aa r X i v : . [ m a t h . NA ] J un DUAL BASES FUNCTIONS IN SUBSPACES
SCOTT N. KERSEY
Abstract.
In this paper we study dual bases functions in subspaces. These are bases which aredual to functionals on larger linear space. Our goal is construct and derive properties of certainbases obtained from the construction, with primary focus on polynomial spaces in B-form. Whenthey exist, our bases are always affine (not convex), and we define a symmetric configuration thatconverges to Lagrange polynomial bases. Because of affineness of our bases, we are able to derivecertain approximation theoretic results involving quasi-interpolation and a Bernstein-type operator.In a broad sense, it is the aim of this paper to present a new way to view approximation problemsin subspaces. In subsequent work, we will apply our results to dual bases in subspaces of spline andmultivariate polynomial spaces, and apply this to the construction of blended function approximantsused for approximation in the sum of certain tensor product spaces. Introduction
Let X be a finite dimensional vector spaces (of dimension n ). Of fundamental importance is thebasis. There are various reasons to choose a particular basis, and each basis has advantages anddisadvantages. Often it is the action of certain functionals that lends importance to a particularbasis. For example, the Lagrange basis is important because it is dual to point evaluation. Ingeneral, any basis Φ n = [Φ , . . . , Φ n ] for X has a dual map Λ n = [ λ , . . . , λ n ] for X ∗ satisfying λ i Φ j = δ ij . That is, Λ nT Φ n = I , the identity matrix. Conversely, any basis Λ n ⊂ X ∗ is dual tosome basis Φ n of X . It is the dual map Λ n that extracts information from the functions that is ofparticular interest here, which arguably plays a more prominent role than the basis Φ n itself. Thisprocess allows one to consider the information we are trying to capture as our primary goal.The above concepts are well-known and well-studied. In this paper, we are interested in investi-gating bases for subspaces Y of X that are dual to subsets of the functionals in the map Λ n . Forexample, suppose that Λ n = [ λ , . . . , λ n ] are linearly independent on the n-dimensional space X , and Y is an m-dimensional subspace of X with m < n . Then, given an injective map s : [1 : m ] → [1 : n ],our questions are:(1) Is the subset Λ n ( s ) = [ λ s (1) , . . . , λ s ( m ) ] of Λ n linearly independent on Y ?(2) If linearly independent, what is the basis D m for Y that is dual to Λ n ( s ) in the sense thatΛ n ( s ) T D m = I .(3) What are the properties of this dual basis. Date : September 11, 2018.1991
Mathematics Subject Classification.
In this paper we consider the action of subsets of functionals for certain basis on subspaces ofthe original space. This allows us to view how the subspace looks according to information on thewhole space. In some sense, we are addressing the question “how do you approximate with less(or not enough) information?” But it is not always possible to construct such bases, because thefunctionals are not always linearly independent on the subspace. For example, the only subset ofΛ = [ δ , δ D, δ D ] that is linearly independent on Ran[1 , ( · )] is Λ = [ δ , δ D ], not Λ = [ δ , δ D ] orΛ = [ δ D, δ D ]. As we will see in this paper, the situation is different in the Bernstein basis.It is the objective of this paper to determine if dual bases exist for certain subspaces of polynomialspaces, and if so to compute and characterize these bases, and then determine their properties. Inparticular, we show that in the Bernstein basis for $ n , any m + 1-selection of the dual functionals arelinearly independent on $ m . Moreover, by choosing a particular symmetric choice of the functionals,we show that the corresponding dual basis converge to the Lagrange polynomial basis. Later, wederive certain approximation results concerning quasi-interpolation and a Bernstein-type operator.2. Dual Bases in Subspaces Y of X As defined above, X is a vector space of dimension n with basis Φ n and dual map Λ n , and Y asubspace of X of dimension m , m < n , with basis Φ m . For y ∈ Y ⊂ X , we have y = Φ m α = Φ n β for some coefficient sequences α ∈ R m and β ∈ R n . Applying the dual basis, we haveΛ nT Φ m α = Λ nT Φ n β = β. Hence, β = Eα with E := Λ nT Φ m . The matrix E embeds the coefficients α of y in the basis for Y to it’s coefficients in the basis for X . By inclusion of Y into X , we can construct the embedding e := Φ n ◦ E ◦ (Φ m ) − of Y into X . And since Φ m α = φ n Eα for all α ∈ R m , it follows that Φ m = Φ n E .All said, this can be visualized as in the following commutative diagram. Y (cid:31) (cid:127) e / / X R m Φ m O O E / / R n Φ n O O Figure 2.1.
Embedding of $ m,y into $ n,x .Moreover, we note thatΛ nT Φ m = Λ nT Φ n E = E = E Λ mT Φ m = (Λ m E T ) T Φ m . Therefore, Λ n = Λ m E T . Hence, we have the following result concerning the action of Λ n on asubspace Y of X . Proposition 2.1.
Let ( Y, Φ m , Λ m ) be an m -dimensional subspace of a vector space ( X, Φ n , Λ n ) , withbasis Φ m dual to Λ m and Φ n dual to Λ n . Then, Y embeds into X by the map e = Φ n ◦ E ◦ (Φ m ) − UAL BASES FUNCTIONS IN SUBSPACES 3 with E = Λ nT Φ m , and the bases transform as Φ m = Φ n E .
Further, the dual basis for X ∗ map to the dual basis for Y ∗ by the map Λ n = Λ m E T . Since Λ n ⊂ X ∗ ⊂ Y ∗ , the functionals λ ni ∈ X ∗ are also functionals on Y . But since dim Y < n ,the functionals in Λ n cannot be linearly independent on Y . However, sinceΛ nT Φ m = Λ nT Φ n E = E is a 1-1 matrix, Λ n does span Y ∗ (moreover, Λ n E is a basis for Y ∗ since (Λ n E ) T Φ m = E T E isinvertible). Therefore, we can trim Λ n to a basis for Y ∗ by removing inessential vectors (this is astandard construction, c.f. [5]). This leaves an m -subvector Λ n ( s ) of Λ n that is linearly independenton Y , where s is an injective map s : [1 : m ] → [1 : n ] (which we call a selection map ). Then, with I the n × n identity matrix, we haveΛ n ( s ) T Φ m = Λ n ( s ) T Φ n E = (Λ n I (: , s )) T Φ n E = I ( s, :)Λ nT Φ n E = I ( s, :) E = E ( s, :) . Therefore, an m -subvector Λ n ( s ) of Λ n is linearly independent on Y iff E ( s, :) is invertible. Moreover,when E ( s, :) is invertible, the basis of Y = Ran(Φ m ) that is dual to Λ n ( s ) is Φ m E ( s, :) − , as followsby computing the change of basis Φ m → Φ m A : I = Λ n ( s ) T Φ m A = E ( s, :) A. Hence, the basis of Y that is dual to Λ n ( s ) is D m := Φ m E ( s, :) − . The following summarizes theabove statements: Proposition 2.2.
