MATHEMATICS

V. S. VIDENSKII

EXTREMAL ESTIMATES OF THE DERIVATIVE OF A TRIGONOMETRIC POLYNOMIAL ON AN INTERVAL SHORTER THAN THE PERIOD

(Presented by Academician S. N. Bernstein on 29 VIII 1959)

Set

\[ t_n(\theta)=-\cos 2n\arccos \frac{\sin(\theta/2)}{\sin(\omega/2)}, \qquad u_n(\theta)=\sin 2n\arccos \frac{\sin(\theta/2)}{\sin(\omega/2)}; \]

it is easy to verify that \(t_n(\theta)\) is a trigonometric polynomial of order \(n\) for \(-\omega \leqslant \theta \leqslant \omega,\ 0<\omega<\pi\).

Theorem. If a trigonometric polynomial \(s_n(\theta)\) of order \(n\) satisfies the inequality

\[ |s_n(\theta)|\leqslant 1,\qquad -\omega\leqslant \theta\leqslant \omega,\qquad 0<\omega<\pi, \tag{1} \]

then

\[ |s'_n(\theta)|\leqslant |t'_n(\theta)+iu'_n(\theta)| =n\cos(\theta/2)\,[\sin^2(\omega/2)-\sin^2(\theta/2)]^{-1/2}, \]

\[ -\omega<\theta<\omega, \tag{2} \]

and for \(n>[3\tg^2(\omega/2)+1]^{1/2}/2\)

\[ |s'_n(\theta)|\leqslant t'_n(\omega)=2n^2\ctg(\omega/2),\qquad -\omega\leqslant \theta\leqslant \omega. \tag{3} \]

In (2) equality is attained only for polynomials \(s_n(\theta)=\gamma t_n(\theta)\), \(|\gamma|=1\), at the \(2n\) points \(\theta_\nu\) that are zeros of \(t_n(\theta)\) on \([-\omega,\omega]\); in (3), for the same polynomials, but only at the points \(\theta=\pm\omega\).

Inequalities (2) and (3) are analogous to the classical inequalities of S. N. Bernstein and A. A. Markov for algebraic polynomials. Inequality (2) refines a known result of I. I. Privalov \((^1)\), who proved that (1) implies the inequality \(|s'_n(\theta)|\leqslant C_{\omega,\varepsilon}n\) for \(-\omega+\varepsilon\leqslant\theta\leqslant\omega-\varepsilon,\ 0<\varepsilon<\omega/2\), where \(C_{\omega,\varepsilon}\) is a constant depending on \(\omega\) and \(\varepsilon\). Inequality (3) correspondingly refines Jackson’s estimate \((^2)\): \(|s'_n(\theta)|\leqslant C_\omega n^2,\ -\omega\leqslant\theta\leqslant\omega\), where \(C_\omega\) is a constant depending on \(\omega\).

After inequality (2) had been established, it could have been derived from a general theorem of N. I. Akhiezer and B. Ya. Levin \((^3)\) by investigating the corresponding conformal mapping; but since that theorem itself is proved rather intricately, another path is chosen here, connected with the author’s previous works and admitting further generalizations. Note that if \(|s_n(\theta)|\leqslant 1\) for \(-\pi\leqslant\theta\leqslant\pi\), then passage to the limit \(\omega\to\pi\) in the right-hand side of (2) gives the classical S. N. Bernstein inequality \(|s'_n(\theta)|\leqslant n,\ -\pi\leqslant\theta\leqslant\pi\). In all subsequent arguments we shall assume the coefficients of the polynomial \(s_n(\theta)\) to be real, since passing to complex coefficients presents no difficulty (see \((^4)\), p. 45).

Lemma. If \(s_n(\theta)\) is a trigonometric polynomial satisfying the inequality \(|s_n(\theta)|<1,\ -\omega\leqslant\theta\leqslant\omega\), then for all real \(\beta\) the function

\[ F(\theta;\beta)=\cos\beta\,t_n(\theta)+\sin\beta\,u_n(\theta)-s_n(\theta) \]

has only simple zeros on \([-\omega,\omega]\).

