Geometric Estimates in Linear Interpolation

on a Cube and a Ball

Mikhail Nevskii

Department of Mathematics, P. G. Demidov Yaroslavl State University,

Sovetskaya str., 14, Yaroslavl, 150003, Russia

February 18, 2024

11footnotetext:  E-mail address<EMAIL_ADDRESS>

Abstract

The paper contains a survey of the results obtained by the author in recent
years. These results concern the application in multivariate polynomial
interpolation of some geometric constructions and methods. In particular, we
give estimates of the projector’s norms through the characteristics of sets
associated with homothety. The known exact values and nowaday best estimates
of these norms are given. Also we formulate some open problems.

MSC: 41A05, 52B55, 52C07

Keywords: polynomial interpolation, projector, norm, estimate, simplex,
homothety, absorption index.

##  Introduction

The results related to minimization the norms of projection operators
(briefly, projectors) are important in Approximation Theory thanks to the
inequality

$E_{m}(f;K)_{C(K)}\leq\|f-Pr(f)\|_{C(K)}\leq(1+\|Pr\|)E_{m}(f;K)_{C(K)}.$

Here $E_{m}(f;K)_{C(K)}$ is the best approximation in the uniform norm of a
continuous function $f$ defined on a closed set $K\subset{\mathbb{R}}^{n}$ by
polynomials of degree at most $m$ (the space $\Pi_{m}({\mathbb{R}}^{n})$),
$Pr:C(K)\to\Pi_{m}({\mathbb{R}}^{n})$ is a bounded projector from $C(K)$ onto
$\Pi_{m}({\mathbb{R}}^{n})$, and $\|Pr\|$ is its operator norm.

The paper contains a survey of the results obtained by the author in recent
years. They are related to geometric aspects of multidimensional polynomial
interpolation. The most deep results concern linear interpolation. In
particular, we give estimates of the projector’s norms through the
characteristics of sets associated with homothety. The known exact values and
nowaday best estimates of these norms are given. The approach related to
linear interpolation can also be used for interpolation by polynomials from
wider spaces. Our main focus is on a cube and a ball in ${\mathbb{R}}^{n}$. A
number of statements we give with proofs, otherwise references to the
literature are given. Also we formulate some open questions.

Numerical characteristics connecting simplices and subsets of
${\mathbb{R}}^{n}$ have applications for obtaining various estimates in
polynomial interpolation. This approach and the corresponding analytic methods
in details were described in the author’s works [16]–[27] including the
monograph [24]. Lately these questions have been managed to study also by
computer methods (see [33], [34], [36], [37], [38], [39], [40], [41], [42],
[44], [54], and also the survey paper [43]). Computer calculations were
carried out by Alexey Ukhalov, whose skill and persistence produced many
impressive results. Various computer math systems, such as Wolfram Mathematica
(see, e. g., [57]), were used significantly along the way. Some of the
numerical results were obtained by our students.

In the early 1970s, at Yaroslavl’ State University (Yaroslavl’, Russia) there
was formed a school on Function Theory, Approximation Theory, Linear Operator
Interpolation Theory, and related areas. The founder and head of this school
for about 20 years was Professor Yuri Abramovich Brudnyi. It was Prof. Brudnyi
who inspired the author to take up Approximation Theory. Under his supervision
the author performed his dissertation research related to piecewise-polynomial
approximation in Orlicz classes.

Throughout their lives, all of Yu. A. Brudnyi’s students have felt great
gratitude, love and respect for their teacher. This text is dedicated to Yuri
Abramovich Brudnyi in connection with the upcoming 90th anniversary of his
birth.

##  1\. Notation and preliminaries

Let us fix some notation. Throughout the paper $n\in{\mathbb{N}}$ is a
positive integer. By $e_{1}$, $\ldots$, $e_{n}$ we denote the standard basis
in ${\mathbb{R}}^{n}$. Given $x=(x_{1},...x_{n})\in{\mathbb{R}}^{n}$, by
$\|x\|$ we denote its standard Euclidean norm

$\|x\|=\sqrt{\langle
x,x\rangle}=\left(\sum\limits_{i=1}^{n}x_{i}^{2}\right)^{1/2}.$

Hereafter, for $x=(x_{1},...x_{n}),y=(y_{1},...y_{n})\in{\mathbb{R}}^{n}$ by
$\langle{x,y}\rangle=x_{1}y_{1}+...+x_{n}y_{n}$ we denote the standard inner
product in ${\mathbb{R}}^{n}$.

Given $x^{(0)}\in{\mathbb{R}}^{n}$ and $R>0$, we let $B(x^{(0)};R)$ denote the
Euclidean ball with center $x^{(0)}$ and radius $R$:

$B\left(x^{(0)};R\right)=\\{x\in{\mathbb{R}}^{n}:\|x-x^{(0)}\|\leq R\\}.$

We also set

$B_{n}=B(0;1),~{}~{}~{}Q_{n}=[0,1]^{n}~{}~{}~{}\text{and}~{}~{}~{}Q_{n}^{\prime}:=[-1,1]^{n}.$

The notation $L(n)\asymp M(n)$ means that there exist absolute constants
$c_{1},c_{2}>0$ such that $c_{1}M(n)\leq L(n)\leq c_{2}M(n)$.

Let $K$ be a convex body in ${\mathbb{R}}^{n}$, i.e., a compact convex subset
of ${\mathbb{R}}^{n}$ with nonempty interior. By $c(K)$ we denote the center
of gravity of $K$. The symbol $\sigma K$ denotes a homothetic copy of $K$ with
the center of homothety in $c(K)$ and the ratio of homothety $\sigma.$

We let ${\rm vol}(K)$ denote the volume of $K$. If $K$ is a convex polytope,
then by ${\rm ver}(K)$ we denote the set of all vertices of $K$. By a
translate we mean the result of a parallel shift.

We say that an $n$-dimensional simplex $S$ is circumscribed around a convex
body $K$ if $K\subset S$ and each $(n-1)$-dimensional face of $S$ contains a
point of $K$. A convex polytope is inscribed into $K$ if every vertex of this
polytope belongs to the boundary of $K$.

Definition 1.1. Let $i\in\\{1,...,n\\}$ and let $d_{i}(K)$ be the maximal
length of a segment contained in $K$ and parallel to the $x_{i}$-axis. We
refer to $d_{i}(K)$ as the $i$th axial diameter of $K$.

The notion of axial diameter of a convex body was introduced by P. Scott
[49],[50].

Definition 1.2. Given convex bodies $K_{1}$, $K_{2}$, by $\xi(K_{1};K_{2})$ we
denote the minimal $\sigma\geq 1$ with the property $K_{1}\subset\sigma
K_{2}$. We call $\xi(K_{1},K_{2})$ the absorption index of $K_{1}$ by $K_{2}$.

By $\alpha(K_{1},K_{2})$ we denote the minimal $\sigma>0$ such that $K_{1}$ is
a subset of a translate of $\sigma K_{2}$.

Note that the equality $\xi(K_{1};K_{2})$ $=1$ is equivalent to the inclusion
$K_{1}\subset K_{2}$. Clearly, $\alpha(K_{1},K_{2})$ $\leq$
$\xi(K_{1},K_{2})$.

Definition 1.3. By $\xi_{n}(K)$ we denote the minimal absorption index of a
convex body $K$ by an inner nondegenerate simplex. In other words,

$\xi_{n}(K)=\min\\{\xi(K;S):\,S\mbox{ is an $n$-dimensional
simplex,}\,S\subset K,\,{\rm vol}(S)\neq 0\\}.$

By $C(K)$ we denote the space of all continuous functions $f:K\to{\mathbb{R}}$
with the uniform norm

$\|f\|_{C(K)}=\max\limits_{x\in K}|f(x)|.$

We let $\Pi_{1}\left({\mathbb{R}}^{n}\right)$ denote the space of polynomials
in $n$ variables of degree at most $1$.

Let $S$ be a nondegenerate simplex in ${\mathbb{R}}^{n}$ with vertices
$x^{(j)}=\left(x_{1}^{(j)},\ldots,x_{n}^{(j)}\right),$ $1\leq j\leq n+1.$ We
define the vertex matrix of this simplex by

${\bf A}=\left(\begin{array}[]{cccc}x_{1}^{(1)}&\ldots&x_{n}^{(1)}&1\\\
x_{1}^{(2)}&\ldots&x_{n}^{(2)}&1\\\ \vdots&\vdots&\vdots&\vdots\\\
x_{1}^{(n+1)}&\ldots&x_{n}^{(n+1)}&1\\\ \end{array}\right).$

Clearly, matrix ${\bf A}$ is nondegenerate and

${\rm vol}(S)=\frac{|\det({\bf A})|}{n!}.$ (1.1)

Let

${\bf A}^{-1}=\left(l_{ij}\right)_{i,j=1}^{n+1}.$ (1.2)

Definition 1.4. Linear polynomials

$\lambda_{j}(x)=l_{1j}x_{1}+\ldots+l_{nj}x_{n}+l_{n+1,j},~{}~{}~{}~{}j=1,...,n+1,$

are called the basic Lagrange polynomials corresponding to $S$.

These polynomials have the following property:

$\lambda_{j}\left(x^{(k)}\right)=\delta_{j}^{k}~{}~{}~{}\text{for
all}~{}~{}~{}j,k=1,...,n+1.$

Here $\delta_{j}^{k}$ is the Kronecker delta. For an arbitrary
$x\in{\mathbb{R}}^{n}$, we have

$x=\sum_{j=1}^{n+1}\lambda_{j}(x)x^{(j)},\quad\sum_{j=1}^{n+1}\lambda_{j}(x)=1.$

Thus, $\lambda_{j}(x)$ are the barycentric coordinates of $x$ with respect to
the simplex $S$. In turn, equations $\lambda_{j}(x)=0$, $j=1,...,n+1$, define
the $(n-1)$-dimensional hyperplanes containing the faces of $S.$ Therefore,

$S=\left\\{x\in{\mathbb{R}}^{n}:\,\lambda_{j}(x)\geq
0,\,j=1,\ldots,n+1\right\\}.$

Also let us note that for every $j=1,...,n+1$ we have

$\lambda_{j}(x)=\frac{\Delta_{j}(x)}{\Delta}.$ (1.3)

Here $\Delta=\det({\bf A})$ and $\Delta_{j}(x)$ is the determinant that
appears from $\Delta$ after replacing the $j$th row by the row $(x_{1}\
\ldots\ x_{n}\ 1).$ For more information on $\lambda_{j}$, see [24], [45].

Definition 1.5. Let $x^{(j)}\in K$, $1\leq j\leq n+1,$ be the vertices of a
nondegenerate simplex $S$. The interpolation projector
$P:C(K)\to\Pi_{1}({\mathbb{R}}^{n})$ with nodes $x^{(j)}$ is determined by the
equalities $Pf\left(x^{(j)}\right)=f_{j}=f\left(x^{(j)}\right)$ , $1\leq j\leq
n+1.$ We say that an interpolation projector
$P:C(K)\to\Pi_{1}({\mathbb{R}}^{n})$ and a nondegenerate simplex $S\subset K$
correspond to each other if the nodes of $P$ coincide with the vertices of
$S$. We use notation $P_{S}$ and $S_{P}$.

For an interpolation projector $P=P_{S}$, the analogue of Lagrange
interpolation formula holds:

$Pf(x)=\sum\limits_{j=1}^{n+1}f\left(x^{(j)}\right)\lambda_{j}(x).$ (1.4)

Here $\lambda_{j}$ are the basic Lagrange polynomials of the simplex $S_{P}$
(see Definition 1.4).

We let $\|P\|_{K}$ denote the norm of $P$ as an operator from $C(K)$ into
$C(K)$. Thanks to (1.4),

$\|P\|_{K}=\sup_{\|f\|_{C(K)}=1}\|Pf\|_{C(K)}=\sup_{-1\leq f_{j}\leq
1}\max_{x\in K}\left|\sum_{j=1}^{n+1}f_{j}\lambda_{j}(x)\right|=$ $=\max_{x\in
K}\sup_{-1\leq f_{j}\leq
1}\left|\sum_{j=1}^{n+1}f_{j}\lambda_{j}(x)\right|=\max_{x\in K}\sup_{-1\leq
f_{j}\leq 1}\sum_{j=1}^{n+1}f_{j}\lambda_{j}(x).$

Because $\sum f_{j}\lambda_{j}(x)$ is linear in $x$ and
$f_{1},\ldots,f_{n+1}$, we have

$\|P\|_{K}=\max_{x\in K}\max_{f_{j}=\pm
1}\sum_{j=1}^{n+1}f_{j}\lambda_{j}(x)=\max_{x\in
K}\sum_{j=1}^{n+1}|\lambda_{j}(x)|.$ (1.5)

If $K$ is a convex polytope in ${\mathbb{R}}^{n}$ (e. g., $K$ is a cube), a
simpler equality

$\|P\|_{K}=\max_{x\in{\rm ver}(K)}\sum_{j=1}^{n+1}|\lambda_{j}(x)|$ (1.6)

holds.

Definition 1.6. We let $\theta_{n}(K)$ denote the minimal value of
$\|P_{S}\|_{K}$ where $S$ runs over all nondegenerate simplices with vertices
in $K$. An interpolation projector
$P:C(K)\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ is called minimal if
$\|P\|_{K}=\theta_{n}(K)$.

Definition 1.7. Let $\mu$ be a positive integer, $1\leq\mu\leq n$. A point
$x\in K$ is called the $\mu$-point of $K$ with respect to a simplex $S\subset
K,$ if the corresponding projector
$P:C(K)\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ satisfies
$\|P\|_{K}=\sum|\lambda_{j}(x)|$ and, in addition, among the numbers
$\lambda_{j}(x)$ there are exactly $\mu$ negatives.

Suppose $K$ is a cube in ${\mathbb{R}}^{n}$. A vertex $x$ of the cube is
called the $\mu$-vertex with respect to $S,$ if $x$ is a $\mu$-point of $K$
with respect to this simplex.

It was shown in [18] that for any interpolation projector
$P:C(K)\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ and the corresponding simplex
$S$ we have

$\frac{n+1}{2n}\Bigl{(}\|P\|_{K}-1\Bigr{)}+1\leq\xi(K;S)\leq\frac{n+1}{2}\Bigl{(}\|P\|_{K}-1\Bigr{)}+1.$
(1.7)

If there exists an $1$-point in $K$ with respect to $S$, then the right-hand
inequality in (1.7) becomes an equality. This statement was proved in [18] in
the equivalent form; the notion of $1$-vertex was introduced later in [19].
Furthermore, it was shown in [19] that if $K=Q$ where $Q$ is a cube in
${\mathbb{R}}^{n}$ then the existence of an $1$-vertex of $Q$ with respect to
a simplex $S\subset Q$ is equivalent to the right-hand equality in (1.7). If
there exists a $\mu$-vertex of $Q$ with respect to $S,$ then

$\frac{n+1}{2\mu}\left(\|P\|_{Q}-1\right)+1\leq\xi(Q;S).$

Thanks to (1.7),

$\frac{n+1}{2n}\Bigl{(}\theta_{n}(K)-1\Bigr{)}+1\leq\xi_{n}(K)\leq\frac{n+1}{2}\Bigl{(}\theta_{n}(K)-1\Bigr{)}+1.$
(1.8)

Obviously, if a projector $P$ satisfies the equality

$\xi_{n}(K)=\frac{n+1}{2}\Bigl{(}\|P\|_{K}-1\Bigr{)}+1,$ (1.9)

then $P$ is minimal and the right-hand relation in (1.8) becomes an equality.

It turns out that calculation of $\xi(K;S)$ and $\alpha(K;S)$ is of particular
interest to us. In [22] we show that

$\xi(K;S)=(n+1)\max_{1\leq k\leq n+1}\max_{x\in
K}(-\lambda_{k}(x))+1\quad(K\not\subset S),$ (1.10)
$\alpha(K;S)=\sum_{j=1}^{n+1}\max_{x\in K}(-\lambda_{j}(x))+1$ (1.11)

(see also [24]). Furthermore, we prove that the equality
$\xi(K;S)=\alpha(K;S)$ holds true if and only if the simplex $\xi(S)S$ is
circumscribed around $K$. This is also equivalent to the relation

$\max_{x\in K}\left(-\lambda_{1}(x)\right)=\ldots=\max_{x\in
K}\left(-\lambda_{n+1}(x)\right).$ (1.12)

If $K$ is a convex polytope, then the maxima on $K$ in (1.10)–(1.12) can also
be taken over $x\in{\rm ver}(K)$. Note that for $K=Q_{n}$ formula (1.10) is
proved in [20].

Occasionally, we will consider the case when $n+1$ is an Hadamard number,
i.e., there exists an Hadamard matrix of order $n+1$.

Definition 1.8. An Hadamard matrix of order $m$ is a square binary matrix $\bf
H$ with entries either $1$ or $-1$ satisfying the equality

${\bf H}^{-1}=\frac{1}{m}\,{\bf H}^{T}.$

An integer $m$, for which an Hadamard matrix of order $m$ exists, is called an
Hadamard number.

Thus, the rows of ${\bf H}$ are pairwise orthogonal with respect to the
standard scalar product on ${\mathbb{R}}^{m}.$

Let us recall some basic facts relating to the theory of Hadamard matrices.
The order of an Hadamard matrix is $1$ or $2$ or some multiple of $4$ (see
[9]). But it is still unknown whether an Hadamard matrix exists for every
order of the form $m=4k$. This is one of the longest lasting open problems in
Mathematics called the Hadamard matrix conjecture. The orders below $1500$ for
which Hadamard matrices are not known are $668,716,892,956,1132,1244,1388$,
and $1436$ (see, e.g., [7], [15]).

Two Hadamard matrices are called equivalent if one can be obtained from
another by performing a finite sequence of the following operations:
multiplication of some rows or columns by $-1$, or permutation of rows or
columns. Up to equivalence, there exists a unique Hadamard matrix of orders
$1,2,4,8$, and $12$. There are $5$ equivalence classes of Hadamard matrices of
order $16,3$ of order $20$, $60$ of order $24$, and $487$ of order $28$. For
orders $32$, $36$, and $40$, the number of equivalence classes is much
greater. For $n=32$ there are at least $3,578,006$ equivalence classes; for
$n=36$, at least $4,745,357$ (see [7]).

It is known that $n+1$ is an Hadamard number if and only if it is possible to
inscribe an $n$-dimensional regular simplex into an $n$-dimensional cube in
such a way that the vertices of the simplex coincide with vertices of the
cube. See [10].

