text:
\section{Introduction}
We study the integral operator $K$ on $L^2(\mathbb{R}_+)$, $\mathbb{R}_+:=\{x\in\mathbb{R}: x>0\}$ given by
\begin{equation}\label{WHO}
(Kg)(x)=\int_0^\infty \textrm{\scriptsize{$\frac{1}{\pi}$}}\operatorname{sinc}(x-y)\,g(y)\d y
\end{equation}

deriving its spectral representation. The sinc function or cardinal sine function $\operatorname{sinc}(x)=\frac{\sin(x)}{x}$ for $x\in\mathbb{R}\setminus \{0\}$, $\operatorname{sinc}(0)=0$   (also called sampling function or slit function)   frequently arises in signal processing and the theory of Fourier transforms. It equals $\operatorname{j}_0$, which is
  the zeroth order spherical Bessel function of the first kind.  $K$ is  an operator of Wiener-Hopf type. It is not  singular  as $\operatorname{sinc}$ is continuous at $0$.
A well-known method  to achieve the  concrete  spectral representation of the operator requires the solution of the corresponding homogeneous integral equation (\ref{HWHE}).
The  original Wiener-Hopf method \cite{WHP31} does not apply   as   $\frac{1}{\pi}\operatorname{sinc}(x)\exp(\lambda|x|)$, $x\in\mathbb{R}$ is not square integrable for any $\lambda>0$.
  Since the kernel $\frac{1}{\pi}\operatorname{sinc}$ is not even integrable and the symbol $1_{[-1,1]}=\sqrt{2\pi}\mathcal{F}^{-1}\big(\frac{1}{\pi}\operatorname{sinc}\big)$ is not continuous,
  more general theories as Krein \cite{K62}, Talenti \cite[Teorema 3.1.2, 3.1.3]{T73} do not apply, either. However, as we will see, one may  successfully  introduce the factor $\exp(-\varepsilon|\cdot|)$, $\varepsilon>0$ to ensure the exponential decay of the kernel, and study the limit $\varepsilon\to 0$.
An alternative approach to the spectral representation of $K$ applies the theory by Rosenblum
 \cite{R61} by turning to an isomorphic Toeplitz operator, see sec.\,\ref{TOIK} for details.\\
\hspace*{6mm}
We came across $K$  when  analyzing   the causal localization  for a Weyl fermion. In sec.\,\ref{CLOWF} it is shown that the localization operator for the unit ball is Hilbert space isomorphic
to the countably infinite orthogonal sum of copies of $K$. In the course of the proof some useful formulae concerning the Bessel functions of the first kind and the related Hankel transformations are deduced. --- Exploiting the Hilbert space isomorphisms transforming Wiener-Hopf operators into Toeplitz operators (\ref{MTRTOWHO}) we obtain in sec.\,\ref{TOIK} the spectral representation of the Toeplitz operators for the indicator functions of the arcs on the unit circle, already derived in \cite{R61}.  The integral transformations effecting the spectral representation are studied in (\ref{GOLBV}), (\ref{SGOLBV}). They map bases of known orthonormal functions onto each other. --- Particularly interesting is the close relationship of $K$ to the finite
Hilbert transformation revealed in sec.\,\ref{MEFHT}, which allows to deduce the spectral representation of the latter \cite{KP59}. In \cite{C20} it is shown that generally a natural isometry exists transforming a self-adjoint Wiener-Hopf operator into a singular integral operator of the well-studied class of Hilbert transformation type operators.
 --- Wiener-Hopf  operators for symbols which are related by  change of variable due to a real fractional linear transformation
are unitarily equivalent. Actually the covariance in (\ref{CVFLTWH}) holds. Using this, in (\ref{HTWHRK}) several Wiener-Hopf operators and singular integral operators are transformed isomorphicly    into $2K-I$.

\section{Spectral Representation of $K$}

Let us call a solution $u$ of the homogeneous Wiener-Hopf equation
\begin{equation}\label{HWHE}
s\,u(x)- \int_0^\infty k(x-y)u(y)\d y=0, \quad x>0
\end{equation}
a generalized eigenfunction or  \textbf{spectral function} of the respective Wiener-Hopf operator for the generalized eigenvalue or  \textbf{spectral value}   $s$.
For the kernel $\frac{1}{\pi}\operatorname{sinc}$ we will show that every $s\in ]0,1[$ is a spectral value and find the corresponding  spectral functions $q_s$. Moreover, these are shown to  give rise to the kernel $q:]0,1[\times \mathbb{R}_+\to \mathbb{C}$, $q(s,x):=n(s)q_s(x)$ for the normalizing factor $n(s)=\frac{\i}{\pi}(2s)^{-1/2}$ such that the integral operator
$(Vh)(x):=\operatorname{l.i.m.}\int_{\downarrow 0}^{\uparrow1}q(s,x)h(s)\d s$ determines a Hilbert space isomorphism from $L^2(0,1)$ onto $L^2(\mathbb{R}_+)$ diagonalizing $K$, i.e.,
$$\big(V^{-1}K\,Vh\big)(s)=s\,h(s)$$
Hence the spectrum of $K$ is $[0,1]$, it is simple,  and  $K$ is absolutely continuous.
As known, the latter  property have all bounded self-adjoint Wiener-Hopf operators.

\subsection{First properties of K}\label{FPOK}
 $\mathcal{F}$ denotes the Fourier transformation on  $L^2(\mathbb{R})$. Recall  $\mathcal{F}f(x)=(2\pi)^{-1/2}\int_\mathbb{R} \operatorname{e}  ^{-\i  xy} f(y) \d y$ for $x\in\mathbb{R}$ if $f\in L^2\cap L^1$ and that in general $\mathcal{F}f$ is the limit $\operatorname{l.i.m.}_{A,A'\to\infty}\mathcal{F}(1_{[-A,A']}f)$  in $L^2$.
For any $\kappa\in L^\infty(\mathbb{R})$ let $M(\kappa)$ denote the multiplication operator by $\kappa$ on $L^2(\mathbb{R})$, which is normal and bounded with norm $\norm{\kappa}_\infty$.  We use the projection $P_+:L^2(\mathbb{R})\to L^2(\mathbb{R}_+)$, $P_+f(x):=f(x)$ and its adjoint, the injection $P_+^*:L^2(\mathbb{R}_+)\to L^2(\mathbb{R})$, $P_+^*g(x)=g(x)$ if $x>0$ and $=0$ otherwise. Note $P_+^*P_+=M(1_{\mathbb{R}_+})$ and $P_+P_+^*=I_{L^2(\mathbb{R}_+)}$.

\begin{Def}
Let $\kappa\in L^\infty(\mathbb{R})$. Then the bounded operator $$W_\kappa:=P_+\mathcal{F}M(\kappa)\mathcal{F}^{-1}P_+^*$$ on $L^2(\mathbb{R}_+)$ is called the Wiener-Hopf operator with symbol $\kappa$. Occasionally  we write $W(\kappa)$ instead of $W_\kappa$.
\end{Def}

Obviously, $\norm{W_\kappa}\le \norm{\kappa}_\infty$, and $W_\kappa$ is self-adjoint for real $\kappa$ with the spectrum contained in $[\operatorname{ess-inf} \kappa, \operatorname{ess-sup}\kappa]$. Actually, as known,  the spectrum equals this interval.

 \begin{Lem}\label{WHMCO}
 Consider the two cases
 \begin{itemize}
 \item[\emph{(a)}] $\kappa\in L^\infty(\mathbb{R})\cap L^2(\mathbb{R})$ and $k:=(2\pi)^{-1/2}\mathcal{F}\kappa$
  \item[\emph{(b)}] $\kappa=\int\operatorname{e}  ^{\i  (\cdot) y}k(y)\d y$ for some $k\in L^1(\mathbb{R})$
 \end{itemize}
 In either case   the  operator $C(k):=\mathcal{F}M(\kappa)\mathcal{F}^{-1}$ on $L^2(\mathbb{R})$
 is bounded with $\norm{C(k)}=\norm{\kappa}_\infty$ and satisfies
 $$\big(C(k)f\big)(x)=\int_{-\infty}^\infty k(x-y)f(y)\d y$$
 In case \emph{(a)} the integral exists  for every $x\in\mathbb{R}$ and determines a continuous function  bounded by $\norm{k}_2\norm{f}_2$.
 \end{Lem}\\
 {\it Proof.} (a) Write $\mathcal{F}^{-1}f=\operatorname{l.i.m.}_{A\to\infty}F_A$ for $F_A(z):=(2\pi)^{-1/2}\int_{-A}^A \operatorname{e}  ^{\i zy} f(y) \d y$. Then  by continuity
$C(k)f=\mathcal{F}M(\kappa)\mathcal{F}^{-1}f=\operatorname{l.i.m.}_{A\to\infty}\mathcal{F}M(\kappa)F_A$. Furthermore,
 $\mathcal{F}M(\kappa)F_A=\operatorname{l.i.m.}_{B\to\infty}G_{A,B}$ with $G_{A,B}(x):= (2\pi)^{-1/2}\int_{-B}^B \kappa(z)\operatorname{e}  ^{-\i xz}\big((2\pi)^{-1/2}\int_{-A}^A \operatorname{e}  ^{\i zy} f(y) \d y\big)\d z$ for $x\in\mathbb{R}$. Then by  Fubini's theorem, $G_{A,B}(x)=\int_{-A}^A K_{x,B}(y)f(y)dy$ for $K_{x,B}(y):=\frac{1}{2\pi}\int_{-B}^B\kappa(z)\operatorname{e}  ^{-\i  z(x-y)}\d z$. As $\operatorname{l.i.m.}_{B\to\infty}K_{x,B}=k\big(x-(\cdot)\big)$, one has $\lim_{B\to\infty}G_{A,B}(x)=\int_{-A}^A k(x-y)f(y)\d y$ for every $x$. Now note $\int|k(x-y)f(y)|\d y\le \norm{k}_2\norm{f}_2$ by Cauchy-Schwarz inequality, whence $\big(C(k)f\big)(x)=\int k(x-y)f(y)\d y$. Moreover, $\phi(x):=\int k(x-y)f(y)\d y$  exists for every $x$ and $2\pi|\phi(x+h)-\phi(x)|^2\le\norm{f}_2^2\int|\big(\operatorname{e}  ^{-\i  xh}-1\big)\kappa(x)|^2\d x\to 0$ for $h\to 0$ by dominated convergence. \\
\hspace*{6mm}  (b) In view of    \cite[(21.31),\,(21.32)]{HS69}   it suffices  to show $\kappa\,\mathcal{F}^{-1}f=\mathcal{F}^{-1}(k\star f)$ for $f\in L^2\cap L^1$. Now,  $k\star f\in L^2\cap L^1$ by \cite[(21.31),\,(21.32)]{HS69}
and hence $\mathcal{F}^{-1}(k\star f)=\kappa\, \mathcal{F}^{-1}f$ by \cite[(21.41)]{HS69}.\qed

In the cases treated in  (\ref{WHMCO}) the Wiener-Hopf operator $W_\kappa$  is the trace  on the subspace $L^2(\mathbb{R}_+)$ of the convolution operator $C(k)$ on $L^2(\mathbb{R})$ and hence has kernel $k$. We apply now (\ref{WHMCO})(a).

\begin{Cor}\label{FPK} As
$\frac{1}{\pi}\operatorname{sinc}=(2\pi)^{-1/2}\mathcal{F}1_{[-1,1]}$ one has
$$K=W(1_{[-1,1]})$$
with its spectrum  contained in $[0,1]$. Note $\norm{\frac{1}{\pi}\operatorname{sinc}}_2=\pi^{-1/2}$.
For $g\in L^2(\mathbb{R}_+)$ the right hand side of \emph{(\ref{WHO})} exists for every $x\in\mathbb{R}$ and $Kg\in L^2(\mathbb{R}_+)$ is bounded continuous  with $\norm{Kg}_2\le\norm{g}_2$ and $\norm{Kg}_\infty\le\pi^{-1/2}\norm{g}_2$.
\end{Cor}

\subsection{An auxiliary Wiener-Hopf operator}\label{AAWHO}
For $\varepsilon>0$   introduce the Wiener-Hopf operator $W_{\kappa_\varepsilon}$ with  kernel $$k_\varepsilon=\textrm{\scriptsize{$\frac{1}{\pi}$}}\operatorname{e}  ^{-\varepsilon |\cdot|}\operatorname{sinc}$$ As $k_\varepsilon$ vanishes exponentially at infinity the Wiener-Hopf method \cite[11.17]{T48} applies and yields a spectral function  of $W_{\kappa_\epsilon}$ for every  value $s\in]0,1[$.  Then the limit $\varepsilon\to 0$ results in a family $q_s$, $s\in]0,1[$ of functions,  which in (\ref{SFKP}) are shown  to be  spectral functions  of $K$.

