# Common Fermi Surface Topology and Nodeless Superconducting Gap in
K0.68Fe1.79Se2 and (Tl0.45K0.34)Fe1.84Se2 Superconductors Revealed from Angle-
Resolved Photoemission Spectroscopy

Lin Zhao1, Daixiang Mou1, Shanyu Liu1, Xiaowen Jia1, Junfeng He1, Yingying
Peng1, Li Yu1, Xu Liu1, Guodong Liu1, Shaolong He1, Xiaoli Dong1, Jun Zhang1,
J. B. He2, D. M. Wang2, G. F. Chen2, J. G. Guo1, X. L. Chen1, Xiaoyang Wang3,
Qinjun Peng3, Zhimin Wang3, Shenjin Zhang3, Feng Yang3, Zuyan Xu3, Chuangtian
Chen3 and X. J. Zhou1,∗
1Beijing National Laboratory for Condensed Matter Physics, Institute of
Physics, Chinese Academy of Sciences, Beijing 100190, China
2Department of Physics, Renmin University of China, Beijing 100872, China
3Technical Institute of Physics and Chemistry, Chinese Academy of Sciences,
Beijing 100190, China

(February 5, 2011)

###### Abstract

We carried out high resolution angle-resolved photoemission measurements on
the electronic structure and superconducting gap of K0.68Fe1.79Se2 (Tc=32 K)
and (Tl0.45K0.34)Fe1.84Se2 (Tc=28 K) superconductors. In addition to the
electron-like Fermi surface near M($\pi$,$\pi$), two electron-like Fermi
pockets are revealed around the zone center $\Gamma$(0,0) in K0.68Fe1.79Se2.
This observation makes the Fermi surface topology of K0.68Fe1.79Se2 consistent
with that of (Tl,Rb)xFe2-ySe2 and (Tl,K)xFe2-ySe2 compounds. A nearly
isotropic superconducting gap ($\Delta$) is observed along the electron-like
Fermi pocket near the M point in K0.68Fe1.79Se2 ($\Delta$$\sim$ 9 meV) and
(Tl0.45K0.34)Fe1.84Se2 ($\Delta$$\sim$ 8 meV). The establishment of a
universal picture on the Fermi surface topology and superconducting gap in the
AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) superconductors will provide important
information in understanding the superconductivity mechanism of the iron-based
superconductors.

###### pacs:

74.70.-b, 74.25.Jb, 79.60.-i, 71.20.-b

The latest discovery of superconductivity with a Tc above 30 K in a new
AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) systemJGGuo ; Switzerland ; Mizuguchi ;
MHFang ; GFChen  has triggered a new wave of broad interest in the iron-based
high temperature superconductorsKamihara ; ZARenSm ; RotterSC ; MKWu11 ;
CQJin111 . A couple of unique characteristics of the AxFe2-ySe2 system provide
new perspectives that ask for rethinking and re-examination of ideas which
have been proposed for other iron-based superconductors, such as the effect of
Fe vacancy and structural modulation on superconductivityMHFang ; GFChen ;
ZWang ; PZavalij , the nature of the underlying parent compoundMHFang ; GFChen
; QMSi ; YiZhou , the role of electron scattering across the bands between the
zone center $\Gamma$(0,0) and zone corner M($\pi$,$\pi$) on superconductivity,
and the pairing symmetry of this new system with a distinct Fermi surface
topologyYiZhou ; FWang . Band structure calculations of AxFe2-ySe2LJZhang ;
IRShein ; XWYan  suggest that the large electron doping in this system leads
to the disappearance of the hole-like Fermi surface pockets around the
$\Gamma$ point that are commonly present in other Fe-based compounds. In this
case, the peculiar Fermi surface topology near $\Gamma$ in the AxFe2-ySe2
superconductors would make it unlikely to have electron scatterings from the
hole-like bands near $\Gamma$ to the electron-like bands near M that are
considered to play an important role in the electron pairing and
superconductivity in the Fe-based superconductors by some theoriesKuroki ;
FeSCMagnetic . Experimental investigations on the electronic structure and the
superconducting gap of AxFe2-ySe2 superconductors are thus crucial for
understanding the physical properties and the pairing mechanism in the iron-
based superconductors.

Figure 1: Fermi surface mapping of K0.68Fe1.79Se2 superconductor (Tc=32 K)(a)
and (Tl0.45K0.34)Fe1.84Se2 superconductor (Tc=28 K) (b) measured by using
h$\nu$=21.2 eV light source. Near the M($\pi$,$\pi$) point, one Fermi surface
sheet is clearly observed which is marked as $\gamma$ (for the sake of
clarity, we refer the four equivalent M points in the first BZ as M1, M2, M3
and M4). Near the $\Gamma$(0,0) point, in addition to a tiny Fermi pocket
observed which is marked as $\alpha$, a weak large Fermi surface sheet (marked
as $\beta$) is also discernable.  Figure 2: Band structure and photoemission
spectra of K0.68Fe1.79Se2 measured along typical high symmetry cuts. (a). Band
structure along the Cut 1 crossing the $\Gamma$ point measured by using
h$\nu$=21.2 eV light source; the location of the cut is shown on the top of
Fig. 2a. (b). Corresponding EDC second derivative image of Fig. 2a. The
$\alpha$ band and two Fermi crossings of the $\beta$ band ($\beta_{L}$ and
$\beta_{R}$) are marked. Two inverse-parabolic GA and GB bands are also
marked. (c). Band structure along the Cut 2 crossing the $\Gamma$ point
measured by using h$\nu$=6.994 eV VUV laser. (d). Corresponding EDC second
derivative image of Fig. 2c. (e). Band structure along the Cut 3 crossing the
M2 point measured by using h$\nu$=21.2 eV. (f). Corresponding EDC second
derivative image of Fig. 2e. Two Fermi crossings of the $\gamma$ band
($\gamma_{L}$ and $\gamma_{R}$) are marked. The photoemission spectra (EDCs)
corresponding to the Cut1, Cut2 and Cut3 are shown in (g), (h) and (i),
respectively.

Angle-resolved photoemission spectroscopy (ARPES) is a powerful tool to
directly measure the electronic structure and superconducting gap of
superconductorsDamascelli . Some initial ARPES measurements on KxFe2-ySe2 did
not observe Fermi surface near $\Gamma$TQian  or observed only a trace of a
tiny electron-like pocket near $\Gamma$YZhang . These seem to be in agreement
with the band structure calculationsLJZhang ; IRShein ; XWYan . However, in
the ARPES measurement on (Tl,Rb)xFe2-ySe2DXMou , two electron-like Fermi
surface sheets are observed near $\Gamma$, with the large one having a similar
size as the one near the electron-like pocket around M. The existence of two
electron-like pockets near $\Gamma$ is also reported in (Tl,K)xFe2-ySe2XPWang
. These results raise an obvious issue on whether the Fermi surface topology
of KxFe2-ySe2 is different from (Tl,Rb,K)xFe2-ySe2; the resolving of this
issue is important for sorting out general electronic structure features in
understanding the Fe-based superconductors.

In this paper, we report the observation of two electron-like Fermi surface
sheets around the zone center $\Gamma$(0,0) in K0.68Fe1.79Se2 superconductor
(Tc=32 K) revealed from our high resolution ARPES measurements. This is
different from the previous ARPES reports that no Fermi pocket or only one
tiny Fermi pocket is present near $\Gamma$ in KxFe2-ySe2TQian ; YZhang . The
observation of two electron-like Fermi pockets near $\Gamma$ makes the Fermi
surface topology of KxFe2-ySe2 consistent with that in (Tl,Rb)xFe2-ySe2DXMou
and (Tl,K)xFe2-ySe2XPWang , thus establishing a coherent picture of Fermi
surface topology in the AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) system. We
observe nearly isotropic superconducting gap ($\Delta$) around the Fermi
pocket near M in K0.68Fe1.79Se2 ($\Delta$$\sim$ 9 meV) and
(Tl0.45K0.34)Fe1.84Se2 ($\Delta$$\sim$ 8 meV). The general picture on the
Fermi surface topology and its associated superconducting gap in the
AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) superconductors will provide key
insights in understanding the iron-based superconductors.

High resolution angle-resolved photoemission (ARPES) measurements were carried
out by using our lab system equipped with a Scienta R4000 electron energy
analyzerGDLiu . We used Helium discharge lamp as the light source which
provides photons with an energy of h$\upsilon$= 21.218 eV (Helium I), as well
as vacuum ultraviolet (VUV) laser which provides h$\upsilon$= 6.994 eV
photons. The energy resolution was set at 10 meV for the Fermi surface mapping
(Fig. 1) and band structure measurements (Fig. 2) and at 4 meV for the
superconducting gap measurements (Figs. 3 and 4). The angular resolution is
$\sim$0.3 degree. The Fermi level is referenced by measuring on a clean
polycrystalline gold that is electrically connected to the sample. The
K0.68Fe1.79Se2 and (Tl0.45K0.34)Fe1.84Se2 single crystals were grown by the
Bridgeman methodGFChen . The composition of the crystals were analyzed by the
energy dispersive X-ray (EDX) spectroscopy. Electrical resistivity and DC
magnetic susceptibility measurements show that the crystals exhibit a sharp
superconducting transition at Tc$\sim$32 K (transition width of $\sim$1 K) for
K0.68Fe1.79Se2 and Tc$\sim$28 K (transition width of $\sim$1 K) for
(Tl0.45K0.34)Fe1.84Se2. The crystal was cleaved in situ and measured in vacuum
with a base pressure better than 5$\times$10-11 Torr.

Fig. 1 shows Fermi surface mapping of K0.68Fe1.79Se2 (Fig. 1a) and
(Tl0.45K0.34)Fe1.84Se2 (Fig. 1b) superconductors. The band structure of
K0.68Fe1.79Se2 along two typical high symmetry cuts are shown in Fig. 2. An
electron-like Fermi surface is clearly observed around M($\pi$,$\pi$), similar
to previous ARPES results on KxFe2-ySe2TQian ; YZhang , (Tl,Rb)xFe2-ySe2DXMou
and (Tl,K)xFe2-ySe2XPWang . Near the $\Gamma$ point, a tiny Fermi pocket
(denoted as $\alpha$) is obvious which is possibly formed by an electron-like
band with its bottom nearly touching the Fermi level. In addition, one can
observe a rather weak but discernable electron-like Fermi surface sheet
(denoted as $\beta$) near $\Gamma$ in both K0.68Fe1.79Se2 (Fig. 1a) and
(Tl0.45K0.34)Fe1.84Se2 (Fig. 1b), with its size being similar to that of the
electron-like pocket near M.

Figure 3: Temperature dependence of the superconducting gap of K0.68Fe1.79Se2
(Tc$\sim$32 K) along the $\gamma$ Fermi pocket near M. (a-e) show
photoemission images taken at different temperatures along a cut near M3; the
location of the cut is marked in the bottom-left inset of (h). (f).
Photoemission spectra measured at different temperatures at the Fermi crossing
kF of the $\gamma$ band, as marked in (a). (g). The corresponding symmetrized
EDCs of (f). (h). Temperature dependence of the measured superconducting gap
(empty red circles). The black dashed line is a curve following the BCS form.

