UDC 517.512.2

MATHEMATICS

G. I. NATANSON

SOME CASES IN WHICH FOURIER SUMS GIVE AN APPROXIMATION OF THE ORDER OF THE BEST

(Presented by Academician V. I. Smirnov on 15 IV 1968)

It is known that for functions of the class \(W^rH_\omega\) (i.e., \(2\pi\)-periodic functions having \(r\)-th derivatives, whose modulus of continuity \(\omega(f^{(r)},\delta)\) does not exceed the true majorant of moduli of continuity \(^*\ \omega(\delta)\)), the deviation of the partial sums of the trigonometric Fourier series has order \(n^{-r}\omega(n^{-1})\ln n\), and for the whole class \(W^rH_\omega\) this order is definitive.

In the present note it is shown that for certain rather broad subclasses of the class \(W^rH_\omega\) the factor \(\ln n\) can be omitted. In other words, for functions from such subclasses the order of the deviation of Fourier sums coincides with the order of the quantity \(\sup\limits_{f\in W^rH_\omega} E_n(f)\), where \(E_n(f)\) is the best approximation of the function \(f\) by polynomials

\[ A_0+\sum_{k=1}^{n}(A_k\cos kx+B_k\sin kx). \]

Lemma 1 (see \((^{1,2})\)). If \(f\in W^rH_\omega\) and \(S_n[f;x]\) is the \(n\)-th partial sum of the trigonometric Fourier series of \(f\), then

\[ S_n[f;x]-f(x)=\frac{n^{1-r}}{2\pi^2}\sum_{k=1}^{m}\frac{1}{k} \int_{y_k}^{y_{k+1}} \left[ f^{(r)}\left(x+\frac{r'\pi}{2n}+t\right) + f^{(r)}\left(x+\frac{r'\pi}{2n}-t\right) \right]\times \]

\[ {}\times \sin nt\,dt+(r+1)O\bigl(n^{-r}\omega(n^{-1})\bigr), \]

where \(m=(n-3)/2,\ y_k=(4k+1)\pi/2n,\ r'\in(-2,2],\ r'\equiv r\pmod 4\); the constant entering the \(O\)-term is absolute.

Lemma 2 \((^3)\). If the majorant \(\omega\) satisfies the condition

\[ \delta\int_{\delta}^{\pi}\frac{\omega(t)}{t^2}\,dt=O(\omega(\delta)), \tag{*} \]

then there exist constants \(A\) and \(\mu\in(0,1)\) such that \(\delta_1^\mu\omega(\delta_2)\le A\delta_2^\mu\omega(\delta_1)\) for \(0<\delta_1<\delta_2\le\pi\).

Theorem 1. Let \(f\in H_\omega=W^0H_\omega\) and be monotone on the intervals \([c_i,c_{i+1}]\), where \(c_0<c_1<\cdots<c_N=c_0+2\pi\). If \(\omega\) satisfies condition \((*)\), then

\[ S_n[f;x]-f(x)=O\bigl(\omega(n^{-1})\ln(N+1)\bigr), \]

where the constant entering the \(O\)-term depends only on \(\omega\).

Proof. On the basis of Lemma 1 we have

\[ S_n[f;x]-f(x)=\frac{n}{2\pi^2}\left\{ \sum_{k=1}^{m}\frac{1}{k} \int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt+ \right. \]

\[ \left. +\sum_{k=1}^{m}\frac{1}{k} \int_{y_k}^{y_{k+1}} f(x-t)\sin nt\,dt \right\} + O\left(\omega\left(\frac{1}{n}\right)\right). \]

\[ \text{}^*\ \omega(\delta)\text{ is a true majorant of moduli of continuity if: 1) }\omega\text{ is defined on }[0,+\infty); \]
\[ \text{2) }\lim_{\delta\to0}\omega(\delta)=\omega(0)=0;\ \text{3) for }0\le\delta_1<\delta_2\text{ one has }0\le\omega(\delta_2)-\omega(\delta_1)\le\omega(\delta_2-\delta_1). \]

Let \(k_0, k_1,\ldots,k_{p+1}\) be the numbers of those intervals \([y_k,y_{k+1}]\) into which at least one of the points \(y_1,y_m,c_i-x\) falls (or a point differing from \(c_i-x\) by an integral multiple of \(2\pi\)). Obviously, \(p\leq N\). Then (as usual, \(\sum_{k=\mu}^{\nu} a_k=0\) if \(\mu>\nu\))

\[ \Sigma_1=\sum_{k=1}^{m}\frac1k\int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt =\sum_{i=0}^{p}\sum_{k=k_i+1}^{k_{i+1}-1}\frac1k\int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt+ \]

\[ +\sum_{i=0}^{p+1}\frac1{k_i}\int_{y_{k_i}}^{y_{k_i+1}} f(x+t)\sin nt\,dt =\Sigma_1'+\Sigma_1''. \]