Following \cite[11.17]{T48} consider the symbol $\kappa_\varepsilon(w)= \int_{-\infty}^\infty\operatorname{e}  ^{\i  wy} k_\varepsilon(y) \d y$ for complex $w:=u+\i  v$, $|v|<\varepsilon$:
\begin{eqnarray*}
2\pi\operatorname{Re} \kappa_\varepsilon(w)&=&\arctan\textrm{\scriptsize{$\left(\frac{u+1}{\varepsilon-v}\right)$}}-\arctan\textrm{\scriptsize{$\left(\frac{u-1}{\varepsilon-v}\right)$}}+\arctan\textrm{\scriptsize{$\left(\frac{u+1}{\varepsilon+v}\right)$}}-\arctan\textrm{\scriptsize{$\left(\frac{u-1}{\varepsilon+v}\right)$}}\\
2\pi\operatorname{Im} \kappa_\varepsilon(w)&=&\textrm{\scriptsize{$\frac{1}{2}$}}\ln\textrm{\scriptsize{$\left(\frac{(u-1)^2+(\varepsilon-v)^2}{(u+1)^2+(\varepsilon-v)^2} \cdot\frac{(u+1)^2+(\varepsilon+v)^2}{(u-1)^2+(\varepsilon+v)^2}\right)$}}
\end{eqnarray*}
In view of the spectrum of $K$ (see (\ref{FPK})) restrict at once $s\in\,]0,1[$. For $\varepsilon>0$ sufficiently small, i.e., $\varepsilon<\cot(\frac{\pi}{2}s)$, the function to be factorized $1-\frac{1}{s}\kappa_\varepsilon$ has  exactly  two  zeros, namely at $\pm\, u_{\varepsilon,s}$ for $u_{\varepsilon,s}:=\big(1-\varepsilon^2+2\varepsilon\cot(\pi s)\big)^{1/2}$, which are real and simple. Put the cut of $\ln$ along the negative real axis. By the factor $R_ {\varepsilon,s}(w):=\frac{w^2+\varepsilon^2}{w^2-u_ {\varepsilon,s}^2}$ the product $S_{\varepsilon,s}:=(1-\frac{1}{s}\kappa_\varepsilon)R_ {\varepsilon,s}$  has no zeros in the strip $\{|v|<\varepsilon\}$ and the variation of $\ln\circ \,S_{\varepsilon,s}$ along the strip is zero. Let
\begin{equation}\label{AWHOF}
T_{\varepsilon,s}(w):=\textrm{\scriptsize{$\frac{1}{2\pi\i  }$}}\int_{-\infty-\i  b}^{\infty-\i  b}\frac{\ln S_{\varepsilon,s}(z)}{z-w}\d z,\quad Q _{\varepsilon,s}(w):=\operatorname{e}  ^{-T_{\varepsilon,s}(w)}\frac{w+\i  \varepsilon}{w^2-u_{\varepsilon,s}^2}
\end{equation}
with $0<b<\varepsilon$ for $v>-b$. Then
\begin{equation}
q _{\varepsilon,s}(x):=(2\pi)^{-1/2}\int_{-\infty+\i  b}^{\infty+\i  b}
\operatorname{e}  ^{-\i  xw}Q _{\varepsilon,s}(w)\d w
\end{equation}
is a spectral function of $W_{\kappa_\epsilon}$ for the spectral value $s\in]0,1[$ (and $q _{\varepsilon,s}(x)=0$ for $x<0$). Moreover,  within the set of  locally integrable functions being $\mathcal{O}(\operatorname{e}  ^{cx})$ for some $c<\varepsilon$ as $x\to \infty$,
$q _{\varepsilon,s}$ is unique up to a complex constant (depending on $s$). Indeed, in case of even kernel a possible polynomial, by which $Q _{\varepsilon,s}$ has to be multiplied, has degree not exceeding $\frac{n}{2}-1$ with $n$ the number of zeros, see \cite[Theorem XVI in Paley,\,Wiener]{WHP31}.

\subsubsection{The limit $\varepsilon\to 0$}
Let $w=u+\i  v\in\mathbb{C}$.
From (\ref{AWHOF}) one obtains $\lim_{\varepsilon\to 0}T_{\varepsilon,s}(w)=\frac{1}{2}\ln\big(\frac{w^2}{w^2-1}\big)-\frac{\i  }{2\pi}\ln(\frac{1}{s}-1)\ln\big(\frac{w-1}{w+1}\big)$ and accordingly we put
\begin{equation}\label{QS}
Q_s(w):=
\textrm{\scriptsize{$\frac{w}{w^2-1}$}}
\exp\textrm{\scriptsize{$\left(-\frac{1}{2}\ln\big(\frac{w^2}{w^2-1}\big)\right)$}}
\exp\textrm{\scriptsize{$\left(\frac{\i  }{2\pi}\ln\big(\frac{1}{s}-1\big)\ln\big(\frac{w-1}{w+1}\big)\right)$}}
\end{equation}

One easily verifies  that $Q_s$ is holomorphic on $\mathbb{C}\setminus[-1,1]$ without zeros, has a cut along $[-1,1]$ and a singularity at $1$ and $-1$, and that
\begin{equation}\label{EQSW}
|Q_s(w)|\le C_s|w^2-1|^{-1/2} \le C_s|u^2-1|^{-1/2} \textrm{\, for \,} C_s:=\big(\textrm{\scriptsize{$\frac{1}{s}-1$}}\big)^{1/2}
\end{equation}
Also one checks that $Q_s^\pm(u):=\lim_{v\to\pm 0}Q_s(u+\i  v)$ exists with
\begin{equation}\label{LQS}
Q_s^\pm(u)=m_s(u)^{\mp 1}|u^2-1|^{-1/2}\exp\textrm{\scriptsize{$\left(\frac{\i  }{2\pi}\ln\big(\frac{1}{s}-1\big)\ln \big|\frac{u-1}{u+1}\big|\right)$}}
\end{equation}
where $m_s(u):=\operatorname{sgn}(u)$ if $|u|>1$ and $m_s(u):=\i  \big(\frac{1}{s}-1\big)^{1/2}$ if $|u|<1$. By (\ref{EQSW}) one has
\begin{equation}\label{LQSI}
Q_s^\pm\in L^{p}(\mathbb{R}) \textrm{ \,for \,} 1<p<2 \textrm{\; \;and \;\;} 1_{[-1,1]}Q_s^\pm\in L^{p}(\mathbb{R})\textrm{ \,for \,} 1\le p<2
\end{equation}

Finally, for $b>0$ and $x\in\mathbb{R}\setminus\{0\}$ define $q_s^\pm(x)\in \mathbb{C}$ by the Cauchy integral
\begin{equation}\label{QSP}
q _s^\pm(x):=(2\pi)^{-1/2}\int_{-\infty\pm\i  b}^{\infty\pm\i  b}
\operatorname{e}  ^{-\i  xw}Q _s(w)\d w
\end{equation}

\subsubsection{Remark on the factorization} Note that (\ref{LQS}) yields
 the factorization
\begin{equation}
1-\textrm{\scriptsize{$\frac{1}{s}$}} 1_{[-1,1]}(u)=Q^-_s(u)/Q^+_s(u), \quad u\in\mathbb{R}\setminus\{-1,1\}
\end{equation}
 for every $s\in]0,1[$, where by (\ref{QS}) $Q^-_s$ and $1/Q^+_s$ is the limit of a  holomorphic function  on the lower and upper half-plane, respectively. Such a factorization of $1-\frac{1}{s}\kappa$ is at the core of  the method for solving (\ref{HWHE}) by Wiener and Hopf and generalized by Krein. In general it is available,   if e.g. $\kappa$ is holomorphic in a strip around the real axis and uniformly square integrable on lines parallel to the real axis (see e.g. \cite[11.17 Lemma]{T48}), or more generally if $\kappa$ is $\alpha$-H\"older continuous ($0<\alpha<1$) \cite[Teorema 2.1.2]{T73}, or if $\kappa$ is the Fourier transform of an integrable function \cite{K62}, see e.g. \cite[Teorema 2.1.4]{T73}.

\subsection{Spectral functions of $K$} We are going to show that $q_s:=q_s^+|_{\mathbb{R}_+}$  are spectral  functions of $K$. We start from their definition  (\ref{QSP}) and forget about their origin in  sec.\,\ref{AAWHO}. For the proof and the ensuing  results it is convenient to consider  also the functions $q_s^-$. Recall $s\in\,]0,1[$ and that $\ln$ denotes the principal branch of the logarithm.

\begin{Pro}\label{MPQS} The integral defining $q_s^\pm(x)$ in \emph{(\ref{QSP})} exists
 in a uniform manner for $s\in[\delta,1[$\,and $x\in]-\infty,-\delta]\cup[\delta,1/\delta]$ for every $\delta\in]0,1[$, and satisfies
\begin{itemize}
\item[\emph{(i)}] $q_s^\pm(x)=0$ for $\pm\, x<0$
\item[\emph{(ii)}] $q_s^+(x)=(2\pi)^{-1/2}\,\frac{1}{s}\int_{-1}^1\operatorname{e}  ^{-\i  xu}Q_s^+(u)\d u$  for $x>0$,\\
$q_s^-(x)=(2\pi)^{-1/2}\,\frac{1}{1-s}\int_{-1}^1\operatorname{e}  ^{-\i  xu}Q_s^-(u)\d u$  for $x<0$
\item[\emph{(iii)}] $q_s^\pm(x)=(2\pi)^{-1/2}\int_{ -\infty}^{ \infty}\operatorname{e}  ^{-\i  xu}Q_s^\pm(u)\d u$  for $x\ne 0$,  more precisely\\ $(2\pi)^{-1/2}\int_{ -A}^A\operatorname{e}  ^{-\i  xu}Q_s^\pm(u)\d u\to q_s^\pm(x)$ as $A\to \infty$ uniformly with respect to $x$ and $s$  in any compact subset of $\mathbb{R}\setminus\{0\}$ and $]0,1[$, respectively
\item[\emph{(iv)}] $\big|q_s^\pm(x)\big|\le(\pi/2)^{1/2}\big(s(1-s)\big)^{-1/2}$ for $x\ne 0$, and
 $q_s^\pm\in L^p(\mathbb{R})$ for $p>2$
\end{itemize}
\end{Pro}
{\it Proof.} We prove the assertions for $q_s^+$. The results for $q_s^-$ follow quite analogously. Put $I(x,s,A):=\int_{-A +\i  b}^{A+\i  b}\operatorname{e}  ^{-\i  xw}Q _s(w)\d w$ for $A>0$.\\

\hspace*{6mm}
(i) By (\ref{EQSW}) one gets $\big|\operatorname{e}  ^{-\i  xw}Q_s(w)\big|  \le C_s \operatorname{e}  ^{xv}(u^2+v^2-1)^{-1/2}$ for $x\in\mathbb{R}$. Now let $x<0$ and consider the contours $L_1,\dots, L_4$ shown in Fig.\,1.

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\vspace{2mm}
Then $\left|\int_{L_2}\operatorname{e}  ^{-\i  xw}Q_s(w)\d w\right|\le C_s \int_0^A \operatorname{e}  ^{xv}(A^2-1)^{-1/2}\d v=C_s(A^2-1)^{-1/2}\frac{1}{x}\big(\operatorname{e}  ^{xA}-1\big)\to 0$ as $A\to\infty$. Similarly, the integral along $L_4$  vanishes for $A\to \infty$. Also $\left|\int_{L_3}\operatorname{e}  ^{-\i  xw}Q_s(w)\d w\right|\le C_s\int_{-A}^A   \operatorname{e}  ^{xA}(u^2+A^2-1)^{-1/2}\d u\le C_s \operatorname{e}  ^{xA}\int_{-A}^A1\d u=2C_sA\operatorname{e}  ^{xA}\to 0$ as $A\to \infty$. Now let $L$ be the closed contour composed by $L_1,\dots, L_4$. Since $\operatorname{e}  ^{-\i  x(\cdot)}Q_s$ is holomorphic in the upper half-plane, one has $0=\int_L\operatorname{e}  ^{-\i  xw}Q_s(w)\d w\to
(2\pi)^{1/2}q _s^+(x)$ as $A\to\infty$. ---
The estimates show that the limit $I(x,s,A)\to 0$ for $A\to\infty$
occurs uniformly with respect to  $x$ and $s$ if $x\le -\delta$ and $\delta\le s<1$ for some $\delta>0$.\\

(ii) Let $x>0$. By holomorphy  the integrals of $\operatorname{e}  ^{-\i  x(\cdot)}Q_s$ along $L$ and $L'$ are equal for every $A>1$. Hence by the same kind of estimations as used in the proof of (i) one obtains
 $I(x,s,A)
\to (2\pi)^{1/2}q _s^+(x)$ for $A\to\infty$ and thus $(2\pi)^{1/2}q _s^+(x)=\int_{L'}\operatorname{e}  ^{-\i  xw}Q_s(w)\d w$. --- Note that the limit occurs uniformly with respect to  $x$ and $s$, for $x$ in a compact set $\subset\mathbb{R}_+$ and $s\in[\delta,1[$ for $\delta>0$.

