Return to Article Details Baskakov-Kantorovich operators reproducing affine functions: inverse results

Baskakov-Kantorovich operators reproducing affine functions: inverse results

Jorge Bustamante $^{*}$

received: June 18, 2022; accepted: June 22, 2022; published online: August 25, 2022.

In a previous paper the author presented a Kantorovich modification of Baskakov operators which reproduce affine functions and he provided an upper estimate for the rate of convergence in polynomial weighted spaces. In this paper, for the same family of operators, a strong inverse inequality is given for the case of approximation in norm.

MSC. 41A35, 41A36, 41A17, 41A37.

Keywords. Baskakov-Kantorovich operators, polynomial weighted spaces, strong inverse results.

$^{*}$ Benemérita Universidad Autónoma de Puebla, Faculty of Physic and Mathematics, Avenida San Claudio y 18 Sur, Colonia San Manuel, Ciudad Universitaria, C.P. 72570, Puebla, Mexico, e-mail: jbusta@fcfm.buap.mx.

1 Introduction

Let $C [0, \infty)$ be the family of all real continuous functions on the semiaxis. Denote $e_{k} (t) = t^{k}$ , $k \geq 0$ .

Throughout the paper, we fix $m \in N$ , $m \geq 2$ , and set

ϱ (x) = \frac{1}{(1 + x)^{m}} a n d φ (x) = \sqrt{x (1 + x)} .

Moreover

C_{ϱ} [0, \infty) = {f \in C [0, \infty) : ‖ f ‖_{ϱ} < \infty},

where $‖ f ‖_{ϱ} = sup_{x \geq 0} | ϱ (x) f (x) |$ .

For a real $λ > 1$ , $f : [0, \infty) \to R$ and $x \geq 0$ , the Baskakov operator is defined by

V_{λ} (f, x) = \sum_{k = 0}^{\infty} f (\frac{k}{λ}) v_{λ, k} (x), v_{λ, k} (x) = (\binom{λ + k - 1}{k}) \frac{x^{k}}{(1 + x)^{λ + k}},

whenever the series converges absolutely.

For some functions $f \in C [0, \infty)$ , a family of Kantorovich-Baskakov type operators reproducing affine functions was introduced in [ 1 ] by setting

M_{λ} (f, x) = λ \sum_{k = 0}^{\infty} Q_{λ, k} (f) v_{λ, k} (x), Q_{λ, k} (f) = \int_{I_{λ, k}} f (a_{k} t) d t,

where

I_{λ, k} = [\frac{k}{λ}, \frac{k + 1}{λ}] a n d a_{k} = \frac{2 k}{2 k + 1}, k \in N_{0} .

Approximation properties of the operators $M_{λ}$ in some weighted spaces were presented in [ 1 ] . The following notations are needed: for $0 \leq β \leq 1$ , set

C_{ϱ, β} [0, \infty) = {h \in C [0, \infty) : h (0) = 0, ‖ h ‖_{ϱ, β} = ‖ φ^{2 β} h ‖_{ϱ} < \infty}

and

K (f, t)_{ϱ, β} = inf {‖ f - g ‖_{ϱ, β} + t ‖ φ^{2} g^{″} ‖_{ϱ, β} : g \in D (ϱ, β)},

where

D (ϱ, β) = {g \in C_{ϱ, β} [0, \infty) : g, g^{'} \in A C_{l o c} : ‖ φ^{2} g^{″} ‖_{ϱ, β} < \infty} .

The following result was proved in [ 1 ] .

Theorem 1

If $β \in [0, 1]$ and $m \geq 2$ , then there exists a constant $C$ such that, for all $λ > 2 (1 + m)$ and every $f \in C_{ϱ, β} [0, \infty)$ ,

‖ M_{λ} (f) - f ‖_{ϱ, β} \leq C K (f, \frac{1}{λ})_{ϱ, β} .

In this paper we present a strong inverse result related with theorem 1.

The work is organized as follows. In section 2 we present notations that will be used throughout the paper, as well as some identities related with the operators $M_{λ}$ and their derivatives. Section 3 is devoted to prove inequalities related with the moments of the operators $M_{λ}$ . In section 4 we collect several inequalities related with the weight $ϱ$ . In section 5 we include some Bernstein type inequalities. A Voronovskaya type theorem is given in section 6. Finally the main result is proved in section 7.

In what follows $C$ and $C_{i}$ ( $i \in N$ ) will denote absolute constants. They may be different on each occurrence. We remark that our arguments allow to obtain bounds for the constants, but not the best.

2 Notations and identities

We will use the notations

M_{λ}^{'} (f, x) = \frac{d}{d x} M_{λ} (f, x), M_{λ}^{″} (f, x) = \frac{d^{2}}{d x^{2}} M_{λ} (f, x)

and

M_{λ}^{‴} (f, x) = \frac{d^{3}}{d x^{3}} M_{λ} (f, x) .

For $λ > 1$ , $k \in N_{0}$ and $x \geq 0$ , we define

\begin{aligned} R_{λ, k} (x) = & (\frac{k}{λ} - x)^{2} - \frac{(1 + 2 x)}{λ} (\frac{k}{λ} - x) - \frac{φ^{2} (x)}{λ}, \\ I_{λ, 1} (x) = & \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{n})} (\frac{k}{λ} - x)^{2}, \\ I_{λ, 2} (x) = & \frac{1 + 2 x}{λ} \sum_{k = 1}^{\infty} | \frac{k}{λ} - x | \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{n})}, \end{aligned}

and

I_{λ, 3} (x) = \frac{φ^{2} (x)}{λ} \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{λ})} .

For $p, q \geq 0$ , set

A_{λ, p, q} (x) = \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ^{p} (\frac{k}{λ}) φ^{q} (\frac{k}{λ})} .

For $f \in C [0, \infty)$ and $k, j \in N$ , we use the notation

J_{λ, k, j} (f) = \int_{I_{λ, k}} \int_{k / λ}^{a_{k + j} (t + j / λ)} f (u) \cdot ((a_{k + j} (t + \frac{j}{λ}) - u) d u d t .

Proposition 1

( [ 1 , Prop. 2.4 ] ) For each $f \in C_{ϱ} [0, \infty)$ , $λ > 1$ and $x > 0$ , one has

\begin{aligned} M_{λ}^{'} (f, x) & = λ^{2} \sum_{k = 0}^{\infty} (Q_{λ, k + 1} (f) - Q_{λ, k} (f)) v_{λ + 1, k} (x) \\ = \frac{λ^{2}}{φ^{2} (x)} \sum_{k = 0}^{\infty} Q_{λ, k} (f) (\frac{k}{λ} - x) v_{λ, k} (x) . \end{aligned}

If $f (0) = 0$ , the term corresponding to $k = 0$ should be omitted.

