MATHEMATICS

V. S. VIDENSKY

ON THE MUTUAL POSITION OF THE ZEROS OF SUCCESSIVE POLYNOMIALS LEAST DEVIATING FROM ZERO

(Presented by Academician S. N. Bernstein, 28 IV 1957)

Among the many remarkable properties possessed by the Chebyshev polynomials
(T_n(x)=\cos n\arccos x), the following is known: the zeros of two successive polynomials (T_n(x)) and (T_{n+1}(x)) are mutually separated; the points of maximum deviation of both polynomials are likewise mutually separated. This assertion is easily verified directly, since for the polynomial (T_n(x)) we know explicit expressions for its zeros (\left(x_{k,n}=\cos\frac{2k-1}{n}\pi,\ 1\leq k\leq n\right)) and for the points of its maximum deviation (\left(\xi_{k,n}=\cos\frac{k\pi}{n},\ 0\leq k\leq n\right)).

In the present note a generalization of the fact under consideration is given under sufficiently broad assumptions concerning the functions introduced below.

Let (f(x)) be a function continuous together with its first derivative on the interval ([a,b]); a zero (x_0) of the function (f(x)) will be called simple if (f'(x_0)\neq0); otherwise the zero (x_0) will be called double.

Theorem. Let on the interval ([a,b]) there be given two functions (t_n(x)) and (t_{n+1}(x)), continuous together with their first derivatives, possessing the following properties:

1) (t_n(x)) has (n) simple zeros in the interval ((a,b)):
[
x_1<x_2<\cdots<x_n;
]
(t_{n+1}(x)) has (n+1) simple zeros in the interval ((a,b)):
[
y_1<y_2<\cdots<y_{n+1};
]

2) every linear combination (\lambda t_n(x)+\mu t_{n+1}(x)) ((\lambda,\mu) real, (\lambda^2+\mu^2\neq0)) has (\leq n+1) zeros on the interval ([a,b]);

3) on the interval ([a,b])
[
|t_n(x)|\leq1,\qquad |t_{n+1}(x)|\leq1,
]
and, moreover, at (n+1) distinct points (\xi_1<\xi_2<\cdots<\xi_{n+1}) and at (n+2) distinct points (\eta_1<\eta_2<\cdots<\eta_{n+2}) the relations
[
t_n(\xi_k)=(-1)^k\qquad (k=1,\ldots,n+1),
\tag{1}
]
[
t_n(\eta_k)=(-1)^k\qquad (k=1,\ldots,n+2).
\tag{2}
]

Then the inequalities
[
a\leq\eta_1\leq\xi_1<\eta_2<\xi_2<\cdots<\eta_{n+1}<\xi_{n+1}\leq\eta_{n+2}\leq b,
\tag{3}
]
[
a<y_1<x_1<y_2<\cdots<x_n<y_{n+1}<b
\tag{4}
]
are valid.

Inequalities (3) are closely connected with S. N. Bernstein’s theorem ((1), p. 87) on the points of maximum deviation of the polynomials least deviating from zero, formed from Chebyshev systems. These inequalities (3) can be established by means of arguments close to those given in the monograph (1). We shall, however, give here a somewhat different proof*, with a view to using it in deriving inequalities (4).

First we shall show that inequalities (3) hold. To this end consider the function

[
\Phi(x)=t_{n+1}^{2}(x)-t_n^{2}(x).
\tag{5}
]

By assumption 2), the function (\Phi(x)) can vanish on ([a,b]) at most (2n+2) times. From equalities (2) it follows that in each interval ((\eta_k,\eta_{k+1})) there lies one zero of the function (t_{n+1}(x)), namely (y_k). Since from (5) it follows that, for all (k),

[
\Phi(\eta_k)\ge 0,\qquad \Phi(\xi_k)\le 0,
\tag{6}
]

then on the segment ([\eta_k,\eta_{k+1}]) we have

[
\Phi(\eta_k)\ge 0,\qquad \Phi(y_k)\le 0,\qquad \Phi(\eta_{k+1})\ge 0
\tag{7}
]

and, consequently, on this segment the function (\Phi(x)) has at least two zeros.

