On the Eigenstructure of the $(\Alpha, Q) $-Bernstein Operator

$On the Eigenstructure of the $(\Alpha, Q) $-Bernstein Operator$

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2 Preliminaries

The deﬁnitions and notations used in this article are adopted from [1, Ch.10]. Let q > 0. The q-integer is deﬁned by

2 n−1 [n]q := 1 + q + q + ··· + q , [0]q := 0 (n = 1, 2,... ), (2.1) the q-factorial of n by

[n]q!:= [1]q[2]q ... [n]q, [0]q!:= 1 (n = 1, 2,... ). From (2.1), one can easily see that

[n]q − [i]q [n − i] = , i = 0, 1,..., n, (2.2) q qi For integers k and n with 0 ≤ k ≤ n, the q− binomial coeﬃcient is n [n] ! = q k q [k]q![n − k]q! The q-shifted products deﬁned by

k−1 ∞ s s (a; q)0 = 1, (a; q)k = (1 − aq ), (a; q)∞ = (1 − aq ). Ys=0 Ys=0

We also need the q-Stirling numbers of the second kind S q(k, r) given by

r 1 i i(i−1) r k = − 2 − S q(k, r) r(r−1)/2 ( 1) q [r i]q. (2.3) [r] ! q i q q Xi=0 Using the induction on k, one can verify that

S q(k + 1, r) = S q(k, r − 1) + [r]qS q(k, r), k ≥ 0, r ≥ 1 (2.4) with S q(0, 0) = 1, S q(k, 0) = 0 for k > 0 and S q(k, r) = 0 for k < r. The α-Bernstein polynomial of f : [0, 1] → R is introduced in [3] as

n i T ( f ; x) = f p(α)(x) (2.5) n,α n n,i Xi=0

2 (α) (α) (α) where pn,i (x) are the α-Bernstein polynomials of degree n given by p1,0(x) = 1 − x, p1,1(x) = x and for n ≥ 2,

n − 2 n − 2 n p(α)(x) = (1 − α)x + (1 − α)(1 − x) + αx(1 − x) xi−1(1 − x)n−1−i. n,i " i ! i − 2! i! #

n n Here i stands for the binomial coeﬃcients and one has i = 0 for i < 0 or i > n. The α-Bernstein operator Tn,α on C[0, 1] is given by

Tn,α : f → Tn,α f

Using the forward diﬀerence operator, (2.5) can also be written as

n n − 1 r n r r Tn,α( f ; x) = (1 − α) ∆ g0 + α ∆ f0 x " r ! r! # Xr=0 where i i i fi = f , gi = 1 − fi + fi+1 n n − 1 n − 1 0 r r−1 r−1 and ∆ fi = fi, ∆ fi = ∆ fi+1 − ∆ fi for r ≥ 1. Lemma 3.1 in [3] states that the higher-order forward diﬀerence of gi can be expressed as

r i r i + r r ∆ gi = 1 − ∆ fi + ∆ fi+1, n − 1 n − 1 The q-analogue of α-Bernstein operators, called (α, q)-Bernstein operators are deﬁned in [18] as Tn,q,α : f → Tn,q,α( f ; .) such that

n [i]q (α) Tn,q,α( f ; x) = f p (x), [n] ! n,q,i Xi=0 q

(α) (α) where pn,q,i(x) are the basis (α, q)-Bernstein polynomials of degree n given by p1,q,0(x) = 1 − x, (α) p1,q,1(x) = x and

n − 2 (1 − α)x n − 2 n (α) = + − n−i−2 + i−1 pn,q,i(x) n−i−1 (1 α)q αx x (x; q)n−i. " i #q 1 − q x " i − 2 #q i q Like α-Bernstein polynomials, the (α, q)-Bernstein polynomials have the representation using the forward diﬀerence operators as

n n − 1 r n r r Tn,q,α( f ; x) = (1 − α) ∆qg0 + α ∆q f0 x (2.6) " r # r q Xr=0 q

3 0 r r−1 r−1 r−1 where ∆q fi = fi, ∆q fi = ∆q fi+1 − q ∆q fi for r ≥ 1. Also, the higher-order forward diﬀerence of gi can be expressed as, see [18, Lemma 2.4],

n−i−1 n−i−1−r r q [i]q r q [i + r]q r ∆qgi = 1 − ∆q fi + ∆q fi+1. (2.7) [n − 1]q ! [n − 1]q It is worth mentioning that when α = 1, the α-Bernstein and (α, q)-Bernstein polynomials become Bernstein and q-Bernstein polynomials, respectively.

