# Onesided, intertwining, positive and copositive polynomial approximation
with interpolatory constraints††thanks: AMS classification: 41A05, 41A10,
41A15, 41A25, 41A29 Keywords and phrases: Intertwining, copositive, positive,
nonnegative, Hermite interpolation, approximation by algebraic polynomials,
exact pointwise estimates

German Dzyubenko Institute of Mathematics NAS of Ukraine (dzyuben@gmail.com).
Supported in part by NAS of Ukraine project 0123U100853, and by Johns Hopkins
University support to U4U. Kirill A. Kopotun Department of Mathematics,
University of Manitoba, Winnipeg, Manitoba, R3T 2N2, Canada
(Crimea_is_Ukraine@shaw.ca). Supported in part by NSERC grant
RGPIN-05678-2020.

###### Abstract

Given $k\in\mathbb{N}$, a nonnegative function $f\in C^{r}[a,b]$, $r\geq 0$,
an arbitrary finite collection of points
$\big{\\{}\alpha_{i}\big{\\}}_{i\in{\mathcal{J}}}\subset[a,b]$, and a
corresponding collection of nonnegative integers
$\big{\\{}m_{i}\big{\\}}_{i\in{\mathcal{J}}}$ with $0\leq m_{i}\leq r$,
$i\in{\mathcal{J}}$, is it true that, for sufficiently large $n\in\mathbb{N}$,
there exists a polynomial $P_{n}$ of degree $n$ such that

  1. (i)

$|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x);[a,b])$,
$x\in[a,b]$, where $\rho_{n}(x):=n^{-1}\sqrt{1-x^{2}}+n^{-2}$ and
${\omega}_{k}$ is the classical $k$-th modulus of smoothness,

  2. (ii)

$P^{(\nu)}(\alpha_{i})=f^{(\nu)}(\alpha_{i})$, for all $0\leq\nu\leq m_{i}$
and all $i\in{\mathcal{J}}$,

  3. and

  4. (iii)

either $P\geq f$ on $[a,b]$ (_onesided_ approximation), or $P\geq 0$ on
$[a,b]$ (_positive_ approximation)?

We provide precise answers not only to this question, but also to similar
questions for more general intertwining and copositive polynomial
approximation. It turns out that many of these answers are quite unexpected.

We also show that, in general, similar questions for $q$-monotone
approximation with $q\geq 1$ have negative answers, i.e., $q$-monotone
approximation with general interpolatory constraints is impossible if $q\geq
1$.

###### Contents

  1. 1 Introduction and Main Results
    1. 1.1 If $q\geq 1$, then $q$-monotone approximation with general interpolatory constraints is impossible
    2. 1.2 Positive, copositive, onesided and intertwining approximation with interpolatory constraints: definitions and a motivational example
    3. 1.3 Summary of main results
  2. 2 Main results
    1. 2.1 Onesided and intertwining approximation with interpolatory constraints
    2. 2.2 Positive and copositive approximation with interpolatory constraints
  3. 3 Negative results
    1. 3.1 Onesided and intertwining approximation with interpolatory constraints
    2. 3.2 Positive and copositive approximation with interpolatory constraints
    3. 3.3 Negative results for copositive approximation with interpolation at the points in $Y_{s}$
  4. 4 Spline approximation with interpolatory constraints
    1. 4.1 Local construction
    2. 4.2 Global construction
  5. 5 Intertwining polynomial approximation of truncated power functions and splines with interpolatory constraints
    1. 5.1 Approximation of truncated power functions
    2. 5.2 Approximation of general splines
  6. 6 Proofs of the main results
  7. 7 Appendix: Exact pointwise estimates

##  1 Introduction and Main Results

In this paper, we are interested in shape preserving approximation (see e.g.
[klps2011] for a brief introduction to this area) with an additional
restriction that approximating polynomials interpolate the function that is
being approximated at a given set of points.

Without any shape restrictions, we can easily do this and also preserve the
rate of best approximation in the uniform norm (it is rather evident that this
is no longer possible in the integral norms). Indeed, it is sufficient to
correct the polynomial of best approximation using an appropriate Lagrange
polynomial obtaining a near best polynomial approximant satisfying the needed
interpolation conditions (the constant will necessarily depend on the minimum
distance among all interpolation points).

Hence, the following question is rather natural: If $f$ has a certain shape
that we wish to preserve while approximating it by polynomials, is
interpolation at a given set of points possible and, if so, what can be said
about errors of approximation of $f$ by such polynomials? Our goal is to
provide answers to this question.

In order to discuss this further, we now recall some standard notations and
definitions (see e.g. [klps2011]). We say that a function $f$ is $q$-monotone
on an interval $J$ and write $f\in\Delta^{(q)}(J)$) if, for all collections of
$q+1$ distinct points $t_{0},t_{1},\dots,t_{q}$ in $J$, the divided
differences $[t_{0},\dots,t_{q};f]$ are nonnegative. In particular,
$\Delta^{(0)}(J)$, $\Delta^{(1)}(J)$ and $\Delta^{(2)}(J)$ are the sets of all
nonnegative, nondecreasing and convex functions on $J$, respectively. If
$q\geq 3$, then functions $f$ from $\Delta^{(q)}(J)$ are $(q-2)$-times
differentiable in $\mathop{\rm int}\nolimits J$ (the interior of $J$) and
$f^{(q-2)}\in\Delta^{(2)}(\mathop{\rm int}\nolimits J)$.

Let $\mathbb{Y}_{s}$, $s\in\mathbb{N}$, be the set of all collections
$Y_{s}:=\big{\\{}y_{i}\big{\\}}_{i=1}^{s}$ of points $y_{i}$, such that
$-1<y_{s}<\dots<y_{1}<1$, and define $y_{0}:=1$ and $y_{s+1}:=-1$. We denote
by $\Delta^{(q)}(Y_{s})$ the set of functions which change their
$q$-monotonicity at the points in $Y_{s}$. More precisely,

$f\in\Delta^{(q)}(Y_{s})\quad\iff\quad(-1)^{i}f\in\Delta^{(q)}([y_{i+1},y_{i}]),\;0\leq
i\leq s.$

Also, if $s=0$, then $\mathbb{Y}_{0}:=\\{\emptyset\\}$ and
$\Delta^{(q)}(Y_{0}):=\Delta^{(q)}(I)$, where $I:=[-1,1]$.

Let $C^{r}:=C^{r}(I)$ denote the space of all $r$-times continuously
differentiable functions on $I$, and let $\mathbb{A}_{p}$, $p\in\mathbb{N}$,
be the set of all collections $A_{p}:=\big{\\{}\alpha_{i}\big{\\}}_{i=1}^{p}$
of points $\alpha_{i}$ such that $-1\leq\alpha_{p}<\dots<\alpha_{1}\leq 1$.
For a given $f\in C^{r}(I)$, $r\in\mathbb{N}_{0}$, we denote

${\mathcal{I}}^{(r)}(f,A_{p}):=\left\\{g\in
C^{r}(I)\;\;\big{|}\;\;g^{(i)}(\alpha)=f^{(i)}(\alpha),\;\text{for all
$\alpha\in A_{p}$ and $0\leq i\leq r$}\right\\}.$

In particular, if $r=0$, we use the notation
${\mathcal{I}}(f,A_{p}):={\mathcal{I}}^{(0)}(f,A_{p})$. For convenience, we
also denote $\mathbb{A}_{0}:=\\{\emptyset\\}$. Obviously, if $f\in C^{r}(I)$,
$r\geq 1$, then
${\mathcal{I}}^{(m+1)}(f,A_{p})\subset{\mathcal{I}}^{(m)}(f,A_{p})$, for all
$0\leq m\leq r-1$.

As usual, $\Pi_{n}$ is the set of all polynomials of degree $\leq n$,
$\left\|\cdot\right\|_{J}$ is the uniform norm on $J$,
${\omega}_{k}(f,\cdot;J)$ is the $k$-th modulus of smoothness of $f$ on $J$,
$\rho_{n}(x):=n^{-1}\sqrt{1-x^{2}}+n^{-2}$ and
$\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$. Throughout this paper, $c(\dots)$
denote positive constants which depend only on the parameters in the
parentheses. Also, the interval $I=[-1,1]$ will be omitted from the notation,
i.e., $\left\|\cdot\right\|:=\left\|\cdot\right\|_{I}$,
${\omega}_{k}(f,\cdot):={\omega}_{k}(f,\cdot;I)$, etc.

Throughout this paper, if sets $Y_{s}$ and $A_{p}$ both appear in a statement,
it is assumed that they have no points in common, i.e., the restriction
$Y_{s}\cap A_{p}=\emptyset$ is always assumed but is not explicitly stated.
Note that this has no influence on the generality of the obtained results and
is done for convenience only.

###  1.1 If $q\geq 1$, then $q$-monotone approximation with general
interpolatory constraints is impossible

We start with the following observation that implies that $q$-monotone
polynomial approximation with interpolatory constraints is, in general, not
possible if $q\geq 1$.

######  Lemma 1.1.

For any $q\in\mathbb{N}$, $s\in\mathbb{N}_{0}$ and $Y_{s}\in\mathbb{Y}_{s}$,
there exists $f\in\Delta^{(q)}(Y_{s})$ and a set $A_{p}\in\mathbb{A}_{p}$,
$p\geq q+2$, such that, for any $n\in\mathbb{N}_{0}$,

$\Delta^{(q)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})\cap\Pi_{n}=\emptyset.$

###### Proof.

Given $q\in\mathbb{N}$, $s\in\mathbb{N}_{0}$ and $Y_{s}\in\mathbb{Y}_{s}$, we
set $\xi:=(1+y_{1})/2$ and pick $A_{p}$, $p\geq q+2$, so that
$y_{1}<\alpha_{q+2}<\dots<\alpha_{2}<\xi<\alpha_{1}$. Then,
$f:=(\cdot-\xi)_{+}^{q}$ is clearly in $\Delta^{(q)}(Y_{s})$. We now recall
the following property of $q$-monotone functions (see [b]*Theorem 5):

> Let $f\in\Delta^{(q)}(J)$, $q\in\mathbb{N}$, and let
> $l_{q-1}:=l_{q-1}(z_{q},\dots,z_{1})\in\Pi_{q-1}$ be the Lagrange polynomial
> interpolating $f$ at $q$ distinct points in $J$:
> $z_{q}<z_{q-1}<\dots<z_{1}$. Then, $(-1)^{i}\left(f(x)-l_{q-1}(x)\right)\geq
> 0$, for $x\in[z_{i+1},z_{i}]$, $1\leq i\leq q$, and $f(x)-l_{q-1}(x)\geq 0$,
> for $x\geq z_{1}$. In other words,
>
> (1.1) $f-l_{q-1}\in\Delta^{(0)}\left(\\{z_{i}\\}_{i=1}^{q}\right).$

Note that this result is a generalization of the property that a convex
function always lies below its chord.

Suppose now that
$p_{n}\in\Delta^{(q)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})\cap\Pi_{n}$. Then,
$p_{n}$ is $q$-monotone on $[y_{1},1]$, $p_{n}(\alpha_{i})=0$, $2\leq i\leq
q+2$ and $p_{n}(\alpha_{1})=f(\alpha_{1})>0$. Now, polynomials
$\ell_{1}:=l_{q-1}(\alpha_{q+2},\dots,\alpha_{3})\equiv 0$ and
$\ell_{2}:=l_{q-1}(\alpha_{q+2},\dots,\alpha_{4},\alpha_{2})\equiv 0$ (in the
case $q=1$, $\ell_{2}:=l_{0}(\alpha_{2})\equiv 0$) interpolate $p_{n}$ at $q$
points in $[y_{1},1]$, and so it follows from (1.1) that
$p_{n}(x)=p_{n}(x)-\ell_{1}(x)\geq 0$, for $x\geq\alpha_{3}$, and
$-p_{n}(x)=(-1)\left(p_{n}(x)-\ell_{2}(x)\right)\geq 0$, for $\alpha_{4}\leq
x\leq\alpha_{2}$. Therefore, $p_{n}(x)=0$, $\alpha_{3}\leq x\leq\alpha_{2}$,
and so $p_{n}\equiv 0$. Hence, $p_{n}(\alpha_{1})>0$ does not hold. This
contradiction implies that such a polynomial $p_{n}$ does not exist. ∎

We remark that, if the number of points $p$ in $A_{p}\in\mathbb{A}_{p}$ is
sufficiently small, then $q$-monotone approximation with interpolatory
constraints may be possible, but it may or may not be as good as $q$-monotone
approximation without interpolation at the points in $A_{p}$. For example, if
$q\geq 1$, then denoting

$E_{n}^{(q)}(f,Y_{s};A_{p}):=\inf\left\\{\left\|f-p_{n}\right\|\;\;\big{|}\;\;p_{n}\in\Delta^{(q)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})\cap\Pi_{n}\right\\},$

we have

$E_{n}^{(q)}(f,Y_{s};A_{1})\leq 2E_{n}^{(q)}(f,Y_{s};\emptyset),\quad\text{for
any $A_{1}=\\{\alpha\\}\subset I$,}$

and

$E_{n}^{(q)}(f,Y_{s};A_{q+1})=1,\quad\text{if $Y_{s}\subset[-1,0)$,
$f(x)=x_{+}^{q}\in\Delta^{(q)}(Y_{s})$ and $A_{q+1}\subset(y_{1},0)$.}$

###  1.2 Positive, copositive, onesided and intertwining approximation with
interpolatory constraints: definitions and a motivational example

Note that the case for $q=0$ (copositive approximation) is not covered by
Lemma 1.1. In fact, as shown below, (co)positive polynomial approximation
requiring interpolation on general sets $A_{p}$ is always possible. However,
its rate of approximation is often worse than the rate of (co)positive
approximation without these additional interpolatory constraints.

Another type of shape preserving approximation that is related to (co)positive
approximation is the so-called onesided and, more generally, intertwining
approximation. We say (see [hky]; as far as we know, this is where
intertwining approximation was first introduced and discussed) that
$\\{P,Q\\}$ is an intertwining pair of functions (polynomials, splines, etc.)
for $f$ with respect to $Y_{s}\in\mathbb{Y}_{s}$ if

$P-f,f-Q\in\Delta^{(0)}(Y_{s}).$

In particular, in the case $s=0$, we have $Q\leq f\leq P$ on $I$, and
approximation of $f$ by $P$ and $Q$ is usually referred to as onesided
approximation.

The error of intertwining/onesided polynomial approximation is usually defined
as the infimum of $\left\|P-Q\right\|$ taken over all intertwining pairs of
polynomials for $f$. In this paper, we also work with an equivalent quantity
(and equivalent corresponding pointwise estimates)

$\widetilde{E}_{n}(f,Y_{s}):=\inf\left\\{\|f-P\|\;\;\big{|}\;\;P\in\Pi_{n}\cap\widetilde{\Delta}(f,Y_{s})\right\\},$

where

$\widetilde{\Delta}(f,Y_{s}):=\left\\{P:I\mapsto\mathbb{R}\;\;\big{|}\;\;P-f\in\Delta^{(0)}(Y_{s})\right\\}.$

The error of onesided approximation is denoted by
$\widetilde{E}_{n}(f):=\widetilde{E}_{n}(f,Y_{0})$.

Estimates for intertwining/onesided approximation immediately imply those for
copositive/positive approximation since, if $f\in\Delta^{(0)}(Y_{s})$ and
$P\in\widetilde{\Delta}(f,Y_{s})$, then $P=(P-f)+f\in\Delta^{(0)}(Y_{s})$.
However, as we show below, there are many cases when copositive (polynomial)
approximation (with or without interpolatory constraints) is possible while
intertwining approximation is not.

It is well known that the ‘correct’ pointwise estimates for polynomial
approximation need to involve the quantity $\rho_{n}(x)$ since, without it,
matching inverse results would not be possible. Namely, the classical Timan-
Dzyadyk-Freud-Brudnyi direct theorem for the approximation by algebraic
polynomials can be stated as follows (see e.g. [kls21] for discussions):

> if $k\in\mathbb{N}$, $r\in\mathbb{N}_{0}$ and $f\in C^{r}$, then for each
> $n\geq k+r-1$ there is a polynomial $p_{n}\in\Pi_{n}$ satisfying
>
> (1.2) $|f(x)-p_{n}(x)|\leq
> c(k,r)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\quad x\in I.$

Hence, estimates of type (1.2) is what we are interested in while discussing
approximation with interpolatory constraints (see also Section 7 for
discussions of how interpolation yields further improvement of these
estimates).

Illustrative example: Suppose that $r\in\mathbb{N}_{0}$ is fixed, and two
functions $f,g\in C^{r}$ are such that $g\in\Delta^{(0)}(I)$ and
$f\in\Delta^{(0)}(\\{0\\})$ (i.e., $g\geq 0$ on $[-1,1]$, and $f\leq 0$ on
$[-1,0]$ and $f\geq 0$ on $[0,1]$), and recall that, for $0\leq m\leq r$,
${\mathcal{I}}^{(m)}(f,\\{\alpha\\})$ denotes the set of functions $h\in
C^{m}$ such that $h^{(i)}(\alpha)=f^{(i)}(\alpha)$ for all $0\leq i\leq m$.

We are interested in precise answers to the following questions:

  1. Q1.

For which $m$ does there exist
$p_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m)}(g,\\{1/2\\})$ such that $p_{n}\geq g$
on $I$ and (1.2) (with $f$ replaced by $g$) holds? (_onesided approximation_)

  2. Q2.

For which $m$ does there exist
$p_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m)}(g,\\{1/2\\})$ such that $p_{n}\geq 0$
on $I$ and (1.2) (with $f$ replaced by $g$) holds? (_positive approximation_)

  3. Q3.

For which $m_{1}$ and $m_{2}$ does there exist
$p_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m_{1})}(f,\\{1/2\\})\cap{\mathcal{I}}^{(m_{2})}(f,\\{0\\})$
such that $p_{n}\leq f$ on $[-1,0]$, $p_{n}\geq f$ on $[0,1]$, and (1.2)
holds? (_intertwining approximation_)

  4. Q4.

For which $m_{1}$ and $m_{2}$ does there exist
$p_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m_{1})}(f,\\{1/2\\})\cap{\mathcal{I}}^{(m_{2})}(f,\\{0\\})$
such that $p_{n}\in\Delta^{(0)}(Y_{s})$ and (1.2) holds? (_copositive
approximation_)

The answers are rather unexpected:

  1. A1.

Such $p_{n}$ does not exist if $r=0$ or $r=1$. If $r\geq 2$ is even, then
$p_{n}$ exists if $m=r-1$ and does not exist if $m=r$. If $r\geq 3$ is odd,
then $p_{n}$ exists if $m=r-2$ and does not exist if $m=r-1$.

  2. A2.

If $r=0$ and $m=0$, then such $p_{n}$ exists but only if $k=2$ ($k$ is the
order of the modulus of smoothness in (1.2)). If $k=3$, then such $p_{n}$ no
longer exists. If $r=1$ and $m=0$, then such $p_{n}$ exists if $k=3$ and does
not exist if $k=4$. If $r=1$ and $m=1$, then such $p_{n}$ does not exist (for
any $k$). If $r\geq 2$, then the answers are the same as in “A1”.

  3. A3.

Such $p_{n}$ does not exist if $r=0$ or $r=1$. If $r\geq 2$ is even, then such
$p_{n}$ exists if $m_{1}=r-1$ and $m_{2}=r-2$, and does not exist if either
$m_{1}=r$ or $m_{2}=r-1$. If $r\geq 3$ is odd, then such $p_{n}$ exists if
$m_{1}=r-2$ and $m_{2}=r-1$, and does not exist if either $m_{1}=r-1$ or
$m_{2}=r$.

  4. A4.

This is the most involved answer.

     * $\mathbf{r=0}$:

If $m_{1}=m_{2}=0$, then such $p_{n}$ exists if $k=2$ and does not exist if
$k=3$.

     * $\mathbf{r=1}$:

If $m_{1}=m_{2}=0$, then such $p_{n}$ exists if $k=3$ and does not exist if
$k=4$. If $m_{1}=1$, then such $p_{n}$ does not exist. If $m_{1}=0$ and
$m_{2}=1$, then such $p_{n}$ exists if $k=2$ and does not exist if $k=3$.

     * $\mathbf{r=2}$:

If $m_{1}=1$ and $m_{2}=0$, then such $p_{n}$ exists (for any $k$). However,
$m_{1}=1$ cannot be increased to $2$. If $m_{1}=1$ and $m_{2}=1$, then such
$p_{n}$ exists if $k=3$ and does not exist if $k=4$. If $m_{2}=2$, then such
$p_{n}$ does not exist.

     * $\mathbf{r\geq 3}$:

The answers are the same as in “A3”.

###  1.3 Summary of main results

We start this section with the summary of all known results involving
pointwise estimates for (co)positive, onesided and intertwining polynomial
approximation without any additional interpolatory restrictions.

Table 1: Approximation without interpolatory constraints Onesided
approximation
---
$f\in C$ |  $\exists$ $P_{n}\geq f$: $|f(x)-P_{n}(x)|\leq c{\omega}_{k}(f,\rho_{n}(x))$ |  [hky]*Theorem 1
Positive approximation
$f\in C$ | $\exists$ $P_{n}\geq 0$: $|f(x)-P_{n}(x)|\leq c{\omega}_{k}(f,\rho_{n}(x))$ |  [dz96]*Theorem 3,
or [hky]*Corollary 2
Intertwining approximation
$f\in C$ | $\widetilde{E}_{n}(f,Y_{s})\not\leq c\|f\|$ |  [hky]*Theorem 13
$f\in C^{1}$ |  $\exists$ $P_{n}\in\widetilde{\Delta}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ |  [hky]*Theorem 5
Copositive approximation
$f\in C$ |  $\exists$ $P_{n}$, copositive with $f$: $|f(x)-P_{n}(x)|\leq c{\omega}_{3}(f,\rho_{n}(x))$ |  [hky]*Theorem 7
${\omega}_{3}(f,\rho_{n}(x))$ cannot be replaced by ${\omega}_{4}(f,n^{-1})$ | Zhou [zhou, zhou-atappl]
$f\in C^{1}$ |  $\exists$ $P_{n}$, copositive with $f$: |  [dz96]*Theorem 1,
$|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ | or [hky]*Corollary 6

The main goal of this paper is to investigate what changes in Table
LABEL:tablewithout are needed if we add the requirement that approximating
polynomials also interpolate the function and possibly its derivatives (i.e.,
Hermite interpolation) at the points in $A_{p}\in\mathbb{A}_{p}$.
Additionally, it is clear that any continuous function from
$\widetilde{\Delta}(f,Y_{s})$ automatically interpolates $f$ at all points in
$Y_{s}\in\mathbb{Y}_{s}$ (i.e.,
$\widetilde{\Delta}(f,Y_{s})\subset{\mathcal{I}}(f,Y_{s})$), and we also
investigate if interpolation of derivatives of $f$ at the points in $Y_{s}$ is
possible as well as its limitations.

