Fast Decreasing Trigonometric Polynomials and Applications

Abstract

We construct trigonometric polynomials that fast decrease towards $\pm \pi$. We apply them to construct a trigonometric polynomial the derivative of which interpolates the derivative of a given $2\pi$-periodic function, at some prescribed distinct points in $[-\pi ,\pi )$, and vanishes at some other prescribed points in that interval. The construction requires that the function possesses derivatives where the interpolation is supposed to take place. Still, we are able to apply the result to trigonometric approximation of a $2\pi$-periodic piecewise algebraic polynomial which is merely continuous, while interpolating its derivative at some points (that, obviously, are not knots).

1 Introduction and the main results

Let $I:=[-\pi ,\pi ]$ and let C[a, b] and $C^l[a,b]$ denote, respectively, the space of continuous functions and of l times continuously differentiable functions, and denote by $\tilde{C}$ the space of $2\pi$-periodic continuous functions. As usual all spaces are equipped with the sup-norm, i.e., $\Vert f\Vert _{[a,b]}:=\max _{x\in [a,b]}|f(x)|$ and $\Vert f\Vert :=\max _{x\in \mathbb R}|f(x)|$, respectively.

For $\beta \in \mathbb N$ and $n\in \mathbb N$, let

$$\begin{aligned} J(x)=J_{n,\beta }(x)=\frac{n}{\gamma _{n,\beta }}\left( \frac{\sin (nx/2)}{n\sin (x/2)}\right) ^{2\beta }, \end{aligned}$$

(1.1)

be a Jackson-type kernel, where $C_*(\beta )\le \gamma _{n,\beta }\le C^*(\beta )$ is a normalizing factor, so that J is a trigonometric polynomial of degree $\beta (n-1)$, and

$$\begin{aligned} \frac{1}{\pi }\int _{-\pi }^\pi J(x)\,dx=1, \end{aligned}$$

(1.2)

(see, e.g., [1, p. 204]). For the asymptotic behavior of $\gamma _{n,\beta }$, see [3, Theorem 1].

Put $h:=\pi /n$. Clearly,

$$\begin{aligned} J(x)\ge \frac{C_1(\beta )}{h},\quad |x|\le h, \end{aligned}$$

(1.3)

and Bernstein’s inequality implies, for all $\nu \in \mathbb N_0$,

$$\begin{aligned} \Vert J^{(\nu )}\Vert \le \frac{C_2(\beta ,\nu )}{h^{\nu +1}}. \end{aligned}$$

(1.4)

In addition, we will show that (see the end of Sect. 2) for $0\le \nu \le 2\beta$,

$$\begin{aligned} |J^{(\nu )}(x)|\le \frac{C_3(\beta )}{h^{\nu +1}}\frac{1}{(n|x|)^{2\beta -\nu }},\quad 0<|x|\le \pi , \end{aligned}$$

(1.5)

where $C_1(\beta )$ and $C_3(\beta )$ depend only on $\beta$, and $C_2(\beta ,\nu )$ may depend also on $\nu$.

We wish to construct a trigonometric polynomial $L_n$, of degree $\beta (n-1)$, which satisfies analogues of (1.3) through (1.5), not only for $h\approx \frac{\pi }{n}$, but rather, for any $\frac{\pi }{n}\le h\le \mathop {\text {const}}\nolimits$.

Theorem 1.1

For each $m\in \mathbb N$, $\beta \in \mathbb N$ and $0<\epsilon \le 1$, there are constants $K_1>0$ and $K_2$, depending only on m, $\beta$ and $\epsilon$, such that if $\frac{\pi }{n}\le h\le \frac{\pi }{m}$, then the trigonometric polynomial

$$\begin{aligned} L_n(x)&:=\frac{1}{h^m}\underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m\text { times}}J(x+t_1+\cdots +t_m)\,dt_1\cdots \,dt_m\\&=:\frac{1}{h^m}W_m(x,h,J), \end{aligned}$$

of degree $\le \beta (n-1)$, satisfies

$$\begin{aligned} \frac{1}{\pi }\int _{-\pi }^{\pi }L_n(t)\,dt=1, \end{aligned}$$

(1.6)

$$\begin{aligned} L_n(x)\ge \frac{K_1}{h},\quad |x|\le \frac{(m-\epsilon )h}{2}, \end{aligned}$$

(1.7)

and for all $0\le \nu \le m-1$,

$$\begin{aligned} \Vert L_n^{(\nu )}\Vert \le 2^\nu \pi h^{-(\nu +1)}, \end{aligned}$$

(1.8)

and

$$\begin{aligned} |L_n^{(\nu )}(x)|\le K_2\frac{h^{-(\nu +1)}}{(n|x|)^{2\beta -1}},\quad \frac{(m+\epsilon )h}{2}\le |x|\le \pi . \end{aligned}$$

(1.9)

Remark 1.2

The above constants may be replaced by $K_1:=C_1\epsilon ^{m-1}$ and $K_2:=C_2\epsilon ^{-2\beta }$, where $C_1$ and $C_2$ depend only on m and $\beta$.

We apply Theorem 1.1 to obtain an interpolation result. Namely,

Theorem 1.3

Given $n,s,\eta \in \mathbb N$ and $0<\epsilon \le 1/2$. For $\frac{\pi }{n}\le h\le \frac{\pi }{s+2}$, denote

$$ \dot{O}:=[-(s+1/2-\epsilon )h,(s+1/2-\epsilon )h]\quad \text {and}\quad \ddot{O}:=(-(s+1/2+\epsilon )h,(s+1/2+\epsilon )h). $$

Given a collection $\{z_i\}_{i=1}^{2s}$, of distinct points in $[-\pi ,\pi )$. Let l, $0\le l\le 2s$, be such that

$$\begin{aligned} z_q\in \dot{O},\quad 1\le q\le l,\quad \text {(no }z_q\text { in }\dot{O}\text {, if }l=0\text {)}, \end{aligned}$$

(1.10)

and

$$\begin{aligned} z_q\in [-\pi ,\pi )\setminus \ddot{O},\quad l+1\le q\le 2s\quad \text {(no }z_q\text { in }[-\pi ,\pi )\setminus \ddot{O}\text {, if }l=2s\text {)}. \end{aligned}$$

(1.11)

Assume that f is defined in $\dot{O}$, and if $l\ge 1$, assume that $f\in C^{l-1}(\dot{O})$ and satisfies,

$$\begin{aligned} |f^{(\nu )}(x)|\le h^{-\nu },\quad x\in \dot{O},\quad \nu =0\dots l-1. \end{aligned}$$

(1.12)

Then, there exists a constant $c=c(s,\eta ,\epsilon )$ and a trigonometric polynomial $D_l$ of degree $\le ([\eta /2]+2s+1)n$, such that

$$\begin{aligned} \Vert D_l\Vert \le ch, \end{aligned}$$

(1.13)

and its derivative $d_l:=D'_l$ satisfies,

$$\begin{aligned} d_l(z_q)=f(z_q),\quad 1\le q\le l, \end{aligned}$$

(1.14)

$$\begin{aligned} d_l(z_q)=0,\quad l+1\le q\le 2s, \end{aligned}$$

(1.15)

and, for all $\nu =0,\dots ,2s$,

$$\begin{aligned} |d_l^{(\nu )}(x)|\le ch^{-\nu }M^\eta (x),\quad x\in [-\pi ,\pi ], \end{aligned}$$

(1.16)

where

$$\begin{aligned} M(x):={\left\{ \begin{array}{ll} 1&{}\quad x\in \ddot{O}\\ \frac{1}{n|x|}&{}\quad x\in [-\pi ,\pi ]\setminus \ddot{O}. \end{array}\right. } \end{aligned}$$

(1.17)

Finally, in Sect. 6, we apply Theorem 1.3 to obtain a trigonometric polynomial which approximates a $2\pi$-periodic continuous piecewise algebraic polynomial, and the derivative of which, interpolates the derivative of the latter at a given collection of points (obviously, not knots).