Taking further into account that if an interior point (\eta_k) of the interval ((a,b)) is a zero of the function (\Phi(x)), then (\eta_k) will be a double zero of this function, we conclude that the function (\Phi(x)) has (2n+2) zeros on the segment ([\eta_1,\eta_{n+2}]). It is proved analogously that on each segment ([\xi_k,\xi_{k+1}]) there lie (\ge 2) zeros and that on the segment ([\xi_1,\xi_{n+1}]) there lie not fewer than (2n) zeros of (\Phi(x)).

We now show that

[
\eta_1\le \xi_1,\qquad \xi_{n+1}\le \eta_{n+2}.
\tag{8}
]

This is obvious if (\eta_1=a,\ \eta_{n+2}=b). Let (\eta_1>a); suppose, contrary to (8), that (a\le \xi_1<\eta_1). Then from (6) it would follow that in the interval ([\xi_1,\eta_1)) there must necessarily lie a zero of (\Phi(x)), but this is impossible, since all (2n+2) zeros of this function lie on the segment ([\eta_1,\eta_{n+2}]). The second of the inequalities is proved analogously.

Let us verify that any segment ([\eta_k,\eta_{k+1}]) ((k=2,\ldots,n)) can contain (\le 1) points (\xi_j). If (\eta_k=\xi_j,\ \eta_{k+1}=\xi_{j+1}), then these points would be double zeros of (\Phi(x)), and since, moreover, (\Phi(x_i)\ge 0,\ \Phi(y_i)\le 0), there would lie (\ge 5) zeros on the segment ([\eta_k,\eta_{k+1}]), whence it follows that on the segment ([\eta_1,\eta_{n+2}]) there would have to be (\ge 2n+3) zeros of (\Phi(x)).

If (\eta_k=\xi_j<\xi_{j+1}<\eta_{k+1}), then the interval ([\eta_k,\eta_{k+1})) contains (\ge 4) zeros, whence it again follows that the total number of zeros is (\ge 2n+3). In exactly the same way, from the assumption (\eta_k<\xi_j<\xi_{j+1}<\eta_{k+1}) it follows that either inside the interval ((\eta_k,\eta_{k+1})) there are (\ge 4) zeros of (\Phi(x)), or on the segment ([\eta_k,\eta_{k+1}]) there are (\ge 5) zeros.

Further we have

[
\xi_1<\eta_2,\qquad \eta_{n+1}<\xi_{n+1}.
\tag{9}
]

[
\underline{\hspace{3cm}}
]

* A similar device was used by me in the note (2).

Indeed, if (\eta_2 \leqslant \xi_1), then, since the segment ([\eta_2,\eta_{n+1}]) can contain (\leqslant n-1) points (\xi_j), we would have (\eta_{n+1}<\xi_n<\xi_{n+1}\leqslant \eta_{n+2}), and in the interval ((\eta_{n+1},\eta_{n+2}]) there would lie (\geqslant 3) zeros of (\Phi(x)), which leads to a contradiction. In exactly the same way we obtain that (\eta_{n+1}<\xi_{n+1}).

Let us also establish the inequalities

[
\eta_2<\xi_2,\qquad \xi_n<\eta_{n+1}.
\tag{10}
]

Indeed, if (\eta_2>\xi_2), then in the interval ([\eta_1,\eta_2)) there would lie (\geqslant 3) zeros of (\Phi(x)), which is impossible. From the hypothesis that (\eta_2=\xi_2), it follows that (\eta_3<\xi_3<\eta_4), whence we conclude that either in the interval ([\eta_1,\eta_4]) there lie (\geqslant 7) zeros, or on the segment ([\eta_1,\eta_4]) there lie (\geqslant 8) zeros of (\Phi(x)), but each of these conclusions leads to a contradiction. Analogously it is proved that (\xi_n<\eta_{n+1}).

Thus we have shown, among other things, that each of the segments ([\eta_k,\eta_{k+1}]) ((k=1,\ldots,n+1)) contains one and only one point (\xi_j).