Remark 1

The proof of proposition 1 (see [ 1 ] ) is a consequence of the identities ( $v_{λ, - 1} = 0$ )

v_{λ, k}^{'} (x) = \frac{k - λ x}{x (1 + x)} v_{λ, k} (x) = λ (v_{λ + 1, k - 1} (x) - v_{λ + 1, k} (x))) .

Here need the analogous of proposition 1 for the second and the third derivatives.

Proposition 2

For each $λ > 1$ , $f \in C_{ϱ} [0, \infty)$ , and $x > 0$ , one has

\begin{aligned} M_{λ} ” (f, x) & = λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (Q_{λ, k + 2} (f) - 2 Q_{λ, k + 1} (f) + Q_{λ, k} (f)) v_{λ + 2, k} (x) \\ = \frac{λ^{3}}{φ^{4} (x)} \sum_{k = 0}^{\infty} Q_{λ, k} (f) R_{λ, k} (x) v_{λ, k} (x), \end{aligned}

where $R_{λ, k}$ was defined in 2.

Moreover, if $g \in C_{ϱ}^{2} (0, \infty)$ and $h \in C_{ϱ}^{3} (0, \infty)$ , then

M_{λ}^{″} (g, x) = λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{λ, k, j} (g^{″})) v_{λ + 2, k} (x)

and

\frac{φ^{2} (x) M_{λ}^{‴} (g, x)}{λ^{2} (λ + 1) (λ + 2)} = \sum_{k = 0}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{λ, k, j} (g^{″})) (\frac{k}{λ + 2} - x) v_{λ + 2, k} (x) .

If $f (0) = 0$ or $g (0) = 0$ the term corresponding to $k = 0$ should be omitted.

Proof â–¼

It follows from proposition 1 and 7 that

\begin{aligned} M_{λ} ” (f, x) = & λ^{2} \sum_{k = 0}^{\infty} (Q_{λ, k + 1} (f) - Q_{λ, k} (f)) v_{λ + 1, k}^{'} (x) \\ = & λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (Q_{λ, k + 1} (f) - Q_{λ, k} (f)) (v_{λ + 2, k - 1} (x) - v_{λ + 2, k} (x))) \\ = & λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (Q_{λ, k + 2} (f) - 2 Q_{λ, k + 1} (f) + Q_{λ, k} (f)) v_{λ + 2, k} (x) . \end{aligned}

On the other hand,

\begin{aligned} \frac{φ^{4} (x)}{λ^{2}} M_{λ} ” (f, x) = & \frac{φ^{2} (x)}{λ} (\frac{φ^{2} (x)}{λ} M_{λ}^{'} (f, x))^{'} - \frac{1 + 2 x}{λ} \frac{φ^{2} (x)}{λ} M_{λ}^{'} (f, x) \\ = & \frac{φ^{2} (x)}{λ} (λ \sum_{k = 0}^{\infty} Q_{λ, k} (f) (\frac{k}{λ} - x) v_{λ, k}^{'} (x) - λ \sum_{k = 0}^{\infty} Q_{λ, k} (f) v_{λ, k} (x)) \\ - (1 + 2 x) \sum_{k = 0}^{\infty} Q_{λ, k} (f) (\frac{k}{λ} - x) v_{λ, k} (x) \\ = & λ \sum_{k = 0}^{\infty} Q_{λ, k} (f) ((\frac{k}{λ} - x)^{2} - \frac{(1 + 2 x)}{λ} (\frac{k}{λ} - x) - \frac{φ^{2} (x)}{λ}) v_{λ, k} (x) . \end{aligned}

Notice that, if

T_{λ} (t, k) = (a_{k + 2} (t + \frac{2}{λ}) - \frac{k}{λ}) - 2 (a_{k + 1} (t + \frac{1}{λ}) - \frac{k}{λ}) + (a_{k} t - \frac{k}{λ}),

then

\begin{aligned} 2 \int_{\frac{k}{λ}}^{\frac{k + 1}{λ}} T_{λ} (t, k) d t = \\ = a_{k + 2} ((\frac{k + 3}{λ})^{2} - (\frac{k + 2}{λ})^{2}) - 2 a_{k + 1} ((\frac{k + 2}{λ})^{2} - (\frac{k + 1}{λ})^{2}) + a_{k} ((\frac{k + 1}{λ})^{2} - (\frac{k}{λ})^{2}) \\ = \frac{1}{λ^{2}} (a_{k + 2} (2 k + 5) - 2 a_{k + 1} (2 k + 3) + a_{k} (2 k + 1)) \\ = \frac{1}{λ^{2}} (2 (k + 2) - 4 (k + 1) + 2 k) = 0. \end{aligned}

Hence, using the representation

g (y) = g (\frac{k}{λ}) + g^{'} (\frac{k}{λ}) (y - \frac{k}{λ}) + \int_{k / λ}^{y} g^{″} (u) (y - u) d u,

one has

\begin{aligned} Q_{λ, k + 2} (g) - 2 Q_{λ, k + 1} (g) + Q_{λ, k} (g) = \\ = & \int_{k / λ}^{(k + 1) / λ} (g (a_{k + 2} (t + 2 / λ)) - 2 g (a_{k + 1} (t + \frac{1}{λ})) + g (a_{k} t)) d t \\ = & \int_{k / λ}^{(k + 1) / λ} (\int_{k / λ}^{a_{k + 2} (t + 2 / λ)} g ” (u) (a_{k + 2} (t + \frac{2}{λ}) - u) \\ - 2 \int_{k / λ}^{a_{k + 1} (t + 2 / λ)} g ” (u) (a_{k + 1} (t + \frac{1}{λ}) - u) + \int_{k / λ}^{a_{k} t} g ” (u) (a_{k} t - u)) d u \\ = & \sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{λ, k, j} (g ”) . \end{aligned}

On the other hand

\begin{aligned} φ^{2} (x) M_{λ} ”^{'} (g, x) = \\ = φ^{2} (x) λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{λ, k, j} (g ”)) v_{λ + 2, k}^{'} (x) \\ = φ^{2} (x) λ^{2} (λ + 1) \sum_{k = 0}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{n, k, j} (g ”)) \frac{k - (λ + 2) x}{x (1 + x)} v_{λ + 2, k} (x) \\ = λ^{2} (λ + 1) (λ + 2) \sum_{k = 0}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) (- 1)^{j} J_{n, k, j} (g ”)) (\frac{k}{λ + 2} - x) v_{λ + 2, k} (x) . \end{aligned}

3 Estimates for the moments

Here, for $q \in N$ , we should consider the absolute moments

V_{λ, q} (x) = V_{λ} (| e_{1} - x |^{q}, x) a n d M_{λ, q} (x) = M_{λ} (| e_{1} - x |^{q}, x) .