Let ( X, Φ n , Λ n ) be a vector space with basis Φ n dual to Λ n , and let ( Y, Φ m , Λ m ) bea subspace with basis Φ m = Φ n E dual to Λ m . Let s : [1 : m ] → [1 : n ] denote a selection (injective)map. (1) Λ n E is linearly independent on Y and of length m (hence a basis for Y ∗ ). (2) There exist (at least one) selections Λ n ( s ) of Λ n linearly independent on Y . Sometimes onlyone selection is linearly independent, and sometimes all selections are linearly independent(this is of particular interest later in this paper). (3) E ( s, :) = Λ n ( s ) T Φ m (4) Λ n ( s ) is linearly independent on Y iff E ( s, :) is invertible. (5) If Λ n ( s ) is linearly independent on Y , then the basis of Y that is dual to Λ n ( s ) in the sensethat Λ n ( s ) T D m = I is D m = Φ m E ( s, :) − . SCOTT N. KERSEY Additional Properties of Dual Bases in Subspaces
In this section we state definitions and properties useful for certain constructions that we considerin the subsequent sections. Firstly, by Proposition 2.2, there always exists some s such that Λ n ( s )is linearly independent on Y , and in this case we have the basis D m = Φ m E ( s, :) − with E =Λ n ( s ) T Φ m . However, this may or may not be true for all s . Hence, we give the following definitionto make this distinction. Definition 3.1.
We say the embedding of ( Y, Φ m , Λ m ) into ( X, Φ n , Λ n ) is complete if Λ n ( s ) islinearly independent on Y for all injective maps (selections) s : [1 : m ] → [1 : n ]. Proposition 3.2.
Let Y = $ m be the space of polynomials of degree at most m , and let X = $ n bethe space of polynomials of degree at most n . (1) Let Φ m and Φ n be power basis for $ m and $ n . Then, the embedding is not complete. Moreover, Λ n ( s ) is linearly independent on Φ m iff s = [0 : m ] . (2) Let Φ m and Φ n be Bernstein bases for $ m and $ n , respectively. Then, the embedding iscomplete.Proof. Part (2) is non-trivial, and will be proved later in this paper. For part (1), recall that thepower basis P n = [1 , ( · ) , . . . , ( · ) n ] dual to Λ n = [ δ , δ D, . . . , δ D n /n !], and ($ m , P m , Λ m ) is thesubspace with m < n . Then P m = P n E for E = I (: , m ), with I is the ( n + 1) × ( n + 1) identitymatrix. It follows by Proposition 2.1 that Λ n = Λ m E T = Λ m I (0 : m, :) on $ m . Let s be a selectionmap. Since I (0 : m, s ) is invertible iff s = 0 : m , it follows that Λ n ( s ) is linearly independent on$ m iff s = 0 : m . That is, on $ m , [ λ n , . . . , λ nm ] = [ λ m , . . . , λ mm ]. Hence, there is only one choice forthe selection s of Λ n to consider, and this choice is Λ m = Λ n ( s ). Hence, the Lagrange basis is notcomplete. (cid:3) The next properties are characteristic of Bernstein bases, which we consider later in this paper.
Definition 3.3. (1) Let Φ n be a basis of finite-dimensional function space. We say Φ n is affine on S if Φ n ( t ) α isan affine combination of α for all t ∈ S . That is, P i Φ ni ( t ) = 1 for all t ∈ S . If it holds forall t ∈ dom(Φ n ), then we say Φ n is affine .(2) Let E be a (real) matrix. We say E is row affine if P j E ( i, j ) = 1 for all i , and columnaffine if P i E ( i, j ) = 1 for all j . Lemma 3.4. (1)
Matrix inversion preserves the property row-affine. That is, the inverse of an invertiblerow-affine matrix is row-affine. (2)
Matrix multiplication preserves the property row-affine. That is, the product of two row-affine matrices is row-affine.
UAL BASES FUNCTIONS IN SUBSPACES 5
Proof.
For the first result, let A be an n × n invertible row-affine matrix. Then P nj =1 A ( i, j ) = 1for i = 1 : n . Since A − A = I ,1 = n X j =1 I ( k, j ) = n X j =1 n X i =1 A − ( k, i ) A ( i, j ) = n X i =1 A − ( k, i ) n X j =1 A ( i, j ) = n X i =1 A − ( k, i ) , for k = 1 : n . Hence, A − is row-affine.For the second result, assume C = AB with A and B row-affine of dimensions m × n and n × p ,respectively. Then, C is of dimension m × p , and p X j =1 C ( k, j ) = p X j =1 n X i =1 A ( k, i ) B ( i, j ) = n X i =1 A ( k, i ) p X j =1 B ( i, j ) = n X i =1 A ( k, i ) = 1 , for k = 1 : m . Therefore, C is row-affine. (cid:3) Theorem 3.5.
