Before proving our main results, the following elementary inequalities will be useful:
Lemma 1
The following estimates hold for
\(a_{k}=\frac{1}{2\left( (n+1)-(n+2)\sin ^2 \theta _{k}/2\right) }=\frac{1}{n+(n+2)\cos \theta _{k}}\)
-
1.
\(|a_{k}|\le \frac{1}{n}, \; 1 \le k \le [n/2], or\;\; 0 \le \theta _{k} \le \pi /2\)
-
2.
\(|a_{k}| \le \left| \frac{1}{n-(n+2)n/k}\right| ,\;\; [n/2]+1\le k \le n-1, \;or\; \pi /2<\theta _k<\pi .\)
Also we shall require.
Lemma 2
The following inequality hold true for all
\(n \ge 1.\)
$$ \frac{n}{10}\cos ^2 \theta _{k}/2\le \left| (n+1)-(n+2) \sin ^2 \theta _{k}/2\right|. $$
The first result in this paper is
Theorem 1
For
\(f(x) \in C^1[-1,1],\)
the estimation
\(| H_{2n+1}(f;x)| =\mathcal{O}(n^2)\)
holds true as
\(n \rightarrow \infty .\)
Proof
By (1), (9), it follows that
$$ \left| H_{2n+1}(f;x)\right| \le \Vert f\Vert _{1} \left\{ \sum _{k=0}^n\left| v_{k}(x)\ell _{k}^2(x)\right| +\sum _{k=0}^{n}\left| (x-x_{k})\ell _{k}^2(x)\right| \right\} = \Vert f\Vert _{1}\{J_{1}+J_{2}\} $$
To estimate \(J_{1}\), by (10) and (11), it follows that
$$\begin{aligned} \sum _{k=0}^n\left| v_{k}(x) \ell ^2_{k}(x)\right|\le &\, \left| v_{0}(x) \ell _{0}^2(x)\right| +\sum _{k=1}^{[n/2]}\left| v_{k}(x)\ell _{k}^2(x)\right| \\&+\,\sum _{k=[n/2]+1}^{n-1}\left| v_{k}(x)\ell _{k}^2(x)\right| +\left| v_{n}(x) \ell _{n}^2(x)\right| \\= & \,J_{1}^{(1)}+J_{1}^{(2)}+J_{1}^{(3)}+J_{1}^{(4)}. \end{aligned}$$
We shall proceed to estimate each of these terms: For \(J_{1}^{(1)}\), by (14), we have
$$\begin{aligned} J_{1}^{(1)}=\,\left| v_{0}(x) \ell _{0}^2(x)\right|\le\, &\, \ell _{0}^2(x)+\left( \frac{2}{3}n(n+1)+1\right) (1-x^2)\frac{(1+x^2)\sin ^2\left( \frac{2n+1}{2}\right) \theta }{4(2n+1)^2\sin ^2 \theta /2}\\\le &\, 1+\frac{1}{12}(2n+1)^2\frac{\left( 4 \sin ^2 \theta /2 \cos ^2 \theta /2\right) \left( 2 \cos ^2 \theta /2\right) \sin ^2\left( \frac{2n+1}{2}\right) \theta }{(2n+1)^2\sin ^2 \theta /2}\le \frac{5}{3}. \end{aligned}$$
For \(J_{1}^{(4)}\), we have
$$\begin{aligned} J_{1}^{(4)}=\left| v_{n}(x) \ell _{n}^2(x)\right|\le\, &\, \ell _{n}^2(x)+\{2n(n+1)+1\}(1+x)\frac{(1-x^2)\sin ^2(2n+1/2)\theta }{4\sin ^2 \theta /2}\\\le &\, 1+\left( 4n^2+4n+2\right) =4n^2+4n+3. \end{aligned}$$
For \(J_{1}^{(2)}\), we have
$$\begin{aligned} J_{1}^{(2)}= \sum _{k=1}^{[n/2]}\left| v_{k}(x)\ell _{k}^2(x)\right|\le &\, \sum _{k=1}^{[n/2]}\left| \ell _{k}^2(x)\right| +\sqrt{2}\sum _{k=1}^{[n/2]}\left| \frac{a_{k}w(x)\left( 2+n(n+3)-\frac{1}{2\sin ^2 \theta _{k}/2}\right) }{2\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) }\right| \\\le &\, 2+2\sqrt{2}\sum _{k=1}^{[n/2]}\left| \frac{2(n+1)(n+2)\sin ^2 \theta _{k}/2-n}{4\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) ^2 \sin ^2\theta _{k}/2}\right| , \end{aligned}$$
