# Certain Approximation Results for Kantorovich Exponential Sampling Series

Shivam Bajpeyi  School of Mathematics, Indian Institute of Science Education
and Research, Thiruvananthapuram, India<EMAIL_ADDRESS>,  A.
Sathish Kumar  Department of Mathematics, Indian Institute of Technology
Madras, Chennai-600036, India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  P. Devaraj  School of Mathematics, Indian Institute of Science Education
and Research, Thiruvananthapuram, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study a strong inverse approximation theorem and saturation
order for the family of Kantorovich exponential sampling operators. The class
of log-uniformly continuous and bounded functions, and class of log-Hölderian
functions are considered to derive these results. We also prove some auxiliary
results including Voronovskaya type theorem, and a relation between the
Kantorovich exponential sampling series and the generalized exponential
sampling series, to achieve the desired plan. Moreover, some examples of
kernels satisfying the conditions, which are assumed in the hypotheses of our
theorems, are discussed.

###### Key words and phrases:

Kantorovich exponential sampling series, Inverse approximation, Saturation
order, Mellin derivative.

###### 2010 Mathematics Subject Classification:

41A35; 30D10; 94A20; 41A25

##  1\. Introduction

The problem of sampling and reconstruction of functions is a fundamental
aspect of approximation theory, with important applications in signal analysis
and image processing ([18, 32]). A significant breakthrough in sampling and
reconstruction theory was collectively achieved by Whittaker-Kotelnikov-
Shannon. They established that any band-limited signal $f$, i.e. the Fourier
transform of $f$ is compactly supported, can be completely recovered using its
regularly spaced sample values (see [19]). This result is widely known as WKS
sampling theorem. Butzer and Stens [21] generalized this result significantly
for not-necessarily band-limited signals. Since then, several mathematicians
have been making significant advancements in this direction, see [22, 8, 39,
29, 1].

The problem of approximating functions with their exponentially-spaced sample
values can be traced back to the work of Ostrowski et.al. [40], Bertero and
Pike [20], and Gori [33]. In order to deal with exponentially-spaced data,
they provided a series representation for the class of Mellin band-limited
functions (defined in Section 2). This reconstruction formula is referred as
the exponential sampling formula and defined as follows. For
$f:\mathbb{R}^{+}\rightarrow\mathbb{C}$ and $c\in\mathbb{R},$ the exponential
sampling formula is given by (see [23])

$(E_{c,T}f)(x):=\sum_{k=-\infty}^{\infty}lin_{\frac{c}{T}}(e^{-k}x^{T})f(e^{\frac{k}{T}})$
(1.1)

where $lin_{c}(x)=\dfrac{x^{-c}}{2\pi i}\dfrac{x^{\pi i}-x^{-\pi i}}{\log
c}=x^{-c}sinc(\log x)$ with continuous extension $lin_{c}(1)=1.$ Moreover, if
$f$ is Mellin band-limited to $[-T,T],$ then $(E_{c,T}f)(x)=f(x)$ for each
$x\in\mathbb{R}^{+}.$

The exponentially spaced data can be observed in various problems emerging in
optical physics and engineering, for example Fraunhofer diffraction,
polydispersity analysis by photon correlation spectroscopy, neuron scattering,
radio astronomy etc (see [27, 40, 20, 33]). Therefore, it became crucial to
examine the extensions and variations of the exponential sampling formula
(1.1). Butzer and Jansche [25] investigated into the exponential sampling
formula, incorporating the analytical tools of Mellin analysis. They
established that the theory of Mellin transform provides a suitable framework
to handle sampling and approximation problem related to exponentially-spaced
data. The foundational work on the Mellin transform theory was initially
undertaken by Mamedov [38]. Subsequently, Butzer and his colleagues made
significant contributions to the field of Mellin theory in [23, 25]. For some
notable developments on Mellin theory, we refer to [10, 11, 12, 13] etc. In
order to approximate a function which is not necessarily Mellin band-limited,
the theory of exponential sampling formula (1.1) was extended in [14] using
generalized kernel satisfying suitable conditions. This gives a method to
approximate the class of log-continuous functions by employing its
exponentially spaced sample values. For $x\in\mathbb{R}^{+}$ and $w>0,$ the
generalized exponential sampling series is given by (see [14])

$(S_{w}^{\chi}f)(x)=\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})f(e^{\frac{k}{w}})$
(1.2)

for any $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ such that the series (1.2)
converges absolutely. Various approximation properties associated with the
family of operators (1.2) can be observed in [7, 15, 16, 35]. The
approximation properties of exponential sampling operators based on artificial
neural network can be found in [5, 6]. In order to approximate integrable
functions, the series (1.2) is not suitable. To overcome with this, the
following Kantorovich type modification of the family (1.2) was studied in
[2]. For $x\in\mathbb{R}^{+},k\in\mathbb{Z}$ and $w>0,$ the Kantorovich
exponential sampling series is defined by

$(I_{w}^{\chi}f)(x):=\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})\
w\int_{\frac{k}{w}}^{\frac{k+1}{w}}f(e^{u})\ du\ \ $ (1.3)

whenever the series (1.3) is absolutely convergent for any locally integrable
function $f:\mathbb{R}^{+}\rightarrow\mathbb{R}.$ Some direct and inverse
approximation results for the family $(I_{w}^{\chi}f)$ have been discussed in
[2, 4] which includes basic convergence theorem, higher order asymptotic
convergence result and quantitative approximation theorem. Also, an inverse
approximation theorem in case of $f\in\mathcal{C}^{(1)}(\mathbb{R}^{+})$ was
proved in [2] under the assumption that the fist order moment vanishes on
$\mathbb{R}^{+}.$ For some recent advancements related to the family (1.3), we
refer to [36, 3, 37, 34].

In the present work, we deduce a strong inverse approximation result for the
family of Kantorovich exponential sampling operator $(I_{w}^{\chi})$ for
$f\in\mathcal{C}(\mathbb{R}^{+}),$ without assuming that first order algebraic
moment vanishes. This not only broadens the underlying class of functions but
also enable the application of our theory to some other kernels, for instance,
the class of Mellin B-spline kernels (see Section 4). We also establish the
saturation order, i.e. the highest order of convergence that can be achieved,
for $(I_{w}^{\chi}f)$ in case of $f\in\mathcal{C}(\mathbb{R}^{+}).$ The
problem of saturation order for the family of operators $(I_{w}^{\chi}),w>0$
is to find a suitable class $\mathcal{F}$ of real valued functions defined on
$\mathbb{R}^{+},$ a subclass $\mathcal{S}$ and a positive non-increasing
function $\rho(w),w>0$ satisfying the following: there exists
$h\in\mathcal{F}\setminus\mathcal{S}$ with
$\|I_{w}^{\chi}h-h\|=\mathcal{O}(\rho(w))$ as $w\rightarrow\infty$ and
whenever $f\in\mathcal{F}$ with $\|I_{w}^{\chi}f-f\|={o}(\rho(w))$ as
$w\rightarrow\infty$ implies that $f\in\mathcal{S}$ and vice versa. Several
authors have investigated the inverse approximation results and saturation
order for various sampling operators significantly, see [28, 17, 29, 30, 31]
etc.

The proposed plan of the paper is as follows. In order to derive these
results, we first define an appropriate average type kernel and derive some
auxiliary results mainly concerned with this new kernel in Section 2. In
Section 3, we establish a relation between the operator (1.3) and the
derivative of the operator (1.2) based on average type kernel. Further, we
derive the asymptotic formula for the operator (1.3) using Mellin Taylor
formula. By using these results, we prove the saturation theorem and inverse
result for the family of sampling operators (1.3). In Section 4, we discuss
some examples of kernels satisfying the conditions, which are assumed in the
hypotheses of the theorems.

##  2\. Preliminaries and Auxiliary Results

Let $\mathbb{R}^{+}$ be the set of positive real numbers and
$L^{p}(\mathbb{R}^{+}),\ 1\leq p<\infty$ consists of all p-integrable
functions in the Lebesgue sense on $\mathbb{R}^{+}$ with usual $p-$norm.
Further $L^{\infty}(\mathbb{R}^{+})$ denotes the class of bounded measurable
functions defined on $\mathbb{R}^{+}$ with $\|.\|_{\infty}$ norm. Let $X_{c}$
be the space of functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ such that
$f(\cdot)(\cdot)^{c-1}\in L^{1}(\mathbb{R}^{+})$ for some $c\in\mathbb{R},$
equipped with following norm

$\|f\|_{X_{c}}=\int_{0}^{\infty}|f(t)|\ t^{c}\frac{dt}{t}.$

For $f\in X_{c},$ the Mellin transform of $f$ is given by

$[f]^{\wedge}_{M}(s):=\int_{0}^{\infty}f(t)\ t^{s}\frac{dt}{t},\ \ \ (s=c+ix,\
x\in\mathbb{R}).$

One can observe that the Mellin transform is well defined in $X_{c}$ as a
Lebesgue integral. Further, for $c,t\in\mathbb{R}$ and $T>0,$ any function
$f\in X_{c}(\mathbb{R}^{+})$ is said to be Mellin band-limited to $[-T,T],$ if
$[f]_{\hat{M}}(c+it)=0$ for $|t|>T.$ For more details on theory of Mellin
transform, we refer to [23, 24]. The point-wise Mellin derivative of the
function $f$ is defined by the following limit