Proposition 3

For each fixed $q \in N$ , there exists a constant $C_{q}$ such that, if $λ > 1$ , $x > 0$ and $λ x \geq \frac{2}{3}$ , then

V_{λ, q} (x) \leq C_{q} \frac{φ^{q} (x)}{λ^{\frac{q}{2}}} .

Proof â–¼

It was proved in [ 3 , Prop. 3.2 ] that, if

q = 2 j

j \in N

, the assertion holds whenever

λ x \geq 1

. But the proof can be modified to include the case

λ x \geq \frac{2}{3}

. Of course, with a different constant.

If $q = 2 j - 1$ , $j \in N$ , then

\begin{aligned} V_{λ} (∣ e_{1} - x ∣^{2 j - 1}, x) & \leq (V_{λ} ((e_{1} - x)^{2 (j - 1)}, x) V_{λ} ((e_{1} - x)^{2 j}, x))^{1 / 2} \\ \leq C (\frac{φ^{2 (j - 1)} (x)}{λ^{j - 1}} \frac{φ^{2 j} (x)}{λ^{j}})^{1 / 2} = C \frac{φ^{2 j - 1} (x)}{λ^{j - 1 / 2}} . \end{aligned}

We need an extension of proposition 3 to the case of the operators $M_{λ}$ , but only for $1 \leq q \leq 6$ .

Proposition 4

For each fixed $q \in N$ , $1 \leq q \leq 6$ , there exists a constant $C_{q}$ such that, if $λ > 1$ , $x > 0$ and $λ x \geq \frac{2}{3}$ , then

M_{λ, q} (x) \leq C_{q} \frac{φ^{q} (x)}{λ^{\frac{q}{2}}} .

Proof â–¼

It was proved in [ 1 , Cor. 2.3 ] that, for each

λ > 1

and

x \geq 0

, one has

M_{λ, 2} (x) \leq \frac{13}{12} \frac{φ^{2} (x)}{λ} a n d M_{λ, 1} (x) \leq \sqrt{\frac{13}{12}} \frac{φ (x)}{\sqrt{λ}} .

Moreover, if $x \geq \frac{1}{2 (λ + 1)}$ , then $M_{λ, 4} (x) \leq \frac{16 φ^{4} (x)}{λ^{2}}$ . Since $\frac{2}{3 λ} \geq \frac{1}{2 (λ + 1)}$ , the inequality holds under the conditions assumed above.

Notice that

M_{λ, 3} (x) \leq \sqrt{M_{λ, 2} (x) M_{λ, 4} (x)} \leq 4 \sqrt{2} \frac{φ^{3} (x)}{λ^{3 / 2}} .

We should present a proof for $M_{λ, 6} (x)$ . For $k \in N$ ,

\begin{aligned} 0 & \leq λ \int_{I_{λ, k}} (a_{k} t - x)^{6} d t = \frac{λ}{7 a_{k}} ((\frac{a_{k} k}{λ} - x + \frac{a_{k}}{λ})^{7} - (\frac{a_{k} k}{λ} - x)^{7}) \\ = \frac{λ}{7 a_{k}} \sum_{i = 0}^{6} (\binom{7}{i}) (\frac{a_{k}}{λ})^{7 - i} (\frac{a_{k} k}{λ} - x)^{i} = \frac{1}{7} \sum_{i = 0}^{6} (\binom{7}{i}) (\frac{a_{k}}{λ})^{6 - i} (\frac{a_{k} k}{λ} - x)^{i} \\ = \frac{1}{7} \sum_{i = 0}^{6} (\binom{7}{i}) (\frac{a_{k}}{λ})^{6 - i} (\frac{k}{λ} - x - \frac{k}{(2 k + 1) λ})^{i} \\ \leq \frac{1}{7} \sum_{i = 0}^{6} (\binom{7}{i}) 2^{i} (\frac{1}{λ})^{6 - i} (| \frac{k}{λ} - x |^{i} + (\frac{k}{(2 k + 1) λ})^{i}) \\ \leq \frac{1}{7 λ^{6}} \sum_{i = 0}^{6} (\binom{7}{i}) + \frac{1}{7} \sum_{i = 0}^{6} (\binom{7}{i}) 2^{i} (\frac{1}{λ})^{6 - i} | \frac{k}{λ} - x |^{i} . \end{aligned}

Therefore

\begin{aligned} λ \sum_{k = 1}^{\infty} Q_{λ, k} ((e_{1} - x)^{6}) v_{λ, k} (x) & \leq \frac{2^{7}}{7 λ^{6}} + \sum_{i = 1}^{6} (\binom{7}{i}) \frac{2^{i}}{λ^{6 - i}} V_{λ} (∣ e_{1} - x ∣^{i}, x) \\ \leq \frac{2^{7}}{7 λ^{6}} + \sum_{i = 1}^{6} (\binom{7}{i}) \frac{2^{i} C_{i}}{λ^{6 - i}} \frac{φ^{i} (x)}{λ^{i / 2}} \\ = \frac{2^{7}}{7 λ^{6}} + \sum_{i = 1}^{6} (\binom{7}{i}) \frac{2^{i} C_{i} φ^{i} (x)}{λ^{3 + (6 - i) / 2}} \\ \leq \frac{3^{3} x^{3}}{2^{3} 7 λ^{3}} + \sum_{i = 1}^{6} (\binom{7}{i}) \frac{2^{i} 3^{(6 - i) / 2} C_{i} φ^{i} (x) x^{(6 - i) / 2}}{2^{(6 - i) / 2} λ^{3}} \\ \leq \frac{φ^{6} (x)}{λ^{3}} + C_{1} \sum_{i = 1}^{6} (\binom{7}{i}) \frac{φ^{i} (x) φ^{6 - i} (x)}{λ^{3}} \leq C_{2} \frac{φ^{6} (x)}{λ^{3}} . \end{aligned}

Moreover, taking into account proposition 3, one has

0 \leq λ \int_{I_{λ, 0}} (a_{0} t - x)^{6} d t v_{λ, 0} = x^{6} v_{λ, 0} \leq V_{λ} ((e_{1} - x)^{6}, x) \leq C_{6} \frac{φ^{6} (x)}{λ^{3}} .

This yields the inequality for $m = 6$ .

