Caputo Fractional Approximation by Sublinear Operators
September 25, 2017. Accepted: October 12, 2018. Published online: February 17, 2019.
\(^\ast \)Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A, e-mail: ganastss@memphis.edu.
Here we consider the approximation of functions by sublinear positive operators with applications to a big variety of Max-Product operators under Caputo fractional differentiability. Our study is based on our general fractional results about positive sublinear operators. We produce Jackson type inequalities under simple initial conditions. So our approach is quantitative by producing inequalities with their right hand sides involving the modulus of continuity of fractional derivative of the function under approximation.
MSC. 26A33, 41A17, 41A25, 41A36.
Keywords. positive sublinear operators, Max-product operators, modulus of continuity, Caputo fractional derivative
1 Introduction
The main motivation here is the monograph by B. Bede, L. Coroianu and S. Gal [ 5 , 2016 ] .
Let \(N\in \mathbb {N}\), the well-known Bernstein polynomials [ 11 ] are positive linear operators, defined by the formula
T. Popoviciu in [ 12 ] (1935), proved for \(f\in C\left( \left[ 0,1\right] \right) \) that
where
is the first modulus of continuity, here \(\left[ a,b\right] =\left[ 0,1\right] \).
G.G. Lorentz in [ 11 , p. 21 ] (1986), proved for \(f\in C^{1}\left( \left[ 0,1\right] \right) \) that
In [ 5 , p. 1 ] , the authors introduced the basic Max-product Bernstein operators,
where \(\bigvee \) stands for maximum, and \(p_{N,k}\left( x\right) =\tbinom Nk x^{k}\left( 1-x\right) ^{N-k}\) and \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}=[0,\infty ).\)
These are nonlinear and piecewise rational operators.
The authors in [ 5 ] studied similar such nonlinear operators such as: the Max-product Favard-Szász-Mirakjan operators and their truncated version, the Max-product Baskakov operators and their truncated version, also many other similar specific operators. The study in [ 5 ] is based on presented there general theory of sublinear operators. These Max-product operators tend to converge faster to the on hand function.
So we mention from [ 5 , p. 30 ] , that for \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) continuous, we have the estimate
Also from [ 5 , p. 36 ] , we mention that for \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) being concave function we get that
a much faster convergence.
In this article we expand the study in [ 5 ] by considering Caputo fractional smoothness of functions. So our inequalities are with respect to \(\omega _{1}\left( D^{\alpha }f,\delta \right) \), \(\delta {\gt}0\), where \(D^{\alpha }f\) with \(\alpha {\gt}0\) is the Caputo fractional derivative.
2 Main Results
We need
Let \(\nu \geq 0\), \(n=\left\lceil \nu \right\rceil \) (\(\left\lceil \cdot \right\rceil \) is the ceiling of the number), \(f\in AC^{n}\left( \left[ a,b\right] \right) \) (space of functions \(f\) with \(f^{\left( n-1\right) }\in AC\left( \left[ a,b\right] \right) \), absolutely continuous functions). We call left Caputo fractional derivative (see [ 7 , p. 49 ] , [ 10 ] , [ 13 ] ) the function
where \(\Gamma \) is the gamma function \(\Gamma \left( v\right) =\int _{0}^{\infty }e^{-t}t^{v-1}dt\), \(v{\gt}0.\)
We set \(D_{\ast a}^{0}f\left( x\right) =f\left( x\right) \), \(\forall \) \(x\in \left[ a,b\right] .\)
[ 2 ] Let \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\), \(n=\left\lceil \nu \right\rceil \), \(f\in C^{n-1}\left( \left[ a,b\right] \right) \) and \(f^{\left( n\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Then \(D_{\ast a}^{\nu }f\left( a\right) =0.\)
We need
(see also [ 1 ] , [ 8 ] , [ 10 ] ) Let \(f\in AC^{m}\left( \left[ a,b\right] \right) ,\) \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0 \). The right Caputo fractional derivative of order \(\alpha {\gt}0\) is given by
We set \(D_{b-}^{0}f\left( x\right) =f\left( x\right) \).
[ 2 ] Let \(f\in C^{m-1}\left( \left[ a,b\right] \right) ,\) \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) ,\) \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\). Then \(D_{b-}^{\alpha }f\left( b\right) =0.\)
We assume that
and
for all \(x,x_{0}\in \left[ a,b\right] .\)
We mention
[ 2 ] Let \(f\in C^{n}\left( \left[ a,b\right] \right) \), \(n=\left\lceil \nu \right\rceil \), \(\nu {\gt}0\). Then \(D_{\ast a}^{\nu }f\left( x\right) \) is continuous in \(x\in \left[ a,b\right] .\)
[ 2 ] Let \(f\in C^{m}\left( \left[ a,b\right] \right) \), \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\). Then \(D_{b-}^{\alpha }f\left( x\right) \) is continuous in \(x\in \left[ a,b\right] .\)
The modulus of continuity \(\omega _{1}\left( f,\delta \right) \) is defined the same way for bounded functions, see (1.3), and it is finite.