where in the last inequality we used \(|w(x) \le 2\). Now using the first part of Lemma 1, and \(\sin ^2\theta _{k}/2=\sin ^2\theta _{k}/(4\cos ^2 \theta _{k}/2)\), we obtain
$$\begin{aligned} J_{1}^{(2)}&= 2+2\sqrt{2}\sum _{k=1}^{[n/2]}\left| \frac{4 \cos ^2 \theta _{k}/2\left( 2(n+1)(n+2)+n\right) }{n^2 \sin ^2 \theta _{k}}\right| \\&\le \, 2+2\sqrt{2}\sum _{k=1}^{[n/2]}\left| \frac{4\left( 2(n+1)(n+2)+n\right) }{n^2\sin ^2\theta _{k}}\right| . \end{aligned}$$
Using the Jordan inequality: \(\sin \theta \ge 2\theta /\pi , \; 0 \le \theta \le \pi /2\), we have
$$\begin{aligned} J_{1}^{(2)}\le & 2+2\sqrt{2}\sum _{k=1}^{[n/2]}\left| \frac{4(2(n+1)(n+2)+n)n^2}{4n^2k^2}\right| \\\le & 2+2\sqrt{2} \left( 2n^2+7n+4\right) \sum _{k=1}^{[n/2]}\frac{1}{k^2}\\\le & 2+2\sqrt{2} \left( 2n^2+7n+4\right) \sum _{k=1}^{\infty }\frac{1}{k^2}\le 2+2\sqrt{2} \left( 2n^2+7n+4\right) \frac{\pi ^2}{3}. \end{aligned}$$
For \(J_{1}^{(3)}\), we have
$$\begin{aligned} J_{1}^{(3)}= & \sum _{\left[ \frac{n}{2}\right] +1}^{n-1}\left| v_{k}(x)\ell _{k}^2(x)\right| \\\le & \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1} \ell _{k}^2(x)+\sqrt{2}\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{a_{k} w(x)\left( 2+n\left( (n+3)-\frac{1}{2\sin ^2 \theta _{k}/2}\right) \right) }{2\left( (n+1)-(n+2) \sin ^2 \theta _{k}/2\right) }\right| \\\le & 2+2 \sqrt{2}\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| \frac{2(n+1)(n+2)\sin ^2 \theta _{k}/2-n}{4\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) ^2 \sin ^2\theta _{k}/2}\right| . \end{aligned}$$
Now by the second part of Lemma 1, and \(\sin ^2\theta _{k}/2=\sin ^2 \theta _{k}/\left( 4 \cos ^2\theta _{k}/2\right) \) we obtain
$$\begin{aligned} J_{1}^{(3)}\le \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{4(2(n+1)(n+2)+n)\cos ^2\theta _{k}/2}{n^2\left( \frac{k-(n+2)}{k}\right) ^2 \sin ^2\theta _{k}}\right| . \end{aligned}$$
Since \(\sin \theta _{k}=\sin \theta _{k-n}\), we can apply Jordan inequality again, so that
$$\begin{aligned} J_{1}^{(3)}\le\, & 2+2\sqrt{2}\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{4\left( 2(n+1)(n+2)+n\right) n^2}{4n^2\left( \frac{k-(n+2)}{k}\right) ^2k^2}\right| \\\le \,& 2+2 \sqrt{2}\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{2(n+1)(n+2)+n}{(k-(n+2))^2}\right| . \end{aligned}$$
But,
$$\begin{aligned} \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\frac{1}{\left( k-(n+2)\right) ^2}=\sum _{k=\left[ \frac{n}{2}\right] +1}^{-3}\frac{1}{k^2} =\sum _{k=3}^{\left[ \frac{n}{2}\right] +1}\frac{1}{k^2}\le \frac{\pi ^2}{6}. \end{aligned}$$
Therefore,
$$ J_{1}^{(3)}\le 2+ \frac{\pi ^2\sqrt{2}}{3}\left( 2n^2+7n+4\right). $$