It turns out that whether optimal estimates are possible depends not only on
the smoothness of $f$, but also on whether the order of the highest derivative
of $f$ is even or odd, and on the location of the interpolation points (i.e.,
on whether $A_{p}$ contains any points in the interior of $I$). Hermite
interpolation at the points from $Y_{s}$ is even more involved.

For clarity, we summarize our main results in the following four tables.

Table 2: Onesided approximation Onesided approximation with interpolatory
constraints
---
$f\in C$ | not possible in general | Lemma 3.1 with $r=0$
$f\in C^{1}$ | not possible if $A_{p}\cap\mathop{\rm int}\nolimits I\neq\emptyset$ | Lemma 3.1(i) with $r=1$
$\exists$ $P_{n}\geq f$ and $P_{n}\in{\mathcal{I}}(f,\\{\pm 1\\})$: | Theorem 2.2 with
$|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ | $(r,m_{1},m_{3})=(1,-1,0)$
${\mathcal{I}}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,\\{\pm 1\\})$ | Lemma 3.1(ii) ($r=1$)
$f\in C^{r}$, $r\geq 2$ is even |  $\exists$ $P_{n}\geq f$ and $P_{n}\in{\mathcal{I}}^{(r-1)}(f,A_{p})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.2 with
$(r,m_{1},m_{3})$
$=(2j,2j-1,2j-1)$
${\mathcal{I}}^{(r-1)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,A_{p})$ | Lemma 3.1
$f\in C^{r}$, $r\geq 3$ is odd |  $\exists$ $P_{n}\geq f$ and $P_{n}\in{\mathcal{I}}^{(r-2)}(f,A_{p})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$
${\mathcal{I}}^{(r-2)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,A_{p})$ unless $A_{p}=\\{\pm 1\\}$ |  Lemma 3.1(i)
$\exists$ $P_{n}\geq f$ and $P_{n}\in{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.2 with
$(r,m_{1},m_{3})$
$=(2j+1,2j-1,2j)$
${\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,\\{\pm 1\\})$ | Lemma 3.1(ii)
Table 3: Positive approximation Positive approximation with interpolatory
constraints ($f\geq 0$)
---
|  $\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}(f,A_{p})$: | Theorem 2.4 with
$f\in C$ | $|f(x)-P_{n}(x)|\leq c{\omega}_{2}(f,\rho_{n}(x))$ | $(r,k,m_{1},m_{3})=(0,2,0,0)$
|  ${\omega}_{2}(f,\rho_{n}(x))$ cannot be replaced by ${\omega}_{3}(f,\rho_{n}(x))$ | Lemma 3.3
$f\in C^{1}$ |  $\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}(f,A_{p})$: | Theorem 2.4 with
$|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{3}(f^{\prime},\rho_{n}(x))$ | $(r,k,m_{1},m_{3})=(1,3,0,0)$
${\omega}_{3}(f^{\prime},\rho_{n}(x))$ cannot be replaced by ${\omega}_{4}(f^{\prime},\rho_{n}(x))$ unless $A_{p}=\\{\pm 1\\}$ | Lemma 3.5
${\mathcal{I}}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,A_{p})$ | Lemma 3.9
$\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}(f,\\{\pm 1\\})$: | Theorem 2.4 with
$|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ | $(r,k,m_{1},m_{3})=(1,k,-1,0)$
${\mathcal{I}}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,\\{\pm 1\\})$ | Lemma 3.9
$f\in C^{r}$, $r\geq 2$ is even |  $\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}^{(r-1)}(f,A_{p})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.4 with
$(r,k,m_{1},m_{3})$
$=(2j,k,2j-1,2j-1)$
${\mathcal{I}}^{(r-1)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,A_{p})$ | Lemma 3.6
$f\in C^{r}$, $r\geq 3$ is odd |  $\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}^{(r-2)}(f,A_{p})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$
${\mathcal{I}}^{(r-2)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,A_{p})$ unless $A_{p}=\\{\pm 1\\}$ | Lemma 3.8
$\exists$ $P_{n}\geq 0$ and $P_{n}\in{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.4 with
$(r,k,m_{1},m_{3})$
$=(2j+1,k,2j-1,2j)$
${\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,\\{\pm 1\\})$ | Lemma 3.9
Table 4: Intertwining approximation Intertwining approximation with
interpolatory constraints
---
$f\in C$ | not possible in general | Lemma 3.1 ($r=0$)
$f\in C^{1}$ | not possible if $A_{p}\cap\mathop{\rm int}\nolimits I\neq\emptyset$ | Lemma 3.1(i) ($r=1$)
$\exists$ $P_{n}\in\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ | Theorem 2.1 with
$(r,m_{1},m_{2},m_{3})$
$=(1,-1,0,0)$
${\mathcal{I}}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,\\{\pm 1\\})$ | Lemma 3.1(ii) ($r=1$)
$\widetilde{\Delta}(f,Y_{s})$ cannot be replaced by $\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(1)}(f,Y_{s})$ | Lemma 3.2 ($r=1$)
$f\in C^{r}$, $r\geq 2$ is even |  $\exists$ $P_{n}\in\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(r-1)}(f,A_{p})\cap{\mathcal{I}}^{(r-2)}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.1 with
$(r,m_{1},m_{2},m_{3})$
$=(2j,2j-1,2j-2,2j-1)$
${\mathcal{I}}^{(r-1)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,A_{p})$ | Lemma 3.1
${\mathcal{I}}^{(r-2)}(f,Y_{s})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,Y_{s})$ | Lemma 3.2
$f\in C^{r}$, $r\geq 3$ is odd |  $\exists$ $P_{n}\in\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,Y_{s})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$
${\mathcal{I}}^{(r-2)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,A_{p})$ unless $A_{p}=\\{\pm 1\\}$ |  Lemma 3.1(i)
${\mathcal{I}}^{(r-1)}(f,Y_{s})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,Y_{s})$ | Lemma 3.2
$\exists$ $P_{n}\in\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,Y_{s})\cap{\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.1 with
$(r,m_{1},m_{2},m_{3})$
$=(2j+1,2j-1,2j,2j)$
${\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,\\{\pm 1\\})$ | Lemma 3.1(ii)
Table 5: Copositive approximation Copositive approximation with interpolatory
constraints ($f\in\Delta^{(0)}(Y_{s})$)
---
$f\in C$ |  $\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})$: $|f(x)-P_{n}(x)|\leq c{\omega}_{2}(f,\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(0,2,0,0,0)$
${\omega}_{2}(f,\rho_{n}(x))$ cannot be replaced by ${\omega}_{3}(f,\rho_{n}(x))$ | Lemma 3.4
$f\in C^{1}$ |  $\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{3}(f^{\prime},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(1,3,0,0,0)$
${\omega}_{3}(f^{\prime},\rho_{n}(x))$ cannot be replaced by ${\omega}_{4}(f^{\prime},\rho_{n}(x))$ unless $A_{p}=\\{\pm 1\\}$ |  Lemma 3.5
${\mathcal{I}}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,A_{p})$ | Lemma 3.9
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{k}(f^{\prime},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(1,k,-1,0,0)$
${\mathcal{I}}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(1)}(f,\\{\pm 1\\})$ | Lemma 3.9
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(1)}(f,Y_{s})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{2}(f^{\prime},\rho_{n}(x))$
${\omega}_{2}(f^{\prime},\rho_{n}(x))$ cannot be replaced by ${\omega}_{3}(f^{\prime},\rho_{n}(x))$ | Lemma 3.10
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}(f,A_{p})\cap{\mathcal{I}}^{(1)}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}(x){\omega}_{2}(f^{\prime},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(1,2,0,1,0)$
$f\in C^{2}$ |  $\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(1)}(f,A_{p})\cap{\mathcal{I}}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{2}(x){\omega}_{k}(f^{\prime\prime},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(2,k,1,0,1)$
${\mathcal{I}}^{(1)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(2)}(f,A_{p})$ | Lemma 3.6
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(1)}(f,Y_{s})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq
c\rho_{n}^{2}(x){\omega}_{3}(f^{\prime\prime},\rho_{n}(x))$
${\omega}_{3}(f^{\prime\prime},\rho_{n}(x))$ cannot be replaced by ${\omega}_{4}(f^{\prime\prime},\rho_{n}(x))$ | Lemma 3.11
${\mathcal{I}}^{(1)}(f,Y_{s})$ cannot be replaced with ${\mathcal{I}}^{(2)}(f,Y_{s})$ | Lemma 3.12
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(1)}(f,A_{p})\cap{\mathcal{I}}^{(1)}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{2}(x){\omega}_{3}(f^{\prime\prime},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(2,3,1,1,1)$
$f\in C^{r}$, $r\geq 3$ is odd |  $\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,Y_{s})$: |  follows from a stronger statement below
$|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$
${\mathcal{I}}^{(r-2)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,A_{p})$ unless $A_{p}=\\{\pm 1\\}$ |  Lemma 3.8
${\mathcal{I}}^{(r-1)}(f,Y_{s})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,Y_{s})$ | Lemma 3.13
$\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(r-2)}(f,A_{p})\cap{\mathcal{I}}^{(r-1)}(f,Y_{s})\cap{\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})$
$=(2j+1,k,2j-1,2j,2j)$
${\mathcal{I}}^{(r-1)}(f,\\{\pm 1\\})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,\\{\pm 1\\})$ | Lemma 3.9
$f\in C^{r}$, $r\geq 4$ is even |  $\exists$ $P_{n}\in\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(r-1)}(f,A_{p})\cap{\mathcal{I}}^{(r-2)}(f,Y_{s})$: $|f(x)-P_{n}(x)|\leq c\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x))$ | Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})=$
$(2j,k,2j-1,2j-2,2j-1)$
${\mathcal{I}}^{(r-1)}(f,A_{p})$ cannot be replaced by ${\mathcal{I}}^{(r)}(f,A_{p})$ | Lemma 3.6
${\mathcal{I}}^{(r-2)}(f,Y_{s})$ cannot be replaced by ${\mathcal{I}}^{(r-1)}(f,Y_{s})$ |  Lemma 3.14

For $Y_{s}\in\mathbb{Y}_{s}$ and $A_{p}\in\mathbb{A}_{p}$, recalling that it
is always assumed that $Y_{s}\cap A_{p}=\emptyset$, it is convenient to denote

(1.3) $Y_{s}\cup A_{p}=:\left\\{\beta_{i}\right\\}_{i=1}^{s+p},\quad\text{with
$-1\leq\beta_{s+p}<\beta_{s+p-1}<\dots<\beta_{2}<\beta_{1}\leq 1$,}$

and

(1.4) $\displaystyle d(Y_{s},A_{p})$
$\displaystyle:=\min\left\\{|u-v|\;\;\big{|}\;\;u,v\in Y_{s}\cup
A_{p}\cup\\{\pm 1\\}\quad\mbox{\rm and}\quad u\neq v\right\\}$
$\displaystyle=\mathop{\rm
pmin}\nolimits\left\\{\beta_{i-1}-\beta_{i}\;\;\big{|}\;\;1\leq i\leq
s+p+1\right\\},$

where $\beta_{0}:=1$, $\beta_{s+p+1}:=-1$, and $\mathop{\rm pmin}\nolimits(S)$
is the smallest positive number from the set $S$ of nonnegative reals. In
particular, $d(Y_{s}):=d(Y_{s},\emptyset)$ and $d(A_{p}):=d(\emptyset,A_{p})$.

It follows from Remark 3.7 as well as Lemmas 4.4 and 4.6 that, in general,
dependence of various constants in theorems below on $d(Y_{s},A_{p})$ cannot
be removed.

##  2 Main results

###  2.1 Onesided and intertwining approximation with interpolatory
constraints

For convenience, we introduce the following notation:

$\Upsilon_{\text{intertwining}}:=\bigcup_{r\geq 1}\Upsilon_{r},$

where

$\displaystyle\Upsilon_{r}:=\left\\{(r,m_{1},m_{2},m_{3})\;\;\big{|}\;\;\right.$
$\displaystyle m_{1}=r-1,\;m_{2}=r-2,\;m_{3}=r-1\;\text{if $r$ is even; }$
$\displaystyle\left.m_{1}=r-2,\;m_{2}=r-1,\;m_{3}=r-1\;\text{if $r$ is
odd}\right\\}.$

Hence,

(2.1) $\displaystyle\Upsilon_{\text{intertwining}}=$
$\displaystyle\left\\{(2j,2j-1,2j-2,2j-1)\;\;\big{|}\;\;j\in\mathbb{N}\right\\}$
$\displaystyle\cup\left\\{(2j+1,2j-1,2j,2j)\;\;\big{|}\;\;j\in\mathbb{N}_{0}\right\\}.$

Note that, when $r=1$,

$\Upsilon_{1}:=\left\\{(1,-1,0,0)\right\\},$

which is the only case when any of $m_{i}$’s can become negative.

The following is our main result (or rather a collection of results combined
in one statement) for intertwining approximation with interpolatory
constraints.

######  Theorem 2.1 (intertwining approximation with interpolatory
constraints).

Let $k,p,s\in\mathbb{N}$, $Y_{s}\in\mathbb{Y}_{s}$, $A_{p}\in\mathbb{A}_{p}$,
and $(r,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}}$, defined in
(2.1). Then, for any $f\in C^{r}$ and any $n\geq c(d(Y_{s},A_{p}))$, there
exists

$P_{n}\in\Pi_{n}\cap\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I,$

where ${\mathcal{I}}^{(-1)}(f,A_{p}):=\left\\{h:I\mapsto\mathbb{R}\right\\}$,
the set of all real valued functions on $I$, i.e., the restriction
$P_{n}\in{\mathcal{I}}^{(-1)}(f,A_{p})$ is vacuous, and so there is no
requirement that $P_{n}$ interpolate $f$ at the points in $A_{p}$ if
$m_{1}=-1$.

The results for onesided approximation are similar and could have been
included in the statement of Theorem 2.1 by allowing $s$ to be $0$ and making
the restriction $P_{n}\in{\mathcal{I}}^{(m_{2})}(f,Y_{0})$ vacuous.
Nevertheless, in order to avoid possible confusion and for readers’ benefit,
we provide an explicit statement.

Let

(2.2) $\displaystyle\Upsilon_{\text{onesided}}:=$
$\displaystyle\left\\{(r,m_{1},m_{3})\;\;\big{|}\;\;(r,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}},\;\text{for
some $m_{2}\geq 0$}\right\\}$ $\displaystyle=$
$\displaystyle\left\\{(2j,2j-1,2j-1)\;\;\big{|}\;\;j\in\mathbb{N}\right\\}\cup\left\\{(2j+1,2j-1,2j)\;\;\big{|}\;\;j\in\mathbb{N}_{0}\right\\}.$

######  Theorem 2.2 (onesided approximation with interpolatory constraints).

Let $k,p\in\mathbb{N}$, $A_{p}\in\mathbb{A}_{p}$, and
$(r,m_{1},m_{3})\in\Upsilon_{\text{onesided}}$, defined in (2.2). Then, for
any $f\in C^{r}$ and any $n\geq c(d(A_{p}))$, there exists

$P_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that $P_{n}\geq f$ on $I$, and

$|f(x)-P_{n}(x)|\leq
c(k,r,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I.$

Here, as in the statement of Theorem 2.1, there is no requirement that $P_{n}$
interpolate $f$ at the points in $A_{p}$ if $m_{1}=-1$.

It follows from Lemmas 3.1 and 3.2 that Theorems 2.1 and 2.2 cannot be
strengthened by increasing any of the parameters $m_{i}$’s (for any
$r\in\mathbb{N}$), or by including the case for $r=0$.

###  2.2 Positive and copositive approximation with interpolatory constraints

Similarly to the previous section, we start with introducing the notation for
the set of indices $\Upsilon^{(0)}_{\text{copositive}}$. However, in order to
capture all statements in Table LABEL:coptable, this notation needs to be more
complicated and involve the order of the modulus $k$. We denote

$\Upsilon^{(0)}_{\text{copositive}}:=\bigcup_{r\geq 0}\Upsilon^{(0)}_{r},$

where

$\displaystyle\Upsilon^{(0)}_{0}$ $\displaystyle:=\\{(0,2,0,0,0)\\},$
$\displaystyle\Upsilon^{(0)}_{1}$
$\displaystyle:=\\{(1,3,0,0,0),(1,2,0,1,0)\\}\cup\left\\{(1,k,-1,0,0)\;\;\big{|}\;\;k\in\mathbb{N}\right\\},$
$\displaystyle\Upsilon^{(0)}_{2}$
$\displaystyle:=\\{(2,3,1,1,1)\\}\cup\left\\{(2,k,1,0,1)\;\;\big{|}\;\;k\in\mathbb{N}\right\\},$

and, for $r\geq 3$,

$\displaystyle\Upsilon^{(0)}_{r}:=\left\\{(r,k,m_{1},m_{2},m_{3})\;\;\big{|}\;\;\right.$
$\displaystyle\left.k\in\mathbb{N}\quad\mbox{\rm
and}\quad(r,m_{1},m_{2},m_{3})\in\Upsilon_{r}\right\\}$
$\displaystyle=\left\\{(r,k,m_{1},m_{2},m_{3})\;\;\big{|}\;\;\right.$
$\displaystyle
k\in\mathbb{N},\;\;\text{and}\;\;m_{1}=r-1,\;m_{2}=r-2,\;m_{3}=r-1\;\text{if
$r$ is even; }$
$\displaystyle\left.m_{1}=r-2,\;m_{2}=r-1,\;m_{3}=r-1\;\text{if $r$ is
odd}\right\\}.$

Hence,

(2.3) $\displaystyle\Upsilon^{(0)}_{\text{copositive}}=$
$\displaystyle\left\\{(0,2,0,0,0),(1,3,0,0,0),(1,2,0,1,0),(2,3,1,1,1)\right\\}$
$\displaystyle\cup\left\\{(2j,k,2j-1,2j-2,2j-1)\;\;\big{|}\;\;j,k\in\mathbb{N}\right\\}$
$\displaystyle\cup\left\\{(2j+1,k,2j-1,2j,2j)\;\;\big{|}\;\;j\in\mathbb{N}_{0},\;k\in\mathbb{N}\right\\}.$

######  Theorem 2.3 (copositive approximation with interpolatory constraints).

Let $p,s\in\mathbb{N}$, $Y_{s}\in\mathbb{Y}_{s}$, $A_{p}\in\mathbb{A}_{p}$,
and $(r,k,m_{1},m_{2},m_{3})\in\Upsilon^{(0)}_{\text{copositive}}$, defined in
(2.3). Then, for any $f\in C^{r}\cap\Delta^{(0)}(Y_{s})$ and any $n\geq
c(d(Y_{s},A_{p}))$, there exists

$P_{n}\in\Pi_{n}\cap\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I.$

Here, as in the statement of Theorem 2.1, there is no requirement that $P_{n}$
interpolate $f$ at the points in $A_{p}$ if $m_{1}=-1$.

As in the case for intertwining and onesided approximation, the results for
positive approximation are similar to those in the copositive case and could
have been included in the statement of Theorem 2.3 by allowing $s$ to be $0$
and making the restriction $P_{n}\in{\mathcal{I}}^{(m_{2})}(f,Y_{0})$ vacuous.
Again, we provide an explicit statement for readers’ convenience.

Let

(2.4) $\displaystyle\Upsilon^{(0)}_{\text{positive}}:=$
$\displaystyle\left\\{(r,k,m_{1},m_{3})\;\;\big{|}\;\;(r,k,m_{1},m_{2},m_{3})\in\Upsilon^{(0)}_{\text{copositive}},\;\text{for
some $m_{2}\geq 0$}\right\\}$ $\displaystyle=$
$\displaystyle\left\\{(0,2,0,0),(1,2,0,0),(1,3,0,0),(2,3,1,1)\right\\}$
$\displaystyle\cup\left\\{(2j,k,2j-1,2j-1)\;\;\big{|}\;\;j,k\in\mathbb{N}\right\\}$
$\displaystyle\cup\left\\{(2j+1,k,2j-1,2j)\;\;\big{|}\;\;j\in\mathbb{N}_{0},\;k\in\mathbb{N}\right\\}.$

######  Theorem 2.4 (positive approximation with interpolatory constraints).

Let $p\in\mathbb{N}$, $A_{p}\in\mathbb{A}_{p}$, and
$(r,k,m_{1},m_{3})\in\Upsilon^{(0)}_{\text{positive}}$, defined in (2.4).
Then, for any $f\in C^{r}$ such that $f\geq 0$ on $I$, and any $n\geq
c(d(A_{p}))$, there exists

$P_{n}\in\Pi_{n}\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that $P_{n}\geq 0$ on $I$, and

$|f(x)-P_{n}(x)|\leq
c(k,r,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I.$

Here, as in the statements above, there is no requirement that $P_{n}$
interpolate $f$ at the points in $A_{p}$ if $m_{1}=-1$.

It follows from lemmas in Sections 3.2 and 3.3 that Theorems 2.3 and 2.4
cannot be strengthened by increasing any of the parameters $k$ or $m_{i}$’s,
for any $r\geq 0$.

