# Optimal parameter choice for regularized Shannon sampling formulas

Melanie Kircheis111Corresponding author<EMAIL_ADDRESS>Chemnitz University of Technology, Faculty of Mathematics, D–09107 Chemnitz,
Germany Daniel<EMAIL_ADDRESS>Chemnitz University of
Technology, Faculty of Mathematics, D–09107 Chemnitz, Germany Manfred
<EMAIL_ADDRESS>University of Rostock, Institute of
Mathematics, D–18051 Rostock, Germany

###### Abstract

The fast reconstruction of a bandlimited function from its sample data is an
essential problem in signal processing. In this paper, we consider the widely
used Gaussian regularized Shannon sampling formula in comparison to
regularized Shannon sampling formulas employing alternative window functions,
including the modified Gaussian function, the $\sinh$-type window function,
and the continuous Kaiser–Bessel window function. It is shown that the
approximation errors of these regularized Shannon sampling formulas possess an
exponential decay with respect to the truncation parameter. The main focus of
this paper is to identify the optimal variance of the (modified) Gaussian
function as well as the optimal shape parameters of the $\sinh$-type window
function and the continuous Kaiser–Bessel window function, with the aim of
achieving the fastest exponential decay of the approximation error. In doing
so, we demonstrate that the decay rate of the $\sinh$-type regularized Shannon
sampling formula is considerably superior to that of the Gaussian regularized
Shannon sampling formula. Additionally, numerical experiments illustrate the
theoretical results.

_Key words_ : Shannon sampling series, optimal regularization, bandlimited
function, approximation error, exponential decay, Gaussian regularized Shannon
sampling formulas, $\sinh$-type regularized Shannon sampling formulas.

AMS _Subject Classifications_ : 62D05, 42A10, 41A25, 94A20.

##  1 Introduction

In signal processing, the fast reconstruction of a bandlimited function from
its sample data is of fundamental importance. A function $f\in
L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is called _bandlimited_ with _bandwidth_
$\delta>0$, if its Fourier transform

$\displaystyle({\mathcal{F}}f)(\omega)={\hat{f}}(\omega)\coloneqq\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}f(t)\,{\mathrm{e}}^{-{\mathrm{i}}t\omega}\,{\mathrm{d}}t\,,\quad\omega\in\mathbb{R}\,,$
(1.1)

vanishes for all $|\omega|\geq\delta$. For such a bandlimited function with
$\delta\in(0,\,\pi]$ the famous Shannon sampling theorem, see [31, 14, 28],
states that

$\displaystyle f(t)=\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{sinc}(t-k)\,,\quad
t\in\mathbb{R}\,,$ (1.2)

where

$\displaystyle\mathrm{sinc}(t)\coloneqq\begin{cases}\frac{\sin(\pi t)}{\pi
t}&\colon t\in\mathbb{R}\setminus\\{0\\}\,,\\\ 1&\colon t=0\,,\end{cases}$
(1.3)

denotes the _cardinal sine function_. It is known that the Shannon sampling
series (1.2) converges absolutely and uniformly on whole $\mathbb{R}$.
However, the practical use of (1.2) is limited, since its evaluation requires
infinitely many samples and its truncated version is not a good approximation
due to the slow decay of the cardinal sine function, see [11]. In addition to
this rather poor convergence, it is known, see [8, 9, 7], that in the presence
of noise in the samples $f(k)$, $k\in\mathbb{Z}$, of a bandlimited function
$f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ the convergence of Shannon sampling
series (1.2) may even break down completely. Therefore, it was proposed to
consider the regularization of the Shannon sampling series with a suitable
window function. Note that many authors such as [6, 17, 26, 19, 29] used
window functions in the frequency domain, but the recent study [13] has shown
that it is much more beneficial to employ a window function in the spatial
domain, cf. [22, 23, 29, 16, 15, 5, 12]. In the following, a _window function_
$\varphi:\,\mathbb{R}\to[0,\,1]$ is an even function in $L^{2}(\mathbb{R})\cap
C(\mathbb{R})$ which decreases on $[0,\,\infty)$ and fulfills $\varphi(0)=1$.
By ${\mathbf{1}}_{[-m,\,m]}$ we denote the _characteristic function_ of the
interval $[-m,\,m]$ with $m\in\mathbb{N}\setminus\\{1\\}$, i. e., the function

$\displaystyle{\mathbf{1}}_{[-m,\,m]}(t)\coloneqq\begin{cases}1&\colon
t\in[-m,\,m]\,,\\\ 0&\colon t\in\mathbb{R}\setminus[-m,\,m]\,.\end{cases}$

In this paper, we assume that the bandwidth $\delta$ of $f$ fulfills the so-
called _oversampling condition_ $0<\delta<\pi$. Then we recover $f$ by the
_$\varphi$ -regularized Shannon sampling formula_

$\displaystyle\big{(}R_{\varphi,m}f\big{)}(t)\coloneqq\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)\,{\mathbf{1}}_{[-m,\,m]}(t-k)\,,\quad
t\in\mathbb{R}\,,$ (1.4)

where $m\in\mathbb{N}\setminus\\{1\\}$ is the so-called _truncation
parameter_. In doing so, we consider the following window functions
$\varphi:\,\mathbb{R}\to[0,\,1]$.

######  Example 1.1.

The most popular window function, see e. g. [22, 25, 27, 30, 15, 5], is the
_Gaussian function_

$\displaystyle\varphi_{\mathrm{Gauss}}(t)\coloneqq\mathrm{e}^{-t^{2}/(2\sigma^{2})}\,,\quad
t\in\mathbb{R}\,,$ (1.5)

with _variance_ $\sigma^{2}>0$. In addition, [25, 24] considered the _modified
Gaussian function_

$\displaystyle\varphi_{\mathrm{modGauss}}(t)\coloneqq\mathrm{e}^{-t^{2}/(2\sigma^{2})}\,\cos(\lambda
t)\,,\quad t\in\mathbb{R}\,,$ (1.6)

with the parameters $\sigma^{2}>0$ and $\lambda\geq 0$. Then the corresponding
expression (1.4) is named the _(modified) Gaussian regularized Shannon
sampling formula_. Note that these two window functions (1.5) and (1.6) are
supported on whole $\mathbb{R}$.

Here we prefer window functions which are compactly supported on the interval
$[-m,\,m]$, as studied in [12, 13]. The $\sinh$-_type window function_ is
defined as

$\displaystyle\varphi_{\sinh}(t)\coloneqq\begin{cases}\frac{1}{\sinh\beta}\,\sinh\Big{(}\beta\,\sqrt{1-\tfrac{t^{2}}{m^{2}}}\,\Big{)}&\colon
t\in[-m,\,m]\,,\\\ 0&\colon t\in\mathbb{R}\setminus[-m,\,m]\,,\end{cases}$
(1.7)

with _shape parameter_ $\beta>0$, see [21]. Then the corresponding expression
(1.4) is termed the $\sinh$-_type regularized Shannon sampling formula_. The
_continuous Kaiser–Bessel window function_ is defined as

$\displaystyle\varphi_{\mathrm{cKB}}(t)\coloneqq\begin{cases}\frac{1}{I_{0}(\beta)-1}\,\big{(}I_{0}(\beta\sqrt{1-t^{2}/m^{2}})-1\big{)}&\colon
t\in[-m,\,m]\,,\\\ 0&\colon t\in\mathbb{R}\setminus[-m,\,m]\,,\end{cases}$
(1.8)

with convenient shape parameter $\beta>0$, see [21]. Then the corresponding
expression (1.4) is called the _continuous Kaiser–Bessel regularized Shannon
sampling formula_. We remark that these two window functions (1.7) and (1.8)
are well-studied in the context of the nonuniform fast Fourier transform
(NFFT), see e. g. [20, Section 6] and [4, 3].

Due to the definition of the cardinal sine function (1.3) we have
$\mathrm{sinc}(n-k)=\delta_{n,k}$ and therefore the regularized Shannon
sampling formula $R_{\varphi,m}f$ in (1.4) has the _interpolation property_

$\displaystyle\big{(}R_{\varphi,m}f\big{)}(n)=f(n)\,,\quad n\in\mathbb{Z}\,.$
(1.9)

Moreover, the use of the characteristic function ${\mathbf{1}}_{[-m,\,m]}$ in
(1.4) leads to _localized sampling_ of $f$, i. e., the computation of
$\big{(}R_{\varphi,m}f\big{)}(t)$ for any $t\in\mathbb{R}\setminus\mathbb{Z}$
requires only $2m$ samples $f(k)$, where $k\in\mathbb{Z}$ fulfills the
condition $|k-t|\leq m$. Especially, for $t\in(0,\,1)$ we obtain the finite
sum

$\displaystyle\big{(}R_{\varphi,m}f\big{)}(t)=\sum_{k=1-m}^{m}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)\,.$

As in many applications, we use _oversampling_ of the given bandlimited
function $f$ with bandwidth $\delta<\pi$, i. e., the function $f$ is sampled
on the integer grid $\mathbb{Z}$.

In this paper, we focus on the $\varphi$-regularized Shannon sampling formulas
(1.4) for the window functions $\varphi$ given in Example 1.1. To compare the
corresponding approaches, we present estimates on the uniform approximation
error

$\displaystyle\|f-R_{\varphi,m}f\|_{C_{0}(\mathbb{R})}\coloneqq\max_{t\in\mathbb{R}}\big{|}f(t)-\big{(}R_{\varphi,m}f\big{)}(t)\big{|}\,,$
(1.10)

where $C_{0}(\mathbb{R})$ denotes the Banach space of continuous functions
$g\colon\\!\mathbb{R}\to\mathbb{C}$ vanishing as $|t|\to\infty$. The main
focus of this paper is to find the optimal variance $\sigma^{2}$ of the
(modified) Gaussian window function (1.5) and (1.6), respectively, as well as
the optimal shape parameter $\beta$ of the $\sinh$-type window function (1.7)
and the continuous Kaiser–Bessel window function (1.8), such that the
exponential decay of the approximation error (1.10) is the fastest.

For this purpose, we initially study the uniform approximation error of
general $\varphi$-regularized Shannon sampling formulas (1.4) in Section 2.
Afterwards, we specify our findings for the window functions $\varphi$
introduced in Example 1.1. In particular, Section 3 deals with the (modified)
Gaussian window function (1.5) and (1.6), respectively, while Section 4 is
concerned with the $\sinh$-type window function (1.7) and Section 5 with the
continuous Kaiser–Bessel window function (1.8).

##  2 Approximation error of regularized Shannon sampling formulas

First we estimate the uniform approximation error of the $\varphi$-regularized
Shannon sampling formula (1.4), analogous to [12, Theorem 3.2] and [13,
Theorem 4.1].

######  Theorem 2.1.

Assume that $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is bandlimited with
bandwidth $\delta\in(0,\,\pi)$. Further let $\varphi:\,\mathbb{R}\to[0,\,1]$
be an even function in $L^{2}(\mathbb{R})\cap C(\mathbb{R})$ which is
decreasing on $[0,\,\infty)$ with $\varphi(0)=1$, and let
$m\in{\mathbb{N}}\setminus\\{1\\}$ be given.
Then the $\varphi$-regularized Shannon sampling formula (1.4) satisfies the
error estimate

$\displaystyle\|f-R_{\varphi,m}f\|_{C_{0}(\mathbb{R})}\leq\big{(}E_{1}(m)+E_{2}(m)\big{)}\,\|f\|_{L^{2}(\mathbb{R})}\,,\quad
m\in\mathbb{N}\setminus\\{1\\}\,,$

with the error constants

$\displaystyle E_{1}(m)$
$\displaystyle\coloneqq\max_{\omega\in[-\delta,\,\delta]}\bigg{|}1-\frac{1}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}{\hat{\varphi}}(\tau)\,{\mathrm{d}}\tau\,\bigg{|}\,,$
(2.1) $\displaystyle E_{2}(m)$
$\displaystyle\coloneqq\frac{\sqrt{2}}{\pi\,m}\,\sqrt{\varphi^{2}(m)+\int_{m}^{\infty}\varphi^{2}(t)\,\mathrm{d}t}\,.$
(2.2)

_Proof_. (i) Initially, we consider only the case $t\in(0,\,1)$, where we
split the approximation error

$\displaystyle
f(t)-\big{(}R_{\varphi,m}f\big{)}(t)=e_{1}(t)+e_{2,0}(t)\,,\quad
t\in(0,\,1)\,,$

into the _regularization error_

$\displaystyle e_{1}(t)\coloneqq
f(t)-\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)\,,\quad
t\in\mathbb{R}\,,$ (2.3)

and the _truncation error_

$\displaystyle e_{2,0}(t)$
$\displaystyle\coloneqq\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)-\big{(}R_{\varphi,m}f\big{)}(t)$
$\displaystyle=\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)\,,\quad
t\in(0,\,1)\,.$ (2.4)

(ii) To estimate the regularization error (2.3), we start our study by
considering the Fourier transform (1.1) of the function
$\varphi\,\mathrm{sinc}$, i. e., the term

$\displaystyle{\mathcal{F}}(\varphi\,\mathrm{sinc})(\omega)=\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}\varphi(t)\,\mathrm{sinc}(t)\,\mathrm{e}^{-\mathrm{i}\omega
t}\,\mathrm{d}t\,.$

Using the convolution property of $\mathcal{F}$ in $L^{2}(\mathbb{R})$ (see
[20, Theorem 2.26]), we have