$\theta_{c}f(t)=\lim_{h\rightarrow
1}\frac{\tau_{h}^{c}f(t)-f(t)}{h-1}=tf^{{}^{\prime}}(t)+cf(t)\ ,$

provided $f^{{}^{\prime}}$ exists, where $\tau_{h}^{c}$ is the Mellin
translation operator $(\tau_{h}^{c}f)(t):=h^{c}f(ht).$ Furthermore, the Mellin
differential operator of order $r$ is given by
$\theta_{c}^{r}:=\theta_{c}(\theta_{c}^{r-1}).$ Throughout this paper, we
consider $\theta_{c}:=\theta_{c}^{1}$ and $\theta f:=\theta_{0}f.$

We now give the definition of recurrent function. We say that a function
$f:\mathbb{R}^{+}\rightarrow\mathbb{C}$ is recurrent if $f(x)=f(e^{a}x),$
$\forall$ $x\in\mathbb{R}^{+}$ and for some $a\in\mathbb{R}$ (see [26]). The
fundamental interval of the above recurrent functions can be taken as
$[1,e^{a}].$

Let $C(\mathbb{R}^{+})$ denotes the space of all uniformly continuous and
bounded functions on $\mathbb{R}^{+}$ with norm
$\|f\|_{\infty}:=\sup_{t\in\mathbb{R}^{+}}|f(t)|.$ For any $\nu\in\mathbb{N},$
$C^{(\nu)}(\mathbb{R}^{+})$ be the subspace of $C(\mathbb{R}^{+})$ such that
$f^{(r)}\in C(\mathbb{R}^{+})$ for each $r\leq\nu,r\in\mathbb{N}.$ Also,
$C_{c}^{\infty}(\mathbb{R}^{+})$ represents the space of all infinitely
differentiable functions which are compactly supported in $\mathbb{R}^{+}.$ A
function $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ is said to be log-uniformly
continuous on $\mathbb{R}^{+}$ if $\forall$ $\epsilon>0,$ there exists a
$\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|\log x-\log y|<\delta,$
for any $x,y\in\mathbb{R}^{+}.$ Further, $\mathcal{C}(\mathbb{R}^{+})$ denotes
the space of all log-uniformly continuous and bounded functions defined on
$\mathbb{R}^{+}.$ Analogous to the classical case, for any $\nu\in\mathbb{N},$
$\mathcal{C}^{(\nu)}(\mathbb{R}^{+})$ be the subspace of
$\mathcal{C}(\mathbb{R}^{+})$ such that
$(\theta^{r}f)\in\mathcal{C}(\mathbb{R}^{+})$ for each
$r\leq\nu,r\in\mathbb{N}.$ For $f\in C^{(r)}(\mathbb{R}^{+}),$ the Mellin’s
Taylor formula is defined by (see [9])

$f(tx)=f(x)+(\theta f)(x)\log
t+\frac{(\theta^{2}f)(x)}{2!}\log^{2}t+\cdots+\frac{(\theta^{n}f)(x)}{n!}\log^{n}t+h(x)\log^{n}t\
,$

where $h:\mathbb{R}^{+}\rightarrow\mathbb{R}$ is bounded and $h(x)\rightarrow
0$ as $x\rightarrow 1.$

A continuous function $\chi:\mathbb{R}^{+}\rightarrow\mathbb{R}$ is said to be
kernel if it fulfils the following conditions:

  * $(\chi_{1})$

For any $u\in\mathbb{R}^{+},\ $
$\displaystyle\sum_{k=-\infty}^{\infty}\chi(e^{-k}u)=1,\hskip
5.69046pt\mbox{uniformly on}\ \mathbb{R}^{+}.$

  * $(\chi_{2})$

$\displaystyle m_{1}(\chi,u):=\sum_{k=-\infty}^{\infty}\chi(e^{-k}u)(k-\log
u)\ :=m_{1}^{\chi}\in\mathbb{R}.$

  * $(\chi_{3})$

For some $\beta\geq 1,$ $\displaystyle
M_{\beta}(\chi):=\sup_{u\in\mathbb{R}^{+}}\sum_{k=-\infty}^{\infty}|\chi(e^{-k}u)||k-\log
u|^{\beta}<\infty.$

  * $(\chi_{4})$

For every $\gamma>0,$ $\displaystyle\lim_{w\rightarrow\infty}\sum_{|w\log
x-k|>w\gamma}|\chi(e^{-k}x^{w})|\ |w\log x-k|=0$ uniformly on
$\mathbb{R}^{+}.$

######  Remark 1.

[2] One can deduce that for $\alpha,\beta\in\mathbb{N}_{0}$ with
$\alpha<\beta,$ $M_{\alpha}(\chi)<\infty$ whenever $M_{\beta}(\chi)<\infty.$

To establish the proposed results for the Kantorovich exponential sampling
operator (1.3), we define the average type kernel as follows:

$\displaystyle\bar{\chi}(t)=\int_{e^{\frac{-1}{2}}}^{e^{\frac{1}{2}}}\chi(tu)\frac{du}{u}=\int_{\frac{-1}{2}}^{\frac{1}{2}}\chi(te^{p})dp,\
\ \ \ t\in\mathbb{R}^{+}.$ (2.1)

In the following lemma, we show that the average type kernel satisfies the
conditions $(\chi_{1})$-$(\chi_{4}).$

######  Lemma 1.

Let $\chi:\mathbb{R}^{+}\rightarrow\mathbb{R}$ be the kernel function
satisfying $(\chi_{1})$ -$(\chi_{4})$ and $\bar{\chi}(t)$ be defined as in
(2.1). Then $\bar{\chi}$ also satisfies $(\chi_{1})$ -$(\chi_{4}).$

###### Proof.

Since $\chi$ is continuous, the averaged type kernel $\bar{\chi}$ is also
continuous. Now for $u\in\mathbb{R}^{+},$ we have

$m_{0}(\bar{\chi})=\sum_{k=-\infty}^{\infty}\bar{\chi}(e^{-k}u)=\sum_{k=-\infty}^{\infty}\int_{\frac{-1}{2}}^{\frac{1}{2}}\chi(e^{-k}ue^{p})dp=\displaystyle\int_{\frac{-1}{2}}^{\frac{1}{2}}dp=1.$

Hence $\overline{\chi}$ satisfies $(\chi_{1}).$ Using the condition
$(\chi_{2}),$ we get

$\displaystyle m_{1}(\bar{\chi},u)$ $\displaystyle=$
$\displaystyle\sum_{k=-\infty}^{\infty}\bar{\chi}(e^{-k}u)(k-\log u)$
$\displaystyle=$
$\displaystyle\sum_{k=-\infty}^{\infty}\int_{\frac{-1}{2}}^{\frac{1}{2}}\chi(e^{-k}ue^{p})(k-\log
u+p-p)\ dp$ $\displaystyle=$
$\displaystyle\int_{\frac{-1}{2}}^{\frac{1}{2}}\sum_{k=-\infty}^{\infty}\chi(e^{-k}ue^{p})(k-\log(ue^{p})+p)\
dp$ $\displaystyle=$ $\displaystyle
m_{1}(\chi,u)\int_{\frac{-1}{2}}^{\frac{1}{2}}dp\ +\
\int_{\frac{-1}{2}}^{\frac{1}{2}}p\ dp=m_{1}^{\chi}.$

We define $\displaystyle
M_{\beta}(\bar{\chi}):=\sup_{u\in\mathbb{R}^{+}}\sum_{k=-\infty}^{\infty}|\bar{\chi}(e^{-k}u)|\
|k-\log u|^{\beta}.$ For $\beta\geq 1,$ it is given that
$M_{\beta}(\chi)<\infty.$ Now we show that $M_{\beta}(\bar{\chi})<\infty.$ In
view of (2.1), we have

$\displaystyle M_{\beta}(\bar{\chi})$ $\displaystyle\leq$
$\displaystyle\sum_{k=-\infty}^{\infty}\int_{\frac{-1}{2}}^{\frac{1}{2}}\left(|\chi(e^{-k}ue^{p})|\
|k-\log u+p-p|^{\beta}\right)dp$ $\displaystyle\leq$
$\displaystyle\int_{\frac{-1}{2}}^{\frac{1}{2}}\left(\sum_{k=-\infty}^{\infty}|\chi(e^{-k}ue^{p})|\
|k-\log(ue^{p})+p|^{\beta}\right)dp$ $\displaystyle\leq$
$\displaystyle\int_{\frac{-1}{2}}^{\frac{1}{2}}\sum_{k=-\infty}^{\infty}|\chi(e^{-k}ue^{p})|\left(2^{\beta-1}|k-\log(ue^{p})|^{\beta}+|p|^{\beta}\right)dp.$

Since $\beta\geq 1,$ then by using the inequality $|a+b|^{\beta}\leq
2^{\beta-1}\left(|a|^{\beta}+|b|^{\beta}\right),$ we obtain

