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

Baskakov-Kantorovich operators reproducing affine functions: inverse results

Jorge Bustamante$^\ast $

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.

$^\ast $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 $m\in \mathbb {N}$, $m\geq 2$, and set

\[ \varrho (x)=\tfrac 1{(1+x)^m} \quad {\rm and } \quad \varphi (x)=\sqrt{x(1+x)}. \]

Moreover

\[ C_{\varrho }[0,\infty )=\big\{ f\in C[0,\infty ): \, \, \Vert f\Vert _\varrho {\lt} \infty \big\} , \]

where $\Vert f\Vert _\varrho =\sup _{x\geq 0}| \varrho (x)f(x)|$.

For a real $\lambda {\gt}1$, $f:[0,\infty )\to \mathbb {R}$ and $x\geq 0$, the Baskakov operator is defined by

\[ V_{\lambda }(f,x)=\sum _{k=0}^\infty f\big( \tfrac {k}{\lambda }\big)v_{\lambda ,k}(x), \quad v_{\lambda ,k}(x)= \tbinom {\lambda +k-1}{k}\tfrac {x^k}{(1+x)^{\lambda +k}}, \]

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

\[ M_{\lambda }(f,x)=\lambda \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)v_{\lambda ,k}(x) , \quad Q_{\lambda ,k}(f)=\int _{I_{\lambda ,k}}f(a_{k}t)dt, \]

where

\[ I_{\lambda ,k}=\big[\tfrac k{\lambda },\tfrac {k+1}{\lambda }\big]\quad \quad {\rm and} \quad \quad a_{k}=\tfrac {2k}{2k+1},\quad k \in \mathbb {N}_0. \]

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

\[ C_{\varrho ,\beta }[0,\infty )=\Big\{ h\in C[0,\infty ): \, h(0)=0, \, \, \, \Vert h\Vert _{\varrho ,\beta }=\Vert \varphi ^{2\beta }h\Vert _\varrho {\lt} \infty \Big\} \]

and

\begin{equation} \label{LaKfun} K(f, t)_{\varrho ,\beta }= \inf \Big\{ \Vert f-g\Vert _{\varrho ,\beta } +\, t \Vert \varphi ^{2}g''\Vert _{\varrho ,\beta } :\, g\in D(\varrho ,\beta )\Big\} , \end{equation}

where

\[ D(\varrho ,\beta )=\Big\{ g\in C_{\varrho ,\beta }[0,\infty )\, :\, g,g'\in AC_{loc}:\, \Vert \varphi ^{2}g''\Vert _{\varrho ,\beta }{\lt}\infty \Big\} . \]

The following result was proved in [ 1 ] .

Theorem 1

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

\[ \Vert M_{\lambda }(f)-f\Vert _{\varrho ,\beta } \leq CK\big(f, \tfrac {1}{\lambda }\big)_{\varrho ,\beta }. \]

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_\lambda $ and their derivatives. Section 3 is devoted to prove inequalities related with the moments of the operators $M_\lambda $. In section 4 we collect several inequalities related with the weight $\varrho $. 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 $C_i$ ($i\in \mathbb {N}$) 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

\[ M_\lambda '(f,x)=\frac{d}{dx}M_\lambda (f,x),\quad M_\lambda ''(f,x)=\frac{d^2}{dx^2}M_\lambda (f,x) \]

and

\[ M_\lambda '''(f,x)=\frac{d^3}{dx^3}M_\lambda (f,x). \]

For $\lambda {\gt}1$, $k\in \mathbb {N}_0$ and $x\geq 0$, we define

\begin{align} \label{LasRlambda} R_{\lambda ,k}(x)=& \Big(\tfrac {k}{\lambda }-x\Big)^2 -\tfrac {(1+2x)}{\lambda }\Big(\tfrac {k}{\lambda }-x\Big) -\tfrac {\varphi ^2(x)}{\lambda }, \\ \label{LasI1} I_{\lambda ,1}( x)=& \sum \limits _{k=1}^\infty \tfrac {v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac kn)} \Big(\tfrac {k}{\lambda }-x\Big)^2, \\ \label{LasI2} I_{\lambda ,2}( x)=& \tfrac {1+2x}{\lambda } \sum \limits _{k=1}^\infty \Big|\tfrac {k}{\lambda }-x\Big| \tfrac {v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac kn)}, \end{align}

and

\begin{equation} \label{LasI3} I_{\lambda ,3}( x)= \tfrac {\varphi ^2(x)}{\lambda }\sum \limits _{k=1}^\infty \tfrac {v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac k{\lambda })}. \end{equation}

For $p,q\geq 0$, set

\begin{equation} \label{losA} A_{\lambda ,p,q}(x)= \sum \limits _{k=1}^\infty \tfrac {v_{\lambda ,k}(x)}{\varrho ^p(\frac k{\lambda })\varphi ^{q}(\frac k{\lambda })}. \end{equation}

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

\[ J_{\lambda ,k,j}(f)=\int _{I_{\lambda ,k}} \int _{k/\lambda }^{a_{k+j}(t+j/\lambda )} f(u)\cdot \Big( (a_{k+j}(t+\tfrac j{\lambda })-u\Big)\, du\, dt. \]

Proposition 1

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

\begin{align*} M_{\lambda }’(f,x)& =\lambda ^2\sum \limits _{k=0}^\infty \Big(Q_{\lambda ,k+1}(f)-Q_{\lambda ,k}(f)\Big)v_{\lambda +1,k}(x) \\ & = \tfrac {\lambda ^2}{\varphi ^2(x)}\sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)\Big(\tfrac {k}{\lambda } -x\Big)v_{\lambda ,k}(x). \end{align*}

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 ($v_{\lambda ,-1}=0$)

\begin{equation} \label{gigante} v_{\lambda ,k}'(x)=\tfrac {k- \lambda x}{x(1+x)}v_{\lambda ,k}(x) =\lambda \big(v_{\lambda +1,k-1}(x)- v_{\lambda +1,k}(x))\big). \end{equation}

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

Proposition 2

For each $\lambda {\gt}1$, $f\in C_\varrho [0,\infty )$, and $x{\gt} 0$, one has

\begin{align} \label{segunda1} M_\lambda ”(f,x) & =\lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(Q_{\lambda ,k+2}(f)-2Q_{\lambda ,k+1}(f)+Q_{\lambda ,k}(f)\Big) v_{\lambda +2,k}(x) \\ \label{segunda1MN} & = \tfrac {\lambda ^3}{\varphi ^4(x)} \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f) R_{\lambda ,k}(x)v_{\lambda ,k}(x), \end{align}

where $R_{\lambda ,k}$ was defined in 2.