We make
[ 2 ] Let \(f\in C^{n-1}\left( \left[ a,b\right] \right) \), \(f^{\left( n\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \), \(n=\left\lceil \nu \right\rceil \), \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\). Then
Similarly, let \(f\in C^{m-1}\left( \left[ a,b\right] \right) \), \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \), \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), then
That is \(\omega _{1}\left( D_{\ast a}^{\nu }f,\delta \right) \), \(\omega _{1}\left( D_{b-}^{\alpha }f,\delta \right) \) are finite.
Clearly, above \(D_{\ast a}^{\nu }f\) and \(D_{b-}^{\alpha }f\) are bounded, from
see [ 2 ] .â–¡
We need
Let \(D_{x_{0}}^{\alpha }f\) denote any of \(D_{x_{0}-}^{\alpha }f\), \(D_{\ast x_{0}}^{\alpha }f\), and \(\delta {\gt}0\). We set
where \(x_{0}\in \left[ a,b\right] .\) Here the moduli of continuity are considered over \(\left[ a,x_{0}\right] \) and \(\left[ x_{0},b\right] \), respectively.
We need
Let \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \) (i.e. \(f^{\left( m-1\right) }\in AC\left( \left[ a,b\right] \right) \), absolutely continuous functions on \(\left[ a,b\right] \)), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\). Then
for all \(a\leq x\leq b.\)
If \(0{\lt}\alpha {\lt}1\), then we do not need initial conditions.
for all \(x_{0}\leq x\leq b.\)
Also from [ 1 ] , using the right Caputo fractional Taylor formula we get
for all \(a\leq x\leq x_{0}.\)
By the assumption \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\), we get
for all \(x_{0}\leq x\leq b.\)
And it holds
for all \(a\leq x\leq x_{0}.\)
Notice that when \(0{\lt}\alpha {\lt}1\), then \(m=1\), and (2.19) and (2.20) are valid without initial conditions.
Since \(D_{x_{0}-}^{\alpha }f\left( x_{0}\right) =D_{\ast x_{0}}^{\alpha }f\left( x_{0}\right) =0\), we get
\(x_{0}\leq x\leq b,\) and
\(a\leq x\leq x_{0}.\)
We have that (\(x_{0}\leq x\leq b\))
We have proved that
\(\delta _{1}{\gt}0\), and \(x_{0}\leq x\leq b.\)
Similarly acting, we get (\(a\leq x\leq x_{0}\))
We have proved that
\(\delta _{2}{\gt}0\), and (\(a\leq x\leq x_{0}\))\(.\) Choosing \(\delta =\delta _{1}=\delta _{2}{\gt}0\), by (2.25) and (2.28), we get (2.16).
We need
Here \(C_{+}\left( \left[ a,b\right] \right) :=\left\{ f:\left[ a,b\right] \rightarrow \mathbb {R}_{+},\ f\text{ continuous}\right\} .\) Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), operators, \(\forall \) \(N\in \mathbb {N}\), such that
(i)
(ii) if \(f,g\in C_{+}\left( \left[ a,b\right] \right) :f\leq g,\) then
(iii)
We call \(\left\{ L_{N}\right\} _{N\in \mathbb {N}}\) positive sublinear operators.
We need a Hölder’s type inequality, see next:
(see [ 3 ] ) Let \(L:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), be a positive sublinear operator and \(f,g\in C_{+}\left( \left[ a,b\right] \right) \), furthermore let \(p,q{\gt}1:\frac{1}{p}+\frac{1}{q}=1\). Assume that \(L\left( \left( f\left( \cdot \right) \right) ^{p}\right) \left( s_{\ast }\right) ,\) \(L\left( \left( g\left( \cdot \right) \right) ^{q}\right) \left( s_{\ast }\right) {\gt}0\) for some \(s_{\ast }\in \left[ a,b\right] \). Then
We make
By [ 5 , p. 17 ] ,we get: let \(f,g\in C_{+}\left( \left[ a,b\right] \right) \), then
Furthermore, we also have that
\(\forall \) \(x\in \left[ a,b\right] \); \(e_{0}\left( t\right) =1.\)
From now on we assume that \(L_{N}\left( 1\right) =1\). Hence it holds
Using Theorem 2.10 and (2.16) with (2.35) we get:
â–¡
We have proved
Let \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Furthermore we assume that
\(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\). Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), \(\forall \) \(N\in \mathbb {N},\) be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Then
\(\delta {\gt}0\), \(\forall \) \(N\in \mathbb {N}\).