Finally the proof in the case $m = 5$ is obtained by using Hölder inequality. $◻$

4 Preparatory computations

Proposition 5

Suppose that $m \geq 2$ , $γ \in [0, 2)$ , $f \in C [0, \infty)$ , $f (0) = 0$ , and $‖ φ^{2 γ} f ‖_{ϱ} < \infty$ . If $λ > 1$ and $k \in N$ , then

∣ Q_{λ, k} (f) ∣\leq \frac{2^{m}}{λ} \frac{‖ φ^{2 γ} f ‖_{ϱ}}{ϱ (\frac{k}{λ}) φ^{2 γ} (\frac{k}{λ})} .

Proof â–¼

It was proved in [ 1 , Prop. 3.4 ] that, if

γ \in [0, 2)

m \geq γ

λ > 1

and

k > 0

, then

\int_{k / λ}^{(k + 1) / λ} \frac{d t}{ϱ (a_{k} t) φ^{2 γ} (a_{k} t)} \leq \frac{2^{m}}{λ} \frac{1}{ϱ (\frac{k}{λ}) φ^{2 γ} (\frac{k}{λ})} .

Proposition 6

Assume $m \in N$ , $p, q \geq 0$ , $m p \geq q \geq 0$ and set $s = 2 (1 + 2 m p - q)$ . There exists a constant $C (p, q)$ such that, if $λ \geq s$ and $x > 0$ , then

A_{λ, p, q} (x) \leq C (p, q) \frac{(1 + x)^{m p}}{φ^{q} (x)},

where $A_{λ, p, q} (x)$ is defined in 6.

Proof â–¼

It is known that (see [ 3 , Prop. 3.8 ] ), if

a > - 1

q \in R

, and

r = max {2 a, 2 | c |, \frac{| c |}{1 + a}}

, then

\sum_{k = 1}^{\infty} (\frac{λ}{k})^{a} (1 + \frac{k}{λ})^{c} v_{λ, k} (x) \leq C \frac{(1 + x)^{c}}{x^{a}}, λ \geq 2 (1 + r) .

We apply this result with $a = \frac{q}{2}$ and $c = m p - \frac{q}{2}$ . Notice that $q \leq 2 m p - q$ . Hence, if $λ \geq 2 (1 + 2 m p - q)$ ,

\begin{aligned} \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ^{p} (\frac{k}{λ}) φ^{q} (\frac{k}{λ})} = & \sum_{k = 1}^{\infty} (\frac{λ}{k})^{\frac{q}{2}} (1 + \frac{k}{λ})^{m p - \frac{q}{2}} v_{λ, k} (x) \\ \leq & C \frac{(1 + x)^{m p - \frac{q}{2}}}{x^{\frac{q}{2}}} = C \frac{(1 + x)^{m p}}{φ^{q} (x)} . \end{aligned}

Proposition 7

Assume $β \in [0, 1]$ and $m \geq 2$ . There exists a constant $C$ such that, if $k > 0$ , $j \in {0, 1, 2}$ , $λ > 2$ , and $g \in D (ϱ, β)$ , then

∣ J_{λ, k, j} (g^{″}) ∣\leq \frac{C}{ϱ (\frac{k}{λ + 2}) φ^{2 + 2 β} (\frac{k}{λ + 2})} \frac{‖ φ^{2} g^{″} ‖_{ϱ, β}}{λ^{3}} .

Proof â–¼

It is sufficient to consider the case

‖ φ^{2} g^{″} ‖_{ϱ, β} = 1

If we set $g (x) = ϱ (x) φ^{2 + 2 β} (x)$ , then

| J_{λ, k, j} (g^{″}) | \leq \int_{I_{λ, k}} \int_{k / λ}^{a_{k + j} (t + \frac{j}{λ})} \frac{((a_{k + j} (t + \frac{j}{λ}) - u)}{g (u)} d u d t .

(A) Let us first consider the case $j \in {1, 2}$ . Since

k (1 + 2 (k + j)) \leq 2 k (k + j) + 2 j (k + j) = 2 (k + j)^{2},

if $t \geq \frac{k}{λ}$ , then

\frac{k}{λ} \leq (\frac{2 (k + j)}{1 + 2 (k + j)} (\frac{k}{λ} + \frac{j}{λ}) \leq a_{k + j} (t + \frac{j}{λ}) .

Moreover, if $u \leq a_{k + j} (t + j / λ)$ and $t \leq (k + 1) / λ$ , then

\begin{aligned} \frac{1}{ϱ (u)} & = (1 + u)^{m} \leq (1 + (a_{k + j} (\frac{k + 1 + j}{λ}))^{m} \leq (1 + (\frac{2 k + 2}{λ}))^{m} \\ \leq (1 + (\frac{4 k}{λ}))^{m} \leq 4^{m} (1 + \frac{k}{λ})^{m} \leq \frac{4^{m} \cdot 2^{m}}{ϱ (\frac{k}{λ + 2})} . \end{aligned}

On the other hand, since $φ^{2 + 2 β} (x)$ increases, if $\frac{k}{λ} \leq u$ , then

\frac{1}{φ^{2 + 2 β} (u)} \leq \frac{1}{φ^{2 + 2 β} (\frac{k}{λ})} \leq \frac{1}{φ^{4 β} (\frac{k}{λ + 2})} .

Therefore,

\begin{aligned} | J_{λ, k, j} (g ”) | \leq & \frac{2^{3 m}}{g (\frac{k}{λ + 2})} \int_{I_{λ, k}} \int_{k / λ}^{a_{k + j} (t + j / λ)} (a_{k + j} (t + \frac{j}{λ}) - u) d u d t \\ = & \frac{2^{3 m}}{2 g (\frac{k}{λ + 2})} \int_{I_{λ, k}} (a_{k + j} (t + \frac{j}{λ}) - \frac{k}{λ})^{2} d t \\ \leq & \frac{2^{3 m}}{2 g (\frac{k}{λ + 2})} \int_{k / λ}^{(k + 1) / λ} ((t + \frac{j}{λ}) - \frac{k}{λ})^{2} d t \\ \leq & \frac{2^{3 m}}{2 g (\frac{k}{λ + 2})} \int_{k / λ}^{(k + 1) / λ} (\frac{1 + j}{λ})^{2} d t \leq \frac{9 \cdot 2^{3 m}}{2 g (\frac{k}{λ + 2})} \frac{1}{λ^{3}} . \end{aligned}

(B) Now assume $j = 0$ . First notice that, if $k \in N$ , then

\frac{2}{3} \leq a_{k} < 1.

In this case, taking into account 11, for $t \geq k / λ$ ,

φ^{2} (a_{k} t) = a_{k} t (1 + a_{k} t) \geq \frac{2}{3} t (1 + \frac{2 t}{3}) \geq \frac{4}{9} φ (\frac{k}{λ}) \geq \frac{4}{9} φ (\frac{k}{λ + 2}) .