Moreover, if $g\in C^2_\varrho (0,\infty )$ and $h\in C^3_\varrho (0,\infty )$, then

\begin{equation} \label{repretercera} M_{\lambda }''(g,x)= \lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(\sum \limits _{j=0}^2 \tbinom {2}{j}(-1)^jJ_{\lambda ,k,j}(g'')\Big)v_{\lambda +2,k}(x) \end{equation}

and

\[ \frac{\varphi ^2(x)M_{\lambda }'''(g,x)}{ \lambda ^2(\lambda +1)(\lambda +2)} =\sum \limits _{k=0}^\infty \Big(\sum \limits _{j=0}^2 \tbinom {2}{j}(-1)^jJ_{\lambda ,k,j}(g'')\Big) \Big(\tfrac {k}{\lambda +2}- x\Big)v_{\lambda +2,k}(x). \]

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{align*} M_{\lambda }”(f,x)= & \lambda ^2\sum \limits _{k=0}^\infty \Big(Q_{\lambda ,k+1}(f)-Q_{\lambda ,k}(f)\Big)v_{\lambda +1,k}’(x) \\ = & \lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(Q_{\lambda ,k+1}(f)-Q_{\lambda ,k}(f)\Big) \big(v_{\lambda +2,k-1}(x)- v_{\lambda +2,k}(x))\big) \\ = & \lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(Q_{\lambda ,k+2}(f)-2Q_{\lambda ,k+1}(f)+Q_{\lambda ,k}(f)\Big) v_{\lambda +2,k}(x). \end{align*}

On the other hand,

\begin{align*} \tfrac {\varphi ^4(x)}{\lambda ^2}M_{\lambda }”(f,x)= & \tfrac {\varphi ^2(x)}{\lambda }\Big(\tfrac {\varphi ^2(x)}{\lambda }M_{\lambda }’(f,x)\Big)’ -\tfrac {1+2x}{\lambda }\tfrac {\varphi ^2(x)}{\lambda }M_{\lambda }’(f,x) \\ = & \tfrac {\varphi ^2(x)}{\lambda }\Big( \lambda \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)\Big(\tfrac {k}{\lambda }-x\Big)v_{\lambda ,k}’(x) -\lambda \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)v_{\lambda ,k}(x) \Big) \\ & -(1+2x) \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)\Big(\tfrac {k}{\lambda } -x\Big)v_{\lambda ,k}(x) \\ = & \lambda \sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)\Big(\Big(\tfrac {k}{\lambda }-x\Big)^2 -\tfrac {(1+2x)}{\lambda }\Big(\tfrac {k}{\lambda }-x\Big) -\tfrac {\varphi ^2(x)}{\lambda }\Big)v_{\lambda ,k}(x). \end{align*}

Notice that, if

\[ T_\lambda (t,k)= \Big( a_{k+2} (t+\tfrac 2{\lambda }) - \tfrac k{\lambda } \Big)-2\Big(a_{k+1} (t+\tfrac 1{\lambda })-\tfrac k{\lambda } \Big)+\Big(a_{k} t -\tfrac k{\lambda }\Big), \]

then

\begin{align*} & 2 \int _{\frac k{\lambda }}^{\frac{k+1}{\lambda }} T_\lambda (t,k)dt = \\ & = a_{k+2} \Big((\tfrac {k+3}{\lambda })^2-(\tfrac {k+2}{\lambda })^2) -2a_{k+1}\Big((\tfrac {k+2}{\lambda })^2-(\tfrac {k+1}{\lambda })^2\Big) +a_{k}\Big((\tfrac {k+1}{\lambda })^2-(\tfrac k{\lambda })^2\Big) \\ & = \tfrac {1}{\lambda ^2}\Big(a_{k+2}(2k+5) -2a_{k+1}(2k+3)+a_{k}(2k+1) \Big) \\ & = \tfrac {1}{\lambda ^2}(2(k+2) -4(k+1)+2k )=0. \end{align*}

Hence, using the representation

\[ g(y)= g(\tfrac k{\lambda })+g' (\tfrac k{\lambda })(y-\tfrac k{\lambda })+\int _{k /\lambda }^y g''(u)(y-u)du, \]

one has

\begin{align*} & Q_{\lambda ,k+2}(g)-2Q_{\lambda ,k+1}(g)+Q_{\lambda ,k}(g)= \\ =& \int _{k/\lambda }^{(k+1)/\lambda }\Big( g\Big( a_{k+2} (t+2/\lambda ) \Big)-2g\Big( a_{k+1} (t+\tfrac 1{\lambda }) \Big)+g(a_{k} t)\Big)dt \\ = & \int _{k/\lambda }^{(k+1)/\lambda }\Big( \int _{k/\lambda }^{a_{k+2} (t+2/\lambda )} g”(u)(a_{k+2} (t+\tfrac 2{\lambda })-u) \\ & -2\int _{k/\lambda }^{a_{k+1} (t+2/\lambda )} g”(u)(a_{k+1} (t+\tfrac 1{\lambda })-u) +\int _{k/\lambda }^{a_{k} t} g”(u)(a_{k}t -u)\Big)du \\ = & \sum \limits _{j=0}^2\tbinom {2}{j}(-1)^jJ_{\lambda ,k,j}(g”). \end{align*}

On the other hand

\begin{align*} & \varphi ^2(x)M_{\lambda }”’(g,x) = \\ & =\varphi ^2(x)\lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(\sum \limits _{j=0}^2 \tbinom {2}{j}(-1)^jJ_{\lambda ,k,j}(g”)\Big)v_{\lambda +2,k}’(x) \\ & = \varphi ^2(x)\lambda ^2(\lambda +1)\sum \limits _{k=0}^\infty \Big(\sum \limits _{j=0}^2 \tbinom {2}{j}(-1)^jJ_{n,k,j}(g”)\Big) \tfrac {k- (\lambda +2) x}{x(1+x)}v_{\lambda +2,k}(x) \\ & = \lambda ^2(\lambda +1)(\lambda +2)\sum \limits _{k=0}^\infty \Big(\sum \limits _{j=0}^2 \tbinom {2}{j}(-1)^jJ_{n,k,j}(g”)\Big) \Big(\tfrac {k}{\lambda +2}- x\Big)v_{\lambda +2,k}(x). \end{align*}

3 Estimates for the moments

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

\[ V_{\lambda ,q}(x)=V_\lambda (| e_1-x|^q,x) \quad {\rm and}\quad M_{\lambda ,q}(x)=M_\lambda (| e_1-x|^q,x). \]

Proposition 3

For each fixed $q\in \mathbb {N}$, there exists a constant $C_q$ such that, if $\lambda {\gt}1$, $ x{\gt}0$ and $\lambda x\geq \frac23$, then

\[ V_{\lambda ,q}(x)\leq C_q\frac{\varphi ^{q}(x)}{\lambda ^{\frac q2}}. \]

Proof â–¼

It was proved in [ 3 , Prop. 3.2 ] that, if $q=2j$, $j\in \mathbb {N}$, the assertion holds whenever $\lambda x\geq 1$. But the proof can be modified to include the case $\lambda x\geq \frac23$. Of course, with a different constant.