From (8), (9), and (10) we have (\eta_1\leqslant \xi_1<\eta_2<\xi_2). Suppose by induction that the inequalities

[
\eta_1\leqslant \xi_1<\eta_2<\xi_2<\cdots<\eta_r<\xi_r
]

have already been proved for some (r) ((2\leqslant r\leqslant n)). We shall show that then

[
\xi_r<\eta_{r+1}<\xi_{r+1}.
]

If (\xi_r\geqslant \eta_{r+1}), then on the segment ([\eta_{r+1},\eta_{n+2}]) there would lie (n+2-r) points (\xi_j); consequently, on at least one of the (n+1-r) segments ([\eta_{r+k},\eta_{r+k+1}]) ((k=1,\ldots,n+1-r)) there would fall 2 points (\xi_j), which cannot occur. In exactly the same way, if (\xi_{r+1}\leqslant \eta_{r+1}), then on the segment ([\eta_1,\eta_{r+1}]) there would fall (r+1) points (\xi_j), and, consequently, on one of the (r) segments ([\eta_k,\eta_{k+1}]) ((k=1,\ldots,r)) there would fall 2 points (\xi_j). Thus, the inequalities (3) are proved.

We pass to the proof of the inequalities (4). We first show that
(y_1<x_1,\ x_n<y_{n+1}).

Assume, to the contrary, that (x_1\leqslant y_1). Then, by virtue of the inequalities

[
\Phi(\eta_1)\geqslant 0,\qquad
\Phi(\xi_1)\leqslant 0,\qquad
\Phi(x_1)\geqslant 0,\qquad
\Phi(y_1)\leqslant 0,\qquad
\Phi(\eta_2)\geqslant 0,
]

either in the interval ([\eta_1,\eta_2)) there will lie (\geqslant 3) zeros of (\Phi(x)), or on the segment ([\eta_1,\eta_2]) there will be (\geqslant 4) zeros; this, as was shown in the derivation of the inequalities (3), leads to a contradiction. Further, from (3) and the obvious inequalities (\xi_k<x_k<\xi_{k+1},\ \eta_k<y_k<\eta_{k+1}), we directly obtain

[
x_{k-2}<\xi_{k-1}<\eta_k<y_k<\eta_{k+1}<\xi_{k+1}<x_{k+1}
\qquad (k=3,\ldots,n-1),
]

i.e. (x_{k-2}<y_k<x_{k+1}). Let us establish that

[
x_{k-1}<y_k<x_k.
]

It follows from (5) that (\Phi(y_k)\leqslant 0). If (\Phi(y_k)=0), then (y_k) is a double zero of the function (\Phi(x)), and hence (y_k) coincides with one of the two points (x_{k-1}) or (x_k). Suppose, for definiteness, that (y_k=x_k); then the inequalities

[
\eta_k<y_k=x_k<\xi_{k+1}<\eta_{k+1},
]

[
\Phi(\eta_k)\geqslant 0,\qquad
\Phi(y_k)=0,\qquad
\Phi(\xi_{k+1})\leqslant 0,\qquad
\Phi(\eta_{k+1})\geqslant 0,
]

from which it follows that on the interval ([x_k,\eta_{k+1}]\subset(\eta_k,\eta_{k+1}]) there must be (\geq 3) zeros if (\Phi(\eta_k)>0), or (\geq 4) zeros if (\Phi(\eta_k)=0). Similarly it is proved that (y_k\ne x_{k-1}). Thus, (\Phi(y_k)<0). If (x_k<y_k), then from the inequalities

[
\eta_k<\xi_k<x_k<y_k<\eta_{k+1},
]

[
\Phi(\eta_k)\geq 0,\qquad
\Phi(\xi_k)\leq 0,\qquad
\Phi(x_k)>0,\qquad
\Phi(y_k)<0,\qquad
\Phi(\eta_{k+1})\geq 0
]

it would follow that in the interval ((\eta_k,\eta_{k+1})) there must lie (\geq 4) zeros. In the same way we obtain the inequality (x_{k-1}