On Some Estimate for the Norm of an Interpolation Projector

Let $Q_n=[0,1]^n$ be the unit cube in ${\mathbb R}^n$ and let $C(Q_n)$ be a space of continuous functions $f:Q_n\to{\mathbb R}$ with the norm $\|f\|_{C(Q_n)}:=\max_{x\in Q_n}|f(x)|.$ By $\Pi_1\left({\mathbb R}^n\right)$ denote a set of polynomials of degree $\leq 1$, i.e., a set of linear functions on ${\mathbb R}^n$. The interpolation projector $P:C(Q_n)\to \Pi_1({\mathbb R}^n)$ with the nodes $x^{(j)}\in Q_n$ is defined by the equalities $Pf\left(x^{(j)}\right)= f\left(x^{(j)}\right)$, $j=1,$ $\ldots,$ $ n+1$. Let $\|P\|_{Q_n}$ be the norm of $P$ as an operator from $C(Q_n)$ to $C(Q_n)$. If $n+1$ is an Hadamard number, then there exists a nondegenerate regular simplex having the vertices at vertices of $Q_n$. We discuss some approaches to get inequalities of the form $||P||_{Q_n}\leq c\sqrt{n}$ for the norm of the corresponding projector $P$.


Introduction
Let K be a convex body in R n .i. e., a compact convex subset of R n with nonempty interior.Denote by C(K) a space of continuous functions f : K → R with the uniform norm By Π 1 (R n ) we mean a set of polynomials in n variables of degree ≤ 1, i. e., of linear functions on R n .For x (0) ∈ R n , R > 0, by B(x (0) ; R) we denote the n-dimensional Euclidean ball given by the inequality x − x (0) ≤ R.Here .
We say that an interpolation projector P : C(K) → Π 1 (R n ) corresponds to a simplex S ⊂ K if the nodes of P coincide with the vertices of S. This projector is defined by the equalities P f x (j) = f x (j) .The following analogue of the Lagrange interpolation formula holds: Denote by P K the norm of P as an operator from C(K) in C(K).From (1), it follows that If K is a convex polytope, then the maxima in (2) also can be taken only over x ∈ ver(K), where ver(K) denotes the vertex set of K.
In this paper, we consider the case when n + 1 is an Hadamard number, i. e., there exists an Hadamard matrix of order n+1.An Hadamard matrix of order m is a square binary matrix H with entries either 1 or −1 which satisfies This means that the rows of H are pairwise orthogonal, with respect to the standard scalar product on R m .The order of an Hadamard matrix is 1 or 2 or some multiple of 4 (see [4]).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 (for links, see, e.g., [5], [8]).
Two Hadamard matrices are called equivalent iff 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 is 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 [5]).
Let Q be an arbitrary n-dimensional cube.If n + 1 is an Hadamard number, then there exists a nondegenerate regular simplex having the vertices coinciding with vertices of Q.In the paper, we discuss some approaches to obtain estimates of the form ||P || Q ≤ c √ n for the norm of the corresponding interpolation projector.For convenience, we will take