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Since the  singularities at $\pm1$ are integrable, the integrals along $L_{\pm,\varepsilon}$ vanish if $\varepsilon\to 0$. Hence $(2\pi)^{1/2}q _s^+(x)=\int_{-1}^1\operatorname{e}  ^{-\i  xu}\big(Q^+_s(u)-Q^-_s(u)\big)\d u$. By (\ref{LQS}) this yields the result. \\

(iii) Let $A>1$. Reasoning as in the proof of (i) one has $(2\pi)^{1/2}q _s^+(x)=\int_{L}\operatorname{e}  ^{-\i  xw}Q_s(w)\d w$

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\vspace{2mm}
(see Fig.\,3) independently of $b>0$. Note
$I(x,s,A)
\to\int_{-A}^{A}\operatorname{e}  ^{-\i  xu}Q^+_s(u)\d u$ if $b\to 0$, since $C_s\operatorname{e}  ^{|x|}\big|u^2-1\big|^{-1/2}$ is integrable on $[-A,A]$ and  dominates $\operatorname{e}  ^{-\i  x(u+\i  b)}Q_s(u+\i  b)$  for $b\le 1$ by (\ref{EQSW}). Remember the uniformity of the limit of $I(x,s,A)$ for $A\to\infty$ in (i) and (ii). This implies the result.

(iv) From (i) and (ii) and due to  (\ref{EQSW}) one gets $|q_s^+(x)| \le \frac{1}{s}(2\pi)^{-1/2}\int_{-1}^1|Q_s^+(u)|\d u=(\pi/2)^{1/2}\big(s(1-s)\big)^{-1/2}$.
 According to   \cite[Theorem 74]{T48} the results (iii)  and   (\ref{LQSI}) imply $q_s^+\in L^p(\mathbb{R})$ for $p>2$.\qed

\begin{Cor} \label{EEQS}
$q_s^\pm|_{\mathbb{R}_\pm}$ has an entire extension.
\end{Cor}\\
{\it Proof.} Recall (\ref{MPQS})(ii) and (\ref{LQSI}). Hence by the Paley-Wiener theorem the Fourier-Laplace transform of $\frac{1}{s}1_{[-1,1]} Q_s^\pm$ is the asserted entire extension of $q_s^\pm|_{\mathbb{R}_\pm}$.\qed

\begin{Lem}\label{IKTQS}  Let $x\in\mathbb{R}$. The function $y\mapsto  \frac{1}{\pi}\operatorname{sinc}(x-y)q_s^\pm(y)$ on $\mathbb{R}$ belongs to $L^r(\mathbb{R})$ for $r>\frac{2}{3}$ and hence  is integrable. The function $x\mapsto \frac{1}{\pi}\int_{-\infty}^{\infty} \operatorname{sinc}(x-y)q_s^\pm(y)\d y$ on $\mathbb{R}$ is continuous.
\end{Lem}\\
{\it Proof.} It suffices to consider the case $q_s^+$. Let $k$ equal $\operatorname{sinc}$ or more generally any $k\in\cap_{p>1} L^p(\mathbb{R})$, as e.g. $k_0:\mathbb{R}\to\mathbb{R}$, $k_0(x):=1$ if $|x|\le 2$ and $k_0(x):=\frac{2}{|x|}$ if $|x|>2$.\\
\hspace*{6mm}
 By H\"older's inequality   $\int |q_s^+(x)|^r|k(x-y)|^r\d y\le \big(\int |q_s^+(x)|^{rp''}\d y\big)^{1/p''} \big(\int |k(x-y)|^{rq''}\d y\big)^{1/q''}$ with $1/p''+1/q''=1$. We choose $\frac{1}{r}<q''<\frac{3}{2}$ for $r>\frac{2}{3}$. Then $rq''>1$ and $1-r<1-\frac{1}{q''}=\frac{1}{p''}<\frac{1}{3}$. The latter implies $rp''>2$. Thus by (\ref{MPQS})(iv)  the right hand side of H\"older's inequality is finite. This implies the first part of the assertion.\\
\hspace*{6mm}
Let $f(x):=\int_{-\infty}^{\infty} \operatorname{sinc}(x-y)q_s^+(y)\d y$. Let $h\in\mathbb{R}$ with $|h|\le 1$. Check $\frac{1}{|t+h|}<\frac{2}{|t|}$ for $|t|>2$. Therefore $|\operatorname{sinc}(x+h-y)|<k_0(x-y)$
for all $x,y$. Hence $f(x+h)\to f(x)$ for $h\to 0$ by dominated convergence, whence the second part of the assertion.\qed

We turn to the main result of this section. Recall $q_s=q_s^+|_{\mathbb{R}_+}$ (see (\ref{QSP})).

\begin{The}\label{SFKP}
Let $s\in\,]0,1[$. Then
$$\int_0^\infty \textrm{\scriptsize{$\frac{1}{\pi}$}}\operatorname{sinc}(x-y)\,q_s(y)\d y\,=\,s\,q_s(x) \quad\forall \;x>0$$
Hence $q_s$ is a spectral function of $K$ for the spectral value $s$.
\end{The} \\
{\it Proof.} By (\ref{MPQS})(i) and (\ref{IKTQS}) the integral exists for every $x>0$.  By (\ref{MPQS})(iv) one has $q_s^+\in L^3(\mathbb{R})$, $Q_s^+\in L^{3/2}(\mathbb{R})$. Obviously, $k:=\frac{1}{\pi}\operatorname{sinc}\in L^{3/2}(\mathbb{R})$.
Because of (\ref{MPQS})(iii) the Fourier transform of $Q_s^+$ according to \cite[(4.1.1)]{T48} coincides a.e. with $q_s^+$. Hence the  Fourier transform of $q_s^+$ in the sense of \cite[(4.1.4]{T48}
coincides a.e. with $Q_s^+$. Recall $k(x)=(2\pi)^{-1/2}\int_{-\infty}^\infty  \operatorname{e}  ^{\i  xu}(2\pi)^{-1/2}1_{[-1,1]}(u)\d u$ for all $x\in\mathbb{R}$. Therefore the Fourier transform of $k$ in the sense of \cite[(4.1.1]{T48}
coincides a.e. with $(2\pi)^{-1/2}1_{[-1,1]}$.\\
\hspace*{6mm}
 Apply \cite[Theorem 76]{T48}. Thus $(2\pi)^{-1/2}\int_{-\infty}^\infty \operatorname{e}  ^{-\i  xu}(2\pi)^{-1/2}1_{[-1,1]}(u)Q_s^+(u)\d u=(2\pi)^{-1/2}\int_{-\infty}^\infty k(x-y)q_s^+(y)\d y$ a.e. Hence by (\ref{MPQS})(i),\,(ii) one has $s\,q_s(x)=\int_{0}^\infty k(x-y)q_s(y)\d y$ for almost all $x>0$. Actually equality holds for all $x>0$ since both sides are continuous functions of $x$ by (\ref{EEQS}) and (\ref{IKTQS}).\qed

\subsection{Hilbert space isomorphism associated with spectral functions of $K$}

As we will prove in (\ref{HSISFK}), the spectral functions $q_s$ of $K$ multiplied by the  normalization constants $n(s):=\frac{\i  }{\pi}(2s)^{-1/2}$ gives rise to a kernel $q$ for an integral operator, which determines an Hilbert space isomorphism $V$ from  $L^2(0,1)$ onto  $L^2(\mathbb{R}_+)$.
The following definition considers both cases $q_s=q_s^+|_{\mathbb{R}_+}$ and $q_s^-|_{\mathbb{R}_-}$.
\begin{equation}\label{KSFK}
q_\pm:]0,1[\times \mathbb{R}_\pm\to \mathbb{C}, \quad q_+(s,x):=n(s)q_s^+(x),\;\;q_-(s,x):=-n(1-s)q_s^-(x)
\end{equation}
By (\ref{MPQS})(ii) one has more explicitly
\begin{equation}\label{EKSFK}
q_+(s,x)=\textrm{\scriptsize{$\frac{1}{2}$}}
\pi^{-3/2}s^{-1}(1-s)^{-1/2}\int_{-1}^1\operatorname{e}  ^{-\i  xu}
(1-u^2)^{-1/2}
\exp\textrm{\scriptsize{$\left(\frac{\i  }{2\pi}\ln\big(\frac{1}{s}-1\big)\ln \big(\frac{1-u}{1+u}\big)\right)$}}\d u
\end{equation}
and $q_-(s,x)=q_+(1-s,-x)$. Note that $q_+$ and $q_-$ are real.

\begin{Lem}\label{SHSI}
\begin{itemize}
\item[\emph{(i)}] $R_\pm: L^2(0,1)\to L^2(\mathbb{R},\mathbb{C}^2),\; (R_\pm h)(x):=(2\pi)^{1/2}\textrm{\scriptsize{$\left(\begin{array}{c} l_\pm(s) \\ \pm\, l_\mp(s) \end{array}\right)$}}\,h(s)$ with $l_+(s):=s\,(1-s)^{1/2},\; l_-(s):=l_+(1-s)$ and $s:=\big(1+\operatorname{e}  ^{-2\pi x}\big)^{-1}$ for $x\in\mathbb{R}$,
 is a linear isometry with range $\{f\in L^2(\mathbb{R},\mathbb{C}^2):f_2(x)=\pm \operatorname{e}  ^{\mp\pi x}f_1(x) \operatorname{a.e.}\}.$ If $f$ is in the range of $R_\pm$, then  $(\mathbb{R}_\pm^{-1}f)(s)=(2\pi)^{-1/2}
 l_\pm(s)^{-1}\,f_1\big(\textrm{\scriptsize{$-\frac{1}{2\pi}$}}\ln(\textrm{\scriptsize{$\frac{1}{s}$}}-1)\big)$. \\

\item[\emph{(ii)}] $R:L^2(]0,1[,\mathbb{C}^2)\to L^2(\mathbb{R},\mathbb{C}^2)$, $R\,h:=R_+h_1+R_-h_2$ is a Hilbert space isomorphism, $(R^{-1}f)(s)=(2\pi)^{-1/2}\textrm{\scriptsize{$\left(\begin{array}{cc}(1-s)^{-1/2} & s^{-1/2} \\ s^{-1/2}  & -(1-s)^{-1/2} \end{array}\right)$}}f\big(\textrm{\scriptsize{$-\frac{1}{2\pi}$}}\ln(\textrm{\scriptsize{$\frac{1}{s}$}}-1)\big)$.\\

\item[\emph{(iii)}] $S: L^2(\mathbb{R})\to L^2(\mathbb{R},\mathbb{C}^2)$, $(Sf)(x):=\sqrt{2}\operatorname{e}  ^{x/2}\textrm{\scriptsize{$\left(\begin{array}{c} (1+\operatorname{e}  ^x)^{-1}f\big(  (1-\operatorname{e}  ^x)  (1+\operatorname{e}  ^x)^{-1}\big)\\\i   \,(1-\operatorname{e}  ^x)^{-1}f\big(  (1+\operatorname{e}  ^x)  (1-\operatorname{e}  ^x)^{-1}\big)\end{array}\right)$}}$ is a Hilbert space isomorphism, $(S^{-1}f)(u)=\sqrt{2}\,|1-u^2|^{-1/2} \big(\i  \operatorname{sgn}(u)\big)^{1-j}f_j\big(\ln\big|\frac{1-u}{1+u}\big|\,\big)$ where $j=1$ if $|u|<1$ and $j=2$ if $|u|>1$.
\end{itemize}
\end{Lem}
{\it Proof.} The Hilbert space transformations all arise from change of variable and are easily checked.\qed

See sec.\,\ref{FPOK} for the projection $P_+$ and define $P_-$ analogously.