The existence of the $\beta$ Fermi pocket near the $\Gamma$ point in
K0.68Fe1.79Se2 can also be identified from the measured band structure (Fig.
2a and Fig. 2b). We note that the feature of the $\beta$ band (Fig. 2a) and
its associated Fermi surface (Fig. 1a) near $\Gamma$ are rather weak in
K0.68Fe1.79Se2, much weaker than in (Tl,Rb)xFe2-ySe2DXMou ; this is probably
why it was not revealed beforeTQian ; YZhang . We also notice that the band
structure of K0.68Fe1.79Se2 near the $\Gamma$ point (Figs. 2a, 2b, 2c and 2d)
presents some new features that were not observed before. As shown in Fig. 2b,
in addition to the electron-like $\alpha$ band and the electron-like $\beta$
band, at least two more bands are clearly present within the measured energy
window. The observation of the hole-like GB band is consistent with other
measurements on KxFe2-ySe2TQian ; YZhang  that is also commonly observed in
(Tl,Rb)xFe2-ySe2DXMou  and (Tl,K)xFe2-ySe2XPWang . However, the presence of a
new GA band is very clear in our measurement (Figs. 2b and 2d) which was not
observed in the previous measurementsTQian ; YZhang . The revelation of this
GA band is important when comparing the experimental results with the band
structure calculations and considering electron scatterings between various
bands.

The observation of two electron-like Fermi pockets, $\alpha$ and $\beta$,
around $\Gamma$ in K0.68Fe1.79Se2 is interesting. It is distinct from other
Fe-based compounds where hole-like Fermi surface sheets are expected around
the $\Gamma$ pointDJSingh1111 ; Kuroki . It is also different from band
structure calculationsLJZhang ; IRShein ; XWYan ; YZhang ; TQian  and previous
ARPES measurementsYZhang ; TQian  on KxFe2-ySe2 that only suggest
disappearance of hole-like Fermi surface sheets near the $\Gamma$ point. It
becomes now consistent with the ARPES measurements on (Tl,Rb)xFe2-ySe2DXMou
and (Tl,K)xFe2-ySe2XPWang  to provide a general picture on the Fermi surface
topology in the AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) superconductors.

Now we turn to investigate the superconducting gap in the K0.68Fe1.79Se2 and
(Tl0.45K0.34)Fe1.84Se2 superconductors. Since the $\beta$ feature near
$\Gamma$ is too weak to give reasonable information on the superconducting
gap, we will focus in this paper on the superconducting gap along the $\gamma$
Fermi surface near M. Figs. 3(a-e) show the photoemission images measured on
K0.68Fe1.79Se2 along a cut near M (its location shown in the bottom-left inset
of Fig. 3h) at different temperatures. The photoemission spectra (energy
distribution curves, EDCs) on the Fermi momentum at different temperatures are
shown in Fig. 3f. To visually inspect possible gap opening and remove the
effect of Fermi distribution function near the Fermi level, these original
EDCs are symmetrized to get spectra in Fig. 3g, following the procedure that
is commonly used in high temperature cuprate superconductorsMNorman . As seen
from Fig. 3g, there is a clear superconducting gap opening below Tc$\sim$ 32 K
which is closed above Tc. The superconducting gap size is extracted from the
peak position of the symmetrized EDCs in this paperMNorman  (Fig. 3g); it is
$\sim$9 meV at 12 K and its temperature dependence roughly follows the BCS-
type form (Fig. 3h).

In order to measure the momentum-dependence of the superconducting gap, we
took high resolution Fermi surface mapping of the $\gamma$ Fermi pocket at M
for K0.68Fe1.79Se2 (Fig. 4a) and (Tl0.45K0.34)Fe1.84Se2 (Fig. 4e)
superconductors. Fig. 4b shows photoemission spectra around the $\gamma$ Fermi
pocket (Fig. 4a) measured in the superconducting state (T= 15 K); the
corresponding symmetrized photoemission spectra are shown in Fig. 4c. The
superconducting gap (Fig. 4d), extracted by picking up the peak position of
the symmetrized EDCs (Fig. 4c), is nearly isotropic with a size of (9$\pm$2)
meV. By the same procedure, the superconducting gap around the $\gamma$ Fermi
pocket near M for the (Tl0.45K0.34)Fe1.84Se2 superconductor (Fig. 4h) is also
nearly isotropic with a size of (8$\pm$2) meV.

Figure 4: Momentum dependent superconducting gap of K0.68Fe1.79Se2
superconductor (Tc=32 K) and (Tl0.45K0.34)Fe1.84Se2 superconductor (Tc=28 K)
measured along the $\gamma$ Fermi surface sheet near M at a temperature of 15
K. (a). High resolution Fermi surface mapping of K0.68Fe1.79Se2 near M3; the
corresponding Fermi crossings are marked by empty black circles. (b) and (c)
show several typical EDCs along the $\gamma$ Fermi surface and their
corresponding symmetrized EDCs, respectively. (d). Momentum dependence of the
superconducting gap along the $\gamma$ Fermi surface sheet (solid red
circles). (e), (f), (g) and (h) show, respectively, the high resolution Fermi
surface mapping near M2, EDCs along the Fermi surface, their corresponding
symmetrized EDCs, and the obtained momentum-dependent superconudtcing gap for
the (Tl0.45K0.34)Fe1.84Se2 superconductor.

In summary, we have identified two electron-like Fermi pockets near the
$\Gamma$ point in K0.68Fe1.79Se2 and (Tl0.45K0.34)Fe1.84Se2 superconductors.
This has established a consistent picture on the Fermi surface topology in the
AxFe2-ySe2 (A=K, Tl, Cs, Rb and etc.) superconductors. The distinct Fermi
surface topology in the AxFe2-ySe2 superconductors definitely asks for re-
evaluation of the pairing mechanisms, based on electron scatterings between
the bands near $\Gamma$ and the bands near M, proposed before for other Fe-
based superconductors Kuroki ; FeSCMagnetic . We have observed nearly
isotropic superconducting gap around the $\gamma$ Fermi pocket near the M
point in K0.68Fe1.79Se2 ($\Delta$$\sim$ 9 meV) and (Tl0.45K0.34)Fe1.84Se2
($\Delta$$\sim$ 8 meV). These are consistent with other ARPES
measurementsYZhang ; DXMou ; XPWang  to build a general picture on an
isotropic superconducting gap along the $\gamma$ Fermi surface near M. These
results, together with the observation of nearly isotropic superconducting gap
along the $\beta$ pocket near $\Gamma$DXMou ; XPWang , indicate that the
AxFe2-ySe2 superconductors are nodeless in its gap structure, a fact that
appears to favor an s-wave symmetry or a nodeless d-wave symmetryYiZhou ;
FWang . These rich information on the Fermi surface topology and the
associated superconducting gap will provide crucial information and
constraints on understanding the superconductivity mechanism in the Fe-based
superconductors.

XJZ thanks the funding support from NSFC (Grant No. 10734120) and the MOST of
China (973 program No: 2011CB921703).

∗Corresponding author<EMAIL_ADDRESS>
## References

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  * (28) M. R. Norman et al., Phys. Rev. B 57, R11093 (1998). The gap size can be obtained by either fitting the symmetrized EDCs using the phenonolological formula as proposed in the paper, or picking up the peak position. We found that, when the selected energy window is large to cover the overall peak, the gap value from the fitting procedure tends to be (2$\sim$3) meV larger than that obtained directly from the peak position. The gap size in this paper is obatined by picking the peak position while it was obtained by fitting the symetrized EDCs over a large energy window in DXMou .


∎

11institutetext: Domingos Djinja 22institutetext: Department of Mathematics
and Informatics, Faculty of Sciences, Eduardo Mondlane University, Box 257,
Maputo, Mozambique. 33institutetext: Division of Mathematics and Physics,
School of Education, Culture and Communication, Mälardalens University, Box
883, 72123 Västerås, Sweden.
33email<EMAIL_ADDRESS><EMAIL_ADDRESS>44institutetext: Sergei Silvestrov 55institutetext: Division of Mathematics
and Physics, School of Education, Culture and Communication, Mälardalens
University, Box 883, 72123 Västerås, Sweden.
55email<EMAIL_ADDRESS>66institutetext: Alex Behakanira Tumwesigye
77institutetext: Department of Mathematics, College of Natural Sciences,
Makerere University, Box 7062, Kampala, Uganda.
77email<EMAIL_ADDRESS>
# Multiplication and linear integral operators on $L_{p}$ spaces representing
polynomial covariance type commutation relations

Domingos Djinja  Sergei Silvestrov  Alex Behakanira Tumwesigye

###### Abstract

Representations of polynomial covariant type commutation relations by pairs of
linear integral operators and multiplication operators on Banach spaces
$L_{p}$ are constructed.
MSC2020 Classification: 47L80, 47L55, 47L65, 47G10

###### Keywords:

multiplication operators, integral operators, covariance commutation relations

##  1 Introduction

Commutation relations of the form

$AB=BF(A)$ (1)

where $A,B$ are elements of an associative algebra and $F$ is a function of
the elements of the algebra, are important in many areas of Mathematics and
applications. Such commutation relations are usually called covariance
relations, crossed product relations or semi-direct product relations. The
pairs $(A,B)$ of elements of an algebra that satisfy (1) are called
representations of this relation in that algebra. Representations of
covariance commutation relations (1) by linear operators are important for
study of actions and induced representations of groups and semigroups, crossed
product operator algebras, dynamical systems, harmonic analysis, wavelets and
fractals analysis and applications in physics and engineering
BratJorgIFSAMSmemo99 ; BratJorgbook ; JorgWavSignFracbook ; JorgOpRepTh88 ;
JorMoore84 ; MACbook1 ; MACbook2 ; MACbook3 ; OstSambook ; Pedbook79 ;
SamoilenkbookSTDST .

The structure of representations for the relations of the form (1) by bounded
and unbounded self-adjoint operators, normal operators, unitary operators,
partial isometries and other linear operators with special involution
conditions on a Hilbert space, have been considered in BratEvansJorg2000 ;
CarlsenSilvExpoMath07 ; CarlsenSilvAAM09 ; CarlsenSilvProcEAS10 ; DutkayJorg3
; DJS12JFASilv ; DLS09 ; DutSilvProcAMS ; DutSilvSV ; JSvT12a ; JSvT12b ;
Mansour16 ; JMusondaPhdth18 ; JMusonda19 ; Musonda20 ; Nazaikinskii96 ;
OstSambook ; PerssonSilvestrov031 ; PerssonSilvestrov032 ;
PersSilv:CommutRelDinSyst ; RST16 ; RSST16 ; SamoilenkbookSTDST ; SaV8894 ;
SilPhD95 ; STomdynsystype1 ; SilWallin96 ; SvSJ07a ; SvSJ07b ; SvSJ07c ; SvT09
; Tomiyama87 ; Tomiama:SeoulLN1992 ; Tomiama:SeoulLN2part2000 ;
TumwesigyePhDThesis2018 ; TumwRiSilv:ComCrPrAlgPieccnstfnctreallineSPAS19v2 ;
VaislebSa90  using reordering formulas for functions of the algebra elements
and operators satisfying covariance commutation relation, functional calculus
and spectral representation of operators AkkLnearOperatorsDST  and interplay
with dynamical systems generated by iteration of the maps involved in the
commutation relations.

In this paper we construct representations of (1) by pairs of linear integral
and multiplication operators on Banach space $L_{p}$. Such representations can
also be viewed as solutions for operator equations $AX=XF(A)$, when $A$ is
specified or $XB=BF(X)$ when $B$ is specified. In contrast to OstSambook ;
SamoilenkbookSTDST ; SaV8894 ; VaislebSa90  devoted to involutive
representations of covariance type relations by operators on Hilbert spaces
using spectral theory of operators on Hilbert spaces, we aim at direct
construction of various classes of representations of covariance type
relations in specific important classes of operators on Banach spaces more
general than Hilbert spaces without imposing any involution conditions and not
using classical spectral theory of operators. This paper is organized in three
sections. After the introduction, we present in Section 2 preliminaries,
notations and basic definitions. In Section 3 we present the main results
about construction of specific representations on Banach function spaces
$L_{p}$.