$\displaystyle M_{\beta}(\overline{\chi})$ $\displaystyle\leq$ $\displaystyle
2^{\beta-1}\int_{\frac{-1}{2}}^{\frac{1}{2}}\left(\sum_{k=-\infty}^{\infty}|\chi(e^{-k}ue^{p})|\
|k-\log(ue^{p})|^{\beta}\right)dp$
$\displaystyle+2^{\beta-1}\int_{\frac{-1}{2}}^{\frac{1}{2}}\left(\sum_{k=-\infty}^{\infty}|\chi(e^{-k}ue^{p})|\
|p|^{\beta}\right)dp$ $\displaystyle\leq$ $\displaystyle
2^{\beta-1}\left(M_{\beta}(\chi)+M_{0}({\chi})\int_{\frac{-1}{2}}^{\frac{1}{2}}|p|^{\beta}dp\right)$
$\displaystyle\leq$ $\displaystyle
2^{\beta-1}M_{\beta}(\chi)+\frac{M_{0}(\chi)}{2^{\beta+1}(\beta+1)}((-1)^{\beta+1}+1).$

Since $M_{\beta}(\chi)<\infty,$ so we have $M_{\beta}(\bar{\chi})<\infty.$

Now we show that for $\gamma>0,$

$\displaystyle\lim_{w\rightarrow\infty}\sum_{|w\log
x-k|>w\gamma}|\bar{\chi}(e^{-k}x^{w})|\ |w\log x-k|=0$

uniformly w.r.t. $x\in\mathbb{R}^{+}.$ For $w>\dfrac{1}{\gamma},$ by using the
definition of $\bar{\chi},$ we can write

$\displaystyle\sum_{|w\log x-k|>w\gamma}|\bar{\chi}(e^{-k}x^{w})|\ |w\log
x-k|$

$\displaystyle\leq$ $\displaystyle\sup_{p\in[-1/2,1/2]}\left(\sum_{|w\log
x-k|>w\gamma}|\chi(e^{-k}x^{w}e^{p})|\ |w\log
x-k+\log(e^{p})-\log(e^{p})|\right)$ $\displaystyle\leq$
$\displaystyle\sup_{p\in[-1/2,1/2]}\left(\sum_{|w\log
x-k+\log(e^{p})|>w\gamma-1/2}|\chi(e^{-k}x^{w}e^{p})|\ (|w\log
x-k+\log(e^{p})|+|\log(e^{p})|)\right)$ $\displaystyle\leq$
$\displaystyle\sup_{y\in\mathbb{R}^{+}}\left(\sum_{|w\log
y-k|>w\gamma-1/2}|\chi(e^{-k}y^{w})|\left(|w\log
y-k|+\frac{1}{2}\right)\right)$ $\displaystyle\leq$
$\displaystyle\sup_{y\in\mathbb{R}^{+}}\left(\sum_{|w\log
y-k|>w\gamma/2}|\chi(e^{-k}y^{w})||w\log y-k|+\frac{1}{2}\sum_{|w\log
y-k|>w\gamma/2}|\chi(e^{-k}y^{w})|\right).$

Using the condition $(\chi_{4}),$ we deduce that

$\displaystyle\lim_{w\rightarrow\infty}\sum_{|w\log
x-k|>w\gamma}|\bar{\chi}(e^{-k}x^{w})|\ |w\log x-k|=0$

uniformly w.r.t. $x\in\mathbb{R}^{+}.$ This concludes the proof. ∎

Next we deduce the following result which will be useful to obtain a relation
between the sampling series (1.2) and (1.3).

######  Lemma 2.

Let $[a,b]\subset\mathbb{R}^{+}$ and $f:[a,b]\rightarrow\mathbb{R}$ be
continuous. If

$F(x)=\int_{0}^{x}f(t)\frac{dt}{t},\ \ \ x\in[a,b].$ (2.2)

Then $F$ is Mellin differentiable and $\ (\theta F)(x)=f(x),\ \forall
x\in[a,b].$

###### Proof.

We have

$\displaystyle F(x)$ $\displaystyle=$
$\displaystyle\int_{a}^{x}f(t)\frac{dt}{t}=\int_{\log a}^{\log x}f(e^{v})\
dv.$

This gives

$\displaystyle F(sx)-F(x)$ $\displaystyle=$ $\displaystyle\int_{\log a}^{\log
sx}f(e^{v})\ dv-\int_{\log a}^{\log x}f(e^{v})\ dv$ $\displaystyle=$
$\displaystyle\begin{cases}{\displaystyle\int_{\log x}^{\log sx}f(e^{v})\
dv}&\quad\text{if}\ \ \ \ {s>1}\\\ {\displaystyle-\int_{\log x}^{\log
sx}f(e^{v})\ dv}&\quad\text{if}\ \ \ \ {s<1.}\\\ \end{cases}$

On applying the mean value theorem for integral calculus, we get

$\int_{\log x}^{\log sx}f(e^{v})\ dv=(\log s)f(e^{\xi}),$

where $\xi\in[\log x,\log sx]$ for $s>1$ and $\xi\in[\log sx,\log x]$ for
$s<1.$ This gives

$F(sx)-F(x)=\log(s)f(e^{\xi}),\ \ \xi\in[\log x,\log sx].$

This implies that

$\frac{F(sx)-F(x)}{\log s}=f(sx).$

Taking limit as $s\rightarrow 1,$ we deduce

$\lim_{s\rightarrow 1}\frac{F(sx)-F(x)}{\log s}=f(x).$

This gives $(\theta F)(x)=f(x),\ \ x\in[a,b].$ Thus, the proof is completed. ∎

##  3\. Main Results

In this section, we derive the inverse approximation result and saturation
order for the Kantorovich exponential sampling series $(I_{w}^{\chi}f).$ First
we establish the relation between $(S_{w}^{\chi}f)$ and $(I_{w}^{\chi}f).$
Using the continuity of $\chi$ and Lemma 2, we obtain

$\theta\bar{\chi}(t)=\chi(te^{\frac{1}{2}})-\chi(te^{\frac{-1}{2}}).$ (3.1)

######  Lemma 3.

Let $f\in C(\mathbb{R}^{+})$ and $F$ be the Mellin anti-derivative of $f.$
Then for $x,w\in\mathbb{R}^{+},$ there holds

$(I_{w}^{\chi}f)(x)=(\theta S^{\bar{\chi}}_{w}F)(xe^{\frac{1}{2w}}).$

###### Proof.

Using (2.2), we can write

$\displaystyle(I_{w}^{\chi}f)(x)$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})\
w\int_{\frac{k}{w}}^{\frac{k+1}{w}}f(e^{u})\ du$ $\displaystyle=$
$\displaystyle\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})\
w\left(F(e^{\frac{k+1}{w}})-F(e^{\frac{k}{w}})\right)$ $\displaystyle=$
$\displaystyle
w\left(\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})F(e^{\frac{k+1}{w}})-\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})F(e^{\frac{k}{w}})\right).$

Setting $\widetilde{k}=k+1$ in the first term of the above expression, we have

$\displaystyle(I_{w}^{\chi}f)(x)$ $\displaystyle=$ $\displaystyle
w\left(\sum_{\widetilde{k}=-\infty}^{\infty}\chi(e^{\widetilde{-k}}x^{w})F(e^{\frac{\widetilde{k}}{w}})-\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})F(e^{\frac{k}{w}})\right)$
$\displaystyle=$ $\displaystyle
w\left(\sum_{\widetilde{k}=-\infty}^{\infty}\chi(e^{\widetilde{-k}}x^{w}e^{\frac{1}{2}}e^{\frac{1}{2}})F(e^{\frac{\widetilde{k}}{w}})\right)-w\left(\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w}e^{\frac{1}{2}}e^{\frac{-1}{2}})F(e^{\frac{k}{w}})\right)$
$\displaystyle=$ $\displaystyle\sum_{k=-\infty}^{\infty}F(e^{\frac{k}{w}})\
w\left(\chi(e^{-k}x^{w}e^{\frac{1}{2}}e^{\frac{1}{2}})-\chi(e^{-k}x^{w}e^{\frac{1}{2}}e^{\frac{-1}{2}})\right).$

From (3.1), we get

$\displaystyle(I_{w}^{\chi}f)(x)=\sum_{k=-\infty}^{\infty}F(e^{\frac{k}{w}})\
w\ (\theta\bar{\chi})(e^{-k}x^{w}e^{\frac{1}{2}}).$

Since $(\theta f)(x)=xf^{{}^{\prime}}(x),$ thus we obtain

$(\theta
S_{w}^{\bar{\chi}}F)(xe^{\frac{1}{2w}})=\sum_{k=-\infty}^{\infty}F(e^{\frac{k}{w}})\
w\ (\theta\bar{\chi})(e^{-k}x^{w}e^{\frac{1}{2}}).$

Hence, the proof is established. ∎

Next we establish the asymptotic formula for the series $(S_{w}^{\chi}f).$
This result is required to derive the saturation order for the Kantorovich
exponential sampling series $(I_{w}^{\chi}f).$

######  Theorem 1.

If $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ be differentiable such that
$(\theta f)$ is log-uniformly continuous and bounded on $\mathbb{R}^{+},$ then

$\displaystyle\lim_{w\to+\infty}w\ ((S_{w}^{\chi}f)(x)-f(x))=m_{1}^{\chi}\
(\theta f)(x),$

uniformly on $\mathbb{R}^{+}.$

###### Proof.

Setting $u=k/w$ in the first order Mellin Taylor formula, we obtain

$f(e^{k/w})=f(x)+\left(\frac{k}{w}-\log x\right)(\theta f)(\xi),\ \ \ \
\xi\in(k/w,\log x).$