Let us prove this statement. Let $Q_{n}^{\prime}=[-1,1]^{n}$. Then there
exists a simple correspondence between Hadamard matrices of order $n+1$ with
the last column consisting of $1$’s and $n$-dimensional regular simplices
whose vertices are at the vertices of $Q_{n}^{\prime}$. Since later on we will
need this correspondence, we will dwell on it in more detail.

If $S$ is a regular simplex inscribed into $Q_{n}^{\prime}$, then its vertex
matrix ${\bf A}$ is an Hadamard matrix of order $n+1$. Indeed, let
$x^{(1)},\ldots,x^{(n+1)}$ be the vertices of $S$. Since $x^{(j)}$,
$j=1,...,n+1,$ coincide with vertices of the cube, the entries of $\bf A$ are
$\pm 1$ and its last column consists of $1$’s. Let us denote by $h^{(j)}$ the
rows of ${\bf A}$ and make sure that these vectors are pairwise orthogonal in
${\mathbb{R}}^{n+1}$. Equalities $||x^{(j)}||=\sqrt{n}$ mean that simplex $S$
is inscribed into an $n$-dimensional ball of radius $\sqrt{n}$. The edge-
length $d$ of a regular simplex and radius $R$ of the circumscribed ball
satisfy the following equality:

$d=R\sqrt{2}\sqrt{\frac{n+1}{n}}$ (1.13)

(see, e. g., [24]). If $R=\sqrt{n},$ then $d=\sqrt{2(n+1)}.$ Hence,

$2(n+1)=||x^{(j)}-x^{(k)}||^{2}=2||x^{(j)}||^{2}-2\langle
x^{(j)},x^{(k)}\rangle=2n-2\langle x^{(j)},x^{(k)}\rangle.$

We see that $\langle x^{(j)},x^{(k)}\rangle=-1.$ Since the rows of $\bf S$ in
the first $n$ columns contain coordinates of vertices of the simplex and the
last element of any row is $1$, we have

$\langle h^{(j)},h^{(k)}\rangle=\langle x^{(j)},x^{(k)}\rangle+1=0.$

Thus, $h^{(j)}$ are pairwise orthogonal in ${\mathbb{R}}^{n+1}$ so that $\bf
A$ is an Hadamard matrix of order $n+1$. Hence,

${\bf A}^{-1}=\frac{1}{n+1}\,{\bf A}^{T}.$ (1.14)

Conversely, let ${\bf H}$ be an Hadamard matrix of order $n+1$ with the last
column consisting of $1$’s. Consider the simplex $S$ whose vertex matrix is
$\bf H$. In other words, the vertices of $S$ are given by the rows of ${\bf
H}$ (excepting the last component). Let us see that the simplex $S$ is
inscribed into the cube $Q_{n}^{\prime},$ and ${\rm ver}(S)\subset{\rm
ver}(Q_{n}^{\prime})$. Indeed, if $h^{(j)}$ are the rows of ${\bf H}$ and
$x^{(j)}$ are the vertices of $S$, then $\langle h^{(j)},h^{(k)}\rangle=0,$
$\langle x^{(j)},x^{(k)}\rangle=-1,$ $\|x^{(j)}\|^{2}=n.$ Hence,
$||x^{(j)}-x^{(k)}||^{2}=2(n+1)$ proving that $S$ is a regular simplex with
the edge-length $\sqrt{2(n+1)}$.

Notice that any Hadamard matrix of order $n+1$ is equivalent to the vertex
matrix of some $n$-dimensional regular simplex inscribed in $Q_{n}^{\prime}$.
This vertex matrix can be obtained from the given Hadamard matrix after
multiplication of some rows by $-1$.

Definition 1.9. By $h_{n}$ we denote the maximum value of a determinant of
order $n$ with entries $0$ or $1$. Let $\nu_{n}$ be the maximum volume of an
$n$-dimensional simplex contained in $Q_{n}$.

The numbers $h_{n}$ and $\nu_{n}$ satisfy the equality $h_{n}=n!\nu_{n}$ (see
[10]). For any $n$, there exists in $Q_{n}$ a maximum volume simplex with some
vertex coinciding with a vertex of the cube. For $n>1$, the following
inequalities hold

$\frac{1}{2}\left(1-\frac{\log(4/3)}{\log n}\right)n\log
n<\log(2^{n-1}h_{n-1})\leq\frac{1}{2}\,n\log n.$ (1.15)

The right-hand inequality in (1.15) was proved by Hadamard [8] and the left-
hand one by Clements and Lindström [4]. Consequently, for all
$n\in{\mathbb{N}}$

$\left(\frac{3}{4}\right)^{(n+1)/2}\,\frac{\left(n+1\right)^{(n+1)/2}}{2^{n}}<h_{n}\leq\frac{\left(n+1\right)^{(n+1)/2}}{2^{n}},$
(1.16)
$\left(\frac{3}{4}\right)^{(n+1)/2}\,\frac{\left(n+1\right)^{(n+1)/2}}{2^{n}n!}<\nu_{n}\leq\frac{\left(n+1\right)^{(n+1)/2}}{2^{n}n!}.$
(1.17)

The right-hand equality in each relation holds if and only if $n+1$ is an
Hadamard number [10]. In some cases the right-hand inequality in (1.17) has
been improved. For instance, if $n$ is even, then

$\nu_{n}\leq\frac{n^{n/2}\sqrt{2n+1}}{2^{n}n!}.$ (1.18)

If $n>1$ and $n\equiv 1({\rm mod}~{}4),$ then

$\nu_{n}\leq\frac{(n-1)^{(n-1)/2}}{2^{n-1}(n-1)!}$ (1.19)

(see [10]). For many $n$, the values of $\nu_{n}$ and $h_{n}$ are known
exactly. The first $12$ numbers $\nu_{n}$ are

$\nu_{1}=1,\quad\nu_{2}=\frac{1}{2},\quad\nu_{3}=\frac{1}{3},\quad\nu_{4}=\frac{1}{8},\quad\nu_{5}=\frac{1}{24},\quad\nu_{6}=\frac{1}{80},\quad\nu_{7}=\frac{2}{315},$
$\nu_{8}=\frac{1}{720},\quad\nu_{9}=\frac{1}{2520},\quad\nu_{10}=\frac{1}{11340},\quad\nu_{11}=\frac{9}{246400},\quad\nu_{12}=\frac{3}{394240}.$

##  2\. The values of $\alpha(Q_{n};S)$ and $\xi(Q_{n};S)$

Recall that $d_{i}(K)$ is the maximal length of a segment contained in $K$ and
parallel to the $x_{i}$-axis (see Definition 1.1). In [22] we have proved the
following statement.

Theorem 2.1. Any convex body $K\subset{\mathbb{R}}^{n}$ contains a translate
of the cube $\sigma Q_{n}$ where

$\sigma=\left(\sum_{i=1}^{n}\frac{1}{d_{i}(K)}\right)^{-1}.$

This result immediately implies the inequality

$\alpha(Q_{n};K)\leq\sum_{i=1}^{n}\frac{1}{d_{i}(K)}.$ (2.1)

See Definition 1.2 for $\alpha(Q_{n};K)$. It was also shown in [22] that if
$K=S$ where $S$ is a nondegenerate simplex, then inequality (2.1) becomes an
equality.

Theorem 2.2. For an arbitrary $n$-dimensional nondegenerate simplex $S$,

$\alpha(Q_{n};S)=\sum_{i=1}^{n}\frac{1}{d_{i}(S)}.$ (2.2)

Earlier, in [20], we have proved that

$\frac{1}{d_{i}(S)}=\frac{1}{2}\sum_{j=1}^{n+1}|l_{ij}|.$ (2.3)

Recall that $l_{ij}$, $i,j=1,...,n+1$, are the entries of the matrix ${\bf
A}^{-1}$, see (1.2). Also we have shown that the only segment of length
$d_{i}(S)$ parallel to the $i$th axis is located in $S$ in such a way that
every $(n-1)$-dimensional face of the simplex contains at least one of its
endpoints. Formulas for the endpoints of this segment are also given in [20].
For a generalization of these results to the case of maximum segments of
arbitrary directions in a simplex, we refer the reader to [27].

Corollary 2.1. We have

$\alpha(Q_{n};S)=\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n+1}|l_{ij}|.$ (2.4)

This is immediate from (2.2) and (2.3). Recall that
$Q_{n}^{\prime}=[-1,1]^{n}.$

Corollary 2.2.

$\alpha(Q_{n}^{\prime};S)=\sum_{i=1}^{n}\sum_{j=1}^{n+1}|l_{ij}|.$ (2.5)

Proof. Obviously, $\alpha(K_{1};K_{2})$ is invariant with respect to shifts,
and $\alpha(\tau K_{1};K_{2})=\tau\alpha(K_{1};K_{2})$ for every $\tau>0$.
Since $Q_{n}^{\prime}=[-1,1]^{n}$ is a translate of $2Q_{n}$, then (2.5) is
the direct consequence of (2.4). $\Box$

We note that formula (2.2) has many useful corollaries which we present in
works [22] and [24]. Let us give some examples.

Corollary 2.3. If $S\subset Q_{n}$, then $\xi(Q_{n};S)\geq n.$ The equality
$\xi(Q_{n};S)=n$ implies $\xi(Q_{n};S)=\alpha(Q_{n};S)$ and $d_{i}(S)=1$ for
every $i=1,...,n$.

Proof. Since $S\subset Q_{n},$ we have $d_{i}(S)\leq 1$. Therefore,

$\xi(Q_{n};S)\geq\alpha(Q_{n};S)=\sum_{i=1}^{n}\frac{1}{d_{i}(S)}\geq n.$
(2.6)

If, in addition, $\xi(Q_{n};S)=n$, then (2.6) gives
$\xi(Q_{n};S)=\alpha(Q_{n};S)=n$ and $d_{i}(S)=1$. $\Box$

Equality (2.2) yields the following property first proven in [20].

Corollary 2.4. If an $n$-dimensional simplex $S$ contains the cube $Q_{n}$,
then for some $i\in\\{1,...,n\\}$ this simplex contains a segment of length
$n$ parallel to the $i$th axis.

Proof. Because $Q_{n}\subset S$, we have $\alpha(Q_{n};S)\leq 1.$ From (2.2),
it follows

$\sum_{i=1}^{n}\frac{1}{d_{i}(S)}\leq 1$

proving the existence of $i$ such that $d_{i}(S)\geq n$. $\Box$

Corollary 2.5. Let $S$ be a nondegenerate simplex and let $D,$ $D^{*}$ be
parallelotopes in ${\mathbb{R}}^{n}.$ Suppose $D^{*}$ is a homothetic copy of
$D$ with ratio $\sigma>1.$ If $D\subset S\subset D^{*},$ then $\sigma\geq n.$

Proof. It is sufficient to consider the case $D=Q_{n}.$ Then $D^{*}$ is a
translate of the cube $\sigma Q_{n}$. Since $S$ is contained in this
translate, we have $d_{i}(S)\leq\sigma.$ Hence, $1/\sigma\leq 1/d_{i}(S)$ for
all $i$. Summing up over $i=1,\ldots,n$, we get

$\frac{n}{\sigma}\leq\sum_{i=1}^{n}\frac{1}{d_{i}(S)}=\alpha(Q_{n};S).$

The inclusion $S\subset Q_{n}$ means that $\alpha(Q_{n};S)\leq 1$ (for
$\alpha(Q_{n};S)$, see Definition 1.2). Consequently, $n/\sigma\leq 1,$ i. e.,
$\sigma\geq n.$ $\Box$

Finally, let us note the result obtained earlier by M. Lassak [13].

Corollary 2.6. If a simplex $S\subset Q_{n}$ has the maximum volume, then
$d_{i}(S)=1$ for every $i=1,...,n$.

Proof. If a simplex $S\subset Q_{n}$ has the maximum volume, then $S\subset-
nS$. In fact, if this is not the case, then some vertex of $S$ can be moved in
$Q_{n}$ in such a way that its distance from the opposite face of the simplex
increases. In this case the volume of $S$ will increase so that this volume is
not maximal, a contradiction. The inclusion $Q_{n}\subset-nS$ means that
$-Q_{n}$ is a subset of a translate of $nS$. Since $-Q_{n}=Q_{n}$, we have
$\alpha(S)\leq n$. But, thanks to (2.6), $\alpha(Q_{n};S)\geq n$ proving that
$\alpha(Q_{n};S)=n$. Let us also note that $d_{i}(S)\leq 1$ because $S\subset
Q_{n}$. From these properties and (2.2) we have $d_{i}(S)=1$ for all
$i=1,...,n,$ proving the statement. $\Box$

Note that this property of a maximum volume simplex was essentially used in
[41].

Corollary 2.7. For any $n$, we have $\xi_{n}(Q_{n})\geq n.$

This follows immediately from Corollary 2.3. The simplicity of obtaining this
estimate is apparent, since both the proofs of formula (2.2) given in [22] are
quite labor-intensive.

Corollary 2.8. For any $n$,

$\theta_{n}(Q_{n})\geq 3-\frac{4}{n+1}.$ (2.7)

Furthermore, if there exists a projector
$P:C(Q_{n})\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ satisfying the equality

$\|P\|_{Q_{n}}=3-\frac{4}{n+1}$ (2.8)

then $P$ is minimal. Moreover, in this case $\xi_{n}(Q_{n})=n.$

Proof. Let us combine the right-hand inequality in (1.7) with the estimate
$\xi_{n}(Q_{n})\geq n$:

$n\leq\xi_{n}(Q_{n})\leq\frac{n+1}{2}\Bigl{(}\theta_{n}(Q_{n})-1\Bigr{)}+1.$
(2.9)

Therefore, for any $n$, we have (2.7). The equality (2.8) means that we have
an equality also in (2.7). This means that the outer parts of (2.9) are equal
to $n$. Hence, in this case also $\xi_{n}(Q_{n})=n$. $\Box$

Let us present a series of results devoted to calculation of the constant
$\xi_{n}(Q_{n})$ for small $n$.

The case $n=1$ is trivial: $\xi_{1}(Q_{1})=1$ and a unique extremal simplex is
the segment $[0,1]$ coinciding with $Q_{1}$. Starting from $n=2$, the problem
of calculating $\xi_{n}(Q_{n})$ is rather difficult, especially, if $n+1$ is
not an Hadamard number. The values $\xi_{2}(Q_{2})$ and $\xi_{3}(Q_{3})$ were
discovered in [17]; see also [24].

For $n=2,$

$\xi_{2}(Q_{2})=\frac{3\sqrt{5}}{5}+1=2.3416...$

Up to rotations, the only extremal simplex is the triangle with vertices
$(0,0),$ $(1,\tau),$ $(\tau,1)$, where $\tau=(3-\sqrt{5})/2=0.3819\ldots$ .
This number satisfies $\tau^{2}-3\tau+1=0$ or

$\frac{\tau}{1-\tau}=\frac{1-\tau}{1}.$

Therefore, $\tau$ delivers the so called golden section of the segment [0,1].
Sharp inequality $\xi(Q_{2};S)\geq 3\sqrt{5}/5+1$ for simplices $S\subset
Q_{2}$ gives new characterization of this classical notion. See Fig. 1.

Figure 1: The case $n=2$. The golden section simplex

Combining our approach with some results of [13], we prove in [17] that
$\xi_{3}(Q_{3})=3.$ Up to a change of coordinates, each extremal simplex in
$Q_{3}$ coincides with the tetrahedron $S^{\prime}$ with vertices

$(1,1,0),~{}~{}~{}(1,0,1),~{}~{}~{}(0,1,1),~{}~{}~{}(0,0,0)$

or the tetrahedron $S^{\prime\prime}$ with vertices

$\left(\frac{1}{2},0,0\right),~{}~{}~{}\left(\frac{1}{2},1,0\right),~{}~{}~{}\left(0,\frac{1}{2},1\right),~{}~{}~{}\left(1,\frac{1}{2},1\right).$

In other words, if $\xi(Q_{3};S)=3,$ then either vertices of $S$ coincide with
vertices of the cube and form a regular tetrahedron or coincide with the
centers of opposite edges of two opposite faces of the cube and does not
belong to a common plane. One can see that all axial diameters of $S^{\prime}$
and $S^{\prime\prime}$ are equal to 1. This also follows from the equality
$\xi_{3}(Q_{3})=3.$ Note that ${\rm vol}(S^{\prime})=\nu_{3}=1/3,$ and ${\rm
vol}(S^{\prime\prime})=1/6$.

In [35] and [45] we show that each extremal simplex for $n=2$ and $n=3$ is
equisecting, i.e., the hyperplanes containing its faces cut off from the cube
in outer direction from the simplex the domains of equal volumes.

The estimate $\xi_{n}(Q_{n})\geq n$ occurs to be exact in order of $n$.
Indeed, let $n>2$ and let $S^{*}$ be the simplex with the zero-vertex and
other $n$ vertices coinciding with the vertices of $Q_{n}$ adjacent to
$(1,\ldots,1)$, i.e., with the vertices

$(0,1,\ldots,1),~{}~{}~{}(1,0,\ldots,1),~{}~{}~{}\ldots,~{}~{}~{}(1,1,\ldots,0),~{}~{}~{}(0,0,\ldots,0)$
(2.10)

(see [21], [24, § 3.2]). If $n\geq 3$, then $S^{*}$ has the following property
(see [10, Lemma 3.3]): replacement of an arbitrary vertex of $S^{*}$ by any
point of $Q_{n}$ decreases the volume of the simplex. For $n=2,3,4$ (and only
in these cases), simplex $S^{*}$ has maximum volume in $Q_{n}$ (see also
[10]). As it is shown in [21], if $n\geq 2$, then $d_{i}(S^{*})=1$ for every
$i=1,...,n$, and therefore, thanks to (2.2), $\alpha(Q_{n};S^{*})=n.$ For this
simplex,

$\xi(Q_{n};S^{*})=\frac{n^{2}-3}{n-1}.$

Hence, for $n>2$,

$\xi_{n}(Q_{n})\leq\frac{n^{2}-3}{n-1}.$ (2.11)

If $n>1$, the right-hand side of (2.11) is strictly smaller than $n+1$. Since

$\xi_{1}(Q_{1})=1,~{}~{}~{}\xi_{2}(Q_{2})=\frac{3\sqrt{5}}{5}+1=2.3416\ldots,$

inequality $\xi_{n}<n+1$ holds true also for $n=1,2$.

We note that if the volume of simplex $S\subset Q_{n}$ is maximal, then
$\xi(Q_{n};S)\leq n+2$, see [24].

Simplices satisfying the inclusions $S\subset Q_{n}\subset nS$ were studied in
[35]. The existence of such simplices for a given $n$ is equivalent to the
equality $\xi_{n}(Q_{n})=n.$ This holds not for each $n$, but nowaday a unique
known dimension for which $\xi_{n}(Q_{n})>n$ is $n=2$.