3 Main Results

Since the α-Bernstein polynomials are obtained from (α, q)-Bernstein polynomials as q → 1, we will present only the results for the latter one. Similar results can be derived for α-Bernstein polynomials by taking the limit as q → 1. It is knownthat (α, q)-Bernstein polynomials possess some properties of q-Bernstein polynomials. For example, see [3] and [18], they have the end-point interpolation property:

Tn,q,α( f ; 0) = f (0), Tn,q,α( f ; 1) = f (1), n = 1, 2,..., q > 0, and leave the functions invariant:

Tn,q,α(at + b; x) = ax + b, n = 1, 2,..., q > 0.

k Moreover, they are degree reducing on polynomials, that is, Tn,q,α(t ; x) is a polynomial of degree min{n, k}. This implies that, for k ≤ n, the subspace Pk of polynomials of degree at most k is invariant under Tn,q,α. To be speciﬁc, it is known (see [18]) that

k k k−1 Tn,q,α(t ; x) = ak x + ak−1 x + ··· + a1 x, where k(k−1)/2 q [n − 2]q! a = (1 − α)[n − k] [n − 1 + k] + α[n] [n − 1] . k [n − k] ![n]k q q q q q q n o The above representation gives only an explicit formula for the leading coeﬃcient. However, in our study, we need all coeﬃcients explicitly which are given in the following lemma. Lemma 3.1. Let f (t) = tk. Then, for r ≤ k andi = 0, 1,..., n, one has

1 r r ∆r = − s s(s−1)/2 + − k . q fi k ( 1) q [i r s]q (3.1) [n] s q q Xs=0 Proof. Take f (t) = tk in [17, formula (2.1)].

4 Lemma 3.2. For n ≥ k ≥ 1, one has

k k r Tn,q,α(t ; x) = a(r, k)x Xr=0 where

r(r−1) q 2 [n − 2] ! = q − − + − + + a(r, k) k (1 α)[n r]q [n r 1]qS q(k 1, r 1) [n]q[n − r]q!

− [r + 1]q[n − 1]qS q(k, r + 1) + α[n]q[n − 1]qS q(k, r) . (3.2) Proof. From (2.6) one has

n − 1 r n r a(r, k) = (1 − α) ∆qg0 + α ∆q f0. " r #q r q For i = 0, (3.1) becomes

1 r r ∆r = − s s(s−1)/2 − k, q f0 k ( 1) q [r s]q [n] s q q Xs=0 and using (2.3), we get

r(r−1)/2 r [r]q!q ∆q f0 = S q(k, r). [n]q!

r+1 r r r As ∆q f0 = ∆q f1 − q ∆q f0, we have

r r+1 r r ∆q f1 = ∆q f0 + q ∆q f0 [r] !qr(r+1)/2 = q + , + + , k ([r 1]qS q(k r 1) S q(k r)) [n]q [r] !qr(r+1)/2 = q + , + k S q(k 1 r 1) [n]q Also, i = 0 in (2.7) results in

n−1−r r r q [r]q r ∆qg0 = ∆q f0 + ∆q f1 [n − 1]q r(r−1) n−1 [r] !q 2 q [r] = q + q + + k S q(k, r) S q(k 1, r 1) [n]q [n − 1]q

