MATHEMATICS

P. P. KOROVKIN

ON THE ORDER OF APPROXIMATION OF FUNCTIONS BY LINEAR POSITIVE OPERATORS

(Presented by Academician V. I. Smirnov, 10 I 1957)

Let \(\mathscr L_n(f;x)\) be a linear positive operator whose value, for any continuous function, is an algebraic or trigonometric polynomial of degree not exceeding \(n\). Put

\[ \Delta_n=\mathscr L_n(f;x)-f(x). \tag{1} \]

E. V. Voronovskaya \((^1)\), studying the order of approximation of functions by S. N. Bernstein polynomials, established that the order of smallness of the quantity \(\Delta_n\) in this case is, generally speaking, no higher than \(1/n\), and proved the equality

\[ \lim_{n\to\infty} n\Delta_n=\frac12 x(1-x)f''(x). \tag{2} \]

In this note it will be shown that the slow order of convergence to zero of the quantity \(\Delta_n\) occurs for all linear and positive polynomial operators. In addition, an analogue of equality (2) will be established for arbitrary sequences of positive operators.

Theorem 1. If \(\{\mathscr L_n(f;x)\}\) is a sequence of linear positive operators such that the value of the operator \(\mathscr L_n(f;x)\) for any function continuous on the interval \(-1\le x\le 1\) is an algebraic polynomial of degree not exceeding \(n\), then at least one of the sequences

\[ \{n^2\|\mathscr L_n(1;x)-1\|\},\qquad \{n^2\|\mathscr L_n(t;x)-x\|\},\qquad \{n^2\|\mathscr L_n(t^2;x)-x^2\|\} \]

does not tend to zero.

Proof. We have

\[ \begin{aligned} \left|\mathscr L_n(|t|;x)-|x|\right| &\le \left|\mathscr L_n(|t|;x)-|x|\,\mathscr L_n(1;x)\right| +|x|\left|\mathscr L_n(1;x)-1\right| \\ &\le \left|\mathscr L_n(|t|-|x|;x)\right| +\left|\mathscr L_n(1;x)-1\right|. \end{aligned} \tag{3} \]

Using the properties of linear positive operators (monotonicity and the Cauchy–Bunyakovsky inequality), we obtain

\[ \begin{aligned} \left|\mathscr L_n(|t|-|x|;x)\right| &\le \mathscr L_n(|t-x|;x) \le \\ &\le \sqrt{\mathscr L_n\{(t-x)^2;x\}\,\mathscr L_n(1;x)}= \\ &= \sqrt{\{\mathscr L_n(t^2;x)-x^2\}-2x\{\mathscr L_n(t;x)-x\} +x^2\{\mathscr L_n(1;x)-1\}}\, \sqrt{\mathscr L_n(1;x)}. \end{aligned} \tag{4} \]

Since \(\mathscr L_n(f;x)\) is a polynomial operator, it follows that

\[ E_n(|x|)\le \|\mathscr L_n(|t|;x)-|x|\|. \tag{5} \]

From (3), (4), and (5) we obtain

\[ nE_n(|x|) \leq n^2 \| \mathscr{L}_n(1;x)-1 \|+ \tag{6} \]

\[ + \sqrt{\|\mathscr{L}_n(1;x)\|}\sqrt{n^2(\|\mathscr{L}_n(t^2;x)-x^2\|+2\|\mathscr{L}_n(t;x)-x\|+\|\mathscr{L}_n(1;x)-1\|)}. \]

If the theorem were not true, then the right-hand side of the last inequality would tend to zero. But this is impossible, since \(nE_n(|x|)\) does not tend to zero.

It follows from the theorem proved that the order of approximation of functions by linear positive and polynomial operators is not higher than \(1/n^2\), even for analytic functions. A similar assertion is easily carried over also to the trigonometric case. In what follows, sequences of linear positive and polynomial operators will be indicated for which the order of approximation \(1/n^2\) holds for any twice differentiable function, \(\|f''(x)\|<\infty\).

Theorem 2. Let \(\varphi_n(x)\), \(n=1,2,\ldots\), be nondecreasing functions on the interval \(a \leq x \leq b\), and

\[ \Phi_n(f)=\int_a^b f(x)\,d\varphi_n(x). \]

Let \(\psi(x)\) be a continuous function on this interval,

\[ \psi(c)=0,\qquad \psi(x)>0,\qquad x\ne c,\qquad a\leq c\leq b. \]

In order that from the equality

\[ \lim_{x\to c}\frac{f(x)}{\psi(x)}=A<\infty \tag{7} \]

where \(f(x)\) is any function for which the last limit exists, there follow the equality

\[ \lim_{n\to\infty}\frac{\Phi_n(f)}{\Phi_n(\psi)}=A, \tag{8} \]

it is necessary and sufficient that for every \(\delta>0\) the equality

\[ \lim_{n\to\infty}\frac{\alpha_n(\delta)}{\Phi_n(\psi)}=0 \tag{9} \]

hold, where

\[ \alpha_n(\delta)=\int_{|x-c|\geq\delta} d\varphi_n(x). \tag{10} \]

Proof. Putting \(f(x)=\psi^2(x)\), we obtain

\[ \Phi_n(f)=\int_a^b f(x)\,d\varphi_n(x)\geq \int_{|x-c|\geq\delta} f(x)\,d\varphi_n(x)\geq m\alpha_n(\delta), \]

where \(m\) is the least value of the function \(f(x)\) on the set \(|x-c|\geq\delta\). Consequently,

\[ \frac{\Phi_n(f)}{\Phi_n(\psi)}\geq m\frac{\alpha_n(\delta)}{\Phi_n(\psi)}. \tag{11} \]