Thus, always $n\leq\xi_{n}(Q_{n})<n+1$, i.e., $\xi_{n}(Q_{n})-n\in[0,1)$.
However, the exact values of the constant $\xi_{n}(Q_{n})$ are currently only
known for $n=2$, $n=5$ and $n=9$, as well as for an infinite set of $n$ for
which there exists an Hadamard matrix of order $n+1$.

In all these cases, except $n=2$, the equality $\xi_{n}(Q_{n})=n$ holds. In
the noted Hadamard case, one can give the proof using the structure of
Hadamard matrix of order $n+1$, see [37]. In [37], we have also discovered the
exact values of $\xi_{n}(Q_{n})$ for $n=5$ and $n=9$ and constructed several
infinite families of extremal simplices for $n=5,7,9$.

We call a nondegenerate simplex $S\subset Q_{n}$ perfect provided
$\xi(Q_{n};S)=\xi_{n}(Q_{n})$ and the cube $Q_{n}$ is inscribed into the
simplex $\xi(Q_{n};S)\,S$. For a long time, only two dimensions were known,
namely $n=1$ and $n=3$, for which such simplices exist. Every three-
dimensional perfect simplex is similar to $S^{\prime\prime}$. In Fig. 2 it can
be seen that all vertices of $Q_{3}$ belong to the boundary of the simplex
$3S^{\prime\prime}$.

Figure 2: The case $n=3$. Simplex $S^{\prime\prime}$

On the contrary, simplex $S^{\prime}$ is not perfect, While $Q_{3}\subset
3S^{\prime}$, only four vertices of the cube lie on the boundary of the
simplex $3S^{\prime}$. See Fig. 3.

Figure 3: The case $n=3$. Simplex $S^{\prime}$

It was shown in [37] that perfect simplices do exist also in
${\mathbb{R}}^{5}$.

As we have mentioned above, nowaday $n=2$ is a unique known dimension for
which $\xi_{n}(Q_{n})>n.$

Let us also note that dimension $n=2$ remains the only even positive integer
where we know the sharp value of the constant $\xi_{n}(Q_{n})$. Also, it is
still unknown whether there exists an even $n$ such that $\xi_{n}(Q_{n})=n.$
The best of known today upper bounds for $n=4$ and $n=6$ are

$\xi_{4}(Q_{4})\leq\frac{19+5\sqrt{13}}{9}=4.1141\ldots,\quad\xi_{6}(Q_{6})<6.0166.$

(see [33], [34]). Using simplices of maximum volume in the cube, A.Yu. Ukhalov
perfomed computer calculations to get as accurate upper bounds for
$\xi_{n}(Q_{n})$ as possible; the results for $n\leq 118$ are given in [36]
and also in the survey paper [43].

Let us note that inequality $\xi_{n}(K)\leq n+2$ holds for an arbitrary convex
body $K\subset{\mathbb{R}}^{n}$. This is immediate from the following
proposition proven by M. Lassak [14]: If $S$ is a maximum volume simplex
contained in $K$, then $\xi(K;S)\leq n+2.$ In other words, we have

$S\subset K\subset(n+2)S.$ (2.12)

Note that this statement easily follows from formula (1.10). In fact, because
$S\subset K$ has maximum volume, we have $|\Delta_{j}(x)|\leq|\Delta|$ for any
$j=1,\ldots,n+1$ and $x\in K.$ It suffices to use the relation (1.1) between
determinants and volumes of simplices. Thanks to (1.3),

$-\lambda_{j}(x)\leq|\lambda(x)|=\frac{|\Delta_{j}(x)|}{|\Delta|}\leq 1,\quad
x\in K.$

Thus, by (1.10), we obtain

$\xi(K;S)=(n+1)\max_{1\leq k\leq n+1}\max_{x\in K}(-\lambda_{k}(x))+1\leq
n+2.$

This approach was used by the author in [24] in the case $K=Q_{n}$.

##  3\. The values of $\alpha(B_{n};S)$ and $\xi(B_{n};S)$

In this section we discuss characteristics associated with the absorption of a
Euclidean ball by a homothetic image of a simplex (with or without
translation). Replacing a cube with a ball makes many questions much more
simple. However, geometric interpretation of general results has a certain
interest also in this particular case. Besides, we will note some new
applications of the basic Lagrange polynomials.

Given an $n$-dimensional simplex $S$, let us introduce the following geometric
characteristics. The inradius of $S$ is the maximum of the radii of balls
contained within $S$. The center of this unique maximum ball is called the
incenter of $S$. The boundary of the maximum ball is a sphere that has a
single common point with each $(n-1)$-dimensional face of $S$. By the
circumradius of S we mean the minimum of the radii of balls containing $S$.
Note that the boundary of this unique minimal ball contains all the vertices
of $S$ if and only if its center lies inside the simplex.

The inradius $r$ and the circumradius $R$ of a simplex $S$ satisfy the so-
called Euler inequality

$R\geq nr.$ (3.1)

Equality in (3.1) takes place if and only if $S$ is a regular simplex. For the
proof of the Euler inequality, its history and various generalizations we
refer the reader to [12], [58] and [56].

Let us note that the Euler inequality is equivalent to the following
statement: Suppose $B$ is a ball with radius $r_{1}$ and $B^{*}$ is a ball
with radius $r_{2}$. If $B\subset S\subset B^{*}$, then $r_{2}\geq nr_{1}.$
Equality takes place if and only if $S$ is a regular simplex inscribed into
$B^{*}$ and $B$ is the ball inscribed into $S$. Another equivalent form of
these propositions is given by Theorem 3.3.

Also we remark that the analog of the above property for parallelotopes is
expressed by Corollary 2.5.

We turn to presentation of some computational formulae for the quantities
$\alpha(B_{n};S)$ and $\xi(B_{n};S)$ proven in [28]. Let $x^{(1)},$ $\ldots,$
$x^{(n+1)}$ be the vertices and $\lambda_{1},$ $\ldots,$ $\lambda_{n+1}$ be
the basic Lagrange polynomials of a nondegenerate simplex
$S\subset{\mathbb{R}}^{n}$. As usual, $l_{ij}$ are the coefficients of
$\lambda_{j}.$ See formulas from Section 1.

Given $j\in\\{1,...,n+1\\}$, let $\Gamma_{j}$ be the $(n-1)$-dimensional
hyperplane determined by the equation $\lambda_{j}(x)=0$, and let $\Sigma_{j}$
be the $(n-1)$-dimensional face of $S$ contained in $\Gamma_{j}$. We let
$h_{j}$ denote the height of $S$ conducted from $x^{(j)}$ onto $\Gamma_{j}$,
and we let $r$ denote the inradius of $S$. By $\sigma_{j}$ we denote
$(n-1)$-dimensional measure of $\Sigma_{j}$. Finally, we put
$\sigma=\sum\limits_{j=1}^{n+1}\sigma_{j}$.

In [28] we show that the value $\alpha(B_{n};S)$ can be calculated in various
ways.

Theorem 3.1. The following equalities hold:

$\alpha(B_{n};S)=\sum_{j=1}^{n+1}\left(\sum_{i=1}^{n}l_{ij}^{2}\right)^{1/2},$
(3.2) $\alpha(B_{n};S)=\sum_{j=1}^{n+1}\frac{1}{h_{j}},$ (3.3)
$\alpha(B_{n};S)=\frac{1}{r},$ (3.4) $\alpha(B_{n};S)=\frac{\sigma}{n\,{\rm
vol}(S)}.$

From (3.3) and (3.4) we get

$\frac{1}{r}=\sum_{j=1}^{n+1}\frac{1}{h_{j}}.$

This geometric relation (which evidently can be obtained in the direct way)
occurs to be equivalent to general formula (1.11) for $\alpha(K;S)$ in the
case when the convex body $K$ coincide with the unit ball.

It is interesting to compare (3.2) with the formula

$\alpha(Q_{n}^{\prime};S)=\sum_{i=1}^{n}\sum_{j=1}^{n+1}|l_{ij}|.$

for the cube $Q_{n}^{\prime}=[-1,1]^{n}$ (see Corollary 2.2). Since
$B_{n}\subset Q_{n}^{\prime}$, it follows that
$\alpha(B_{n};S)\leq\alpha(Q_{n}^{\prime};S)$. Of course, analytically, this
inequality is immediate from (3.2) and the above formula for
$\alpha(Q_{n}^{\prime};S)$.

In turn, the inradius $r$ and the incenter $z$ of a simplex $S$ can be
calculated as follows:

$r=\frac{1}{\sum\limits_{j=1}^{n+1}\left(\sum\limits_{i=1}^{n}l_{ij}^{2}\right)^{1/2}},\quad
z=\frac{1}{\sum\limits_{j=1}^{n+1}\left(\sum\limits_{i=1}^{n}l_{ij}^{2}\right)^{1/2}}\sum\limits_{j=1}^{n+1}\left(\sum\limits_{i=1}^{n}l_{ij}^{2}\right)^{1/2}x^{(j)}.$

For arbitrary $x^{(0)}$ and $\varrho>0$, the quantity
$\alpha\left(B(x^{(0)};\varrho);S\right)$ one can calculate using the equality
$\alpha(B(x^{(0)};\varrho);S)$ $=$ $\varrho\alpha(B_{n};S)$.

The general formula (1.10) for $\xi(K;S)$ in the case
$K=B\left(x^{(0)};\varrho\right)$ provides the following (see [28]).

Theorem 3.2. If $B\left(x^{(0)};\varrho\right)\not\subset S$, we have

$\xi\left(B\left(x^{(0)};\varrho\right);S\right)=(n+1)\max_{1\leq j\leq
n+1}\left[\varrho\left(\sum_{i=1}^{n}l_{ij}^{2}\right)^{1/2}-\sum_{i=1}^{n}l_{ij}x_{i}^{(0)}-l_{n+1,j}\right]+1.$
(3.5)

In particular, in the case $B_{n}\not\subset S$

$\xi(B_{n};S)=(n+1)\max_{1\leq j\leq
n+1}\left[\left(\sum_{i=1}^{n}l_{ij}^{2}\right)^{1/2}-l_{n+1,j}\right]+1.$
(3.6)

Now we turn to lower bounds for $\alpha(B_{n};S)$ and $\xi(B_{n};S)$ for
$S\subset B_{n}.$ It is easy to show that the Euler inequality is equivalent
to the following statement.

Theorem 3.3. If $S\subset B_{n}$, then $\alpha(B_{n};S)\geq n.$ The equality
$\alpha(B_{n};S)=n$ holds true if and only if $S$ is a regular simplex
inscribed into $B_{n}$.

Recall that the minimum value of $\alpha(Q_{n};S)$ for $S\subset Q_{n}$ is
also equal to $n$ (see (2.6)). This value corresponds to those and only those
$S\subset Q_{n}$ for which every axial diameter $d_{i}(S)$ is equal to $1$.
The latter property is fulfilled for the maximum volume simplices in $Q_{n}$,
but, if $n>2$, not for the only ones.

If $S\subset B_{n}$, then $\xi(B_{n};S)\geq n.$ The equality $\xi(B_{n};S)=n$
takes place if and only if $S$ is a regular simplex inscribed into $B_{n}$.
These statements immediately follow from Theorem 3.3 and the inequality
$\xi(B_{n};S)\geq\alpha(B_{n};S)$. Paper [28] contains also the direct
arguments without applying the Euler inequality that was used in [28] to
obtain the estimate $\alpha(B_{n};S)\geq n$.

Now, let us consider an analog of the quantity $\xi_{n}(Q_{n})$, the quantity
$\xi_{n}(B_{n})$. In accordance with Definition 1.3,

$\xi_{n}(B_{n})=\min\\{\xi(B_{n};S):\,S\mbox{ is an $n$-dimensional
simplex,}\,S\subset B_{n},\,{\rm vol}(S)\neq 0\\}.$

While many problems concerning $\xi_{n}(Q_{n})$ yet have not been solved (see
Section 2), the problem on numbers $\xi_{n}(B_{n})$ turns out to be simple.
For any $n$, we have $\xi_{n}(B_{n})=n$. The only simplex $S\subset B_{n}$
extremal with respect to $\xi_{n}(B_{n})$ is an arbitrary regular simplex
inscribed into $B_{n}$.

Let us note that, thanks to (1.7), for every interpolation projector
$P:C(B_{n})\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ and the corresponding
simplex $S\subset B_{n}$, the following inequalities

$\frac{n+1}{2n}\Bigl{(}\|P\|_{B_{n}}-1\Bigr{)}+1\leq\xi(B_{n};S)\leq\frac{n+1}{2}\Bigl{(}\|P\|_{B_{n}}-1\Bigr{)}+1$
(3.7)

hold. Furthermore, if there exists an $1$-point in $B_{n}$ with respect to $S$
(see Section 1), then the right-hand inequality in (3.7) becomes an equality.
Hence,

$\frac{n+1}{2n}\Bigl{(}\theta_{n}(B_{n})-1\Bigr{)}+1\leq\xi_{n}(B_{n})=n\leq\frac{n+1}{2}\Bigl{(}\theta_{n}(B_{n})-1\Bigr{)}+1.$
(3.8)

Starting from $n=5$, the right-hand inequality in (3.7) is strict, while for
$1\leq n\leq 4$ it turns into equality. This fact is noted in Section 7. Here
is another argument, but also using some of the properties below. Since we
know the precise value of $\theta_{n}(B_{n})$ (see Section 9), the cases
$1\leq n\leq 5$ are cheking directly. The result for $n\geq 6$ follows from
the estimate $\theta_{n}(B_{n})\geq\sqrt{n}$, since then

$n<\frac{n+1}{2}\Bigl{(}\sqrt{n}-1\Bigr{)}+1.$

##  4\. Estimates for $\theta_{n}(Q_{n})$

Despite the apparent simplicity of formulation, the problem of finding exact
values of $\theta_{n}(Q_{n})$ is very difficult. Since 2006, these values are
known only for $n=1,2,3,$ and $7$. Namely,

$\theta_{1}(Q_{1})=1,\quad\theta_{2}(Q_{2})=\frac{2\sqrt{5}}{5}+1=1.8944\ldots,\quad\theta_{3}(Q_{3})=2,\quad\theta_{7}(Q_{7})=\frac{5}{2}.$

Note that for each $n=1,2,3,7$ the right-hand relation in (1.8) becomes an
equality:

$\xi_{n}(Q_{n})=\frac{n+1}{2}\left(\theta_{n}(Q_{n})-1\right)+1.$ (4.1)

If $n=1,3,$ or $7$, then an equality takes place also in (2.7), i. e.,

$\theta_{n}(Q_{n})=3-\frac{4}{n+1}.$

For $1\leq n\leq 3$, the simplices corresponding to minimal projectors are
just the same that the extremal simplices with respect to $\xi_{n}(Q_{n})$,
see Section 2. In the seven-dimensional case, the nodes of minimal projector
appear from a unique (up to equivalence) Hadamard matrix of order $8$. The
proofs are given in original paper [19] and also in [24] and [46].

Clearly, $\theta_{1}(Q_{1})=1$. This value can be found also from (1.8). For
any projector, there is a $1$-vertex $Q_{1}$ relative to the corresponding
simplex (which is a segment). Hence, we have (4.1). Now $\xi_{1}(Q_{1})=1$
implies $\theta_{1}(Q_{1})=1.$ The nodes of a unique minimal projector are 0
and 1. For the most complicated cases $n=2,3,$ see [24].

Let us turn to the case $n=7$. Since $8$ is an Hadamard number, there exists a
seven-dimensional regular simplex having vertices at vertices of the cube. We
can take the simplex with the vertices

$(1,1,1,1,1,1,1),~{}~{}~{}(0,1,0,1,0,1,0),~{}~{}~{}(0,0,1,1,0,0,1),~{}~{}~{}(1,0,0,1,1,0,0),$
$(0,0,0,0,1,1,1),~{}~{}~{}(1,0,1,0,0,1,0),~{}~{}~{}(1,1,0,0,0,0,1),~{}~{}~{}(0,1,1,0,1,0,0).$

The equality $\xi_{7}(Q_{7})=7$ (see Section 2) and inequality

$\xi_{n}(Q_{n})\leq\frac{n+1}{2}\left(\theta_{n}(Q_{n})-1\right)+1$

imply $\theta_{7}(Q_{7})\geq\frac{5}{2}$. But for the corresponding projector,
$\|P\|_{Q_{7}}=\frac{5}{2}$. Therefore, $\theta_{7}=\frac{5}{2}$, and this
projector is minimal.

Calculations seems to be more easy if we make use of similarity reasons. Let
us take the cube $Q_{7}^{\prime}=[-1,1]$ and consider the projector
$P:C(Q_{7}^{\prime})\to\Pi_{1}\left({\mathbb{R}}^{7}\right)$ constructed from
the Hadamard matrix

${\bf H}_{8}=\left(\begin{array}[]{cccccccc}1&1&1&1&1&1&1&1\\\
-1&1&-1&1&-1&1&-1&1\\\ -1&-1&1&1&-1&-1&1&1\\\ 1&-1&-1&1&1&-1&-1&1\\\
-1&-1&-1&-1&1&1&1&1\\\ 1&-1&1&-1&-1&1&-1&1\\\ 1&1&-1&-1&-1&-1&1&1\\\
-1&1&1&-1&1&-1&-1&1\end{array}\right).$

(This matrix is unique up to the equivalence of Hadamard matrix of order 8.)
Suppose the nodes of $P$ are written in the rows of this matrix except the
last column. Since ${\bf H}_{8}^{-1}$ $=$ $({1}/{8}){\bf H}_{8}^{T},$
coefficients of $\lambda_{j}$ form the $j$th row of $({1}/{8}){\bf H}_{8}:$

$\lambda_{1}(x)=\frac{1}{8}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+1\right),$
$\lambda_{2}(x)=\frac{1}{8}\left(-x_{1}+x_{2}-x_{3}+x_{4}-x_{5}+x_{6}-x_{7}+1\right),$

and so on. Using (1.6), we get $\|P\|_{Q_{7}^{\prime}}={5}/{2}$, and, thanks
to the above approach, $\theta_{7}(Q^{\prime}_{7})={5}/{2}.$

However, $n=7$ is the biggest $n$ such that (i) $n+1$ is an Hadamard number
and (ii) (2.8) holds provided $P$ corresponds to a regular simplex inscribed
into the cube.