$\displaystyle{\mathcal{F}}(\varphi\,\mathrm{sinc})(\omega)=\big{(}{\hat{\varphi}}\star({\mathcal{F}}\mathrm{sinc})\big{)}(\omega)=\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}\hat{\varphi}(\omega-\tau)\,(\mathcal{F}\mathrm{sinc})(\tau)\,\mathrm{d}\tau\,,$

and hence by

$\displaystyle(\mathcal{F}\mathrm{sinc})(\tau)=\frac{1}{\sqrt{2\pi}}\,{\mathbf{1}}_{[-\pi,\pi]}(\tau)$

we obtain

$\displaystyle{\mathcal{F}}(\varphi\,\mathrm{sinc})(\omega)=\frac{1}{2\pi}\,\int_{\omega-\pi}^{\omega+\pi}\hat{\varphi}(\tau)\,\,\mathrm{d}\tau\,.$

Consequently, using the shifting property of $\mathcal{F}$, the Fourier
transform (1.1) of the shifted function $\varphi(t-k)\,\mathrm{sinc}(t-k)$
with $k\in\mathbb{Z}$ reads as

$\displaystyle\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}\varphi(t-k)\,\mathrm{sinc}(t-k)\,\mathrm{e}^{-\mathrm{i}\omega
t}\,\mathrm{d}t=\mathrm{e}^{-\mathrm{i}\omega
k}\,{\mathcal{F}}(\varphi\,\mathrm{sinc})(\omega)=\frac{1}{2\pi}\,\mathrm{e}^{-\mathrm{i}\omega
k}\,\int_{\omega-\pi}^{\omega+\pi}\hat{\varphi}(\tau)\,\,\mathrm{d}\tau\,.$

Therefore, the Fourier transform of the regularization error $e_{1}$ in (2.3)
has the form

$\displaystyle{\hat{e}}_{1}(\omega)={\hat{f}}(\omega)-\bigg{(}\frac{1}{2\pi}\,\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{e}^{-\mathrm{i}\omega
k}\bigg{)}\,\int_{\omega-\pi}^{\omega+\pi}\hat{\varphi}(\tau)\,\,\mathrm{d}\tau\,.$
(2.5)

Note that since the set of shifted cardinal sine functions
$\mathrm{sinc}(\cdot-k)$ with $k\in\mathbb{Z}$ forms an orthonormal system in
$L^{2}(\mathbb{R})$, i. e.

$\displaystyle\int_{\mathbb{R}}\mathrm{sinc}(t-k)\,\mathrm{sinc}(t-\ell)\,\mathrm{d}t=\delta_{k,\ell}\,,\quad
k,\,\ell\in\mathbb{Z}\,,$

and the given function $f$ can be represented by the Shannon sampling series
(1.2), we obtain that

$\displaystyle\sum_{k\in\mathbb{Z}}|f(k)|^{2}$
$\displaystyle=\sum_{k\in\mathbb{Z}}\sum_{\ell\in\mathbb{Z}}f(k)\,\overline{f(\ell)}\int_{\mathbb{R}}\mathrm{sinc}(t-k)\,\mathrm{sinc}(t-\ell)\,\mathrm{d}t$
$\displaystyle=\int_{\mathbb{R}}f(t)\,\overline{f(t)}\,\mathrm{d}t=\|f\|_{L^{2}(\mathbb{R})}^{2}<\infty\,,$
(2.6)

and thus the series

$\displaystyle\sum_{k\in\mathbb{Z}}f(k)\,{\mathrm{e}}^{-{\mathrm{i}}\omega k}$

converges in $L^{2}([-\pi,\pi])$. Moreover, since $f$ is bandlimited with
bandwidth $\delta\in(0,\,\pi)$, we have ${\hat{f}}(\omega)=0$ for all
$\omega\in\mathbb{R}\setminus[-\delta,\,\delta]$, and thereby the restricted
function $\hat{f}\big{|}_{[-\pi,\,\pi]}$ belongs to $L^{2}([-\pi,\pi])$.
Hence, this restricted function possesses the $2\pi$-periodic Fourier
expansion

$\displaystyle{\hat{f}}(\omega)=\sum_{k\in\mathbb{Z}}c_{k}(\hat{f})\,\mathrm{e}^{-\mathrm{i}\omega
k}\,,\quad\omega\in[-\pi,\,\pi]\,,$

with the Fourier coefficients

$\displaystyle c_{k}(\hat{f})$
$\displaystyle=\frac{1}{2\pi}\,\int_{-\pi}^{\pi}{\hat{f}}(\tau)\,\mathrm{e}^{\mathrm{i}k\tau}\,\mathrm{d}\tau=\frac{1}{2\pi}\,\int_{\mathbb{R}}{\hat{f}}(\tau)\,\mathrm{e}^{\mathrm{i}k\tau}\,\mathrm{d}\tau=\frac{1}{\sqrt{2\pi}}\,f(k)\,,\quad
k\in\mathbb{Z}\,,$

by inverse Fourier transform. In other words, the function $\hat{f}$ can be
represented in the form

$\displaystyle\hat{f}(\omega)={\hat{f}}(\omega)\,{\mathbf{1}}_{[-\delta,\delta]}(\omega)=\frac{1}{\sqrt{2\pi}}\,\bigg{(}\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{e}^{-\mathrm{i}k\omega}\bigg{)}\,{\mathbf{1}}_{[-\delta,\delta]}(\omega)\,,\quad\omega\in\mathbb{R}\,.$
(2.7)

Introducing the auxiliary function

$\displaystyle\Delta_{\varphi}(\omega)\coloneqq{\mathbf{1}}_{[-\delta,\delta]}(\omega)-\frac{1}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}\hat{\varphi}(\tau)\,\,\mathrm{d}\tau\,,\quad\omega\in\mathbb{R}\,,$

we see by inserting (2.7) into (2.5) that

$\displaystyle{\hat{e}}_{1}(\omega)={\hat{f}}(\omega)\,\Delta_{\varphi}(\omega)\,,\quad\omega\in\mathbb{R}\,,$

and thereby

$\displaystyle\big{|}{\hat{e}}_{1}(\omega)\big{|}=\big{|}{\hat{f}}(\omega)\big{|}\,\big{|}\Delta_{\varphi}(\omega)\big{|}\,,\quad\omega\in\mathbb{R}\,.$

Thus, inverse Fourier transform and the definition (2.1) yields

$\displaystyle|e_{1}(t)|$
$\displaystyle\leq\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}\big{|}{\hat{e}}_{1}(\omega)\big{|}\,\mathrm{d}\omega=\frac{1}{\sqrt{2\pi}}\,\int_{\delta}^{\delta}\big{|}{\hat{f}}(\omega)\big{|}\,\big{|}\Delta_{\varphi}(\omega)\big{|}\,\mathrm{d}\omega$
$\displaystyle\leq\frac{1}{\sqrt{2\pi}}\,\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\varphi}(\omega)\big{|}\;\int_{-\delta}^{\delta}\big{|}{\hat{f}}(\omega)\big{|}\,\mathrm{d}\omega=\frac{1}{\sqrt{2\pi}}\,E_{1}(m)\,\int_{-\delta}^{\delta}\big{|}{\hat{f}}(\omega)\big{|}\,\mathrm{d}\omega\,.$

By the Cauchy–Schwarz inequality and the Parseval equality
$\|\hat{f}\|_{L^{2}(\mathbb{R}}=\|f\|_{L^{2}(\mathbb{R})}$ we obtain

$\displaystyle\int_{-\delta}^{\delta}\big{|}1\cdot{\hat{f}}(\omega)\big{|}\,\mathrm{d}\omega$
$\displaystyle\leq\bigg{(}\int_{-\delta}^{\delta}1^{2}\,\mathrm{d}\omega\bigg{)}^{1/2}\,\bigg{(}\int_{-\delta}^{\delta}\big{|}{\hat{f}}(\omega)\big{|}^{2}\,\mathrm{d}\omega\bigg{)}^{1/2}=\sqrt{2\delta}\,\|\hat{f}\|_{L^{2}(\mathbb{R}}\leq\sqrt{2\pi}\,\|f\|_{L^{2}(\mathbb{R})}\,.$

Consequently, we receive the estimate

$\displaystyle|e_{1}(t)|\leq E_{1}(m)\,\|f\|_{L^{2}(\mathbb{R})}\,,\quad
t\in\mathbb{R}\,,$

and hence

$\displaystyle\max_{t\in\mathbb{R}}|e_{1}(t)|\leq
E_{1}(m)\,\|f\|_{L^{2}(\mathbb{R})}\,.$

(iii) Now we estimate the truncation error $e_{2,0}(t)$ for $t\in(0,\,1)$. By
(2) and $\varphi(t)\geq 0$, we obtain

$\displaystyle|e_{2,0}(t)|\leq\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}|f(k)|\,|\mathrm{sinc}(t-k)|\,\varphi(t-k)\,,\quad
t\in(0,\,1)\,.$

For $t\in(0,\,1)$ and $k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}$, we estimate

$\displaystyle|\mathrm{sinc}(t-k)|\leq\frac{1}{\pi\,|t-k|}\leq\frac{1}{\pi
m}\,,$

such that

$\displaystyle|e_{2,0}(t)|\leq\frac{1}{\pi
m}\,\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}|f(k)|\,\varphi(t-k)\,,\quad
t\in(0,\,1)\,.$

Then the Cauchy–Schwarz inequality implies

$\displaystyle|e_{2,0}(t)|\leq\frac{1}{\pi
m}\,\bigg{(}\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}|f(k)|^{2}\bigg{)}^{1/2}\,\bigg{(}\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}\varphi^{2}(t-k)\bigg{)}^{1/2}\,,\quad
t\in(0,\,1)\,.$

From (2) it follows that

$\displaystyle|e_{2,0}(t)|\leq\frac{1}{\pi
m}\,\|f\|_{L^{2}(\mathbb{R})}\,\bigg{(}\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}\varphi^{2}(t-k)\bigg{)}^{1/2}\,,\quad
t\in(0,\,1)\,.$

Since by assumption the window function $\varphi$ is even and
$\varphi\big{|}_{[0,\infty)}$ decreases, we can estimate the series

$\displaystyle\sum_{k\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}\varphi^{2}(t-k)$
$\displaystyle=\sum_{k=-\infty}^{-m}\varphi^{2}(t-k)+\sum_{k=m+1}^{\infty}\varphi^{2}(t-k)$
$\displaystyle=\sum_{k=m}^{\infty}\varphi^{2}(t+k)+\sum_{k=m+1}^{\infty}\varphi^{2}(k-t)$
$\displaystyle\leq\sum_{k=m}^{\infty}\varphi^{2}(k)+\sum_{k=m+1}^{\infty}\varphi^{2}(k-1)=2\,\sum_{k=m}^{\infty}\varphi^{2}(k)\,,\quad
t\in(0,\,1)\,.$

Applying the integral test for convergence of series, we obtain that

$\displaystyle
2\,\sum_{k=m}^{\infty}\varphi^{2}(k)=2\,\varphi^{2}(m)+2\,\sum_{k=m+1}^{\infty}\varphi^{2}(k)<2\,\varphi^{2}(m)+2\,\int_{m}^{\infty}\varphi^{2}(t)\,\mathrm{d}t\,.$

Thus, for each $t\in(0,\,1)$ we have by definition (2.2) that

$\displaystyle|e_{2,0}(t)|\leq\frac{\sqrt{2}}{\pi
m}\,\bigg{(}\varphi^{2}(m)+\int_{m}^{\infty}\varphi^{2}(t)\,\mathrm{d}t\bigg{)}^{1/2}\,\|f\|_{L^{2}(\mathbb{R})}=E_{2}(m)\,\|f\|_{L^{2}(\mathbb{R})}<\infty\,.$

Furthermore, by the interpolation property (1.9) of $R_{\varphi,m}f$ we have
$e_{2,0}(0)=e_{2,0}(1)=0$, such that

$\displaystyle\max_{t\in[0,1]}|e_{2,0}(t)|\leq
E_{2}(m)\,\|f\|_{L^{2}(\mathbb{R})}\,.$

(iv) By the same technique, the error estimate

$\displaystyle\max_{t\in[n,n+1]}\big{|}f(t)-\big{(}R_{\varphi,m}f\big{)}(t)\big{|}\leq\big{(}E_{1}(m)+E_{2}(m)\big{)}\,\|f\|_{L^{2}(\mathbb{R})}$

can be shown for the interval $[n,\,n+1]$ with arbitrary $n\in\mathbb{Z}$. On
the open interval $(n,\,n+1)$, we decompose the approximation error as

$\displaystyle
f(t+n)-\big{(}R_{\varphi,m}f\big{)}(t)(t+n)=e_{1}(t+n)+e_{2,n}(t)\,,\quad
t\in(0,\,1)\,,$

with

$\displaystyle e_{1}(t+n)$
$\displaystyle=f(t+n)-\sum_{k\in\mathbb{Z}}f(k)\,\mathrm{sinc}\big{(}t-(k-n)\big{)}\,\varphi\big{(}t-(k-n)\big{)}$
$\displaystyle=f(t+n)-\sum_{\ell\in\mathbb{Z}}f(\ell+n)\,\mathrm{sinc}(t-\ell)\,\varphi(t-\ell)\,,$
$\displaystyle e_{2,n}(t)$
$\displaystyle\coloneqq\sum_{\ell\in\mathbb{Z}\setminus\\{1-m,\ldots,m\\}}f(\ell+n)\,\mathrm{sinc}(t-\ell)\,\varphi(t-\ell)\,.$

As shown in steps (ii) and (iii), we have

$\displaystyle\|e_{1}(\cdot+n)\|_{C_{0}(\mathbb{R})}$
$\displaystyle=\|e_{1}\|_{C_{0}(\mathbb{R})}\,,$ $\displaystyle|e_{2,n}(t)|$
$\displaystyle\leq E_{2}(m)\,\|f\|_{L^{2}(\mathbb{R})}\,,\quad t\in(0,\,1)\,.$

Furthermore, by the interpolation property (1.9) of $R_{\varphi,m}f$, we have
$e_{2,n}(0)=e_{2,n}(1)=0$ for each $n\in\mathbb{Z}$ and thus

$\displaystyle\max_{t\in[n,n+1]}\big{|}e_{2,n}(t)\big{|}\leq
E_{2}(m)\,\|f\|_{L^{2}(\mathbb{R})}\,.$

Hence, it follows that

$\displaystyle\max_{t\in[n,n+1]}\big{|}f(t)-\big{(}R_{\varphi,m}f\big{)}(t)\big{|}$
$\displaystyle\leq\|e_{1}\|_{C_{0}(\mathbb{R})}+\max_{t\in[n,n+1]}\big{|}e_{2,n}(t)\big{|}$
$\displaystyle\leq\big{(}E_{1}(m)+E_{2}(m)\big{)}\,\|f\|_{L^{2}(\mathbb{R})}\,,$

which completes the proof.