If (9) does not hold for some \(\delta>0\), then the right-hand side, and hence the left-hand side, of the last inequality do not tend to zero, although

\[ \lim_{x\to c}\frac{f(x)}{\psi(x)}=\lim_{x\to c}\psi(x)=0. \]

To prove the sufficiency of the conditions of the theorem, note that from (7) there follows the inequality

\[ |f(x)-A\psi(x)|<\varepsilon\psi(x),\qquad |x-c|<\delta . \tag{12} \]

Now we have

\[ \Phi_n(f)-A\Phi_n(\psi)=\int_a^b\{f(x)-A\psi(x)\}\,d\varphi_n(x)= \]

\[ =\int_{|x-c|\leqslant\delta}\{f(x)-A\psi(x)\}\,d\varphi_n(x) +\int_{|x-c|\geqslant\delta}\{f(x)-A\psi(x)\}\,d\varphi_n(x). \]

But

\[ \int_{|x-c|\geqslant\delta}|f(x)-A\psi(x)|\,d\varphi_n(x)\leqslant M\alpha_n(\delta), \qquad M=\|f(x)-A\psi(x)\|, \]

\[ \int_{|x-c|<\delta}|f(x)-A\psi(x)|\,d\varphi_n(x) <\varepsilon\int_{|x-c|<\delta}\psi(x)\,d\varphi_n(x) \leqslant \varepsilon\Phi_n(\psi). \]

Consequently,

\[ |\Phi_n(f)-A\Phi_n(\psi)|<\varepsilon\Phi_n(\psi)+M\alpha_n(\delta). \]

From this inequality, (9), and the arbitrariness of \(\varepsilon>0\), the theorem follows.

To obtain Voronovskaya’s theorem from this theorem, it is necessary to note that

\[ B_n(1;x)=1,\qquad B_n\{(t-x);x\}=0,\qquad B_n\{(t-x)^2;x\}=\frac{x(1-x)}{n}, \]

\[ \alpha_n(\delta)= \sum_{|k/n-x|\geqslant\delta} C_n^k x^k(1-x)^{n-k} =o\!\left(\frac1n\right). \]

Since

\[ \alpha_n(\delta)=o\bigl(B_n\{(t-x)^2;x\}\bigr),\qquad 0<x<1, \]

by Theorem 2, from the equality

\[ \lim_{t\to x} \frac{f(t)-f(x)-(t-x)f'(x)}{(t-x)^2} =\frac12 f''(x) \]

there follows the equality

\[ \lim_{n\to\infty} \frac{B_n(f;x)-f(x)B_n(1;x)-f'(x)B_n(t-x;x)} {B_n\{(t-x)^2;x\}} = \]

\[ =\lim_{n\to\infty} \frac{B_n(f;x)-f(x)} {\dfrac{x(1-x)}{n}} = \frac12 f''(x), \]

i.e. Voronovskaya’s theorem.

Let us also apply Theorem 2 to Jackson’s operators

\[ D_n(f;x)=\frac{3}{2n(2n^2+1)}\frac1\pi \int_{-\pi}^{\pi} f(t+x) \left(\frac{\sin(nt/2)}{\sin(t/2)}\right)^4\,dt . \]

We have

\[ \alpha_n(\delta)= \frac{3}{2n(2n^2+1)}\frac1\pi \left\{ \int_{-\pi}^{-\delta}+\int_{\delta}^{\pi} \left(\frac{\sin(nt/2)}{\sin(t/2)}\right)^4\,dt \right\} = O\!\left(\frac1{n^3}\right). \]

On the other hand,

\[ D_n\left(\sin^2 \frac{t-x}{2}; x\right) = \frac{3}{2n(2n^2+1)}\frac{1}{\pi} \int_{-\pi}^{\pi} \left(\frac12-\frac12\cos nt\right) \frac{\sin^2(nt/2)}{\sin^2(t/2)}\,dt = \]

\[ = \frac{3}{2(2n^2+1)} = O\left(\frac{1}{n^2}\right). \]

Applying Theorem 2, from the equality

\[ \lim_{t\to x} \frac{f(t)-f(x)-f'(x)\sin(t-x)} {\sin^2\frac{t-x}{2}} = 2f''(x) \]

we obtain

\[ \lim_{n\to\infty} \frac{D_n(f;x)-f(x)} {D_n\left(\sin^2\frac{t-x}{2};x\right)} = \lim_{n\to\infty} \frac{4}{3}n^2\{D_n(f;x)-f(x)\} = 2f''(x). \]

Consequently,

\[ \lim_{n\to\infty}4n^2\bigl(D_n(f;x)-f(x)\bigr)=6f''(x). \]

It would not be difficult to show that the last relation holds uniformly for all \(x\), if the periodic function \(f(x)\) has a continuous second derivative. Since the linear and positive operator \(D_n(f,x)\) is a trigonometric polynomial of order not exceeding \(2n\), the remark to Theorem 1 concerning the existence of operators realizing the order of approximation \(1/n^2\) for good functions has also been proved.

Moscow City Pedagogical Institute named after V. P. Potemkin

Received
16 XII 1956

REFERENCES

1. E. V. Voronovskaya, DAN, A, 79 (1932).