##  2 Preliminaries and notations

In this section we present some preliminaries, basic definitions and
notations. For more details please read AdamsGDST ; DudleyDST ; FollandRADST ;
KantarovitchAkilovFunctionalAnalysis ; KolmogorovFominElemTheorFuncFuncAnal ;
RudinRCADST ; RynneLFADST .

Let $S\subseteq\mathbb{R}$, ($\mathbb{R}$ is the set of real numbers), be a
Lebesgue measurable set and let $(S,\Sigma,\tilde{m})$ be a $\sigma$-finite
measure space, that is, $S$ is a nonempty set, $\Sigma$ is a $\sigma-$algebra
with subsets of $S$, where $S$ can be cover with at most countable many
disjoint sets $E_{1},E_{2},E_{3},\ldots$ such that
$E_{i}\in\Sigma,\,\tilde{m}(E_{i})<\infty$, $i=1,2,\ldots$ and $\tilde{m}$ is
the Lebesgue measure. For $1\leqslant p<\infty,$ we denote by $L_{p}(S)$, the
set of all classes of equivalent measurable functions $f:S\to\mathbb{R}$ such
that $\int\limits_{S}|f(t)|^{p}dt<\infty.$ This is a Banach space (Hilbert
space when $p=2$) with norm
$\|f\|_{p}=\left(\int\limits_{S}|f(t)|^{p}dt\right)^{\frac{1}{p}}.$ We denote
by $L_{\infty}(S)$ the set of all classes of equivalent measurable functions
$f:S\to\mathbb{R}$ such that there is a constant $\lambda>0$,
$|f(t)|\leq\lambda$ almost everywhere. This is a Banach space with norm
$\displaystyle\|f\|_{\infty}=\mathop{\operatorname{ess\;sup}}_{t\in S}|f(t)|.$

##  3 Operator representations of covariance commutation relations

Before we proceed with constructions of more complicated operator
representations of commutation relations (1) on more complicated Banach
spaces, we wish to mention the following two observations that, while being
elementary, nevertheless explicitly indicate differences in how the different
operator representations of commutation relations (1) interact with the
function $F$.

###### Proposition 3.1

Let $A:E\to E$ and $B:E\to E$, $B\not=0$, be linear operators on a linear
space $E$, such that any composition among them is well defined and consider
$F:\mathbb{R}\to\mathbb{R}$ a polynomial. If $A=\alpha I$, then $AB=BF(A)$ if
and only if $F(\alpha)=\alpha$.

###### Proof

If $A=\alpha I$, then

$\displaystyle AB=\alpha IB=\alpha B,$ $\displaystyle BF(A)=BF(\alpha
I)=BF(\alpha)I=F(\alpha)B.$

We have then $AB=BF(A)$, $B\neq 0$ if and only if $F(\alpha)=\alpha$. ∎

###### Proposition 3.2

Let $A:E\to E$ and $B:E\to E$ be linear operators such that any composition
among them is well defined and consider a polynomial
$F:\mathbb{R}\to\mathbb{R}$. If $B=\alpha I$, where $\alpha\not=0$, then
$AB=BF(A)$ if and only if $F$ satisfies $F(A)=A$.

###### Proof

If $B=\alpha I$ then

$\displaystyle AB=A(\alpha I)=\alpha A,$ $\displaystyle BF(A)=\alpha
IF(A)=\alpha F(A).$

We have then $AB=BF(A)$ if and only if $F(A)=A$. ∎

###  3.1 Representations of covariance commutation relations by integral and
multiplication operators on $L_{p}$ spaces

We consider first a useful lemma for integral operators.

###### Lemma 3.1

Let $f:[\alpha_{1},\beta_{1}]\to\mathbb{R}$,
$g:[\alpha_{2},\beta_{2}]\to\mathbb{R}$ be two measurable functions such that
for all $x\in L_{p}(\mathbb{R})$, $1\leq p\leq\infty$,

$\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)x(t)dt,\quad\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)x(t)dt,$

exist and are finite, where
$\alpha_{1},\beta_{1},\alpha_{2},\beta_{2}\in\mathbb{R}$,
$\alpha_{1}<\beta_{1}$ and $\alpha_{2}<\beta_{2}$. Set
$G=[\alpha_{1},\beta_{1}]\cap[\alpha_{2},\beta_{2}].$ Then the following
statements are equivalent.

  1. 1)

For all $x\in L_{p}(\mathbb{R})$, where $1\leq p\leq\infty$, the following
holds

$\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)x(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)x(t)dt.$

  2. 2)

The following conditions hold:

    1. a)

for almost every $t\in G$, $f(t)=g(t)$;

    2. b)

for almost every $t\in[\alpha_{1},\beta_{1}]\setminus G,\ f(t)=0;$

    3. c)

for almost every $t\in[\alpha_{2},\beta_{2}]\setminus G,\ g(t)=0.$

###### Proof

$2.\,\Rightarrow\,1.$ follows from direct computation.

Suppose that 1. is true. Take $x(t)=I_{G_{1}}(t)$ the indicator function of
the set $G_{1}=[\alpha_{1},\beta_{1}]\cup[\alpha_{2},\beta_{2}]$. For this
function we have,

$\displaystyle\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)x(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)x(t)dt=\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)dt=\eta,$

$\eta$ is a constant. Now by taking $x(t)=I_{[\alpha_{1},\beta_{1}]\setminus
G}(t)$ we get

$\displaystyle\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)x(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)x(t)dt=\int\limits_{[\alpha_{1},\beta_{1}]\setminus
G}f(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)\cdot 0dt=0.$

Then $\int\limits_{[\alpha_{1},\beta_{1}]\setminus G}f(t)dt=0.$ If instead
$x(t)=I_{[\alpha_{2},\beta_{2}]\setminus G}(t)$, then
$\int\limits_{[\alpha_{2},\beta_{2}]\setminus G}g(t)dt=0.$ We claim that
$f(t)=0$ for almost every $t\in[\alpha_{1},\beta_{1}]\setminus G$ and $g(t)=0$
for almost every $t\in[\alpha_{2},\beta_{2}]\setminus G$. We take a partition
$\bigcup S_{i}$ of the set $[\alpha_{1},\beta_{1}]\setminus G$ such that
$S_{i}\cap S_{j}\neq\emptyset$ for $i\neq j$ and each set $S_{i}$ has positive
measure. For each $x(t)=I_{S_{i}}(t)$, we have

$\displaystyle\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)x(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)x(t)dt=\int\limits_{S_{i}}f(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)\cdot
0dt=0.$

Thus, $\int\limits_{S_{i}}f(t)dt=0.$ Since we can choose arbitrary partition
with positive measure on each of its elements we have

$\displaystyle f(t)=0\quad\mbox{ for almost every
}t\in[\alpha_{1},\beta_{1}]\setminus G.$

Analogously, $g(t)=0$ for almost every $t\in[\alpha_{2},\beta_{2}]\setminus
G.$ Then,

$\eta=\int\limits_{\alpha_{1}}^{\beta_{1}}f(t)dt=\int\limits_{\alpha_{2}}^{\beta_{2}}g(t)dt=\int\limits_{G}f(t)dt=\int\limits_{G}g(t)dt.$

Then, for all function $x\in L_{p}(\mathbb{R})$ we have

$\displaystyle\int\limits_{G}f(t)x(t)dt=\int\limits_{G}g(t)x(t)dt\Longleftrightarrow\int\limits_{G}[f(t)-g(t)]x(t)dt=0.$

By taking $x(t)=\left\\{\begin{array}[]{cc}1,&\mbox{if }f(t)-g(t)>0,\\\
-1,&\mbox{ if }f(t)-g(t)<0,\end{array}\right.$ for almost every $t\in G$ and
$x(t)=0$ for almost every $t\in\mathbb{R}\setminus G$, we get
$\int\limits_{G}|f(t)-g(t)|dt=0$. This implies that $f(t)=g(t)$ for almost
every $t\in G$. ∎

###### Remark 3.1

When operators are given in abstract form, we use the notation
$A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$ meaning that operator $A$ is well
defined from $L_{p}(\mathbb{R})$ to $L_{p}(\mathbb{R})$ without discussing
sufficient conditions for it to be satisfied. For instance, for the following
integral operator

$(Ax)(t)=\int\limits_{\mathbb{R}}k(t,s)x(s)ds$

there are sufficient conditions on the kernel $k(\cdot,\cdot)$ such that
operator $A$ is well defined as a linear operator from $L_{p}(\mathbb{R})$ to
$L_{p}(\mathbb{R})$, and bounded ConwayFunctionalAnalysisDST ; FollandRADST .
For instance, (FollandRADST, , Theorem 6.18) states the following: if $1\leq
p\leq\infty$ and $k:\mathbb{R}\times[\alpha,\beta]\to\mathbb{R}$ is a
measurable function, $\alpha,\beta\in\mathbb{R}$, $\alpha<\beta$, and there is
a constant $\lambda>0$ such that

$\mathop{\operatorname{ess\;sup}}_{s\in[\alpha,\beta]}\int\limits_{\mathbb{R}}|k(t,s)|dt\leq\lambda,\quad\mathop{\operatorname{ess\;sup}}_{t\in\mathbb{R}}\int\limits_{\alpha}^{\beta}|k(t,s)|ds\leq\lambda,$

then $A$ is well defined from $L_{p}(\mathbb{R})$ to $L_{p}(\mathbb{R})$,
$1\leq p\leq\infty$ and bounded.