##  3 Negative results

Recall that, for functions $g:I\mapsto\mathbb{R}$ with absolutely continuous
$(r-1)$st derivatives on $I$, the following (Taylor’s formula) holds:

$g(x)=\sum_{i=0}^{r-1}\frac{1}{i!}g^{(i)}(a)(x-a)^{i}+\frac{1}{(r-1)!}\int_{a}^{x}(x-t)^{r-1}g^{(r)}(t)dt,\quad
x\in I.$

We often use this elementary identity without explicitly mentioning it.

###  3.1 Onesided and intertwining approximation with interpolatory
constraints

The purpose of the following lemmas is to establish general negative results
for onesided and intertwining approximation with interpolatory constraints.

######  Lemma 3.1 (local onesided approximation with interpolatory
constraints: negative result for $f\in C^{r}$ with $r\geq 0$).

Let $r\in\mathbb{N}_{0}$. Then,

  1. (i)

if $\lambda$ is an interior point of $I$ (i.e., $\lambda\in(-1,1)$), then
there exists a function $f\in C^{r}$ such that $f\equiv 0$ on $[-1,\lambda]$,
and

${\mathcal{I}}^{({\tilde{r}})}(f,\\{\lambda\\})\cap\left\\{g\in
C^{r+1}\;\;\big{|}\;\;\text{$g\geq f$ on
$(\lambda-\varepsilon,\lambda+\varepsilon)$, for some
$\varepsilon>0$}\right\\}=\emptyset,$

where

${\tilde{r}}:=2\lfloor r/2\rfloor=\begin{cases}r-1\,,&\text{if $r$ is odd,}\\\
r\,,&\text{if $r$ is even.}\end{cases}$

  2. (ii)

if $\lambda$ is an endpoint of $I$ (i.e., $\lambda=\pm 1$), then there exists
$f\in C^{r}$ such that

${\mathcal{I}}^{(r)}(f,\\{\lambda\\})\cap\left\\{g\in
C^{r+1}\;\;\big{|}\;\;\text{$g\geq f$ on
$(\lambda-\varepsilon,\lambda+\varepsilon)\cap I$, for some
$\varepsilon>0$}\right\\}=\emptyset.$

###### Proof.

If $\lambda$ is an interior point of $I$ then, without loss of generality, we
assume that $\lambda=0$ and define

$f(x)=\frac{1}{(r-1)!}\int_{0}^{x}(x-t)^{r-1}f^{(r)}(t)dt,\quad\text{where}\quad
f^{(r)}(x):=\begin{cases}-x\ln x,&\text{if $0<x\leq 1$,}\\\ 0,&\text{if
$-1\leq x\leq 0$.}\end{cases}$

Clearly, $f\in C^{r}$ and $f\equiv 0$ on $[-1,0]$. Now, suppose that, for some
$\varepsilon>0$, there exists $g\in C^{r+1}$ satisfying $g(x)\geq f(x)$,
$x\in(-\varepsilon,\varepsilon)$, and such that $g^{(i)}(0)=f^{(i)}(0)=0$,
$0\leq i\leq{\tilde{r}}$. Then, since ${\tilde{r}}\geq r-1$,

$\displaystyle g(x)-f(x)$ $\displaystyle=$
$\displaystyle\frac{1}{(r-1)!}\int_{0}^{x}(x-t)^{r-1}\left[g^{(r)}(t)-f^{(r)}(t)\right]dt.$

If $r$ is even, then ${\tilde{r}}=r$, and so $g^{(r)}(0)=0$, and we will now
show that this equality also holds if $r$ is odd. Indeed, for
$-\varepsilon<x<0$,

$0\leq g(x)-f(x)=\frac{(-1)^{r}}{(r-1)!}\int_{x}^{0}(t-x)^{r-1}g^{(r)}(t)dt,$

and so, by continuity of $g^{(r)}$, we must have $(-1)^{r}g^{(r)}(0)\geq 0$.
Similarly, because $g\geq f$ on $(0,\varepsilon)$, we should have
$g^{(r)}(0)\geq f^{(r)}(0)=0$. Hence, if $r$ is odd, this implies that
$g^{(r)}(0)=0$.

Thus, for $x>0$, we have

$\displaystyle g(x)-f(x)$ $\displaystyle=$
$\displaystyle\frac{1}{(r-1)!}\int_{0}^{x}(x-t)^{r-1}\left[g^{(r)}(t)+t\ln
t\right]dt$ $\displaystyle=$
$\displaystyle\frac{1}{(r-1)!}\int_{0}^{x}(x-t)^{r-1}\int_{0}^{t}\left(g^{(r+1)}(u)+\ln
u+1\right)\,du\,dt.$

Since $g\in C^{r+1}$ and $\lim_{u\to 0^{+}}\ln u=-\infty$, there exists
$\delta>0$ such that $g^{(r+1)}(u)+\ln u+1<0$, for all $0<u<\delta$. Hence,
$g(x)-f(x)<0$, for $0<x<\delta$, which is a contradiction.

If $\lambda$ is an endpoint of $I$, then the same proof as above works with no
changes (for example, one can consider $[0,1]$ instead of $I$, $\lambda=0$,
and then arrive at a contradiction using (3.1)). ∎

The same proof can be used to verify validity of the following lemma that is
used for negative results involving Hermite interpolation at the points in
$Y_{s}$.

######  Lemma 3.2 (local intertwining approximation with interpolation at
$Y_{s}$: negative result for $f\in C^{r}$ with $r\geq 1$).

Let $r\in\mathbb{N}$ and suppose that $\lambda\in(-1,1)$. Then there exists a
function $f\in C^{r}$ such that $f\equiv 0$ on $[-1,\lambda]$, and

${\mathcal{I}}^{({\hat{r}})}(f,\\{\lambda\\})\cap\left\\{g\in
C^{r+1}\;\;\big{|}\;\;\text{$g\leq f$ on $(\lambda-\varepsilon,\lambda]$, and
$g\geq f$ on $[\lambda,\lambda+\varepsilon)$, for some
$\varepsilon>0$}\right\\}=\emptyset,$

where

${\hat{r}}:=2\lceil r/2\rceil-1=\begin{cases}r\,,&\text{if $r$ is odd,}\\\
r-1\,,&\text{if $r$ is even.}\end{cases}$

We remark that in all cases that are not covered by the statements of Lemmas
3.1 and 3.2 onesided/intertwining polynomial approximation with interpolatory
constraints is possible (see Theorems 2.1 and 2.2).

Note that the only reason for including the property that $f\equiv 0$ on
$[-1,\lambda]$ in the statements of Lemma 3.1 (if $\lambda$ is an interior
point of $I$) and Lemma 3.2 is to indicate that these lemmas are still valid
if $f$ is assumed to be from the class $\Delta^{(0)}(Y_{s})$ (with
$y_{1}\leq\lambda$). In particular, this is important for the illustrative
example in Section 1.2.

###  3.2 Positive and copositive approximation with interpolatory constraints

The following is the first lemma in this paper that provides a negative result
for polynomial approximation with interpolatory constraints. Even though, in
some sense, it is less general than Lemma 3.4 that follows it, its proof is
rather simple and illustrates the main ideas that are used in the proofs of
more complicated negative results below but without the technical details
required there.

######  Lemma 3.3 (positive approximation with interpolatory constraints:
negative result for $f\in C$).

For any $n\in\mathbb{N}$, $A>0$ and $\lambda\in I$, there exists $f\in C$
which is nonnegative on $I$ and such that, for any nonnegative on $I$
polynomial $P_{n}\in\Pi_{n}$ satisfying $P_{n}(\lambda)=f(\lambda)$, we have

$\left\|f-P_{n}\right\|>A{\omega}_{3}(f,1).$

###### Proof.

Let $n\in\mathbb{N}$ and $A>0$ be fixed, and without loss of generality
suppose that $\lambda\in[0,1]$. We now let $\varepsilon\in(0,\lambda/2)$, and
pick
$f(x):=\max\left\\{(x-\lambda)(x-\lambda+\varepsilon)/\varepsilon^{2},0\right\\}$
and suppose that $P_{n}\in\Pi_{n}$ is nonnegative on $I$, satisfies
$P_{n}(\lambda)=f(\lambda)=0$ (this necessarily implies that
$P_{n}^{\prime}(\lambda)\leq 0$, and actually $P_{n}^{\prime}(\lambda)=0$ if
$\lambda\neq 1$) and such that $\left\|f-P_{n}\right\|\leq
A{\omega}_{3}(f,1)$. If
$Q(x):=(x-\lambda)(x-\lambda+\varepsilon)/\varepsilon^{2}$, then
$\left\|f-Q\right\|=\left\|Q\right\|_{[\lambda-\varepsilon,\lambda]}=1/4$ and

${\omega}_{3}(f,1)={\omega}_{3}(f-Q,1)\leq 8\left\|f-Q\right\|=2.$

Hence,

$\left\|P_{n}-Q\right\|\leq\left\|P_{n}-f\right\|+\left\|f-Q\right\|\leq
2A+1/4,$

and so, by Markov’s inequality,

$1/\varepsilon=Q^{\prime}(\lambda)\leq
Q^{\prime}(\lambda)-P_{n}^{\prime}(\lambda)\leq\left\|Q^{\prime}-P_{n}^{\prime}\right\|\leq(2A+1/4)(\max\\{n,2\\})^{2}.$

We now get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

######  Lemma 3.4 (copositive approximation with interpolatory constraints:
negative result for $f\in C$).

For any $A>1$, $\lambda\in[0,1]$, $n_{0}\in\mathbb{N}$, and any positive
sequence $(\alpha_{n})_{n\in\mathbb{N}}$ converging to $0$, there exist
$n\in\mathbb{N}$, $n\geq n_{0}$, and $f=f_{n}\in C$ such that $f\geq 0$ on
$I$, $f\equiv 0$ on $[-1,-1/2]$, and for any polynomial $P_{n}\in\Pi_{n}$
which is nonnegative in some nonempty neighborhood of $\lambda$ and satisfies
$P_{n}(\lambda)=f(\lambda)$, we have

$\left\|\frac{f-P_{n}}{{\omega}_{3}(f,\rho_{n})}\right\|>A\,\alpha_{n}\sqrt{n}.$

###### Proof.

Recall that the classical “glue function” is

${\widetilde{G}}(x):={\widetilde{G}}(x;[a,b]):=\frac{G(x;[a,b])}{G(1;[a,b])},\quad\text{where}\quad
G(x;[a,b]):=\int_{-1}^{x}g(t;(a+b)/2,(b-a)/2)\,dt,$
$g(x;x_{0},d):=\begin{cases}\exp(d^{2}/((x-x_{0})^{2}-d^{2})),&x\in(x_{0}-d,x_{0}+d),\\\
0,&\text{otherwise,}\end{cases}$

and $[a,b]\subset[-1,1]$. Then, ${\widetilde{G}}\in C^{\infty}$,
${\widetilde{G}}(x)=0$ if $x\leq a$, ${\widetilde{G}}(x)=1$ if $x\geq b$, and
$0\leq{\widetilde{G}}(x)\leq 1$ if $a\leq x\leq b$.

Now, let $\varepsilon:=\rho_{n}^{3/2}(\lambda)$,
$Q(x):=(x-\lambda)(x-\lambda+\varepsilon)$, and define

$f(x):=\begin{cases}0,&x\in[\lambda-\varepsilon,\lambda],\\\
Q(x),&x\in[-1/4,1]\setminus[\lambda-\varepsilon,\lambda],\\\
Q(x)\,{\widetilde{G}}(x;[-1/2,-1/4]),&x\in(-1/2,-1/4),\\\
0,&x\in[-1,-1/2].\end{cases}$

(Here, we assume that $n$ is sufficiently large so that everything is well
defined.)

Suppose now that $P_{n}\in\Pi_{n}$ is nonnegative near $\lambda$, satisfies
$P_{n}(\lambda)=f(\lambda)=0$ (this necessarily implies that
$P_{n}^{\prime}(\lambda)\leq 0$, and actually $P_{n}^{\prime}(\lambda)=0$ if
$\lambda\neq 1$) and such that $|f(x)-P_{n}(x)|\leq
A\alpha_{n}\sqrt{n}\,{\omega}_{3}(f,\rho_{n}(x))$, $x\in I$.

Clearly, $f\in C^{\infty}[-1,-1/8]$, and it is not difficult to see that
$\left\|f^{(3)}\right\|_{[-1,-1/8]}\leq c_{0}$, for some absolute constant
$c_{0}$ (it is convenient to assume that $c_{0}\geq 2$). Hence,

${\omega}_{3}(f,t;[-1,-1/8])\leq t^{3}\left\|f^{(3)}\right\|_{[-1,-1/8]}\leq
c_{0}t^{3},\quad t>0.$

Since

$\left\|f-Q\right\|_{[-1/4,1]}=\left\|Q\right\|_{[\lambda-\varepsilon,\lambda]}=\varepsilon^{2}/4,$

we also have

${\omega}_{3}(f,t;[-1/4,1])={\omega}_{3}(f-Q,t;[-1/4,1])\leq
8\left\|f-Q\right\|_{[-1/4,1]}=2\varepsilon^{2}.$

Hence,

${\omega}_{3}(f,\rho_{n}(x))\leq
c_{0}\max\\{\rho_{n}^{3}(x),\rho_{n}^{3}(\lambda)\\},\quad x\in I.$

For $x\in[-1/4,1]$, we now have

(3.2)
$|P_{n}(x)-Q(x)|\leq|P_{n}(x)-f(x)|+|f(x)-Q(x)|\leq(c_{0}A\alpha_{n}\sqrt{n}+1)\max\\{\rho_{n}^{3}(x),\rho_{n}^{3}(\lambda)\\}.$

We now consider two cases.

Case 1: $\lambda<1$
Since $\rho_{n}(x)\leq 2n^{-1}$, $x\in I$, using (3.2) and Bernstein’s
inequality on $[-1/4,1]$, we have

$\displaystyle(1-\lambda^{2})^{3/4}n^{-3/2}$
$\displaystyle\leq\varepsilon=Q^{\prime}(\lambda)=|Q^{\prime}(\lambda)-P_{n}^{\prime}(\lambda)|\leq(1-\lambda)^{-1/2}(\lambda+1/4)^{-1/2}n\left\|Q-P_{n}\right\|_{[-1/4,1]}$
$\displaystyle\leq
16(1-\lambda)^{-1/2}n^{-2}\left(c_{0}A\alpha_{n}\sqrt{n}+1\right),$

and we arrive at a contradiction by taking $n$ to be sufficiently large.

Case 2: $\lambda=1$
We need the following Dzyadyk inequality (see e.g. [dzya]*p. 386 or
[kls21]*Lemma 2.5):

> If $m\in\mathbb{N}$, $x_{0}\in I$, $R_{n}\in\Pi_{n}$, and
>
> $|R_{n}(x)|\leq(|x-x_{0}|+\rho_{n}(x))^{m},\quad x\in I,$
>
> then
>
> $|R_{n}^{\prime}(x_{0})|\leq c(m)\rho_{n}^{m-1}(x_{0}).$

We use this result with $m=3$, $x_{0}=\lambda=1$ and $R_{n}=B^{-1}(P_{n}-Q)$,
where $B:=c_{0}A\alpha_{n}\sqrt{n}+1$. Suppose that we verified that

(3.3) $|P_{n}(x)-Q(x)|\leq B(|x-1|+\rho_{n}(x))^{3},\quad x\in I.$

Then, by Dzyadyk’s inequality,

$n^{-3}=\varepsilon=Q^{\prime}(1)\leq Q^{\prime}(1)-P_{n}^{\prime}(1)\leq
cB\rho_{n}^{2}(1)=c(c_{0}A\alpha_{n}\sqrt{n}+1)n^{-4},$

and we arrive at a contradiction by taking $n$ to be sufficiently large.

It remains to verify (3.3). Note that (3.3) immediately follows from (3.2) for
$x\in[-1/4,1]$ and, for $x\in[-1,-1/4]$, we have

$|Q(x)|\leq x(x-1)\leq(1-x)^{3},$

and so

$\displaystyle|P_{n}(x)-Q(x)|$
$\displaystyle\leq|P_{n}(x)-f(x)|+|f(x)-Q(x)|\leq|P_{n}(x)-f(x)|+|Q(x)|$
$\displaystyle\leq c_{0}A\alpha_{n}\sqrt{n}\rho_{n}^{3}(x)+|x-1|^{3}\leq
B(|x-1|+\rho_{n}(x))^{3}.$

The proof is now complete. ∎

######  Lemma 3.5 (copositive approximation with interpolatory constraints:
negative result for $f\in C^{1}$ and $\lambda\neq 1$).

For any $A>1$, $\lambda\in[0,1)$, $n_{0}\in\mathbb{N}$, and any positive
sequence $(\alpha_{n})_{n\in\mathbb{N}}$ converging to $0$, there exist
$n\in\mathbb{N}$, $n\geq n_{0}$, and $f=f_{n}\in C^{1}$ such that $f\geq 0$ on
$I$, $f\equiv 0$ on $[-1,-1/2]$, and for any polynomial $P_{n}\in\Pi_{n}$
which is nonnegative on $(\lambda-1/n,\lambda+1/n)$ and satisfies
$P_{n}(\lambda)=f(\lambda)$, we have

(3.4)
$\left\|f-P_{n}\right\|>A\,\alpha_{n}n^{-2/3}{\omega}_{4}(f^{\prime},n^{-1}).$

We remark that inequality (3.4) implies that

$\left\|\frac{f-P_{n}}{\rho_{n}{\omega}_{4}(f^{\prime},\rho_{n})}\right\|>cA\,\alpha_{n}n^{1/3},$

where $c>0$ is some absolute constant. Also, note that the statement of this
lemma is no longer valid if $\lambda=1$, i.e., if interpolation takes place at
an endpoint of $I$ (see Theorem 2.3 with
$(r,k,m_{1},m_{2},m_{3})=(1,k,-1,0,0)$).

###### Proof of Lemma 3.5.

Suppose that $n\geq 25(1-\lambda)^{-1}$, let $\varepsilon:=n^{-4/3}$ and
define

$g(x):=\begin{cases}Q(x),&\text{if $x<\lambda$,}\\\ 0,&\text{if $\lambda\leq
x\leq\lambda+\varepsilon$,}\\\ Q(x)+\varepsilon^{4},&\text{if
$x\geq\lambda+\varepsilon$.}\end{cases}$

where $Q(x):=3(x-\lambda)^{4}-4\varepsilon(x-\lambda)^{3}$. Note that $g\in
C^{1}$ and

$g^{\prime}(x)=\begin{cases}Q^{\prime}(x),&\text{if $x<\lambda$ or
$x>\lambda+\varepsilon$,}\\\ 0,&\text{if $\lambda\leq
x\leq\lambda+\varepsilon$,}\end{cases}$

with $Q^{\prime}(x)=12(x-\lambda)^{2}(x-\lambda-\varepsilon)$.

Now, define

$f(x):=\begin{cases}g(x),&x\in[-1/4,1],\\\
g(x)\,{\widetilde{G}}(x;[-1/2,-1/4]),&x\in(-1/2,-1/4),\\\
0,&x\in[-1,-1/2],\end{cases}$

where ${\widetilde{G}}$ is the glue function from the proof of Lemma 3.4.

Suppose that $P_{n}\in\Pi_{n}$ is nonnegative on $(\lambda-1/n,\lambda+1/n)$ ,
satisfies $P_{n}(\lambda)=f(\lambda)=0$ (this necessarily implies that
$P_{n}^{\prime}(\lambda)=0$) and such that $\left\|f-P_{n}\right\|\leq
A\alpha_{n}n^{-2/3}\,{\omega}_{4}(f^{\prime},1/n)$.

Clearly, $f\in C^{\infty}[-1,-1/8]$, and it is not difficult to see that
$\left\|f^{(5)}\right\|_{[-1,-1/8]}\leq c_{0}$, for some absolute constant
$c_{0}\geq 32$. Hence,

${\omega}_{4}(f^{\prime},n^{-1};[-1,-1/8])\leq
n^{-4}\left\|f^{(5)}\right\|_{[-1,-1/8]}\leq c_{0}n^{-4}.$

Since

$\left\|f^{\prime}-Q^{\prime}\right\|_{[-1/4,1]}=\left\|Q^{\prime}\right\|_{[\lambda,\lambda+\varepsilon]}\leq
2\varepsilon^{3},$

we also have

${\omega}_{4}(f^{\prime},n^{-1};[-1/4,1])={\omega}_{4}(f^{\prime}-Q^{\prime},n^{-1};[-1/4,1])\leq
16\left\|f^{\prime}-Q^{\prime}\right\|_{[-1/4,1]}\leq 32\varepsilon^{3}.$

Hence,

${\omega}_{4}(f^{\prime},n^{-1})\leq\max\\{c_{0}n^{-4},32\varepsilon^{3}\\}=c_{0}n^{-4},$

and so

$\displaystyle\left\|P_{n}-Q\right\|_{[-1/4,1]}$
$\displaystyle\leq\left\|P_{n}-f\right\|_{[-1/4,1]}+\left\|g-Q\right\|_{[-1/4,1]}$
$\displaystyle\leq
A\alpha_{n}n^{-2/3}\,{\omega}_{4}(f^{\prime},1/n)+\max\left\\{\varepsilon^{4},\left\|Q\right\|_{[\lambda,\lambda+\varepsilon]}\right\\}$
$\displaystyle\leq n^{-14/3}\left(c_{0}A\alpha_{n}+n^{-2/3}\right).$

By the Markov-Bernstein inequality on $[-1/4,1]$,

$\left\|P_{n}^{\prime\prime}-Q^{\prime\prime}\right\|_{[\lambda,\lambda+\varepsilon]}\leq
c(1-\lambda)^{-1}n^{-8/3}\left(c_{0}A\alpha_{n}+n^{-2/3}\right),$

and taking into account that $P_{n}(\lambda+\varepsilon)\geq 0$ and
$P_{n}^{(i)}(\lambda)=Q^{(i)}(\lambda)=0$, $i=0,1$, we have

$\displaystyle 1$
$\displaystyle\leq\varepsilon^{-4}\left[P_{n}(\lambda+\varepsilon)-Q(\lambda+\varepsilon)\right]=\varepsilon^{-4}\int_{\lambda}^{\lambda+\varepsilon}(\lambda+\varepsilon-t)\left[P_{n}^{\prime\prime}(t)-Q^{\prime\prime}(t)\right]dt$
$\displaystyle\leq
c\varepsilon^{-2}(1-\lambda)^{-1}n^{-8/3}\left(c_{0}A\alpha_{n}+n^{-2/3}\right)$
$\displaystyle\leq c(1-\lambda)^{-1}\left(c_{0}A\alpha_{n}+n^{-2/3}\right),$

and we arrive at a contradiction by taking $n$ to be sufficiently large. ∎

######  Lemma 3.6 (copositive approximation with interpolatory constraints:
negative result for $f\in C^{r}$ with even $r\geq 2$).

