# Optimal Polynomial Approximants in $L^{p}$

Raymond Centner  Department of Mathematics and Statistics, University of South
Florida, 4202 E. Fowler Avenue, Tampa, Florida 33620-5700, USA.
<EMAIL_ADDRESS>
###### Abstract.

Over the past several years, optimal polynomial approximants (OPAs) have been
studied in many different function spaces. In these settings, numerous papers
have been devoted to studying the properties of their zeros. In this paper, we
introduce the notion of optimal polynomial approximant in the space $L^{p}$,
$1\leq p\leq\infty$. We begin our treatment by showing existence and
uniqueness for $1<p<\infty$. For the extreme cases of $p=1$ and $p=\infty$, we
show that uniqueness does not necessarily hold. We continue our development by
elaborating on the special case of $L^{2}$. Here, we create a test to
determine whether or not a given 1st degree OPA is zero-free in
$\overline{\mathbb{D}}$. Afterward, we shed light on an orthogonality
condition in $L^{p}$. This allows us to study OPAs in $L^{p}$ with the
additional tools from the $L^{2}$ setting. Throughout this paper, we focus
many of our discussions on the zeros of OPAs. In particular, we show that if
$1<p<\infty$, $f\in H^{p}$, and $f(0)\neq 0$, then there exists a disk,
centered at the origin, in which all the associated OPAs are zero-free. Toward
the end of this paper, we use the orthogonality condition to compute the
coefficients of some OPAs in $L^{p}$. To inspire further research in the
general theory, we pose several open questions throughout our discussions.

###### 2020 Mathematics Subject Classification:

Primary 30E10; Secondary 46E30. Key words: optimal polynomial approximant,
$L^{p}$ space, Hardy space, digital filter, Shanks’ Conjecture

###### Contents

  1. 1 Introduction
  2. 2 Existence and Uniqueness
  3. 3 OPAs in $L^{2}$
  4. 4 Orthogonality Condition
  5. 5 Location of Zeros
  6. 6 Computation of Coefficients
  7. 7 Concluding Remarks

##  1\. Introduction

The methods of least-squares approximation have been used in many areas of
engineering since the latter half of the 20th century. In 1963, Robinson [19]
studied these methods in the context of digital signal processing and
geophysical studies. His goal was to obtain a finite-length wavelet whose
convolution with a given finite-length wavelet best approximates the unit
spike. In other words, given a finite sequence of real numbers
$b:=(b_{0},b_{1},\dots,b_{m})$ and a particular value of $n\in\mathbb{N}$, he
was trying to find another finite sequence of real numbers
$a:=(a_{0},a_{1},\dots,a_{n})$ such that the difference between $a*b$ and the
unit spike $(1,0,\dots,0)$ has the smallest $l^{2}$-norm. This sequence $a$
was referred to as a least-squares approximate inverse.

In the 1970s, least-squares approximation appeared in connection with 2D
recursive digital filter design. Applications of 2D digital filtering include
the processing of medical pictures, satellite photographs, seismic data
mappings, gravity waves data, magnetic recordings, and radar and sonar maps
[15]. In 1972, Justice, Shanks, and Treitel [17] applied the methods of least-
squares approximation in an effort to ensure 2D filter stability. Generally
speaking, a filter is called stable if it corresponds to a system in which
bounded input yields bounded output. In their studies, they considered the
following problem: given a polynomial $f(z,w)$ of two complex variables, find
another polynomial $q(z,w)$ such that the difference between $qf$ and $1$ has
the smallest $H^{2}(\mathbb{D}^{2})$-norm. The polynomial $q$ was referred to
as a planar least-squares inverse (PLSI) polynomial. It is one of the earliest
examples of an optimal polynomial approximant.

In 1980, Chui [9] formulated the problem of Robinson in the
$H^{2}(\mathbb{D})$ setting. More specifically, he considered the following
problem: given $n\in\mathbb{N}$ and a polynomial
$f(z):=\sum_{k=0}^{m}b_{k}z^{k}$ with $f(0)\neq 0$, find an $n$th degree
polynomial $q(z):=\sum_{k=0}^{n}a_{k}z^{k}$ such that the difference between
$qf$ and $1$ has the smallest $H^{2}$-norm. The polynomial $q$ was referred to
as a least-squares inverse polynomial. For brevity, we’ll refer to these as
LSI polynomials. Chui’s formulation of the problem ties in well with signal
processing since digital filters are often represented by quotients of
polynomials. Now, in many contexts of $1$D recursive digital filter design, it
is important to ensure that the poles of the filter are outside of
$\overline{\mathbb{D}}$. If this is the case, then the filter will be stable.
In 1982, Chui and Chan [10] proposed a design method that guaranteed filter
stability. In the build-up to their method, they reformulated the problem of
Robinson to include arbitrary $f\in H^{2}(\mathbb{D})$ with $f(0)\neq 0$. They
proceeded to show that for this general class of functions, the LSI
polynomials are zero-free in $\overline{\mathbb{D}}$.

Since the 1980s, least-squares inverse polynomials have been studied in many
different function spaces. In the mathematics community, these polynomials
became known as optimal polynomial approximants (OPAs). In 2015, OPAs were
introduced in the Dirichlet-type spaces $\mathcal{D}_{\alpha}$,
$\alpha\in\mathbb{R}$ (see [3] for details). Since then, several papers have
been devoted to investigating the properties of these polynomials (see, e.g.,
[5, 6, 7]). Such properties include boundary behavior, connections to
reproducing kernel functions, and location of zeros. In [5], it was shown that
if $\alpha\geq 0$, $f\in\mathcal{D}_{\alpha}$, and $f(0)\neq 0$, then the
zeros of the associated OPA lie outside of $\overline{\mathbb{D}}$; in more
recent literature, optimal polynomial approximants and their zeros have been
studied in the context of $\ell^{p}_{A}$ (see [8] for details).

The primary motivation for writing this paper is to investigate the zeros of
OPAs in the context of $H^{p}:=H^{p}(\mathbb{D})$, $1\leq p\leq\infty$. In
particular, we would like to know if the nontrivial OPAs are zero-free in the
closed disk $\overline{\mathbb{D}}$, which is the case in the Hardy space
$H^{2}$ and the Dirichlet-type spaces $\mathcal{D}_{\alpha}$, $\alpha\geq 0$.
Recall that for $1\leq p<\infty$, $H^{p}$ is the space of analytic functions
$f$ in the open unit disk $\mathbb{D}$ such that

$\sup_{0\leq r<1}\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(re^{it})|^{p}dt<\infty.$

It is well known that $H^{p}$ is a Banach space with norm

$\|f\|_{p}:=\bigg{\\{}\sup_{0\leq
r<1}\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(re^{it})|^{p}dt\bigg{\\}}^{\frac{1}{p}}.$

On the other hand, $H^{\infty}$ is the space of bounded analytic functions in
$\mathbb{D}$; this is a Banach space with norm

$\|f\|_{\infty}:=\sup_{z\in\mathbb{D}}|f(z)|.$

To facilitate our study in the $H^{p}$ setting, we introduce the notion of
optimal polynomial approximant in the space $L^{p}:=L^{p}(\mathbb{T})$; in
Section 4, we will see how OPAs in this context relate to OPAs in the more
structured space $L^{2}$. For $1\leq p<\infty$, $L^{p}$ is the space of
Lebesgue measurable $\mathbb{C}$-valued functions $f$ on the unit circle
$\mathbb{T}$ such that

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(e^{it})|^{p}dt<\infty.$

It is well known that $L^{p}$ is a Banach space with norm

$\|f\|_{p}:=\bigg{\\{}\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(e^{it})|^{p}dt\bigg{\\}}^{\frac{1}{p}}.$

On the other hand, $L^{\infty}$ is the space of essentially bounded Lebesgue
measurable $\mathbb{C}$-valued functions $f$ on $\mathbb{T}$; this is a Banach
space with norm

$\|f\|_{\infty}:=\operatorname*{ess\,sup}_{t\in[-\pi,\pi]}|f(e^{it})|.$

It can be shown that for any $f\in H^{p}$, $1\leq p\leq\infty$, the
nontangential limits $f^{*}(e^{it})$ exist almost everywhere on $\mathbb{T}$
and $f^{*}\in L^{p}$ (see, e.g., [12]). Futhermore, $\|f^{*}\|_{p}=\|f\|_{p}$.
This shows that $H^{p}$ is isometrically imbedded in $L^{p}$. Therefore, we
regard any $f\in H^{p}$ as a function in $L^{p}$ with the understanding that
we are identifying $f$ with its nontangential limit function $f^{*}$.
Accordingly, any result we obtain for functions in $L^{p}$ will automatically
hold for functions in $H^{p}$.