\begin{The}\label{HSISFK} $V_\pm:L^2(0,1)\to L^2(\mathbb{R}_\pm)$, $$V_\pm h:=\operatorname{l.i.m.}_{0<\alpha,\alpha'\to 0}\int_{\alpha}^{1-\alpha'}q_\pm(s,\cdot)h(s)\d s$$
 are Hilbert space isomorphisms with $$V_+^{-1}g=\operatorname{l.i.m.}_{A\to\infty}\int_0^Aq_+(\cdot,x)g(x)\d x, \quad V_-^{-1}g=\operatorname{l.i.m.}_{A\to\infty}\int_{-A}^0q_-(\cdot,x)g(x)\d x$$
 Moreover, the  formula
$P_\pm^*\,V_\pm= \mathcal{F}^{-1}S^{-1} \big(\mathcal{F}^{-1}\oplus  \mathcal{F}^{-1}\big)R_\pm$ holds.\end{The}\\
{\it Proof.} Put $W_\pm:=\mathcal{F}^{-1}S^{-1} \big(\mathcal{F}^{-1}\oplus  \mathcal{F}^{-1}\big)R_\pm$ and  $f_{A,A'}:=(2\pi)^{-1/2}\int_{-A}^{A'}\operatorname{e}  ^{\i  (\cdot)y}\big(R_+h\big)(y)\d y$ for $h\in L^2(0,1)$ and $A,A'\ge 0$. Then $W_+h=\operatorname{l.i.m.}_{A,A'\to \infty}\mathcal{F}^{-1}S^{-1}f_{A,A'}$. \\
\hspace*{6mm}
For every $x\in\mathbb{R}$, one has
$f_{A,A'}(x)=(2\pi)^{-1}\int_{\alpha}^{1-\alpha'}\textrm{\scriptsize{$\operatorname{e}  ^{-\i  \frac{x}{2\pi}\ln\big(\frac{1}{s}-1\big)}$}}\textrm{\scriptsize{$\left(\begin{array}{c} (1-s)^{-1/2} \\  s^{-1/2} \end{array}\right)$}}\,h(s)\d s$ by the change of variable $s:=\big(1+\operatorname{e}  ^{-2\pi y}\big)^{-1}$. Therefore, $G_{\alpha,\alpha'}(u):=\big(S^{-1}f_{A,A'}\big)(u)=\int_{\alpha}^{1-\alpha'}\overline{n(s)}\overline{Q_s^+(u)}\,h(s)\d s$ for every $u\in\mathbb{R}$. Put $J(x,s,B):=(2\pi)^{-1/2}\int_{-B}^B \operatorname{e}  ^{\i  xu}\overline{Q_s^+(u)}\d u$.\\
\hspace*{6mm}
Then $\mathcal{F}^{-1}G_{\alpha,\alpha'}=\operatorname{l.i.m.}_{B\to\infty}\int_{\alpha}^{1-\alpha'}\overline{n(s)}h(s)
J(\cdot,s,B)\d s$,
where the order of integration has been inverted by Fubini's theorem since $\big|h(s) n(s)Q_s^+(u)\big|\le s^{-1}|h(s)||u^2-1|^{-1/2}$ (see (\ref{EQSW})) is integrable on $[\alpha,1-\alpha']\times[-B,B]$.\\
\hspace*{6mm}
It follows from (\ref{MPQS})(iii),(iv) that for $x\ne 0$ there is $B_x>0$ such that $|J(x,s,B)|\le |q_s^+(x)|+1\le (\pi/2)^{1/2}\big(s(1-s)\big)^{-1/2}+1$ for all $B\ge B_x$ and $s\in[\alpha,1-\alpha']$ and that $\lim_{B\to\infty}J(x,s,B)=\overline{q_s^+(x)}$. Thus
$(\mathcal{F}^{-1}G_{\alpha,\alpha'})(x)=\int_{\alpha}^{1-\alpha'}\overline{n(s)}\overline{q_s^+(x)}h(s)\d s$ a.e. by dominated convergence. Recall (\ref{MPQS})(i) and that $q_+$ is real. Then this proves  $W_+=P_+^*\,V_+$.
 In the same way one proves $W_-=P_-^*\,V_-$.\\
\hspace*{6mm}
By (\ref{SHSI})(ii), $W:L^2(]0,1[,\mathbb{C}^2)\to L^2(\mathbb{R})$, $W\,h:=W_+h_1+W_-h_2$ is an isomorphism. This implies that $V_+$ and $V_-$ are isomorphisms, too.\\
\hspace*{6mm}
It remains to verify the integral representation for $V_+^{-1}$ and similarly that for $V_-^{-1}$.
By the foregoing result on $W$ one has the formula $V_+^{-1}g=\big(W^{-1}P_+^*g\big)_1=R_+^{-1}\mathcal{F}^{(2)}S\mathcal{F}P_+^*g$ for $g\in L^2(\mathbb{R}_+)$, which is evaluated now. Let $A>0$ and $G_A:=(2\pi)^{-1/2}\int_0^A\operatorname{e}  ^{-\i  (\cdot)x}g(x)\d x$. Since $\int_0^A|g(x)|\d x<\infty$, $G_A$ is bounded on $\mathbb{R}$. Clearly $(SG_A)_1\in L^2(\mathbb{R})$. Moreover,
$(SG_A)_1(y)=\sqrt{2}\operatorname{e}  ^{y/2} (1+\operatorname{e}  ^y)^{-1}G_A\big(  (1-\operatorname{e}  ^y)  (1+\operatorname{e}  ^y)^{-1}\big)$, whence $\big|(SG_A)_1(y)\big|\le C\operatorname{e}  ^{y/2} (1+\operatorname{e}  ^y)^{-1}$ and hence $(SG_A)_1
\in L^1(\mathbb{R})$.
So $(2\pi)^{1/2}\big(\mathcal{F}(SG_A)_1\big)(x')=\int \operatorname{e}  ^{-\i  yx'}(SG_A)_1(y)\d y=\sqrt{2}\int_{-1}^1
\exp\textrm{\scriptsize{$\left(-\i  x'\ln \big(\frac{1-u}{1+u}\big)\right)$}}(1-u^2)^{-1/2}G_A(u)\d u$ by the new variable $u:=(1-\operatorname{e}  ^y)  (1+\operatorname{e}  ^y)^{-1}$. Now, Fubini's theorem obviously applies and, doing first the integral on $u$, one obtains at $x'=-(2\pi)^{-1}\ln\big(\frac{1}{s}-1\big)$ using (\ref{EKSFK}) the expression $\big(2\pi s^2(1-s)\big)^{-1/2}\big(\mathcal{F}(SG_A)_1\big)(x') =\int_0^Aq_+(s,x)g(x)\d x$, which equals  $\big(R_+^{-1}\mathcal{F}^{(2)}S\mathcal{F}G_A\big)(s)$
by  (\ref{SHSI})(i). Finally perform the limit in the mean for $A\to \infty$ on both sides.\qed

\hspace*{6mm}
See also (\ref{SGOLBV}) for a determining property of $V_+$.

\subsection{Spectral representation of $K$}\label{SRK}

In the following put  $V:=V_+$.

\begin{The}
$\big(V^{-1}K\,V\,h\big)(s) = s \,h(s) \textrm{ \;a.e.\;}$ for every $h\in L^2(0,1)$.
\end{The}\\
{\it Proof.} Let $h\in L^2(0,1)$ vanish outside some interval $[\alpha,1-\alpha]\subset]0,1[$. The set of these functions is dense in   $L^2(0,1)$. By (\ref{FPK}),\,(\ref{HSISFK}), the proof of (\ref{HSISFK}), and (\ref{MPQS})(i) one has $KVh=P_+\mathcal{F}^{-1}M(1_{[-1,1]})\mathcal{F}P_+^*P_+\mathcal{F}^{-1}G_{\alpha,\alpha}=P_+\mathcal{F}^{-1}M(1_{[-1,1]})G_{\alpha,\alpha}$. Recall that $q_+$ is real. By  (\ref{EQSW}) Fubini's theorem applies so that by (\ref{MPQS})(ii)  $\big(KVh\big)(x)=\int_0^1q_+(s,x)\,s\,h(s)\d s$ for almost all $x>0$. Hence $\big(V^{-1}K\,V\,h\big)(s) = s \,h(s)$ a.e. by (\ref{HSISFK}). The result extends to all $h\in L^2(0,1)$ by continuity.\qed

So $K$ is Hilbert space isomorphic to the multiplication operator $M(\operatorname{id}_{[0,1]})$ by the identity  $\operatorname{id}_{[0,1]}$ on $L^2(0,1)$:
\begin{equation}\label{KHSIJ}
 K\,=\,VM(\operatorname{id}_{[0,1]})V^{-1}
 \end{equation}

Recall  the spectral theorem (see e.g. \cite[12.23 Theorem]{R74}) for a self-adjoint operator $A$, by which  there exists a unique  spectral measure, i.e., a projection valued measure $E_A$ on the Borel sets of $\mathbb{R}$, satisfying $A=\int \lambda \d E_A(\lambda)$. Obviously, $E_{M(\operatorname{id}_{[0,1]})}(\Delta)h=1_{\Delta\cap [0,1]}h$, $h\in L^2(0,1)$, $\Delta\subset \mathbb{R}$ Borel set. So  $E_K(\Delta)=VE_{M(\operatorname{id}_{[0,1]})}(\Delta)V^{-1}$.

\section{Some Operators related to $K$}

\subsection{Causal localization operator}\label{CLOWF} We came across $K$ (\ref{WHO})  when  analyzing    the causal localization  for a Weyl fermion, namely the localization operator $T^{\chi,\eta}(B)$ for the unit ball $B=\{x\in\mathbb{R}^3:|x|\le 1\}$, see (\ref{ALOWF}). One recalls that, as shown in \cite[sections 16, 23]{C18},  the two Weyl fermions ($\eta=+$) and their antiparticles ($\eta=-$) are the only massless relativistic quantum systems which satisfy the conditions imposed by causality. Here
$\chi\in\{+,-\}$ indicates the handedness of the particle, whence $m:=\chi\eta\frac{1}{2}$ its helicity.\\
\hspace*{6mm}
In the course of  the following considerations some useful formulae concerning  the Bessel functions of the first kind $\operatorname{J}_\nu$, $\nu\in\mathbb{C}$ and the related Hankel transformations $\mathcal{H}_\nu$, $\nu\in[-\frac{1}{2},\infty[$ are deduced, see (ii),\,(iii) and ($\alpha$),\,($\beta$) in (\ref{FAGKO}).

In position representation on $L^2(\mathbb{R}^3,\mathbb{C}^2)$ with $\mathbb{R}^3$ the coordinate space $T^{\chi,\eta}$ is the trace of the canonical projection valued measure  on the carrier space of the representation for the respective Weyl fermion, which in energy representation is simply $L^2(\mathbb{R}^2)\oplus\{0\}$ for $m=\frac{1}{2}$
and $\{0\}\oplus L^2(\mathbb{R}^2)$ for $m=-\frac{1}{2}$.\\
\hspace*{6mm}
 Let $T_m$ on $L^2(\mathbb{R}^3)$ denote $T^{\chi,\eta}(B)$ in the energy representation. For $\varphi\in L^2\cap L^1$  one has explicitly
\begin{equation}\label{CLOUB}
\big(T_m\varphi\big)(p)=(2\pi)^{-3/2}\int_{\mathbb{R}^3}D^{(1/2)}_{mm}(B_p^{-1}B_{p'})\big(\mathcal{F}1_B\big)(p-p')\varphi(p')\d p'
\end{equation}
Here $D^{(1/2)}$ is the identity representation of $SU(2)$ and $B_p\in SU(2)$ the helicity cross section satisfying $|p|B_pe_3=p$, $p\in\mathbb{R}^3$. Note $B_p=B_{p/|p|}$ for $p\ne 0$.     In the sequel identify $L^2(\mathbb{R}^3)$  by $p=s\,w$  with $L^2_\rho(\mathbb{R}_+)\otimes L^2_\omega(S^2)$, $\d \rho(s):=4\pi s^2\d s$, $\omega$ the normalized rotational invariant measure on the sphere $S^2$. Moreover consider the Hilbert space isomorphism $$\delta: L^2_\rho(\mathbb{R}_+)\to L^2(\mathbb{R}_+),\quad (\delta g)(s):=\sqrt{4\pi} s g(s)$$
 Recall that by the Peter-Weyl theorem the functions $w\to (2j+1)^{1/2}D^{(j)}_{m m'}(B^{-1}_w)$,  for $j\in |m|+\mathbb{N}_0$ and $m'\in\{-j,-j+1,\dots,j-1,j\}$, form an  ONB for $L^2_\omega(S^2)$. The orthogonal projection onto the subspace determined by fixed $j$ is given by
$$(P^j_mf)(w)=(2j+1)\int_{S^2}D^{(j)}_{mm}(B^{-1}_wB_{w'})f(w')\d \omega(w')$$
Finally by \cite[7.14.2\,(32)]{E53} for $i\in\{0,1\}$ the functions $s\to (4n+2i+1)^{1/2}s^{-1/2}\operatorname{J}_{2n+i+\frac{1}{2}}(s)$, $n\in\mathbb{N}_0$ form an ONS in $L^2(\mathbb{R}_+)$. Let $P(2n+i+\frac{1}{2})$ denote the orthogonal projection onto the one-dimensional subspace spanned by each of these functions.

