Traces of certain integral operators related to the Riemann hypothesis

1.   Introduction

Let $B(H)$ be the algebra of all bounded linear operators on a separable complex Hilbert space $H$. The adjoint of an operator $T\in B(H)$ is denoted by $T^{*}$ and the symbol $I$ stands for identity maps. Denote by $\{s_n(T)\}_{n\geq 1}$ the sequence of singular values of a compact operator $T\in B(H)$. If $0 < p < \infty$ we say that $T\in S^{p}(H)$ if ${\sum^{\infty}_{n = 1}s^{p}_{n}(T) < \infty}$; the set $S^{p}(H)$ is a two-side ideal in $B(H)$. The sets $S^{1}(H)$ and $S^{2}(H)$ will denote, respectively, the set of nuclear and Hilbert-Schmidt operators. By an ideal we mean a two-sided ideal in $B(H)$. A linear functional $\tau$ from the ideal $J$ into $\mathbb{C}$ is said to be a trace if:

$i)$ $\tau(U^{*}TU) = \tau(T)$ for every $T\in J$ and $U\in B(H)$ unitary. Equivalently, $\tau(ST) = \tau(TS)$ for every $T\in J$ and $S\in B(H)$.

$ii)$ $\tau(T)\geq 0$ for every $T\in J$ with $T$ a non-negative operator. We denote by $T\geq 0$ when $T$ is a non-negative operator.

Then a trace is a positive unitarily invariant linear functional.

Let $T\in S^{1}(H)$, and let $\{\varphi_n\}_{n\geq 1}$ be an orthonormal system in $H$. By [13, p. 56], we have

Therefore,

is well-defined, since the right-hand series converges absolutely, and its value does not depend on the choice of the orthonormal basis $\{\varphi_n\}_{n\geq 1}$. Clearly, this linear functional is a trace on the ideal $S^{1}(H)$. There is also the description of Tr as the sum of eigenvalues,

where $\{\lambda_n(T)\}_{n\geq 1}$ is the sequence of nonzero eigenvalues of $T$, ordered in such a way that $\left\lvert{\lambda_n(T)}\right\rvert\geq \left\lvert{\lambda_{n+1}(T)}\right\rvert$, $\forall\, n\in \mathbb{N}$ and each one of them being counted according to its algebraic multiplicity. This result was shown by Von Neumann in ^[26] for self-adjoint operators and by Lidskii in ^[19] in general case. Formula (1.1) is called the Lidskii formula.

A natural question concerning the extension of the Lidskii formula to other ideals and traces on these ideals has been addressed in ^{[6,13,20,21,22]}. Thus, the notion of spectral traces arises. A trace $\tau$ on an ideal $J$ is called spectral if for every $T\in J$, the value of $\tau(T)$ depends only on the eigenvalues of $T$ and their multiplicities. For example, the classical trace Tr on the ideal $S^{1}(H)$ is spectral. Motivated by the problem of identifying spectral traces, it is shown in [12, Corollary 2.4] that every trace on a geometrically stable ideal is spectral. This result has been generalized in the setting of ideals closed with respect to the logarithmic submajorization (see ^[24]).

A trace $\tau$ on an ideal $J$ will be called singular if it vanishes on the set $\mathcal{F}(H)$ of finite rank operators. This definition makes sense, since by the Calkin Theorem ^[9], each proper ideal in $B(H)$ contains the finite rank operators and is contained in the ideal $K(H)$ of the compact linear operators on $H$.

In 1966, J. Dixmier proved the existence of singular traces ^[11]. These traces are called Dixmier traces, and its importance is due to their applications in noncommutative geometry ^[10]. Other examples of singular traces appeared in ^[1,17,25].

The question whether an operator belongs to the domain of some singular trace was answered in ^[15]. Motivated by this, we give a nontrivial singular trace taking the value of zero on certain integral operator related to the Riemann hypothesis.

In ^[2], J. Alcántara-Bode has reformulated the Riemann hypothesis as a problem of functional analysis by means of the following theorem.

Theorem 1.1. Let ${(A_\rho f)(\theta) = \int^{1}_{0}\rho\left(\frac{\theta}{x}\right)f(x)dx}$, where $\rho$ is the fractional part function, be considered as an operator on $L^{2}(0, 1)$. Then the Riemann hypothesis holds if and only if ${\rm{Ker}}(A_\rho) = \{0\}$, or if and only if $h\notin {\rm{Ran}}\, (A_\rho)$ where $h(x) = x$.