Throughout this paper, we assume that $\mathbb{N}:=\\{0,1,2,\dots\\}$ and
denote $\mathcal{P}_{n}$ as the space of analytic polynomials of degree at
most $n\in\mathbb{N}$; for any $f\in L^{p}$, we let
$f\mathcal{P}_{n}:=\\{fq\colon q\in\mathcal{P}_{n}\\}$. We begin the paper by
discussing existence and uniqueness of optimal polynomial approximants in
$L^{p}$, $1\leq p\leq\infty$; for the extreme cases of $p=1$ and $p=\infty$,
there seems to be a lot of fertile ground for research. We then introduce OPAs
in the Hilbert space $L^{2}$; this is a generalization of the familiar theory
in $H^{2}$ (see [2] for a survey of OPAs in $\mathcal{D}_{\alpha}$). In
Section 4, we use an orthogonality condition to characterize OPAs in $L^{p}$,
$1<p<\infty$; we later use this condition to compute some OPAs for several
values of $p$. Throughout this paper, we discuss what we know and pose several
open questions about the zeros of optimal polynomial approximants in $L^{p}$.

##  2\. Existence and Uniqueness

The context of this paper will be the space $L^{p}$, $1\leq p\leq\infty$; we
would like to have an adequate structure to support the theory, so it seems
natural to consider only the values of $p$ for which $L^{p}$ is a Banach
space. In the context of $L^{p}$, existence and uniqueness of best
approximations can be a delicate issue. However, if $Y$ is a finite
dimensional subspace of $L^{p}$, then there exists a best approximation to any
$g\in L^{p}$ out of $Y$ (see, e.g., [18, Theorem 6.1-1]); this allows us to
define the notion of optimal polynomial approximant.

######  Proposition 2.1 (Existence).

Let $1\leq p\leq\infty$ and $n\in\mathbb{N}$. For $f,g\in L^{p}$, $f\not\equiv
0$, there exists a polynomial $Q\in\mathcal{P}_{n}$ such that

$\inf_{q\in\mathcal{P}_{n}}\|qf-g\|_{p}=\|Qf-g\|_{p}.$

###### Proof.

Since $f\mathcal{P}_{n}$ is a finite dimensional subspace of $L^{p}$, the
result is immediate. ∎

Proposition 2.1 states that there is a best approximation to $g$ out of the
space $f\mathcal{P}_{n}$. If $fQ$ is such a best approximation, then the
polynomial $Q\in\mathcal{P}_{n}$ is referred to as an optimal polynomial
approximant in $L^{p}$. Now, it’s important to note that best approximations
in the Banach space setting are not always unique. However, $L^{p}$ is a
strictly convex for $1<p<\infty$. Therefore, for any given subspace $Y$ of
$L^{p}$, there is at most one best approximation to any $g\in L^{p}$ out of
$Y$ (see, e.g., [18, Theorem 6.2-3]). This guarantees that optimal polynomial
approximants in $L^{p}$, $1<p<\infty$, are unique. (See [20] for a similar
discussion of uniqueness of OPAs in $\ell^{p}_{A}$.)

######  Proposition 2.2 (Uniqueness).

Let $1<p<\infty$ and $n\in\mathbb{N}$. For $f,g\in L^{p}$, $f\not\equiv 0$,
there exists a unique polynomial $Q\in\mathcal{P}_{n}$ such that

$\inf_{q\in\mathcal{P}_{n}}\|qf-g\|_{p}=\|Qf-g\|_{p}.$

###### Proof.

It is well-known that $L^{p}$ is strictly convex for $1<p<\infty$. Since $g$
has a best approximation out of $f\mathcal{P}_{n}$, it must be unique. ∎

In the case of $L^{1}$, Proposition 2.1 implies that the set of best
approximations to a given function $g$ out of $f\mathcal{P}_{n}$ is nonempty.
Now, since the set of best approximations must be convex (see, e.g., [18,
Lemma 6.2-1]), there is either a unique best approximation or an infinite
number of best approximations. The following proposition shows that the former
does not need to hold.

######  Proposition 2.3.

Optimal polynomial approximants in $L^{1}$ are not necessarily unique.

###### Proof.

Let $f\equiv 1$, and consider the characteristic function
$g(e^{it}):=\chi_{[-\pi,0)}(t)$. We show that

$\inf_{q\in\mathcal{P}_{0}}\|qf-g\|_{1}=\|af-g\|_{1}$

for all $a\in[0,1]$. Let $\Lambda:=\\{h\in L^{\infty}\colon\|h\|_{\infty}\leq
1\;\text{and}\;\frac{1}{2\pi}\int_{-\pi}^{\pi}hdt=0\\}$. By Hahn-Banach
duality, it follows that

$\inf_{q\in\mathcal{P}_{0}}\|qf-g\|_{1}=\sup_{h\in\Lambda}\bigg{|}\frac{1}{2\pi}\int_{-\pi}^{\pi}ghdt\bigg{|}.$

If we let $h_{0}(e^{it}):=\chi_{[-\pi,0)}(t)-\chi_{[0,\pi)}(t)$, then we see
that $h_{0}\in\Lambda$. Consequently,

$\inf_{q\in\mathcal{P}_{0}}\|qf-g\|_{1}\geq\bigg{|}\frac{1}{2\pi}\int_{-\pi}^{\pi}gh_{0}dt\bigg{|}=\frac{1}{2}.$
(1)

Now, suppose that $a\in[0,1]$. Then

$\displaystyle\|af-g\|_{1}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}|a-g|dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{0}(1-a)dt+\frac{1}{2\pi}\int_{0}^{\pi}adt$
$\displaystyle=\frac{1}{2}.$ (2)

From (1) and (2), we conclude that

$\inf_{q\in\mathcal{P}_{0}}\|qf-g\|_{1}=\|af-g\|_{1}$

whenever $a\in[0,1]$. ∎

On the other hand, consider the case of $L^{\infty}$. From Proposition 2.1 and
our previous discussion, the set of best approximations to a given function
$g$ out of $f\mathcal{P}_{n}$ consists of either a single element or an
infinite number of elements. As in the case of $L^{1}$, the former does not
need to hold.

######  Proposition 2.4.

Optimal polynomial approximants in $L^{\infty}$ are not necessarily unique.

###### Proof.

Let $f(z)=1-z$. We show that

$\inf_{q\in\mathcal{P}_{0}}\|qf-1\|_{\infty}=\|af-1\|_{\infty}$

for all $a\in[0,1]$. For any $a\in\mathbb{C}$,

$\|af-1\|_{\infty}=\|(a-1)-az\|_{\infty}\leq|a-1|+|a|.$ (3)

If $a=0$, then equality in (3) clearly holds. If $a\neq 0$, let

$w:=-\frac{|a|}{a}e^{i\arg(a-1)}.$

Note that

$\displaystyle|af(w)-1|$ $\displaystyle=|(a-1)-aw|$
$\displaystyle=\big{|}|a-1|e^{i\arg(a-1)}+|a|e^{i\arg(a-1)}\big{|}$
$\displaystyle=|a-1|+|a|.$

Thus, equality in (3) holds, and it follows that

$\|af-1\|_{\infty}=|a+1|+|a|\geq(1-|a|)+|a|=1$

for all $a\in\mathbb{C}$. Consequently,

$\inf_{q\in\mathcal{P}_{0}}\|qf-1\|_{\infty}=1,$ (4)

where the infimum is attained for all $a\in\mathbb{C}$ such that
$|a-1|+|a|=1$.

If $a\in\mathbb{C}$ satisfies $|a-1|+|a|=1$, then

$|1-a|+|a|=|(1-a)+a|.$

Hence, $a=\alpha(1-a)$ for some positive real $\alpha$. It follows that $a$ is
a real number in $[0,1]$. Conversely, if $a\in[0,1]$, then

$|a-1|+|a|=(1-a)+a=1.$

Therefore, we conclude that $|a-1|+|a|=1$ if and only if $a\in[0,1]$. It
follows that

$\|af-1\|_{\infty}=1\quad\text{for all}\quad a\in[0,1].$ (5)

By combining (4) and (5), the result follows. ∎

Although optimal polynomial approximants in $L^{p}$ are not necessarily unique
for $p=1$ or $p=\infty$, there are still interesting questions that one can
ask about best approximations in these spaces. For example, when there are
infinitely many best approximations to $g$ out of $f\mathcal{P}_{n}$, is there
an optimal polynomial approximant that has norm strictly smaller than all the
others? In the case of Proposition 2.3 (or Proposition 2.4), this optimal
polynomial approximant is clearly the constant $0$. As another example, is
there a way to characterize the functions $f$ such that the constant function
$0$ is a best approximation to $1$ out of $f\mathcal{P}_{n}$? In Proposition
5.2, we will see an analogous characterization in the context of $H^{p}$,
$1<p<\infty$. Nevertheless, a best approximation to $g$ out of
$f\mathcal{P}_{n}$ is unique whenever $1<p<\infty$. This leads us to the
following definition.