\begin{The}\label{ALOWF} Let $m=\pm \frac{1}{2}$. Then
$$T_m\simeq K\oplus K\oplus K\oplus\dots$$
i.e., $T_m$ is Hilbert space isomorphic to the countably infinite orthogonal sum of copies of $K$. More precisely,
$$T_m=\bigoplus_{j\in \mathbb{N}_0+1/2} C_j\;\otimes \,P_m^j $$
where
$\delta C_j \delta^{-1}$ equals $K-\textrm{\scriptsize{$\frac{1}{2}$}}\sum_{k=0}^{j-1/2}P(k+\textrm{\scriptsize{$\frac{1}{2}$}})$, which is unitarily equivalent to $K$.
\end{The}

{\it Proof.}  The ingredients for the proof are
\begin{itemize}
\item[(i)] $\big(\mathcal{F}1_B\big)(p-p')=(2/\pi)^{1/2}\int_0^1r^2\operatorname{sinc}(r|p-p'|)\d r$
\item[(ii)] $\operatorname{sinc}(r|p-p'|)=\frac{\pi}{2}\sum_{l=0}^\infty(2l+1)(rsrs')^{-1/2}\operatorname{J}_{l+1/2}(rs) \operatorname{J}_{l+1/2}(rs')\,\operatorname{P}_l\big(\cos\sphericalangle(p,p')
\big)$
\item[(iii)] $(2l+1)\operatorname{P}_l=l\operatorname{P}_{l-1}^{(0,1)}+(l+1)\operatorname{P}_l^{(0,1)}$
\item[(iv)] $D^{(n+1/2)}_{mm}(B^{-1}_wB_{w'})=D^{(1/2)}_{mm}(B^{-1}_wB_{w'})\,\operatorname{P}_n^{(0,1)}\big(\cos\sphericalangle(w,w')\big)$
\item[(v)] $\mathcal{H}_\nu: L^2(\mathbb{R}_+)\to L^2(\mathbb{R}_+)$, $\nu\in[-\frac{1}{2},\infty[$ the unitary involutory Hankel transformation given by $(\mathcal{H}_\nu g)(s)=\int_{\mathbb{R}_+}(ss')^{1/2}\operatorname{J}_\nu(ss')g(s')\d s'$ if $g\in L^2\cap L^1$ and  in general $\mathcal{H}_\nu g=\operatorname{l.i.m.}_{A\to\infty}\mathcal{H}_\nu (1_{[0,A]}g)$.
\end{itemize}
The first (i) is easily checked. For (ii) see Gegenbauer's addition theorem \cite[7.15(30)]{E53} for $\nu=1/2$.
 Item (iii) is \cite[10.8(36)]{E53} for $\alpha=0$, $\beta=1$, since $\operatorname{P}_l=\operatorname{P}_l^{(0,0)}$.
 As to (iv) one recalls that the matrix elements $D^{(j)}_{\sigma\sigma'}(B)$ can be expressed by the Jacobi polynomials $\operatorname{P}^{(\alpha,\beta)}_n$. In particular
$D^{(j)}_{\sigma\sigma}(B)$ equals $B_{11}^{2|\sigma|}\operatorname{P}^{(0,2|\sigma|)}_{j-|\sigma|}\big(2|B_{11}|^2-1\big)$ or  its complex conjugate depending on whether $\sigma\ge 0$ or $\sigma\le 0$. By this formula one obtains (iv). \\
\hspace*{6mm}
It suffices to evaluate (\ref{CLOUB}) for $\varphi=g\times f $ with integrable $g\in L^2_\rho(\mathbb{R}_+)$, $f\in L^2_\omega(S^2)$. Employing (ii), (iii), and (iv),  (\ref{CLOUB}) becomes
$\big(T_m\varphi\big)(p)=\int_0^\infty\d s'\int_{S^2}\d \omega(w')\int_0^1\d r \times \sum_{l=0}^\infty\, r^2\, F_l(rs,rs')\,\big(l\,D^{(l-1/2)}_{mm}(B^{-1}_wB_{w'})\,+\,(l+1)\,D^{(l+1/2)}_{mm}(B^{-1}_wB_{w'})\big)\,s'^2\,g(s')\,f(w')$, where
$$ F_l(x,x'):=(xx')^{-1/2}\operatorname{J}_{l+1/2}(x)\operatorname{J}_{l+1/2}(x'), \quad x,x'\in\mathbb{R}_+$$
The integrand is dominated by $\sum_{l=0}^\infty (2l+1)|F_l(rs,rs')|\,1_{[0,1]}(r)\,s'^2g(s')f(w')$. Formula (ii) yields the special case $2/\pi=\sum_{l=0}^\infty (2l+1)F_l(x,x)$,
whence $\sum_{l=0}^\infty (2l+1)|F_l(rs,rs')|\le 2/\pi$ independent of $r,s,s'$. Hence summation and  integrations may be interchanged. Doing first the integral for $\omega$ one obtains $\big(T_m\varphi\big)(p)=\sum_{l=0}^\infty\int_0^1\d r\int_0^\infty\d s' r^2\frac{1}{2}\big(F_l(rs,rs')+F_{l+1}(rs,rs')\big)s'^2g(s')\,(P^{l+1/2}_mf)(w)$. By (iv)  the remaining integrations yield
$T_mg\times f=\sum_{l=0}^\infty \delta^{-1}K_{l+1/2}\delta \,g\otimes P^{l+1/2}_mf$ with
\begin{equation}\label{GOK}
K_{l+1/2}:=\textrm{\scriptsize{$\frac{1}{2}$}}\big(\mathcal{H}_{l+1/2}M(1_{[0,1]})\mathcal{H}_{l+1/2}+\mathcal{H}_{l+3/2}M(1_{[0,1]})\mathcal{H}_{l+3/2}\big)
\end{equation}
Here $M(1_{[0,1]})$ denotes the multiplication operator in $L^2(\mathbb{R}_+)$.\\
\hspace*{6mm}
 $K_{l+1/2}$, $l\in\mathbb{N}\cup\{-1,0\}$ is an integral operator on $L^2(\mathbb{R}_+)$ with kernel
$$k_{l+1/2}(r,t)=\textrm{\scriptsize{$\frac{1}{2}$}}\frac{(rt)^{1/2}}{r-t}\big(\operatorname{J}_{l+3/2}(r)\operatorname{J}_{l+1/2}(t)-\operatorname{J}_{l+1/2}(r)\operatorname{J}_{l+3/2}(t)\big)$$
The latter follows evaluating $(rt)^{1/2}\int_0^1s\operatorname{J}_j(rs)\operatorname{J}_j(st)\d s$ for $j=l+1/2$ and $j=l+3/2$ by \cite[7.14.1(8)]{E53}. Note that $(r-t)k_{l+1/2}(r,t)$ is bounded on $\mathbb{R}_+\times\mathbb{R}_+$ due to \cite[7.2(2)]{E53} for small and \cite[7.13.1(3)]{E53} for large variable $s$ of $s^{1/2}\operatorname{J}_{l+1/2}(s)$. So by \cite[Theorem 6.17.1]{Pu67} the operator
$K_{l+1/2}$ is absolutely continuous. Now (\ref{FAGKO})(i),(iii) will show $K=K_{-1/2}$ and
\begin{equation}\label{FDPOK}
K_{l+1/2}=K-\textrm{\scriptsize{$\frac{1}{2}$}}\sum_{k=0}^{l}P(k+\textrm{\scriptsize{$\frac{1}{2}$}})
\end{equation}
which is a a finite-dimensional perturbation of $K$. Hence by \cite[Theorem 6.20.1]{Pu67} all $\delta C_j \delta^{-1}=K_{l+1/2}$ are unitarily equivalent to $K$.\qed

Recall  that $\mathcal{H}_{-1/2}=\mathcal{F}_c$ and  $\mathcal{H}_{1/2} =\mathcal{F}_s$ are the involutory Fourier cosine and  Fourier sine transformation.

\begin{Lem}\label{FAGKO}
Let  $n\in\mathbb{N}_0$, $i\in\{0,1\}$. Then
\begin{itemize}
\item[\emph{(i)}] $K=\textrm{\scriptsize{$\frac{1}{2}$}}\big(\mathcal{F}_cM(1_{[0,1]})\mathcal{F}_c+\mathcal{F}_sM(1_{[0,1]})\mathcal{F}_s\big)$
\item[\emph{(ii)}] $\mathcal{F}_cM(1_{[0,1]})\mathcal{F}_c=\bigoplus_{n\in\mathbb{N}_0} P(2n+1/2)$,\quad $\mathcal{F}_sM(1_{[0,1]})\mathcal{F}_s=\bigoplus_{n\in\mathbb{N}_0} P(2n+3/2)$
\item[\emph{(iii)}] $\mathcal{H}_{2n+i+1/2}M(1_{[0,1]})\mathcal{H}_{2n+i+1/2}=\mathcal{H}_{1/2-i}M(1_{[0,1]})\mathcal{H}_{1/2-i}-\bigoplus_{k=0}^{n-1+i}P(2k-i+3/2)$
\end{itemize}
\end{Lem}
{\it Proof.} (i) is the special case $K_{-1/2}=K$ of (\ref{GOK}) for $l=-1$ as $k_{-1/2}(r,t)=\frac{1}{\pi}\operatorname{sinc}(r-t)$. As to (ii) specialize the formula (ii) in the proof of (\ref{ALOWF}) to the case of parallel and antiparallel $p,p'$. Note $\operatorname{P}_l(1)=1$, $\operatorname{P}_l(-1)=(-1)^l$. One obtains \begin{itemize}
\item $\frac{1}{\pi}\big(\operatorname{sinc}(x-x')+\operatorname{sinc}(x+x')\big)=\sum_{l=0}^\infty(4l+1)F_{2l}(x,x')$\\
\item $\frac{1}{\pi}\big(\operatorname{sinc}(x-x')-\operatorname{sinc}(x+x')\big)=\sum_{l=0}^\infty(4l+3)F_{2l+1}(x,x')$
\end{itemize}
(for $F_l$ see the proof of (\ref{ALOWF})).
This is (ii) for the  kernels of the relative integral operators.

\hspace*{6mm}
 For the proof of (iii) we first deduce the  formula
 \begin{itemize}
 \item[($\alpha$)] $\frac{\d }{\d z}\sum_{k=n}^{n'}2(\nu+2k)(ab)^{-1}\operatorname{J}_{\nu+2k}(az)\operatorname{J}_{\nu+2k}(bz)=$
 \item[] \begin{flushright}
 $z\big(\operatorname{J}_{\nu+2n-1}(az)\operatorname{J}_{\nu+2n-1}(bz)-\operatorname{J}_{\nu+2n'+1}(az)\operatorname{J}_{\nu+2n'+1}(bz)\big)$
 \end{flushright}
 \end{itemize}
valid for $\nu, z, a,b\in\mathbb{C}$, $ab\ne 0$, $n,n'\in\mathbb{Z}$, $n\le n'$ using the relation
 \begin{itemize}
 \item[($\beta$)] $2\nu\big(z_1\operatorname{J}_{\nu}'(z_1)\operatorname{J}_{\nu}(z_2)+z_2\operatorname{J}_{\nu}(z_1)\operatorname{J}_{\nu}'(z_2)\big)=
 z_1z_2\big(\operatorname{J}_{\nu-1}(z_1)\operatorname{J}_{\nu-1}(z_2)-\operatorname{J}_{\nu+1}(z_1)\operatorname{J}_{\nu+1}(z_2)\big)$
 \end{itemize}
which follows from \cite[7.2.8(54),(55),(56)]{E53} as $2\nu\big(z_1\operatorname{J}_{\nu}'(z_1)\operatorname{J}_{\nu}(z_2)+z_2\operatorname{J}_{\nu}(z_1)\operatorname{J}_{\nu}'(z_2)\big)=2\nu\Big(\big(z_1\operatorname{J}_{\nu-1}(z_1)-\nu\operatorname{J}_{\nu}(z_1)\big)\operatorname{J}_{\nu}(z_2)+\operatorname{J}_{\nu}(z_1)\big(-z_2\operatorname{J}_{\nu+1}(z_2)+\nu\operatorname{J}_{\nu}(z_2)\big)\Big)=2\nu\big(z_1\operatorname{J}_{\nu-1}(z_1)\operatorname{J}_{\nu}(z_2)-z_2\operatorname{J}_{\nu}(z_1)\operatorname{J}_{\nu+1}(z_2)\big)=z_1z_2\Big(\operatorname{J}_{\nu-1}(z_1)\big(\operatorname{J}_{\nu-1}(z_2)+\operatorname{J}_{\nu+1}(z_2)\big)-\big(\operatorname{J}_{\nu-1}(z_1)+\operatorname{J}_{\nu+1}(z_1)\big)\operatorname{J}_{\nu+1}(z_2)\Big)=z_1z_2\big(\operatorname{J}_{\nu-1}(z_1)\operatorname{J}_{\nu-1}(z_2)-\operatorname{J}_{\nu+1}(z_1)\operatorname{J}_{\nu+1}(z_2)\big)$. Now, to the left hand side of ($\alpha$) apply the product rule of differentiation and subsequently use ($\beta$).\\
\hspace*{6mm}
In ($\alpha$) choose $n=0$, $n'=n-1+i$, $\nu=3/2-i$, $a=r$, $b=t$, $z=s$. Integrate ($\alpha$) with respect to $s\in[0,1]$ and note $\operatorname{J}_{\nu}(0)=0$ for $\nu\in\mathbb{R}_+$ according to \cite[7.2.1(2)]{E53}. One obtains the formula (iii) for the  kernels of the relative integral operators.\qed

\subsection{Toeplitz operators isomorphic to $K$}\label{TOIK} Using well-known transformations like that indicated in  \cite[chap.\,9]{BS90} (cf.\,also \cite[sec.\,4]{R65}) we obtain   in (\ref{HSITO}) Toeplitz operators, which are  Hilbert space   isomorphic to $K$, and  in (\ref{SRTO}) their spectral representation. The latter has been  already derived in \cite{R61} for the corresponding discrete Wiener-Hopf operators. In (\ref{GOLBV}) the integral transformation effecting the spectral representation is shown to map the standard basis of the Hardy space
onto the orthonormal basis (\ref{MPPCV}) related to the Meixner-Pollaczek polynomials by a change of variable.