Operating $\displaystyle\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})$ on both
sides of the above expression, we get

$(S_{w}^{\chi}f)(x)=f(x)+\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})(\theta
f)(\xi)\left(\frac{k}{w}-\log x\right):=f(x)+R_{w}(x).$

The above expression implies

$w\left((S_{w}^{\chi}f)(x)-f(x)\right)-m_{1}^{\chi}(\theta f)(x)=w\
R_{w}(x)-m_{1}^{\chi}(\theta f)(x).$

For $\delta>0,$ we can write

$w\ R_{w}(x)-m_{1}^{\chi}\ (\theta f)(x)$

$\displaystyle=$
$\displaystyle\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})(k-w\log x)((\theta
f)(\xi)-(\theta f)(x))$ $\displaystyle=$
$\displaystyle\left(\sum_{\left|\frac{k}{w}-\log
x\right|<\delta}+\sum_{\left|\frac{k}{w}-\log
x\right|\geq\delta}\right)\chi(e^{-k}x^{w})(k-w\log x)((\theta f)(\xi)-(\theta
f)(x))$ $\displaystyle:=$ $\displaystyle I_{1}+I_{2}.$

Since $(\theta f)$ is log-uniformly continuous, $\forall\epsilon>0,$
$\exists\,\ \delta>0$ such that $\left|(\theta f)(\xi)-(\theta
f)(x)\right|<\epsilon,$ for $\left|\dfrac{k}{w}-\log x\right|<\delta.$ This
gives

$\displaystyle|I_{1}|$ $\displaystyle\leq$
$\displaystyle\epsilon\sum_{\left|\frac{k}{w}-\log
x\right|<\delta}|\chi(e^{-k}x^{w})||k-w\log x|$ $\displaystyle\leq$
$\displaystyle\epsilon\ M_{1}(\chi).$

Since $\epsilon>0$ is arbitrary, we obtain $I_{1}\rightarrow 0$ as
$w\rightarrow\infty.$ In view of boundedness of $(\theta f),$ we get

$\displaystyle|I_{2}|$ $\displaystyle\leq$ $\displaystyle 2\|\theta
f\|_{\infty}\sum_{|{k}-w\log x|\geq w\delta}|\chi(e^{-k}x^{w})||k-w\log x|.$

Using the condition $(\chi_{4}),$ we obtain $I_{2}\rightarrow 0$ as
$w\rightarrow\infty.$ On collecting the estimates $I_{1}-I_{2},$ we obtain

$\lim_{w\to+\infty}w\left[(S_{w}^{\chi}f)(x)-f(x)\right]=m_{1}^{\chi}\ (\theta
f)(x).$

Thus, the proof is completed. ∎

This gives the following corollary.

######  Corollary 1.

Let $f:\mathbb{R}^{+}\to\mathbb{R}$ be such that $(\theta
f)\in\mathcal{C}(\mathbb{R}^{+}).$ Then for any $x\in\mathbb{R}^{+}$ and
$w>0,$ we have

$\lim_{w\to+\infty}w[(S_{w}^{\chi}f)(xe^{1/2w})-f(x)]=(2m_{1}^{\chi}+1)\frac{(\theta
f)(x)}{2},$

uniformly on $\mathbb{R}^{+}.$

###### Proof.

Using Theorem 1, we write

$\Big{|}w\left[(S_{w}^{\chi}f)(u)-f(u)\right]-m_{1}^{\chi}\ (\theta
f)(u)\Big{|}<\epsilon\ ,$

hold for any $u\in\mathbb{R}^{+}$ and for every fixed $\epsilon>0.$ If we take
$u=xe^{1/2w},$ then from the above estimate, we obtain

$\Big{|}w\left[(S_{w}^{\chi}f)(xe^{1/2w})-f(xe^{1/2w})\right]-m_{1}^{\chi}\
(\theta f)(xe^{1/2w})\Big{|}<\epsilon.$

Thus, we have

$\Big{|}w\left[(S_{w}^{\chi}f)(xe^{1/2w})-f(x)\right]-(2m_{1}^{\chi}+1)\dfrac{(\theta
f)(x)}{2}\Big{|}$

$\displaystyle<$
$\displaystyle\epsilon+\Bigg{|}w(f(xe^{1/2w})-f(x))-\frac{(\theta
f)(x)}{2}\Bigg{|}+\Big{|}m_{1}^{\chi}\left((\theta f)(x)-(\theta
f)(xe^{\frac{1}{2w}})\right)\Big{|}$ $\displaystyle<$
$\displaystyle\epsilon+\frac{1}{2}\Bigg{|}\frac{f(xe^{1/2w})-f(x)}{1/2w}-\frac{(\theta
f)(x)}{2}\Bigg{|}+\Big{|}m_{1}^{\chi}\left((\theta f)(x)-(\theta
f)(xe^{\frac{1}{2w}})\right)\Big{|}$ $\displaystyle:=$
$\displaystyle\epsilon+I_{3}+I_{4}.$

Since
$\displaystyle\lim_{w\rightarrow\infty}2w\left(f(xe^{1/2w})-f(x)\right)=\frac{(\theta
f)(x)}{2},$ we deduce that $I_{3}\rightarrow 0$ as $w\rightarrow\infty.$ Since
$\theta f$ is log-uniformly continuous, we have

$|\theta f(x)-\theta f(xe^{1/2w})|<\epsilon,$

whenever $|\log x-\log(xe^{1/2w})|<\delta.$ Therefore, we obtain

$\Bigg{|}w\left[(S_{w}^{\chi}f)(xe^{1/2w})-f(x)\right]-2(m_{1}^{\chi}+1)\frac{(\theta
f)(x)}{2}\Bigg{|}<\epsilon.$

Hence we get the desired result. ∎

In what follows, we shall define the class of log-Hölderian functions as

$L_{\alpha}:=\\{f:I\rightarrow\mathbb{R}\ :\exists\ \mbox{K $>$ 0 \ s.t.}\ \
|f(x)-f(y)|\leq K|\log x-\log y|^{\alpha};\hskip 3.1298ptx,y\in I\\},$

with $I\subseteq\mathbb{R}^{+}$ and $0<\alpha\leq 1.$ Now we prove the
following direct approximation result.

######  Theorem 2.

Let $\chi$ be a kernel function and $f\in L_{\alpha}.$ Then the following
holds

$\|I_{w}^{\chi}f-f\|_{\infty}=\mathcal{O}(w^{-\alpha})~{}~{}~{}\mbox{as}\ \
w\rightarrow\infty.$

###### Proof.

Consider $|(I_{w}^{\chi}f)(x)-f(x)|$

$\displaystyle=$
$\displaystyle\Big{|}\sum_{k=-\infty}^{\infty}\chi(e^{-k}x^{w})w\int_{k/w}^{{k+1}/w}[f(e^{u})-f(x)]\
du\Big{|}$ $\displaystyle\leq$
$\displaystyle\sum_{k=-\infty}^{\infty}|\chi(e^{-k}x^{w})|w\int_{k/w}^{{k+1}/w}|f(e^{u})-f(x)|\
du.$

Since $f\in L_{\alpha},$ so we obtain

$\displaystyle|(I_{w}^{\chi}f)(x)-f(x)|$ $\displaystyle\leq$
$\displaystyle\sum_{k=-\infty}^{\infty}|\chi(e^{-k}x^{w})|w\int_{k/w}^{{k+1}/w}|u-\log
x|^{\alpha}du$ $\displaystyle\leq$
$\displaystyle\frac{w}{(\alpha+1)}\sum_{k=-\infty}^{\infty}|\chi(e^{-k}x^{w})|\left[\
\Big{|}\frac{k+1}{w}-\log x\Big{|}^{\alpha+1}-\ \Big{|}\frac{k}{w}-\log
x\Big{|}^{\alpha+1}\right].$

Since $|a+b|^{\alpha+1}\leq 2^{\alpha}\left(|a|^{\alpha}+|b|^{\alpha}\right)$
for $\alpha>0,$ we can write

$|(I_{w}^{\chi}f)(x)-f(x)|$

$\displaystyle\leq$
$\displaystyle\frac{w}{(\alpha+1)}\sum_{k=-\infty}^{\infty}|\chi(e^{-k}x^{w})|\left[\
2^{\alpha}\left(\Big{|}\frac{k}{w}-\log
x\Big{|}^{\alpha+1}+\frac{1}{w^{\alpha+1}}\right)-\ \Big{|}\frac{k}{w}-\log
x\Big{|}^{\alpha+1}\right]$ $\displaystyle\leq$
$\displaystyle\frac{w}{(\alpha+1)}\sum_{k=-\infty}^{\infty}|\chi(e^{-k}x^{w})|\
\left[(2^{\alpha}-1)\Big{|}\frac{k}{w}-\log
x\Big{|}^{\alpha+1}+\frac{1}{w^{\alpha+1}}\right]$ $\displaystyle\leq$
$\displaystyle\frac{w^{-\alpha}}{(\alpha+1)}\left((2^{\alpha}-1)M_{\alpha+1}(\chi)+M_{0}(\chi)\right).$

This completes the proof. ∎

We derive the saturation order for the Kantorovich exponential sampling series
as follows.