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

\begin{align*} V_\lambda (\mid e_1-x\mid ^{2j-1},x) & \leq \Big( V_\lambda ((e_1-x)^{2(j-1)},x) V_\lambda ((e_1-x)^{2j},x)\Big)^{1/2} \\ & \leq C \Big( \tfrac {\varphi ^{2(j-1)}(x)}{\lambda ^{j-1}} \tfrac {\varphi ^{2j}(x)}{\lambda ^{j}}\Big)^{1/2}= C\tfrac {\varphi ^{2j-1}(x)}{\lambda ^{j-1/2}}. \end{align*}

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

Proposition 4

For each fixed $q\in \mathbb {N}$, $1\leq q\leq 6$, there exists a constant $C_q$ such that, if $\lambda {\gt}1$, $ x{\gt}0$ and $\lambda x\geq \frac23$, then

\[ M_{\lambda ,q}(x)\leq C_q\frac{\varphi ^{q}(x)}{\lambda ^{\frac q2}}. \]

Proof â–¼

It was proved in [ 1 , Cor. 2.3 ] that, for each $\lambda {\gt}1$ and $x\geq 0$, one has

\[ M_{\lambda ,2}(x)\leq \tfrac {13}{12}\tfrac {\varphi ^2(x)}{\lambda }\quad \quad {\rm and} \quad \quad M_{\lambda ,1}(x)\leq \sqrt{\tfrac {13}{12}}\tfrac {\varphi (x)}{\sqrt{\lambda }}. \]

Moreover, if $x\geq \frac1{2(\lambda +1)}$, then $M_{\lambda ,4}(x) \leq \frac{16 \varphi ^4(x)}{\lambda ^2}$. Since $\frac2{3\lambda }\geq \frac1{2(\lambda +1)}$, the inequality holds under the conditions assumed above.

Notice that

\[ M_{\lambda ,3}(x)\leq \sqrt{M_{\lambda ,2}(x)M_{\lambda ,4}(x)} \leq 4\sqrt{2} \tfrac {\varphi ^{3}(x)}{\lambda ^{3/2}}. \]

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

\begin{align*} 0 & \leq \lambda \int _{I_{\lambda ,k}}(a_kt-x)^6dt=\tfrac {\lambda }{7a_k}\Big( \Big(\tfrac {a_k k}{\lambda } -x+\tfrac {a_k}{\lambda } \Big)^7 -\Big(\tfrac {a_k k}{\lambda }-x \Big)^7 \Big) \\ & =\tfrac {\lambda }{7a_k}\sum _{i=0}^6\tbinom {7}{i}\Big(\tfrac {a_k}{\lambda }\Big)^{7-i} \Big(\tfrac {a_k k}{\lambda } -x\Big)^i =\tfrac {1}{7}\sum _{i=0}^6\tbinom {7}{i}\Big(\tfrac {a_k }{\lambda }\Big)^{6-i} \Big(\tfrac {a_k k}{\lambda } -x\Big)^i \\ & =\tfrac {1}{7}\sum _{i=0}^6\tbinom {7}{i}\Big(\tfrac {a_k }{\lambda }\Big)^{6-i} \Big(\tfrac { k}{\lambda } -x-\tfrac {k}{(2k+1)\lambda }\Big)^i \\ & \leq \tfrac {1}{7}\sum _{i=0}^6\tbinom {7}{i}2^i\Big(\tfrac {1 }{\lambda }\Big)^{6-i} \Big(\Big|\tfrac { k}{\lambda } -x\Big|^i+\Big(\tfrac {k}{(2k+1)\lambda }\Big)^i\Big) \\ & \leq \tfrac {1}{7\lambda ^6}\sum _{i=0}^6\tbinom {7}{i} +\tfrac {1}{7}\sum _{i=0}^6\tbinom {7}{i}2^i\Big(\tfrac {1 }{\lambda }\Big)^{6-i} \Big|\tfrac { k}{\lambda } -x\Big|^i. \end{align*}

Therefore

\begin{align*} \lambda \sum \limits _{k=1}^\infty Q_{\lambda ,k}((e_1-x)^6)v_{\lambda ,k}(x) & \leq \tfrac {2^7}{7\lambda ^6}+\sum _{i=1}^6\tbinom {7}{i}\tfrac {2^i }{\lambda ^{6-i}} V_\lambda (\mid e_1-x\mid ^i,x) \\ & \leq \tfrac {2^7}{7\lambda ^6}+\sum _{i=1}^6\tbinom {7}{i}\tfrac {2^iC_i }{\lambda ^{6-i}} \tfrac {\varphi ^i(x)}{\lambda ^{i/2}} \\ & = \tfrac {2^7}{7\lambda ^6}+\sum _{i=1}^6\tbinom {7}{i}\tfrac {2^iC_i\varphi ^i(x) }{\lambda ^{3+(6-i)/2}} \\ & \leq \tfrac {3^3x^{3}}{2^37\lambda ^{3}}+\sum _{i=1}^6\tbinom {7}{i}\tfrac {2^i3^{(6-i)/2}C_i\varphi ^i(x) x^{(6-i)/2} }{ 2^{(6-i)/2}\lambda ^{3}} \\ & \leq \tfrac {\varphi ^{6}(x)}{\lambda ^{3}}+C_1\sum _{i=1}^6\tbinom {7}{i}\tfrac {\varphi ^i(x) \varphi ^{6-i}(x) }{\lambda ^{3}} \leq C_2\tfrac {\varphi ^6(x)}{\lambda ^{3}}. \end{align*}

Moreover, taking into account proposition 3, one has

\[ 0\leq \lambda \int _{I_{\lambda ,0}}(a_0t-x)^6dt v_{\lambda ,0}= x^6 v_{\lambda ,0} \leq V_{\lambda }((e_1-x)^6,x)\leq C_6\tfrac {\varphi ^6(x)}{\lambda ^3}. \]

This yields the inequality for $m=6$.

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

4 Preparatory computations

Proposition 5

Suppose that $m\geq 2$, $\gamma \in [0,2)$, $f\in C[0,\infty )$, $f(0)=0$, and $\Vert \varphi ^{2\gamma } f\Vert _\varrho {\lt}\infty $. If $\lambda {\gt}1$ and $k\in \mathbb {N}$, then

\[ \mid Q_{\lambda ,k}(f)\mid \leq \frac{2^m}{\lambda }\frac{\Vert \varphi ^{2\gamma } f\Vert _\varrho }{\varrho (\frac k{\lambda })\varphi ^{2\gamma }(\frac k{\lambda })}. \]

Proof â–¼

It was proved in [ 1 , Prop. 3.4 ] that, if $\gamma \in [0,2)$, $m\geq \gamma $, $\lambda {\gt} 1$ and $k{\gt}0$, then

\[ \int _{k/\lambda } ^{(k+1)/\lambda } \tfrac {dt}{\varrho (a_kt)\varphi ^{2\gamma }(a_{k}t)} \leq \tfrac {2^m}{\lambda }\tfrac {1}{\varrho (\frac k{\lambda })\varphi ^{2\gamma }(\frac k{\lambda })}. \]