##  3 Optimal regularization with the (modified) Gaussian function

In this section we consider the Gaussian function (1.5) with variance
$\sigma^{2}>0$, analogous to [12, Theorem 4.1]. In order to achieve fast
convergence of the Gaussian regularized Shannon sampling formula, we put
special emphasis on the optimal choice of this variance $\sigma^{2}$,
comparable to [5].

######  Theorem 3.1.

Assume that $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is bandlimited with
bandwidth $\delta\in(0,\,\pi)$. Further let $\varphi_{\mathrm{Gauss}}$ be the
Gaussian function (1.5) with variance $\sigma^{2}=\frac{m}{\pi-\delta}$ and
let $m\in{\mathbb{N}}\setminus\\{1\\}$ be given.
Then the Gaussian regularized Shannon sampling formula satisfies the error
estimate

$\displaystyle\big{\|}f-R_{\mathrm{Gauss},m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\frac{2\sqrt{2}}{\sqrt{\pi
m\,(\pi-\delta)}}\,\mathrm{e}^{-m\,(\pi-\delta)/2}\,\|f\|_{L^{2}(\mathbb{R})}\,.$
(3.1)

_Proof_. (i) At first, we estimate the regularization error constant (2.1) for
the Gaussian function (1.5). Since the Fourier transform of
$\varphi_{\mathrm{Gauss}}$ reads as

$\displaystyle{\hat{\varphi}}_{\mathrm{Gauss}}(\omega)=\frac{1}{\sqrt{2\pi}}\,\int_{\mathbb{R}}\varphi_{\mathrm{Gauss}}(t)\,\mathrm{e}^{-\mathrm{i}\,t\omega}\,\mathrm{d}t=\sigma\,\mathrm{e}^{-\omega^{2}\sigma^{2}/2}\,,\quad\omega\in\mathbb{R}\,,$

cf. [20, Example 2.6], we have

$\displaystyle
E_{1}(m)=\max_{\omega\in[-\delta,\delta]}\bigg{|}1-\frac{\sigma}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}\mathrm{e}^{-\tau^{2}\sigma^{2}/2}\,\mathrm{d}\tau\bigg{|}\,.$

Substituting $s=\tau\sigma/\sqrt{2}$ and using the integral

$\displaystyle\int_{\mathbb{R}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s=\sqrt{\pi}\,,$
(3.2)

we obtain for $\omega\in[-\delta,\,\delta]$ with $\delta\in(0,\,\pi)$ that

$\displaystyle\Delta_{\mathrm{Gauss}}(\omega)$ $\displaystyle\coloneqq
1-\frac{1}{\sqrt{\pi}}\,\int_{(\omega-\pi)\sigma/\sqrt{2}}^{(\omega+\pi)\sigma/\sqrt{2}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s$
$\displaystyle=\frac{1}{\sqrt{\pi}}\,\bigg{(}\int_{\mathbb{R}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s-\int_{(\omega-\pi)\sigma/\sqrt{2}}^{(\omega+\pi)\sigma/\sqrt{2}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\bigg{)}$
$\displaystyle=\frac{1}{\sqrt{\pi}}\,\bigg{(}\int_{-\infty}^{(\omega-\pi)\sigma/\sqrt{2}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s+\int_{(\omega+\pi)\sigma/\sqrt{2}}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\bigg{)}$
$\displaystyle=\frac{1}{\sqrt{\pi}}\,\bigg{(}\int_{(\pi-\omega)\sigma/\sqrt{2}}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s+\int_{(\omega+\pi)\sigma/\sqrt{2}}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\bigg{)}\,.$

Since $\Delta_{\mathrm{Gauss}}$ is even, we consider only the case
$\omega\in[0,\,\delta]$. Applying the inequality

$\displaystyle\int_{a}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s=\int_{0}^{\infty}\mathrm{e}^{-(t+a)^{2}}\,\mathrm{d}t\leq\mathrm{e}^{-a^{2}}\,\int_{0}^{\infty}\mathrm{e}^{-2at}\,\mathrm{d}t=\frac{1}{2a}\,\mathrm{e}^{-a^{2}}\,,\quad
a>0\,,$ (3.3)

we obtain

$\displaystyle 0\leq\Delta_{\mathrm{Gauss}}(\omega)$
$\displaystyle\leq\frac{1}{\sqrt{2\pi}}\,\bigg{(}\frac{\mathrm{e}^{-(\pi-\omega)^{2}\sigma^{2}/2}}{(\pi-\omega)\,\sigma}+\frac{\mathrm{e}^{-(\pi+\omega)^{2}\sigma^{2}/2}}{(\pi+\omega)\,\sigma}\bigg{)}\leq\sqrt{\frac{2}{\pi}}\,\frac{\mathrm{e}^{-(\pi-\omega)^{2}\sigma^{2}/2}}{(\pi-\omega)\,\sigma}\,,\quad\omega\in[0,\,\delta]\,.$

Consequently, we have for all $\omega\in[-\delta,\,\delta]$ that

$\displaystyle
0\leq\Delta_{\mathrm{Gauss}}(\omega)\leq\sqrt{\frac{2}{\pi}}\,\frac{\mathrm{e}^{-(\pi-|\omega|)^{2}\sigma^{2}/2}}{(\pi-|\omega|)\,\sigma}$

and hence

$\displaystyle
E_{1}(m)\leq\sqrt{\frac{2}{\pi}}\,\frac{\mathrm{e}^{-(\pi-\delta)^{2}\sigma^{2}/2}}{(\pi-\delta)\,\sigma}\,.$
(3.4)

(ii) Now we examine the truncation error constant (2.2) for the Gaussian
function (1.5). By
$\varphi_{\mathrm{Gauss}}^{2}(m)=\mathrm{e}^{-m^{2}/\sigma^{2}}$ and the
inequality

$\displaystyle\int_{m}^{\infty}\varphi_{\mathrm{Gauss}}^{2}(t)\,\mathrm{d}t=\sigma\,\int_{m/\sigma}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\leq\frac{\sigma^{2}}{2m}\,\mathrm{e}^{-m^{2}/\sigma^{2}}$

we obtain

$\displaystyle E_{2}(m)$ $\displaystyle\leq\frac{\sqrt{2}}{\pi
m}\,\sqrt{\mathrm{e}^{-m^{2}/\sigma^{2}}+\frac{\sigma^{2}}{2m}\,\mathrm{e}^{-m^{2}/\sigma^{2}}}=\frac{\sqrt{2}}{\pi
m}\,\sqrt{1+\frac{\sigma^{2}}{2m}}\,{\mathrm{e}}^{-m^{2}/(2\sigma^{2})}\,.$
(3.5)

(iii) Finally, we say that the variance $\sigma^{2}$ of the Gaussian function
(1.5) is _optimal_ , if $E_{1}(m)$ and $E_{2}(m)$ possess the same exponential
decay with respect to $m$. From (3.4) and (3.5) it follows that

$\displaystyle\sigma_{\mathrm{opt}}^{2}\coloneqq\frac{m}{\pi-\delta}$ (3.6)

is the optimal variance with

$\displaystyle E_{1}(m)$
$\displaystyle\leq\sqrt{\frac{2}{\pi}}\,\frac{1}{\sqrt{m\,(\pi-\delta)}}\,\mathrm{e}^{-m\,(\pi-\delta)/2}\,,$
$\displaystyle E_{2}(m)$ $\displaystyle\leq\frac{\sqrt{2}}{\pi
m}\,\sqrt{1+\frac{1}{2\,(\pi-\delta)}}\,\mathrm{e}^{-m\,(\pi-\delta)/2}\,.$

Note that since $m\in\mathbb{N}\setminus\\{1\\}$ and $\delta\in(0,\,\pi)$, we
have

$\displaystyle\left(\frac{\sqrt{2}}{\sqrt{\pi
m\,(\pi-\delta)}}\right)^{-1}\cdot\frac{\sqrt{2}}{\pi
m}\,\sqrt{1+\frac{1}{2\,(\pi-\delta)}}$
$\displaystyle=\sqrt{\frac{2(\pi-\delta)+1}{2\pi
m}}\leq\sqrt{\frac{2\pi+1}{4\pi}}<1$

and therefore

$\displaystyle E_{2}(m)\leq\frac{\sqrt{2}}{\sqrt{\pi
m\,(\pi-\delta)}}\,\mathrm{e}^{-m\,(\pi-\delta)/2}\,.$

Thus, the Gaussian regularized Shannon sampling formula with the optimal
variance (3.6) fulfills the error estimate (3.1). This completes the proof.

Note that already in [12, Theorem 4.1] bounds on the approximation error of
the Shannon sampling formula (1.4) were shown for the Gaussian function (1.5)
with suitably chosen variance $\sigma^{2}$, which is basically the same as the
one in Theorem 3.1, only looking slightly different due to the different
setting considered in [12].

######  Remark 3.2.

We remark that in [5] a different optimal variance
$\sigma^{2}=\frac{m-1}{\pi-\delta}$ is presented for the Gaussian regularizer
(1.5). However, by Theorem 3.1 we see that the choice (3.6) is optimal for the
Shannon sampling formula (1.4) with the Gaussian function (1.5), while a
slightly different truncation than in (1.4) was considered in [5].
Nevertheless, both results, Theorem 3.1 and [5, Theorem 1.1], possess the same
asymptotic behavior.

Additionally, it should be noted that in [5] the approximation error is
estimated only up to an unknown constant, while our error estimate of the
Gaussian regularized Shannon sampling formula contains relatively small
explicit constants, which is more favorable for practical applications.
Moreover, we estimate the approximation error differently by splitting it into
the regularization error (2.3) and the truncation error (2), which seems more
intuitive than the rather artificial analysis presented in [5, Theorem 1.1].

######  Example 3.3.

Now we visualize the optimality of the variance (3.6) for the Gaussian
regularized Shannon sampling formula shown in Theorem 3.1. For this purpose,
for a given bandlimited function $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$
with bandwidth $\delta\in(0,\,\pi]$ we consider the regularized Shannon
sampling formula (1.4) with the Gaussian function $\varphi_{\mathrm{Gauss}}$
in (1.5) and compute the corresponding approximation error

$\displaystyle\max_{t\in[-1,\,1]}\big{|}f(t)-\big{(}R_{\varphi,m}f\big{)}(t)\big{|}\,,$
(3.7)

cf. (1.10). The error (3.7) shall here be approximated by evaluating a given
function $f$ and its approximation $R_{\varphi,m}f$ at equidistant points
$t_{s}\in[-1,\,1]$, $s=1,\dots,S$, with $S=10^{5}$. Note that by the
definition of the regularized Shannon sampling formula (1.4) we have

$\displaystyle\big{(}R_{\varphi,m}f\big{)}(t)=\sum_{k=-m-1}^{m+1}f(k)\,\mathrm{sinc}(t-k)\,\varphi(t-k)\,,\quad
t\in[-1,\,1]\,.$

Analogous to [18, Section IV, C] we study the bandlimited function

$\displaystyle
f(t)=\frac{2\delta}{\sqrt{5\pi\delta+4\pi\sin\delta}}\left[\mathrm{sinc}\bigg{(}\frac{\delta
t}{\pi}\bigg{)}+\frac{1}{2}\,\mathrm{sinc}\bigg{(}\frac{\delta(t-1)}{\pi}\bigg{)}\right]\,,\quad
t\in\mathbb{R}\,,$ (3.8)

with $\|f\|_{L^{2}(\mathbb{R})}=1$, for several bandwidth parameters
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$,
i. e., several oversampling rates $\frac{\pi}{\delta}>1$. To compare with the
optimal variance $\sigma_{\mathrm{opt}}$ in (3.6), we choose the parameter of
the Gaussian function (1.5) as
$\sigma\in\big{\\{}\frac{1}{2}\sigma_{\mathrm{opt}},\sigma_{\mathrm{opt}},2\sigma_{\mathrm{opt}}\big{\\}}$.

The corresponding results for different truncation parameters
$m\in\\{2,3,\ldots,10\\}$ are displayed in Figure 3.1. It can clearly be seen
that both an increase and a decrease of the variance in (3.6) cause worsened
error decay rates with respect to $m$. Thus, the numerical results confirm
that the variance (3.6) of Theorem 3.1 is optimal, and that this fact can
already be observed for very small truncation parameters
$m\in\mathbb{N}\setminus\\{1\\}$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 3.1: Maximum approximation error (3.7) using the Gaussian function
$\varphi_{\mathrm{Gauss}}$ in (1.5) with different variances
$\sigma^{2}\in\big{\\{}\frac{m}{4(\pi-\delta)},\frac{m}{\pi-\delta},\frac{4m}{\pi-\delta}\big{\\}}$,
for the bandlimited function (3.8) with bandwidths
$\delta\in\big{\\{}\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4}\big{\\}}$ and
truncation parameters $m\in\\{2,3,\ldots,10\\}$.

