On Berman’s phenomenon
for (0,1,2) Hermite–Fejér interpolation

Graeme J. Byrne$^\ast $ Simon J. Smith$^\bullet $

October 7, 2018. Accepted: February 4, 2019. Published online: October 10, 2019.

$^\ast $Department of Mathematics and Statistics, La Trobe University, P. O. Box 199, Bendigo, VIC 3552, Australia, e-mail: g.byrne@latrobe.edu.au.

$^\bullet $54 Broad Parade, Bendigo, VIC 3550, Australia, e-mail: sjsmith3550@gmail.com. The work of this author has been supported by La Trobe University.

Given $f\in C[-1,1]$ and $n$ points (nodes) in $[-1,1]$, the Hermite–Fejér interpolation (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 Berman’s phenomenon. 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.

MSC. Primary 41A05; Secondary 41A10

Keywords. interpolation, polynomial interpolation, Hermite–Fejér interpolation, Chebyshev nodes, Berman’s phenomenon

1 Introduction

Suppose $f\in C[-1,1]$ and let

\begin{equation} X=\{ x_{k,n}:k=0,1,2,\ldots ,n-1;\ n=1,2,3,\ldots \} \label{X1} \end{equation}

be an infinite triangular matrix of nodes such that, for all $n$,

\begin{equation} 1\geq x_{0,n}>x_{1,n}>\ldots >x_{n-1,n}\geq -1.\label{X2} \end{equation}

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

\[ L_{n}(X,f,x_{k,n})=f(x_{k,n}),\qquad 0\leq k\leq n-1. \]

A classic result due to Faber [ 6 ] 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

\[ T=\left\{ x_{k,n}=\cos \big(\tfrac {2k+1}{2n}\pi \big):k=0,1,2,\ldots ,n-1;\ n=1,2,3,\ldots \right\} \]

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

\[ \omega (\delta ;f)=\max \big\{ |f(s)-f(t)|:-1\leq s,t\leq 1, |s-t|\leq \delta \big\} , \]

then $L_{n}(T,f)$ converges uniformly to $f$ under the quite mild restriction

\[ \omega (\tfrac 1n;f)\log n\rightarrow 0,\ \mbox{as}\ n\rightarrow \infty \]

(see Rivlin [ 11 , Chapter 4 ] for details and references).

A generalization of Lagrange interpolation is provided by Hermite–Fejér interpolation (HFI). Given a non-negative integer $m$ and nodes $X$ defined by (1) and (2), 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

\[ \left\{ \begin{array}{ll} H_{m,n}(X,f,x_{k,n})=f(x_{k,n}), & 0\leq k\leq n-1, \\[0.25 cm] H^{(r)}_{m,n}(X,f,x_{k,n})=0, & 1\leq r\leq m,\ 0\leq k\leq n-1. \end{array} \right. \]

Note that $H_{0,n}(X,f)=L_{n}(X,f)$. The original motivation for studying HFI was provided by Fejér [ 7 ] , 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.

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 Berman’s phenomenon, 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

\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}

Thus, for each $n$, the $x_{k,n+2}$ are the zeros of $(1-x^{2})T_{n}(x)$.

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 [ 1 ] was able to show that if $f(x)=|x|$, then $H_{1,n}(T_{a},f,0)$ diverges, while later [ 2 ] 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.

Theorem 1 Bojanic

[ 3 ] . 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 $f_{L}'(1)=f_{R}'(-1)=0$.

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 not occur. This follows from the representation

\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*}

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 (??). Thus $L_{n}(T_{a},f)\rightarrow f$ uniformly on $[-1,1]$ whenever $L_{n}(T,f)\rightarrow f$ uniformly on $[-1,1]$.

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 [ 5 ] 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 [ 5 ] that the term Berman’s phenomenon was first used.) The result of Cook and Mills was later extended by Maky [ 9 ] 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 [ 8 ] that $H_{3,n}(T,f)$ converges uniformly to $f$ on $[-1,1]$ for any $f\in C[-1,1]$.

