$$ \frac{f_n(t)}{t^\alpha} = \frac{f_n(t)}{t^M} \cdot \frac{t^M}{t^\alpha} = g_n(t) \cdot \frac{t^M}{t^\alpha}. $$

So

$$ \left| \frac{f_n(t)}{t^\alpha} \right|^2 = |g_n(t)|^2 \cdot t^{2M-2\alpha}. $$

Since $g_n$ is continuous on $[0, 1]$ and $\alpha < M + \frac{1}{2}$, thus $2M - 2\alpha > -1$, it follows that $f_n(t)/t^\alpha$ belongs to $L^2(0, 1)$. Then, for all positive integers $n, m$ which are not in the set $A$, we have

$$ \langle \frac{f_n}{t^\alpha}, t_\alpha r_m \rangle = \int_0^1 \left( e^{i\lambda_n t} + a_{n,1}e^{i\Lambda_1 t} + a_{n,2}e^{i\Lambda_2 t} + \dots + a_{n,M}e^{i\Lambda_M t} \right) \cdot \overline{r_m} dt = \begin{cases} 1, & n = m \\ 0, & n \neq m \end{cases}. $$
Thus, the family ${\frac{f_n(t)}{t^\alpha}}_{n \in \mathbb{Z} \setminus A}$ is biorthogonal to the sequence ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus A}$. This means that the system (1.3) is minimal.

So, it remains to verify the existence of the numbers $a_{n,j}$ so that (3.4) is true. Now, if a trigonometric polynomial $f_n$ as in (3.3) vanishes at the point $z = 0$ at least $M$ times, then

$$0 = f_n(0) = f'_n(0) = f''_n(0) = \cdots f^{(M-1)}(0) \qquad (3.5)$$

where $f_n^{(k)}(z)$ is the $k$th derivative of $f_n$ with respect to $z$. From (3.3) and (3.5) we get

$$\begin{aligned} 0 &= 1 + a_{n,1} + a_{n,2} + a_{n,3} + \cdots + a_{n,M} \\ &= \lambda_n + a_{n,1} \cdot \Lambda_1 + a_{n,2} \cdot \Lambda_2 + a_{n,3} \cdot \Lambda_3 + \cdots + a_{n,M} \cdot \Lambda_M \\ &= \lambda_n^2 + a_{n,1} \cdot \Lambda_1^2 + a_{n,2} \cdot \Lambda_2^2 + a_{n,3} \cdot \Lambda_3^2 + \cdots + a_{n,M} \cdot \Lambda_M^2 \end{aligned}$$

and so on, until the last equation $(0 = f_n^{(M-1)}(0))$

$$0 = \lambda_n^{M-1} + a_{n,1} \cdot \Lambda_1^{M-1} + a_{n,2} \cdot \Lambda_2^{M-1} + a_{n,3} \cdot \Lambda_3^{M-1} + \cdots + a_{n,M} \cdot \Lambda_M^{M-1}.$$

We rewrite this system of equations as

$$\begin{pmatrix} -1 \\ -\lambda_n \\ -\lambda_n^2 \\ \vdots \\ -\lambda_n^{M-1} \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ \Lambda_1 & \Lambda_2 & \Lambda_3 & \cdots & \Lambda_M \\ \Lambda_1^2 & \Lambda_2^2 & \Lambda_3^2 & \cdots & \Lambda_M^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \Lambda_1^{M-1} & \Lambda_2^{M-1} & \Lambda_3^{M-1} & \cdots & \Lambda_M^{M-1} \end{pmatrix} \cdot \begin{pmatrix} a_{n,1} \\ a_{n,2} \\ a_{n,3} \\ \vdots \\ a_{n,M} \end{pmatrix}. \qquad (3.6)$$

Since the $\Lambda_i$'s are all different, then the above square Vandermonde matrix is invertible. Therefore, real numbers $a_{n,j}$, $j = 1, 2, \cdots, M$ do exist so that $f_n$ vanishes at $z = 0$ at least $M$ times. Thus

$$f_n(z) = z^M \cdot g_n(z) \quad \text{for some entire function } g_n(z),$$

hence (3.4) holds.