Make the substitution \(x\tg(\omega/2)=\tg(\theta/2)\) and, in addition, put \(a=\tg(\omega/2)\), \(\alpha=\sqrt{1+a^2}\). Then

\[ s_n(\theta)=(1+a^2x^2)^{-n}P_{2n}(x), \]

\[ F(\theta;\beta)=(1+a^2x^2)^{-n}\left[\cos\beta\,M_{2n}(x)+\sin\beta\sqrt{1-x^2}\,N_{2n-1}(x)-P_{2n}(x)\right], \]

where \(P_{2n}(x)\) is an algebraic polynomial of degree \(2n\), while \(M_{2n}(x)\) and \(N_{2n-1}(x)\) are algebraic polynomials of degrees \(2n\) and \(2n-1\), respectively, which are determined by the formulas

\[ M_{2n}(x)=\Re\left(\alpha x+i\sqrt{1-x^2}\right)^{2n},\qquad \sqrt{1-x^2}\,N_{2n-1}(x)=\Im\left(\alpha x+i\sqrt{1-x^2}\right)^{2n}. \]

All the zeros of \(M_{2n}(x)\) and \(N_{2n-1}(x)\) lie in the interval \((-1,1)\) and mutually interlace (see, for example, \((^5)\)). From (1) there follows the inequality

\[ |P_{2n}(x)|\leqslant (1+a^2x^2)^n =\left|M_{2n}(x)+i\sqrt{1-x^2}\,N_{2n-1}(x)\right|,\qquad -1\leqslant x\leqslant 1. \tag{4} \]

But, as can be shown by following the method of my note \((^6)\), it follows from (4) that \(\cos\beta\,M_{2n}(x)+\sin\beta\sqrt{1-x^2}N_{2n-1}(x)-P_{2n}(x)\) cannot have multiple zeros on the interval \(-1\leqslant x\leqslant 1\). Hence we conclude that the function \(F(\theta;\beta)\) has only simple zeros on the interval \(-\omega\leqslant\theta\leqslant\omega\), since \(dx/d\theta=(1+a^2x^2)/2a\) does not vanish. The lemma is proved.

Suppose that at some point \(\theta_0\), \(-\omega<\theta_0<\omega\), inequality (2) is violated, i.e.

\[ s'_n(\theta_0)\geqslant |t'_n(\theta_0)+iu'_n(\theta_0)|\qquad (s_n(\theta)\ne \pm t_n(\theta)). \tag{5} \]

Choose the parameter \(\beta\) occurring in the function \(F(\theta;\beta)\) according to the condition

\[ [\cos\beta\,t_n(\theta_0)+\sin\beta\,u_n(\theta_0)]s'_n(\theta_0) -[\cos\beta\,t'_n(\theta_0)+\sin\beta\,u'_n(\theta_0)]s_n(\theta_0)=0, \tag{6} \]

which, of course, is always possible. Next choose the parameter \(\lambda\) according to the condition

\[ \cos\beta\,t'_n(\theta_0)+\sin\beta\,u'_n(\theta_0)-\lambda s'_n(\theta_0)=0. \tag{7} \]

From (6) and (7) it follows that

\[ \cos\beta\,t_n(\theta_0)+\sin\beta\,u_n(\theta_0)-\lambda s_n(\theta_0)=0. \tag{8} \]

Moreover, from (7) and (5) it follows that \(-1\leqslant\lambda\leqslant 1\), since by the Cauchy–Bunyakovsky inequality we have

\[ |\lambda s'_0(\theta_0)|\leqslant [t_n'^2(\theta_0)+u_n'^2(\theta_0)]^{1/2}. \]

Consequently, the polynomial \(\lambda s_n(\theta)\) satisfies the inequality \(|\lambda s_n(\theta)|<1\), and the corresponding function \(F(\theta;\beta)\) has, at the point \(\theta_0\), in view of (7) and (8), a double zero, which contradicts the lemma. Inequality (2) is established.

We pass to the proof of inequality (3). Let \(\xi_1<\xi_2<\cdots<\xi_{2n-1}\) be the zeros of the polynomial \(t'_n(\theta)\) on \([-\omega,\omega]\). First of all, let us show that on \([\xi_{2n-1},\omega]\) the polynomial \(t'_n(\theta)\) increases monotonically. In view of the oddness of the polynomial \(t'_n(\theta)\), it has a zero at the point \(\theta=\pi\). By Rolle’s theorem, in the interval \(\xi_{2n-1}<\theta<\pi\) there lies one and only one zero of its derivative \(t''_n(\theta)\). It is easy to verify that this zero is located to the right of the point \(\omega\). Indeed, since at the point \(\xi_{2n-1}\) the polynomial \(t_{2n}(\theta)\) has a minimum, \(t''_n(\xi_{2n-1})>0\), while on the other hand a direct calculation gives \(t''_n(\omega)>0\) for \(2n>[3\tg^2(\omega/2)+1]^{1/2}\). Consequently, \(t''_n(\theta)>0\) on \([\xi_{2n-1},\omega]\), and \(t'_n(\theta)\) increases monotonically.