In particular 2.37 is true for \(\alpha {\gt}1\), \(\alpha \notin \mathbb {N}\).
Let \(0{\lt}\alpha {\lt}1\), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ a,b\right] \right) \). Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), \(\forall \) \(N\in \mathbb {N},\) be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Then 2.37 is valid.
We give
Let \(0{\lt}\alpha {\lt}1\), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ a,b\right] \right) \). Let \(L_{N}\) from \(C_{+}\left( \left[ a,b\right] \right) \) into itself be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Assume that \(L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) {\gt}0\), \(\forall \) \(N\in \mathbb {N}\). Then
Choose
i.e. \(\delta ^{\alpha +1}=L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) .\)
By (2.37) we obtain
proving (2.38).
3 Applications
I) Case \(0{\lt}\alpha {\lt}1.\)
Here we apply Theorem 2.16 to well known Max-product operators.
We make
The Max-product Bernstein operators \(B_{N}^{\left( M\right) }\left( f\right) \left( x\right) \) are defined by (1.5), see also [ 5 , p. 10 ] ; here \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) is a continuous function.
We have \(B_{N}^{\left( M\right) }\left( 1\right) =1\), and
see [ 5 , p. 31 ] .
\(B_{N}^{\left( M\right) }\) are positive sublinear operators and thus they possess the monotonicity property, also since \(\left| \cdot -x\right| \leq 1\), then \(\left| \cdot -x\right| ^{\beta }\leq 1\), \(\forall \) \(x\in \left[ 0,1\right] \), \(\forall \) \(\beta {\gt}0.\)
Therefore it holds
Furthermore, clearly it holds that
The operator \(B_{N}^{\left( M\right) }\) maps \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. â–¡
We present
Let \(0{\lt}\alpha {\lt}1\), any \(x\in \left( 0,1\right) \), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then
\(\forall \) \(N\in \mathbb {N}.\)
As \(N\rightarrow +\infty \), we get \(B_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), any \(x\in \left( 0,1\right) . \)
We continue with
The truncated Favard-Szász-Mirakjan operators are given by
\(s_{N,k}\left( x\right) =\frac{\left( Nx\right) ^{k}}{k!}\), see also [ 5 , p. 11 ] .
By [ 5 , p. 178-179 ] ,we get that
Clearly it holds
The operators \(T_{N}^{\left( M\right) }\) are positive sublinear operators mapping \(C_{+}\left( \left[ 0,1\right] \right) \) into itself, with \(T_{N}^{\left( M\right) }\left( 1\right) =1\).
Furthermore it holds
â–¡
We give
Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1]\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then
As \(N\rightarrow +\infty \), we get \(T_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1].\)
We make
Next we study the truncated Max-product Baskakov operators (see [ 5 , p. 11 ] )
where
From [ 5 , pp. 217-218 ] , we get (\(x\in \left[ 0,1\right] \))
Let \(\lambda \geq 1\), clearly then it holds
Also it holds \(U_{N}^{\left( M\right) }\left( 1\right) =1\), and \(U_{N}^{\left( M\right) }\) are positive sublinear operators from \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. Furthermore it holds
â–¡
We give
Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1]\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then
\(\forall \text{ }N\geq 2\text{, }N\in \mathbb {N}\).
As \(N\rightarrow +\infty \), we get \(U_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1].\)
We continue with
Here we study the Max-product Meyer-Köning and Zeller operators (see [ 5 , p. 11 ] ) defined by
\(s_{N,k}\left( x\right) = \begin{array}[c]{c}\binom {N+k} {k} \end{array} x^{k}\), \(x\in \left[ 0,1\right] \).
By [ 5 , p. 253 ] , we get that
As before we get that (for \(\lambda \geq 1\))
\(\forall \) \(x\in \left[ 0,1\right] \), \(N\geq 4\), \(N\in \mathbb {N}.\)
Also it holds \(Z_{N}^{\left( M\right) }\left( 1\right) =1\), and \(Z_{N}^{\left( M\right) }\) are positive sublinear operators from \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. Also it holds
We give
Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1)\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then
\(\forall \) \(N\geq 4\), \(N\in \mathbb {N}.\)
As \(N\rightarrow +\infty \), we get \(Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1).\)
We continue with
Here we deal with the Max-product truncated sampling operators (see [ 5 , p. 13 ] ) defined by
and
\(\forall \) \(x\in \left[ 0,\pi \right] \), \(f:\left[ 0,\pi \right] \rightarrow \mathbb {R}_{+}\) a continuous function.