######  Theorem 3.

Let $\chi$ be a kernel function such that $m_{1}^{\chi}\neq-1/2$ and let
$f\in\mathcal{C}(\mathbb{R}^{+})$ be such that

$\|I_{w}^{\chi}f-f\|_{\infty}=o(w^{-1})~{}~{}~{}~{}as~{}~{}~{}w\to+\infty.$

Then $f$ is constant on $\mathbb{R}^{+}.$

###### Proof.

Let $\phi\in C_{c}^{\infty}(\mathbb{R}^{+})$ be fixed. We define

$G_{f}(\phi):=w\int_{\mathbb{R}^{+}}\big{[(I_{w}^{\chi}f)-f(x)\big{]}}\phi(x)\frac{dx}{x},\
\ \ \ w>0.$ (3.2)

We assume that $[a,b]\subset\mathbb{R}^{+}$ be such that the compact support
of $\phi$ is properly contained in $[a,b],$ i.e. $supp(\phi)\subset[a,b].$
This gives $\phi(a)=0=\phi(b).$ So, the integral (3.2) reduces to

$G_{f}(\phi)=w\int_{a}^{b}\big{[(I_{w}^{\chi}f)-f(x)\big{]}}\phi(x)\frac{dx}{x},\
\ \ \ w>0.$

Using Lemma 3, we obtain

$\displaystyle G_{f}(\phi)$ $\displaystyle=w\int_{a}^{b}\big{[(\theta
S_{w}^{\bar{\chi}}F)(xe^{1/2w})-f(x)\big{]}}\phi(x)\frac{dx}{x}$
$\displaystyle=w\int_{a}^{b}\big{[(\theta
S_{w}^{\bar{\chi}}F)(xe^{1/2w})-(\theta F)(x)\big{]}}\phi(x)\frac{dx}{x},$

where $F$ is the Mellin anti-derivative of $f,$ i.e. $\displaystyle
F(x)=\int_{0}^{x}f(t)\frac{dt}{t},~{}~{}x\in\mathbb{R}^{+}.$ Using integration
by parts in the Mellin-sense, we obtain

$\displaystyle G_{f}(\phi)=w\left\\{\phi(x)\int\big{[(\theta
S_{w}^{\bar{\chi}}F)(xe^{1/2w})-(\theta
F)(x)\big{]}}\frac{dx}{x}\right\\}_{a}^{b}$
$\displaystyle-w\int_{a}^{b}\big{[(S_{w}^{\bar{\chi}}F)(xe^{1/2w})-F(x)\big{]}}(\theta\phi)(x)\frac{dx}{x}.$

Since $\phi(a)=0=\phi(b),$ we have

$G_{f}(\phi)=-w\int_{a}^{b}\big{[(S_{w}^{\bar{\chi}}F)(xe^{1/2w})-F(x)\big{]}}(\theta\phi)(x)\
\frac{dx}{x}.$

Applying the limit $w\rightarrow\infty$ on both sides of the above equation
and using Vitali’s convergence theorem, we get

$\lim\limits_{w\to\infty}G_{f}(\phi)=-(m_{1}^{\chi}+1/2)\int_{a}^{b}(\theta\phi)(x)(\theta
F)(x)\ \frac{dx}{x}.$ (3.3)

Since $(I_{w}^{\chi}f)$ converges to $f$ uniformly as $w\to\infty,$ we obtain

$0=-(m_{1}^{\chi}+1/2)\int_{a}^{b}(\theta\phi)(x)(\theta F)(x)\frac{dx}{x}.$

As $(\theta F)(x)=f(x),$ we have

$0=-(m_{1}^{\chi}+1/2)\int_{a}^{b}(\theta\phi)(x)f(x)\frac{dx}{x},$

Since $\phi\in C_{c}^{\infty}(\mathbb{R}^{+})$ is arbitrary function, $f$ is
constant in $\mathbb{R}^{+}.$ Hence, the highest order of convergence that
$(I_{w}^{\chi}f)$ can achieve for $f\in\mathcal{C}(\mathbb{R}^{+})$ is one,
provided $m_{1}(\chi,u)\neq-1/2.$ Hence, the result is proved. ∎

For $f\in\mathcal{C}(\mathbb{R}^{+}),$ the logarithmic modulus of continuity
is defined as

$\omega(f,\delta):=\sup\\{|f(u)-f(v)|:\ |\log u-\log v|\leq\delta,\ \
\delta\in\mathbb{R}^{+}\\}.$

For every $\delta>0$ and $u,v\in\mathbb{R}^{+},$ $\omega$ has the following
properties:

  * (a)

$\omega(f,\delta)\rightarrow 0$ as $\delta\rightarrow 0.$

  * (b)

$\displaystyle|f(u)-f(v)|\leq\omega(f,\delta)\left(1+\frac{|\log u-\log
v|}{\delta}\right).$

For more details on modulus of continuity, we refer to [38, 11].

Now we prove the proposed inverse approximation result for the Kantorovich
exponential sampling series $I_{w}^{\chi}.$

######  Theorem 4.

Let $\chi$ be differentiable and $M_{1}(\theta\chi)<+\infty$ and
$f\in\mathcal{C}(\mathbb{R}^{+})$ be such that

$\|I_{w}^{\chi}f-f\|_{\infty}=\mathcal{O}(w^{-\alpha})~{}~{}~{}as~{}~{}~{}w\to\infty$

with $0<\alpha\leq 1.$ Then $f$ belongs to $L_{\alpha}.$

###### Proof.

Since
$\|I_{w}^{\chi}f-f\|_{\infty}=\mathcal{O}(w^{-\alpha})~{}~{}~{}as~{}~{}~{}w\to\infty,$
there exist $K$ and $w_{0}$ such that

$\|I_{w}^{\chi}f-f\|_{\infty}\leq Kw^{-\alpha},\ \ \ \mbox{for every}\ \
w>w_{0}.$

Now for every fixed $x,y\in\mathbb{R}^{+}$ and for $x\neq y,$ we can write

$\displaystyle|f(x)-f(y)|$
$\displaystyle=|f(x)-(I_{w}^{\chi}f)(x)+(I_{w}^{\chi}f)(x)-f(y)+(I_{w}^{\chi}f)(y)-(I_{w}^{\chi}f)(y)|$
$\displaystyle\leq|f(y)-(I_{w}^{\chi}f)(y)|+|f(x)-(I_{w}^{\chi}f)(x)|+|(I_{w}^{\chi}f)(x)-(I_{w}^{\chi}f)(y)|$
$\displaystyle\leq
Kw^{-\alpha}+|(I_{w}^{\chi}f)(x)-(I_{w}^{\chi}f)(y)|+Kw^{-\alpha}$
$\displaystyle\leq 2Kw^{-\alpha}+|(I_{w}^{\chi}f)(x)-(I_{w}^{\chi}f)(y)|$
$\displaystyle\leq 2Kw^{-\alpha}+\Bigg{|}\int_{y}^{x}(\theta
I_{w}^{\chi}f)(t)\frac{dt}{t}\Bigg{|}=:I_{1}+I_{2}.$

Now we estimate $I_{2}.$ It is easy to see that

$(\theta
I_{w}^{\chi}f)(x)=\sum_{k=-\infty}^{\infty}(\theta{\chi})(e^{-k}x^{w})\
w^{2}\int_{k/w}^{{k+1}/w}f(e^{u})\ du.$ (3.4)

Let 1 represents the constant function such that $\textbf{1}(x)=1,\ \forall
x\in\mathbb{R}^{+}.$ We can easily observe that
$(I_{w}^{\chi}\textbf{1})(x)=1,\ \forall x\in\mathbb{R}^{+}.$ This gives

$(\theta
I_{w}^{\chi}\textbf{1})(x)=w\sum_{k=-\infty}^{\infty}\chi^{{}^{\prime}}(e^{-k}x^{w})\
(e^{-k}x^{w})=0.$

Again we have

$\displaystyle|(\theta I_{w}^{\chi}f)(x)|$ $\displaystyle=$
$\displaystyle\big{|}(\theta I_{w}^{\chi}f)(x)-f(x)\ (\theta
I_{w}^{\chi}\textbf{1})(x)\big{|}$ $\displaystyle=$
$\displaystyle\Big{|}w\sum_{k=-\infty}^{\infty}(\theta\chi)(e^{-k}x^{w})\
w\int_{k/w}^{{k+1}/w}f(e^{u})\ du-
wf(x)\sum_{k=-\infty}^{\infty}(\theta\chi)(e^{-k}x^{w})\big{|}$
$\displaystyle=$ $\displaystyle
w^{2}\Big{|}\sum_{k=-\infty}^{\infty}(\theta\chi)(e^{-k}x^{w})\int_{k/w}^{{k+1}/w}(f(e^{u})-f(x))\
du\Big{|}$ $\displaystyle\leq$ $\displaystyle
w^{2}\sum_{k=-\infty}^{\infty}|(\theta\chi)(e^{-k}x^{w})|\int_{k/w}^{{k+1}/w}|f(e^{u})-f(x)|\
du.$

