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<title>On Berman’s phenomenon <br />for (0,1,2) Hermite–Fejér interpolation: On Berman’s phenomenon <br />for (0,1,2) Hermite–Fejér interpolation</title>
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<h1>On Berman’s phenomenon <br />for (0,1,2) Hermite–Fejér interpolation</h1>
<p class="authors">
<span class="author">Graeme J. Byrne\(^\ast \) Simon J. Smith\(^\bullet \)</span>
</p>
<p class="date">October 7, 2018. Accepted: February 4, 2019. Published online: October 10, 2019.</p>
</div>
<p>\(^\ast \)Department of Mathematics and Statistics, La&#160;Trobe University, P. O. Box 199, Bendigo, VIC 3552, Australia, e-mail: <span class="tt">g.byrne@latrobe.edu.au</span>. </p>
<p>\(^\bullet \)54 Broad Parade, Bendigo, VIC 3550, Australia, e-mail: <span class="tt">sjsmith3550@gmail.com</span>. The work of this author has been supported by La&#160;Trobe University. </p>
<div class="abstract"><p> Given \(f\in C[-1,1]\) and \(n\) points (nodes) in \([-1,1]\), the <i class="it">Hermite–Fejér interpolation</i> (HFI) polynomial is the polynomial of degree at most \(2n-1\) which agrees with \(f\) and has zero derivative at each of the nodes. In 1916, L. Fejér showed that if the nodes are chosen to be the zeros of \(T_{n}(x)\), the \(n\)th Chebyshev polynomial of the first kind, then the HFI polynomials converge uniformly to \(f\) as \(n\rightarrow \infty \). Later, D. L. Berman established the rather surprising result that this convergence property is no longer true for all \(f\) if the Chebyshev nodes are augmented by including the endpoints \(-1\) and 1 as additional nodes. This behaviour has become known as <em>Berman’s phenomenon</em>. The aim of this paper is to investigate Berman’s phenomenon in the setting of \((0,1,2)\) HFI, where the interpolation polynomial agrees with \(f\) and has vanishing first and second derivatives at each node. The principal result provides simple necessary and sufficient conditions, in terms of the (one-sided) derivatives of \(f\) at \(\pm 1\), for pointwise and uniform convergence of \((0,1,2)\) HFI on the augmented Chebyshev nodes if \(f\in C^{4}[-1,1]\), and confirms that Berman’s phenomenon occurs for \((0,1,2)\) HFI. </p>
<p><b class="bf">MSC.</b> Primary 41A05; Secondary 41A10 </p>
<p><b class="bf">Keywords.</b> interpolation, polynomial interpolation, Hermite–Fejér interpolation, Chebyshev nodes, Berman’s phenomenon </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Suppose \(f\in C[-1,1]\) and let </p>
<div class="equation" id="X1">
<p>
  <div class="equation_content">
    \begin{equation}  X=\{ x_{k,n}:k=0,1,2,\ldots ,n-1;\  n=1,2,3,\ldots \} \label{X1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> be an infinite triangular matrix of nodes such that, for all \(n\), </p>
<div class="equation" id="X2">
<p>
  <div class="equation_content">
    \begin{equation}  1\geq x_{0,n}>x_{1,n}>\ldots >x_{n-1,n}\geq -1.\label{X2} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> The well-known Lagrange interpolation polynomial of \(f\) is the polynomial \(L_{n}(X,f)(x)=L_{n}(X,f,x)\) of degree at most \(n-1\) which satisfies </p>
<div class="displaymath" id="a0000000003">
  \[ L_{n}(X,f,x_{k,n})=f(x_{k,n}),\qquad 0\leq k\leq n-1. \]
</div>
<p>A classic result due to Faber&#160;<span class="cite">
	[
	<a href="#Faber" >6</a>
	]
</span> states that for any \(X\) there exists \(f\in C[-1,1]\) so that \(L_{n}(X,f)\) does not converge uniformly to \(f\) on \([-1,1]\) as \(n\rightarrow \infty \). On the other hand, a more positive result occurs for the matrix of Chebyshev nodes </p>
<div class="displaymath" id="a0000000004">
  \[ T=\left\{ x_{k,n}=\cos \big(\tfrac {2k+1}{2n}\pi \big):k=0,1,2,\ldots ,n-1;\  n=1,2,3,\ldots \right\}  \]
</div>
<p> where, for each \(n\), the \(x_{k,n}\) are the zeros of the \(n\)th Chebyshev polynomial \(T_{n}(x)=\cos (n\arccos x)\), \(-1\leq x\leq 1\). This result states that if the modulus of continuity \(\omega (\delta ;f)\) of \(f\) is defined by </p>
<div class="displaymath" id="a0000000005">
  \[ \omega (\delta ;f)=\max \big\{ |f(s)-f(t)|:-1\leq s,t\leq 1, |s-t|\leq \delta \big\} , \]
</div>
<p> then \(L_{n}(T,f)\) converges uniformly to \(f\) under the quite mild restriction </p>
<div class="displaymath" id="a0000000006">
  \[ \omega (\tfrac 1n;f)\log n\rightarrow 0,\  \mbox{as}\  n\rightarrow \infty  \]
</div>
<p> (see Rivlin&#160;<span class="cite">
	[
	<a href="#RivlinApproxofFunctions" >11</a>
	, 
	Chapter
	
	4
	]
</span> for details and references). </p>
<p>A generalization of Lagrange interpolation is provided by Hermite–Fejér interpolation (HFI). Given a non-negative integer \(m\) and nodes \(X\) defined by&#160;(<a href="#X1">1</a>) and&#160;(<a href="#X2">2</a>), the \((0,1,\ldots ,m)\) HFI polynomial \(H_{m,n}(X,f)(x)=H_{m,n}(X,f,x)\) of \(f\) is the unique polynomial of degree at most \((m+1)n-1\) which satisfies the \((m+1)n\) conditions </p>
<div class="displaymath" id="a0000000007">
  \[ \left\{ \begin{array}{ll} H_{m,n}(X,f,x_{k,n})=f(x_{k,n}), &  0\leq k\leq n-1, \\[0.25 cm] H^{(r)}_{m,n}(X,f,x_{k,n})=0, &  1\leq r\leq m,\  0\leq k\leq n-1. \end{array} \right.  \]
</div>
<p> Note that \(H_{0,n}(X,f)=L_{n}(X,f)\). The original motivation for studying HFI was provided by Fejér&#160;<span class="cite">
	[
	<a href="#Fejer" >7</a>
	]
</span>, who in 1916 showed that if \(f\in C[-1,1]\), then \(\| H_{1,n}(T,f)-f\| \rightarrow 0\) as \(n\rightarrow \infty \) (here and subsequently, \(\| \cdot \| \) denotes the uniform norm on \([-1,1]\)). Thus on the Chebyshev nodes, \((0,1)\) HFI succeeds where Lagrange interpolation may fail. </p>
<p>In the years since Fejér’s work, \((0,1,\ldots ,m)\) HFI has been much studied by many authors. In this paper our focus is on an aspect of HFI that has become known as <em>Berman’s phenomenon</em>, which occurs if the Chebyshev nodes are augmented by the end points of the interval \([-1,1]\). In other words, we will be studying \((0,1,\ldots ,m)\) HFI on the nodes \(T_{a}=\{ x_{k,n+2}:0\leq k\leq n+1, n=1,2,3,\ldots \} \), where </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996BE6450>">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}{l} x_{0,n+2}=1,\quad x_{n+1,n+2}=-1, \\[0.25 cm] x_{k,n+2}=\cos \big(\tfrac {(2k-1)\pi }{2n}\big),\quad 1\leq k\leq n. \end{array} \right. \label{augmented_Chebyshev nodes} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> Thus, for each \(n\), the \(x_{k,n+2}\) are the zeros of \((1-x^{2})T_{n}(x)\). </p>