In this paper our focus will be on $(0,1,2)$ HFI. Here it was shown by Szabados and Varma [ 13 ] 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. [ 4 , Theorem 1 ] that if $\omega (1/n;f)\log n\rightarrow 0$ as $n\rightarrow \infty $, then $H_{2,n}(T,f)$ converges uniformly to $f$.

To investigate $(0,1,2)$ HFI on $T_{a}$ it proves helpful to follow the approach of Bojanic [ 3 ] , 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 (??) 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

\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}

and define the polynomial $R_{2,n+2}(T_{a},f)$ of degree at most $3n+3$ by the $3n+4$ conditions

\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}

Also, for future reference, recall that $H_{2,n+2}(T_{a},f)$ is defined by the $3n+6$ conditions

\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}

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.

Theorem 2

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 (??)–(??), where the interpolation nodes $x_{k}=x_{k,n+2}$ are defined by (??).

If $f\in C[-1,1]$ and ${\displaystyle \lim _{n\rightarrow \infty }\| H_{2,n}(T,f)-f\| =0}$, then

\[ \lim _{n\rightarrow \infty }\| Q_{2,n+2}(T_{a},f)-f\| =0. \]
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 $0$, 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 $f'(1)=f'(-1)$.
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

\begin{equation} f'(1)=f'(-1)=f''(1)=f''(-1)=0.\label{Theorem_conditions} \end{equation}
7

If $0{\lt}|x|{\lt}1$ and (??) 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 0, then

\begin{equation} \limsup _{n\rightarrow \infty }\tfrac {1}{n^{2}}|H_{2,n+2}(T_{a},f,x)|>0. \label{limsup} \end{equation}
8

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

\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}
9

Theorem 2 confirms that Berman’s phenomenon occurs for (0,1,2) HFI. In particular, the following results hold.

Corollary 3

$(0,1,2)$ HFI provides the following illustrations of Berman’s phenomenon.

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 ${\displaystyle \lim _{n\rightarrow \infty }H_{2,n+2}(T_{a},f,0)=0}$.
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 $(-1,1)$.

The theorem will be proved via a sequence of lemmas in the next section.

2 Proof of the main result (Theorem 2)

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

\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}

\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}

and

\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}

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 (??)) of the polynomial on the left-hand side. Also note (Byrne et al. [ 4 , Section 1 ] ) that $H_{2,n}(T,f)$ has the explicit formula

\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}

where, with $x_{k}=x_{k,n+2}$ defined by (??),

\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}

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 (??).

Lemma 4

For $x_{k}=x_{k,n+2}$ defined by (??) and $f\in C[-1,1]$, the following summation formulas hold.

\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}

\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}

\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}

\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}

Proof â–¼

The formula (14) for Lagrange interpolation on the Chebyshev nodes is well-known (see, for example, Rivlin [ 12 , Section 1.3 ] ). The remaining formulas (15)–(17) follow by successive differentiation of (14).

Lemma 5

For $x_{k}=x_{k,n+2}$ defined by (??), the following summation formulas hold.

\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}

Proof â–¼

If $f(x)\equiv 1$, then $L_{n}(T,f,x)\equiv 1$, so (14) gives

\[ \sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{x-x_{k}}=\tfrac {n}{T_{n}(x)}. \]

Putting $x=1$ gives (18). The formulas (19)–(21) follow from (15)–(17) 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 [ 12 , p. 38 ] )

\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}

Finally, (22) is a straightforward consequence of the symmetry of the Chebyshev nodes (i.e. $x_{k,n+2}=-x_{n-k+1,n+2}$ for $1\leq k\leq n$).

Proof â–¼

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 (??).