#### Representations when $A$ is integral operator and $B$ is multiplication
operator

###### Proposition 3.3

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1\leq p\leq\infty$, be defined as follows, for almost all
$t\in\mathbb{R}$,

$\displaystyle(Ax)(t)=\int\limits_{\alpha}^{\beta}k(t,s)x(s)ds,\quad(Bx)(t)=b(t)x(t),\quad\alpha,\beta\in\mathbb{R},\
\alpha<\beta,$

where $k:\mathbb{R}\times[\alpha,\beta]\to\mathbb{R}$ is a measurable
function, and $b:\mathbb{R}\to\mathbb{R}$ is a measurable function. Consider a
polynomial $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$, where
$\delta_{0},\ldots,\delta_{n}\in\mathbb{R}$. We set

$\displaystyle k_{0}(t,s)=k(t,s),\quad\quad
k_{m}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k_{m-1}(\tau,s)d\tau,\quad
m\in\\{1,\ldots,n\\}$ $\displaystyle F_{0}(k(t,s))=0,\quad
F_{n}(k(t,s))=\sum_{j=1}^{n}\delta_{j}k_{j-1}(t,s),\quad n\geq 1.$ (2)

Then $AB=BF(A)$ if and only if

$\forall\ x\in L_{p}(\mathbb{R}):\quad
b(t)\delta_{0}x(t)+b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds.$
(3)

If $\delta_{0}=0$, that is, $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, then
the condition (3) reduces to the following: for almost every $(t,s)$ in
$\mathbb{R}\times[\alpha,\beta]$,

$\displaystyle b(t)F_{n}(k(t,s))=k(t,s)b(s).$ (4)

###### Proof

By applying Fubini Theorem from AdamsGDST  and iterative kernels from
KrasnolskZabreykoPuSoIntegralopssummablefuncs , We have

$\displaystyle(A^{2}x)(t)$ $\displaystyle=$
$\displaystyle\int\limits_{\alpha}^{\beta}k(t,s)(Ax)(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)\left(\int\limits_{\alpha}^{\beta}k(s,\tau)x(\tau)d\tau\right)ds$
$\displaystyle=$
$\displaystyle\int\limits_{\alpha}^{\beta}\left(\int\limits_{\alpha}^{\beta}k(t,s)k(s,\tau)ds\right)x(\tau)d\tau=\int\limits_{\alpha}^{\beta}k_{1}(t,\tau)x(\tau)d\tau,$

where $k_{1}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k(\tau,s)d\tau.$ In the
same way,

$\displaystyle(A^{3}x)(t)=\int\limits_{\alpha}^{\beta}k(t,s)(A^{2}x)(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)\left(\int\limits_{\alpha}^{\beta}k_{1}(s,\tau)x(\tau)d\tau\right)ds$
$\displaystyle=\int\limits_{\alpha}^{\beta}k_{2}(t,s)x(s)ds,$

where $k_{2}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k_{1}(\tau,s)d\tau.$
For every $m\geq 1$,

$(A^{m}x)(t)=\int\limits_{\alpha}^{\beta}k_{m-1}(t,s)x(s)ds,$

where $k_{m}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k_{m-1}(\tau,s)d\tau,\
m=1,\ldots,n,\ k_{0}(t,s)=k(t,s).$ Thus,

$\displaystyle(F(A)x)(t)$ $\displaystyle=$
$\displaystyle\delta_{0}x(t)+\sum\limits_{j=1}^{n}\delta_{j}(A^{j}x)(t)=\delta_{0}x(t)+\sum\limits_{j=1}^{n}\delta_{j}\int\limits_{\alpha}^{\beta}k_{j-1}(t,s)x(s)ds$
$\displaystyle=$
$\displaystyle\delta_{0}x(t)+\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds,$

where $F_{n}(k(t,s))=\sum\limits_{j=1}^{n}\delta_{j}k_{j-1}(t,s),$ for $n\geq
1,$ and $F_{0}(k(t,s))=0.$ So, we can compute $BF(A)x$ and $(AB)x$ as follows:

$\displaystyle(BF(A)x)(t)$
$\displaystyle=b(t)(F(A)x)(t)=b(t)\delta_{0}x(t)+b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds,$
$\displaystyle(ABx)(t)$
$\displaystyle=A(Bx)(t)=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds.$

It follows that $ABx=BF(A)x$ if and only if condition (3) holds.

If $\delta_{0}=0$ then condition (3) reduces to the following:

$\displaystyle\forall\ x\in
L_{p}(\mathbb{R}):\quad\int\limits_{\alpha}^{\beta}b(t)F_{n}(k(t,s))x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds.$

Let $f(t,s)=b(t)F_{n}(k(t,s))-k(t,s)b(s)$. By applying Lemma 3.1 we have for
almost every $t\in\mathbb{R}$ that $f(t,\cdot)=0$ almost everywhere. Since the
set $N=\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
f(t,s)\not=0\\}\subset\mathbb{R}^{2}$ is measurable and almost all sections
$N_{t}=\\{s\in[\alpha,\beta]:\ (t,s)\in N\\}$ of the plane has Lebesgue
measure zero, by the reciprocal Fubini Theorem OxtobyDST , the set $N$ has
Lebesgue measure zero on the plane $\mathbb{R}^{2}$. ∎

###### Corollary 3.4

For $M_{1},M_{2}\in\mathbb{R}$, $M_{1}<M_{2}$ and $1\leq p\leq\infty$, let
$A:L_{p}([M_{1},M_{2}])\to L_{p}([M_{1},M_{2}])$ and
$B:L_{p}([M_{1},M_{2}])\to L_{p}([M_{1},M_{2}])$ be nonzero operators defined,
for almost all $t$, by

$\displaystyle(Ax)(t)=\int\limits_{\alpha}^{\beta}k(t,s)x(s)ds,\quad(Bx)(t)=b(t)x(t),\quad\alpha,\beta\in\mathbb{R},\
\alpha<\beta,$

where $[M_{1},M_{2}]\supseteq[\alpha,\beta]$, and
$k(\cdot,\cdot):[M_{1},M_{2}]\times[\alpha,\beta]\to\mathbb{R}$,
$b:[M_{1},M_{2}]\to\mathbb{R}$ are given by

$\displaystyle k(t,s)=a_{0}+a_{1}t+c_{1}s,\qquad
b(t)=\sum\limits_{j=0}^{n}b_{j}t^{j},$

where $n$ is non-negative integer, $a_{0},\,a_{1},\,c_{1},\,b_{j}$ are real
numbers for $j=0,\ldots,n$. Consider a polynomial defined by
$F(z)=\delta_{0}+\delta_{1}z+\delta_{2}z^{2}$, where
$\delta_{0},\,\delta_{1},\,\delta_{2}\in\mathbb{R}$.

Then, $AB=BF(A)$ if and only if

$\displaystyle\forall\ x\in L_{p}([M_{1},M_{2}]):\quad
b(t)\delta_{0}x(t)+b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds,$

where $F_{n}(k(t,s))$ is given by (2).

If $\delta_{0}=0$, that is, $F(z)=\delta_{1}z+\delta_{2}z^{2}$ then the last
condition reduces to the condition that for almost every $(t,s)$ in
$[M_{1},M_{2}]\times[\alpha,\beta]$

$b(t)F_{2}(k(t,s))=k(t,s)b(s).$ (5)

Condition (5) is equivalent to that $b(\cdot)\equiv b_{0}\neq 0$ is a nonzero
constant $(b_{j}=0,$ $j=1,\ldots,n)$ and one of the following cases holds:

  1. 1)

if $\delta_{2}=0$, $\delta_{1}=1$, then $a_{0},a_{1},c_{1}\in\mathbb{R}$ can
be arbitrary;

  2. 2)

if $\delta_{2}\not=0$, $\delta_{1}=1$, $a_{1}\not=0$, $c_{1}=0$, then

$a_{0}=-\frac{\beta+\alpha}{2}a_{1};$

  3. 3)

if $\delta_{2}\not=0$, $\delta_{1}=1$, $a_{1}=0$, $c_{1}\not=0$, then

$a_{0}=-\frac{\beta+\alpha}{2}c_{1};$

  4. 4)

if $\delta_{2}\not=0$, $\delta_{1}\not=1$, $a_{1}\not=0$, $c_{1}=0$, then

$a_{0}=\frac{2-2\delta_{1}-\delta_{2}(\beta^{2}-\alpha^{2})a_{1}}{2\delta_{2}(\beta-\alpha)};$

  5. 5)

if $\delta_{2}\not=0$, $\delta_{1}\not=1$, $c_{1}\not=0$, $a_{1}=0$, then

$a_{0}=\frac{2-2\delta_{1}-\delta_{2}(\beta^{2}-\alpha^{2})c_{1}}{2\delta_{2}(\beta-\alpha)};$

  6. 6)

if $\delta_{2}\not=0$, $\delta_{1}\not=1$, $a_{1}=0$ and $c_{1}=0$, then

$a_{0}=\frac{1-\delta_{1}}{\delta_{2}(\beta-\alpha)}.$

###### Proof

Operator $A$ is well defined on $L_{p}[M_{1},M_{2}]$, $1\leq p\leq\infty$. In
fact, it follows by (HutsonPymDST, , Theorem 3.4.10). Moreover, kernel
$k(\cdot,\cdot)$ is continuous on a closed and bounded set
$[-M,M]\times[\alpha,\beta]$ and $b(\cdot)$ is continuous in $[M_{1},M_{2}]$,
so these functions are measurable. By applying Proposition 3.3 we just need to
check when the condition (4) is satisfied for $k(\cdot,\cdot)$ and $b(\cdot)$.
We compute

$\displaystyle k_{1}(t,s)=$
$\displaystyle\int\limits_{\alpha}^{\beta}k(t,\tau)k(\tau,s)d\tau=\int\limits_{\alpha}^{\beta}(a_{0}+a_{1}t+c_{1}\tau)(a_{0}+a_{1}\tau+c_{1}s)d\tau$
$\displaystyle=$
$\displaystyle\int\limits_{\alpha}^{\beta}[(a_{0}^{2}+a_{0}a_{1}t+a_{0}c_{1}s+a_{1}c_{1}ts)$
$\displaystyle+(a_{0}a_{1}+a_{0}c_{1}+a_{1}^{2}t+c_{1}^{2}s)\tau+a_{1}c_{1}\tau^{2}]d\tau$
$\displaystyle=$
$\displaystyle(\beta-\alpha)(a_{0}^{2}+a_{0}a_{1}t+a_{0}c_{1}s+a_{1}c_{1}ts)$
$\displaystyle+\frac{\beta^{2}-\alpha^{2}}{2}\cdot(a_{0}a_{1}+a_{0}c_{1}+a_{1}^{2}t+c_{1}^{2}s)$
$\displaystyle+\frac{\beta^{3}-\alpha^{3}}{3}a_{1}c_{1}=\nu_{0}+\nu_{1}t+\nu_{2}s+\nu_{3}ts,$
(6)

where

$\displaystyle\begin{array}[]{cclccl}\nu_{0}&=&a_{0}^{2}(\beta-\alpha)+\frac{\beta^{2}-\alpha^{2}}{2}a_{0}(a_{1}+c_{1})+a_{1}c_{1}\frac{\beta^{3}-\alpha^{3}}{3},&\nu_{2}&=&a_{0}c_{1}(\beta-\alpha)+c_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2},\\\
\nu_{1}&=&a_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+a_{1}a_{0}(\beta-\alpha),&\nu_{3}&=&a_{1}c_{1}(\beta-\alpha).\end{array}$

Then, we have

$\displaystyle
b(t)F_{2}(k(t,s))=b(t)(\delta_{1}k(t,s)+\delta_{2}k_{1}(t,s))=(a_{0}\delta_{1}+\delta_{2}\nu_{0})\sum_{j=0}^{n}b_{j}t^{j}$
$\displaystyle+(a_{1}\delta_{1}+\delta_{2}\nu_{1})\sum_{j=0}^{n}b_{j}t^{j+1}+(c_{1}\delta_{1}+\delta_{2}\nu_{2})\sum_{j=0}^{n}b_{j}t^{j}s+\nu_{3}\delta_{2}\sum_{j=0}^{n}b_{j}t^{j+1}s$
$\displaystyle+\sum_{j=1}^{n}((c_{1}\delta_{1}+\nu_{2}\delta_{2})b_{j}+\nu_{3}\delta_{2}b_{j-1})t^{j}s+(\delta_{1}a_{1}+\delta_{2}\nu_{1})b_{n}t^{n+1}+\nu_{3}\delta_{2}b_{n}t^{n+1}s$
$\displaystyle
k(t,s)b(s)=a_{0}\sum_{j=0}^{n}b_{j}s^{j}+a_{1}\sum_{j=0}^{n}b_{j}s^{j}t+c_{1}\sum_{j=0}^{n}b_{j}s^{j+1}=a_{0}b_{0}+a_{1}b_{0}t$
$\displaystyle+\sum_{j=1}^{n}(a_{0}b_{j}+c_{1}b_{j-1})s^{j}+\sum_{j=1}^{n}a_{1}b_{j}s^{j}t+c_{1}b_{n}s^{n+1}.$