\subsubsection{Toeplitz operators}
 Let $\mathbb{T}\subset \mathbb{C}$ denote the unit circle endowed with the normalized Lebesgue measure. The Hardy space $H^2(\mathbb{T})$ is the closed subspace of $L^2(\mathbb{T})$ with orthonormal basis $e_n,\ n\in\mathbb{N}_0$, where $e_n(z):=z^n$. Let $Q_+:L^2(\mathbb{T})\to H^2(\mathbb{T})$ be the orthogonal projection. The adjoint $Q_+^*$ is the inclusion $H^2(\mathbb{T})\hookrightarrow L^2(\mathbb{T})$. Then for any $\omega\in L^\infty(\mathbb{T})$ consider the multiplication operator
$M(\omega)$ by $\omega$ in $L^2(\mathbb{T})$ and define the Toeplitz operator
\begin{equation}\label{TOETS}
T_\omega:=Q_+\,M(\omega)\,Q_+^*
\end{equation}
in $H^2(\mathbb{T})$. Occasionally  we write $T(\omega)$ instead of $T_\omega$.

\subsubsection{Hilbert space isomorphisms transforming Wiener-Hopf operators into Toeplitz operators}
The M\"{o}bius transformations mapping $\mathbb{R}$ to  $\mathbb{T}$ are $\chi(x):=\operatorname{e}  ^{\i  \alpha}\frac{x+\overline{d}}{x+d}$ for $\alpha\in\mathbb{R}$, $d\in\mathbb{C}\setminus\mathbb{R}$, as e.g.\,the Cayley transformation for $\alpha=0$ and $d=\i  $. Each transformation gives rise to a Hilbert space isomorphism $X:L^2(\mathbb{R})\to L^2(\mathbb{T})$,
\begin{equation}\label{HSIRHT}
(Xf)(z):=\,(2\pi)^{1/2}\operatorname{e}  ^{\i  \alpha}\,\textrm{\scriptsize{$\frac{\sqrt{2|\operatorname{Im} d|}}{-z+\operatorname{e}  ^{\i  \alpha}}$}} f\big(\chi^{-1}(z)\big)
\end{equation}
 Note $\chi^{-1}(z)=\frac{dz-
\operatorname{e}  ^{\i  \alpha}\overline{d}}{-z+\operatorname{e}  ^{\i  \alpha}}$. Its inverse  is
$(X^{-1} h)(x)=(2\pi)^{-1/2}\i  \operatorname{sgn}(\operatorname{Im} d)\,\frac{\sqrt{2|\operatorname{Im} d|}}{x+d}\,h\big(\chi(x)\big)$.    It is easy to verify
$$XM(\kappa)X^{-1}=M(\kappa\circ\chi^{-1})$$
for $\kappa\in L^\infty(\mathbb{R})$.

 Recall that $P:=P^*_+P_+=M(1_{[0,\infty[})$ is the orthogonal projection in $L^2(\mathbb{R})$ with range $L^2(\mathbb{R}_+)$.
 Similarly $Q:=Q^*_+Q_+$ is the orthogonal projection in $L^2(\mathbb{T})$ with range $H^2(\mathbb{T})$.

As to  (\ref{MTRTOWHO}) see \cite[9.1(e),\,(f), 9.2(d), 9.5(e)]{BS90} regarding $U=-\i  X^{-1}$ for $\alpha=\pi$, $d=\i  $.

\begin{The}\label{MTRTOWHO}  Let $\operatorname{Im} d>0$. Then $X^{-1}QX=\mathcal{F}^{-1}P\mathcal{F}$ holds and $X_+: L^2(\mathbb{R}_+)\to H^2(\mathbb{T})$,  $X_+:=Q_+X\mathcal{F}^{-1}P_+^*$
is a Hilbert space isomorphism satisfying $$T(\kappa\circ \chi^{-1})=X_+W_\kappa X_+^{-1}$$
for $\kappa\in L^{\infty}(\mathbb{R})$.
\end{The}\\
{\it Proof.} Put $f_n:=X^{-1}e_n$ for $n\in\mathbb{Z}$. Then  $X^{-1}QXf_n$ equals $f_n$ if $n\ge 0$ and is $0$ for $n<0$. We are going to show that $\mathcal{F}f_n$ vanishes on $]-\infty,0[$ for  $n\ge 0$ and vanishes on $]0,\infty[$ for $n< 0$, thus implying the first part of the assertion. Note $f_n(x)= \frac{C}{x+d}\chi(x)^n$ with $C$ constant. Hence $\big(\mathcal{F}f_n\big)(x)$ equals the Cauchy integral $C(2\pi)^{-1/2}\int_{-\infty}^\infty \operatorname{e}  ^{-\i  xy} \frac{1}{y+d}\chi(y)^n\d y$, which after a linear change of variable becomes $C'\int_{-\infty}^\infty \operatorname{e}  ^{-\i  \delta xy} \frac{1}{y+i}\big(\frac{y-\i  }{y+\i  }\big)^n\d y$ with $\delta:=\operatorname{Im} d>0$ and $C':=(2\pi)^{-1/2}\operatorname{e}  ^{\i  (n\alpha+x\gamma)}$, $\gamma:=
\operatorname{Re} d$. Note that for $n\ge 0$ there are no residues in the upper half-plane and  that for $n<0$ there are none in the lower-half plane. For $x<0$ the integral can be evaluated closing the path along the real line from $-A$ to $A$ by the half-circle in the upper half-plane and performing the limit $A\to\infty$. Similarly for $x>0$ one closes the path  by the half-circle in the lower half-plane. So the result holds by the residue theorem.\\
\hspace*{6mm}
Hence one has $QX\mathcal{F}^{-1}=X\mathcal{F}^{-1}P$, whence  $X_+^*X_+=I_{L^2(\mathbb{R}_+)}$,  $X_+X_+^*=I_{H^2(\mathbb{T}}$  and the last part of the assertion.\qed

Recall that $l_q(x):= (1-q)^{-1}\exp\big(-\frac{1}{2}\frac{1+q}{1-q}\, x\big)=\sum_{n=0}^\infty \operatorname{l}_n(x)q^n$, $|q|<1$ is  the generating function of the Laguerre functions $l_n$, $n\ge 0$, which form an orthonormal basis in $L^2(\mathbb{R}_+)$. Similarly $e_q(z):=\frac{1}{1-qz}$  generates  the standard basis $(e_n)_{n\ge 0}$ of $H^2(\mathbb{T})$.

As to (\ref{ILFUIXP}) cf.\,\cite[sec.\,4,\,iv)]{R65} regarding $X_+$ for $\alpha=0$, $d=\frac{\i  }{2}$.

\begin{Lem}\label{ILFUIXP} Consider $X$ in \emph{(\ref{HSIRHT})} for $\operatorname{Im} d>0$. If $\alpha=0$, $d=\frac{\i  }{2}$
one has  $X_+l_n=e_n$, $n\ge 0$. Generally $$X_+l_n(z)=\i  \operatorname{e}  ^{\i  \alpha}\sqrt{2\operatorname{Im} d}\,\big((d-\textrm{\scriptsize{$\frac{\i  }{2}$}})z-\operatorname{e}  ^{\i  \alpha}(\overline{d}-\textrm{\scriptsize{$\frac{\i  }{2}$}})\big)^{-1}e_n\big(\mu^{-1}(z)\big)$$ holds, where the M\"obius transformation $\mu(z):=\operatorname{e}  ^{\i  \alpha}\frac{(\overline{d}-\textrm{\scriptsize{$\frac{\i  }{2}$}})z-\overline{d}-\textrm{\scriptsize{$\frac{\i  }{2}$}}}{(d-\textrm{\scriptsize{$\frac{\i  }{2}$}})z-d-\textrm{\scriptsize{$\frac{\i  }{2}$}}}$ maps $\mathbb{T}$ onto itself.
\end{Lem}\\
{\it Proof.} One easily computes $\mathcal{F}^{-1}P_+^*l_q(x)=(2\pi)^{-1/2}\frac{\i  }{x+\frac{\i  }{2}}\,e_q\Big(\frac{x-\frac{\i  }{2}}{x+\frac{\i  }{2}}\Big)=X_0^{-1}e_q(x)$, where the subscript $0$ denotes the case $\alpha=0, d=\frac{\i  }{2}$.
Therefore $X_+l_n=Q_+X\mathcal{F}^{-1}P_+^*l_n=Q_+XX_0^{-1}e_n$, $n\ge 0$, whence the result.\qed

\subsubsection{Toeplitz operators isomorphic to $K$ and their spectral representations}\label{TOITK} Let $A$  denote the  arc $\{\operatorname{e}  ^{\i  \varphi}:\alpha+\beta\le\varphi\le\alpha+2\pi-\beta\}$
for  $\alpha\in [0,2\pi[$, $\beta\in]0,\pi[$.
One easily verifies $1_A=1_{[-1,1]}\circ\chi^{-1}$ for $\chi(x)=\operatorname{e}  ^{\i  \alpha}\frac{x-\i  \tan(\beta/2)}{x+\i  \tan(\beta/2)}$, \,$\chi^{-1}(z)= \i  \tan(\beta/2)\frac{z+\operatorname{e}  ^{\i  \alpha}}{-z+\operatorname{e}  ^{\i  \alpha}}$,
whence by (\ref{MTRTOWHO})
\begin{equation}\label{HSITO}
 T(1_A)=X_+\,K\,X_+^{-1}
\end{equation}

Recall the Hilbert space isomorphism $V=V_+$ in (\ref{HSISFK}) providing the spectral representation  of $K$  (\ref{KHSIJ}).
Then obviously the Hilbert space isomorphism $Y:=X_+V: L^2(0,1)\to H^2(\mathbb{T})$ yields the spectral representation
\begin{equation}\label{SRTO}
 T(1_A)=Y\,M(\operatorname{id}_{[0,1]})\,Y^{-1}
\end{equation}

\hspace*{6mm} For $n\in\mathbb{N}_0$ let $h_n:]0,1[\to \mathbb{R}$
 \begin{equation}\label{MPPCV}
h_n(s):=\operatorname{e}  ^{\i  n\alpha}\Big(\textrm{\scriptsize{$\frac{\sin\beta}{\pi}$}}\Big)^{1/2}\big(\textrm{\scriptsize{$\frac{1}{s}$}}-1\big)^{\beta/(2\pi)}(1-s)^{-1/2}\operatorname{P}_n^{(1/2)}\Big(\textrm{\scriptsize{$\frac{1}{2\pi}$}}\ln\big(\textrm{\scriptsize{$\frac{1}{s}$}}-1\big);\beta\Big)
\end{equation}
where $\operatorname{P}_n^{(1/2)}(x;\beta)$ is the $n$th Meixner-Pollaczek polynomial of order $1/2$.  The generating function of the  polynomials is $p_q(x):=(1-q\operatorname{e}  ^{\i  \beta})^{-1/2 + \i  x}(1-q\operatorname{e}  ^{-\i  \beta})^{-1/2 - \i  x}$. Orthogonality   and completeness of the polynomials $\operatorname{P}_n^{(1/2)}$ on $\mathbb{R}$ carry over  to the functions $h_n$ by the change of variable $x=\frac{1}{2\pi}\ln\big(\frac{1}{s}-1\big)$ so that the latter form an orthonormal basis of $L^2(0,1)$.