5 Therefore,

r(r−1) n−1 [r] !q 2 n − 1 q [r] n = q − + q + + + a(r, k) k (1 α) S q(k, r) S q(k 1, r 1) α S q(k, r) [n]q " r #q [n − 1]q ! r q r(r−1) n−1 q 2 [n − 1] ! q [r] [n] ! = − q + q + + + q k (1 α) S q(k, r) S q(k 1, r 1) α S q(k, r) . [n]q [n − r − 1]q! [n − 1]q ! [n − r]q! Using (2.4), the last equality becomes

r(r−1) q 2 [n − 2] ! = q − − − + + − + + a(r, k) k (1 α)[n 1]q[n r]q S q(k 1, r 1) [r 1]qS q(k, r 1) [n]q[n − r]q! n−1 q [r]q + S q(k + 1, r + 1) + α[n]q[n − 1]qS q(k, r) [n − 1]q r(r−1) q 2 [n − 2] ! = q − α − − + n−1 + , + k (1 )[n r]q [n 1]q q [r]q S q(k 1 r 1) [n] [n − r]q! q − [r + 1]q[n − 1]qS q(k, r + 1) + α[n]q[n − 1]qS q(k, r) r(r−1) q 2 [n − 2] ! = q − α − + − + , + k (1 )[n r]q [n r 1]qS q(k 1 r 1) [n]q[n − r]q!

− [r + 1]q[n − 1]qS q(k, r + 1) + α[n]q[n − 1]qS q(k, r) which is (3.2) as claimed. Lemma 3.3. The numbers

k(k−1) q 2 [n − 2]q! λ(α,n) = ( 1 − α)[n − k] [n − 1 + k] + α[n] [n − 1] k,q [n − k] ![n]k q q q q q q are distinct for α ∈ [0, 1] and k = 2,..., n. Proof. One can write

k−1 [n − k]q[n + k − 1]q [m]q λ(α,n) = α + (1 − α) 1 − . (3.3) k,q [n] [n − 1] ! [n] ! q q Ym=1 q

(α,n) Dividing (3.3) by λk− j,q for j = 1, 2,..., k − 1, we get

(α,n) λ (1 − α)[n − k] [n + k − 1] + α[n] [n − 1] k−1 [m] k,q = q q q q − q (α,n) 1 . (1 − α)[n − k + j]q[n + k − j − 1]q + α[n]q[n − 1]q ! [n]q ! λk− j,q mY=k− j

6 Obviously k−1 [m]q 1 − < 1. [n] ! mY=k− j q To complete the proof, note that

[n − k]q[n + k − 1]q ≤ [n − k + j]q[n + k − j − 1]q ⇔(1 − qn−k)(1 − qn+k−1) ≤ (1 − qn−k+ j)(1 − qn+k− j−1) ⇔(q j − 1)(q2k− j−1 − 1) ≥ 0 for all q > 0 and j = 1, 2,..., k − 1.

(α,n) k Remark 3.1. It is worth mentioning that the numbers λk,q are the leading coeﬃcients of Tn,q,α(t ; x). (α,n) That is, λk,q = a(k, k), and hence

k (α,n) k (α,n) Tn,q,α(t , x) = λk,q x + Pk−1 (x)

(α,n) where Pk−1 (x) is a polynomial of degree k − 1.

4 Eigenvalues and eigenvectors of Tn,q,α

In this part, the eigenvalues and the corresponding eigenvectors of Tn,q,α are found. The coeﬃcients of the eigenvectors are given recursively. For some speciﬁc values of n, the eigenvalues are plotted.

Lemma 4.1. For all q > 0 and α ∈ [0, 1], the operator Tn,q,α has n + 1 linearly independent monic (α,n) (α,n) (α,n) eigenvectors pk,q (x) of degree k = 0, 1,..., n corresponding to the eigenvalues λ0,q = λ1,q = 1 and

k(k−1) q 2 [n − 2]q! λ(α,n) = ( 1 − α)[n − k] [n − 1 + k] + α[n] [n − 1] k,q [n − k] ![n]k q q q q q q for k = 2, 3,....

Proof. The proof is clear for k = 0 and k = 1. For k = 2, 3,..., we have

k (α,n) k (α,n) Tn,q,α(t ; x) = λk,q x + Pk−1 (x),

(α,n) where Pk−1 (x) is a polynomial of degree k − 1. Let

(α,n) k k−1 pk (x) = x + βk−1 x + ... + β1 x

7 (α,n) stand for the monic eigenvector of Tn,q,α corresponding to λk,q , that is,

(α,n) (α,n) (α,n) Tn,q,α(pk (t); x) = λk,q pk (x).

