UDC 517.512.6

MATHEMATICS

D. L. BERMAN

ON EVERYWHERE DIVERGENT EXTENDED HERMITE–FEJÉR INTERPOLATION PROCESSES

(Presented by Academician L. V. Kantorovich on 29 XII 1969)

1°. Introduction.

In 1916 L. Fejér (¹) proved the following important theorem.

Let \(f(x)\) be an arbitrary continuous function on the segment \([-1,1]\), and let \(H_n(f,x)\) be the polynomial of degree \((2n-1)\), uniquely determined by the conditions

\[ H_n\bigl(f,x_k^{(n)}\bigr)=f\bigl(x_k^{(n)}\bigr),\qquad H'_n\bigl(f,x_k^{(n)}\bigr)=0,\qquad k=1,2,\ldots,n, \]

\[ x_k^{(n)}=\cos \frac{2k-1}{2n}\,\pi . \tag{1} \]

Then the relation

\[ H_n(f,x)\to f(x),\qquad n\to\infty \]

holds uniformly on \([-1,1]\).

Let us now consider the so-called extended Hermite–Fejér interpolation polynomial \(F_n(f,x)\) of degree \((2n+3)\), which is uniquely determined by the equalities

\[ F_n(f,1)=f(1),\qquad F_n(f,-1)=f(-1),\qquad F'_n(f,1)=0, \tag{2} \]

\[ F'_n(f,-1)=0,\qquad F_n\bigl(f,x_k^{(n)}\bigr)=f\bigl(x_k^{(n)}\bigr),\qquad F'_n\bigl(f,x_k^{(n)}\bigr)=0,\qquad k=1,2,\ldots,n, \]

where \(\{x_k^{(n)}\}_{k=1}^n\) are the Chebyshev nodes (1). In (², ³) it was proved that the process constructed even for such a simple function as \(|x|\) diverges at \(x=0\). Since the node matrix of the process \(\{F_n\}_{n=1}^{\infty}\) is obtained by extending the node matrix of the process \(\{H_n\}_{n=1}^{\infty}\) by adding the points \(\pm1\) as nodes, this result, in view of the aforementioned theorem of L. Fejér, is unexpected. We now weaken conditions (2); namely, we consider the polynomial \(A_n(f)\) of degree \((2n+2)\), which is uniquely determined by the equalities

\[ A_n(f,1)=f(1),\qquad A_n(f,-1)=f(-1),\qquad A'_n(f,1)=0, \]

\[ A_n\bigl(f,x_k^{(n)}\bigr)=f\bigl(x_k^{(n)}\bigr), \tag{3} \]

\[ A'_n\bigl(f,x_k^{(n)}\bigr)=0,\qquad k=1,2,\ldots,n, \]

where \(x_k^{(n)}\) are, as before, the nodes (1). The polynomials \(A_n\) are in a certain sense closer to the polynomials \(H_n\) than are the polynomials \(F_n\). Therefore the question of whether the process \(\{A_n(f)\}_{n=1}^{\infty}\) will converge uniformly for every function continuous on \([-1,1]\) is of definite interest. Recently R. B. Saxena (⁴), with the aid of results from (³), proved simply that the process \(\{A_n(f)\}_{n=1}^{\infty}\) diverges at \(x=0\) if \(f(x)=|x|\). Thus here the same situation occurs as in the case of the process \(\{F_n(f)\}_{n=1}^{\infty}\).

In connection with the above, the following questions arise:

1. Does there exist a function continuous on \([-1,1]\) for which the process \(\{A_n(f,x)\}_{n=1}^{\infty}\) diverges at all points of \((-1,1)\)?

2. Does there exist a function continuous on \([-1,1]\) for which the process \(\{F_n(f,x)\}_{n=1}^{\infty}\) diverges at all points of \((-1,1)\)?

In this note only the first question is considered. The second question is resolved in an analogous way.

2°. Formulation and proof of the theorem.

Theorem 1. The interpolation process \(\{A_n(f,x)\}_{n=1}^{\infty}\), constructed at the nodes (1) for \(f(x)=1-x^2\), diverges at all points of the interval \((-1,1)\).

Proof. We need the following

Lemma. For any \(\theta\in[0,\pi/2]\) one can find a sequence of natural numbers \(n_1<n_2<\cdots,\ n_k\to\infty,\ k\to\infty\), such that the equality
\[ \lim_{k\to\infty}\sin^2 n_k\theta=0 \]
holds.