Following [ 5 , p. 343 ] , and making the convention \(\frac{\sin \left( 0\right) }{0}=1\) and denoting \(s_{N,k}\left( x\right) =\frac{\sin \left( Nx-k\pi \right) }{Nx-k\pi }\), we get that \(s_{N,k}\left( \frac{k\pi }{N}\right) =1\), and \(s_{N,k}\left( \frac{j\pi }{N}\right) =0\), if \(k\neq j\), furthermore \(W_{N}^{\left( M\right) }\left( f\right) ( \frac{j\pi }{N}) =f( \frac{j\pi }{N}) \), for all \(j\in \left\{ 0,...,N\right\} .\)
Clearly \(W_{N}^{\left( M\right) }\left( f\right) \) is a well-defined function for all \(x\in \left[ 0,\pi \right] \), and it is continuous on \(\left[ 0,\pi \right] \), also \(W_{N}^{\left( M\right) }\left( 1\right) =1.\)
By [ 5 , p. 344 ] , \(W_{N}^{\left( M\right) }\) are positive sublinear operators.
Call \(I_{N}^{+}\left( x\right) =\left\{ k\in \left\{ 0,1,...,N\right\} ;s_{N,k}\left( x\right) {\gt}0\right\} \), and set \(x_{N,k}:=\frac{k\pi }{N}\), \(k\in \left\{ 0,1,...,N\right\} \).
We see that
By [ 5 , p. 346 ] ,we have
Notice also \(\left| x_{N,k}-x\right| \leq \pi \), \(\forall \) \(x\in \left[ 0,\pi \right] .\)
Therefore (\(\lambda \geq 1\)) it holds
If \(x\in \left( \tfrac {j\pi }{N},\tfrac {\left( j+1\right) \pi }{N}\right) \), with \(j\in \left\{ 0,1,...,N\right\} \), we obtain \(nx-j\pi \in \left( 0,\pi \right) \) and thus \(s_{N,j}\left( x\right) =\tfrac {\sin \left( Nx-j\pi \right) }{Nx-j\pi }{\gt}0\), see [ 5 , pp. 343-344 ] .
Consequently it holds (\(\lambda \geq 1\))
such that \(x\neq x_{N,k}\), for any \(k\in \left\{ 0,1,...,N\right\} .\) â–¡
We give
Let \(0{\lt}\alpha {\lt}1\), any \(x\in \left[ 0,\pi \right] \) be such that \(x\neq \tfrac {k\pi }{N}\), \(k\in \left\{ 0,1,...,N\right\} \), \(\forall \) \(N\in \mathbb {N}\); \(f\in AC\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Then
As \(N\rightarrow +\infty \), we get \(W_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \).
We make
Here we continue with the Max-product truncated sampling operators (see [ 5 , p. 13 ] ) defined by
\(\forall \) \(x\in \left[ 0,\pi \right] \), \(f:\left[ 0,\pi \right] \rightarrow \mathbb {R}_{+}\) a continuous function.
Following [ 5 , p. 350 ] , and making the convention \(\tfrac {\sin \left( 0\right) }{0}=1\) and denoting \(s_{N,k}\left( x\right) =\tfrac {\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}\), we get that \(s_{N,k}\left( \tfrac {k\pi }{N}\right) =1\), and \(s_{N,k}\left( \tfrac {j\pi }{N}\right) =0\), if \(k\neq j\), furthermore \(K_{N}^{\left( M\right) }\left( f\right) \left( \tfrac {j\pi }{N}\right) =f\left( \tfrac {j\pi }{N}\right) \), for all \(j\in \left\{ 0,...,N\right\} .\)
Since \(s_{N,j}\left( \tfrac {j\pi }{N}\right) =1\) it follows that \(\bigvee _{k=0}^{N}s_{N,k}\left( \tfrac {j\pi }{N}\right) \geq 1{\gt}0,\) for all \(j\in \left\{ 0,1,...,N\right\} \). Hence \(K_{N}^{\left( M\right) }\left( f\right) \) is well-defined function for all \(x\in \left[ 0,\pi \right] \), and it is continuous on \(\left[ 0,\pi \right] \), also \(K_{N}^{\left( M\right) }\left( 1\right) =1.\) By [ 5 , p. 350 ] , \(K_{N}^{\left( M\right) } \) are positive sublinear operators.
Denote \(x_{N,k}:=\frac{k\pi }{N}\), \(k\in \left\{ 0,1,...,N\right\} \).