It follows from Theorem 1.1 that the problem of verifying the condition $h\in{\mbox{Ran}}\, (A_\rho)$ is ill posed in the sense of Hadamard[16, Definition 2.1.2]. This leads to regularizing the ill posed problem replacing $A_\rho$ by

Observe that $A_\rho(1) = A_{\rho}$. Since $A_\rho$ is nonnuclear (see [2, Theorem 6]), it was proved in [23, Theorem 4.3] that there exists a nontrivial singular trace $\tau$ with domain a geometrically stable ideal $\mathcal{J}$ such that $A_\rho\in \mathcal{J}$ and $\tau(A_\rho) = 0$. It follows from [24, Lemma 35] that $\mathcal{J}$ is closed with respect to the logarithmic submajorization.

Let $\mathcal{J}$ be as in the previous paragraph. We show that if $0 < \alpha\leq 1$ then $A_\rho(\alpha)\in \mathcal{J}$, and that each nontrivial singular trace on $\mathcal{J}$ takes the value of zero on $A_\rho(\alpha)$. More precisely, our first main result of the paper is the following theorem.

Theorem 1.2. If $0 < \alpha\leq 1$ then $\tau(A_\rho(\alpha)) = 0$ for every $\tau$ nontrivial singular trace on $\mathcal{J}$, where $\mathcal{J}$ is the geometrically stable ideal in the above paragraph.

The approach used to prove Theorem 1.2 is based on spectral traces and the fact that the operators $\alpha(A_\rho+Q_{f_1})$ and $A_\rho+Q_{f_\alpha}$, where $\alpha\in ]0, 1[$, have the same nonzero eigenvalues with the same algebraic and geometric multiplicities. Here $Q_f g = {\langle {g, f}\rangle}h$, $h(x) = x$, $f_\alpha(x) = -\alpha^{-1}\rho\left(\dfrac{\alpha}{x}\right)$.

Finally, our second main result is a recursion formula to calculate the traces ${\mbox{Tr}}\, (A^{r}_{\rho}(\alpha))$, $r\in \mathbb{N}$, $r\geq 2$.

Theorem 1.3. If $0 < \alpha\leq 1$ then for every $r\in \mathbb{N}$ with $r\geq 2$ we have

where

In order to prove Theorem 1.3, we use the modified Fredholm determinant of $I-uA_\rho(\alpha)$ (see ^[3]).

2.   Singular traces

Let $l^{\infty}$ the space of all bounded sequence of complex numbers and $w$ a dilation invariant extended limit on $l^{\infty}$, that is, $w$ is an extended limit on $l^{\infty}$ and

The Dixmier trace of $T\in M^{1, \infty}(H)$ with $T\geq 0$ is the number

where

It was shown in ^[11] that the weight ${\mbox{Tr}}_w$ defines a positive, unitarily invariant, additive and positive homogeneous function on the positive cone of $M^{1, \infty}(H)$, that can uniquely be extended to a singular trace on all of $M^{1\infty}(H)$, i.e., for an arbitrary $T\in M^{1, \infty}(H)$, its Dixmier trace is defined by

where $T = T_1-T_2+iT_3-iT_4$, $0\leq T_j\in M^{1, \infty}(H)$, $j = 1, 2, 3, 4$. In addition to this, the Dixmier trace vanishes on the ideal $S^{1}(H)$ and is continuous in the norm $\left\lVert{\cdot}\right\rVert_{1, \infty}$.

The existence of a singular trace which is nontrivial on a compact operator $T$, i.e., on the two-sided ideal generated by $T$,

was studied by J. Varga ^[25], and it has been completely characterized in ^[1]. This leads to study irregular, eccentric and generalized eccentric operators.

Definition 2.1. We say that a compact operator $T\in B(H)$ is

$a)$ regular if ${\sum\limits^{n}_{k = 1}s_k(T) = \mathcal{O}(ns_n(T))}$ ($n\to \infty$);

$b)$ irregular if it is not regular;

$c)$ eccentric if it is irregular but not nuclear;

$d)$ generalized eccentric if 1 is a limit point of the sequence $\left \{{\dfrac{S_{2n}(T)}{S_n(T)}}\right \}_{n\geq 1}$, where

Remark 2.2. By [25, Lemma 1], the class of generalized eccentric operators which are not nuclear coincides with the class of eccentric operators.