Combining the estimates of \(J_{1}^{(1)},J_{1}^{(2)},J_{1}^{(3)}\) and \(J_{1}^{(4)}\), there is a constant \(K_{1}\) such that
$$ J_{1}\le K_{1}\left( 11+22n+5n^2\right). $$
(17)
To estimate \(J_{2}\), by (13), (15) and Lemma 2, we have
$$\begin{aligned} \sum _{k=1}^{n-1}\left| (x-x_{k})\ell _{k}^2(x)\right|= & \sum _{k=1}^{n-1}\sqrt{2}\left[ \frac{w(x)}{2(n+1)-(n+2)\sin ^2\theta _{k}/2}\right] \\\le & \sum _{k=1}^{n-1}\left| \frac{10 \sqrt{2}}{n \cos ^2 \theta _{k}/2}\right| \le 40 \sqrt{2} \sum _{k=1}^{n-1}\frac{1}{n \sin ^2 \theta _{k}}. \end{aligned}$$
Because
$$\begin{aligned} \sum _{k=1}^{n-1}\frac{1}{n\sin ^2 \theta _{k}}=\sum _{k=1}^{\left[ \frac{n}{2}\right] }\frac{1}{\sin ^2 \theta _{k}} +\sum _{k=1}^{n-\left[ \frac{n}{2}\right] -1}\frac{1}{\sin ^2 \theta _{n-k}} \end{aligned}$$
and \(\sin \theta _{n-k}=\sin \theta _{k}\), using the Jordan inequality: \(\sin \theta \ge 2\theta /\pi \) for \(0 \le \theta \le \pi /2\), we obtain
$$\begin{aligned} \sum _{k=1}^{n-1}\frac{1}{n\sin ^2 \theta _{k}}\le \sum _{k=1}^{\left[ \frac{n}{2}\right] }\frac{n}{4k^2}+\sum _{k=1}^{n-\left[ \frac{n}{2}\right] -1}\frac{n}{4k^2} <\frac{n}{2}\sum _{k=1}^{\infty }\frac{1}{k^2}=\frac{\pi ^2n}{12}. \end{aligned}$$
Therefore,
$$ \sum _{k=1}^{n-1}\left| (x-x_{k})\ell _{k}^2(x)\right| \le \frac{\sqrt{2}\pi ^2}{3}n. $$
Again, by (13), we have \(|(x-1)\ell _{0}^2(x)|+|(x+1)\ell _{n}^2(x)|\le |x-1|+|x+1|=2, \; (-1<x<1)\). Therefore,
$$ J_{2} \le 50 n+2. $$
(18)
From this and (17), we obtain the desired result. \(\square \)
Theorem 2
For
\(f(x) \in C^1[-1,1]\)
, the estimation
\(|H^{\prime }_{2n+1}(f;x)| =\mathcal{O}(n^2)\)
holds true as
\(n\rightarrow \infty \).
Proof
By (9) and \(\frac{\hbox {d}}{\hbox {d}x}\sum _{k=0}^n (v_{k}(x) \ell _{k}^2(x))=0\), we have
$$\begin{aligned} H^{\prime }_{2n+1}(f;x)= & \sum _{k=0}^{n}\left( f(x_{k})-f(x)\right) \left( v_{k}(x)\ell _{k}^2(x)\right) ^{\prime }+\sum _{k=0}^{n}f^{\prime }(x_{k})\left( (x-x_{k})\ell _{k}^2(x)\right) ^{\prime }\\= & \sum _{k=0}^n\left( f(x_{k})-f(x)\right) \left( v^{\prime }_{k}(x)\ell ^2_{k}(x)+2v_{k}(x)\ell _{k}(x)\ell ^{\prime }_{k}(x)\right) \\&+\,\sum _{k=0}^{n}f^{\prime }(x_{k})\left( \ell ^2_{k}(x) +2(x-x_{k})\ell _{k}(x)\ell ^{\prime }_{k}(x)\right) . \end{aligned}$$
Since \(|f(x)-f(x_{k})|\le \max _{|x|\le 1}|f^{\prime }(x)||x-x_{k}|\), the above formula implies that
$$\begin{aligned} \left| H^{\prime }_{2n+1}(f;x)\right|\le & \max _{|x|\le 1} |f^{\prime }(x)|\left\{ \sum _{k=0}^n \left| (x-x_{k})v^{\prime }_{k}(x)\ell ^2_{k}(x)\right| +\sum _{k=0}^n \ell ^2_{k}(x)\right. \\&\left. +\,2\sum _{k=0}^n\left| (x-x_{k})v_{k}(x)\ell _{k}(x)\ell ^{\prime }_{k}(x)\right| +2\sum _{k=0}^n\left| (x-x_{k})\ell _{k}(x)\ell ^{\prime }_{k}(x)\right| \right\} \\= & \max _{|x|\le 1} |f^{\prime }(x)|\left\{ I_{1}+I_{2}+2I_{3}+2I_{4}\right\} . \end{aligned}$$
