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\begin{document}

\title[On the Lebesgue constants]{On the Lebesgue constants for interpolation of analytic functions}
\author{K. Yu.\ Osipenko}




\maketitle


Let $\tau:=(t_1,\ldots,t_n)$ be a system of distinct points in the interval $[-1,1]$ and let $k\in[0,1)$. We put
$$L_n(\tau,p,k)=\max_{t\in[-1,1]}\|D(t)\|_p,$$
where $D(t)=(D_1(t),\ldots,D_n(t))$,
\begin{gather*}
D_j(t)=\frac{\omega_j(t)}{\omega_j(t_j)}\left(1-kW_j^2(t)\right),\quad\omega_j(t)=
\prod_{m\ne j}W_m(t),\\
W_j(t)=\frac{t-t_j}{1-kt_jt},\\
\|a\|_p=\begin{cases}\displaystyle\left(\sum_{j=1}^n|a_j|^p\right)^{1/p},&1\le p<\infty,\\
\displaystyle\max_{1\le j\le n}|a_j|,&p=\infty.\end{cases}
\end{gather*}

For $k=0$
$$L_n(\tau,p,0)=\Lambda_n(\tau,p),$$
where
\begin{gather*}
\Lambda_n(\tau,p)=\max_{t\in[-1,1]}\|l(t)\|_p,\\
l(t)=(l_1(t),\ldots,l_n(t)),\quad l_j(t)=\prod_{\substack{m=1\\m\ne j}}^n\frac{t-t_m}{t_j-t_m}.
\end{gather*}

Thus, $\Lambda_n(\tau,p)$ is the Lebesgue constant occurring in the theory of approximation by Lagrange interpolating polynomials. Lebesgue constants (mainly for $p=1$) are studied in a rather great number of papers (see, for example, \cite{2,3,4,5}, \cite{10,11,12} and the references cited there).

The quantity $L_n(\tau,p,k)$ is linked up with the problems of interpolation of analytic functions. We denote by $B_k$ the class of functions analytic in the circle
$$D_k=\{\,z\in\mathbb C:|z|<1/\sqrt k\,\},\quad k\in)0,1),$$
and bounded there in absolute value by $1$. We call a method $S_0$ the best method  of reconstruction of the function value $f\in B_k$ at a point $t\in D_k$ on the basis of the
information $I(f)=(f(t_1),\ldots,f(t_n))$ if
$$E(t,\tau,k)\equiv\infp_S\sup_{f\in B_k}|f(t)-S(t,I(f))|=\sup_{f\in B_k}|f(t)-S_0(t,I(f))|,$$
where the infimum is taken with respect to all possible functions $S(t,I(f))$. It follows from \cite{8} that for any point $t\in[-l,1]$ the best method is given by
$$S_0(t,I(f))=\sum_{j=1}^nD_j(t)f(t_j),$$
where
$$E(t,\tau,k)=k^{n/2}|W(t)|,\quad W(t)=\prod_{j=1}^n\frac{t-t_j}{1-kt_jt}.$$

Now let us use, instead of $I(f)$, an information on approximate values $\widetilde I=(f_1,\ldots,f_n)$ such that for every $f\in B_k$
$$\|I(f)-\widetilde I\|_q\le\delta.$$
We put
\begin{gather*}
R_q(t,\tau,k,\delta)=\sup_{f\in B_k\vphantom{\widetilde I}}\sup_{\widetilde I:\|I(f)-\widetilde I\|_q\le\delta}|f(t)-S_0(t,\widetilde I)|,\\
F_q(\tau,k,\delta)=\sup_{t\in[-1,1]}R_q(t,\tau,k,\delta).
\end{gather*}

\begin{theorem}\label{T1}
For all $t\in [-1,1]$ we have
$$R_q(t,\tau,k,\delta)=E(t,\tau,k)+\delta\|D(t)\|_p\quad\left(\frac1p+\frac1q=1\right).$$
\end{theorem}