<p>Initially it might be thought that augmenting the Chebyshev nodes with \(\pm 1\) will have little effect on the convergence behaviour of interpolation polynomials when compared with interpolation on the Chebyshev nodes alone. However, D. L. Berman&#160;<span class="cite">
	[
	<a href="#Berman1965" >1</a>
	]
</span> was able to show that if \(f(x)=|x|\), then \(H_{1,n}(T_{a},f,0)\) diverges, while later&#160;<span class="cite">
	[
	<a href="#Berman1969" >2</a>
	]
</span> he showed that for \(g(x)=x^{2}\), \(H_{1,n}(T_{a},g,x)\) does not converge to \(x^{2}\) at any point of \((-1,1)\). (This result for \(x^{2}\) does not extend to \([-1,1]\) because \(\pm 1\) are interpolation nodes for all \(n\).) An explanation for Berman’s phenomenon was provided by Bojanic, as follows. </p>
<p><div class="theorem_thmwrapper " id="Bojanic's_Theorem">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
    <span class="theorem_thmtitle">Bojanic</span>
  </div>
  <div class="theorem_thmcontent">
  <p>  <span class="cite">
	[
	<a href="#Bojanic" >3</a>
	]
</span>. If \(f\in C[-1,1]\) has left and right derivatives \(f_{L}'(1)\) and \(f_{R}'(-1)\) at 1 and \(-1\), respectively, then \(H_{1,n}(T_{a},f)\) converges uniformly to \(f\) on \([-1,1]\) if and only if \(f_{L}'(1)=f_{R}'(-1)=0\). </p>

  </div>
</div> </p>
<p>At this point it is natural to ask whether Berman’s phenomenon occurs for \((0,1,\ldots ,m)\) HFI if \(m\neq 1\). When \(m=0\) (Lagrange interpolation) it is straightforward to show that it does <em>not</em> occur. This follows from the representation </p>
<div class="displaymath" id="a0000000008">
  \begin{align*}  L_{n+2}(T_{a},f,x)=& L_{n}(T,f,x) +\tfrac {1}{2}T_{n}(x)\big\{ (1\! +\! x)[f(1)-L_{n-1}(T,f,1)]\\ & \hspace{3.3cm}+(-1)^{n}(1\! -\! x)[f(-1)-L_{n-1}(T,f,-1)]\big\} , \end{align*}
</div>
<p> which can be verified by observing that both sides of the equation are polynomials of degree at most \(n+1\) which agree at the \(n+2\) nodes \(x_{k,n+2}\) given by&#160;(??). Thus \(L_{n}(T_{a},f)\rightarrow f\) uniformly on \([-1,1]\) whenever \(L_{n}(T,f)\rightarrow f\) uniformly on \([-1,1]\). </p>
<p>The question of whether Berman’s phenomenon occurs for \((0,1,\ldots ,m)\) HFI for any \(m{\gt}1\) was answered in the affirmative by Cook and Mills&#160;<span class="cite">
	[
	<a href="#CookMills" >5</a>
	]
</span> in 1975, who showed that if \(h(x)=(1-x^{2})^{3}\), then \(H_{3,n}(T_{a},h,0)\) diverges. (Incidentally, it was in&#160;<span class="cite">
	[
	<a href="#CookMills" >5</a>
	]
</span> that the term <em>Berman’s phenomenon</em> was first used.) The result of Cook and Mills was later extended by Maky&#160;<span class="cite">
	[
	<a href="#Maky" >9</a>
	]
</span> who showed that \(H_{3,n}(T_{a},h,x)\) diverges at each point in \((-1,1)\). These findings for \((0,1,2,3)\) HFI on \(T_{a}\) contrast with the earlier result of Krylov and Steuermann&#160;<span class="cite">
	[
	<a href="#KrylovSteuermann" >8</a>
	]
</span> that \(H_{3,n}(T,f)\) converges uniformly to \(f\) on \([-1,1]\) for <em>any</em> \(f\in C[-1,1]\). </p>
<p>In this paper our focus will be on \((0,1,2)\) HFI. Here it was shown by Szabados and Varma&#160;<span class="cite">
	[
	<a href="#SzabadosVarma" >13</a>
	]
</span> that, as with Lagrange interpolation, for any matrix of nodes \(X\) there exists \(f\in C[-1,1]\) so that \(H_{2,n}(X,f)\) does not converge uniformly to \(f\) on \([-1,1]\). On the other hand, and again like Lagrange interpolation, it follows from Byrne et al.&#160;<span class="cite">
	[
	<a href="#ByrneMillsSmith" >4</a>
	, 
	Theorem
	
	1
	]
</span> that if \(\omega (1/n;f)\log n\rightarrow 0\) as \(n\rightarrow \infty \), then \(H_{2,n}(T,f)\) converges uniformly to \(f\). </p>
<p>To investigate \((0,1,2)\) HFI on \(T_{a}\) it proves helpful to follow the approach of Bojanic&#160;<span class="cite">
	[
	<a href="#Bojanic" >3</a>
	]
</span>, and introduce incremental modifications to the \((0,1,2)\) HFI process on \(T\). To this end, with nodes \(x_{k}=x_{k,n+2}\) defined by&#160;(??) and \(f\in C[-1,1]\), define the polynomial \(Q_{2,n+2}(T_{a},f)\) of degree at most \(3n+1\) by the \(3n+2\) conditions </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996C535F0>">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}{ll} Q_{2,n+2}(T_{a},f,x_{k})=f(x_{k}), &  0\leq k\leq n+1, \\[0.25 cm] Q_{2,n+2}’(T_{a},f,x_{k})=Q_{2,n+2}”(T_{a},f,x_{k})=0,\  \  \  &  1\leq k\leq n, \end{array} \right.\label{Defn_of_Q2,n+2} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> and define the polynomial \(R_{2,n+2}(T_{a},f)\) of degree at most \(3n+3\) by the \(3n+4\) conditions </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996CD1C70>">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}{ll} R_{2,n+2}(T_{a},f,x_{k})=f(x_{k}), R_{2,n+2}’(T_{a},f,x_{k})=0,\  \  \  &  0\leq k\leq n+1, \\[0.25 cm] R_{2,n+2}”(T_{a},f,x_{k})=0, &  1\leq k\leq n. \end{array} \right.\label{Defn_of_R2,n+2} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> Also, for future reference, recall that \(H_{2,n+2}(T_{a},f)\) is defined by the \(3n+6\) conditions </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996CD2CF0>">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}{ll} H_{2,n+2}(T_{a},f,x_{k})=f(x_{k}), & \\[0.25 cm] H_{2,n+2}’(T_{a},f,x_{k})=H_{2,n+2}”(T_{a},f,x_{k})=0,\  \  \  &  0\leq k\leq n+1. \end{array} \right.\label{Defn_of_H2,n+2} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>Our principal results are presented in the following theorem. Note that throughout this paper we are concerned with functions defined on \([-1,1]\), and so any derivative evaluated at 1 or \(-1\) is assumed to be the appropriate one-sided derivative. </p>