Throughout the paper we will have positive constants c and C that may differ from one another on different occurrences even if they appear in the same line.

2 Pointwise Bernstein Inequality

We extend the well known Bernstein inequality

$$ \Vert (T_n^{r})^{(\nu )}\Vert \le (rn)^{\nu }\Vert T_n^{r}\Vert , $$

which is valid for all trigonometric polynomials $T_n$, of degree $\le n$, and for all $r\in \mathbb N$ and $\nu \in \mathbb N$, into a pointwise version. Namely,

Lemma 2.1

For arbitrary trigonometric polynomial $T_n$ of degree $\le n$, any $r\in \mathbb N$ and each natural $\nu \le r$ the inequality

$$\begin{aligned} |(T_n^r)^{(\nu )}(x)|\le (rn)^{\nu }\Vert T_n\Vert ^\nu |T_n^{r-\nu }(x)|, \quad x\in \mathbb R, \end{aligned}$$

(2.1)

holds.

Proof

Without loss of generality assume that $\Vert T_n\Vert =1$, so we have to prove the inequality

$$\begin{aligned} |(T_n^r)^{(\nu )}(x)|\le (rn)^{\nu }|T_n^{r-\nu }(x)|,\quad x\in \mathbb R. \end{aligned}$$

(2.2)

Then Bernstein inequality implies $\Vert T_n^{(\nu )}\Vert \le n^\nu$ and

$$ |(T_n^r)'(x)|=r|T_n^{r-1}(x)T_n'(x)|\le r n|T_n^{r-1}(x)|,\quad r\in \mathbb N. $$

Assuming by induction, that for some $\nu -1<r$ the inequality (2.1) holds for all $r\in \mathbb N$, we get

$$\begin{aligned} |(T_n^r)^{(\nu )}(x)|&=r\left| \left( T_n^{r-1}(x)T_n'(x)\right) ^{(\nu -1)}\right| \\&=r\left| \sum _{j=0}^{\nu -1}\left( {\begin{array}{c}\nu -1\\ j\end{array}}\right) (T_n^{r-1})^{(\nu -1-j)}(x)(T_n)^{(j+1)}(x)\right| \\&\le r\sum _{j=0}^{\nu -1}\left( {\begin{array}{c}\nu -1\\ j\end{array}}\right) ((r-1)n)^{\nu -1-j}|T^{r-\nu +j}_n(x)|n^{j+1}\\&\le rn^\nu |T_n^{r-\nu }(x)|\sum _{j=0}^{\nu -1}\left( {\begin{array}{c}\nu -1\\ j\end{array}}\right) (r-1)^{\nu -1-j}=(rn)^{\nu }|T_n^{r-\nu }(x)|, \end{aligned}$$

which is (2.2). $\square$

Applying Lemma 2.1 to the polynomial $T_n(u)=\frac{\sin {nu}}{n\sin u}$, we readily obtain (1.5).

3 Fast Decreasing Trigonometric Polynomials

Proof of Theorem 1.1

Since J is a trigonometric polynomial of degree $<\beta n$, $L_n$ is also a trigonometric polynomial of degree $<\beta n$.

First, we have

$$\begin{aligned} \int _{-\pi }^{\pi }L_n(x)\,dx&=\frac{1}{h^m}\int _{-\pi }^{\pi }\left( \underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m\text { times}}J(x+t_1+\cdots +t_m)\,dt_1\cdots \,dt_m\right) \,dx\\&=\frac{1}{h^m}\underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m\text { times}}\left( \int _{-\pi }^\pi J(x+t_1+\cdots +t_m)\, dx\right) \,dt_1\cdots \,dt_m\\&=\frac{1}{h^m}\underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m\text { times}}\pi \,dt_1\cdots \,dt_m=\pi , \end{aligned}$$

and (1.6) is proved.

Evidently, for every $\nu =1,\dots , m-1$, there is a $\theta =\theta _\nu \in [x-\nu h/2,x+\nu h/2]$, such that

$$\begin{aligned} |W_m^{(\nu )}(x,h,f)|\le 2^\nu |W_{m-\nu }(\theta ,h,f)|. \end{aligned}$$

(3.1)

Now, for any $a\in \mathbb R$,

$$ \int _{-h/2}^{h/2}J(a+t)\,dt\le \int _{-\pi }^{\pi }J(a+t)\,dt=\pi . $$

Thus, (3.1) implies,

$$ |L_n^{(\nu )}(x)|\le \frac{2^\nu }{h^m}W_{m-\nu }(\theta ,h,J)\le \frac{2^\nu \pi }{h^m} \underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m-\nu -1\text { times}}\,dt_1\cdots \,dt_{m-\nu -1}=2^\nu \pi h^{-(\nu +1)}, $$

which is (1.8).

In order to prove (1.9), take $\frac{(m+\epsilon )h}{2}\le |x|\le \pi$ and $|\theta -x|\le \nu h/2$. If

$$ |t_j|\le h/2,\quad 1\le j\le m-\nu ,\quad \text {and}\quad \Theta :=\theta +t_1+\cdots +t_{m-\nu }\in I, $$

then

$$ |\Theta |\ge |x|-(m-\nu +\nu )\frac{h}{2}=|x|-\frac{mh}{2},$$

which implies,

$$\begin{aligned} J(\Theta )\le \frac{1}{\gamma _{n,\beta }}\frac{n}{\bigl (n\sin {\Theta /2}\bigr )^{2\beta }}\le \frac{Cn}{\left( n(|x|-\frac{mh}{2})\right) ^{2\beta }}. \end{aligned}$$

(3.2)

Here and in the rest of the proof C and $C^*$ depend only on $\beta$.

If, on the other hand, $|\Theta |>\pi$, then $\pi <|\Theta |\le \frac{3\pi }{2}$, which implies

$$ |\sin {\Theta /2}|\ge \sin {\pi /4}\ge \sin \frac{|x|-\frac{mh}{2}}{4}\ge \frac{|x|-\frac{mh}{2}}{2\pi }, $$

so that (3.2) is valid in this case too.

Combined, (3.1) and (3.2) yield

$$\begin{aligned} |L_n^{(\nu )}(x)|&\le \frac{2^\nu }{h^m}W_{m-\nu }(\theta ,h,J)\le \frac{2^\nu }{h^m}\frac{Cn}{\left( n(|x|-\frac{mh}{2})\right) ^{2\beta }} \underbrace{\int _{-h/2}^{h/2}\dots \int _{-h/2}^{h/2}}_{m-\nu \text { times}}\,dt_1\cdots \,dt_{m-\nu }\\&=C\frac{2^\nu h^{-\nu }}{\left( n(|x|-\frac{mh}{2})\right) ^{2\beta -1}}\frac{1}{|x|-\frac{mh}{2}}\le C\frac{2^\nu h^{-\nu }(m+1)^{2\beta -1}}{\left( \epsilon n|x|\right) ^{2\beta -1}}\frac{2}{\epsilon h}\\&\le K_2\frac{h^{-(\nu +1)}}{(n|x|)^{2\beta -1}}, \end{aligned}$$

where, for $|x|\ge \frac{(m+\epsilon )h}{2}$, we applied the inequalities $|x|-\frac{mh}{2}\ge \frac{\epsilon |x|}{m+1}$ and $|x|-\frac{mh}{2}\ge \frac{\epsilon h}{2}$. Thus, we obtain (1.9).