### 3.3 The system (1.3) is Exact if and only if $M - \frac{1}{2} \le \alpha < M + \frac{1}{2}$

If $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$, then the system (1.3) is both complete and minimal in $L^2(0, 1)$, hence it is exact.

Assume now that the system (1.3) is exact: we will show that $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$.

Suppose first that it is exact but $\alpha < M - \frac{1}{2}$, say $M - \frac{3}{2} \le \alpha < M - \frac{1}{2}$. Then, if $B \subset \mathbb{Z}$ such that $B = A \setminus \{\tau\}$ where $\tau$ is an integer not in the set $A$, then the cardinality of $B$ equals $M - 1$. Hence the system ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus B}$ is exact. So the system ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus A}$ will have excess equal to $-1$ instead of zero. Similarly for other positive values of $\alpha$ when $\alpha < M - \frac{3}{2}$. So $\alpha$ cannot be less than $M - \frac{1}{2}$.

Suppose now that it is exact but $\alpha \ge M + \frac{1}{2}$, say $M + \frac{1}{2} \le \alpha < M + \frac{3}{2}$. Then, if $C \subset \mathbb{Z}$ such that $C = A \cup \{\tau\}$ where $\tau$ is an integer not in the set $A$, then the cardinality of $C$ equals $M + 1$. Hence the system ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus C}$ is exact. So the system ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus A}$ will have excess equal to $+1$ instead of zero. Similarly for other positive values of $\alpha$ when $\alpha \ge M + \frac{3}{2}$. So $\alpha$ cannot be greater than or equal to $M + \frac{1}{2}$.

In this way we conclude that if the system (1.3) is exact, then $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$.

### 3.4 The exact system (1.3) is not a Riesz Basis for $L^2(0, 1)$

Suppose that for $\alpha \in [M - \frac{1}{2}, M + \frac{1}{2})$ the exact system ${t^\alpha \cdot r_n(t)}_{n \in \mathbb{Z} \setminus A}$ is a Riesz Basis for $L^2(0, 1)$ and hence a Schauder Basis as well. Let $f_n$ be as in (3.3): since the family ${f_n/t_\alpha}_{n \in \mathbb{Z} \setminus A}$ is biorthogonal
to the Schauder Basis (1.3), then {$f_n/t_\alpha$}$_{n \in \mathbb{Z} \setminus A}$ is the dual Schauder basis, hence for any $f \in L^2(0, 1)$, we have

$$f(t) = \sum_{n \in \mathbb{Z} \setminus A} \langle f, \frac{f_n}{t_\alpha} \rangle \cdot t^\alpha r_n(t),$$

with convergence in the $L^2(0, 1)$ norm. This implies that $|\langle f, \frac{f_n}{t_\alpha} \rangle| \cdot ||t_\alpha r_n|| \to 0$ as $n \to \infty$.

In particular, for each integer $m$ belonging to the set $A$ which has been excluded from $\mathbb{Z}$, we have

$$t^{\alpha} \cdot r_m(t) = \sum_{n \in \mathbb{Z} \setminus A} \langle t_{\alpha} r_m, \frac{f_n}{t_{\alpha}} \rangle \cdot t^{\alpha} r_n(t) \quad (3.7)$$

with convergence in the $L^2(0, 1)$ norm. Fix such an $m \in A$. Then

$$|\langle t_{\alpha} r_m, \frac{f_n}{t_{\alpha}} \rangle| \cdot ||t_{\alpha} r_n|| \to 0 \quad \text{as} \quad n \to \infty. \quad (3.8)$$

