MATHEMATICS

G. I. NATANSON

ON A THEOREM OF S. M. LOZINSKII

(Presented by Academician V. I. Smirnov on 23 V 1957)

1. S. M. Lozinskii proved ((^1)) the following theorem.

Theorem. Let the matrix of multipliers ({\rho_m^{(n)}}) ((n=0, 1, 2, \ldots;\; m=0, 1, 2, \ldots, n)) be such that for every (f(x)\in C_{2\pi}), uniformly on ((-\infty,+\infty)), one has

[
\lim_{n\to\infty}\left[\rho_0^{(n)}a_0+
\sum_{m=1}^{n}\rho_m^{(n)}(a_m\cos mx+b_m\sin mx)\right]=f(x)
]

((a_m, b_m) are the Fourier coefficients of (f(x))). Then, if (f(x)\in C_{2\pi}) and

[
T(x)=a_0^{(n)}+\sum_{m=1}^{n}\left(a_m^{(n)}\cos mx+b_m^{(n)}\sin mx\right)
]

is a polynomial coinciding with (f(x)) at the nodes (x_k=\dfrac{2k\pi}{2n+1}) ((k=0,1,2,\ldots,2n)), then

[
\lim_{n\to\infty}\left[\rho_0^{(n)}a_0^{(n)}+
\sum_{m=1}^{n}\rho_m^{(n)}\left(a_m^{(n)}\cos mx+b_m^{(n)}\sin mx\right)\right]=f(x)
]

uniformly on ((-\infty,+\infty)).

1. This theorem carries over to the theory of series in ultraspherical polynomials. Namely, the following theorem is valid:

Theorem. Let (-\frac{1}{2}\leq \alpha \leq \frac{1}{2}) and let (J_n(x)\equiv J_n^{(\alpha)}(x)) be polynomials orthonormal on ([-1,1]) with weight (p(x)=(1-x^2)^\alpha). Let the matrix ({\rho_m^{(n)}}) ((n=0,1,2,\ldots;\; m=0,1,2,\ldots,n)) be such that for every (f(x)\in C([-1,1])) one has

[
S_n^{(\rho)}[f;x]=\sum_{m=0}^{n}\rho_m^{(n)}a_mJ_m(x)\longrightarrow f(x)
\quad n\to\infty
]

uniformly on ([-1+h,\,1-h]), where

[
a_m=\int_{-1}^{1} f(t)J_m(t)p(t)\,dt,\qquad h\in[0,1).
]

For the function (f(x)\in C([-1,1])), form the interpolation polynomial (L_n[f;x]), coinciding with (f(x)) at the roots (x_k^{(n)}) of the polynomial (J_n(x)). Expand (L_n[f;x]) in the polynomials (J_m(x)):

[
L_n[f;x]=\sum_{m=0}^{n-1}a_m^{(n)}J_m(x)
]

and form

[
L_n^{(\rho)}[f;x]=\sum_{m=0}^{n-1}\rho_m^{(n)}a_m^{(n)}J_m(x).
]

Then

[
\lim_{n\to\infty} L_n^{(\rho)}[f;x]=f(x)
]

uniformly on ([-1+h,\,1-h]).

3. Lemma. If (T(x)) is an even trigonometric polynomial of degree not exceeding (n) and (\delta>0), then

[
\int_0^\pi |T(x)|\,dx \leq 6(\pi\gamma)^\delta n^\delta \int_0^\pi \sin^\delta x\,|T(x)|\,dx,
]

where (\gamma=[\ln(4/3)]^{-1}).

Proof. Obviously,

[
T(x)=T\left(x+\frac{1}{2\gamma n}\right)
-\sum_{k=1}^{\infty}\frac{T^{(k)}(x)}{(2\gamma n)^k k!},
]

[
T(x)=T\left(x-\frac{1}{2\gamma n}\right)
-\sum_{k=1}^{\infty}\frac{(-1)^k T^{(k)}(x)}{(2\gamma n)^k k!}.
]

Consequently,

[
\int_0^{\pi/2}|T(x)|\,dx
\leq
\int_0^{\pi/2}\left|T\left(x+\frac{1}{2\gamma n}\right)\right|\,dx
+\sum_{k=1}^{\infty}\frac{1}{(2\gamma n)^k k!}
\int_0^\pi |T^{(k)}(x)|\,dx,
\tag{1}
]

[
\int_{\pi/2}^{\pi}|T(x)|\,dx
\leq
\int_{\pi/2}^{\pi}\left|T\left(x-\frac{1}{2\gamma n}\right)\right|\,dx
+\sum_{k=1}^{\infty}\frac{1}{(2\gamma n)^k k!}
\int_0^\pi |T^{(k)}(x)|\,dx.
\tag{2}
]

Applying (k) times Zygmund’s inequality (see, for example, ((^2)), p. 566), we find

[
\int_0^\pi |T^{(k)}(x)|\,dx
\leq
(2n)^k\int_0^\pi |T(x)|\,dx.
]