######  Definition 2.5 ($n$th OPA).

Let $1<p<\infty$ and $n\in\mathbb{N}$. For $f,g\in L^{p}$, $f\not\equiv 0$,
the unique polynomial that minimizes the norm $\|qf-g\|_{p}$, where $q$ varies
over $\mathcal{P}_{n}$, is denoted by $q_{n,p}[f,g]$ and is called the $n$th
OPA of $g/f$ in $L^{p}$.

In the next section, we develop the theory of OPAs in $L^{2}$; the inner
product structure gives us an efficient way to compute the coefficients of
OPAs; it also gives us an easy way to verify whether or not the OPA
$q_{1,2}[f,g]$ is zero-free in $\overline{\mathbb{D}}$. As we will see in
Section 4, it is possible to relate certain OPAs in $L^{p}$, $p\geq 2$, to
OPAs in $L^{2}$.

##  3\. OPAs in $L^{2}$

In 1980, Chui [9] presented the idea of least-squares inverse polynomials in
the context of $H^{2}$. The LSI polynomials being studied can be viewed as
OPAs of the form $q_{n,2}[f,1]$, where $f$ is a polynomial. In 1982, Chui and
Chan [10] presented a filter design method that extended this study to include
arbitrary functions $f\in H^{2}$, $f(0)\neq 0$. The reason for considering
more general $f$ was to approximate certain outer functions that were defined
in conjunction with an ideal digital filter. One of the selling points of this
method was that the coefficients of the LSI polynomial could be expressed as
the solution to a system of linear equations. Moreover, the associated matrix
is positive definite and Toeplitz. (In fact, it’s a Gram matrix.) Therefore,
its inverse can be computed quickly with any of the available algorithms (see
[1] for details).

In this section, we’ll see that some of the most important OPA properties in
$H^{2}$ (particularly the fact that the coefficients satisfy a system of
linear equations) generalize naturally to the $L^{2}$ setting. Before we
begin, let’s note that the following proposition follows immediately from the
notion of orthogonal projection.

######  Proposition 3.1 (Linearity).

Let $f,g,h\in L^{2}$, $f\not\equiv 0$, $n\in\mathbb{N}$, and
$\alpha,\beta\in\mathbb{C}$. Then

$q_{n,2}[f,\alpha g+\beta h]=\alpha q_{n,2}[f,g]+\beta q_{n,2}[f,h].$

###### Proof.

By definition, the OPA $q_{n,2}[f,g]$ minimizes the norm $\|qf-g\|_{2}$, where
$q$ varies over $\mathcal{P}_{n}$. Since $L^{2}$ is a Hilbert space and
$f\mathcal{P}_{n}$ is a closed subspace, it follows that $q_{n,2}[f,g]f$ is
the orthogonal projection of $g$ onto $f\mathcal{P}_{n}$. The proposition
follows by linearity of the projection. ∎

In the following result, we characterize OPAs of the form $q_{n,2}[f,g]$,
where $f,g\in L^{2}$.

######  Proposition 3.2.

Let $f,g\in L^{2}$, $f\not\equiv 0$, $n\in\mathbb{N}$, and
$Q\in\mathcal{P}_{n}$. Then

$\langle Qf-g,z^{k}f\rangle=0$

for $k=0,\dots,n$ if and only if $Q=q_{n,2}[f,g]$.

###### Proof.

Let $q_{n,2}:=q_{n,2}[f,g]$. If $\langle Qf-g,z^{k}f\rangle=0$ for
$k=0,\dots,n$, then $(Qf-g)\perp f\mathcal{P}_{n}$. Since $g=Qf+(g-Qf)$, it
follows that $Qf$ is the projection of $g$ onto $f\mathcal{P}_{n}$, i.e.,
$Q=q_{n,2}$. Conversely, since $q_{n,2}f$ is the projection of $g$ onto
$f\mathcal{P}_{n}$, and since $g=q_{n,2}f+(g-q_{n,2}f)$, it follows that
$(q_{n,2}f-g)\in L^{2}\ominus f\mathcal{P}_{n}$. In particular, $\langle
q_{n,2}f-g,z^{k}f\rangle=0$ for $k=0,\dots,n$. ∎

A valuable aspect of Proposition 3.2 is that it allows us to compute the
coefficients of $q_{n,2}[f,g]$ in an efficient way; as we will see, the
coefficients are expressed as a solution to a system of linear equations. In
fact, the associated matrix has the same properties as the one that Chui and
Chan were considering in [10]. (For a recent discussion of the computation of
coefficients in the context of analytic function spaces, see [4, 7].)

######  Theorem 3.3 (Coefficients in $L^{2}$).

Let $f,g\in L^{2}$, $f\not\equiv 0$, and $n\in\mathbb{N}$. Set
$q_{n,2}[f,g](z)=\sum_{j=0}^{n}a_{j}z^{j}$. Then the coefficients of
$q_{n,2}[f,g]$ satisfy

$\begin{bmatrix}\langle f,f\rangle&\langle zf,f\rangle&\dots&\langle
z^{n}f,f\rangle\\\ \langle f,zf\rangle&\langle zf,zf\rangle&\dots&\langle
z^{n}f,zf\rangle\\\ \vdots&\vdots&&\vdots\\\ \langle f,z^{n}f\rangle&\langle
zf,z^{n}f\rangle&\dots&\langle
z^{n}f,z^{n}f\rangle\end{bmatrix}\begin{bmatrix}a_{0}\\\ a_{1}\\\ \vdots\\\
a_{n}\end{bmatrix}=\begin{bmatrix}\langle g,f\rangle\\\ \langle g,zf\rangle\\\
\vdots\\\ \langle g,z^{n}f\rangle\end{bmatrix}.$

###### Proof.

From Proposition 3.2, it follows that

$\sum_{j=0}^{n}\langle z^{j}f,z^{k}f\rangle a_{j}=\langle
g,z^{k}f\rangle\quad\text{for}\quad k=0,\dots,n.$ (6)

Let $a$ denote the vector with components $a_{j}$, $j=0,\dots,n$, and $y$
denote the vector with components $\langle g,z^{k}f\rangle$, $k=0,\dots,n$. If
$B$ is the $(n+1)\times(n+1)$ matrix with entries $b_{kj}=\langle
z^{j}f,z^{k}f\rangle$, $0\leq k,j\leq n$, it is easy to see that (6) is
equivalent to $Ba=y$. ∎

Let’s look at an example to demonstrate the computation of coefficients in
$L^{2}$.

######  Example 3.4.

Let $f(z)=1-z$. Then $q_{1,2}[f,1](z)=\frac{1}{3}z+\frac{2}{3}$.

###### Proof.

Set $q_{1,2}[f,1](z)=a_{1}z+a_{0}$. By Theorem 3.3, the coefficients of
$q_{1,2}[f,1](z)$ satisfy

$\begin{bmatrix}2&-1\\\ -1&2\end{bmatrix}\begin{bmatrix}a_{0}\\\
a_{1}\end{bmatrix}=\begin{bmatrix}1\\\ 0\end{bmatrix}.$

Therefore, it easily follows that $q_{1,2}[f,1](z)=\frac{1}{3}z+\frac{2}{3}$.
∎

As previously suggested, the design method of Chui and Chan was motivated by
the fact that all nontrivial LSI polynomials are zero-free in
$\overline{\mathbb{D}}$. Since we are extending these polynomials to the
$L^{2}$ setting, it’s natural to question if OPAs of the form $q_{n,2}[f,g]$,
where $f,g\in L^{2}$, are zero-free in $\overline{\mathbb{D}}$. It turns out
that this is not the case; the following theorem shows that the function $g$
plays a prominent role.