Proposition 6

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

\[ A_{\lambda ,p,q}(x)\leq C(p,q)\tfrac {(1+x)^{mp}}{\varphi ^{q}(x)}, \]

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

Proof â–¼

It is known that (see [ 3 , Prop. 3.8 ] ), if $a{\gt}-1$, $q\in \mathbb {R}$, and $r=\max \Big\{ 2a, 2|c|, \frac{|c|}{1+a}\Big\} $, then

\[ \sum \limits _{k=1}^\infty \Big(\tfrac {\lambda }{k}\Big)^a \Big(1+\tfrac {k}{\lambda }\Big)^cv_{\lambda ,k}(x)\leq C \tfrac {(1+x)^c}{x^a}, \quad \quad \lambda \geq 2(1+r). \]

We apply this result with $a=\frac q2$ and $c=mp-\frac q2$. Notice that $q\leq 2mp-q$. Hence, if $\lambda \geq 2(1+2mp-q)$,

\begin{align*} \sum \limits _{k=1}^\infty \frac{v_{\lambda ,k}(x)}{\varrho ^p(\frac k{\lambda })\varphi ^{q}(\frac k{\lambda })} = & \sum \limits _{k=1}^\infty \Big(\tfrac {\lambda }{k}\Big)^{\frac q2} \Big(1+\tfrac {k}{\lambda }\Big)^{mp-\frac q2}v_{\lambda ,k}(x) \\ \leq & C\frac{(1+x)^{mp-\frac q2}}{x^{\frac q2}} = C\frac{(1+x)^{mp}}{\varphi ^{q}(x)}. \end{align*}

Proposition 7

Assume $\beta \in [0,1]$ and $m\geq 2$. There exists a constant $C$ such that, if $k{\gt}0$, $j\in \{ 0,1,2\} $, $\lambda {\gt}2$, and $g\in D(\varrho ,\beta )$, then

\[ \mid J_{\lambda ,k,j}(g'')\mid \leq \frac{C}{ \varrho (\frac k{\lambda +2})\varphi ^{2+2\beta }(\frac k{\lambda +2})}\frac{\Vert \varphi ^{2} g''\Vert _{\varrho ,\beta }}{\lambda ^3}. \]

Proof â–¼

It is sufficient to consider the case $\Vert \varphi ^{2} g''\Vert _{\varrho ,\beta }=1$.

If we set $g(x)= \varrho (x)\varphi ^{2+2\beta }(x)$, then

\[ |J_{\lambda ,k,j}(g'')| \leq \int _{I_{\lambda ,k}} \int _{k/\lambda }^{a_{k+j}(t+\frac j{\lambda })} \frac{( (a_{k+j}(t+\tfrac j{\lambda })-u)}{g(u)}\, du\, dt. \]

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

\[ k(1+2(k+j))\leq 2k(k+j)+2j(k+j)=2(k+j)^2, \]

if $t\geq \frac k{\lambda }$, then

\[ \tfrac {k}{\lambda }\leq \Big( \tfrac {2(k+j)}{1+2(k+j)}\Big(\tfrac {k}{\lambda }+\tfrac {j}{\lambda }\Big)\leq a_{k+j}(t+\tfrac j{\lambda }). \]

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

\begin{align*} \frac{1}{\varrho (u)}& =(1+u)^m\leq (1+(a_{k+j}(\tfrac {k+1+j}{\lambda }))^m \leq (1+(\tfrac {2k+2}{\lambda }))^m \\ & \leq (1+(\tfrac {4k}{\lambda }))^m\leq 4^m(1+\tfrac k{\lambda })^m\leq \frac{4^m\cdot 2^m}{\varrho (\frac k{\lambda +2})}. \end{align*}

On the other hand, since $\varphi ^{2+2\beta }(x)$ increases, if $\frac k{\lambda } \leq u$, then

\[ \frac{1}{\varphi ^{2+2\beta }(u)}\leq \frac{1}{\varphi ^{2+2\beta }(\frac k{\lambda })} \leq \frac{1}{\varphi ^{4\beta }(\frac k{\lambda +2})}. \]

Therefore,

\begin{align*} | J_{\lambda ,k,j}(g”)| \leq & \tfrac {2^{3m}}{g(\frac k{\lambda +2})}\int _{I_{\lambda ,k}} \int _{k/\lambda }^{a_{k+j}(t+j/\lambda )} \big(a_{k+j}(t+\tfrac j{\lambda })-u\big) du dt \\ = & \tfrac {2^{3m}}{2\, g(\frac k{\lambda +2})}\int _{I_{\lambda ,k}} \big(a_{k+j}(t+\tfrac j{\lambda })-\tfrac k{\lambda }\big)^2 dt \\ \leq & \tfrac {2^{3m}}{2\, g(\frac k{\lambda +2})}\int _{k/\lambda }^{(k+1)/\lambda } \big((t+\tfrac j{\lambda })-\tfrac k{\lambda }\big)^2 dt \\ \leq & \tfrac {2^{3m}}{2\, g(\frac k{\lambda +2})} \int _{k/\lambda }^{(k+1)/\lambda } \big(\tfrac {1+j}{\lambda }\big)^2dt \leq \tfrac {9\cdot 2^{3m}}{2\, g(\frac k{\lambda +2})}\tfrac {1}{\lambda ^3}. \end{align*}

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

\begin{equation} \label{acotaak} \tfrac {2}{3}\leq a_{k} <1. \end{equation}

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

\[ \varphi ^2(a_{k}t)=a_{k}t(1+a_{k}t)\geq \tfrac {2}{3}t(1+\tfrac {2t}3) \geq \tfrac {4}{9}\varphi (\tfrac k{\lambda }) \geq \tfrac {4}{9}\varphi (\tfrac k{\lambda +2}). \]

On the other hand, since

\[ 1+\tfrac {k}{\lambda }\leq 2+\tfrac {k}{\lambda +2} \tfrac {\lambda +2}{\lambda } \leq 2\Big(1+\tfrac {k}{\lambda +2} \Big), \]

one has

\begin{equation} \label{cambialask} \frac{1}{\varrho (\tfrac k{\lambda })}\leq \frac{2^m}{\varrho (\frac k{\lambda +2})}. \end{equation}

Moreover

\[ (k+1) a_k= \tfrac {2k(k+1)}{2k+1}\geq \tfrac {k(2k+1)}{2k+1}=k. \]