Since

\[ \int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt=O\bigl(n^{-1}\omega(n^{-1})\bigr), \]

we have

\[ \Sigma_1''=O\!\left(\frac1n\,\omega\!\left(\frac1n\right)\right) \sum_{i=0}^{p+1}\frac1{k_i} =O\!\left(\frac1n\,\omega\!\left(\frac1n\right)\right) \sum_{i=1}^{N+2}\frac1i =O\!\left(\frac{\ln(N+1)}{n}\,\omega\!\left(\frac1n\right)\right). \]

Since \(f(x+\cdot)\) is monotone on \([y_{k_i+1},y_{k_{i+1}}]\), for \(k_i<k<k_{i+1}\) we have

\[ \left|\int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt\right| \leq \frac{2\pi}{n}\left|\Delta_{2\pi/n}f(x+y_k)\right|, \]

whence

\[ |\Sigma_1'|\leq \frac{2\pi}{n}\sum_{i=0}^{p} \left|\sum_{k=k_i+1}^{k_{i+1}-1}\frac1k\Delta_{2\pi/n}f(x+y_k)\right|. \]

But

\[ \sum_{k=k_i+1}^{k_{i+1}-1}\frac1k\Delta_{2\pi/n}f(x+y_k) =\sum_{k=k_i+1}^{k_{i+1}-1} \frac{f(x+y_k)-f(x+y_{k_i+1})}{(k-1)k} + \]

\[ +\frac{f(x+y_{k_i+1})-f(x+y_{k_{i+1}})}{k_{i+1}-1}, \]

\[ |f(x+y_k)-f(x+y_{k_i+1})| \leq \omega\!\left(\frac{2(k-1)}{n}\pi\right), \]

\[ |f(x+y_{k_i+1})-f(x+y_{k_{i+1}})| \leq \omega\!\left(\frac{2(k_{i+1}-1-k_i)}{n}\pi\right). \]

Thus,

\[ |\Sigma_1'|\leq \frac{4\pi}{n}\sum_{i=1}^{p} \left[ \sum_{k=k_i+1}^{k_{i+1}-1} \frac{\omega((k-1)\pi n^{-1})}{(k-1)k} + \frac{\omega((k_{i+1}-1-k_i)\pi n^{-1})}{k_{i+1}-1} \right]\leq \]

\[ \leq \frac{8\pi}{n}\sum_{k=2}^{m}\frac{\omega((k-1)\pi n^{-1})}{k^2} +\frac{4\pi}{n}\sum_{i=0}^{p} \frac{\omega((k_{i+1}-1-k_i)\pi n^{-1})}{k_{i+1}-1}. \]

The first of the resulting sums, by virtue of \((*)\), is
\(O(n^{-1}\omega(n^{-1}))\).
To estimate the second sum we apply Lemma 2:

\[ \frac{4\pi}{n}\sum_{i=0}^{p} \frac{\omega((k_{i+1}-1-k_i)\pi n^{-1})}{k_{i+1}-1} = O\!\left(\frac1n\,\omega\!\left(\frac1n\right)\right) \sum_{i=0}^{p}\frac{(k_{i+1}-1-k_i)^{\mu}}{k_{i+1}-1}. \]

Denote by \(l\) such a number that
\(k_l\leq (p+1)^{1/(1-\mu)}+1<k_{l+1}\) (if all the numbers
\(k_1,k_2,\ldots,k_{p+1}\) lie on one side of \((p+1)^{1/(1-\mu)}+1\), then

below one of the sums vanishes). Then

\[ \sum_{i=0}^{p} \frac{(k_{i+1}-1-k_i)^\mu}{k_{i+1}-1} \ll \sum_{i=0}^{l-1} \frac{k_{i+1}-1-k_i}{k_{i+1}-1} + \sum_{i=l}^{p} (k_{i+1}-1)^{\mu-1} < \]

\[ < \sum_{i=0}^{l-1} \sum_{k=k_i}^{k_{i+1}-1} \frac{1}{k} + \sum_{i=0}^{p} [(p+1)^{1/(1-\mu)}]^{\mu-1} \ll \frac{1}{1-\mu}\ln(p+1)+2. \]

Thus,

\[ \Sigma_1' = O\left(\frac{1}{n}\omega\left(\frac{1}{n}\right)\right) \left(1+\frac{1}{1-\mu}\ln(p+1)+2\right) = O\left(\frac{\ln(N+1)}{n}\omega\left(\frac{1}{n}\right)\right). \]

Consequently,

\[ \Sigma_1 = O\left(\ln(N+1)\omega n^{-1}/n\right). \]

Similarly,

\[ \Sigma_2 = \sum_{k=1}^{m}\frac{1}{k} \int_{y_k}^{y_{k+1}} f(x-t)\sin nt\,dt = O\left(\frac{\ln(N+1)}{n}\omega\left(\frac{1}{n}\right)\right). \]

The theorem is proved.

