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\begin{document}
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\title{$n$-Widths and Kolmogorov's Inequality in Hardy--Sobolev
Classes}
\author{K. Yu. Osipenko (Moscow)}

\maketitle

Let $H^r_{\infty,\beta}$, $r\ge0$ be the class of all real-valued on $\mathbb R$ and analytic on the strip $S_\beta:=\{\,z\in\mathbb C:|\IM z|<\beta\,\}$ functions which satisfy the restriction $|f^{(r)}(z)|\le1$, $z\in S_\beta$. Denote by $\widetilde H^r_{\infty,\beta}$ $2\pi$-periodic functions from $H^r_{\infty,\beta}$. Let $\Lambda$ and $\Lambda'$ be the complete
elliptic integrals of the first kind with moduli $\lambda$ and $\lambda'=
\sqrt{1-\lambda^2}$. For every $\nu\in(0,\infty)$ determine $\lambda=\lambda
(\nu)$ by the equation
$$\frac{\Lambda'}\Lambda=\frac{4\beta\nu}\pi.$$
For $r\ge-1$ put
$$\Phi^\beta_{\nu,r}(z):=\frac\pi{\sqrt\lambda\Lambda\nu^r}\sum_{s=0}^\infty
\frac{\sin((2s+1)\nu z-\pi r/2)}{(2s+1)^r\sinh((2s+1)2\nu\beta)}.$$
The functions $\Phi^\beta_{\nu,r}$ play a role which is analogous to that
played by the Euler perfect splines in the case of the Sobolev classes
of functions. It can be shown that
$$\|\Phi^\beta_{\nu,r}\|_\infty:=\frac\pi{\sqrt\lambda\Lambda\nu^r}\sum_{s=0}
^\infty\frac{(-1)^{s(r+1)}}{(2s+1)^r\sinh((2s+1)2\nu\beta)}$$
where $\|\cdot\|_\infty$ is the norm in $L_\infty(\mathbb R)$.

Denote by $d_n$, $\lambda_n$ and $d^n$ the Kolmogorov, linear and Gel'fand
$n$-widths, respectively. The following two results are considered.

\begin{theorem}
For all integer $r\ge0$ and $L_\infty:=L_\infty[0,2\pi]$
$$d_{2n}\big(\widetilde H^r_{\infty,\beta},L_\infty\big)=\lambda_{2n}\big(
\widetilde H^r_{\infty,\beta},L_\infty\big)=d^{2n}\big(\widetilde H^r_{
\infty,\beta},L_\infty\big)=\|\Phi^\beta_{n,r}\|_\infty.$$
\end{theorem}

\begin{theorem} 
Suppose that $\delta\in(0,\infty)$ if $r\ge1$, and $\delta\in(0,1)$ if $r=0$. Then for all $1\le k\le r+1$
$$\sup_{\substack{f\in H^r_{\infty,\beta}\\\|f\|_\infty\le
\delta}}\|f^{(k)}\|_\infty=\|\Phi^\beta_{\nu,r-k}\|_\infty$$
where $\nu$ is determined by the equation
$$\|\Phi^\beta_{\nu,r}\|_\infty=\delta.$$
\end{theorem}
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