Therefore

\begin{align*} & \int _{I_{\lambda ,k}}\left| \int _{k/\lambda }^{a_{k}t} \tfrac {a_{k}t-u} {g(u)}du \right| dt = \\ & =\int _{k/\lambda }^{k/(\lambda a_{k})} \int ^{k/\lambda }_{a_{k}t} \tfrac {u-a_{k}t} {g(u)}du dt +\int _{k/(\lambda a_{k})}^{(k+1)/\lambda } \int _{k/\lambda }^{a_{k}t} \tfrac {a_{k}t-u} {g(u)}du dt \\ & \leq \tfrac {C_1}{g(k/(\lambda +2))} \Big(\int _{k/n}^{k/(\lambda a_{k})} \int ^{k/\lambda }_{a_{k}t} (u-a_{k}t)du dt +\int _{k/(na_{k})}^{(k+1)/\lambda } \int _{k/\lambda }^{a_{k}t} (a_{k}t-u) du dt\Big) \\ & = \tfrac {C_1}{2g(\frac k{\lambda +2})} \Big(\int _{k/\lambda }^{k/(\lambda a_{k})} (\tfrac k{\lambda } -a_{k}t)^2 dt +\int _{k/(\lambda a_{k})}^{(k+1)/\lambda } (a_{k}t-\tfrac k{\lambda } )^2 dt\Big) \\ & = \tfrac {C_1}{6a_k\, g(\frac k{\lambda +2}) } \Big(\tfrac {k^3}{\lambda ^3}(1-a_{k})^3 + \Big(\tfrac {a_{k}(k+1)}{\lambda }-\tfrac {k}{\lambda }\Big)^3 \Big) \\ & \leq \tfrac {C_1}{4\, g(\frac k{\lambda +2})} \Big(\tfrac {k^3}{\lambda ^3}\tfrac {1}{(2k+1)^3} + \Big(\tfrac {k+1}{\lambda }-\tfrac {k}{\lambda }\Big)^3 \Big) \leq \tfrac {C_2}{ g(\frac k{\lambda +2} )}\tfrac {1}{\lambda ^3}, \end{align*}

where we use 12.â–¡

Proposition 8

If $\lambda \geq 2$, $\beta \in [0,1)$, there exists a constant $C$ such that, if $x{\gt}0$, $k\in \{ 1,2\} $, and $\lambda x{\lt} 2/3$, then

\[ \varphi ^{2\beta }(x)v_{\lambda ,k}(x)\lambda \int _{k/\lambda }^{(k+1)/\lambda } \left| \int _x^{a_kt}\frac{(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}\right|dt\leq \tfrac {C(k+1)^2}{\lambda ^{2}}. \]

Proof â–¼

Since $x{\lt}2/(3\lambda )\leq a_k/\lambda \leq a_kk/\lambda $ and $0{\lt}x{\lt}1$,

\begin{align*} & \varphi ^{2\beta }(x)v_{\lambda ,k}(x)\lambda \int _{k/\lambda }^{(k+1)/\lambda } \int _x^{a_kt}\tfrac {(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)} dt \leq \\ & \leq \tfrac {x^k}{\varphi ^{2}(x)\sqrt{1+x}}\, \lambda \int _{k/\lambda }^{(k+1)/\lambda } (1+t)^m (t-x)^{2} \int _x^{t} \tfrac {ds}{s^{1/2}} dt \\ & =\tfrac {2x^k}{\varphi ^{2}(x)\sqrt{1+x}}\lambda \int _{k/\lambda }^{(k+1)/\lambda } (1+t)^m (t-x)^{2} (\sqrt{t}-\sqrt{x}) dt \\ & \leq C_1x^{k-1}\Big(1+\tfrac {k+1}{\lambda }\Big)^m \lambda \int _{k/\lambda }^{(k+1)/\lambda } (t-x)^{2} dt \leq C_2 \Big(\tfrac {k+1}{\lambda }\Big)^2. \end{align*}

Proposition 9

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

\[ I_{\lambda ,i}(x)\leq C \frac{(1+x)^{m}}{\varphi ^{2\beta }(x)}\frac{\varphi ^2(x)}{\lambda }. \]

where we use the notations in 3–5.

Proof â–¼

We will use the notation in 6. Notice that $4\beta {\lt}4\leq 2m$, thus we can apply proposition 6 with $\lambda \geq 2(1+2m)\geq 2(1+2m-4\beta )$.

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

\begin{align*} I_1(\lambda , x) & = \sum \limits _{k=1}^\infty \frac{v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac kn)} \Big(\tfrac {k}{\lambda }-x\Big)^2 \\ & \leq \sqrt{A_{\lambda ,2,4\beta }(x)} \Big( \sum \limits _{k=1}^\infty v_{\lambda ,k}(x)\Big(\tfrac {k}{\lambda }-x\Big)^4\Big)^{1/2} \leq C_1\frac{(1+x)^{m}}{\varphi ^{2\beta }(x)}\frac{\varphi ^2(x)}{\lambda }. \end{align*}

On the other hand

\begin{align*} I_2(\lambda , x)= & \tfrac {1+2x}{\lambda } \sum \limits _{k=1}^\infty \Big|\tfrac {k}{\lambda }-x\Big| \tfrac {v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac kn)} \\ \leq & \tfrac {1+2x}{\lambda } \sqrt{A_{\lambda ,2,4\beta }(x)} \Big( \sum \limits _{k=1}^\infty v_{\lambda ,k}(x)\big(\tfrac {k}{n}-x\big)^2\Big)^{1/2} \\ \leq & C_2\tfrac {(1+x)^{m}}{\varphi ^{2\beta }(x)}\tfrac {(1+x)}{\lambda }\sqrt{\tfrac {x(1+x)}{\lambda }} = C_2\tfrac {(1+x)^{m}}{\varphi ^{2\beta }(x)}\tfrac {x(1+x)}{\lambda }\sqrt{\Big(1+\tfrac {1}{x}\Big)\tfrac {1}{\lambda }} \\ \leq & C_2\tfrac {(1+x)^{m}}{\varphi ^{2\beta }(x)}\tfrac {\varphi ^2(x)}{\lambda }\sqrt{\tfrac {1+\lambda }{\lambda }} \leq \sqrt{2}C_2\tfrac {(1+x)^{m}}{\varphi ^{2\beta }(x)}\tfrac {\varphi ^2(x)}{\lambda }. \end{align*}

Finally

\[ \hspace{7mm} \frac{\lambda }{\varphi ^2(x)} I_{\lambda ,3}(x)= \sum \limits _{k=1}^\infty \frac{v_{\lambda ,k}(x)}{\varrho (\frac k{\lambda })\varphi ^{2\beta }(\frac k{\lambda })} =A_{\lambda ,1,2\beta }(x) \leq C_3 \frac{(1+x)^{m}}{\varphi ^{2\beta }(x)}. \]

5 Bernstein type inequalities

Theorem 2

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

\[ \Vert \varphi ^{2} M_\lambda ''(f)\Vert _{\varrho ,\beta }\leq C\, \lambda \, \Vert f\Vert _{\varrho ,\beta }. \]

Proof â–¼

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

\begin{align*} \varrho (x)\varphi ^{2+2\beta }(x) | M_\lambda ”(f,x)| & =\frac{\varrho (x)\lambda ^3}{\varphi ^{2(1-\beta )}(x)} \Big|\sum \limits _{k=0}^\infty Q_{\lambda ,k}(f)R_{\lambda ,k}(x)v_{\lambda ,k}(x)\Big| \\ & \leq \frac{\varrho (x)\lambda ^2\Vert f\Vert _{\varrho ,\beta }}{\varphi ^{2(1-\beta )}(x)} \Big(I_1(\lambda ,x)+I_2(\lambda ,x)+I_3(\lambda ,x)\Big) \\ & \leq C\lambda \, \Vert f\Vert _{\varrho ,\beta }. \end{align*}

