∎

11institutetext: Domingos Djinja 22institutetext: Department of Mathematics
and Informatics, Faculty of Sciences, Eduardo Mondlane University, Box 257,
Maputo, Mozambique. Division of Mathematics and Physics, Division of
Mathematics and Physics, School of Education, Culture and Communication,
Mälardalens University, Mälardalens University, Box 883, Västerȧs, Sweden.
22email<EMAIL_ADDRESS><EMAIL_ADDRESS>33institutetext: Sergei Silvestrov 44institutetext: Division of Mathematics
and Physics, School of Education, Culture and Communication, Mälardalens
University, Box 883, Västerȧs, Sweden.
44email<EMAIL_ADDRESS>55institutetext: Alex Behakanira Tumwesigye
66institutetext: Department of Mathematics, College of Natural Sciences,
Makerere University, Box 7062, Kampala, Uganda.
66email<EMAIL_ADDRESS>
# Representations of polynomial covariance type commutation relations by
piecewise function multiplication and composition operators

Domingos Djinja  Sergei Silvestrov  Alex Behakanira Tumwesigye

###### Abstract

Representations of polynomial covariance type commutation relations are
constructed on Banach spaces $L_{p}$ and $C[\alpha,\beta]\
\alpha,\beta\in\mathbb{R}$. Representations involve operators with piecewise
functions, multiplication operators and inner superposition operators.
MSC2020 Classification: 47B33, 47L80, 47L65

###### Keywords:

piecewise function multiplication operator, composition operator, covariance
commutation relations

##  1 Introduction

In many areas of applications, there can be found relations of the form

$\displaystyle ST=F(TS)$ (1)

where $S,T$ are elements of an associative algebra and
$F:\mathbb{R}\to\mathbb{R}$ is a function satisfying certain conditions. For
example, if $F(z)=z$, then $S$ and $T$ commute. If $F(z)=-z$, then $S$ and $T$
anti-commute. If $F(z)=\delta_{0}+\delta_{1}z$, then $S$ and $T$ satisfy the
relation

$\displaystyle ST-\delta_{1}TS=\delta_{0}I,$ (2)

where $\delta_{0},\delta_{1}$ are constants and $I$ is the identity element.
This relation is known as the deformed Heisenberg commutation relation
HellstromSSDST2 . If $\delta_{0}=\delta_{1}=1$, then (2) reduces to the
canonical Heisenberg commutation relation $ST-TS=I$ which is important in
differential and integral calculus and quantum physics. If $\delta_{0}=0$ then
(2) reduces to the quantum plane relation $ST=\delta_{1}TS.$ Elements of an
algebra that satisfy a commutation relation are called a representation of
this commutation 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 ; DaubechiesIngridDST2 ; JorgWavSignFracbook ; JorgOpRepTh88 ;
JorMoore84 ; MACbook1 ; MACbook2 ; MACbook3 ; OstSambookDST2 ; Pedbook79 ;
SamoilenkobookDST2 . The structure of representations for the relations of the
form (1), (3), (4), 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 ; OstSambookDST2 ; PerssonSilvestrov031 ;
PerssonSilvestrov032 ; PersSilv:CommutRelDinSystDST2 ; RST16 ; RSST16 ;
SamoilenkobookDST2 ; SaV8894 ; SilPhD95 ; STomdynsystype1 ; SilWallin96 ;
SvSJ07a ; SvSJ07b ; SvSJ07c ; SvT09 ; Tomiyama87 ; Tomiama:SeoulLN1992 ;
Tomiama:SeoulLN2part2000 ; AlexThesis2018DST2 ;
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 and interplay with dynamical systems generated by
iteration of the maps involved in the commutation relations.

In case of $*$-algebras, when $S=X$ and $T=X^{*}$ where $X$ is an element in
the algebra, the relation (1) reduces to $XX^{*}=F(X^{*}X)$. This relation
often can be transformed to relations of the form

$\displaystyle AB=BF(A),$ (3) $\displaystyle BA=F(A)B$ (4)

for some other elements $A,B$ of the $*$-algebra obtained from $X$ and $X^{*}$
using some transformations or factorizations in an appropriate functional
calculus (see for example OstSambookDST2 ; PersSilv:CommutRelDinSystDST2 ;
SamoilenkobookDST2 ; SaV8894 ; VaislebSa90 ; AlexThesis2018DST2  and
references cited their). A description of the structure of representations of
relation (3) by bounded and unbounded self-adjoint linear operators on Hilbert
space by using spectral representation AkkLnearOperatorsDST2  of such
operators is given in SamoilenkobookDST2 , where also more general families of
commuting self-adjoint operators satisfying relation (3) with other operators
on Hilbert spaces are considered using spectral theory and non-commutative
analysis for bounded and unbounded operators on Hilbert spaces.

In this paper we construct representations of relation (3) and (4) by linear
operators acting on Banach spaces $L_{p}$ and $C[\alpha,\beta]$ for
$\alpha,\beta\in\mathbb{R}$, and $F$ is a polynomial. 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. We consider representations of
(3) involving linear operators with piecewise function multiplication
operators, multiplication operators and inner superposition operators. We
derive conditions on the parameters or functional coefficients of operators so
that they satisfy (3) for a polynomial $F$. In contrast to OstSambookDST2 ;
SamoilenkobookDST2 ; SaV8894 ; VaislebSa90  devoted to involutive
representations of covariance type relations by operators on Hilbert spaces
using spectral theory of operators on Hilbert spaces, aiming 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 five sections, after the introduction, we present
in Section 2 preliminaries, notations and basic definitions. In Section 3, we
present representations involving piecewise function multiplication operators
acting on $L_{p}$ for $1<p<\infty$. In Section 4, we construct representations
involving inner superposition operators. These operators are important in
wavelets analysis for instance. In Section 5 we construct representations by
multiplication operators acting on $L_{p}$ for $1<p<\infty$ and the space of
continuous functions.

##  2 Preliminaries and notations

We use the following basic standard definitions and notations (see for example
AdamsGDST2 ; BrezisFASobolevSpacesDST2 ; ConwayFunctionalAnalysisDST2 ;
DudleyDST2 ; FollandRADST2 ; KantarovitchAkilovFunctionalAnalysisDST2 ;
KolmogorovDST2 ; RudinRCADST2 ). 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 countably 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 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 exists a constant $\lambda>0$ for which
$|f(t)|\leq\lambda$ for almost every $t$. This is a Banach space with norm
$\|f\|_{\infty}=\mathop{\operatorname{ess\;sup}}_{t\in S}|f(t)|.$ We denote by
$C[\alpha,\beta]$ the set of all continuous functions
$f:[\alpha,\beta]\to\mathbb{R}$. This is a Banach space with norm
$\|f\|=\max_{t\in[\alpha,\beta]}|f(t)|.$

##  3 Representations by operators involving piecewise functions

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 that to be satisfied.

###### Proposition 3.1

Let $(\mathbb{R},\Sigma,\tilde{m})$ be the standard Lebesgue measure space in
the real line. 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

$\displaystyle(Ax)(t)=\sum_{i=1}^{l}\alpha_{i}a(t)I_{G_{i}}(t)x(t),\qquad(Bx)(t)=\sum_{i=1}^{m}\beta_{i}b(t)I_{H_{i}}(t)x(t),$
(5)

where $a,b:\mathbb{R}\to\mathbb{R}$ are measurable functions, $G_{i}\in\Sigma$
for $i=1,\ldots,l$, $\tilde{m}(G_{i}\cap G_{j})=0$ if $i\not=j$,
$H_{i}\in\Sigma$ for $i=1,\ldots,m$, $\tilde{m}(H_{i}\cap H_{j})=0$ if
$i\not=j$, and $I_{G_{k}}$ is the indicator function of the set $G_{k}$. Then,
for a polynomial $F_{1}(z)=\delta_{0}+F(z)$ with
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$ and $\delta_{j}\in\mathbb{R}$ for
$j=0,\dots,n$, the commutation relation $AB=BF_{1}(A)$ is satisfied if and
only if, for almost every $t$,

$\sum_{i=1}^{l}\sum_{j=1}^{m}\alpha_{i}\beta_{j}a(t)b(t)I_{G_{i}\cap
H_{j}}(t)=\sum_{j=1}^{m}\delta_{0}\beta_{j}b(t)I_{H_{j}}(t)+\sum_{i=1}^{l}\sum_{j=1}^{m}\beta_{j}b(t)I_{G_{i}\cap
H_{j}}(t)F(a(t)\alpha_{i}).$