Using property (b) of $\omega$ and choosing $\delta=1/w$ to get

$\displaystyle|(\theta I_{w}^{\chi}f)(x)|$ $\displaystyle\leq$ $\displaystyle
w^{2}\sum_{k=-\infty}^{\infty}|(\theta\chi)(e^{-k}x^{w})|\int_{k/w}^{{k+1}/w}\omega(f,1/w)\
(1+w\ |u-\log x|)\ du$ $\displaystyle\leq$ $\displaystyle w^{3}\
\omega(f,1/w)\
\sum_{k=-\infty}^{\infty}|(\theta\chi)(e^{-k}x^{w})|\int_{k/w}^{{k+1}/w}|u-\log
x|\ du$ $\displaystyle+w\ \omega(f,1/w)\
\sum_{k=-\infty}^{\infty}|(\theta\chi)(e^{-k}x^{w})|$ $\displaystyle\leq$
$\displaystyle w\ \omega(f,1/w)\ M_{0}(\theta\chi)+\frac{w}{2}\ \omega(f,1/w)\
(M_{0}(\theta\chi)+2M_{1}(\theta\chi))$ $\displaystyle\leq$
$\displaystyle\frac{w}{2}\
\omega(f,1/w)\left(3M_{0}(\theta\chi)+2M_{1}(\theta\chi)\right).$

This gives

$\displaystyle|f(x)-f(y)|$ $\displaystyle\leq$ $\displaystyle
2Kw^{-\alpha}+\frac{w}{2}\
\omega(f,1/w)\int_{y}^{x}\left(3M_{0}(\theta\chi)+2M_{1}(\theta\chi)\right)\frac{dt}{t}$
$\displaystyle\leq$ $\displaystyle 2Kw^{-\alpha}+\frac{w}{2}\
\omega(f,1/w)\left(3M_{0}(\theta\chi)+2M_{1}(\theta\chi)\right)|\log x-\log
y|.$

Since $M_{0}(\theta\chi)$ and $M_{1}(\theta\chi)$ are finite, we define
$N:=\mbox{max}\\{M_{0}(\theta\chi),M_{1}(\theta\chi)\\}$ to obtain

$\omega(f,\delta)\leq 2Kw^{-\alpha}+w\delta\
\omega(f,1/w)\left(\frac{5N}{2}\right).$

For any fixed $\delta>0,$ there exists sufficiently large $w$ such that
$\frac{1}{w}<\delta.$ Now, in view of monotonicity of $\omega,$ we write

$\omega(f,1/w)<\omega(f,\delta).$

This gives $\omega(f,\delta)\leq H\ w^{-\alpha},$ where $\displaystyle
H:=\frac{4K}{5-2N}.$ Hence, we conclude that $f\in L_{\alpha},\ 0<\alpha\leq
1.$ Thus, we obtain the desired result. ∎

##  4\. Examples of kernel

In this section, we discuss some examples of kernel functions satisfying the
assumptions of our theory.

###  4.1. Mellin B-spline kernel

We start with the Mellin B-spline kernel of order $n\in\mathbb{N}$ which are
defined by (see [14])

$\overline{{B}_{n}}(t):=\frac{1}{(n-1)!}\sum_{j=0}^{n}(-1)^{j}{n\choose
j}\bigg{(}\frac{n}{2}+\log t-j\bigg{)}_{+}^{n-1},\,\,\,\ t\in\mathbb{R}^{+},$

where $(x)_{+}:=\max\\{x,0\\},$ $x\in\mathbb{R}.$ We observe that for
$t\in\mathbb{R}^{+},$ $B_{n}(\log t)=\overline{{B}_{n}}(t),$ where

${B}_{n}(x):=\frac{1}{(n-1)!}\sum_{j=0}^{n}(-1)^{j}{n\choose
j}\bigg{(}\frac{n}{2}+x-j\bigg{)}_{+}^{n-1},\,\,\,\ x\in\mathbb{R},$

denotes the classical $B$-splines of order $n\in\mathbb{N}.$ Also, we noted
$B_{n}(x)=\overline{{B}_{n}}(e^{x}),$ $x\in\mathbb{R}.$ Now using the
definition of the Mellin transform, we have

$[\overline{{B}_{n}}]^{\wedge}_{M}(is)=\int_{0}^{\infty}\overline{{B}_{n}}(t)\
t^{is}\ \frac{dt}{t},\ \ \ s\in\mathbb{R}.$

Subtituting $u=\log t,$ we easily obtain

$\displaystyle[\overline{{B}_{n}}]^{\wedge}_{M}(is)$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\overline{{B}_{n}}(e^{u})\ e^{isu}\
du=\int_{-\infty}^{\infty}{B}_{n}(u)\ e^{isu}\ du=\widehat{B_{n}}(-s),$

where
$\widehat{f}(u):=\displaystyle\int_{-\infty}^{\infty}f(x)e^{-iux}dx,u\in\mathbb{R}$
denotes the Fourier transform of the function $f.$ Now, let
$f(x):=\displaystyle\sum_{k=-\infty}^{\infty}\overline{{B}_{n}}(e^{k}x).$ This
$f$ is a recurrent function with fundamental interval $[1,e].$ That is
$f(ex)=f(x),$ $\forall x\in\mathbb{R}^{+}.$ The Mellin-Fourier cofficient
$m_{k}(f)$ of $f$ is given by

$\displaystyle m_{k}(f):=\displaystyle\int_{1}^{e}f(x)x^{2k\pi
i}\frac{dx}{x}=\displaystyle\int_{1}^{e}\sum_{j=-\infty}^{\infty}\overline{{B}_{n}}(e^{j}x)x^{2k\pi
i}\frac{dx}{x}=\sum_{j=-\infty}^{\infty}\displaystyle\int_{1}^{e}\overline{{B}_{n}}(e^{j}x)x^{2k\pi
i}\frac{dx}{x}.$

Substituting $u=e^{j}x,$ we obtain

$\displaystyle
m_{k}(f)=\sum_{j=-\infty}^{\infty}\displaystyle\int_{e^{j}}^{e^{j+1}}\overline{{B}_{n}}(u){\left(\frac{u}{e^{j}}\right)}^{2k\pi
i}\frac{du}{u}=\displaystyle\int_{0}^{\infty}\overline{{B}_{n}}(u){u}^{2k\pi
i}\frac{du}{u}=[\overline{{B}_{n}}]^{\wedge}_{M}(2k\pi i).$

Therefore, we get the following Mellin Poisson summation formula

$\displaystyle\displaystyle\sum_{k=-\infty}^{\infty}\overline{{B}_{n}}(e^{k}x)=\sum_{k=-\infty}^{\infty}[\overline{{B}_{n}}]^{\wedge}_{M}(2k\pi
i)x^{-2k\pi i}.$ (4.1)

It is easy to see that

$\displaystyle[\overline{{B}_{n}}]^{\wedge}_{M}(c+is)=\bigg{(}\frac{\sin(\frac{s}{2})}{(\frac{s}{2})}\Bigg{)}^{n},\
\ \hskip 14.22636pts\neq 0,c=0.$ (4.2)

Therefore, we have

$[\overline{{B}_{n}}]^{\wedge}_{M}(2k\pi i)=\widehat{B_{n}}(-2k\pi
i)=\begin{cases}{1,}&\quad\text{if}\ \ {k=0}\\\ {0,}&\quad\ \
{\text{otherwise.}}\\\ \end{cases}$ (4.3)

Using the Mellin Poisson summation formula (4.1), we get

$\displaystyle\sum_{k=-\infty}^{\infty}\overline{{B}_{n}}(e^{k}x)=1,\ \
\forall x\in\mathbb{R}^{+}.$

Hence $\overline{{B}_{n}}$ satisfies the condition $(\chi_{1}).$

Now we establish the condition $(\chi_{2}).$ Again using the Mellin transform
and by differentiating under the sign of the integral, we get

$\frac{d}{ds}\left([\overline{{B}_{n}}]^{\wedge}_{M}(is)\right)=\frac{d}{ds}\left(\displaystyle\int_{0}^{\infty}\overline{{B}_{n}}(t){t}^{is}\frac{dt}{t}\right)=\int_{0}^{\infty}\overline{{B}_{n}}(t){t}^{is}(i\log
t)\frac{dt}{t}.$

Similarly, we can easily obtain

$\frac{d^{j}}{ds^{j}}\left([\overline{{B}_{n}}]^{\wedge}_{M}(is)\right)=\int_{0}^{\infty}\overline{{B}_{n}}(t){t}^{is}(i\log
t)^{j}\frac{dt}{t}=(i)^{j}[\overline{f_{j}}]^{\wedge}_{M}(is),$

where $f_{j}(t)=\overline{{B}_{n}}(t)(\log t)^{j}.$ Thus, we get

$\displaystyle\displaystyle\sum_{k=-\infty}^{\infty}\overline{{B}_{n}}(e^{-k}x)(k-\log
x)^{j}=\sum_{k=-\infty}^{\infty}(-i)^{j}\frac{d^{j}}{ds^{j}}\overline{{B}_{n}}]^{\wedge}_{M}(2k\pi
i)x^{-2k\pi i}.$ (4.4)

Again from (4.2), we obtain

$\displaystyle\dfrac{d}{ds}\left([\overline{{B}_{n}}]^{\wedge}_{M}(is)\right)=n\bigg{(}\frac{\sin(\frac{s}{2})}{(\frac{s}{2})}\Bigg{)}^{n-1}\left(\dfrac{s\cos(s/2)-2\sin(s/2)}{s^{2}}\right),\
\ \hskip 14.22636pts\neq 0,$

which in-turn gives

$\displaystyle\dfrac{d}{ds}\left([\overline{{B}_{n}}]^{\wedge}_{M}(is)\right)(2k\pi
i)=0,\ \ k\in\mathbb{Z}.$

Therefore, we obtain $m_{1}(\overline{{B}_{n}},x)=0,\ \forall n\in\mathbb{N},$
by using (4.4). Hence the condition $(\chi_{2})$ is also satisfied.

