Reports of the Academy of Sciences of the USSR
1964, Volume 159, No. 5

MATHEMATICS

G. M. MIRAKYAN

APPROXIMATION OF CONTINUOUS FUNCTIONS BY S. N. BERNSTEIN POLYNOMIALS

(Presented by Academician S. N. Bernstein on 27 I 1964)

For estimating the approximation of an arbitrary function \(f(x)\), continuous on the segment \(0 \leqslant x \leqslant 1\), by means of the S. N. Bernstein polynomials

\[ B_n(x)=\sum_{k=0}^{n} f\left(\frac{k}{n}\right)b_k(x) \quad \left(b_k(x)=C_n^k x^k(1-x)^{n-k}\right) \tag{1} \]

the estimate, valid on this segment for any natural \(n\), is

\[ |f(x)-B_n(x)|\leqslant \frac{5}{4}\,\omega\left(\frac{1}{\sqrt n}\right), \tag{2} \]

where here and below \(\omega(\delta)\) denotes the modulus of continuity of \(f(x)\). Inequality (2) was established by Popoviciu \((^1)\). Under the same assumptions, Li Wen-ching obtained the more precise estimate \((^2)\)

\[ |f(x)-B_n(x)|\leqslant \frac{19}{16}\,\omega\left(\frac{1}{\sqrt n}\right). \tag{3} \]

Below an upper estimate is obtained for the best asymptotic (as \(n\to\infty\)) value \(S_n\) for

\[ \max_{[0,1]} |f(x)-B_n(x)| \sim S_n . \]

Keeping in mind the identity

\[ \sum_{k=0}^{n} b_k(x) \equiv 1, \tag{4} \]

we successively find

\[ |f(x)-B_n(x)| \leqslant \sum_{k=0}^{n} \left|f(x)-f\left(\frac{k}{n}\right)\right| b_k(k) \leqslant \sum_{k=0}^{n} \omega\left(\left|x-\frac{k}{n}\right|\right)b_k(x). \tag{5} \]

Denote by \(v\) the integer part of the number \(\dfrac{|x-k/n|}{\delta}\), where \(\delta\) denotes an arbitrary positive number,

\[ v=\mathrm{E}\left(\frac{|x-k/n|}{\delta}\right). \tag{6} \]

It follows that

\[ v \leqslant \frac{|x-k/n|}{\delta} < v+1 . \tag{7} \]

From (7) we find
\[ |x-k/n|<(v+1)\delta, \]
and, by the property of \(\omega(\delta)\), we have
\[ \omega(|x-k/n|)\leqslant \omega[(v+1)\delta]\leqslant (v+1)\omega(\delta). \]

With the aid of (5) and the last inequality we find
\[ |f(x)-B_n(x)|\leqslant \omega(\delta)\sum_{k=0}^{n}\left[1-v\left(x,\frac{k}{n}\right)\right]b_k(x). \]

The last inequality, using (4), can be put in the form
\[ |f(x)-B_n(x)|\leqslant \omega(\delta)\left[1+\sum_{k=0}^{n}v\left(x,\frac{k}{y}\right)b_k(x)\right]. \tag{8} \]

Next we estimate
\[ \sum_{k=0}^{n}v\left(x,\frac{k}{n}\right)b_k(x). \tag{9} \]

From the left-hand side of the double inequality (7) we obtain
\[ v\left(x,\frac{k}{n}\right)\leqslant \frac{|x-k/n|^\alpha}{\delta^\alpha}, \tag{10} \]
where \(\alpha\geqslant 1\). From (9) and (10) we have
\[ \sum_{k=0}^{n}v\left(x,\frac{k}{n}\right)b_k(x) < \frac{1}{\delta^\alpha}\sum_{k=0}^{n}\left|x-\frac{k}{n}\right|^\alpha b_k(x). \tag{11} \]

Let us find the asymptotic value of the right-hand side of (11) as \(n\to\infty\). For this purpose it is convenient to represent the sum on the right-hand side of (11) in the form of an integral over a simple piecewise-smooth closed contour \(C\), which in the complex \(z\)-plane encloses the segment \([0,1]\) of the real axis \(Ox\):
\[ \sum_{k=0}^{n}\left|x-\frac{k}{n}\right|^\alpha b_k(x) = \frac{n!}{2\pi i}\int_C |z-x|^\alpha \frac{(x-1)^{n(1-z)}x^{nz}} {z(nz-1)(nz-2)\ldots(nz-n)}\,dz. \]

As a result of applying Stirling’s formula we obtain
\[ \sum_{k=0}^{n}\left|x-\frac{k}{n}\right|^\alpha b_k(x) = \sqrt{\frac{n}{2\pi}}\int_C (1+\rho_n')|z-x|^\alpha \frac{1}{\sqrt{z(z-1)}} \left[\left(\frac{x-1}{z-1}\right)^{1-z}\left(\frac{x}{z}\right)^z\right]^n dz, \tag{12} \]
where here \(n\rho_n'\) remains bounded as \(n\to\infty\), and, moreover, we assume \(0<x<1\) \(\left({}^{3}\right)\).