###### Proof

Consider a monomial $M(z)=\delta z^{r}$, where $r$ is a positive integer, and
$\delta\in\mathbb{R}$. We compute $AB$, $\delta A^{r}$ and $\delta BA^{r}$ as
follows:

$\displaystyle(ABx)(t)=\sum_{i=1}^{l}\alpha_{i}a(t)I_{G_{i}}(t)\left(\sum_{j=1}^{m}\beta_{j}b(t)I_{H_{j}}(t)x(t)\right)$
$\displaystyle=\sum_{i=1}^{l}\sum_{j=1}^{m}\alpha_{i}\beta_{j}a(t)b(t)I_{G_{i}\cap
H_{j}}(t)x(t),$
$\displaystyle(A^{2}x)(t)=\sum_{i=1}^{l}\sum_{j=1}^{l}\alpha_{i}a(t)\alpha_{j}a(t)I_{G_{i}\cap
G_{j}}(t)x(t)=\sum_{i=1}^{l}\alpha^{2}_{i}a(t)^{2}I_{G_{i}}(t)x(t)$

for almost all $t$. In general, for almost all $t\in\mathbb{R}$,

$\displaystyle(\delta A^{r}x)(t)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{l}\delta\alpha^{r}_{i}a(t)^{r}I_{G_{i}}(t)x(t),$
$\displaystyle(B\delta A^{r}x)(t)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{m}\sum_{i=1}^{l}\beta_{j}b(t)\delta\alpha^{r}_{i}a(t)^{r}I_{G_{i}}(t)^{r}I_{H_{j}}(t)x(t)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{m}\sum_{i=1}^{l}\beta_{j}b(t)\delta\alpha^{r}_{i}a(t)^{r}I_{G_{i}\cap
H_{j}}(t)x(t).$

Suppose now that $F_{1}(z)=\delta_{0}+F(z)$, where
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$,
$j=0,\dots,n$. Then, for almost every $t\in\mathbb{R}$ we have

$\displaystyle(F_{1}(A)x)(t)$ $\displaystyle=$
$\displaystyle\delta_{0}x(t)+(F(A)x)(t)=\delta_{0}x(t)+\sum_{i=1}^{l}F(\alpha_{i}a(t)I_{G_{i}}(t))x(t)$
$\displaystyle B(F_{1}(A)x)(t)$ $\displaystyle=$
$\displaystyle\delta_{0}(Bx)(t)+B(F(A)x)(t)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{m}\delta_{0}\beta_{j}b(t)I_{H_{j}}(t)x(t)+\sum_{i=1}^{l}\sum_{j=1}^{m}b(t)F(\alpha_{i}a(t))\beta_{j}I_{G_{i}\cap
H_{j}}(t)x(t).$

Then $AB=BF_{1}(A)$ on $L_{p}(\mathbb{R})$ if and only if, for almost every
$t\in\mathbb{R}$,

$\displaystyle\sum_{i=1}^{l}\sum_{j=1}^{m}\alpha_{i}a(t)b(t)\beta_{j}I_{G_{i}\cap
H_{j}}(t)$
$\displaystyle=\sum_{j=1}^{m}\delta_{0}b(t)\beta_{j}I_{H_{j}}(t)+\sum_{i=1}^{l}\sum_{j=1}^{m}F(\alpha_{i}a(t))b(t)\beta_{j}I_{G_{i}\cap
H_{j}}(t),$

since if this equality holds for almost every $t\in\mathbb{R}$, then
multiplying both sides of this equality by $x(t)$ for any $x\in
L_{p}(\mathbb{R})$ yields $ABx=BF_{1}(A)x$ on $L_{p}(\mathbb{R})$, and if the
equality does not hold on a set $S$ of finite positive measure
$0<\tilde{m}(S)<\infty$, then $ABI_{S}\neq BF_{1}(A)I_{S}$ and so $AB\neq
BF_{1}(A)$ as linear operators on $L_{p}(\mathbb{R})$. ∎

###### Example 3.2

Consider the operators

$A:L_{p}([1,3])\to L_{p}([1,3]),\ B:L_{p}([1,3])\to L_{p}([1,3]),\ 1\leq
p\leq\infty$

defined as follows, for almost every $t$,

$\displaystyle(Ax)(t)=\sum_{i=1}^{3}\alpha_{i}I_{G_{i}}(t)x(t),\quad(Ax)(t)=\sum_{i=1}^{3}\beta_{i}I_{H_{i}}(t)x(t),$

where

$\begin{array}[]{l}G_{1}=[1,\frac{3}{2}[,\ G_{2}=[\frac{3}{2},2],\
G_{3}=]2,3],\\\ H_{1}=[1,2],\ H_{2}=]2,\frac{5}{2}],\
H_{3}=]\frac{5}{2},3],\end{array}$ (6)

$\alpha_{i},$ $\beta_{i}\in\mathbb{R}$, $i=1,2,3$. Let
$F(z)=\delta_{0}+\delta_{3}z^{3}$, where $\delta_{0},\delta_{3}\in\mathbb{R}$.
By applying Proposition 5.1 we have $AB=B\delta_{0}+\delta_{3}BA^{3}$ if and
only if

$\displaystyle\sum_{i=1}^{3}\sum_{j=1}^{3}\alpha_{i}\beta_{j}I_{G_{i}\cap
H_{j}}(t)=\delta_{0}\sum_{j=1}^{3}\beta_{j}I_{H_{j}}(t)+\sum_{i=1}^{3}\sum_{j=1}^{3}\delta_{3}\alpha_{i}^{3}\beta_{j}I_{G_{i}\cap
H_{j}}(t).$

By simplifying this we have

$\displaystyle\beta_{1}\sum_{i=1}^{2}\alpha_{i}I_{G_{i}}(t)+\alpha_{3}\sum_{i=2}^{3}\beta_{i}I_{H_{i}}(t)$
$\displaystyle=\sum_{i=1}^{3}\delta_{0}\beta_{i}I_{H_{i}}(t)+\beta_{1}\sum_{i=1}^{2}\delta_{3}\alpha_{i}^{3}I_{G_{i}}(t)+\alpha_{3}^{3}\delta_{3}\sum_{i=2}^{3}\beta_{i}I_{H_{i}}(t).$

This is equivalent to

$\displaystyle\left\\{\begin{array}[]{c}\beta_{1}\alpha_{1}=\delta_{0}\beta_{1}+\beta_{1}\delta_{3}\alpha_{1}^{3}\\\
\beta_{1}\alpha_{2}=\delta_{0}\beta_{1}+\beta_{1}\delta_{3}\alpha_{2}^{3}\\\
\beta_{2}\alpha_{3}=\delta_{0}\beta_{2}+\beta_{2}\delta_{3}\alpha_{3}^{3}\\\
\beta_{3}\alpha_{3}=\delta_{0}\beta_{3}+\beta_{3}\delta_{3}\alpha_{3}^{3}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{cc}F(\alpha_{i})\beta_{1}=\alpha_{i}\beta_{1},\quad
i=1,2,\\\ F(\alpha_{3})\beta_{i}=\alpha_{3}\beta_{i},\quad
i=2,3.\end{array}\right.$

This is equivalent to the following:

$\displaystyle\left\\{\begin{array}[]{c}\beta_{1}=0,\ \text{ or }\
F(\alpha_{i})=\alpha_{i},\ i=1,2,\\\ F(\alpha_{3})=\alpha_{3},\ \text{ or }\
\beta_{j}=0,\ j=2,3.\end{array}\right.$

That is, one of the following cases hold:

  1. 1)

$(\beta_{1},\beta_{2},\beta_{3})\in(\\{0\\}\times(\mathbb{R}\setminus\\{0\\})\times\mathbb{R})\cup(\\{0\\}\times\mathbb{R}\times(\mathbb{R}\setminus{0}))$,
$(\alpha_{1},\alpha_{2},\alpha_{3})\in\mathbb{R}\times\mathbb{R}\times\mbox{Fix}(F)$.

  2. 2)

$(\beta_{1},\beta_{2},\beta_{3})\in((\mathbb{R}\setminus\\{0\\})\times(\mathbb{R}\setminus\\{0\\})\times\mathbb{R})\cup((\mathbb{R}\setminus\\{0\\})\times\mathbb{R}\times(\mathbb{R}\setminus{0}))$,
$(\alpha_{1},\alpha_{2},\alpha_{3})\in\mbox{Fix}(F)\times\mbox{Fix}(F)\times\mbox{Fix}(F)$.