Let Φ m and Φ n be affine bases of Y and X , respectively, with Y a subspace of X . (1) If Φ m = Φ n E for some matrix E , then E is row affine. (2) If, moreover, E ( s, :) is invertible for some selection s : [1 : m ] → [1 : n ] , then the dual basis D m = Φ m E ( s, :) − exists and is affine.Proof. For (1), it follows by affineness of the two bases that1 = X i Φ mi ( t ) = X i X j Φ nj ( t ) E ( j, i ) = X j Φ j ( t ) X i E ( j, i ) . Let λ nk be the functional on X such that λ k Φ nj = δ ij . Then, λ nk (1) = λ nk ( X j Φ nj ) = X j λ nk (Φ nj ) = λ nk (Φ nk ) = 1 , and so 1 = λ nk (1) = λ nk ( X j Φ j ( t ) X i E ( j, i )) = X i E ( k, i ) . This establishes (1).For (2), we recall that D m = Φ m E ( s, :) − is the basis for Y dual to Λ n ( s ) with E = Λ nT Φ m ,and hence E ( s, :) = Λ n ( s ) T Φ m . By part (1) of this theorem, E ( s, :) is row affine, and by Lemma 3, A := ( E ( s, :)) − is row affine as well. Therefore, n X i =0 D mi = m X i =0 X j =0: m Φ mj A ( j, i ) = m X j =0 B mj X i =0: m A ( j, i ) = m X j =0 B mj = 1 . And so D m is an affine basis: (cid:3) These results regarding affine bases will concern the Bernstein basis, which we investigate in theremaining sections, and in particular we show that dual bases are affine. We remark here thatthe same is not true of convexity. That is, the dual bases constructed are not convex. Indeed,matrix inversion does not preserve convexity, as it does affineness. Morever, as it turns out, certainapproximation properties do not actually require convexity. I.e., affineness is enough. Hence, our
SCOTT N. KERSEY dual bases D m will enjoy many of the same properties as B m do, in the Bernstein-basis setting, justnot convexity.In our construction we derive dual bases in terms of data maps that are dual on a large space.However, it is possible to have different data maps that are both dual to the same bases. Forexample, in terms of the point-evaluation function δ x f = f ( x ) and derivative operator D , bothmaps Λ = [ δ , δ D ] and ˜Λ = [ δ , δ − δ ] are dual to the power basis Φ = [1 , ( · )], as can be seenby Λ T Φ = ˜Λ T Φ = I . However, even with different data maps, dual bases are the same, as shownnext. This idea is important in constructing approximation operators where some data maps mayapply and not others. More to the point, we will derive dual basis in this paper with respect todual data maps that involve differentiation, hence do not apply on C ([0 , C ([0 , Proposition 3.6.
Let Λ n and ˜Λ n be data maps on X both dual to the basis Φ n . Then both mapsare equivalent on X , and the dual bases in a subspace Y are invariant of which dual map is used.That is, if D m and ˜ D m are bases for the subspace Y that are dual to Φ m with respect to the selection s and data maps Λ n and ˜Λ n , respectively. Then, D m = ˜ D m .Proof. To show that Λ n and ˜Λ n are equivalent on X , let f = Φ n α ∈ X . Then, since both maps aredual to Φ n , we have Λ nT Φ n α = α and ˜Λ nT Φ n α = α . Hence, they are equivalent on X .The basis D m is dual to Λ n ( s ) in the sense that Λ nT D m = I . With this basis represented D m = Φ m A = Φ n EA , for some embedding, we getΛ nT D m = Λ nT Φ n EA = EA.
Therefore, I = Λ nT ( s ) D m = E ( s, :) A, and so A = E ( s, :) − , which is depends only on the embedding E and not the data map Λ n . Hence, if ˜ A is the transformation matrix for ˜Λ n , then ˜ A = A , and so D m = ˜ D m . (cid:3) Dual Bernstein Bases in the subspace $ m of $ n Let $ n be the space of polynomials of degree at most n and $ m the subspace of polynomialsof degree at most m , for m < n . In this section we derive bases for $ m that are dual to subsetsof the dual Bernstein functionals on $ n . We begin with some basic formulas involving Bernsteinpolynomials that will be used. • Bernstein basis for $ n : B n = [ B n , . . . , B nn ] with B ni = (cid:0) ni (cid:1) (1 − · ) n − i ( · ) i , P ni =0 B ni = 1 and B ni ≥ , , B n α is aconvex combination of α for all t ∈ [0 , • Degree Elevation of Bernstein Basis (See [1, 6]): B m = B n E with E the ( n + 1) × ( m + 1)matrix with entries E ( i, j ) = (cid:0) n − im − j (cid:1)(cid:0) ij (cid:1)(cid:0) nm (cid:1) = (cid:0) n − mi − j (cid:1)(cid:0) mj (cid:1)(cid:0) ni (cid:1) UAL BASES FUNCTIONS IN SUBSPACES 7 for 0 ≤ i ≤ n and 0 ≤ j ≤ m , with E ( i, j ) = 0 if i < j or n − i < m − j . That is, B mj = n X i =0 E ( i, j ) B ni = n X i =0 (cid:0) n − im − j (cid:1)(cid:0) ij (cid:1)(cid:0) nm (cid:1) B ni . • Dual Bernstein Functionals: Λ n = [ λ n , . . . , λ nn ] with λ nk = k X j =0 (cid:0) kj (cid:1)(cid:0) nj (cid:1) j ! δ D j = n − k X j =0 ( − j (cid:0) n − kj (cid:1)(cid:0) nj (cid:1) j ! δ D j . Therefore, λ nk B ni = δ ki and Λ nT B n = I . (Note that the two forms are equivalent on $ n , butnot on all spaces.) • Reduction of Dual Bernstein Basis: On $ m , Λ n = Λ m E T (by Proposition 2.1). Hence, λ nk = m X j =0 E T ( j, k ) λ mj = m X j =0 (cid:0) n − km − j (cid:1)(cid:0) kj (cid:1)(cid:0) nm (cid:1) λ mj . Hence, in the context of this paper, we have the following:
Theorem 4.1.
