Extending a Chebyshev Subspace to a Weak Chebyshev Subspace of Higher Dimension and Related Results

Let G={g1,...,gn} be an n-dimensional Chebyshev sub-space of C[a, b] such that 1∉G and U=(u0, u1,...,un) be an (n+1)-dimensional subspace of C[a, b] where u0=1, ui=gi, i=1..... n. Under certain restriction on G, we proved that U is a Chebyshev subspace if and only if it is a Weak Chebyshev subspace. In addition, some other related results are established. Citation: Alyazidi-Asiry M (2017) Extending a Chebyshev Subspace to a Weak Chebyshev Subspace of Higher Dimension and Related Results. J Appl Computat Math 6: 347. doi: 10.4172/2168-9679.1000347


Introduction
The finite set of functions {g 1 ,….,g n } and C[a, b] is called a Chebyshev system on [a, b] if it is linearly independent and for all 1 {x } n j j= such that a ≤ x 1 ,<x 2 <….<x n ≤ b, and the n-dimensional subspace G=[g 1 ,…,g n ] of C[a, b] will be called a Chebyshev subspace [1][2][3][4]. Using the continuity of the determinant, it can be shown that the sign of the determinant is constant [5], so we will assume that the sign of the determinant is always positive through this paper (replace g 1 by -g 1 if necessary). And the finite set of functions {g 1,…. g n} and C[a, b] is called a Weak Chebyshev system on [a, b] if it is linearly independent and for all 1 {x } n j j= such that a ≤ x 1 , < x 2 <….< x n ≤ b and the n-dimensional subspace G={g 1,…… g n} of C[a, b] will be called a weak Chebyshev subspace, C[a, b] is the space of all real-valued continuous functions. Extending an n-dimensional Chebyshev subspace which does not contain a constant function to an (n+1)-dimensional Chebyshev subspace containing a constant function was investigated [4]. In what follows is the statement of the problem considered in this paper: Let G=[g 1,……. g n ] be a Chebyshev subspace of C[a, b] such that 1 G ∉ and U={u 0, u 1,…… u n } be an (n+1)dimensional subspace of C[a, b] where u 0 =1, u i =g i , i=1,….,n [6][7][8]. Our main purpose is to prove that, under certain restriction on G, U is a Chebyshev subspace of C[a, b] if and only if it is a Weak Chebyshev subspace of C[a, b]. An example illustrating that the preceding assertion is not true in general is presented and some related results are given at the end of the last section.

Preliminary
We start this section by the following well known theorem [3,5].

Theorem
For an n-dimensional subspace G of C[a, b], the following statements are equivalent.
(i) G is a Chebyshev subspace.
(ii) Every nontrivial function g ∈ G has at most n-1 distinct zeros in [a, b].
(iii) For all points a=t 0 ≤ t 1 < …. < t n-1 ≤ t n =b, there exists a function g ∈G such that We need the following definitions: and f ∈U such that f(x)=0. We call x an essential zero of f with respect to U, if and only if there is a g ∈U with g(x) ≠ 0.
If no confusion arises, the term "with respect to U "will be omitted.
The following theorem is a version of theorem 1 of Stockenberg [6].

Theorem
Let G be an n-dimensional Weak Chebyshev subspace of C[a, b]. Then the following statements hold.
1. If there is a g ∈ G with n separated, essential zeros a ≤ t 1 <…< t n ≤ b, then g(t)=0 for all t with t ≤ t 1 ≥ t n .
2. No g ∈ G has more than n separated, essential zeros.

The Main Result
We start this section with the following lemma.

Lemma
Let G={g 1 ,…,g n } be an n -dimensional Chebyshev subspace of C[a, b] such that 1 ∉ and U={u 0 ,u 1 ,….,u n } be an (n+1)-dimensional subspace of C[a,b] where u 0 =1, u i =g i , I=1,…..,n. If there are two nontrivial functions h, k∈ U and a set of n points with x< y there is appoint z, x<z<y such v that f (z) ≠ f (x).

Lemma
Let G=(g 1 ,…., g n ) be an n-dimensional Chebyshev subspace of C[a, b] such that 1∈G and U=(u 0 ,u 1 ,…, u n ) be an (n+1)-dimensional subspace of C[a, b] where u 0 =1, u i =g i , i=1,…, n. If G satisfies Assumption A, then the zeros of each nontrivial function h ∈ U are separated and essential.
Proof: Let h be a nontrivial element of U such that h(x)=h(y)=0 for some x, y with a ≤ x < y ≤ b. If h ∈ G, then n ≤ 3, for otherwise h ≡ 0, and since G is an n-dimensional Chebyshev subspace of C[a, b], there is a point z ∈ (x, y) such that h(z) ≠ 0. If h∉ G, then h=α+g, where α ≠ 0 and g∈G, hence g(x)=g(y)=-α, but G satisfies Assumption A, so there is a point z∈ (x, y ) such that g(z) ≠-α that is h (z)≠ 0, this shows that the zeros of h are separated. For the second part of the assertion of the lemma, it is clear that each zero of any nontrivial element of u is an essential zero, that is because 1∈ U.