  3. 3)

$(\beta_{1},\beta_{2},\beta_{3})\in((\mathbb{R}\setminus\\{0\\})\times\\{0\\}\times\\{0\\})$,$(\alpha_{1},\alpha_{2},\alpha_{3})\in\mbox{Fix}(F)\times\mbox{Fix}(F)\times\mathbb{R}$,

  4. 4)

$(\beta_{1},\beta_{2},\beta_{3})\in\\{0\\}\times\\{0\\}\times\\{0\\}$,
$(\alpha_{1},\alpha_{2},\alpha_{3})\in\mathbb{R}^{3}$.

where Fix$(F)=\\{\alpha\in\mathbb{R}:\ F(\alpha)=\alpha\\}$. Note that
Fix$(F)\not=\emptyset$, since $deg(F(z)-z)=1$ if $\delta_{3}=0$,
$deg(F(z)-z)=3$ if $\delta_{3}\not=0$, and all coefficients of
$F(z)-z=\delta_{3}z^{3}-z+\delta_{0}$ are real.

In the following corollary we consider a case where both operators $A$ and $B$
are considered on the same partition.

###### Corollary 3.3

Let $(\mathbb{R},\Sigma,\tilde{m})$ be the standard Lebesgue measure space in
the real line. 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

$\displaystyle(Ax)(t)=\sum_{i=1}^{l}\alpha_{i}a(t)I_{G_{i}}(t)x(t),\qquad(Bx)(t)=\sum_{i=1}^{l}\beta_{i}b(t)I_{G_{i}}(t)x(t),$

where $G_{i}\in\Sigma$ for $i=1,\ldots,l$, $\tilde{m}(G_{i}\cap G_{j})=0$ for
$i\not=j$, $I_{G_{k}}$ is the indicator function of the set $G_{k}$, and
$a,b:\mathbb{R}\to\mathbb{R}$ are measurable functions. Then, for a polynomial
$F_{1}(z)=\delta_{0}+F(z)$, where $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$
and $\delta_{j}\in\mathbb{R}$ for $j=0,\dots,n$, the commutation relation
$AB=BF_{1}(A)$ is satisfied if and only if

$\displaystyle\sum_{i=1}^{l}\alpha_{i}\beta_{i}a(t)b(t)I_{G_{i}}(t)=\sum_{i=1}^{l}b(t)F_{1}(a(t)\alpha_{i})\beta_{i}I_{G_{i}}(t)\quad\mbox{for
almost every $t$}.$

###### Proof

By applying Proposition 3.1 we have $AB=BF_{1}(A)$ if and only if

$\displaystyle\sum_{i,j=1}^{l}\alpha_{i}a(t)b(t)\beta_{j}I_{G_{i}\cap
G_{j}}(t)=\sum_{j=1}^{l}b(t)\delta_{0}\beta_{j}I_{G_{j}}(t)+\sum_{j,i=1}^{l}b(t)F(\alpha_{i}a(t))\beta_{j}I_{G_{i}\cap
G_{j}}(t)$

for almost every $t$. Since $G_{i}\cap G_{j}=\emptyset$ when $i\not=j$, the
last condition becomes

$\displaystyle\sum_{i=1}^{l}\alpha_{i}a(t)b(t)\beta_{i}I_{G_{i}}(t)=\sum_{i=1}^{l}b(t)\delta_{0}\beta_{i}I_{G_{i}}(t)+\sum_{i=1}^{l}b(t)F(\alpha_{i}a(t))\beta_{i}I_{G_{i}}(t)$
$\displaystyle=\sum_{i=1}^{l}b(t)\beta_{i}I_{G_{i}}(t)(\delta_{0}+F(\alpha_{i}a(t)))=\sum_{i=1}^{l}b(t)\beta_{i}I_{G_{i}}(t)F_{1}(\alpha_{i}a(t)).$
∎

###### Corollary 3.4

Let $(\mathbb{R},\Sigma,\tilde{m})$ be the standard Lebesgue measure space in
the real line. Consider the operators $A$ and $B$ defined in (5) where
$\beta_{j}\not=0$ for all $j\in\\{1,\ldots,m\\}$. Then, for a constant
monomial $F_{1}(z)=\delta_{0}$, we have $AB=BF_{1}(A)$ if and only if for all
$i\in\\{1,\ldots,l\\}$, $j\in\\{1,\ldots,m\\}$ the set

$\displaystyle\operatorname{\rm
supp\,}(\alpha_{i}a(t)-\delta_{0})\cap\operatorname{\rm
supp\,}b(t)\cap(G_{i}\cap H_{j})$

has measure zero.

###### Proof

We have $(F_{1}(A)x)(t)=\delta_{0}x(t),$ for almost all $t$. Moreover, for
almost every $t$

$\displaystyle(ABx)(t)=\sum_{i=1}^{l}\alpha_{i}a(t)I_{G_{i}}(t)\left(\sum_{j=1}^{m}\beta_{j}b(t)I_{H_{j}}(t)x(t)\right)$
$\displaystyle=\sum_{i=1}^{l}\sum_{j=1}^{m}\alpha_{i}\beta_{j}a(t)b(t)I_{G_{i}\cap
H_{j}}x(t)$
$\displaystyle(BF_{1}(A)x)(t)=\sum_{j=1}^{m}\delta_{0}\beta_{j}b(t)I_{H_{j}}(t)x(t),$

Then, for almost every $t\in\mathbb{R}$ we have $AB=BF_{1}(A)$ if and only if

$\sum_{i=1}^{l}\sum_{j=1}^{m}\alpha_{i}a(t)\beta_{j}b(t)I_{G_{i}\cap
H_{j}}(t)=\sum_{j=1}^{m}\delta_{0}\beta_{j}b(t)I_{H_{j}}(t),$

which is equivalent to

$\displaystyle\sum_{i=1}^{l}\alpha_{i}a(t)\beta_{j}b(t)I_{G_{i}\cap
H_{j}}(t)=\delta_{0}\beta_{j}b(t)I_{H_{j}}(t)$ (7)
$\displaystyle\quad\text{for almost every}\ t\in\mathbb{R},\
j\in\\{1,\ldots,m\\}$

Since $\beta_{j}\not=0$ for all $j$, (7) is equivalent to the following: for
$t\in G_{i}\cap H_{j}$, $1\leq i\leq l$, $1\leq j\leq m$,
$(\alpha_{i}a(t)-\delta_{0})b(t)=0$, which is equivalent to that the set

$\displaystyle\operatorname{\rm
supp\,}(\alpha_{i}a(t)-\delta_{0})\cap\operatorname{\rm
supp\,}b(t)\cap(G_{i}\cap H_{j})$

has measure zero, for all $i\in\\{1,\ldots,l\\}$, $j\in\\{1,\ldots,m\\}$. ∎

In the following corollary we consider a case of the operator $B$
corresponding to one of the subsets of the partition defining the operator
$A$.

###### Corollary 3.5

Let $(\mathbb{R},\Sigma,\tilde{m})$ the standard Lebesgue measure space in the
real line. 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 every $t$,

$\displaystyle
Ax(t)=\sum\limits_{i=1}^{l}\alpha_{i}a(t)I_{G_{i}}(t)x(t),\qquad(Bx)(t)=I_{G_{k}}(t)b(t)x(t),\quad
1\leq k\leq l$

where $G_{i}\in\Sigma$ for each $i=1,\ldots,l$, $\tilde{m}(G_{i}\cap G_{j})=0$
for $i\not=j$, $I_{G_{k}}(t)$ is the indicator function for the set $G_{k}$,
$a,b:\mathbb{R}\to\mathbb{R}$ are measurable functions. For a polynomial
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$ for
$j=1,\dots,n$, the commutation relation $AB=BF(A)$ is satisfied if and only if

$\displaystyle\operatorname{\rm supp\,}b\cap\operatorname{\rm
supp\,}\left(\alpha_{k}a-F(\alpha_{k}a)\right)\cap G_{k}$

has measure zero.