By [1, Lemma 2.6], that an operator is generalized eccentric can be reformulated as follows.

Lemma 2.3. Let $T\in B(H)$ be a compact operator. Then $T$ is generalized eccentric if and only if there exists an increasing sequence of natural numbers $\{p_n\}_{n\geq 1}$ such that ${\lim\limits_{k\to +\infty}\frac{S_{kp_k}(T)}{S_{p_k}(T)} = 1}$.

In this context, the main result in ^[1] is the following.

Theorem 2.4. Let $T\in B(H)$ be a compact operator. Then the following are equivalent:

a) There exists a singular trace $\tau$ such that $0 < \tau(\left\lvert{T}\right\rvert) < \infty$.

b) $T$ is generalized eccentric.

The process to construct the singular trace given by $a)$ is as follows:

We introduce a triple $\Omega = (T, w, \{n_k\}_{k\geq 1})$, where $T$ is a generalized eccentric operator, $w$ is an extended limit and $n_k = kp_k$, $k\in \mathbb{N}$, where $\{p_k\}_{k\geq 1}$ is the sequence given in Lemma 2.3. Associated with the triple $\Omega$, on the positive part of the ideal $(T)$, we defined the functional

and by [1, Theorem 2.11], this functional extends linearly to a singular trace on the ideal $(T)$.

Remark 2.5.

$a)$ By Theorem 2.4, finite rank operators cannot by generalized eccentric.

$b)$ The question whether an operator belongs to the domain of some singular trace is treated in ^[15]. By [15, Theorem 3.1 (i)], every compact operator $A$ is in the kernel (hence in the domain) of some singular trace. The main idea for proving this theorem is the existence of a generalized eccentric operator $B$ such that $A\in (B)_0\subset (B)$. Here $(B)_0$ denotes the kernel of $(B)$ (see ^[15]).

Example 2.6. Let $V:L^{2}(0, 1)\to L^{2}(0, 1)$ be the integral operator

By [14, p. 250], ${s_n(V) = \frac{4}{(2n-1)\pi}}$, $n = 1, 2, \ldots$. Therefore, by Remark 2.2, $V$ is a generalized eccentric operator. Let ${\mbox{Re}}(V)$ and ${\mbox{Im}}(V)$ be the real and imaginary parts of $V$, respectively. Then

and

By [14, p. 172], ${s_n({\rm{Re}}(V)) = \frac{2}{(2n-1)\pi}}$, $n = 1, 2, \ldots$, it follows that ${\rm{Re}}(V)$ is a generalized eccentric operator. However, the operator ${\rm{Im}}(V)$ has rank one, and by Remark 2.5 (a), ${\rm{Im}}(V)$ is not an eccentric operator.

3.   Commutator subspace

Now we concentrate on the commutator subspace of geometrically stable ideals and ideals closed with respect to the logarithmic submajorization, terminologies used in ^[18,24], respectively.

Definition 3.1. Let $J$ an ideal in $B(H)$. The subspace

where $[A, B] = AB-BA$, is called the commutator subspace of $J$.

Definition 3.2. An ideal $J$ of $B(H)$ is called geometrically stable if a diagonal operator ${\mbox{diag}}\{s_1, s_2, \ldots\}\in J$, where $s_1\geq s_2\geq \ldots\geq 0$, then ${\mbox{diag}}\{u_1, u_2, \ldots\}\in J$, where $u_n = (s_1s_2\ldots s_n)^{1/n}$.

The following theorem [18, Theorem 3.3] characterizes in terms of arithmetic means the commutator subspace of geometrically stable ideals.

Theorem 3.3. Suppose that $J$ is a geometrically stable ideal of $B(H)$. Then $T\in {\rm{Com}}(J)$ if and only if ${\rm{diag}}\{\frac{1}{n}(\lambda_1 (T)+\ldots +\lambda_n(T))\}\in J$.

In order to extend the previous result, ideals closed with respect to the logarithmic submajorization are introduced in ^[24].