Lemma 6

For $x_{k}=x_{k,n+2}$ defined by (??) and $f\in C[-1,1]$,

\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}

and

\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}

where

\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}

Proof â–¼

For convenience, denote the RHS of (24) by $r(x)$. To establish (24) it is sufficient to show $r(x)$ satisfies the $3n+4$ defining conditions (??) 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 (??) that (upon using $T_{n}'(1)=n^{2}$)

\[ r'(1)=H_{2,n}'(T,f,1)+3n^{2}\left(f(1)-H_{2,n}(T,f,1)\right)-A_{1,n}. \]

However, by (12) and (13),

\[ H_{2,n}'(T,f,1)=3n^{2}H_{2,n}(T,f,1)+\tfrac {1}{n^{3}}S_{n}'(f,1), \]

where

\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}

Thus

\[ r'(1)=3n^{2}f(1)+\tfrac {1}{n^{3}}S_{n}'(f,1)-A_{1,n}, \]

and this is zero by (19), (20) and (26). The result $r'(-1)=0$ is established by similar means.

Again for convenience, denote the RHS of (25) by $h(x)$. By the same arguments as above, it is evident that (25) will be proved if it can be shown that $h''(1)=h''(-1)=0$. To show $h''(1)=0$, begin by noting that from (24) and (25),

\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}

However, by (12) and (13),

\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}

where

\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}

By substituting (26), (28), (30), (32) and (33) into (31), and employing summation formulas (19)–(21), it follows (after somewhat tedious calculations) that $h''(1)=0$. The result $h''(-1)=0$ is proved in similar fashion.

Proof â–¼

The next result characterizes the quantities $A_{i,n}$ and $B_{i,n}$ of Lemma 6 in terms of values of Lagrange interpolation polynomials and their derivatives.

Lemma 7

For $i=1,2$, the quantities $A_{i,n}$ and $B_{i,n}$ that are defined by (26)–(29) can be written as

\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}

\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}

\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}

\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}

Proof â–¼

To obtain (??), begin by writing (26) as

\[ A_{1,n}=\tfrac {3}{n^{3}}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{3}}g(x_{k}) +\tfrac {n^{2}-1}{2n^{3}}\sum _{k=1}^{n}\tfrac {(-1)^{k-1}\sqrt{1-x_{k}^{2}}}{(1-x_{k})^{2}}g(x_{k}), \]

where $g(x)=f(1)-f(x)$. Now apply the identities (15) and (16) at $x=1$, noting that $L_{n}(T,g,x)=f(1)-L_{n}(T,f,x)$ and using the result (??) for $T_{n}^{(r)}(1)$. To obtain (??), apply (15) and (16) at $x=-1$ to (27) with $g(x)=f(-1)-f(x)$, and use $T_{n}^{(r)}(-1)=(-1)^{n+r}T_{n}^{(r)}(1)$. The remaining expressions (??) and (??) are proved in a similar fashion.

Proof â–¼

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 [ 10 , Satz 2, p. 168 ] :

\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}

Lemma 8

Suppose $A_{i,n}$ and $B_{i,n}$ are defined by (26)–(29). If $f\in C^{2}[-1,1]$, then

\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}

If $f\in C^{4}[-1,1]$, then

\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}

Proof â–¼

Suppose $f\in C^{2}[-1,1]$. We work with the expression (??) 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

\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*}

Thus, by (??),

\begin{equation} \lim _{n\rightarrow \infty }n^{2}\left[f(1)-L_{n}(T,f,1)\right]=0. \label{partial_resultA} \end{equation}

Also, if $\mathcal{P}_{n}$ denotes the set of polynomials of degree at most $n$, let $q$ be the best uniform approximation to $f'$ in $\mathcal{P}_{n-3}$. Then, by Markov’s inequality,

\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*}

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 [ 14 , p. 284 ] ), so

\[ \tfrac {1}{n^{2}}|L_{n}''(T,f,1)|\leq \| L_{n}'(T,f)-f'\| +\| f'-q\| +c\, \omega (\tfrac 1{n-3};f'). \]

Since $f\in C^{2}[-1,1]$, it follows from (??) and the Weierstrass approximation theorem that

\begin{equation} \lim _{n\rightarrow \infty }\tfrac {1}{n^{2}}L_{n}''(T,f,1)=0. \label{partial_resultB} \end{equation}

Substituting (??) and (??) into (??) and using (??) then gives the result (??). The result (??) is established from (??) in near-identical fashion.