In (3.3) we have for every $n \in \mathbb{Z} \setminus A$, $f_n(t) = e^{i\lambda_n t} + a_{n,1}e^{i\Lambda_1 t} + a_{n,2}e^{i\Lambda_2 t} + \dots + a_{n,M}e^{i\Lambda_M t}$. Since $m \in A$ and $\{\lambda_i : i \in A\}$ is rewritten as $\{\Lambda_1, \Lambda_2, \dots, \Lambda_M\}$ such that $\Lambda_1 < \Lambda_2 < \dots < \Lambda_M$, then $\lambda_m = \Lambda_j$ for some $j \in \{1, 2, \dots, M\}$, and $e^{i\lambda_m t} = e^{i\Lambda_j t}$. Since the family $\{r_n\}_{n \in \mathbb{Z}}$ is biorthogonal to the family $\{e^{i\lambda_n t}\}_{n \in \mathbb{Z}}$, we get

$$\langle t_{\alpha} r_m, \frac{f_n}{t_{\alpha}} \rangle = \overline{a_{n,j}} \quad \text{for all } n \in \mathbb{Z} \setminus A.$$

Replacing into (3.8) we have

$$|a_{n,j}| \cdot ||t_{\alpha} r_n|| \to 0 \quad \text{as} \quad n \to \infty. \quad (3.9)$$

Now, if the system (1.3) is a Riesz basis for $L^2(0, 1)$, then there are positive constants $p$ and $q$ so that $p < ||t_\alpha r_n|| < q$ for all $n \in \mathbb{Z} \setminus A$ ([1, Lemma 3.6.9]). We will show that there is a positive number $\delta$, independent of $n \in \mathbb{Z} \setminus A$ and $j \in \{1, 2, \dots, M\}$, so that

$$\delta \cdot |\lambda_n|^{M-1} \le |a_{n,j}| \quad \text{finally for all } n \in \mathbb{Z} \setminus A, \text{ and } j \in \{1, 2, \dots, M\}. \quad (3.10)$$

But then, combining the above with $p < ||t_\alpha r_n|| < q$ for all $n \in \mathbb{Z} \setminus A$, gives

$$p \cdot \delta \cdot |\lambda_n|^{M-1} \le |a_{n,j}| \cdot ||t_\alpha r_n|| \quad \text{for all } n \in \mathbb{Z} \setminus A, \text{ and } j \in \{1, 2, \dots, M\}.$$

However, $M - 1 \ge 0$ and $|\lambda_n| \to \infty$ as $n \to \infty$, thus the above shows that (3.9) is not true. Hence the exact system (1.3) is not a Riesz Basis for $L^2(0, 1)$.

So it remains to verify (3.10). Returning to the relation (3.6), consider the invertible Vandermonde matrix

$$\Lambda := \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ \Lambda_1 & \Lambda_2 & \Lambda_3 & \cdots & \Lambda_M \\ \Lambda_1^2 & \Lambda_2^2 & \Lambda_3^2 & \cdots & \Lambda_M^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \Lambda_1^{M-1} & \Lambda_2^{M-1} & \Lambda_3^{M-1} & \cdots & \Lambda_M^{M-1} \end{pmatrix}$$

and denote by $\Lambda^{-1}$, $det(\Lambda)$, and $adj(\Lambda)$, the inverse, determinant and adjoint of $\Lambda$ respectively. Then we have

$$\begin{pmatrix} -1 \\ -\lambda_n \\ -\lambda_n^2 \\ \vdots \\ -\lambda_n^{M-1} \end{pmatrix} = \Lambda \cdot \begin{pmatrix} a_{n,1} \\ a_{n,2} \\ a_{n,3} \\ \vdots \\ a_{n,M} \end{pmatrix}$$
and

$$ 
\begin{pmatrix} a_{n,1} \\ a_{n,2} \\ a_{n,3} \\ \vdots \\ a_{n,M} \end{pmatrix} = \Lambda^{-1} \cdot \begin{pmatrix} -1 \\ -\lambda_n \\ -\lambda_n^2 \\ \vdots \\ -\lambda_n^{M-1} \end{pmatrix} \\
= \frac{1}{\det(\Lambda)} \cdot \mathrm{adj}(\Lambda) \cdot \begin{pmatrix} -1 \\ -\lambda_n \\ -\lambda_n^2 \\ \vdots \\ -\lambda_n^{M-1} \end{pmatrix}.
$$