Theorem 3

Suppose $\beta \in [0,1)$ and $m\geq 2$. There exist a constant $\Lambda _1$ such that, if $\lambda \geq 2(1+2m)$ and $g \in D(\varrho ,\beta )$, then

\[ \Vert \varphi ^{3} M_{\lambda }'''(g)\Vert _{\varrho ,\beta } \leq \Lambda _1\, \sqrt{\lambda } \, \Vert \varphi ^{2} g''\Vert _{\varrho ,\beta }. \]

Proof â–¼

Set $Z(f)=\Vert \varphi ^{2} g''\Vert _{\varrho ,\beta }$. From ?? we obtain

\begin{align*} & (\varrho \varphi ^{3+2\beta })(x)| M_{\lambda }”’(g,x)| \leq \\ & \leq C_1\, Z(f)\, (\varrho \varphi ^{1+2\beta })(x)\lambda ^4 \sum \limits _{k=1}^\infty \Big(\sum \limits _{j=0}^2\tbinom {2}{j}\mid J_{n,k,j}\mid \Big|\tfrac {k}{\lambda +2}- x\Big|v_{\lambda +2,k}(x) \\ & \leq C_2 Z(f) (\varrho \varphi ^{1+2\beta })(x) \, \lambda \, \sum \limits _{k=1}^\infty \Big|\tfrac {k}{\lambda +2}- x\Big| \tfrac {v_{\lambda +2,k}(x)}{\varrho (\frac{k}{\lambda +2})\varphi ^{2+2\beta }(\frac{k}{\lambda +2})} \\ & \leq C_2 \, Z(f)\, (\varrho \varphi ^{1+2\beta })(x)\lambda \sqrt{A_{\lambda +2,2,2(1+\beta )}(x)} \sqrt{ V_{\lambda +2}((e_1-x)^2,x)} \\ & \leq C_3\, Z(f)\, (\varrho \varphi ^{1+2\beta })(x) \tfrac {(1+x)^{m}}{\varphi ^{2+2\beta }(x)}\lambda \tfrac {\varphi (x)}{\sqrt{\lambda +2}}\leq C_4\, Z(f)\, \sqrt{\lambda }. \end{align*}

6 A Voronovskaya type theorem

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

Theorem 4

If $\beta \in [0,1]$ and $m\geq 2$, there exists a constant $\Lambda _2$ such that, for all $ \lambda {\gt} 2(1+m)$ and every $ f\in C_{\varrho ,\beta }[0,\infty )$, one has

\[ \Vert M_\lambda (f)\Vert _{\varrho ,\beta } \leq \Lambda _2 \Vert f\Vert _{\varrho ,\beta }. \]

Let

\[ C^3_{\varrho ,\beta }[0,\infty )=\Big\{ g\in C_{\varrho ,\beta }[0,\infty )\, :\, g,g',g''\in AC_{loc},\, \, \, \Vert \varphi ^{3}g'''\Vert _{\varrho ,\beta }{\lt}\infty \Big\} . \]

Theorem 5

Suppose $\beta \in [0,1)$ and $m\geq 3$ $($or $m=2$ and $\beta \in [0,1/2])$. There exists a constant $\Lambda _3$ such that, if $g\in C^3_{\varrho ,\beta }[0,\infty )$ and $\lambda \geq 2(1+2m)$, then

\[ \Big\Vert M_\lambda (g)-g-\tfrac {g”}{2}M_{\lambda ,2}\Big\Vert _{\varrho ,\beta } \leq \, \tfrac {\Lambda _3}{\, \lambda ^{3/2}}\, \Vert \varphi ^{3}g'''\Vert _{\varrho .\beta }\, . \]

Proof â–¼

Let us denote

\[ R_\lambda (g,x)=M_\lambda (g,x)-g(x)-\tfrac {g”(x)}{2}M_\lambda ((t-x)^2,x). \]

By Taylor’s expansion

\[ g(t)=g(x)+g'(x)(x-t)+\tfrac {1}{2}g''(x)(t-x)^2+\tfrac {1}{2}\int _x^t g'''(s)(t-s)^2ds, \]

one has

\begin{align*} & \varphi ^{2\beta }(x)\varrho (x)\mid R_\lambda (g,x)\mid = \\ & =\varphi ^{2\beta }(x)\varrho (x)\left| \tfrac {\lambda }{2}\sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x)\Big(\int _{k/\lambda }^{(k+1)/\lambda } \int _x^{a_kt}\left(a_kt-s\right)^{2}g”’(s)ds \Big)dt\right| \\ & \leq \, \varphi ^{2\beta }(x)\varrho (x)\Vert \varphi ^{3}g”’\Vert _{\varrho ,\beta } \tfrac {\lambda }{2} \sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\left| \int _x^{a_kt}\tfrac {(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}\right|dt \\ & =\tfrac {1}{2}\, \Vert \varphi ^{3}g”’\Vert _{\varrho ,\beta }\, F_n(x), \end{align*}

where

\[ F_n(x)= \varphi ^{2\beta }(x)\varrho (x)\lambda \sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\left| \int _x^{a_kt}\tfrac {(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}\right|dt. \]

Case 1. Assume $\lambda x {\lt}2/3.$

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

Since $a_0=0$, one has

\begin{align*} \tfrac {(\varrho \varphi ^{2\beta })(x)}{(1+x)^\lambda }\lambda \int _{0}^{1/\lambda } \int _0^x\tfrac {s^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}dt & \leq \varrho (x)x^\beta \int _0^x s^{(1-2\beta )/2}(1+s)^{m}ds \\ & \leq \varrho (x) x^\beta (1+x)^m \int _0^x s^{(1-2\beta )/2}ds \\ & = \tfrac {2x^{\beta }x^{3/2-\beta }}{(3-2\beta )} \leq \tfrac {2}{(3-2\beta )}\tfrac {1}{\lambda ^{3/2}}. \end{align*}

On the other hand, it follows from proposition 8 that

\[ (\varrho \varphi ^{2\beta })(x)\lambda \sum _{k=1}^2 v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda } \left| \int _x^{a_kt}\tfrac {(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}\right|dt \leq \tfrac {C_1(2^2+3^2)}{\lambda ^{2}}. \]

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

\[ \tfrac {k}{\lambda }v_{\lambda ,k}(x)=x v_{\lambda +1,k-1}(x). \]

In particular

\[ \tfrac {k}{\lambda }\tfrac {k-1}{\lambda +1}\tfrac {k-2}{\lambda +2}v_{\lambda ,k}(x)=x^3 v_{\lambda +3,k-3}(x). \]