Hence

[
\sum_{k=1}^{\infty}\frac{1}{(2\gamma n)^k k!}
\int_0^\pi |T^{(k)}(x)|\,dx
\leq
(e^{1/\gamma}-1)\int_0^\pi |T(x)|\,dx
=
\frac13\int_0^\pi |T(x)|\,dx.
\tag{3}
]

Adding (1) and (2) and using (3), we obtain

[
\int_0^\pi |T(x)|\,dx
-\frac23\int_0^\pi |T(x)|\,dx
\leq
\int_{1/2\gamma n}^{\pi/2+1/2\gamma n}|T(x)|\,dx
+
\int_{\pi/2-1/2\gamma n}^{\pi-1/2\gamma n} |T(x)|\,dx,
]

or

[
\int_0^\pi |T(x)|\,dx
\leq
6\int_{1/2\gamma n}^{\pi-1/2\gamma n}|T(x)|\,dx.
]

For (x\in\left[\dfrac{1}{2\gamma n},\,\pi-\dfrac{1}{2\gamma n}\right]) we have (\sin x\geq \dfrac{1}{\pi\gamma n}). Therefore

[
\int_0^\pi |T(x)|\,dx
\leq
6\int_{1/2\gamma n}^{1/2\gamma n}(\sin x\cdot \pi\gamma n)^\delta |T(x)|\,dx,
]

whence the lemma follows.

4. Passing to the proof of the theorem, let us note that (S_n^{(\rho)}[f;x]) is a linear operator from the space (C([-1,1])) into the space (C([-1+h,\,1-h])) with norm

[
|S_n^{(\rho)}|=
\max_{x\in[-1+h,\,1-h]}
\int_{-1}^{1}
\left|
\sum_{m=0}^{n}\rho_m^{(n)}J_m(x)J_m(t)
\right|
\rho(t)\,dt.
\tag{4}
]

By the well-known Banach theorem,

[
\left| S_n^{(\rho)} \right| = O(1).
\tag{5}
]

Put

[
l_k^{(n)}(x)=\frac{J_n(x)}{(x-x_k^{(n)})J_n'(x_k^{(n)})}.
]

Then (\bigl(J_n(x_k^{(n)})=0\bigr))

[
l_k^{(n)}(x)
=
-\left[J_{n+1}(x_k^{(n)})J_n'(x_k^{(n)})\sqrt{\lambda_{n+1}}\right]^{-1}
\sqrt{\lambda_{n+1}}\,
\frac{J_n(x)J_{n+1}(x_k^{(n)})-J_{n+1}(x)J_n(x_k^{(n)})}{x_k^{(n)}-x},
]

where

[
\lambda_{n+1}
=
\frac{(n+1)(2\alpha+n+1)}{(2\alpha+2n+1)(2\alpha+2n+3)}
=
\frac14+O(n^{-1}).
]

Denoting

[
\left[J_{n+1}(x_k^{(n)})J_n'(x_k^{(n)})\sqrt{\lambda_{n+1}}\right]^{-1}
=
q_k^{(n)},
]

with the aid of the Christoffel–Darboux formula we obtain

[
l_k^{(n)}(x)
=
-q_k^{(n)}\sum_{m=0}^{n}J_m(x_k^{(n)})J_m(x).
]

Consequently,

[
L_n^{(\rho)}[f;x]
=
-\sum_{k=1}^{n} q_k^{(n)} f(x_k^{(n)})
\sum_{m=0}^{n}\rho_m^{(n)}J_m(x_k^{(n)})J_m(x),
]

[
\left|L_n^{(\rho)}\right|
=
\max_{x\in[-1+h,\,1-h]}
\sum_{k=1}^{n}
\left|
q_k^{(n)}
\sum_{m=0}^{n}\rho_m^{(n)}J_m(x_k^{(n)})J_m(x)
\right|.
\tag{6}
]

For ultraspherical polynomials the following formulas hold (see, for example, (3), pp. 83 and 232):

[
(n+1)\sqrt{\frac{(n+2\alpha+1)(2n+2\alpha+1)}{(n+1)(2n+2\alpha+3)}}\,J_{n+1}(x)
=
]

[

(n+2\alpha+1)xJ_n(x)-(1-x^2)J_n'(x),
]

[
\left|J_n'(x_k^{(n)})\right|^{-1}
=
O\left(\frac{k^{\alpha+3/2}}{n^{\alpha+5/2}}\right)
\qquad
\left(0\le x_{[n/2]}^{(n)}<\cdots<x_2^{(n)}<x_1^{(n)}<1\right).
]

Hence, putting (x_k^{(n)}=\cos\theta_k), we find

[
q_k^{(n)}
=
\frac{O(1)\,k^{2\alpha+3}}{n^{2\alpha+4}\sin^2\theta_k}.
]

For (\theta_k) the Markov–Stieltjes inequalities are known (see (4), p. 40):

[
\frac{2k-1}{2n+1}\pi
\le
\theta_k
\le
\frac{2k}{2n+1}\pi.
]

Thus, for (1\le k\le [n/2]),

[
q_k^{(n)}
=
O(n^{-1})\sin^{2\alpha+1}\theta_k.
\tag{7}
]