\begin{proof}
Let
$$f\in B_k\quad\mbox{and}\quad\|I(f)-\widetilde I\|_q\le\delta.$$
Then
\begin{multline*}
|f(t)-S_0(t,\widetilde I)|\le|f(t)-S_0(t,I(f))|+\sum_{j=1}^n|D_j(t)|f_j-f(t_j)|\\
\le E(t,\tau,k)+\delta\|D(t)\|_p.
\end{multline*}
Hence
$$R_q(t,\tau,k,\delta)\le E(t,\tau,k)+\delta\|D(t)\|_p.$$
For $1\le p<\infty$, we set
$$f_j=-\delta\frac{|D_j(t)|^{p-1}\sign D_j(t)}{\|D(t)\|_p^{p-1}},\quad j=1,\ldots,n;$$
while for $p=\infty$ and $\displaystyle\max_{1\le j\le n}|D_j(t)|=|D_m(t)|$ let
$$f_j=\begin{cases}-\delta\sign D_m(t),&j=m,\\
0,&j\ne m.\end{cases}$$
We consider the function
$$f(z)=k^{n/2}W(z)\sign W(t).$$
By virtue of the facts that $f\in B_k$, $I(f)=0$, and $\|\widetilde I\|_q=\delta$, we have
$$R_q(t,\tau,k,\delta)\ge|f(t)-S_0(t,\widetilde I)|=E(t,\tau,k)+\delta\|D(t)\|_p.$$
The theorem is proved.
\end{proof}

Thus,
\begin{equation}\label{1}
F_q(\tau,k,\delta)\le\max_{t\in[-1,1]}E(t,\tau,k)+\delta L_n(\tau,p,k).
\end{equation}

It follows from \cite{7} that
$$\infp_\tau\max_{t\in[-1,1]}E(t,\tau,k)=\max_{t\in[-1,1]}E(t,Z,k),$$
where
$$Z=\left\{\sn\left[\left(\frac{2j-1}n-1\right)K,k\right]\right\}_1^n;$$
here and in the sequel, $K$, $L$ and $M$, denote the complete elliptic integrals of the first
kind for the moduli, $k$, $l$ and $\mu_n$, respectively, while $K'$, $L'$ and $M_n'$ denote those for the complemented moduli; concerning the notations $\sn x$, $\cn x$, $\dn x$, etc. we refer
to \cite[p.~240]{13}. The point system $Z$ plays a role which is analogous to that played by the Chebyshev system
$$T=\left\{\cos\frac{2j-1}{2n}\pi\right\}_1^n$$
in the case of interpolation by Lagrange polynomials. In addition, for $k=0$ the system $Z$ coincides with the Chebyshev system, due to the equalities
$$K=\pi/2\quad\mbox{and}\quad\sn(x,0)=\sin x.$$
By this token,
\begin{equation}\label{2}
L_n(Z,p,0)=\Lambda_n(T,p).
\end{equation}

We consider some estimates of the quantity $L_n(Z,p,k)$.

\begin{theorem}\label{T2}
For all $n\ge1$, $2\le p\le\infty$, and $k\in[0,1)$ we have
\begin{equation}\label{3}
L_n(Z,p,k)\le\sqrt{1+\sn\left(\frac{n-1}nK,k\right)}<\sqrt2.
\end{equation}
\end{theorem}

