UDC 517.512.6+517.537

MATHEMATICS

D. L. BERMAN

ON A GENERALIZED FORMULA OF M. RIESZ

(Presented by Academician S. N. Bernstein on 4 III 1968)

1°. Denote by \(\Pi_n\) the set of all trigonometric polynomials of order \(\leqslant n\). For a given polynomial
\[ \Phi_n(t)=\sum_{k=0}^{n} r_k\sin(kt+\alpha_k) \]
construct the polynomial
\[ \widetilde{\Phi}_n(t)=r_n+2\sum_{k=0}^{n-1} r_k\cos[(n-k)t+\alpha_n-\alpha_k], \]
which we shall call associated with the polynomial \(\Phi_n\). In \((^1)\) the following theorem was established:

If \(t\in\Pi_n\), then the identity
\[ \int_0^{2\pi} t(x+\theta)\Phi_n(\theta)\,d\theta = \frac{\pi}{2n}\sum_{r=1}^{2n} t(x+\psi_r)(-1)^{r-1}\widetilde{\Phi}_n(\psi_r), \tag{1} \]
holds, where
\[ \psi_r=\frac{2r-1}{2n}\pi-\frac{\alpha_n}{n}. \]

It is interesting that if
\[ \Phi_n(t)=\frac{1}{\pi}\left(\frac{\sin(n+1/2)}{2\sin t/2}\right)', \]
then
\[ \widetilde{\Phi}_n(t)=\frac{\sin^2 nt/2}{\pi\sin^2 t/2}. \]
Therefore, with this choice of the polynomial \(\Phi_n\), equality (1) becomes the well-known identity of M. Riesz \((^2)\)
\[ t'(x)=\frac{1}{2n}\sum_{r=1}^{2n} t(x+\varphi_r)\frac{(-1)^{r-1}}{2\sin^2\varphi_r/2}, \qquad t\in\Pi_n,\quad \varphi_r=\frac{(2r-1)\pi}{2n}. \tag{2} \]

The present note is devoted to extending identity (1) to the complex domain and to the case of semi-entire trigonometric polynomials.

2°. Let \(D\) be an arbitrary finite domain with simply connected complement \(D_1\) and rectifiable boundary \(\Gamma\). Denote by \(w=\Phi(z)\) the function mapping \(D_1\) conformally onto the domain \(|w|\geqslant 1\) of the \(w\)-plane under the condition \(\Phi(\infty)=\infty\). Let \(z=\psi(w)\) be the inverse function. Denote by \(A(D)\) the set of all functions \(f(z)\) continuous in the closed domain \(\overline{D}\) and analytic at every interior point of \(D\). Put
\[ \|f\|=\max_{z\in\overline{D}} |f(z)|. \]
Define the shift \(f_t\) of a function \(f\in A(D)\) by the real number \(t\), \(-\infty<t<\infty\), by means of the equality
\[ f_t(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f\{\psi[\Phi(\zeta)e^{it}]\}}{\zeta-z}\,d\zeta. \tag{3} \]
The transformation (3) has been well studied in the works of V. K. Dzyadyk \((^{3,4})\). For fixed \(\Phi_n\in\Pi_n\) introduce the operators
\[ \sigma(f)=\sigma(f,z)=\int_{-\pi}^{\pi} f_t(z)\Phi_n(t)\,dt,\qquad f\in A(D), \]
\[ U(f,z)=\frac{\pi}{2n}\sum_{r=1}^{2n} f_{\psi_r}(z)(-1)^{r-1}\widetilde{\Phi}_n(\psi_r). \]

Theorem 1. If \(f\) is a polynomial of degree \(\leqslant n\), then the identity
\[ \sigma(f)=U(f) \]
holds.

We outline the proof. Denote by \(F_n\) the Faber polynomial of degree \(n\) generated by the domain \(D\). It is known that any polynomial of degree \(n\) is expressed linearly in terms of the polynomials \(\{F_k\}\), \(0\leq k\leq n\). Therefore it suffices to prove the identity only for the polynomials \(F_k\), \(0\leq k\leq n\).

The Faber polynomials have the important property that

\[ (F_k)_t=F_k e^{ikt}. \]

Therefore

\[ \sigma(F_k)=F_k\int_{-\pi}^{\pi}\Phi_n(t)e^{ikt}\,dt, \tag{4} \]

\[ U(F_k)=F_k\frac{\pi}{2n}\sum_{r=1}^{2n}e^{ik\psi_r}(-1)^{r-1}\widetilde{\Phi}_n(\psi_r). \tag{5} \]

By virtue of identity (1), valid for any trigonometric polynomial of order \(n\),

\[ \int_{-\pi}^{\pi}\Phi_n(t)e^{ikt}\,dt = \frac{\pi}{2n}\sum_{r=1}^{2n}e^{ik\psi_r}(-1)^{r-1}\widetilde{\Phi}_n(\psi_r). \]

Therefore it follows from (4) and (5) that \(\sigma(F_k)=U(F_k)\).