If $r\geq 2$ is even, then for any $n\geq r$, $A>0$ and $\lambda\in[0,1]$,
there exists $f\in C^{r}$ which is nonnegative on $I$, is identically $0$ on
$[-1,-1/2]$, and such that, for any nonnegative on $(\lambda-1/n,\lambda)$
polynomial $P_{n}\in\Pi_{n}$ satisfying
$P_{n}^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r$, we have

(3.5) $\left\|f-P_{n}\right\|>A\left\|f^{(r)}\right\|.$

###### Proof.

Let $n\geq r$ and $A\geq 1$ be fixed, and define $f\in C^{r}$ as

$f(x)=\frac{1}{(r-1)!}\int_{\lambda-\varepsilon}^{x}(x-t)^{r-1}f^{(r)}(t)dt,$

where

$f^{(r)}(x):=\begin{cases}0,&\text{if $x\leq\lambda-\varepsilon$ or
$x\geq\lambda$,}\\\
-(x-\lambda)(x-\lambda+\varepsilon)/\varepsilon^{2},&\text{if
$\lambda-\varepsilon<x<\lambda$,}\end{cases}$

and $\varepsilon\in(0,1/(2n))$. Then, $f\equiv 0$ on
$[-1,\lambda-\varepsilon]$, $f\uparrow$ on $[\lambda-\varepsilon,1]$ (and so
$f$ is nonnegative on $I$), $\left\|f^{(r)}\right\|=1/4$ and, for
$x\geq\lambda$,

(3.6)
$f(x)=Q(x):=\frac{1}{(r-1)!}\cdot\frac{1}{\varepsilon^{2}}\int_{\lambda-\varepsilon}^{\lambda}(x-t)^{r-1}(\lambda-t)(t-\lambda+\varepsilon)dt.$

Note that $Q$ is a polynomial of degree $\leq r-1$ and

$Q(\lambda-\varepsilon)=\frac{(-1)^{r-1}}{(r-1)!}\cdot\frac{1}{\varepsilon^{2}}\int_{\lambda-\varepsilon}^{\lambda}(t-\lambda+\varepsilon)^{r}(\lambda-t)dt=(-1)^{r-1}\frac{r}{(r+2)!}\varepsilon^{r}.$

Now, suppose that $P_{n}\in\Pi_{n}$ is such that $P_{n}\geq 0$ on
$(\lambda-1/n,\lambda)$,
$P_{n}^{(i)}(\lambda)=f^{(i)}(\lambda)=Q^{(i)}(\lambda)$, $0\leq i\leq r$, and
$\left\|f-P_{n}\right\|\leq A\left\|f^{(r)}\right\|=A/4$. Hence,

$\left\|P_{n}\right\|_{[-1,\lambda]}\leq\left\|P_{n}-f\right\|_{[-1,\lambda]}+\left\|f\right\|_{[-1,\lambda]}\leq
A\left\|f^{(r)}\right\|+\left\|f\right\|_{[\lambda-\varepsilon,\lambda]}\leq(A+\varepsilon^{r}/r!)\left\|f^{(r)}\right\|\leq
A.$

Now, by Markov’s inequality,

$\left\|P_{n}^{(r+1)}\right\|_{[-1,\lambda]}\leq\left(\frac{2n^{2}}{\lambda+1}\right)^{r+1}\left\|P_{n}\right\|_{[-1,\lambda]}\leq(2n^{2})^{r+1}A,$

and so, since $Q^{(r+1)}\equiv 0$ and $r$ is even,

$\displaystyle\frac{r}{(r+2)!}\varepsilon^{r}$
$\displaystyle=-Q(\lambda-\varepsilon)\leq
P_{n}(\lambda-\varepsilon)-Q(\lambda-\varepsilon)$
$\displaystyle=\frac{1}{r!}\int_{\lambda}^{\lambda-\varepsilon}(\lambda-\varepsilon-t)^{r}\left(P_{n}^{(r+1)}(t)-Q^{(r+1)}(t)\right)dt$
$\displaystyle\leq\frac{1}{(r+1)!}\varepsilon^{r+1}\left\|P_{n}^{(r+1)}\right\|_{[\lambda-\varepsilon,\lambda]}\leq\frac{(2n^{2})^{r+1}A}{(r+1)!}\varepsilon^{r+1},$

and we get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

######  Remark 3.7.

The same proof can be used to show that the estimate (3.5) holds if
$\lambda\in[0,1)$ and $P_{n}\in\Pi_{n}$, such that $P_{n}\geq 0$ on
$(\lambda-1/n,\lambda)$, is assumed to satisfy e.g.
$P_{n}^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r-1$, and
$P_{n}(\lambda+\varepsilon)=f(\lambda+\varepsilon)$. In particular, this
implies that constants $c(d(A_{p}))$ in Theorems 2.3 and 2.4 (and, hence, in
Theorems 2.1 and 2.2) cannot be made independent of the minimal distance among
points in $A_{p}$.

######  Lemma 3.8 (copositive approximation with interpolatory constraints:
negative result for $f\in C^{r}$ with odd $r\geq 3$).

If $r\geq 3$ is odd, then for any $n\geq r-1$ and $A>0$, there exists $f\in
C^{r}$ which is identically $0$ on $[-1,-1/2]$, is nonnegative on $I$ and such
that, for any nonnegative on $[-1/2,1/2]$ polynomial $P_{n}\in\Pi_{n}$
satisfying $P_{n}^{(i)}(0)=f^{(i)}(0)$, $0\leq i\leq r-1$, we have

$\left\|f-P_{n}\right\|>A\left\|f^{(r)}\right\|.$

We note that an analog of this lemma is not valid if we require that $P_{n}$
be nonnegative on $[-1/2,1]$ and $P_{n}^{(i)}(1)=f^{(i)}(1)$, $0\leq i\leq
r-1$ (i.e., this negative result does not hold if $A_{p}=\\{\pm 1\\}$), see
e.g. Theorem 2.3 with $(r,k,m_{1},m_{2},m_{3})=(2j+1,k,2j-1,2j,2j)$,
$j\in\mathbb{N}$.

###### Proof.

Let $n\geq r-1$ and $A>0$ be fixed, and define $f\in C^{r}$ as

$f(x)=\frac{1}{(r-2)!}\int_{-\varepsilon}^{x}(x-t)^{r-2}f^{(r-1)}(t)dt,$

where

$f^{(r-1)}(x):=\begin{cases}0,&\text{if $x<-\varepsilon$ or $x>0$,}\\\
x^{2}(x+\varepsilon)^{2}/\varepsilon^{3},&\text{if $-\varepsilon\leq x\leq
0$,}\end{cases}$

and $\varepsilon\in(0,1/2)$. Then, $\left\|f^{(r)}\right\|\leq 1$, $f\equiv 0$
on $[-1,-\varepsilon]$, and $f$ is clearly nonnegative on $I$. Now, for $x\geq
0$, we have

(3.7)
$f(x)=Q(x):=\frac{1}{(r-2)!}\cdot\frac{1}{\varepsilon^{3}}\int_{-\varepsilon}^{0}(x-t)^{r-2}t^{2}(t+\varepsilon)^{2}dt.$

Note that $Q\in\Pi_{r-2}$,

(3.8)
$\left\|Q\right\|\leq\frac{2^{r-2}}{(r-2)!}\cdot\frac{1}{\varepsilon^{3}}\int_{-\varepsilon}^{0}t^{2}(t+\varepsilon)^{2}dt\leq\frac{2^{r-2}}{(r-2)!}\varepsilon^{2}=:c_{0}\varepsilon^{2},$

and

$Q(-\varepsilon)=(-1)^{r}{2(r^{2}-r)\over(r+3)!}\varepsilon^{r}=:-c_{1}\varepsilon^{r},\quad
c_{1}=c_{1}(r)>0,$

since $r$ is odd.

Suppose that there exists a polynomial $P_{n}\in\Pi_{n}$ which is nonnegative
on $[-1/2,1/2]$, satisfies $P_{n}^{(i)}(0)=f^{(i)}(0)=Q^{(i)}(0)$, $0\leq
i\leq r-1$, and such that

(3.9) $\left\|f-P_{n}\right\|\leq A\left\|f^{(r)}\right\|\leq A.$

Then, $P_{n}$ can be written as

$P_{n}(x)=Q(x)+a_{r}x^{r}+a_{r+1}x^{r+1}+\dots+a_{n}x^{n},$

and taking into account that

$\left\|P_{n}-Q\right\|_{[0,1]}=\left\|P_{n}-f\right\|_{[0,1]}\leq A,$

by Markov’s inequality, we conclude that

$|a_{i}|=|P_{n}^{(i)}(0)|/i!=|P_{n}^{(i)}(0)-Q^{(i)}(0)|/i!\leq K,\quad r\leq
i\leq n,$

where the constant $K$ depends only on $A$, $r$ and $n$.

Since $r+1$ is even, for any $x\in I$, we have the following estimate

$\displaystyle 0\leq P_{n}(x)$ $\displaystyle\leq
Q(x)+a_{r}x^{r}+|x|^{r+1}\sum_{i=r+1}^{n}|a_{i}||x|^{i-r-1}$
$\displaystyle\leq Q(x)+a_{r}x^{r}+Bx^{r+1},$

where $B:=nK$. We also note that $a_{r}$ satisfies $|a_{r}|\leq K<B$.

Now, let $x_{0}=-a_{r}/(2B)$. Then, $x_{0}\in(-1/2,1/2)$ and

$0\leq P_{n}(x_{0})\leq
Q(x_{0})+(-1)^{r}\frac{a_{r}^{r+1}}{2^{r+1}B^{r}}=Q(x_{0})-\frac{a_{r}^{r+1}}{2^{r+1}B^{r}},$

and so by (3.8) (recall that $r+1$ is even)

$a_{r}^{r+1}\leq 2^{r+1}B^{r}Q(x_{0})\leq
c_{0}2^{r+1}B^{r}\varepsilon^{2}\quad\Rightarrow\quad|a_{r}|\leq
D\varepsilon^{2/(r+1)},$

where $D$ is a positive constant that depends only on $A$, $r$ and $n$, and is
independent of $\varepsilon$.

Now, we have

$\displaystyle 0\leq P_{n}(-\varepsilon)$ $\displaystyle\leq
Q(-\varepsilon)+D\varepsilon^{2/(r+1)}\varepsilon^{r}+B\varepsilon^{r+1}$
$\displaystyle=\varepsilon^{r}\left(-c_{1}+O(\varepsilon^{2/(r+1)})\right)<0\quad\text{as
$\varepsilon\to 0^{+}$},$

which is a contradiction. ∎

The following lemma closes a few gaps left in Lemma 3.8. It is applicable in
the case $r=1$ as well as for establishing negative results for interpolation
at the endpoints of $I$ (the case for $\lambda=1$). However, if $r\geq 3$ and
$\lambda\neq 1$, then this lemma is weaker than Lemma 3.8.

######  Lemma 3.9 (copositive approximation with interpolatory constraints:
negative result for $f\in C^{r}$ with odd $r\geq 1$).

If $r\in\mathbb{N}$ is odd, then for any $n\geq r$, $A>0$ and
$\lambda\in[0,1]$, there exists $f\in C^{r}$ which is identically $0$ on
$[-1,-1/2]$, is nonnegative on $I$ and such that, for any nonnegative on
$(\lambda-1/n,\lambda)$ polynomial $P_{n}\in\Pi_{n}$ satisfying
$P_{n}^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r$, we have

$\left\|f-P_{n}\right\|>A\left\|f^{(r)}\right\|.$

###### Proof.

Let $n\geq r$ and $A\geq 1$ be fixed, and define $f\in C^{r}$ as

$f(x)=\frac{1}{(r-1)!}\int_{\lambda-\varepsilon}^{x}(x-t)^{r-1}f^{(r)}(t)dt,$

where

$f^{(r)}(x):=\begin{cases}0,&\text{if $x<\lambda-\varepsilon$,}\\\
(x-\lambda+\varepsilon)/\varepsilon,&\text{if $\lambda-\varepsilon\leq
x\leq\lambda$,}\\\ 1,&\text{if $x>\lambda$,}\end{cases}$

and $\varepsilon\in(0,1/(2n))$. Note that $f$ is nonnegative on $I$, $f\equiv
0$ on $[-1,\lambda-\varepsilon]$, $\left\|f^{(r)}\right\|=1$ and, for
$x\geq\lambda$,

$\displaystyle f(x)=Q(x)$
$\displaystyle:=\frac{1}{(r+1)!}\cdot\frac{1}{\varepsilon}\left[(x-\lambda+\varepsilon)^{r+1}-(x-\lambda)^{r+1}\right].$

Now, suppose that $P_{n}\in\Pi_{n}$ is such that $P_{n}\geq 0$ on
$(\lambda-1/n,\lambda)$,
$P_{n}^{(i)}(\lambda)=f^{(i)}(\lambda)=Q^{(i)}(\lambda)$, $0\leq i\leq r$, and
$\left\|f-P_{n}\right\|\leq A\left\|f^{(r)}\right\|=A$. Then,

$\left\|P_{n}\right\|_{[-1,\lambda]}\leq\left\|P_{n}-f\right\|_{[-1,\lambda]}+\left\|f\right\|_{[-1,\lambda]}\leq
A\left\|f^{(r)}\right\|+\left\|f\right\|_{[\lambda-\varepsilon,\lambda]}\leq(A+\varepsilon^{r}/r!)\left\|f^{(r)}\right\|\leq
2A.$

Now, by Markov’s inequality,

$\left\|P_{n}^{(r+1)}\right\|_{[-1,\lambda]}\leq\left(\frac{2n^{2}}{\lambda+1}\right)^{r+1}\left\|P_{n}\right\|_{[-1,\lambda]}\leq
2(2n^{2})^{r+1}A,$

and so, since $Q^{(r+1)}\equiv 0$,

$\displaystyle\frac{1}{(r+1)!}\varepsilon^{r}$
$\displaystyle=-Q(\lambda-\varepsilon)\leq
P_{n}(\lambda-\varepsilon)-Q(\lambda-\varepsilon)$
$\displaystyle=\frac{1}{r!}\int_{\lambda}^{\lambda-\varepsilon}(\lambda-\varepsilon-t)^{r}\left(P_{n}^{(r+1)}(t)-Q^{(r+1)}(t)\right)dt$
$\displaystyle\leq\frac{1}{(r+1)!}\varepsilon^{r+1}\left\|P_{n}^{(r+1)}\right\|_{[\lambda-\varepsilon,\lambda]}\leq\frac{2(2n^{2})^{r+1}A}{(r+1)!}\varepsilon^{r+1},$

and we get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

###  3.3 Negative results for copositive approximation with interpolation at
the points in $Y_{s}$

######  Lemma 3.10 (copositive approximation with interpolation at $Y_{s}$:
negative result for $f\in C^{1}$ and $Y_{s}=\\{0\\}$).

For any $n\in\mathbb{N}$ and $A>0$, there exists $f\in
C^{1}\cap\Delta^{(0)}(\\{0\\})$ such that, for any polynomial
$P_{n}\in\Pi_{n}$ such that $P_{n}\geq 0$ on $(0,1/n)$ and satisfies
$P_{n}^{(i)}(0)=f^{(i)}(0)$, $i=0,1$, we have

$\left\|f-P_{n}\right\|>A{\omega}_{3}(f^{\prime},1).$

###### Proof.

Let $n\in\mathbb{N}$ and $A>0$ be fixed (without loss of generality, we can
assume that $n\geq 3$), and define $f\in C^{1}$ as

$f(x)=\int_{0}^{x}f^{\prime}(t)dt,\quad\text{ where
}\;f^{\prime}(x):=\begin{cases}x(x-\varepsilon)/\varepsilon^{2},&\text{if
$x<0$ or $x>\varepsilon$,}\\\ 0,&\text{if $0\leq
x\leq\varepsilon$,}\end{cases}$

and $\varepsilon\in(0,1/(2n))$. Note that $f\in\Delta^{(0)}(\\{0\\})$,
$f^{(i)}(0)=0$, $i=0,1$, and denoting
$q(x):=x(x-\varepsilon)/\varepsilon^{2}$, we have

${\omega}_{3}(f^{\prime},1)={\omega}_{3}(f^{\prime}-q,1)\leq
8\left\|f^{\prime}-q\right\|=8\left\|q\right\|_{[0,\varepsilon]}=2.$

Then, for $x\leq 0$, we have

$f(x)=Q(x):=(2x^{3}-3\varepsilon x^{2})/(6\varepsilon^{2}).$

Suppose that $P_{n}\in\Pi_{n}$ is nonnegative on $(0,1/n)$, satisfies
$P_{n}^{(i)}(0)=f^{(i)}(0)=Q^{(i)}(0)=0$, $i=0,1$, and such that

(3.10) $\left\|f-P_{n}\right\|\leq A{\omega}_{3}(f^{\prime},1)\leq 2A.$

Taking into account that

$\left\|P_{n}-Q\right\|_{[-1,\varepsilon]}\leq\left\|P_{n}-f\right\|_{[-1,\varepsilon]}+\left\|f-Q\right\|_{[-1,\varepsilon]}\leq\left\|P_{n}-f\right\|+\left\|f-Q\right\|_{[0,\varepsilon]}\leq
2A+1,$

by Markov’s inequality, we conclude that

$\left\|P_{n}^{\prime\prime}-Q^{\prime\prime}\right\|_{[-1,\varepsilon]}\leq\left(\frac{2n^{2}}{\varepsilon+1}\right)^{2}\left\|P_{n}-Q\right\|_{[-1,\varepsilon]}\leq
4n^{4}(2A+1).$

Now,

$\displaystyle\varepsilon/6$ $\displaystyle=-Q(\varepsilon)\leq
P_{n}(\varepsilon)-Q(\varepsilon)=\int_{0}^{\varepsilon}(\varepsilon-t)\left(P_{n}^{\prime\prime}(t)-Q^{\prime\prime}(t)\right)dt\leq
2n^{4}(2A+1)\varepsilon^{2},$

and we get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

######  Lemma 3.11 (copositive approximation with interpolation at $Y_{s}$:
negative result for $f\in C^{2}$ and $Y_{s}=\\{0\\}$).

For any $n\in\mathbb{N}$ and $A>0$, there exists $f\in
C^{2}\cap\Delta^{(0)}(\\{0\\})$ such that, for any polynomial
$P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ such that
$P_{n}^{(i)}(0)=f^{(i)}(0)$, $i=0,1$, we have

$\left\|f-P_{n}\right\|>A{\omega}_{4}(f^{\prime\prime},1).$

###### Proof.

Let $n\in\mathbb{N}$ and $A>0$ be fixed (without loss of generality, we can
assume that $n\geq 5$), and define $f\in C^{2}$ as

$f(x)=\int_{0}^{x}(x-t)f^{\prime\prime}(t)dt,\quad\text{ where
}\;f^{\prime\prime}(x):=\begin{cases}x(x^{2}-\varepsilon^{2})/\varepsilon^{3},&\text{if
$|x|>\varepsilon$,}\\\ 0,&\text{if $|x|\leq\varepsilon$,}\end{cases}$

and $\varepsilon\in(0,1/2)$. Note that $f\in\Delta^{(0)}(\\{0\\})$,
$f^{(i)}(0)=0$, $i=0,1$, and denoting
$q(x):=x(x^{2}-\varepsilon^{2})/\varepsilon^{3}$, we have

${\omega}_{4}(f^{\prime\prime},1)={\omega}_{4}(f^{\prime\prime}-q,1)\leq
16\left\|f^{\prime\prime}-q\right\|=16\left\|q\right\|_{[-\varepsilon,\varepsilon]}\leq
7.$

Then, for $x\geq\varepsilon$, we have

$f(x)=Q(x):=\int_{\varepsilon}^{x}(x-t)q(t)dt.$

Suppose that $P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ satisfies
$P_{n}^{(i)}(0)=f^{(i)}(0)=0$, $i=0,1$, and such that

$\left\|f-P_{n}\right\|\leq A{\omega}_{4}(f^{\prime\prime},1)\leq 7A.$

Taking into account that

$\left\|P_{n}-Q\right\|_{[0,1]}\leq\left\|P_{n}-f\right\|_{[0,1]}+\left\|f-Q\right\|_{[0,1]}\leq\left\|P_{n}-f\right\|+\left\|Q\right\|_{[0,\varepsilon]}\leq
7A+1,$

by Markov’s inequality, we conclude that

$\left\|P_{n}^{\prime\prime\prime}-Q^{\prime\prime\prime}\right\|_{[0,1]}\leq\left(2n^{2}\right)^{3}\left\|P_{n}-Q\right\|_{[0,1]}\leq
8n^{6}(7A+1).$

Since $P_{n}\in\Delta^{(0)}(\\{0\\})$ and $P_{n}(0)=P_{n}^{\prime}(0)=0$, we
have $P_{n}^{\prime\prime}(0)=0$ and $P_{n}^{\prime\prime\prime}(0)\geq 0$.
Hence,

$\displaystyle 1/\varepsilon$
$\displaystyle=-q^{\prime}(0)=-Q^{\prime\prime\prime}(0)\leq
P_{n}^{\prime\prime\prime}(0)-Q^{\prime\prime\prime}(0)\leq 8n^{6}(7A+1),$

and we get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

######  Lemma 3.12 (copositive approximation with interpolation at $Y_{s}$:
negative result for $f\in C^{2}$ and $Y_{s}=\\{0\\}$).