\begin{proof}
Let $0<k<1$. Substituting
\begin{gather*}
\nu_1=\ldots=\nu_n=2,\quad g(\xi)=1,\quad x=\sqrt kt,\\
z_j=\sqrt kt_j,\quad j=1,\ldots,n,
\end{gather*}
in \cite[Lemma~2.1]{9}, we have
\begin{equation}\label{4}
1-\sum_{j=1}^nD_j^2(t)h_j(t)=k^{2n}W^4(t),
\end{equation}
where
$$h_j(t)=\frac{1+kW_j^2(t)}{1-kW_j^2(t)}-\frac{2\omega_j'(t_j)W_j(t)}{\omega_j(t_j)W_j'(t_j)
(1-kW_j^2(t))}.$$
In the case $k=0$, equality \eqref{4} is of the form
\begin{gather*}
1-\sum_{j=1}^nl_j^2(t)v_j(t)=0,\\
v_j(t)=1-\frac{\omega''(t_j)}{\omega'(t_j)}(t-t_j),\quad\omega(t)=\prod_{j=1}^n(t-t_j),
\end{gather*}
and remains valid, since it is the expansion of the unity by the interpolating Hermite polynomial (see \cite{4}, \cite{11}). For the system $Z$ of nodes the function $W(t)$ can be written by means of the first basic transformation of $n$th degree of elliptic functions (see, for example \cite[p.~284]{1}) in the parametric form
\begin{equation}\label{5}
W(t)=(-1)^{n+1}a_n\sn\left[(nx-1)M_n,\mu_n\right],\quad t=\sn[(x-1)K,k],
\end{equation}
where
$$a_n=\prod_{j=1}^{[n/2]}\sn^2\left(\frac{2j-1}nK,k\right),\quad\mu_n=k^na_n^2.$$
Since $W(t)=W_j(t)\omega_j(t)$, we have
$$\omega_j(t_j)=\frac{W'(t_j)}{W_j'(t_j)},\quad\omega_j'(t_j)=\frac{W''(t_j)W_j'(t_j)-W'(t_j)
W_j''(t_j)}{2\left[W_j'(t_j)\right]^2}.$$
Computing the values $W'(t_j)$ and $W''(t_j)$ from representation \eqref{5}, we get
$$h_j(t)=\frac{1+kW_j^2(t)}{1-kW_j^2(t)}-\frac{(1-k)^2(1+kt_j^2)(t-t_j)(1-kt_jt)t_j}
{(1-t_j)^2(1-k^2t_j^2)(1-kt_j^2)(1-kt^2)}.$$
It is not hard to see that for all $t\in[-1,1]$ and $k\in[0,1)$ the inequalities
\begin{gather*}
(1-k)^2\le(1-kt^2)(1-kt_j^2),\quad(1+kt_j^2)\le1+k|t_j|,\\
t_j(t-t_j)(1-kt_jt)\le|t_j|(1-|t_j|)(1-k|t_j|)
\end{gather*}
are satisfied. Hence
$$h_j(t)\ge1-\frac{|t_j|}{1+|t_j|}\ge\left[1+\sn\left(\frac{n-1}nK,k\right)\right]^{-1}.$$
Thus,
\begin{multline*}
\left[1+\sn\left(\frac{n-1}nK,k\right)\right]^{-1}\sum_{j=1}^nD_j^2(t)\le
\sum_{j=1}^nD_j^2(t)h_j(t)\\=1-k^{2n}W^4(t)\le1.
\end{multline*}
Consequently,
$$\|D(t)\|_2\le\sqrt{1+\sn\left(\frac{n-1}nK,k\right)}<\sqrt2.$$
Now inequality \eqref{3} follows from the fact that $\|\cdot\|_p\le\|\cdot\|_2 $ for $2\le p\le\infty$. The theorem is proved.
\end{proof}

By virtue of \eqref{2} and \eqref{3}, for $2\le p\le\infty$ we obtain the inequality
$$\Lambda_n(T,p)\le\sqrt{1+\cos\frac\pi{2n}}<\sqrt2,$$
which was obtained in \cite{4} for the case $p=2$.

Taking \eqref{1} and \eqref{5} into account, Theorem~\ref{T2} implies the following

\begin{corollary}\label{C1}
For all $1\le q\le2$, $k\in(0,1)$, and $\delta\ge0$ we have
$$F_q(Z,k,\delta)\le\sqrt{\mu_n}+\delta\sqrt2.$$
\end{corollary}

We consider the case $p=1$.

\begin{theorem}\label{T3}
For all $n\ge1$ and $k\in[0,1)$ we have
\begin{equation}\label{6}
L_n(Z,1,k)=[1+\alpha_n(k)]\frac L{nM_n}\sum_{j=1}^n\ctn\frac{2j-1}{2n}L,
\end{equation}
where
$$0\le\alpha_n(k)\le\frac{2kK^2}{(1-k)^2n^2},\quad l=\frac{2\sqrt k}{1+k},\quad\ctn x=\frac{\cn x\dn x}{\sn x}$$
$($here and in the sequel, we do not indicate the dependence of the elliptic functions upon the modulus if the latter equals $l$$)$. In addition, $\alpha_n(k)=0$ for
\begin{equation}\label{7}
n\ge\frac{8k}{(1-k)^2}+1.
\end{equation}
\end{theorem}