Now let us note some upper bounds for $\theta_{n}(Q_{n})$. For $n\geq 3$, the
projector with nodes (2.10) satisfies $\|P\|_{Q_{n}}\leq({n+1})/{2}$ with the
equality for odd $n$ ([16], [24]). Therefore, if $n\neq 2$, then

$\theta_{n}(Q_{n})\leq\frac{n+1}{2}.$

Recall that $h_{n}$ denotes the maximum value of a determinant of order $n$
with entries $0$ or $1$ (see Definition 1.9). Suppose $P$ corresponds to a
maximum volume simplex $S\subset Q_{n}$ with some vertex coinciding with a
vertex of the cube. Then

$\|P\|_{Q_{n}}\leq\frac{2h_{n+1}}{h_{n}}+1.$ (4.2)

Hence,

$\theta_{n}(Q_{n})\leq\frac{2h_{n+1}}{h_{n}}+1$

for all $n$ and also $\theta_{n}(Q_{n})\leq\sqrt{2n+3}+1$, provided $n+1$ is
an Hadamard number (see [31]). Earlier (see [16] and [24]) it was proved that
in the Hadamard case $\xi_{n}(Q_{n})\leq\sqrt{n+1}$. Let us give here the
proof from [31] essentially making use of the structure of an Hadamard matrix.

Theorem 4.1. Let $n+1$ be an Hadamard number, and let $S$ be an
$n$-dimensional regular simplex having the vertices at vertices of
$Q_{n}^{\prime}$. Then, for the corresponding interpolation projector
$P:C(Q_{n}^{\prime})\to\Pi_{1}({\mathbb{R}}^{n})$, we have

$\|P\|_{Q_{n}^{\prime}}\leq\sqrt{n+1}.$ (4.3)

Proof. Let $x^{(1)},\ldots,x^{(n+1)}$ be the vertices and let
$\lambda_{1},\ldots,\lambda_{n+1}$ be the basic Lagrange polynomials of the
simplex. Thanks to the theorem’s hypothesis, the vertex matrix $\bf S$ is an
Hadamard matrix of order $n+1$ with the last column consisting of $1$’s.

Let us show that

$\sum_{j=1}^{n+1}\lambda_{j}(x)^{2}=\frac{||x||^{2}+1}{n+1}~{}~{}~{}\text{for
every}~{}~{}~{}x\in{\mathbb{R}}^{n}.$ (4.4)

Let $y=(x_{1},\ldots,x_{n},1)\in{\mathbb{R}}^{n+1}.$ The coefficients of
$\lambda_{j}$ form the $j$th column of ${\bf S}^{-1}$. Since $\bf S$ is an
Hadamard matrix, it satisfies (1.14), whence

$\lambda_{j}(x)=\frac{1}{n+1}\langle h^{(j)},y\rangle.$

Further, ${h^{(j)}}/{\sqrt{n+1}}$ form an orthonormalized basis of
${\mathbb{R}}^{n+1},$ therefore,

$y=\sum_{j=1}^{m}\frac{\langle
h^{(j)},y\rangle}{\sqrt{n+1}}\frac{h^{(j)}}{\sqrt{n+1}},\quad\langle
y,y\rangle=\sum_{j=1}^{n+1}\frac{\langle h^{(j)},y\rangle^{2}}{n+1}.$

From this, we have

$||x||^{2}+1=\langle y,y\rangle=\sum_{j=1}^{n+1}\frac{\langle
h^{(j)},y\rangle^{2}}{n+1}=(n+1)\sum_{j=1}^{n+1}\lambda_{j}(x)^{2}$

proving (4.4).

If $x$ is a vertex of $Q_{n}^{\prime}$, then $||x||^{2}=n$ and (4.4) implies
$\sum\lambda_{j}(x)^{2}=1.$ Applying Cauchy inequality, for each $x\in{\rm
ver}(Q_{n}^{\prime})$, we have

$\sum_{j=1}^{n+1}|\lambda_{j}(x)|\leq\left(\sum_{j=1}^{n+1}\lambda_{j}(x)^{2}\right)^{\frac{1}{2}}\cdot\sqrt{n+1}=\sqrt{n+1}.$

Now, let $P:C(Q_{n}^{\prime})\to\Pi_{1}({\mathbb{R}}^{n})$ be an interpolation
projector corresponding to $S$. Then, thanks to (1.6),

$\|P\|_{Q_{n}^{\prime}}=\max_{x\in{\rm
ver}(Q_{n}^{\prime})}\sum_{j=1}^{n+1}|\lambda_{j}(x)|\leq\sqrt{n+1}$

completing the proof of the theorem. $\Box$

By similarity reasons, Theorem 4.1 overcomes to any $n$-dimensional cube, e.
g., to the cube $Q_{n}=[0,1]^{n}$. Thus, if $n+1$ is an Hadamard number, then
$\theta_{n}(Q_{n})\leq\sqrt{n+1}.$

We can slightly improve this estimate by using interpolation on a ball.

Let $p_{n}=\max\\{\psi(a),\psi(a+1)\\}$ where

$\psi(t):=\frac{2\sqrt{n}}{n+1}\Bigl{(}t(n+1-t)\Bigr{)}^{1/2}+\left|1-\frac{2t}{n+1}\right|,\quad
a=\left\lfloor\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}\right\rfloor.$

Now, let us assume that $n+1$ is an Hadamard number. Consider an interpolation
projector $P$ with the nodes at those vertices of the cube $Q_{n}^{\prime}$
that form a regular simplex $S$. Since $Q_{n}^{\prime}$ is inscribed into the
unit ball $B_{n}$, simplex $S$ is also inscribed into $B_{n}$. It remains to
apply formula (1.5) for the projector’s norm both on the cube and on the ball:

$\|P\|_{Q_{n}^{\prime}}=\max_{x\in
Q_{n}^{\prime}}\sum_{j=1}^{n+1}|\lambda_{j}(x)|\leq\max_{x\in
B_{n}}\sum_{j=1}^{n+1}|\lambda_{j}(x)|=\|P\|_{B_{n}}=p_{n}\leq\sqrt{n+1}.$

(For the equality $\|P\|_{B_{n}}=p_{n}$ and the final estimate, we refer the
reader to Section 7.) Hence,

$\theta_{n}(Q_{n})\leq p_{n}\leq\sqrt{n+1}.$

The equality $\|P\|_{Q_{n}^{\prime}}=\sqrt{n+1}$ may hold as for all regular
simplices having vertices at vertices of the cube $(n=1,n=3)$, as for some of
them $(n=15),$ or may not be executed at all.

As it is shown in Section 6, for each $n$,
$\theta_{n}(Q_{n})\geq\sqrt{n-1}/e.$ Therefore, if $n+1$ is an Hadamard
number, then

$\frac{\sqrt{n-1}}{e}\leq\theta_{n}(Q_{n})\leq\sqrt{n+1}.$

The upper bounds of $\theta_{n}(Q_{n})$ for special $n$ were improved by A.
Ukhalov and his students with the help of computer methods. In particular,
simplices of maximum volume in the cube were considered. In all situations
where $n+1$ is an Hadamard number, the full set of Hadamard matrices of the
corresponding order was used. In particular, to obtain an estimate for
$\theta_{23}$, all existing $60$ Hadamard matrices of order $24$ were
considered. To estimate $\theta_{27}$, we have to consider $487$ Hadamard
matrices of order $28$. Known nowaday upper estimates for $1\leq n\leq 27$ are
given in [46]. (Here, for brevity, $\theta_{n}=\theta_{n}(Q_{n})$.)

$\theta_{1}=1,\quad\theta_{2}=\frac{2\sqrt{5}}{5}+1,\quad\theta_{3}=2,\quad\theta_{4}\leq\frac{3(4+\sqrt{2})}{7},\quad\theta_{5}\leq
2.448804,$ $\theta_{6}\leq
2.6000\ldots,\quad\theta_{7}=\frac{5}{2},\quad\theta_{8}\leq\frac{22}{7},\quad\theta_{9}\leq
3,\quad\theta_{10}\leq\frac{19}{5},\quad\theta_{11}\leq 3,$
$\theta_{12}\leq\frac{17}{5},\quad\theta_{13}\leq\frac{49}{13},\quad\theta_{14}\leq\frac{21}{5},\quad\theta_{15}\leq\frac{7}{2},\quad\theta_{16}\leq\frac{21}{5},\quad\theta_{17}\leq\frac{139}{34},$
$\theta_{18}\leq 5.1400\ldots,\quad\theta_{19}\leq 4,\quad\theta_{20}\leq
4.68879\ldots,\quad\theta_{21}\leq\frac{251}{50},\quad\theta_{22}\leq\frac{1817}{335},$
$\theta_{23}\leq\frac{9}{2},\quad\theta_{24}\leq\frac{103}{21},\quad\theta_{25}\leq
5,\quad\theta_{26}\leq\frac{474}{91},\quad\theta_{27}\leq 5.$

Note that $n$-dimensional regular simplices with vertices at the vertices of
the cube can be located differently with respect to the vertices and faces of
the cube. This is quite evident when the norms of the corresponding projectors
are different. But this is also possible if regular simplices generate the
same norms. In [31] we describe an approach based on comparison of
$\mu$-vertices of the cube with respect to various simplices.

Recall that a $\mu$-vertex of the cube $Q$ with respect to a simplex $S\subset
Q$ is such a vertex $x$ of the cube that for the corresponding projector
$P=P_{S}$ we have $\|P\|_{Q}=\sum|\lambda_{j}(x)|$ and exactly $\mu$ numbers
$\lambda_{j}(x)$ are negative (see Definition 1.7). Of course, simplices which
have different sets of $\mu$-vertices with respect to the containing cube, are
differently located in the cube, even if they have the same projector’s norms.
These simplices are non-equivalent in the following sense: one of them cannot
be mapped into another by an orthogonal transform which maps the cube into
itself. Let us give some examples for the case $Q=Q_{n}^{\prime}.$

While obtaining estimates for minimal norms of projectors, in [38] various
$n$-dimensional regular simplices arising from different Hadamard matrices of
order $n+1$ were calculated. For $n=15$, the order of matrices is equal to
$16$. Up to equivalence, there are exactly five Hadamard matrices. They
correspond to five simplices described in Table 1.

Table 1:  Regular simplices for $n=15$

$\begin{array}[]{|c|c|c|c|}\hline\cr S&{\|P\|}_{Q_{15}^{\prime}}&\mbox{Values
}\ \mu&\mbox{ Number of}\ \mu\mbox{-vertices}\\\ \hline\cr
S_{1}&4&6&m_{6}=448\\\ S_{2}&4&6&m_{6}=192\\\ S_{3}&4&6&m_{6}=64\\\
S_{4}&\frac{7}{2}&4,5,6,8&m_{4}=896,m_{5}=1344,m_{6}=5376,m_{8}=1344\\\
S_{5}&\frac{7}{2}&4,5,6,8&m_{4}=896,m_{5}=1344,m_{6}=5376,m_{8}=1344\\\
\hline\cr\end{array}$

By $m_{\mu}$ we denote a number of $\mu$-vertices of the cube $Q_{n}^{\prime}$
with respect to every simplex. For the rest $1\leq\mu\leq 15$, excepting the
given in Table 1, values $m_{\mu}$ are zero. Every simplex $S_{1}$, $S_{2}$,
and $S_{3}$ generates the same projector’s norm and has only $6$-vertices. But
the numbers of $6$-vertices for them are different; hence, these simplices are
pairwise non-equivalent. Each of them also is non-equivalent as to $S_{4}$, so
as to $S_{5}.$ The latter simplices generates equal norms and have the same
sets of $\mu$-vertices. Also we have $\theta_{15}\leq{7}/{2}=3.5$. This is
more exact than the estimate $\theta_{15}(Q_{15})\leq 4$ provided by (4.3).

Another example given in [31] is related to $n=23.$ Regular inscribed
simplices were built from the available 60 pairwise non-equivalent Hadamard
matrices of order $24$. For all simplices, excepting ones with the numbers 16,
53, 59, and 60, the projector norm equals ${14}/{3}=4.6666\ldots$, while for
each of these four simplices the projector norm is ${9}/{2}=4.5$. In
particular, we have $\theta_{23}(Q_{23})\leq 4.5$ (which was also noted in
[38]). This inequality is more exact than the estimate
$\theta_{23}(Q_{23})\leq\sqrt{24}=4.8989\ldots$. Each of the four exceptional
simplices is not equivalent to any of the 56 others.

Despite the possible differs, for all regular simplices with vertices at the
vertices of the cube, inequality (4.3) holds. The inscribed regular simplices
satisfying $\|P\|_{Q_{n}^{\prime}}=\sqrt{n+1}$ exist at least for $n=1$,
$n=3$, and $n=15.$ The problem of full description of dimensions $n$ with such
property is still open.

The best nowaday known lower bound of $\theta_{n}(Q_{n})$ has the form

$\theta_{n}(Q_{n})\geq\max\left[3-\frac{4}{n+1},\,\chi_{n}^{-1}\left(\frac{1}{\nu_{n}}\right)\right].$
(4.5)

Here $\chi_{n}$ is the standardized Legendre polynomial of degree $n$, see
Section 5.

The values of the right-hand side of (4.5) for $1\leq n\leq 54$ are given in
[46]. Inequality

$\chi_{n}^{-1}\left(\frac{1}{\nu_{n}}\right)>3-\frac{4}{n+1}$

takes place for $n\geq 53$. On estimates via the function $\chi_{n}^{-1}$ see
Section 6.

As noted in Section 1 (see the right-hand relation in (1.7)), the inequality

$\xi_{n}(Q_{n})\leq\frac{n+1}{2}\left(\theta_{n}(Q_{n})-1\right)+1$ (4.6)

is true for any $n$. So far, we know only four values of $n$ for which this
relation becomes an equality: $n=1,2,3$, and $7$. These are exactly the cases
in which we know the exact values of $\theta_{n}(Q_{n})$ and $\xi_{n}(Q_{n})$.
In [33] the authors conjectured that the minimum of $n$ for which inequality
(4.6) is strict is $4.$ This is still an open problem.

We note that, thanks to the equivalence $\xi_{n}(Q_{n})\asymp n$ and
inequality $\theta_{n}\geq c\sqrt{n}$, for all sufficiently large $n$, we have

$\xi_{n}(Q_{n})<\frac{n+1}{2}\left(\theta_{n}(Q_{n})-1\right)+1.$ (4.7)

Let $n_{0}$ be the minimal natural number such that for all $n\geq n_{0}$
inequality (4.7) holds. The problem about the exact value of $n_{0}$ is very
difficult. The known lower and upper bounds differ quite significantly. From
the preceding, we have the estimate $n_{0}\geq 8.$ In 2009 we proved that
$n_{0}\leq 57$ (see [19], [24, § 3.7]). A sufficient condition for the
validity of (4.7) for $n>2$ is the inequality

$\chi_{n}\left(\frac{3n-5}{n-1}\right)\cdot\nu_{n}<1.$ (4.8)

It was proved in [19] that (4.8) is satisfied for $n\leq 57$. Later
calculations allowed the upper bound of $n_{0}$ to be slightly lowered. These
results are described in [38] where it is noted that $n_{0}\leq 53$. In other
words, inequality (4.7) is satisfied at least starting from $n=53$. Note that
a better estimate from above for $n_{0}$ is an open problem.

##  5\. Legendre polynomials and the measure of $E_{n,\gamma}$

The standardized Legendre polynomial of degree $n$ is the function

$\chi_{n}(t)=\frac{1}{2^{n}n!}\left[(t^{2}-1)^{n}\right]^{(n)}$

(Rodrigues’ formula). For properties of $\chi_{n}$ see, e.g., [52],[53].
Legendre polynomials are orthogonal on the segment $[-1,1]$ with the weight
$w(t)\equiv 1.$ The first standardized Legendre polynomials are

$\chi_{0}(t)=1,\quad\chi_{1}(t)=t,\quad\chi_{2}(t)=\frac{1}{2}\left(3t^{2}-1\right),\quad\chi_{3}(t)=\frac{1}{2}\left(5t^{3}-3t\right),$
$\chi_{4}(t)=\frac{1}{8}\left(35t^{4}-30t^{2}+3\right),\quad\chi_{5}(t)=\frac{1}{8}\left(63t^{5}-70t^{3}+15t\right).$

We have

$\chi_{n+1}(t)=\frac{2n+1}{n+1}\,t\chi_{n}(t)-\frac{n}{n+1}\,\chi_{n-1}(t).$
(5.1)

This implies $\chi_{n}(1)=1$. If $n\geq 1$, then $\chi_{n}(t)$ increases as
$t\geq 1$. The latter properties also easily follow from (5.2).

We let $\chi_{n}^{-1}$ denote the function inverse to $\chi_{n}$ on the semi-
axis $[1,+\infty)$.

The appearance of Legendre polynomials in the circle of our questions is due
to some their property. For $\gamma\geq 1$, let us define the set

$E_{n,\gamma}:=\left\\{x\in{\mathbb{R}}^{n}:\sum_{j=1}^{n}|x_{j}|+\left|1-\sum_{j=1}^{n}x_{j}\right|\leq\gamma\right\\}.$

In [16] we proved such a statement (the proof is also given in [24] and [46]).

