AN EXAMPLE OF A SEQUENCE OF LINEAR POSITIVE OPERATORS IN THE SPACE OF CONTINUOUS FUNCTIONS
Consider a sequence of functions
Submitted 1957-01-01 | RussiaRxiv: ru-195701.96472 | Translated from Russian

Abstract

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MATHEMATICS

V. A. BASKAKOV

AN EXAMPLE OF A SEQUENCE OF LINEAR POSITIVE OPERATORS IN THE SPACE OF CONTINUOUS FUNCTIONS

(Presented by Academician V. I. Smirnov, 8 X 1956)

Consider a sequence of functions

[
\varphi_n(y), \qquad n=1,2,\ldots,
]

each of which has the following properties:

a) it is analytic in the closed disk of radius (R>0) with center at the point (M(R;0));

b) (\varphi_n(0)=1);

c) ((-1)^k\varphi_n^{(k)}(x)\geq 0,\ k=0,1,2,\ldots,\ x\in[0,R]);

d)

[
-\varphi_n^{(k)}(x)=n\varphi_n^{(k-1)}(x)\,(1+\alpha_{kn}(x)), \qquad k=1,2,\ldots,
]

where (x\in[0,R]) and (\alpha_{kn}(x)) tends to zero uniformly with respect to (k) and (x) as (n) tends to infinity;

e)

[
\lim_{n\to\infty}\frac{n}{n_n}=1.
]

Expand the function (\varphi_n(y)) in a Taylor series in powers of ((y-x)) ((x\in[0,R])):

[
\varphi_n(y)=\sum_{k=0}^{\infty}\frac{\varphi_n^{(k)}(x)}{k!}(y-x)^k.
]

From conditions a) and b) it follows that

[
1=\sum_{k=0}^{\infty}(-1)^k\frac{\varphi_n^{(k)}(x)}{k!}x^k.
\tag{1}
]

Consider the sequence of linear operators

[
L_n(f;x)=\sum_{k=0}^{\infty}(-1)^k\frac{\varphi_n^{(k)}(x)}{k!}x^k f!\left(\frac{k}{n}\right), \qquad n=1,2,\ldots,
\tag{2}
]

which will be positive on the basis of condition c).

We shall show that the sequence (2) converges uniformly on the interval ([0,R]) for each of the three functions (1,x), and (x^2).

Uniform convergence to the function (f(x)=1) follows from identity (1). For the function (f(x)=x) we have

[
L_n(y;x)=\sum_{k=0}^{\infty}(-1)^k\frac{\varphi_n^{(k)}(x)}{k!}x^k\frac{k}{n}.
]

On the basis of condition d),

[
L_n(y;x)=x\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\varphi_n^{(k-1)}(x)}{(k-1)!}x^{k-1}(1-\alpha_{kn}(x))=
]

[
= x\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\varphi_n^{(k-1)}(x)}{(k-1)!}x^{k-1}
+ x\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\varphi_n^{(k-1)}(x)}{(k-1)!}x^{k-1}\alpha_{kn}(x).
]

Since

[
\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\varphi_{m_n}^{(k-1)}(x)}{(k-1)!}\,x^{k-1}
=\varphi_{m_n}(0)=1,
\tag{3}
]

and the sum

[
\sum_{k=1}^{\infty}(-1)^k
\frac{\varphi_{m_n}^{(k-1)}(x)}{(k-1)!}\,x^{k-1}\alpha_{kn}(x),
]

by virtue of the conditions imposed on (\alpha_{kn}(x)), tends uniformly to zero, it follows that (L_n(y;x)) converges uniformly to (x).

For the function (f(x)=x^2) we have

[
\begin{aligned}
L_n(y^2;x)
&=\sum_{k=0}^{\infty}(-1)^k\frac{\varphi_n^{(k)}(x)}{k!}\,x^k\left(\frac{k}{n}\right)^2 \
&=x\sum_{k=1}^{\infty}(-1)^{k-1}
\frac{\varphi_{m_n}^{(k-1)}(x)}{(k-1)!}\,x^{k-1}(1+\alpha_{kn}(x))\frac{k}{n} \
&=x^2\frac{m_n}{n}\sum_{k=2}^{\infty}(-1)^{k-2}
\frac{\varphi_{m_nm_n}^{(k-2)}(x)}{(k-2)!}\,x^{k-2}
(1+\alpha_{kn}(x))(1+\alpha_{k-1\,m_n}(x)) \
&\quad+\frac{x}{n}\sum_{k=1}^{\infty}(-1)^{k-1}
\frac{\varphi_{m_n}^{(k-1)}(x)}{(k-1)!}\,x^{k-1}(1+\alpha_{kn}(x)).
\end{aligned}
]

The uniform convergence of (L_n(y^2;x)) to (x^2) follows from condition d), (3), from the fact that

[
\sum_{k=2}^{\infty}(-1)^{k-2}
\frac{\varphi_{m_nm_n}^{(k-2)}(x)}{(k-2)!}\,x^{k-2}
=\varphi_{m_nm_n}(0)=1,
]

and the sums

[
1)\quad
\sum_{k=2}^{\infty}(-1)^{k-2}
\frac{\varphi_{m_nm_n}^{(k-2)}(x)}{(k-2)!}\,x^{k-2}
\bigl(\alpha_{kn}(x)+\alpha_{k-1\,m_n}(x)+\alpha_{kn}(x)\alpha_{k-1\,m_n}(x)\bigr),
]

[
2)\quad
\sum_{k=1}^{\infty}(-1)^{k-1}
\frac{\varphi_{m_n}^{(k-1)}(x)}{(k-1)!}\,x^{k-1}\alpha_{kn}(x)
]

converge uniformly to zero.