Since Tn,q,α is linear, this equality becomes

k k−1 (α,n) k k−1 Tn,q,α(t ; x) + βk−1Tn,q,α(t ; x) + ··· + β1Tn,q,α(t; x) = λm (x + βk−1 x + ··· + β1 x).

m Comparing the coeﬃcients of x , m = 1, 2,..., k − 1, we get the system in the unknowns β1,β2,..., βk−1, whose coeﬃcient matrix is

(α,n) (α,n) λk,q − λk−1,q 0 0 ··· 0 (α,n) (α,n)  ∗ λk,q − λk−2,q 0 ··· 0  Λ =  .  .  ∗ ∗ λ(α,n) − λ(α,n) .. 0   k,q k−3,q   (α,n) (α,n)  ∗ ∗ ∗ ... λ − λ   k,q 1,q    Clearly (α,n) (α,n) (α,n) (α,n) (α,n) (α,n) , det(Λ) = (λk,q − λk−1,q)(λk,q − λk−2,q) ··· (λk,q − λ1,q ) 0 (α,n) by Lemma 3.3. Thus, there exist unique numbers β1,...,βk−1 and hence pk (x) is an eigenvector of (α,n) Tn,q,α corresponding to the eigenvalue λk,q .

(α,n) (α,n) Theorem 4.2. The monic eigenvector pk,q (x) of Tn,q,α associated with λk,q is a polynomial of degree k given by

k (α,n) j pk,q (x) = cn,q( j, k)x , Xj=0

(α,n) (α,n) where p0,q (x) = 1, p1,q (x) = x and

j− 1 1 − = − − − cn,q(k j, k) (α,n) (α,n) cn,q(k i, k)an,q(k j, k i) λk,q − λk− j,q Xi=0 for j = 1, 2,..., k, k = 2, 3,....

(α,n) Proof. Let us write the eigenvector pk,q (x) of Tn,q,α in the form

k (α,n) r pk,q (x) = cn,q(r, k)x . Xr=0

8 By the assumption that cn,q(k, k) = 1, the relation

(α,n) (α,n) (α,n) T(pk,q ; x) = λk,q pk,q (x) implies

k k (α,n) s r λk,q cn,q(s, k)x = cn,q(r, k)T(t ; x) Xs=0 Xr=0 k r i = cn,q(r, k) an,q(i, r)x Xr=0 Xi=0 k k s = cn,q(r, k)an,q(s, r)x , Xs=0 Xr=s which leads to

k (α,n) λk,q cn,q(s, k) = cn,q(r, k)an,q(s, r). (4.1) Xr=s Substituting s = k − j and r = k − i on (4.1), we obtain

j (α,n) λk,q cn,q(k − j, k) = cn,q(k − i, k)an,q(k − j, k − i) Xi=0 j−1

= cn,q(k − j, k)an,q(k − j, k − j) + cn,q(k − i, k)an,q(k − j, k − i). (4.2) Xi=0 It is seen from (4.2) that

j− 1 1 − = − − − cn,q(k j, k) (α,n) (α,n) cn,q(k i, k)an,q(k j, k i). λk,q − λk− j,q Xi=0 The proof of Theorem 4.2 is completed

(α,n) (α,n) Example 4.1. As mentioned before p0,q (x) = 1 and p1,q (x) = x for all n. Also, by the endpoint (α,n) 2 interpolation, one can easily derive that p2,q (x) = x − x for all n. For n = 3 and k = 3, one has

(α,3) 3 2 p3,q (x) = x + a2 x + a1 x where (1 − α)q4 + (2 − α)q3 + 3q2 + (2α + 1)q + 2 a = − 2 (1 − α)q4 + q3 + 2q2 + (1 + α)q + 1

9 (1 − α)q3 + q2 + αq + 1 a = . 1 (1 − α)q4 + q3 + 2q2 + (1 + α)q + 1

(α,3) The graph of p3,q (x) for α = 0.4 and various values q is plotted on Figure 1. Also, the graph of (α,3) p3,q (x) for ﬁxed q and variable α is depicted on Figure 2.

