Abstract
Full Text
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}) 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}^{\infty}) is obtained by extending the node matrix of the process ({H_n}_{n=1}^{\infty
[
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)}).}^{\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:
-
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))?
-
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,}^{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\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).