By [ 5 , p. 352 ] , we have
Notice also \(\left| x_{N,k}-x\right| \leq \pi \), \(\forall \) \(x\in \left[ 0,\pi \right] .\)
Therefore (\(\lambda \geq 1\)) it holds
If \(x\in \left( \tfrac {j\pi }{N},\tfrac {\left( j+1\right) \pi }{N}\right) \), with \(j\in \left\{ 0,1,...,N\right\} \), we obtain \(nx-j\pi \in \left( 0,\pi \right) \) and thus \(s_{N,j}\left( x\right) =\tfrac {\sin ^{2}\left( Nx-j\pi \right) }{\left( Nx-j\pi \right) ^{2}}{\gt}0\), see [ 5 , pp. 350 ] .
Consequently it holds (\(\lambda \geq 1\))
such that \(x\neq x_{N,k}\), for any \(k\in \left\{ 0,1,...,N\right\} .\) â–¡
We give
Let \(0{\lt}\alpha {\lt}1\), \(x\in \left[ 0,\pi \right] \) be such that \(x\neq \frac{k\pi }{N}\), \(k\in \left\{ 0,1,...,N\right\} \), \(\forall \) \(N\in \mathbb {N}\); \(f\in AC\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Then
As \(N\rightarrow +\infty \), we get \(K_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \).
When \(\alpha =\frac{1}{2}\) we get:
Let \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then
Due to lack of space we avoid to give other applications when \(\alpha =\frac{1}{2}\) from the other Max-product operators.
II) Case \(\alpha {\gt}1\), \(\alpha \notin \mathbb {N}\).
Here we apply Theorem 2.14 to well known Max-product operators.
We present
Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\) \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,1\right] \), \(f\in AC^{m}\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then
We get \(\underset {N\rightarrow +\infty }{\lim }B_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
Choose \(\delta =\left( \tfrac {6}{\sqrt{N+1}}\right) ^{\frac{1}{\alpha +1}}\), then \(\delta ^{\alpha +1}=\frac{6}{\sqrt{N+1}}\), and apply it to (3.76). Clearly we derive (3.75).
We continue with
Same assumptions as in Theorem 3.14. Then
\(\forall \text{ }N\in \mathbb {N}.\)
We get \(\underset {N\rightarrow +\infty }{\lim }T_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We give
Same assumptions as in Theorem 3.14. Then
\(\forall \text{ }N\in \mathbb {N},\text{ }N\geq 2.\)
We get \(\underset {N\rightarrow +\infty }{\lim }U_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We give
Same assumptions as in Theorem 3.14. Then
\(\forall \text{ }N\in \mathbb {N}, \text{ }N\geq 4.\)
We get \(\underset {N\rightarrow +\infty }{\lim }Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) ,\) where \(\rho \left( x\right) \) is as in 3.59.
We continue with
Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,\pi \right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right)\) \( =0\), \(k=1,...,m-1\). Then
\(\forall \text{ }N\in \mathbb {N}.\)
We have that \(\underset {N\rightarrow +\infty }{\lim }W_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
Choose \(\delta =\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{1}{\alpha +1}}\), i.e. \(\delta ^{\alpha +1}=\tfrac {\pi ^{\alpha +1}}{2N}\), and \(\delta ^{\alpha }=\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}}\). We use the last into (3.81) and we obtain (3.80).
We also have
Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,\pi \right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Furthermore we assume that
\(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then
\( \forall \text{ }N\in \mathbb {N}.\)
We have that \(\underset {N\rightarrow +\infty }{\lim }K_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We make
We mention the interpolation Hermite-Fejer polynomials on Chebyshev knots of the first kind (see [ 5 , p. 4 ] ): Let \(f:\left[ -1,1\right] \rightarrow \mathbb {R}\) and based on the knots \(x_{N,k}=\cos \left( \tfrac {\left( 2\left( N-k\right) +1\right) }{2\left( N+1\right) }\pi \right) \in \left( -1,1\right) \), \(k\in \left\{ 0,...,N\right\} \), \(-1{\lt}x_{N,0}{\lt}x_{N,1}{\lt}...{\lt}x_{N,N}{\lt}1\), which are the roots of the first kind Chebyshev polynomial \(T_{N+1}\left( x\right) =\cos \left( \left( N+1\right) \arccos x\right) \), we define (see Fejér [ 9 ] )
where
the fundamental interpolation polynomials.
The Max-product interpolation Hermite-Fejér operators on Chebyshev
knots of the first kind (see
[
5
,
p.
12
]
) are defined by
where \(f:\left[ -1,1\right] \rightarrow \mathbb {R}_{+}\) is continuous.