In order to prove (1.7), we observe that if $|a|\le h/2$, then

$$\begin{aligned} \int _{-h/2}^{h/2}J(t+a)\,dt\ge \int _0^{h/2}J(t)\,dt\ge \int _0^{\pi /(2n)}J(t)\,dt>C^*. \end{aligned}$$

(3.3)

For $m=1$, this readily yields (1.7). Thus, let $m>1$ and denote

$$ H:=\left[ -\frac{h}{2},\frac{h}{2}\right] \bigcap \left[ \frac{-x}{m-1}-\frac{h}{2(m-1)},\frac{-x}{m-1}+\frac{h}{2(m-1)}\right] . $$

Note that if $t_j\in H$, $1\le j\le m-1$, then $-h/2\le a:=x+t_1+\cdots +t_{m-1}\le h/2$.

Hence, for $|x|\le \frac{(m-1)h}{2}$,

$$\begin{aligned} L_n(x)&\ge \frac{1}{h^m}\underbrace{\int _{H}\dots \int _{H}}_{m-1\text { times}}\left( \int _{-h/2}^{h/2}J(x+t_1+\cdots +t_m)\,dt_m\right) \, dt_1\cdots \,dt_{m-1}\\&\ge \frac{1}{h^m}\underbrace{\int _{H}\dots \int _{H}}_{m-1\text { times}}C^*\, dt_1\cdots \,dt_{m-1}=\frac{1}{h^m}C^*|H|^{m-1}\ge \frac{C}{h},\nonumber \end{aligned}$$

(3.4)

where we used the fact that $|H|\ge \frac{h}{2(m-1)}$.

Finally, if $\frac{(m-1)h}{2}<|x|\le \frac{(m-\epsilon )h}{2}$, then

$$ |H|=\frac{h}{2}-\left( \frac{|x|}{m-1}-\frac{h}{2(m-1)}\right) \ge \frac{h}{2}-\left( \frac{(m-\epsilon )h}{2(m-1)}-\frac{h}{2(m-1)}\right) =\frac{\epsilon h}{2(m-1)}. $$

Substituting in (3.4), completes the proof. $\square$

4 Interpolating Trigonometric Polynomials

Proof of Theorem 1.3

If $l=0$, then (1.14) is empty, so we may take $D_0(x)\equiv 0$.

We proceed by induction. By the induction assumption, there is a polynomial $D_{l-1}$, $1\le l\le 2s$, satisfying (1.14) through (1.16) with $l-1$ instead of l, and with any $\tilde{z}_l\in \ddot{O}$, $\tilde{z}_l\notin \{z_i\}_{i=l+1}^{2s}$, instead of the given $z_l\in \dot{O}$.

We will construct the derivative $d_l$ and then put $D_l(x):=\int _{-\pi }^x d_l(t)\,dt$.

To construct $d_l$ we first note that for the polynomial $L_n$, defined by Theorem 1.1 with $\beta =[\eta /2]+2s+1$ and $m=2s+1$, we have for $x\in [-\pi ,\pi ]$,

$$\begin{aligned} |L_n^{(\nu )}(x)|\le ch^{-(\nu +1)}M^{2\beta -1}(x),\quad 0\le \nu \le 2s. \end{aligned}$$

(4.1)

Here and in the rest of the proof $c=c(s,\eta ,\epsilon )$.

We will show that the desired polynomial $d_l$ may be taken in the form

$$ d_l=d_{l-1}+\hat{d}_l-\frac{\hat{B}_l}{\breve{B}_l}\breve{d}_l, $$

where

$$\begin{aligned} \hat{d}_l(x)&:=\underbrace{\frac{f(z_l)-d_{l-1}(z_l)}{\prod _{q=1}^{l-1}\sin \bigl ((z_l-z_q)/2\bigr )}}_{=:F_l}\underbrace{\prod _{q=1}^{l-1} \sin \bigl ((x-z_q)/2\bigr )\frac{\sin \bigl ((x-\tilde{z}_l)/2\bigr )}{\sin \bigl ((z_l-\tilde{z}_l)/2\bigr )}\prod _{q=l+1}^{2s}\frac{\sin \bigl ((x-z_q)/2\bigr )}{\sin \bigl ((z_l-z_q)/2\bigr )}}_{=:\hat{I}(x)}\\ \qquad&\times \frac{L_n(x)}{L_n(z_l)} =F_l\hat{I}(x)\frac{L_n(x)}{L_n(z_l)}, \end{aligned}$$

$$ \breve{d}_l(x):=h^{1-2\,l}L_n(x)\underbrace{\prod _{q=1}^l\sin ^2\bigl ((x-z_q)/2\bigr ) \prod _{q=l+1}^{2\,s}\frac{\sin ^2\bigl ((x-z_q)/2\bigr )}{\sin ^2\bigl ((z_l-z_q)/2\bigr )}}_{=:\breve{I}(x)}=h^{1-2\,l}L_n(x)\breve{I}(x), $$

$$ \hat{B}_l:=\int _{-\pi }^\pi \hat{d}_l(x)\,dx\quad \text {and}\quad \breve{B}_l:=\int _{-\pi }^\pi \breve{d}_l(x)\,dx. $$

If $l=1$, we mean $\prod _{q=1}^0=1$, and recall that $d_{l-1}=d_0\equiv 0$. Similarly, $\prod _{q=2s+1}^{2s}=1$.

Evidently, $d_l$ is a trigonometric polynomial of degree $\le \beta (n-1)+2s$, and (1.14) and (1.15) hold.

We first estimate the polynomials $\hat{d}_l$ and $\breve{d}_l$, and their derivatives.

By the induction assumption, (1.12) and (1.16) imply,

$$ |f^{(l-1)}(x)- d_{l-1}^{(l-1)}(x)|\le ch^{1-l},\quad x\in \dot{O}, $$

while (1.14) yields,

$$ f(z_q)- d_{l-1}(z_q)=0,\quad q<l. $$

Thus, we have

$$\begin{aligned} |F_l|=\left| \frac{f(z_l)-d_{l-1}(z_l)}{\prod _{q=1}^{l-1}(z_l-z_q)}\right|&=\bigl |[z_1,\dots ,z_l;f-d_{l-1}]|\\&=\left| \frac{f^{(l-1)}(\theta )-d_{k-1}^{(l-1)}(\theta )}{(l-1)!}\right| \le ch^{1-l},\nonumber \end{aligned}$$

(4.2)

where the middle term is the divided difference of $f-d_{l-1}$, and in the last inequality we used the fact that $\theta \in \dot{O}$.