is attained on this interval. From this we shall derive that \(|t'_n(\theta)| \leqslant t'(\omega)=2n^2\operatorname{ctg}(\omega/2)\) for \(-\omega \leqslant \theta \leqslant \omega\). Indeed, this follows directly for the intervals \(-\omega \leqslant \theta \leqslant \xi_1\) and \(\xi_{2n-1} \leqslant \theta \leqslant \omega\) from the monotonicity of \(t'_n(\theta)\) on these intervals. If \(0 \leqslant \theta \leqslant \theta_{2n}\), \(\theta_{2n}\) being the rightmost zero of \(t_n(\theta)\) on \([-\omega,\omega]\), then, observing that the zeros of \(t_n(\theta)\) coincide with the zeros of \(u'_n(\theta)\) and that \(|t'_n(\theta)+iu'_n(\theta)|\) increases monotonically on \([0,\theta_{2n}]\), from (2) we conclude that
\[ |t'_n(\theta)| \leqslant |t'_n(\theta_{2n})+iu'_n(\theta_{2n})|=t'(\theta_{2n}), \]
but \(t'(\theta_{2n}) \leqslant t'_n(\omega)\), since \(\xi_{2n-1}<\theta_{2n}<\omega\).

The proof of inequality (3) is equivalent to proving that, among the trigonometric polynomials \(s_n(\theta)\) of order \(n\) with the given value
\[ \max |s'_n(\theta)|=2n^2\operatorname{ctg}(\omega/2) \]
on the interval \(-\omega \leqslant \theta \leqslant \omega\), the polynomial least deviating from zero on this interval is the polynomial \(t_n(\theta)\) (for \(2n>[3\operatorname{tg}^2(\omega/2)+1]^{1/2}\)). Suppose that this is not so and that the extremal polynomial is \(\sigma_n(\theta)\ne \pm t_n(\theta)\). Clearly, \(|\sigma_n(\theta)|<1\) on \([-\omega,\omega]\), hence
\[ |\sigma'_n(\theta)|<|t'_n(\theta)+iu'_n(\theta)|. \]
If \(\theta_1 \leqslant \theta \leqslant \theta_{2n}\) (\(\theta_1,\theta_{2n}\) are the extreme zeros of \(t_n(\theta)\) on \([-\omega,\omega]\)), then
\[ |\sigma'_n(\theta)|<|t'_n(\theta)+iu_n(\theta)|\leqslant |t'_n(\theta_{2n})+iu'_n(\theta_{2n})|=t'_n(\theta_{2n}). \]
Therefore the point \(\theta_0\), at which \(\max|\sigma'_n(\theta)|=|\sigma'_n(\theta_0)|\) is attained, must lie in one of the intervals \((-\omega,\theta_1)\), \((\theta_{2n},\omega)\). Suppose, for definiteness, that \(\theta_{2n}<\theta_0<\omega\). The polynomial \(\sigma_n(\theta)\) must have on \([-\omega,\omega]\) at least \(2n\) points of maximum deviation with successively opposite signs. Suppose this is not so and that the number of the indicated points of maximum deviation is \(2n-1\); denote them by \(\varphi_1,\ldots,\varphi_{2n-1}\). Then one can construct such a trigonometric polynomial \(S_n(\theta)\) of order \(\leqslant n\) which, at the points \(\varphi_\nu\), would have successively opposite signs and for which \(S'_n(\theta_0)=0\). Put
\[ R_{2n}(x)=\sum_{\nu=1}^{2n-1}\frac{(-1)^\nu Q(x)}{(x-x_\nu)Q'(x_\nu)}+C(x-\mu)Q(x),\qquad Q(x)=\prod_{\nu=1}^{2n-1}(x-x_\nu), \]
where \(ax_\nu=\operatorname{tg}(\varphi_\nu/2)\), \(a=\operatorname{tg}(\omega/2)\), and \(C\) and \(\mu\) are certain constants, which we shall now choose in a suitable way. Since \(Q(x)\) has no multiple zeros, the parameter \(\mu\) can be chosen so that
\[ (1+a^2x_0^2)[(x_0-\mu)Q'(x_0)+Q(x_0)]-2na^2x_0(x_0-\mu)Q(x_0)\ne 0, \tag{9} \]
where \(ax_0=\operatorname{tg}(\theta_0/2)\). Now put
\[ S_n(\theta)=(1+a^2x^2)^{-n}R_{2n}(x),\qquad ax=\operatorname{tg}(\theta/2), \]
then
\[ 2aS'_n(\theta)=(1+a^2x^2)^{-n}\{(1+a^2x^2)R'_{2n}(x)-2na^2xR_{2n}(x)\}, \]
and, thanks to (9), we can choose the parameter \(C\) so that \(S'_n(\theta_0)=0\). On the other hand, \(S_n(\varphi_\nu)S_n(\varphi_{\nu+1})<0\). Consequently, there is such an \(\varepsilon\) that the polynomial \(\sigma_n(\theta)+\varepsilon S_n(\theta)\) will deviate from zero on \([-\omega,\omega]\) less than \(\sigma_n(\theta)\).