Definition 3.4. If $A, C\in B(H)$, then the operator $C$ is logarithmically submajorized by the operator $A$ (written $C\prec\prec_{\log} A$) if

Definition 3.5. An ideal $J$ is said to be closed with respect to the logarithmic submajorization if $C\prec\prec_{\log}A\in J$ implies $C\in J$.

Remark 3.6. By [24, Lemma 35], every geometrically stable ideal is closed with respect to the logarithmic submajorization. However, by [24, Theorem 36 (c)], the converse assertion fails.

The following theorem [24, Theorem 7] extends Theorem 3.3 to ideals closed with respect to the logarithmic submajorization.

Theorem 3.7. Let an ideal $J$ be closed with respect to the logarithmic submajorization and let $T\in J$. Then $T\in {\rm{Com}}(J)$ if and only if ${\rm{diag}}\{\frac{1}{n}(\lambda_1(T)+\ldots +\lambda_n(T))\}\in J$.

A Lidskii-type formula holds for every trace defined on ideals closed with respect to the logarithmic submajorization (see [24, Theorem 8]), the statement is the following.

Theorem 3.8. Let $J$ be an ideal in $B(H)$ and let $\tau$ be a trace on $J$. If $J$ is closed with respect to the logarithmic submajorization, then $\tau$ is a spectral trace.

Example 3.9.

$a)$ Let $\mathcal{L}_{1, \infty}(H)$ be the ideal

Since $\mathcal{L}_{1, \infty}(H)$ is a quasi-Banach ideal with the complete quasi-norm ${\left\lVert{T}\right\rVert_{\mathcal{L}_{1, \infty}(H)} = \sup_{n\geq 1}\{ns_n(T)\}}$ and every quasi-Banach ideal is geometrically stable (see [18, Proposition 3.2]), then by Remark 3.6, $\mathcal{L}_{1, \infty}(H)$ is closed with respect to the logarithmic submajorization.

$b)$ It can be shown that the integral operator $V:L^{2}(0, 1)\to L^{2}(0, 1)$, ${(Vf)(t) = 2i\int^{t}_{0}f(s)ds}$ has no eigenvalues ([14, p. 178]). Since $V\in \mathcal{L}_{1, \infty}(L^{2}(0, 1))$, then by Theorem 3.7, $V\in {\mbox{Com}}(\mathcal{L}_{1, \infty}(L^{2}(0, 1)))$. Hence $\tau(V) = 0$ for every $\tau$ trace on $\mathcal{L}_{1, \infty}(L^{2}(0, 1))$.

4.   Modified Fredholm determinant

If $T\in S^{2}(H)$ and $\{\lambda_n(T)\}_{n\geq 1}$ is the sequence of non-zero eigenvalues of $T$, each repeated according to its algebraic multiplicity and ordered in such a way that $\left\lvert{\lambda_n(T)}\right\rvert\geq \left\lvert{\lambda_{n+1}(A)}\right\rvert$, $\forall n\in \mathbb{N}$ then

is an entire function, known as the modified Fredholm determinant of $I-\lambda T$. By the formula of Plemelj-Smithies [13, p. 166], it follows that

where $d_0 = 1$ and

and ${\sigma_n(T) = {\mbox{Tr}}\, (T^{n}) = \sum\limits^{\infty}_{j = 1}\lambda^{n}_{j}(T)}$, $n\geq 2$.

In particular, for Hilbert-Schmidt integral operators of the form

where $k$ is a measurable function and ${\int^{b}_{a}\int^{b}_{a}\left\lvert{k(t, s)}\right\rvert^{2}dtds < \infty}$, the modified Fredholm determinant of $I-\lambda T$ is equal to its Hilbert-Carleman determinant (see [13, p. 176]). More precisely,

where

5.   The integral operator $A_\rho$

To study the Riemann hypothesis, J. Alcántara-Bode ^[2] introduced the integral operator ${A_\rho:L^{2}(0, 1)\to L^{2}(0, 1)}$, ${(A_\rho f)(\theta) = \int^{1}_{0}\rho\left(\frac{\theta}{x}\right)f(x)dx}$. By Theorem 1.1, the Riemann hypothesis holds if and only if ${\mbox{Ker}}(A_\rho) = \{0\}$, or if and only if $h\notin {\mbox{Ran}}(A_\rho)$ where $h(x) = x$.

As we have seen in the introduction, the operators $A_\rho(\alpha)$ arise from the attempt to verify the condition $h\in{\mbox{Ran}}\, (A_\rho)$.