\begin{proof}
We put
$$t(u)=\left(\frac2{1+k}u-1\right)\left(1-\frac{2k}{1+k}u\right)^{-1}.$$
Changing variable $t=t(u)$, we get that
$$L_n(Z,1,k)=\max_{u\in[0,1]}\sum_{j=1}^n|\widetilde D_j(u)|,$$
where
$$\widetilde D_j(u)=\frac{\widetilde W(u)(1-l^2u^2)}{\widetilde W'(u_j)(u-u_j)(1-l^2u_ju)},\quad
\widetilde W(u)=\prod_{j=1}^n\frac{u-u_j}{1-i^2u_ju},$$
while $t(u_j)=t_j$. By the Gauss transformation of elliptic functions \cite[p.~134]{1}, one
can show that
$$u_j=\sn^2\frac{2j-1}{2n}L,\quad j=1,\ldots,n.$$
Applying the first basic transformation of $2n$-th degree, we can write the function
$\widetilde W(u)$ in the form
\begin{gather*}
\widetilde W(u)=(-1)^n\widetilde a_n\sn\left[\left(\frac{2nt}L+1\right)M_n,\mu_m\right],\quad u=\sn^2t,\\
\widetilde a_n=\prod_{j=1}^n\sn^2\frac{2j-1}{2n}L.
\end{gather*}
Hence, finding $\widetilde W'(u_j)$ and using identities well-known in the theory of elliptic
functions, we get that
\begin{multline}\label{8}
\widetilde D_j(\sn^2t)=(-1)^{j+1}\frac L{2nM_n}\sn\left[\left(\frac{2nt}L+1\right)M_n,\mu_m\right]\\
\times\left[\ctn(t_j+t)+\ctn(t_j-t)\right].
\end{multline}
From the inequality
\begin{multline*}
\sum_{j=1}^n|\widetilde D_j(\sn^2t)|\le\frac L{2nM_n}\left|\sn\left[\left(\frac{2nt}L+1\right)M_n,\mu_m\right]\right|\\
\times\sum_{j=1}^n\left[|\ctn(t_j+t)|+|\ctn(t_j-t)|\right],
\end{multline*}
which turns into equality for $t\in\left[-L/(2n),L/(2n)\right]$, furthermore, from the evenness and periodicity, with period $L/n$, of the function occurring on the right-hand side of this inequality it follows that
$$\max_{u\in[0,1]}\sum_{j=1}^n|\widetilde D_j(u)|=\max_{t\in\left[0,L/(2n)\right]}\sum_{j=1}^n|\widetilde D_j(\sn^2t)|=\max_{u\in[0,u_1]}\sum_{j=1}^n|\widetilde D_j(u)|.$$
Since for $u\in[0,u_1]$
$$|\widetilde D_j(u)|=\frac{\widetilde\omega_j(u)}{|\widetilde W'(u_j)|}\frac{1-l^2u^2}{(1-l^2u_ju)^2},\quad \widetilde\omega_j(u)=\prod_{\substack{m=1\\m\ne j}}^n\frac{u_m-u}{1-i^2u_mu},$$
and since each of the factors $(u_m-u)/(1-l^2u_ju)$ is monotone decreasing, while the function $(1-l^2u^2)/(1-l^2u_ju)^2$ is increasing, we can infer that
$$\max_{u\in[0,u_1]}\sum_{j=1}^n|\widetilde D_j(u)|\le\frac{\widetilde\omega_j(0)}{|\widetilde W'(u_j)|}\frac{1-l^2u_1^2}{(1-l^2u_ju_1)^2}=|\widetilde D_j(0)|\frac{1-l^2u_1^2}{(1-l^2u_ju_1)^2}.$$
By virtue of the inequalities
\begin{gather*}
\frac{1-l^2u_1^2}{(1-l^2u_ju_1)^2}\le1+\frac{2l^2u_nu_1}{(1-l^2u_nu_1)^2},\\
l^2u_nu_1=l^2\sn^2\left(L-\frac L{2n}\right)\sn^2\frac L{2n}\le l^2\sn^4\frac L2=\frac{l^2}{(1+\sqrt{1-l^2})^2}=k,
\end{gather*}
as well as
$$l^2u_nu_1\le l^2\sn^2\frac L{2n}\le\frac{l^2L^2}{4n^2}=\frac{kK^2}{n^2},$$
we have
$$\sum_{j=1}^n|\widetilde D_j(0)|\le\max_{u\in[0,u_1]}\sum_{j=1}^n|\widetilde D_j(u)|\le\left[1+\frac{2kK^2}{(1-k)^2n^2}\right]\sum_{j=1}^n|\widetilde D_j(0)|.$$
Now equality \eqref{6} follows from \eqref{8} for $t=0$.