Corollary. If the kernel \(\Phi_n\) is such that

\[ \widetilde{\Phi}_n(\psi_r)\geq 0,\qquad r=1,2,\ldots,2n, \tag{6} \]

then for any polynomial \(f\) of degree \(\leq n\) the inequality

\[ \left\|\int_{-\pi}^{\pi} f_t(z)\Phi_n(t)\,dt\right\| \leq \pi r_n \max_{r=1,2,\ldots,2n}\|f_{\psi_r}(z)\| \tag{7} \]

holds.

Indeed, by Theorem 1 and (6) we have

\[ \left\|\int_{-\pi}^{\pi} f_t(z)\Phi_n(t)\,dt\right\| \leq \frac{\pi}{2n} \max_{r=1,2,\ldots,2n}\|f_{\psi_r}(z)\| \sum_{r=1}^{2n}\widetilde{\Phi}_n(\psi_r). \tag{8} \]

One can verify that the last sum is equal to \(2nr_n\). Therefore (7) follows from (8).

Inequality (7) is a generalization of the classical inequality of S. N. Bernstein on the maximum modulus of the derivative of a trigonometric polynomial.

\(3^\circ\). As is known, an expression of the form

\[ \tau_n(x)=\sum_{k=0}^{n}\left(a_k\cos (k+{}^{1}/{}_{2})x+b_k\sin (k+{}^{1}/{}_{2})x\right), \]

where \(\{a_k\}_{k=0}^{n}\), \(\{b_k\}_{k=0}^{n}\) are constants, is called a half-integer trigonometric polynomial of order \(n\). We denote the set of all such polynomials by \(\Pi_{n+1/2}\). It is obvious that any polynomial \(\Phi\in\Pi_{n+1/2}\) can be written in the form

\[ \Phi(x)=\sum_{k=0}^{n} r_k\sin\left((k+{}^{1}/{}_{2})x+\alpha_k\right), \tag{9} \]

where \(r_k\) and \(\alpha_k\) are real numbers, with \(r_k\geq 0\), \(0\leq k\leq n\).

Let \(\Phi\) be a fixed polynomial from \(\Pi_{n+1/2}\). Associate with it the polynomial

\[ \widetilde{\Phi}(x)=2\sum_{k=0}^{n} r_k\cos\left[(n+1)x-(k+{}^{1}/{}_{2})x+\alpha_n-\alpha_k\right], \]

which we shall call the polynomial associated with the polynomial (9).

Theorem 2. Let \(t_n \in \Pi_{n+1/2}\); then the identity holds

\[ \int_{0}^{2\pi} t_n(x+\theta)\Phi(\theta)\,d\theta = \frac{\pi}{2n+2}\sum_{r=1}^{2n+2} t_n(x+\delta_r)(-1)^{r-1}\widetilde{\Phi}(\delta_r), \tag{10} \]

where

\[ \delta_r=\frac{(2r-1)\pi}{2n+2}-\frac{\alpha_n}{n+1}. \]

Proof. Obviously, it is enough to prove equality (10) only for the functions
\(\{\cos (k+1/2)x\}_{k=0}^n,\ \{\sin (k+1/2)x\}_{k=0}^n\). Let us verify, for example, identity (10) for
\(c_k(x)=\cos (k+1/2)x,\ 0\leq k\leq n\). After simple calculations we obtain that

\[ \int_{0}^{2\pi} c_k(x+\theta)\Phi(\theta)\,d\theta = -\pi r_k \sin\bigl[(k+1/2)x-\alpha_k\bigr]. \tag{11} \]

Denote the right-hand side of (10) by \(U(t_n,x)\). Then, by virtue of (11), it is necessary to prove that

\[ U(c_k,x)=-\pi r_k\sin\bigl[(k+1/2)x-\alpha_k\bigr]. \]

This equality follows from the equalities

\[ \sum_{r=1}^{2n+2}(-1)^{r-1}\cos p\delta_r = \sum_{r=1}^{2n+2}(-1)^{r-1}\sin p\delta_r = 0 \qquad 0\leq p<n+1; \]

\[ \sum_{r=1}^{2n+2}(-1)^{r-1}\cos (n+1)\delta_r = (2n+2)\sin\alpha_n; \]

\[ \sum_{r=1}^{2n+2}(-1)^{r-1}\sin (n+1)\delta_r = (2n+2)\cos\alpha_n. \]

Let us give some applications of Theorem 2. Let

\[ \Phi(x)=\sum_{k=0}^{n}\frac{2k+1}{2}\sin (k+1/2)x. \tag{12} \]

It is easy to check that in this case

\[ \widetilde{\Phi}(x)= \frac{\sin^2 (n+1)x/2}{\sin^2 x/2}\cos\frac{x}{2}, \qquad \frac{1}{\pi}\int_{0}^{2\pi} t(x+\theta)\Phi(\theta)\,d\theta=t'(x) \tag{13} \]

for any \(t\in\Pi_{n+1/2}\). Therefore, if one takes into account that
\(\sin^2 \frac{n+1}{2}\delta_r=1/2\), then by Theorem 2 we obtain