On the other hand, since

1 + \frac{k}{λ} \leq 2 + \frac{k}{λ + 2} \frac{λ + 2}{λ} \leq 2 (1 + \frac{k}{λ + 2}),

one has

\frac{1}{ϱ (\frac{k}{λ})} \leq \frac{2^{m}}{ϱ (\frac{k}{λ + 2})} .

Moreover

(k + 1) a_{k} = \frac{2 k (k + 1)}{2 k + 1} \geq \frac{k (2 k + 1)}{2 k + 1} = k .

Therefore

\begin{aligned} \int_{I_{λ, k}} | \int_{k / λ}^{a_{k} t} \frac{a_{k} t - u}{g (u)} d u | d t = \\ = \int_{k / λ}^{k / (λ a_{k})} \int_{a_{k} t}^{k / λ} \frac{u - a_{k} t}{g (u)} d u d t + \int_{k / (λ a_{k})}^{(k + 1) / λ} \int_{k / λ}^{a_{k} t} \frac{a_{k} t - u}{g (u)} d u d t \\ \leq \frac{C_{1}}{g (k / (λ + 2))} (\int_{k / n}^{k / (λ a_{k})} \int_{a_{k} t}^{k / λ} (u - a_{k} t) d u d t + \int_{k / (n a_{k})}^{(k + 1) / λ} \int_{k / λ}^{a_{k} t} (a_{k} t - u) d u d t) \\ = \frac{C_{1}}{2 g (\frac{k}{λ + 2})} (\int_{k / λ}^{k / (λ a_{k})} (\frac{k}{λ} - a_{k} t)^{2} d t + \int_{k / (λ a_{k})}^{(k + 1) / λ} (a_{k} t - \frac{k}{λ})^{2} d t) \\ = \frac{C_{1}}{6 a_{k} g (\frac{k}{λ + 2})} (\frac{k^{3}}{λ^{3}} (1 - a_{k})^{3} + (\frac{a_{k} (k + 1)}{λ} - \frac{k}{λ})^{3}) \\ \leq \frac{C_{1}}{4 g (\frac{k}{λ + 2})} (\frac{k^{3}}{λ^{3}} \frac{1}{(2 k + 1)^{3}} + (\frac{k + 1}{λ} - \frac{k}{λ})^{3}) \leq \frac{C_{2}}{g (\frac{k}{λ + 2})} \frac{1}{λ^{3}}, \end{aligned}

where we use 12.â–¡

Proposition 8

If $λ \geq 2$ , $β \in [0, 1)$ , there exists a constant $C$ such that, if $x > 0$ , $k \in {1, 2}$ , and $λ x < 2 / 3$ , then

φ^{2 β} (x) v_{λ, k} (x) λ \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} | d t \leq \frac{C (k + 1)^{2}}{λ^{2}} .

Proof â–¼

Since

x < 2 / (3 λ) \leq a_{k} / λ \leq a_{k} k / λ

and

0 < x < 1

\begin{aligned} φ^{2 β} (x) v_{λ, k} (x) λ \int_{k / λ}^{(k + 1) / λ} \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} d t \leq \\ \leq \frac{x^{k}}{φ^{2} (x) \sqrt{1 + x}} λ \int_{k / λ}^{(k + 1) / λ} (1 + t)^{m} (t - x)^{2} \int_{x}^{t} \frac{d s}{s^{1 / 2}} d t \\ = \frac{2 x^{k}}{φ^{2} (x) \sqrt{1 + x}} λ \int_{k / λ}^{(k + 1) / λ} (1 + t)^{m} (t - x)^{2} (\sqrt{t} - \sqrt{x}) d t \\ \leq C_{1} x^{k - 1} (1 + \frac{k + 1}{λ})^{m} λ \int_{k / λ}^{(k + 1) / λ} (t - x)^{2} d t \leq C_{2} (\frac{k + 1}{λ})^{2} . \end{aligned}

Proposition 9

If $m \geq 2$ and $β \in [0, 1)$ , there exists a constant $C$ such that for $i \in {1, 2, 3}$ , $λ \geq 2 (1 + 2 m)$ , and $λ x \geq 2 / 3$ ,

I_{λ, i} (x) \leq C \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{φ^{2} (x)}{λ} .

where we use the notations in 3–5.

Proof â–¼

We will use the notation in 6. Notice that

4 β < 4 \leq 2 m

, thus we can apply proposition 6 with

λ \geq 2 (1 + 2 m) \geq 2 (1 + 2 m - 4 β)

Taking into account Hölder’s inequality and using proposition 6 and proposition 3, we have

\begin{aligned} I_{1} (λ, x) & = \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{n})} (\frac{k}{λ} - x)^{2} \\ \leq \sqrt{A_{λ, 2, 4 β} (x)} (\sum_{k = 1}^{\infty} v_{λ, k} (x) (\frac{k}{λ} - x)^{4})^{1 / 2} \leq C_{1} \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{φ^{2} (x)}{λ} . \end{aligned}

On the other hand

\begin{aligned} I_{2} (λ, x) = & \frac{1 + 2 x}{λ} \sum_{k = 1}^{\infty} | \frac{k}{λ} - x | \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{n})} \\ \leq & \frac{1 + 2 x}{λ} \sqrt{A_{λ, 2, 4 β} (x)} (\sum_{k = 1}^{\infty} v_{λ, k} (x) (\frac{k}{n} - x)^{2})^{1 / 2} \\ \leq & C_{2} \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{(1 + x)}{λ} \sqrt{\frac{x (1 + x)}{λ}} = C_{2} \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{x (1 + x)}{λ} \sqrt{(1 + \frac{1}{x}) \frac{1}{λ}} \\ \leq & C_{2} \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{φ^{2} (x)}{λ} \sqrt{\frac{1 + λ}{λ}} \leq \sqrt{2} C_{2} \frac{(1 + x)^{m}}{φ^{2 β} (x)} \frac{φ^{2} (x)}{λ} . \end{aligned}

Finally

\frac{λ}{φ^{2} (x)} I_{λ, 3} (x) = \sum_{k = 1}^{\infty} \frac{v_{λ, k} (x)}{ϱ (\frac{k}{λ}) φ^{2 β} (\frac{k}{λ})} = A_{λ, 1, 2 β} (x) \leq C_{3} \frac{(1 + x)^{m}}{φ^{2 β} (x)} .

5 Bernstein type inequalities

Theorem 2

Suppose $β \in [0, 1)$ and $m \geq 2$ . There exists a constant $C$ such that, if $λ \geq 2 (1 + 2 m)$ and $f \in C_{ϱ, β} [0, \infty)$ , then

‖ φ^{2} M_{λ}^{″} (f) ‖_{ϱ, β} \leq C λ ‖ f ‖_{ϱ, β} .