In order to estimate $\hat{I}(x)$ and $\breve{I}(x)$, and their derivatives, we observe that

$$ |z|<(s+1)h,\quad z\in \dot{O}, $$

and

$$ |z-z_l|\ge 2h\epsilon ,\quad z\in I{\setminus }\ddot{O}. $$

Also

$$ |x|M(x)\le (s+1)h,\quad x\in I. $$

Hence, for $x\in I$ and $z\in \dot{O}$, we have

$$ \frac{|\sin \bigl ((x-z)/2\bigr )|}{h}M(x)\le \frac{|x|}{2\,h}M(x)+\frac{|z|}{2\,h}M(x)\le \frac{|x|}{2\,h}M(x)+\frac{s+1}{2}\le s+1\le c, $$

and for $x\in I$ and $z\in I\setminus \ddot{O}$, we have

$$\begin{aligned} \frac{2}{\pi }\left| \frac{\sin \bigl ((x-z)/2\bigr )}{\sin \bigl ((z_l-z)/2\bigr )}\right| M(x)&\le \frac{|x-z|}{|z_l-z|}M(x)\le 1+\frac{|x|+|z_l|}{|z_l-z|}M(x)\\&\le 1+\frac{|x|+|z_l|}{2h\epsilon }M(x)\le c. \end{aligned}$$

Therefore, we write $\hat{I}(x)=h^{l-1}\prod _{q=1}^{2s}\alpha _q(x)$, $x\in I$, where for each $1\le q\le 2s$,

$$ |\alpha _q(x)|\le \frac{c}{M(x)}\quad \text {and}\quad |\alpha _q^{(\nu )}(x)|\le \frac{c}{h},\quad \nu \in \mathbb N,\quad x\in I,. $$

This, in turn, yields for each $0\le \nu \le 2s$,

$$ |\hat{I}^{(\nu )}(x)|\le ch^{l-1}\frac{h^{-\nu }}{M^{2\,s}(x)},\quad x\in I. $$

Combining with (4.2), (1.7) and (4.1), we obtain for each $0\le \nu \le 2s$,

$$\begin{aligned} |\hat{d}_l^{(\nu )}(x)|&\le \frac{2^\nu |F_l|}{|L_n(z_l)|}\sum _{\mu =0}^{\nu }|L_n^{(\mu )}(x)\hat{I}^{(\nu -\mu )}(x)|\\&\le c\frac{ h^{1-l}}{1/h}h^{l-1}\sum _{\mu =0}^{\nu }\frac{M^{2\beta -1}(x)}{h^{\mu +1}}\frac{h^{\mu -\nu }}{M^{2s}(x)}\nonumber \\&=c(\nu +1)h^{-\nu }M^{2\beta -2s-1}(x)\le c h^{-\nu }M^{\eta }(x),\quad x\in I.\nonumber \end{aligned}$$

(4.3)

Similarly,

$$ |\breve{I}^{(\nu )}(x)|\le ch^{2\,l-\nu }\frac{1}{M^{4\,s}(x)},\quad x\in I, $$

whence,

$$\begin{aligned} |\breve{d}_l^{(\nu )}(x)|\le c(\nu +1)h^{-\nu }M^{2\beta -4s-1}(x) \le ch^{-\nu }M^{\eta }(x),\quad x\in I. \end{aligned}$$

(4.4)

It follows by (1.13) with $l-1$, that

$$ \int _{-\pi }^\pi d_l(x)\,dx=0. $$

By virtue of (4.3) and (4.4), we obtain

$$ \int _{-\pi }^x |\hat{d}_l(t)|\,dt\le ch\quad \text {and}\quad \int _{-\pi }^x\breve{d}_l(t)\,dt\le ch,\quad x\in I, $$

and, in particular,

$$ |\hat{B}_l|\le ch\quad \text {and}\quad \breve{B}_l\le ch. $$

So, in order to complete the proof of (1.13) and (1.16), we will prove that

$$\begin{aligned} \breve{B}_l\ge ch. \end{aligned}$$

(4.5)

To this end, we note that if $x\in [-sh,sh]\subset \dot{O}$, then

$$ L_n(x)>\frac{c}{h}, $$

and

$$ \left| \frac{\sin \bigl ((x-z)/2\bigr )}{\sin \bigl ((z_l-z)/2\bigr )}\right| >c,\quad z\in I{\setminus }\ddot{O}. $$

Hence,

$$ \breve{B}_l\ge \int _{-sh}^{sh}\breve{d}_l(x)\,dx \ge \frac{c}{h^{2\,l}}\int _{-sh}^{sh}\prod _{q=1}^l(x-z_q)^2\,dx. $$

Now, the algebraic polynomial,

$$ Q(t):=\int _{-sh}^t\prod _{q=1}^l(x-z_q)^2\,dx,\quad -sh\le t\le sh, $$

of degree $2l+1$, satisfies, by Markov’s inequality,

$$ \Vert Q\Vert _{[-sh,sh]}\ge (sh)^{2\,l+1}c\Vert Q^{(2\,l+1)}\Vert _{[-sh,sh]}=c(2\,l)!(sh)^{2\,l+1}=ch^{2\,l+1}, $$

and (4.5) is proved. Thus, the proofs of (1.13) and (1.16) are complete. $\square$

5 An Auxiliary Lemma

For $j\in \mathbb Z$, let

$$ x_j:=\frac{j\pi }{n},\quad I_j:=[x_j,x_{j+1}],\quad \text {and}\quad |I_j|=\frac{\pi }{n}. $$

Denote by $\mathbb P_k$, the space of algebraic polynomials of degree $<k$, and by $\widetilde{\Sigma }_{k,n}$, the space of $2\pi$-periodic continuous piecewise algebraic polynomials S, of degree $<k$, with knots $x_j$, that is,

$$ S\left| _{I_j}\right. =p_j,\quad p_j\in \mathbb P_k,\quad j\in \mathbb Z. $$

Given $S\in \widetilde{\Sigma }_{k,n}$, $n_1, \beta \in \mathbb N$ and $J_{n_1}:=J_{n_1,\beta }$, let $\nu \in \mathbb N_0$ and denote

$$\begin{aligned} B_{\nu ,n_1}(x):=\frac{d^\nu }{dx^\nu }\int _{-\pi }^\pi S(x+\sigma t)J_{n_1}(t)\,dt-\int _{-\pi }^\pi S^{(\nu )}(x+\sigma t)J_{n_1}(t)\,dt. \end{aligned}$$

(5.1)

Lemma 5.1

Let $S\in \widetilde{\Sigma }_{k,n}$and let $\sigma \in \mathbb N$. For each $\nu \in \mathbb N$ we have

$$\begin{aligned} B_{\nu ,n_1}(x)=\frac{1}{\sigma }\sum _{l=1}^\nu \sum _{j=-n\sigma +1}^{n\sigma }\left( p_j^{(l-1)}(x_j)- p_{j-1}^{(l-1)}(x_j)\right) \frac{d^{\nu -l}}{dx^{\nu -l}}J_{n_1}\left( \frac{x-x_j}{\sigma }\right) .\nonumber \\ \end{aligned}$$

(5.2)

Proof

We first prove that for each $\nu \in \mathbb N$,

$$\begin{aligned} B_{\nu ,n_1}(x)= & {} \frac{1}{\sigma }\sum _{j=-n\sigma +1}^{n\sigma }\left( p^{(\nu -1)}_j(x_j)-p^{(\nu -1)}_{j-1}(x_j)\right) J_{n_1}\left( \frac{x-x_j}{\sigma }\right) +\frac{d}{dx}B_{\nu -1,n_1}(x).\nonumber \\ \end{aligned}$$

(5.3)

To this end, we observe that since S is differentiable of any degree $l\in \mathbb N$, except at a final number of points in any compact interval, the following integrals exist, for all $l\in \mathbb N_0$, and are equal.

$$\begin{aligned} \int _{-\pi }^\pi S^{(l)}(x+\sigma t)J_{n_1}(t)\,dt&=\frac{1}{\sigma }\int _{x-\sigma \pi }^{x+\sigma \pi }S^{(l)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \,du\\&=\frac{1}{\sigma }\int _{-\sigma \pi }^{\sigma \pi } S^{(l)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \,du\nonumber \\&=\frac{1}{\sigma }\sum _{j=-n\sigma }^{n\sigma -1}\int _{x_j}^{x_{j+1}}p_j^{(l)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \,du,\nonumber \end{aligned}$$

(5.4)

where for the second equation we used the fact that the integrand is $2\pi \sigma$-periodic.