Call
Then by [ 5 , p. 287 ] , we obtain that
For \(m{\gt}1\), we get
Hence it holds
Furthermore we have
and \(H_{2N+1}^{\left( M\right) }\) maps continuous functions to continuous functions over \(\left[ -1,1\right] \) and for any \(x\in \mathbb {R}\) we have \(\bigvee _{k=0}^{N}h_{N,k}\left( x\right) {\gt}0\).
We also have \(h_{N,k}\left( x_{N,k}\right) =1\), and \(h_{N,k}\left( x_{N,j}\right) =0\), if \(k\neq j\), furthermore it holds \(H_{2N+1}^{\left( M\right) }\left( f\right) \left( x_{N,j}\right) =f\left( x_{N,j}\right) \), for all \(j\in \left\{ 0,...,N\right\} \), see [ 5 , p. 282 ] .
\(H_{2N+1}^{\left( M\right) }\) are positive sublinear operators, [ 5 , p. 282 ] . â–¡
We give
Let \(\alpha \! \! {\gt}\! \! 1,\) \(\alpha \notin \mathbb {N}\), \(m\! \! =\! \! \left\lceil \alpha \right\rceil \), \(x\in \left[ \! -\! 1,1\right] \), \(f\in AC^{m}\left( \left[ \! -\! 1,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ -1,1\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then
\(\forall \text{ }N\in \mathbb {N}.\)
Furthermore it holds \(\underset {N\rightarrow +\infty }{\lim }H_{2N+1}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We continue with
Here we deal with Lagrange interpolation polynomials on Chebyshev knots of second kind plus the endpoints \(\pm 1\) (see [ 5 , p. 5 ] ). These polynomials are linear operators attached to \(f:\left[ -1,1\right] \rightarrow \mathbb {R}\) and to the knots \(x_{N,k}=\cos \left( \left( \tfrac {N-k}{N-1}\right) \pi \right) \in \left[ -1,1\right] \), \(k=1,...,N,\) \(N\in \mathbb {N}\), which are the roots of \(\omega _{N}\left( x\right) =\sin \left( N-1\right) t\sin t\), \(x=\cos t\). Notice that \(x_{N,1}=-1\) and \(x_{N,N}=1.\) Their formula is given by [ 5 , p. 377 ]
where
\(N\geq 2\), \(k=1,...,N\), and \(\omega _{N}\left( x\right) =\prod _{k=1}^{N}\left( x-x_{N,k}\right) \) and \(\delta _{i,j}\) denotes the Kronecher’s symbol, that is \(\delta _{i,j}=1\), if \(i=j\), and \(\delta _{i,j}=0 \), if \(i\neq j\).
The Max-product Lagrange interpolation operators on Chebyshev knots of second kind, plus the endpoints \(\pm 1\), are defined by [ 5 , p. 12 ]
where \(f:\left[ -1,1\right] \rightarrow \mathbb {R}_{+}\) continuous.
First we see that \(L_{N}^{\left( M\right) }\left( f\right) \left( x\right) \) is well defined and continuous for any \(x\in \left[ -1,1\right] \). Following [ 5 , p. 289 ] , because \(\sum _{k=1}^{N}l_{N,k}\left( x\right) =1\), \(\forall \) \(x\in \mathbb {R}\), for any \(x\) there exists \(k\in \left\{ 1,...,N\right\} :l_{N,k}\left( x\right) {\gt}0\), hence \(\bigvee _{k=1}^{N}l_{N,k}\left( x\right) {\gt}0\). We have that \(l_{N,k}\left( x_{N,k}\right) =1\), and \(l_{N,k}\left( x_{N,j}\right) =0\), if \(k\neq j\). Furthermore it holds \(L_{N}^{\left( M\right) }\left( f\right) \left( x_{N,j}\right) =f\left( x_{N,j}\right) \), all \(j\in \left\{ 1,...,N\right\} ,\) and \(L_{N}^{\left( M\right) }\left( 1\right) =1\).
Call \(I_{N}^{+}\left( x\right) =\left\{ k\in \left\{ 1,...,N\right\} ;l_{N,k}\left( x\right) {\gt}0\right\} \), then \(I_{N}^{+}\left( x\right) \neq \emptyset \).
So for \(f\in C_{+}\left( \left[ -1,1\right] \right) \) we get
Notice here that \(\left| x_{N,k}-x\right| \leq 2\), \(\forall \) \(x\in \left[ -1,1\right] .\)
By [ 5 , p. 297 ] , we get that
\(N\geq 3\), \(\forall \) \(x\in \left( -1,1\right) \), \(N\) is odd.