######  Theorem 3.5.

Let $f,g\in L^{2}$ and $f\not\equiv 0$. The OPA $q_{1,2}[f,g]$ is zero-free in
$\overline{\mathbb{D}}$ if and only if

$|\langle g,f\rangle|>|\langle g,zf\rangle|.$

###### Proof.

Set $q_{1,2}[f,g](z)=a_{1}z+a_{0}$. From Proposition 3.2, we have that

$a_{0}\langle f,z^{k}f\rangle+a_{1}\langle zf,z^{k}f\rangle=\langle
g,z^{k}f\rangle\quad\text{for}\quad k=0,1.$ (7)

This is equivalent to

$\begin{bmatrix}\langle f,f\rangle&\langle zf,f\rangle\\\ \langle
f,zf\rangle&\langle zf,zf\rangle\end{bmatrix}\begin{bmatrix}a_{0}\\\
a_{1}\end{bmatrix}=\begin{bmatrix}\langle g,f\rangle\\\ \langle
g,zf\rangle\end{bmatrix}.$ (8)

Let $A$ denote the $2\times 2$ matrix in (8). Since $|\langle
f,f\rangle|>|\langle f,zf\rangle|$, and since the shift operator is an
isometry on $H^{2}$, it follows that $\det A\neq 0$. Therefore,

$a_{0}=\frac{\langle zf,zf\rangle\langle g,f\rangle-\langle zf,f\rangle\langle
g,zf\rangle}{\det A}$

and

$a_{1}=\frac{\langle f,f\rangle\langle g,zf\rangle-\langle f,zf\rangle\langle
g,f\rangle}{\det A}.$

If $q_{1,2}[f,g]$ is zero-free in $\overline{\mathbb{D}}$, then $a_{0}\neq 0$.
It follows from (7) that

$\displaystyle|\langle g,f\rangle|=|\langle a_{0}f,f\rangle|>|\langle
a_{0}f,zf\rangle|=|\langle g,zf\rangle|.$

Conversely, suppose that $|\langle g,f\rangle|>|\langle g,zf\rangle|$. Then
$\langle g,f\rangle$ and $\langle g,zf\rangle$ cannot simultaneously be zero.
From $(\ref{eq2})$, it follows that $a_{0}\neq 0$, i.e., $q_{1,2}[f,g]$ is
zero-free in $\overline{\mathbb{D}}$.

Let $q_{1,2}[f,g](z_{0})=0$. Then

$\displaystyle z_{0}=-\frac{a_{0}}{a_{1}}=\frac{\langle zf,f\rangle\langle
g,zf\rangle-\langle zf,zf\rangle\langle g,f\rangle}{\langle f,f\rangle\langle
g,zf\rangle-\langle f,zf\rangle\langle g,f\rangle}.$ (9)

If $\langle g,zf\rangle=0$, then it follows from (8) that $\langle
g,f\rangle\neq 0$. From (9), we see that

$|z_{0}|=\frac{|\langle zf,zf\rangle|}{|\langle f,zf\rangle|}>1.$

Therefore, both implications in the theorem are trivially true. We thus assume
that $\langle g,zf\rangle\neq 0$. Let $w:=\frac{\langle g,f\rangle}{\langle
g,zf\rangle}$ and $\alpha:=\frac{\langle zf,f\rangle}{\langle zf,zf\rangle}$.
From (9), it follows that

$|z_{0}|=\bigg{|}\frac{w-\alpha}{1-\overline{\alpha}w}\bigg{|}.$

Since $\alpha\in\mathbb{D}$, $z_{0}\in\overline{\mathbb{D}}$ if and only if
$w\in\overline{\mathbb{D}}$. Equivalently, $q_{1,2}[f,g]$ is zero-free in
$\overline{\mathbb{D}}$ if and only if $|\langle g,f\rangle|>|\langle
g,zf\rangle|$. ∎

The following example demonstrates the relative ease of applying Theorem 3.5.

######  Example 3.6 (Zero in $\overline{\mathbb{D}}$).

Let $f(z)=z^{2}+1$ and $g(z)=2z-1$. Note that

$|\langle g,f\rangle|=1<2=|\langle g,zf\rangle|.$

By Theorem 3.5, $q_{1,2}[f,g]$ has its zero in $\overline{\mathbb{D}}$.

It’s worth noting that the results in this section can be formulated in the
context of $L^{2}(\mu)$, where $d\mu:=\frac{1}{2\pi}|f|^{2}dt$. At any rate,
in the case where $n$ is arbitrary, the following is still unknown.

######  Open Question 1.

Let $n\in\mathbb{N}$. For which functions $f,g\in L^{2}$ can we guarantee that
$q_{n,2}[f,g]$ is zero-free in $\overline{\mathbb{D}}$?

##  4\. Orthogonality Condition

Our main motivation for writing this paper is to investigate the zeros of
optimal polynomial approximants in the context of $H^{p}$. For the Hilbert
space $H^{2}$, we already know that OPAs of the form $q_{n,2}[f,1]$, where
$f\in H^{2}$ and $f(0)\neq 0$, are zero-free in $\overline{\mathbb{D}}$; one
proof of this follows from a special case of Proposition 3.2. Now, in an
effort to understand the zeros of OPAs in $H^{p}$, we generalize Proposition
3.2 to functions in $L^{p}$. As a result, this allows us to relate OPAs in
$L^{p}$ to OPAs in the more structured space $L^{2}$.

Recall that in a normed space $(X,\|\cdot\|)$, an element $x\in X$ is said to
be orthogonal to a subspace $Y$ if $\|x\|\leq\|x+y\|$ for all $y\in Y$. In the
case where $Y$ is a subspace of $L^{p}$, $1<p<\infty$, Shapiro [21, Theorem
4.2.1] characterized the functions $f\in L^{p}$ that are orthogonal to $Y$.
Now, it follows from the definition of OPA that $Qf-g$ is orthogonal to
$f\mathcal{P}_{n}$ if and only if $Q=q_{n,p}[f,g]$. This observation, along
with Shapiro’s result, allows us to generalize Proposition 3.2. We refer to
this generalization as the orthogonality condition in $L^{p}$.

######  Proposition 4.1 (Orthogonality Condition).

Let $1<p<\infty$, $f,g\in L^{p}$, $f\not\equiv 0$, $n\in\mathbb{N}$, and
$Q\in\mathcal{P}_{n}$. Then

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-g|^{p-1}\operatorname{sgn}(Qf-g)e^{-ikt}\overline{f}dt=0$

for $k=0,\dots,n$ if and only if $Q=q_{n,p}[f,g]$.

###### Proof.

Let $F:=Qf-g$. From [21, Theorem 4.2.1],

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|F|^{p-1}\overline{\operatorname{sgn}(F)}qfdt=0$
(10)

for all $q\in\mathcal{P}_{n}$ if and only if $F$ is orthogonal to
$f\mathcal{P}_{n}$. From our previous discussion, it follows that (10) is true
for all $q\in\mathcal{P}_{n}$ if and only if $Q=q_{n,p}[f,g]$. This is
equivalent to our assertion. ∎

One of the primary reasons Proposition 3.2 is useful is that it allows us to
easily compute the coefficients of OPAs in $L^{2}$; since Proposition 4.1 is a
generalization of Proposition 3.2, it’s natural to wonder if Proposition 4.1
can be used to readily compute the coefficients of OPAs in $L^{p}$. By
imposing certain conditions on $f$ and $p$, we will see in Section 6 that this
is the case.

On a different note, Proposition 4.1 can give us insight into OPAs of the form
$q_{n,p}[f,1]$, where $2\leq p<\infty$. Before we look into the details,
consider the following remark.

######  Remark 4.2.

Let $2\leq p<\infty$, $f\in L^{p}\setminus\\{0\\}$, and $n\in\mathbb{N}$.
Define the function $g:=\\{q_{n,p}[f,1]f-1\\}^{\frac{p-2}{2}}$. If $g\equiv
0$, then $q_{n,p}[f,1]$ is zero-free on $\mathbb{T}$. Moreover, if $f\in
H^{p}$ and $g\equiv 0$, then $q_{n,p}[f,1]$ is zero-free in
$\overline{\mathbb{D}}$.

We will reference this remark when we study the location of zeros in the next
section. Nevertheless, perhaps the most valuable insight into the zeros of
OPAs can be obtained by the following theorem.