Let B m be the Bernstein basis for $ m , and let B n be the Bernstein basis for $ n withdual map Λ n given above, m ≤ n . Let E be the degree elevation matrix. Then, (1) The embedding e of ($ m , B m , Λ m ) into ($ n , B n , Λ n ) is complete. (2) The basis for $ m dual to Λ n ( s ) can be represented D m = B m E ( s, :) − , with D m = B m when m = n .Proof. For (1), we need to show that Λ n ( s ) is linearly independent on $ m for any selection s : [0 : m ] → [0 : n ]. The conversion from between the power basis P n and Bernstein basis B n for $ n canbe expressed P n D n = B n T n with D n := Diag([ (cid:0) nj (cid:1) : j = 0 : n ]) and T n Pascal’s (lower triangular) matrix T n := [ (cid:18) ij (cid:19) : 0 ≤ i, j ≤ n ] = · · ·
01 1 0 · · ·
01 2 1 · · · (cid:0) n (cid:1) (cid:0) n (cid:1) (cid:0) n (cid:1) · · · (cid:0) nn (cid:1) . This follows directly from the identity (cid:18) nj (cid:19) t j = n X i =0 (cid:18) ij (cid:19) B ni SCOTT N. KERSEY (see [6], section 2.8, for a short proof). Let d n := ( D n ) − = [1 / (cid:0) nj (cid:1) : j = 0 : n ]. Then,Λ n ( s ) T P m = (Λ n I (: , s )) T P n I (: , m )= I ( s, :)Λ nT B n T n d n I (: , m )= I ( s, :) T n d n I (: , m )= T n ( s, m ) d n (0 : m, m )In [3] it was shown that the truncated Pascal matrix T n ( s, m ) is invertible for any m + 1 selection s : [0 : m ] → [0 : n ] of the rows of T n . Therefore, Λ n ( s ) T P m is invertible, and so Λ n ( s ) is linearlyindependent on $ m .For (2), we have B m = B n E with E the degree elevation matrix. Therefore, we are exactly inthe framework of Proposition 2.2, with Φ m = B m and Φ n = B n . Since the embedding is complete,we have that E ( s, :) is invertible for any selection map s , Therefore, the dual basis of $ m dual toΛ n ( s ) exists for any selection map s , and can bre represented D m = D m = B m E ( s, :) − . In the case m = n , there is only one selection s = [1 : n ], and so Λ n ( s ) = Λ n . Hence, E ( s, :) = E is the identitymatrix, and D m = B m = B n are therefore dual to Λ n . (cid:3) Since, as is well known, B m and B n are affine bases, the next result follows directly from Theorem3.5. Instead, we give a more direct proof below using properties of the degree elevation matrix E . Theorem 4.2.
For any selection (injective) map s : [0 : m ] → [0 : n ] , D m := B m A is an affinebasis of $ m dual to Λ n ( s ) , with A := E ( s, :) − .Proof. We note that by the previous lemma Λ n ( s ) is linearly independent on $ m . Therefore,Λ n ( s ) T B m is invertible. By Proposition 2.2, Λ n ( s ) T B m = E ( s, :) and D m = B m E ( s, :) − is thebasis for $ m dual to Λ n ( s ). Recall that E ( i, j ) = (cid:0) n − im − j (cid:1)(cid:0) ij (cid:1)(cid:0) nm (cid:1) . By the Chu-Vandermonde identity (cid:18) nm (cid:19) m X j =0 E ( i, j ) = m X j =0 (cid:18) n − im − j (cid:19)(cid:18) ij (cid:19) = m X j =0 (cid:18) n − ij (cid:19)(cid:18) im − j (cid:19) = (cid:18) nm (cid:19) , and so E ( s, :) is row affine. By Lemma 3, ( E ( s, :)) − is row affine as well. Therefore, n X i =0 D mi = m X i =0 X j =0: m B mj A ( j, i ) = m X j =0 B mj X i =0: m A ( j, i ) = m X j =0 B mj = 1 . And so D m is an affine basis: (cid:3) The next property is a generalization of the following property of Bernstein polynomials: x = B m ( x ) v m = m X i =0 ξ mi B mi ( x ) UAL BASES FUNCTIONS IN SUBSPACES 9 for all x , with ξ mi := im . Theorem 4.3.
Let D m = B m E ( s, :) − be the dual basis for some selection map s . (1) Eξ m = ξ n (2) ξ m = E ( s, :) − ξ n ( s ) . (3) x = B m ξ m = D m ( x ) ξ n ( s ) = P mi =0 ξ ns ( i ) D mi ( x ) .Proof. Recall that the Bernstein functions on $ n can be represented λ nk = k X j =0 (cid:0) kj (cid:1)(cid:0) nj (cid:1) j ! δ D j . In particular, λ n x = δ x = 0. For k >
0, the sum of the terms for λ nk x are zero except when j = 1.Hence, λ nk x = n X j =0 (cid:0) kj (cid:1)(cid:0) nj (cid:1) j ! δ D j x = (cid:0) k (cid:1)(cid:0) n (cid:1) δ D x = kn . Therefore, Λ nT x = ξ n . From this we get, ξ n = Λ nT x = Λ nT B m ( x ) ξ m = Λ nT B n ( x ) Eξ m = Eξ m . Thereofore, ξ n = Eξ m . Hence, ξ n ( s ) = E ( s, :) ξ m implies ξ m = E ( s, :) − ξ n ( s ), and D m ( x ) ξ n ( s ) = B m ( x ) E ( s, :) − ξ n ( s ) = B m ( x ) ξ m = x. (cid:3) Plotting Polynomials in the Dual Bases
Since dual bases D m are bases for the subspace Y , and function in Y can be represented by thisbasis. In particular, in the Berntstein setup above, any polynomial p ∈ S m can be represented as p = D m α for some α ∈ R n +1 . And so, this basis can be used in computation with functions in thispolynomial space.Now, since D m = B m A with A = E ( s, :) − , we can write p = B m ( Aα ). Hence, one can transformthe coefficients α by the matrix A , and then use B-form techniques in computation. In particular, toplot the curves, one can use DeCasteljau’s algortihm on the control polygon with points ( im , ( Aα )( i )),for i = 0 : m .This is depicted in Figure 5.1. The control polygon for the coefficients α is displayed in solidbroken line and the transformed control polygon is dashed. Figure 5.1.
Control polygon and transformed polygon for deCasteljau’s algorithmwith dual Bernstein subspace bases.6.