Note that in [25, 24] the modified Gaussian function (1.6) was used for the
regularization of Shannon sampling formulas, using a slightly different
notation. By the same techniques as in Theorem 3.1 one can determine the
optimal parameter $\sigma^{2}$ of (1.6) subject to $\lambda$.

######  Theorem 3.4.

Assume that $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is bandlimited with
bandwidth $\delta\in(0,\,\pi)$. Further let $\varphi_{\mathrm{modGauss}}$ be
the modified Gaussian function (1.6) with parameter
$\sigma^{2}=\frac{m}{\pi-\lambda-\delta}$, $0\leq\lambda<\pi-\delta$, and let
$m\in{\mathbb{N}}\setminus\\{1\\}$ be given.
Then the modified Gaussian regularized Shannon sampling formula satisfies the
error estimate

$\displaystyle\big{\|}f-R_{\mathrm{modGauss},m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\frac{2\sqrt{2}}{\sqrt{\pi
m\,(\pi-\lambda-\delta)}}\,\mathrm{e}^{-m\,(\pi-\lambda-\delta)/2}\,\|f\|_{L^{2}(\mathbb{R})}\,.$
(3.9)

_Proof_. (i) At first, we estimate the regularization error constant (2.1) for
the modified Gaussian function (1.6). By [18, p. 21, 5.24] the Fourier
transform of $\varphi_{\mathrm{modGauss}}$ reads as

$\displaystyle{\hat{\varphi}}_{\mathrm{modGauss}}(\omega)$
$\displaystyle=\sigma\,\mathrm{e}^{-\sigma^{2}\,(\lambda^{2}+\omega^{2})/2}\,\cosh(\lambda\sigma^{2}\omega)=\frac{\sigma}{2}\left(\mathrm{e}^{-\sigma^{2}\,(\omega+\lambda)^{2}/2}+\mathrm{e}^{-\sigma^{2}\,(\omega-\lambda)^{2}/2}\right)\,,\quad\omega\in\mathbb{R}\,.$

Therefore, we obtain for $\omega\in[-\delta,\,\delta]$ with
$\delta\in(0,\,\pi)$ that

$\displaystyle\Delta_{\mathrm{modGauss}}(\omega)$ $\displaystyle\coloneqq
1-\frac{1}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}{\hat{\varphi}}_{\mathrm{modGauss}}(\tau)\,\mathrm{d}\tau$
$\displaystyle=\frac{1}{2}\bigg{(}1-\frac{\sigma}{\sqrt{2\pi}}\int_{\omega-\pi}^{\omega+\pi}\mathrm{e}^{-\sigma^{2}(\tau+\lambda)^{2}/2}\,\mathrm{d}\tau\bigg{)}+\frac{1}{2}\bigg{(}1-\frac{\sigma}{\sqrt{2\pi}}\int_{\omega-\pi}^{\omega+\pi}\mathrm{e}^{-\sigma^{2}(\tau-\lambda)^{2}/2}\,\mathrm{d}\tau\bigg{)}\,.$

Substituting $s=\sigma(\tau+\lambda)/\sqrt{2}$ in the first term and
$s=\sigma(\tau-\lambda)/\sqrt{2}$ in the second term, as well as using the
integral (3.2), it follows that

$\displaystyle\Delta_{\mathrm{modGauss}}(\omega)$
$\displaystyle=\frac{1}{2\sqrt{\pi}}\bigg{(}\int_{\mathbb{R}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s-\int_{(\omega-\pi+\lambda)\sigma/\sqrt{2}}^{(\omega+\pi+\lambda)\sigma/\sqrt{2}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\bigg{)}$
$\displaystyle\qquad+\,\frac{1}{2\sqrt{\pi}}\bigg{(}\int_{\mathbb{R}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s-\int_{(\omega-\pi-\lambda)\sigma/\sqrt{2}}^{(\omega+\pi-\lambda)\sigma/\sqrt{2}}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\bigg{)}$
$\displaystyle=\frac{1}{2\sqrt{\pi}}\bigg{(}\int_{-\infty}^{(\omega-\pi+\lambda)\sigma/\sqrt{2}}+\int_{(\omega+\pi+\lambda)\sigma/\sqrt{2}}^{\infty}$
$\displaystyle\qquad+\,\int_{-\infty}^{(\omega-\pi-\lambda)\sigma/\sqrt{2}}+\int_{(\omega+\pi-\lambda)\sigma/\sqrt{2}}^{\infty}\bigg{)}\,\mathrm{e}^{-s^{2}}\,\mathrm{d}s$
$\displaystyle=\frac{1}{2\sqrt{\pi}}\bigg{(}\int_{(\pi-\lambda-\omega)\sigma/\sqrt{2}}^{\infty}+\int_{(\pi+\lambda+\omega)\sigma/\sqrt{2}}^{\infty}$
$\displaystyle\qquad+\,\int_{(\pi+\lambda-\omega)\sigma/\sqrt{2}}^{\infty}+\int_{(\pi-\lambda+\omega)\sigma/\sqrt{2}}^{\infty}\bigg{)}\,\mathrm{e}^{-s^{2}}\,\mathrm{d}s\,.$

Since $\Delta_{\mathrm{modGauss}}$ is even, we consider only the case
$\omega\in[0,\delta]$. By (3.3) we obtain

$\displaystyle 0\leq\Delta_{\mathrm{modGauss}}(\omega)$
$\displaystyle\leq\frac{1}{4\sqrt{\pi}}\bigg{(}\frac{\sqrt{2}}{(\pi-\lambda-\omega)\sigma}\,\mathrm{e}^{(\pi-\lambda-\omega)^{2}\sigma^{2}/2}+\frac{\sqrt{2}}{(\pi+\lambda+\omega)\sigma}\,\mathrm{e}^{(\pi+\lambda+\omega)^{2}\sigma^{2}/2}$
$\displaystyle\qquad+\frac{\sqrt{2}}{(\pi+\lambda-\omega)\sigma}\,\mathrm{e}^{(\pi+\lambda-\omega)^{2}\sigma^{2}/2}+\frac{\sqrt{2}}{(\pi-\lambda+\omega)\sigma}\,\mathrm{e}^{(\pi-\lambda+\omega)^{2}\sigma^{2}/2}\bigg{)}$
$\displaystyle\leq\frac{1}{\sqrt{\pi}}\,\frac{\sqrt{2}}{(\pi-\lambda-\omega)\sigma}\,\mathrm{e}^{-(\pi-\lambda-\omega)^{2}\sigma^{2}/2}\,,\quad\omega\in[0,\delta]\,.$

Consequently, we have for all $\omega\in[-\delta,\,\delta]$ that

$\displaystyle
0\leq\Delta_{\mathrm{modGauss}}(\omega)\leq\sqrt{\frac{2}{\pi}}\,\frac{\mathrm{e}^{-(\pi-\lambda-|\omega|)^{2}\sigma^{2}/2}}{(\pi-\lambda-|\omega|)\,\sigma}$

and hence

$\displaystyle
E_{1}(m)\leq\sqrt{\frac{2}{\pi}}\,\frac{\mathrm{e}^{-(\pi-\lambda-\delta)^{2}\sigma^{2}/2}}{(\pi-\lambda-\delta)\,\sigma}\,.$
(3.10)

(ii) For the truncation error constant (2.2) for the modified Gaussian
function (1.6), we observe that by the inequalities
$\varphi_{\mathrm{modGauss}}^{2}(m)\leq\mathrm{e}^{-m^{2}/\sigma^{2}}$ and

$\displaystyle\int_{m}^{\infty}\varphi_{\mathrm{modGauss}}^{2}(t)\,\mathrm{d}t\leq\sigma\,\int_{m/\sigma}^{\infty}\mathrm{e}^{-s^{2}}\,\mathrm{d}s\leq\frac{\sigma^{2}}{2m}\,\mathrm{e}^{-m^{2}/\sigma^{2}}$

we also have (3.5).
(iii) Finally, we say that the parameter $\sigma^{2}$ of the modified Gaussian
function (1.6) is _optimal_ , if $E_{1}(m)$ and $E_{2}(m)$ possess the same
exponential decay with respect to $m$. From (3.10) and (3.5) it follows that

$\displaystyle\sigma^{2}=\frac{m}{\pi-\lambda-\delta}$ (3.11)

is the optimal parameter with

$\displaystyle E_{1}(m)$
$\displaystyle\leq\sqrt{\frac{2}{\pi}}\,\frac{1}{\sqrt{m\,(\pi-\lambda-\delta)}}\,\mathrm{e}^{-m\,(\pi-\lambda-\delta)/2}\,,$
$\displaystyle E_{2}(m)$ $\displaystyle\leq\frac{\sqrt{2}}{\pi
m}\,\sqrt{1+\frac{1}{2\,(\pi-\lambda-\delta)}}\,\mathrm{e}^{-m\,(\pi-\lambda-\delta)/2}\,.$

where we have $\pi-\lambda-\delta>0$ by the assumption
$\lambda\in[0,\,\pi-\delta)$.

Note that since $m\in\mathbb{N}\setminus\\{1\\}$, $\delta\in(0,\,\pi)$, and
$\lambda\in[0,\,\pi-\delta)$, we have

$\displaystyle\left(\frac{\sqrt{2}}{\sqrt{\pi
m\,(\pi-\lambda-\delta)}}\right)^{-1}\cdot\frac{\sqrt{2}}{\pi
m}\,\sqrt{1+\frac{1}{2\,(\pi-\lambda-\delta)}}$
$\displaystyle=\sqrt{\frac{2(\pi-\lambda-\delta)+1}{2\pi
m}}\leq\sqrt{\frac{2\pi+1}{4\pi}}<1$

and therefore

$\displaystyle E_{2}(m)\leq\frac{\sqrt{2}}{\sqrt{\pi
m\,(\pi-\lambda-\delta)}}\,\mathrm{e}^{-m\,(\pi-\lambda-\delta)/2}\,.$

Thus, the modified Gaussian regularized Shannon sampling formula with the
optimal parameter (3.11) fulfills the error estimate (3.9). This completes the
proof.

Thereby, Theorem 3.4 shows that the approximation error of the regularized
Shannon sampling formula with the modified Gaussian function (1.6) has the
best exponential decay in the case $\lambda=0$. In other words, the Gaussian
function $\varphi_{\mathrm{Gauss}}$ in (1.5) is much more favorable than the
modified Gaussian function $\varphi_{\mathrm{modGauss}}$ in (1.6).

##  4 Optimal regularization with the sinh-type window function

In this section, we consider the $\sinh$-type window function (1.7) with shape
parameter $\beta>0$, analogous to [12, Theorem 6.1] and [13, Theorem 4.2]. In
order to achieve fast convergence of the $\sinh$-type regularized Shannon
sampling formula, we put special emphasis on the optimal choice of this shape
parameter $\beta$. Moreover, we demonstrate that the exponential decay with
respect to the truncation parameter $m\in\mathbb{N}\setminus\\{1\\}$ is much
better for the uniform approximation error
$\big{\|}f-R_{\sinh,m}f\big{\|}_{C_{0}(\mathbb{R})}$ than for the
approximation error
$\big{\|}f-R_{\mathrm{Gauss},m}f\big{\|}_{C_{0}(\mathbb{R})}$ in Theorem 3.1.