As $\overline{{B}_{n}}$ is compactly supported, there exists $\nu>0$ such that
$supp(\overline{{B}_{n}})\subseteq[e^{-\nu},e^{\nu}].$ Thus, we get
$|\\{k:e^{-\nu}\leq e^{-k}u\leq e^{\nu}\\}|\leq 2[\nu]+1,$ $\forall
u\in\mathbb{R}^{+},$ where $[.]$ denotes the integer part. Hence we obtain

$\sum_{k=-\infty}^{\infty}|\overline{{B}_{n}}(e^{-k}u)|\ |k-\log
u|^{\beta}\leq(2[\nu]+1)\left(\sup_{u\in\mathbb{R}^{+}}\overline{{B}_{n}}(u)\right)\nu^{\beta}<\infty.$

This establishes the condition $(\chi_{3}).$

Again using the fact that
$supp(\overline{{B}_{n}})\subseteq[e^{-\nu},e^{\nu}],$ for some $\nu>0.$
Choosing $w\gamma>\nu,$ we get $\displaystyle\sum_{|k-w\log
x|>w\gamma}|\overline{{B}_{n}}(e^{-k}x^{w})|\ |k-w\log x|=0,$ $\forall
x\in\mathbb{R}^{+}.$ Therefore, the condition $(\chi_{4})$ is satisfied. To
verify the conditions which are used in the hypothesis of the theorem, we
consider the third order Mellin $B$-spline kernel by

$\overline{{B}_{3}}(x)=\begin{cases}{-\frac{1}{2}\left(\frac{3}{2}+\log
x\right)^{2},}&\quad\text{}\ \ {\text{$e^{-3/2}<x<e^{-1/2},$}}\\\
{\frac{3}{4}-\log^{2}x,}&\quad\text{}\ \ {e^{-1/2}<x<e^{1/2},}\\\
{-\frac{1}{2}\left(\frac{3}{2}-\log x\right)^{2},}&\quad\text{}\ \
{\text{$e^{1/2}<x<e^{3/2},$}}\\\ {0,}&\quad\text{}\ \ {\text{otherwise}.}\\\
\end{cases}$

The Mellin derivative of $\overline{{B}_{3}}(x)$ is given by

$(\theta\overline{{B}_{3}})(x)=\begin{cases}{-\left(\frac{3}{2}+\log
x\right),}&\quad\text{}\ \ {\text{$e^{-3/2}<x<e^{-1/2},$}}\\\ {-2\log
x,}&\quad\text{}\ \ {e^{-1/2}<x<e^{1/2},}\\\ {\frac{3}{2}-\log
x,}&\quad\text{}\ \ {\text{$e^{1/2}<x<e^{3/2},$}}\\\ {0,}&\quad\text{}\ \ \
{\text{otherwise}.}\\\ \end{cases}$

Evidently,
$supp(\theta\overline{{B}_{3}})\subseteq\left(e^{-3/2},e^{3/2}\right).$ Hence,
$M_{\beta}(\theta\overline{{B}_{3}})$ are finite. So it satisfies the
assumption of Theorem 4. Moreover, we have
$m_{1}^{\overline{{B}_{3}}}:=m_{1}(\overline{{B}_{3}},x)=0,\ \forall
x\in\mathbb{R}^{+}.$ Thus, $\overline{{B}_{3}}(x)$ also satisfies the
assumption of Theorem 3.

###  4.2. Mellin Jackson kernel

Next we consider Mellin Jackson kernel. For $c\in\mathbb{R},\alpha\geq
1,n\in\mathbb{N}$ and $x\in\mathbb{R}^{+},$ the Mellin Jackson kernel is
defined by (see [14])

$\overline{J_{\alpha,n}}(x):=C_{\alpha,n}\
x^{-c}\textit{sinc}^{2n}\left(\frac{\log x}{2\alpha n\pi}\right),$

where $\displaystyle
C^{-1}_{\alpha,n}:=\int_{0}^{\infty}\textit{sinc}^{2n}\left(\frac{\log
x}{2\alpha n\pi}\right)\frac{dx}{x}$ and $sinc(u)=\begin{cases}{\dfrac{\sin\pi
u}{\pi u},}&\quad\text{}\ \ {u\neq 0}\\\ {1,}&\quad\text{}\ \ {u=0}.\\\
\end{cases}$

It is evident that $\overline{J_{\alpha,n}}(x)=x^{-c}{J_{\alpha,n}}(\log x),$
where ${J_{\alpha,n}}$ represents the generalized Jackson kernel (see [8])
given by

${J_{\alpha,n}}(x):=c_{\alpha,n}\textit{sinc}^{2n}\left(\frac{x}{2\alpha
n\pi}\right),$

where $\displaystyle
c^{-1}_{\alpha,n}:=\int_{-\infty}^{\infty}\textit{sinc}^{2n}\left(\frac{x}{2\alpha
n\pi}\right)dx\ $ is constant. Now we have

$\displaystyle\|\overline{J_{\alpha,n}}\|_{X_{c}}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}|\overline{J_{\alpha,n}}(x)|x^{c}\
\frac{dx}{x}=\int_{0}^{\infty}|{J_{\alpha,n}}(\log x)|\frac{dx}{x}=1.$

Hence $\overline{J_{\alpha,n}}(x)\in X_{c}$ and therefore its Mellin transform
is well defined. Now, we obtain

$\displaystyle[\overline{J_{\alpha,n}}]^{\wedge}_{M}(c+iv)=\int_{0}^{\infty}\overline{J_{\alpha,n}}(x)x^{c+iv}\frac{dx}{x}=C_{\alpha,n}\int_{0}^{\infty}x^{iv}\frac{\textit{sin}^{2n}\left(\frac{\log
x}{2\alpha n}\right)}{\left(\frac{\log x}{2\alpha
n}\right)^{2n}}\frac{dx}{x}.$

On substituting $u=\log x,$ we obtain

$\displaystyle[\overline{J_{\alpha,n}}]^{\wedge}_{M}(c+iv)$ $\displaystyle=$
$\displaystyle\frac{C_{\alpha,n}}{(n\alpha)^{2n}}\int_{-\infty}^{\infty}e^{iuv}\left(\widehat{\chi}_{[-\frac{1}{2n\alpha},\frac{1}{2n\alpha}]}(u)\right)^{2n}du$
$\displaystyle=$
$\displaystyle\frac{C_{\alpha,n}}{(n\alpha)^{2n}}\left({\chi}_{[-\frac{1}{2n\alpha},\frac{1}{2n\alpha}]}\ast{\chi}_{[-\frac{1}{2n\alpha},\frac{1}{2n\alpha}]}\ast...\ast{\chi}_{[-\frac{1}{2n\alpha},\frac{1}{2n\alpha}]}\right)(u),\
\ (2n\ \ times),$

where $\ast$ denotes the convolution. From the above relation, we see that
$supp[\overline{J_{\alpha,n}}]^{\wedge}_{M}\subseteq[-\frac{1}{\alpha},\frac{1}{\alpha}],$
hence we get $[\overline{J_{\alpha,n}}]^{\wedge}_{M}(c+iv)=0,$ for
$|v|>\frac{1}{\alpha}.$ Thus $\overline{J_{\alpha,n}}$ is Mellin band-limited.
Further, we observe that $[\overline{J_{\alpha,n}}]^{\wedge}_{M}(0)=1$ and
$[\overline{J_{\alpha,n}}]^{\wedge}_{M}(2k\pi i)=0,$ for $k\neq 0.$ Hence
using the Mellin-Poisson summation formula, we get

$\displaystyle\displaystyle\sum_{k=-\infty}^{\infty}\overline{J_{\alpha,n}}(e^{k}x)=\sum_{k=-\infty}^{\infty}[\overline{J_{\alpha,n}}]^{\wedge}_{M}(2k\pi
i)x^{-2k\pi i}=1.$

Hence $(\chi_{1})$ is satisfied.

In view of the Mellin transform, we obtain

$\displaystyle\frac{d}{dv}\left([\overline{J_{\alpha,n}}]^{\wedge}_{M}(0)\right)$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\overline{J_{\alpha,n}}(u)(i\log
u)\frac{du}{u}=iC_{\alpha,n}\int_{0}^{\infty}\frac{\textit{sin}^{2n}\left(\frac{\log
u}{2\alpha n}\right)}{\left(\frac{\log u}{2\alpha n}\right)}\log
u\frac{du}{u}.$

Hence, we obtain
$\dfrac{d}{dv}\left([{J_{\alpha},n}]^{\wedge}_{M}(0)\right)=0.$ Thus from the
Mellin-Poisson summation formula, we get $m_{1}(\overline{J_{\alpha,n}},u)=0.$
Therfore, the condition $(\chi_{2})$ is established. Further, we see that
$m_{1}^{\overline{J_{\alpha,n}}}\neq-\dfrac{1}{2},$ the condition of Theorem 3
is fulfilled.