<p><div class="theorem_thmwrapper " id="newtheorem">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Suppose that the polynomials \(Q_{2,n+2}(T_{a},f)\), \(R_{2,n+2}(T_{a},f)\) and \(H_{2,n+2}(T_{a},f)\) are defined by <span class="rm">(??)</span>–<span class="rm">(??)</span>, where the interpolation nodes \(x_{k}=x_{k,n+2}\) are defined by&#160;<span class="rm">(??)</span>. </p>
<ul class="itemize">
  <li><p>If \(f\in C[-1,1]\) and \({\displaystyle \lim _{n\rightarrow \infty }\| H_{2,n}(T,f)-f\| =0}\), then </p>
<div class="displaymath" id="a0000000009">
  \[ \lim _{n\rightarrow \infty }\| Q_{2,n+2}(T_{a},f)-f\| =0. \]
</div>
</li>
  <li><p>If \(f\in C^{2}[-1,1]\), then \({\displaystyle \lim _{n\rightarrow \infty }\| R_{2,n+2}(T_{a},f)-f\| =0}\) if and only if \(f'(1)=f'(-1)=0\). Furthermore, if \(f'(1)\) and \(f'(-1)\) are not both&#160;\(0\), then \(R_{2,n+2}(T_{a},f,x)\) is divergent if \(0{\lt}|x|{\lt}1\), and \(R_{2,n+2}(T_{a},f,0)\) converges to \(f(0)\) if and only if \(f'(1)=f'(-1)\). </p>
</li>
  <li><p>If \(f\in C^{4}[-1,1]\), then \({\displaystyle \lim _{n\rightarrow \infty }\| H_{2,n+2}(T_{a},f)-f\| =0}\) if and only if </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996D0A930>">
<p>
  <div class="equation_content">
    \begin{equation}  f'(1)=f'(-1)=f''(1)=f''(-1)=0.\label{Theorem_conditions} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> If \(0{\lt}|x|{\lt}1\) and <span class="rm">(??)</span> does not hold, then \(H_{2,n+2}(T_{a},f,x)\) is divergent; in particular, if \(f'(1)\) and \(f'(-1)\) are not both&#160;0, then </p>
<div class="equation" id="limsup">
<p>
  <div class="equation_content">
    \begin{equation}  \limsup _{n\rightarrow \infty }\tfrac {1}{n^{2}}|H_{2,n+2}(T_{a},f,x)|>0. \label{limsup} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> Furthermore, \(H_{2,n+2}(T_{a},f,0)\) converges to \(f(0)\) if and only if \(f'(1)=f'(-1)\) and \(f''(1)=-f''(-1)\) (which occurs, for example, if \(f\) is odd). If \(f'(1)\neq f'(-1)\), then </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996D0BBF0>">
<p>
  <div class="equation_content">
    \begin{equation}  \limsup _{n\rightarrow \infty }\tfrac {1}{n^{2}}|H_{2,n+2}(T_{a},f,0)|>0. \label{limsup_at_0} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
</li>
</ul>

  </div>
</div> </p>
<p>Theorem&#160;<a href="#newtheorem">2</a> confirms that Berman’s phenomenon occurs for (0,1,2) HFI. In particular, the following results hold. <div class="corollary_thmwrapper " id="a0000000010">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">3</span>
  </div>
  <div class="corollary_thmcontent">
  <p>\((0,1,2)\) HFI provides the following illustrations of Berman’s phenomenon. </p>
<ul class="itemize">
  <li><p>If \(f(x)=x\), then \({\displaystyle \limsup _{n\rightarrow \infty }\tfrac {1}{n^{2}}|H_{2,n+2}(T_{a},f,x)|{\gt}0}\) for \(0{\lt}|x|{\lt}1\) and \({\displaystyle \lim _{n\rightarrow \infty }H_{2,n+2}(T_{a},f,0)=0}\). </p>
</li>
  <li><p>If \(g(x)=x^{2}\), then \({\displaystyle \limsup _{n\rightarrow \infty }\tfrac {1}{n^{2}}|H_{2,n+2}(T_{a},g,x)|{\gt}0}\) for all \(x\) in \((-1,1)\). </p>
</li>
</ul>

  </div>
</div> </p>
<p>The theorem will be proved <i class="it">via</i> a sequence of lemmas in the next section. </p>
<h1 id="a0000000011">2 Proof of the main result (Theorem&#160;2)</h1>
<p>We begin by noting that the polynomials \(Q_{2,n+2}(T_{a},f)\), \(R_{2,n+2}(T_{a},f)\) and \(H_{2,n+2}(T_{a},f)\), defined by (??)–(??), are related to each other and to \(H_{2,n}(T,f)\) according to the formulas </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996D3DD90>">
  \begin{align}  \label{Formula_for_Q2,n+2} & Q_{2,n+2}(T_{a},f,x)=\\ \nonumber & =H_{2,n}(T,f,x)+\tfrac {1}{2}\left(T_{n}(x)\right)^{3}\\ \nonumber & \quad \times \left(1+x)\left(f(1)-H_{2,n}(T,f,1)\right)+(-1)^{n}(1-x)\left(f(-1)-H_{2,n}(T,f,-1)\right)\right],\nonumber \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996BAF2F0>">
  \begin{align}  \label{Formula_for_R2,n+2} & R_{2,n+2}(T_{a},f,x)=\\ \nonumber & =Q_{2,n+2}(T_{a},f,x)+\tfrac {1}{4}\left(T_{n}(x)\right)^{3}(1-x^{2})\\ \nonumber & \quad \times \big[(1+x)Q_{2,n+2}’(T_{a},f,1)-(-1)^{n}(1-x)Q_{2,n+2}’(T_{a},f,-1)\big] \end{align}
</div>
<p> and </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996CE8CB0>">
  \begin{align}  \label{Formula_for_H2,n+2} & H_{2,n+2}(T_{a},f,x)=\\ \nonumber & =R_{2,n+2}(T_{a},f,x)-\tfrac {1}{16}\left(T_{n}(x)\right)^{3}(1-x^{2})^{2}\\ \nonumber & \quad \times \big[(1+x)R_{2,n+2}”(T_{a},f,1)+(-1)^{n}(1-x)R_{2,n+2}”(T_{a},f,-1)\big]. \end{align}
</div>
<p> Each of (??)–(??) can be verified by simply checking that the polynomial on the right-hand side satisfies the defining conditions (in terms of degree and values at the interpolation nodes&#160;(??)) of the polynomial on the left-hand side. Also note (Byrne <i class="it">et al.</i>&#160;<span class="cite">
	[
	<a href="#ByrneMillsSmith" >4</a>
	, 
	Section
	
	1
	]
</span>) that \(H_{2,n}(T,f)\) has the explicit formula </p>
<div class="equation" id="H2,n">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2,n}(T,f,x)=\tfrac {1}{n^{3}}\left(T_{n}(x)\right)^{3}S_{n}(f,x) \label{H2,n} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> where, with \(x_{k}=x_{k,n+2}\) defined by&#160;(??), </p>
<div class="equation" id="Sn">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n}(f,x)=\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {1-xx_{k}}{(x-x_{k})^{3}} -\tfrac {x_{k}}{2(x-x_{k})^{2}}+\tfrac {n^{2}-1}{2(x-x_{k})}\right)f(x_{k}). \label{Sn} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>The results of the following two lemmas will be used to develop alternative representations for \(R_{2,n+2}(T_{a},f)\) and \(H_{2,n+2}(T_{a},f)\) to those in (??) and (??). </p>
<p><div class="lemma_thmwrapper " id="a0000000012">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
  </div>
  <div class="lemma_thmcontent">
  <p>For \(x_{k}=x_{k,n+2}\) defined by&#160;<span class="rm">(??)</span> and \(f\in C[-1,1]\), the following summation formulas hold. </p>