Remark 1. If \(\omega(\delta)=\delta^\mu\), where \(0<\mu<1\), then

\[ S_n[f;x]-f(x)=O\left(\frac{\ln(N+1)}{(1-\mu)n^\mu}\right), \]

and the constant entering the \(O\)-term is absolute.

Remark 2. The condition of piecewise monotonicity of \(f\) cannot be replaced by the condition of bounded variation.

Remark 3. The condition \((*)\) imposed on the majorant \(\omega\) is essential.

Lemma 3. If \(f\) is convex upward on \([y_k,y_{k+1}]\), then

\[ \frac{4}{\pi n}\Delta_{\pi/n}^{2} f(y_k) \ll \int_{y_k}^{y_{k+1}} f(t)\sin nt\,dt \ll 0. \]

Proof. We shall show that

\[ -\frac{8}{\pi}g(\pi) \ll \int_{0}^{2\pi} g(z)\cos z\,dz \ll 0, \]

if \(g\) is convex upward on \([0,2\pi]\), \(g(0)=g(2\pi)=0\). Let

\[ h(z)=[g(z)+g(2\pi-z)]/2,\qquad l(z)=2h(\pi/2)z/\pi. \]

Then

\[ \int_{0}^{2\pi} g(z)\cos z\,dz = 2\int_{0}^{\pi} h(z)\cos z\,dz \gg \]

\[ \gg 2\int_{0}^{\pi} l(z)\cos z\,dz = -\frac{8}{\pi}h\left(\frac{\pi}{2}\right) \gg -\frac{8}{\pi}g(\pi). \]

The inequality

\[ \int_{0}^{2\pi} g(z)\cos z\,dz\ll 0 \]

is obvious. To complete the proof it is enough to put

\[ f\left(\frac{z}{n}+y_k\right) - \frac{z}{2\pi}\,[f(y_{k+1})-f(y_k)] - f(y_k) = g(z) \]

and to note that

\[ g(\pi)=-\frac{1}{2}\Delta_{\pi/n}^{2} f(y_k). \]

Theorem 2. If \(f\in H_\omega\) and is convex (upward or downward) on each of the intervals \([c_i,c_{i+1}]\), where \(c_0<c_1<\cdots<c_N=c_0+2\pi\), then

\[ S_n[f;x]-f(x)=O\bigl(\omega(n^{-1})\ln(N+1)\bigr), \]

where the constant entering the \(O\)-term is absolute.

Proof. With the preceding notation it is enough to show that

\[ \Sigma_1' = O\left(\ln(N+1)\omega n^{-1}/n\right). \]

The function \(f(x+\cdot)\) preserves the direction of convexity on

\[ [y_{k_i+1},y_{k_{i+1}}]. \]

Hence, by Lemma 3,

\[ |\Sigma_1'| \ll \sum_{i=0}^{p} \left| \sum_{k=k_i+1}^{k_{i+1}-1} \frac{1}{k} \int_{y_k}^{y_{k+1}} f(x+t)\sin nt\,dt \right| \ll \frac{4}{\pi n} \sum_{i=0}^{p} \left| \sum_{k=k_i+1}^{k_{i+1}-1} \frac{1}{k} \Delta_{\pi/n}^{2} f(y_k) \right|. \]

Moreover, \(y_{k+1/2}=(4k+3)\pi/2n\),

\[ |\Sigma'_1|\leq \frac{8}{\pi n}\sum_{i=0}^{p} \left|\sum_{\nu=2k_i+2}^{2k_{i+1}-2} \frac{1}{\nu}\Delta_{\pi/n}^{2}f(y_{\nu/2})\right|. \]

Next

\[ \sum_{\nu=2k_i+2}^{2k_{i+1}-2}\frac{1}{\nu}\Delta_{\pi/n}^{2}f(y_{\nu/2}) = \sum_{\nu=2k_i+2}^{2k_{i+1}-2} \frac{\Delta_{\pi/n}f(y_{\nu/2})}{(\nu-1)\nu} - \frac{\Delta_{\pi/n}f(y_{k_i+1})}{2k_i+1} + \frac{\Delta_{\pi/n}f(y_{k_{i+1}-1/2})}{2k_{i+1}-2} \]

and \(|\Delta_{\pi/n}f(x)|\leq \omega(\pi/n)\). Therefore

\[ \left| \sum_{\nu=2k_i+2}^{2k_{i+1}-2} \frac{1}{\nu}\Delta_{\pi/n}^{2}f(y_{\nu/2}) \right| \leq \omega\left(\frac{\pi}{n}\right) \left[ \sum_{\nu=2k_i+1}^{2k_{i+1}} \frac{1}{(\nu-1)\nu} + \frac{1}{k_i} \right]. \]