By symmetry of (J_n(x)), (7) is also valid for ([n/2]\le k\le n). Let the sum on the right-hand side of (6) attain its maximum at (x=x_0\in[-1+h,\,1-h]). Introducing the notation

[
\sum_{m=0}^{n}\rho_m^{(n)}J_m(\cos\theta)J_m(x_0)=T(\theta)
]

((T(\theta)) is an even trigonometric polynomial of order (n)), we have

[
\left|L_n^{(\rho)}\right|
=
O(n^{-1})\sum_{k=1}^{n}\sin^{2\alpha+1}\theta_k\,|T(\theta_k)|
=
O(1)\sum_{k=-n}^{n}
\frac{\left|\sin^{2\alpha+1}\theta_k\,T(\theta_k)\right|}{2n+1},
]

where (\theta_{-k}=-\theta_k).

If (s_n(\theta)) is a trigonometric polynomial of best approximation to (\sin^{2\alpha+1}\theta) of order not exceeding (n), then

[
\sin^{2\alpha+1}\theta=s_n(\theta)+O\left(n^{-2\alpha-1}\right),
\tag{8}
]

[
\left|L_n^{(\rho)}\right|
=
O(1)\sum_{k=-n}^{n}\frac{\left|s_n(\theta_k)T(\theta_k)\right|}{2n+1}
+
O\left(n^{-2\alpha-1}\right)
\sum_{k=-n}^{n}\frac{\left|T(\theta_k)\right|}{2n+1}.
]

Exactly as in ((^{2})), p. 568, it is proved that

[
\sum_{k=-n}^{n}\frac{\left|s_n(\theta_k)T(\theta_k)\right|}{2n+1}
=
O(1)\int_{-\pi}^{\pi}\left|s_n(\theta)T(\theta)\right|\,d\theta,
]

[
\sum_{k=-n}^{n}\frac{\left|T(\theta_k)\right|}{2n+1}
=
O(1)\int_{-\pi}^{\pi}\left|T(\theta)\right|\,d\theta.
]

Therefore

[
\left|L_n^{(\rho)}\right|
=
O(1)\int_{-\pi}^{\pi}\left|s_n(\theta)T(\theta)\right|\,d\theta
+
O\left(n^{-2\alpha-1}\right)
\int_{-\pi}^{\pi}\left|T(\theta)\right|\,d\theta,
]

and, by virtue of (8) and the evenness of (T(\theta)),

[
\left|L_n^{(\rho)}\right|
=
O(1)\int_{0}^{\pi}\sin^{2\alpha+1}\theta\,\left|T(\theta)\right|\,d\theta
+
O\left(n^{-2\alpha-1}\right)
\int_{0}^{\pi}\left|T(\theta)\right|\,d\theta.
]

Applying the lemma, we find

[
\left|L_n^{(\rho)}\right|
=
O(1)\int_{0}^{\pi}\left|T(\theta)\right|\sin^{2\alpha+1}\theta\,d\theta
=
O(1)\int_{-1}^{1}
\left|\sum_{m=0}^{n}\rho_m^{(n)}J_m(t)J_m(x_0)\right|
\rho(t)\,dt.
]

Thus, by virtue of (4) and (5), the norms of the linear operators (L_n^{(\rho)}[f;x]) are uniformly bounded. It remains to observe that, if (P(x)) is an algebraic polynomial, then
(\lim\limits_{n\to\infty}L_n^{(\rho)}[P;x]=P(x)), since for (n) greater than the degree of (P(x)),

[
L_n^{(\rho)}[P;x]=S_n^{(\rho)}[P;x].
]

Starting from estimates for the deviation of (S_n^{(\rho)}[f;x]) from (f(x)), one can obtain convergent estimates also for the deviation of (L_n^{(\rho)}[f;x]) from (f(x)). We formulate the simplest result of this kind.

Let, for every (f(x)\in C([-1,1])),

[
\left|f(x)-S_n^{(\rho)}[f;x]\right|
<
A\omega_f(\varphi(n))
\qquad
(-1+h\leq x\leq 1-h),
]

where (\omega_f) is the modulus of continuity of (f(x)) on ([-1,1]); the quantity (\dfrac{1}{n\varphi(n)}) is bounded; (\lim\limits_{n\to\infty}\varphi(n)=0); and (A) and (\varphi) do not depend on (n), (x), and (f(x)). Then

[
\left|f(x)-L_n^{(\rho)}[f;x]\right|
<
B\omega_f(\varphi(n))
\qquad
(-1+h\leq x\leq 1-h),
]

where (B) likewise does not depend on (n), (x), or (f(x)).

There is also a trigonometric analogue of this fact.

Leningrad State Pedagogical Institute
named after A. I. Herzen

Received
21 V 1957

CITED LITERATURE

1. S. M. Lozinskii, Mat. sbornik, 14 (56), 175 (1944).
2. I. P. Natanson, Constructive Theory of Functions, Moscow–Leningrad, 1949.
3. G. Szegő, Orthogonal Polynomials, Am. Math. Soc. Coll. Publ., 23, 1939.
4. Ya. L. Geronimus, Theory of Orthogonal Polynomials, Moscow–Leningrad, 1950.