######  Theorem 4.1.

Assume that $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is bandlimited with
bandwidth $\delta\in(0,\,\pi)$. Further let $\varphi_{\sinh}$ be the
$\sinh$-type window function (1.7) with shape parameter
$\beta=m\,(\pi-\delta)$ and let $m\in{\mathbb{N}}\setminus\\{1\\}$ be given.
Then the $\sinh$-type regularized Shannon sampling formula satisfies the error
estimate

$\displaystyle\big{\|}f-R_{\sinh,m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\mathrm{e}^{-m\,(\pi-\delta)}\,\|f\|_{L^{2}(\mathbb{R})}\,.$
(4.1)

_Proof_. (i) Since $\varphi_{\sinh}$ in (1.7) is compactly supported on
$[-m,\,m]$ and $\varphi_{\sinh}(m)=0$, we have $E_{2}(m)=0$. Thus, according
to Theorem 2.1, the approximation error can be estimated by

$\displaystyle\big{\|}f-R_{\sinh,m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\|f\|_{L^{2}(\mathbb{R})}\,\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\sinh}(\omega)\big{|}\,,$

where

$\displaystyle\Delta_{\sinh}(\omega)\coloneqq
1-\frac{1}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}{\hat{\varphi}}_{\sinh}(\tau)\,\mathrm{d}\tau\,,\quad\omega\in[-\delta,\,\delta]\,.$
(4.2)

Following [18, p. 38, 7.58], the Fourier transform of (1.7) has the form

$\displaystyle{\hat{\varphi}}_{\sinh}(\tau)=\frac{m\,\sqrt{\pi}}{\sqrt{2}\,\sinh\beta}\cdot\begin{cases}(1-\nu^{2})^{-1/2}\,I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}&\colon|\nu|<1\,,\\\\[4.30554pt]
(\nu^{2}-1)^{-1/2}\,J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}&\colon|\nu|>1\,,\end{cases}$
(4.3)

with the scaled frequency $\nu=\frac{m}{\beta}\,\tau$, where $J_{1}$ denotes
the Bessel function and $I_{1}$ the modified Bessel function of first order.
Substituting $\tau=\frac{\beta}{m}\,\nu$ in the integral in (4.2), the
function $\Delta_{\sinh}$ reads as

$\displaystyle\Delta_{\sinh}(\omega)\coloneqq
1-\frac{\beta}{\sqrt{2\pi}\,m}\,\int_{-\nu_{1}(-\omega)}^{\nu_{1}(\omega)}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,,\quad\omega\in[-\delta,\,\delta]\,,$
(4.4)

with the increasing linear function

$\displaystyle\nu_{1}(\omega)\coloneqq\frac{m}{\beta}\,(\omega+\pi)\,,\quad\omega\in[-\delta,\,\delta]\,.$
(4.5)

(ii) Now we choose the shape parameter of (1.7) in the special form
$\beta=m\,(\pi-\delta)$. Thus, we have

$\displaystyle
1=\nu_{1}(-\delta)\leq\nu_{1}(\omega)=\frac{\omega+\pi}{\pi-\delta}\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\pi-\delta}\,,\quad\omega\in[-\delta,\,\delta]\,.$

In view of (4.3) we split (4.4) in the form
$\Delta_{\sinh}(\omega)=\Delta_{\sinh,1}-\Delta_{\sinh,2}(\omega)$ with

$\displaystyle\Delta_{\sinh,1}$ $\displaystyle\coloneqq
1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,,$
$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle\coloneqq\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1}^{\nu_{1}(-\omega)}+\int_{1}^{\nu_{1}(\omega)}\bigg{)}\,\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,\mathrm{d}\nu\,.$

Using [10, 6.681–11] and [1, 10.2.13], we get

$\displaystyle\int_{0}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu=\int_{0}^{\pi/2}I_{1}(\beta\,\cos\sigma)\,\mathrm{d}\sigma=\frac{\pi}{2}\,\bigg{(}I_{1/2}\bigg{(}\frac{\beta}{2}\bigg{)}\bigg{)}^{2}=\frac{2}{\beta}\left(\sinh\frac{\beta}{2}\right)^{2}$
(4.6)

and hence

$\displaystyle\Delta_{\sinh,1}=1-\frac{2\,\big{(}\sinh\frac{\beta}{2}\big{)}^{2}}{\sinh\beta}=\frac{2\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}\,.$
(4.7)

By [10, 6.645–1] we have

$\displaystyle\int_{1}^{\infty}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,\mathrm{d}\nu=I_{1/2}\bigg{(}\frac{\beta}{2}\bigg{)}\,K_{1/2}\bigg{(}\frac{\beta}{2}\bigg{)}=\frac{1-\mathrm{e}^{-\beta}}{\mathrm{e}^{\beta}}\,,$

where $I_{1/2}$ and $K_{1/2}$ are modified Bessel functions of half order (see
[1, 10.2.13, 10.2.14, and 10.2.17]. Numerical experiments, cf. [12], have
shown that for all $W>1$ we have

$\displaystyle
0<\int_{1}^{W}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,\mathrm{d}\nu\leq\frac{3\,(1-\mathrm{e}^{-\beta})}{2\,\beta}\,,$
(4.8)

such that

$\displaystyle
0\leq\Delta_{\sinh,2}(\omega)\leq\frac{\beta}{2\,\sinh\beta}\,\frac{3\,(1-\mathrm{e}^{-\beta})}{\beta}=\frac{3\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}\,,\quad\omega\in[-\delta,\,\delta]\,.$
(4.9)

Thereby, it follows from (4.7) and (4.9) that the expressions in (4.12) have
the same exponential decay $m\,(\pi-\delta)$ and that

$\displaystyle\big{|}\Delta_{\sinh}(\omega)\big{|}=\big{|}\Delta_{\sinh,1}-\Delta_{\sinh,2}(\omega)\big{|}\leq\frac{\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}<\mathrm{e}^{-\beta}\,,\quad\omega\in[-\delta,\,\delta]\,.$

Thus, the $\sinh$-type regularized Shannon sampling formula with the chosen
shape parameter $\beta=m\,(\pi-\delta)$ fulfills the error estimate (4.1).
This completes the proof.

Now we show that the choice of the shape parameter $\beta=m\,(\pi-\delta)$ of
(1.7) is optimal in a certain sense. To this end, let the parameters
$\alpha>0$, $m\in\mathbb{N}\setminus\\{1\\}$, and $\delta\in(0,\pi)$ be given,
and consider shape parameters of the form $\beta=\alpha\,m\,(\pi-\delta)$.
Then the increasing linear function (4.5) fulfills

$\displaystyle\frac{1}{\alpha}=\nu_{1}(-\delta)\leq\nu_{1}(\omega)=\frac{\omega+\pi}{\alpha\,(\pi-\delta)}\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\alpha\,(\pi-\delta)}\,,\quad\omega\in[-\delta,\,\delta]\,.$

Therefore, we split (4.4) as
$\Delta_{\sinh}(\omega)=\Delta_{\sinh,1}-\Delta_{\sinh,2}(\omega)$,
$\omega\in[-\delta,\,\delta]$, with

$\displaystyle\Delta_{\sinh,1}$ $\displaystyle\coloneqq
1-\frac{\beta}{m\,\sqrt{2\pi}}\,\int_{-1/\alpha}^{1/\alpha}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,,$
(4.10) and $\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle\coloneqq\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{-\nu_{1}(-\omega)}^{-1/\alpha}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,.$
(4.11)

Introducing the terms

$\displaystyle D_{1}(m)\coloneqq\big{|}\Delta_{\sinh,1}\big{|}\,,\quad
D_{2}(m)\coloneqq\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\sinh,2}(\omega)\big{|}\,,$
(4.12)

it is known by Theorem 4.1 that for $\alpha=1$ both expressions in (4.12)
possess the same exponential decay $m\,(\pi-\delta)$. In the following, we
discuss the other cases $0<\alpha<1$ and $\alpha>1$. More precisely, we show
in Theorem 4.2 that for $\alpha\neq 1$ both expressions in (4.12) have the
same exponential decay smaller than $m\,(\pi-\delta)$. In this sense, it
follows immediately that the shape parameter $\beta=m\,(\pi-\delta)$ of the
$\sinh$-type window function (1.7) is _optimal_ , since both expressions in
(4.12) tend to zero as $m\to\infty$ with the same maximum exponential decay.

######  Theorem 4.2.

For $\delta\in(0,\,\pi)$, let $\varphi_{\sinh}$ be the $\sinh$-type window
function (1.7) with the shape parameter $\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha>0$, $\alpha\not=1$, and $m\in{\mathbb{N}}\setminus\\{1\\}$.

  1. a)

In the case $\alpha\in(0,\,1)$, both expressions in (4.12) tend to zero as
$m\to\infty$ with the same exponential decay $\alpha\,m\,(\pi-\delta)$.

  2. b)

In the case $\alpha>1$, both expressions in (4.12) tend to zero as
$m\to\infty$ with exponential decay smaller than $m\,(\pi-\delta)$.

_Proof_. a) First we consider the shape parameter
$\beta=\alpha\,m\,(\pi-\delta)$ with $\alpha\in(0,\,1)$. Then we have by
(4.10), (4.3) and (4.6) that

$\displaystyle\Delta_{\sinh,1}$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\bigg{(}\int_{0}^{1}+\int_{1}^{1/\alpha}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\bigg{(}1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,{\mathrm{d}}\nu\bigg{)}-\frac{\beta}{\sinh\beta}\,\int_{1}^{1/\alpha}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,{\mathrm{d}}\nu$
$\displaystyle=\bigg{(}1-\frac{2\,\big{(}\sinh\frac{\beta}{2}\big{)}^{2}}{\sinh\beta}\bigg{)}-\frac{\beta}{\sinh\beta}\,\int_{1}^{1/\alpha}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{2\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}-\frac{\beta}{\sinh\beta}\,\int_{1}^{1/\alpha}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,{\mathrm{d}}\nu\,.$

Hence, by (4.8) this yields

$\displaystyle\big{|}\Delta_{\sinh,1}\big{|}\leq\frac{2\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}+\frac{\beta}{\sinh\beta}\,\frac{3\,(1-\mathrm{e}^{-\beta})}{2\,\beta}=\frac{5\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}\,.$
(4.13)

For the second term (4.11) we have by (4.3) that

$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\,\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1}^{\nu_{1}(-\omega)}+\int_{1}^{\nu_{1}(\omega)}-2\,\int_{1}^{1/\alpha}\bigg{)}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,{\mathrm{d}}\nu\,,\quad\omega\in[-\delta,\,\delta]\,,$

such that (4.8) implies

$\displaystyle\big{|}\Delta_{\sinh,2}(\omega)\big{|}\leq\frac{3\,\big{(}1-\mathrm{e}^{-\beta}\big{)}}{\sinh\beta}=\frac{6\,\mathrm{e}^{-\beta}}{1+\mathrm{e}^{-\beta}}\,,\quad\omega\in[-\delta,\,\delta]\,.$
(4.14)

Thus, by (4.13) and (4.14) the quantities (4.12) tend to zero as $m\to\infty$
with the same exponential decay $\alpha\,m\,(\pi-\delta)$, which is smaller
than $m\,(\pi-\delta)$ as $\alpha\in(0,\,1)$.

b) Now we investigate the shape parameter $\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha>1$.
(i) By (4.10) and (4.3) we obtain

$\displaystyle\Delta_{\sinh,1}$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu=1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1/\alpha}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,.$

Numerical experiments, cf. Figure 4.1, have shown that

$\displaystyle\Delta_{\sinh,1}>1-\frac{m\,(\pi-\delta)}{\sinh\big{(}m\,(\pi-\delta)\big{)}}\,\int_{0}^{1}\frac{I_{1}\big{(}m\,(\pi-\delta)\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,.$

Hence, by (4.6) we have

$\displaystyle\Delta_{\sinh,1}>1-\frac{2\,\big{(}\sinh\big{(}\frac{m\,(\pi-\delta)}{2}\big{)}\big{)}^{2}}{\sinh\big{(}m\,(\pi-\delta)\big{)}}=\frac{2\,\mathrm{e}^{-m\,(\pi-\delta)}}{1+\mathrm{e}^{-m\,(\pi-\delta)}}>0\,,$

i. e., $D_{1}(m)=\Delta_{\sinh,1}$ tends to zero as $m\to\infty$ with
exponential decay smaller than $m\,(\pi-\delta)$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 4.1: Semilogarithmic plots of the term
$1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1/\alpha}\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(ii) On the one hand, we consider the expression (4.11) in the case

$\displaystyle 1<\frac{\pi+\delta}{\pi-\delta}\leq\alpha\,,$

i. e., for (4.5) we have
$\nu_{1}(\delta)=\frac{\pi+\delta}{\alpha\,(\pi-\delta)}\leq 1$, such that

$\displaystyle\frac{1}{\alpha}\leq\nu_{1}(\pm\omega)\leq
1\,,\quad\omega\in[-\delta,\,\delta]\,.$ (4.15)

Then by (4.11) and (4.3) we obtain

$\displaystyle 0<\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,,\quad\omega\in[-\delta,\,\delta]\,.$

Note that by (4.5) and (4.15) we have
$\frac{1}{\alpha}<\nu_{1}(0)=\frac{\pi}{\alpha\,(\pi-\delta)}<1$, and

$\displaystyle\nu_{1}(\omega)=\nu_{1}(0)+\frac{\omega}{\alpha\,(\pi-\delta)}\,,\quad\nu_{1}(-\omega)=\nu_{1}(0)-\frac{\omega}{\alpha\,(\pi-\delta)}\,.$
(4.16)

Hence, for $\omega\in[0\,,\delta]$ it follows that

$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{\nu_{1}(0)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle\qquad+\frac{\beta}{2\,\sinh\beta}\,\Bigg{(}\int_{\nu_{1}(0)}^{\nu_{1}(0)+\frac{\omega}{\alpha(\pi-\delta)}}-\int_{\nu_{1}(0)-\frac{\omega}{\alpha(\pi-\delta)}}^{\nu_{1}(0)}\Bigg{)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,.$
(4.17)

An analogous decomposition of $\Delta_{\sinh,2}(\omega)$ also applies for
$\omega\in[-\delta\,,0]$. By the power series expansion of the modified Bessel
function $I_{1}$, the integrand

$\displaystyle\frac{I_{1}(\beta\,\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}=\frac{\beta}{2}\,\sum_{k=0}^{\infty}\frac{(1-\nu^{2})^{k}}{2^{2k}k!(k+1)!}\,,\quad\nu\in(-1,\,1)\,,$
(4.18)

is positive for $\nu\in(-1,1)$. Since the integrand
$\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}$ is also monotonously
decreasing on $[0,1)$, the second term in (4) is negative for
$\omega\in(0\,,\delta]$ as we have two integration intervals of the same
length by (4.16), and therefore

$\displaystyle
D_{2}(m)=\max_{\omega\in[-\delta,\delta]}\Delta_{\sinh,2}(\omega)=\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{\nu_{1}(0)}\frac{I_{1}(\beta\,\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,.$
(4.19)

Numerical experiments, cf. Figure 4.2, have shown that (4.19), tends to zero
as $m\to\infty$ with exponential decay smaller than $m\,(\pi-\delta)$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 4.2: Semilogarithmic plots of the term
$\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{v_{1}(0)}\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1.1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(iii) On the other hand, we consider the expression (4.11) in the case

$\displaystyle 1<\alpha<\frac{\pi+\delta}{\pi-\delta}\,.$

Then by (4.5) we have $\nu_{1}(\omega_{1})=1$ for
$\omega_{1}\coloneqq\alpha\,(\pi-\delta)-\pi$. Without loss of generality, we
can assume that $\omega_{1}\geq 0$. In the case $\omega_{1}>0$, we split the
interval $[-\delta,\delta]$ into the three subintervals
$[-\delta,-\omega_{1}]$, $[-\omega_{1},\omega_{1}]$, and
$[\omega_{1},\delta]$. In the case $\omega_{1}=0$, the interval
$[-\delta,\delta]$ is decomposed into $[-\delta,0]$ and $[0,\delta]$. In the
following, we discuss only the case $\omega_{1}>0$.