Proof â–¼

Taking into account the notations 3–5, it follows from 9, proposition 5, and proposition 9 that

\begin{aligned} ϱ (x) φ^{2 + 2 β} (x) | M_{λ} ” (f, x) | & = \frac{ϱ (x) λ^{3}}{φ^{2 (1 - β)} (x)} | \sum_{k = 0}^{\infty} Q_{λ, k} (f) R_{λ, k} (x) v_{λ, k} (x) | \\ \leq \frac{ϱ (x) λ^{2} ‖ f ‖_{ϱ, β}}{φ^{2 (1 - β)} (x)} (I_{1} (λ, x) + I_{2} (λ, x) + I_{3} (λ, x)) \\ \leq C λ ‖ f ‖_{ϱ, β} . \end{aligned}

Theorem 3

Suppose $β \in [0, 1)$ and $m \geq 2$ . There exist a constant $Λ_{1}$ such that, if $λ \geq 2 (1 + 2 m)$ and $g \in D (ϱ, β)$ , then

‖ φ^{3} M_{λ}^{‴} (g) ‖_{ϱ, β} \leq Λ_{1} \sqrt{λ} ‖ φ^{2} g^{″} ‖_{ϱ, β} .

Proof â–¼

Set

Z (f) = ‖ φ^{2} g^{″} ‖_{ϱ, β}

. From ?? we obtain

\begin{aligned} (ϱ φ^{3 + 2 β}) (x) | M_{λ} ”^{'} (g, x) | \leq \\ \leq C_{1} Z (f) (ϱ φ^{1 + 2 β}) (x) λ^{4} \sum_{k = 1}^{\infty} (\sum_{j = 0}^{2} (\binom{2}{j}) ∣ J_{n, k, j} ∣ | \frac{k}{λ + 2} - x | v_{λ + 2, k} (x) \\ \leq C_{2} Z (f) (ϱ φ^{1 + 2 β}) (x) λ \sum_{k = 1}^{\infty} | \frac{k}{λ + 2} - x | \frac{v_{λ + 2, k} (x)}{ϱ (\frac{k}{λ + 2}) φ^{2 + 2 β} (\frac{k}{λ + 2})} \\ \leq C_{2} Z (f) (ϱ φ^{1 + 2 β}) (x) λ \sqrt{A_{λ + 2, 2, 2 (1 + β)} (x)} \sqrt{V_{λ + 2} ((e_{1} - x)^{2}, x)} \\ \leq C_{3} Z (f) (ϱ φ^{1 + 2 β}) (x) \frac{(1 + x)^{m}}{φ^{2 + 2 β} (x)} λ \frac{φ (x)}{\sqrt{λ + 2}} \leq C_{4} Z (f) \sqrt{λ} . \end{aligned}

6 A Voronovskaya type theorem

We need a result given in Theorem 5.1 of [ 1 ] .

Theorem 4

If $β \in [0, 1]$ and $m \geq 2$ , there exists a constant $Λ_{2}$ such that, for all $λ > 2 (1 + m)$ and every $f \in C_{ϱ, β} [0, \infty)$ , one has

‖ M_{λ} (f) ‖_{ϱ, β} \leq Λ_{2} ‖ f ‖_{ϱ, β} .

Let

C_{ϱ, β}^{3} [0, \infty) = {g \in C_{ϱ, β} [0, \infty) : g, g^{'}, g^{″} \in A C_{l o c}, ‖ φ^{3} g^{‴} ‖_{ϱ, β} < \infty} .

Theorem 5

Suppose $β \in [0, 1)$ and $m \geq 3$ $($ or $m = 2$ and $β \in [0, 1 / 2])$ . There exists a constant $Λ_{3}$ such that, if $g \in C_{ϱ, β}^{3} [0, \infty)$ and $λ \geq 2 (1 + 2 m)$ , then

‖ M_{λ} (g) - g - \frac{g ”}{2} M_{λ, 2} ‖_{ϱ, β} \leq \frac{Λ_{3}}{λ^{3 / 2}} ‖ φ^{3} g^{‴} ‖_{ϱ . β} .

Proof â–¼

Let us denote

R_{λ} (g, x) = M_{λ} (g, x) - g (x) - \frac{g ” (x)}{2} M_{λ} ((t - x)^{2}, x) .

By Taylor’s expansion

g (t) = g (x) + g^{'} (x) (x - t) + \frac{1}{2} g^{″} (x) (t - x)^{2} + \frac{1}{2} \int_{x}^{t} g^{‴} (s) (t - s)^{2} d s,

one has

\begin{aligned} φ^{2 β} (x) ϱ (x) ∣ R_{λ} (g, x) ∣= \\ = φ^{2 β} (x) ϱ (x) | \frac{λ}{2} \sum_{k = 0}^{\infty} v_{λ, k} (x) (\int_{k / λ}^{(k + 1) / λ} \int_{x}^{a_{k} t} {(a_{k} t - s)}^{2} g ”^{'} (s) d s) d t | \\ \leq φ^{2 β} (x) ϱ (x) ‖ φ^{3} g ”^{'} ‖_{ϱ, β} \frac{λ}{2} \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} | d t \\ = \frac{1}{2} ‖ φ^{3} g ”^{'} ‖_{ϱ, β} F_{n} (x), \end{aligned}

where

F_{n} (x) = φ^{2 β} (x) ϱ (x) λ \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} | d t .

Case 1. Assume $λ x < 2 / 3.$

First we estimate the terms corresponding to $k = 0, 1, 2$ .

Since $a_{0} = 0$ , one has

\begin{aligned} \frac{(ϱ φ^{2 β}) (x)}{(1 + x)^{λ}} λ \int_{0}^{1 / λ} \int_{0}^{x} \frac{s^{2} d s}{(ϱ φ^{3 + 2 β} (s)} d t & \leq ϱ (x) x^{β} \int_{0}^{x} s^{(1 - 2 β) / 2} (1 + s)^{m} d s \\ \leq ϱ (x) x^{β} (1 + x)^{m} \int_{0}^{x} s^{(1 - 2 β) / 2} d s \\ = \frac{2 x^{β} x^{3 / 2 - β}}{(3 - 2 β)} \leq \frac{2}{(3 - 2 β)} \frac{1}{λ^{3 / 2}} . \end{aligned}

On the other hand, it follows from proposition 8 that

(ϱ φ^{2 β}) (x) λ \sum_{k = 1}^{2} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} | d t \leq \frac{C_{1} (2^{2} + 3^{2})}{λ^{2}} .

Now we consider the tail of the series. It is known that (see [ 3 ] )

\frac{k}{λ} v_{λ, k} (x) = x v_{λ + 1, k - 1} (x) .