Corollary. For any polynomial \(t\in\Pi_{n+1/2}\) the identity holds

\[ t'(x)= \frac{1}{4n+4}\sum_{r=1}^{2n+2} t(x+\varphi_r)(-1)^{r-1} \frac{\cos \varphi_r/2}{\sin^2 \varphi_r/2}, \qquad \varphi_r=\frac{2r-1}{2n}\pi. \tag{14} \]

Remark. Identity (14) can also be derived from M. Riesz’s identity (2). Applying identity (14) to the case where
\(t_n(x)=\sin(n+1/2)x\), and putting \(x=0\), we obtain

\[ \sum_{r=1}^{2n+2}\operatorname{ctg}^2\frac{2r-1}{4n+4}\pi = (2n+1)(2n+2). \]

Theorem 3. If the kernel \(\Phi_n\) is such that

\[ \widetilde{\Phi}_n(\delta_r)\geq 0, \qquad r=1,2,\ldots,(2n+2), \tag{15} \]

then for any \(t\in\Pi_{n+1/2}\) the inequality holds

\[ \left\| \int_{0}^{2\pi} t(x+\theta)\Phi_n(\theta)\,d\theta \right\| \leq \left( \frac{\pi}{n+1} \sum_{j=0}^{n} r_j \frac{ \sin\bigl[\alpha_n(n-j+1/2)/(n+1)+\alpha_j-\alpha_n\bigr] }{ \sin\bigl[(n-j+1/2)\pi/(2n+2)\bigr] } \right)\|t\|. \tag{16} \]

Inequality (16) is sharp in the sense that for

\[ \Phi_n(\theta)=\sum_{k=0}^{n}\cos (k+1/2)\theta \tag{17} \]

for every \(t\in \Pi_{n+1/2}\) equality holds in (16).

We outline the proof. From Theorem 2 it follows that for every \(t\in \Pi_{n+1/2}\)

\[ \left\|\int_{0}^{2\pi} t(x+\theta)\Phi_n(\theta)\,d\theta\right\| \le \left(\frac{\pi}{2n+2}\sum_{r=1}^{2n+2}|\widetilde{\Phi}_n(\delta_r)|\right)\|t\|. \tag{18} \]

It can be verified that if inequalities (15) are satisfied, then

\[ \sum_{r=1}^{2n+2}|\widetilde{\Phi}_n(\delta_r)| = 2\sum_{j=0}^{n} r_j \frac{\sin[\alpha_n(n-j+1/2)/(n+1)+\alpha_j-\alpha_n]} {\sin[(n-j+1/2)\pi/(2n+2)]}. \]

Therefore (16) follows from (18). Let \(\Phi_n\) be defined according to (17). By simple calculations we obtain that in this case

\[ \widetilde{\Phi}_n(x)=\frac{\sin^2(n+1/2)t}{\sin t/2}. \tag{19} \]

Thus, inequalities (15) are satisfied. Hence Theorem 3 is applicable. But in fact in this case equality holds in (16), since it is directly seen that the right-hand side of (16) is equal to \(\pi\|t\|\). It can be verified that if \(\Phi_n\) is defined according to (17), then for every \(t\in \Pi_{n+1/2}\)

\[ \int_{0}^{2\pi} t(x+\theta)\Phi_n(\theta)\,d\theta=\pi t(x). \]

Therefore the left-hand side of (16) is also equal to \(\pi\|t\|\).

Theorem 4. If the kernel \(\Phi_n\) is such that the inequalities

\[ \Phi_n(\delta_r)\ge 0,\qquad r=1,2,\ldots,n, \tag{20} \]

\[ \Phi_n(\delta_r)\le 0,\qquad r=n+1,n+2,\ldots,2n+2, \tag{21} \]

are satisfied, then for every \(t\in \Pi_{n+1/2}\)

\[ \left\|\int_{0}^{2\pi} t(x+\theta)\Phi_n(\theta)\,d\theta\right\| \le \left( \frac{2\pi}{n+1} \sum_{j=0}^{n} r_j \frac{(-1)^{\,n-j}}{\cos(2j+1)\pi/(2n+2)\,2} \right)\|t\|. \]

From equality (13) it is seen that if \(\Phi_n\) is defined according to (12), then inequalities (20), (21) are satisfied, and hence Theorem 4 is applicable in this case.

Received
21 II 1968

REFERENCES

1. D. L. Berman, DAN, 163, No. 3 (1965).
2. M. Riesz, C. R. 158, 1152 (1914).
3. V. K. Dzyadyk, Izv. AN SSSR, ser. matem., 26, 797 (1962).
4. V. K. Dzyadyk, in: Contemporary Problems of the Theory of Analytic Functions, “Nauka,” 1966.