As to (\ref{GOLBV}) cf.\;the result in \cite[Example 3]{R61}.

\begin{The}\label{GOLBV}\quad  $Y^{-1}e_n=h_n$,   $n\in\mathbb{N}_0$.
\end{The}\\
{\it Proof.} Note $Y^{-1}=V^{-1}X_+^{-1}$. Use the abbreviation $y:=\frac{1}{2\pi}\ln\big(\textrm{\scriptsize{$\frac{1}{s}$}}-1\big)$ and put $t(u):=\ln\big(\frac{1-u}{1+u}\big)$. Then by (\ref{KSFK}), (\ref{EKSFK}), and due to (\ref{MPQS})(i) one has $\sqrt{2}\,\pi s\sqrt{1-s}\,(Y^{-1}h)(s)=(2\pi)^{-1/2}\operatorname{l.i.m.}\int_{-\infty}^{\uparrow\infty}\Big(\int_{-1}^1\operatorname{e}  ^{-\i  xu}(1-u^2)^{-1/2}\operatorname{e}  ^{\i  yt(u)}\d u\Big)(\mathcal{F}X^{-1}h)(x)\d x$.  Let the transform $X^{-1}h$ of $h\in H^2(\mathbb{T})$ have compact support. Then the integrations can be interchanged. Use $\mathcal{F}^2f(u)=f(-u)$. It follows  for all $h\in H^2(\mathbb{T})$
\begin{equation*}(Y^{-1}h)(s)=\big(\sqrt{ 2}\,\pi \,s\sqrt{1-s}\big)^{-1}  \operatorname{l.i.m.} \textrm{\scriptsize{$\int_{\downarrow-1}^{\uparrow 1}$}}(1-u^2)^{-1/2}\operatorname{e}  ^{-\i  yt(u)}(X^{-1}h)(u)\d u \tag{$\star$}
\end{equation*}
Now we evaluate ($\star$) for $h=e_q$.  We treat the case $\alpha=0$. For the general case simply substitute  $q$ with $q\operatorname{e}  ^{\i  \alpha}$. Put $\delta:=\tan\frac{\beta}{2}>0$ and $C(s):=\i  \sqrt{2\delta}\,\big( 2\,\pi^{3/2} \,s\sqrt{1-s}\big)^{-1}$. Perform the change of variable $t=t(u)$. One obtains $Y^{-1}e_q(s)=C(s)\int_{-\infty}^\infty \operatorname{e}  ^{-\i  yt}\frac{\operatorname{e}  ^{t/2}}{c_q-\overline{c}_q\operatorname{e}  ^t}\d t$ for $c_q:=1-q+\i  \delta(1+q)$ and $q\in]-1,1[$.  Note $\overline{c}_q/c_q=\operatorname{e}  ^{\i  \gamma}$ for some $\gamma\in]-\pi,0[$. Hence the residue theorem yields for all $y\in\mathbb{R}$
$$\int_{-\infty}^\infty \operatorname{e}  ^{-\i  yt}\frac{\operatorname{e}  ^{t/2}}{1-\operatorname{e}  ^{\i  \gamma}\operatorname{e}  ^t}\d t=-2\pi\i  \frac{\operatorname{e}  ^{-\gamma(y+\i  /2)}}{1+\operatorname{e}  	^{2\pi y}}$$
Simple computations show
$c_q=\sec\beta/2\,\operatorname{e}  ^{\i  \beta/2}\big(1-q\operatorname{e}  ^{-\i  \beta}\big)$, $\operatorname{e}  ^{\i  \gamma}=\operatorname{e}  ^{-\i  \beta}
\frac{1-q\operatorname{e}  ^{\i  \beta}}{1-q\operatorname{e}  ^{-\i  \beta}}$, whence
$\operatorname{e}  ^{-\gamma(y+\i  /2)}=\Big(\frac{1-q\operatorname{e}  ^{\i  \beta}}{1-q\operatorname{e}  ^{-\i  \beta}}\Big)^{-1/2+\i  y}\operatorname{e}  ^{\i  \beta/2}\operatorname{e}  ^{\beta y}$. Also note that  $|\operatorname{arg}(1+w)|\le\frac{\pi}{2}$ if $|w|<1$. Therefore putting together the terms one obtains $Y^{-1}e_q(s)=\big(\frac{\sin\beta}{\pi}\big)^{1/2}\big(\frac{1}{s}-1\big)^{\beta/(2\pi)}(1-s)^{-1/2}p_q\big(\frac{1}{2\pi}\ln(\frac{1}{s}-1)\big)$, whence the result.\qed\\

 Performing in ($\star$) the change of variable $z=\chi(u)$  one gets the integral transformation
 \begin{equation}
 (Y^{-1}h)(s)=\i  (2\tan\beta/2)^{1/2} (\pi s)^{-1/2}\int_A\frac{Q\big(1-2s,\chi^{-1}(z)\big)}{-z+\operatorname{e}  ^{\i  \alpha}}h(z)\d z
 \end{equation}
with $Q$ from (\ref{IKFHT}) for $h\in H^2(\mathbb{T})$ with compact support within the open arc. Because of (\ref{SRTO}) the kernel of this transformation yields  normalized spectral functions of $T(1_A)$ for every spectral value $s\in]0,1[$.\\

\hspace*{6mm}
Consider the special case $\alpha=0$ and $\tan\frac{\beta}{2}=\frac{1}{2}$. Then (\ref{GOLBV}) yields by (\ref{ILFUIXP})
\begin{equation}\label{SGOLBV}
V^{-1}\operatorname{l}_n=h_n
\end{equation}
and, more precisely,
$ \int_0^\infty q_+(s,x)\operatorname{l}_n(x)\d x=h_n(s) \quad\forall \;s\in]0,1[,\; n\in\mathbb{N}_0$.

\subsection{Finite Hilbert transformation}\label{MEFHT} The following consideration show
a close relationship  between  $K$ and the finite Hilbert transformation on $[-1,1]$.

\subsubsection{Hilbert transformation} The spectral representation of the  finite Hilbert transformation on $L^2(a,b)$, $-\infty<a<b<\infty$
\begin{equation}\label{FHT}
(H_{[a,b]}\,g)(x):=\textrm{\scriptsize{$\frac{1}{\i  \pi}$}}\int_{a}^b\frac{g(y)}{y-x}\d y
\end{equation}
(in the sense of the principal value at $x$) is given in  \cite{KP59}.  See also \cite[Theorem 6.20.1]{Pu67} for another approach.
  In \cite[Lemma 2.1]{KP59}   spectral functions of $H_{[a,b]}$ are found by solving a related boundary value problem. Then, by the Hilbert space isomorphism $U_{a,b}$  associated to the former obtained in \cite[Theorem 3.1]{KP59},   the representation
\begin{equation}\label{SRFHT}
H_{[a,b]}=\,U_{a,b}\,M(\operatorname{id}_{[-1,1]})\,U_{a,b}^{-1}
\end{equation}
is shown in \cite[Theorem 3.2]{KP59}). Here $M(\operatorname{id}_{[-1,1]})$ denotes the multiplication operator on $L^2(-1,1)$ by the identity.\\
\hspace*{6mm}
Recall  the well-known  representation of the  Hilbert transformation  $H$ ($a=-\infty$, $b=\infty$)
\begin{equation}\label{SPHT}
H=\mathcal{F}^{-1}\,  M(\operatorname{sgn})\,\mathcal{F}= -\mathcal{F}\,  M(\operatorname{sgn})\,\mathcal{F}^{-1}
\end{equation}
where $M(\operatorname{sgn})$ denotes the multiplication operator on $L^2(\mathbb{R})$ by the signum function. See e.g.\,\cite[Theorems 91,\,95]{T48} or \cite[Lemma 1.35]{D79}, and for more details \cite[Teorema 1.1.1]{T73}. Hence
\begin{equation}\label{FHTFR}
H_{[a,b]}=\,P_{[a,b]}\,\mathcal{F}^{-1}\,M(\operatorname{sgn})\,\mathcal{F}\,P_{[a,b]}^*
\end{equation}
where $P_{[a,b]}:L^2(\mathbb{R})\to L^2(a,b)$ denotes the projection $\big(P_{[a,b]}f\big)(u):=f(u)$. Its adjoint $P_{[a,b]}^*$ is the injection $\big(P_{[a,b]}^*k\big)(x)=k(x)$ if $a\le x\le b$ and $0$ otherwise. Analogous formulae hold for the semi-finite transformations  $H_{]-\infty,b]}$ and $H_{[a,\infty[}$. \\
 \hspace*{6mm}
In the following $a=-1$, $b=1$ is assumed. This is no restriction as $H_{[a,b]}$  is obtained by an obvious linear change of variable. Even the semi-finite transformations are Hilbert space isomorphic to the finite one using a fractional linear change of variable as shown in \cite[sec.\,4]{KP59}. Indeed, apply  the Hilbert space isomorphism $\Gamma:L^2(a,b)\to L^2(-1,1)$, $(\Gamma g)(t):=\sqrt{|\gamma'(t)|}\,g(\gamma(t))$, $\gamma:]-1,1[\to ]a,b[$ with $\gamma(t):= \frac{b-a}{2}t+  \frac{b+a}{2}$ if $a,b$ are finite, and $\gamma(t):=b+\frac{t-1}{t+1}$ if $a=-\infty$, $\gamma(t):=a+\frac{1+t}{1-t}$ if $b=\infty$.

\subsubsection{Spectral analysis of $H_{[-1,1]}$}

  We are going to derive the spectral analysis of $H_{[-1,1]}$ from  the foregoing analysis of $K$, comprising  the spectral representation
 (\ref{SRFHT}), which is  shown in (\ref{PSRFHT}). \\
 \hspace*{6mm}
Introduce $A:=P_+\mathcal{F}P_{[-1,1]}^*$, $A^*=P_{[-1,1]}\mathcal{F}^{-1}P_+^*$. Using (\ref{FPK}), (\ref{FHTFR}), and $M(1_{[0,\infty[})=\frac{1}{2}\big(1+M(\operatorname{sgn})\big)$
one easily checks \begin{itemize}
\item[(a)] $A\,A^*=K$
\item[(b)] $A^*A=\frac{1}{2}(I+H_{[-1,1]})$
\end{itemize}

The relation (a), (b) is not accidental and as shown in \cite{C20} it relates isomorphicly
 self-adjoint Wiener-Hopf operators to singular integral operators with kernels of Cauchy type. \\
 \hspace*{6mm}
The following computations are valid in a  distributional sense.  Spectral functions $Q_s$ for $\frac{1}{2}(I+H_{[-1,1]})$  are given by $A^*q_s =P_{[-1,1]}\mathcal{F}^{-1}P_+^*q_s$ since
 $(A^*A)\,A^*q_s=A^*(AA^*)q_s=sA^*q_s$. By  (\ref{MPQS})(i),\,(iii),\,(i) one gets explicitly
 \begin{equation}\label{SFIPHP}
Q_s=\frac{n(s)}{\sqrt{s}}\; Q^+_s|_{[-1,1]}, \quad s\in]0,1[
\end{equation}
Here $\frac{n(s)}{\sqrt{s}}=\i  (\sqrt{2}\,\pi s)^{-1}$ is the unique (up to a phase) normalization constant such that $Q(s,x):=Q_s(x)$ is  the kernel for a Hilbert space isomorphism  $W:L^2(0,1)\to L^2(-1,1)$ satisfying
\begin{equation}\label{HSIDIPHP}
 Wh=\operatorname{l.i.m.}\int_ {\downarrow 0}^{\uparrow 1} Q(s,\cdot)h(s)\d s
\end{equation}
 The additional factor $\frac{1}{\sqrt{s}}$ regarding the normalization  of $Q_s$  is suggested heuristically by $\langle A^*q_s, A^*q_s\rangle=\langle q_s,AA^*q_s\rangle=s\langle q_s,q_s\rangle$. Indeed,
 (\ref{HSIDIPHP}) follow immediately from
 \begin{equation}\label{ADHSIDIPHP}
W= \Gamma\,\mathcal{F}^{-1}\Gamma^{-1}\Lambda
\end{equation}
 for the Hilbert space isomorphisms
$\Lambda:L^2(0,1)\to L^2(-1,1)$, $\Lambda h\,(t):=\frac{1}{\sqrt{2}} h\big(\frac{1+t}{2}\big)$
due to the change of variable $\lambda(t):=(1+t)/2$, and $\Gamma: L^2(\mathbb{R})\to L^2(-1,1)$,  $(\Gamma f)(x):=\sqrt{|\gamma'(x)|}f(\gamma(x))$ due to the change of variable $\gamma:]-1,1[\to\mathbb{R}$, $\gamma(x):=\sqrt{\frac{1}{2\pi}} \ln\big(\frac{1+x}{1-x}\big)$, $\gamma'(x)=\big(\frac{2}{\pi}\big)^{1/2}(1-x^2)^{-1}$. In order to verify  (\ref{ADHSIDIPHP}) one easily checks that the unitary transformation
\begin{equation}
U:=\Gamma\,\mathcal{F}^{-1}\Gamma^{-1}
\end{equation}
satisfies
\begin{equation}
Uk=\operatorname{l.i.m.}_{0<\alpha,\alpha'\to 0}\int_{-1+\alpha}^{1-\alpha'}Q'(t,\cdot)k(t)\d t
\end{equation}
for $Q'(t,u):=\frac{1}{\sqrt{2}}Q\big(\frac{1+t}{2}\big)$. By (\ref{SFIPHP}),\,(\ref{LQS}) one has explicitly
\begin{equation}\label{IKFHT}
Q'(t,u)=\textrm{\scriptsize{$\frac{1}{\pi}$}}(1-t^2)^{-1/2}(1-u^2)^{-1/2}\exp\textrm{\scriptsize{$\left(\frac{\i  }{2\pi}\ln\Big(\frac{1-t}{1+t}\Big) \ln \Big(\frac{1-u}{1+u}\Big)\right)$}}
\end{equation}