Symmetric Bernstein Class of Dual Bases
As shown in the previous section, for Bernstein functionals the matrix E ( s, :) = Λ( s ) nT B m isinvertible for any selection map s , and we can therefore find dual bases for any selection map.Moreover, these bases are affine. In this section we will use this idea to produce a certain “sym-metric” class of bases, and show that these converge to the Lagrange polynomial basis. We alsoprovide an estimate for the rate of convergence.For k ∈ N , let s ( i ) = ik for i = 0 : m . Hence, s = 0 : k : n with n = k ∗ m . For example: • If k = 1 then m = m and we get s = [0 : m ]. • If m = 4 and k = 3, then n = 12 and s = [0 , , , , D mk = B m A m,k with A m,k = E ( s, :) − . The goal in the remainderof this section is to show that symmetric dual bases converge to point evaluation in a certain sense.To get the most general result, we extend the Bernstein functional to allow k = x to be any realnumber: λ nx := ⌊| x |⌋ X j =0 (cid:0) xj (cid:1)(cid:0) nj (cid:1) j ! δ D j . Here, the factorials in the binomial coefficients will involve the gamma function when x is a non-integer. The functionals reduce to the above formulation when k := x is a non-negative integerwith k ≤ n . For the following, we use the falling factorial notation( x ) j := x · ( x − · · · ( x − j + 1) . Lemma 6.1.
For x ∈ R and j ≥ , lim n →∞ (cid:0) xnj (cid:1)(cid:0) nj (cid:1) = x j , with ( · )! := Γ( · + 1) for non-integer factorials. UAL BASES FUNCTIONS IN SUBSPACES 11
Proof.
By the well-known property Γ( z + 1) = z Γ( z ), we get that j ! (cid:18) xnj (cid:19) = Γ( xn + 1)Γ( xn − j + 1) = ( xn ) j Γ( xn − j + 1)Γ( xn − j + 1) = ( xn ) j . Therefore, lim n →∞ (cid:0) xnj (cid:1)(cid:0) nj (cid:1) = lim n →∞ xnn xn − n − · · · xn − j + 1 n − j + 1 = x j . (cid:3) Proposition 6.2.
Let x ∈ R and p ∈ $ m for some m . The dual functionals converge to pointevaluation in the following sense lim n →∞ λ nxn p = p ( x ) . In particular, with n = mk and x = im , lim k →∞ λ kmik p = p ( im ) . Proof.
Let p ( x ) = P mj =0 α j x j in $ m . Then, p ( j ) (0) = j ! α j if j ≤ m and 0 otherwise, and so by theprevious lemma we havelim n →∞ λ nxn p = lim n →∞ ⌊| xn |⌋ X j =0 (cid:0) xnj (cid:1)(cid:0) nj (cid:1) j ! p ( j ) (0) = lim n →∞ m X j =0 (cid:0) xnj (cid:1)(cid:0) nj (cid:1) α j = m X j =0 α j lim n →∞ (cid:0) xnj (cid:1)(cid:0) nj (cid:1) = m X j =0 α j x j = p ( x ) . (cid:3) Corollary 6.3.
Let s = 0 : k : km and D mk = B m A k with A k = E ( s, :) − . Then, D mk → L m with L m the Lagrange basis. In figure 6.1, we display the dual bases for various degree polynomial spaces and level of refinementto illustrate convergence to the Lagrange basis.
Figure 6.1.
Dual Bases D mk of degree m = 2, 3 and 4, for k = 1 (solid) and 10(dotted), respectively. Note that k = 1 is the Bernstein basis, and k = 10 is close tothe Lagrange basis. Rate of Convergence of Symmetric Configuration to Lagrange Interpolation
In this section we determine the rate of convergence for this symmetric configuration to Lagrangeinterpolation, which moreover provides an alternate proof or convergence to Lagrange interpolation.Recall that D mk = B m A k is the basis for $ m that is dual to Λ mk ( s k ) with s ki = ik for i = 0 : m , and A k = E ( s k , :) − with E = (Λ mk ) T B m . Let L m be the Lagrange basis for $ m dual to point evaluationat im for i = 0 : m , and let A be the matrix such that L m = B m A . Then, we have the following: Lemma 7.1.
For h small: (1) ( α − ℓh ) · · · ( α − nh ) = α n − ℓ +1 − α n − ℓ ( n + ℓ )( n + 1 − ℓ ) h + O ( h ) . (2) a − bh + O ( h ) c − dh + O ( h ) = ac + ad − bcc h + O ( h ) .Proof. Part (2) follows by a Maclaurin’s expansion. For (1), there are n − ℓ + 1 terms. On expandingin powers of h , we have( α − ℓh ) · · · ( α − nh ) = α n − ℓ +1 − α n − ℓ ( ℓ + · · · + n ) h + O ( h )= α n − ℓ +1 − α n − ℓ h(cid:18) n + 12 (cid:19) − (cid:18) ℓ (cid:19)i h + O ( h )= α n − ℓ +1 − α n − ℓ ( n + 1) n − ℓ ( ℓ − h + O ( h )= α n − ℓ +1 − α n − ℓ ( n + ℓ )( n + 1 − ℓ ) h + O ( h ) . (cid:3) Lemma 7.2.