Theorem 5.1. The following equalities hold:

${\rm mes_{n}}(E_{n,\gamma})=\frac{1}{2^{n}n!}\sum_{i=0}^{n}{n\choose
i}^{2}(\gamma-1)^{n-i}(\gamma+1)^{i}=\frac{\chi_{n}(\gamma)}{n!}.$ (5.2)

Proof. First let us establish the left-hand equality in (5.2). Let

$E^{(1)}=\\{x\in E_{n,\gamma}:\sum x_{i}>1\\},~{}~{}~{}~{}E^{(2)}=\\{x\in
E_{n,\gamma}:\sum x_{i}\leq 1\\}.$

Let us find sequentially $m_{1}={\rm mes_{n}}(E^{(1)})$ and $m_{2}={\rm
mes_{n}}(E^{(2)}).$

Temporarily fix $k$, $1\leq k\leq n$, and consider a non-empty subset
$G\subset E^{(1)}$ consisting of all points $x=(x_{1},...,x_{n})$ such that
$x_{1},\ldots,x_{k}\geq 0$ and $x_{k+1},\ldots,x_{n}<0$. Let $y_{i}=x_{i}$ for
$i=1,...,k$, $y_{i}=-x_{i}$ for $i=k+1,...,n$, and let $y=(y_{1},...,y_{n})$.
Then

$G=\left\\{y:1+y_{k+1}+\ldots+y_{n}\leq
y_{1}+\ldots+y_{k}\leq\frac{\gamma+1}{2},\ y_{i}\geq 0\right\\},$

hence

${\rm mes_{n}}(G)=\int\limits_{1}^{\alpha}\,dy_{1}\int\limits_{1}^{\alpha-
y_{1}}\,dy_{2}\ldots\int\limits_{1}^{\alpha-y_{1}-\ldots-
y_{k-1}}\,dy_{k}\cdot$
$\cdot\int\limits_{0}^{y_{1}+\ldots+y_{k}-1}\,dy_{k+1}\int\limits_{0}^{y_{1}+\ldots+y_{k}-1-y_{k+1}}\,dy_{k+2}\ldots\int\limits_{0}^{y_{1}+\ldots+y_{k}-1-y_{k+1}-\ldots-
y_{n-1}}\,dy_{n}.$

Throughout the proof, $\alpha=(\gamma+1)/2.$ If $b>0$, then

$\int\limits_{0}^{b}\,dz_{1}\int\limits_{0}^{b-z_{1}}\,dz_{2}\ldots\int\limits_{0}^{b-z_{1}-\ldots-
z_{l-1}}\,dz_{l}=\frac{b^{l}}{l!},$

so

${\rm mes_{n}}(G)=$ $=\int\limits_{1}^{\alpha}\,dy_{1}\int\limits_{1}^{\alpha-
y_{1}}\,dy_{2}\ldots\int\limits_{1}^{\alpha-y_{1}-\ldots-
y_{k-1}}\frac{1}{(n-k)!}{\left(y_{1}+\ldots+y_{k}-1\right)}^{n-k}\,dy_{k}=$
$=\left(\int\limits_{\
y_{1}+\ldots+y_{k}\leq\alpha}-\int\limits_{y_{1}+\ldots+y_{k}\leq
1}\right)\frac{1}{(n-k)!}{(y_{1}+\ldots+y_{k}-1)}^{n-k}\,dy_{1}\ldots\,dy_{k}=$
$=J_{1}-J_{2}.$

The first integral equals

$J_{1}=\sum_{j=1}^{k}(-1)^{j+1}\frac{(\alpha-1)^{n-k+j}}{(n-k+j)!}\frac{\alpha^{k-j}}{(k-j)!}+\frac{(-1)^{n+k}}{n!}.$

The value of $J_{2}$ appears from this expression if instead of $\alpha$ we
take 1. Consequently,

${\rm
mes_{n}}(G)=J_{1}-J_{2}=\sum_{j=1}^{k}(-1)^{j+1}\frac{(\alpha-1)^{n-k+j}}{(n-k+j)!}\frac{\alpha^{k-j}}{(k-j)!}=$
$=\frac{(-1)^{k+1}}{n!}\sum_{i=0}^{k-1}{n\choose
i}(\alpha-1)^{n-i}(-\alpha)^{i}.$

Clearly, the set $E^{(1)}$ is the union of all pairwise disjoint sets $G$ with
various $k=1,\ldots,n,$. Therefore, the measure of $E^{(1)}$ is equal to

$m_{1}=\sum_{k=1}^{n}{n\choose
k}\frac{(-1)^{k+1}}{n!}\sum_{i=0}^{k-1}{n\choose
i}(\alpha-1)^{n-i}(-\alpha)^{i}.$

Changing the order of summation and using the identity

$\sum_{k=0}^{i}(-1)^{k}{n\choose k}=(-1)^{i}{{n-1}\choose i}$ (5.3)

(see, e. g., [48]) we get

$m_{1}=\frac{1}{n!}\sum_{i=0}^{n-1}{n\choose
i}(\alpha-1)^{n-i}(-\alpha)^{i}\sum_{k=0}^{i}(-1)^{k}{n\choose
k}=\frac{1}{n!}\sum_{i=0}^{n-1}{n\choose i}{{n-1}\choose
i}(\alpha-1)^{n-i}\alpha^{i}.$ (5.4)

Now let us turn to $E^{(2)}.$ First, note that $E^{(2)}$ contains the domain
$S:=\\{x_{i}\geq 0,$ $\sum x_{i}\leq 1\\},$ with the measure $1/n!$. Next, fix
$k\in\\{1,...,n\\}$ and consider the subset $G^{\prime}\subset E^{(2)}$
corresponding to the inequalities $x_{1},\ldots,x_{k}<0;$
$x_{k+1},\ldots,x_{n}\geq 0.$ Put $y_{1}=-x_{1},\ldots,y_{k}=-x_{k};$
$y_{k+1}=x_{k+1},\ldots,y_{n}=x_{n}.$ Then

$G^{\prime}=\\{y:y_{k+1}+\ldots+y_{n}\leq
1+y_{1}+\ldots+y_{k}\leq\frac{\gamma-1}{2},\ y_{i}\geq 0\\}.$

Let $\beta=({\gamma-1})/{2}.$ Then the following equalities hold:

${\rm
mes_{n}}(G^{\prime})=\int\limits_{0}^{\beta}\,dy_{1}\int\limits_{0}^{\beta-
y_{1}}\,dy_{2}\ldots\int\limits_{0}^{\beta-y_{1}-\ldots-y_{k-1}}\,dy_{k}\cdot$
$\cdot\int\limits_{0}^{1+y_{1}+\ldots+y_{k}}\,dy_{k+1}\int\limits_{0}^{1+y_{1}+\ldots+y_{k}-y_{k+1}}\,dy_{k+2}\ldots\int\limits_{0}^{1+y_{1}+\ldots+y_{k}-y_{k+1}-\ldots-
y_{n-1}}\,dy_{n}=$ $=\int\limits_{0}^{\beta}\,dy_{1}\int\limits_{0}^{\beta-
y_{1}}\,dy_{2}\ldots\int\limits_{0}^{\beta-y_{1}-\ldots-
y_{k-1}}\frac{(1+y_{1}+\ldots+y_{k})^{n-k}}{(n-k)!}\,dy_{k}=$
$=\sum_{j=0}^{k-1}(-1)^{k-1-j}\frac{(1+\beta)^{n-j}\beta^{j}}{(n-j)!j!}+\frac{(-1)^{k}}{n!}=$
$=\frac{(-1)^{k+1}}{n!}\Bigl{(}\sum_{j=0}^{k-1}{n\choose
j}(1+\beta)^{n-j}(-\beta)^{j}-1\Bigr{)}.$

Clearly, the set $E^{(2)}$ is the union of all such sets $G^{\prime}$
corresponding to various $k=1,\ldots,n,$ and the simplex $S.$ Therefore,

$m_{2}={\rm mes_{n}}(E^{(2)})=$
$=\frac{1}{n!}\Bigl{(}\sum_{k=1}^{n}(-1)^{k+1}{n\choose
k}\Bigl{(}\sum_{j=0}^{k-1}{n\choose
j}(1+\beta)^{n-j}(-\beta)^{j}-1\Bigr{)}+1\Bigr{)}.$

Remark that

$\displaystyle
1+\beta=\frac{\gamma+1}{2}=\alpha,\quad\beta=\frac{\gamma-1}{2}=\alpha-1.$

Using the substitution $i=n-j$ in the internal sum we obtain

$m_{2}=$ $=\frac{1}{n!}\Bigl{(}1+\sum_{k=1}^{n}(-1)^{k+1}{n\choose
k}\Bigl{(}(-1)^{n}\sum_{i=n-k+1}^{n}{n\choose
i}(\alpha-1)^{n-i}(-\alpha)^{i}-1\Bigr{)}\Bigr{)}=$
$=\frac{(-1)^{n}}{n!}\sum_{k=1}^{n}(-1)^{k+1}{n\choose
k}\sum_{i=n-k+1}^{n}{n\choose i}(\alpha-1)^{n-i}(-\alpha)^{i}.$

We took into account that

$\sum_{k=0}^{n}(-1)^{k}{n\choose k}=\sum_{k=1}^{n}(-1)^{k}{n\choose k}+1=0.$

Changing the order of summation, we obtain

$m_{2}=\frac{(-1)^{n}}{n!}\sum_{i=1}^{n}{n\choose
i}(\alpha-1)^{n-i}(-\alpha)^{i}\sum_{k=n+1-i}^{n}(-1)^{k+1}{n\choose k}.$

Using (5.3), we can write

$\sum_{k=n+1-i}^{n}(-1)^{k+1}{n\choose
k}=\sum_{k=n+1-i}^{n}(-1)^{k+1}{n\choose{n-k}}=\sum_{j=0}^{i-1}(-1)^{n-j+1}{n\choose
j}=(-1)^{n+i}{{n-1}\choose{i-1}}.$

Thus we have

$m_{2}=\frac{1}{n!}\sum_{i=1}^{n}{n\choose
i}{{n-1}\choose{i-1}}(\alpha-1)^{n-i}\alpha^{i}.$ (5.5)

Thanks to equalities (5.4) and (5.5), we have

${\rm mes_{n}}(E_{n,\gamma})=m_{1}+m_{2}=$
$=\frac{1}{n!}\sum_{i=0}^{n-1}{n\choose i}{{n-1}\choose
i}(\alpha-1)^{n-i}\alpha^{i}+\frac{1}{n!}\sum_{i=1}^{n}{n\choose
i}{{n-1}\choose{i-1}}(\alpha-1)^{n-i}\alpha^{i}=$
$=\frac{1}{n!}\sum_{i=1}^{n-1}{n\choose i}\left({{n-1}\choose
i}+{{n-1}\choose{i-1}}\right)(\alpha-1)^{n-i}\alpha^{i}+$
$+\frac{1}{n!}\Bigl{(}(\alpha-1)^{n}+\alpha^{n}\Bigr{)}=\frac{1}{n!}\sum_{i=0}^{n}{n\choose
i}^{2}(\alpha-1)^{n-i}\alpha^{i}=$ $=\frac{1}{2^{n}n!}\sum_{i=0}^{n}{n\choose
i}^{2}(\gamma-1)^{n-i}(\gamma+1)^{i}.$

We took into account that

$\displaystyle{{n-1}\choose i}+{{n-1}\choose{i-1}}={n\choose i}$

completing the proof of the left-hand equality in (5.2).

The right-hand equality in (5.2) follows from the identity

$\sum_{i=0}^{n}{n\choose
i}^{2}t^{i}=(1-t)^{n}\chi_{n}\left(\frac{1+t}{1-t}\right)$

(see [48]). Let us put $t=\frac{\gamma-1}{\gamma+1},$ then

$(1-t)^{n}=2^{n}(\gamma+1)^{-n},\quad\frac{1+t}{1-t}=\gamma.$

Hence,

${\rm mes_{n}}(G)=\frac{1}{2^{n}n!}\sum_{i=0}^{n}{n\choose
i}^{2}(\gamma-1)^{n-i}(\gamma+1)^{i}=$
$=\frac{1}{2^{n}n!}\sum_{i=0}^{n}{n\choose
i}^{2}(\gamma+1)^{n-i}(\gamma-1)^{i}=$
$=\frac{1}{2^{n}n!}(\gamma+1)^{n}\sum_{i=0}^{n}{n\choose
i}^{2}\Bigl{(}\frac{\gamma-1}{\gamma+1}\Bigr{)}^{i}=\frac{\chi_{n}(\gamma)}{n!}.$

The proof of the theorem is complete. $\Box$

Let us give some simple examples. The set

$E_{1,2}=\\{x\in{\mathbb{R}}:|x|+|1-x|\leq 2\\}$

is the segment $\left[-\frac{1}{2},\frac{3}{2}\right]$ with the length

${\rm mes_{1}}(E_{1,2})=\frac{\chi_{1}(2)}{1!}=2.$

The two-dimensional set
$E_{2,2}=\\{x\in{\mathbb{R}}^{2}:|x_{1}|+|x_{2}|+|1-x_{1}-x_{2}|\leq 2\\}$ is
shown in Fig. 4.

Figure 4: The set $E_{2,2}$

The area of this set is equal to

${\rm mes_{2}}(E_{2,2})=\frac{\chi_{2}(2)}{2!}=\frac{11}{4}.$

The three-dimensional domain

$E_{3,2}=\\{x\in{\mathbb{R}}^{3}:|x_{1}|+|x_{2}|+|x_{3}|+|1-x_{1}-x_{2}-x_{3}|\leq
2\\}$

is shown in Fig. 5. It has the volume

${\rm mes_{3}}(E_{3,2})=\frac{\chi_{3}(2)}{3!}=\frac{17}{6}.$

Figure 5: The set $E_{3,2}$

There is an interesting open problem related to the above-mentioned equality
(5.2). Essentially, this equality along with Rodrigues’ formula and other
well-known relations also gives a characterization of Legendre polynomials.
This characterization is written via the volumes of some convex polyhedra.
Namely, for $t\geq 1$

$\chi_{n}(t)=n!\,{\rm mes}_{n}(E_{n,t}),$ (5.6)

where

$E_{n,t}=\left\\{x\in{\mathbb{R}}^{n}:\sum_{j=1}^{n}\left|x_{j}\right|+\left|1-\sum_{j=1}^{n}x_{j}\right|\leq
t\right\\}.$

The question arises about analogues of (5.6) for other classes of orthogonal
polynomials, such as Chebyshev polynomials or, more generally, Jacobi
polynomials: Is the equality (5.6) a particular case of a more general
pattern?

The author would be grateful for any information on this matter.

Let ua note that, from (5.6) and (5.1), we have

${\rm mes}_{n+1}(E_{n+1,t})=\frac{2n+1}{(n+1)^{2}}\,t\,{\rm
mes}_{n}(E_{n,t})-\frac{1}{(n+1)^{2}}{\rm mes}_{n-1}(E_{n-1,t}).$

The direct establishing this recurrence relation for the measures of the set
$E_{n,t}$ could provide a proof of Theorem 5.1 different from the above.

##  6\. Inequalities $\theta_{n}(Q_{n})>c\sqrt{n}$ and
$\theta_{n}(B_{n})>c\sqrt{n}$

Based on Theorem 5.1, in this section we obtain lower bounds for the norms of
the projector associated with linear interpolation on the cube and the ball.

First let us consider linear interpolation on the unit cube $Q_{n}=[0,1]^{n}$.

Theorem 6.1. Suppose $P:C(Q_{n})\to\Pi_{1}\left({\mathbb{R}}^{n}\right)$ is an
arbitrary interpolation projector. Then for the corresponding simplex
$S\subset Q_{n}$ and the node matrix ${\bf A}$ we have

$\|P\|_{Q_{n}}\geq\chi_{n}^{-1}\left(\frac{n!}{|\det({\bf
A})}\right)=\chi_{n}^{-1}\left(\frac{1}{{\rm vol}(S)}\right).$ (6.1)

Proof. For each $i=1,$ $\ldots,$ $n$, let us subtract from the $i$th row of
${\bf A}$ its $(n+1)$th row. Denote by ${\bf B}$ the submatrix of order $n$
which stands in the first $n$ rows and columns of the result matrix. Then

$|\det({\bf B})|=|\det({\bf A})|=n!{\rm vol}(S)\leq n!\nu_{n}$

so that

$\frac{|\det({\bf B})|}{n!\nu_{n}}\leq 1.$ (6.2)

Let $x^{(j)}$ be the vertices and $\lambda_{j}$ be the basic Lagrange
polynomials of simplex $S$. Since $\lambda_{1}(x),$ $\ldots$ ,
$\lambda_{n+1}(x)$ are the barycentric coordinates of a point $x$, we have

$\|P\|_{Q_{n}}=\max_{x\in
Q_{n}}\sum_{j=1}^{n+1}|\lambda_{j}(x)|=\max\left\\{\sum_{j=1}^{n+1}|\beta_{j}|:\,\sum_{j=1}^{n+1}\beta_{j}=1,\,\sum_{j=1}^{n+1}\beta_{j}x^{(j)}\in
Q_{n}\right\\}.$

Let us replace $\beta_{n+1}$ with the equal value
$1-\sum\limits_{j=1}^{n}\beta_{j}.$ The condition
$\sum\limits_{j=1}^{n+1}\beta_{j}x^{(j)}$ $\in Q_{n}$ is equivalent to

$\sum\limits_{j=1}^{n}\beta_{j}(x^{(j)}-x^{(n+1)})\in
Q^{\prime}:=Q_{n}-x^{(n+1)}.$

Hence,

$\|P\|_{Q_{n}}=\max\left\\{\sum_{j=1}^{n}\left|\beta_{j}\right|+\left|1-\sum_{j=1}^{n}\beta_{j}\right|\right\\}$
(6.3)

where the maximum is taken over all $\beta_{j}$ such that
$\sum\limits_{j=1}^{n}\beta_{j}(x^{(j)}-x^{(n+1)})\in Q^{\prime}.$ Clearly,

${\rm vol}(Q^{\prime})={\rm vol}(Q_{n})=1.$

Let us consider the nondegenerate linear operator
$F:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n},$ which maps the point
$\beta=(\beta_{1},\ldots,\beta_{n})$ to the point $x=F(\beta)$ according to
the rule

$x=\sum_{j=1}^{n}\beta_{j}\left(x^{(j)}-x^{(n+1)}\right).$

We have the matrix equality $F(\beta)=(\beta_{1},\ldots,\beta_{n}){\bf B},$
where $\bf B$ is the above introduced matrix of order $n$ with elements
$b_{ij}=x_{j}^{(i)}-x_{j}^{(n+1)}.$ Put

$\gamma^{*}:=\chi_{n}^{-1}\left(\displaystyle\frac{n!}{|\det{\bf B}|}\right).$

This definition is correct since by (6.2)

$\displaystyle\frac{n!}{|\det({\bf B})|}\geq\frac{1}{\nu_{n}}\geq 1.$

Note that

$\chi_{n}(\gamma^{*})=\frac{n!}{|\det({\bf B})|}.$

Given $\gamma\geq 1$, let us introduce a set

$E_{n,\gamma}:=\left\\{\beta=(\beta_{1},\ldots,\beta_{n})\in{\mathbb{R}}^{n}:\sum_{j=1}^{n}|\beta_{j}|+\left|1-\sum_{j=1}^{n}\beta_{j}\right|\leq\gamma\right\\}.$

Let us show that $Q^{\prime}\not\subset F(E_{n,\gamma})$ provided
$\gamma<\gamma^{*}.$ Indeed, this is immediate from Theorem 5.1 because,
thanks to this theorem,

${\rm mes_{n}}(F(E_{n,\gamma}))<{\rm mes_{n}}(F(E_{n,\gamma^{*}}))=|\det{\bf
B}|\cdot{\rm mes_{n}}(E_{n,\gamma^{*}})=$ $=|\det{\bf
B}|\cdot\frac{\chi_{n}(\gamma^{*})}{n!}=1={\rm mes_{n}}(Q^{\prime}).$

Thus, for every $\varepsilon>0$ there exists a point $x^{(\varepsilon)}$ with
the properties:

$x^{(\varepsilon)}\sum\beta_{j}^{(\varepsilon)}\left(x^{(j)}-x^{(n+1)}\right)\in
Q^{\prime},\quad\left|\sum\beta_{j}^{(\varepsilon)}\right|+\left|1-\sum\beta_{j}^{(\varepsilon)}\right|\geq\gamma^{*}-\varepsilon.$

In view of (6.3) this gives $\|P\|_{Q_{n}}\geq\gamma^{*}-\varepsilon.$ Since
$\varepsilon>0$ is an arbitrary, we obtain

$\|P\|_{Q_{n}}\geq\gamma^{*}=\chi_{n}^{-1}\left(\frac{n!}{|\det({\bf
B})|}\right)=\chi_{n}^{-1}\left(\frac{n!}{|\det({\bf
A})|}\right)=\chi_{n}^{-1}\left(\frac{1}{{\rm vol}(S)}\right).$

The theorem is proved. $\Box$

Corollary 6.1. For each $n$,

$\theta_{n}(Q_{n})\geq\chi_{n}^{-1}\left(\displaystyle\frac{1}{\nu_{n}}\right).$
(6.4)

The Stirling formula (see, e. g., [6])

$n!=\sqrt{2\pi
n}\left(\frac{n}{e}\right)^{n}e^{\frac{\zeta_{n}}{12n}},~{}~{}~{}0<\zeta_{n}<1,$

yields

$\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}<n!<\sqrt{2\pi
n}\left(\frac{n}{e}\right)^{n}e^{\frac{1}{12n}}.$ (6.5)

Let us see that for every $n>1$ the following inequality

$\chi_{n}^{-1}(s)>\left(\frac{s}{{n\choose{\lfloor n/2\rfloor}}}\right)^{1/n}$
(6.6)

holds. In fact, if $t\geq 1$ and $n>1$, then, according to (5.2),

$\chi_{n}(t)=\frac{1}{2^{n}}\sum_{i=0}^{n}{n\choose
i}^{2}(t-1)^{n-i}(t+1)^{i}<$ $<{n\choose{\lfloor
n/2\rfloor}}\cdot\frac{1}{2^{n}}\sum_{i=0}^{n}{n\choose
i}(t-1)^{n-i}(t+1)^{i}=$ $={n\choose{\lfloor n/2\rfloor}}\cdot(2t)^{n}\cdot
2^{-n}={n\choose{\lfloor n/2\rfloor}}\,t^{n}$

proving (6.6).