We now claim that all the entries of the far right column of $adj(\Lambda)$ are non-zero real numbers. So we need to check for the cofactors of the entries of the bottom row of $\Lambda$. For example, for the entry $\Lambda_2^{M-1}$, its cofactor is equal to $(-1)^{2+M}$ times the determinant of the matrix that remains when we delete the second column and the last row. That matrix is

$$ 
\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ \Lambda_1 & \Lambda_3 & \Lambda_4 & \cdots & \Lambda_M \\ \Lambda_1^2 & \Lambda_3^2 & \Lambda_4^2 & \cdots & \Lambda_M^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \Lambda_1^{M-2} & \Lambda_3^{M-2} & \Lambda_4^{M-2} & \cdots & \Lambda_M^{M-2} \end{pmatrix}.
$$

Clearly this is again an invertible Vandermonde matrix. The same is true for the cofactors of the other elements of the last row of $\Lambda$, thus we conclude that all the entries of the far right column of $adj(\Lambda)$ are non-zero real numbers. This means that for all $n \in \mathbb{Z} \setminus A$ and all $j = 1, 2, \dots, M$, we have

$$ a_{n,j} = \sum_{k=1}^{M-1} c_{j,k} (-\lambda_n)^{k-1} + c_{j,M} (-\lambda_n)^{M-1} $$

where all the $c_{j,k}$ are real numbers, and especially $c_{j,M} \neq 0$ for all $j = 1, 2, \dots, M$. This implies the existence of some positive $\delta$ so that (3.10) holds.

## 3.5 The exact system (1.1) is not a Schauder Basis for $L^2(0, 1)$

If the exponential Schauder basis for $L^2(0, 1)$ is the family $\{e^{i2\pi nt}\}_{n\mathbb{Z}}$, then its dual Schauder basis is the same family, thus $r_n = e^{i2\pi nt}$ for all $n \in \mathbb{Z}$. In this case, we have $||t_\alpha r_n|| = 1$, thus (3.9) becomes

$$ |a_{n,j}| \to 0 \quad \text{as} \quad n \to \infty. $$

Clearly this is false due to (3.10), therefore the exact system (1.1) is not a Schauder Basis for $L^2(0, 1)$. The proof of Theorem 1.1 is now finished.

## References

[1] O. Christensen, An introduction to Frames and Riesz Bases, Second Edition, Applied and Numerical Harmonic Analysis, Birkhäuser, Boston, (2016).

[2] C. A. Heil, G. J. Yoon, Duals of weighted exponential systems, Acta Appl. Math. **119** (2012), 97–112.

[3] C. A. Heil, Basis theory primer, Expanded Edition, Applied and Numerical Harmonic Analysis, Birkhäuser/Springer, New York, (2011).
[4] K. S. Kazarian, On the multiplicative completion of some incomplete orthonormal systems to bases in $L^p$, $1 \le p < \infty$, Analysis Math. **4** no. 1 (1978), 37–52.

[5] K. S. Kazarian, Summability of generalized Fourier series and Dirichlet's problem in $L^p(d\mu)$ and weighted $H^p$-spaces ($p > 1$), Analysis Math. **13** (1987) 173–197.

[6] A. S. Shukurov, Basis of the properties of weighted exponential systems with excess, Vestn. Samar. Univ. Estestvennonauchn. Ser. **24** no. 1 (2018), 14–19.

[7] R. M. Young, An introduction to Nonharmonic Fourier Series, Revised first edition. Academic Press, Inc., San Diego, CA, (2001) xiv+234 pp. ISBN: 0-12-772955-0.