Making Use of the Numbers h n
Denote by h n the maximum value of a determinant of order n with entries 0 or 1.Let ν n be the maximum volume of an n-dimensional simplex contained in Q n .These numbers satisfy the equality h n = n!ν n (see [6]).For any n, there exists in Q n a maximum volume simplex with some vertex coinciding with a vertex of the cube.For such simplices, we have the following theorem.
Theorem 1. Suppose S ⊂ Q n is a maximum volume simplex with some vertex coinciding with a vertex of the cube, P is the corresponding interpolation projector.Then Proof.Let x (1) , . . ., x (n+1) be the vertices of S. We may assume that x (n+1) = 0, i. e., the vertex matrix of the simplex has the form From definition of ∆ j (x) and properties of determinant, Here we mark the jth row.Consequently, there exist numbers Define n-dimensional binary vectors v and w as follows.If u j = 1, then v j = 1 and w j = 0; if u j = −1, then v j = 0 and w j = 1.We have u = v − w, hence Denote these determinants of order n+1 by d 1 and d 2 .Since their entries are in [0, 1], It remains to include the case j = n + 1. Recall that therefore, for any vertex of the cube, We obtain Corollary 1.
Corollary 2. There exist c, c 1 > 0 does not depend on n such that Proof.The author proved that θ n ≥ γ √ n, γ > 0 not depends on n (see [12]).Hence, the left-hand inequality in (5) follows from (4).Since h n = n!ν n , the righthand inequality appears from the left-hand one.
Let us note more explicit forms of (4).For any n, Consequently, .
The above proof of Theorem 1 follows the approach of [10].Unfortunately, later it was noticed that the arguments in [10] perfomed for an arbitrary n have a lacuna.Yet in the case when n + 1 is an Hadamard number Theorem 1 allows to obtain for θ n upper bounds which cannot be improved in order of n.To show this, let us utilize the estimates The equality in (6) takes place if and only if n + 1 is an Hadamard number.Relation (6) was proved by Hadamard [3]; inequality (7) was obtained by Barba [1] (see [6]).Corollary 3. Suppose n + 1 is an Hadamard number.If S is a regular simplex with the vertices coinciding with vertices of Q n , then for the corresponding interpolation projector P P Qn ≤ √ 2n + 3 + 1.
Proof.If n + 1 is an Hadamard number, then the considered simplex S maximum volume among all simplices contained in Q n .Hence, we can apply to S Theorem 1: As it was noted above, in this case Since n + 1 is even, we can apply to h n+1 inequality (7): .
From this, it follows Corollary 4. If n + 1 is an Hadamard number, then θ n ≤ √ 2n + 3 + 1. Upper bounds given in Corollaries 3 and 4 are unimprovable in order of n, but can be made more precise.We will show this in next two sections.