The polynomial \(A_n(f,x)\), uniquely determined by the conditions (2), has the form
\[ \begin{aligned} A_n(f,x)=&\left[\left(\frac{1-x}{2}\right)^2 f(-1) +\frac{1+x}{2}\left\{1+\frac{4n^2+1}{2}(1-x)\right\}f(1)\right]T_n^2(x)+\\ &+\sum_{\nu=1}^{n} f(x_\nu^{(n)}) \frac{(1-x^2)(1-x)}{(1-x_\nu^2)(1-x_\nu)} \frac{(1-x_\nu^2)+(1+2x_\nu)(x-x_\nu)} {n^2(x-x_\nu)^2}\,T_n^2(x), \end{aligned} \]
\[ T_n(x)=\cos n\arccos x. \tag{4} \]

Therefore, for \(f(x)=1-x^2\) we have:
\[ A_n(f,x)= \frac{(1-x^2)(1-x)T_n^2(x)}{n^2} \left[ \sum_{\nu=1}^{n}\frac{1+x_\nu}{(x-x_\nu)^2} + \sum_{\nu=1}^{n}\frac{1+2x_\nu}{(1-x_\nu)(x-x_\nu)} \right]. \tag{5} \]

Let us note that
\[ \frac{1+2x_\nu}{(1-x_\nu)(x-x_\nu)} = \frac{1+2x}{(1-x)(x-x_\nu)} + \frac{3}{(x-1)(1-x_\nu)}. \]

Consequently, (5) can be written in the form
\[ \begin{aligned} A_n(f,x)=\frac{(1-x^2)(1-x)T_n^2(x)}{n^2} \Bigg[& \sum_{\nu=1}^{n}\frac{1}{(x-x_\nu)^2} +\sum_{\nu=1}^{n}\frac{x}{(x-x_\nu)^2} -\sum_{\nu=1}^{n}\frac{1}{x-x_\nu} \\ &+\frac{1+2x}{1-x}\sum_{\nu=1}^{n}\frac{1}{x-x_\nu} +\frac{3}{x-1}\sum_{\nu=1}^{n}\frac{1}{1-x_\nu} \Bigg]. \end{aligned} \tag{6} \]

It is known that
\[ \frac{T'_n(x)}{T_n(x)} = \sum_{\nu=1}^{n}\frac{1}{x-x_\nu}, \qquad \frac{(1-x^2)T_n^2(x)}{n^2} \sum_{\nu=1}^{n}\frac{1}{(x-x_\nu)^2} = 1-\frac{\sin 2n\theta\cos\theta}{2n\sin\theta}, \quad x=\cos\theta, \]
\[ \tag{7} \sum_{\nu=1}^{n}\frac{1}{1-x_\nu}=n^2. \]

Therefore from (6) we derive
\[ A_n(f,x) = (1-x^2)+\frac{\sin 2n\theta\sin 2\theta}{2n} -3\sin^2\theta\cos^2 n\theta, \qquad x=\cos\theta. \tag{8} \]

Consider first the case when \(0\le x<1\). Suppose that, for some \(x\) in \([0,1)\), convergence of the process \(\{A_n(f,x)\}_{n=1}^{\infty}\) takes place. Then, by (8), we have
\[ \sin 2n\theta\,\sin 2\theta/2n - 3\sin^2\theta\cos^2 n\theta \to 0,\qquad n\to\infty. \tag{9} \]

This equality is equivalent to the equality
\[ \lim_{n\to\infty}\cos^2 n\theta=0, \qquad \theta\in(0,\pi/2], \]
which contradicts the lemma. Consequently, in \([0,1)\) the process diverges.

Consider now the interval \((-1,0)\). If at some point \(\tilde x \in (-1,0)\) the process \(\{A_n\}_{n=1}^{\infty}\) converged, then, according to the preceding, \(\lim_{n\to\infty}\cos^2 n\bar\theta=0\) (9), where \(\tilde x=\cos\bar\theta\). Put \(\theta=\pi-\bar\theta\), \(\pi/2<\bar\theta<\pi\). Thus \(0<\theta<\pi/2\). Since

\[ \cos^2 n\bar\theta=\cos^2 n\theta, \]

it follows from (9) that \(\lim_{n\to\infty}\cos n\theta=0\). This again contradicts the lemma. Thus the process \(\{A_n(f)\}_{n=1}^{\infty}\) also diverges on the interval \((-1,0)\). The theorem is proved.

Along with the polynomial \(A_n(f,x)\), consider the polynomial \(B_n(f,x)\), uniquely determined by the conditions

\[ B_n(f,1)=f(1), \qquad B_n(f,-1)=f(-1), \qquad B'_n(f,-1)=0, \]

\[ B_n(f,x_k^{(n)})=f(x_k^{(n)}), \qquad B'_n(f,x_k^{(n)})=0, \qquad k=1,2,\ldots,n. \]

By an almost verbatim repetition of the arguments in the proof of Theorem 1, we obtain Theorem 2.

