Asymptotic Properties of Derivatives of the Stieltjes Polynomials

Theorem 2.1. Let 0 < λ < 1 and r⩾1 be a positive integer. Then, for all $$,

Moreover, one has and especially one has, for $$,

Theorem 2.2. Let 0 < λ < 1 and r⩾1 be a positive integer. Then, for all $$,

Moreover, one has and especially, for $$,

Theorem 2.3. Let 0 < λ < 1 and r⩾2 be an even integer. For 1 ⩽ μ ⩽ n + 1, one has

Theorem 2.4. Let 0 < λ < 1 and r⩾2 be an even integer. For 1 ⩽ ν ⩽ 2n + 1, one has

Theorem 2.5. Let 0 < λ < 1 and 0 < ɛ < 1. Suppose $$. Then,

• (a)
  $$ ()

• (b)
  $$ ()
  In addition,
  $$ ()

Theorem 2.6. Let 0 < λ < 1 and 0 < ɛ < 1. Suppose $$. Then,

• (a)
  $$ ()

• (b)
  $$ ()
  In addition,
  $$ ()

Theorem 2.7. Let 0 < λ < 1 and 0 < ɛ < 1. Suppose $$. Then, one has, for a positive integer ℓ⩾1,

where c_ℓ = 4^ℓ − 2^ℓ−1.

Proposition 3.1. (a) Let λ > −1/2. Then, P_λ,n(x) satisfies the second-order differential equation as follows:

(b) Let λ > −1/2. Then,

(c) Let λ > −1/2. Then, for 1 ⩽ μ ⩽ n + 1,

(d) Let λ > −1/2. Then, for 1 ⩽ ν ⩽ n,

(e) Let λ > −1/2. Then, for x ∈ (−1,1) and r⩾0,

(f) Let 0 ⩽ λ ⩽ 1. Then, for 1 ⩽ ν ⩽ n,

(g) Let λ > −1/2 and r⩾0. Then, P_λ,n(x) satisfies the higher-order differential equation as follows:

Proof. (a) It is from [9, (4.2.1)]. (b) It is from [9, (4.7.14)]. (c) It is from [6, Lemma 3.4]. (d) It is from [9, (8.9.7)]. (e) For r = 0, it follows from [9, (7.33.5)], and, for r⩾1, it comes from (b) and the case of r = 0. (f) It is from [6, Lemma 3.3 (3.23)]. (g) Equation (3.7) comes from (a).

Proposition 3.2 (see [4].)Let 0 < λ < 1. Let $$, μ = 1, …, n + 1 and $$, ν = 1, …, 2n + 1. Then, for μ = 0,1, …, n + 2 and ν = 0,1, …, 2n + 2,

where $$ and $$.

Proposition 3.3 ([6, Proposition  2.3]). Let 0 < λ < 1. Then, for all x ∈ [−1,1],

where Then I_λ,n(x) is a polynomial of degree n − 1 satisfying

Proposition 3.4 ([4, Theorem  2.1]). Let 0 < λ < 1. Then, for n⩾0,

Furthermore, E_λ,n+1(1) ≳ 1.

Proposition 3.5 ([6, Theorem  2.5]). Let 0 < λ < 1.

• (a)

  For all $$,

  $$ ()
  Moreover, one has, for $$,
  $$ ()

• (b)

  For all $$,

  $$ ()
  Moreover, one has, for $$,
  $$ ()

Proposition 3.6 ([6, Corollary  2.6]). Let 0 < λ < 1. Then, for all x ∈ [−1,1],

Here, J_λ,n(x) is a polynomial of degree of 2n + 1 defined in (4.37) such that, for $$, and, for $$,

Proposition 3.7 ([6, Corollary  2.7]). Let 0 < λ < 1.

• (a)

  For all $$,

  $$ ()
  Moreover, one has, for $$,
  $$ ()

• (b)

  For all $$,

  $$ ()
  Moreover, one has, for $$,
  $$ ()

Proposition 3.8 ([4, Lemma  5.5]). Let 0 < λ < 1. Then, for μ = 1,2, …, n + 1,

and, for ν = 1,2, …, 2n + 1,

Proposition 3.9 ([6, Theorem  2.9]). Let 0 < λ < 1. Then, for μ = 1,2, …, n + 1,

and, for ν = 1,2, …, 2n + 1,

Lemma 4.1. Let 0 < λ < 1. Then, for all x ∈ [−1,1] and r⩾2,

and, for $$, Here, I_λ,n(x) is a polynomial of degree n − 1 defined in (3.10); such that