We briefly summarize properties of $A_\rho(\alpha)$ established in ^[2,3].

$i)$ $A_\rho(\alpha)$, $0 < \alpha\leq 1$, is Hilbert-Schmidt, but neither nuclear, nor normal, nor monotone.

$ii)$ $\lambda\in \sigma(A_\rho(\alpha))\setminus \{0\}$ ($\sigma(A_\rho(\alpha))$ is the spectrum of $A_\rho(\alpha)$), $0 < \alpha\leq 1$, if and only if $T_\alpha(\lambda^{-1}) = 0$ where

iii) The modified Fredholm determinant of $I-uA_\rho(\alpha)$ is

6.   Main results

For a given compact operator $T\notin S^{1}(H)$, where $H$ is a separable complex Hilbert space, the following theorem shows the existence of a nontrivial singular trace defined on an ideal closed with respect to the logarithmic submajorization taking the value of zero on $T$.

Theorem 6.1. For every compact operator $T\notin S^{1}(H)$ there exists an ideal $\mathcal{I}$ closed with respect to the logarithmic submajorization and a nontrivial singular trace $\tau$ on $\mathcal{I}$, such that $T\in \mathcal{I}$ and $\tau(T) = 0$.

Proof. By Remark 2.5, there exists a generalized eccentric operator $B$ such that $T\in (B)_0\subset (B)$. As we explained in the construction of the singular trace in Theorem 2.4, we can take the triple $\Omega = (B, w, \{n_k\}_{k\geq 1})$. Associated with $\Omega$, on the positive part of $(B)$, we have the functional

that extends linearly to a singular trace on the ideal $(B)$. We also denote this extension by $t_\Omega$. Clearly $t_\Omega(\left\lvert{B}\right\rvert) = 1$. Since $T\notin S^{1}(H)$, it follows that $B\notin S^{1}(H)$ and then

It follows from (6.1) that $t_\Omega$ is bounded with respect to the norm $\left\lVert{\cdot}\right\rVert_B$, where

so it extends by continuity to a singular trace $\tilde{t}_{\Omega}:\overline{(B)}^{\left\lVert{\cdot}\right\rVert_{B}}\to \mathbb{C}$. Here $\overline{(B)}^{\left\lVert{\cdot}\right\rVert_{B}}$ denotes the closure of $(B)$ with respect to $\left\lVert{\cdot}\right\rVert_{B}$. By [20, Lemma 5.5.9] and Remark 3.6 the ideal $\overline{(B)}^{\left\lVert{\cdot}\right\rVert_{B}}$ is closed with respect to the logarithmic submajorization. Finally, it follows from [15, Proposition 2.7] that $\tilde{\tau}_{\Omega}(T) = 0$. □

Since $A_\rho\notin S^{1}(L^{2}(0, 1))$, it follows from Theorem 6.1 that there exists an ideal $\mathcal{J}$ of $B(L^{2}(0, 1))$ closed with respect to the logarithmic submajorization that contains $A_\rho$ and a nontrivial singular trace $\tau$ on $\mathcal{J}$ such that $\tau(A_\rho) = 0$. The following theorem shows that every nontrivial singular trace on $\mathcal{J}$ take the value of zero on $A_\rho$.

Theorem 6.2. $\tau(A_\rho) = 0$ for every $\tau$ nontrivial singular trace on $\mathcal{J}$.

Proof. Let $\tau$ be a nontrivial singular trace on $\mathcal{J}$ (since $A_\rho\notin S^{1}(L^{2}(0, 1))$, the existence of $\mathcal{J}$ and a nontrivial singular trace on $\mathcal{J}$ are guaranteed by Theorem 6.1. If $0 < \alpha < 1$, by ^[5], the operators $\alpha(A_\rho +Q_{f_1})$ and $A_\rho+Q_{f_\alpha}$ have the same nonzero eigenvalues with the same algebraic and geometric multiplicities, where $Q_f(g) = {\langle {g, f}\rangle}h$, $f_1(x) = -\rho(\frac{1}{x})$ and $f_\alpha(x) = -\alpha^{-1}\rho(\frac{\alpha}{x})$. It follows from Theorem 3.8 that $\tau$ is a spectral trace and, therefore,

Hence, by definition of singular trace, it follows from (6.2) that $\tau(A_\rho) = 0$. □

Remark 6.3.