For \(I_{1}\), by (12), we know that
$$\begin{aligned} I_{1}&= \, \left| (x-1)\left( \frac{2}{3}n(n+1)+1\right) \ell _{0}^2(x)\right| +\left| (x+1)(2n(n+1)+1)\ell _{n}^2(x)\right| \\&+\,\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| (x-x_{k})\left( \frac{2+n\left( (n+3)-\frac{1}{2\sin ^2\theta _{k}/2}\right) }{2\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) }\right) \ell ^2_{k}(x)\right| \\&+\,\sum _{k={\left[ \frac{n}{2}\right] }+1}^{n-1}\left| (x-x_{k})\left( \frac{2+n\left( (n+3)-\frac{1}{2\sin ^2\theta _{k}/2}\right) }{2\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) }\right) \ell ^2_{k}(x)\right| \\&= \, I_{1}^{(1)}+I_{1}^{(2)}+I_{1}^{(3)}+I_{1}^{(4)}. \end{aligned}$$
Again by (13), we have
$$\begin{aligned} I_{1}^{(1)}= &\, \left| (1-x)\left( \frac{2}{3}n(n+1)+1\right) +\frac{(1+x)^2\sin ^2\left( \frac{2n+1}{2}\right) \theta }{4(2n+1)^2\sin ^2\theta /2}\right| \\\le & \left| \frac{1}{3}(2n^2+2n+1)\frac{2}{(2n+1)^2}\right| \le \frac{1}{3}. \end{aligned}$$
Similarly, we can show that \(I_{1}^{(2)}\le 4n^2+4n+2.\) For \(I_{1}^{(3)}\), with the aid of the first part of Lemma 1, we have
$$\begin{aligned} I_{1}^{(3)}= &\, \sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| (x-x_{k})\left( \frac{2(n+1)(n+2)\sin ^2\theta _{k}/2-n}{4\left( (n+1)- (n+2)\sin ^2\theta _{k}/2\right) \sin ^2\theta _{k}/2}\right) \ell ^2_{k}(x)\right| \\\le & \sqrt{2}\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| \frac{2(n+1)(n+2)+n}{4\left( (n+1)- (n+2)\sin ^2\theta _{k}/2\right) ^2\sin ^2\theta _{k}/2}\right| \\\le & \sqrt{2}\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| \frac{2(n+1)(n+2)+n}{n^2\sin ^2\theta _{k}/2}\right| \le \sqrt{2}\left( 2n^2+7n+4\right) \sum _{k=1}^{\infty }\frac{1}{k^2}\\\le & \sqrt{2}\left( 2n^2+7n+4\right) \frac{\pi ^2}{6}. \end{aligned}$$
For \(I_{1}^{(4)}\), with the aid of the second part of Lemma 1, we have
$$\begin{aligned} I_{1}^{(4)}\le & \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{2(n+1)(n+2)+n}{4\left( (n+1)-(n+2)\sin ^2\theta _{k}\right) ^2\sin ^2\theta _{k}/2}\right| \\\le & \sqrt{2}\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{4(2(n+1)(n+2)+n)\cos ^2\theta _{k}/2}{n^2\sin ^2\theta _{k}\left( \frac{k-(n+2)}{k}\right) ^2}\right| . \end{aligned}$$
Since \(\sin \theta _{k}=\sin \theta _{k-n}\), we can apply Jordan inequality to obtain
$$\begin{aligned} I_{1}^{(4)}\le & \sqrt{2}\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{2(n+1)(n+2)+n}{(k-(n+2))^2}\right| \\= &\, \sqrt{2}(2(n+1)(n+2)+n)\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\frac{1}{(k-(n+2))^2} =\sqrt{2}(2(n+1)(n+2)+n)\sum _{k=3}^{\left[ \frac{n}{2}\right] +1}\frac{1}{k^2}\\\le & \sqrt{2}(2(n+1)(n+2)+n) \frac{\pi ^2}{6}. \end{aligned}$$