Thus we have $\displaystyle k(t,s)b(s)=b(t)F_{2}(k(t,s))$ for all
$(t,s)\in[M_{1},M_{2}]\times[\alpha,\beta]$ if and only if

$\displaystyle a_{0}b_{0}$ $\displaystyle=$
$\displaystyle(a_{0}\delta_{1}+\delta_{2}\nu_{0})b_{0}$ $\displaystyle
a_{1}b_{0}$ $\displaystyle=$
$\displaystyle(a_{0}\delta_{1}+\delta_{2}\nu_{0})b_{1}+(a_{1}\delta_{1}+\delta_{2}\nu_{1})b_{0}$
$\displaystyle a_{0}b_{1}+c_{1}b_{0}$ $\displaystyle=$
$\displaystyle(c_{1}\delta_{1}+\delta_{2}\nu_{2})b_{0}$ (7) $\displaystyle
a_{1}b_{1}$ $\displaystyle=$
$\displaystyle(c_{1}\delta_{1}+\delta_{2}\nu_{2})b_{1}+\delta_{2}\nu_{3}b_{0}$
(8) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
a_{0}b_{j}+c_{1}b_{j-1},\quad 2\leq j\leq n$ (9) $\displaystyle 0$
$\displaystyle=$
$\displaystyle(a_{0}\delta_{1}+\delta_{2}\nu_{0})b_{j}+(a_{1}\delta_{1}+\delta_{2}\nu_{1})b_{j-1},\quad
2\leq j\leq n$ $\displaystyle a_{1}b_{j}$ $\displaystyle=$ $\displaystyle
0,\quad 2\leq j\leq n$ (10) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
c_{1}\delta_{1}b_{j}+\delta_{2}\nu_{3}b_{j-1}+\delta_{2}\nu_{2}b_{j}\quad
2\leq j\leq n$ $\displaystyle 0$ $\displaystyle=$ $\displaystyle
a_{1}\delta_{1}b_{n}+\delta_{2}\nu_{1}b_{n},\quad\mbox{ if }n\geq 1$
$\displaystyle c_{1}b_{n}$ $\displaystyle=$ $\displaystyle 0,\quad\mbox{ if
}n\geq 1$ (11) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\delta_{2}\nu_{3}b_{n},\quad\mbox{ if }n\geq 1.$

Suppose that $n\geq 1$. We proceed by induction to prove that $b_{j}=0$, for
all $j=1,\ldots,n$. For $i=0$, we suppose that $b_{n}=b_{n-i}\not=0$. Then
from the equation (10) we have $a_{1}b_{n}=0$ and thus $a_{1}=0$. From the
equation (11) we have $c_{1}b_{n}=0$ and thus $c_{1}=0$. From the equation (9)
we have $0=a_{0}b_{n}+c_{1}b_{n-1}=a_{0}b_{n}$ and thus $a_{0}=0$. This
implies that $k(t,s)\equiv 0$, that is, $A=0$. So for $i=0$,
$b_{n}=b_{n-i}\not=0$ implies $A=0$. Hence, $b_{n}=0$. Let $1<m\leq n-2$ and
suppose that $b_{n-i}=0$ for all $i=1,\ldots,m-1$. Let us show that then
$b_{n-m}=0$. If $b_{n-m}\not=0$, then from equation (10) we have
$a_{1}b_{n-m}=0$ which implies $a_{1}=0$. From equation (9) and for $j=n-m+1$
by induction assumption $a_{0}b_{n-m+1}+c_{1}b_{n-m}=c_{1}b_{n-m}=0$ which
implies $c_{1}=0$. Therefore from (9) and for $j=n-m$ we have $a_{0}b_{n-m}=0$
which implies $a_{0}=0$. Then $k(t,s)\equiv 0$, that is $A=0$. So we must have
$b_{n-m}=0$. If $m=n-1$, then let us show that $b_{n-m}=b_{1}=0$. If
$b_{n-m}\not=0$ then (9) gives $c_{1}b_{n-m}=c_{1}b_{1}=0$ when $j=n-m+1=2$.
Then $c_{1}=0$ and by (8), since $\nu_{2}=\nu_{3}=0$ we get $a_{1}b_{1}=0$
which yields $a_{1}=0$. Therefore, (7) gives $a_{0}b_{1}=0$ which yields
$a_{0}=0$. Thus $A=0$. Since $A\not=0$, $b_{1}=0$ is proved. Thus
$b(\cdot)=b_{0}$ is proved.

Since $B\neq 0$ and $B=b_{0}I$ (multiple of identity operator), $b_{0}\neq 0$
and the commutation relation is equivalent to $F(A)=A$. By (4) we have
$F_{2}(k(t,s))=k(t,s)$ which can be written as follows

$\delta_{1}k(t,s)+\delta_{2}k_{1}(t,s)=k(t,s),$ (12)

where $k(t,s)=a_{0}+a_{1}t+c_{1}s$ and
$k_{1}(t,s)=\nu_{0}+\nu_{1}t+\nu_{2}s+\nu_{3}ts$,

$\displaystyle\begin{array}[]{cclccl}\nu_{0}&=&a_{0}^{2}(\beta-\alpha)+\frac{\beta^{2}-\alpha^{2}}{2}a_{0}(a_{1}+c_{1})+a_{1}c_{1}\frac{\beta^{3}-\alpha^{3}}{3},&\nu_{2}&=&a_{0}c_{1}(\beta-\alpha)+c_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2},\\\
\nu_{1}&=&a_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+a_{1}a_{0}(\beta-\alpha),&\nu_{3}&=&a_{1}c_{1}(\beta-\alpha).\end{array}$

If $\delta_{2}=0$, then (12) becomes $(\delta_{1}-1)k(\cdot,\cdot)=0$ and
$A\not=0$ yields $\delta_{1}=1$. Thus, if $\delta_{2}=0$ and $\delta_{1}=1$,
then (12) is satisfied for any $a_{0},a_{1},c_{1}\in\mathbb{R}$.

If $\delta_{2}\not=0$ and $\delta_{1}=1$ then (12) becomes
$k_{1}(\cdot,\cdot)=0$, that is, $\nu_{0}=\nu_{1}=\nu_{2}=\nu_{3}=0$, where

$\displaystyle\begin{array}[]{cclccl}\nu_{0}&=&a_{0}^{2}(\beta-\alpha)+\frac{\beta^{2}-\alpha^{2}}{2}a_{0}(a_{1}+c_{1})+a_{1}c_{1}\frac{\beta^{3}-\alpha^{3}}{3},&\nu_{2}&=&a_{0}c_{1}(\beta-\alpha)+c_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2},\\\
\nu_{1}&=&a_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+a_{1}a_{0}(\beta-\alpha),&\nu_{3}&=&a_{1}c_{1}(\beta-\alpha).\end{array}$

Since $\alpha<\beta$, $a_{1}c_{1}(\beta-\alpha)=0$ is equivalent to either
$a_{1}=0$ or $c_{1}=0$. If $a_{1}\neq 0$, $c_{1}=0$, then

$\displaystyle\left\\{\begin{array}[]{c}\nu_{0}=0\\\ \nu_{1}=0\\\ \nu_{2}=0\\\
\nu_{3}=0\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{c}(\beta-\alpha)a_{0}^{2}+\frac{\beta^{2}-\alpha^{2}}{2}a_{0}a_{1}=0\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt(\beta-\alpha)a_{1}a_{0}+\frac{\beta^{2}-\alpha^{2}}{2}a^{2}_{1}=0\end{array}\right.\Leftrightarrow
a_{0}+\frac{\beta+\alpha}{2}a_{1}=0,$

which is equivalent to $a_{0}=-\frac{\beta+\alpha}{2}a_{1}$. If $a_{1}=0$,
$c_{1}\neq 0$, then

$\displaystyle\left\\{\begin{array}[]{c}\nu_{0}=0\\\ \nu_{1}=0\\\ \nu_{2}=0\\\
\nu_{3}=0\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{c}(\beta-\alpha)a_{0}^{2}+\frac{\beta^{2}-\alpha^{2}}{2}a_{0}c_{1}=0\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt(\beta-\alpha)c_{1}a_{0}+\frac{\beta^{2}-\alpha^{2}}{2}c^{2}_{1}=0\end{array}\right.\Leftrightarrow
a_{0}+\frac{\beta+\alpha}{2}c_{1}=0,$

which is equivalent to $a_{0}=-\frac{\beta+\alpha}{2}c_{1}$. If $a_{1}=0$,
$c_{1}=0$, then $\nu_{0}=\nu_{1}=\nu_{2}=\nu_{3}=0$ is equivalent to
$a_{0}^{2}(\beta-\alpha)=0$, that is, $a_{0}=0$. This implies $A=0$.
Therefore, $\delta_{2}\not=0$, $\delta_{1}=1$, $a_{1}=c_{1}=0$ yields $A=0$.

Consider $\delta_{2}\not=0$ and $\delta_{1}\not=1$, and note that (12) is
equivalent to:

$\displaystyle\left\\{\begin{array}[]{cl}a_{0}&=\delta_{1}a_{0}+\delta_{2}a_{0}^{2}(\beta-\alpha)+\delta_{2}\frac{\beta^{2}-\alpha^{2}}{2}a_{0}(a_{1}+c_{1})+\delta_{2}a_{1}c_{1}\frac{\beta^{3}-\alpha^{3}}{3}\\\
a_{1}&=\delta_{1}a_{1}+\delta_{2}a_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+\delta_{2}a_{1}a_{0}(\beta-\alpha)\\\
c_{1}&=\delta_{1}c_{1}+\delta_{2}a_{0}c_{1}(\beta-\alpha)+\delta_{2}c_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}\\\
0&=\delta_{2}a_{1}c_{1}(\beta-\alpha).\end{array}\right.$ (17)

Since $\alpha<\beta$ and $\delta_{2}\not=0$, equation
$\delta_{2}a_{1}c_{1}(\beta-\alpha)=0$ implies that either $a_{1}=0$ or
$c_{1}=0$. If $\delta_{2}\not=0$, $\delta_{1}\not=1$, $a_{1}\not=0$ and
$c_{1}=0$, then (17) becomes

$\displaystyle a_{0}$ $\displaystyle=$
$\displaystyle\delta_{1}a_{0}+\delta_{2}a_{0}^{2}(\beta-\alpha)+\delta_{2}\frac{\beta^{2}-\alpha^{2}}{2}a_{0}a_{1}$
$\displaystyle a_{1}$ $\displaystyle=$
$\displaystyle\delta_{1}a_{1}+\delta_{2}a_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+\delta_{2}a_{1}a_{0}(\beta-\alpha)$

which is equivalent to
$1=\delta_{1}+\delta_{2}(\beta-\alpha)a_{0}+\delta_{2}\frac{\beta^{2}-\alpha^{2}}{2}a_{1}$.
Then,

$a_{0}=\frac{2-2\delta_{1}-\delta_{2}(\beta^{2}-\alpha^{2})a_{1}}{2\delta_{2}(\beta-\alpha)}.$

If $\delta_{2}\not=0$, $\delta_{1}\not=1$, $a_{1}=0$ and $c_{1}\neq 0$, then
(17) becomes