Let inequality \eqref{7} be satisfied. In order to prove that $\alpha_n(k)=0$ it is enough to
prove the decrease of the functions
$$\varphi_j(u)=\widetilde\omega_j(u)\frac{1-l^2u^2}{(1-l^2u_ju)^2}$$
for $u\in[0,u_1]$. From the relations
\begin{multline*}
(1-l^2u^2)\frac{\varphi_j'(u)}{\varphi_j(u)}=2l^2\frac{u_j-u}{1-l^2u_ju}-(1-l^2u^2)\sum_{
\substack{m=1\\m\ne j}}^n\frac{1-l^2u_m^2}{(u_m-u)(1-l^2u_mu)}\\
\le2l^2u_j-\sum_{\substack{m=1\\m\ne j}}^n\frac{1-l^2u_m^2}{u_m}\le2l^2-(n-1)(1-l^2)
\end{multline*}
it follows that $\varphi_j'(u)\le0$ for $u\in[0,u_1]$ and
$$n\ge\frac{2l^2}{1-l^2}+1=\frac{8k}{(1-k)^2}+1.$$
The theorem is proved.
\end{proof}

\begin{corollary}\label{C2}
If \eqref{7} is satisfied, then we have
$$F_\infty(Z,k,\delta)=\sqrt{\mu_n}+\delta L_n(Z,1,k).$$
\end{corollary}

\begin{proof}
It follows from the proof of Theorem~\ref{T3} that if condition \eqref{7} is satisfied,
then
\begin{equation}\label{9}
L_n(Z,1,k)=\sum_{j=1}^n|D_j(1)|=\frac L{nM_n}\sum_{j=1}^n\ctn\frac{2j-1}{2n}L.
\end{equation}
By virtue of representation \eqref{5},
$$\max_{t\in[-1,1]}E(t,Z,k)=E(1,Z,k)=\sqrt{\mu_n}.$$
Hence and by Theorem~\ref{T1},
\begin{multline*}
F_\infty(Z,k,\delta)\ge R_\infty(1,Z,k,\delta)=E(1,Z,k)+\delta\|D(1)\|_1\\
=\max_{t\in[-1,1]}E(t,Z,k)+\delta L_n(Z,1,k).
\end{multline*}
The lower estimate follows from inequality \eqref{1}. The corollary is proved.
\end{proof}

We note that for $l=0$ we have $\ctn x=\cot x$. Therefore, taking \eqref{2} into account for $k=0$, equalities \eqref{9} turn into the following:
$$\Lambda_n(T,1)=\sum_{j=1}^n|l_j(1)|=\frac1n\sum_{j=1}^n\cot\frac{2j-1}{4n}\pi,$$
which were proved in \cite{3}, \cite{10}.

\begin{theorem}\label{T4}
If \eqref{7} is satisfied, then we have
\begin{multline*}
L_n(Z,1,k)=\frac2\pi\ln n+A(k)+\sum_{s=1}^{m-1}\frac{a_s}{n^{2s}}\left[B_{2s}\left(2^{2s-1}-1\right)-4sH_{2s-1}(k)\right]\\
+\frac{a_m}{n^{2m}}\rho_m(n,k)-h^{2n}\left[\frac8\pi\ln n+B(k)\right]\theta_n(k),
\end{multline*}
where
$$A(k)=\frac2\pi\left(\gamma+\ln\frac4{(1+k)K}\right),$$
$\gamma=0.5772\ldots$ is the Euler constant, $B_{2s}$ are the Bernoulli numbers,
\begin{gather*}
a_s=4\frac{|B_{2s}|\left(2^{2s-1}-1\right)\pi^{2s-1}}{2^{2s}s(2s)!},\\
H_s(k)=\sum_{r=1}^\infty\frac{h^{2r-1}}{1+h^{2r-1}}(2r-1)^s,\quad h=\exp\left(-\frac{\pi K'}{2K}\right),\\
-2\pi mH_{2m}(k)\le(-1)^{m+1}\rho_m(n,k)\le2\pi mH_{2m}(k)+|B_{2m}|\left(2^{2m-1}-1\right),\\
B(k)=4A(k)+\frac{2\pi^2}3H_2(k)+\frac\pi{18},\quad0\le\theta_n(k)\le1.
\end{gather*}
\end{theorem}