Now,

$$\begin{aligned}&\int _{x_j}^{x_{j+1}}p_j^{(l)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \,du -\frac{d}{dx}\int _{x_j}^{x_{j+1}}p_j^{(l-1)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \,du\\&\qquad =\int _{x_j}^{x_{j+1}}\frac{\partial }{\partial u}\left( p_j^{(l-1)}(u)J_{n_1}\left( \frac{u-x}{\sigma }\right) \right) \,du\\&\qquad =p_j^{(l-1)}(x_{j+1})J_{n_1}\left( \frac{x-x_{j+1}}{\sigma }\right) - p_j^{(l-1)}(x_{j})J_{n_1}\left( \frac{x-x_{j}}{\sigma }\right) . \end{aligned}$$

Hence,

$$\begin{aligned}&B_{\nu ,n_1}(x)-\frac{d}{dx}B_{\nu -1,n_1}(x)\\&=-\int _{-\pi }^\pi S^{(\nu )}(x+\sigma t)J_{n_1}(t)\,dt+\frac{d}{dx}\int _{-\pi }^\pi S^{(\nu -1)}(x+\sigma t)J_{n_1}(t)\,dt\\&=\frac{1}{\sigma }\sum _{j=-n\sigma }^{n\sigma -1}\left( p_j^{(\nu -1)}(x_j)J_{n_1}\left( \frac{x-x_j}{\sigma }\right) - p_j^{(\nu -1)}(x_{j+1})J_{n_1}\left( \frac{x-x_{j+1}}{\sigma }\right) \right) \\&=\frac{1}{\sigma }\sum _{j=-n\sigma +1}^{n\sigma }\left( p^{(\nu -1)}_j(x_j)-p^{(\nu -1)}_{j-1}(x_j)\right) J_{n_1}\left( \frac{x-x_j}{\sigma }\right) , \end{aligned}$$

where for the last equation we used the fact that $p^{(l)}_{-n\sigma }(x_{-n\sigma })=p^{(l)}_{n\sigma }(x_{n\sigma })$.

Thus (5.3) is proved.

Since $B_{0,n_1}(x)\equiv 0$, the lemma follows by induction. $\square$

Remark 5.2

Since S is a continuous function, the $(l=1)$-term of (5.2) vanishes, so that the sum begins with $l=2$.

6 Approximating a Piecewise Polynomial

For $f\in C[a,b]$, let

$$ \Delta _h(f,x)=\Delta ^1_h(f,x):={\left\{ \begin{array}{ll}f(x+h)-f(x),&{}\quad x,x+h\in [a,b]\\ 0,&{}\quad \text {otherwise}, \end{array}\right. } $$

and, for $k>1$, let

$$ \Delta ^k_h(f,x):=\Delta _h\left( \Delta ^{k-1}_h(f,\cdot ),x\right) . $$

Denote by

$$ \omega _k(f,t;[a,b]):=\sup _{\begin{array}{c} 0\le h\le t\\ x\in [a,b] \end{array}}\left| \Delta ^k_h(f,x)\right| ,\quad k\ge 1, $$

the kth modulus of smoothness of f. (See [1, Chapter 2, Section 7] for properties of moduli of smoothness.)

Similarly, for $f\in \widetilde{C}$, the space of $2\pi$-periodic functions on $\mathbb R$, let $\Delta _h(f,x)=\Delta ^1_h(f,x):=f(x+h)-f(x)$, and for $k>1$, let $\Delta ^k_h(f,x):=\Delta _h\left( \Delta ^{k-1}_h(f,\cdot ),x\right)$.

Finally, denote by

$$ \omega _k(f,t):=\sup _{\begin{array}{c} 0\le h\le t\\ x\in \mathbb R \end{array}}\left| \Delta ^k_h(f,x)\right| ,\quad k\ge 1, $$

the kth modulus of smoothness of f. Note that for such f, $\omega _k(f,t)=\omega _k(f,t;[-2\pi ,2\pi ])$ for $0<t<2\pi /k$.

We call a closed interval E a proper interval, if $E=[x_{j_*},x_{j^*}]$ for some indices $j_*$ and $j^*$, and $x_{j^*}-x_{j_*}<2\pi$.

Let $Y_s:=\{y_i\}_{i\in \mathbb Z}$, $s\ge 1$, be a set of points, such that $y_i<y_{i+1}$ and $y_{i+2s}=y_i+2\pi ,\quad i\in \mathbb Z$.

For each $i\in \mathbb Z$, let $j_i$ be the index such that $y_i\in [x_{j_i},x_{j_i+1})$. We denote by O the interior of the union

$$ \cup _{i\in \mathbb Z}[x_{j_i-1},x_{j_i+2}]. $$

We will write $S\in \widetilde{\Sigma }_{k,n}(Y_s)$, if $S\in \widetilde{\Sigma }_{k,n}$ and $p_{j\pm 1}\equiv p_j$ for $x_j\in O$.

For $x\in E$ and $\eta \in \mathbb N$, denote

$$\begin{aligned} A(x,E):=\omega _k(S,1/n;E)+\omega _k(S,1/n)\left( \frac{1}{n_1\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)}\right) ^\eta , \end{aligned}$$

(6.1)

and, finally, let

$$\begin{aligned} \pi (t):=\prod _{i=1}^{2s}\frac{|\sin \frac{1}{2}(t-y_i)|}{|\sin \frac{1}{2}(t-y_i)|+1/n}. \end{aligned}$$

(6.2)

We devote this section to proving,

Theorem 6.1

Let $n_1\ge n$ and $S\in \widetilde{\Sigma }_{k,n}(Y_s)$. Then there is a trigonometric polynomial $\mathcal T$ of degree $<cn_1$, such that

$$\begin{aligned} \Vert S-\mathcal T\Vert \le c\omega _k(S,1/n), \end{aligned}$$

(6.3)

and if E is a proper interval, then

$$\begin{aligned} |S'(x)-\mathcal T'(x)|\le cn\pi (x)A(x,E),\quad x\in E. \end{aligned}$$

(6.4)

Here and in the sequel c and C denote constants which depend only on some or all the parameters k, s and $\eta$.

Let

$$\begin{aligned} O_\nu =(x_{\nu ^-}, x_{\nu ^+}),\quad \nu \in \mathbb Z, \end{aligned}$$

(6.5)

be the connected components of the set O, enumerated from right to left, and set

$$\begin{aligned} \tilde{O}_\nu =(x_{\nu ^-}+\pi /(2n),x_{\nu ^+}-\pi /(2n)). \end{aligned}$$

(6.6)

We need a few lemmas.

Lemma 6.2

Let $n_1\ge n$, $1\le \nu \le 2s+1$, $1\le \sigma \le k$, $S\in \tilde{\Sigma }_{k,n}(Y_s)$ $\eta \ge 2s$ and $J_{n_1}:=J_{n_1,\eta }$. If E is a proper interval and $O_\mu \subset E$, then for all $x\in \tilde{O}_\mu$,

$$\begin{aligned} |B_{\nu ,n_1}(x)|\le Cn^\nu A(x,E). \end{aligned}$$

(6.7)

Proof

Clearly, for any $j^0\in \mathbb Z$, we may rewrite (5.2) as

$$ B_{\nu ,n_1}(x)=\frac{1}{\sigma }\sum _{l=2}^\nu \sum _{\begin{array}{c} j=-n\sigma +j^0+1\\ x_j\notin O \end{array}}^{n\sigma + j^0}\left( p^{(l-1)}_j(x_j)-p^{(l-1)}_{j-1}(x_j)\right) \frac{d^{\nu -l}}{dx^{\nu -l}} J_{n_1}\left( \frac{x-x_j}{\sigma }\right) , $$

where we use the fact that S is continuous and $2\pi$-periodic, so that $p_j(x_j)=p_{j-1}(x_j)$, $j\in \mathbb Z$, and $p_j\equiv p_{j-1}$ for all j such that $x_j\in O$, and that $J_{n_1}(t/\sigma )$ is $2\pi \sigma$-periodic.