Now we give another proof of  \cite[Theorem 3.2]{KP59}).

\begin{Cor}\label{PSRFHT} The  spectral representation
$H_{[-1,1]}=U\,M(\operatorname{id}_{[-1,1]})\,U^{-1}$ holds.
\end{Cor}\\
{\it Proof.}  It suffices to show $P_{[-1,1]}\mathcal{F}^{-1}\,M(1_{[0,\infty[})\,\mathcal{F}P_{[-1,1]}^*Wh=W\,M(\operatorname{id}_{[0,1]})h$ for  $h$ with compact support in $]0,1[$. For $A>1$,
$\int_{-A}^A\operatorname{e}  ^{-\i  ux}\big(\int_{0}^11_{]-1,1[}(u)Q(s,u)h(s)\d s\big)\d u=\int_{0}^1\big(\int_{-1}^1\operatorname{e}  ^{-\i  ux}Q(s,u)\d u\big)h(s)\d s$ for every $x\in\mathbb{R}$ by Fubini's theorem since $|Q(s,u)h(s)|$ is integrable by (\ref{LQS}). By (\ref{MPQS})(ii) the double integral becomes $\i  \pi^{-1/2}\int_{0}^1q_s^+(x)h(s)\d s$. The latter is zero if $x<0$ by (\ref{MPQS})(i), and due to (\ref{MPQS})(iii) equals $\int_{-\infty}^\infty\operatorname{e}  ^{-\i  ux}\Big(\int_{0}^1Q(s,u)sh(s) \d s\Big) \d u$. This implies the assertion.\qed\\

\subsection{Covariance under real fractional linear transformations} One starts from the following simple observation. Recall that $H$ denotes the Hilbert transformation.

\begin{Lem}\label{CVUT} Let $F:L^2(\mathbb{R})\to L^2(\mathbb{R})$ be unitary.  Put $F':=P_+\mathcal{F}F\mathcal{F}^{-1}P_+^*$.  Suppose
\begin{itemize}
\item[\emph{($\alpha$)}] $FH=HF$
\item[\emph{($\beta$)}]
For every $\kappa\in L^\infty(\mathbb{R})$ there is $ \kappa'\in L^\infty(\mathbb{R})$ such that $FM(\kappa)F^{-1}=M(\kappa')$
\end{itemize}
Then $F'$ is unitary and $F'W_\kappa F'^{-1}=W(\kappa')$
holds.
\end{Lem}\\
{\it Proof.} By (\ref{SPHT}) and ($\alpha$),  $M(1_{[0,\infty[})$ commutes with $\mathcal{F}F\mathcal{F}^{-1}$ reducing it to  $F'$. Hence $F'P_+=P_+\mathcal{F}F\mathcal{F}^{-1}$, whence the result by ($\beta$).\qed\\

 With $A=\big(\textrm{\scriptsize{$ \begin{array}{cc} a & b\\ c & d \end{array}$}}\big) \in SL(2,\mathbb{R})$
  associate the real fractional linear transformation $A\cdot x:=\frac{ax+b}{cx+d}$ for almost all  $x\in\mathbb{R}$.
 Actually this is a group action as $A\cdot (A'\cdot x)= (AA' )\cdot x$ and $I_2\cdot x=x$. Note that $A'\cdot x=A\cdot x$ for almost all $x$  if and only if $A'\in\{A,-A\}$.

\begin{The}\label{CVFLTWH}  For $A\in SL(2,\mathbb{R})$, $f\in L^2(\mathbb{R})$ let $\big(F_Af\big)(x):=\frac{1}{-cx+a}f(A^{-1}\cdot x)$. Then $F$ is a unitary representation of $SL(2,\mathbb{R})$ in $L^2(\mathbb{R})$ commuting with $H$. Let $F'$ denote the subrepresentation of $\mathcal{F}F\mathcal{F}^{-1}$ according to \emph{(\ref{CVUT})}. One has the covariance
$$W(A\cdot \kappa)\,=\,F'_A\,W_\kappa\, F_A'^{-1}$$
where $\big(A\cdot \kappa\big)(x):=\kappa(A^{-1}\cdot x)$ for $\kappa\in L^\infty(\mathbb{R})$.
 \end{The}\\
 {\it Proof.} The result follows by elementary computations. Verify $F_AH=HF_A$ and $M(A\cdot \kappa)=F_AM(\kappa)F_A^{-1}$, and apply (\ref{CVUT}).\qed\\

 In particular   $W(1_J)$ for $J$ a bounded or semi-bounded interval is unitarily equivalent to $K$ by some $F'_A$. For instance $A:=\frac{1}{\sqrt{2}}\big(\textrm{\scriptsize{$ \begin{array}{cc} 1-\gamma & 1+\gamma\\ -1 & 1 \end{array}$}}\big)$ provides for the isomorphism of $K$ to $W(1_{[\gamma,\infty[})$, which has the distribution kernel $k(x)=\frac{1}{2}\delta(x)+ \frac{1}{2\pi \i  }\frac{\exp(-\i  \gamma x)}{-x}$, whence $W(1_{[0,\infty[})=\frac{1}{2}(I+H_{[0,\infty[})$ if $\gamma=0$ (cf.\,(\ref{FHTFR})). More generally, $W(a1_{]-\infty,\gamma[}+b1_{[\gamma,\infty[})$
 has the kernel $k(x)=\frac{b+a}{2}\delta(x)+ \frac{b-a}{2\pi \i  }\frac{\exp(-\i  \gamma x)}{-x}$.
 \\

For (\ref{TLWHHT})($\beta$) see the remark on  (0.3) in \cite{D79}.

\begin{Lem} \label{TLWHHT} One has
\begin{itemize}
\item[\emph{($\alpha$)}] $H_{[0,\infty[}=W_{-\operatorname{sgn}}$, where
$W_{-\operatorname{sgn}}$ has the kernel $-\frac{1}{x}$
\item[\emph{($\beta$)}] $H_{[0,1]}=\Gamma \,W_{-\tanh}\Gamma^{-1}$, where $W_{-\tanh}$ has the kernel $\frac{1}{2\i  \sinh(\pi x/2)}$
\end{itemize}
Here  $\Gamma:L^2(\mathbb{R}_+)\to L^2(0,1)$, $(\Gamma g)(t):=(\pi t)^{-1/2}g\big(-\frac{1}{\pi}\ln(t)\big)$ is the Hilbert space isomorphism due to the change of variable $\gamma(t):=-\frac{1}{\pi}\ln(t)$.
\end{Lem}

{\it Proof.}  For ($\alpha$) see (\ref{FHTFR}). As to ($\beta$), check $(\Gamma^{-1}k)(x)=\sqrt{\pi}\operatorname{e}  ^{-\pi x/2}k(\operatorname{e}  ^{-\pi x})$ and  that $\Gamma^{-1}H_{[0,1]}\Gamma$ equals the Wiener-Hopf operator with the indicated kernel. Then its symbol $-\tanh(x)$ is the principal value of $\int \frac{\exp(\i  xy)}{2\i  \sinh(\pi y/2)}\d y$ at $y=0$, which is evaluated calculating the residues.\qed

We use these results in an obvious way  to relate explicitly $K$ to several other operators. Note that $2K-I=W_{\sigma}$ with $\sigma(x)=1$ if $|x|\le1$ and $=-1$ otherwise.

\begin{Cor}\label{HTWHRK} The finite and semi-finite Hilbert transformations  as well the Wiener-Hopf operators $W_\kappa$ for $\kappa(x)=-\tanh(\frac{ax+b}{cx+d})$ with $ad-bc=1$ are Hilbert space isomorphic to $2K-I$. More precisely
$$H_{[a,b]}=\Lambda\,(2K-I)\,\Lambda^{-1}, \quad W_\kappa=\Lambda'\,(2K-I)\,\Lambda'^{-1}$$
Here $\Lambda= \Gamma\, F'_B$, where  $B:=\frac{1}{\sqrt{2}}\big(\textrm{\scriptsize{$ \begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}$}}\big)$ and $\Gamma:L^2(0,\infty)\to L^2(a,b)$ due to the change of variable $\gamma(t):=\frac{t-a}{-t+b}$ if $a,b$ are finite, $\gamma(t):=t-a$ if $b=\infty$, $\gamma(t):=\frac{1}{-t+b}$ if $a=-\infty$. Further
 $\Lambda':=F_A'^{-1}\Gamma^{-1}\Lambda$ with $A=\big(\textrm{\scriptsize{$ \begin{array}{cc} a & b\\ c & d \end{array}$}}\big)$, $\Gamma$ from \emph{(\ref{TLWHHT})($\beta$)}, and $\Lambda$ relating $H_{[0,1]}$ to $2K-I$.
 \end{Cor}\\
 {\it Proof.} Put $F:=F_B$. Check $B\cdot (21_{]-1,1[}-1)=-\operatorname{sgn}$, whence $F'(2K+I)F'^{-1}=H_{[0,\infty[}$ by (\ref{TLWHHT})($\alpha$),\,(\ref{CVFLTWH}). Further check $\Gamma H_{[0,\infty[}\Gamma^{-1}=H_{[a,b]}$. Hence the first part of the assertion holds. The remainder follows from this result for $H_{[a,b]}=H_{[0,1]}$ and by (\ref{TLWHHT})($\beta$),  (\ref{CVFLTWH}).\qed\\

 In particular (\ref{HTWHRK}) yields
\begin{equation} \label{CHSIHK}
H_{[-1,1]}\,=\,\Lambda (2K-I) \Lambda^{-1}\; \textrm{ with }  \Lambda=P_{[-1,1]}F_B\mathcal{F}F_B\mathcal{F}^{-1}P_+^*
 \end{equation}

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Q: how to have a menu across top with links left aligned and right aligned this may seem simple. I have a menu across the top of the page currently right aligned.
I want one link to be to the left of the page. How do it?

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css
    ul#bottomnavlistl li
{
    display: inline;
    padding: 10px 0px 5px 0px;
    border-right: 1px dotted #8A8575;
    float:right;
}

A: Use float:left
http://htmldog.com/reference/cssproperties/float/

A: Your css should have somthing like
.a{float:left;}
.first{float:right}

A demo here

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text:
Q: Passing XML data into a file I have an interface which takes two videos and plays them simultaneously side by side, at any point the user can add comments to the video in the 3 text boxes shown above. What I want to do is save the text kept in these boxes with the time stamp. I have a class to create a object of saved - which takes the 3 text inputs and position of the video on the click of a save button. What i'm not sure of is how to save the input once I have passed it back to my class.
Here is my code to call the class on button click:
Save.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent i) {

            Saved saved = new Saved();
            saved.setPosition(mediaPlayerComponent2.getMediaPlayer().getPosition() * 1000);
            saved.setGap(textArea.getText());
            saved.setForces(textArea1.getText());
            saved.setFindings(textArea2.getText());

            try {
                File file = new File("text.xml");
                JAXBContext jaxbContext = JAXBContext.newInstance(Saved.class);
                Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

                // output printed
                jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

                jaxbMarshaller.marshal(saved, file);
                jaxbMarshaller.marshal(saved, System.out);

            } catch (JAXBException e) {
                e.printStackTrace();
            }
        }

    });

Here is my class for creating a Saved object:
@XmlRootElement
public class Saved {
    float position;
    String gap;
    String forces;
    String findings;

    public float getPosition(){
        return position;
    }

    @XmlAttribute
    public void setPosition(float position){
        this.position = position;

    }

    public String getgap() {
    return gap;
}

@XmlElement
public void setGap(String gap) {
    this.gap = gap;
}

public String forces() {
    return forces;
}

@XmlElement
public void setForces(String forces) {
    this.forces = forces;
}

    public String findings() {
    return forces;
}

@XmlElement
public void setFindings(String findings) {
    this.findings = findings;
    }

}

A: Based on the comments:
First of all you try to marshall a non-static inner class, this could be solved by change it to a static nested class or an "outer" class.
Then JAXB doesn't recognize the getters for gap, forces and findings.
So change getgap(), forces(), findings to getGap(), getForces(), getFindings().

Also note that findings() returns the value of forces.