Proof. For r⩾2, (4.3) is obtained by r − 2 times differentiation of (3.9). Equation (4.6) follows by (3.11) and the use of Markov-Bernstein inequality. Now, we prove (4.4). We know that the Chebyshev polynomial T_n(x) satisfies the second-order differential equation

so we have, by r − 2 times differentiation of (4.7), Let for a nonnegative integer j⩾0, Observe that in the view of Szegö’s result (cf. [1]) Then, since $$ (note (4.10)), we will prove that, for $$ and j⩾0, instead of (4.4). Since, from the proof of [6, Proposition  2.3], and, for x = cos θ, we obtain that I_λ,n,0(x)≲n² and I_λ,n,1(x)≲n³φ⁻¹(x). Using (4.8), we have, for 2 ⩽ j ⩽ n, Therefore, (4.11) is proved by the mathematical induction on j. Consequently, we have (4.4).

Lemma 4.2. Let 0 < λ < 1 and r⩾2. Let $$. If r is even, then one has

and, if r is odd, then one has

Proof. Let r⩾2. From (4.3) and (4.4), we have, for $$,

and especially that is, we have (4.15) for r = 2. From Proposition 3.2, we see $$, $$, so we have, for $$, Then, from (4.17) with r = 3 and (4.18), we know that that is, we have (4.16) for r = 3. Assume that (4.15) and (4.16) hold for 3,4, …, r − 1 times differentiation. Let r be an even number. Then, we have, from (4.17), (4.19), and the assumptions for r − 1 and r − 2, that is, we have (4.15). Similarly, we also have (4.16) for an odd r.

Lemma 4.3. Let −1 < x₁ < x₂ < ⋯<x_n < 1 and

Then, P^′(x) is a polynomial of degree of n − 1 and has distinct real n − 1 zeros in (−1,1). Moreover, if one lets −1 < y₁ < y₂ < ⋯<y_n−1 < 1 be the zeros of P^′(x), then $$ is interlaced with the zeros of P(x) that is, x_i < y_i < x_i+1, i = 1,2, …, n − 1.

Proof. Since the sign of P^′(x_i) is (−1) ⁿ⁻ⁱ, it is proved.

Lemma 4.4. Let r be a nonnegative integer. Then, $$ has distinct n + 1 − r real zeros on (−1,1). If one lets $$ be the zeros of the polynomial $$ with

then one has, for 1 ⩽ r ⩽ n and k = 1, …, n + 1 − r,

Proof. From Lemma 4.3, we know that $$ has distinct real n + 1 − r zeros on (−1,1). By the interlaced zeros property of Lemma 4.3, we see that, for k = 1, …, n + 1 − r,

Thus, (4.24) is proved.

Proof of Theorem 2.1. Let r⩾1. Equation (2.2) comes from (4.15), (4.16), (3.12), and (3.13). From Propositions 3.4, 3.5, and (4.19), we have

Hence, using the Markov-Bernstein inequality, we have (2.3). To prove (2.4), we will use the mathematical induction. We use (3.14). The formula (2.3) holds for r = 1 from (3.14). We suppose that, for $$ and r⩾2, Then, by Lemma 4.4 and (3.8), we have, for $$ and r⩾2, Here, the last inequality is obtained by Proposition 3.2, that is, Therefore, we have, for $$ and r⩾2, Hence, from (2.3), we have (2.4). For $$, the proof is similar.

Lemma 4.5. Let ℓ be a nonnegative integer and $$. Then,

Proof. (a) When ℓ = 0, it is obvious from (3.5) and (3.12). Now, suppose ℓ⩾1. From (3.12), (2.2), and (3.5), we have

(b) From (4.4) and (3.5), we have

(c) Similarly to the proof of (a), we have, from (2.2) and (3.5),

Lemma 4.6. Let 0 < λ < 1. Then, for all x ∈ [−1,1] and r⩾2,

and, for $$, Here, J_λ,n(x) is a polynomial of degree of 2n + 1 defined as follows: Furthermore, one has

Proof. Similarly to the proof of Lemma 4.1, (4.35) is obtained by r − 2 times differentiation of the second-order differential relation with respect to F_λ,2n+1(x), that is, (3.17). So it is sufficient to prove (4.36) and (4.38). From (4.37), we know that

By Lemma 4.5 (a) and (b), we have, for $$, From (4.19) and Lemma 4.5 (c), we have, for $$ and r⩾4, When r = 2,3, we can similarly obtain that Therefore, we have (4.36). On the other hand, from (2.4), (4.6), and (3.2), we know that, for a nonnegative integer ℓ, Then, similarly to the proof of Lemma 4.5, we obtain that Therefore, we have (4.38).