For any $n\geq 2$ and $A>0$, there exists $f\in
C^{2}\cap\Delta^{(0)}(\\{0\\})$ such that, for any polynomial
$P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ such that
$P_{n}^{(i)}(0)=f^{(i)}(0)$, $0\leq i\leq 2$, we have

$\left\|f-P_{n}\right\|>A\left\|f^{\prime\prime}\right\|.$

###### Proof.

Let $n\geq 2$ and $A\geq 1$ be fixed, and define $f\in C^{2}$ as

$f(x)=\int_{0}^{x}\int_{\varepsilon}^{t}f^{\prime\prime}(u)dudt,\quad\text{
where }\;f^{\prime\prime}(x):=\begin{cases}-1,&\text{if $x<0$,}\\\ 0,&\text{if
$x>\varepsilon$,}\\\ x/\varepsilon-1,&\text{if $0\leq
x\leq\varepsilon$,}\end{cases}$

and $\varepsilon\in(0,1/2)$. Note that $f\in\Delta^{(0)}(\\{0\\})$,
$\left\|f^{\prime\prime}\right\|=1$ and, for $x\leq 0$,

$f(x)=Q(x):=-\frac{1}{2}x^{2}+\frac{1}{2}\varepsilon x.$

Suppose that $P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ satisfies
$P_{n}^{(i)}(0)=f^{(i)}(0)=Q^{(i)}(0)$, $0\leq i\leq 2$, and such that

$\left\|f-P_{n}\right\|\leq A\left\|f^{\prime\prime}\right\|=A.$

Then,

$P_{n}(x)=Q(x)+a_{3}x^{3}+\dots+a_{n}x^{n},$

and since

$\left\|P_{n}-Q\right\|_{[-1,0]}=\left\|P_{n}-f\right\|_{[-1,0]}\leq A,$

by Markov’s inequality, we conclude that

$|a_{i}|=|P_{n}^{(i)}(0)-Q^{(i)}(0)|/i!\leq K,\quad 3\leq i\leq n,$

where the constant $K$ depends only on $A$ and $n$. Hence,

$0\leq P_{n}(2\varepsilon)\leq
Q(2\varepsilon)+nK(2\varepsilon)^{3}=\varepsilon^{2}(-1+8nK\varepsilon)<0\quad\text{as
$\varepsilon\to 0^{+}$,}$

which is a contradiction. ∎

######  Lemma 3.13 (copositive approximation with interpolation at $Y_{s}$:
negative result for $f\in C^{r}$ with odd $r\geq 3$ and $Y_{s}=\\{0\\}$).

If $r\geq 3$ is odd, then for any $n\geq r$ and $A>0$, there exists $f\in
C^{r}\cap\Delta^{(0)}(\\{0\\})$ such that, for any polynomial
$P_{n}\in\Pi_{n}$ such that $P_{n}\leq 0$ on $(-1/n,0)$ and satisfies
$P_{n}^{(i)}(0)=f^{(i)}(0)$, $0\leq i\leq r$, we have

(3.11) $\left\|f-P_{n}\right\|>A\left\|f^{(r)}\right\|.$

###### Proof.

Let $n\geq r$ and $A\geq 1$ be fixed, and let $f$ and $Q$ be the same as in
the proof of Lemma 3.6 with $\lambda=0$ (but $r$ is odd now). Then, for $x\geq
0$ (see (3.6)),

$\displaystyle f(x)$
$\displaystyle=Q(x):=\frac{1}{(r-1)!}\cdot\frac{1}{\varepsilon^{2}}\int_{-\varepsilon}^{0}(x-t)^{r-1}(-t)(t+\varepsilon)dt.$

Note that $Q(0)=r\varepsilon^{r}/(r+2)!$ and

$\displaystyle Q(-2\varepsilon)$
$\displaystyle=\frac{1}{(r-1)!}\cdot\frac{1}{\varepsilon^{2}}\int_{0}^{\varepsilon}(2\varepsilon-t)^{r-1}t(\varepsilon-t)dt=\frac{1}{(r+2)!}\left(2^{r+1}(r-2)+r+4\right)\varepsilon^{r}.$

Now, define

${\widetilde{f}}(x):=f(x)-Q(0)\quad\mbox{\rm
and}\quad{\widetilde{Q}}(x):=Q(x)-Q(0).$

Then, ${\widetilde{f}}\in C^{r}\cap\Delta^{(0)}(\\{0\\})$, and we now suppose
that $P_{n}\in\Pi_{n}$ is such that $P_{n}\leq 0$ on $(-1/n,0)$,
$P_{n}^{(i)}(0)={\widetilde{f}}^{(i)}(0)={\widetilde{Q}}^{(i)}(0)$, $0\leq
i\leq r$, and $\left\|{\widetilde{f}}-P_{n}\right\|\leq
A\left\|{\widetilde{f}}^{(r)}\right\|=A/4$. As in the proof of Lemma 3.6,

$\left\|P_{n}\right\|_{[-1,0]}\leq\left\|P_{n}-{\widetilde{f}}\right\|_{[-1,0]}+\left\|{\widetilde{f}}\right\|_{[-1,0]}\leq
A\left\|{\widetilde{f}}^{(r)}\right\|+\left\|f\right\|_{[-\varepsilon,0]}+|Q(0)|\leq
A,$

and, by Markov’s inequality,

$\left\|P_{n}^{(r+1)}\right\|_{[-1,0]}\leq\left(2n^{2}\right)^{r+1}\left\|P_{n}\right\|_{[-1,0]}\leq(2n^{2})^{r+1}A.$

Finally, since ${\widetilde{Q}}^{(r+1)}\equiv 0$ and $P_{n}(-2\varepsilon)\leq
0$,

$\displaystyle\frac{4}{(r+2)!}\varepsilon^{r}$ $\displaystyle\leq
Q(-2\varepsilon)-Q(0)={\widetilde{Q}}(-2\varepsilon)\leq{\widetilde{Q}}(-2\varepsilon)-P_{n}(-2\varepsilon)$
$\displaystyle=\frac{1}{r!}\int_{0}^{-2\varepsilon}(-2\varepsilon-t)^{r}\left({\widetilde{Q}}^{(r+1)}(t)-P_{n}^{(r+1)}(t)\right)dt$
$\displaystyle\leq\frac{1}{(r+1)!}(2\varepsilon)^{r+1}\left\|P_{n}^{(r+1)}\right\|_{[-2\varepsilon,0]}\leq\frac{(4n^{2})^{r+1}A}{(r+1)!}\varepsilon^{r+1},$

and we get a contradiction by letting $\varepsilon\to 0^{+}$. ∎

######  Lemma 3.14 (copositive approximation with interpolation at $Y_{s}$:
negative result for $f\in C^{r}$ with even $r\geq 4$ and $Y_{s}=\\{0\\}$).

If $r\geq 4$ is even, then for any $n\geq r-1$ and $A>0$, there exists $f\in
C^{r}\cap\Delta^{(0)}(\\{0\\})$ such that, for any polynomial
$P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ satisfying
$P_{n}^{(i)}(0)=f^{(i)}(0)$, $0\leq i\leq r-1$, we have

$\left\|f-P_{n}\right\|>A\left\|f^{(r)}\right\|.$

###### Proof.

Let $n\geq r$ and $A\geq 1$ be fixed, and let $f$ and $Q$ be the same as in
the proof of Lemma 3.8 (but $r$ is even now). Recall (see (3.7)) that, for
$x\geq 0$,

$f(x)=Q(x):=\frac{1}{(r-2)!}\cdot\frac{1}{\varepsilon^{3}}\int_{-\varepsilon}^{0}(x-t)^{r-2}t^{2}(t+\varepsilon)^{2}dt$

and, in particular,

$f(0)=Q(0)={2(r^{2}-r)\over(r+3)!}\varepsilon^{r}=c_{1}\varepsilon^{r}.$

Define

${\widetilde{f}}(x):=f(x)-Q(0)\quad\mbox{\rm
and}\quad{\widetilde{Q}}(x):=Q(x)-Q(0).$

Then, ${\widetilde{f}}\in C^{r}\cap\Delta^{(0)}(\\{0\\})$,
${\widetilde{Q}}\in\Pi_{r-2}$, and (3.8) implies that

$\left\|{\widetilde{Q}}\right\|\leq\left\|Q\right\|+|Q(0)|\leq
c_{0}\varepsilon^{2}+c_{1}\varepsilon^{r}\leq(c_{0}+c_{1})\varepsilon^{2}.$

We now suppose that $P_{n}\in\Pi_{n}\cap\Delta^{(0)}(\\{0\\})$ is such that
$P_{n}^{(i)}(0)={\widetilde{f}}^{(i)}(0)={\widetilde{Q}}^{(i)}(0)$, $0\leq
i\leq r-1$, and $\left\|{\widetilde{f}}-P_{n}\right\|\leq
A\left\|{\widetilde{f}}^{(r)}\right\|\leq A$. As in the proof of Lemma 3.8, we
conclude that $P_{n}$ can be written as

$P_{n}(x)={\widetilde{Q}}(x)+a_{r}x^{r}+a_{r+1}x^{r+1}+\dots+a_{n}x^{n},$

where

$|a_{i}|\leq K,\quad r\leq i\leq n,$

where the constant $K$ depends only on $A$, $r$ and $n$. This implies that

(3.12) $|P_{n}(x)-{\widetilde{Q}}(x)-a_{r}x^{r}|\leq B|x|^{r+1},\quad x\in I,$

where $B=nK$. We now have

$\displaystyle{\widetilde{Q}}(-3\varepsilon)$
$\displaystyle=Q(-3\varepsilon)-Q(0)=\frac{1}{(r-2)!}\cdot\frac{1}{\varepsilon^{3}}\int_{-\varepsilon}^{0}\left[(3\varepsilon+t)^{r-2}-t^{r-2}\right]t^{2}(t+\varepsilon)^{2}dt$
$\displaystyle\geq\frac{2^{r-2}-1}{(r-2)!}\cdot\varepsilon^{r-5}\int_{-\varepsilon}^{0}t^{2}(t+\varepsilon)^{2}dt=\frac{2^{r-2}-1}{30(r-2)!}\cdot\varepsilon^{r}=:c_{2}\varepsilon^{r},$

and, since $P_{n}(-3\varepsilon)\leq 0$,

(3.13) $0\leq-
P_{n}(-3\varepsilon)\leq-{\widetilde{Q}}(-3\varepsilon)-a_{r}(-3\varepsilon)^{r}+B(3\varepsilon)^{r+1}\leq\varepsilon^{r}\left(-c_{2}-3^{r}a_{r}+3^{r+1}B\varepsilon\right).$

If $a_{r}\geq 0$, then we immediately get a contradiction by letting
$\varepsilon\to 0^{+}$.

Suppose now that $a_{r}<0$. Then, $x_{0}=-a_{r}/(2B)$ is positive,
$P_{n}(x_{0})\geq 0$, and so (3.12) implies

$0\leq
P_{n}(x_{0})\leq{\widetilde{Q}}(x_{0})+a_{r}x_{0}^{r}+Bx_{0}^{r+1}={\widetilde{Q}}(x_{0})+(-1)^{r}\frac{a_{r}^{r+1}}{2^{r+1}B^{r}}={\widetilde{Q}}(x_{0})+\frac{a_{r}^{r+1}}{2^{r+1}B^{r}}.$

Hence,

$-a_{r}^{r+1}\leq 2^{r+1}B{\widetilde{Q}}(x_{0})\leq
2^{r+1}B(c_{0}+c_{1})\varepsilon^{2}$

and, since $r+1$ is odd and $a_{r}<0$, we have

$|a_{r}|=-a_{r}\leq\left[2^{r+1}B(c_{0}+c_{1})\varepsilon^{2}\right]^{1/(r+1)}.$

Therefore, (3.13) implies

$0\leq\varepsilon^{r}\left(-c_{2}+O(\varepsilon^{2/(r+1)})\right)<0\quad\text{as
$\varepsilon\to 0^{+}$,}$

and we again arrive at a contradiction. ∎

##  4 Spline approximation with interpolatory constraints

Given a collection of $q$ points $Y=\\{\upsilon_{j}\\}_{j=0}^{q-1}$ with
possible repetitions,
$\upsilon_{0}\leq\upsilon_{1}\leq\dots\leq\upsilon_{q-1}$, we recall that, for
each $j$, the multiplicity $m_{j}$ of $\upsilon_{j}$ is the number of
$\upsilon_{i}$ such that $\upsilon_{i}=\upsilon_{j}$. Also, we let $l_{j}$ be
the number of $\upsilon_{i}=\upsilon_{j}$ with $i\leq j$. Suppose that a
function $f$ is defined at all points in $Y$ and, moreover, for each
$\upsilon_{j}\in Y$, $f^{(l_{j}-1)}(\upsilon_{j})$ is defined as well. In
other words, $f$ has $m_{j}-1$ derivatives at each point that has multiplicity
$m_{j}$. Then, there is a unique Lagrange-Hermite polynomial
$L_{q-1}(\cdot;f,Y)$ of degree $\leq q-1$ that satisfies

$L_{q-1}^{(l_{j}-1)}(\upsilon_{j};f,Y)=f^{(l_{j}-1)}(\upsilon_{j}),\quad\text{for
all }\;0\leq j\leq q-1.$

The following lemma is a generalization of the classical Whitney’s theorem. It
is used for construction of local polynomial pieces near interpolation points
in Lemmas 4.5 and 4.7.

######  Lemma 4.1 (see [kls-umzh]*Theorem 5.2 and Lemma 3.1 or [kls21]*Lemma
1.12).

Let $r\in\mathbb{N}_{0}$ and $q\in\mathbb{N}$ be such that $q\geq r+1$, and
suppose that a set $V=\\{\upsilon_{j}\\}_{j=0}^{q-1}\subset[a,b]$ is such
that, if $q\geq r+2$, then $\Lambda_{r}(V)\geq\lambda(b-a)$, where
$0<\lambda\leq 1$ and

$\Lambda_{r}(V):=\min_{0\leq j\leq
q-r-2}(\upsilon_{j+r+1}-\upsilon_{j}),\quad\;q\geq r+2.$

(If $q=r+1$, this condition is not needed.) If $f\in C^{r}[a,b]$ then, for all
$x\in[a,b]$,

$|f(x)-L_{q-1}(x;f,V)|\leq
c(q,\lambda)(b-a)^{r}{\omega}_{q-r}(f^{(r)},b-a;[a,b]).$

###  4.1 Local construction

######  Lemma 4.2 (onesided and intertwining approximation with interpolatory
constraints: $f\in C^{r}$, $r\geq 1$).

Suppose that $r\in\mathbb{N}$, $f\in C^{r}[a,b]$ and $\lambda\in[a,b]$. Then,
there exists $p:=p(\cdot;f,k,r,\lambda)\in\Pi_{k+r-1}$ such that
$p^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r-1$,

(4.1) $\left\|f-p\right\|_{[a,b]}\leq
c(b-a)^{r}{\omega}_{k}(f^{(r)},b-a;[a,b]),\quad\text{where $c=c(k,r)$,}$

and

  1. (i)

if $\lambda\in(a,b)$, then $p-f\geq 0$ on $[\lambda,b]$, and
$(-1)^{r}\left(p-f\right)\geq 0$ on $[a,\lambda]$,

  2. (ii)

if $\lambda=a$, then $p-f\geq 0$ on $[a,b]$,

  3. (iii)

if $\lambda=b$, then $(-1)^{r}\left(p-f\right)\geq 0$ on $[a,b]$.

###### Proof.

Since $f^{(r)}\in C$, there exists a polynomial $q\in\Pi_{k-1}$ such that

$\left\|f^{(r)}-q\right\|_{[a,b]}\leq
c{\omega}_{k}(f^{(r)},b-a;[a,b])\quad\mbox{\rm and}\quad q(x)\geq
f^{(r)}(x),\;x\in[a,b].$

We now define

$p(x):=\frac{1}{(r-1)!}\int_{\lambda}^{x}(x-t)^{r-1}q(t)dt+\sum_{i=0}^{r-1}\frac{1}{i!}f^{(i)}(\lambda)(x-\lambda)^{i}$

and note that $p^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r-1$, and

$p(x)-f(x)=\frac{1}{(r-1)!}\int_{\lambda}^{x}(x-t)^{r-1}\left(q(t)-f^{(r)}(t)\right)dt,\quad
x\in[a,b].$

This immediately implies (4.1), and all other properties of $p$ are easily
verified. ∎

######  Lemma 4.3 (positive approximation with interpolatory constraints:
$f\in C$).

Suppose that $f\in C[a,b]$ is nonnegative on $[a,b]$, and $\lambda\in[a,b]$ is
such that either $\lambda$ is one of the endpoints of $[a,b]$, or
$\min\\{|a-\lambda|,|b-\lambda|\\}\geq\gamma|b-a|$, $\gamma>0$. Then, there
exists $l\in\Pi_{1}$ such that $l$ is nonnegative on $[a,b]$,
$l(\lambda)=f(\lambda)$ and

(4.2) $\left\|f-l\right\|_{[a,b]}\leq c{\omega}_{2}(f,b-a),$

where $c$ is an absolute constant if $\lambda$ is an endpoint of $[a,b]$, or
$c=c(\gamma)$ if $\lambda\in(a,b)$.

###### Proof.

Let $l_{1},l_{2}\in\Pi_{1}$ be linear polynomials interpolating $f$ at $a$ and
$\lambda$, and at $\lambda$ and $b$, respectively. Then, by Whitney’s
inequality, $\left\|f-l_{i}\right\|_{[a,b]}\leq c{\omega}_{2}(f,b-a)$,
$i=1,2$, and if at least one of them is nonnegative on $[a,b]$, then we let
$l$ to be that nonnegative linear polynomial (this is obviously the case if
$\lambda$ is an endpoint of $[a,b]$). Otherwise, $l_{1}(b)<0$ and
$l_{2}(a)<0$, and so

$0\leq f(a)<f(a)-l_{2}(a)\leq c{\omega}_{2}(f,b-a)\quad\mbox{\rm and}\quad
0\leq f(b)<f(b)-l_{1}(b)\leq c{\omega}_{2}(f,b-a).$

Also, since in this case $f(\lambda)<\min\\{f(a),f(b)\\}$, we have $0\leq
f(\lambda)\leq c{\omega}_{2}(f,b-a)$. This implies that
$\left\|l_{i}\right\|_{[a,b]}\leq c{\omega}_{2}(f,b-a)$, $i=1,2$, and so
$\left\|f\right\|_{[a,b]}\leq c{\omega}_{2}(f,b-a)$. Hence, we can set
$l\equiv f(\lambda)$ on $[a,b]$. ∎

It follows from the following lemma that the constant $c$ in (4.2) cannot be
made independent of $\gamma$. At the same time, it is rather obvious that, if
${\omega}_{2}$ in the estimate (4.2) is replaced by ${\omega}$, then $c$ no
longer depends on $\gamma$.

######  Lemma 4.4 (positive approximation with interpolatory constraints:
negative result for $f\in C$ and $\lambda$ near the endpoints).

For any $n\in\mathbb{N}$ and $A>0$, there exist $f\in C$ which is nonnegative
on $I$ and $\lambda\in(-1,1)$ such that, for any nonnegative on $I$ polynomial
$P_{n}\in\Pi_{n}$ satisfying $P_{n}(\lambda)=f(\lambda)$, we have

$\left\|f-P_{n}\right\|>A{\omega}_{2}(f,1).$

###### Proof.

Let $n\in\mathbb{N}$ and $A>0$ be fixed, pick
$f(x):=\max\left\\{(x-\lambda)/(\lambda+1),0\right\\}$ and suppose that
$P_{n}\in\Pi_{n}$ is nonnegative on $I$, satisfies
$P_{n}(\lambda)=f(\lambda)=0$ (this necessarily implies that
$P_{n}^{\prime}(\lambda)=0$) and such that $\left\|f-P_{n}\right\|\leq
A{\omega}_{2}(f,1)$. If $Q(x):=(x-\lambda)/(\lambda+1)$, then
$\left\|f-Q\right\|=\left\|Q\right\|_{[-1,\lambda]}=1$ and

${\omega}_{2}(f,1)={\omega}_{2}(f-Q,1)\leq 4\left\|f-Q\right\|=4.$

Hence,

$\left\|P_{n}-Q\right\|\leq\left\|P_{n}-f\right\|+\left\|f-Q\right\|\leq
4A+1,$

and so, by Markov’s inequality,

$1/(\lambda+1)=Q^{\prime}(\lambda)=Q^{\prime}(\lambda)-P_{n}^{\prime}(\lambda)\leq\left\|Q^{\prime}-P_{n}^{\prime}\right\|\leq
n^{2}\left(4A+1\right).$

We now get a contradiction by letting $\lambda\to-1^{+}$. ∎

######  Lemma 4.5 (positive approximation with interpolatory constraints:
$f\in C^{1}$).

Suppose that $f\in C^{1}[a,b]$ is nonnegative on $[a,b]$, and
$\lambda\in[a,b]$ is such that
$\min\\{|a-\lambda|,|b-\lambda|\\}\geq\gamma|b-a|$, $\gamma>0$. Then, there
exists $p\in\Pi_{3}$ such that $p\geq 0$ on $[a,b]$, $p(\lambda)=f(\lambda)$
and

(4.3) $\left\|f-p\right\|_{[a,b]}\leq
c(b-a){\omega}_{3}(f^{\prime},b-a;[a,b]),\quad\text{where $c=c(\gamma)$.}$

###### Proof.

Assume, for convenience, that $[a,b]=I$ and $\lambda=0$.