In particular

\frac{k}{λ} \frac{k - 1}{λ + 1} \frac{k - 2}{λ + 2} v_{λ, k} (x) = x^{3} v_{λ + 3, k - 3} (x) .

From this we obtain

\begin{aligned} (ϱ φ^{2 β}) (x) λ \sum_{k = 3}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} d s}{(ϱ φ^{3 + 2 β} (s)} | d t = \\ = (ϱ φ^{2 β}) (x) λ \sum_{k = 3}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} \int_{x}^{a_{k} t} \frac{(a_{k} t - s)^{2} (1 + s)^{m} d s}{φ^{3 + 2 β} (s)} d t \\ \leq \frac{λ ϱ (x)}{φ^{3} (x)} \sum_{k = 3}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} \int_{x}^{t} (t - s)^{2} (1 + s)^{m} d s d t \\ \leq \frac{λ ϱ (x)}{φ^{3} (x)} \sum_{k = 1}^{\infty} (1 + \frac{k + 1}{λ})^{m} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} t^{3} d t \\ \leq \frac{C_{2} ϱ (x)}{φ^{3} (x)} \sum_{k = 3}^{\infty} (1 + \frac{k}{λ})^{m} \frac{k^{3}}{λ^{3}} v_{λ, k} (x) \\ \leq \frac{C_{3} ϱ (x)}{φ^{3} (x)} \sum_{k = 3}^{\infty} v_{λ, k} (x) (1 + \frac{k}{λ + 3})^{m} \frac{k}{λ} \frac{k - 1}{λ + 1} \frac{k - 2}{λ + 2} \\ \leq C_{4} ϱ (x) x^{3 / 2} \sum_{k = 3}^{\infty} v_{λ + 3, k - 3} (x) (1 + \frac{k - 3}{λ + 3})^{m} \leq C_{5} x^{3 / 2} \leq \frac{C_{6}}{λ^{3 / 2}}, \end{aligned}

where we use proposition 6, with $q = 0$ and $p = 1$ .

Case 2. Assume $λ x \geq 2 / 3$ and set $c = m - 3 / 2 - β$ .

From Proposition 3.3 of [ 3 ] we know that

\begin{aligned} | \int_{x}^{a_{k} t} \frac{(a_{k} t - u)^{2} d u}{(ϱ φ^{3 + 2 β} (u)} | & = | \int_{x}^{a_{k} t} \frac{(a_{k} t - u)^{2} (1 + u)^{c} d u}{u^{(3 + 2 β) / 2}} | \\ \leq \frac{| a_{k} t - x |^{3}}{(3 - (1 + 2 β) / 2) x^{(3 + 2 β) / 2}} ((1 + x)^{c} + (1 + t)^{c}) \end{aligned}

If $λ x \geq 2 / 3$ , since $3 / 2 + β < 5 / 2 < 3$ , one has

\begin{aligned} F_{n} (x) = & (ϱ φ^{2 β}) (x) λ \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | \int_{x}^{a_{k} t} \frac{(a_{k} t - u)^{2} d u}{(ϱ φ^{3 + 2 β} (u)} | d t \\ \leq & \frac{C_{1} φ^{2 β} (x) ϱ (x) λ}{x^{(3 + 2 β) / 2}} \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} ∣ a_{k} t - x ∣^{3} ((1 + x)^{c} + (1 + t)^{c}) d t \\ = & \frac{C_{1} φ^{2 β} (x) ϱ (x) (1 + x)^{c}}{x^{(3 + 2 β) / 2}} M_{λ, 3} (x) \\ + \frac{C_{1} φ^{2 β} (x) ϱ (x)}{x^{(3 + 2 β) / 2}} λ \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | a_{k} t - x |^{3} (1 + t)^{c} d t \\ = & \frac{C_{1}}{φ^{3} (x)} M_{λ, 3} (x) \\ + \frac{C_{1} (1 + x)^{β}}{x^{3 / 2} (1 + x)^{m}} λ \sum_{k = 0}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} | a_{k} t - x |^{3} (1 + t)^{c} d t . \end{aligned}

It follows from proposition 4 that

\frac{1}{φ^{3} (x)} M_{λ} (∣ t - x ∣^{3}, x) \leq \frac{C_{2}}{λ^{3 / 2}} .

For the other terms we first estimate the case $k = 0$ . Notice that, for $t \in (0, \frac{1}{λ})$ ,

(1 + t)^{c} \leq (1 + \frac{1}{λ})^{c} \leq (1 + \frac{3 x}{2})^{c} \leq 2^{c} (1 + x)^{c} .

Here the condition $m \geq 3$ was used. Therefore

\begin{aligned} \frac{(1 + x)^{β}}{x^{3 / 2} (1 + x)^{m}} λ v_{λ, 0} (x) \int_{0}^{1 / λ} x^{3} (1 + t)^{c} d t & \leq \frac{2^{c} (1 + x)^{β} x^{3}}{x^{3 / 2} (1 + x)^{m}} v_{λ, 0} (x) (1 + x)^{c} \\ = \frac{2^{c} x^{3}}{x^{3 / 2} (1 + x)^{3 / 2}} v_{λ, 0} (x) \\ \leq \frac{2^{c}}{φ^{3} (x)} V_{λ} (∣ t - x ∣^{3}, x) \leq \frac{C_{3}}{λ^{3 / 2}} . \end{aligned}

On the other hand, since

\begin{aligned} \int_{k / λ}^{(k + 1) / λ} ∣ a_{k} t - x ∣^{3} (1 + t)^{c} d t \leq \\ \leq (\int_{k / λ}^{(k + 1) / λ} (a_{k} t - x)^{6} d t)^{1 / 2} (\int_{k / λ}^{(k + 1) / λ} (1 + t)^{2 c} d t)^{1 / 2}, \end{aligned}

from theorem 4 and proposition 4 we obtain

\begin{aligned} \frac{(1 + x)^{β}}{x^{3 / 2} (1 + x)^{m}} λ \sum_{k = 1}^{\infty} v_{λ, k} (x) \int_{k / λ}^{(k + 1) / λ} ∣ a_{k} t - x ∣^{3} (1 + t)^{c} d t \leq \\ \leq \frac{(1 + x)^{β}}{x^{3 / 2} (1 + x)^{m}} \sqrt{M_{λ} ((1 + t)^{2 c}, x)} \sqrt{M_{λ} ((e_{1} - x)^{6}, x)} \\ \leq \frac{C_{4} (1 + x)^{β}}{x^{3 / 2} (1 + x)^{m}} (1 + x)^{m - 3 / 2 - β} \frac{φ^{3} (x)}{λ^{3 / 2}} = \frac{C_{4}}{λ^{3 / 2}} . \end{aligned}

This completes the proof.â–¡

Remark 2

We do not know if theorem 5 holds $m = 2$ and $β \in (\frac{1}{2}, 1)$ .