Summarizing these estimates for \(I_{1}\), we get \(I_{1}\le \mathcal{O}(n^2), (n \rightarrow \infty )\). For \(I_{2}\), using the fact that \(I_{2} =\sum _{k=1}^{n-1} \ell _{k}^2(x) \le 2\) and (2.12) we get \(I_{2} \le 4\). For \(I_{3}\),
$$\begin{aligned} I_{3}&= \, \sum _{k=0}^{n} \left| (x-x_{k}) v_{k}(x) \ell _{k}(x) \ell ^{\prime }_{k}(x)\right| \\\le & \sum _{k=0}^n\left| v_{k}(x) \ell _{k}(x)\left\{ a_{k} w^{\prime }(x)-\ell _{k}(x)\right\} \right| \\\le & \left| v_{0}(x) \ell _{0}(x) a_{0} w^{\prime }(x)\right| +\left| v_{n}(x) \ell _{n}(x) a_{n} w^{\prime }(x)\right| +\left| v_{0}(x)\ell ^2_{0}(x)\right| +\left| v_{n}(x)\ell ^2_{n}(x)\right| \\&+ \,\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| v_{k}(x) \ell _{k}(x) a_{k} w^{\prime }(x)\right| +\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| v_{k}(x) \ell _{k}(x) a_{k} w^{\prime }(x) \right| +\sum _{k=1}^{\left[ \frac{n}{2}\right] }\left| \ell _{k}^2(x) v_{k}(x) \right| \\&+\,\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \ell _{k}^2(x) v_{k}(x) \right| \\&= \, I_{3}^{(1)}+I_{3}^{(2)}+I_{3}^{(3)}+I_{3}^{(4)}+I_{3}^{(5)}+I_{3}^{(6)}+I_{3}^{(7)}+I_{3}^{(8)}. \end{aligned}$$
Note that \(|a_{0} w^{\prime }(x)|\le 1\), so that
$$\begin{aligned} I_{3}^{(1)}\le & \left| \ell _{0}(x)\left( 1-\left( \frac{2}{3}n(n+1)+1\right) \right) (x-1)\right| \\\le & \left| \ell _{0}(x) \right| +\left| \left( \frac{2}{3}n(n+1)+1\right) \frac{\left( 1-x^2\right) \sin \left( \frac{2n+1}{2}\right) \theta }{2(2n+1)\sin \theta /2}\right| \le 1+\frac{1}{3}(n+1), \end{aligned}$$
and
$$\begin{aligned} I_{3}^{(2)}\le & \sqrt{2}(2n+1)+\left| (2n(n+1)+1)(x+1)\frac{(1-x^2)(1+x)\sin ^2\left( \frac{2n+1}{2}\right) \theta }{2(1+x)\sin \theta /2} \right| \\\le \,& 2n^2+7n+3. \end{aligned}$$
For \(I_{3}^{(3)}\), we have,
$$\begin{aligned} I_{3}^{(3)}\le \ell _{0}^2(x)+\left( \frac{2}{3}n(n+1)+1\right) (1-x)\frac{(1+x)^2\sin ^2\left( \frac{2n+1}{2}\right) \theta }{2(1+x)\sin \theta /2}\le \frac{5}{3}. \end{aligned}$$
Similar to the estimate in \(I_{3}^{(2)}\), we can show that \(I_{3}^{(4)}=|v_{n}(x) \ell _{n}^2(x)|\le 4n^2+4n+3.\) For \(I_{3}^{(5)}\): By using \(|w^{\prime }(x)|\le 2(2n+1)\) and part one of Lemma 1, we obtain