From this we obtain

\begin{align*} & (\varrho \varphi ^{2\beta })(x)\, \lambda \, \sum \limits _{k=3}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\left| \int _x^{a_kt}\tfrac {(a_kt-s)^{2}ds}{(\varrho \varphi ^{3+2\beta } (s)}\right|dt = \\ & =(\varrho \varphi ^{2\beta })(x)\, \lambda \, \sum \limits _{k=3}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\int _x^{a_kt}\tfrac {(a_kt-s)^{2}(1+s)^m\, ds}{ \varphi ^{3+2\beta }(s)}dt \\ & \leq \tfrac {\lambda \, \varrho (x)}{\varphi ^{3} (x)} \sum \limits _{k=3}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\int _x^{t}(t-s)^{2}(1+s)^{m}\, ds\, dt \\ & \leq \tfrac {\lambda \, \varrho (x)}{\varphi ^{3}(x)} \sum \limits _{k=1}^{\infty }\Big(1+\tfrac {k+1}{\lambda }\Big)^m v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\, t^{3}dt \\ & \leq \tfrac {C_2\, \varrho (x)}{\varphi ^{3} (x)} \sum \limits _{k=3}^{\infty } \Big(1+\tfrac {k}{\lambda }\Big)^m\tfrac {k^3}{\lambda ^3} v_{\lambda ,k}(x) \\ & \leq \tfrac {C_3\, \varrho (x)}{\varphi ^{3} (x)} \sum \limits _{k=3}^{\infty } v_{\lambda ,k}(x)\Big(1+\tfrac {k}{\lambda +3}\Big)^m \tfrac {k}{\lambda } \tfrac {k-1}{\lambda +1} \tfrac {k-2}{\lambda +2} \\ & \leq C_4\, \varrho (x)\, x^{3/2}\, \sum \limits _{k=3}^{\infty } v_{\lambda +3,k-3}(x)\Big(1+\tfrac {k-3}{\lambda +3}\Big)^m \leq C_5\, x^{3/2} \leq \tfrac {C_6}{\lambda ^{3/2}}, \end{align*}

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

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

From Proposition 3.3 of [ 3 ] we know that

\begin{align*} \left| \int _x^{a_kt} \frac{(a_kt-u)^{2}du}{(\varrho \varphi ^{3+2\beta } (u)}\right| & = \left| \int _x^{a_kt} \tfrac {(a_kt-u)^{2}(1+u)^{c}du}{u^{(3+2\beta )/2} }\right| \\ & \leq \tfrac {| a_kt-x|^3}{(3-(1+2\beta )/2)x^{(3+2\beta )/2}}\Big((1+x)^c+(1+t)^c\Big) \end{align*}

If $\lambda x\geq 2/3$, since $ 3/2+\beta {\lt} 5/2 {\lt} 3$, one has

\begin{align*} F_n(x)= & (\varrho \varphi ^{2\beta })(x)\, \lambda \, \sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda }\left| \int _x^{a_kt} \tfrac {(a_kt-u)^{2}du}{(\varrho \varphi ^{3+2\beta } (u)}\right|dt \\ \leq & \tfrac {C_1\varphi ^{2\beta }(x)\varrho (x)\lambda }{x^{(3+2\beta )/2}}\sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x)\int _{k/\lambda }^{(k+1)/\lambda } \mid a_kt-x\mid ^3 \Big((1+x)^c+(1+t)^c\Big)dt \\ = & \tfrac {C_1\varphi ^{2\beta }(x)\varrho (x)(1+x)^c}{x^{(3+2\beta )/2}}M_{\lambda ,3}(x) \\ & + \tfrac {C_1\varphi ^{2\beta }(x)\varrho (x)}{x^{(3+2\beta )/2}}\, \lambda \sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x) \int _{k/\lambda }^{(k+1)/\lambda } | a_kt-x|^3 (1+t)^c dt \\ = & \tfrac {C_1}{\varphi ^{3}(x)}M_{\lambda ,3}(x) \\ & + \tfrac {C_1(1+x)^{\beta }}{x^{3/2}(1+x)^m}\, \lambda \sum \limits _{k=0}^{\infty } v_{\lambda ,k}(x) \int _{k/\lambda }^{(k+1)/\lambda } | a_kt-x|^3 (1+t)^c dt. \end{align*}

It follows from proposition 4 that

\[ \tfrac {1}{\varphi ^{3}(x)}M_\lambda (\mid t-x\mid ^3,x) \leq \tfrac {C_2}{\lambda ^{3/2}}. \]

For the other terms we first estimate the case $k=0$. Notice that, for $t\in (0,\frac1{\lambda })$,

\[ (1+t)^c\leq (1+\tfrac 1{\lambda })^c\leq (1+\tfrac {3x}2)^c\leq 2^c(1+x)^c. \]

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

\begin{align*} \tfrac {(1+x)^{\beta }}{x^{3/2}(1+x)^m}\, \lambda v_{\lambda ,0}(x) \int _{0}^{1/\lambda } x^3 (1+t)^c dt & \leq \tfrac {2^c(1+x)^{\beta }x^3}{x^{3/2}(1+x)^m}\, v_{\lambda ,0}(x) (1+x)^c \\ & =\tfrac {2^cx^3}{x^{3/2}(1+x)^{3/2}}\, v_{\lambda ,0}(x) \\ & \leq \tfrac {2^c}{\varphi ^{3}(x)}V_\lambda (\mid t-x\mid ^3,x) \leq \tfrac {C_3}{\lambda ^{3/2}}. \end{align*}

On the other hand, since

\begin{align*} & \int _{k/\lambda }^{(k+1)/\lambda } \mid a_kt-x\mid ^3 (1+t)^c dt \leq \\ & \leq \Big(\int _{k/\lambda }^{(k+1)/\lambda } ( a_kt-x)^6 dt\Big)^{1/2}\Big( \int _{k/\lambda }^{(k+1)/\lambda } (1+t)^{2c} dt\Big)^{1/2}, \end{align*}

from theorem 4 and proposition 4 we obtain

\begin{align*} & \tfrac {(1+x)^{\beta }}{x^{3/2}(1+x)^m}\, \lambda \sum \limits _{k=1}^{\infty } v_{\lambda ,k}(x) \int _{k/\lambda }^{(k+1)/\lambda } \mid a_kt-x\mid ^3 (1+t)^c dt \leq \\ & \leq \tfrac {(1+x)^{\beta }}{x^{3/2}(1+x)^m}\sqrt{M_\lambda ((1+t)^{2c},x)} \sqrt{M_\lambda ((e_1-x)^6,x)} \\ & \leq \tfrac {C_4(1+x)^{\beta }}{x^{3/2}(1+x)^m}\, (1+x)^{m-3/2-\beta } \tfrac {\varphi ^3(x)}{\lambda ^{3/2}} =\tfrac {C_4}{\lambda ^{3/2}}. \end{align*}

This completes the proof.â–¡

Remark 2

We do not know if theorem 5 holds $m=2$ and $\beta \in (\frac12,1)$.