Lemma 4.7. Let 0 < λ < 1. Then, for r⩾2, if r is even, one has, for $$,

and, if r is odd, one has

Proof. Using (4.35) and (4.36), we obtain the result similarly to the proof of Lemma 4.2.

Proof of Theorem 2.2. Equation (2.5) comes from Lemma 4.7 and Proposition 3.7. We will show (2.6) and (2.7). From Proposition 3.7, we see

Hence, using Markov-Bernstein inequality for $$, we have (2.6). Now, we show (2.7). By Proposition 3.7 (a), it is true for r = 1. We suppose that, for r⩾2, As the proof of Theorem 2.1, we have, for r⩾2 and for $$, Therefore, we see that by induction with Proposition 3.7, (2.7) holds for every r = 1,2, 3, ….

Corollary 4.8. Let 0 < λ < 1 and r⩾2. Then, for $$,

Proof. Corollary 4.8 comes from (4.35), (2.7), and (3.8).

Proof of Theorem 2.3. Equation (2.8) comes from (4.15) and (3.24).

Lemma 4.9. For 1 ⩽ μ ⩽ n + 1 and 1 ⩽ ν ⩽ 2n + 1,

Proof. Since we know from (3.24) and (3.25) that

(4.51) is obviously proved. Similarly, since we have, from (3.4) and (3.25), (4.52) is obtained.

Lemma 4.10. Let 0 < λ < 1 and r⩾1. Let r be an odd integer.

• (a)

  For 1 ⩽ μ ⩽ n + 1,

  $$ ()

• (b)

  For 1 ⩽ ν ⩽ n,

  $$ ()

Proof. (a) We know, from (3.3),

So (4.55) holds for k = 1. Assume that, for k = 1,2, …, ℓ, Then, we have from (3.5), (3.7), and (4.58) that Therefore, we have the result using the mathematical induction.

(b) For an odd integer r⩾1, we have from (3.6), (4.16), and (4.52)

Lemma 4.11. Let r⩾2. If r is even, then

Proof. It is easily proved from (3.4) and (3.7).

Lemma 4.12. Let k be a positive integer. Then, one has, for 1 ⩽ ν ⩽ 2n + 1,

Proof. From (4.37), we know that

From Lemma 4.5 (a) and (b), we know that for $$ ($$) On the other hand, we estimate $$ splitting into three terms as follows: Here, from Lemma 4.5 (c), we have for $$ For the first term, we also split into two terms as follows: From (2.2) and (4.55), we know that for even q and r Also, we know from (4.56) and (3.5) that for even q and r Then, we have Similarly, for odd q and r, we have, by (2.8) and (3.5), and, by (2.2) and (4.61), Thus, we have Therefore, we have the result.

Proof of Theorem 2.4. When r = 2, (2.9) holds from (3.27). Let even r > 2, and suppose that (2.9) holds for r − 2. Since we know by (4.35), (2.5), (4.62), and (4.19)

we obtain, using mathematical induction, Therefore, (2.9) is proved.

Proof of Theorem 2.5. (a) From (4.3), we know that

Therefore, we have, by (2.2), (3.24), and (3.26), Suppose that, for an integer ℓ⩾2, Then, from (4.3), we obtain Therefore, by (2.2), (2.8), and (4.4), we have

(b) Similarly to the proof of (a), by (4.51), (3.3), and (3.7), we can obtain

In addition, we see that, from (4.51),

Proof of Theorem 2.6. (a) From (3.9), we know that

Therefore, by (4.4) and (4.56), we have Then, we obtain from (4.3), (2.2), and (4.56) that Therefore, we have the result inductively.

(b) From (3.7), we know that

Therefore, by (3.4) and (4.61), we have Suppose that, for an integer ℓ⩾2, Then from (3.5), (3.7), and (4.61) Here, we see that for ℓ⩾2 Hence, we obtain, from (3.4),

Lemma 4.13. Let 0 < ɛ < 1 and $$. Then, for a nonnegative integer ℓ⩾0,

Proof. Let $$. From (4.37) and Lemma 4.5, we see that

Here, we let $$ when ℓ = 0. To estimate the first term, we split it into two terms as follows: Then, using $$, from (2.13) and (2.15), we obtain and, from (4.56) and from (4.61), Thus, we have, for $$, Similarly, noting $$, from (2.10) and (2.12) and, from (2.8) and (4.55), Then, we have, for $$, Therefore, we have, for $$, Thus, we have, for $$,

Proof of Theorem 2.7. From (4.35), (3.25), (3.27), and Lemma 4.13, we have

Suppose that Then, we obtain from (4.35) Therefore, (2.16) is proved.