###### Proof

We write the operator $B$ as follows

$\displaystyle(Bx)(t)=I_{G_{k}}(t)b(t)x(t)=\sum_{j=1}^{l}\beta_{j}b(t)I_{G_{j}}(t)x(t),$

almost everywhere, where $\beta_{j}=\left\\{\begin{array}[]{l}\,1,\quad j=k\\\
\,0,\ \mbox{ otherwise }\end{array}\right..$ By using Corollary 3.3 we have
$AB=BF(A)$ if and only if, for almost very $t$,

$\displaystyle\sum\limits_{i=1}^{l}\alpha_{i}\beta_{i}a(t)b(t)I_{G_{i}}(t)=\sum\limits_{i=1}^{l}b(t)F(a(t)\alpha_{i})\beta_{i}I_{G_{i}}(t).$

We can simplify the last relation as follows

$\displaystyle\alpha_{k}a(t)b(t)I_{G_{k}}(t)=b(t)F(a(t)\alpha_{k})I_{G_{k}}(t)\
\Leftrightarrow\ b(t)(a(t)\alpha_{k}-F(a(t)\alpha_{k}))=0,$

for almost every $t\in G_{k}$, which is equivalent to that the set

$\displaystyle\operatorname{\rm supp\,}b\cap\operatorname{\rm
supp\,}\left(\alpha_{k}a-F(\alpha_{k}a)\right)\cap G_{k}$

has measure zero. ∎

###### Example 3.6

Let $A:L_{p}[0,1]\to L_{p}[0,1]$, $B:L_{p}[0,1]\to L_{p}[0,1]$, $1\leq
p\leq\infty$ be defined as follows

$\displaystyle(Ax)(t)=\alpha I_{[0,1/3]}(t)x(t)+\beta
I_{]1/3,1/2[}(t)x(t)+\gamma I_{]1/2,1]}(t)x(t),$
$\displaystyle(Bx)(t)=I_{]1/3,1/2[}(t)x(t),$

where $\alpha,\,\beta$ are constants and $I_{E}$ is the indicator function of
the set $E$. For a monomial $F(z)=z^{n}$, where $n$ is a positive integer, we
have

$AB=BF(A)\quad(\mbox{that is}\ AB=BA^{n})$

if and only if $F(\beta)=\beta.$ In fact, taking a partition $G_{1}\cup
G_{2}\cup G_{3}$ where

$G_{1}=[0,\frac{1}{3}],\quad G_{2}=]\frac{1}{3},\frac{1}{2}[,\quad
G_{3}=[\frac{1}{2},1],$

we have $(Ax)(t)=(\alpha I_{G_{1}}(t)+\beta I_{G_{2}}(t)+\gamma
I_{G_{3}}(t))x(t)$ and $(Bx)(t)=I_{G_{2}}(t)x(t)$. Note that
$a(\cdot)=b(\cdot)=1$. By Corollary 3.5, we have $AB=BF(A)$ if and only if

$\operatorname{\rm supp\,}b\cap\operatorname{\rm
supp\,}\left(\alpha_{k}a-F(\alpha_{k}a)\right)\cap
G_{k}=]\frac{1}{3},\frac{1}{2}[\ \cap\ \operatorname{\rm
supp\,}(\beta-F(\beta))$

has measure zero which holds if and only if $F(\beta)=\beta$.

In this case, we can also get the same result by the following direct
computation. For almost every $t\in[0,1]$,

$\displaystyle(ABx)(t)=A(Bx)(t)=(\alpha I_{[0,1/3]}(t)+\beta
I_{]1/3,1/2[}(t)+\gamma I_{]1/2,1]}(t))(Bx)(t)$ $\displaystyle=(\alpha
I_{[0,1/3]}(t)+\beta I_{]1/3,1/2[}(t)+\gamma
I_{]1/2,1]}(t))I_{]1/3,1/2[(t)}x(t)=\beta I_{]1/3,1/2[(t)}x(t)$
$\displaystyle(A^{2}x)(t)=A(Ax)(t)=(\alpha I_{[0,1/3]}+\beta
I_{]1/3,1/2[}+\gamma I_{]1/2,1]})(Ax)(t)$ $\displaystyle=(\alpha
I_{[0,1/3]}(t)+\beta I_{]1/3,1/2[}(t)+\gamma I_{]1/2,1]}(t))$
$\displaystyle\hskip 56.9055pt\cdot(\alpha I_{[0,1/3]}(t)+\beta
I_{]1/3,1/2[}(t)+\gamma I_{]1/2,1]}(t))x(t)$
$\displaystyle=(\alpha^{2}I_{[0,1/3]}(t)+\beta^{2}I_{]1/3,1/2[}(t)+\gamma^{2}I_{]1/2,1]}(t))x(t).$

In general, for almost every $t$,

$\displaystyle(A^{n}x)(t)=(\alpha^{n}I_{[0,1/3]}(t)+\beta^{n}I_{]1/3,1/2[}(t)+\gamma^{n}I_{[1/2,1]}(t))x(t).$

Thus, for almost every $t$,

$\displaystyle(BF(A)x)(t)$ $\displaystyle=$
$\displaystyle(BA^{n}x)(t)=B(A^{n}x)(t)=I_{]1/3,1/2[}(t)(A^{n}x)(t)$
$\displaystyle=$
$\displaystyle\beta^{n}I_{]1/3,1/2[}(t)x(t)=F(\beta)I_{]1/3,1/2[}(t)x(t).$

##  4 Representations involving inner superposition operators

In this section we will look at the commutation relation $BA=F(A)B$ which has
important applications in wavelets analysis for instance. When $B=0$, the
relation $BA=F(A)B$ is trivially satisfied for any $A$. If $A=0$ then the
relation $BA=F(A)B$ reduces to $F(0)B=0$. This implies either $B=0$, or
$F(0)=0$ with $B$ being any well defined operator. Thus, we focus on
construction and properties of non-zero representations of $BA=F(A)B$.

###### Proposition 4.1

Let $(\mathbb{R},\Sigma,\tilde{m})$ the standard Lebesgue measure space in the
real line. Let $A:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$,
$B:L_{p}(\mathbb{R})\to L_{p}(\mathbb{R})$, $p\geq 1$ be defined as follows,
for almost every $t\in\mathbb{R}$,

$\displaystyle(Ax)(t)=\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(t)x(t-1),\qquad(Bx)(t)=\beta
x(\gamma t),$

where $G_{i}\in\Sigma$ for each $i\in\mathbb{Z}$, $\tilde{m}(G_{i}\cap
G_{j})=0$ for $i\not=j$, $I_{G_{k}}(t)$ is the indicator function for the set
$G_{k}$, $\beta,\gamma\in\mathbb{R}\setminus\\{0\\}$ and $\gamma>0$,
$\alpha_{i}\in\mathbb{R}$, for all $i\in\mathbb{Z}$. For a monomial
$F(z)=\delta z^{m}$, where $\delta\in\mathbb{R}\setminus\\{0\\}$ and $m>0$ is
a positive integer, $(BAx)(t)=(F(A)Bx)(t)$ is satisfied for almost every $t$
if and only if

$\displaystyle\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(\gamma
t)x(\gamma
t-1)=\delta\sum\limits_{i=-\infty}^{\infty}\alpha^{m}_{i}I_{G_{i}}(t)x(\gamma
t-\gamma m),$ (8)

for almost every $t$.

###### Proof

For almost every $t\in\mathbb{R}$ we have

$\displaystyle(A^{2}x)(t)$ $\displaystyle=$ $\displaystyle
A\left(\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(t)x(t-1)\right)=\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(t)(Ax)(t-1)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(t)\sum\limits_{j=-\infty}^{\infty}\alpha_{j}I_{G_{j}}(t)x(t-2)=\sum\limits_{i=-\infty}^{\infty}\alpha^{2}_{i}I_{G_{i}}(t)x(t-2).$

We suppose that for almost every $t\in\mathbb{R}$,

$\displaystyle(A^{m}x)(t)=\sum\limits_{i=-\infty}^{\infty}\alpha^{m}_{i}I_{G_{i}}(t)x(t-m),\quad\text{for}\
m\geq 1$

Then we have, for almost every $t$,

$\displaystyle(A^{m+1}x)(t)$ $\displaystyle=$ $\displaystyle
A(A^{m}x)(t)=A\left(\sum\limits_{i=-\infty}^{\infty}\alpha^{m}_{i}I_{G_{i}}(t)x(t-m)\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(t)\sum\limits_{j=-\infty}^{\infty}\alpha^{m}_{j}I_{G_{j}}(t)x(t-m-1)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=-\infty}^{\infty}\alpha^{m+1}_{i}I_{G_{i}}(t)x(t-(m+1)).$

Moreover, we have for almost every $t\in\mathbb{R}$,

$\displaystyle(BAx)(t)$ $\displaystyle=$
$\displaystyle\beta\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(\gamma
t)x(\gamma t-1)$ $\displaystyle\delta(A^{m}Bx)(t)$ $\displaystyle=$
$\displaystyle\delta\beta\sum\limits_{i=-\infty}^{\infty}\alpha^{m}_{i}I_{G_{i}}(t)x(\gamma
t-\gamma m).$