On the basis of P. P. Korovkin’s theorem ((^1)), the sequence of operators (2) converges uniformly on the interval ([0,R]) to the function (f(x)), continuous on this interval, continuous at the point (R) from the right, and bounded on the set (x\ge 0).

Let us consider some special cases.

A. Taking (\varphi_n(y)=(1-y)^n), we obtain the Bernstein polynomials

[
B_n(f;x)=\sum_{k=1}^{n} f\left(\frac{k}{n}\right) C_n^k x^k(1-x)^{n-k}
\tag{4}
]

and his theorem.

B. Taking (\varphi_n(y)=e^{-ny}), we obtain the sequence of operators

[
L_n(f;x)=e^{-nx}\sum_{k=0}^{\infty} f\left(\frac{k}{n}\right)\frac{n^k}{k!}x^k,
\tag{5}
]

which converges uniformly on the interval ([0,R]) ((R>0)) to the function (f(x)), continuous on ([0,R]), continuous from the right at the point (R), and bounded on ([0,\infty)). These operators have been considered by many authors, for example G. M. Mirakyan ((^2)), Favard ((^4)), Szasz ((^4)), Băcer ((^6)), and others.

c. Taking (\varphi_n=\dfrac{1}{(1+y)^n}), we obtain the sequence of operators

[
L_n(f;y)=\frac{1}{(1+x)^n}\sum_{k=0}^{\infty} f!\left(\frac{k}{n}\right)
\frac{n(n+1)\ldots(n+k-1)}{k!}
\left(\frac{x}{1+x}\right)^k,
\tag{6}
]

converging uniformly on the interval ([0,R]) to the function (f(x)), continuous on ([0,R]), continuous at the point (R) from the right, and bounded on ([0,\infty)).

The following theorems describe the character of convergence of the operators (2).

Theorem 1. If the following conditions are satisfied: 1) (f(x)) is continuous on ([0,R]); 2) (f(x)=f(R)), if (x>R); 3) (\alpha_{kn}(x)=\alpha_n), then

[
\left|L_n(f;x)-f(x)\right|\leq
]

[
\leq \omega!\left(\frac{1}{\sqrt n}\right)
\left[
\sqrt n\,
\sqrt{
x^2\frac{m_n}{n}(1+\alpha_n)(1+\alpha_{m_n})
+\frac{x}{n}(1+\alpha_n)-2x^2(1+\alpha_n)+x^2+1
}
\right],
]

where (\omega(\delta)) is the modulus of continuity of (f(x)).

Theorem 2. If the following conditions are satisfied: 1) the function (f(x)) is continuous on ([0,R]), continuous at the point (R) from the right, bounded on the set (x>0), and has a finite second derivative at the point (x\in[0,R]); 2) (m_n=n+c), where (c) is an integer; 3) (\alpha_{kn}(x)\equiv0), then

[
L_n(f;x)=f(x)+\frac{f''(x)}{2!}\frac{x+cx^2}{n}+o!\left(\frac{1}{n}\right).
]

Corollary 1. For the polynomials of S. N. Bernstein (4), for which (c=-1), we have

[
B_n(f;x)=f(x)+\frac{f''(x)}{2!}\frac{x(1-x)}{n}+o!\left(\frac{1}{n}\right).
]

This proves the theorem of E. V. Voronovskaya (³).

Corollary 2. For the operators (5), (c=0) and, consequently,

[
L_n(f;x)=f(x)+\frac{f''(x)}{2!}\frac{x}{n}+o!\left(\frac{1}{n}\right).
]

Corollary 3. For the operators (6), (c=1) and, consequently,

[
L_n(f;x)=f(x)+\frac{f''(x)}{2!}\frac{x(x+1)}{n}+o!\left(\frac{1}{n}\right).
]

Kaluga State
Pedagogical Institute

Received
26 IX 1956

REFERENCES CITED

¹ P. P. Korovkin, DAN, 90, No. 6 (1953).
² G. M. Mirakyan, DAN, 31, No. 3 (1941).
³ E. V. Voronovskaya, DAN (A), 79 (1932).
⁴ M. J. Favard, J. math. pures et appl., 23, 219 (1944).
⁵ O. Szasz, J. Res. Nat. Bur. Stand., 45, No. 3, 239 (1950).
⁶ P. L. Butzer, Proc. Am. Math. Soc., 5, No. 4, 547 (1954).

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AN EXAMPLE OF A SEQUENCE OF LINEAR POSITIVE OPERATORS IN THE SPACE OF CONTINUOUS FUNCTIONS