(A) Since $\nu_{1}(\omega)$ in (4.5) is an increasing linear function, we have
for $\omega\in[-\delta,-\omega_{1}]$ that

$\displaystyle\frac{1}{\alpha}\leq\nu_{1}(\omega)\leq\nu_{1}(-\omega_{1})<1\,,\quad
1\leq\nu_{1}(-\omega)\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\alpha\,(\pi-\delta)}\,.$
(4.20)

Then from (4.11) and (4.3) it follows that

$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{1}+\int_{1}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1/\alpha}^{1}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle\qquad+\,\frac{\beta}{2\,\sinh\beta}\,\int_{1}^{\nu_{1}(-\omega)}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,\mathrm{d}\nu\,,\quad\omega\in[-\delta,-\omega_{1}]\,.$

Since the integrand $\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}$ is
nonnegative by (4.18), using (4.20) and (4.8) implies

$\displaystyle\frac{\beta}{2\,\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle<\Delta_{\sinh,2}(\omega)$
$\displaystyle<\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu+\frac{\beta}{2\,\sinh\beta}\,\frac{3\,\big{(}1-\mathrm{e}^{-\beta}\big{)}}{2\,\beta}$
$\displaystyle=\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu+\frac{3\,\mathrm{e}^{-\beta}}{2\,(1+\mathrm{e}^{-\beta})}\,.$

Numerical experiments, cf. Figure 4.3, have shown that

$\displaystyle\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,,\quad\alpha>1\,,$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$. Therefore, we obtain that

$\displaystyle\max_{\omega\in[-\delta,-\omega_{1}]}\Delta_{\sinh,2}(\omega)$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 4.3: Semilogarithmic plots of the term
$\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1.1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(B) For $\omega\in[-\omega_{1},\omega_{1}]$ we have

$\displaystyle\frac{1}{\alpha}<\nu_{1}(-\omega_{1})\leq\nu_{1}(\pm\omega)\leq
1\,.$ (4.21)

Then from (4.11) and (4.3) it follows that

$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\,,\quad\omega\in[-\omega_{1},\omega_{1}]\,.$

Note that by (4.5) and (4.21) we have again (4.16) and therefore (4) holds for
$\omega\in[0\,,\omega_{1}]$. Analogous to (4.19) this implies

$\displaystyle\max_{\omega\in[-\omega_{1},\omega_{1}]}\Delta_{\sinh,2}(\omega)=\frac{\beta}{\sinh\beta}\int_{1/\alpha}^{\nu_{1}(0)}\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}\mathrm{d}\nu\,.$

Hence, by the numerical experiments in Figure 4.2 we see that

$\displaystyle\max_{\omega\in[-\omega_{1},\omega_{1}]}\Delta_{\sinh,2}(\omega)$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

(C) For $\omega\in[\omega_{1},\delta]$ we have

$\displaystyle
1\leq\nu_{1}(\omega)\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\alpha\,(\pi-\delta)}\,,\quad\frac{1}{\alpha}\leq\nu_{1}(-\omega)\leq\nu_{1}(-\omega_{1})<1\,.$
(4.22)

Then from (4.11) and (4.3) it follows that

$\displaystyle\Delta_{\sinh,2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{1}+\int_{1}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\sinh}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{2\,\sinh\beta}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{1}\bigg{)}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle\qquad+\,\frac{\beta}{2\,\sinh\beta}\,\int_{1}^{\nu_{1}(\omega)}\frac{J_{1}\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\sqrt{\nu^{2}-1}}\,\mathrm{d}\nu\,.$

Since the integrand $\frac{I_{1}(\beta\sqrt{1-\nu^{2}})}{\sqrt{1-\nu^{2}}}$ is
nonnegative by (4.18), using (4.22) and (4.8) implies

$\displaystyle\frac{\beta}{2\,\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle<\Delta_{\sinh,2}(\omega)$
$\displaystyle<\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu+\frac{\beta}{2\,\sinh\beta}\,\frac{3\,\big{(}1-\mathrm{e}^{-\beta}\big{)}}{2\,\beta}$
$\displaystyle=\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu+\frac{3\,\mathrm{e}^{-\beta}}{2\,(1+\mathrm{e}^{-\beta})}\,.$

Hence, by the numerical experiments in Figure 4.3 we obtain that

$\displaystyle\max_{\omega\in[\omega_{1},\delta]}\Delta_{\sinh,2}(\omega)$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

In summary,

$\displaystyle
D_{2}(m)=\max_{\omega\in[-\delta,\delta]}\Delta_{\sinh,2}(\omega)$

tends to zero for $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

######  Remark 4.3.

Note that the similarity between Figures 4.1 and 4.2 can be easily explained,
since using (4.6) and (4.7) we have

$\displaystyle\bigg{(}1$
$\displaystyle-\frac{\beta}{\sinh\beta}\,\int_{0}^{1/\alpha}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu\bigg{)}-\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu$
$\displaystyle=1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu=1-\frac{2\,\big{(}\sinh\frac{\beta}{2}\big{)}^{2}}{\sinh\beta}=\frac{2\,{\mathrm{e}}^{-\beta}}{1+{\mathrm{e}}^{-\beta}}$

and hence

$\displaystyle
1-\frac{\beta}{\sinh\beta}\,\int_{0}^{1/\alpha}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu=\frac{\beta}{\sinh\beta}\,\int_{1/\alpha}^{1}\frac{I_{1}\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu+\frac{2\,{\mathrm{e}}^{-\beta}}{1+{\mathrm{e}}^{-\beta}}\,,$

where the term $\frac{2\,{\mathrm{e}}^{-\beta}}{1+{\mathrm{e}}^{-\beta}}$ is
very small.

######  Example 4.4.

Analogous to Example 3.3 we now visualize the optimality of the shape
parameter $\beta=m\,(\pi-\delta)$ for the $\sinh$-type regularized Shannon
sampling formula shown in Theorems 4.1 and 4.2. More precisely, for the
bandlimited function (3.8) with several bandwidth parameters
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$,
i. e., several oversampling rates $\frac{\pi}{\delta}>1$, we consider the
regularized Shannon sampling formula (1.4) with the $\sinh$-type window
function $\varphi_{\sinh}$ in (1.7). The corresponding approximation error
(3.7) shall again be approximated by evaluating the given function $f$ and its
approximation $R_{\varphi,m}f$ at equidistant points $t_{s}\in[-1,\,1]$,
$s=1,\dots,S$, with $S=10^{5}$. To compare with the optimal parameter, we
choose the shape parameter of the $\sinh$-type window function (1.7) as
$\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha\in\big{\\{}\frac{1}{2},1,2\big{\\}}$.

The results for different truncation parameters $m\in\\{2,3,\ldots,10\\}$ are
depicted in Figure 4.4. As stated in Theorem 4.2, it can clearly be seen that
the choice of $\alpha\neq 1$ causes worsened error decay rates with respect to
$m$. Thus, the numerical results confirm that the shape parameter
$\beta=m\,(\pi-\delta)$ of Theorem 4.1 is optimal, and that this fact can
already be observed for very small truncation parameters
$m\in\mathbb{N}\setminus\\{1\\}$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 4.4: Maximum approximation error (3.7) using the $\sinh$-type window
function $\varphi_{\mathrm{sinh}}$ in (1.7) with different shape parameters
$\beta=\alpha\,m(\pi-\delta)$, $\alpha\in\big{\\{}\frac{1}{2},1,2\big{\\}}$,
for the bandlimited function (3.8) with bandwidths
$\delta\in\big{\\{}\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4}\big{\\}}$ and
truncation parameters $m\in\\{2,3,\ldots,10\\}$.

We further remark that already in [12, Theorem 6.1] and [13, Theorem 4.2]
bounds on the approximation error of the Shannon sampling formula (1.4) were
shown for the $\sinh$-type window function (1.7) with suitably chosen shape
parameter $\beta$. However, in these previous works the optimal parameter was
only conjectured by numerical testing, whereas the proof of the optimality was
still an open problem. Note that although the respective parameters look
different than the one in Theorem 4.1, they are basically the same, only
adapted to the slightly different settings considered in [12, 13]. Therefore,
the newly proposed Theorem 4.2 provides not only a proof for the optimality of
the parameter choice in Theorem 4.1 but also for the parameter choice of [12,
13].

##  5 Optimal regularization with the continuous Kaiser–Bessel window function

In this section, we consider the continuous Kaiser–Bessel window function
(1.8) with shape parameter $\beta>0$, analogous to [13, Theorem 4.3]. In order
to achieve fast convergence of the continuous Kaiser–Bessel regularized
Shannon sampling formula, we again put special emphasis on the optimal choice
of this shape parameter $\beta$. Furthermore, we show that the exponential
decay with respect to the truncation parameter
$m\in\mathbb{N}\setminus\\{1\\}$ for the uniform approximation error
$\big{\|}f-R_{\mathrm{cKB},m}f\big{\|}_{C_{0}(\mathbb{R})}$ is similar to the
approximation error $\big{\|}f-R_{\sinh,m}f\big{\|}_{C_{0}(\mathbb{R})}$ in
Theorem 4.1.

######  Theorem 5.1.

Assume that $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ is bandlimited with
bandwidth $\delta\in\big{(}0,\,\frac{m-1}{m}\pi\big{]}$. Further let
$\varphi_{\mathrm{cKB}}$ be the continuous Kaiser–Bessel window function (1.8)
with shape parameter $\beta=m\,(\pi-\delta)$ and let
$m\in{\mathbb{N}}\setminus\\{1\\}$ be given.
Then the continuous Kaiser–Bessel regularized Shannon sampling formula
satisfies the error estimate

$\displaystyle\big{\|}f-R_{\mathrm{cKB},m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\bigg{(}\frac{7}{8}\,m\,(\pi-\delta)+\frac{7}{\pi}\,m^{2}(\pi-\delta)^{2}\bigg{)}\,\mathrm{e}^{-m\,(\pi-\delta)}\,\|f\|_{L^{2}(\mathbb{R})}\,.$
(5.1)

_Proof_. (i) Since $\varphi_{\mathrm{cKB}}$ in (1.8) is compactly supported on
$[-m,\,m]$ and $\varphi_{\mathrm{cKB}}(m)=0$, we have $E_{2}(m)=0$. Thus,
according to Theorem 2.1, the approximation error can be estimated by

$\displaystyle\big{\|}f-R_{\mathrm{cKB},m}f\big{\|}_{C_{0}(\mathbb{R})}\leq\|f\|_{L^{2}(\mathbb{R})}\,\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\mathrm{cKB}}(\omega)\big{|}$

where

$\displaystyle\Delta_{\mathrm{cKB}}(\omega)\coloneqq
1-\frac{1}{\sqrt{2\pi}}\,\int_{\omega-\pi}^{\omega+\pi}{\hat{\varphi}}_{\mathrm{cKB}}(\tau)\,\mathrm{d}\tau\,,\quad\omega\in[-\delta,\,\delta]\,.$
(5.2)

Following [18, p. 3, 1.1, and p. 95, 18.31], the Fourier transform of (1.8)
has the form

$\displaystyle{\hat{\varphi}}_{\mathrm{cKB}}(\tau)=\frac{m\,\sqrt{2}}{(I_{0}(\beta)-1)\,\sqrt{\pi}}\cdot\begin{cases}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}&\colon|\nu|<1\,,\\\\[8.61108pt]
\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}&\colon|\nu|>1\,,\end{cases}$
(5.3)

with the scaled frequency $\nu=\frac{m}{\beta}\,\tau$. Substituting
$\tau=\frac{\beta}{m}\,\nu$ in the integral in (5.2), the function
$\Delta_{\mathrm{cKB}}$ reads as

$\displaystyle\Delta_{\mathrm{cKB}}(\omega)=1-\frac{\beta}{m\,\sqrt{2\pi}}\,\int_{-\nu_{1}(-\omega)}^{\nu_{1}(\omega)}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,,\quad\omega\in[-\delta,\,\delta]\,,$
(5.4)

with the increasing linear function (4.5).