Let $L_{0}(x)$ be the Lagrange-Hermite polynomial of degree $\leq 3$ such that
$L_{0}^{(i)}(0)=f^{(i)}(0)$, $i=0,1$ and $L_{0}(\pm 1)=f(\pm 1)$:

$L_{0}(x)=f(0)+f^{\prime}(0)\,x+\left[\frac{f(-1)+f(1)}{2}-f(0)\right]\,x^{2}+\left[\frac{f(1)-f(-1)}{2}-f^{\prime}(0)\right]\,x^{3}.$

It follows from Lemma 4.1 with $q=4$ and $r=1$ that

$\left\|f-L_{0}\right\|\leq c{\omega}_{3}(f^{\prime},1).$

This also implies that

(4.4) $\left\|f^{\prime}-L_{0}^{\prime}\right\|\leq
A{\omega}_{3}(f^{\prime},1),$

where $A$ is some positive constant that we now consider fixed.

Consider the following two cases:

Case 1: $|f^{\prime}(0)|\leq 2A{\omega}_{3}(f^{\prime},1)$
We define

$p(x):=f(0)+\left[\frac{f(-1)+f(1)}{2}-f(0)\right]\,x^{2}+\left[\frac{f(1)-f(-1)}{2}\right]\,x^{3}=:f(0)+ax^{2}+bx^{3}$

and note that $\left\|L_{0}-p\right\|\leq|f^{\prime}(0)|$ which implies that
$\left\|f-p\right\|\leq c{\omega}_{3}(f^{\prime},1)$. Clearly, $p(0)=f(0)$,
and so it remains to show that $p$ is nonnegative on $I$.

Note that $p(1)=f(1)$ and $p(-1)=f(-1)$. If $b=0$, then $p$ is clearly
nonnegative on $I$, and so suppose that $b\neq 0$ and, in fact, without loss
of generality, we can assume that $b>0$.

Clearly, $p^{\prime}(x)=0$ iff $x=0$ or $x=-2a/(3b)=:x_{0}$. Now, if $a\geq
0$, then $x_{0}\leq 0$, and $p\downarrow$ on $[x_{0},0]$ and $p\uparrow$,
otherwise. This implies that $p\geq 0$ on $I$.

If $a<0$, then $x_{0}>0$, and $p\downarrow$ on $[0,x_{0}]$ and $p\uparrow$,
otherwise. If $x_{0}\geq 1$, then $p$ is clearly nonnegative on $I$ . Hence,
it remains to show that $p(x_{0})\geq 0$ in the case $0<x_{0}<1$ (i.e., if
$0<-2a<3b$). Indeed, we have

$p(x_{0})=f(0)+\frac{4}{27}\cdot\frac{a^{3}}{b^{2}}>f(0)-\frac{b}{2}>\frac{f(-1)+f(1)}{2}-\frac{f(1)-f(-1)}{4}=\frac{3f(-1)+f(1)}{4}\geq
0.$

Case 2: $|f^{\prime}(0)|>2A{\omega}_{3}(f^{\prime},1)$
Without loss of generality, we assume that $f^{\prime}(0)<0$. Define

$p(x):=L_{0}(x)+A(x+x^{2}){\omega}_{3}(f^{\prime},1)$

and note that $\left\|f-p\right\|\leq c{\omega}_{3}(f^{\prime},1)$ and
$p(0)=f(0)$, and so we only need to show that $p$ is nonnegative on $I$. It
follows from (4.4) that, for $x\in[0,1]$,

$p^{\prime}(x)=L_{0}^{\prime}(x)+A(1+2x){\omega}_{3}(f^{\prime},1)\geq
L_{0}^{\prime}(x)+A{\omega}_{3}(f^{\prime},1)\geq f^{\prime}(x),$

and so

$p(x)-f(x)=\int_{0}^{x}[p^{\prime}(u)-f^{\prime}(u)]\,du\geq 0,\quad 0\leq
x\leq 1.$

For convenience, we denote

$p(x)=f(0)+ax+bx^{2}+cx^{3},$

and note that

$a=p^{\prime}(0)=L_{0}^{\prime}(0)+A{\omega}_{3}(f^{\prime},1)=f^{\prime}(0)+A{\omega}_{3}(f^{\prime},1)<-A{\omega}_{3}(f^{\prime},1)<0.$

If $c\leq 0$, then, for any $x\in[-1,0)$, we have

$p(x)=p(-x)+2ax+2cx^{3}\geq p(-x)\geq 0.$

Hence, it remains to consider the case $c>0$.

If $p^{\prime}\leq 0$ on $[-1,0]$, then $p\downarrow$ on $[-1,0]$ and so is
nonnegative there. Hence, we can assume that $p^{\prime}$ is positive
somewhere in $[-1,0]$. Since $c>0$, this implies that there exists
$\alpha_{0}\in(-1,0)$ such that $p^{\prime}>0$ on $[-1,\alpha_{0})$, and
$p^{\prime}<0$ on $(\alpha_{0},0]$. Hence, $p\uparrow$ on $[-1,\alpha_{0}]$
and $p\downarrow$ on $[\alpha_{0},0]$. This implies that we only need to
verify that $p(-1)\geq 0$ in order to conclude that $p$ is nonnegative on the
whole interval $I$. We have

$p(-1)=L_{0}(-1)=f(-1)\geq 0,$

and the proof is now complete. ∎

We remark that if $\lambda$ in the statement of Lemma 4.5 is one of the
endpoints of $[a,b]$, then a much stronger result is valid (see Lemma 4.2
which implies an analogous result for positive approximation). Also, it
follows from the following lemma that the constant $c$ in (4.3) cannot be made
independent of $\gamma$.

######  Lemma 4.6 (positive approximation with interpolatory constraints:
negative result for $f\in C^{1}$ and $\lambda$ near the endpoints).

For any $n\in\mathbb{N}$ and $A>0$, there exist $f\in C^{1}$ which is
nonnegative on $I$ and $\lambda\in(-1,1)$ such that, for any nonnegative on
$I$ polynomial $P_{n}\in\Pi_{n}$ satisfying $P_{n}(\lambda)=f(\lambda)$, we
have

$\left\|f-P_{n}\right\|>A{\omega}_{3}(f^{\prime},1).$

###### Proof.

Let $n\in\mathbb{N}$ and $A>0$ be fixed, pick
$f(x):=\max\left\\{(x-\lambda)^{3}/(\lambda+1)^{2},0\right\\}$ and suppose
that $P_{n}\in\Pi_{n}$ is nonnegative on $I$, satisfies
$P_{n}(\lambda)=f(\lambda)=0$ (this necessarily implies that
$P_{n}^{\prime}(\lambda)=0$) and such that $\left\|f-P_{n}\right\|\leq
A{\omega}_{3}(f^{\prime},1)$. If $Q(x):=(x-\lambda)^{3}/(\lambda+1)^{2}$, then

$\left\|f^{\prime}-Q^{\prime}\right\|=\left\|Q^{\prime}\right\|_{[-1,\lambda]}=3,\quad\left\|f-Q\right\|=\left\|Q\right\|_{[-1,\lambda]}=\lambda+1,$

and

${\omega}_{3}(f^{\prime},1)={\omega}_{3}(f^{\prime}-Q^{\prime},1)\leq
8\left\|f^{\prime}-Q^{\prime}\right\|=24.$

Hence,

$\left\|P_{n}-Q\right\|\leq\left\|P_{n}-f\right\|+\left\|f-Q\right\|\leq
24A+2,$

and so, by Markov’s inequality,

$\left\|P_{n}^{\prime\prime}-Q^{\prime\prime}\right\|\leq n^{4}(24A+2)=:K,$

and so, since $P_{n}^{(i)}(\lambda)=Q^{(i)}(\lambda)$, $i=0,1$,

$\lambda+1=-Q(-1)\leq
P_{n}(-1)-Q(-1)=\int_{\lambda}^{-1}(-1-t)\left[P_{n}^{\prime\prime}(t)-Q^{\prime\prime}(t)\right]\,dt\leq
K(\lambda+1)^{2}/2,$

and we get a contradiction by letting $\lambda\to-1^{+}$. ∎

######  Lemma 4.7 (copositive approximation with interpolation at $Y_{s}$:
$f\in C^{1}$).

Suppose that $f\in C^{1}[a,b]$ is such that $(x-\lambda)f(x)\geq 0$,
$x\in[a,b]$, where $\lambda\in[a,b]$ is such that
$\min\\{|a-\lambda|,|b-\lambda|\\}\geq\gamma|b-a|$, $\gamma>0$. Then, there
exists $p\in\Pi_{2}$ such that $(x-\lambda)p(x)\geq 0$, $x\in[a,b]$,
$p^{(i)}(\lambda)=f^{(i)}(\lambda)$, $i=0,1$, and

(4.5) $\left\|f-p\right\|_{[a,b]}\leq
c(b-a){\omega}_{2}(f^{\prime},b-a;[a,b]),\quad\text{where $c=c(\gamma)$.}$

###### Proof.

Assume, for convenience, that $[a,b]=I$ and $\lambda=0$. Then,
$f\in\Delta^{(0)}(\\{0\\})$, and we need to construct
$p\in\Delta^{(0)}(\\{0\\})\cap\Pi_{2}$ such that $p^{(i)}(0)=f^{(i)}(0)$,
$i=0,1$, and $\left\|f-p\right\|\leq c{\omega}_{2}(f^{\prime},1)$.

Let $L_{1}$ and $L_{2}$ be the quadratic Lagrange-Hermite polynomials such
that $L_{1}^{(i)}(0)=L_{2}^{(i)}(0)=f^{(i)}(0)$, $i=0,1$, $L_{1}(-1)=f(-1)$
and $L_{2}(1)=f(1)$. Note that $f(0)=0$ and $f^{\prime}(0)\geq 0$. Hence,

$L_{1}(x)=f^{\prime}(0)\,x+\left[f(-1)+f^{\prime}(0)\right]\,x^{2}$

and

$L_{2}(x)=f^{\prime}(0)\,x+\left[f(1)-f^{\prime}(0)\right]\,x^{2}.$

It follows from Lemma 4.1 with $q=3$ and $r=1$ that

$\left\|f-L_{j}\right\|\leq c{\omega}_{2}(f^{\prime},1),\quad j=1,2.$

Now, note that $L_{1}\in\Delta^{(0)}(\\{0\\})$ iff $L_{1}(1)\geq 0$, and
$L_{2}\in\Delta^{(0)}(\\{0\\})$ iff $L_{2}(-1)\leq 0$. So, if $L_{1}(1)\geq
0$, we set $p:=L_{1}$ and, if $L_{2}(-1)\leq 0$, then we set $p:=L_{2}$, and
the proof is complete. Suppose now that

$L_{1}(1)=2f^{\prime}(0)+f(-1)<0\quad\mbox{\rm and}\quad
L_{2}(-1)=f(1)-2f^{\prime}(0)>0.$

Then,

$0\leq f(1)\leq f(1)-L_{1}(1)\leq c{\omega}_{2}(f^{\prime},1)\quad\mbox{\rm
and}\quad 0\leq-f(-1)\leq L_{2}(-1)-f(-1)\leq c{\omega}_{2}(f^{\prime},1).$

We now set

$p(x):=f^{\prime}(0)\,x,$

and note that $p\in\Delta^{(0)}(\\{0\\})$, $p^{(i)}(0)=f^{(i)}(0)$, $i=0,1$,
and

$\left\|f-p\right\|\leq\left\|f-L_{1}\right\|+|f(-1)|+|f^{\prime}(0)|\leq
c{\omega}_{2}(f^{\prime},1),$

since $0\leq f^{\prime}(0)<f(1)/2\leq c{\omega}_{2}(f^{\prime},1)$. ∎

######  Lemma 4.8 (copositive approximation with interpolation at $Y_{s}$:
$f\in C^{2}$).

Suppose that $f\in C^{2}[a,b]$ is such that $(x-\lambda)f(x)\geq 0$,
$x\in[a,b]$, where $\lambda\in[a,b]$ is such that
$\min\\{|a-\lambda|,|b-\lambda|\\}\geq\gamma|b-a|$, $\gamma>0$. Then, there
exists $P\in\Pi_{4}$ such that $(x-\lambda)P(x)\geq 0$, $x\in[a,b]$,
$P^{(i)}(\lambda)=f^{(i)}(\lambda)$, $i=0,1$, and

(4.6) $\left\|f-P\right\|_{[a,b]}\leq
c(b-a)^{2}{\omega}_{3}(f^{\prime\prime},b-a;[a,b]),\quad\text{where
$c=c(\gamma)$.}$

###### Proof.

Assume, for convenience, that $[a,b]=I$ and $\lambda=0$. Then,
$f\in\Delta^{(0)}(\\{0\\})$, and we need to construct
$P\in\Pi_{4}\cap\Delta^{(0)}(\\{0\\})$ such that $P^{(i)}(0)=f^{(i)}(0)$,
$i=0,1$, and $\left\|f-P\right\|\leq c{\omega}_{3}(f^{\prime\prime},1)$.

Note that $f\in\Delta^{(0)}(\\{0\\})$ implies that $f(0)=0$ and
$f^{\prime}(0)\geq 0$.

Let $p\in\Pi_{2}$ be any quadratic polynomial such that

$\left\|f^{\prime\prime}-p\right\|\leq c{\omega}_{3}(f^{\prime\prime},1)$

(e.g. $p$ is the quadratic polynomial of best approximation of
$f^{\prime\prime}$ on $I$). We now shift $p$ so that it is below/above
$f^{\prime\prime}$, i.e., let

$p_{1}:=p-\left\|f^{\prime\prime}-p\right\|\leq f^{\prime\prime}\quad\mbox{\rm
and}\quad p_{2}:=p+\left\|f^{\prime\prime}-p\right\|\geq f^{\prime\prime},$

and suppose that $p_{i}(x)=ax^{2}+bx+c_{i}$, $i=1,2$. If
$f^{\prime\prime}\equiv p$, then there is nothing to prove, so assume that
$f^{\prime\prime}\not\in\Pi_{2}$. Then,
$0<c_{2}-c_{1}=2\left\|f^{\prime\prime}-p\right\|\leq
c{\omega}_{3}(f^{\prime\prime},1)$. Now, for $i=1,2$, we define

$P_{i}(x):=f^{\prime}(0)x+\int_{0}^{x}(x-t)p_{i}(t)dt=f^{\prime}(0)x+\frac{a}{12}x^{4}+\frac{b}{6}x^{3}+\frac{c_{i}}{2}x^{2}$

and note that $P_{1}\leq f\leq P_{2}$ on $I$. Since
$f\in\Delta^{(0)}(\\{0\\})$, this implies that $P_{2}\geq 0$ on $[0,1]$, and
$P_{1}\leq 0$ on $[-1,0]$:

(4.7) $\displaystyle Q_{2}(x):=P_{2}(x)/x$
$\displaystyle=f^{\prime}(0)+\frac{a}{12}x^{3}+\frac{b}{6}x^{2}+\frac{c_{2}}{2}x\geq
0,\quad x\in[0,1],$ $\displaystyle Q_{1}(x):=P_{1}(x)/x$
$\displaystyle=f^{\prime}(0)+\frac{a}{12}x^{3}+\frac{b}{6}x^{2}+\frac{c_{1}}{2}x\geq
0,\quad x\in[-1,0].$

We now denote $\varepsilon:=(c_{2}-c_{1})/2$ and define

$P(x):=\begin{cases}P_{1}(x)+\varepsilon x^{3},&\text{if $c_{2}>c_{1}\geq
0$,}\\\ P_{2}(x)+\varepsilon x^{3},&\text{if $c_{1}<c_{2}\leq 0$,}\\\
f^{\prime}(0)x+ax^{4}/12+(b/6+\varepsilon)x^{3},&\text{if
$c_{1}<0<c_{2}$.}\end{cases}$

It is obvious that $P^{(i)}(0)=f^{(i)}(0)$, $i=0,1$, and
$\left\|f-P\right\|\leq c{\omega}_{3}(f^{\prime\prime},1)$ (in the third case,
one uses the fact that $c_{2}-c_{1}=|c_{1}|+|c_{2}|\leq
c{\omega}_{3}(f^{\prime\prime},1)$), and it only remains to check that
$P\in\Delta^{(0)}(\\{0\\})$. We denote $Q(x):=P(x)/x$ and note that
$P\in\Delta^{(0)}(\\{0\\})$ iff $Q\geq 0$ on $I$.

  1. (i)

$c_{2}>c_{1}\geq 0$.

In this case, $P\leq P_{1}\leq f\leq 0$ on $[-1,0]$, and so we need to show
that

$Q(x)=f^{\prime}(0)+\frac{a}{12}x^{3}+\left(\frac{b}{6}+\varepsilon\right)x^{2}+\frac{c_{1}}{2}x\geq
0,\quad x\in[0,1].$

If $a\geq 0$, then

$\displaystyle Q(x)$ $\displaystyle=Q_{1}(-x)+\frac{a}{6}x^{3}+\varepsilon
x^{2}+c_{1}x\geq 0,\quad x\in[0,1],$

since, by (4.7), $Q_{1}(-x)\geq 0$, $x\in[0,1]$, and $a,\varepsilon,c_{1}\geq
0$.

If $a<0$, then $Q^{\prime}$ is a concave down parabola, and
$Q^{\prime}(0)=c_{1}/2\geq 0$. Hence, either $Q^{\prime}\geq 0$ ($Q\uparrow$)
on $[0,1]$, or there exists $x_{0}\in[0,1]$ such that $Q^{\prime}\geq 0$
($Q\uparrow$) on $[0,x_{0}]$, and $Q^{\prime}\leq 0$ ($Q\downarrow$) on
$[x_{0},1]$. Therefore, using (4.7) again, we have

$\displaystyle\min_{x\in[0,1]}Q(x)$
$\displaystyle=\min\\{Q(0),Q(1)\\}=\min\left\\{f^{\prime}(0),Q_{2}(1)+\varepsilon-(c_{2}-c_{1})/2\right\\}$
$\displaystyle=\min\left\\{f^{\prime}(0),Q_{2}(1)\right\\}\geq 0.$

  2. (ii)

$c_{1}<c_{2}\leq 0$.

This is analogous to the previous case. Indeed, $P\geq P_{2}\geq f\geq 0$ on
$[0,1]$, and we need to verify that

$Q(x)=f^{\prime}(0)+\frac{a}{12}x^{3}+\left(\frac{b}{6}+\varepsilon\right)x^{2}+\frac{c_{2}}{2}x\geq
0,\quad x\in[-1,0].$

If $a\leq 0$, then

$\displaystyle Q(x)$ $\displaystyle=Q_{2}(-x)+\frac{a}{6}x^{3}+\varepsilon
x^{2}+c_{2}x\geq 0,\quad x\in[-1,0],$

since, by (4.7), $Q_{2}(-x)\geq 0$, $x\in[-1,0]$, and $a,c_{2}\leq 0$ and
$\varepsilon>0$.

If $a>0$, then $Q^{\prime}$ is a concave up parabola, and
$Q^{\prime}(0)=c_{2}/2\leq 0$. Hence, either $Q^{\prime}\leq 0$
($Q\downarrow$) on $[-1,0]$, or there exists $x_{0}\in[-1,0]$ such that
$Q^{\prime}\geq 0$ ($Q\uparrow$) on $[-1,x_{0}]$, and $Q^{\prime}\leq 0$
($Q\downarrow$) on $[x_{0},0]$. Therefore, using (4.7) again, we have

$\displaystyle\min_{x\in[-1,0]}Q(x)$
$\displaystyle=\min\\{Q(0),Q(-1)\\}=\min\left\\{f^{\prime}(0),Q_{1}(-1)+\varepsilon-(c_{2}-c_{1})/2\right\\}$
$\displaystyle=\min\left\\{f^{\prime}(0),Q_{1}(-1)\right\\}\geq 0.$

  3. (iii)

$c_{1}<0<c_{2}$. We need to check that

$Q(x)=f^{\prime}(0)+\frac{a}{12}x^{3}+\left(\frac{b}{6}+\varepsilon\right)x^{2}\geq
0,\quad x\in I.$

Denote

${\widetilde{Q}}(x):=f^{\prime}(0)-\frac{|a|}{12}x^{3}+\left(\frac{b}{6}+\varepsilon\right)x^{2},$

and note that $Q\geq 0$ on $I$ iff ${\widetilde{Q}}\geq 0$ on $[0,1]$.

Now, since ${\widetilde{Q}}^{\prime}$ is a concave down parabola (or a linear
function) with ${\widetilde{Q}}^{\prime}(0)=0$, either
${\widetilde{Q}}^{\prime}$ does not change its sign (and so ${\widetilde{Q}}$
is either $\uparrow$ or $\downarrow$) on $[0,1]$, or there exists
$x_{0}\in[0,1]$, such that ${\widetilde{Q}}^{\prime}\geq 0$
(${\widetilde{Q}}\uparrow$) on $[0,x_{0}]$ and ${\widetilde{Q}}^{\prime}\leq
0$ (${\widetilde{Q}}\downarrow$) on $[x_{0},1]$. This implies that
${\widetilde{Q}}$ achieves its minimum value on $[0,1]$ at $0$ or $1$, and
since ${\widetilde{Q}}(0)=f^{\prime}(0)\geq 0$, it remains to verify that
${\widetilde{Q}}(1)\geq 0$.

Using (4.7), if $a\geq 0$, we have

${\widetilde{Q}}(1)=f^{\prime}(0)-\frac{a}{12}+\frac{b}{6}+\varepsilon=Q_{1}(-1)+\varepsilon+\frac{c_{1}}{2}=Q_{1}(-1)+\frac{c_{2}}{2}\geq
0,$

and if $a<0$, then

${\widetilde{Q}}(1)=f^{\prime}(0)+\frac{a}{12}+\frac{b}{6}+\varepsilon=Q_{2}(1)+\varepsilon-\frac{c_{2}}{2}=Q_{2}(1)-\frac{c_{1}}{2}\geq
0.$

The proof is now complete. ∎

###  4.2 Global construction

Let $\mathbf{Z}_{n}:=\\{z_{j}\\}_{j=0}^{n}$, where
$-1:=z_{n}<z_{n-1}<\dots<z_{1}<z_{0}:=1$, $n\in\mathbb{N}$, be a given knot
sequence on $I$, and set $z_{i}:=-1$, $i>n$, $z_{i}:=1$, $i<0$, and
$J_{i}:=[z_{i},z_{i-1}]$.