Remark 1:
Note that if 1 ∉ U and 0 ∉ f ∈U, f(x)=0, then since G is a Chebyshev space, x is an essential zero for f. Indeed, there is an element g ∈G such that g(x) ≠ 0.

Theorem
Let G=(g 1 ,… g n ) be an n-dimensional Chebyshev subspace of C[a, b] such that 1 ∈ G and U=(u 0, u 1,…., u n) be an Proof: One direction is trivial.
For the other direction, suppose U=(u 0 , u 1 , ….. u n ) is an (n+1)dimensional Weak Chesbyshev subspace of C [a, b] where u 0= 1, u i =g i , I=1, n and G=(g 1 ,… g n ) is an n-dimensional Chesbyshev subspace of C[a, b] satisfying Assumption A. Let ū be a nontrivial element of U such that we (x ) 0,...i 1,...d, If d>n+1, then by lemma (2) together with theorem (2) we must have u ≡ 0, so d ≥ n+1, if d≤ n, then is nothing to prove, so to this end, we will assume that d=n+1 and again from lemma (2) and theorem (2)  and there is a point s ∈ (y n , b], such that w(s)=0, so w has at least n distinct zeros in [a,b]. A similar argument as in case A1 shows that u must be identically zero.
Case A3: u(x)< 0 for all x ∈ (x n, b) and v(x)<0 for all x ∈ [x n , y n ), then v(x) > 0 for all x∈ (y n , b], taking w=u-v, we have w(y n )=-v(x n ) > 0 and w (y n )=u (y n )< 0, and continuing exactly as in case A1, we conclude that u must be identically zero.
Case A4: u(x) < 0 for all x ∈ (x n , b) and v(x) > 0 for all x∈ [x n , y n ), then v(x) <0 for all x∈ (y n ,b], taking w=u-v, we have w (y n )=u (y n ) < 0 and w( b)=-v (b) >0 and an argument similar to that of case A2 shows be identically zero.
Case B: a< x 1 and x n+1 =b As in case A, for any q ∈ (x n , b) there is a function such that v(t)=0,then by theorem (2) we must have t=b or t=a.
If t=b, then u and v are two nontrivial elements of U such that U(x n+ 1 )=v(x n+1 )=0, u(x i )=v(x i )=0, I=1, …n-1 and by lemma 1 there is a nonzero constant λ such that u=λ v, this implies that where t i =x i , i=1, …..n, t n+ 1 =y n and t n+2 =x n+1 =b so u has at least n+2 separated zeros in[a, b] which implies that u=v ≡ 0 and this is a contradiction. so t ≠ b and the situation becomes exactly as in case A, proceedings as in case A we conclude that u must be identically zero.
Case C: a=x 1 and x n+1 < b The proof of this case requires that n ≥ 2 and the proof for n=1 will be given in remark (2). Now, for any point p ∈ (a,x 1 ) there is a function such that v(t)=0, then by therome (2)we must have t=a or t=b.
If t=a, then u and v are two nontrivial elements of U such that A similar argument to that of the other cases leads to a contradiction.
So t ≠ a and on the interval [a, x 2 ] we have u(a)=u(x 2 )=0, u(x)≠ 0, x ∈ (a,x 2 ) and v(y 1 )=0, y 1 ∈ (a,x 2 ),v(t) ≠ 0 for every t ∈ [a,x 2 ]\{y 1 }. If x ∈ [a,y 1 ), y ∈ (y 1 ,x 2 ] then sign v(x)=-sign v(y), and as in the other cases we are presented with four different subcases. In each case, a similar argument to that of the cases in A can be used to show that the function u-v in G has at least n zeros which leads to the conclusion that u must be identically zero. Hence U is a Chebyshev subspace of C [a,b].

Remark 2:
The following is the proof for theorem (3) when n=1 which is somehow more direct: This shows that U is not a weak Chebyshev subspace and the theorem is proved.
The followings example illustrates that theorem (3) is not true in general is proved.

Example 1
That is U is a 2-dimensional weak Chebyshev Subspace of C[0,2] but not a Chebyshev Subspace.
If H is n-dimensional subspace of C[a, b], then it is possible that H is a Chebyshev subspace on one of the intervals (a, b] or [a, b) but not on the closed interval [a,b] as illustrated in the following example.

Example 2
Let H=(sin x, cos x), it can be easily checked that H is a Chebyshev subspace of dimension 2 on each of the intervals (0,π] of dimension 2. In next result we give a necessary and sufficient condition for an ndimensional Chebyshev H on (a,b] or [a,b) to be a chebyshev subspace on the closed interval [a,b].

Theorem
Let H be an n-dimensional subspace of C [a, b] on [a, b], there is a non-trivial element u ∈ U such that u(z 1 )≠ 0, u (z i )=0, i=2, …n.
Since H is a Chebyshev subspace on (a, b], there is a subset E={h 1 ,…. h n } and H such that Using a similar argument when I=[a, b) leads to a contradiction and the theorem is proved.