For even $n$, $\displaystyle{n\choose{\lfloor
n/2\rfloor}}={n\choose{n/2}}=(n!)/((n/2)!)^{2},$ therefore,

$\chi_{n}^{-1}(s)>\left(\frac{s\left((n/2)!\right)^{2}}{n!}\right)^{1/n}.$
(6.7)

If $n$ is odd, then

${n\choose{\lfloor n/2\rfloor}}=\frac{n!}{\frac{n+1}{2}!\frac{n-1}{2}!}$

and (6.6) has the form

$\chi_{n}^{-1}(s)>\left(\frac{s\frac{n+1}{2}!\frac{n-1}{2}!}{n!}\right)^{1/n}.$
(6.8)

Theorem 6.2. For all $n\in{\mathbb{N}}$,

$\theta_{n}(Q_{n})>\frac{\sqrt{n-1}}{e}.$ (6.9)

Proof. The case $n=1$ is trivial. If $n>1,$ let us apply (6.6)–(6.8) and the
Hadamard inequality (see (1.17))

$\nu_{n}\leq\frac{\left(n+1\right)^{(n+1)/2}}{2^{n}n!}.$

For even $n$, from (6.4), the left-hand inequality in (6.5), and (6.7), we get

$\theta_{n}(Q_{n})\geq\chi_{n}^{-1}\left(\frac{1}{\nu_{n}}\right)\geq\chi_{n}^{-1}\left(\frac{2^{n}n!}{(n+1)^{(n+1)/2}}\right)>2\left(\frac{[(n/2)!]^{2}}{(n+1)^{(n+1)/2}}\right)^{1/n}>$
$>\frac{2}{(n+1)^{1/2+1/(2n)}}\left(\sqrt{\pi
n}\left(\frac{n}{2e}\right)^{n/2}\right)^{2/n}=\frac{\left(\pi
n\right)^{1/n}n}{e(n+1)^{1/2+1/(2n)}}>\frac{\sqrt{n-1}}{e}.$

If $n$ is odd, then, thanks to (6.8),

$\theta_{n}(Q_{n})\geq\chi_{n}^{-1}\left(\frac{1}{\nu_{n}}\right)\geq\chi_{n}^{-1}\left(\frac{2^{n}n!}{(n+1)^{(n+1)/2}}\right)>\left(\frac{2^{n}\frac{n+1}{2}!\frac{n-1}{2}!}{(n+1)^{(n+1)/2}}\right)^{1/n}>$
$>2\left(\frac{\pi\sqrt{n^{2}-1}\left(n^{2}-1\right)^{(n-1)/2}(n+1)}{(2e)^{n}}\right)^{1/n}=$
$=\frac{1}{e}\pi^{1/n}\sqrt{n-1}(n+1)^{1/(2n)}>\frac{\sqrt{n-1}}{e}$

proving that (6.9) holds for all $n$. $\Box$

In some situations, the estimates of Theorem 2.4 can be improved [24]. In
particular, we prove that $\theta_{n}>\sqrt{n}/{e}$ if $n$ is even, and
$\displaystyle\theta_{n}>n/(e\sqrt{n-1})$ provided $n>1$ and $n\equiv 1({\rm
mod}~{}4)$.

Let us also note that, thanks to (6.9), the inequality
$\theta_{n}(Q_{n})>c\sqrt{n}$ holds with some $c$, $0<c<1$. A suitable
estimate is

$\theta_{n}(Q_{n})>\frac{2\sqrt{2}}{3e}\,\sqrt{n}.$ (6.10)

Indeed, if $n\leq 8$, then the right-hand side of (6.10) is less than 1, while
for $n\geq 9$

$\theta_{n}(Q_{n})>\frac{\sqrt{n-1}}{e}\geq\frac{2\sqrt{2}}{3e}\,\sqrt{n}.$

Notice that $\displaystyle\frac{2\sqrt{2}}{3e}=0.3468\ldots$

Later this approach was extended to linear interpolation on the unit ball
$B_{n}$, see [29]. Let us present some results obtained in this direction.

A regular simplex inscribed into an $n$-dimensional ball has the maximal
volume among all simplices contained in this ball. Furthermore, there are no
other simplices with this property (see [5], [51], and [55]).

Let $\varkappa_{n}={\rm vol}(B_{n})$, and let $\sigma_{n}$ be the volume of a
regular simplex inscribed into $B_{n}$.

Theorem 6.3. Let $P:C(B_{n})\to\Pi_{1}({\mathbb{R}}^{n})$ be an interpolation
projector. Then, for the corresponding simplex $S\subset B_{n}$ and the node
matrix ${\bf A}$, we have

$\|P\|_{B_{n}}\geq\chi_{n}^{-1}\left(\frac{n!\varkappa_{n}}{|\det({\bf
A})|}\right)=\chi_{n}^{-1}\left(\frac{\varkappa_{n}}{{\rm vol}(S)}\right).$
(6.11)

Corollary 6.2. For each $n$,

$\theta_{n}(B_{n})\geq\chi_{n}^{-1}\left(\frac{\varkappa_{n}}{\sigma_{n}}\right).$
(6.12)

Proof. Let $P:C(B_{n})\to\Pi_{1}({\mathbb{R}}^{n})$ be an arbitrary
interpolation projector. The corresponding simplex $S\subset B_{n}$ satisfies
${\rm vol}(S)\leq\sigma_{n}$. Then, thanks to inequality (6.11),

$\|P\|_{B_{n}}\geq\chi_{n}^{-1}\left(\frac{\varkappa_{n}}{{\rm
vol}(S)}\right)\geq\chi_{n}^{-1}\left(\frac{\varkappa_{n}}{\sigma_{n}}\right)$

proving (6.12). $\Box$

It is known that

$\varkappa_{n}=\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)},\qquad\sigma_{n}=\frac{1}{n!}\sqrt{n+1}\left(\frac{n+1}{n}\right)^{\frac{n}{2}},$
(6.13)
$\varkappa_{2k}=\frac{\pi^{k}}{k!},\qquad\varkappa_{2k+1}=\frac{2^{k+1}\pi^{k}}{(2k+1)!!}=\frac{2(k!)(4\pi)^{k}}{(2k+1)!}.$
(6.14)

These equalities and inequality (6.12) imply the following low bound for the
constant $\theta_{n}(B_{n})$.

Corollary 6.3. For every $n$,

$\theta_{n}(B_{n})\geq\chi_{n}^{-1}\left(\frac{\pi^{\frac{n}{2}}n!}{\Gamma\left(\frac{n}{2}+1\right)\sqrt{n+1}\left(\frac{n+1}{n}\right)^{\frac{n}{2}}}\right).$
(6.15)

If $n=2k$, then (6.15) is equivalent to the inequality

$\theta_{2k}(B_{2k})\geq\chi_{2k}^{-1}\left(\frac{\pi^{k}(2k)!}{k!\sqrt{2k+1}\left(\frac{2k+1}{2k}\right)^{k}}\right).$
(6.16)

For $n=2k+1$ we have

$\theta_{2k+1}(B_{2k+1})\geq\chi_{2k+1}^{-1}\left(\frac{2(k!)(4\pi)^{k}}{\sqrt{2k+2}\left(\frac{2k+2}{2k+1}\right)^{\frac{2k+1}{2}}}\right).$
(6.17)

Proof. It is sufficient to apply (6.12), (6.13), and (6.14). $\Box$

Also we will need the following estimates which were proved in [24, Section
3.4.2]:

$\chi_{2k}^{-1}(s)>\left(\frac{(k!)^{2}s}{(2k)!}\right)^{\frac{1}{2k}},\qquad\chi_{2k+1}^{-1}(s)>\left(\frac{(k+1)!k!s}{(2k+1)!}\right)^{\frac{1}{2k+1}}.$
(6.18)

Theorem 6.4. There exists an absolute constant $c>0$ such that

$\theta_{n}(B_{n})>c\sqrt{n}.$ (6.19)

In [30] we have shown that the inequality (6.19) takes place with the constant

$c=\frac{\sqrt[3]{\pi}}{\sqrt{12e}\cdot\sqrt[6]{3}}=0.2135...\,.$ (6.20)

Corollary 6.4. $\theta_{n}(B_{n})\asymp\sqrt{n}$.

Proof. By the results of [40], $\theta_{n}(B_{n})\leq\sqrt{n+1}.$
Consequently, the lower estimate $\theta_{n}(B_{n})>c\sqrt{n}$ is precise with
respect to dimension $n$. $\Box$

Corollary 6.5. Let $P:C(B_{n})\to\Pi_{1}({\mathbb{R}}^{n})$ be the
interpolation projector whose nodes coincide with vertices of a regular
simplex being inscribed into the boundary sphere $\|x\|=1$. Then
$\|P\|_{B_{n}}\asymp\theta_{n}(B_{n})$.

Proof. Since $\sqrt{n}\leq\|P\|_{B_{n}}\ \leq\sqrt{n+1}$, it remains to
utilize the previous corollary. $\Box$

Until 2023, the equality $\|P\|_{B_{n}}=\theta_{n}(B_{n})$, where $P$ is the
projector from Corollary 6.5, remained proven only for $1\leq n\leq 4$. As it
turned out later, this equality is true for every $n$. Moreover, if $n>1$,
then, in addition to the inequality (6.19) with $c$ defined by (6.20), we
proved that $\theta_{n}(B_{n})>\sqrt{n}$. However, the latter only became
clear after finding the exact value of $\theta_{n}(B_{n})$. See Section 9.

The approach proposed above can be also applied to the case of an arbitrary
convex body $K\subset{\mathbb{R}}^{n}.$

Theorem 6.5. For an arbitrary interpolation projector
$P:C(K)\to\Pi_{1}({\mathbb{R}}^{n})$,

$\|P\|_{K}\geq\chi_{n}^{-1}\left(\frac{n!{\rm vol}(K)}{|\det({\bf
A})|}\right)=\chi_{n}^{-1}\left(\frac{{\rm vol}(K)}{{\rm vol}(S)}\right).$
(6.21)

Here $S\subset K$ is the corresponding simplex and ${\bf A}$ is the
corresponding node matrix.

We let ${\rm simp_{n}}(K)$ denote the maximum volume of a nondegenerate
simplex $S$ with vertices in $K$. Thus, in these settings, $\nu_{n}={\rm
simp_{n}}(Q_{n})$ and $\sigma_{n}={\rm simp_{n}}(B_{n})$.

Corollary 6.6. Let $K$ be an arbitrary convex body in ${\mathbb{R}}^{n}$. Then

$\theta_{n}(K)\geq\chi_{n}^{-1}\left(\frac{{\rm vol}(K)}{{\rm
simp_{n}}(K)}\right).$ (6.22)

This inequality is immediate from (6.21).

For a simplex $S$ of the maximum volume in $K$, we have $K\subset(n+2)S,$ see
(2.12). Therefore, the ratio ${\rm vol}(K)/{\rm simp_{n}}(K)$ in (6.22) is
bounded by $(n+2)^{n}.$

Note that the approach given in this section works also for an arbitrary (not
necessarily convex) compact.

##  7\. The norm $\|P\|_{B}$ for an inscribed regular simplex

Let $B=B(x^{(0)};R)$ and let $P:C(B)\to\Pi_{1}({\mathbb{R}}^{n})$ be the
interpolation projector with the nodes $x^{(j)}\in B$, $j=1,...,n+1$. Let $S$
be the simplex with vertices $x^{(j)}$ and let
$\lambda_{j}(x)=l_{1j}x_{1}+\ldots+l_{nj}x_{n}+l_{n+1,j}$ be the basic
Lagrange polynomials of $S$. In this case

$\|P\|_{B}=\max_{x\in B}\sum_{j=1}^{n+1}|\lambda_{j}(x)|$

(see (1.5) for $K=B$). In [40] another formula for $\|P\|_{B}$ was obtained:

$\|P\|_{B}=\max\limits_{f_{j}=\pm
1}\left[R\left(\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}\right)^{1/2}+\left|\sum_{j=1}^{n+1}f_{j}\left(\sum_{i=1}^{n}l_{ij}x_{i}^{(0)}+l_{n+1,j}\right)\right|\right]=$
$=\max\limits_{f_{j}=\pm
1}\left[R\left(\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}\right)^{1/2}+\left|\sum_{j=1}^{n+1}f_{j}\lambda_{j}(x^{(0)})\right|\right].$
(7.1)

If $c(S)=x^{(0)}$, then $\lambda_{j}(x^{(0)})=1/(n+1)$ which implies the
following simpler formula for $\|P\|_{B}$:

$\|P\|_{B}=\max\limits_{f_{j}=\pm
1}\left[R\left(\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}\right)^{1/2}+\frac{1}{n+1}\left|\sum_{j=1}^{n+1}f_{j}\right|\right].$
(7.2)

Now, suppose that $S$ is a regular simplex inscribed into the $n$-dimensional
ball $B=B(x^{(0)};R)$ and $P:C(B)\to\Pi_{1}({\mathbb{R}}^{n})$ is the
corresponding interpolation projector. Clearly, $\|P\|_{B}$ does not depend on
the center $x^{(0)}$ and the radius $R$ of the ball and on the choice of a
regular simplex inscribed into that ball. In other words, $\|P\|_{B}$ is a
function of only dimension $n$. The exact expression for $\|P\|_{B}$ which we
present below was found in [40].

We let $\psi:[0,n+1]\to{\mathbb{R}}$ denote a function defined by

$\psi(t):=\frac{2\sqrt{n}}{n+1}\Bigl{(}t(n+1-t)\Bigr{)}^{1/2}+\left|1-\frac{2t}{n+1}\right|.$
(7.3)

Let

$a=a_{n}=\left\lfloor\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}\right\rfloor.$ (7.4)

Theorem 7.1. The following relations hold:

$\|P\|_{B}=\max\\{\psi(a),\psi(a+1)\\},$ (7.5)
$\sqrt{n}\leq\|P\|_{B}\leq\sqrt{n+1}.$ (7.6)

The equality $\|P\|_{B}=\sqrt{n}$ is true only for $n=1$. The equality
$\|P\|_{B}=\sqrt{n+1}$ holds if and only if $\sqrt{n+1}$ is an integer.

Proof. First we prove (7.5). If $n=1$, then $\psi(t)=\sqrt{t(2-t)}+|1-t|$,
$a=0$, $\psi(a)=\psi(a+1)=1.$ Since $\|P\|_{B}=1$, the equality (7.5) is true.

Let $n\geq 2$. Consider the simplex $S$ with vertices

$x^{(1)}=e_{1},\ \ \ldots,\ \ x^{(n)}=e_{n},\quad
x^{(n+1)}=\left(\frac{1-\sqrt{n+1}}{n},\ldots,\frac{1-\sqrt{n+1}}{n}\right).$

Since the length of any edge of $S$ is equal to $\sqrt{2}$, this simplex is
regular. The simplex is inscribed into the ball $B=B(x^{(0)};R)$, where

$x^{(0)}=c(S)=\left(\frac{1-\sqrt{\frac{1}{n+1}}}{n},\ldots,\frac{1-\sqrt{\frac{1}{n+1}}}{n}\right),\quad
R=\sqrt{\frac{n}{n+1}}.$

Note that the $(n+1)$th vertex of $S$ is obtained by shifting the zero-vertex
of the simplex $x_{i}\geq 0$, $\sum x_{i}\leq 1$ in direction from the
hyperplane $\sum x_{i}=1$. It is important that $S$ is invariant with respect
to changing the order of coordinates. It is enough to find $\|P\|_{B}$ for
this simplex.

The corresponding matrices ${\bf A}$ and ${\bf A}^{-1}$ are

${\bf A}=\left(\begin{array}[]{ccccc}1&0&\ldots&0&1\\\ 0&1&\ldots&0&1\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&\ldots&1&1\\\
-\tau&-\tau&\ldots&-\tau&1\\\ \end{array}\right),\quad{\bf
A}^{-1}=\frac{1}{\sqrt{n+1}}\left(\begin{array}[]{ccccc}\sigma&-\tau&\ldots&-\tau&-1\\\
-\tau&\sigma&\ldots&-\tau&-1\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\
-\tau&-\tau&\ldots&\sigma&-1\\\ \tau&\tau&\ldots&\tau&1\\\
\end{array}\right).$ (7.7)

Here

$\sigma:=\frac{(n-1)\sqrt{n+1}+1}{n},\quad\tau:=\frac{\sqrt{n+1}-1}{n}.$ (7.8)

Since $c(S)=x^{(0)}$, for calculation $\|P\|_{B}$ we can use (7.2) with
$R=\sqrt{n/(n+1)}$:

$\|P\|_{B}=\max\limits_{f_{j}=\pm
1}\left[\sqrt{\frac{n}{n+1}}\left(\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}\right)^{1/2}+\frac{1}{n+1}\left|\sum_{j=1}^{n+1}f_{j}\right|\right].$

Here $l_{ij}$ are the elements of ${\bf A}^{-1}$ (see (7.7)).