0.15

0.1

0.05

-0.05 0 0.2 0.4 0.6 0.8 1

(α,3) Figure 1: The eigenvector p3,q (x) for α = 0.4 and diﬀerent values of q.

4.1 The limit behavior of the eigenvalues and eigenvectors of Tn,q,α Lemma 4.3. For q ∈ (0, 1) and k = 2, 3,..., one has:

r(r−1) k−r (i) lim an,q(r, k) = q 2 (1 − q) S q(k, r), r = 0, 1,..., k n→∞

(ii) lim λ(α,n) = qk(k−1)/2. n→∞ k,q

Proof. The proof follows, immediately, from the fact that limn→∞[n]q = 1/(1−q), for all q ∈ (0, 1).

10 0.08

0.06

0.04

0.02

-0.02

-0.04 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

(α,3) Figure 2: The eigenvector p3,q (x) for q = 0.6 diﬀerent values of α.

(α,n) Theorem 4.4. Let 0 < q < 1 and pk,q (x), k = 0, 1,..., n be the monic eigenvectors of Tn,q,α given in Theorem 4.2. Then

lim cn,q( j, k) = bq( j, k) n→∞ holds for 0 ≤ j ≤ k, k = 0, 1,... , where 1, j = k,  0, j = k − 1 = 0, bq( j, k) =  k − i− j  (1 q) S q(i, k)  bq(i, k), otherwise.  q(k− j)(k+ j−1)/2 − 1  iX= j+1  In other words, when q ∈ (0,1), we have

k (α,n) j lim p (x) = pk(x) = bq( j, k)x n→∞ k,q Xj=0 uniformly on [0, 1].

11 (n) (n) Proof. p0 (x) = p0(x) = 1 and p1 (x) = p1(x) = x, by Theorem 4.2. So it is enough to consider the case k ≥ 2. Suppose that limn→∞ cn,q(k − i, k) = bq(k − i, k), i = 0,..., j − 1, where 0 ≤ j ≤ k. One can easily see that

j−i an,q(k − j, k − i) (1 − q) S q(k − i, k − j) lim = . n→∞ (α,n) (α,n) q j(2k− j−1)/2 − 1 λk,q − λk− j,q Then

j−1 j−i (1 − q) S q(k − i, k − j) lim cn,q(k − j, k) = cn,q(k − i, k). n→∞ q j(2k− j−1)/2 − 1 Xi=0 Substituting j by k − j and i by k − i, we obtain

k (1 − q)i− jS (i, k) b ( j, k) = q b (i, k) q q(k− j)(k+ j−1)/2 − 1 q iX= j+1 which completes the proof. Lemma 4.5. For q > 1, one has: (i) lim λ(α,n) = 1. n→∞ k,q

an,q(k − j, k − i) (ii) lim = 0 for i < j − 1, j = 1, 2,..., k − 1. n→∞ (α,n) (α,n) λk,q − λk− j,q (iii) a (k − j, k − j + 1) S (k − j + 1, k − j) + (1 − α)q j−k[k − j] [k − j + 1] n,q = − q q q lim (α,n) (α,n) n→∞ [k − 1]q + [k − 2]q + ··· + [k − j]q λk,q − λk− j,q for j = 1, 2,..., k − 1. Proof. The proof of (i) is obvious. For the proof of (ii) and (iii), one can write

r(r−1) q 2 [n] ! [n − r] [n − r + 1] = q − q q + + an,q(r, k) k (1 α) S q(k 1, r 1) [n]q[n − r]q! [n]q[n − 1]q [n − r]q − (1 − α) [r + 1]qS q(k, r + 1) + αS q(k, r) [n]q and

k(k−1) q 2 [n] ! [n − k] [n + k − 1] (α,n) = q − q q + λk,q k (1 α) α [n]q[n − k]q! [n]q[n − 1]q