<div class="equation" id="summation1">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {T_{n}(x)}{n}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{x-x_{k}}f(x_{k})=L_{n}(T,f,x) \label{summation1} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<div class="equation" id="summation2">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {T_{n}(x)}{n}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(x-x_{k})^{2}}f(x_{k})=\tfrac {T_{n}’(x)}{T_{n}(x)}L_{n}(T,f,x)-L_{n}'(T,f,x) \label{summation2} \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<div class="displaymath" id="summation3">
  \begin{align}  \tfrac {T_{n}(x)}{n}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(x-x_{k})^{3}}f(x_{k})=& \left[\left(\tfrac {T_{n}’(x)}{T_{n}(x)}\right)^{\! \! 2}-\tfrac {1}{2}\tfrac {T_{n}”(x)}{T_{n}(x)}\right]L_{n}(T,f,x)\label{summation3} \\ & -\tfrac {T_{n}’(x)}{T_{n}(x)}L_{n}’(T,f,x)+\tfrac {1}{2}L_{n}”(T,f,x)\nonumber \end{align}
</div>
<div class="displaymath" id="summation4">
  \begin{align}  \label{summation4} & \tfrac {T_{n}(x)}{n}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(x-x_{k})^{4}}f(x_{k})=\\ \nonumber & =\left[\tfrac {1}{6}\tfrac {T_{n}”’(x)}{T_{n}(x)}-\tfrac {T_{n}’(x)T_{n}”(x)}{(T_{n}(x))^{2}}+\left(\tfrac {T_{n}’(x)}{T_{n}(x)}\right)^{\! \! 3\, }\right]L_{n}(T,f,x)\\ \nonumber & \quad +\left[\tfrac {1}{2}\tfrac {T_{n}”(x)}{T_{n}(x)}-\left(\tfrac {T_{n}’(x)}{T_{n}(x)}\right)^{\! \! 2\, }\right]L_{n}’(T,f,x)+\tfrac {1}{2}\tfrac {T_{n}’(x)}{T_{n}(x)}L_{n}”(T,f,x)-\tfrac {1}{6}L_{n}”’(T,f,x) \end{align}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>The formula&#160;(<a href="#summation1">14</a>) for Lagrange interpolation on the Chebyshev nodes is well-known (see, for example, Rivlin&#160;<span class="cite">
	[
	<a href="#RivlinChebyshevPolynomials" >12</a>
	, 
	Section
	
	1.3
	]
</span>). The remaining formulas (<a href="#summation2">15</a>)–(<a href="#summation4">17</a>) follow by successive differentiation of &#160;(<a href="#summation1">14</a>). </p>
<p><div class="lemma_thmwrapper " id="a0000000014">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p>For \(x_{k}=x_{k,n+2}\) defined by&#160;<span class="rm">(??)</span>, the following summation formulas hold. </p>
<div class="displaymath" id="summation5">
  \begin{align}  \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{1-x_{k}}&  =n \label{summation5} \\ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{2}}&  =n^{3} \label{summation6} \\ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{3}}&  =\tfrac {1}{6}n^{3}(5n^{2}+1) \label{summation7} \\ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{4}}&  =\tfrac {1}{90}n^{3}(61n^{4}+25n^{2}+4) \label{summation8} \\ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1+x_{k})^{r}}&  =(-1)^{n+1}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{r}},\quad r=1,2,3,\ldots \label{summation9} \end{align}
</div>
<p> <div class="proof_wrapper" id="a0000000015">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> If \(f(x)\equiv 1\), then \(L_{n}(T,f,x)\equiv 1\), so&#160;(<a href="#summation1">14</a>) gives </p>
<div class="displaymath" id="a0000000016">
  \[ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{x-x_{k}}=\tfrac {n}{T_{n}(x)}. \]
</div>
<p> Putting \(x=1\) gives&#160;(<a href="#summation5">18</a>). The formulas (<a href="#summation6">19</a>)–(<a href="#summation8">21</a>) follow from (<a href="#summation2">15</a>)–(<a href="#summation4">17</a>) in similar fashion after noting that if \(f(x)\equiv 1\), then \(L_{n}^{(r)}(T,f,x)\equiv 0\) for \(r\geq 1\), and (Rivlin&#160;<span class="cite">
	[
	<a href="#RivlinChebyshevPolynomials" >12</a>
	, 
	p.
	 
	38
	]
</span>) </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996BF71D0>">
<p>
  <div class="equation_content">
    \begin{equation}  T_{n}^{(r)}(1)=\tfrac {n^{2}(n^{2}-1^{2})(n^{2}-2^{2})\ldots (n^{2}-(r-1)^{2})}{1\, .\, 3\, .\, 5\ldots (2r-1)}. \label{Chebyshev_derivatives_at_1} \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> Finally,&#160;(<a href="#summation9">22</a>) is a straightforward consequence of the symmetry of the Chebyshev nodes (i.e. \(x_{k,n+2}=-x_{n-k+1,n+2}\) for \(1\leq k\leq n\)). <div class="proof_wrapper" id="a0000000017">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p>We now obtain alternative representations for the interpolation polynomials \(R_{2,n+2}(T_{a},f)\) and \(H_{2,n+2}(T_{a},f)\) to those in (??) and (??). </p>
<p><div class="lemma_thmwrapper " id="AltRepns">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">6</span>
  </div>
  <div class="lemma_thmcontent">
  <p> For \(x_{k}=x_{k,n+2}\) defined by&#160;<span class="rm">(??)</span> and \(f\in C[-1,1]\), </p>
<div class="displaymath" id="r(x)">
  \begin{align}  R_{2,n+2}(T_{a},f,x)=& H_{2,n}(T,f,x)\label{r(x)} \\ & +\tfrac {1}{4}(T_{n}(x))^{3}\Big\{ (1+x)^{2}(2-x)\left[f(1)-H_{2,n}(T,f,1)\right]\nonumber \\ & +(-1)^{n}(1-x)^{2}(2+x)\left[f(-1)-H_{2,n}(T,f,-1)\right]\nonumber \\ & +(1-x^{2})\left[(1+x)A_{1,n}-(1-x)B_{1,n}\right]\Big\} \nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="h(x)">
  \begin{align}  H_{2,n+2}(T_{a},f,x)=& R_{2,n+2}(T_{a},f,x)+\tfrac {1}{16}(T_{n}(x))^{3}(1-x^{2})^{2}\label{h(x)}\times \\ & \times \! \Big[3x\big\{ \! \left[f(1)\! \! -\! \! H_{2,n}(T,f,1)\right]\! -\! (\! -\! 1)^{n}\left[f(\! -\! 1)\! -\! H_{2,n}(T,f,\! -\! 1)\right]\! \big\} \nonumber \\ & +(1+x)(A_{2,n}+B_{1,n})-(1-x)(B_{2,n}+A_{1,n})\Big] \nonumber \end{align}
</div>
<p> where </p>
<div class="displaymath" id="A1,n">
  \begin{align}  A_{1,n}&  =\tfrac {1}{n^{3}}\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {3}{(1-x_{k})^{2}}+\tfrac {n^{2}-1}{2(1-x_{k})}\right)\tfrac {f(1)-f(x_{k})}{1-x_{k}}, \label{A1,n} \\ B_{1,n}&  =\tfrac {1}{n^{3}}\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {3}{(1+x_{k})^{2}}+\tfrac {n^{2}-1}{2(1+x_{k})}\right)\tfrac {f(-1)-f(x_{k})}{1+x_{k}}, \label{B1,n} \\ A_{2,n}&  =\tfrac {1}{n^{3}}\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {15}{(1-x_{k})^{3}}+\tfrac {n^{2}+2}{(1-x_{k})^{2}}+\tfrac {n^{2}-1}{1-x_{k}}\right)\tfrac {f(1)-f(x_{k})}{1-x_{k}}, \label{A2,n} \\ B_{2,n}&  =\tfrac {1}{n^{3}}\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {15}{(1+x_{k})^{3}}+\tfrac {n^{2}+2}{(1+x_{k})^{2}}+\tfrac {n^{2}-1}{1+x_{k}}\right)\tfrac {f(-1)-f(x_{k})}{1+x_{k}}. \label{B2,n} \end{align}
</div>