For the symmetric configuration, A − ( i, j ) − A − k ( i, j ) = 0 if i = 0 or m . Otherwise, A − ( i, j ) − A − k ( i, j ) = C ij k + O ( 1 k ) as k → ∞ , with C ij = 12 B mj (cid:16) im (cid:17)h ( j − j ( m − i ) im | {z } if j> + ( m − j )(2 mj − im + i − ij ) m ( m − i ) | {z } if j The Lagrange basis is dual to the point evaluation map ∆ := [ δ im : i = 0 : m ], giving I = ∆ T L m = ∆ T B m A . Therefore, A − = ∆ T B m = [ B mj ( im )] . We also have A − k = E k ( s k , :) with E the degree elevation matrix. Hence, A − ( i, j ) − A − k ( i, j ) = B mj ( im ) − E ( ik, j ) = B mj ( im ) − (cid:0) mk − mik − j (cid:1)(cid:0) mj (cid:1)(cid:0) mkik (cid:1) . UAL BASES FUNCTIONS IN SUBSPACES 13 It is easily checked that this vanishes for i = 0 or i = m (provided k > < i < m . In this case, ik > j and mk − m ≥ ik − j for k ≥ m , which, since k → ∞ , wecan assume as well. Factoring out (cid:0) mj (cid:1) , we can rewrite the second term as1 (cid:0) mj (cid:1) E ( ik, j ) = (cid:0) mk − mik − j (cid:1)(cid:0) mkik (cid:1) = ( ik )!( ik − j )! ( mk − m )!( mk )! ( mk − ik )!( mk − ik − m + j )!= ( ik ) · · · ( ik − j + 1)( mk ) · · · ( mk − j + 1) · ( mk − ik ) · · · ( mk − ik − m + j + 1)( mk − j ) · · · ( mk − m + 1)= i · · · ( i − j − k ) m · · · ( m − j − k ) | {z } appears if j> · ( m − i ) · · · ( m − i − m − j − k )( m − jk ) · · · ( m − m − k ) | {z } appears if j 0, we have i · · · ( i − j − k ) m · · · ( m − j − k ) = i j − i j − ( j − j k + O ( k ) m j − m j − ( j − j k + O ( k )= i j m j + i j m j − ( j − j − m j i j − ( j − j ( m j ) k + O ( 1 k )= (cid:16) im (cid:17) j − (cid:16) im (cid:17) j ( j − j ( m − i ) im k + O ( 1 k )= (cid:16) im (cid:17) j h − 12 ( j − j ( m − i ) im k i + O ( 1 k ) . Likewise, when j < m ,( m − i ) · · · ( m − i − m − j − k )( m − jk ) · · · ( m − m − k ) = ( m − i ) m − j − ( m − i ) m − j − ( m − j − m − j ) k + O ( k ) m m − j − m m − j − ( m + j − m − j ) k + O ( k )= (cid:16) m − im (cid:17) m − j + ( m − i ) m − j m m − j − ( m + j − m − j ) − m m − j ( m − i ) m − j − ( m − j − m − j ) m m − j ) k + O ( 1 k )= (cid:16) − im (cid:17) m − j + 12 h(cid:0) m − im (cid:1) m − j ( m + j − m − j ) m − (cid:0) m − im (cid:1) m − j ( m − j )( m − j − m − i i k + O ( 1 k )= (cid:16) − im (cid:17) m − j + 12 (cid:0) m − im (cid:1) m − j ( m − j ) h m + j − m − m − j − m − i i k + O ( 1 k )= (cid:16) − im (cid:17) m − j h − 12 ( m − j )(2 mj − im + i − ij ) m ( m − i ) 1 k i + O ( 1 k ) . Multiplying these two terms together and by (cid:0) mj (cid:1) gives us, when 0 < j < m , E ( ik, j ) = (cid:18) mj (cid:19) i · · · ( i − j − k ) m · · · ( m − j − k ) · ( m − i ) · · · ( m − i − m − j − k )( m − jk ) · · · ( m − m − k )= B mj (cid:16) im (cid:17)h − 12 ( j − j ( m − i ) im k ih − 12 ( m − j )(2 mj − im + i − ij ) m ( m − i ) 1 k i + O ( 1 k )= B mj (cid:16) im (cid:17) − B mj (cid:16) im (cid:17)h ( j − j ( m − i ) im + ( m − j )(2 mj − im + i − ij ) m ( m − i ) i k + O ( 1 k ) . For j = 0, E ( ik, 0) = B m (cid:16) im (cid:17) + 12 B m (cid:16) im (cid:17) ( m − im − i k + O ( 1 k ) . For j = m , E ( ik, m ) = B mm (cid:16) im (cid:17) − B mm (cid:16) im (cid:17) ( m − m − i ) i k + O ( 1 k ) . Putting this all together, we get: A − ( i, j ) − A − k ( i, j ) = C ij k + O ( 1 k )with C ij = 12 B mj (cid:16) im (cid:17)h ( j − j ( m − i ) im | {z } if j> + ( m − j )(2 mj − im + i − ij ) m ( m − i ) | {z } if j Let p = L m α and p k = D m,k α with L m = B m A the Lagrange basis and D m,k = B m A k with A k = E ( s k , :) − the dual basis for the symmetric configuration given in the previous section.Then, || p − p k || [0 , = (cid:16) || A || ∞ || C || ∞ k + O ( 1 k ) (cid:17) || α || , and || L i ( t ) − D m,ki || [0 , = || A || ∞ || C || ∞ k + O ( 1 k ) . Proof. From A k − A = A ( A − − A − k ) A k , we get || A k − A || ≤ || A || || A − − A − k || || A k || . Since, as we proved in the previous section, that D m,k converges to the Lagrange basis L m , we knowthat A k → A . Hence, lim sup k || A k || ∞ || | | A ∞ . We also know that || A − − A − k || ∞ = || C || ∞ k + O ( 1 k ) . Therefore, || A k − A || ∞ = || A || ∞ || C || ∞ k + O ( 1 k . Now, let p = L m α = B m Aα and p k = D m,k α = B m A k α for α ∈ R m +1 . Then, || p − p k || [0 , = || B m ( A − A k ) α || [0 , = max t ∈ [0 , || B m ( t )( A − A k ) α ||≤ max t ∈ [0 , || B m ( t ) || ∞ || A − A k || ∞ || α || ∞ = || A − A k || ∞ || α || ∞ = (cid:16) || A || ∞ || C || ∞ k + O ( 1 k ) (cid:17) || α || ∞ . UAL BASES FUNCTIONS IN SUBSPACES 15 In