Interpolation by Linear Functions on an $n$-Dimensional Ball

By $B=B(x^{(0)};R)$ we denote the Euclidean ball in ${\mathbb R}^n$ given by the inequality $\|x-x^{(0)}\|\leq R$. Here $x^{(0)}\in{\mathbb R}^n, R>0$, $\|x\|:=\left(\sum_{i=1}^n x_i^2\right)^{1/2}$. We mean by $C(B)$ the space of continuous functions $f:B\to{\mathbb R}$ with the norm $\|f\|_{C(B)}:=\max_{x\in B}|f(x)|$ and by $\Pi_1\left({\mathbb R}^n\right)$ the set of polynomials in $n$ variables of degree $\leq 1$, i.e., linear functions on ${\mathbb R}^n$. Let $x^{(1)}, \ldots, x^{(n+1)}$ be the vertices of $n$-dimensional nondegenerate simplex $S\subset B$. The interpolation projector $P:C(B)\to \Pi_1({\mathbb R}^n)$ corresponding to $S$ is defined by the equalities $Pf\left(x^{(j)}\right)=f\left(x^{(j)}\right).$ We obtain the formula to compute the norm of $P$ as an operator from $C(B)$ into $C(B)$ via $x^{(0)}$, $R$ and coefficients of basic Lagrange polynomials of $S$. In more details we study the case when $S$ is a regular simplex inscribed into $B_n=B(0,1)$.


Introduction
We begin with basic definitions and notations. Everywhere further n ∈ N. An element x ∈ R n we present in the form x = (x 1 , . . . , x n ). Denote by e 1 , . . ., e n the standard basis in R n . Let us put Everywhere further S is a nondegenerate simplex in R n . By c(S) denote the center of gravity of S. The notation τ S means the result of the homotety of S with homothetic center c(S) and coefficient τ . We denote ver(G) the set of vertices of the convex polyhedron G. Let Ω ⊂ R n be the closed bounded set. Consider the value ξ(Ω; S) := min{σ ≥ 1 : Ω ⊂ σS}. We call this number the absorption index of S with respect to Ω.
Let x (j) = x (j) 1 , . . . , x (j) n , 1 ≤ j ≤ n + 1 be the vertices of the simplex S. The vertex matrix is non-degenerate. Put A −1 = (l ij ). We call linear polinomials λ j (x) = l 1j x 1 + . . . + l nj x n + l n+1,j the basic Lagrange polynomials of S. Note that the coefficients of λ j form the columns of A −1 . These polinomials have the property λ j x (k) = δ k j . The numbers λ j (x) are barycentric coordinates of the point x ∈ R n with respect to S. The simplex S is given by the system of linear inequalities λ j (x) ≥ 0. More on polinomials λ j see in [4,Ch. 1].
Let us introduce the value ξ n (Ω) := min{ξ(Ω; S) : S n-dimensional simplex, ver(S) ⊂ Ω, vol(S) = 0}, ξ n := ξ n (Q n ). Various estimates of numbers ξ n were obtained in papers [2], [3], [5], [6], [7], [8], [13] and in monograph [4]. Here let us note that n ≤ ξ n < n+1. The exact values of ξ n are known only for n = 2, n = 5, n = 9, and for infinite numer of such n that there exists a Hadamard matrix of order n + 1. In all these cases, except n = 2, holds ξ n = n, while ξ 2 = 3 √ 5 5 + 1 = 2.34 . . . Up to present time n = 2 is the only even n for which the exact value of ξ n is known. The number ξ n for a Euclidean ball, i. e., exact value of ξ n (B n ), was found in [10]. In particular, for any S ⊂ B n holds ξ(B n ; S) ≥ n with the equality only for a regular simplex S inscribed into the border sphere. Further, in such a case, we say that the simplex is inscribed into the ball. Therefore, always ξ n (B n ) = n.
Denote by C(Ω) the space of continuous functions f : Ω → R with the uniform norm By Π 1 (R n ) we mean the set of polynomials in n variables of degree ≤ 1 (or the set of linear functions in n variables). Let x (j) ∈ Ω, 1 ≤ j ≤ n + 1, be the vertices of n-dimensional non-degenerate simplex S. We say that interpolation projector P : C(Ω) → Π 1 (R n ) corresponds to S if nodes of interpolation coincide with the points x (j) . This projector is determined by the equalities P f x (j) = f j := f x (j) . The analogue of the Lagrange interpolation formula holds true. Denote by P Ω the norm of P as an operator from C(Ω) into C(Ω). It follows from (1) that .