<p> <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> For convenience, denote the RHS of&#160;(<a href="#r(x)">24</a>) by \(r(x)\). To establish&#160;(<a href="#r(x)">24</a>) it is sufficient to show \(r(x)\) satisfies the \(3n+4\) defining conditions&#160;(??) of \(R_{2,n+2}(T_{a},f)\). In fact, apart from \(r'(1)=r'(-1)=0\), the conditions are verified easily. To show \(r'(1)=0\), firstly observe from&#160;(??) that (upon using \(T_{n}'(1)=n^{2}\)) </p>
<div class="displaymath" id="a0000000019">
  \[ r'(1)=H_{2,n}'(T,f,1)+3n^{2}\left(f(1)-H_{2,n}(T,f,1)\right)-A_{1,n}. \]
</div>
<p> However, by&#160;(<a href="#H2,n">12</a>) and&#160;(<a href="#Sn">13</a>), </p>
<div class="displaymath" id="a0000000020">
  \[ H_{2,n}'(T,f,1)=3n^{2}H_{2,n}(T,f,1)+\tfrac {1}{n^{3}}S_{n}'(f,1), \]
</div>
<p> where </p>
<div class="equation" id="Sn'(1)">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n}'(f,1)=-\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {3}{(1-x_{k})^{3}} +\tfrac {n^{2}-1}{2(1-x_{k})^{2}}\right)f(x_{k}). \label{Sn'(1)} \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p> Thus </p>
<div class="displaymath" id="a0000000021">
  \[ r'(1)=3n^{2}f(1)+\tfrac {1}{n^{3}}S_{n}'(f,1)-A_{1,n}, \]
</div>
<p> and this is zero by (<a href="#summation6">19</a>), (<a href="#summation7">20</a>) and&#160;(<a href="#A1,n">26</a>). The result \(r'(-1)=0\) is established by similar means. </p>
<p>Again for convenience, denote the RHS of&#160;(<a href="#h(x)">25</a>) by \(h(x)\). By the same arguments as above, it is evident that&#160;(<a href="#h(x)">25</a>) will be proved if it can be shown that \(h''(1)=h''(-1)=0\). To show \(h''(1)=0\), begin by noting that from&#160;(<a href="#r(x)">24</a>) and&#160;(<a href="#h(x)">25</a>), </p>
<div class="displaymath" id="h''(1)">
  \begin{align}  h”(1)=& H_{2,n}”(T,f,1)+(7n^{4}-n^{2})\left(f(1)-H_{2,n}(T,f,1)\right)\label{h''(1)} -(6n^{2}+2)A_{1,n}+A_{2,n}. \end{align}
</div>
<p> However, by&#160;(<a href="#H2,n">12</a>) and&#160;(<a href="#Sn">13</a>), </p>
<div class="equation" id="H2,n''(1)">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2,n}''(T,f,1)=(7n^{4}-n^{2})H_{2,n}(T,f,1)+\tfrac {6}{n}S_{n}'(f,1)+\tfrac {1}{n^{3}}S_{n}''(f,1), \label{H2,n''(1)} \end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p> where </p>
<div class="equation" id="Sn''(1)">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n}''(f,1)=\sum _{k=1}^{n}(-1)^{k-1}\sqrt{1-x_{k}^{2}}\left(\tfrac {3x_{k}+12}{(1-x_{k})^{4}} +\tfrac {n^{2}-1}{(1-x_{k})^{3}}\right)f(x_{k}). \label{Sn''(1)} \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> By substituting (<a href="#A1,n">26</a>), (<a href="#A2,n">28</a>), (<a href="#Sn'(1)">30</a>), (<a href="#H2,n''(1)">32</a>) and&#160;(<a href="#Sn''(1)">33</a>) into&#160;(<a href="#h''(1)">31</a>), and employing summation formulas (<a href="#summation6">19</a>)–(<a href="#summation8">21</a>), it follows (after somewhat tedious calculations) that \(h''(1)=0\). The result \(h''(-1)=0\) is proved in similar fashion. <div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p>The next result characterizes the quantities \(A_{i,n}\) and \(B_{i,n}\) of Lemma&#160;<a href="#AltRepns">6</a> in terms of values of Lagrange interpolation polynomials and their derivatives. </p>
<p><div class="lemma_thmwrapper " id="a0000000023">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">7</span>
  </div>
  <div class="lemma_thmcontent">
  <p>For \(i=1,2\), the quantities \(A_{i,n}\) and \(B_{i,n}\) that are defined by <span class="rm">(<a href="#A1,n">26</a>)</span>–<span class="rm">(<a href="#B2,n">29</a>)</span> can be written as </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996DD5C10>">
  \begin{align}  A_{1,n}=3n^{2}\left[f(1)-L_{n}(T,f,1)\right]+\tfrac {7n^{2}-1}{2n^{2}}L_{n}’(T,f,1)-\tfrac {3}{2n^{2}}L_{n}”(T,f,1), \label{A1,n_alt} \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996DD6690>">
  \begin{align}  B_{1,n}=& (-1)^{n+1}\big[3n^{2}\left[f(-1)-L_{n}(T,f,-1)\right]\label{B1,n_alt} \\ & -\tfrac {7n^{2}-1}{2n^{2}}L_{n}’(T,f,-1)-\tfrac {3}{2n^{2}}L_{n}”(T,f,-1)\big],\nonumber \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996DD6210>">
  \begin{align}  A_{2,n}=& (11n^{4}+7n^{2})\left[f(1)-L_{n}(T,f,1)\right]+\tfrac {27n^{4}+11n^{2}-2}{2n^{2}}L_{n}’(T,f,1)\label{A2,n_alt} \\ & -\tfrac {8n^{2}+1}{n^{2}}L_{n}”(T,f,1)+\tfrac {5}{2n^{2}}L_{n}”’(T,f,1),\nonumber \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996DD7E90>">
  \begin{align}  \label{B2,n_alt} B_{2,n}=& (-1)^{n+1}\big[(11n^{4}+7n^{2})\left[f(-1)-L_{n}(T,f,-1)\right]\\ \nonumber & -\tfrac {27n^{4}+11n^{2}-2}{2n^{2}}L_{n}’(T,f,\! -\! 1)\! -\! \tfrac {8n^{2}+1}{n^{2}}L_{n}”(T,f,\! -\! 1)\! -\! \tfrac {5}{2n^{2}}L_{n}”’(T,f,\! -\! 1)\big]. \end{align}
</div>
<p> <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> To obtain&#160;(??), begin by writing&#160;(<a href="#A1,n">26</a>) as </p>
<div class="displaymath" id="a0000000025">
  \[ A_{1,n}=\tfrac {3}{n^{3}}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{3}}g(x_{k}) +\tfrac {n^{2}-1}{2n^{3}}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{2}}g(x_{k}), \]
</div>
<p> where \(g(x)=f(1)-f(x)\). Now apply the identities&#160;(<a href="#summation2">15</a>) and&#160;(<a href="#summation3">16</a>) at \(x=1\), noting that \(L_{n}(T,g,x)=f(1)-L_{n}(T,f,x)\) and using the result&#160;(??) for \(T_{n}^{(r)}(1)\). To obtain&#160;(??), apply&#160;(<a href="#summation2">15</a>) and&#160;(<a href="#summation3">16</a>) at \(x=-1\) to&#160;(<a href="#B1,n">27</a>) with \(g(x)=f(-1)-f(x)\), and use \(T_{n}^{(r)}(-1)=(-1)^{n+r}T_{n}^{(r)}(1)\). The remaining expressions&#160;(??) and&#160;(??) are proved in a similar fashion. <div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p>For a function \(f\) that has continuous derivatives on \([-1,1]\), the following lemma characterizes the limiting behaviour of the \(A_{i,n}\) and \(B_{i,n}\) as \(n\rightarrow \infty \) in terms of these derivatives. The proof of the lemma uses the following result which is a special case of a more general theorem concerning Lagrange interpolation on Jacobi nodes that is due to Neckermann and Runck&#160;<span class="cite">
	[
	<a href="#NeckermannRunck" >10</a>
	, 
	Satz
	
	2,
	 
	p.