Denote by $\Sigma_{k}(\mathbf{Z}_{n})$ the collection of all continuous
piecewise polynomials of degree $\leq k-1$ (of order $k$) on the knot sequence
$\mathbf{Z}_{n}$. Functions from $\Sigma_{k}(\mathbf{Z}_{n})\cap C^{k-2}$ are
usually called splines (or splines with minimal defect) of order $k$ with
knots $\mathbf{Z}_{n}$.

We need the following result which is usually referred to as Beatson’s lemma.

######  Lemma 4.9 (Beatson [bea]*Lemma 3.2).

Let $m\geq 2$ be an integer and $d=2(m-1)^{2}$. Let
$T=\\{t_{i}\\}_{i=-\infty}^{\infty}$ be a strictly increasing knot sequence
with $t_{0}=a$ and $t_{d}=b$. If $p_{1},p_{2}\in\Pi_{m-1}$, then there exists
a spline $S\in\Sigma_{m}(T)\cap C^{m-2}$ such that

  1. (i)

$S(x)$ is a number between $p_{1}(x)$ and $p_{2}(x)$ for each $x\in[a,b]$,

  2. (ii)

$S\equiv p_{1}$ on $(-\infty,a]$ and $S\equiv p_{2}$ on $[b,\infty)$.

The following theorem is our main result on intertwining approximation with
interpolatory constraints by splines.

######  Theorem 4.10 (intertwining spline approximation with interpolatory
constraints).

Let $k,p\in\mathbb{N}$, $s\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$,
$A_{p}\in\mathbb{A}_{p}$,
$(r,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}}$, defined in (2.1),
and suppose that $f\in C^{r}$. Also, let $\mathbf{Z}_{n}$ be a given knot
sequence such that there are at least $4(k+r-1)^{2}$ knots in each nonempty
open interval $(\beta_{j},\beta_{j-1})$, $1\leq j\leq s+p+1$, where
$\\{\beta_{j}\\}_{j=1}^{s+p}=Y_{s}\cup A_{p}$, $\beta_{0}=1$ and
$\beta_{s+p+1}=-1$ (see (1.3)). Then, there exists

(4.8) $S\in\Sigma_{k+r}(\mathbf{Z}_{n})\cap
C^{k+r-2}\cap\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that, for $0\leq i\leq n-1$,

(4.9) $\left\|f-S\right\|_{J_{i}}\leq c|J_{i}|^{r}{\omega}_{k}(f^{(r)},|{\cal
J}_{i}|;{\cal J}_{i}),\quad c=c(k,r,\rho),$

where $\rho:=\rho(\mathbf{Z}_{n}):=\max\left\\{|J_{i\pm
1}|/|J_{i}|\;\;\big{|}\;\;1\leq i\leq n\right\\}$, and ${\cal J}_{i}$ is an
interval such that $J_{i}\subset{\cal
J}_{i}\subseteq\bigl{[}z_{i+6(k+r-1)^{2}},\,z_{i-6(k+r-1)^{2}}\bigr{]}$. Here,
if $m_{1}=-1$, then the restriction $S\in{\mathcal{I}}^{(-1)}(f,A_{p})$ is
vacuous, i.e., there is no requirement that $S$ interpolate $f$ at the points
in $A_{p}$.

We recall that, if $s=0$, then $\mathbb{Y}_{0}=\emptyset$, and so Theorem 4.10
becomes a theorem on onesided spline approximation with interpolatory
constraints.

###### Proof.

The idea of the proof is quite similar to that of [hky]*Theorem 3. The only
difference is that, near the points from $Y_{s}\cup A_{p}\cup\\{\pm 1\\}$, we
use Lemma 4.2 to construct an appropriate polynomial piece of $S$. For
completeness, we provide details.

Without loss of generality, we may assume that $\pm 1\in A_{p}$. We now denote
$m:=k+r$, $d:=2(m-1)^{2}$, $N:=\lceil n/d\rceil$ and $\bar{z}_{i}:=z_{di}$,
$i\in\mathbb{Z}$. Note that $\bar{z}_{i}=1$ for $i\leq 0$ and $\bar{z}_{i}=-1$
for $i\geq N$.

We now construct overlapping polynomial pieces of degree $\leq m-1$ on the
coarser partition $\overline{\mathbf{Z}}_{N}:=\\{\bar{z}_{i}\\}_{i=0}^{N}$. We
say that the interval $\bar{I}_{i}:=[\bar{z}_{i},\,\bar{z}_{i-1}]$ is
“$Y_{s}$-contaminated” if $\bar{z}_{i}\leq y_{j}<\bar{z}_{i-1}$ for some point
$y_{j}\in Y_{s}$, and it is “$A_{p}$-contaminated” if
$\bar{z}_{i}\leq\alpha_{j}<\bar{z}_{i-1}$, $i\geq 2$, or
$\bar{z}_{1}\leq\alpha_{j}\leq\bar{z}_{0}=1$, for some point $\alpha_{j}\in
A_{p}$. By assumption in the statement of the theorem, there exists exactly
one $\beta_{j}$ in each of the contaminated intervals $\bar{I}_{\mu_{j}}$,
$1\leq j\leq s+p$, and there is at least one non-contaminated interval between
$\bar{I}_{\mu_{j}}$ and $\bar{I}_{\mu_{j+1}}$, i.e.,

(4.10) $\mu_{j}+2\leq\mu_{j+1},\qquad 1\leq j\leq s+p-1.$

Note that $\mu_{1}=1$ and $\mu_{s+p}=N$ by our assumption that $\pm 1\in
A_{p}$.

Also, if an interval $J\subset I$ does not contain any points from $Y_{s}$,
then $J\subset[y_{j},y_{j-1}]$, for some $1\leq j\leq s+1$, and we say that
$J$ is “positive” if $j$ is odd, and it is “negative” if $j$ is even. Any
function $S\in\widetilde{\Delta}(f,Y_{s})$ satisfies $S\geq f$ on all positive
intervals, and $S\leq f$ on all negative intervals.

If $\mu_{j+1}>\mu_{j}+2$, on each of the intervals
$[\bar{z}_{i},\,\bar{z}_{i-2}]$, $i=\mu_{j}+2,\dots,\mu_{j+1}-1$, by Whitney’s
inequality there exist two polynomials $P_{i}$ and $Q_{i}$ of degree $<m$ such
that

$P_{i}(x)\geq f(x)\geq Q_{i}(x),\quad x\in[\bar{z}_{i},\,\bar{z}_{i-2}],$

and

$\left\|P_{i}-Q_{i}\right\|_{[\bar{z}_{i},\,\bar{z}_{i-2}]}\leq
c{\omega}_{m}(f,|\bar{I}_{i}|;[\bar{z}_{i},\,\bar{z}_{i-2}])\leq
c|\bar{I}_{i}|^{r}{\omega}_{k}(f^{(r)},|\bar{I}_{i}|;[\bar{z}_{i},\,\bar{z}_{i-2}]),\quad
c=c(m,\rho).$

We now define $p_{i}$ on $[\bar{z}_{i},\,\bar{z}_{i-2}]$ as

$p_{i}:=\begin{cases}P_{i},&\text{if $(\bar{z}_{i},\,\bar{z}_{i-2})$ is
positive,}\\\ Q_{i},&\text{if $(\bar{z}_{i},\,\bar{z}_{i-2})$ is
negative.}\end{cases}$

Then,

(4.11) $\left\|f-p_{i}\right\|_{[\bar{z}_{i},\,\bar{z}_{i-2}]}\leq
c|\bar{I}_{i}|^{r}{\omega}_{k}(f^{(r)},|\bar{I}_{i}|;[\bar{z}_{i},\,\bar{z}_{i-2}]),\quad
c=c(k,r,\rho).$

If $\mu_{j+1}=\mu_{j}+2$ (i.e., if there is only one non-contaminated interval
between $\bar{I}_{\mu_{j}}$ and $\bar{I}_{\mu_{j+1}}$), then the above
construction is not needed.

We now construct local polynomial pieces near each contaminated interval
$\bar{I}_{\mu_{j}}$, $1\leq j\leq s+p$. Denote
$\widetilde{I}_{\mu_{j}}:=[\bar{z}_{\mu_{j}+1},\bar{z}_{\mu_{j}-2}]$ and use
Lemma 4.2 with $[a,b]=\widetilde{I}_{\mu_{j}}$ to define $p_{\mu_{j}}$ on
$\widetilde{I}_{\mu_{j}}$. We recall that $p(\cdot;f,k,r,\lambda)$ denotes the
polynomialfrom the statement of Lemma 4.2 and consider three cases.

Case 1: $\bar{I}_{\mu_{j}}$ is $\\{\pm 1\\}$-contaminated (i.e., $j=1$ or
$j=s+p$).

  1. (i)

If $j=1$ and $r$ is even, then $p_{\mu_{1}}=p_{1}:=p(\cdot;f,k,r,1)$.

  2. (ii)

If $j=1$ and $r$ is odd, then $p_{\mu_{1}}=p_{1}:=-p(\cdot;-f,k,r,1)$.

  3. (iii)

If $j=s+p$ and $\bar{I}_{\mu_{s+p}}=\bar{I}_{N}$ is positive, then
$p_{\mu_{s+p}}=p_{N}:=p(\cdot;f,k,r,-1)$.

  4. (iv)

If $j=s+p$ and $\bar{I}_{\mu_{s+p}}=\bar{I}_{N}$ is negative, then
$p_{\mu_{s+p}}=p_{N}:=-p(\cdot;-f,k,r,-1)$.

Case 2: $1<j<s+p$ and $\bar{I}_{\mu_{j}}$ is $A_{p}$-contaminated. (Recall
that, in this case, $r=1$ is excluded, and so $r\geq 2$.)

  1. (i)

If $r$ is even and $\bar{I}_{\mu_{j}}$ is positive, then
$p_{\mu_{j}}=p(\cdot;f,k,r,\beta_{j})$.

  2. (ii)

If $r$ is even and $\bar{I}_{\mu_{j}}$ is negative, then
$p_{\mu_{j}}=-p(\cdot;-f,k,r,\beta_{j})$.

  3. (iii)

If $r$ is odd and $\bar{I}_{\mu_{j}}$ is positive, then
$p_{\mu_{j}}=p(\cdot;f,k+1,r-1,\beta_{j})$.

  4. (iv)

If $r$ is odd and $\bar{I}_{\mu_{j}}$ is negative, then
$p_{\mu_{j}}=-p(\cdot;-f,k+1,r-1,\beta_{j})$.

Case 3: $\bar{I}_{\mu_{j}}$ is $Y_{s}$-contaminated (and so $\beta_{j}=y_{i}$,
for some $1\leq i\leq s$).

  1. (i)

If $r$ is odd and $i$ is odd, then $p_{\mu_{j}}=p(\cdot;f,k,r,\beta_{j})$.

  2. (ii)

If $r$ is odd and $i$ is even, then $p_{\mu_{j}}=-p(\cdot;-f,k,r,\beta_{j})$.

  3. (iii)

If $r$ is even and $i$ is odd, then
$p_{\mu_{j}}=p(\cdot;f,k+1,r-1,\beta_{j})$.

  4. (iv)

If $r$ is even and $i$ is even, then
$p_{\mu_{j}}=-p(\cdot;-f,k+1,r-1,\beta_{j})$.

We remark that, if $r\geq 2$ and $p_{\mu_{j}}=p(\cdot;f,k+1,r-1,\lambda)$,
then $p_{\mu_{j}}^{(i)}(\lambda)=f^{(i)}(\lambda)$, $0\leq i\leq r-2$, and

$\left\|f-p_{\mu_{j}}\right\|_{\widetilde{I}_{\mu_{j}}}\leq
c|\widetilde{I}_{\mu_{j}}|^{r-1}{\omega}_{k+1}(f^{(r-1)},|\widetilde{I}_{\mu_{j}}|;\widetilde{I}_{\mu_{j}})\leq
c|\widetilde{I}_{\mu_{j}}|^{r}{\omega}_{k}(f^{(r)},|\widetilde{I}_{\mu_{j}}|;\widetilde{I}_{\mu_{j}}).$

All overlapping polynomial pieces with all the right properties have now been
constructed, and it remains to blend them together to obtain a smooth spline
$S$ on the original knot sequence $\mathbf{Z}_{n}$ with the same properties.

If both $\bar{I}_{i}$ and $\bar{I}_{i+1}$ are non-contaminated, then $p_{i}$
and $p_{i+1}$ overlap on $\bar{I}_{i}$, which contains $d-1$ interior knots
from $\mathbf{Z}_{n}$. By Lemma 4.9 there exists a spline $S_{i}$ of order $m$
on $\bar{I}_{i}$ on these knots that connects with $p_{i+1}$ and $p_{i}$ in a
$C^{m-2}$ manner at $\bar{z}_{i}=z_{di}$ and $\bar{z}_{i-1}=z_{d(i-1)}$,
respectively. Moreover, the graph of $S_{i}$ lies between those of $p_{i+1}$
and $p_{i}$, and, hence, $\mathop{\rm
sgn}\nolimits(p_{i+1}(x)-f(x))=\mathop{\rm
sgn}\nolimits(p_{i}(x)-f(x))=\mathop{\rm sgn}\nolimits(S_{i}(x)-f(x))$,
$x\in\bar{I}_{i}$. Additionally, it follows from (4.11) that

$\left\|f-S_{i}\right\|_{[\bar{z}_{i},\,\bar{z}_{i-2}]}\leq
c|\bar{I}_{i}|^{r}{\omega}_{k}(f^{(r)},|\bar{I}_{i}|;[\bar{z}_{i+1},\,\bar{z}_{i-2}]).$

The blending of overlapping polynomial pieces involving contaminated intervals
is done in exactly the same way. The only difference is that the spline pieces
$S_{i}$ thus produced satisfy the estimate above with a slightly larger
interval in place of $[\bar{z}_{i+1},\,\bar{z}_{i-2}]$ on the right-hand side,
($[\bar{z}_{i+2},\,\bar{z}_{i-3}]$ at worst), which will make no difference in
the rest of the proof.

We define the final spline $S$ on each $\bar{I}_{i}$ as follows: if there is
only one local polynomial $p_{i}$ over $\bar{I}_{i}$, set $S$ to this
polynomial; if there are two polynomials overlapping on $\bar{I}_{i}$, then
there must be a blending local spline $S_{i}$, and so set $S$ to $S_{i}$. It
is clear from this construction that $S\in\Sigma_{k+r}(\mathbf{Z}_{n})\cap
C^{k+r-2}\cap\widetilde{\Delta}(f,Y_{s})$, and $S$ also satisfies
interpolatory conditions stated in (4.8).

Finally, all neighboring intervals $I_{i}:=[z_{i},\,z_{i-1}]$ in the original
partition $\mathbf{Z}_{n}$ are comparable in size and each interval
$\bar{I}_{i}=[z_{di},\,z_{d(i-1)}]$ contains no more than $d$ such intervals.
Therefore, (4.9) is satisfied. ∎

######  Theorem 4.11 (copositive spline approximation with interpolatory
constraints).

Let $p\in\mathbb{N}$, $s\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$,
$A_{p}\in\mathbb{A}_{p}$, and
$(r,k,m_{1},m_{2},m_{3})\in\Upsilon^{(0)}_{\text{copositive}}$, defined in
(2.3), and suppose that $f\in C^{r}\cap\Delta^{(0)}(Y_{s})$. Also, let
$\mathbf{Z}_{n}$ be a given knot sequence such that there are at least
$4(k+r-1)^{2}$ knots in each nonempty open interval $(\beta_{j},\beta_{j-1})$,
$1\leq j\leq s+p+1$, where $\\{\beta_{j}\\}_{j=1}^{s+p}=Y_{s}\cup A_{p}$,
$\beta_{0}=1$ and $\beta_{s+p+1}=-1$ (see (1.3)). Then, there exists

(4.12) $S\in\Sigma_{k+r}(\mathbf{Z}_{n})\cap
C^{k+r-2}\cap\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that, for $0\leq i\leq n-1$,

(4.13) $\left\|f-S\right\|_{J_{i}}\leq c|J_{i}|^{r}{\omega}_{k}(f^{(r)},|{\cal
J}_{i}|;{\cal J}_{i}),\quad c=c(k,r,\rho),$

where $\rho$ and ${\cal J}_{i}$ are the same as in the statement of Theorem
4.10. Here, if $m_{1}=-1$, then the restriction
$S\in{\mathcal{I}}^{(-1)}(f,A_{p})$ is vacuous, i.e., there is no requirement
that $S$ interpolate $f$ at the points in $A_{p}$.

###### Proof.

First, we note that Theorem 4.11 immediately follows from Theorem 4.10 if

$\displaystyle(r,k,m_{1},m_{2},m_{3})\in$
$\displaystyle\left\\{(1,k,-1,0,0),(2,k,1,0,1)\;\;\big{|}\;\;k\in\mathbb{N}\right\\}$
$\displaystyle\cup\left\\{(r,k,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}}\;\;\big{|}\;\;r\geq
3\right\\}.$

In the remaining cases, its proof is exactly the same as that of Theorem 4.10
with the only difference that, while constructing local polynomial pieces near
the points from $Y_{s}\cup A_{p}\cup\\{\pm 1\\}$, instead of Lemma 4.2, we
need to use:

  * •

Lemma 4.3 near $A_{p}$ and a (trivial) linear interpolant near $Y_{s}$, if
$r=0$ and $k=2$,

  * •

Lemma 4.5 near $A_{p}$, if $r=1$ and $k=3$,

  * •

Lemma 4.7 near $Y_{s}$ and Lemma 4.5 near $A_{p}$, if $r=1$, $k=2$ and
$m_{2}=1$,

  * •

Lemma 4.8 near $Y_{s}$, if $r=2$, $k=3$ and $m_{2}=1$.

We omit details. ∎

##  5 Intertwining polynomial approximation of truncated power functions and
splines with interpolatory constraints

We reserve the special notation $\mathbf{T}_{n}$ for the Chebyshev partition
of $I$, i.e., $\mathbf{T}_{n}:=\\{x_{j}\\}_{j=0}^{n}$, where
$x_{j}:=x_{j,n}:=\cos\left(j\pi/n\right)$, $0\leq j\leq n$. Also, for $1\leq
j\leq n$, we denote

$I_{j}:=I_{j,n}:=[x_{j},x_{j-1}],\quad|I_{j}|=x_{j-1}-x_{j}\quad\mbox{\rm
and}\quad\psi_{j}(x):=\psi_{j,n}(x):=\frac{|I_{j}|}{|x-x_{j}|+|I_{j}|}.$

Note that $|I_{j\pm 1}|<3|I_{j}|$ and $\rho_{n}(x)<|I_{j}|<5\rho_{n}(x)$,
$x\in I_{j}$, and so $\psi_{j\pm 1}(x)\sim\psi_{j}(x)$, $x\in I$.

Additionally, we denote

$(x-\lambda)_{+}^{m}:=(x-\lambda)^{m}\chi[\lambda,1](x),\quad m\geq 0,$

where

$\chi[a,\,b](x):=\left\\{\begin{array}[]{ll}1,&\mbox{\rm if }x\in[a,\,b],\\\
0,&\mbox{\rm otherwise.}\end{array}\right.$

It is also convenient to denote $x_{j}:=1$ for $j<0$, and $x_{j}:=-1$ for
$j>n$.

###  5.1 Approximation of truncated power functions

######  Lemma 5.1 (intertwining approximation of truncated powers with
interpolatory constraints).

Let $s,p,r,m\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$ and
$A_{p}\in\mathbb{A}_{p}$, and suppose that $j,n\in\mathbb{N}$ are such that
the interval $(x_{j+1},x_{j-2})$ does not contain any points from $Y_{s}\cup
A_{p}\cup\\{\pm 1\\}$, and let $\lambda\in[{x_{j}},{x_{j-1}}]$ and
$g_{m,\lambda}(x):=(x-\lambda)_{+}^{m}$.

Then, there exists a sufficiently large constant $\mu=\mu(s,p,r,m)$ and
polynomials $R_{1}$ and $R_{2}$ of degree $\leq c(\mu)n$ such that

(5.1) $R_{1},R_{2}\in{\mathcal{I}}^{(r)}(g_{m,\lambda},A_{p}\cup
Y_{s}\cup\\{\pm 1\\}),$ (5.2)
$\left\\{R_{1}-g_{m,\lambda},g_{m,\lambda}-R_{2}\right\\}\subset\Delta^{(0)}(Y_{s}),$

i.e., $\\{R_{1},R_{2}\\}$ is an intertwining pair of polynomials for
$g_{m,\lambda}$ with respect to $Y_{s}$, and

(5.3) $\left|R_{i}(x)-g_{m,\lambda}(x)\right|\leq
c(\mu)\psi_{j}(x)^{\mu-m}|I_{j}|^{m},\quad\text{for all $x\in I$ and
$i=1,2$.}$

###### Proof.

In order to avoid possible confusion, we note that, in this proof, the prime
notation is used to denote new parameters/functions and is not used to denote
any derivatives. It is also clear that, without loss of generality, we can
assume that $\pm 1\in A_{p}$.

Case 1: even $m$. For even $m$ and without (5.1), Lemma 5.1 is [hky]*Lemma 19
with

$\\{a_{1},\dots,a_{k}\\}=\left\\{y\in
Y_{s}\;\;\big{|}\;\;y<x_{j}\right\\}\quad\mbox{\rm
and}\quad\\{b_{1},\dots,b_{l}\\}=\left\\{y\in
Y_{s}\;\;\big{|}\;\;y>x_{j}\right\\}.$

For even $m$, the same construction also works if points $a_{i}$’s and
$b_{i}$’s are allowed to coalesce, which implies Hermite interpolation of
$g_{m,\lambda}$ by $R_{i}$’s at these points. Hence, it is sufficient to set

$\\{a_{1},\dots,a_{k}\\}:=\bigcup_{y\in
Y_{s},\;y<x_{j}}\big{\\{}\underbrace{y,\dots,y}_{r^{\prime}}\big{\\}}\cup\bigcup_{\alpha\in
A_{p},\;\alpha<x_{j}}\big{\\{}\underbrace{\alpha,\dots,\alpha}_{r^{\prime\prime}}\big{\\}}$

and

$\\{b_{1},\dots,b_{l}\\}:=\bigcup_{y\in
Y_{s},\;y>x_{j}}\big{\\{}\underbrace{y,\dots,y}_{r^{\prime}}\big{\\}}\cup\bigcup_{\alpha\in
A_{p},\;\alpha>x_{j}}\big{\\{}\underbrace{\alpha,\dots,\alpha}_{r^{\prime\prime}}\big{\\}},$

where $r^{\prime}$ and $r^{\prime\prime}$ are, respectively, odd and even
integers which are not less than $r+1$ (for example, we can set
$r^{\prime}:=2r+1$ and $r^{\prime\prime}:=2(r+1)$), so that
$R_{i}-g_{m,\lambda}$, $i=1,2$, changes its sign at the points in $Y_{s}$ and
does not change it at the points in $A_{p}$.