(ii) Now we choose the shape parameter of (1.8) in the special form
$\beta=m\,(\pi-\delta)$. Thus, we have

$\displaystyle
1=\nu_{1}(-\delta)\leq\nu_{1}(\omega)=\frac{\omega+\pi}{\pi-\delta}\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\pi-\delta}\,,\quad\omega\in[-\delta,\,\delta]\,.$

In view of (5.3) we split (5.4) in the form
$\Delta_{\mathrm{cKB}}(\omega)=\Delta_{\mathrm{cKB},1}-\Delta_{\mathrm{cKB},2}(\omega)$
with

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle=1-\frac{2\beta}{\pi\,(I_{0}(\beta)-1)}\,\int_{0}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,,$
$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{\pi\,(I_{0}(\beta)-1)}\,\bigg{(}\int_{1}^{\nu_{1}(-\omega)}+\int_{1}^{\nu_{1}(\omega)}\bigg{)}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,.$
(5.5)

Using [10, 3.997–1] we have

$\displaystyle\int_{0}^{1}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}\,\mathrm{d}\nu=\frac{1}{\beta}\,\int_{0}^{\pi/2}\sinh(\beta\,\cos
s)\,\mathrm{d}s=\frac{\pi}{2\beta}\,{\textbf{L}}_{0}(\beta)\,,$

where ${\textbf{L}}_{0}$ denotes the _modified Struve function_ (see [1,
12.2.1])

$\displaystyle{\textbf{L}}_{0}(x)\coloneqq\sum_{k=0}^{\infty}\frac{(x/2)^{2k+1}}{\big{(}\Gamma\big{(}k+\frac{3}{2}\big{)}\big{)}^{2}}=\frac{2x}{\pi}\,\sum_{k=0}^{\infty}\frac{x^{2k}}{\big{(}(2k+1)!!\big{)}^{2}}\,,\quad
x\in\mathbb{R}\,.$

Additionally, by the definition of the _sine integral function_

$\displaystyle\mathrm{Si}(x)\coloneqq\int_{0}^{x}\frac{\sin
v}{v}\,{\mathrm{d}}v\,,\quad x\in\mathbb{R}\,,$

we have

$\displaystyle\int_{0}^{1}\frac{\sin(\beta v)}{\beta
v}\,{\mathrm{d}}v=\frac{1}{\beta}\,\mathrm{Si}(\beta)\,,$

such that we obtain

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle=1-\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\frac{\pi}{2\beta}\,{\textbf{L}}_{0}(\beta)-\frac{1}{\beta}\,\mathrm{Si}(\beta)\bigg{)}$
$\displaystyle=\frac{1}{I_{0}(\beta)-1}\,\bigg{(}I_{0}(\beta)-{\textbf{L}}_{0}(\beta)-1+\frac{2}{\pi}\,\mathrm{Si}(\beta)\bigg{)}\,.$
(5.6)

Note that by [2, Theorem 1] the function $I_{0}(x)-{\textbf{L}}_{0}(x)$ is
completely monotonic on $[0,\,\infty)$ and tends to zero as $x\to\infty$.
Moreover, by a numerical test (see [13, Figure 4.2]) we see that for
$\beta=m\,(\pi-\delta)>0$ we have

$\displaystyle 0\leq
I_{0}(\beta)-{\textbf{L}}_{0}(\beta)-1+\frac{2}{\pi}\,\mathrm{Si}(\beta)\leq\frac{1}{2}\,.$
(5.7)

Since additionally $I_{0}(\beta)>1$ holds for $\beta=m\,(\pi-\delta)>0$, this
yields

$\displaystyle
0\leq\Delta_{\mathrm{cKB},1}\leq\frac{1}{2\,\big{(}I_{0}(\beta)-1\big{)}}\,.$

Now we estimate$\Delta_{\mathrm{cKB},2}(\omega)$ in (5.5) for
$\omega\in[-\delta,\,\delta]$ by the triangle inequality as

$\displaystyle\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}\leq\frac{\beta}{\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1}^{\nu_{1}(-\omega)}+\int_{1}^{\nu_{1}(\omega)}\bigg{)}\bigg{|}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{|}\,\mathrm{d}\nu\,.$

By [21, Lemma 4] we have

$\displaystyle\bigg{|}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{|}\leq\frac{2}{\nu^{2}}\,,\quad\nu\geq
1\,,$ (5.8)

and therefore

$\displaystyle|\Delta_{\mathrm{cKB},2}(\omega)|\leq\frac{4\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\infty}\frac{1}{\nu^{2}}\,{\mathrm{d}}\nu=\frac{4\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,.$

Thus, we conclude that

$\displaystyle|\Delta_{\mathrm{cKB}}(\omega)|\leq\Delta_{\mathrm{cKB},1}+|\Delta_{\mathrm{cKB}2}(\omega)|\leq\frac{1}{I_{0}(\beta)-1}\,\bigg{(}\frac{1}{2}+\frac{4\beta}{\pi}\bigg{)}\,,\quad\omega\in[-\delta,\,\delta]\,.$

Since numerical experiments have shown that
$\frac{{\mathrm{e}}^{x}}{x\,(I_{0}(x)-1)}$ is strictly decreasing on
$[1,\infty)$ and by the assumption $0<\delta\leq\frac{m-1}{m}\,\pi$ we have
$\beta=m\,(\pi-\delta)\geq\pi$ for $m\in\mathbb{N}\setminus\\{1\\}$, it
follows that

$\displaystyle\frac{{\mathrm{e}}^{\beta}}{\beta\,(I_{0}(\beta)-1)}\leq\frac{{\mathrm{e}}^{\pi}}{\pi\,(I_{0}(\pi)-1)}=1.644967\ldots<\frac{7}{4}\,.$
(5.9)

Hence, this yields

$\displaystyle\frac{1}{I_{0}(\beta)-1}\,\bigg{(}\frac{1}{2}+\frac{4\beta}{\pi}\bigg{)}$
$\displaystyle<\frac{7\beta}{4}\,\bigg{(}\frac{1}{2}+\frac{4\beta}{\pi}\bigg{)}\,{\mathrm{e}}^{-\beta}=\bigg{(}\frac{7}{8}\,\beta+\frac{7}{\pi}\,\beta^{2}\bigg{)}\,{\mathrm{e}}^{-\beta}\,.$
(5.10)

Thus, the continuous Kaiser–Bessel regularized Shannon sampling formula with
the chosen shape parameter $\beta=m\,(\pi-\delta)$ fulfills the error estimate
(5.1). This completes the proof.

Now we show that the choice of the shape parameter $\beta=m\,(\pi-\delta)$ of
(1.8) is optimal in a certain sense. To this end, let the parameters
$\alpha\geq\frac{1}{\pi}$, $m\in\mathbb{N}\setminus\\{1\\}$, and
$\delta\in\big{(}0,\,\frac{m-1}{m}\,\pi\big{)}$ be given, and consider shape
parameters of the form $\beta=\alpha\,m\,(\pi-\delta)$. Then the increasing
linear function (4.5) fulfills

$\displaystyle\frac{1}{\alpha}=\nu_{1}(-\delta)\leq\nu_{1}(\omega)=\frac{\omega+\pi}{\alpha\,(\pi-\delta)}\leq\nu_{1}(\delta)=\frac{\pi+\delta}{\alpha\,(\pi-\delta)}\,,\quad\omega\in[-\delta,\,\delta]\,.$

Therefore, we split (5.4) as
$\Delta_{\mathrm{cKB}}(\omega)=\Delta_{\mathrm{cKB},1}-\Delta_{\mathrm{cKB},2}(\omega)$,
$\omega\in[-\delta,\,\delta]$, with

$\displaystyle\Delta_{\mathrm{cKB},1}$ $\displaystyle\coloneqq
1-\frac{\beta}{m\,\sqrt{2\pi}}\,\int_{-1/\alpha}^{1/\alpha}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,,$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,,$
(5.11) $\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle\coloneqq\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{-\nu_{1}(-\omega)}^{-1/\alpha}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,\mathrm{d}\nu\,.$
(5.12)

Introducing the terms

$\displaystyle D_{1}(m)\coloneqq\big{|}\Delta_{\mathrm{cKB},1}\big{|}\,,\quad
D_{2}(m)\coloneqq\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}\,,$
(5.13)

it is known by Theorem 5.1 that for $\alpha=1$ both expressions in (5.13)
possess the same exponential decay $m\,(\pi-\delta)$. In the following, we
discuss the other cases $0<\alpha<1$ and $\alpha>1$. More precisely, we show
in Theorem 5.2 that for $\alpha\neq 1$ both expressions in (5.13) have the
same exponential decay smaller than $m\,(\pi-\delta)$. In this sense, it
follows immediately that the shape parameter $\beta=m\,(\pi-\delta)$ of the
continuous Kaiser–Bessel window function (1.8) is _optimal_ , since both
expressions in (5.13) tend to zero as $m\to\infty$ with the same maximum
exponential decay.

######  Theorem 5.2.

For $\delta\in\big{(}0,\,\frac{m-1}{m}\,\pi\big{]}$, let
$\varphi_{\mathrm{cKB}}$ be the continuous Kaiser–Bessel window function (1.8)
with the shape parameter $\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha\geq\frac{1}{\pi}$, $\alpha\not=1$, and
$m\in{\mathbb{N}}\setminus\\{1\\}$.

  1. a)

In the case $\alpha\in\big{[}\frac{1}{\pi},\,1\big{)}$, both expressions in
(5.13) tend to zero as $m\to\infty$ with the same exponential decay
$\alpha\,m\,(\pi-\delta)$.

  2. b)

In the case $\alpha>1$, both expressions in (5.13) tend to zero as
$m\to\infty$ with exponential decay smaller than $m\,(\pi-\delta)$.

_Proof_. a) First we consider the shape parameter
$\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha\in\big{[}\frac{1}{\pi},\,1\big{)}$. Then we have by (5.11), (5.3) and
(5.6) that

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\bigg{(}\int_{0}^{1}+\int_{1}^{1/\alpha}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\bigg{(}1-\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{0}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,{\mathrm{d}}\nu\bigg{)}$
$\displaystyle\qquad-\,\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{1/\alpha}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{1}{I_{0}(\beta)-1}\,\bigg{(}I_{0}(\beta)-{\textbf{L}}_{0}(\beta)-1+\frac{2}{\pi}\,\mathrm{Si}(\beta)\bigg{)}$
$\displaystyle\qquad-\,\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{1/\alpha}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,{\mathrm{d}}\nu\,.$

Using (5.7), (5.8), and (5.10), it follows that

$\displaystyle\big{|}\Delta_{\mathrm{cKB},1}\big{|}\leq\frac{1}{2\,\big{(}I_{0}(\beta)-1\big{)}}+\frac{4\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}<\bigg{(}\frac{7}{8}\,\beta+\frac{7}{\pi}\,\beta^{2}\bigg{)}\,\mathrm{e}^{-\beta}\,.$
(5.14)

For the second term (5.12) have by (5.3) that

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\,\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,{\mathrm{d}}\nu\,,$

such that (5.8) in connection with $\frac{1}{\alpha}>1$ and the monotonicity
of $\frac{1}{\nu^{2}}$ implies

$\displaystyle\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}\leq\frac{4\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\infty}\frac{1}{\nu^{2}}\,\mathrm{d}\nu=\frac{4\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,.$

Since numerical experiments have shown that
$\frac{{\mathrm{e}}^{x}}{x\,(I_{0}(x)-1)}$ is strictly decreasing on
$[1,\infty)$ and by the assumption $0\leq\delta\leq\frac{m-1}{m}\,\pi$ and
$\frac{1}{\pi}\leq\alpha<1$ we have
$\beta=\alpha\,m\,(\pi-\delta)\geq\alpha\,\pi\geq 1$, it follows that

$\displaystyle\frac{\mathrm{e}^{\beta}}{\beta\,\big{(}I_{0}(\beta)-1\big{)}}\leq\frac{\mathrm{e}}{I_{0}(1)-1}=10.216574\ldots<11$

and therefore

$\displaystyle\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}\leq\frac{44\beta^{2}}{\pi}\,\mathrm{e}^{-\beta}\,,\quad\omega\in[-\delta,\,\delta]\,.$
(5.15)

Thus, by (5.14) and (5.15) the quantities (5.13) tend to zero as $m\to\infty$
with the same exponential decay $\alpha\,m\,(\pi-\delta)$, which is smaller
than $m\,(\pi-\delta)$ as $\alpha\in\big{[}\frac{1}{\pi},\,1\big{)}$.

b) Now we investigate the shape parameter $\beta=\alpha m\,(\pi-\delta)$ with
$\alpha>1$.
(i) By (5.11) and (5.3) we obtain

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle=1-\frac{\beta\,\sqrt{2}}{m\,\sqrt{\pi}}\,\int_{0}^{1/\alpha}{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=1-\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{0}^{1/\alpha}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Numerical experiments, cf. Figure 5.1, have shown that

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle>1-\frac{2m\,(\pi-\delta)}{\pi\,\big{(}I_{0}(m\,(\pi-\delta))-1\big{)}}\,\int_{0}^{1}\bigg{(}\frac{\sinh\big{(}m\,(\pi-\delta)\,\sqrt{1-\nu^{2}}\big{)}}{m\,(\pi-\delta)\,\sqrt{1-\nu^{2}}}-\frac{\sin\big{(}m\,(\pi-\delta)\,\nu\big{)}}{m\,(\pi-\delta)\,\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Hence, by (5.6) we have

$\displaystyle\Delta_{\mathrm{cKB},1}$
$\displaystyle>\frac{1}{I_{0}(m\,(\pi-\delta))-1}\,\bigg{(}I_{0}(m\,(\pi-\delta))-{\textbf{L}}_{0}(m\,(\pi-\delta))-1+\frac{2}{\pi}\,\mathrm{Si}(m\,(\pi-\delta))\bigg{)}>0\,,$

i. e., $D_{1}(m)=\Delta_{\mathrm{cKB},1}$ tends to zero as $m\to\infty$ with
exponential decay smaller than $m\,(\pi-\delta)$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 5.1: Semilogarithmic plots of the term
$1-\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{0}^{1/\alpha}\Big{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\Big{)}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(ii) On the one hand, we consider the expression (5.12) in the case

$\displaystyle 1<\frac{\pi+\delta}{\pi-\delta}\leq\alpha\,,$

where we have (4.15). Then by (5.12) and (5.3) we obtain for
$\omega\in[-\delta,\,\delta]$ that

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Note that we have again (4.16). Hence, for $\omega\in[0,\,\delta]$ it follows
that

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)=\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{\nu_{1}(0)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\bigg{)}\,\mathrm{d}\nu$
(5.16)
$\displaystyle\quad+\,\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\Bigg{(}\int_{\nu_{1}(0)}^{\nu_{1}(0)+\frac{\omega}{\alpha\,(\pi-\delta)}}-\int_{\nu_{1}(0)-\frac{\omega}{\alpha\,(\pi-\delta)}}^{\nu_{1}(0)}\Bigg{)}\,\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\bigg{)}\,\mathrm{d}\nu\,.$

An analogous decomposition of $\Delta_{\mathrm{cKB},2}(\omega)$ also applies
for $\omega\in[-\delta,\,0]$. By Figure 5.2 the even integrand

$\displaystyle\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\,,\quad\nu\in(-1,\,1)\,,$
(5.17)

is positive and monotonously decreasing on $[0,\,1)$. Thus, the second term in
(5.16) is negative for $\omega\in(0\,,\delta]$ as we have two integration
intervals of the same length by (5.16), and therefore

$\displaystyle
D_{2}(m)=\max_{\omega\in[-\delta,\,\delta]}\Delta_{\mathrm{cKB},2}(\omega)=\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{\nu_{1}(0)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Numerical experiments, cf. Figure 5.3, demonstrate that $D_{2}(m)$ tends to
zero as $m\to\infty$ with exponential decay smaller than $m\,(\pi-\delta)$.