	 
	168
	]
</span>: </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996E05B50>">
<p>
  <div class="equation_content">
    \begin{equation}  \mbox{If $m>0$ and $f^{(2m)}\in C[-1,1]$, then\  }\lim _{n\rightarrow \infty }\| L_{n}^{(m)}(T,f)-f^{(m)}\| =0. \label{Neckermann_and_Runck's_result} \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p><div class="lemma_thmwrapper " id="a0000000027">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">8</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Suppose \(A_{i,n}\) and \(B_{i,n}\) are defined by <span class="rm">(<a href="#A1,n">26</a></span>)–<span class="rm">(<a href="#B2,n">29</a>)</span>. If \(f\in C^{2}[-1,1]\), then </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996E06810>">
  \begin{align}  \lim _{n\rightarrow \infty }A_{1,n}&  =\tfrac {7}{2}f’(1),\label{A1,n_limit} \\ \lim _{n\rightarrow \infty }(-1)^{n}B_{1,n}&  =\tfrac {7}{2}f’(-1).\label{B1,n_limit} \end{align}
</div>
<p> If \(f\in C^{4}[-1,1]\), then </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000012996E07B30>">
  \begin{align}  \lim _{n\rightarrow \infty }\left(A_{2,n}-\tfrac {27n^{2}}{2}f’(1)\right)&  =\tfrac {11}{2}f’(1)-8f”(1),\label{A2,n_limit} \\ \lim _{n\rightarrow \infty }\left((-1)^{n}B_{2,n}-\tfrac {27n^{2}}{2}f’(-1)\right)&  =\tfrac {11}{2}f’(-1)+8f”(-1).\label{B2,n_limit} \end{align}
</div>
<p> <div class="proof_wrapper" id="a0000000028">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Suppose \(f\in C^{2}[-1,1]\). We work with the expression&#160;(??) for \(A_{1,n}\). Firstly, with \(x_{1}=\cos (\pi /(2n))\), it follows from the Mean Value Theorem that there exists \(\alpha \in (x_{1},1)\) so that </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  \Big|f(1)-L_{n}(T,f,1)\Big|&  =\Big|\left(f(1)-L_{n}(T,f,1)\right)-\left(f(x_{1})-L_{n}(T,f,x_{1})\right)\Big| \\ &  =\Big|(1-x_{1})\left(f’(\alpha )-L_{n}’(T,f,\alpha )\right)\Big| \\ &  \leq 2\sin ^{2}\! \left(\tfrac {\pi }{4n}\right)\| f’-L_{n}’(T,f)\| . \end{align*}
</div>
<p> Thus, by&#160;(??), </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996DF3230>">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }n^{2}\left[f(1)-L_{n}(T,f,1)\right]=0. \label{partial_resultA} \end{equation}
  </div>
  <span class="equation_label">43</span>
</p>
</div>
<p> Also, if \(\mathcal{P}_{n}\) denotes the set of polynomials of degree at most&#160;\(n\), let \(q\) be the best uniform approximation to \(f'\) in \(\mathcal{P}_{n-3}\). Then, by Markov’s inequality, </p>
<div class="displaymath" id="a0000000030">
  \begin{align*}  |L_{n}”(T,f,1)|&  \leq \| L_{n}”(T,f)-q’\| +\| q’\|  \\ &  \leq (n-2)^{2}\| L_{n}’(T,f)-q\| +\| q’\|  \\ &  \leq (n-2)^{2}\left(\| L_{n}’(T,f)-f’\| +\| f’-q\| \right)+\| q’\| . \end{align*}
</div>
<p> Now, there exists an absolute constant \(c\) so that \(\| q'\| \leq c(n-3)^{2}\omega (\tfrac 1{n-3};f')\) (see Szabados and Vértesi&#160;<span class="cite">
	[
	<a href="#SzabadosVertesi" >14</a>
	, 
	p.
