October 02, 2025
In this note we show that the only result of [Rocky Mountain J. Math. 54 (2024), no. 4, 995–1004] is nothing more than a misformulated version of an exercise from classical texts, presented with a flawed proof. To place the matter on firmer ground, we provide instead a direct solution of a more general problem.
In [Rocky Mountain J. Math. 54 (2024), no. 4, 995–1004] the authors claim to have proved the following result (see [1]):
Theorem 3.2. Let \(\mu\) be a finite measure defined on the Borelian \(\sigma\)-algebra of \(\mathbb{C}\) such that it contains an infinite number of points and let \((L_n(z))\) be the system of monic orthogonal polynomials with respect to \(\mu\). Let \(\xi\) be a fixed complex number. Let \(k\) be a positive integer. All the zeros of \(Q_{n;k}(z;\xi)\) are contained in the closed disk \(\overline{D\!\left(0,\,|\xi|+(k+1)(1+|\xi|)\right)}\).
Here \(\overline{D\!\left(0,\,|\xi|+(k+1)(1+|\xi|)\right)}\) denotes the closed disk in the complex plane, centered at the origin, with radius \(|\xi|+(k+1)(1+|\xi|)\). The rest of the undefined elements in [1] are provided by the authors in [1]:
Definition 1.3. Let \(\mu\) be a finite measure defined on the Borelian \(\sigma\)-algebra of \(\mathbb{C}\) such that it contains an infinite number of points and let \((L_n(z))\) the system of monic orthogonal polynomials with respect to \(\mu\). Let \(\xi\) be a fixed complex number. Let \(k\) be a positive integer. The \(k\)-polar polynomial related to \(\mu\), which will be denoted by \(Q_{n;k}(z;\xi)\), is the polynomial solution of degree \(n\) of the \(k\)-th order linear differential equation 1 \[\frac{\mathrm{d}^k}{\mathrm{d}z^k}(z-\xi)^k P(z) = (n+1)\cdots(n+k)L_n(z).\]
The authors established the connection with the orthogonal polynomials on the unit circle (in short, OPUC) in their work as follows:
Barry Simon [2] proved that if \(\mu\) is a finite positive measure defined on the Borelian \(\sigma\)-algebra of \(\mathbb{C}\), \(\mu\) is absolutely continuous with respect to the Lebesgue measure \(d\theta/(2\pi)\) on \([-\pi,\pi]\).
Although the hypotheses of [1] say nothing about the support of the measure, so that \((L_n)_{n\geq 0}\) cannot be regarded as a sequence of OPUC2, let us nevertheless make an imaginative effort, following the previous statement, to assume that the authors have in mind a measure supported on the unit circle. That said, it is worth emphasizing that the preceding claim is neither due to Simon nor could it be, since it is false; indeed, we know that in the case at hand \(\mu = \mu_{\mathrm{ac}} + \mu_{\mathrm{s}},\) where \(\mu_{\mathrm{ac}}\) denotes the part absolutely continuous with respect to the Lebesgue measure on the unit circle and \(\mu_{\mathrm{s}}\) the singular part [2]. Two illustrative counterexamples to the above claim, each with a singular part of a different nature and vanishing absolutely continuous part, are given by the Cantor measure on the unit circle and \[\mu=\sum_{j=1}^\infty 2^{-j}\,\delta_{e^{i/j}}.\]
A reading of [1] restricted to its definitions and results suggests that everything revolves around a certain measure defined on a particular \(\sigma\)-algebra of \(\mathbb{C}\) or on OPUC. The truth, however, is that it serves no real purpose other than that of a conjurer’s artifice—an illusion, embellished with a long list of superfluous known results, a contrived connection with Jacobi polynomials, redundant examples, and an artificial link to Sendov’s conjecture, all devised to divert attention from what is truly essential. In essence, the entirety of [1] reduces to the simple step of choosing a polynomial \(L_n\) (in the notation of [1]) whose zeros all lie in the open unit disk (indeed, OPUC have their zeros there). What may be even more striking is that [1] is nothing more than Exercise E.8 on page 24 of the classical book by Borwein and Erdélyi [3]. But if this reference appears recent, we recommend that readers consult instead Exercise \(21\) on page \(74\) of Marden’s book [4], in its second edition of 1966. To deepen the historical perspective on the origin of these results, the reader will find useful hints in the books cited. For a more modern view, and to appreciate the general scope of the problem, our suggestion is to begin with the works of Borcea and Brändén [5], [6].