In particular we may take $j^0$ where $x\in I_{j^0}$. Thus, without loss of generality, we may assume that $x\in I_0$ and $j^0=0$. Then, for each j, $-n\sigma +1\le j\le n\sigma$, we have $\frac{|x-x_j|}{\sigma }\le \pi$.

By virtue of (1.5) we obtain for each $\hat{j}$, $-n\sigma +1\le \hat{j}\le \mu ^-$,

$$\begin{aligned}&\sum _{j=-n\sigma +1}^{\hat{j}}\left| J^{(\nu -l)}_{n_1}\left( \frac{x-x_j}{\sigma }\right) \right| \le C \sum _{j=-n\sigma +1}^{\hat{j}}\frac{n_1^{\nu -l+1}}{(n_1(x-x_j))^{2\eta -\nu +l}}\\&\quad \le C\sum _{j=-n\sigma +1}^{\hat{j}}\frac{n_1n^{\nu -l}}{(n_1(x-x_j))^{\eta +1}}\le C\frac{n^{\nu -l}}{n_1(n_1(x- x_{\hat{j}})^{\eta -1}} \sum _{j=-n\sigma +1}^{\hat{j}}\frac{1}{(x-x_j)^2}\\&\quad \le C\frac{n^{\nu -l+1}}{(n_1(x-x_{\hat{j}}))^\eta }. \end{aligned}$$

Similarly, (1.5) implies for each $\breve{j}$, $\mu ^+\le \breve{j}\le n\sigma$,

$$ \sum _{j=\breve{j}}^{n\sigma }\left| J^{(\nu -l)}_{n_1}\left( \frac{x-x_j}{\sigma }\right) \right| \le C\frac{n^{\nu -l+1}}{(n_1(x_{\breve{j}}-x))^\eta }. $$

Let $E=[x_{j_*},x_{j^*}]$. By Markov’s inequality $|p^{(l)}_j(x_j)-p^{(l)}_{j-1}(x_j)|\le cn^l\omega _k(S,1/n;E)$, if either $j_*<j\le \mu ^-$, or $\mu ^+\le j<j^*$. Also $|p^{(l)}_j(x_j)-p^{(l)}_{j-1}(x_j)|\le cn^l\omega _k(S,1/n)$ for all $j\in \mathbb Z$.

We have to separate the proof for $\sigma =1$ and $\sigma \ge 2$. We begin with the latter, so that

$$ -n\sigma +1\le j_*\le \mu ^-<\mu ^+\le j^*\le n\sigma . $$

Hence,

$$\begin{aligned}&\sum _{j=-n\sigma +1}^{j_*}\left| \left( p^{(l-1)}_{j-1}(x_j)-p^{(l-1)}_j(x_j)\right) J^{(\nu -l)}_{n_1}\left( \frac{x-x_j}{\sigma }\right) \right| \\&\quad \le C\omega _k(S,1/n)\frac{n^{l-1}n^{\nu -l+1}}{(n_1(x-x_{ j_*}))^\eta } \le C\omega _k(S,1/n)n^\nu \left( \frac{1}{n_1\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)}\right) ^{\eta },\nonumber \end{aligned}$$

(6.8)

and

$$\begin{aligned}&\sum _{j=j_*+1}^{\mu ^-}\left| \left( p^{(l-1)}_{j-1}(x_j)-p^{(l-1)}_j(x_j)\right) J^{(\nu -l)}_{n_1}\left( \frac{x-x_j}{\sigma }\right) \right| \\&\quad \le C\omega _k(S,1/n;E)\frac{n^{l-1}n^{\nu -l+1}}{(n_1(x-x_{ j_{\mu ^-}}))^\eta } \le C\omega _k(S,1/n;E)n^\nu .\nonumber \end{aligned}$$

(6.9)

Similarly,

$$ \sum _{j=\mu ^+}^{n\sigma }\left| \left( p^{(l-1)}_j(x_j)-p^{(l-1)}_{j-1}(x_j)\right) J^{(\nu -l)}_{n_1}\left( \frac{x-x_j}{\sigma }\right) \right| \le Cn^\nu A(x,E). $$

If $\sigma =1$, then we may have $-n+1\le j_*<j^*\le n$, and the proof follows verbatim as above. Otherwise, we may have that $j_*\le -n$, so that (6.8) is irrelevant, while in the summation in (6.9) we replace $j=-j_*+1$ by $j=-n+1$, where we note that $\mu ^->-n$; or we may have $j^*>n$, so that we replace the upper end of the summation $j^*$ with n, where we note that $\mu ^+<n$. This completes the proof. $\square$

Lemma 6.3

Let $n_1\ge n$, $1\le \nu \le 2s+1$, $S\in \tilde{\Sigma }_{k,n}(Y_s)$ $\eta \ge 2s$ and $J_{n_1}:=J_{n_1,\eta }$, and let

$$ T(x):=\frac{1}{\pi }\int _{-\pi }^\pi (S(x)-(-1)^k\Delta _t^k(S,x))J_{n_1}(t)\,dt, $$

be the trigonometric polynomial of degree $<\eta n_1$. Then

$$\begin{aligned} \Vert S-T\Vert \le c\omega _k(S,1/n). \end{aligned}$$

(6.10)

If E is a proper interval, then

$$\begin{aligned} |S'(x)-T'(x)|\le cnA(x,E),\quad x\in E, \end{aligned}$$

(6.11)

where A(x, E) was defined in (6.1).

Moreover, if $O_\mu \subset E$, then for all $x\in \tilde{O}_\mu$ and $1\le \nu \le 2s+1$,

$$\begin{aligned} |S^{(\nu )}(x)-T^{(\nu )}(x)|\le cn^\nu A(x,E),\quad x\in \tilde{O}_\mu . \end{aligned}$$

(6.12)

Proof

The inequality $\Vert S-T\Vert \le c\omega _k(S,1/n_1)\le c\omega _k(S,1/n)$ is well-known.

Since S is a piecewise algebraic polynomial, it possesses all left- and right-hand derivatives at each $x_j$, $j\in \mathbb Z$. Thus, if it happens that $x+\nu t=x_j$ for some $j\in \mathbb Z$, then $\Delta _t^k(S^{(\nu )},x)$ may not be well defined. But, for each fixed x, this may happen only for finitely many values of t, and would not influence the integration below. However, when we wish to estimate $\Delta _t^k(S^{(\nu )},x)$ we must consider the collection of all (finitely many) possible values we may have by assigning the various appropriate left- or right-hand values that may occur.