We get that (\(m{\gt}1\))
\(N\geq 3\) odd, \(\forall \) \(x\in \left( -1,1\right) .\)
\(L_{N}^{\left( M\right) }\) are positive sublinear operators, [ 5 , p. 290 ] . â–¡
We give
Same assumptions as in Theorem 3.21. Then
\(\forall \text{ }N\in \mathbb {N}:N\geq 3\text{, odd.}\)
It holds \(\underset {N\rightarrow +\infty }{\lim }L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We make
Let \(f\in C_{+}\left( \left[ -1,1\right] \right) \), \(N\geq 4\), \(N\in \mathbb {N}\), \(N\) even.
By [ 5 , p. 298 ] , we get
Hence (\(m{\gt}1\))
â–¡
We present
Same assumptions as in Theorem 3.21. Then
\(\forall \text{ }N\in \mathbb {N},\text{ }N\geq 4\text{, }N\) is even.
It holds \(\underset {N\rightarrow +\infty }{\lim }L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
We need
( [ 6 , p. 41 ] ). Let \(I\subset \mathbb {R}\) be an interval of finite or infinite length, and \(f:I\rightarrow \mathbb {R}\) a bounded or uniformly continuous function. We define the first modulus of continuity
Clearly, it holds \(\omega _{1}\left( f,\delta \right) _{I}{\lt}+\infty .\)
We also have
Let a real number \(m{\gt}1\), from now on we assume that \(D_{x_{0}-}^{m}f\) is either bounded or uniformly continuous function on \((-\infty ,x_{0}]\), similarly from now on we assume that \(D_{\ast x_{0}}^{m}f\) is either bounded or uniformly continuous function on \([x_{0},+\infty ).\)
We need
Let \(D_{x_{0}}^{m}f\) (real number \(m{\gt}1\)) denote any of \(D_{x_{0}-}^{m}f\), \(D_{\ast x_{0}}^{m}f\) and \(\delta {\gt}0\). We set
where \(x_{0}\in \mathbb {R}\). Notice that \(\omega _{1}\left( D_{x_{0}}^{m}f,\delta \right) _{\mathbb {R}}{\lt}+\infty .\)
We will use
Let the real number \(m{\gt}0\), \(m\notin \mathbb {N}\), \(\lambda =\left\lceil m\right\rceil \), \(x_{0}\in \mathbb {R}\), \(f\in AC^{\lambda }\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \) (i.e. \(f^{\left( \lambda -1\right) }\in AC\left[ a,b\right] ,\) absolutely continuous functions on \(\left[ a,b\right] \)), \(\forall \) \(\left[ a,b\right] \subset \mathbb {R}\), and \(f^{\left( \lambda \right) }\in L_{\infty }\left( \mathbb {R}\right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,\lambda -1\). The Convention 3.27 is imposed here. Then
for all \(x\in \mathbb {R}\).
If \(0{\lt}m{\lt}1\), then we do not need initial conditions.
We continue with
Let \(b:\mathbb {R}\rightarrow \mathbb {R}_{+}\) be a centered (it takes a global maximum at \(0\)) bell-shaped function, with compact support \(\left[ -T,T\right] \), \(T{\gt}0\) (that is \(b\left( x\right) {\gt}0\) for all \(x\in \left( -T,T\right) \)) and \(I=\int _{-T}^{T}b\left( x\right) dx{\gt}0.\)
The Cardaliaguet-Euvrard neural network operators are defined by (see [ 4 ] )
\(0{\lt}\alpha {\lt}1\), \(N\in \mathbb {N}\) and typically here \(f:\mathbb {R}\rightarrow \mathbb {R}\) is continuous and bounded or uniformly continuous on \(\mathbb {R}\).
\(CB\left( \mathbb {R}\right) \) denotes the continuous and bounded function on \(\mathbb {R}\), and
The corresponding max-product Cardaliaguet-Euvrard neural network operators will be given by
\(x\in \mathbb {R}\), typically here \(f\in CB_{+}\left( \mathbb {R}\right) \), see also [ 4 ] .
Next we follow [ 4 ] .
For any \(x\in \mathbb {R}\), denoting
we can write
\(x\in \mathbb {R}\), \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} \), where \(J_{T,N}\left( x\right) \neq \emptyset \). Indeed, we have \(\bigvee _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) {\gt}0\), \(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} .\)
We have that \(C_{N,\alpha }^{\left( M\right) }\left( 1\right) \left( x\right) =1\), \(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} \). â–¡
See in [ 4 , Lemma 2.1, Corollary 2.2 and Remarks ] .