Therefore, $(BAx)(t)=(F(A)Bx)(t)$ for almost every $t$ if and only if

$\displaystyle\sum\limits_{i=-\infty}^{\infty}\alpha_{i}I_{G_{i}}(\gamma
t)x(\gamma
t-1)=\delta\sum\limits_{i=-\infty}^{\infty}\alpha^{m}_{i}I_{G_{i}}(t)x(\gamma
t-\gamma m),$ $\displaystyle\hskip 113.81102pt\mbox{for almost every}\ t,\
i\in\mathbb{Z}.$ ∎

###### Proposition 4.2

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,
$(Ax)(\cdot)=\alpha x(\upsilon(\cdot)),$ $(Bx)(\cdot)=\beta x(\sigma(\cdot)),$
where $\upsilon,\sigma:\mathbb{R}\to\mathbb{R}$ are continuous functions and
$\alpha,\beta\in\mathbb{R}\setminus\\{0\\}$. For a monomial $F(z)=\delta
z^{m}$ where $\delta\in\mathbb{R}\setminus\\{0\\}$,
$m\in\mathbb{Z}_{>0}=\\{1,2,\ldots\\}$, for $x\in L_{p}(\mathbb{R})$ and for
almost every $t$,

$\displaystyle(BAx)(t)=(F(A)Bx)(t),$

if and only if
$x(\upsilon(\sigma(t)))=\delta\alpha^{m-1}x(\sigma(\upsilon^{\circ(m)})(t))$.

###### Proof

We have for almost every $t$

$\displaystyle(BAx)(t)=\alpha\beta x(\upsilon(\sigma(t))),$
$\displaystyle(A^{2}x)(t)=\alpha^{2}x(\upsilon(\upsilon(t))).$

In the same way we have for almost every $t$

$\displaystyle(A^{m}x)(t)=\alpha^{m}x(\upsilon^{\circ(m)}(t)),$
$\displaystyle\delta(A^{m}Bx)(t)=\delta\alpha^{m}\beta
x(\sigma(\upsilon^{\circ(m)})(t)).$

Then, for all $L_{p}(\mathbb{R})$, $1\leq p\leq\infty$, $BAx=(F(A)B)x$ if and
only if, for almost every $t$,

$\alpha\beta x(\upsilon(\sigma(t)))=\delta\alpha^{m}\beta
x(\sigma(\upsilon^{\circ(m)})(t)).$ ∎

###### Example 4.3

Consider operators $A:L_{2}(\mathbb{R})\to L_{2}(\mathbb{R})$,
$B:L_{2}(\mathbb{R})\to L_{2}(\mathbb{R})$ be defined as follows, for almost
every $t$,

$(Ax)(t)=x(t-1),\quad(Bx)(t)=\gamma^{1/2}x(\gamma t),\quad\gamma>0.$

These operators are particular case of the corresponding ones of Proposition
4.2 when $\upsilon(t)=t-1$, $\sigma(t)={\gamma t}$, $\alpha=1$,
$\beta=\gamma^{\frac{1}{2}}$. We get

$\upsilon(\sigma(t))={\gamma t-1},\quad\sigma(\upsilon^{\circ(m)})(t)=\gamma
t-\gamma m,\quad m\in\mathbb{Z}_{>0}=\\{1,2,\ldots\\}$

Then $BAx=\delta A^{m}Bx$ for $x\in L_{p}(\mathbb{R})$, for almost every $t$,
if and only

$x(\gamma t-1)=\delta x(\gamma t-\gamma m).$

If $\gamma=\frac{1}{m}$ and $\delta=1$, then we get on both sides
$x(\frac{t}{m}-1)$ and thus $BA=A^{m}B$ on $L_{p}(\mathbb{R})$. For example,
when $\gamma=1/2$ and $m=2$, we have operators well known operators important
for example in wavelets theory,

$(Ax)(t)=x\left(t-1\right),\quad(Bx)(t)=\frac{1}{2^{1/2}}x\left(\frac{1}{2}t\right),$

satisfying $BA=A^{2}B$.

##  5 Representations involving weighted composition operators

In this section we consider pairs of operators $(A,B)$ which involve weighted
composition operators, and the multiplication operator composed with the point
evaluation functional $\upsilon_{\gamma}:x(t)\mapsto x(\gamma)$ (”boundary
value” operator, ”one-dimensional range” operator). These operators act on
some spaces of real-valued functions $x(t)$ of one real variable by the
formulas

$(T_{w,\sigma}x)(t)=w(t)x(\sigma(t)),\quad
T_{w,\upsilon_{\gamma}}x(t)=w(t)x(\gamma),\quad t,\gamma\in\mathbb{R}.$

Note that the multiplication operator composed with the point evaluation
functional is a special case of weighted composition operator in these
notations since

$T_{w,\upsilon_{\gamma}}=T_{w,\sigma},\quad\mbox{for}\quad\sigma=\upsilon_{\gamma}.$

###### Proposition 5.1

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 for measurable functions
$a,b:\mathbb{R}\to\mathbb{R}$ by

$(Ax)(\cdot)=a(\cdot)x(\cdot),\quad(Bx)(\cdot)=b(\cdot)x(\cdot).$

For a polynomial $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$ with
$\delta_{0},\ldots,\delta_{n}\in\mathbb{R}$,

$AB=BF(A),$

if and only if the set $\operatorname{\rm supp\,}b\ \cap\ \operatorname{\rm
supp\,}(a-F(a))$ has measure zero.

###### Proof

For almost every $t\in\mathbb{R}$,

$\displaystyle(ABx)(t)=A(Bx)(t)=a(t)b(t)x(t),$
$\displaystyle(A^{2}x)(t)=A(Ax)(t)=a(t)(Ax)(t)=a(t)^{2}x(t)=a(t)^{2}x(t),$
$\displaystyle(A^{3}x)(t)=A(A^{2}x)(t)=a(t)^{3}x(t).$

Therefore, for $m\geq 1$ and for almost every $t$,

$\displaystyle(A^{m}x)(t)=a(t)^{m}x(t),\
(BA^{m}x)(t)=B(A^{m}x)(t)=b(t)(A^{m}x)(t)=b(t)a(t)^{m}x(t).$

Thus, we have for almost every $t$,

$(BF(A)x)(t)=\sum_{k=0}^{n}\delta_{k}(B(A^{k}x))(t)=\sum_{k=0}^{n}\delta_{k}b(t)a(t)^{k}x(t)=b(t)F(a(t))x(t).$

Then $AB=BF(A)$ if and only if for almost every $t$,

$b(t)F(a(t))=a(t)b(t).$

This is equivalent to the set $\operatorname{\rm supp\,}b\ \cap\
\operatorname{\rm supp\,}(a-F(a))$ having measure zero. ∎

###  5.1 Representations by operators on $C[\alpha,\beta]$

In this subsection we consider pairs of operators $(A,B)$ which involve
weighted composition operators, and the multiplication operator composed with
the point evaluation functional $\upsilon_{\gamma}:x(t)\mapsto x(\gamma)$
(”boundary value” operator, ”one-dimensional range” operator). These operators
act on some spaces of continuous real-valued functions of one real variable
$x(t)$ by the formulas:

$(T_{w,\sigma}x)(t)=w(t)x(\sigma(t)),\quad
T_{w,\upsilon_{\gamma}}x(t)=w(t)x(\gamma),\quad t,\gamma\in\mathbb{R}.$

###### Proposition 5.2

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$, $1\leq p\leq\infty$ defined as follows

$(Ax)(t)=a(t)x(\upsilon(t)),\quad(Bx)(t)=b(t)x(\sigma(t)),$ (9)

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$a,b,\upsilon,\sigma:[\alpha,\beta]\to[\alpha,\beta]$ are continuous
functions. For a monomial $F(z)=\delta z^{m}$ where
$\delta\in\mathbb{R}\setminus\\{0\\}$, $m\in\mathbb{Z}_{>0}=\\{1,2,\ldots\\}$,
and for $x\in C[\alpha,\beta]$ and for each $t\in[\alpha,\beta]$,

$(ABx)(t)=(BF(A)x)(t),$

if and only if

$a(t)b(\upsilon(t))x(\sigma(\upsilon(t)))\\\ =\delta b(t)a(\sigma(t))\cdot
a(\upsilon(\sigma(t)))\cdot\ldots\cdot
a(\upsilon^{\circ(m-1)}(\sigma(t)))x(\upsilon^{\circ(m)}(\sigma(t))).$