12 then

(k− j)(k− j−1) q 2 [n]q! A a (k − j, k − i) [n]k−i[n−k+ j] ! n n,q = q q (n,α) (n,α) k(k−1) (k− j)(k− j−1) λk,q − λk− j,q q 2 [n]q! q 2 [n]q! k Bn − k− j Cn [n]q[n−k]q! [n]q [n−k+ j]q! i [n] An = q , j(2k− j−1)/2 j [n − k + j]q ··· [n − k + 1]qq Bn − [n]qCn where

[n − k + j]q[n + k − j − 1]q An = (1 − α) S q(k − i + 1, k − j + 1) [n]q[n − 1]q [n − k + j]q − (1 − α) [k − j + 1]qS q(k − i, k − j + 1) + αS q(k − i, k − j), [n]q [n − k]q[n + k − 1]q Bn = (1 − α) + α [n]q[n − 1]q and

[n − k + j]q[n + k − j − 1]q Cn = (1 − α) + α. [n]q[n − 1]q Using (2.2), one can write

i a (k − j, k − i) [n] An n,q = q (α,n) (α,n) j λk,q − λk− j,q ([n]q − [k − j]q)([n]q − [k − j − 1]q) ··· ([n]q − [k − 1]q)Bn − [n]qCn i [n] An = q j j−1 j−2 , [n]q(Bn − Cn) − [n]q ([k − 1]q + [k − 2]q + ··· + [k − j]q)Bn + O([n]q ) Note that

−k+ j lim An = (1 − α) S q(k − i + 1, k − j + 1) − q [k − j + 1]S q(k − i, k − j + 1) n→∞ + αS q(k − i, k − j) −k+ j = S q(k − j + 1, k − j) + (1 − α)q [k − j]q[k − j + 1]q where (2.4) is used. Also, limn→∞ Bn = 1 and limn→∞ Cn = 1. It is obvious now that if i < j − 1, then

an,q(k − j, k − i) lim = 0 n→∞ (α,n) (α,n) λk,q − λk− j,q

13 which is the claim in (ii). Moreover, if i = j − 1, then we have

a (k − j, k − i) S (k − j + 1, k − j) + (1 − α)q−k+ j[k − j] [k − j + 1] n,q = − q q q lim (α,n) (α,n) n→∞ [k − 1]q + [k − 2]q + ··· + [k − j]q λk,q − λk− j,q which completes the proof.

Theorem 4.6. For q > 1 and 0 ≤ j ≤ k, we have

lim cn,q( j, k) = dq( j, k), n→∞ where

k− j −k+i S q(k − i + 1, k − i) + (1 − α)q [k − i]q[k − i + 1]q d (0, 1) = 0, d ( j, k) = − q q [k − 1] + ··· + [k − i] Yi=1 q q Proof. For k = 0 and k = 1, there is nothing to prove. Assume that k ≥ 2, and use strong induction on j. Since cn,q(k, k) = 1, the statement is true for j = k. Assume that limn→∞ cn,q(k − i, k) = dq(k − i, k) for i = 0, 1,..., j − 1. Then using Lemma 4.5 (i) and (ii), we obtain

j−1 an,q(k − j, k − i) lim cn,q(k − j, k) = dq(k − i, k) lim n→∞ n→∞ (α,n) (α,n) Xi=0 λk,q − λk,q −k+ j S q(k − j + 1, k − j) + (1 − α)q [k − j]q[k − j + 1]q = − dq(k − j + 1, k) [k − 1]q + ··· + [k − i]q From the fact that

j−1 −k+i S q(k − i + 1, k − i) + (1 − α)q [k − i]q[k − i + 1]q d (k − j + 1, k) = − q [k − 1] + ··· + [k − i] Yi=1 q q then we get

k− j S (k − i + 1, k − i) + (1 − α)q−k+i[k − i] [k − i + 1] d ( j, k) = − q q q q [k − 1] + ··· + [k − i] Yi=1 q q which completes the the induction. Remark 4.1. Note that when q > 1, the limiting coeﬃcients depend on α unlike the case 0 < q < 1.

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