We need
[ 4 ] . Let \(b\left( x\right) \) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\), \(0{\lt}\alpha {\lt}1\) and \(C_{N,\alpha }^{\left( M\right) }\) be defined as in 3.107.
(i) If \(\left| f\left( x\right) \right| \leq c\) for all \(x\in \mathbb {R}\) then \(\left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \right| \leq c\), for all \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} \) and \(C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \) is continuous at any point \(x\in \mathbb {R}\), for all \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} ;\)
(ii) If \(f,g\in CB_{+}\left( \mathbb {R}\right) \) satisfy \(f\left( x\right) \leq g\left( x\right) \) for all \(x\in \mathbb {R}\), then \(C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \leq C_{N,\alpha }^{\left( M\right) }\left( g\right) \left( x\right) \) for all \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} ;\)
(iii) \(C_{N,\alpha }^{\left( M\right) }\left( f+g\right) \left( x\right) \leq C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) +C_{N,\alpha }^{\left( M\right) }\left( g\right) \left( x\right) \) for all \(f,g\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} ;\)
(iv) For all \(f,g\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} \), we have
(v) \(C_{N,\alpha }^{\left( M\right) }\) is positive homogeneous, that is \(C_{N,\alpha }^{\left( M\right) }\left( \lambda f\right) \left( x\right) =\lambda C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \) for all \(\lambda \geq 0\), \(x\in \mathbb {R}\), \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} \) and \(f\in CB_{+}\left( \mathbb {R}\right) .\)
We make
We have that
\(\forall \) \(x\in \mathbb {R}\), and \(N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} .\) â–¡
We mention from [ 4 ] the following:
[ 4 ] . Let \(b\left( x\right) \) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\) and \(0{\lt}\alpha {\lt}1\). In addition, suppose that the following requirements are fulfilled:
(i) There exist \(0{\lt}m_{1}\leq M_{1}{\lt}\infty \) such that \(m_{1}\left( T-x\right) \leq b\left( x\right) \leq M_{1}\left( T-x\right) \), \(\forall \) \(x\in \left[ 0,T\right] ;\)
(ii) There exist \(0{\lt}m_{2}\leq M_{2}{\lt}\infty \) such that \(m_{2}\left( x+T\right) \leq b\left( x\right) \leq M_{2}\left( x+T\right) \), \(\forall \) \(x\in \left[ -T,0\right] \).
Then for all \(f\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and for all \(N\in \mathbb {N}\) satisying \(N{\gt}\max \left\{ T+\left| x\right| ,\left( \frac{2}{T}\right) ^{\frac{1}{\alpha }}\right\} \), we have the estimate
where
and
We make
In [ 4 ] , was proved that
That is
From (3.109) we have that \(\left| x-\frac{k}{N}\right| \leq \frac{T}{N^{1-\alpha }}.\)
Hence (\(m{\gt}1\)) (\(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \Big\{ T+\left| x\right| ,\big( \frac{2}{T}\big) ^{\frac{1}{\alpha }}\Big\} \))
Then (\(m{\gt}1\)) it holds
Call
Consequently (\(m{\gt}1\)) we derive
â–¡
We need
All here as in Theorem 3.29, where \(x=x_{0}\in \mathbb {R}\) is fixed. Let \(b\) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\), \(0{\lt}\alpha {\lt}1\) and \(C_{N,\alpha }^{\left( M\right) }\) be defined as in 3.107. Then
\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} .\)
true over \(\mathbb {R}\).
As in Theorem 3.31 and using similar reasoning and \(C_{N,\alpha }^{\left( M\right) }\left( 1\right) =1\), we get
\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{ T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\} .\)
We continue with
Here all as in Theorem 3.29, where \(x=x_{0}\in \mathbb {R}\) is fixed and \(m{\gt}1\). Also the same assumptions as in Theorem 3.33. Then
\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{ T+\left| x\right| ,\big( \frac{2}{T}\big) ^{\frac{1}{\alpha }}\right\} .\)
We have that \(\underset {N\rightarrow +\infty }{\lim }C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\)
thus \(\delta ^{m+1}=\frac{\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\), and
Therefore we have
\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{ T+\left\vert x\right\vert ,\left( \frac{2}{T}\right) ^{\frac{1}{\alpha }}\right\} , \) proving the inequality (3.119).
We finish with (case of \(\alpha =1.5\))
Let \(x\in \left[ 0,1\right] \), \(f\in AC^{2}\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \) and \(f^{\left( 2\right) }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Assume that \(f^{\prime }\left( x\right) =0\). Then
\(\forall \text{ }N\in \mathbb {N}\text{.}\)
Due to lack of space we do not give other example applications.
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