In particular, if $x\in C[\alpha,\beta]$ satisfies
$x(\sigma(\upsilon(\cdot))))=x(\upsilon^{\circ(m)}(\sigma(\cdot))))$, then
$ABx=BF(A)x$ if and only if

$a(t)b(\upsilon(t))=\delta b(t)a(\sigma(t))\cdot
a(\upsilon(\sigma(t)))\cdot\ldots\cdot a(\upsilon^{\circ(m-1)}(\sigma(t)))$

for all $t\in\operatorname{\rm
supp\,}x(\sigma(\upsilon(\cdot))))\bigcap[\alpha,\beta]$

###### Proof

For each $t\in[\alpha,\beta]$,

$\displaystyle(ABx)(t)$ $\displaystyle=$ $\displaystyle
a(t)b(\upsilon(t))x(\sigma(\upsilon(t))),$ $\displaystyle(A^{2}x)(t)$
$\displaystyle=$ $\displaystyle a(t)a(\upsilon(t))x(\upsilon(\upsilon(t)))).$

In the same way we have for $n\geq 1$ and for each $t\in[\alpha,\beta]$,

$\displaystyle(A^{n}x)(t)=a(t)a(\upsilon(t))a(\upsilon^{\circ(2)}(t))\ldots
a(\upsilon^{\circ(n-1)}(t))x(\upsilon^{\circ(n)}(t)),$
$\displaystyle\delta(BA^{m}x)(t)=\delta
b(t)a(\sigma(t))a(\upsilon(\sigma(t)))\cdot\ldots\cdot
a(\upsilon^{\circ(m-1)}(\sigma(t)))x(\upsilon^{\circ(m)}(\sigma(t))).$

Then $ABx=BF(A)x$ for all $x\in C[\alpha,\beta]$ if and only if for all
$t\in[\alpha,\beta]$,

$a(t)b(\upsilon(t))x(\sigma(\upsilon(t)))\\\ =\delta b(t)a(\sigma(t))\cdot
a(\upsilon(\sigma(t)))\cdot\ldots\cdot
a(\upsilon^{\circ(m-1)}(\sigma(t)))x(\upsilon^{\circ(m)}(\sigma(t))).$

If $x(\sigma(\upsilon(\cdot)))=x(\upsilon^{\circ(m)}(\sigma(\cdot)))$, then
$ABx=BF(A)x$ if and only if

$\displaystyle a(t)b(\upsilon(t))=\delta b(t)a(\sigma(t))\cdot
a(\upsilon(\sigma(t)))\cdot\ldots\cdot a(\upsilon^{\circ(m-1)}(\sigma(t))).$

for $t\in[\alpha,\beta]$ such that $x(\sigma(\upsilon(t)))\neq 0$, since
$(ABx)(t)=0=(BF(A)x)(t)$ for $t\in[\alpha,\beta]$ such that
$x(\sigma(\upsilon(t)))=0$. ∎

#### Representations when $B$ is the multiplication operator

Let us consider the operators (9) with constant $\upsilon(\cdot)=\gamma$ and
with $\sigma_{B}(t)=t,t\in\mathbb{R}$.

###### Lemma 5.3

Let $a,b\in C[\alpha,\beta]$, $\alpha,\beta\in\mathbb{R}$, $\alpha<\beta$. If
the set $\Omega=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm supp\,}b$
has Lebesgue measure zero, then it is empty.

###### Proof

Without loss of generality we suppose that there exists
$\alpha_{0}\in]\alpha,\beta[$ such that $\alpha_{0}\in\Omega$, that is,
$\Omega$ is not empty. Then, $a(\alpha_{0})\not=0$ and $b(\alpha_{0})\not=0$.
Since $a,b$ are continuous functions, there is an open interval
$V_{\alpha_{0}}=]\alpha_{0}-\varepsilon,\alpha_{0}+\varepsilon[\ \subseteq\
]\alpha,\beta[$, $\varepsilon>0$ such that $a(t)\not=0$ and $b(t)\not=0$ for
all $t\in V_{\alpha_{0}}$. Then the Lebesgue measure of $\Omega$ is positive.
But this contradicts the hypothesis. Then the set $\Omega$ must be empty. ∎

###### Proposition 5.4

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ defined by

$(Ax)(t)=a(t)x(\gamma),\quad(Bx)(t)=b(t)x(t),$

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$\gamma\in[\alpha,\beta]$ and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous
functions. Let $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$,
$\delta_{j}\in\mathbb{R}$, $j=0,\ldots,n$. For $x\in C[\alpha,\beta]$ and
$t\in[\alpha,\beta]$,

$(ABx)(t)=(BF(A)x)(t)$

if and only if,

$a(t)b(\gamma)x(\gamma)=\delta_{0}b(t)x(t)+b(t)a(t)x(\gamma)\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}.$
(10)

###### Proof

We have

$\displaystyle(ABx)(t)=A(Bx)(t)=a(t)(Bx)(\gamma)=a(t)b(\gamma)x(\gamma)$
$\displaystyle(A^{m}x)(t)=a(t)a(\gamma)^{m-1}x(\gamma),\quad m=1,2,\ldots$
$\displaystyle((BA^{m})x)(t)=(BA^{m}x)(t)=b(t)a(t)a(\gamma)^{m-1}x(\gamma),\
m=1,2,\ldots$
$\displaystyle(BF(A)x)(t)=\delta_{0}b(t)x(t)+\sum\limits_{j=1}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).$

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

$a(t)b(\gamma)x(\gamma)=\delta_{0}b(t)x(t)+\sum\limits_{j=1}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).\
\qed$

###### Corollary 5.5

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(\gamma),\quad(Bx)(t)=b(t)x(t),$

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$\gamma\in[\alpha,\beta]$ and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous
functions such that $a(\gamma)\not=0$, $b(\gamma)\not=0$. Let
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, where $\delta_{j}\in\mathbb{R}$,
$j=1,\ldots,n$. The following statements hold.

  1. 1)

If $b(\cdot)$ is not constant in any open interval included in
$[\alpha,\beta]$, then

$AB\not=BF(A).$

  2. 2)

If $b(\cdot)$ is constant in some open interval included in $[\alpha,\beta]$,
then

$AB=BF(A)$

if and only if $\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}\left(b(\gamma)-b\cdot k_{1}\right)=\emptyset,$ where
$k_{1}=\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}.$

In particular, if $b(\cdot)=b(\gamma)$ is identically non-zero constant in
$[\alpha,\beta]$ then $AB=BF(A)$ if and only if $k_{1}=1$.

###### Proof

Suppose that there exist two continuous functions $a,\ b$, $a(\gamma)\not=0$,
$b(\gamma)\not=0$ and a polynomial $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$
such that $ABx=BF(A)x$ for all $x\in C[\alpha,\beta]$. By applying Proposition
5.4 we have $AB=BF(A)$ if and only if for all $x\in C[\alpha,\beta]$

$a(t)b(\gamma)x(\gamma)=\sum\limits_{j=1}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).$

This is equivalent to

$a(t)b(\gamma)=\sum\limits_{j=1}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}\Leftrightarrow
a(t)\left(b(\gamma)-b(t)\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}\right)=0,$

which is equivalent to the set

$\Omega=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}\left(b(\gamma)-b\cdot\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}\right)$

has measure zero. Since $a,b$ are continuous functions, by applying Lemma 5.3
we conclude that the set $\Omega$ is empty. We consider the following cases.

  1. Case 1: If $b(\cdot)$ is not constant in any open interval included in $[\alpha,\beta]$ and

$k_{1}=\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}\not=1,$

then $\gamma\in\Omega$. This contradicts the fact that $\Omega$ is empty.

  2. Case 2: Suppose $k_{1}=1$ and $b(\cdot)$ is not constant in any open interval included in $[\alpha,\beta]$ and without loss of generality suppose that $\gamma\in]\alpha,\beta[$. Since the functions $a$ and $b(\gamma)-k_{1}b(t)=b(\gamma)-b(t)$ are continuous, then for some positive $\varepsilon$, we can find an open interval $]\gamma-\varepsilon,\ \gamma+\varepsilon[\subseteq]\alpha,\beta[$ such that the set

$\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}\left(b(\gamma)-b\right)\,\cap\,]\gamma-\varepsilon,\
\gamma+\varepsilon[\not=\emptyset,$

which is a contradiction.