We now show that $(\chi_{4})$ is satisfied, i.e.
$\displaystyle\lim_{w\rightarrow\infty}\sum_{|k-w\log
x|>w\gamma}|\overline{J_{\alpha,n}}(e^{-k}x^{w})|\ |k-w\log x|=0$ uniformly on
$\mathbb{R}^{+}.$ Let $\epsilon>0$ and $n>1.$ Then, there exists
$N\in\mathbb{Z}$ such that
$\displaystyle\sum_{k>N}\frac{1}{k^{2n-1}}<\epsilon.$ For $w\gamma>N,$ we can
write

$\displaystyle\sum_{|k-w\log
x|>w\gamma}|\overline{J_{\alpha,n}}(e^{-k}x^{w})|\ |k-w\log x|$

$\displaystyle=$ $\displaystyle\left(\sum_{k-w\log x>w\gamma}+\sum_{k-w\log
x<-w\gamma}\right)|\overline{J_{\alpha,n}}(e^{-k}x^{w})|\ |k-w\log x|$
$\displaystyle:=$ $\displaystyle S_{1}+S_{2}.$

First we estimate $S_{1}.$ We have

$\displaystyle S_{1}$ $\displaystyle=$ $\displaystyle\sum_{k-w\log
x>w\gamma}C_{\alpha,2n}sinc\left(\frac{|w\log x-k|}{2\alpha
n\pi}\right)^{2n}|k-w\log x|$ $\displaystyle\leq$ $\displaystyle\sum_{k-w\log
x>w\gamma}C_{\alpha,2n}(2\alpha n)^{2n}\frac{1}{|k-w\log x|^{2n-1}}$
$\displaystyle\leq$ $\displaystyle C_{\alpha,2n}(2\alpha
n)^{2n}\sum_{k>N}\frac{1}{k^{2n-1}}<C_{\alpha,2n}(2\alpha n)^{2n}\epsilon.$

Similarly, for $S_{2},$ we obtain

$\displaystyle S_{2}$ $\displaystyle\leq$ $\displaystyle\sum_{k-w\log
x<-w\gamma}C_{\alpha,2n}(2\alpha n)^{2n}\frac{1}{|k-w\log x|^{2n-1}}$
$\displaystyle\leq$ $\displaystyle C_{\alpha,2n}(2\alpha
n)^{2n}\sum_{k=1}^{\infty}\frac{1}{(N+k)^{2n-1}}$ $\displaystyle<$
$\displaystyle C_{\alpha,2n}(2\alpha n)^{2n}\epsilon.$

On combining $S_{1}-S_{2},$ we get $S<2C_{\alpha,2n}(2\alpha n)^{2n}\epsilon,$
for $n>1.$ Hence $(\chi_{4})$ is satisfied for $n>1.$

Now we establish the condition $(\chi_{3}).$ Let $u\in\mathbb{R}^{+}$ and
$\beta<2n-1,c=0.$ Then $\exists$ $k_{0}\in\mathbb{Z}$ such that $k_{0}\leq\log
u<k_{0}+1.$ This gives $|\log u-k|\geq|k-k_{0}|,$ if $k<k_{0}$ and $|\log
u-k|>|k-(k_{0}+1)|,$ if $k>k_{0}+1.$ Now, we have

$\displaystyle\sum_{k=-\infty}^{\infty}|{\overline{J_{\alpha,n}}}(e^{-k}u)|\
|k-\log u|^{\beta}$ $\displaystyle=$
$\displaystyle\left(\sum_{k<k_{0}}+\sum_{k=k_{0},k_{0}+1}+\sum_{k>k_{0}}\right)|\overline{J_{\alpha,n}}(e^{-k}u)|\
|k-\log u|^{\beta}$ $\displaystyle:=$ $\displaystyle
S_{1}^{{}^{\prime}}+S_{2}^{{}^{\prime}}+S_{3}{{}^{\prime}}.$

Using definition of $\overline{J_{\alpha,n}},$ $S_{1}^{{}^{\prime}}$ is
estimated by

$\displaystyle S_{1}^{{}^{\prime}}$ $\displaystyle=$ $\displaystyle
C_{\alpha,n}\sum_{k<k_{0}}\sin^{2n}\left(\frac{|k-\log
u|}{2n\alpha}\right)(2\alpha n)^{2n}\frac{1}{|\log u-k|^{2n-\beta}}$
$\displaystyle\leq$ $\displaystyle C_{\alpha,n}(2\alpha
n)^{2n}\sum_{k<k_{0}}\frac{1}{|\log u-k|^{2n-\beta}}$ $\displaystyle\leq$
$\displaystyle C_{\alpha,n}(2\alpha
n)^{2n}\sum_{k<k_{0}}\frac{1}{|k-k_{0}|^{2n-\beta}}\leq C_{\alpha,n}(2\alpha
n)^{2n}\sum_{k=1}^{\infty}\frac{1}{k^{2n-\beta}}.$

The above sum is finite if $\beta<2n-1.$ Similarly, for $S_{3}^{{}^{\prime}},$
we obtain

$\displaystyle S_{3}^{{}^{\prime}}$ $\displaystyle\leq$ $\displaystyle
C_{\alpha,n}(2\alpha n)^{2n}\sum_{k>k_{0}+1}\frac{1}{|\log u-k|^{2n-\beta}}$
$\displaystyle\leq$ $\displaystyle C_{\alpha,n}(2\alpha
n)^{2n}\sum_{k>k_{0}+1}\frac{1}{|k-(k_{0}+1)|^{2n-\beta}}\leq
C_{\alpha,n}(2\alpha n)^{2n}\sum_{k=1}^{\infty}\frac{1}{k^{2n-\beta}}.$

This gives $S_{3}^{{}^{\prime}}<\infty$ for $\beta<2n-1.$ Finally
$S_{2}^{{}^{\prime}}$ is estimated as

$\displaystyle S_{2}^{{}^{\prime}}$ $\displaystyle=$ $\displaystyle
C_{\alpha,n}\left(\frac{\sin(\frac{|\log u-k_{0}|}{2n\alpha})}{\frac{|\log
u-k_{0}|}{2n\alpha}}\right)^{2n}|\log
u-k_{0}|^{\beta}+C_{\alpha,n}\left(\frac{\sin(\frac{|\log
u-k_{0}-1|}{2n\alpha})}{\frac{|\log u-k_{0}-1|}{2n\alpha}}\right)^{2n}|\log
u-(k_{0}-1)|^{\beta}$ $\displaystyle\leq$ $\displaystyle
C_{\alpha,n}\left(\frac{\sin(\frac{|\log u-k_{0}|}{2n\alpha})}{\frac{|\log
u-k_{0}|}{2n\alpha}}\right)^{2n}+C_{\alpha,n}\left(\frac{\sin(\frac{|\log
u-k_{0}-1|}{2n\alpha})}{\frac{|\log u-k_{0}-1|}{2n\alpha}}\right)^{2n}$
$\displaystyle\leq$ $\displaystyle 2C_{\alpha,n}\sup_{0\leq u\leq
1}\left(\frac{\sin(\frac{u}{2n\alpha})}{\frac{u}{2n\alpha}}\right)^{2n}\leq
2C_{\alpha,n}.$

Combining the estimates
$S_{1}^{{}^{\prime}},S_{2}^{{}^{\prime}},S_{3}^{{}^{\prime}},$ we get

$\sup_{u\in\mathbb{R}^{+}}\sum_{k=-\infty}^{\infty}|{\overline{J_{\alpha,n}}}(e^{-k}u)|\
|k-\log u|^{\beta}<\infty$

and hence $(\chi_{3})$ is verified.

Putting $c=0$ in the Mellin-Jackson kernel $\overline{J_{\alpha,n}}$ and
taking the Mellin derivative of the kernel, we get

$(\theta\overline{J_{\alpha,n}})(x)=\frac{C_{\alpha,n}}{\log^{2n}x}\left[\frac{2n}{\rho^{2n-1}}\sin^{2n-1}(\rho\log
x)\left(\cos(\rho\log x)-\frac{\sin(\rho\log x)}{\rho\log x}\right)\right].$

Proceeding along the lines of the estimate of
$S_{1}^{{}^{\prime}},S_{2}^{{}^{\prime}},S_{3}^{{}^{\prime}}$, it follows that
$M_{1}(\theta\overline{J_{\alpha,n}})<\infty,$ and thus
$\overline{J_{\alpha,n}}$ satisfies the assumptions of Theorem 4 for $n>1$ and
$\beta<2n-1$.

##  5\. Acknowledgement

S. Bajpeyi gratefully thank Indian Institute of Science Education and Research
(IISER) Thiruvananthapuram for the postdoctoral fellowship to carry out this
research work. A. Sathish Kumar acknowledges DST-SERB, India Research Grant
MTR/2021/000428 for the financial support and NFIG Grant, IIT Madras, Grant
No. RF/22-23/0984/MA/NFIG/009017.

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