Figure 5.2: The integrand (5.17) for $m=5$, $\alpha=1$, and
$\delta=\frac{\pi}{4}$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 5.3: Semilogarithmic plots of the term
$\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{\nu_{1}(0)}\Big{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\Big{)}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1.1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(iii) On the other hand, we consider the expression (5.12) in the case

$\displaystyle 1<\alpha<\frac{\pi+\delta}{\pi-\delta}\,.$

Then by (4.5) we have $\nu_{1}(\omega_{1})=1$ for
$\omega_{1}\coloneqq\alpha\,(\pi-\delta)-\pi$. Without loss of generality, we
can assume that $\omega_{1}\geq 0$. In the case $\omega_{1}>0$, we split the
interval $[-\delta,\,\delta]$ into three subintervals
$[-\delta,\,-\omega_{1}]$, $[-\omega_{1},\,\omega_{1}]$, and
$[\omega_{1},\,\delta]$. In the case $\omega_{1}=0$, the interval
$[-\delta,\,\delta]$ is decomposed into $[-\delta,\,0]$ and $[0,\,\delta]$. In
the following, we discuss only the case $\omega_{1}>0$.

(A) For $\omega\in[-\delta,\,-\omega_{1}]$ we have again (4.20). Then from
(5.12) and (5.3) it follows that

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{1}+\int_{1}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1/\alpha}^{1}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\,\nu)}{\beta\,\nu}\bigg{)}\,\mathrm{d}\nu$
$\displaystyle\qquad+\,\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\nu_{1}(-\omega)}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\,\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,,\quad\omega\in[-\delta,-\omega_{1}]\,.$

Since the integrand (5.17) is nonnegative by Figure 5.2, using (4.20) and
(5.8) implies that

$\displaystyle\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu<\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}$
$\displaystyle\qquad<\,\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu$
$\displaystyle\qquad\qquad+\,\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\nu_{1}(-\omega)}\bigg{|}\frac{\sin\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{|}\,\mathrm{d}\nu$
$\displaystyle\qquad\leq\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu+\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,.$

We remark that by $\beta=\alpha\,m\,(\pi-\delta)$ with $\alpha>1$ and
$0<\delta\leq\frac{m-1}{m}\,\delta$ we have $\beta>\pi$, and hence by (5.9)
this yields

$\displaystyle\frac{1}{I_{0}(\beta)-1}<\frac{7}{4}\,\beta\,\mathrm{e}^{-\beta}\,.$

Numerical experiments, cf. Figure 5.4, have shown that

$\displaystyle\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,,\quad\alpha>1\,,$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$. Therefore, we obtain that

$\displaystyle\max_{\omega\in[-\delta,-\omega_{1}]}\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 5.4: Semilogarithmic plots of the term
$\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\Big{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\Big{)}\,\mathrm{d}\nu$
for $m=1,\ldots,20$, $\alpha\in\\{1.1,\,1.2,\,1.5,\,2,\,3\\}$, and
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$.

(B) For $\omega\in[-\omega_{1},\,\omega_{1}]$ we have again (4.21). Then from
(5.12) and (5.3) it follows that for $\omega\in[-\omega_{1},\,\omega_{1}]$ we
have

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{\nu_{1}(\omega)}\bigg{)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Note that by (4.5) and (4.21) we have again (4.16) and therefore (5.16) holds
for $\omega\in[0\,,\omega_{1}]$. Analogous to step (ii) this implies

$\displaystyle\max_{\omega\in[-\omega_{1},\omega_{1}]}\Delta_{\mathrm{cKB},2}(\omega)=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{\nu_{1}(0)}\bigg{(}\frac{\sinh(\beta\sqrt{1-\nu^{2}})}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Hence, by the numerical experiments in Figure 5.3 we see that

$\displaystyle\max_{\omega\in[-\omega_{1},\omega_{1}]}\Delta_{\mathrm{cKB},2}(\omega)$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

(C) For $\omega\in[\omega_{1},\,\delta]$ we have again (4.22). Then from
(5.12) and (5.3) it follows that

$\displaystyle\Delta_{\mathrm{cKB},2}(\omega)$
$\displaystyle=\frac{\beta}{m\,\sqrt{2\pi}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{1}+\int_{1}^{\nu_{1}(\omega)}\bigg{)}\,{\hat{\varphi}}_{\mathrm{cKB}}\big{(}\tfrac{\beta}{m}\,\nu\big{)}\,{\mathrm{d}}\nu$
$\displaystyle=\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\bigg{(}\int_{1/\alpha}^{\nu_{1}(-\omega)}+\int_{1/\alpha}^{1}\bigg{)}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu$
$\displaystyle\qquad+\,\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\nu_{1}(\omega)}\bigg{(}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu\,.$

Since the integrand (5.17) is nonnegative by Figure 5.2, using (4.22), (5.8)
and (5.9) for the shape parameter
$\beta=\alpha\,m\,(\pi-\delta)>\alpha\,\pi>\pi$ implies

$\displaystyle\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu<\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}$
$\displaystyle\qquad<\,\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu$
$\displaystyle\qquad\qquad+\frac{\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1}^{\nu_{1}(\omega)}\bigg{|}\frac{\sin\big{(}\beta\,\sqrt{\nu^{2}-1}\big{)}}{\beta\,\sqrt{\nu^{2}-1}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{|}\,\mathrm{d}\nu$
$\displaystyle\qquad<\,\frac{2\beta}{\pi\,\big{(}I_{0}(\beta)-1\big{)}}\,\int_{1/\alpha}^{1}\bigg{(}\frac{\sinh\big{(}\beta\,\sqrt{1-\nu^{2}}\big{)}}{\beta\,\sqrt{1-\nu^{2}}}-\frac{\sin(\beta\nu)}{\beta\nu}\bigg{)}\,\mathrm{d}\nu+\frac{7}{2\pi}\,\beta^{2}\,\mathrm{e}^{-\beta}\,.$

Hence, by the numerical experiments in Figure 5.4 we obtain that

$\displaystyle\max_{\omega\in[\omega_{1},\delta]}\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}$

tends to zero as $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

In summary,

$\displaystyle
D_{2}(m)=\max_{\omega\in[-\delta,\delta]}\big{|}\Delta_{\mathrm{cKB},2}(\omega)\big{|}$

tends to zero for $m\to\infty$ with exponential decay smaller than
$m\,(\pi-\delta)$.

######  Example 5.3.

Analogous to Example 4.4 we now visualize the optimality of the shape
parameter $\beta=m\,(\pi-\delta)$ for the continuous Kaiser–Bessel regularized
Shannon sampling formula shown in Theorems 5.1 and 5.2. More precisely, for
the bandlimited function (3.8) with several bandwidth parameters
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$,
i. e., several oversampling rates $\frac{\pi}{\delta}>1$, we consider the
regularized Shannon sampling formula (1.4) with the continuous Kaiser–Bessel
window function $\varphi_{\mathrm{cKB}}$ in (1.8). The corresponding
approximation error (3.7) shall again be approximated by evaluating the given
function $f$ and its approximation $R_{\varphi,m}f$ at equidistant points
$t_{s}\in[-1,\,1]$, $s=1,\dots,S$, with $S=10^{5}$. To compare with the
optimal parameter, we choose the shape parameter of the continuous
Kaiser–Bessel window function (1.8) as $\beta=\alpha\,m\,(\pi-\delta)$ with
$\alpha\in\big{\\{}\frac{1}{2},1,2\big{\\}}$.

The results for different truncation parameters $m\in\\{2,3,\ldots,10\\}$ are
depicted in Figure 5.5. As stated in Theorem 5.2, it can clearly be seen that
the choice of $\alpha\neq 1$ causes worsened error decay rates with respect to
$m$. Thus, the numerical results confirm that the shape parameter
$\beta=m\,(\pi-\delta)$ of Theorem 5.1 is optimal, and that this fact can
already be observed for very small truncation parameters
$m\in\mathbb{N}\setminus\\{1\\}$.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 5.5: Maximum approximation error (3.7) using the continuous
Kaiser–Bessel window function $\varphi_{\mathrm{cKB}}$ in (1.8) with different
shape parameters $\beta=\alpha\,m(\pi-\delta)$,
$\alpha\in\big{\\{}\frac{1}{2},1,2\big{\\}}$, for the bandlimited function
(3.8) with bandwidths
$\delta\in\big{\\{}\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4}\big{\\}}$ and
truncation parameters $m\in\\{2,3,\ldots,10\\}$.

We further remark that already in [13, Theorem 4.3] bounds on the
approximation error of the Shannon sampling formula (1.4) were shown for the
continuous Kaiser–Bessel window function (1.8) with suitably chosen shape
parameter $\beta$. However, in this previous work the optimal parameter was
only conjectured by numerical testing, whereas the proof of the optimality was
still an open problem. Note that although the respective parameter looks
different than the one in Theorem 5.1, it is basically the same, only adapted
to the slightly different setting considered in [13]. Therefore, the newly
proposed Theorem 5.2 provides not only a proof for the optimality of the
parameter choice in Theorem 5.1 but also for the parameter choice of [13].

######  Remark 5.4.

Note that the code files for this and all the other experiments are available
on https://github.com/melaniekircheis/Optimal-parameter-choice-for-
regularized-Shannon-sampling-formulas.

##  6 Conclusion

In this paper, we have studied the regularized Shannon sampling formula (1.4)
for the widely used Gaussian function (1.5), the modified Gaussian function
(1.6), the $\sinh$-type window function (1.7), and the continuous
Kaiser–Bessel window function (1.8). More precisely, for an arbitrary
bandlimited function $f\in L^{2}(\mathbb{R})\cap C(\mathbb{R})$ with bandwidth
$\delta\in(0,\,\pi)$ we have shown that the uniform approximation error (1.10)
of the regularized Shannon sampling formulas of $f$ possess an exponential
decay with respect to the truncation parameter $m$. In doing so, we have
demonstrated that the decay rate $m\,(\pi-\delta)$ of the $\sinh$-type
regularized Shannon sampling formula, see Theorem 4.1, and the continuous
Kaiser–Bessel regularized Shannon sampling formula, see Theorem 5.1, is much
better than the decay rate $m\,(\pi-\delta)/2$ of the Gaussian regularized
Shannon sampling formula, see Theorem 3.1. Note that the $\sinh$-type
regularized Shannon sampling formula is even better than the continuous
Kaiser–Bessel regularized Shannon sampling formula due to the constant factors
in (4.1) and (5.1), see also Figure 6.1.

Moreover, we found that the exponential decay of the approximation error of
the regularized Shannon sampling formula (1.4) strongly depends on the shape
parameter of the corresponding window function. Namely, the optimal choice of
the variance $\sigma^{2}$ of the (modified) Gaussian function and of the shape
parameter $\beta$ of the $\sinh$-type window function and the continuous
Kaiser–Bessel function is a crucial ingredient for a fast and accurate
reconstruction of $f$. Therefore, the main focus of this paper was to
determine the optimal variances $\sigma^{2}$ in Theorems 3.1 and 3.4 as well
as the optimal shape parameters $\beta$ in Theorems 4.2 and 5.2, such that the
exponential decay of the approximation error (1.10) is as large as possible.
These results further emphasize the superiority of the $\sinh$-type
regularized Shannon sampling formula of $f$, since the approximation errors of
the regularized Shannon sampling formulas were compared for the optimal shape
parameters each.

(a) $\delta=\frac{\pi}{4}$

(b) $\delta=\frac{\pi}{2}$

(c) $\delta=\frac{3\pi}{4}$

Figure 6.1: Maximum approximation error (3.7) (solid) and error constants
(dashed) using
$\varphi\in\\{\varphi_{\mathrm{Gauss}},\varphi_{\mathrm{mGauss}},\varphi_{\sinh},\varphi_{\mathrm{cKB}}\\}$,
see (1.5), (1.6), (1.7), and (1.8), for the bandlimited function (3.8) with
$\delta\in\big{\\{}\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\pi}{4}\big{\\}}$
and $m\in\\{2,3,\ldots,10\\}$.

## Acknowledgments

Melanie Kircheis gratefully acknowledges the support from the BMBF grant
01$\mid$S20053A (project SA$\ell$E).

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