Let $k$ be the number of $f_{j}$ equal to $-1$. Then the number of $f_{j}$
equal to $1$ is $n+1-k$. The simplex $S$ does not change with renumerating of
coordinates, so, we may assume that $f_{1}=\ldots=f_{k}=-1,$
$f_{k+1}=\ldots=f_{n+1}=1$. Since the function being maximized does not change
when the signs of $f_{j}$ change simultaneously, we can consider only the
interval $1\leq k\leq\frac{n+1}{2}$. Thus,

$\|P\|_{B}=\max\limits_{1\leq
k\leq\frac{n+1}{2}}\left[\sqrt{\frac{n}{n+1}}\left(\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}\right)^{1/2}+\frac{n+1-2k}{n+1}\right],$
(7.9)

where $f_{j}=-1$ for $1\leq j\leq k$ and $f_{j}=1$ for all other $j$. The
number $n+1-2k$ is equal to the sum under the sign of absolute value. Taking
into account the multiplier $1/\sqrt{n+1}$ in the equality for ${\bf A}^{-1}$,
let us rewrite the value

$W=(n+1)\sum_{i=1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}.$

Utilizing the explicit expressions for $l_{ij}$ we present this sum in the
form

$W=(n+1)\sum_{i=1}^{k}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}+(n+1)\sum_{i=k+1}^{n}\left(\sum_{j=1}^{n+1}f_{j}l_{ij}\right)^{2}=W_{1}+W_{2}.$

From (7.7) and the described distribution of values $f_{j}$, we get

$W_{1}=\sum_{i=1}^{k}(-\sigma+(k-1)\tau-(n-k)\tau-1)^{2}=k(2k\tau-\alpha)^{2},$
$W_{2}=\sum_{i=k+1}^{n}(k\tau+\sigma-(n-1-k)\tau-1)^{2}=(n-k)(2k\tau+\beta)^{2}.$

Here $\alpha=\sigma+(n+1)\tau+1$, $\beta=\sigma-(n-1)\tau-1$. It follows from
(7.8) that $\alpha=2\sqrt{n+1},$ $\beta=0$, therefore,

$W=4k(k\tau-\sqrt{n+1})^{2}+(n-k)\cdot
4k^{2}\tau^{2}=k^{2}(-8\sqrt{n+1}\tau+4n\tau^{2})+4k(n+1)=$
$=-4k^{2}+4k(n+1)=4k(n+1-k).$

This gives

$\|P\|_{B}=\max\limits_{1\leq
k\leq\frac{n+1}{2}}\left[\sqrt{\frac{n}{n+1}}\left(\frac{1}{n+1}W\right)^{1/2}+\frac{n+1-2k}{n+1}\right]=$
$=\max\limits_{1\leq
k\leq\frac{n+1}{2}}\left[\frac{2\sqrt{n}}{n+1}\bigl{(}k(n+1-k)\bigr{)}^{1/2}+1-\frac{2k}{n+1}\right].$

Recalling (7.3), we obtain

$\|P\|_{B}=\max\limits_{1\leq k\leq\frac{n+1}{2}}\psi(k).$ (7.10)

It remains to show that the last maximum is equal to the largest of numbers
$\psi(a)$ and $\psi(a+1)$, where
$a=\left\lfloor\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}\right\rfloor.$ To do this,
we analyze the behavior of $\psi(t)$ over the whole interval $[0,n+1]$.

The function

$\psi(t)=\frac{2\sqrt{n}}{n+1}\Bigl{(}t(n+1-t)\Bigr{)}^{1/2}+\left|1-\frac{2t}{n+1}\right|,\quad
0\leq t\leq n+1,$

has the following properties:

$\psi(t)>0,~{}~{}~{}\psi(0)=\psi(n+1)=1,~{}~{}~{}\psi\left(\frac{n+1}{2}\right)=\sqrt{n},~{}~{}~{}\psi(n+1-t)=\psi(t).$

The graph of $\psi(t)$ is symmetric with respect to the straight line
$t=\frac{n+1}{2}$. On each half of $[0,n+1]$ the function $\psi(t)$ is concave
as a sum of two concave functions. Indeed, for $0\leq t\leq\frac{n+1}{2}$,

$\psi(t)=\varphi_{1}(t)+\varphi_{2}(t),\quad\varphi_{1}(t)=\frac{2\sqrt{n}}{n+1}\Bigl{(}t(n+1-t)\Bigr{)}^{1/2},\quad\varphi_{2}(t)=1-\frac{2t}{n+1},$

where $\varphi_{1}(t)$ is concave as a superposition of the concave function
$t(n+1-t)$ and the increasing concave function $\sqrt{t}$, while
$\varphi_{2}(t)$ is a linear function. The derivation $\psi^{\prime}(t)$ is
equal to zero only in two points

$t_{-}=\frac{n+1}{2}-\frac{\sqrt{n+1}}{2},\quad
t_{+}=\frac{n+1}{2}+\frac{\sqrt{n+1}}{2}.$

These points lie strictly inside the segments $\left[0,\frac{n+1}{2}\right]$
and $\left[\frac{n+1}{2},n+1\right]$ respectively. From the concavity of
$\psi(t)$ on each of these segments, it follows that

$\max_{0\leq\psi(t)\leq n+1}\psi(t)=\psi(t_{-})=\psi(t_{+})=\sqrt{n+1}.$

Moreover, $t_{-}$ and $t_{+}$ are the only maximum points on the left and
right halves of $[0,n+1]$. Hence, $\psi(t)$ increases for $0\leq t\leq t_{-}$
and decreases for $t_{-}\leq t\leq\frac{n+1}{2}$. On the left half $\psi(t)$
behaves symmetrically: it increases for $\frac{n+1}{2}\leq t\leq t_{+}$ and
decreases for $t_{+}\leq t\leq n+1.$

Let us now consider only the segment $\left[0,\frac{n+1}{2}\right]$. Since
$a=\left\lfloor t_{-}\right\rfloor$, then always $a\leq t_{-}<a+1$. Let $k$ be
a whole number, $1\leq k\leq\frac{n+1}{2}$. If $k<a$, then $\psi(k)<\psi(a)$,
while $\psi(k)<\psi(a+1)$ provided $k>a+1$. Taking into account (7.10), we
have

$\|P\|_{B}=\max\limits_{1\leq
k\leq\frac{n+1}{2}}\psi(k)=\max\\{\psi(a),\psi(a+1)\\}.$

The equality (7.5) is proved.

We now turn to inequalities (7.6). We have already obtained the right one:

$\|P\|_{B}=\max_{k\in{\mathbb{Z}}:\,1\leq
k\leq\frac{n+1}{2}}\psi(k)\leq\max_{1\leq
t\leq\frac{n+1}{2}}\psi(t)=\psi(t_{-})=\sqrt{n+1}.$

Let us describe such dimensions $n$ that $\|P\|_{B}=\sqrt{n+1}$. These
dimensions are characterized by the fact that the non-strict inequality in the
last relation becomes an equality, i.e., the number $t_{-}$ is integer.

Note that both $t_{-}$ and $t_{+}$ are integer if and only if $\sqrt{n+1}$ is
an integer. For example, assume that

$t_{-}=\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}=d\in{\mathbb{Z}}.$

Then $\sqrt{n+1}=n+1-2d$ is an integer. On the contrary, if
$\sqrt{n+1}=m\in{\mathbb{Z}}$, then $m(m-1)$ is an even number and
$t_{-}=\frac{m(m-1)}{2}$ is an integer. The statement for $t_{+}$ can be
proved similarly. Also it can be deduced from the previous statement, since
$t_{+}-t_{-}=\sqrt{n+1}$.

We have obtained that for any dimension of the form $n=m^{2}-1$, $m$ is an
integer, and only in these situations, the number $t_{-}$ is integer.
Consequently,

$\|P\|_{B}=\max_{k\in{\mathbb{Z}}:\,1\leq
k\leq\frac{n+1}{2}}\psi(k)=\max_{1\leq
t\leq\frac{n+1}{2}}\psi(t)=\psi(t_{-})=\sqrt{n+1}=m.$

These equalities are equivalent to (7.5), since in the cases considered
$a=\left\lfloor t_{-}\right\rfloor=t_{-}$ and $\|P\|_{B}=\psi(a)$. For all
other $n$, holds $\|P\|<\sqrt{n+1}.$ Indeed, if $n\neq m^{2}-1$, then
$a<t_{-}<a+1$ and the maximum in (7.10) is reached either for $k=a$ or for
$k=a+1$. For any $n\neq m^{2}-1$ the norm of $P$, i.e., the maximum $\psi(k)$
for integer $k\in\left[1,\frac{n+1}{2}\right]$, is strictly less than maximum
of $\psi(t)$ over all this interval.

It remains to show that always $\|P\|_{B}\geq\sqrt{n}$ and for $n>1$ this
equality is a strict one. If $n>3$, then $\sqrt{n+1}>2$, hence,

$a+1=\left\lfloor\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}\right\rfloor+1\leq\frac{n+1}{2}-\frac{\sqrt{n+1}}{2}+1<\frac{n+1}{2}.$

From (7.5) and properties of $\psi(t)$, we have

$\|P\|_{B}=\max\\{\psi(a),\psi(a+1)\\}\geq\psi(a+1)>\psi\left(\frac{n+1}{2}\right)=\sqrt{n}\qquad(n>3).$

Also, $\|P\|_{B}>\sqrt{n}$ provided $n=2$ and $n=3$. Therefore,
$\|P\|=\sqrt{n}$ only for $n=1$, and the proof of the theorem is complete.
$\Box$

Figure 6: The graph of $\psi(t)$ for $n=5$. Here $n+1=5$, $a=1$,
$t_{-}=(6-\sqrt{6})/2$

The graph of the function $\psi(t)$ for $n=5$ is given in Fig.6. In this case
$n+1=5$, $a=1$, and $t_{-}=(6-\sqrt{6})/2$.

Obviously, $\theta_{n}(B)\leq||P||_{B}\leq\sqrt{n+1}$. Combining these
inequalities with the inequality $\theta_{n}(B_{n})\geq c\sqrt{n}$ (see
(6.19)), we obtain the exact order of the value of $\theta_{n}(B_{n})$ related
to dimension $n$, namely, $\theta_{n}(B_{n})\asymp\sqrt{n}.$

Another corollary of Theorem 7.1 provides the following sharp values of
$\theta_{n}(B_{n})$ for $1\leq n\leq 4$:

$\theta_{1}(B_{1})=1,\quad\theta_{2}(B_{2})=\frac{5}{3},\quad\theta_{3}(B_{3})=2,\quad\theta_{4}(B_{4})=\frac{11}{5}.$
(7.11)

Of course, $\theta_{1}(B_{1})=1$, but this case also fits into the general
scheme. For proving, let us calculate the norm of the projector corresponding
to a regular simplex inscribed into $B_{n}$. By Theorem 7.1,

$\|P\|_{B_{1}}=1,\quad\|P\|_{B_{2}}=\frac{5}{3},\quad\|P\|_{B_{3}}=2,\quad\|P\|_{B_{4}}=\frac{11}{5}.$

Note that for each $n=1,2,3,4,$ we have

$\|P\|_{B_{n}}=3-\frac{4}{n+1}.$

But $\theta_{n}(B_{n})\geq 3-4/(n+1)$ for every positive integer $n$ so that,
if $1\leq n\leq 4$, then

$\theta_{n}(B_{n})=3-\frac{4}{n+1}.$ (7.12)

This is equivalent to (7.11). If (7.12) is true, then any projector having
minimum norm corresponds to a regular simplex inscribed into $B_{n}$. Really,
thanks to (1.7), for any projector $P$ and the simplex $S=S_{P}$

$\frac{n+1}{2}\Bigl{(}\|P\|_{B_{n}}-1\Bigr{)}+1\geq\xi(B_{n};S)\geq\xi_{n}(B_{n})=n,$

i. e.,

$\|P\|_{B_{n}}\geq 3-\frac{4}{n+1}.$

If for some projector we have here an equality, then equalities take place in
the above chain too. In this case, the corresponding simplex satisfies
$\xi(B_{n};S)=\xi_{n}(B_{n})=n.$ This way $S$ is a regular simplex inscribed
in the boundary sphere (see Section 3). Remark that the same approach provides
the exact values of $\theta_{n}(Q_{n})$ for dimensions $n=1,2,3,7.$

Let us define an integer $k=k_{n}$ equal to that of the numbers $a$ or $a+1$,
for which $\psi(t)$ takes a larger value. As it shown in [40], if $k=1$, then
in $B_{n}$ there exists an $1$-point with respect to the simplex corresponding
to $P$. For such $n$,

$\xi(B_{n};S)=\frac{n+1}{2}\Bigl{(}\|P\|_{B_{n}}-1\Bigr{)}+1$ (7.13)

(see (1.7)). Since $S$ is a regular simplex inscribed into $B_{n}$, then
$\xi(B_{n};S)=n$ and (7.13) is equivalent to the equality
$\|P\|_{B_{n}}=3-\frac{4}{n+1}$. However, $k>1$ provided $n\geq 5$. This
corresponds to the fact that (7.12) and (7.13) hold only for $1\leq n\leq 4$.
Initially, this effect was discovered in the course of computer experiments
perfomed by Alexey Ukhalov. Later the author gave an analytical solution to
this problem. See [40] for details.

Numbers $k_{n}$ increase with $n$, but not strictly monotonously. For example,

$k_{1}=k_{2}=k_{3}=k_{4}=1,\quad k_{5}=k_{6}=2,\quad k_{7}=k_{8}=k_{9}=3,\quad
k_{10}=k_{11}=4,$ $k_{12}=k_{13}=5,\quad k_{14}=k_{15}=6,\quad k_{50}=22,\quad
k_{100}=45,\quad k_{1000}=485.$

Furthermore, $k_{n}\leq n/2$ provided $n\geq 2$.

##  8\. A theorem on a simplex and its minimal ellipsoid

In 1948, F. John [11] proved that any convex body in ${\mathbb{R}}^{n}$
contains a unique ellipsoid of the maximum volume. Also he gave a
characterization of those convex bodies for which the maximal ellipsoid is the
unit Euclidean ball $B_{n}$ (see, e.g., [2], [3] for details). John’s theorem
implies the analogous statement which characterizes a unique minimum volume
ellipsoid containing a given convex body.

We shall consider the minimum volume ellipsoid containing a given
nondegenerate simplex. For brevity, such an ellipsoid will be called the
minimal ellipsoid. Obviously, the minimal ellipsoid of a simplex is
circumscribed around this simplex. The center of the ellipsoid coincides with
the center of gravity of the simplex. The minimal ellipsoid of a simplex is a
Euclidean ball if and only if this simplex is regular. This is equivalent to
the well-known fact that the volume of a simplex contained in a ball is
maximal iff this simplex is regular and inscribed into the ball (see, e. g.,
[5], [51], [55]).

Let $S$ be a nondegenerate $n$-dimensional simplex in ${\mathbb{R}}^{n}$ and
let $E$ be the minimum volume ellipsoid containing $S$. Let $m$, $1\leq m\leq
n$, be a positive integer. To any $m$-point set of vertices of $S$, let us
assign a point $y\in E$ defined as follows. Let $g$ be the center of gravity
of the $(m-1)$-dimensional face of $S$ which contains the chosen vertices, and
let $h$ be the center of gravity of the $(n-m)$-dimensional face which
contains the rest $n+1-m$ vertices. Then $y$ is the intersection point of the
straight line $(gh)$ with the boundary of the ellipsoid in direction from $g$
to $h$.

Theorem 8.1. ([32]) For every nondegenerate simplex $S\subset B_{n}$, there
exists such a set of $m$ vertices for which $y\in B_{n}$.

Proof. Let $x^{(j)}$, $j=1,...,n+1$, be the vertices of $S$. The center of
gravity of the simplex and also the center of gravity of its minimal ellipsoid
$E$ are the point $c=({1}/({n+1}))\sum x^{(j)}.$ We let $r$ denote the ratio
of the distance between the center of gravity of a regular simplex and the
center of gravity of its $(m-1)$-dimensional face to the radius of the
circumscribed sphere. It is easy to see that

$r=\frac{1}{m}\sqrt{m-\frac{m(m-1)}{n}}.$

Let $N$ be the number of $(m-1)$-dimensional faces of an $n$-dimensional
simplex; thus, $N={n+1\choose m}$.