The polynomial \(\sigma_n(\theta)\) cannot have in the open interval \((-\omega,\omega)\) \(2n\) points of maximum deviation with successively opposite signs, since then we would have
\[ \sigma_n(\theta)=2n\operatorname{ctg}(\omega/2)\cos(n\theta-\beta), \]
\[ \max|\sigma_n(\theta)|=2n\operatorname{ctg}(\omega/2)>1=\max|t_n(\theta)|. \]
The set of points of maximum deviation of the polynomial \(\sigma_n(\theta)\) also cannot consist of \(2n-1\) points lying inside the interval \((-\omega,\omega)\) and one point coinciding with one of the endpoints \(\pm\omega\), since then the po

the polynomial \(\sigma_n(\theta+\delta)\), for sufficiently small \(\delta\), would have only \(2n-1\) points of maximum deviation on \([-\omega,\omega]\), and moreover
\[ \max |\sigma_n(\theta)|=\max |\sigma_n(\theta+\delta)|. \]
But the polynomial \(\sigma_n(\theta+\delta)\), by the preceding, cannot be the least deviating from zero.

Finally, suppose that the polynomial \(\sigma_n(\theta)\) has \(2n-2\) points of maximum deviation inside \((-\omega,\omega)\), and that both endpoints \(\pm\omega\) are such points. Let
\[ \max_{-\omega\le \theta\le \omega}|\sigma_n(\theta)|=L<1 \]
and
\[ \sigma_n(\theta)=(1+a^2x^2)^{-n}\Phi_{2n}(x),\qquad ax=\operatorname{tg}(\theta/2), \tag{10} \]
where \(\Phi_{2n}(x)\) is the corresponding algebraic polynomial of degree \(2n\), satisfying the inequality (see (4))
\[ |\Phi_{2n}(x)|\le L|M_{2n}(x)+i\sqrt{1-x^2}N_{2n-1}(x)|,\qquad -1\le x\le 1. \tag{11} \]
From (11) follows (7), that \(|\Phi'_{2n}(1)|<LM'_{2n}(1)\). From (10) we have \(\Phi_{2n}(1)=LM_{2n}(1)\). Direct calculation gives
\[ t'_n(\omega)-\sigma'_n(\omega)= \]
\[ =\frac{1}{2a}(1+a^2)^{-n}\{(1+a^2)[M'_{2n}(1)-\Phi'_{2n}(1)]-2na^2[M_{2n}(1)-\Phi_{2n}(1)]\}= \]
\[ =\frac{1}{2a}(1+a^2)^{-n}\{(1+a^2)[LM'_{2n}(1)-\Phi'_{2n}(1)]+ \]
\[ +(1-L)[(1+a^2)M'_{2n}(1)-2na^2M_{2n}(1)]\}= \]
\[ =\frac{1}{2a}(1+a^2)^{-n+1}[LM'_{2n}(1)-\Phi'_{2n}(1)]+(1-L)2n^2\operatorname{ctg}(\omega/2)>0. \]

In addition, by virtue of our assumptions,
\[ t'_n(\theta_{2n})-\sigma'_n(\theta_{2n})>0,\qquad t'_n(\theta_0)-\sigma'_n(\theta_0)<0. \]
Consequently, the polynomial \(t'_n(\theta)-\sigma'_n(\theta)\) of order \(n\) has at least 2 zeros in the interval \((\theta_{2n},\omega)\), and, by virtue of the inequalities
\[ |\sigma'_n(\theta_\nu)|<|t'_n(\theta_\nu)+iu'_n(\theta_\nu)|=|t'_n(\theta_\nu)|, \]
at least \(2n-1\) zeros in the interval \((\theta_1,\theta_{2n})\), i.e. \(\ge 2n+1\) zeros in the interval \((\theta_1,\omega)\) of length \(<2\pi\), which is impossible. Thus, the extremal polynomial is the polynomial having \(2n+1\) points of maximum deviation with successively alternating signs, i.e. the polynomial \(\pm t_n(\theta)\). The theorem is completely proved.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
28 VIII 1959

REFERENCES

1. I. I. Privalov, Cauchy Integral, Saratov, 1919.
2. D. Jackson, Bull. Am. Math. Soc., 37, 883 (1931).
3. N. I. Akhiezer, B. Ya. Levin, DAN, 117, No. 5, 735 (1957).
4. S. Bernstein, Leçons sur les propriétés extrémales, Paris, 1926.
5. S. N. Bernstein, Collected Works, 1, 1952, article No. 42, pp. 452–467.
6. V. S. Videnskii, DAN, 67, No. 5, 777 (1949).
7. V. S. Videnskii, DAN, 73, 257, No. 2 (1950).