Now suppose $f\in C^{4}[-1,1]$. Here we work with the expression (??) 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

\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*}

and so, by (??),

\begin{equation} \lim _{n\rightarrow \infty }n^{4}\left[f(1)-L_{n}(T,f,1)\right]=0. \label{partial_resultC} \end{equation}

Similarly,

\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*}

\begin{equation} \lim _{n\rightarrow \infty }n^{2}\left[f'(1)-L_{n}'(T,f,1)\right]=0. \label{partial_resultD} \end{equation}

Now let $r$ be the best uniform approximation to $f''$ in $\mathcal{P}_{n-4}$. By the method used above to derive (??) it follows that

\[ \tfrac {1}{n^{2}}|L_{n}'''(T,f,1)|\leq \| L_{n}''(T,f)-f''\| +\| f''-r\| +c\, \omega (\tfrac 1{n-4};f''), \]

and so

\begin{equation} \lim _{n\rightarrow \infty }\tfrac {1}{n^{2}}L_{n}'''(T,f,1)=0. \label{partial_resultE} \end{equation}

Substituting (??)–(??) into (??) and using (??) then gives (??). The result (??) is established similarly.

Proof â–¼

We have now developed all the preliminary results needed to establish our theorem.

Proof â–¼

[Proof of Theorem 2] Firstly observe that a) is an immediate consequence of the representation (??) for $Q_{2,n+2}(T_{a},f)$.

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 (24), $R_{2,n+2}(T_{a},f,x)$ converges (pointwise or uniformly) to $f(x)$ if and only if

\[ \lim _{n\rightarrow \infty }(T_{n}(x))^{3}\left[(1+x)A_{1,n}-(1-x)B_{1,n}\right]=0, \]

which (by (??) and (??)) is equivalent to

\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}

Clearly, then, $R_{2,n+2}(T_{a},f)$ converges uniformly to $f$ if $f'(1)=f'(-1)=0$.

On the other hand, since $T_{n}(0)=0$ for $n$ odd and $|T_{n}(0)|=1$ for $n$ even, (48) 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 $f'(1)=f'(-1)$.

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 (48) 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 $f'(1)=f'(-1)=0$.

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 (24) and (25),

\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*}

where, here and subsequently, the $o(1)$ term is uniform in $x$. Then by (??)-(??),

\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}

It is obvious from this expression that $H_{2,n+2}(T_{a},f)$ converges uniformly to $f$ if $f'(1)=f'(-1)=f''(1)=f''(-1)=0$.

On the other hand, (49) shows that $H_{2,n+2}(T_{a},f,0)\rightarrow f(0)$ if and only if

\[ \lim _{\shortstack {$\scriptstyle n\rightarrow \infty $\\ $\scriptstyle n\textup{\ even}$}}(27n^{2}+32)\big[f'(1)-f'(-1)\big]-16\big[f''(1)+f''(-1)\big]=0, \]

which is equivalent to $f'(1)=f'(-1)$ and $f''(1)=-f''(-1)$. Furthermore, if $f'(1)\neq f'(-1)$, (??) follows from (49).

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 (49) that

\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}

Because $\{ m_{r}\} $ has infinitely many values of each parity, (52) 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 0, (8) is a consequence of (49) and the definition of $m_{r}$.

Proof â–¼

Remark

Theorem 1 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 2 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.

Acknowledgements

The authors thank Professor Terry Mills for helpful discussions during the preparation of this paper.

Bibliography

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On Berman’s phenomenon for (0,1,2) Hermite–Fejér interpolation

1 Introduction

2 Proof of the main result (Theorem 2)

Bibliography

On Berman’s phenomenon
for (0,1,2) Hermite–Fejér interpolation