Applying to the right-hand side of (12) Perron’s method \(\left({}^{4}\right)\) for finding the asymptotic value of integrals, we obtain, as \(n\to\infty\) and \(0<x<1\),
\[ \sum_{k=0}^{n}\left|x-\frac{k}{n}\right|^\alpha b_k(x) \sim \frac{1}{\sqrt{\pi}}[2x(1-x)]^{\alpha/2}\Gamma\left(\frac{\alpha+1}{2}\right)\frac{1}{n^{\alpha/2}}. \]

Hence we conclude that, in the interval \(0<x<1\),
\[ \max_{(0,1)}\sum_{k=0}^{n}\left|x-\frac{k}{n}\right|^\alpha b_k(x) \sim \frac{1}{\sqrt{\pi}}\frac{\Gamma((\alpha+1)/2)}{2^{\alpha/2}}\frac{1}{n^{\alpha/2}}. \tag{13} \]

Thus, from (8), (11), and (13) we obtain on the segment \(0\leqslant x\leqslant 1\)

\[ \max_{[0,1]} |f(x)-B_n(x)|\sim S_n\leqslant \left[1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\alpha+1)/2)}{2^{\alpha/2}}\, \frac{1}{(\delta\sqrt n)^\alpha}\right]\omega(\delta), \]

since \(f(0)-B_n(0)=0\) and \(f(1)-B_n(1)=0\).

Putting here \(\delta=1/\sqrt n\), we obtain

\[ \max_{[0,1]} |f(x)-B_n(x)|\sim S_n\leqslant \left[1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\alpha+1)/2)}{2^{\alpha/2}}\right] \omega\left(\frac{1}{\sqrt n}\right). \tag{14} \]

Put in (14) \(\alpha=2\); then

\[ \max_{[0,1]} |f(x)-B_n(x)|\sim \frac{5}{4}\,\omega\left(\frac{1}{\sqrt n}\right) =1.25\,\omega\left(\frac{1}{\sqrt n}\right); \]

this is a result due to Popoviciu. If, however, we put \(\alpha=4\), then we obtain

\[ \max_{[0,1]} |f(x)-B_n(x)|\sim \frac{19}{16}\,\omega\left(\frac{1}{\sqrt n}\right) =1.1875\,\omega\left(\frac{1}{\sqrt n}\right); \]

this is the estimate obtained by Li Wen-ching.

Having in the right-hand side of (14) a general expression for arbitrary \(\alpha\geqslant 1\), it is not difficult to obtain for the constant

\[ 1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\alpha+1)/2)}{2^{\alpha/2}} \tag{15} \]

the best value. For this it is necessary to find such a number \(\alpha=\xi\) for which expression (15) is the least. This number \(\xi\) is the unique positive root of the equation

\[ \frac{\Gamma'((\xi+1)/2)}{\Gamma((\xi+1)/2)}-\ln 2=0. \tag{16} \]

Solving this equation, we find

\[ \xi=2.958\ldots . \]

Consequently, we have the upper estimate of the best asymptotic (as \(n\to\infty\)) value \(S_n\)

\[ \max_{[0,1]} |f(x)-B_n(x)|\sim S_n\leqslant \left[1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\xi+1)/2)}{2^{\xi/2}}+o(1)\right] \omega\left(\frac{1}{\sqrt n}\right)= \]

\[ =\left[1.1761\ldots+o(1)\right]\omega\left(\frac{1}{\sqrt n}\right). \tag{17} \]

One may suppose that

\[ S_n\approx \left[1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\xi+1)/2)}{2^{\xi/2}}\right] \omega\left(\frac{1}{\sqrt n}\right) \]

and, moreover,

\[ |f(x)-B_n(x)|\leqslant \left[1+\frac{1}{\sqrt{\pi}}\, \frac{\Gamma((\xi+1)/2)}{2^{\xi/2}}\right] \omega\left(\frac{1}{\sqrt n}\right) \]

is fulfilled for \(0\leqslant x\leqslant 1\) for all natural \(n\).

Odessa Higher Engineering
Marine School

Received
24 I 1964

CITED LITERATURE

\({}^{1}\) T. Popoviciu, Matematica (Cluj), 10, 49 (1935).
\({}^{2}\) Li Wen-ching, Shuxue jinzhhan, 4, No. 4, 567 (1958).
\({}^{3}\) S. N. Bernstein, Collected Works, 2, 1954, p. 312.
\({}^{4}\) V. I. Smirnov, Course of Higher Mathematics, 7th ed., vol. 2, part 2, 1958, pp. 286–295.
\({}^{5}\) G. G. Lorentz, Bernstein Polynomials, Toronto, 1953.
\({}^{6}\) G. M. Mirakyan, Fifth All-Union Conference on Function Theory, Erevan, 1960, p. 74.