  3. Case 3: If $b(\cdot)$ is constant in an open interval contained in $[\alpha,\beta]$ then $AB=BF(A)$ if and only if

$\Omega=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}\left(b(\gamma)-b\cdot\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}\right)=\emptyset.$

  4. Case 4:  If $b(\cdot)$ is identically constant in $[\alpha,\beta]$ and $k_{1}=1$ then $\Omega$ is empty. If $k_{1}\not=1$ then $\Omega=\operatorname{\rm supp\,}a$ must be empty. This implies $a(\cdot)\equiv 0$, but, $a(\gamma)\not=0$. That is a contradiction. Therefore, the condition of Proposition 5.4 is fulfilled. ∎

###### Remark 5.1

Note that if operators $A$ and $B$ defined in Corollary 5.5 are such that
$AB=BF(A)$ for some polynomial $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j},$
then $k_{1}=\sum\limits_{j=1}^{n}\delta_{j}a(\gamma)^{j-1}=1$. In fact, if
$k_{1}\not=1$, then $\gamma\in\Omega$, thus by Proposition 5.4,
$AB\not=BF(A)$.

###### Example 5.6

Let $A:C[0,3]\to C[0,3]$, $B:C[0,3]\to C[0,3]$ be defined as follows

$(Ax)(t)=a(t)x(1),\quad(Bx)(t)=b(t)x(t),$

where $a(t)=\left\\{\begin{array}[]{rr}t(2-t),&\mbox{ if }\ 0\leq t\leq 2\\\
0,&\mbox{ otherwise, }\end{array}\right.$ and
$b(t)=\left\\{\begin{array}[]{rr}1,&\mbox{ if }\ 0\leq t\leq 2\\\
(t-1)^{3},&\mbox{ if }\ 2<t\leq 3\end{array}.\right.$ Consider a polynomial
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$, $\delta_{j}\in\mathbb{R}$,
$j=1,\ldots,n$. If $k_{1}=\sum\limits_{j=1}^{n}\delta_{j}=1,$ then these
operators satisfy $AB=BF(A)$. Indeed, by Corollary 5.5 we have

$\Omega=\operatorname{\rm supp\,}a\,\cap\,(b(1)-b)=\emptyset.$

###### Corollary 5.7

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(\gamma),\quad(Bx)(t)=b(t)x(t),$

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$\gamma\in[\alpha,\beta]$ and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous
functions such that $b(\gamma)\not=0$. Consider a polynomial
$F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$ for
$j=0,\ldots,n$. If $a(\gamma)=0$, then $AB=BF(A)$ if and only if
$\delta_{0}=0$ and

$\Omega=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}(b(\gamma)-\delta_{1}b)=\emptyset.$

###### Proof

By applying Proposition 5.4 we have $(ABx)(t)=(BF(A)x)(t)$ if and only if, for
all $x\in C[\alpha,\beta]$,

$a(t)b(\gamma)x(\gamma)=\delta_{0}b(t)x(t)+\delta_{1}b(t)a(t)x(\gamma)+\sum\limits_{j=2}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).$

By taking $t=\gamma$ and if $a(\gamma)=0$ we get
$\delta_{0}b(\gamma)x(\gamma)=0$ for all $x\in C[\alpha,\beta]$. Since
$b(\gamma)\not=0$ then $\delta_{0}$ must be zero. By using this, we remain
with the equation

$a(\cdot)b(\gamma)x(\gamma)=\delta_{1}a(\cdot)b(\cdot)x(\gamma)$

for all $x\in C[\alpha,\beta]$. This is equivalent to
$a(\cdot)(b(\gamma)-\delta_{1}b(\cdot))=0,$ which itself is equivalent to the
set $\Omega=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm
supp\,}(b(\gamma)-\delta_{1}b)$ having measure zero. Since the functions
involved are continuous then the set $\Omega$ is empty. ∎

###### Remark 5.2

According to Corollary 5.7, if $a(\gamma)=0$ the role of the polynomial
$F(\cdot)$ is only played by the coefficient $\delta_{1}$. In particular,
Corollary 5.7 establishes conditions for representations of the quantum plane
relation $AB=\delta_{1}BA$, for a real constant $\delta_{1}$.

###### Corollary 5.8

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(\gamma),\quad(Bx)(t)=b(t)x(t),$

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$\gamma\in[\alpha,\beta]$ and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous
functions such that $a(\gamma)\not=0$. Consider a polynomial
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$ for
$j=1,\ldots,n$. Let $k_{1}=\sum\limits_{j=1}^{n}\delta_{j}(a(\gamma))^{j-1}.$
If $b(\gamma)=0$, then $AB=BF(A)$ if and only if $k_{1}=0$ or
$\operatorname{\rm supp\,}a\,\cap\operatorname{\rm supp\,}b=\emptyset.$

###### Proof

By applying Proposition 5.4 we have $(ABx)(t)=(BF(A)x)(t)$ if and only if for
all $x\in C[\alpha,\beta]$ and for all $t\in[\alpha,\beta]$,

$a(t)b(\gamma)x(\gamma)=\sum\limits_{j=1}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).$

By using the hypothesis, this reduces to the following:

$\forall\,x\in C[\alpha,\beta]:\ k_{1}a(t)b(t)x(\gamma)=0.$

This is equivalent to $k_{1}a(\cdot)b(\cdot)=0$ which is equivalent to
$k_{1}=0$ or to the set

$\Omega_{1}=\operatorname{\rm supp\,}a\,\cap\,\operatorname{\rm supp\,}b$

having measure zero. Since $a,b$ are continuous functions, the set
$\Omega_{1}$ is empty. ∎

###### Example 5.9

Let $A:C[0,2]\to C[0,2]$, $B:C[0,2]\to C[0,2]$ be defined as follows

$(Ax)(t)=a(t)x(\frac{1}{2}),\quad(Bx)(t)=b(t)x(t),$

where $a(t)=\left\\{\begin{array}[]{rr}t(1-t),&\mbox{ if }\ 0\leq t\leq 1\\\
0,&\mbox{ otherwise, }\end{array}\right.$ and
$b(t)=\left\\{\begin{array}[]{cc}0,&\mbox{ if }\ 0\leq t\leq 1\\\
(1-t)(t-2),&\mbox{ if }1<t\leq 2\end{array}\right..$ These operators satisfy
$AB=BF(A)$ for any polynomial $F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$,
$\delta_{j}\in\mathbb{R}$, $j=1,\ldots,n$. Indeed this follows by Corollary
5.8.

###### Example 5.10

Let $A:C[-1,1]\to C[-1,1]$, $B:C[-1,1]\to C[-1,1]$ be defined by

$(Ax)(t)=a(t)x(0),\quad(Bx)(t)=b(t)x(t),$

where $a(t)=1+t^{2}$, $b(t)=t$. Consider a polynomial
$F(z)=\sum\limits_{j=1}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$,
$j=1,\ldots,n$. If $k_{1}=\delta_{1}+\ldots+\delta_{n}=0$, then these
operators satisfy $AB=BF(A)$. Indeed this follows by Corollary 5.8.

###### Corollary 5.11

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(\gamma),\quad(Bx)(t)=b(t)x(t),$

where $\alpha,\beta$ are real numbers, $\alpha<\beta$,
$\gamma\in[\alpha,\beta]$ and $a,b:[\alpha,\beta]\to\mathbb{R}$ are non-zero
continuous functions. Consider a polynomial
$F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$, $\delta_{j}\in\mathbb{R}$,
$j=0,\ldots,n$. If $a(\gamma)=b(\gamma)=0$ then $AB=BF(A)$ if and only if
$\delta_{0}=0$ and either $\delta_{1}=0$ or the set $\Omega=\operatorname{\rm
supp\,}a\,\cap\,\operatorname{\rm supp\,}b=\emptyset.$

###### Proof

By applying Proposition 5.4 we have $(ABx)(t)=(BF(A)x)(t)$ if and only if for
all $x\in C[\alpha,\beta]$ and for all $t\in[\alpha,\beta]$,

$a(t)b(\gamma)x(\gamma)=\delta_{0}b(t)x(t)+\delta_{1}b(t)a(t)x(\gamma)+\sum\limits_{j=2}^{n}\delta_{j}b(t)a(t)a(\gamma)^{j-1}x(\gamma).$

By hypothesis, this reduces to the condition

$\forall\,x\in C[\alpha,\beta]:\
\delta_{0}b(\cdot)x(\cdot)+\delta_{1}a(\cdot)b(\cdot)x(\gamma)=0.$ (11)