$\displaystyle a_{0}$ $\displaystyle=$
$\displaystyle\delta_{1}a_{0}+\delta_{2}a_{0}^{2}(\beta-\alpha)+\delta_{2}\frac{\beta^{2}-\alpha^{2}}{2}a_{0}c_{1}$
$\displaystyle c_{1}$ $\displaystyle=$
$\displaystyle\delta_{1}c_{1}+\delta_{2}c_{1}^{2}\frac{\beta^{2}-\alpha^{2}}{2}+\delta_{2}c_{1}a_{0}(\beta-\alpha)$

which is equivalent to
$1=\delta_{1}+\delta_{2}(\beta-\alpha)a_{0}+\delta_{2}\frac{\beta^{2}-\alpha^{2}}{2}c_{1}$.
Then,

$a_{0}=\frac{2-2\delta_{1}-\delta_{2}(\beta^{2}-\alpha^{2})c_{1}}{2\delta_{2}(\beta-\alpha)}.$

If $\delta_{2}\not=0$, $\delta_{1}\not=1$, $a_{1}=0$ and $c_{1}=0$, then
$A\not=0$ yields $a_{0}\not=0$ and (17) becomes

$\displaystyle a_{0}=\delta_{1}a_{0}+\delta_{2}a_{0}^{2}(\beta-\alpha)$

which is equivalent to $a_{0}=\frac{1-\delta_{1}}{\delta_{2}(\beta-\alpha)}.$
∎

###### Remark 3.2

The integral operator given by
$(Ax)(t)=\int\limits_{\alpha_{1}}^{\beta_{1}}k(t,s)x(s)ds$ for almost all $t$,
where $k:[\alpha_{1},\beta_{1}]\times[\alpha_{1},\beta_{1}]\to\mathbb{R}$ is a
measurable function that satisfies

$\int\limits_{\alpha_{1}}^{\beta_{1}}\left(\int\limits_{\alpha_{1}}^{\beta_{1}}|k(t,s)|^{q}ds\right)^{\frac{p}{q}}dt<\infty,$

by (HutsonPymDST, , Theorem 3.4.10) is a well defined and bounded linear
operator from $L_{p}{[\alpha_{1},\beta_{1}]}$ to
$L_{p}{[\alpha_{1},\beta_{1}]}$, where $1<p<\infty$, $1<q<\infty$,
$\frac{1}{p}+\frac{1}{q}=1$.

###### Remark 3.3

If in the Corollary 3.4 when $0\not\in[M_{1},M_{2}]$, one takes $b(t)$ to be a
Laurent polynomial with only negative powers of $t$ then there is no non-zero
kernel $k(t,s)=a_{0}+a_{1}t+c_{1}s$ (there is no $A\not=0$ with such kernels)
such that $AB=BF(A)$. In fact, let $n$ be a positive integer and consider
$b(t)=\sum\limits_{j=1}^{n}b_{j}t^{-j}$, where $t\in[M_{1},M_{2}]$,
$b_{j}\in\mathbb{R}$ for $j=1,\ldots,n$ and $b_{n}\not=0$. We set $k_{1}(t,s)$
as defined by (6). Then we have

$\displaystyle
b(t)F_{2}(k(t,s))=b(t)(\delta_{1}k(t,s)+\delta_{2}k_{1}(t,s))=(a_{0}\delta_{1}+\delta_{2}\nu_{0})\sum_{j=1}^{n}b_{j}t^{-j}$
$\displaystyle+(a_{1}\delta_{1}+\delta_{2}\nu_{1})\sum_{j=1}^{n}b_{j}t^{-j+1}+(c_{1}\delta_{1}+\delta_{2}\nu_{2})\sum_{j=1}^{n}b_{j}t^{-j}s+\nu_{3}\delta_{2}\sum_{j=1}^{n}b_{j}t^{-j+1}s$
$\displaystyle=(a_{1}\delta_{1}+\delta_{2}\nu_{1})b_{1}+\nu_{3}\delta_{2}b_{1}s+\sum_{j=1}^{n-1}((a_{0}\delta_{1}+\delta_{2}\nu_{0})b_{j}+(a_{1}\delta_{1}+\delta_{2}\nu_{1})b_{j+1})t^{-j}$
$\displaystyle
k(t,s)b(s)=\,a_{0}\sum\limits_{j=1}^{n}b_{j}s^{-j}+a_{1}\sum\limits_{j=1}^{n}b_{j}s^{-j}t+c_{1}\sum\limits_{j=1}^{n}b_{j}s^{-j+1}$
$\displaystyle=$
$\displaystyle\,c_{1}b_{1}+\sum\limits_{j=1}^{n-1}(a_{0}b_{j}+c_{1}b_{j+1})s^{-j}+\sum\limits_{j=1}^{n}a_{1}b_{j}s^{-j}t+a_{0}b_{n}s^{-n}.$

Thus we have $k(t,s)b(s)=b(t)F_{2}(k(t,s))$ for almost every
$(t,s)\in[M_{1},M_{2}]\times[\alpha,\beta]$ if and only if

$\displaystyle c_{1}b_{1}$ $\displaystyle=$ $\displaystyle
a_{1}\delta_{1}b_{1}+\delta_{2}\nu_{1}b_{1},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle\delta_{2}\nu_{3}b_{1},$ $\displaystyle 0$
$\displaystyle=$
$\displaystyle(a_{0}\delta_{1}+\delta_{2}\nu_{0})b_{j}+(\delta_{1}a_{1}+\delta_{2}\nu_{1})b_{j+1},\quad
1\leq j\leq n-1,$ $\displaystyle a_{0}b_{j}+c_{1}b_{j+1}$ $\displaystyle=$
$\displaystyle 0,\quad 1\leq j\leq n-1,$ (18) $\displaystyle 0$
$\displaystyle=$ $\displaystyle
c_{1}\delta_{1}b_{j}+\delta_{2}\nu_{2}b_{j}+\delta_{2}\nu_{3}b_{j+1},\quad
1\leq j\leq n-1,$ $\displaystyle a_{1}b_{j}$ $\displaystyle=$ $\displaystyle
0,\quad 1\leq j\leq n,$ (19) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
a_{0}\delta_{1}b_{n}+\delta_{2}\nu_{0}b_{n},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle a_{0}b_{n},$ (20) $\displaystyle 0$
$\displaystyle=$ $\displaystyle c_{1}\delta_{1}b_{n}+\delta_{2}\nu_{3}b_{n}.$

Since $b_{n}\not=0$ then from the equation (20) we have $a_{0}b_{n}=0$ and
thus $a_{0}=0$. From the equation (18) for $j=n-1$ we get $c_{1}b_{n}=0$ and
thus $c_{1}=0$. Finally from the equation (19) we have $0=a_{1}b_{j}$ for
$j=n$ and thus $a_{1}=0$. This implies that $k(t,s)\equiv 0$, that is, $A=0$.
So $b_{n}\not=0$ implies $A=0$.

###### Corollary 3.5

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$, be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=\int\limits_{\alpha}^{\beta}k(t,s)x(s)ds,\quad(Bx)(t)=b(t)x(t),\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where $k(t,s):\mathbb{R}\times[\alpha,\beta]\to\mathbb{R}$ is a measurable
function, and $b\in L_{\infty}(\mathbb{R})$ is a nonzero function such that
the set $\operatorname{\rm supp\,}b(t)\cap[\alpha,\beta]$ has measure zero.
Consider a polynomial $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$, where
$\delta_{0},\ldots,\delta_{n}\in\mathbb{R}$. Let

$\displaystyle k_{0}(t,s)=k(t,s),\quad\quad
k_{m}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k_{m-1}(\tau,s)d\tau,\quad
m=1,\ldots,n,$ $\displaystyle F_{0}(k(t,s))=0,\quad
F_{n}(k(t,s))=\sum_{j=1}^{n}\delta_{j}k_{j-1}(t,s),\quad n\geq 1.$

Then $AB=BF(A)$ if and only if $\delta_{0}=0$ and the set

$\displaystyle\operatorname{\rm supp\,}b(t)\,\cap\,\operatorname{\rm
supp\,}F_{n}(k(t,s))$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$.

###### Proof

Suppose that the set $\operatorname{\rm supp\,}b\cap[\alpha,\beta]$ has
measure zero. By Proposition 3.3 we have $AB=BF(A)$ if and only if condition
(3) holds, that is,

$\displaystyle\forall\ x\in L_{p}(\mathbb{R}):\quad
b(t)\delta_{0}x(t)+b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds,$

almost everywhere. By taking $x(\cdot)=I_{[M_{1},M_{2}]}(\cdot)b(\cdot)$,
where $M_{1},M_{2}\in\mathbb{R}$, $M_{1}<M_{2}$,
$[M_{1},M_{2}]\supset[\alpha,\beta]$,
$\tilde{m}([M_{1},M_{2}]\setminus[\alpha,\beta])>0$, $I_{E}(\cdot)$ is the
indicator function of the set $E$, the condition (3) reduces to

$I_{[M_{1},M_{2}]}(\cdot)b(\cdot)^{2}\delta_{0}=0.$

Since $b$ has support with positive measure (otherwise $B\equiv 0$), then
$\delta_{0}=0$. By using this, condition (3) reduces to the following

$\displaystyle\forall\ x\in L_{p}(\mathbb{R}):\quad
b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)b(s)x(s)ds.$

By hypothesis the right hand side is equal zero. Then condition (3) reduces to

$\displaystyle\forall\ x\in L_{p}(\mathbb{R}):\quad
b(t)\int\limits_{\alpha}^{\beta}F_{n}(k(t,s))x(s)ds=0.$

This is equivalent to

$\displaystyle b(t)F_{n}(k(t,s))=0\quad\mbox{ for almost every }\
s\in[\alpha,\beta].$ (21)

By applying a similar argument as in the proof of Proposition 3.3 we conclude
that condition (21) is equivalent to that the set

$\displaystyle\operatorname{\rm supp\,}b(t)\,\cap\,\operatorname{\rm
supp\,}F_{n}(k(t,s))$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$. ∎

###### Corollary 3.6

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1\leq p\leq\infty$, be defined as follows, for almost all
$t$,

$\displaystyle(Ax)(t)=\int\limits_{\alpha}^{\beta}a(t)c(s)x(s)ds,\quad(Bx)(t)=b(t)x(t),\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where $a:\mathbb{R}\to\mathbb{R}$, $c:[\alpha,\beta]\to\mathbb{R}$,
$b:\mathbb{R}\to\mathbb{R}$ are measurable functions. Consider a polynomial
defined by $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, where
$\delta_{1},\ldots,\delta_{n}\in\mathbb{R}$. Let
$\mu=\int\limits_{\alpha}^{\beta}a(s)c(s)ds.$ Then, $AB=BF(A)$ if and only if
the set

$\displaystyle\operatorname{\rm supp\,}(a(t)c(s))\cap\operatorname{\rm
supp\,}\left(b(t)\sum\limits_{j=1}^{n}\delta_{j}\mu^{j-1}-b(s)\right)$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$.