$$\begin{aligned} I_{3}^{(5)}\le \,& 4\sum _{k=1}^{\left[ \frac{n}{2}\right] }2(2n+1)\left( \frac{2+n(n+3)-\frac{n}{\sin ^2 \theta _{k}/2}}{8((n+1)-(n+2)\sin ^2 \theta _{k}/2)^3}+\ell _{k}(x)\right) \\\le\, & 2(2n+1)\left( n\left( \sum _{k=1}^{\left[ \frac{n}{2}\right] }\ell _{k}^2(x)\right) ^{1/2}+\sum _{k=1}^{\left[ \frac{n}{2}\right] }\frac{2(n+1)(n+2)+n}{n^3\sin ^2\theta _{k}/2}\right) \\\le \,& 2(2n+1)\left( \sqrt{2}n +\sum _{k=1}^{\left[ \frac{n}{2}\right] }\frac{n^2(2(n+1)(n+2)+n)}{n^3k^2}\right) \\\le \,& \left( 4\sqrt{2}+4\pi ^2/3\right) n^2 +\left( 2\sqrt{2}+4\pi ^2\right) n +5 \pi ^2 /3. \end{aligned}$$
For \(I_{3}^{(6)}\), we have
$$\begin{aligned} I_{3}^{(6)}\le & \,2(2n+1)\left( \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left( \frac{2(n+1)(n+2)\sin ^2 \theta _{k}/2-n}{8((n+1)-(n+2)\sin ^2\theta _{k}/2)^3 \sin ^2\theta _{k}/2}+\ell _{k}(x)\right) \right) \\\le &\, 2(2n+1)\left( n\left( \sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\ell _{k}(x)\right) ^{1/2}+\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{2(n+1)(n+2)\sin ^2\theta _{k}/2-n}{8\left( (n+1)-(n+2)\sin ^2\theta _{k}/2\right) ^3\sin ^2\theta _{k}/2}\right| \right) \\\le &\, 2(2n+1)\left( \sqrt{2}n+\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\left| \frac{20(n+1)(n+2)\sin ^2\theta _{k}/2-10n}{n\cos ^2\theta _{k}/2 \sin ^2\theta _{k}/2(n-n(n+2)/k)^2}\right| \right) \\\le &\, 2(2n+1)\left( \sqrt{2}n+\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\frac{40k^2(2(n+1)(n+2)\sin ^2\theta _{k}/2-n)}{n^3\sin ^2\theta _{k}(k-(n+2))^2}\right) . \end{aligned}$$
Since \(\sin {\theta _{k}}=\sin \theta _{k-n}\), we can apply Jordan inequality to obtain
$$ I_{3}^{(6)}\le 2(2n+1)\left( \sqrt{2}n+\sum _{k=\left[ \frac{n}{2}\right] +1}^{n-1}\frac{10\left( 2n^2+7n+3\right) }{\pi ^2 n(k-(n+2))^2}\right) \le 2(2n+1)\left( \sqrt{2}n+\frac{20n+70}{6}\right). $$
It is easy to show that \(I_{3}^{(7)}\le 2+ \frac{\pi ^2\sqrt{2}}{3}(2n^2+7n+4)\) and \(I_{3}^{(8)}\le 2 +K_{2}(2n^2+7n+4)\) for some constant \(K_{2}\). Combining all the estimates for \(I_{3}^{(1)}\) till \(I_{3}^{(8)}\), we observe that \(\parallel I_{3} \parallel ={{\mathcal {O}}}(n^2)\). For \(I_{4}\):
$$ I_{4}\le \sum _{n=0}^n \ell _{k}^2(x)+\sum _{k=0}^n |a_{k}w^{\prime }(x)\ell _{k}(x)|. $$
In view of \(\sum _{k=0}^n \ell _{k}^2(x) \le 4\); and using Lemma 2, we have
$$\begin{aligned} \sum _{k=0}^n |a_{k}w^{\prime }(x)\ell _{k}(x)|\le &\, |a_{0}w'(x) \ell _{0}(x)|+\sum _{k=0}^{n-1}|a_{k}w'(x)\ell _{k}(x)|+|a_{n}w'(x)\ell _{n}(x)|\\\le & \,1+\sqrt{2}\sum _{k=1}^{n-1}\frac{2(2n+1)}{2((n+1)-(n+2)\sin ^2\theta _{k}/2)}+2n+1\\\le &\, 2+2n+2\sqrt{2}(2n+1)\sum _{k=1}^{n-1}\frac{5}{n\cos ^2 \theta _{k}/2}\\\le & 2+2n+40 \sqrt{2}(2n+1)\left( \frac{n \pi ^2}{12}\right) . \end{aligned}$$
Therefore,
$$ I_{4}\le 6+\left( 2+\frac{10 \sqrt{2}\pi ^2}{3}\right) n+\left( \frac{20\sqrt{2}\pi ^2}{3}\right) n^2. $$
Combining all the estimates of \(I_{1}, I_{2}, I_{3}, I_{4}\), we complete the proof of Theorem 2. \(\square \)