A byproduct of the construction in [hky]*Lemma 19 was the fact that
$R_{i}(\lambda)=0$, $i=1,2$ in the case $m>0$. This makes it impossible for
the same construction to work for odd $m$. For example, if $m=1$, then it is
clear that there does not exist a polynomial $R\in\Pi_{n}$ such that
$R(\lambda)=0$ and, at the same time, $R\geq g_{1,\lambda}$ on
$(\lambda-\varepsilon,\lambda+\varepsilon)$.

Case 2: odd $m$. Suppose that $m$ is odd, and let $n^{\prime}:=2n$,
$j^{\prime}:=2j$ and $Y_{s}^{\prime}:=Y_{s}\cup\\{y^{\prime}\\}$, where
$y^{\prime}:=x_{j^{\prime}+1,n^{\prime}}$.

Note that

$x_{j^{\prime},n^{\prime}}=x_{j},\quad
I_{j^{\prime}+1,n^{\prime}}=[x_{j^{\prime}+1,n^{\prime}},x_{j^{\prime},n^{\prime}}]\subset
I_{j}\quad\mbox{\rm
and}\quad\operatorname{dist}\\{y^{\prime},\\{x_{j},x_{j+1}\\}\geq|I_{j+1}|/4.$

It now follows from the statement of the lemma for even $m$ (Case 1 above)
that there exist polynomials $R_{1}^{\prime}$ and $R_{2}^{\prime}$ such that

(5.4)
$R_{1}^{\prime},R_{2}^{\prime}\in{\mathcal{I}}^{(r)}(g_{m-1,\lambda},A_{p}\cup
Y_{s}^{\prime}),$ (5.5)
$\left\\{R_{1}^{\prime}-g_{m-1,\lambda},g_{m-1,\lambda}-R_{2}^{\prime}\right\\}\subset\Delta^{(0)}(Y_{s}^{\prime}),$

and

(5.6) $\left|R_{i}^{\prime}(x)-g_{m-1,\lambda}(x)\right|\leq
c(\mu)\psi_{j}(x)^{\mu-m+1}|I_{j}|^{m-1},\quad\text{for all $x\in I$ and
$i=1,2$.}$

We define

$R_{i}^{\prime\prime}(x):=(x-\lambda)R_{i}^{\prime}(x),\quad i=1,2,$

and note that
$R_{1}^{\prime\prime},R_{2}^{\prime\prime}\in{\mathcal{I}}^{(r)}(g_{m,\lambda},A_{p}\cup
Y_{s})$ and, for all $x\in I$ and $i=1,2$,

(5.7) $\left|R_{i}^{\prime\prime}(x)-g_{m,\lambda}(x)\right|\leq
c(\mu)|x-\lambda|\psi_{j}(x)^{\mu-m+1}|I_{j}|^{m-1}\leq
c(\mu)\psi_{j}(x)^{\mu-m}|I_{j}|^{m},$

since

$|x-\lambda|\leq|x-x_{j}|+|I_{j}|=\psi_{j}(x)^{-1}|I_{j}|.$

Let $J$ be the index such that $y_{i}\in[x_{j-2},1]$ for $1\leq i\leq J$, and
$y_{i}\in[-1,x_{j+1}]$ for $J+1\leq i\leq s$, and note that, if $J=0$ or
$J=s$, then there are no $y_{i}$’s in the intervals $[x_{j-2},1]$ or
$[-1,x_{j+1}]$, respectively.

Now, letting $h$ be either $R_{1}^{\prime}-g_{m-1,\lambda}$ or
$g_{m-1,\lambda}-R_{2}^{\prime}$, we note that

$h\in\Delta^{(0)}(Y_{s}^{\prime})\quad\iff\quad\begin{cases}(-1)^{i}h\geq
0&\text{on $[y_{i+1},y_{i}]$, for $0\leq i\leq J-1$,}\\\ (-1)^{J}h\geq
0&\text{on $[y^{\prime},y_{J}]$,}\\\ (-1)^{J+1}h\geq 0&\text{on
$[y_{J+1},y^{\prime}]$,}\\\ (-1)^{i+1}h\geq 0&\text{on $[y_{i+1},y_{i}]$, for
$J+1\leq i\leq s+1$,}\end{cases}$

and so, for $h^{\prime}:=(\cdot-\lambda)h$, we have

$\begin{cases}(-1)^{i}h^{\prime}\geq 0&\text{on $[y_{i+1},y_{i}]$, for $0\leq
i\leq J-1$,}\\\ (-1)^{J}h^{\prime}\geq 0&\text{on $[\lambda,y_{J}]$,}\\\
(-1)^{J+1}h^{\prime}\geq 0&\text{on $[y^{\prime},\lambda]$,}\\\
(-1)^{J}h^{\prime}\geq 0&\text{on $[y_{J+1},y^{\prime}]$,}\\\
(-1)^{i}h^{\prime}\geq 0&\text{on $[y_{i+1},y_{i}]$, for $J+1\leq i\leq
s+1$.}\end{cases}$

In other words, all the right inequalities for
$h^{\prime}\in\Delta^{(0)}(Y_{s})$ are satisfied except that the restriction
$(-1)^{J}h^{\prime}\geq 0$ that we need on $[y_{J+1},y_{J}]$ fails on its
small subinterval $[y^{\prime},\lambda]$. Hence, the polynomials
$R_{i}^{\prime\prime}$, $i=1,2$, need to be corrected on
$[y^{\prime},\lambda]$ in order to guarantee that
$\widetilde{R}_{1}^{\prime\prime}-g_{m,\lambda},g_{m,\lambda}-\widetilde{R}_{2}^{\prime\prime}\in\Delta^{(0)}(Y_{s})$,
where $\widetilde{R}_{i}^{\prime\prime}$, $i=1,2$, are these corrected
polynomials.

We note that it follows from (5.7) that

$\left\|h^{\prime}\right\|\leq
c(\mu)\left\|\psi_{j}^{\mu-m}\right\|\cdot|I_{j}|^{m}\leq
c(\mu)|I_{j}|^{m}:=c_{1}|I_{j}|^{m}.$

We now recall (see e.g. [dzya]) that the algebraic polynomial of degree $\leq
4n-2$,

$t_{j}(x):=(x-x_{j}^{0})^{-2}\cos^{2}2n\arccos
x+(x-\bar{x}_{j})^{-2}\sin^{2}2n\arccos x,$

where

$\bar{x}_{j}:=\cos\left(\frac{j\pi}{n}-\frac{\pi}{2n}\right)\quad\mbox{\rm
and}\quad
x_{j}^{0}:=\begin{cases}\cos\left(\frac{j\pi}{n}-\frac{\pi}{4n}\right),&\text{if
$j<n/2$,}\\\ \cos\left(\frac{j\pi}{n}-\frac{3\pi}{4n}\right),&\text{if $j\geq
n/2$,}\\\ \end{cases}$

satisfies

$|I_{j}|^{-2}\psi_{j}(x)^{2}\leq t_{j}(x)\leq
4000|I_{j}|^{-2}\psi_{j}(x)^{2},\quad\text{$x\in I$ and $1\leq j\leq n-1$.}$

We now define

$H(x):=t_{j}(x)^{\mu^{\prime}}|I_{j}|^{2\mu^{\prime}}\prod_{y\in
Y_{s}}\frac{(x-y)^{2r+1}}{|x_{j}-y|^{2r+1}}\prod_{\alpha\in
A_{p}}\frac{(x-\alpha)^{2r+2}}{|x_{j}-\alpha|^{2r+2}}$

with $\mu^{\prime}:=\mu+(2r+2)(s+p)$, for example, and note that

$H\in{\mathcal{I}}^{(r)}(0,A_{p}\cup Y_{s})\cap\Delta^{(0)}(Y_{s})$

(in particular, $(-1)^{J}H\geq 0$ on $[y_{J+1},y_{J}]$),

$\displaystyle|H(x)|$ $\displaystyle\leq
c\psi_{j}(x)^{2\mu^{\prime}}\prod_{y\in
Y_{s}}\frac{|x-y|^{2r+1}}{|x_{j}-y|^{2r+1}}\prod_{\alpha\in
A_{p}}\frac{|x-\alpha|^{2r+2}}{|x_{j}-\alpha|^{2r+2}}$ $\displaystyle\leq
c\psi_{j}(x)^{2\mu^{\prime}-(2r+2)(s+p)}\leq c\psi_{j}(x)^{\mu},\quad x\in I,$

since, for $\beta\in Y_{s}\cup A_{p}$,

$\displaystyle\frac{|x-\beta|}{|x_{j}-\beta|}$ $\displaystyle\leq
1+\frac{|x-x_{j}|}{|x_{j}-\beta|}\leq
1+\frac{|x-x_{j}|+|I_{j}|}{|I_{j}|}\cdot\frac{|I_{j}|}{|x_{j}-\beta|}$
$\displaystyle\leq
1+\psi_{j}(x)^{-1}\cdot\frac{|I_{j}|}{\min\\{|I_{j-1}|,|I_{j+1}|\\}}\leq
1+3\psi_{j}(x)^{-1}\leq 4\psi_{j}(x)^{-1}.$

Also, for $x\in[y^{\prime},\lambda]$, we have

$t_{j}(x)\geq(|x-x_{j}|+|I_{j}|)^{-2}\geq\left(\max\\{|y^{\prime}-x_{j}|+|I_{j}|,|\lambda-
x_{j}|+|I_{j}|\\}\right)^{-2}\geq 16^{-1}|I_{j}|^{-2}$

and

$\frac{|x-\beta|}{|x_{j}-\beta|}\geq\min\left\\{\frac{|x_{j+1}-y^{\prime}|}{|x_{j+1}-x_{j}|},\frac{|x_{j-2}-\lambda|}{|x_{j-2}-x_{j}|}\right\\}\geq\frac{1}{4}.$

Hence,

$(-1)^{J}H(x)=|H(x)|\geq 4^{-2\mu^{\prime}-(2r+1)s-(2r+2)p}:=c_{2},\quad
x\in[y^{\prime},\lambda].$

This implies that, for the function

$h^{\prime\prime}:=h^{\prime}+\widetilde{H},\quad\text{where
$\widetilde{H}(x):=(c_{1}/c_{2})H(x)|I_{j}|^{m}$,}$

we have

$(-1)^{J}h^{\prime\prime}(x)\geq-\left\|h^{\prime}\right\|+(c_{1}/c_{2})(-1)^{J}H(x)|I_{j}|^{m}\geq
0,\quad x\in[y^{\prime},\lambda],$

and so $h^{\prime\prime}\in\Delta^{(0)}(Y_{s})$.

It remains to notice that polynomials

$\widetilde{R}_{1}^{\prime\prime}:=R_{1}^{\prime\prime}+\widetilde{H}\quad\mbox{\rm
and}\quad\widetilde{R}_{2}^{\prime\prime}:=R_{2}^{\prime\prime}-\widetilde{H}$

satisfy (5.1)-(5.3), and the proof is now complete. ∎

###  5.2 Approximation of general splines

If $V_{m}=\\{v_{i}\\}_{i=1}^{m}$ is some finite collection of points in $I$,
then we denote by $\Sigma_{k}(\mathbf{Z}_{n},V_{m})$ the subset of
$\Sigma_{k}(\mathbf{Z}_{n})$ consisting of those continuous piecewise
polynomials $S$ that do not have any knots “too close” to the points in
$V_{m}$. More precisely, if $j_{i}$, $1\leq i\leq m$, are chosen so that
$v_{i}\in[z_{j_{i}},z_{j_{i}-1})$ (or $v_{i}\in[z_{j_{i}},z_{j_{i}-1}]$ if
$v_{i}=1$, i.e., $j_{1}:=1$ in this case), then $S$ is in
$\Sigma_{k}(\mathbf{Z}_{n},V_{m})$ if and only if
$S\in\Sigma_{k}(\mathbf{Z}_{n})$ and, for every $1\leq i\leq m$, the
restriction of $S$ to $(z_{j_{i}+1},z_{j_{i}-2})$ is a polynomial.

With this notation, the following lemma which is a corollary of a more general
result [hky]*Theorem 4, can be stated as follows.

######  Lemma 5.2 (see [hky]*Theorem 4).

Let $s,m\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$, $\mu\geq 2m+30$, and let
$S\in\Sigma_{2m+1}(\mathbf{T}_{n},Y_{s})\cap C^{2m-1}$, where $n>c(Y_{s})$ is
such that there are at least $4$ knots $x_{j}$ in each interval
$(y_{i+1},y_{i})$, $0\leq i\leq s$.

Then, there exists an intertwining pair of polynomials
$\\{P_{1},P_{2}\\}\subset\Pi_{c(\mu)n}$ for $S$ with respect to $Y_{s}$ such
that

$|P_{1}(x)-P_{2}(x)|\leq
c(m,\mu,s)\sum_{j=1}^{n-1}\psi_{j}(x)^{\mu}E_{2m}(S,I_{j}\cup I_{j+1}),$

where $E_{n}(f,[a,b]):=\inf_{P_{n}\in\Pi_{n}}\|f-P_{n}\|_{C[a,b]}$.

Exactly the same proof also works with an additional condition that
polynomials $P_{1}$ and $P_{2}$ (Hermite) interpolate $S$ at the points in
$Y_{s}\cup A_{p}\cup\\{\pm 1\\}$. Moreover, in view of Lemma 5.1, the
restriction that the order of the spline $S$ is odd is no longer required, and
we arrive at the following result.

######  Lemma 5.3.

Let $k\in\mathbb{N}$, $p,s,r\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$,
$A_{p}\in\mathbb{A}_{p}$, $\mu\geq k+30$, and let
$S\in\Sigma_{k}(\mathbf{T}_{n},Y_{s}\cup A_{p}\cup\\{\pm 1\\})\cap C^{k-2}$,
where $n>c(d(Y_{s},A_{p}))$ is such that there are at least $4$ knots $x_{j}$
in each nonempty open interval $(\beta_{j},\beta_{j-1})$, $1\leq j\leq s+p+1$,
where $\\{\beta_{j}\\}_{j=1}^{s+p}=Y_{s}\cup A_{p}$, $\beta_{0}=1$ and
$\beta_{s+p+1}=-1$ (see (1.3)).

Then, there exists an intertwining pair of polynomials
$\\{P_{1},P_{2}\\}\subset\Pi_{c(\mu)n}$ for $S$ with respect to $Y_{s}$ such
that

$P_{1},P_{2}\in{\mathcal{I}}^{(r)}(S,A_{p}\cup Y_{s}\cup\\{\pm 1\\})$

and

$|P_{1}(x)-P_{2}(x)|\leq
c(k,r,\mu,s,p)\sum_{j=1}^{n-1}\psi_{j}(x)^{\mu}E_{k-1}(S,I_{j}\cup I_{j+1}).$

##  6 Proofs of the main results

Theorems 2.1/2.2 and 2.3/2.4 immediately follow from Lemma 5.3 and,
respectively, Theorems 4.10 and 4.11, using the same sequence of estimates as
in [hky]*Section 3.

##  7 Appendix: Exact pointwise estimates

It is clear that if a polynomial $P_{n}$ not only approximates $f$ but also
interpolates it (and perhaps its derivatives) then the rate of approximation
(1.2) can be improved near all interpolation points.

The following theorem follows from the results obtained in [kls21].

######  Theorem 7.1 ([kls21]*Theorems 1.5, 1.6 and 1.7).

Let $k\in\mathbb{N}$, $x_{0}\in I$, $r,n\in\mathbb{N}_{0}$, $f\in C^{r}$, and
suppose that $m\in\mathbb{N}_{0}$ is such that $m\leq r$. If $P_{n}\in\Pi_{n}$
is such that

(7.1) $|f(x)-P_{n}(x)|\leq
A\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\quad x\in I,$

and

(7.2) $P_{n}^{(j)}(x_{0})=f^{(j)}(x_{0}),\quad\text{for }\;0\leq j\leq m,$

then, for all $1\leq\ell\leq k$ and $x\in I$, we have

(7.3) $\displaystyle|$ $\displaystyle f(x)-P_{n}(x)|$ $\displaystyle\leq
c(k,r)A\begin{cases}|x-x_{0}|^{m+1}\rho_{n}^{r-m-1}(x)\omega_{\ell}(f^{(r)},\rho_{n}(x)),&\text{if
}\;m\leq r-1,\\\
|x-x_{0}|^{r}\omega_{\ell}(f^{(r)},|x-x_{0}|^{1/\ell}\rho_{n}^{1-1/\ell}(x)),&\text{if
}\;m=r.\end{cases}$

Moreover, estimates in (7.3) cannot be improved in the sense that none of the
powers of $|x-x_{0}|$ can be increased.

The following result immediately follows from Theorems 2.1, 2.2 and 7.1
observing that, for any
$(r,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}}$, we have $m_{i}<r$,
$i=1,2,3$, and so the case $m=r$ in the statement of Theorem 7.1 does not
hold. Note also that we are incorporating the onesided case into a general
result for intertwining approximation by allowing $s$ to be $0$, in which case
the restriction $P_{n}\in{\mathcal{I}}^{(m_{2})}(f,Y_{0})$ becomes vacuous
(recall that the restriction $P_{n}\in{\mathcal{I}}^{(-1)}(f,A_{p})$ is also
vacuous by definition).

######  Corollary 7.2 (intertwining approximation with interpolatory
constraints).

Let $k,p\in\mathbb{N}$, $s\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$,
$A_{p}\in\mathbb{A}_{p}$, and
$(r,m_{1},m_{2},m_{3})\in\Upsilon_{\text{intertwining}}$, defined in (2.1).
Then, for any $f\in C^{r}$ and any $n\geq c(d(Y_{s},A_{p}))$, there exists

$P_{n}\in\Pi_{n}\cap\widetilde{\Delta}(f,Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I.$

Moreover, for all $1\leq\ell\leq k$ and $\lambda\in\mathbb{Y}_{s}\cup
A_{p}\cup\\{\pm 1\\}$, we also have

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)|x-\lambda|^{m+1}\rho_{n}^{r-m-1}(x){\omega}_{\ell}(f^{(r)},\rho_{n}(x)),\quad
x\in I,$

where

$m:=\begin{cases}m_{1},&\text{if $\lambda\in A_{p}$,}\\\ m_{2},&\text{if
$\lambda\in Y_{s}$,}\\\ m_{3},&\text{if $\lambda=\pm 1$.}\end{cases}$

We also have a similar result for (co)positive approximation with
interpolatory constraints which follows from Theorems 2.3, 2.4 and 7.1. The
only difference is that there are a few cases for $r=0$ and $r=1$ when
$m_{i}$’s can be equal to $r$, and so the estimates need to be modified
accordingly.

######  Corollary 7.3 (copositive approximation with interpolatory
constraints).

Let $p\in\mathbb{N}$, $s\in\mathbb{N}_{0}$, $Y_{s}\in\mathbb{Y}_{s}$,
$A_{p}\in\mathbb{A}_{p}$, and
$(r,k,m_{1},m_{2},m_{3})\in\Upsilon^{(0)}_{\text{copositive}}$, defined in
(2.3). Then, for any $f\in C^{r}\cap\Delta^{(0)}(Y_{s})$ and any $n\geq
c(d(Y_{s},A_{p}))$, there exists

$P_{n}\in\Pi_{n}\cap\Delta^{(0)}(Y_{s})\cap{\mathcal{I}}^{(m_{1})}(f,A_{p})\cap{\mathcal{I}}^{(m_{2})}(f,Y_{s})\cap{\mathcal{I}}^{(m_{3})}(f,\\{\pm
1\\})$

such that

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)\rho_{n}^{r}(x){\omega}_{k}(f^{(r)},\rho_{n}(x)),\qquad x\in I.$

Moreover, if $\lambda\in Y_{s}\cup A_{p}\cup\\{\pm 1\\}$ and

$m:=\begin{cases}m_{1},&\text{if $\lambda\in A_{p}$,}\\\ m_{2},&\text{if
$\lambda\in Y_{s}$,}\\\ m_{3},&\text{if $\lambda=\pm 1$.}\end{cases}$

is such that $m<r$, then for all for all $1\leq\ell\leq k$, we also have

$|f(x)-P_{n}(x)|\leq
c(k,r,s,p)|x-\lambda|^{m+1}\rho_{n}^{r-m-1}(x){\omega}_{\ell}(f^{(r)},\rho_{n}(x)),\quad
x\in I.$

Additionally, if $r=1$ and $\lambda\in Y_{s}$ then, for $\ell=1$ or $\ell=2$,

$|f(x)-P_{n}(x)|\leq
c(s,p)|x-\lambda|{\omega}_{\ell}(f^{\prime},|x-\lambda|^{1/\ell}\rho_{n}^{1-1/\ell}(x)),\quad
x\in I,$

and, if $r=0$ and $\lambda\in Y_{s}\cup A_{p}\cup\\{\pm 1\\}$ then, for
$\ell=1$ or $\ell=2$,

$|f(x)-P_{n}(x)|\leq
c(s,p){\omega}_{\ell}(f,|x-\lambda|^{1/\ell}\rho_{n}^{1-1/\ell}(x)),\quad x\in
I.$

We note that, for $r=0$, $A_{p}=\\{-1,1\\}$ and $k=\ell=2$, Corollary 7.3 was
proved in [dz22].

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