\begin{proof}
It was proved in \cite{12} (see also \cite{5}) that for any function $f\in C_{[0,2n]}^{2m}$ we
have
\begin{multline}\label{10}
\sum_{j=1}^nf(2j-1)=\frac12\int_0^{2n}f(x)\,dx\\
+\sum_{s=1}^{m-1}\frac{B_{2s}}{(2s)!}\left(1-2^{2s-1}\right)\left(f^{(2s-1)}(2n)-f^{(2s-1)}(0)
\right)+r_m(n),
\end{multline}
where
\begin{gather*}
r_m(n)=-\frac{2^{2m}}{(2m)!}\int_0^nf^{(2m)}(2x)y_m(x)\,dx,\\ y_m(x)=B_{2m}^*\left(x+\frac12\right)-B_{2m}^*\left(\frac12\right),
\end{gather*}
while $B_{2m}^*(x)$ are the periodic Bernoulli functions with period $1$. In addition, if $f\in C_{[0,2n]}^{2m+2}$ and
$$f^{(2m+2)}(x)f^{(2m)}(x)\ge0\quad\mbox{for all}\quad x\in[0,2n],$$
then
\begin{equation}\label{11}
r_m(n)=\frac{B_{2m}}{(2m)!}\left(1-2^{2m-1}\right)\left[f^{(2m-1)}(2n)-f^{(2m-1)}(0)
\right]\theta,\quad0\le\theta\le1.
\end{equation}
Using the representation of the function $\sn u$ by means of the theta functions and
their expansions into infinite products \cite[pp.~214, 205, 209]{6}, after simple computations we arrive at the equality
\begin{equation}\label{12}
\ctn u=\frac d{du}\ln\sn u=\frac\pi{2L}\cot\frac{\pi u}{2L}-\frac{2\pi}L\sum_{r=1}^\infty\frac{h^r}{1+h^r}\sin\frac{\pi ru}L,\quad|\IM u|<L',
\end{equation}
where
$$h=\exp\left(-\frac{\pi L'}L\right)=\exp\left(-\frac{\pi K'}{2K}\right).$$
We set
$$f(x)=\ctn\frac{xL}{2n}-\frac{2n}{xL}.$$
Then $f(x)=f_1(x)+f_2(x)$, where
\begin{gather}\label{13}
f_1(x)=\frac\pi{2L}\cot\frac{\pi x}{4n}-\frac{2n}{xL}=-\frac1L\sum_{r=1}^\infty\frac{|B_{2r}|\pi^{2r}}{2^{2r-1}(2r)!n^{2r-1}}
x^{2r-1},\quad|x|<4n,\\
f_2(x)=-\frac{2\pi}L\sum_{r=1}^\infty\frac{h^r}{1+h^r}\sin\frac{\pi rx}{2n}.\notag
\end{gather}
Consequently,
$$f^{(2s-1)}(0)=-\frac{|B_{2s}|\pi^{2s}}{L2^{2s}sn^{2s-1}}-
(-1)^{s-1}\frac{4\pi^{2s}}{L2^{2s}n^{2s-1}}\sum_{r=1}^\infty\frac{h^r}{1+h^r}r^{2s-1}.$$
We set
$$g(x)=f_1(x+2n)=-\dfrac\pi{2L}\tan\frac{\pi x}{4n}-\frac{2n}{L(x+2n)}.$$