The expression
f j λ j (x) is linear in x and in f 1 , . . . , f n+1 , therefore, If Ω is a convex polyhedron in R n (for instance, if Ω = Q n ), we have simpler equality We say that the point x ∈ Ω is an 1-point of Ω with respect to S, if for projector P : C(Ω) → Π 1 (R n ) with nodes in vertices of S holds P = λ j (x)| and only one of numbers λ j (x) is negative. In [1, Theorem 2.1] it was established that for any P and corresponding S holds If there exists an 1-point in Ω with respect to S, then the last inequality in (3) becomes an equality. (The last statement was proved in [1] in the equivalent form; the notion of 1-vertex of the cube was introduced later in [2].) By θ n (Ω) denote the minimal value P Ω where x (j) ∈ Ω. It was proved by the first author that for numbers θ n := θ n (Q n ) the asymptotic equality θ n √ n takes place (various estimates were systematized in [4]). Later on some estimates were improved by M. V. Nevskii and A. Yu. Ukhalov, and by students of the second author (see [5], [6], [9] and references in these works). By the present time the exact values of θ n are known only for n = 1, 2, 3, and 7. Specifically, θ 1 = 1, θ 2 = 2 √ 5 5 + 1 = 1.89 . . ., θ 3 = 2, θ 7 = 5 2 . For all these cases ξ n = n + 1 2 θ n − 1 + 1.
In this paper we consider the case Ω = B n . Since for S ⊂ B n holds ξ(B n ; S) ≥ n, then for the corresponding projector P the last inequality in (3) gives So, always θ n (B n ) ≥ 3 − 4 n+1 . If for some interpolation projector P : C(B n ) → Π 1 (R n ) in (4) takes place the equality, then the corresponding simplex S is regular and inscribed into B n . Indeed, in this case n + 1 2 P Bn − 1 + 1 = n, and (3) gives ξ(B n ; S) ≤ n. Hence ξ(B n ; S) = n. This is possible only for a regular simplex inscribed into B n . Therefore, it follows from the equality θ n (B n ) = 3 − 4 n+1 that the simplex corresponding to the minimal projector has the described form.
If S is a regular simplex inscribed into B n and there exists an 1-point of B n with respect to S, then the relation (4) is an equality. When S is not a regular or not inscribed into the ball, (4) is a strict inequality. As it will be demonstrated, an 1-point of the ball with respect to an inscribed regular simplex exists only for 1 ≤ n ≤ 4. Beginning from n = 5 the equality in (4) doesn't hold true for all n and P , therefore, for n ≥ 5 holds θ n (B n ) > 3 − 4 n+1 . This paper includes proofs of this and some other results for linear interpolation on the ball. In particular we will obtain the exact formula for the norm of the interpolation projector with nodes in vertices of an inscribed regular simplex. Utilizing this formula we will show that θ n (B n ) ≤ √ n + 1 and the equality here is possible only for n = m 2 − 1, m ≥ 2.
2 Reduction in the problem of minimal projector An interpolation projector P : C(B n ) → Π 1 (R n ) is called a minimal projector if P Bn = θ n (B n ). The existence of a minimal projector follows from continuity of P as a function of nodes that is defined on closed bounded subset of R m , m = n(n + 1), given by the conditions x (j) ∈ B n , det A ≥ ε n > 0. Let us show that the minimal projector can be determined by the set of nodes belonging to the bounding sphere x = 1.
Assume P is an interpolation projector with the nodes x (j) ∈ B n and λ j are the basic Lagrange polynomials of the corresponding simplex S. Suppose not all x (j) belong to the bounding sphere. We will construct a projector P such that its number of nodes on the border of B n is greater by 1 than for P and such that P ≤ P . Assume the node x (i) of P lies strictly inside B n . Let z be the ortogonal projection of x (i) on the hyperplane λ i (x) = 0 and y be the point of intersection of the sphere and the straight line (zx (i) ) in the same halfspace as x (i) . Hence Denote by T the compression of R n with coefficient σ with respect to hyperplane λ i (x) = 0 in the orthogonal direction. Consider the projector P 1 with the same nodes but acting on ellipsoid T (B n ). The node x (i) belongs to the border of T (B n ) and the rest of nodes belong to the hyperplane λ i (x) = 0. So, the number of nodes on the border of T (B n ) in comparision with the number of original nodes on the border of B n will be greater by 1. Now, let us consider the inverse transformation T −1 . This nondegenerate affine transformation maps the ellipsoid T (B n ) into the ball B n and the simplex S into the simplex S = T −1 (S). Obviously, the ith vertex of S coincides with y while all other vertices of S are the same as those of S. Let P be the interpolation projector with nodes in vertices of S that again is being considered on B n . It remains to compare the norms of the projectors. Since T (B n ) ⊂ B n , we have Because the spaces C(T (B n )) and C(B n ) are isometric, holds P Bn = P 1 T (Bn) . Therefore, P Bn ≤ P Bn .
Applying the described procedure for necessary number of times, we can move all the nodes to the boundary of B n without increasing the projector norm. It means that there exists a minimal projector with all the nodes on the sphere. In particular, this property can be utilized for numerical minimization of the projector norm.