######  Theorem 4.3.

Let $2\leq p<\infty$, $f\in L^{p}\setminus\\{0\\}$, $n\in\mathbb{N}$, and
$g:=\\{q_{n,p}[f,1]f-1\\}^{\frac{p-2}{2}}$. If $g\not\equiv 0$, then

$q_{n,p}[f,1]=q_{n,2}[fg,g].$

###### Proof.

Let $Q:=q_{n,p}[f,1]$. By Proposition 4.1, we have that

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt=0$

for $k=0,\dots,n$. Now, we can write

$\displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|$
${}^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-2}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}(Qf-1)^{\frac{p-2}{2}}(\overline{Qf}-1)^{\frac{p-2}{2}}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}g(Qf-1)e^{-ikt}\overline{fg}dt$
$\displaystyle=\langle Qfg-g,z^{k}fg\rangle.$

Thus, $\langle Qfg-g,z^{k}fg\rangle=0$ for $k=0,\dots,n$. By Proposition 3.2,
we conclude that $Q=q_{n,2}[fg,g]$. ∎

Theorem 4.3 suggests that we can gain a better understanding of OPAs in
$L^{p}$ by studying OPAs of the form $q_{n,2}[fg,g]$, where $f$ and $g$ are
arbitrary functions in $L^{2}$ such that $fg\in L^{2}\setminus\\{0\\}$; this
amounts to studying the unique polynomial that minimizes the norm
$\|\\{qf-1\\}g\|_{2}$, where $q$ varies over $\mathcal{P}_{n}$. As a
particular point of interest, we would like to know the necessary conditions
to impose on $g$ in order for $q_{n,2}[fg,g]$ to be zero-free in
$\overline{\mathbb{D}}$. The following example shows that $g$ cannot be
arbitrary.

######  Example 4.4 (Zero in $\mathbb{D}$).

Let $f(z)=2z+1$ and $g(z)=1-z$. Then
$q_{1,2}[fg,g]=-\frac{6}{35}z-\frac{1}{35}$.

To better align with our main motivation, let’s consider the case where $f\in
H^{p}$. If we assume that $2\leq p<\infty$, then it’s easy to see that the
function $g$ in Theorem 4.3 belongs to $L^{2}$. However, it’s not at all clear
if $g$ belongs to $H^{2}$. Nevertheless, the following proposition shows that
the OPA $q_{n,2}[fg,g]$ is equal to an OPA obtained by replacing $g$ with a
particular function in $H^{2}$.

######  Theorem 4.5.

Let $2\leq p<\infty$, $f\in H^{p}\setminus\\{0\\}$, $n\in\mathbb{N}$, and
$g:=\\{q_{n,p}[f,1]f-1\\}^{\frac{p-2}{2}}$. If $g\not\equiv 0$, then there
exists a function $h\in H^{2}$, which is zero-free in $\mathbb{D}$, such that

$q_{n,p}[f,1]=q_{n,2}[fh,h].$

###### Proof.

Define the function $g:=\\{q_{n,p}[f,1]f-1\\}^{\frac{p-2}{2}}$. Note that we
can write $q_{n,p}[f,1]f-1=u\Phi$, where $u$ is an inner function and $\Phi$
is an outer function. Let $h:=\Phi^{\frac{p-2}{2}}$. Since $\Phi$ is zero-free
in $\mathbb{D}$, it follows that $h\in H^{2}$. Furthermore, $h$ is zero-free
in $\mathbb{D}$. Now, since $g=u^{\frac{p-2}{2}}h$, we see that

$\displaystyle\|q_{n,2}[fh,h]fg-g\|_{2}$
$\displaystyle=\|q_{n,2}[fh,h]fh-h\|_{2}$
$\displaystyle=\inf_{q\in\mathcal{P}_{n}}\|qfh-h\|_{2}$
$\displaystyle=\inf_{q\in\mathcal{P}_{n}}\|qfg-g\|_{2}$
$\displaystyle=\|q_{n,2}[fg,g]fg-g\|_{2}.$

By uniqueness of OPAs and Theorem 4.3, the result follows.

∎

Theorem 4.5 indicates that it’s worth studying OPAs of the form
$q_{n,2}[fg,g]$, where $f$ and $g$ are arbitrary functions in $H^{2}$ such
that $fg\in H^{2}\setminus\\{0\\}$. Accordingly, we would like to address a
modified version of Open Question 1 .

######  Open Question 2.

Let $n\in\mathbb{N}$. For which functions $f,g\in H^{2}$ can we guarantee that
$q_{n,2}[fg,g]$ is zero-free in $\overline{\mathbb{D}}$?

##  5\. Location of Zeros

Our motivation to study the zeros of optimal polynomial approximants comes
from the early application of PLSI polynomials to digital filter design. One
of the most important aspects to designing a recursive digital filter is
ensuring filter stability; recall that a filter is stable if it corresponds to
a system in which bounded input yields bounded output. In 2D signal
processing, a stable filter can be obtained by a rational function of two
complex variables such that the denominator is zero-free in
$\overline{\mathbb{D}^{2}}$. In 1972, Justice, Shanks, and Treitel [17]
proposed that such a rational function can be created with the use of planar
least-squares inverse polynomials. Accordingly, they made the following
conjecture: for any polynomial $f(z,w)\not\equiv 0$, the associated PLSI
polynomial is zero-free in $\overline{\mathbb{D}^{2}}$. This conjecture became
known in the engineering community as Shanks’ Conjecture.

In 1975, Genin and Kamp [13] gave a counterexample to Shanks’ Conjecture. In
particular, they discovered a polynomial with zeros in $\mathbb{D}^{2}$ such
that the associated PLSI polynomial has zeros in $\mathbb{D}^{2}$. A few years
later, they developed a method [14] to construct polynomials with which the
associated PLSI polynomial is guaranteed to have zeros in $\mathbb{D}^{2}$. At
this point, it was unclear whether it was possible to impose conditions on the
polynomial $f(z,w)$ to guarantee that the associated PLSI polynomial is zero-
free in $\overline{\mathbb{D}^{2}}$. In 1980, Delsarte, Genin, and Kamp [11]
made the following conjecture: for any polynomial $f(z,w)$ that doesn’t vanish
in $\overline{\mathbb{D}^{2}}$, the associated PLSI polynomial is zero-free in
$\overline{\mathbb{D}^{2}}$. Although there have been several attempts to
prove this conjecture, it is still unresolved. This conjecture is referred to
as Weak Shanks’ Conjecture.

Since our primary motivation for writing this paper is to understand the zeros
of OPAs in the context of $H^{p}$, it’s natural for us to wonder if “Shanks-
type” results hold in this setting. In fact, we already know that such a
result holds in $H^{2}$. More specifically, we know that the OPA
$q_{n,2}[f,1]$ is zero-free in $\overline{\mathbb{D}}$ whenever $f\in H^{2}$
and $f(0)\neq 0$. With this in mind, we can start by asking if the OPA
$q_{n,p}[f,1]$ is zero-free in $\overline{\mathbb{D}}$ whenever $f\in H^{p}$
and $f(0)\neq 0$. Moreover, we can ask if it’s necessary to impose the
condition $f(0)\neq 0$. On a different note, if there are values of $n$ for
which the OPAs have a zero in $\mathbb{D}$, then we should ask if there’s a
limit to how far in $\mathbb{D}$ the zeros can lie. To begin our
investigation, we establish the following proposition.

######  Proposition 5.1.

Let $1<p<\infty$, $f\in H^{p}\setminus\\{0\\}$, and $n\in\mathbb{N}$. If there
exists some $z_{0}\in\mathbb{D}$ such that $q_{n,p}[f,1](z_{0})=0$, then

$\sqrt{1-\|q_{n,p}[f,1]f-1\|_{p}^{p}}\leq|z_{0}|.$

###### Proof.

Let $Q:=q_{n,p}[f,1]$. Suppose there exists some $z_{0}\in\mathbb{D}$ such
that $Q(z_{0})=0$. Define the analytic functions

$\varphi(z):=\frac{z_{0}-z}{1-\overline{z_{0}}z}\quad\text{and}\quad\psi(z):=\bigg{\\{}\frac{1-|z_{0}|^{2}}{(1-\overline{z_{0}}z)^{2}}\bigg{\\}}^{\frac{1}{p}}.$

Let $\psi C_{\varphi}:H^{p}\rightarrow H^{p}$ denote the weighted composition
operator $h(z)\mapsto\psi(z)h\big{(}\varphi(z)\big{)}$. Since $\big{|}\psi
C_{\varphi}(Qf-1)\big{|}^{p}$ is subharmonic in $\mathbb{D}$, it follows that

$\displaystyle 1-|z_{0}|^{2}$ $\displaystyle=\big{|}\psi
C_{\varphi}(Qf-1)(0)\big{|}^{p}$ $\displaystyle\leq\frac{1}{2\pi
i}\int_{\mathbb{T}}\Big{|}Q\Big{(}\frac{z_{0}-z}{1-\overline{z_{0}}z}\Big{)}f\Big{(}\frac{z_{0}-z}{1-\overline{z_{0}}z}\Big{)}-1\Big{|}^{p}\frac{1-|z_{0}|^{2}}{|1-\overline{z_{0}}z|^{2}}\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi i}\int_{\mathbb{T}}|Q(w)f(w)-1|^{p}\frac{dw}{w}$
$\displaystyle=\|Qf-1\|_{p}^{p}.$

Equivalently, we have that $\sqrt{1-\|Qf-1\|_{p}^{p}}\leq|z_{0}|$. ∎

Proposition 5.1 shows that if the OPA has a zero in $\mathbb{D}$ and the
quantity $\|q_{n,p}[f,1]f-1\|_{p}$ is small, then we can expect the zero to be
close to $\mathbb{T}$. Now, it’s still interesting to know which functions
$f\in H^{p}$ correspond to a quantity $\|q_{n,p}[f,1]f-1\|_{p}$ that’s close
to $1$. Of course, this quantity is equal to $1$ whenever $q_{n,p}[f,1]\equiv
0$. In the case where $p=2$, it is well-known that $q_{n,p}[f,1]\equiv 0$ if
and only if $f(0)=0$ (see, e.g., [5]). Interestingly enough, the following
proposition shows that this is true for all $1<p<\infty$.