Theorem 2. The interpolation process \(\{B_n(f,x)\}_{n=1}^{\infty}\), constructed at the Chebyshev nodes (1) for \(f(x)=1-x^2\), diverges at all points of the interval \((-1,1)\).

Corollary. The interpolation process \(\{A_n(f)\}_{n=1}^{\infty}\), constructed at the Chebyshev nodes (1) for \(f(x)=x^2\), diverges at all points of the interval \((-1,1)\).

Indeed, for \(f\equiv 1\), \(A_n(f)\equiv 1\), hence

\[ A_n(1-z^2,x)=1-A_n(z^2,x). \tag{10} \]

If, for \(x\in(-1,1)\), \(A_n(z^2,x)\to x^2\), \(n\to\infty\), then, according to (10), \(A_n(1-z^2,x)\to 1-x^2\), \(n\to\infty\), which contradicts Theorem 1.

Theorem 3. The interpolation process \(\{A_n(f)\}_{n=1}^{\infty}\), constructed at the Chebyshev nodes (1) for \(f(x)=x\), diverges at all points of the interval \((-1,1)\).

Proof. Since \(A_n(1,x)\equiv 1\), according to (4),

\[ 1-A_n(z,x)=T_n^2(x)\left[\frac{(x-1)^2}{2}+\frac{(1-x)(1-x^2)}{n^2} \sum_{\nu=1}^{n} \frac{(1-x_\nu^2)+(1+2x_\nu)(x-x_\nu)} {(1-x_\nu^2)(x-x_\nu)^2} \right]. \]

Hence, after simple transformations, we obtain

\[ \begin{aligned} 1-A_n(z,x)=T_n^2(x)\Bigg[& \frac{(x-1)^2}{2}+\frac{(1-x)(1-x^2)}{n^2} \Bigg( \frac{3}{2(x-1)}\sum_{\nu=1}^{n}\frac{1}{1-x_\nu} \\ &-\sum_{\nu=1}^{n}\frac{1}{2(x+1)(1+x_\nu)} +\frac{2x+1}{1-x^2}\sum_{\nu=1}^{n}\frac{1}{x-x_\nu} +\sum_{\nu=1}^{n}\frac{1}{(x-x_\nu)^2} \Bigg)\Bigg]. \end{aligned} \tag{11} \]

Expression (11), after application of identity (7) and the identity

\[ \sum_{\nu=1}^{n} 1/(1+x_\nu)=n^2 \]

takes the form

\[ \begin{aligned} 1-A_n(z,x)={}&(1-x) -\frac{(1-x)\sin 2n\theta\cos\theta}{2n\sin\theta} +\frac{3(x^2+1)\cos^2 n\theta}{2} \\ &+\frac{(2x+1)(1-x)\cos n\theta\sin n\theta}{n\sin\theta}, \qquad x=\cos\theta. \end{aligned} \tag{12} \]

If \(A_n(z,x)\to x\), \(n\to\infty\), then from (12) it would follow that \(\lim_{n\to\infty}\cos^2 n\theta=0\), and this contradicts the lemma.

Remark. In connection with Theorems 1 and 3 it is curious that if the process \(\{A_n(f)\}_{n=1}^{\infty}\) is constructed for \(f(x)=(1-x^2)(1-x)\), then it converges uniformly. Moreover,

\[ |A_n(f)-f|\leq 12/n,\qquad |x|\leq 1,\qquad f(x)=(1-x^2)(1-x). \tag{13} \]

Indeed, in the present case

\[ A_n(f,x)= \frac{(1-x)(1-x^2)T_n^2(x)}{n^2} \sum_{\nu=1}^{n} \left( \frac{1-x^2}{(x-x_\nu)^2} -\frac{4x+1}{x-x_\nu} -3 \right). \]

Hence, with the aid of the identities (7), we obtain

\[ A_n(f,x)=(1-x)(1-x^2) -\frac{(1-x)\sin 2n\theta\sin 2\theta}{4n} -\frac{(4x+1)(1-x)\sin 2n\theta\sin\theta}{2n} -\frac{3(1-x)\cos^2 n\theta\sin^2\theta}{n}. \]

Thus, (13) holds.

Leningrad Higher Engineering Naval School
named after Admiral S. O. Makarov

Received
26 XII 1969

CITED LITERATURE

\(^{1}\) L. Fejér, Gött. Nachr., 66 (1916).
\(^{2}\) D. L. Berman, DAN, 163, No. 3 (1965).
\(^{3}\) D. L. Berman, Izv. higher educational institutions, mathematics, No. 1 (1967).
\(^{4}\) R. B. Saxena, Rend. Semin. Mat. Univ. e Politech. Torino, 27, 223 (1967—1968).