Before briefly presenting and solving a more general exercise, in order to prevent future authors from revisiting the same issue, let us make one final remark on [1]. In the last two lines of their proof, the authors attempt to confine the zeros of the polynomial \[\sum_{j=0}^n \binom{n+k}{j+k}\,\omega^j\] to a disk of radius \(k+1\) by means of a faulty argument, citing moreover a work entirely unrelated to the matter. Yet it would have been enough simply to follow the hint provided by Borwein and Erdélyi or Marden in their exercise, which is precisely this. The behavior of the zeros of this polynomial is, of course, already well understood.
After formulating it correctly, if the reader wishes to perceive the weakness of [1], it suffices to consider the following example.
Example 1. (Free case [2])In the notation of [1], let \((L_n)_{n\geq 0}\) denote the sequence of OPUC with respect to the Lebesgue measure on the unit circle; that is, \(L_n(z) = z^n\). Assuming \(\xi = 0\), it follows that \(Q_{n;k}(z;0) = z^n\). Hence, as \(k\) increases, the disk \(D(0,k+1)\) expands to cover the whole complex plane \(\mathbb{C}\), while the zeros of \(Q_{n;k}\) remain fixed at \(0\).
All that follows relies on the following variant of Grace’s theorem, originally due to Szegő (see [7]):
Theorem 1. Let \[P(z) = \sum_{j=0}^n \binom{n}{j} a_j z^j \quad \text{and} \quad Q(z) = \sum_{j=0}^n \binom{n}{j} b_j z^j\] be polynomials of degree \(n\), and suppose that \(K \subset \mathbb{C}\) is an open or closed disk or half-plane, or the open or closed exterior of a disk, that contains all zeros of \(P\). If \(Q(0) \neq 0\), then each zero \(\zeta\) of \[\left(P \star_{\mathrm{G}} Q\right)(z)= \sum_{j=0}^n \binom{n}{j} a_j b_j z^j\] is of the form \(\zeta = -\alpha\beta\) with \(\alpha \in K\) and \(Q(\beta)=0\). If \(Q(0)=0\), then this continues to hold as long as \(K\) is not the open or closed exterior of a disk.
The following exercise, whose proof is included for completeness, constitutes a correctly formulated extension of the main result of [1]. To the best of our knowledge, this statement does not appear in the published literature; we present it here in the form of an exercise, since it may well have remained in private notes without ever being regarded as publishable.
Exercise 2.
Let \(k \in \mathbb{N}\) and let \(P\) be a non-constant monic polynomial of degree \(n\).
a) Prove that there exists a unique monic polynomial \(Q\) of degree exactly \(n\) such that \[\frac{\mathrm{d}^k}{\mathrm{d}z^k}\left(R(z) Q(z)\right)= (n+1)_k P(z),\] where \(R\) is a monic polynomial of degree \(k\) and \((n+1)_k\) denotes the rising factorial \((\)Pochhammer symbol\()\).
b) Let \(\xi \in \mathbb{C}\). Let \(K \subset \mathbb{C}\) an open or closed disk or half-plane, or the open or closed exterior of a disk, that contains all zeros of the shifted polynomial \(P(\xi+\omega)\) in the variable \(\omega\), where \(z=\xi+\omega\). Consider \(R(z)=(z-\xi)^k\). Show that every zero \(\zeta\) of the shifted polynomial \(Q(\xi+\omega)\) can be expressed as \(\zeta=-\alpha\beta\) with \(\alpha \in K\) and \(S(\beta)=0\), where \[S(\omega)=\sum_{j=0}^n \binom{n+k}{j+k}\,\omega^j.\] Equivalently, in the original variable \(z=\xi+\omega\) one has \[Z(Q) \subset \xi - K \cdot Z(S),\] where \(Z(Q)\) and \(Z(S)\) denote the zero sets of \(Q\) and \(S\), respectively.
c) Consider b) in the case where \(R\) is a monic polynomial of degree \(k\), whose zeros are not necessarily simple nor all equal to \(\xi\).
Solution. a) Let \[P(z) = \sum_{j=0}^n a_j z^j \quad \text{and} \quad Q(z) = \sum_{j=0}^n b_j z^j,\] with \(a_n=1\). Let \(\mathbb{C}[\cdot]_{\leq n}\) denote the vector space of all complex polynomials of degree at most \(n\). Define the linear operator \[T_R : \mathbb{C}[\cdot]_{\leq n} \longrightarrow \mathbb{C}[\cdot]_{\leq n}, \qquad Q \longmapsto \frac{\mathrm{d}^k}{\mathrm{d}z^k}\!\left(R(z)\,Q(z)\right),\] and consider the canonical basis \(\mathcal{B}=\{1,z,\dots,z^n\}\). A direct computation shows that \[T_R(z^{j}) = (j+1)_k z^j + \text{(lower degree terms)}.\] Therefore, with respect to the basis \(\mathcal{B}\), the matrix of \(T_R\) is lower triangular, and \[\det T_R = \prod_{j=0}^n (j+1)_k \neq 0.\] Thus \(T_R\) is invertible on \(\mathbb{C}[\cdot]_{\leq n}\). Comparing the coefficients of \(z^n\) in the identity \(T_R(Q)=(n+1)_k P\) yields \((n+1)_k b_n=(n+1)_k a_n\), hence \(b_n=a_n=1\).