######  Proposition 5.2.

Let $1<p<\infty$, $f\in H^{p}\setminus\\{0\\}$, and $n\in\mathbb{N}$. The
following are equivalent:

  1. (i)

$f(0)=0$

  2. (ii)

$q_{n,p}[f,1]\equiv 0$

  3. (iii)

$q_{n,p}[f,1](0)=0$

###### Proof.

Let $q_{n,p}:=q_{n,p}[f,1]$. We first suppose that $f(0)=0$. Since
$|q_{n,p}f-1|^{p}$ is subharmonic in $\mathbb{D}$, it follows that

$\displaystyle 1$ $\displaystyle=|q_{n,p}(0)f(0)-1|^{p}$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}|q_{n,p}f-1|^{p}dt$
$\displaystyle=\|q_{n,p}f-1\|_{p}^{p}.$

Since $\|q_{n,p}f-1\|_{p}\leq 1$, we have that $\|q_{n,p}f-1\|_{p}=1$. By
uniqueness of OPAs, it follows that $q_{n,p}\equiv 0$. This shows that $(i)$
implies $(ii)$.

To prove that $(ii)$ implies $(iii)$ is trivial. Now, suppose that
$q_{n,p}(0)=0$. In a similar argument as above, it follows that $q_{n,p}\equiv
0$. Therefore, Proposition 4.1 implies that

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(e^{it})}dt=0.$ (11)

By conjugating both sides of (11), we see that

$f(0)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{it})dt=0.$

This shows that $(iii)$ implies $(i)$. ∎

As a result of the previous two propositions, we are able to address the
question about how far in $\mathbb{D}$ the zeros of OPAs can lie.

######  Theorem 5.3.

Let $1<p<\infty$, $f\in H^{p}$, $f(0)\neq 0$, and $n\in\mathbb{N}$. Then
$q_{n,p}[f,1](z)$ is zero-free for $|z|<\sqrt{1-\|q_{0,p}[f,1]f-1\|_{p}^{p}}$.

###### Proof.

Note that since $f(0)\neq 0$, it follows from Proposition 5.2 that
$\|q_{0,p}[f,1]f-1\|_{p}<1$, i.e.,
$\big{\\{}z\in\mathbb{C}\colon|z|<\sqrt{1-\|q_{0,p}[f,1]f-1\|_{p}^{p}}\big{\\}}$
is nonempty. Now, suppose $q_{n,p}[f,1](z_{0})=0$ for some
$z_{0}\in\mathbb{D}$. As a result of Proposition 5.1 and the fact that
$\big{\\{}\|q_{k,p}[f,1]f-1\|_{p}\big{\\}}_{k=0}^{\infty}$ is a decreasing
sequence of nonnegative numbers, we conclude that

$\displaystyle\sqrt{1-\|q_{0,p}[f,1]f-1\|_{p}^{p}}$
$\displaystyle=\sqrt{1-\sup_{k\in\mathbb{N}}\|q_{k,p}[f,1]f-1\|_{p}^{p}}$
$\displaystyle=\inf_{k\in\mathbb{N}}\sqrt{1-\|q_{k,p}[f,1]f-1\|_{p}^{p}}$
$\displaystyle\leq\sqrt{1-\|q_{n,p}[f,1]f-1\|_{p}^{p}}$
$\displaystyle\leq|z_{0}|.$

Therefore, $q_{n,p}[f,1](z)$ is zero-free for
$|z|<\sqrt{1-\|q_{0,p}[f,1]f-1\|_{p}^{p}}$. ∎

For a function $f\in H^{p}$ with $f(0)\neq 0$, Theorem 5.3 gives us
considerable insight into the zeros of $q_{n,p}[f,1]$. In fact, it provides us
with the radius of a disk in which $q_{n,p}[f,1]$ is zero-free for every
$n\in\mathbb{N}$. Now, the existence of this disk is partially due to
Proposition 5.2: since $f(0)\neq 0$, we have that $q_{0,p}[f,1]\not\equiv 0$
and hence $\sqrt{1-\|q_{0,p}[f,1]f-1\|_{p}^{p}}>0$. Nonetheless, the following
result (which was noted in [5] for a particular class of weighted Hilbert
spaces of analytic functions) demonstrates another interesting consequence of
Proposition 5.2.

######  Proposition 5.4.

Let $1<p<\infty$. For each $n\in\mathbb{N}\setminus\\{0\\}$, let

$M_{n}:=\inf\\{|z|\colon q_{n,p}[f,1](z)=0\ \text{for at least one}\ f\in
H^{p}\ \text{with}\ f(0)\neq 0\\}.$

Then $M_{1}\leq M_{n}$.

###### Proof.

For each $n\in\mathbb{N}\setminus\\{0\\}$, define the set

$E_{n}:=\\{|z|\colon q_{n,p}[f,1](z)=0\ \text{for at least one}\ f\in H^{p}\
\text{with}\ f(0)\neq 0\\}.$

If $|z_{0}|\in E_{n}$, then there exists some $f\in H^{p}$, with $f(0)\neq 0$,
such that
$q_{n,p}[f,1](z_{0})=0$. Define the function

$h(z):=\frac{q_{n,p}[f,1](z)f(z)}{z-z_{0}},$

and note that

$\|q_{n,p}[f,1]f-1\|_{p}=\|(z-z_{0})h-1\|_{p}.$ (12)

Now, suppose there exists some $q\in\mathcal{P}_{1}$ such that

$\|(z-z_{0})h-1\|_{p}>\|qh-1\|_{p}.$

From (12), it follows that

$\|q_{n,p}[f,1]f-1\|_{p}>\|qh-1\|_{p}=\bigg{\|}q\frac{q_{n,p}[f,1]}{z-z_{0}}f-1\bigg{\|}_{p},$

a contradiction since $q\frac{q_{n,p}[f,1]}{z-z_{0}}$ is a polynomial of
degree at most $n$. We conclude that $q_{1,p}[h,1](z)=z-z_{0}$. Now, since
$f(0)\neq 0$, it follows from Proposition 5.2 that $q_{n,p}[f,1](0)\neq 0$.
This implies that $h(0)\neq 0$. We then see that $|z_{0}|\in E_{1}$, and hence
$E_{n}\subset E_{1}$. We conclude that $M_{1}\leq M_{n}$. ∎

In an effort to find a lower bound for the zeros of $q_{n,p}[f,1]$, where
$f\in H^{p}$ and $f(0)\neq 0$, Proposition 5.4 suggests that we should focus
our attention to the case where $n=1$. Now, by making the additional
assumption that $2\leq p<\infty$, we can gain insight into the zeros of all
the OPAs by implementing the test from Theorem 3.5. With this in mind, we pose
the following question.

######  Open Question 3.

Let $2\leq p<\infty$, $f\in H^{p}$, $f(0)\neq 0$,
$g:=\\{q_{1,p}[f,1]f-1\\}^{\frac{p-2}{2}}$, and $g\not\equiv 0$. Is it true
that $|\langle g,fg\rangle|>|\langle g,zfg\rangle|$?