The Proof Based on Hadamard Matrices
Recall (see [6]) that n + 1 is an Hadamard number iff 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.This can be easily shown for the cube Q There is a simple correspondence between Hadamard matrices of order n + 1 with the last column consisting of 1's and ndimensional regular simplices whose vertices are at the vertices of Q ′ n .Since we need this correspondence, we will dwell on it in more detail.
If S is a regular simplex of the specified type, then its vertex matrix S is an Hadamard matrix of order n + 1.Indeed, let x (1) , . . ., x (n+1) be the vertices of S. Since x (j) coincide with vertices of the cube, the entries of S are ±1 and its last column consists of 1's.Denote by h (j) the rows of S and make sure that these vectors are pairwise orthogonal in R n+1 .Equalities ||x (j) || = √ n mean that simplex S is inscribed into an n-dimensional ball of radius √ n.The edge-length d of a regular simplex and radius R of the circumscribed ball satisfy the relation (see, e. g., [12]).If R = √ n, then d = 2(n + 1).Hence, We see that (x (j) , x (k) ) = −1.Since the rows of S in the first n columns contain coordinates of vertices of the simplex and the last element of any row is 1, then Thus, h (j) are pairwise orthogonal in R n+1 .This means that S is an Hadamard matrix of order n + 1.We have Conversely, let 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 H.In other words, the vertices of S are given by the rows of H (excepting the last component).The simplex S is inscribed into the cube Q ′ n , moreover, ver(S) ⊂ ver(Q ′ n ).If h (j) are the rows of H and x (j) are the vertices of S, then (h (j) , h (k) ) = 0, (x (j) , x (k) ) = −1, x (j) 2 = n.It follows that ||x (j) − x (k) || 2 = 2(n + 1).So, S is a regular simplex with the edge-length 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 .This vertex matrix can be obtained from the given Hadamard matrix after multiplication of some rows by −1.
Theorem 2. Suppose n + 1 is an Hadamard number, S is an n-dimensional regular simplex having the vertices at vertices of Q ′ n .Then for the corresponding interpolation projector P : Proof.Let x (1) , . . ., x (n+1) be the vertices and let λ 1 , . . ., λ n+1 be the basic Lagrange polynomials of the simplex.Under our propositions, the vertex matrix S is an Hadamard matrix of order n + 1 with the last column consisting of 1's.
is the interpolation projector corresponding to S. Then we have Theorem is proved.By similarity reasons, Theorem 2 overcomes to any n-dimensional cube, for instance, to the cube Corollary 5. Suppose n + 1 is an Hadamard number.Then θ n ≤ √ n + 1.The result of Corollary 5 is known (see [9], [12]), but the given proof is more clearly related to Hadamard matrices.
Note that n-dimensional regular simplices with vertices at vertices of the cube can be located differently with respect to 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 more detail, we will describe an approach based on comparison of µ-vertices of the cube with respect to various simplices.
A notion µ-vertex of the cube Q n with respect to a simplex S ⊂ Q n was introduced by the author in [11].
If in place of Q n we consider an arbitraty n-dimensional cube Q, we obtain equivalent results.Assume 1 ≤ µ ≤ n.We say that x ∈ ver(Q) is a µ-vertex of a cube Q with respect to a simplex S ⊂ Q, iff for the interpolation projector P : with the nodes at vertices of S holds Here σS denotes the homothetic copy of S with the center of homothety at the center of gravity of S and the ratio of homothety σ.We call ξ(Q; S) the absorption index of the cube Q by the simplex S. For a projector P : C(Q) → Π 1 (R n ) and the corresponding simplex S, it is proved in [11] that The right-hand equality in (12) takes place if and only if there exists an 1-vertex of Q with respect to S. If for some µ there is a µ-vertex of Q with respect to S, then Of course, simplices which have different sets of µ-vertices with respect to the containing cube, are differently located in the cube, even if have the same projectors' 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 concerning Q = Q ′ n .While obtaining estimates for minimal norms of projectors, in [13] 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 these Hadamard matrices.They correspond to five simplices described in Table 1.The results of calculations were kindly delivered to the author by A. Yu.Ukhalov.
By m µ we denote a number of µ-vertices of the cube Q ′ n with respect to every simplex.For the rest 1 ≤ µ ≤ 15, excepting the given in Table 1, values m µ are zero.Every simplex S 1 , S 2 , and S 3 generates the same projector 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 µ-vertices.Also we have θ 15 ≤ 7 2 = 3.5.This is more exact than the estimate θ 15 ≤ 4 of Corollary 5 for n = 15.
Another example is related to n = 23.In [7], the results for 60 regular simplices inscribed into Q ′ 23 were mentioned.The simplices were built from the available 60 pairwise non-equivalent Hadamard matrices of order 24.For all simplices,  [13]).This inequality is more exact than the estimate θ 23 ≤ √ 24 = 4.8989 . . . of Corollary 5 for n = 23.Each of the four exceptional simplices is not equivalent to any of the 56 others.
Despite the possible differs, for all regular simplices with verices at vertices of the cube, inequality (10)

Connection with Interpolation on a Ball
A regular simplex inscribed into an n-dimensional ball has maximal volume among all simplices contained in this ball.There are no another simplices having this property (see [2], [15], and [16]).In the case when n + 1 is an Hadamard number, the similar property with respect to an n-dimensional cube holds for a regular simplex inscribed into the cube.Furthermore, the upper bound for projectors' norm corresponding to these simplices both on a cube and on a ball is equal to √ n + 1.In this section, we give another proof of Theorem 2. It seems to be shorter than previous one from Section 3, but is based on some relations which were obtained in [14] in a rather complicated way.This approach is related with interpolation on a Euclidean ball.
At first, suppose S is a regular simplex inscribed into an n-dimensional ball B = B(x (0) ; R) and P : C(B) → Π 1 (R n ) is the corresponding interpolation projector.Obviously, P B does not depend neither on the center x (0) and the radius R of the ball, nor on the chosen regular simplex inscribed into this ball.In other words, P B depends only on dimension n.For 0 ≤ t ≤ n + 1, consider the function
holds.The inscribed regular simplices satisfying P Q ′ n = √ 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.