The norm of projector for interpolation on a ball
Let B = B(x (0) ; R) and let P : C(B) → Π 1 (R n ) be the interpolation projector with the nodes x (j) ∈ B. Suppose S is the simplex with vertices x (j) and λ j (x) = l 1j x 1 + . . . + l nj x n + l n+1,j are the basic Lagrange polynomials of S. In this case we have (see (2) for Ω = B). Let us obtain another formula for the projector norm.
Theorem 1. The following equality holds: Proof. Let us write the double maximum in the formula for P B in a different order than while obtaining (5): For fixed f j , the function The function L(x) := Λ(x) − Λ(0) is additive and homogeneous, therefore, Note that Now, the equality (7) gives

Taking into account that
The theorem is proved.
In the case when the center of gravity of the simplex coincides with the center of the ball formula (6) becomes more simple.

The projector norm for a regular simplex inscribed into the ball
Suppose S is a regular simplex inscribed into the n-dimensional ball B = B(x (0) ; R) and P : C(B) → Π 1 (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 dimension n. In this section, we will obtain exact representation of this function and establish its estimates.
For 0 ≤ t ≤ n + 1, consider the function , where s is the integer part of s.
Theorem 2. The following relations hold: The equality P B = √ n is true only for n = 1. The equality P B = √ n + 1 holds if and only if √ n + 1 is an integer.
Proof. We first prove (10). If n = 1, then ψ(t) = t(2 − t) + |1 − t|, a = 0, ψ(a) = ψ(a + 1) = 1. Since P B = 1, the equality (10) is true. Let n ≥ 2. Consider the simplex S with vertices This simplex is regular because the length of any edge is equal to √ 2. Simplex S is inscribed into the ball B = B(x (0) ; R), where Note that the (n + 1)th vertex of S is obtained by shifting of the zero vertex of the simplex x i ≥ 0, x i ≤ 1 in the direction from the hyperplane 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.

This gives
Recalling (9), we obtain It remains to show that the last maximum is equal to the largest of numbers ψ(a) and To do this, we analyze the behavior of ψ(t) over the whole interval [0, n + 1].
The function has following properties: The graph of ψ(t) is symmetric with respect to the straight line t = n+1 2 . On each half of [0, n + 1] function ψ(t) is concave as a sum of two concave functions. Indeed, for where ϕ 1 (t) is concave as a superposition of the concave function t(n + 1 − t) and the increasing concave function √ t, while ϕ 2 (t) is a linear function. The derivation ψ (t) is equal to zero only in two points These points lie strictly inside in 0, n+1 2 and n+1 2 , n + 1 correspondingly. From the concavity of ψ(t) on each of these segments, it follows that max 0≤ψ(t)≤n+1 Moreover, t − and t + are the only maximum points on the left and right halves of [0, n+1]. Hence, ψ(t) increases for 0 ≤ t ≤ t − and decreases for t − ≤ t ≤ n+1 2 . On the left half ψ(t) behaves symmetrically: it increases for n+1 2 ≤ t ≤ t + and decreases for t + ≤ t ≤ n + 1. Let us now consider only the segment 0, n+1 2 . Since a = t − , then always a ≤ t − < a + 1. Let k be a whole number, 1 ≤ k ≤ n+1 2 . If k < a, then ψ(k) < ψ(a), while if k > a + 1, then ψ(k) < ψ(a + 1). Taking into account (15), we have  The equality (10) is proved. We now turn to inequalities (11). We have already obtained the right one: Let us describe such dimensions n that P B = √ 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 √ n + 1 is an integer. Assume for example Then √ n + 1 = n + 1 − 2d is an integer. On the contrary, if √ n + 1 = m ∈ Z, then m(m − 1) is an even number and t − = 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 − = √ 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, These equalities are equivalent to (10), since in the cases considered a = t − = t − and P B = ψ(a). For all other n, holds P < √ n + 1. Indeed, if n = m 2 − 1, then a < t − < a + 1 and the maximum in (15) is reached either for k = a or for k = a + 1. For any n = m 2 − 1 the norm of P , i.e., the maximum ψ(k) for integer k ∈ 1, n+1 2 , is strictly less than maximum of ψ(t) over all this interval.
It remains to show that always P B ≥ √ n and for n > 1 this equality is a strict one. If n > 3, then √ n + 1 > 2, hence, From (10) and properties of ψ(t), we get For n = 2 and n = 3 also holds P B > √ n. Therefore, P = √ n only for n = 1. The theorem is proved.
If n + 1 is an Hadamard number, i. e., there exists an Hadamard matrix of order n + 1 (see [11], [12]), the estimate P B ≥ c √ n can be obtained from previous results of the first author. In 2006, he have proved that for all n θ n (Q) ≥ χ −1 Here Q is an n-dimensional cube, χ n is the normalized Legendre polynomial of degree n and ν n is the maximum volume of a simplex contained in the unit cube Q n = [0, 1] n . For references, definitions and proofs, see [4]. Suppose additionally that n + 1 is an Hadamard number. Then there exists a regular simplex with vertices in the vertices of a cube (see [12]). Let Q be a cube inscribed in the Euclidean ball B. Then this regular simplex will be inscribed also in the ball. Since Q ⊂ B, for the corresponding projector P we have The exact values θ n (B n ) for 1 ≤ n ≤ 4 Utilizing Theorem 2 and properties noted in Introduction, it is possible to obtain the exact values of θ n (B n ) and to describe minimal projectors for n = 1, 2, 3, 4. Of course, θ 1 (B 1 ) = 1, but this case also fits into the general scheme.
Theorem 3. The following equalities hold: For 1 ≤ n ≤ 4 the equality P Bn = θ n (B n ) holds only for projectors corresponding to regular simpleces inscribed into B n .