By the known expansion of $\tan z$, we obtain that
$$f_1^{(2s-1)}(2n)=g^{(2s-1)}(0)=-\frac{|B_{2s}|\pi^{2s}(2^{2s}-1)}{L2^{2s}sn^{2s-1}}+
\frac{(2s-1)!}{L2^{2s-1}n^{2s-1}}.$$
In this way,
\begin{multline*}
f^{(2s-1)}(2n)=-\frac{|B_{2s}|\pi^{2s}(2^{2s}-1)}{L2^{2s}sn^{2s-1}}+
\frac{(2s-1)!}{L2^{2s-1}n^{2s-1}}\\
-(-1)^{s+1}\frac{4\pi^{2s}}{L2^{2s}n^{2s-1}}\sum_{r=1}^\infty(-1)^r\frac{h^r}{1+h^r}r^{2s-1}.
\end{multline*}
Now from \eqref{10} and the fact that
\begin{multline*}
\frac12\int_0^{2n}\left(\ctn\frac{xL}{2n}-\frac{2n}{xL}\right)\,dx=\frac nL\int_0^L\left(\ctn u-\frac1u\right)\,du=\frac nL\log\frac{\sn u}u\bigg|_0^L\\
=\frac nL\ln\frac1L,
\end{multline*}
it follows that
\begin{multline*}
\frac Ln\sum_{j=1}^n\left[\ctn\frac{2j-1}{2n}L-\frac{2n}{(2j-1)L}\right]=\ln\frac1L\\
+\frac\pi2\sum_{s=1}^{m-1}\frac{a_s}{n^{2s}}\left[B_{2s}\left(2^{2s-1}-1\right)-
4sH_{2s-1}(k)\right]-\sum_{s=1}^{m-1}\frac{B_{2s}\left(2^{2s-1}-1\right)}{2^{2s}sn^{2s}}\\
+r_m^{(1)}(n)+r_m^{(2)}(n),
\end{multline*}
where
$$r_m^{(j)}(n)=-\frac{L2^{2m}}{n(2m)!}\int_0^nf_j^{(2m)}(2x)y_m(x)\,dx,\quad j=1,2.$$
Since
$$f_1^{(s)}(x)\le0\quad\mbox{for}\quad x\in[0,2n]$$
holds for an arbitrary $s$ (cf.\ \eqref{13}), by \eqref{11} we have
$$0\le(-1)^{m+1}r_m^{(1)}(n)\le\frac\pi2\frac{a_m}{n^{2m}}|B_{2m}|\left(2^{2m-1}-1\right)-
\frac{|B_{2m}|\left(2^{2m-1}-1\right)}{2^{2m}mn^{2m}}.$$
To estimate the quantity $r_m^{(2)}(n)$, first we note that for any natural number $r$
\begin{equation}\label{14}
\int_0^ny_m(x)\sin\frac{2\pi rx}n\,dx=0.
\end{equation}
This equality follows from the periodicity of the integrand with period $n$ and its oddness. Thus,
$$f_2^{(2m)}(2x)=(-1)^{m+1}\frac{\pi^{2m+1}}{L2^{2m-1}n^{2m}}\sum_{r=1}^\infty\frac{h^r}{1+h^r}
r^{2m}\sin\frac{\pi rx}n,$$
which together with \eqref{14} gives
\begin{multline*}
|r_m^{(2)}(n)|\\=\frac{2\pi^{2m+1}}{(2m)!n^{2m+1}}\biggl|\sum_{r=1}^\infty\frac{h^{2r-1}}{1+h^{2r-1}}
(2r-1)^{2m}\int_0^ny_m(x)\sin\frac{\pi(2r-1)x}n\,dx\biggr|\\
\le\frac{2\pi^{2m+1}}{(2m)!n^{2m+1}}H_{2m}(k)\int_0^n|y_m(x)|\,dx.
\end{multline*}
From well-known properties of the Bernoulli polynomials it follows that
$$(-1)^{m+1}y_m(x)\ge0\quad\mbox{and}\quad\int_0^ny_m(x)\,dx=n\frac{B_{2m}\left(2^{2m-1}-1\right)}
{2^{2m-1}}.$$
A fortiori,
$$|r_m^{(2)}(n)|\le\pi^2m\frac{a_m}{n^{2m}}H_{2m}(k).$$