7 Inverse result

Theorem 6

Suppose $\beta \in [0,1)$ and $m\geq 3$ $($or $m=2$ and $\beta \in [0,1/2])$. There exist positive constants $\kappa $ and $\Lambda _4$ such that, if $f\in C_\varrho [0,\infty )$ and $\lambda \geq 2(1+2m)$, then

\[ \tfrac {1}{\lambda }\Big\Vert \varphi ^{2} (M_\lambda ^2(f))''\Big\Vert _{\varrho ,\beta } \leq \Lambda _3\Big( \Vert M_\lambda (f)- f\Vert _{\varrho ,\beta }+\Vert M_{\kappa \, \lambda }(f)- f \Vert _{\varrho ,\beta }\Big), \]

where $M_\lambda ^2(f)=M_\lambda (M_\lambda (f))$.

Proof â–¼

Let $\Lambda _1$ and $\Lambda _3$ be constants such that the inequalities in theorem 3 and theorem 5 hold for $\lambda \geq 2(1+2m)$. Set

\[ \mu =16(\Lambda _1\Lambda _3)^2\, \lambda . \]

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

\begin{align*} & \tfrac {1}{2\mu } \Big\Vert \varphi ^{2} (M^2_\lambda )” f\Big\Vert _{\varrho ,\beta } \leq \\ \leq & \tfrac {1}{2} \Vert M_{\mu ,2}(M^2_\lambda )” f\Vert _{\varrho ,\beta } \\ \leq & \Big|\Big| M_\mu (M_\lambda ^2 f)-M^2_\lambda f\Big|\Big|_{\varrho ,\beta } + \Big|\Big| M_\mu (M_\lambda ^2 f)-M^2_\lambda f -\tfrac {M_{\mu ,2}}{2} (M_\lambda ^2f)”\Big|\Big|_{\varrho ,\beta } \\ \leq & \Vert M_\mu (M_\lambda ^2f \! -\! M_\lambda f)+ M_\mu (M_\lambda f \! -\! f)+M_\mu f \! - \! f \! + \! f \! -M_\lambda f\! +M_\lambda f -M_\lambda ^2 f\Vert _{\varrho ,\beta } \\ & + \tfrac {\Lambda _3}{\, \mu ^{3/2}}\, \Vert \varphi ^{3}(M_\lambda ^2f)”’\Vert _{\varrho , \beta } \\ \leq & C_1 \Big( \Vert M_\mu f - f \Vert _{\varrho ,\beta } +\Vert M_\lambda f -f \Vert _{\varrho ,\beta }\Big) + \tfrac {\Lambda _1\Lambda _3\sqrt{\lambda }}{\, \mu ^{3/2}}\, \Vert \varphi ^{2}M_\lambda ”(f)\Vert _{\varrho ,\beta } \\ \leq & C_1 \Big( \Vert M_\mu f - f \Vert _{\varrho ,\beta } +\Vert M_\lambda f -f \Vert _{\varrho ,\beta }\Big) \\ & + \tfrac {\Lambda _1\Lambda _3\sqrt{\lambda }}{\, \mu ^{3/2}}\, \Vert \varphi ^{2}M_\lambda ”(f-M_\lambda (f))\Vert _{\varrho ,\beta } +\tfrac {\Lambda _1\Lambda _3\sqrt{\lambda }}{\, \mu ^{3/2}}\, \Vert \varphi ^{2}M_\lambda ”(M_\lambda (f))\Vert _{\varrho ,\beta } \\ \leq & C_1 \Big( \Vert M_\mu f - f \Vert _{\varrho ,\beta } +\Vert M_\lambda f -f \Vert _{\varrho ,\beta }\Big) \\ & + \tfrac {C\Lambda _1\Lambda _3\, \lambda ^{3/2}}{\, \mu ^{3/2}}\, \Vert f-M_\lambda (f)\Vert _{\varrho ,\beta } +\tfrac {1}{4\, \mu }\, \Vert \varphi ^{2}M_\lambda ”(M_\lambda (f))\Vert _{\varrho ,\beta }. \end{align*}

Therefore

\[ \tfrac {1}{4\mu } \Vert \varphi ^{2} (M^2_\lambda )'' f\Vert _{\varrho ,\beta } \leq C_2 \Big( \Vert M_\mu f - f \Vert _{\varrho ,\beta } +\Vert M_\lambda f -f \Vert _{\varrho ,\beta }\Big), \]

and it is sufficient to prove the result, because

\[ \tfrac {1}{\lambda }\left|\left| \varphi ^{2} (M_\lambda ^2(f))''\right|\right|_{\varrho , \beta } =\tfrac {2^6(\Lambda _1\Lambda _2)^2}{4\mu }\left|\left| \varphi ^{2} (M_\lambda ^2(f))''\right|\right|_{\varrho ,\beta }. \]

Theorem 7

Suppose $\beta \in [0,1)$, $m\geq 3$ $($or $m=2$ and $\beta \in [0,\frac12])$, and $\kappa $ is given as in theorem 6. There exists a constant $C$ such that, if $f\in C_\varrho [0,\infty )$ and $\lambda \geq 2(1+2m)$, then

\[ K_\beta \left(f,\tfrac {1}{\lambda }\right)_\varrho \leq C\Big( \| M_\lambda (f)- f \| _{\varrho ,\beta }+\| M_{\kappa \lambda }(f)- f\| _{\varrho ,\beta }\Big), \]

where $K_\beta (f,t)_\varrho $ is defined as in 1.

Proof â–¼

Fix $f\in C_\varrho [0,\infty )$ and denote $g=M_\lambda ^2(f)=M_\lambda (M_\lambda (f))$. From theorem 4 we know that $g\in C_{\varrho ,b}[0,\infty )$.

From the definition of the $K$-functional $K_\beta (f,t)_\varrho $, theorem 4 and theorem 6 we know that

\begin{align*} K_\beta \left(f,\tfrac {1}{\lambda }\right)_\varrho \leq & \| f-M_\lambda ^2(f)\| _{\varrho ,\beta } +\tfrac {1}{\lambda }\, \| \varphi ^{2}( M_\lambda ^2(f))”\| _{\varrho ,\beta } \\ \leq & \| f-M_\lambda (f)\| _{\varrho ,\beta }+\| M_\lambda (f-M_\lambda (f))\| _{\varrho ,\beta } +\tfrac {1}{\lambda }\Big\| \varphi ^{2} (M_\lambda ^2(f))”\Big\| _{\varrho ,\beta } \\ \leq & (1+\Lambda _2) \big\| f-M_\lambda (f)\big\| _{\varrho ,\beta }+\tfrac {1}{\lambda }\Big\| \varphi ^2 (M_\lambda ^2(f))” \Big\| _{\varrho ,\beta } \\ \leq & C\Big( \| M_\lambda (f)- f \| _{\varrho ,\beta }+\| M_{\kappa ,\lambda }(f)- f\| _{\varrho ,\beta }\Big). \end{align*}

Remark 3

In the case $\beta =0$ 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–¡