$i)$ It follows from Theorem 3.8 that the singular trace given in Theorem 6.1 is spectral.

$ii)$ If $0 < \alpha < 1$, by ^[5], the kernel of the operators $\alpha(A_\rho +Q_{f_1})$ and $A_\rho+Q_{f_\alpha}$ is trivial if and only if the Riemann hypothesis is true.

$iii)$ It is easily checked that if $0 < \alpha\leq 1$ and $V_\alpha:L^{2}(0, 1)\to L^{2}(0, 1)$, $(V_\alpha f)(x) = f(\frac{x}{\alpha})\chi_{[0, \alpha]}(x)$, then

The operator $V_\alpha$ was introduced in ^[4].

We have from (6.4) that $A_\rho(\alpha)\in \mathcal{J}$ for every $\alpha\in \mathcal{J}$.

We are now ready to prove the first main result of the paper.

Proof of Theorem 1.2. First, we prove that if $0 < \alpha, \beta\leq 1$ then

where $h(x) = x$. Indeed, by the Müntz-Szasz Theorem [8, Theorem 2.2], it is sufficient to verify (6.5) for $h^{r}$ with $r\in \mathbb{N}$. To this end, we use the identity [7, p. 312]

Evaluating the right-hand side of (6.5) we have

It follows from (6.6) that

Hence, (6.5) is true for $h^{r}$.

Let $\tau$ be a nontrivial singular trace on $\mathcal{J}$. Applying $\tau$ in (6.5), we obtain

It follows from the definition of trace and (6.3) that

Thus, for $0 < \alpha < 1$, we have $\tau(A_\rho V^{*}_{\alpha}) = 0$ and hence $\tau(A_\rho(\alpha)) = 0$. Since $\tau(A_\rho)\! = \!0$ and $A_\rho(1)\! = \!A_\rho$, then $\tau(A_\rho(\alpha))\! = \!0$ for every $0\! < \!\alpha\!\leq\! 1$. □

From now on, we concentrate on a recursion formula from which we can evaluate all the traces ${\mbox{Tr}}\, (A^{r}_{\rho}(\alpha))$, $r\in \mathbb{N}$, $r\geq 2$ and $0 < \alpha\leq 1$. First, we require a preliminary lemma.

Lemma 6.4. ${{\rm{Tr}}(A^{2}_{\rho}(\alpha)) = \alpha^{2}-\frac{\zeta(2)}{2}\alpha^{3}}.$

Proof. It follows from (4.1) that

Since

and

we get that

It follows from (6.8) that

Using (6.7) we obtain that

Finally, by (6.6) we conclude that

□

Proof of Theorem 1.3. By ^[3], we have

where

is an entire function with an infinite number of zeros. It follows from [27, p. 349] that for $u$ sufficiently small we have

Replacing (6.9) in (6.10), we get that

Moreover, we obtain from (6.11) and the Cauchy product formula that for $r\geq 2$ we have

where

This proves the assertion. □

Remark 6.5. Formula (1.2) is a recursion formula from which we can evaluate all the traces ${\mbox{Tr}}\, (A^{r}_{\rho}(\alpha))$, $r\geq 2$. For example, it follows from Lemma 6.4 and Theorem 1.3 that

7.   Conclusions

In this article we have shown the existence of a nontrivial singular trace $\tau$ defined on an ideal $\mathcal{J}$ closed with respect to the logarithmic submajorization such that $\tau(A_\rho(\alpha)) = 0$, where $A_\rho(\alpha):L^{2}(0, 1)\to L^{2}(0, 1)$, ${[A_\rho(\alpha)f](\theta) = \int^{1}_{0}\rho(\alpha\theta/x)f(x)dx}$, $0 < \alpha\leq 1$. Based on the theory of spectral traces and the results of ^[5], we have also shown that $\tau(A_\rho(\alpha)) = 0$ for every $\tau$ nontrivial singular trace on $\mathcal{J}$.

Finally, using the modified Fredholm determinant of $I-uA_\rho(\alpha)$, a recursion formula is presented to calculate all the traces ${\mbox{Tr}}\, (A^{r}_{\rho}(\alpha))$, $r\geq 2$.

Use of AI tools declaration

The author declares he/she has not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The author declares no conflicts of interest.