We apply the equality
$$\sum_{j=1}^n\frac2{2j-1}=\ln n+2\ln2+\gamma+\sum_{s=1}^{m-1}\frac{B_{2s}\left(2^{2s-1}-1\right)}{s2^{2s}n^{2s}}+\delta_m(n),$$
(see \cite{12}), where
$$0\le(-1)^{m+1}\delta_m(n)\le\frac{|B_{2m}|\left(2^{2m-1}-1\right)}
{m2^{2m}n^{2m}}.$$
We have
\begin{multline}\label{15}
\frac Ln\sum_{j=1}^n\ctn\frac{2j-1}{2n}L=\frac Ln\sum_{j=1}^n\left[\ctn\frac{2j-1}{2n}L-\frac{2n}{(2j-1)L}\right]+\sum_{j=1}^n\frac2{2j-1}\\
=\ln n+\frac\pi2A(k)+\frac\pi2\sum_{s=1}^{m-1}\frac{a_s}{n^{2s}}\left[B_{2s}\left(2^{2s-1}-1\right)-
4sH_{2s-1}(k)\right]\\
+\frac\pi2\frac{a_m}{n^{2m}}\rho_m(n,k).
\end{multline}
Denoting by $A_n(k)$ the left part in \eqref{15}, for $m=1$ we get that
\begin{equation}\label{16}
A_n(k)=\ln n+\frac\pi2A(k)+\frac\pi2\frac{a_1}{n^2}\rho_1(n,k)\le\ln n+\frac\pi8B(k).
\end{equation}
Theorem~\ref{T3} implies that, under \eqref{7},
\begin{equation}\label{17}
L_n(Z,1,k)=M_n^{-1}A_n(k).
\end{equation}
For the complete elliptic integral $M_n$ of the first kind the following equalities are
valid \cite[pp.~277, 284]{1} :
$$\sqrt{\frac{2M_n}\pi}=1+2\sum_{r=1}^mh_1^{r^2},\quad h_1=\exp\left(-\frac{\pi M_n'}{M_n}\right)=h^{2n}.$$
Consequently,
$$1\le\sqrt{\frac{2M_n}\pi}\le1+\frac{2h_1}{1-h_1}.$$
Hence it follows easily that
\begin{equation}\label{18}
M_n^{-1}=\frac2\pi-\frac8\pi h^{2n}\sigma_n(k),\quad0\le\sigma_n(k)\le1.
\end{equation}
Now \eqref{15}--\eqref{18} imply the statement of the theorem. The theorem is proved.
\end{proof}

For $k=0$, from Theorem~\ref{T4} a representation of $\Lambda_n(T,1)$ follows, which was
obtained in \cite{12} (see also \cite{2}):
\begin{multline*}
\Lambda_n(T,1)=\frac2\pi\ln n+\frac2\pi\left(\gamma+\ln\frac8\pi\right)+
\sum_{s=1}^{m-1}\frac{a_s}{n^{2s}}B_{2s}\left(2^{2s-1}-1\right)\\
+\frac{a_m}{n^{2m}}\rho_m(n,0),\quad0\le(-1)^{m+1}\rho_m(n,0)\le|B_{2m}|\left(2^{2m-1}-1\right).
\end{multline*}

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\bibitem{4} {\sc L.~Fejer,} Lagrangesche Interpolation und die zugehorigen konjugierten Punkte, {\it Math. Ann.,} {\bf106} (1932), 1--55.
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\bibitem{5} {\sc R.~G\"unter,} Evaluation of Lebesgue constants, {\it SIAM J. Numer. Anal.,} {\bf17} (1980), 512--520.
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\selectlanguage{russian}

\begin{center}
{\bf О константах Лебега при интерполяции аналитических функций}\\
\bigskip
\footnotesize К.~Ю.~ОСИПЕНКО
\end{center}

\bigskip 
\bigskip

Для системы различных точек $\tau=(t_1,\ldots,t_n)$ из отрезка $[-1,1]$ и $k\in[0,1)$ вводится величина
$$L_n(\tau,p,k)=\max_{t\in[-1,1]}\biggl(\sum_{j=1}^n|D_j(t)|^p\biggr)^{1/p},$$
где
\begin{gather*}
D_j(t)=\frac{\omega_j(t)}{\omega_j(t_j)}\left(1-kW_j^2(t)\right),\quad\omega_j(t)=
\prod_{m\ne j}W_m(t),\\ 
W_m(t)=\frac{t-t_m}{1-kt_mt}.
\end{gather*}

При $k=0$ она совпадает с константой Лебега, связанной с интерполяцией многочленом Лагранжа. Показана связь величины $L_n(\tau,p,k)$ с задачами интерполяции аналитических функций. Для системы
$$Z=\left\{\sn\left[\left(\frac{2j-1}n-1\right)K,k\right]\right\}_1^n,$$
являющейся аналогом чебышевской системы, получены оценки $L_n(Z,p,k)$ при $p\ge2$ и $p=1$.

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\noindent\footnotesize{СССР, МОСКВА 103767\\
ПЕТРОВКА 27\\
МОСКОВСКИЙ АВИАЦИОННЫЙ ТЕХНОЛОГИЧЕСКИЙ ИНСТИТУТ\\
ИМЕНИ К.~Э.~ЦИОЛКОВСКОГО}
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