###### Proof

We set $k(t,s)=a(t)c(s)$, so we have

$\displaystyle k_{0}(t,s)=k(t,s)=a(t)c(s),\quad$ $\displaystyle
k_{m}(t,s)=\int\limits_{\alpha}^{\beta}k(t,\tau)k_{m-1}(\tau,s)d\tau=a(t)c(s)\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{m},\
m=1,\ldots,n$ $\displaystyle
F_{n}(k(t,s))=\sum_{j=1}^{n}\delta_{j}k_{j-1}(t,s)=\sum_{j=1}^{n}\delta_{j}a(t)c(s)\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{j-1},\
\quad n\geq 1.$

By applying Proposition 3.3 we have $AB=BF(A)$ if and only if

$\displaystyle
b(t)\sum\limits_{j=1}^{n}\delta_{j}a(t)c(s)\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{j-1}=a(t)c(s)b(s)\
\Longleftrightarrow$ $\displaystyle
a(t)c(s)\left(b(t)\sum\limits_{j=1}^{n}\delta_{j}\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{j-1}-b(s)\right)=0$

for almost every $(t,s)$ in $\mathbb{R}\times[\alpha,\beta]$. The last
condition is equivalent to the set

$\displaystyle\operatorname{\rm supp\,}(a(t)c(s))\cap\operatorname{\rm
supp\,}\left(b(t)\sum\limits_{j=1}^{n}\delta_{j}\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{j-1}-b(s)\right)$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$. We complete the proof by
noticing that the corresponding set can be written as

$\displaystyle\operatorname{\rm supp\,}(a(t)c(s))\cap\operatorname{\rm
supp\,}\left(b(t)\sum\limits_{j=1}^{n}\delta_{j}\mu^{j-1}-b(s)\right),$

where $\displaystyle\mu=\int\limits_{\alpha}^{\beta}a(s)c(s)ds$. ∎

###### Example 3.7

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$(Ax)(t)=\int\limits_{0}^{2}a(t)c(s)x(s)ds,\ (Bx)(t)=b(t)x(t),$

where $a(t)=2tI_{[0,2]}(t)$, $c(s)=I_{[0,1]}(s)$, $b(t)=I_{[1,2]}(t)t^{2}$.
Since kernel has compact support, we can apply (HutsonPymDST, , Theorem
3.4.10) and, we conclude that operators $A$ is well defined and bounded. Since
function $b$ has $4$ as an upper bound then $\|B\|_{L_{p}}\leq 4$. Hence
operator $B$ is well defined and bounded. Consider a polynomial defined by
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, where
$\delta_{1},\ldots,\delta_{n}\in\mathbb{R}$. Then, the above operators satisfy
the relation $AB=BF(A)$ if and only if $\sum\limits_{j=1}^{n}\delta_{j}=0$. In
fact, by applying Corollary 3.6 we have

$\displaystyle\mu=\int\limits_{0}^{2}a(s)c(s)ds=1.$

Hence, $\operatorname{\rm supp\,}\\{b(t)\cdot
0-b(s)\\}=\mathbb{R}\times[1,2].$ Moreover, $\operatorname{\rm
supp\,}a(t)c(s)=[0,2]\times[0,1].$ The set $\operatorname{\rm
supp\,}(a(t)c(s))\cap\operatorname{\rm supp\,}\left(-b(s)\right),$ has measure
zero in $\mathbb{R}\times[0,2]$.

###### Example 3.8

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=\int\limits_{0}^{2}a(t)c(s)x(s)ds,\quad(Bx)(t)=b(t)x(t),$

where $a(t)=I_{[0,2]}(t)\sin(\pi t)$, $c(s)=I_{[0,1]}(s)$,
$b(t)=I_{[1,2]}(t)t^{2}$. Since $a\in L_{p}(\mathbb{R})$ and $c\in
L_{q}[0,2]$, $1<q<\infty$, $\frac{1}{p}+\frac{1}{q}=1$, by applying Hölder
inequality we have that operator $A$ is well defined and bounded. The function
$b\in L_{\infty}$, so $B$ is well defined and bounded because
$\|B\|_{L_{p}}\leq\|b\|_{L_{\infty}}$ we conclude that operator $B$ is well
defined and bounded. Consider a polynomial defined by $F(z)=\delta z^{d}$,
where $\delta\not=0$ is a real constant and $d$ is a positive integer $d\geq
2$. Then, the above operators satisfy the relation $AB=\delta BA^{d}.$ In
fact, by applying Corollary 3.6 we have $\mu=\int\limits_{0}^{2}a(s)c(s)ds=0.$
Hence, $\operatorname{\rm supp\,}\\{b(t)\cdot
0-b(s)\\}=\mathbb{R}\times[1,2].$ Moreover, $\operatorname{\rm
supp\,}a(t)c(s)=[0,2]\times[0,1].$ The set $\operatorname{\rm
supp\,}(a(t)c(s))\cap\operatorname{\rm supp\,}\left(-b(s)\right),$ has measure
zero in $\mathbb{R}\times[0,2]$.

###### Example 3.9

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$, be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=\int\limits_{\alpha}^{\beta}I_{[\alpha,\beta]}(t)x(s)ds,\quad(Bx)(t)=I_{[\alpha,\beta]}(t)x(t),\quad\alpha,\beta\in\mathbb{R},\alpha<\beta.$

Since kernel has compact support, we can apply (HutsonPymDST, , Theorem
3.4.10) and, we conclude that operator $A$ is well defined and bounded. Since
$\|B\|_{L_{p}}\leq 1$ then operator $B$ is well defined and bounded. Consider
a polynomial defined by $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, where
$\delta_{1},\ldots,\delta_{n}\in\mathbb{R}$. Then, the above operators satisfy
the relation $AB=BF(A)$ if and only if
$\sum\limits_{j=1}^{n}\delta_{j}(\beta-\alpha)^{j-1}=1.$ Indeed, if
$a(t)=b(t)=I_{[\alpha,\beta]}(t)$, $c(s)=1$ and

$\displaystyle\lambda=\sum\limits_{j=1}^{n}\delta_{j}\left(\int\limits_{\alpha}^{\beta}a(s)c(s)ds\right)^{j-1}=\sum\limits_{j=1}^{n}\delta_{j}(\beta-\alpha)^{j-1},$

then from Corollary 3.6 we have the following:

  * •

If $\lambda\not=0$, $\lambda\not=1$,

$\displaystyle\operatorname{\rm
supp\,}(b(t)\lambda-b(s)))=\left\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
\lambda I_{[\alpha,\beta]}(t)\not=1\right\\}=\mathbb{R}\times[\alpha,\beta],$
$\displaystyle\operatorname{\rm
supp\,}a(t)c(s)=\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
I_{[\alpha,\beta]}(t)\not=0\\}=[\alpha,\beta]\times[\alpha,\beta].$

The set $\operatorname{\rm supp\,}(\lambda b(t)-b(s))\cap\operatorname{\rm
supp\,}(a(t)c(s))=[\alpha,\beta]\times[\alpha,\beta]$ has positive measure.

  * •

If $\lambda=1$,

$\displaystyle\operatorname{\rm supp\,}(b(t)-b(s))$ $\displaystyle=$
$\displaystyle\left\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
I_{[\alpha,\beta]}(t)\not=1\right\\}$ $\displaystyle=$
$\displaystyle(\mathbb{R}\setminus[\alpha,\beta])\times[\alpha,\beta].$

The set $\operatorname{\rm supp\,}(b(t)-b(s))\cap\operatorname{\rm
supp\,}(a(t)c(s))$ has measure zero in $\mathbb{R}\times[\alpha,\beta]$.

  * •

If $\lambda=0$,

$\displaystyle\operatorname{\rm supp\,}(\lambda b(t)-b(s))$ $\displaystyle=$
$\displaystyle\operatorname{\rm
supp\,}b(s)=\left\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
I_{[\alpha,\beta]}(s)\not=0\right\\}$ $\displaystyle=$
$\displaystyle\left\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\ \alpha\leq
s\leq\beta\right\\}.$

The set $\operatorname{\rm supp\,}b(s)\cap\operatorname{\rm
supp\,}(a(t)c(s))=[\alpha,\beta]\times[\alpha,\beta]$ has measure
$(\beta-\alpha)^{2}$.

The conditions in the Corollary 3.6 are fulfilled only in the second case,
that is, when $\lambda=1$.

#### Representations when $A$ is multiplication operator and $B$ is integral
operator

###### Proposition 3.10

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=a(t)x(t),\quad(Bx)(t)=\int\limits_{\alpha}^{\beta}k(t,s)x(s)ds,\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where $a:\mathbb{R}\to\mathbb{R}$,
$k:\mathbb{R}\times[\alpha,\beta]\to\mathbb{R}$ are measurable functions.
Consider a polynomial defined by $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$,
where $\delta_{0},\ldots,\delta_{n}\in\mathbb{R}$. Then

$\displaystyle AB=BF(A)$

if and only if the set

$\displaystyle\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}k(t,s)$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$.

###### Proof

We have for almost every $t\in\mathbb{R}$

$\displaystyle(ABx)(t)=\int\limits_{\alpha}^{\beta}a(t)k(t,s)x(s)ds$
$\displaystyle(A^{n}x)(t)=[a(t)]^{n}x(t)$
$\displaystyle(F(A)x)(t)=\sum_{i=0}^{n}\delta_{i}(A^{i}x)(t)=\left(\sum_{i=0}^{n}\delta_{i}[a(t)]^{i}\right)x(t)=F(a(t))x(t)$
$\displaystyle(BF(A)x)(t)=\int\limits_{\alpha}^{\beta}k(t,s))F(a(s))x(s)ds.$

Then we have $ABx=BF(A)x$ if and only if

$\displaystyle\int\limits_{\alpha}^{\beta}a(t)k(t,s)x(s)ds=\int\limits_{\alpha}^{\beta}k(t,s)F(a(s))x(s)ds.$
(22)

almost everywhere. By using Lemma 3.1 and by applying the same argument as in
the final steps on the proof of Proposition 3.3, the condition (22) is
equivalent to

$\displaystyle a(t)k(t,s)=k(t,s)F(a(s))\Longleftrightarrow
k(t,s)(a(t)-F(a(s)))=0$

for almost every $(t,s)$ in $\mathbb{R}\times[\alpha,\beta]$.

Since the variables $t$ and $s$ are independent, this is true if and only if
the set

$\displaystyle\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}k(t,s)$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$. ∎

###### Example 3.11

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=I_{[\alpha,\beta]}(t)x(t),\quad(Bx)(t)=\int\limits_{\alpha}^{\beta}I_{[\alpha,\beta]^{2}}(t,s)x(s)ds,\quad\alpha,\beta\in\mathbb{R},\alpha<\beta$

By using properties of norm and (HutsonPymDST, , Theorem 3.4.10),
respectively, for operators $A$ and $B$, we conclude that operators $A$ and
$B$ are well defined and bounded. For a monomial defined by $F(z)=z^{n}$,
$n=1,2,\ldots$, the above operators satisfy the relation $AB=BF(A).$ In fact,
by setting $a(t)=I_{[\alpha,\beta]}(t)$, $k(t,s)=I_{[\alpha,\beta]^{2}}(t,s)$
we have

$\displaystyle\operatorname{\rm
supp\,}(a(t)-F(a(s)))=\left\\{(t,s)\in\mathbb{R}\times[\alpha,\beta]:\
I_{[\alpha,\beta]}(t)\not=1\right\\}=(\mathbb{R}\setminus[\alpha,\beta])\times[\alpha,\beta],$
$\displaystyle\operatorname{\rm
supp\,}k(t,s)=[\alpha,\beta]\times[\alpha,\beta].$

The set $\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}(k(t,s))$ has measure zero in $\mathbb{R}\times[\alpha,\beta]$. So the
result follows from Proposition 3.10.