	 
	284
	]
</span>), so </p>
<div class="displaymath" id="a0000000031">
  \[ \tfrac {1}{n^{2}}|L_{n}''(T,f,1)|\leq \| L_{n}'(T,f)-f'\| +\| f'-q\| +c\, \omega (\tfrac 1{n-3};f'). \]
</div>
<p> Since \(f\in C^{2}[-1,1]\), it follows from&#160;(??) and the Weierstrass approximation theorem that </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996E3D670>">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }\tfrac {1}{n^{2}}L_{n}''(T,f,1)=0. \label{partial_resultB} \end{equation}
  </div>
  <span class="equation_label">44</span>
</p>
</div>
<p> Substituting&#160;(??) and&#160;(??) into&#160;(??) and using&#160;(??) then gives the result&#160;(??). The result&#160;(??) is established from&#160;(??) in near-identical fashion. </p>
<p>Now suppose \(f\in C^{4}[-1,1]\). Here we work with the expression&#160;(??) for \(A_{2,n}\). With \(x_{k}=\cos ((2k-1)\pi /(2n))\), by the Mean Value Theorem there exists \(\beta \in (x_{2},x_{1})\) so that \(f'(\beta )-L_{n}'(T,f,\beta )=0\). Thus, with \(\alpha \in (x_{1},1)\) as above, there exists \(\gamma \in (\beta ,\alpha )\) such that </p>
<div class="displaymath" id="a0000000032">
  \begin{align*}  \left|f(1)-L_{n}(T,f,1)\right| =& \big|(1-x_{1})\left(f’(\alpha )-L_{n}’(T,f,\alpha )\right)\big| \\ \hspace{-2.5em}=& \Big|(1-x_{1})\left[\left(f’(\alpha )-L_{n}’(T,f,\alpha )\right)-\left(f’(\beta )-L_{n}’(T,f,\beta )\right)\right]\Big| \\ \hspace{-2.5em}=& \big|(1-x_{1})(\alpha -\beta )\left(f”(\gamma )-L_{n}”(T,f,\gamma )\right)\big| \\ \hspace{-2.5em}\leq & 4\sin ^{2}\! \left(\tfrac {\pi }{4n}\right)\sin ^{2}\! \left(\tfrac {3\pi }{4n}\right)\| f”-L_{n}”(T,f)\| , \end{align*}
</div>
<p> and so, by&#160;(??), </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996E7D850>">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }n^{4}\left[f(1)-L_{n}(T,f,1)\right]=0. \label{partial_resultC} \end{equation}
  </div>
  <span class="equation_label">45</span>
</p>
</div>
<p> Similarly, </p>
<div class="displaymath" id="a0000000033">
  \begin{align*}  \left|f’(1)-L_{n}’(T,f,1)\right|&  =\big|\left(f’(1)-L_{n}’(T,f,1)\right)-\left(f’(\beta )-L_{n}’(T,f,\beta )\right)\big| \\ &  \leq 2\sin ^{2}\! \left(\tfrac {3\pi }{4n}\right)\| f”-L_{n}”(T,f)\| , \end{align*}
</div>
<p> so </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996E7F170>">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }n^{2}\left[f'(1)-L_{n}'(T,f,1)\right]=0. \label{partial_resultD} \end{equation}
  </div>
  <span class="equation_label">46</span>
</p>
</div>
<p> Now let \(r\) be the best uniform approximation to \(f''\) in \(\mathcal{P}_{n-4}\). By the method used above to derive&#160;(??) it follows that </p>
<div class="displaymath" id="a0000000034">
  \[ \tfrac {1}{n^{2}}|L_{n}'''(T,f,1)|\leq \| L_{n}''(T,f)-f''\| +\| f''-r\| +c\, \omega (\tfrac 1{n-4};f''), \]
</div>
<p> and so </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000012996E7FE90>">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }\tfrac {1}{n^{2}}L_{n}'''(T,f,1)=0. \label{partial_resultE} \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> Substituting (??)–(??) into&#160;(??) and using&#160;(??) then gives&#160;(??). The result&#160;(??) is established similarly. <div class="proof_wrapper" id="a0000000035">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p>We have now developed all the preliminary results needed to establish our theorem. </p>
<p><div class="proof_wrapper" id="a0000000036">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>[Proof of Theorem&#160;<a href="#newtheorem">2</a>] Firstly observe that a) is an immediate consequence of the representation&#160;(??) for \(Q_{2,n+2}(T_{a},f)\). </p>
<p>b) Suppose \(f\in C^{2}[-1,1]\) and \(x\in (-1,1)\). Then \(H_{2,n}(T,f)\) converges uniformly to \(f\), so by&#160;(<a href="#r(x)">24</a>), \(R_{2,n+2}(T_{a},f,x)\) converges (pointwise or uniformly) to \(f(x)\) if and only if </p>
<div class="displaymath" id="a0000000037">
  \[ \lim _{n\rightarrow \infty }(T_{n}(x))^{3}\left[(1+x)A_{1,n}-(1-x)B_{1,n}\right]=0, \]
</div>
<p> which (by&#160;(??) and&#160;(??)) is equivalent to </p>
<div class="equation" id="proof1">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }(T_{n}(x))^{3}\left[(1+x)f'(1)+(-1)^{n+1}(1-x)f'(-1)\right]=0. \label{proof1} \end{equation}
  </div>
  <span class="equation_label">48</span>
</p>
</div>
<p> Clearly, then, \(R_{2,n+2}(T_{a},f)\) converges uniformly to \(f\) if \(f'(1)=f'(-1)=0\). </p>
<p>On the other hand, since \(T_{n}(0)=0\) for \(n\) odd and \(|T_{n}(0)|=1\) for \(n\) even, (<a href="#proof1">48</a>)&#160;holds true at \(x=0\) if and only if \(f'(1)=f'(-1)\). Thus \(R_{2,n+2}(T_{a},f,0)\rightarrow f(0)\) if and only if \(f'(1)=f'(-1)\). </p>
<p>For fixed \(x\) with \(0{\lt}|x|{\lt}1\), it is well-known that there exists a subsequence \(\{ n_{r}\} _{r=1}^{\infty }\) of natural numbers so that \(|T_{n_{r}}(x)|\rightarrow 1\). Since \(|T_{n_{r}+1}(x)|\rightarrow |x|\), it follows that there is a subsequence \(\{ m_{r}\} _{r=1}^{\infty }\) of natural numbers that contains infinitely many odd values and infinitely many even values and satisfies \(|T_{m_{r}}(x)|\geq |x|/2\) for all \(r\). Thus&#160;(<a href="#proof1">48</a>) holds true if and only if \((1+x)f'(1)+(1-x)f'(-1)=0\) and \((1+x)f'(1)-(1-x)f'(-1)=0\), which is equivalent to \(f'(1)=f'(-1)=0\). </p>
<p>c) Suppose \(f\in C^{4}[-1,1]\) and \(x\in (-1,1)\). Again \(H_{2,n}(T,f)\) converges uniformly to \(f\), so by&#160;(<a href="#r(x)">24</a>) and&#160;(<a href="#h(x)">25</a>), </p>
<div class="displaymath" id="a0000000038">
  \begin{align*} & H_{2,n+2}(T_{a},f,x)=\\ & =f(x)+\tfrac {1}{16}(T_{n}(x))^{3}(1-x^{2})\big[(1+x)^{2}(3-x)A_{1,n} \\ & \quad -(1-x)^{2}(3+x)B_{1,n}+(1+x)^{2}(1-x)A_{2,n}-(1-x)^{2}(1+x)B_{2,n}\big]+o(1) \end{align*}
</div>
<p> where, here and subsequently, the \(o(1)\) term is uniform in \(x\). Then by (??)-(??), </p>
<div class="displaymath" id="proof2">
  \begin{align}  \label{proof2} & H_{2,n+2}(T_{a},f,x)=\\ \nonumber & =f(x)+\tfrac {1}{32}(T_{n}(x))^{3}(1-x^{2})\Big[7(1+x)^{2}(3-x)f’(1) \\ \nonumber & \quad \! -\! 7(\! -\! 1)^{n}(1\! -\! x)^{2}(3\! +\! x)f’(\! -\! 1)\! +\! (1+x)^{2}(1\! -\! x)\big[(27n^{2}\! +\! 11)f’(1)\! -\! 16f”(1)\big] \\ \nonumber & \quad -(-1)^{n}(1-x)^{2}(1+x)\big[(27n^{2}+11)f’(-1)+16f”(-1)\big]\Big]+o(1). \end{align}
</div>
<p> It is obvious from this expression that \(H_{2,n+2}(T_{a},f)\) converges uniformly to \(f\) if \(f'(1)=f'(-1)=f''(1)=f''(-1)=0\). </p>
<p>On the other hand, (<a href="#proof2">49</a>)&#160;shows that \(H_{2,n+2}(T_{a},f,0)\rightarrow f(0)\) if and only if </p>
<div class="displaymath" id="a0000000039">