particular, for α = e i the standard unit vector with 1 in the i-th slot, we get || L i ( t ) − D m,ki || [0 , = || A || ∞ || C || ∞ k + O ( 1 k ) . (cid:3) Basis Transformations for Symmetric Class Recall that for the symmetric configuration introduced in the previous section the dual basis isrepresented in terms of the Bernstein basis D m,k = B m A m,k for some m × m matrices A m,k . It seemsrather challenging to explicitly characterize all these transformation matrices for arbitrary m and k ,however, for the convenience of the reader we’ll list out the first several here. To make things morecompact, we define the following notation: km := k − m , nkm := nk − m , pk nkm = pk − nk + m ,and qk pk nkm = qk − pk + nk − m . A ,k = 12! k k − ( k 1) 2(2 k − ( k k .A ,k = 12 · k k − k k 1) 6(2 k k − k k 1) 2( k k k k − k k 1) 6(2 k k − k k k .A ,k = 13 · k k − k (cid:0) k k (cid:1) k k k − k k k 1) 12( k k k − k k k k (cid:0) k k (cid:1) − k k k 1) 24(4 k (cid:0) k k (cid:1) − k k k 1) 4( k (cid:0) k k (cid:1) − k k k 1) 12( k k k − k k k 1) 12(2 k k k − k (cid:0) k k (cid:1) k .A ,k = 14 · k k − k k (cid:0) k k (cid:1) k k k k − k k k k 1) 40( k k k k − k k k k 1) 4( k k k k k (cid:0) k k k (cid:1) − k k k k 11) 20(5 k (cid:0) k k k (cid:1) − k k k k k k (cid:0) k k (cid:1) − k k (cid:0) k k (cid:1) − k k (cid:0) k k (cid:1) k k (cid:0) k k (cid:1) − k k k k 11) 20(5 k (cid:0) k k k (cid:1) − k k k k 11) 2( k (cid:0) k k k (cid:1) k k k k − k k k k 1) 40( k k k k − k k k k k k k k − k k (cid:0) k k (cid:1) k . In particular, for m = 2 and k = 2 : 5: − − 10 0 4 , − − 20 0 6 , − − 30 0 8 , 10 0 0 − − 40 0 10 . For m = 3 and k = 2 : 5: 48 0 0 0 − 18 90 − 30 66 − 30 90 − 180 0 0 48 , 108 0 0 0 − 56 240 − 96 2020 − 96 240 − 560 0 0 108 , 192 0 0 0 − 114 462 − 198 4242 − 198 462 − , 300 0 0 0 − 192 756 − 336 7272 − 336 756 − . For m = 4 and k = 2 : 4: 576 0 0 0 0 − 261 1260 − 630 252 − − 672 1680 − 672 120 − 45 252 − 630 1260 − , − − − − − − 240 1320 − − , − − − − − − 693 3780 − − . For m = 5 and k = 2 : 3: − − − − − − − − − − − − , − − − − − − − − − − − − Quasi-Interpolation In this section we derive some approximation results for our dual basis functions, similar to whatis done in [7] in the multivariate setting. To do so, we will redefine the Bernstein basis and dualfunctionals over a general interval. • Let B n with B ni = (cid:18) ni (cid:19)(cid:16) b − · b − a (cid:17) n − i (cid:16) · − ab − a (cid:17) i be the Bernstein basis over [ a, b ], with dual functionals λ nk = k X j =0 (cid:0) kj (cid:1)(cid:0) nj (cid:1) ( b − a ) j j ! δ a D j . Hence, Λ nT B n = I . UAL BASES FUNCTIONS IN SUBSPACES 17 • Let D m be the basis for $ m dual to Λ n ( s ) for some selection s . That is, Λ nT ( s ) D m = I .Then, D m = B m A with A = E ( s, :) − where B n = B m E . • Let L m be the Lagrange basis at the points ξ m = [ a + im ( b − a ) : i = 0 : m ] with dual map∆ m = [ δ ξ m , . . . , δ ξ mm ]. Hence, ∆ mT L m = I . Then, for B m = L m M m with M m = ∆ m T B m , isthe basis transformation. Let ˜Λ m = ∆ m M − Tm .The purpose of the map ˜Λ m is two-fold. First, it is dual to B m , as we show next. Second, it doesnot involve any derivative evaluations, which makes it suitable for the approximation of functionsin C [ a, b ]. Lemma 9.1. ˜Λ n is dual to B n .Proof. ˜Λ nT B n = (cid:0) ∆ n M − T (cid:1) T B n = M − ∆ nT B n = M − M = I. (cid:3) Theorem 9.2. (Stability) Let p = D m α ∈ $ m . Then, || M − m || ∞ || α || ∞ ≤ || p || [ a,b ] ≤ || A || ∞ || α || ∞ . Proof. For the upper bound: || p || [0 , = || D m α || [0 , = || B m Aα || [0 , = || m X i =0 ( Aα ) i B mi || [0 , ≤ || Aα || ∞ || m X i =0 B mi || [0 , = || Aα || ∞ ≤ || A || ∞ || α || ∞ . For the lower bound, note that∆ mT p = ∆ mT D m α = ∆ mT B m Aα = M m Aα, and so α = A − M − m ∆ mT p. Then, || α || ∞ ≤ || A − || ∞ || M − m || ∞ || ∆ mT p || ∞ ≤ || A − || ∞ || M − m || ∞ || p || [0 , . Since A − = E ( s, :) is row-affine, || A − || ∞ = 1, and so we get1 || M − m || ∞ || α || ∞ ≤ || p || [0 , . (cid:3) For this we recall from Proposition 3.6 that the dual basis is unaffected by which map is used.Hence, we restate this fact in the following lemma: Lemma 9.3. Suppose that data maps Λ n and ˜Λ n are both dual to the basis B n for $ n . Then, thesedata maps are equivalent on $ n , and dual bases D m of $ m for m < n are identical for both maps. Lemma 9.4. For f ∈ C ([ a, b ]) , | ˜ λ nj f | ≤ || f || [ a,b ] || M − || ∞ . Proof. Let C := M − T . Then, ˜Λ m = ∆ C , and so | ˜ λ mj f | = | ∆ C (: , j ) f | = | n X i =0 C ( i, j ) δ in f | = | n X i =0 C ( i, j ) f ( in ) |≤ || f || [ a,b ] n X i =0 | C ( i, j ) | ≤ || f || [ a,b ] || M − T || = || f || [ a,b ] || M − || ∞ . (cid:3) For any selection s , let Q s : C ([ a, b ]) → R : f D m ˜Λ n ( s ) T f. Then, we have following: Lemma 9.5. Q s is a linear projector of C ([ a, b ]) onto $ m .Proof. Linearity is immediate from Q s ( αf + βg ) = D m ˜Λ n ( s ) T ( αf + βg ) = αD m ˜Λ n ( s ) T f + βD m ˜Λ n ( s ) T g = αQ s f + βQ s g, and idempotency follows from Q s = ( D m ˜Λ n ( s ) T ) = D m (cid:16) ˜Λ n ( s ) T D m (cid:17) ˜Λ n ( s ) T = D m ˜Λ n ( s ) T = Q s . since ˜Λ n ( s ) T D m = I . Therefore, Q s is a linear projector. (cid:3) Theorem 9.6. Let f ∈ C [ a, b ] . Then, • || Q s f || [ a,b ] ≤ || A || ∞ || M − m || ∞ || f || [ a,b ] , • || f − Q s f || [ a,b ] ≤ (1 + || Q s || ) d ( f, $ m ) [ a,b ] .Proof. For t ∈ [ a, b ], | Q s f ( t ) | = | ( D m ˜Λ n ( s ) T f )( t ) | = | ( B m A ˜Λ n ( s ) T f )( t ) |≤ || A || ∞ || Λ n ( s ) T f || ∞ m X i =0 B mi ( t ) = || A || ∞ || Λ n ( s ) T f || ∞ ≤ || A || ∞ || M − || ∞ || f || [ a,b ] . UAL BASES FUNCTIONS IN SUBSPACES 19 This establishes the first result. For the second result, take an arbitrary p ∈ $ m . Then, || f − Q s f || [ a,b ] ≤ || f − p || [ a,b ] + || p − Q s p || [ a,b ] | {z } + || Q s ( p − f ) || [ a,b ] ≤ (1 + || Q s || [ a,b ] ) || f − p || [ a,b ] ≤ (1 + || Q s || [ a,b ] ) d ( f, $ m ) [ a,b ] . (cid:3) Bernstein-like Operator Let D m f := D m ∆ nT ( s ) f = m X i =0 f ( ξ ns ( i ) ) D mi be a Bernstein-like operator for our Dual functions, with ξ nj := a + jn ( b − a ).Let || f || k, [ a,b ] := max {| f ( k ) ( x ) | : x ∈ [ a, b ] } , with || f || [ a,b ] := || f || , [ a,b ] , and ω ( f, h ) := max {| f ( x ) − f ( y ) | : x, y ∈ [ a, b ] , | x − y | ≤ h } , the uniform modulus of continuity relative to the interval [ a, b ]. Then, we have the following: Theorem 10.1. || f − D m f || [ a,b ] ≤ || A || ∞ ω ( f, b − a ) , f ∈ C ([ a, b ]);( b − a ) || A || ∞ || f ′ || [ a,b ] , f ∈ C ([ a, b ]); ( b − a ) || A || ∞ || f ′′ || [ a,b ] , f ∈ C ([ a, b ]) . Proof. By affineness of the basis D m = B m A and convexity of B m for all x ∈ [ a, b ], | f ( x ) − D m f ( x ) | = | m X i =0 ( f ( x ) − f ( ξ ns ( i ) ) D mi ( x ) |≤ ω ( f, b − a ) m X i =0 | D mi ( x ) | = ω ( f, b − a ) | m X j =0 B mj ( x ) m X i =0 A ( j, i ) |≤ ω ( f, b − a ) || A || ∞ m X j =0 | B mj ( x ) | = || A || ∞ ω ( f, b − a ) . This gives the first case. The second case follows from the first case and the estimate ω ( f, h ) = max | x − y |≤ h | f ( x ) − f ( y ) | = max | x − y |≤ h | f ( x ) − f ( y ) || x − y | | x − y |≤ max | x − y |≤ h | f ( x ) − f ( y ) || x − y | h ≤ h || f ′ || [ a,b ] . Then, for some η between ξ ns ( i ) and x , | f ( x ) − D m f ( x ) | = | f ( x ) − m X i =0 f ( ξ ns ( i ) ) D mi | = | f ( x ) − m X i =0 h f ( x ) + f ′ ( x )( ξ ns ( i ) − x ) + 12 f ′′ ( η )( ξ ns ( i ) − x ) i D mi | = | f ( x ) − m X i =0 f ( x ) D mi + m X i =0 f ′ ( x )( ξ ns ( i ) − x ) D mi + 12 m X i =0 f ′′ ( η )( ξ ns ( i ) − x ) D mi | = | m X i =0 f ′ ( x )( ξ ns ( i ) − x ) D mi + 12 m X i =0 f ′′ ( η )( ξ ns ( i ) − x ) D mi | = | f ′ ( x ) m X i =0 ξ ns ( i ) D mi − f ′ ( x ) x + 12 m X i =0 f ′′ ( η )( ξ ns ( i ) − x ) D mi | By Theorem 4.3 (tranformed to the interval [ a, b ]), P mi =0 ξ ns ( i ) D mi = x , and so | f ( x ) − D m f ( x ) | = | f ′ ( x ) x − f ′ ( x ) x + 12 m X i =0 f ′′ ( η )( ξ ns ( i ) − x ) D mi | = | m X i =0 f ′′ ( η )( ξ ns ( i ) − x ) D mi |≤ || f ′′ || [ a,b ] ( b − a ) m X i =0 | D mi ( x ) |≤ || f ′′ || [ a,b ] ( b − a ) || A || . (cid:3) References 1. G. Farin, Curves and Surfaces for Computer Aided Geometric Design, 1988.2. R. Goldman, Dual Polynomial Bases, JAT 79(3), 1994, 311–346.3. S. Kersey, Invertibility of submatrices of Pascal’s matrix, arXiv:1303.6159 (2013). UAL BASES FUNCTIONS IN SUBSPACES 21 4. S. Kersey, Dual basis functions in subspaces of inner product spaces, AMC 219 (2013) 10012–10024.5. Y. Katznelson and Y. Katznelson, A (Terse) Introduction to Linear Algebra, AMS (Student MathematicalLibrary) (44), (2008).6. H. Prautzsch, W. Boehm and M. Paluszny, B´ezier and B-spline techniques, Springer (New York) (2002).7. L. Schumaker, M. Lai, Spline Functions on Triangulations, Encylopedia of Mathematics and its Applications,Volume 110, Cambridge University Press (Cambridge, UK), (2007). Current address : Department of Mathematical Science, Georgia Southern University, Statesboro, GA 30460-8093 E-mail address ::