Note that if $g\equiv 0$, then it follows from Remark 4.2 that $q_{1,p}[f,1]$
is zero-free in $\overline{\mathbb{D}}$. Otherwise, it follows from Theorems
3.5 and 4.3 that the OPA $q_{1,p}[f,1]$ is zero-free in
$\overline{\mathbb{D}}$ if and only if $|\langle g,fg\rangle|>|\langle
g,zfg\rangle|$. Therefore, if the answer to Open Question 3 is “Yes”, then we
could deduce from Propositions 5.2 and 5.4 that $q_{n,p}[f,1]$ is zero-free in
$\overline{\mathbb{D}}$ for every $n\in\mathbb{N}$ and every $f\in H^{p}$ with
$f(0)\neq 0$.

Although many of our discussions have been focusing on functions in $H^{p}$,
there are other functions in $L^{p}$ that are worthy of our attention, e.g.,
the modulus of any nontrivial function in $H^{p}$. In the following
proposition, we use Theorem 3.5 to gain insight into the zeros of OPAs
associated with some of these functions.

######  Proposition 5.5.

Let $2\leq p<\infty$. If $f$ is a real-valued nonnegative function in
$L^{p}\setminus\\{0\\}$, then $q_{1,p}[f,1]$ is zero-free in
$\overline{\mathbb{D}}$.

###### Proof.

Let $g:=\\{q_{1,p}[f,1]f-1\\}^{\frac{p-2}{2}}$. If $g\equiv 0$, then
$q_{1,p}[f,1]$ must be a nonzero constant, i.e., $q_{1,p}[f,1]$ is zero-free
in $\overline{\mathbb{D}}$. If $g\not\equiv 0$, then it follows from Theorem
4.3 that $q_{1,p}[f,1]=q_{1,2}[fg,g]$. Since

$\displaystyle|\langle g,zfg\rangle|$
$\displaystyle=\big{|}\big{\langle}g,zf^{\frac{1}{2}}f^{\frac{1}{2}}g\big{\rangle}\big{|}$
$\displaystyle=\big{|}\big{\langle}gf^{\frac{1}{2}},zf^{\frac{1}{2}}g\big{\rangle}\big{|}$
$\displaystyle<\big{\|}gf^{\frac{1}{2}}\big{\|}_{2}^{2}$
$\displaystyle=\big{|}\big{\langle}1,|g|^{2}f\big{\rangle}\big{|}$
$\displaystyle=|\langle g,fg\rangle|,$

we conclude from Theorem 3.5 that $q_{1,p}[f,1]$ is zero-free in
$\overline{\mathbb{D}}$. ∎

######  Example 5.6.

Let $f(z)=|1-z|^{2}$. From Proposition 5.5, it follows that $q_{1,2}[f,1]$ is
zero-free in $\overline{\mathbb{D}}$. More specifically, since
$f(z)=-z+2-z^{-1}$ on $\mathbb{T}$, we can easily show that
$q_{1,2}[f,1](z)=\frac{1}{10}z+\frac{2}{5}$.

As of this point in the paper, our computations of OPAs have only pertained to
the case where $p=2$. In the following section, we develop a method that
allows us to compute the coefficients of OPAs for many different values of
$p$; this method is based on the orthogonality condition that was presented in
Section 4.

##  6\. Computation of Coefficients

As we saw in Section 3, the coefficients of OPAs in $L^{2}$ can be computed by
using Proposition 3.2. In fact, these coefficients can be expressed as the
solution to a system of linear equations for which the associated matrix can
easily be inverted. Now, recall that Proposition 4.1 is a generalization of
Proposition 3.2 to the $L^{p}$ setting. Therefore, one would hope that this
result could be used to compute the coefficients of OPAs in $L^{p}$; after
making restrictions on the functions and considering specific values of $p$,
we find that we are able to solve for the coefficients numerically; in light
of Proposition 5.4, we will focus most of our computations on the case where
$n=1$.

Before we begin the numerical computations, let’s consider an example in which
$n=0$.

######  Example 6.1.

Let $f(z)=1-z$. Then $q_{0,p}[f,1]\equiv\frac{1}{2}$ for all integers $p>1$.

###### Proof.

Let $Q:=\frac{1}{2}$. Note that

$\displaystyle\frac{1}{2\pi}$
$\displaystyle\int_{-\pi}^{\pi}|Qf-1|^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-2}(Qf-1)\overline{f}dt$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\bigg{|}-\frac{1}{2}z-\frac{1}{2}\bigg{|}^{p-2}\bigg{(}-\frac{1}{2}z-\frac{1}{2}\bigg{)}(1-\overline{z})\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\bigg{(}-\frac{1}{2}z-\frac{1}{2}\bigg{)}^{\frac{p-2}{2}}\bigg{(}-\frac{1}{2}\overline{z}-\frac{1}{2}\bigg{)}^{\frac{p-2}{2}}\bigg{(}-\frac{1}{2}z-\frac{1}{2}\bigg{)}(1-\overline{z})\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\bigg{(}-\frac{1}{2}z-\frac{1}{2}\bigg{)}^{p-1}\frac{z-1}{z^{\frac{p}{2}}}\frac{dz}{z}$
$\displaystyle=\bigg{(}-\frac{1}{2}\bigg{)}^{p-1}\frac{1}{2\pi
i}\int_{\mathbb{T}}\Bigg{[}\sum_{j=0}^{p-1}\binom{p-1}{j}z^{j}\Bigg{]}\frac{z-1}{z^{\frac{p}{2}}}\frac{dz}{z}$
$\displaystyle=\bigg{(}-\frac{1}{2}\bigg{)}^{p-1}\frac{1}{2\pi
i}\sum_{j=0}^{p-1}\binom{p-1}{j}\bigg{[}\int_{\mathbb{T}}z^{j-(\frac{p-2}{2})}\frac{dz}{z}-\int_{\mathbb{T}}z^{j-\frac{p}{2}}\frac{dz}{z}\bigg{]}.$
(13)

If $p$ is odd, then each integral in (6) is zero for every $j$. In this case,

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt=0.$

If $p$ is even, then the first integral in (6) is nonzero only when
$j=\frac{p-2}{2}$. The second integral is nonzero only when $j=\frac{p}{2}$.
Since

$\binom{p-1}{\frac{p-2}{2}}=\frac{(p-1)!}{\big{(}\frac{p}{2}\big{)}!\,\big{(}\frac{p-2}{2}\big{)}!}=\binom{p-1}{\frac{p}{2}},$

it follows that

$\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt=0.$

In either case, we conclude from Proposition 4.1 that

$Q=q_{0,p}[f,1]\equiv\frac{1}{2}.$

∎

Let us now begin to formulate a general setting to carry out our computations;
for simplicity, we will only consider OPAs of the form $q_{n,p}[f,1]$. We will
see that it works to our advantage to make certain assumptions on the function
$f$. One of our assumptions will be based on the observation that conjugating
a function’s Fourier coefficients does not change its norm. More specifically,
if $f$ belongs to $L^{p}$ and has Fourier series
$\sum_{k=-\infty}^{\infty}c_{k}z^{k}$, then the function $\tilde{f}$ with
Fourier series $\sum_{k=-\infty}^{\infty}\overline{c}_{k}z^{k}$ belongs to
$L^{p}$ and satisfies $\|\tilde{f}\|_{p}=\|f\|_{p}$. This allows us to deduce
the following result.

######  Proposition 6.2.

Let $1<p<\infty$, $f\in L^{p}\setminus\\{0\\}$, and $n\in\mathbb{N}$. If the
Fourier coefficients of $f$ are all real, then the coefficients of
$q_{n,p}[f,1]$ are all real.

###### Proof.

Let $q_{n,p}:=q_{n,p}[f,1]$. Then we see that

$\displaystyle\inf_{q\in\mathcal{P}_{n}}\|qf-1\|_{p}$
$\displaystyle=\|q_{n,p}f-1\|_{p}$ $\displaystyle=\|(q_{n,p}f-1)^{\sim}\|_{p}$
$\displaystyle=\|\tilde{q}_{n,p}f-1\|_{p}.$

By uniqueness of OPAs, we have that $q_{n,p}=\tilde{q}_{n,p}$. ∎

In the computations that later follow, we will make the assumption that the
Fourier coefficients of the function $f$ are all real. Under this assumption,
Proposition 6.2 will drastically simplify the equations involved in solving
for the OPA coefficients. Nonetheless, the following theorem gives us a method
for computing the coefficients of OPAs whenever $f$ is a polynomial and $p$ is
even.