b) The very formulation of the exercise is nothing more than \[Q(\omega+\xi) = P(\omega+\xi)\,\star_{\mathrm{G}}\,S(\omega),\] where the convolution is taken in the variable \(\omega\). (In any case, a reader not familiar with the topic will better understand this point upon seeing the solution of the next part.) Since \[S(0) =\binom{n+k}{k}\neq 0,\] the result follows directly from Theorem 1.
c) Write the shifted polynomials in the binomial basis \[P(\xi+\omega)=\sum_{j=0}^n \binom{n}{j} \alpha_j\,\omega^j, \quad Q(\xi+\omega)=\sum_{j=0}^n \binom{n}{j} \beta_j\,\omega^j.\] Assume that, for every \(0\le j\le n\), the condition \(\alpha_j=0\) necessarily entails that \(\beta_j=0\). Let \(Q\) be the unique solution of a). Define \(c_j =\beta_j/\alpha_j \,\mathbf{1}_{\{\alpha_j \neq 0\}},\) and set \[S_R(\omega)=\sum_{j=0}^n \binom{n}{j}\,c_j\,\omega^j.\] By construction, \[\begin{align} \label{factor} P(\omega+\xi)\,\star_{\mathrm{G}}\, S_R(\omega) =\sum_{j=0}^n \binom{n}{j} \alpha_j c_j\,\omega^j =\sum_{j=0}^n \binom{n}{j} \beta_j\,\omega^j=Q(\xi+\omega), \end{align}\tag{1}\] and therefore Theorem 1 applies, yielding the desired zero localization. For \(R(z)=(z-\xi)^k\) one has \[\beta_j=\frac{(n+1)_k}{(j+1)_k}\alpha_j.\] Therefore, \[S_R(\omega) =\sum_{j=0}^n \binom{n+k}{j+k}\,\omega^j = S(\omega).\] If the condition “\(\alpha_j=0\) necessarily entails \(\beta_j=0\)” is not satisfied, the factorization 1 may fail. For instance, let \(n=2\), \(\xi=0\), and \(k=1\). Then \[P(\omega)=\omega^2\] corresponds to \(\alpha_0=\alpha_1=0\), \(\alpha_2=1\), while \[Q(\omega)=\omega^2+\omega\] has \(\beta_0=0\), \(\beta_1=1/2\), \(\beta_2=1\). Here \(\alpha_1=0\) but \(\beta_1\neq 0\), so the assumption is violated. In this case, for any \[S_R(\omega)=\sum_{j=0}^2 \binom{2}{j}c_j\,\omega^j\] one finds \((P\star_{\mathrm G} S_R)(\omega)=c_2\,\omega^2,\) which cannot reproduce the linear term in \(Q\). Thus no such factorization exists. ◻
Remark 3. Note that what the authors attempted to establish in [1] amounts to nothing more than taking \(K\) as the open unit disk centered at the origin and invoking the fact (see [3] or [4]) that \[\max_{\gamma \in Z(S)} |\gamma| < k+1.\] From Exercise 2 b) it then follows that \[\begin{align} \max_{\gamma \in Z(Q)} |\gamma| &\leq |\xi| + (|\xi|+1)\,\max_{\gamma \in Z(S)} |\gamma| \\[6pt] &= |\xi| + (|\xi|+1)(k+1). \end{align}\]
Let us revisit Example 1 to see how the previous exercise provides a more convenient way to localize the zeros of the polynomial \(Q\) without computing it explicitly.
Example 2. (Free case, revisited) Since in this case \(K=\{0\}\), it follows at once that all zeros of \(Q\) collapse to the origin.
This work was partially supported by the Centre for Mathematics of the University of Coimbra, funded by the Portuguese Government through FCT/MCTES (DOI: 10.54499/ UIDB/00324/2020). RAN was partially supported by PID2024-155593NB-C21 (FEDER(EU)/ Ministerio de Ciencia, Innovación y Universidades-Agencia Estatal de Investigación) and FQM-415 (Junta de Andalucía). KC acknowledges financial support from FCT under the grant DOI: 10.54499/2022.00143.CEECIND/CP1714/CT0002.