7 Inverse result

Theorem 6

Suppose $β \in [0, 1)$ and $m \geq 3$ $($ or $m = 2$ and $β \in [0, 1 / 2])$ . There exist positive constants $κ$ and $Λ_{4}$ such that, if $f \in C_{ϱ} [0, \infty)$ and $λ \geq 2 (1 + 2 m)$ , then

\frac{1}{λ} ‖ φ^{2} (M_{λ}^{2} (f))^{″} ‖_{ϱ, β} \leq Λ_{3} (‖ M_{λ} (f) - f ‖_{ϱ, β} + ‖ M_{κ λ} (f) - f ‖_{ϱ, β}),

where $M_{λ}^{2} (f) = M_{λ} (M_{λ} (f))$ .

Proof â–¼

Let

Λ_{1}

and

Λ_{3}

be constants such that the inequalities in theorem 3 and theorem 5 hold for

λ \geq 2 (1 + 2 m)

. Set

μ = 16 (Λ_{1} Λ_{3})^{2} λ .

If $λ \geq 2 (1 + 2 m)$ and $g = M_{λ}^{2} (f)$ , from ?? we know that

\begin{aligned} \frac{1}{2 μ} ‖ φ^{2} (M_{λ}^{2}) ” f ‖_{ϱ, β} \leq \\ \leq & \frac{1}{2} ‖ M_{μ, 2} (M_{λ}^{2}) ” f ‖_{ϱ, β} \\ \leq & | | M_{μ} (M_{λ}^{2} f) - M_{λ}^{2} f | |_{ϱ, β} + | | M_{μ} (M_{λ}^{2} f) - M_{λ}^{2} f - \frac{M_{μ, 2}}{2} (M_{λ}^{2} f) ” | |_{ϱ, β} \\ \leq & ‖ M_{μ} (M_{λ}^{2} f - M_{λ} f) + M_{μ} (M_{λ} f - f) + M_{μ} f - f + f - M_{λ} f + M_{λ} f - M_{λ}^{2} f ‖_{ϱ, β} \\ + \frac{Λ_{3}}{μ^{3 / 2}} ‖ φ^{3} (M_{λ}^{2} f) ”^{'} ‖_{ϱ, β} \\ \leq & C_{1} (‖ M_{μ} f - f ‖_{ϱ, β} + ‖ M_{λ} f - f ‖_{ϱ, β}) + \frac{Λ_{1} Λ_{3} \sqrt{λ}}{μ^{3 / 2}} ‖ φ^{2} M_{λ} ” (f) ‖_{ϱ, β} \\ \leq & C_{1} (‖ M_{μ} f - f ‖_{ϱ, β} + ‖ M_{λ} f - f ‖_{ϱ, β}) \\ + \frac{Λ_{1} Λ_{3} \sqrt{λ}}{μ^{3 / 2}} ‖ φ^{2} M_{λ} ” (f - M_{λ} (f)) ‖_{ϱ, β} + \frac{Λ_{1} Λ_{3} \sqrt{λ}}{μ^{3 / 2}} ‖ φ^{2} M_{λ} ” (M_{λ} (f)) ‖_{ϱ, β} \\ \leq & C_{1} (‖ M_{μ} f - f ‖_{ϱ, β} + ‖ M_{λ} f - f ‖_{ϱ, β}) \\ + \frac{C Λ_{1} Λ_{3} λ^{3 / 2}}{μ^{3 / 2}} ‖ f - M_{λ} (f) ‖_{ϱ, β} + \frac{1}{4 μ} ‖ φ^{2} M_{λ} ” (M_{λ} (f)) ‖_{ϱ, β} . \end{aligned}

Therefore

\frac{1}{4 μ} ‖ φ^{2} (M_{λ}^{2})^{″} f ‖_{ϱ, β} \leq C_{2} (‖ M_{μ} f - f ‖_{ϱ, β} + ‖ M_{λ} f - f ‖_{ϱ, β}),

and it is sufficient to prove the result, because

\frac{1}{λ} {| | φ^{2} (M_{λ}^{2} (f))^{″} | |}_{ϱ, β} = \frac{2^{6} (Λ_{1} Λ_{2})^{2}}{4 μ} {| | φ^{2} (M_{λ}^{2} (f))^{″} | |}_{ϱ, β} .

Theorem 7

Suppose $β \in [0, 1)$ , $m \geq 3$ $($ or $m = 2$ and $β \in [0, \frac{1}{2}])$ , and $κ$ is given as in theorem 6. There exists a constant $C$ such that, if $f \in C_{ϱ} [0, \infty)$ and $λ \geq 2 (1 + 2 m)$ , then

K_{β} {(f, \frac{1}{λ})}_{ϱ} \leq C (‖ M_{λ} (f) - f ‖_{ϱ, β} + ‖ M_{κ λ} (f) - f ‖_{ϱ, β}),

where $K_{β} (f, t)_{ϱ}$ is defined as in 1.

Proof â–¼

Fix

f \in C_{ϱ} [0, \infty)

and denote

g = M_{λ}^{2} (f) = M_{λ} (M_{λ} (f))

. From theorem 4 we know that

g \in C_{ϱ, b} [0, \infty)

From the definition of the $K$ -functional $K_{β} (f, t)_{ϱ}$ , theorem 4 and theorem 6 we know that

\begin{aligned} K_{β} {(f, \frac{1}{λ})}_{ϱ} \leq & ‖ f - M_{λ}^{2} (f) ‖_{ϱ, β} + \frac{1}{λ} ‖ φ^{2} (M_{λ}^{2} (f)) ” ‖_{ϱ, β} \\ \leq & ‖ f - M_{λ} (f) ‖_{ϱ, β} + ‖ M_{λ} (f - M_{λ} (f)) ‖_{ϱ, β} + \frac{1}{λ} ‖ φ^{2} (M_{λ}^{2} (f)) ” ‖_{ϱ, β} \\ \leq & (1 + Λ_{2}) ‖ f - M_{λ} (f) ‖_{ϱ, β} + \frac{1}{λ} ‖ φ^{2} (M_{λ}^{2} (f)) ” ‖_{ϱ, β} \\ \leq & C (‖ M_{λ} (f) - f ‖_{ϱ, β} + ‖ M_{κ, λ} (f) - f ‖_{ϱ, β}) . \end{aligned}

Remark 3

In the case $β = 0$ the $K$ -functional can be replaced by a weighted modulus of smoothness as in [ 4 ] . For the classical Baskakov operators $V_{n}$ related results were given [ 6 ] and [ 5 ] . â–¡