######  Theorem 6.3.

Let $n,N\in\mathbb{N}$ and $Q\in\mathcal{P}_{n}$. Suppose that $f$ is a
polynomial of degree at most $N$ and $p\geq 2$ is even. Define the polynomials
$G(z):=Q(z)f(z)-1$,
$R(z):=z^{n+N}\overline{G\big{(}\frac{1}{\overline{z}}\big{)}}$, and
$P(z):=z^{N}G(z)^{\frac{p}{2}}R(z)^{\frac{p-2}{2}}\overline{f\big{(}\frac{1}{\overline{z}}\big{)}}$.
Then

$\left.\frac{d^{k}}{dz^{k}}P(z)\right|_{z=0}=0$

for $\frac{p}{2}(n+N)-n\leq k\leq\frac{p}{2}(n+N)$ if and only if
$Q=q_{n,p}[f,1]$.

###### Proof.

Note that

$\displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|$
${}^{p-1}\operatorname{sgn}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}|Qf-1|^{p-2}(Qf-1)e^{-ikt}\overline{f}dt$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\frac{(Qf-1)^{\frac{p}{2}}(\overline{Qf}-1)^{\frac{p-2}{2}}\overline{f}}{z^{k}}\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\frac{z^{N}G(z)^{\frac{p}{2}}R(z)^{\frac{p-2}{2}}\overline{f\big{(}\frac{1}{\overline{z}}\big{)}}}{z^{k+\frac{p}{2}(n+N)-n}}\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{T}}\frac{P(z)}{z^{k+\frac{p}{2}(n+N)-n}}\frac{dz}{z}.$

Therefore, it follows from Proposition 4.1 that

$\frac{1}{2\pi
i}\int_{\mathbb{T}}\frac{P(z)}{z^{k+\frac{p}{2}(n+N)-n}}\frac{dz}{z}=0\quad\text{for}\quad
k=0,\dots,n$ (14)

if and only if $Q=q_{n,p}[f,1]$. Since $P$ is a polynomial, we can easily see
that (14) is equivalent to

$\left.\frac{d^{k}}{dz^{k}}P(z)\right|_{z=0}=0\quad\text{for}\quad\frac{p}{2}(n+N)-n\leq
k\leq\frac{p}{2}(n+N).$ (15)

Hence, (15) is true if and only if $Q=q_{n,p}[f,1]$. ∎

Let’s have a look at an example to demonstrate this method.

######  Example 6.4.

Let $f(z)=1-z$; let $z_{0}$ be the zero of $q_{1,p}[f,1]$.

  1. (i)

$q_{1,2}[f,1](z)\approx 0.3333333333z+0.6666666667,\quad z_{0}\approx-2.00000$

  2. (ii)

$q_{1,4}[f,1](z)\approx 0.3734388420z+0.6265611579,\quad z_{0}\approx-1.67781$

  3. (iii)

$q_{1,6}[f,1](z)\approx 0.3964823122z+0.6035176878,\quad z_{0}\approx-1.52218$

  4. (iv)

$q_{1,8}[f,1](z)\approx 0.4117075962z+0.5882924038,\quad z_{0}\approx-1.42891$

  5. (v)

$q_{1,10}[f,1](z)\approx 0.4226290585z+0.5773709415,\quad
z_{0}\approx-1.36614$

  6. (vi)

$q_{1,20}[f,1](z)\approx 0.4336619002z+0.5313377953,\quad
z_{0}\approx-1.22524$

  7. (vii)

$q_{1,30}[f,1](z)\approx 0.2117474393z+0.2701449218,\quad
z_{0}\approx-1.27578$

###### Proof.

We only show the work for (ii). However, note the similarity between the OPA
in (i) and the OPA in Example 3.4. Let $Q:=q_{1,4}[f,1]$. Define the
polynomials $G(z)$, $R(z)$, and $P(z)$ as in Theorem 6.3. By setting
$q_{1,4}[f,1]=a_{1}z+a_{0}$, it is easy to see that

$G(z)=-a_{1}z^{2}+(a_{1}-a_{0})z+(a_{0}-1).$

Moreover, we have that

$z\overline{f\bigg{(}\frac{1}{\overline{z}}\bigg{)}}=z-1.$

Now, it follows from Proposition 6.2 that $a_{0}$ and $a_{1}$ are real.
Therefore,

$\displaystyle R(z)$
$\displaystyle=z^{2}\overline{G\bigg{(}\frac{1}{\overline{z}}\bigg{)}}$
$\displaystyle=z^{2}G\bigg{(}\frac{1}{z}\bigg{)}$
$\displaystyle=(a_{0}-1)z^{2}+(a_{1}-a_{0})z-a_{1}.$

We thus find that

$P(z)=\big{\\{}-a_{1}z^{2}+(a_{1}-a_{0})z+(a_{0}-1)\big{\\}}^{2}\big{\\{}(a_{0}-1)z^{2}+(a_{1}-a_{0})z-a_{1}\big{\\}}(z-1).$

To solve for $a_{0}$ and $a_{1}$, we set

$\left.\frac{d^{k}}{dz^{k}}P(z)\right|_{z=0}=0$

for $k=3$ and $k=4$. This gives us the nonlinear equations

$\displaystyle-24a_{1}^{3}+78a_{1}^{2}a_{0}$
$\displaystyle-72a_{1}a_{0}^{2}+36a_{0}^{3}$
$\displaystyle-36a_{1}^{2}+60a_{1}a_{0}-54a_{0}^{2}-12a_{1}+30a_{0}-6=0$

and

$\displaystyle 144a_{1}^{3}-288a_{1}^{2}a_{0}$
$\displaystyle+312a_{1}a_{0}^{2}-96a_{0}^{3}$
$\displaystyle+72a_{1}^{2}-288a_{1}a_{0}+120a_{0}^{2}+96a_{1}-48a_{0}=0.$

Through the use of numerical methods, we find that $a_{1}\approx 0.3734388420$
and $a_{0}\approx 0.6265611579$.

∎

In the above example, it seems plausible that the zeros of $q_{1,p}[f,1]$
remain outside of $\overline{\mathbb{D}}$ for all $p$. Moreover, the zeros
seem to converge as $p\rightarrow\infty$. As a general question, for a
polynomial $f$ with $f(0)\neq 0$, do the zeros of $q_{1,p}[f,1]$ converge as
$p\rightarrow\infty$? Under these conditions, it would be interesting to
determine the smallest disk that contains all the zeros. Perhaps this disk is
disjoint from the closed unit disk $\overline{\mathbb{D}}$.

##  7\. Concluding Remarks

The polynomials obtained by least-squares approximation have been a topic of
interest to engineers and mathematicians over the last several decades. In the
engineering community, many have studied these polynomials in reference to the
Hardy spaces $H^{2}(\mathbb{D}^{2})$ and $H^{2}(\mathbb{D})$. In the
mathematics community, many have studied these polynomials in reference to the
Dirichlet-type spaces $\mathcal{D}_{\alpha}$, $\alpha\in\mathbb{R}$. In this
paper, we introduced an analogous version of these polynomials in the space
$L^{p}$, $1\leq p\leq\infty$. Our main motivation for doing so was to try and
extend “Shanks-type” results to a larger collection of functions. As a result
of our many discussions, we are convinced that such a result holds for
functions in $H^{p}$. Accordingly, we make the following conjecture.

######  Conjecture 7.1.

Let $1<p<\infty$, $f\in H^{p}$, and $n\in\mathbb{N}$. If $f(0)\neq 0$, then
$q_{n,p}[f,1]$ is zero-free in $\overline{\mathbb{D}}$.

For the case where $2\leq p<\infty$, we can approach this conjecture through
the lens of the Hilbert space $L^{2}$; this is because we can express the OPA
$q_{n,p}[f,1]$ in terms of an OPA in $L^{2}$. In particular, one could
formulate the theory through the use of Moore-Penrose inverses (see [16] for a
discussion in the $H^{2}$ setting). More specifically, for any $f\in
L^{\infty}\setminus\\{0\\}$, $g\in L^{2}$, and $n\in\mathbb{N}$, one could
write $q_{n,2}[f,g]=(M_{f}E_{n})^{\dagger}(g)$, where $M_{f}:L^{2}\rightarrow
L^{2}$ is the multiplication operator defined by $h\mapsto fh$,
$E_{n}:L^{2}\rightarrow\mathcal{P}_{n}$ is the orthogonal projection of
$L^{2}$ onto $\mathcal{P}_{n}$, and $(M_{f}E_{n})^{\dagger}$ is the Moore-
Penrose inverse of $M_{f}E_{n}$. Accordingly, one could take an operator
theoretic approach toward gaining insight into the OPA.

Acknowledgments. I would like to thank Catherine Bénéteau for the many
enlightening conversations that we had about these topics. Additionally, I
would like to thank her, Dmitry Khavinson, and the referees for reviewing the
draft of this paper and providing valuable feedback.

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