Recall that in [2, Lemma 5.5], it was proved for a proper interval E, that if $x,x+kt\in E$ and $(x+\ell t)\notin \{x_j\}_{j=-\infty }^\infty$, $0\le \ell \le k$, then

$$\begin{aligned} |\Delta _t^k(S^{(\nu )},x)|\le cn^\nu (1+n|t|)^k\omega _k(S,1/n;E). \end{aligned}$$

(6.13)

Closely observing the proof of [2, Lemma 5.5], we see that (6.13) is valid also with our above relaxation. In addition, the restriction on the length of E, is irrelevant for the next inequality. Thus,we obtain,

$$\begin{aligned} |\Delta _t^k(S^{(\nu )},x)|\le cn^\nu (1+n|t|)^k\omega _k(S,1/n),\quad |t|\le \pi . \end{aligned}$$

(6.14)

If $x\in E$, and $d:=\text {dist}(x,\mathbb R\setminus E)\ge \pi /(2n)$, then

$$\begin{aligned}&\left| \int _{\pi }^\pi \Delta _{t}^k(S^{(\nu )},x)J_{n_1}(t)\,dt\right| \\&\quad \le \left( \int _{|t|\le \frac{\pi }{2kn}}+\int _{\frac{\pi }{2kn}\le |t|\le \frac{d}{k} }+\int _{\frac{d}{k}\le |t|\le \pi }\right) |\Delta _{t}^k(S^{(\nu )},x)|J_{n_1}(t)\,dt\nonumber \\&\quad \le cn^\nu \omega _k(S,1/n;E)\int _IJ_{n_1}(t)\, dt+c\frac{n^{\nu +k}}{n_1^{2\beta -1}}\omega _k(S,1/n;E)\int _{\frac{\pi }{2kn}}^\infty \frac{t^k}{t^{2\beta }}\,dt\nonumber \\&\quad +c\frac{n^{\nu +k}}{n_1^{2\beta -1}}\omega _k(S,1/n)\int _{\frac{d}{k}}^\infty \frac{t^k}{t^{2\beta }}\,dt\nonumber \\&\quad \le cn^\nu A(x,E).\nonumber \end{aligned}$$

(6.15)

In particular, this is the case when $x\in \tilde{O}_\mu$.

Similarly, if $0<d<\pi /(2n)$, then

$$ |S'(x)-T_n'(x)|=\frac{1}{\pi }\left| \int _{-\pi }^\pi \Delta _{t}^k(S',x)J_{n_1}(t)\,dt\right| \le cnA(x,E). $$

Thus, (6.11) is proved.

Finally, if $x\in \tilde{O}_\mu \subset E$, then by (6.15) and (6.7),

$$\begin{aligned}&\pi |S^{(\nu )}(x)-T^{(\nu )}(x)|=\left| \frac{d^\nu }{dx^\nu }\int _{\pi }^\pi \Delta _{t}(S,x)J_{n_1}(t)\,dt\right| \\&\quad \le \left| \int _{\pi }^\pi \Delta _{t}^k(S^{(\nu )},x)J_{n_1}(t)\,dt\right| \\&\quad +\left| \frac{d^\nu }{dx^\nu }\int _{\pi }^\pi \Delta _{t}(S,x)J_{n_1}(t)\,dt-\int _{\pi }^\pi \Delta _{t}^k(S^{(\nu )},x)J_{n_1}(t)\, dt\right| \\&\quad \le cn^\nu A(x,E)+\sum _{\sigma =1}^k\left( {\begin{array}{c}k\\ \sigma \end{array}}\right) \left| \frac{d^\nu }{dx^\nu }\int _{\pi }^\pi S(x+\sigma t)J_{n_1}(t)\,dt-\int _{\pi }^\pi S^{(\nu )}(x+\sigma t)J_{n_1}(t)\,dt\right| \\&\quad \le cn^\nu A(x,E), \end{aligned}$$

and (6.12) follows. $\square$

If a proper interval E is such that its endpoints are not in O, we will call it a $Y_s$-proper interval.

For each $\mu \in \mathbb Z$, let $x_{\mu ^\circ }:=\frac{1}{2}(x_{\mu ^-}+x_{\mu ^-})$ be the midpoint of $O_\mu =(x_{\mu ^-},x_{\mu ^-})$, and for each $Y_s$-proper interval, such that $O_\mu \subset E$, let

$$ A_\mu (E):=A(x_{\mu ^\circ },E). $$

Since $\mathop {\text {dist}}\nolimits (x_{\mu ^\circ },\mathbb R\setminus E)\le C\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)$, for all $x\in \tilde{O}_\mu$, and $\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)\le C\mathop {\text {dist}}\nolimits (x_{\mu ^\circ },\mathbb R\setminus E)$, for all $x\in O_\mu$, it follows that

$$\begin{aligned} A(x,E)\le c A_\mu (E),\quad x\in \tilde{O}_\mu . \end{aligned}$$

(6.16)

and

$$\begin{aligned} A_\mu (E)\le cA(x,E),\quad x\in O_\mu . \end{aligned}$$

(6.17)

Define

$$\begin{aligned} A_\mu :=\min _{E:O_\mu \subset E} A_\mu (E). \end{aligned}$$

(6.18)

Finally, denote $J_\mu :=[x_{\mu ^\circ }-\pi ,x_{\mu ^\circ }+\pi ]$, and let $M_\mu$ be the $2\pi$-periodic function, defined on $J_\mu$ by

$$ M_\mu (x)={\left\{ \begin{array}{ll} 1,\quad &{}x\in \tilde{O}_\mu ,\\ \frac{1}{n_1|x-x_{\mu ^\circ }|}\quad &{}x\in J_\mu {\setminus }\tilde{O}_\mu . \end{array}\right. } $$

Lemma 6.4

Let $\mu \in \mathbb Z$ and $n_1\ge n$. Then, for every $Y_s$-proper interval E and each $x\in E$, we have

$$\begin{aligned} A_\mu M_\mu ^\eta (x)\le CA(x,E). \end{aligned}$$

(6.19)

Proof

It is sufficient to prove (6.19) for $x\in J_\mu$, and we let a $Y_s$-proper interval E be such that $x\in E$.

First, assume that $O_\mu \nsubseteq E$. Thus, there is an endpoint of E, say $\gamma$, lying between x and $x_{\mu ^\circ }$. Then $\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)\le |x-\gamma |\le |x-x_{\mu ^\circ }|$.

Hence,

$$\begin{aligned}&\frac{1}{2}A_{\mu }M_\mu ^\eta (x)\le \omega _k(S,1/n)M_\mu ^{\eta }(x)=\omega _k(S,1/n)\left( \frac{1}{n_1|x-x_{\mu ^\circ }|}\right) ^{\eta }\\&\quad \le \omega _k(S,1/n)\left( \frac{1}{n_1\mathop {\text {dist}}\nolimits (x,\mathbb R\setminus E)}\right) ^{\eta }\le A(x,E). \end{aligned}$$

Otherwise, $O_\mu \subset E$.

If $x\in O_\mu$, then (6.19) is trivial, since $\Vert M_\mu \Vert =1$, and by (6.17)

$$ A_{\mu }M_\mu ^\eta (x)\le A_{\mu }\le A_\mu (E)\le CA(x,E). $$

Similarly, if $x\in E\setminus O_\mu$ and $|x-\gamma |\le |x_{\mu ^\circ }-\gamma |$, where now $\gamma$ is the endpoint of E, closest to $x_{\mu ^\circ }$, then

$$ A_{\mu }M_\mu ^\eta (x)\le A_{\mu }\le A_\mu (E)\le A(x,E), $$

that yields (6.19).