Let $J$ be an $m$-element subset of the set $\\{1,...,n+1\\}$. The center of
gravity of the $(m-1)$-dimensional face with vertices $x^{(j)},j\in J,$ is the
point $g_{J}=(1/m)\sum_{j\in J}x^{(j)}$. Suppose $y_{J}$ coincides with the
point $y$ for this set of vertices. Then

$y_{J}=c+\frac{1}{r}\left(c-g_{J}\right),$

i.e., $y_{J}=(1/r)((r+1)c-g_{J})$. Summing up over all sets $J$, we have

$\sum_{J}\|y_{J}\|^{2}=\sum_{J}\langle y_{J},y_{J}\rangle=$
$=\frac{1}{r^{2}}\left(N\|c\|^{2}(r+1)^{2}+\sum_{J}\|g_{J}\|^{2}-2(r+1)\langle\sum_{J}g_{J},c\rangle\right)=$
$=\frac{1}{r^{2}}\left(N\|c\|^{2}(r^{2}-1)+\sum_{J}\|g_{J}\|^{2}\right).$
(8.1)

We took into account the equality $\sum g_{J}=Nc.$ Further, we claim that

$\sum_{J}\|g_{J}\|^{2}=\frac{1}{mn}{n\choose
m}\sum_{j=1}^{n+1}\|x^{(j)}\|^{2}+\frac{(m-1)(n+1)}{mn}{n+1\choose
m}\|c\|^{2}.$ (8.2)

To obtain (8.2), let us remark the following. The value $\sum\|g_{J}\|^{2}$
contains, for every $j$, exactly ${n\choose m-1}$ numbers $\|x^{(j)}\|^{2}$
taken with coefficient $1/m^{2}$, and exactly ${n-1\choose m-2}$ numbers
$2\langle x^{(i)},x^{(j)}\rangle$, for each $i\neq j$, with the same
coefficient $1/m^{2}$. (The latter ones exist for $m\geq 2$; in case $m=1$
there are no pairwise products.) The expression $\|c\|^{2}$ contains all
numbers $\|x^{(j)}\|^{2}$, for any $j$, and numbers $2\langle
x^{(i)},x^{(j)}\rangle$, for $i\neq j$, with multiplier $1/(n+1)^{2}.$ The
coefficient at $\|c\|^{2}$ in the right-hand part of (8.2) guarantees the
equality of expressions with pairwise products $\langle
x^{(i)},x^{(j)}\rangle$; the coefficient at $\sum\|x^{(j)}\|^{2}$ is chosen so
that the difference of the left-hand part of (8.2) and the second item in the
right-hand part is equal to the first item. Since

$N(r^{2}-1)=-\frac{(m-1)(n+1)}{mn}{n+1\choose m},$

after replacing $\sum\|g_{J}\|^{2}$ in (8.1) with the right-hand part of
(8.2), we notice that items with $\|c\|^{2}$ are compensated. After these
transformations, we get

$\frac{1}{N}\sum_{J}\|y_{J}\|^{2}=\frac{1}{n+1}\sum_{j=1}^{n+1}\|x^{(j)}\|^{2}.$
(8.3)

The inclusion $S\subset B_{n}$ means that $\|x^{(j)}\|\leq 1.$ Therefore, the
mean value of numbers $\|y_{J}\|^{2}$ is also $\leq 1.$ Consequently, for some
set $J$ we have $\|y_{J}\|\leq 1,$ i.e., this point $y_{J}$ lies in $B_{n}.$
Theorem 8.1 is proved. $\Box$

The approach using in this proof was suggested and kindly communicated to the
author by Arseniy Akopyan.

As a conjecture, Theorem 8.1 was formulated by the author in [30] . Also in
[30], the statement of the theorem for $m=1$ was proved in a simpler way. In
the case $m=1$ the points $y$ have the form $y^{(j)}=2c-x^{(j)},$
$j=1,\ldots,n+1,$ where $x^{(j)}\in B_{n}$ are the vertices of $S$ and
$c=c(S)=c(E).$ We need to show that there exists a vertex $x$ of the simplex
such that $\|2c-x\|\leq 1$. Since simplex $S$ is nondegenerate, for some
vertex $x$ holds $\langle c,x-c\rangle\geq 0$. This implies

$\|2c-x\|^{2}=\langle 2c-x,2c-x\rangle=4\langle
c,c-x\rangle+\|x\|^{2}\leq\|x\|^{2}\leq 1,$

i.e., the vertex $x$ is suitable.

##  9\. The value of $\theta_{n}(B_{n})$

In this section we present the main result of the work [32] devoted to
calculation of the exact value of the constant $\theta_{n}(B_{n})$.

Theorem 9.1. Let $\psi$ and $a_{n}$ be the function and the number defined by
the equalities (7.3) and (7.4) respectively. Then, for every
$n\in{\mathbb{N}}$ the following equality

$\theta_{n}(B_{n})=\max\\{\psi(a_{n}),\psi(a_{n}+1)\\}$ (9.1)

holds. Furthermore, an interpolation projector
$P:C(B_{n})\to\Pi_{1}({\mathbb{R}}^{n})$ satisfies the equality
$\|P\|_{B_{n}}=\theta_{n}(B_{n})$ if and only if its interpolation nodes
coincide with the vertices of a regular simplex inscribed into the boundary
sphere.

Proof. First, let $P$ be an interpolation projector corresponding to a regular
inscribed simplex $S$, and let $\lambda_{j}(x)$ be the basic Lagrange
polynomials for this simplex. Let $p_{n}=\|P\|_{B_{n}}$. Theorem 7.1 tells us
that $p_{n}=\max\\{\psi(a_{n}),\psi(a_{n}+1)\\}.$ The points $y\in B_{n}$
satisfying

$\sum_{j=1}^{n+1}|\lambda_{j}(y)|=p_{n}$ (9.2)

were found in [30]. Namely, assume $k_{n}$ coincides with that of the numbers
$a_{n}$ and $a_{n}+1$ which delivers to $\psi(t)$ the maximum value. Let us
fix $m=k_{n}$. Then (9.2) takes place for every point $y$ which is constructed
as in Theorem 8.1 in case $E=B_{n}$. The number of these points $y$ is
$N={n+1\choose m}$.

Now, let $P:C(B_{n})\to\Pi_{1}({\mathbb{R}}^{n})$ be an arbitrary projector
with the nodes $x^{(j)}\in B_{n}$, $S$ be the simplex with the vertices
$x^{(j)}$, and $E$ be the minimum volume ellipsoid contained $S$. We can also
consider $P$ as a projector acting from $C(E)$. By arguments related to affine
equivalence, the norm $\|P\|_{E}$ coincides with the
$(C(B_{n})$-$C(B_{n}))$-norm of a projector corresponding to a regular simplex
inscribed into $B_{n}$, i.e., $\|P\|_{E}=p_{n}.$ Moreover,

$\|P\|_{E}=\sum_{j=1}^{n+1}|\lambda_{j}(y)|$

for $m=k_{n}$ and for each of the points $y$ noted at the beginning of the
previous section (now with respect to the minimal ellipsoid $E$).

Theorem 8.1 tells us that at least one of these points, say $y^{*}$, belongs
to $B_{n}$. Therefore,

$\|P\|_{B_{n}}=\max_{x\in
B_{n}}\sum_{j=1}^{n+1}|\lambda_{j}(x)|\geq\sum_{j=1}^{n+1}|\lambda_{j}(y^{*})|=\|P\|_{E}=p_{n}.$

Thus,
$\theta_{n}(B_{n})=p_{n}=\max\\{\psi(a_{n}),\psi(a_{n}+1)\\}=\psi(k_{n})$.

Let us show that if $\|P\|_{B_{n}}=\theta_{n}(B_{n})$, then the simplex with
vertices at the interpolation nodes is inscribed into $B_{n}$ and regular. For
this simplex $S$, some point $y\in E$, constructed for $m=k_{n}$, falls on the
boundary sphere; otherwise we have $\|P\|_{B_{n}}>p_{n}=\theta_{n}(B_{n})$.
This is due to the fact that $\sum|\lambda_{j}(x)|$ increases monotonously as
$x$ moves in a straight line in direction from $c$ to $y$.

Since $\|y\|=1,$ the mean value of $\|y_{J}\|$ for the rest $y_{J}$ is also
$\leq 1$ (see (8.3)). Hence, another such point also lies in the boundary of
the ball. Thus we obtain $\|y_{J}\|=1$ for all sets $J$ consisting of
$m=k_{n}$ indices $j$. Now (8.3) yields $\|x^{(j)}\|=1,$ i.e., the simplex is
inscribed into the ball. Since all the points $y_{J}$ belong to the boundary
sphere, the function $\sum|\lambda_{j}(x)|$ has maximum upon the ball at
$N={n+1\choose m}$ different points. Consequently, the simplex $S$ is regular,
and the proof of the theorem is complete. $\Box$

For $1\leq n\leq 4$, equality (9.1) and characterization of minimal projectors
were obtained in [40] and [30] by another methods.

Corollary 9.1. For every $n\in{\mathbb{N}}$, the following inequalities

$\sqrt{n}\leq\theta_{n}(B_{n})\leq\sqrt{n+1}$

hold. Moreover, $\theta_{n}(B_{n})=\sqrt{n}$ only for $n=1$ and
$\theta_{n}(B_{n})=\sqrt{n+1}$ if and only if $\sqrt{n+1}$ is an integer.

In fact, it was proved in [40] that the above relations are satisfied for
$p_{n}=\max\\{\psi(a_{n}),\psi(a_{n}+1)\\}$. Combining this property with the
result of Theorem 9.1, we obtain Corollary 9.1.

##  10\. Interpolation by wider polynomial spaces

Let us briefly demonstrate how some of the above methods can be applied to
interpolation by wider polynomial spaces. This approach has been realized in
[18], [19] and in details described in [24].

Let $\Omega$ be a closed bounded subset of ${\mathbb{R}}^{n}$. Let $d\geq
n+1,$ and let $\\{\varphi_{1}(x),\ldots,\varphi_{d}(x)\\}$ be a family of
pairwise different monomials $x^{\alpha}=x_{1}^{\alpha_{1}}\ldots
x_{n}^{\alpha_{n}}.$ Here $x=(x_{1},\ldots,x_{n})\in{\mathbb{R}}^{n}$,
$\alpha=(\alpha_{1},\ldots,\alpha_{n})$ $\in{\mathbb{Z}}_{+}^{n}$. We suppose
that $\varphi_{1}(x)\equiv 1,$ $\varphi_{2}(x)=x_{1},\ \ldots,\
\varphi_{n+1}(x)=x_{n}.$

By the $d$-dimensional space of polynomials in $n$ variables we mean the set
$\Pi={\rm lin}(\varphi_{1},\ldots,\varphi_{d})$, i.e., the linear span of the
family $\\{\varphi_{1}(x),\ldots,\varphi_{d}(x)\\}$. Let us note the following
important cases: $\Pi=\Pi_{k}\left({\mathbb{R}}^{n}\right)$ – the space of
polynomials of degree at most $k$ $(k\in{\mathbb{N}})$, and
$\Pi=\Pi_{\alpha}\left({\mathbb{R}}^{n}\right)$ – the space of polynomials of
degree $\leq\alpha_{i}$ in $x_{i}$ $(\alpha\in{\mathbb{N}}^{n}).$

A collection of points $x^{(1)},\ldots,x^{(d)}\in\Omega$ is called an
admissible set of nodes  for interpolation of functions from $C(\Omega)$ with
the use of polynomials from $\Pi$ provided $\det({\bf A})\neq 0.$ This time
$\bf A$ is $(d\times d)$-matrix

${\bf
A}=\left(\begin{array}[]{cccc}1&\varphi_{2}\left(x^{(1)}\right)&\ldots&\varphi_{d}\left(x^{(1)}\right)\\\
\vdots&\vdots&\vdots&\vdots\\\
1&\varphi_{2}\left(x^{(d)}\right)&\ldots&\varphi_{d}\left(x^{(d)}\right)\end{array}\right).$

Assume that $\Omega$ contains such a set of nodes. For the interpolation
projector $P:C(\Omega)\to\Pi$ with these nodes, we have

$Pf(x)=\sum_{j=1}^{d}f\left(x^{(j)}\right)\lambda_{j}(x),\quad\|P\|_{\Omega}=\max\limits_{x\in\Omega}\sum_{j=1}^{d}|\lambda_{j}(x)|.$
(10.1)

Here $\lambda_{j}\in\Pi$ are the polynomials having the property
$\lambda_{j}\left(x^{(k)}\right)$ $=$ $\delta_{j}^{k}.$ Their coefficients
with respect to the basis $\varphi_{1},$ $\ldots,$ $\varphi_{d}$ form the
columns of ${\bf A}^{-1}.$

Let us introduce the mapping $T:{\mathbb{R}}^{n}\to{\mathbb{R}}^{d-1}$ defined
in the way

$y=T(x)=(\varphi_{2}(x),\ldots,\varphi_{d}(x))=(x_{1},\ldots,x_{n},\varphi_{n+1}(x),\ldots,\varphi_{d}(x)).$

We will consider $T$ on the set $\Omega.$ The choice of the first monomials
$\varphi_{j}(x)$ implies the invertibility of $T.$ Let
$y^{(j)}=T\left(x^{(j)}\right)$. Then the numbers $\lambda_{j}(x)$ are the
barycentric coordinates of the point $y=T(x)$ with respect to the
$(d-1)$-dimensional simplex with vertices $y^{(j)}$.

We set

$\theta(\Pi;\Omega)=\min\limits_{x^{(j)}\in\Omega}\|P\|_{\Omega},$

and

$\xi_{n}(\Omega)=\min\left\\{\xi(\Omega;S):\,S~{}\mbox{is an $n$-dimensional
simplex,}\,{\rm ver}(S)\subset\Omega,\,{\rm vol}(S)\neq 0\right\\}.$

The value $\xi(\Omega;S)$ is defined as for a convex $\Omega$.

If $x^{(1)},\ldots,x^{(d)}\in\Omega$ is an admissible set of nodes to
interpolate functions from $C(\Omega)$ by polynomials from $\Pi$, then
$y^{(j)}=T\left(x^{(j)}\right)$ is an admissible set of nodes to interpolate
functions from $C(T(\Omega))$ by polynomials from
$\Pi_{1}\left({\mathbb{R}}^{d-1}\right).$ For the projector
$\overline{P}:C(T(\Omega))\to\Pi_{1}\left({\mathbb{R}}^{d-1}\right)$ with the
nodes $y^{(1)},\ldots,y^{(d)},$ we have
$\|\overline{P}\|_{T(\Omega)}=\|P\|_{\Omega}.$ By this, when estimating the
norm $\|P\|_{\Omega}$, it turns out to be possible to apply geometric
inequalities for the norm of projector $\overline{P}$ under linear
interpolation on the $(d-1)$ dimensional set $T(\Omega)$. Thus we obtain
estimates of the norms $\|P\|_{\Omega}$ through the absorption coefficients
$\xi(T(\Omega);S)$ or through the function $\chi_{d-1}^{-1}$ (modifications of
(6.22) for $K={\rm conv}(T(\Omega))$) and some others.

For the projector $P:C(\Omega)\to\Pi$ with nodes $x^{(j)}$, we have

$\frac{1}{2}\left(1+\frac{1}{d-1}\right)\left(\|P\|_{\Omega}-1\right)+1\leq\xi(T(\Omega);S)\leq\frac{d}{2}\left(\|P\|_{\Omega}-1\right)+1,$
(10.2)

where $S$ is the $(d-1)$-dimensional simplex with the vertices
$T\left(x^{(j)}\right).$ Also we obtain the following inequalities:

$\frac{1}{2}\left(1+\frac{1}{d-1}\right)\left(\theta(\Pi;\Omega)-1\right)+1\leq\xi_{d-1}(T(\Omega))\leq\frac{d}{2}\left(\theta(\Pi;\Omega)-1\right)+1.$
(10.3)

The right-hand inequality in (10.2) becomes an equality, when there exists a
$1$-point $y=T(x)\in T(\Omega)$ with respect to the simplex $S={\rm
conv}\left(y^{(1)},\ldots,y^{(d)}\right)$. This means that for the
corresponding $x$ simultaneously
$\|P\|_{\Omega}=\sum_{j=1}^{d}|\lambda_{j}(x)|$ and among the numbers
$\lambda_{j}(x)$ there is the only one negative.

For the case $\Pi=\Pi_{1}\left({\mathbb{R}}^{n}\right)$ and $\Omega=Q_{n}$ or
$B_{n}$ see the previous text. Some other examples are given in [24]. The
simplest case of nonlinear interpolation is quadratic interpolation on a
segment. An analytical solution of the minimal projector problem with the
indicated geometric approach is given in [18] (see also [24] and [44]). Let us
consider this case as an illustrative example.

It is well known (see, e. g., [47]) that the minimal norm of an interpolation
projector is attained for regular nodes and equals ${5}/{4}$. We claim that
this result can be also obtained with the use of inequalities (10.2) and
(10.3). In addition, it turns out that there are infinitely many minimal
projectors.

If $\Pi=\Pi_{2}\left({\mathbb{R}}^{1}\right)$, then $d=\dim\Pi=3,$ i.e.,
$k=d-1=2.$ The mapping $T$ has the form $x\longmapsto(x,x^{2})$ and the set
$T(\Omega)=T([-1,1])=\\{(x,x^{2})\in{\mathbb{R}}^{2}:\,-1\leq x\leq 1\\}$ is a
piece of parabola. For the interpolation nodes $-1\leq r<s<t\leq 1$ the
simplex $S$ is the triangle with the vertices
$(r,r^{2}),\,(s,s^{2}),\,(t,t^{2})$ lying on this part of parabola. The
absorption of the parabolic sector by this triangle is shown in Fig. 7 .

Figure 7: The absorption of the parabolic sector by a triangle

The convexity of the function $\psi(x)=x^{2}$ implies that a $1$-point of the
set $T([-1,1])$ with respect to $S$ exists for any nodes. Hence, we have an
equality in the right-hand part of (10.2):

$\xi(T(\Omega);S)=\frac{3}{2}\left(\|P\|_{\Omega}-1\right)+1=\frac{3\|P\|_{\Omega}-1}{2}.$
(10.4)

Since (10.4) holds for an arbitrary projector
$P:C[-1,1]\to\Pi_{2}({\mathbb{R}}^{1})$, we have the right-hand equality also
in (10.3):

$\xi_{2}(T(\Omega))=\frac{3\theta\left(\Pi;\Omega\right)-1}{2}.$

So, finding the minimum norm of a projector $\theta\left(\Pi;\Omega\right)$ is
equivalent to calculating $\xi_{2}(T(\Omega))$. This problem is reducible to a
triangle $S$ with vertices $(-r,r^{2}),$ $(0,0),$ $(r,r^{2}),$ $0<r\leq 1.$
For this triangle,

$\xi(T(\Omega);S)=\max\left(\frac{11}{8},\frac{3}{r^{2}}-2\right),\quad\|P\|_{\Omega}=\max\left(\frac{5}{4},\frac{2}{r^{2}}-1\right).$

If $2\sqrt{2}/3=0.9428\ldots\leq r\leq 1,$ then these values do not depend on
$r$ and are equal correspondingly to $11/8$ and $5/4$ ; these are the minimal
possible values. Thus,

$\xi_{2}(T(\Omega))=\frac{11}{8}~{}~{}~{}\text{and}~{}~{}~{}\theta\left(\Pi;\Omega\right)=\frac{5}{4}.$

The minimal is an arbitrary projector with vertices $-r,$ $0,$ $r$ where
$r\in\left[\frac{2\sqrt{2}}{3},1\right]$. There are no other minimal
projectors.

In [44], we give estimates of
$\theta\left(\Pi_{k}\left({\mathbb{R}}^{1}\right);[-1,1]\right)$ and
$\xi_{k}\left(T([-1,1])\right)$ for $1\leq k\leq 10$. In this paper we have
studied the fulfillment of equality in the right-hand side of (10.2) for
simplices delivering the found upper bounds of $\xi_{k}\left(T([-1,1])\right)$
and also for uniform and Chebyshev nodes.

This work is intended to be continued in the multidimensional case.

Acknowledgements

I am very thankful to Pavel Shvartsman for useful suggestions and remarks.

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