Thus,

\[ |\Sigma'_1|\leq \frac{8}{\pi n}\omega\left(\frac{\pi}{n}\right) \left[ \sum_{i=0}^{p}\sum_{\nu=2k_i+1}^{2k_{i+1}} \frac{1}{(\nu-1)\nu} + \sum_{i=0}^{p}\frac{1}{k_i} \right] = O\left(\frac{\ln(N+1)}{n}\omega\left(\frac{1}{n}\right)\right), \]

which was required to be proved.

We shall call a function \(f\) \(q\)-monotone on \([a,b]\) if, for all \(x\) and \(h>0\) such that \(x,x+qh\in [a,b]\),

\[ \Delta_h^{q}f(x)=\sum_{\nu=0}^{q}(-1)^{q+\nu}C_q^\nu f(x+\nu h)\geq 0\;(\leq 0). \]

It is not difficult to verify that: a) if \(f\) is \(q\)-monotone on \([a,b]\) and \(f'\) exists, then \(f'\) is \((q-1)\)-monotone on \([a,b]\); b) if a continuous \(f\) is \(q\)-monotone on \([a,b]\), then either \(f\) is \((q-1)\)-monotone on \([a,b]\), or there exists a \(c\in(a,b)\) such that \(f\) is \((q-1)\)-monotone on \([a,c]\) and \([c,b]\).

Theorem 3. Let \(f\in WrH_\omega\) and let \(f\) be \(q\)-monotone on the intervals \([c_i,c_{i+1}]\), where \(c_0<c_1<\cdots<c_N=c_0+2\pi\). Suppose, further, that \(r\leq q-1\), and for \(r=q-1\) condition \((*)\) is satisfied for \(\omega\). Then
\[ S_n[f;x]-f(x)=O\bigl((q+\ln(N+1))\omega(n^{-1})n^{-r}\bigr), \]
where the constant entering the \(O\)-term is absolute for \(r<q-1\) and depends only on \(\omega\) for \(r=q-1\).

Proof. For \(r=q-1\) the function \(f^{(r)}\) is monotone on the intervals \([c_i,c_{i+1}]\). The proof of Theorem 1 (with insignificant modifications) gives the required result. If \(r<q-1\), then \(f^{(r)}\), being \((q-r)\)-monotone on \([c_i,c_{i+1}]\), is convex on the intervals \([d_i,d_{i+1}]\), where \(d_0<d_1<\cdots<d_M=d_0+2\pi\) and \(M\leq 2^{q-r-2}N\). Slightly modifying the proof of Theorem 2, here also we obtain the needed estimate.

Assertions analogous to those proved are apparently also valid for Fourier series in Jacobi polynomials \(P_n^{(\alpha,\beta)}(x)\). We confine ourselves to formulating the following result obtained by us.

Let \(0<\mu\leq 1\), \(-1/2<\alpha,\beta<1/2\), and let \(S_n^{(\alpha,\beta)}[f;x]\) be the \(n\)-th partial sum of the Fourier–Jacobi series of a function \(f\) defined on \([-1,1]\). Then, as shown in \((^4)\),

\[ \sup_{f\in \operatorname{Lip}_1\mu} \left|S_n^{(\alpha,\beta)}[f;1]-f(1)\right| \asymp n^{\alpha+1/2-\mu}. \]

If, however, \(f\in \operatorname{Lip}_1\mu\) and the function \(f(\cos\cdot)\) is convex on the intervals \([c_i,c_{i+1}]\), where \(0=c_0<c_1<\cdots<c_N=\pi\), then

\[ S_n^{(\alpha,\beta)}[f;1]-f(1) = O\left(N n^{\alpha-1/2-\mu}+N^{\alpha+1/2+\mu}n^{-2\mu}+n^{-\mu}\right). \]

Leningrad State University
named after A. A. Zhdanov

Received
9 IV 1968

CITED LITERATURE

\(^{1}\) A. V. Efimov, Izv. AN SSSR, ser. matem., 24, 243 (1960).
\(^{2}\) G. I. Natanson, Vestn. LGU, No. 19, 20 (1966).
\(^{3}\) N. K. Bari, S. B. Stechkin, Tr. Mosk. matem. obshch., 5, 483 (1956).
\(^{4}\) S. A. Atakhanov, G. I. Natanson, DAN, 166, 9 (1966).