Degree of Approximation by Hybrid Operators

Lemma 1 (cf. [7]). Let α, m, n, and r be integers with m ≥ 0, r ≥ 1, and n + α > m:

Then one has

• (i)

  R_n,0,r(α; x) = 1 and R_n,1,r(α; x) = ((−α + 1)x + r + 1)/(n + α − 1),

• (ii)

  for m⩾1

  $$ ()

• (iii)
  $$ ()

where g_n,m,r(α; x) is a polynomial of degree ⩽m such that the coefficients of g_n,m,r(α; x) are bounded independently of n.

Proof. Let R_n,m,r(x): = R_n,m,r(α; x). Then (i)

Using we see that (ii) Using $$, we obtain Since we know that we have Then substituting (12) into (10), we consider the following: Then since we see we have Here the last equation follows from integration by parts. Furthermore, we easily see Therefore, we conclude Consequently, (ii) is proved.

(iii) For m = 1, (6) holds. Let us assume (6) for m⩾1. We note

So, we have, by the assumption of induction, Here, if m is even, then and if m is odd, then Hence we have and here we see that g_n,m+1,r(x) is a polynomial of degree ⩽m + 1 such that the coefficients of g_n,m+1,r(x) are bounded independently of n.

Lemma 2 (cf. [7]). Let n, β, and r be integers with r ≥ 0. Let f ∈ C^(r)[0, ∞) satisfy for a positive integer δ

Then one has, for n + β − r > δ, where

Proof. Using

we have

Theorem 3. Let 0 < p ≤ ∞, and let δ and r be nonnegative integers. Let n and β be integers with n + β − r > δ. Let f ∈ C^(r+1)[0, ∞) satisfy

Then one has uniformly, for f and n,

Proof. Let |t − x| < ε and x < ξ < t. By the second inequality of (28),

Let ε = n^−ν,   0 < ν < 1. Then using Lemma 2 and we have From (30) and Lemma 1, we have Next, we estimate E₂. By the use of the first inequality in (28), we have Now using $$and the notation we have Then, with ε = n^−ν, Here for i ≥ 1, we get because Finally we get From (32), If we put ν = 1/3, then we get

Theorem 4. Let r and γ be nonnegative integers. Let n and β be integers with n + β − r > 2γ + 1. Let f ∈ C^(r+2)[0, ∞) satisfy

Then one has uniformly, for f and n,

Proof. For f ∈ C^(r+2)[0, ∞), we have

From (45), (46), and Lemma 2, we get Using (1 + t) ^2γ ≤ C((t − x) ^2γ + (1 + x) ^2γ), we obtain Therefore, we have For x ∈ [0, ∞), we have |g_n,1,r(β − r; x)|ϕ(x) ≤ C, (1 + x) ^2γ|g_n,2,r(β − r; x)|ϕ^2(γ+1)(x) ≤ C, and |g_n,2(γ+1),r(β − r; x)|ϕ^2(γ+1)(x) ≤ C. Hence

Theorem 5. Let r and γ be nonnegative integers. Let n and β be integers with n + β − r > 2γ + 2. Then one has, for f ∈ C^r([0, ∞)),

Theorem 6. Let r and γ be nonnegative integers. Let n and β be integers with n + β − r > 2γ. Let f ∈ C^(r)([0, ∞)) satisfy

Then one has uniformly, for n, f, and x ∈ [0, ∞),

Proof. Using (1 + y) ^2γ ⩽ C((1 + x) ^2γ + (y − x) ^2γ), we have

Therefore, by Lemma 1 (6), we have Since |g_n,2γ,r(β − r; x)ϕ^2γ(x)| is uniformly bounded on [0, ∞), we have with Lemma 2 and (59) Therefore, we have the result.

Lemma 7 (see [8], Lemma 2.4.)Let f(x) ∈ C([0, ∞)), and let η(x) be a positive and nonincreasing function on [0, ∞). Then

• (i)

  [f] _h(x) ∈ C²([0, ∞)),

• (ii)
  $$ ()

• (iii)
  $$ ()

• (iv)
  $$ ()

Proof of Theorem 5. We know that, for f(x) ∈ C^r([0, ∞)),

Then first, we split it as follows: Then for the first term, we have, using Theorem 6, (62), and (65), Here, we suppose 0 < h ⩽ 1, and then we know that For the second term, from Theorem 4, (65), (63), and (64) of Lemma 7, Therefore, we have If we let $$, then because $$.