Finally, if $x\in E\setminus O_\mu$ and $|x-\gamma |>|x_{\mu ^\circ }-\gamma |$, then assume, without loss of generality, that $x_{\mu ^\circ }<\gamma$. Then, it follows that $x+3\pi /(2n)\le x_{\mu ^\circ }\le \gamma -3\pi /(2n)$. Since $g(u):=\frac{\gamma -u}{x_{\mu ^\circ }-u}$ is an increasing function for $u<x_{\mu ^\circ }$, we have,

$$\begin{aligned} \frac{\gamma -x}{x_{\mu ^\circ }-x}&\le \frac{\gamma -(x_{\mu ^\circ }-3\pi /(2n)}{x_{\mu ^\circ }-(x_{\mu ^\circ }-3\pi /(2n)} =\frac{2n}{3\pi }(\gamma -(x_{\mu ^\circ }-3\pi /(2n))\\&\le \frac{2n}{3\pi }2(\gamma -x_{\mu ^\circ })<n(\gamma -x_{\mu ^\circ }). \end{aligned}$$

Hence,

$$ \frac{1}{n_1^2|x_{\mu ^\circ }-\gamma ||x-x_{\mu ^\circ }|}<\frac{1}{n_1|x-\gamma |}. $$

Therefore, $A_{\mu }M_\mu ^\eta (x)\le A(x,E)$. $\square$

We are ready to prove Theorem 6.1. It is easy to show that if an endpoint of a proper interval E belongs to O, say, to its connected component $O_\mu$, then $\omega _k(S,1/n;\overline{E\cup O_\mu })\le c\omega _k(S,1/n;E)$, whence $A(x,\overline{E\cup O_\mu })\le CA(x,E)$, $x\in E$. Therefore, we prove Theorem 6.1 for $Y_s$-proper intervals E and, without loss of generality, we assume that $\eta \ge 2s$.

Proof of Theorem 6.1

We apply Theorem 1.3 for each fixed $\mu \in \mathbb Z$, with $\epsilon =1/12$, and $n_1$ instead of n and h such that,

$$ \dot{O}_\mu :=[x_{\mu ^\circ }-(s+5/(12))h,x_{\mu ^\circ }+(s+5/(12))h]:=[x_{\mu ^-}+\pi /n,x_{\mu ^+}-\pi /n]. $$

Thus,

$$ (2\,s+5/6)h=|\dot{O}_\mu |=|O_\mu |-2\pi /n. $$

Since $\frac{3\pi }{n}\le |O_\mu |\le \frac{6s\pi }{n}$, we conclude that

$$ \frac{\pi }{2(s+1)n}<h<\frac{6\,s\pi }{2sn}=3\pi /n, $$

and

$$ (2\,s+7/6)h=|O_\mu |-2\pi /n+h/3<|O_\mu |-\pi /n. $$

Hence,

$$ \ddot{O}_\mu :=[x_{\mu ^\circ }-(s+7/(12))h,x_{\mu ^\circ }+(s+7/(12))h]\subset [x_{\mu ^-}+\pi /(2n),x_{\mu ^+}-\pi /(2n)]=\tilde{O}_\mu . $$

Note that all points $y_i\in J_\mu$ lie either in $\dot{O}_\mu$ or outside $\ddot{O}_\mu$. Let l be the number of points $y_i\in \dot{O}_\mu$.

Let T be the polynomial, guaranteed by Lemma 6.3, and denote

$$ R_\mu :=\max _{1\le i\le l} h^{i-1}\Vert S^{(i)}-T^{(i)}\Vert _{\dot{O}_\mu }\quad \text {and}\quad f(x):=\frac{S'(x)-T'(x)}{R_\mu }, $$

so that, for all $0\le \nu \le l-1$, we have

$$ |f^{(\nu )}(x)|\le \frac{|S^{(\nu +1)}(x)-T^{(\nu +1)}(x)|}{R_\mu }\le \frac{\Vert S^{(\nu +1)}-T^{(\nu +1)}|_{\dot{O}_\mu }}{h^{\nu }\Vert S^{(\nu +1)}-T^{(\nu +1)}\Vert _{\dot{O}_\mu }}=h^{-\nu },\quad x\in \dot{O}_\mu . $$

Thus, f satisfies (1.12). Hence, (1.14) through (1.17), imply the existence of a polynomial $d_l$, of degree $<cn_1$, such that

$$ d_l(y_i)=f(y_i),\quad y_i\in \dot{O}_\mu ,\quad d_l(y_i)=0,\quad y_i\in J_\mu {\setminus }\ddot{O}_\mu ,\quad \left| \int _{-\pi }^x d_l(t)\,dt\right| \le ch,\quad x\in \mathbb R, $$

and for all $0\le \nu \le 2s$,

$$ |d_l^{(\nu )}(x)|\le ch^{-\nu }M_\mu ^\eta (x),\quad x\in \mathbb R. $$

By (6.12), $R_\mu \le ch^{-1}A_\mu$. Therefore, the polynomial

$$ \tau _\mu :=R_\mu \int _{-\pi }^x d_l(t)\,dt $$

satisfies

$$\begin{aligned} \Vert \tau _\mu \Vert \le c\omega _k(S,1/n), \end{aligned}$$

(6.20)

$$ \tau _\mu '(y_i)=S'(y_i)-T'(y_i),\quad y_i\in O_\mu , $$

$$ \tau _\mu '(y_i)=0,\quad y_i\in J_\mu {\setminus } O_\mu , $$

and for all $1\le \nu \le 2s+1$,

$$ |\tau _\mu ^{(\nu )}(x)|\le cn^\nu A_\mu M_\mu ^\eta (x),\quad x\in \mathbb R, $$

where in the last inequality we used the fact that $\dot{O}_\mu \subset \tilde{O}_\mu$.

Finally, Lemma 6.4 implies that for every $Y_s$-proper interval E, we have

$$\begin{aligned} |\tau _\mu ^{(\nu )}(x)|\le Cn^\nu A(x,E)\quad x\in E. \end{aligned}$$

(6.21)

We will prove that the desired polynomial $\mathcal T$ may be taken in the form

$$ \mathcal T:= T+\sum _{\begin{array}{c} \mu \,\,\text {s.t.}\\ x_{\mu ^\circ }\in [-\pi ,\pi ) \end{array}}\tau _\mu . $$

Indeed, (6.3) readily follows by (6.10) and (6.20).

We observe that,

$$ c\le \pi (t)\le 1,\quad t\notin \tilde{O},\quad \text { where}\quad \tilde{O}:=\cup _{\mu \in \mathbb Z}\tilde{O}_\mu , $$

where $\pi (t)$ was defined in (6.2), and combined with (6.11) and (6.21) with $\nu =1$, we obtain (6.4) for $x\in E\setminus \tilde{O}$.

On the other hand, if $x\in \tilde{O}_{\mu ^*}\subset E$, for some $\mu ^*\in \mathbb Z$, then let $y_{i_\ell }\in O_{\mu ^*}$, $1\le \ell \le l$, and note that $y_{i_\ell }\in \tilde{O}_{\mu ^*}$. Evidently, $S'(y_{i_\ell })=\mathcal T'(y_{i_\ell })$, $1\le \ell \le l$.

Applying (6.12) and (6.21) for all $\mu$, all with $\nu =l+1$, we obtain

$$ |S^{(l+1)}(x)-\mathcal T^{(l+1)}(x)|\le cn^{l+1}A(x,E),\quad x\in \tilde{O}_{\mu _*}. $$

Hence, for $x\in \tilde{O}_{\mu ^*}$,

$$ \frac{|S'(x)-\mathcal T'(x)|}{\prod _{\ell =1}^l|x-y_{i_\ell }|}=[x,y_{i_1},\dots ,y_{i_l};S'- \mathcal T']=\frac{|f^{(l+1)}(\theta )|}{l!}\le Cn^{l+1}A(\theta ,E)\le Cn^{l+1}A_{\mu ^*}(E). $$

Thus, by (6.17), we conclude that

$$ |S'(x)-\mathcal T'(x)|\le cn^{l+1}A_{\mu _*}(E)\prod _{\ell =1}^l|x-y_{i_\ell }|\le cn\pi (x)A_{\mu _*}(E)\le cn\pi (x)A(x,E),\quad x\in \tilde{O}_{\mu _*}. $$

This completes the proof. $\square$