  \[ \lim _{\shortstack {$\scriptstyle n\rightarrow \infty $\\ $\scriptstyle n\textup{\  even}$}}(27n^{2}+32)\big[f'(1)-f'(-1)\big]-16\big[f''(1)+f''(-1)\big]=0, \]
</div>
<p> which is equivalent to \(f'(1)=f'(-1)\) and \(f''(1)=-f''(-1)\). Furthermore, if \(f'(1)\neq f'(-1)\), (??)&#160;follows from&#160;(<a href="#proof2">49</a>). </p>
<p>For fixed \(x\) with \(0{\lt}|x|{\lt}1\), suppose \(H_{2,n+2}(T_{a},f,x)\rightarrow f(x)\). Then, with \(\{ m_{r}\} \) defined as in (b) above, it follows from&#160;(<a href="#proof2">49</a>) that </p>
<div class="displaymath" id="a0000000040">
  \begin{multline}  \lim _{r\rightarrow \infty }\Big[7(1+x)^{2}(3-x)f’(1)-7(-1)^{m_{r}}(1-x)^{2}(3+x)f’(-1) \\ +(1+x)^{2}(1-x)\big[(27m_{r}^{2}+11)f’(1)-16f”(1)\big] \\ -(-1)^{m_{r}}(1-x)^{2}(1+x)\big[(27m_{r}^{2}+11)f’(-1)+16f”(-1)\big]\Big]=0. \label{proof3} \end{multline}
</div>
<p> Because \(\{ m_{r}\} \) has infinitely many values of each parity, (<a href="#proof3">52</a>)&#160;implies that \((1+x)f'(1)+(1-x)f'(-1)=0\) and \((1+x)f'(1)-(1-x)f'(-1)=0\), which means \(f'(1)=f'(-1)=0\). A similar argument then shows that \(f''(1)=f''(-1)=0\). Finally, if \(f'(1)\) and \(f'(-1)\) are not both&#160;0, (<a href="#limsup">8</a>)&#160;is a consequence of&#160;(<a href="#proof2">49</a>) and the definition of \(m_{r}\). <div class="proof_wrapper" id="a0000000041">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="unremark_thmwrapper " id="a0000000042">
  <div class="unremark_thmheading">
    <span class="unremark_thmcaption">
    Remark
    </span>
  </div>
  <div class="unremark_thmcontent">
  <p>Theorem&#160;<a href="#Bojanic's_Theorem">1</a> for \((0,1)\) HFI does not place any conditions on the derivatives of \(f\), other than the existence of \(f'\) at \(\pm 1\). By contrast, our Theorem&#160;<a href="#newtheorem">2</a> for \((0,1,2)\) HFI imposes quite stringent conditions on the derivatives of \(f\). However, it seems difficult to fully extend Bojanic’s methods to \((0,1,2)\) HFI, essentially because the positivity of terms that occur when working with \((0,1)\) HFI no longer applies to the corresponding terms in \((0,1,2)\) HFI. Nevertheless, it would be of interest to know whether the conditions of our theorem can be weakened. </p>

  </div>
</div> </p>
<p><div class="acknowledgement_thmwrapper " id="a0000000043">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgements
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors thank Professor Terry Mills for helpful discussions during the preparation of this paper. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
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  <dd><p><i class="sc">D.L. Berman</i>, <i class="it">On the theory of interpolation</i>, Dokl. Akad. Nauk SSSR, <b class="bf">163</b> (1965), pp.&#160;551–554 (in Russian). [Soviet Math. Dokl., <b class="bf">6</b> (1965), pp.&#160;945–948] </p>
</dd>
  <dt><a name="Berman1969">2</a></dt>
  <dd><p><i class="sc">D.L. Berman</i>, <i class="it">An investigation of the Hermite–Fejér interpolation process</i>, Dokl. Akad. Nauk SSSR, <b class="bf">187</b> (1969), pp.&#160;241–244 (in Russian). [Soviet Math. Dokl., <b class="bf">10</b> (1969), pp.&#160;813–816] </p>
</dd>
  <dt><a name="Bojanic">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/bf01898143"><i class="sc">R. Bojanic</i>, <i class="it">Necessary and sufficient conditions for the convergence of the extended Hermite–Fejér interpolation process</i>, Acta Math. Acad. Sci. Hungar., <b class="bf">36</b> (1980), pp.&#160;271–279. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="ByrneMillsSmith">4</a></dt>
  <dd><p><a href ="https://doi.org/10.1017/s0004972700012223"> <i class="sc">G.J. Byrne, T.M. Mills</i> and <i class="sc">S.J. Smith</i>, <i class="it">On Hermite–Fejér type interpolation on the Chebyshev nodes</i>, Bull. Austral. Math. Soc., <b class="bf">47</b> (1993), pp.&#160;13–24. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="CookMills">5</a></dt>
  <dd><p><a href ="https://doi.org/10.1017/s0004972700024114"> <i class="sc">W.L. Cook</i> and <i class="sc">T.M. Mills</i>, <i class="it">On Berman’s phenomenon in interpolation theory</i>, Bull. Austral. Math. Soc., <b class="bf">12</b> (1975), pp.&#160;457–465. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Faber">6</a></dt>
  <dd><p><i class="sc">G. Faber</i>, <i class="it">Über die interpolatorische Darstellung stetiger Funktionen</i>, Jahresber. Deutsch. Math.-Verein., <b class="bf">23</b> (1914), pp.&#160;190–210. </p>
</dd>
  <dt><a name="Fejer">7</a></dt>
  <dd><p><i class="sc">L. Fejér</i>, <i class="it">Über interpolation</i>, Göttinger Nachrichten, (1916), pp.&#160;66–91. </p>
</dd>
  <dt><a name="KrylovSteuermann">8</a></dt>
  <dd><p><i class="sc">N. Krylov</i> and <i class="sc">E. Steuermann</i>, <i class="it">Sur quelques formules d’interpolation convergentes pour toute fonction continue</i>, Bull. Sci. Phys. Math. Acad. Sci. Ukraine, <b class="bf">1</b> (1922), <br />pp.&#160;13–16. </p>
</dd>
  <dt><a name="Maky">9</a></dt>
  <dd><p><i class="sc">M. Maky</i>, <i class="it">On an extended Hermite–Fejer interpolation process</i>, Journal of al-qadisiyah for pure science (quarterly) [Al-Qadisiyah University, Iraq], <b class="bf">13</b> (2008) no. 2, pp.&#160;1–8. </p>
</dd>
  <dt><a name="NeckermannRunck">10</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/bf02162908"> <i class="sc">L. Neckermann</i> and <i class="sc">P.O. Runck</i>, <i class="it">Über Approximationseigenschaften differenzierter Lagrangescher Interpolationspolynome mit Jacobischen Abszissen</i>, Numer. Math., <b class="bf">12</b> (1968), pp.&#160;159–169. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="RivlinApproxofFunctions">11</a></dt>
  <dd><p><i class="sc">T.J. Rivlin</i>, <i class="it">An introduction to the approximation of functions</i>, Dover, New York, 1981. </p>
</dd>
  <dt><a name="RivlinChebyshevPolynomials">12</a></dt>
  <dd><p><i class="sc">T.J. Rivlin</i>, <i class="it">Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory</i>, 2nd&#160;ed., Wiley, New York, 1990. </p>
</dd>
  <dt><a name="SzabadosVarma">13</a></dt>
  <dd><p><a href ="https://doi.org/10.1090/s0002-9939-1990-1013983-x"> <i class="sc">J. Szabados</i> and <i class="sc">A.K. Varma</i>, <i class="it">On \((0,1,2)\) interpolation in uniform metric</i>, Proc. Amer. Math. Soc., <b class="bf">109</b> (1990), pp.&#160;975–979. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="SzabadosVertesi">14</a></dt>
  <dd><p><i class="sc">J. Szabados</i> and <i class="sc">P. Vértesi</i>, <i class="it">Interpolation of Functions</i>, World Scientific, Singapore, 1990. </p>
</dd>
</dl>


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