Concluding remarks
The inequality (11) of Theorem 2 implies that θ n (B n ) (the minimum norm of the projector for linear interpolation on the unit ball x ≤ 1) satisfies the inequality At least for those n, when √ n + 1 is not integer, i. e. n = m 2 − 1, this inequality is strict. If a projector haves the minimal norm and its nodes are the vertices of a regular simplex inscribed into the sphere x = 1, then θ n (B n ) = max{ψ(a), ψ(a + 1)}.
Questions concerning accuracy of (18) and correctness of (19) for n > 4 will be discussed in a further paper. In conclusion, we present some illustrations and the results of numerical analysis. Graphs of the function , 0 ≤ t ≤ n + 1, for n = 3, 4, 5, 15 are shown in Fig. 1-4. The maximum points of of this function and a + 1 are denoted. One of these two points maximizes ψ(k) for integer 1 ≤ k ≤ n+1 2 . In Fig. 5  i. e., for dimensions of the form m 2 − 1. It is at these points that d n = 0, as it can be seen in Fig. 5. The dashed line denotes the graph of the linear interpolation spline l constructed by the nodes n = m 2 − 2, n = m 2 and by the values of d n in these nodes. Always d n ≤ l(n) and the equality is achieved only for n = m 2 − 2 and n = m 2 . The function l(n) decreases and l(n) → 0 as n → ∞. Thus, the following two-side estimate takes place: √ n + 1 − l(n) ≤ P Bn ≤ √ n + 1. Both the right and the left inequalities turn into equalities for the infinite set of n's. This estimate is more accurate than (11). Table 1 presents the values of P Bn for a regular simplex inscribed into the ball. We use notations from Section 4 The value k * is equal to the number of −1's in the extremal set of f j corresponding to the maximum in (14). Since this maximum is equal to max{ψ(a), ψ(a + 1)}, the integer k * is equal to that of numbers a or a + 1, for which ψ(t) takes a larger value (see the proof of Theorem 2). If k * = 1, then in B n there is an 1-point with respect to the simplex S corresponding to P . For such n, ξ(B n ; S) = n + 1 2 P Bn − 1 + 1 (see (3)). Since S is a regular simplex inscribed into B n , then ξ(B n ; S) = n and (20) is equivalent to P Bn = 3 − 4 n+1 . However, starting from n = 5, always k * > 1. Moreover, k * increases along with n. This corresponds to the fact that (20) and the equality P Bn = 3 − 4 n+1 take place only for 1 ≤ n ≤ 4. Initially, this effect was discovered in the course of computer experiments. Later the analytical solution of the problem was found.
If n + 1 is an Hadamard number, the validity of equality similar to (20) can also be considered for a cube and the inscribed regular simplex. It is interesting to note that an 1-point of Q n with respect to such simplex S exists not only for n = 1 and n = 3, but also for n = 7. In the specified cases ξ(Q n ; S) = n + 1 2 P Qn − 1 + 1, Table 1: Norm of P for a regular simplex inscribed into B n n t − a a + 1 ψ(a) ψ(a + 1) k * P Bn P Bn