###### Example 3.12

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ defined as follows, for almost all $t$,

$(Ax)(t)=(\gamma_{1}I_{[0,1/2)}(t)-\gamma_{2}I_{[1/2,1]}(t))x(t),\quad(Bx)(t)=\int\limits_{0}^{1}k(t,s)x(s)ds$

$k:\mathbb{R}\times[0,1]\to\mathbb{R}$ is a Lebesgue measurable function such
that $B$ is well defined. The operator $A$ is well defined and bounded.
Consider a polynomial defined by $F(z)=\delta_{0}+\delta_{1}z$, where
$\delta_{0}$, $\delta_{1}$, $\gamma_{1}$, $\gamma_{2}$ are constants such that

$\displaystyle|\delta_{0}|+|\delta_{1}|+|\gamma_{1}|+|\gamma_{2}|\not=0.$

If $k(\cdot,\cdot)$ is a measurable function such that one of the following is
fulfilled:

  1. 1)

$\delta_{0}=-\delta_{1}\gamma_{1}$ and $\operatorname{\rm supp\,}\
k(t,s)\subseteq(\mathbb{R}\setminus[0,1])\times[0,1/2]$;

  2. 2)

$\delta_{0}=\delta_{1}\gamma_{2}$ and $\operatorname{\rm supp\,}\
k(t,s)\subseteq(\mathbb{R}\setminus[0,1])\times[1/2,1]$;

  3. 3)

$\delta_{0}+\delta_{1}\gamma_{1}-\gamma_{1}=0$ and $\operatorname{\rm
supp\,}k(t,s)\subseteq[0,1/2]\times[0,1/2]$;

  4. 4)

$\delta_{0}+\delta_{1}\gamma_{1}+\gamma_{2}=0$ and $\operatorname{\rm
supp\,}k(t,s)\subseteq[1/2,1]\times[0,1/2]$;

  5. 5)

$\delta_{0}-\delta_{1}\gamma_{2}-\gamma_{1}=0$ and $\operatorname{\rm
supp\,}k(t,s)\subseteq[0,1/2]\times[1/2,1]$;

  6. 6)

$\delta_{0}-\delta_{1}\gamma_{2}+\gamma_{2}=0$ and $\operatorname{\rm
supp\,}k(t,s)\subseteq[1/2,1]\times[1/2,1]$,

then the above operators satisfy the relation $AB=BF(A)$.

In fact, putting $a(t)=\gamma_{1}I_{[0,1/2)}(t)-\gamma_{2}I_{[1/2,1]}(t)$ we
have

$\displaystyle[a(t)-F(a(s))]=\left\\{\begin{array}[]{cccc}0,&\mbox{ if
}\delta_{0}=-\delta_{1}\gamma_{1},&t\not\in[0,1],&s\in[0,1/2)\\\ 0,&\mbox{ if
}\delta_{0}=\delta_{1}\gamma_{2},&t\not\in[0,1],&s\in[1/2,1]\\\ 0,&\mbox{ if
}\delta_{0}+\delta_{1}\gamma_{1}-\gamma_{1}=0,&t\in[0,1/2),&s\in[0,1/2)\\\
0,&\mbox{ if
}\delta_{0}+\delta_{1}\gamma_{1}+\gamma_{2}=0,&t\in[1/2,1),&s\in[0,1/2]\\\
0,&\mbox{ if
}\delta_{0}-\delta_{1}\gamma_{2}-\gamma_{1}=0,&t\in[0,1/2],&s\in[1/2,1]\\\
0,&\mbox{ if
}\delta_{0}-\delta_{1}\gamma_{2}+\gamma_{2}=0,&t\in[1/2,1],&s\in[1/2,1]\\\
\gamma_{3},&\mbox{ otherwise }&&\\\ \end{array}\right.$

where $\gamma_{3}$ can be different from zero depending on the constants
involved. Thus, in each condition we can choose $k(t,s)=I_{S}(t,s)$, where
$S=\\{(t,s)\in\mathbb{R}\times[0,1]:\ a(t)-F(a(s))=0\\}$ and with a positive
measure. Or for instance we can take:

  1. 1)

$k(t,s)=I_{[2,3]\times[0,1/2]}(t,s)$ if $\delta_{0}=-\delta_{1}\gamma_{1}$;

  2. 2)

$k(t,s)=I_{[2,3]\times[1/2,1]}(t,s)$ if $\delta_{0}=\delta_{1}\gamma_{2}$;

  3. 3)

$k(t,s)=I_{[0,1/3]\times[1/3,1/2]}(t,s)$ if
$\delta_{0}+\delta_{1}\gamma_{1}-\gamma_{1}=0$;

  4. 4)

$k(t,s)=I_{[2/3,1/2]\times[0,1/2]}(t,s)$ if
$\delta_{0}+\delta_{1}\gamma_{1}+\gamma_{2}=0$;

  5. 5)

$k(t,s)=I_{[0,1/3]\times[2/3,1]}(t,s)$ if
$\delta_{0}-\delta_{1}\gamma_{2}-\gamma_{1}=0$;

  6. 6)

$k(t,s)=I_{[2/3,1]\times[2/3,1]}(t,s)$ if
$\delta_{0}-\delta_{1}\gamma_{2}+\gamma_{2}$.

According to the definition, in all above cases the set

$\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm supp\,}\
(k(t,s))$

has measure zero in $\mathbb{R}\times[0,1]$. So the result follows from
Proposition 3.10.

###### Corollary 3.13

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=a(t)x(t),\quad(Bx)(t)=\int\limits_{\alpha}^{\beta}b(t)c(s)x(s)ds,\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where $a:\mathbb{R}\to\mathbb{R}$, $b:\mathbb{R}\to\mathbb{R}$,
$c:[\alpha,\beta]\to\mathbb{R}$ are measurable functions. For a polynomial
defined by $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$, where
$\delta_{0},\ldots,\delta_{n}\in\mathbb{R}$, we have

$\displaystyle AB=BF(A)$

if and only if the set

$\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}(b(t)c(s))$

has measure zero in $\mathbb{R}\times[\alpha,\beta]$.

###### Proof

This follows by Proposition 3.10. ∎

###### Example 3.14

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$(Ax)(t)=a(t)x(t),\quad(Bx)(t)=\int\limits_{\alpha}^{\beta}b(t)c(s)x(s)ds,\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where $a(t)=-1+I_{[\alpha,\beta]}(t)$, $b(t)=I_{[\alpha-2,\alpha-1]}(t)$,
$c(s)=1$. We have that $a\in L_{\infty}(\mathbb{R})$ and so
$\|A\|_{L_{p}}\leq\|a\|_{L_{\infty}}$. Therefore, $A$ is well defined and
bounded. Since kernel has compact support in $\mathbb{R}\times[\alpha,\beta]$,
we can apply (HutsonPymDST, , Theorem 3.4.10) and, we conclude that operators
$B$ is well defined and bounded. Consider a polynomial defined by
$F(z)=-1+\delta_{1}z$, where $\delta_{1}$ is a real constant. Then the above
operators satisfy the relation $AB=BF(A).$ In fact, for
$(t,s)\in\mathbb{R}\times[\alpha,\beta]$ we have

$\displaystyle F(a(s))-a(t)$ $\displaystyle=$
$\displaystyle-\delta_{1}+\delta_{1}I_{[\alpha,\beta]}(s)-I_{[\alpha,\beta]}(t)=-I_{[\alpha,\beta]}(t).$

Therefore, we have

$\displaystyle\operatorname{\rm
supp\,}(a(t)-F(a(s)))=[\alpha,\beta]\times[\alpha,\beta],$
$\displaystyle\operatorname{\rm supp\,}b(t)c(s)=\operatorname{\rm supp\,}\
I_{[\alpha-2,\alpha-1]}(t)I_{[\alpha,\beta]}(s)=[\alpha-2,\alpha-1]\times[\alpha,\beta].$

The set $\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}(I_{[\alpha-2,\alpha-1]}(t)I_{[\alpha,\beta]}(s))$ has measure zero. So
the result follows from Corollary 3.13.

###### Example 3.15

Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $B:L_{p}(\mathbb{R})\to
L_{p}(\mathbb{R})$, $1<p<\infty$ be defined as follows, for almost all $t$,

$\displaystyle(Ax)(t)=a(t)x(t),\quad(Bx)(t)=\int\limits_{\alpha}^{\beta}b(t)c(s)x(s)ds,\quad\alpha,\beta\in\mathbb{R},\alpha<\beta,$

where
$a(t)=\gamma_{0}+I_{\left[\alpha,\frac{\alpha+\beta}{2}\right]}(t)t^{2}$,
$\gamma_{0}$ is a real number, $b(t)=(1+t^{2})I_{[\beta+1,\beta+2]}(t)$,
$c(s)=I_{\left[\frac{\alpha+\beta}{2},\beta\right]}(s)(1+s^{4})$. Consider a
polynomial defined by $F(z)=\delta_{0}+\delta_{1}z$, where
$\delta_{0},\delta_{1}$ are real constants and $\delta_{1}\not=0$. If
$\delta_{0}=\gamma_{0}-\delta_{1}\gamma_{0}$ then the above operators satisfy
the relation

$\displaystyle AB-\delta_{1}BA=\delta_{0}B.$

In fact, $A$ is well defined, bounded since $a\in L_{\infty}$ and this implies
$\|A\|_{L_{p}}\leq\|a\|_{L_{\infty}}$. Operator $B$ is well defined, bounded
since $k(t,s)=b(t)c(s)$, $(t,s)\in\mathbb{R}\times[\alpha,\beta]$ has compact
support and satisfies conditions of (HutsonPymDST, , Theorem 3.4.10). If
$\delta_{0}=\gamma_{0}-\delta_{1}\gamma_{0}$ then we have

$\displaystyle F(a(s))-a(t)$ $\displaystyle=$
$\displaystyle\delta_{0}+\gamma_{0}\delta_{1}+\delta_{1}I_{\left[\alpha,\frac{\alpha+\beta}{2}\right]}(s)s^{2}-\gamma_{0}-I_{\left[\alpha,\frac{\alpha+\beta}{2}\right]}(t)t^{2}$
$\displaystyle=$
$\displaystyle\delta_{1}I_{\left[\alpha,\frac{\alpha+\beta}{2}\right]}(s)s^{2}-I_{\left[\alpha,\frac{\alpha+\beta}{2}\right]}(t)t^{2}.$

Then we have

$\operatorname{\rm
supp\,}[a(t)-F(a(s))]=\left(\mathbb{R}\times\left[\alpha,\frac{\alpha+\beta}{2}\right]\cup\left[\alpha,\frac{\alpha+\beta}{2}\right]\times\left[\frac{\alpha+\beta}{2},\beta\right]\right)\setminus
W,$

where $W\subseteq\mathbb{R}\times[\alpha,\beta]$ is a set with Lebesgue
measure zero, and

$\displaystyle\operatorname{\rm supp\,}\ b(t)c(s)$ $\displaystyle=$
$\displaystyle\operatorname{\rm supp\,}\
(1+t^{2})I_{[\beta+1,\beta+2]}(t)I_{\left[\frac{\alpha+\beta}{2},\beta\right]}(s)(1+s^{4})$
$\displaystyle=$
$\displaystyle[\beta+1,\beta+2]\times\left[\frac{\alpha+\beta}{2},\beta\right].$

The set $\operatorname{\rm supp\,}(a(t)-F(a(s)))\cap\operatorname{\rm
supp\,}(b(t)c(s))$ has measure zero. So the result follows from Corollary
3.13.

###### Acknowledgements.

This work was supported by the Swedish International Development Cooperation
Agency (Sida) bilateral program with Mozambique. Domingos Djinja is grateful
to the research environment Mathematics and Applied Mathematics (MAM),
Division of Mathematics and Physics, School of Education, Culture and
Communication, Mälardalen University for excellent environment for research in
Mathematics. Partial support from Swedish Royal Academy of Sciences is also
gratefully acknowledged.

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