If $\delta_{0}\not=0$ and $t_{0}\in\operatorname{\rm supp\,}b$, then in an
open interval $]t_{0}-\varepsilon,t_{0}+\varepsilon[\subset[\alpha,\beta]$ we
have $x(t)=-\frac{\delta_{1}}{\delta_{0}}a(t)\zeta,\
t\in]t_{0}-\varepsilon,t_{0}+\varepsilon[,$ where $\zeta$ is a constant. Since
$x(\cdot)$ is continuous in $]t_{0}-\varepsilon,t_{0}+\varepsilon[$, thus
$1+x(\cdot)^{2}$ is also continuous in
$]t_{0}-\varepsilon,t_{0}+\varepsilon[$, but the identity (11) is not valid
for this function. This contradicts the condition (11). Since $b(\cdot)$ is
not identically zero, this implies that $\delta_{0}=0$. By using this, the
condition (11) reduces to the following condition: $\forall\,x\in
C[\alpha,\beta]:\ \delta_{1}a(t)b(t)x(\gamma)=0,$ which is equivalent to
$\delta_{1}=0$ or to $\Omega=\operatorname{\rm
supp\,}a\,\cap\,\operatorname{\rm supp\,}b=\emptyset.$ ∎

###### Example 5.12

Let $A:C[0,2]\to C[0,2]$, $B:C[0,2]\to C[0,2]$ be defined by

$(Ax)(t)=a(t)x(1),\quad(Bx)(t)=b(t)x(t),$

where $a(t)=\sin(\pi t)$, $b(t)=t^{2}-1$ and $\gamma=1$. By Corollary 5.11,
these operators satisfy $AB=BF(A)$ for any polynomial
$F(z)=\sum\limits_{j=2}^{n}\delta_{j}z^{j}$ with $\delta_{j}\in\mathbb{R}$,
$j=2,\ldots,n$.

#### Representations when $A$ is multiplication operator

###### Proposition 5.13

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(t),\quad(Bx)(t)=b(t)x(\gamma),$

where $\alpha,\beta\in\mathbb{R}$, $\alpha<\beta$, $\gamma\in[\alpha,\beta]$
and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous functions. Consider a
polynomial $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$,
$\delta_{j}\in\mathbb{R}$, $j=0,\ldots,n$. Then, $AB=BF(A)$ if and only if
$\operatorname{\rm supp\,}(a-F(a(\gamma)))\,\cap\operatorname{\rm
supp\,}b=\emptyset.$

###### Proof

We have

$\displaystyle(ABx)(t)=A(Bx)(t)=a(t)(Bx)(t)=a(t)b(t)x(\gamma)$
$\displaystyle(A^{m}x)(t)=a(t)^{m}x(t),\quad m=0,\ldots,n$
$\displaystyle(BA^{m})x)(t)=(BA^{m}x)(t)=B(A^{m}x)(t)=\left\\{\begin{array}[]{rc}b(t)x(\gamma),&m=0\\\
b(t)a(\gamma)^{m}x(\gamma),&\quad m=1,\ldots,n\end{array}\right.$
$\displaystyle(BF(A)x)(t)=\delta_{0}b(t)x(\gamma)+\sum\limits_{j=1}^{n}\delta_{j}b(t)a(\gamma)^{j}x(\gamma)=b(t)F(a(\gamma))x(\gamma).$

Then $(ABx)(t)=(BF(A)x)(t)$ if and only if for all $x\in C[\alpha,\beta]$ and
for all $t\in[\alpha,\beta]$,

$a(t)b(t)x(\gamma)=b(t)F(a(\gamma))x(\gamma).$

This is equivalent to the equation $a(\cdot)b(\cdot)=b(\cdot)F(a(\gamma))$,
which is equivalent to

$\operatorname{\rm supp\,}\,(a-F(a(\gamma)))\,\cap\operatorname{\rm
supp\,}\,b=\emptyset,$

by Lemma 5.3. ∎

###### Example 5.14

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(t),\quad(Bx)(t)=b(t)x(\gamma),$

where $\alpha,\beta\in\mathbb{R}$, $\alpha<\beta$, $\gamma\in[\alpha,\beta]$
and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous functions. Then, for a
real constant $\zeta$ and a positive integer $d$,

$AB=\zeta BA^{d}$

if and only if $\operatorname{\rm supp\,}(a-\zeta
a(\gamma)^{d})\,\cap\operatorname{\rm supp\,}b=\emptyset.$ This follows by
Proposition 5.13.

###### Example 5.15

Let $A=A_{\nu}:C[0,2]\to C[0,2]$, $B:C[0,2]\to C[0,2]$ be defined by

$(Ax)(t)=A_{\nu}x(t)=a(t)x(t),\quad(Bx)(t)=b(t)x(1/2),$

where $a(t)=I_{[0,1]}(t)\sin(\pi t)+\nu$, $b(t)=I_{[1,2]}(t)\sin(\pi t)$,
$\nu\in\mathbb{R}$, and $I_{[\alpha_{1},\beta_{1}]}(t)$ is the indicator
function of the interval $[\alpha_{1},\beta_{1}]$. Consider a polynomial
$F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$ with $\delta_{i}\in\mathbb{R}$ for
$i=0,\ldots,n$. If $\nu=F(1+\nu),$ then, by Proposition 5.13, these operators
satisfy $AB=BF(A)$. Note that the condition $\nu=F(1+\nu)$ is equivalent to
$\tilde{F}(1+\nu)=1+\nu$, where $\tilde{F}(z)=F(z)+1$, that is to $1+\nu$
being a fixed point of $\tilde{F}(z)$, or equivalently to $1+\nu$ being the
root of $\tilde{F}(z)-z=0$, which in terms of $F$ is the same as $1+\nu$ being
a root of $F(z)-z+1=0$. If this equation has roots in $\mathbb{R}$, then for
such roots the corresponding operators $A$ and $B$ satisfy $AB=BF(A)$. If
$F(z)=z-1$, then $1+\nu$ is a root of $F(z)-z+1=0$ for any real number $\nu$,
since the equation becomes the equality $0=0$. Thus, for any
$\nu\in\mathbb{R}$, the operators $A$ and $B$ satisfy $AB=B(A-I)$ which is
equivalent to $AB-BA=-B$ and to $BA-AB=B$. This can be also checked by direct
computation. In fact, the operator $A_{\nu}$ corresponding to the parameter
$\nu$ can be expressed as $A_{\nu}=A_{0}+\nu I$. For $\nu=0$, $a$ and $b$ have
non-overlapping supports and hence $(A_{0}Bx)(t)=x(\frac{1}{2})a(t)b(t)=0$,
and
$(BA_{0}x)(t)=b(t)(A_{0}x)(\frac{1}{2})=b(t)a(\frac{1}{2})x(\frac{1}{2})=b(t)x(\frac{1}{2})=B$,
and thus $BA-AB=B$, and hence, for any $\nu\in\mathbb{R}$,
$BA_{\nu}-A_{\nu}B=B(A_{0}+\nu I)-(A_{0}+\nu I)B=BA_{0}+\nu B-A_{0}B-\nu
B=BA_{0}-A_{0}B=B$.

###### Corollary 5.16

Let $A:C[\alpha,\beta]\to C[\alpha,\beta]$, $B:C[\alpha,\beta]\to
C[\alpha,\beta]$ be defined by

$(Ax)(t)=a(t)x(t),\quad(Bx)(t)=b(t)x(\gamma),$

where $\alpha,\beta\in\mathbb{R}$, $\alpha<\beta$, $\gamma\in[\alpha,\beta]$
and $a,b:[\alpha,\beta]\to\mathbb{R}$ are continuous functions. Consider a
polynomial $F(z)=\sum\limits_{j=0}^{n}\delta_{j}z^{j}$,
$\delta_{j}\in\mathbb{R}$, $j=0,\ldots,n$. If $a(\gamma)=0$ then $AB=BF(A)$ if
and only if $\operatorname{\rm supp\,}(a-\delta_{0})\,\cap\operatorname{\rm
supp\,}b=\emptyset.$ Furthermore, if $\delta_{0}\not=0$ then $AB=BF(A)$ yields
$b(\gamma)=0$.

###### Proof

This follows by Proposition 5.13. ∎

###### 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 at Mälardalen University, School of
Education, Culture and Communication, Mälardalens University for excellent
environment for research in Mathematics. Partial support from Swedish Royal
Academy of Sciences is also gratefully acknowledged.

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