Multiplication and Linear Integral Operators on L_p Spaces Representing Polynomial Covariant Type Commutation Relations

Abstract

Representations of polynomial covariant type commutation relations by pairs of linear integral operators and multiplication operators on Banach spaces $L_p$ are constructed.

9.1 Introduction

Commutation relations of the form

$$\begin{aligned} AB=B F(A) \end{aligned}$$

(9.1)

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. Elements of an algebra that satisfy (9.1) are called a representation of this relation in that algebra. Representations of covariance commutation relations (9.1) by linear operators are important for the study of actions and induced representations of groups and semigroups, crossed product operator algebras, dynamical systems, harmonic analysis, wavelets and fractals analysis and have applications in physics and engineering [4, 5, 20,21,22, 26,27,28, 34, 36, 45]. A description of the structure of representations for the relation (9.1) and more general families of self-adjoint operators satisfying such relations by bounded and unbounded self-adjoint linear operators on a Hilbert space use 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 involved in the commutation relations [3, 6,7,8, 10, 11, 13,14,15,16,17, 29,30,34, 37,38,41, 45,46,58]. Algebraic properties of the commutation relation (9.1) are important in description of properties of its representations. For instance, there is a well-known link between linear operators satisfying the commutation relation (9.1) and spectral theory [44]. A description of the structure of representations for the relation (9.1) by bounded and unbounded self-adjoint linear operators on a Hilbert space, using spectral representation [2] of such operators, is given in [44] devoted to more general cases of families of commuting self-adjoint operators satisfying relations of the form (9.1).

In this paper we construct representations of (9.1) by pairs of linear integral and multiplication operators on Banach spaces $L_p$. 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 [34, 45, 46, 58] 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 Sect. 9.2 preliminaries, notations and basic definitions. In Sect. 9.3 we present the main results about construction of specific representations on Banach function spaces $L_p$.

9.2 Preliminaries and Notations

  In this section we present some preliminaries, basic definitions and notations. For more details please read [1, 12, 18, 23, 24, 42, 43].

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 covered 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 $f:S\rightarrow \mathbb {R}$ such that $\int \limits _{S} |f(t)|^p dt < \infty .$ This is a Banach space (Hilbert space when $p=2$) with norm $\Vert f\Vert _p= \left( \int \limits _{S} |f(t)|^p dt \right) ^{\frac{1}{p}}.$ We denote by $L_\infty (S)$ the set of all classes of equivalent measurable functions $f:S\rightarrow \mathbb {R}$ such that there is a constant $\lambda >0$, $|f(t)|\le \lambda$ almost everywhere. This is a Banach space with norm $\Vert f\Vert _{\infty }=\mathop {{{\,\mathrm{ess\;sup}\,}}}_{t\in S} |f(t)|.$

9.3 Operator Representations of Covariance Commutation Relations

Before we proceed with constructions of more complicated operator representations of commutation relations (9.1) 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 (9.1) interact with the function F.

Proposition 9.3.1

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

Proof

If $A=\alpha I$, then

$$\begin{aligned}\begin{gathered} AB=\alpha IB=\alpha B, \\ BF(A)=BF(\alpha I)=BF(\alpha ) I=F(\alpha )B. \end{gathered}\end{aligned}$$

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

Proposition 9.3.2

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

Proof

If $B=\alpha I$ then

$$\begin{aligned}\begin{gathered} AB= A(\alpha I)=\alpha A, \\ BF(A)=\alpha IF(A)=\alpha F(A) . \end{gathered}\end{aligned}$$

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

9.3.1 Representations of Covariance Commutation Relations by Integral and Multiplication Operators on $L_p$ Spaces

We consider first a useful lemma for integral operators.

Lemma 9.3.1

Let $f:[\alpha _1,\beta _1]\rightarrow \mathbb {R}$, $g: [\alpha _2,\beta _2]\rightarrow \mathbb {R}$ be two measurable functions such that for all $x\in L_p(\mathbb {R})$, $1\le p \le \infty$,

$$\begin{aligned} \int \limits _{\alpha _1}^{\beta _1} f(t)x(t)dt<\infty ,\quad \int \limits _{\alpha _2}^{\beta _2} g(t)x(t)dt<\infty , \end{aligned}$$

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:

• (i) For all $x\in L_p(\mathbb {R})$, where $1\le p\le \infty$, the following holds

  $$\begin{aligned} \int \limits _{\alpha _1}^{\beta _1} f(t)x(t)dt=\int \limits _{\alpha _2}^{\beta _2} g(t)x(t)dt. \end{aligned}$$

• (ii) The following conditions hold:

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

  • b) for almost every $t \in [\alpha _1,\beta _1]\setminus G,\ f(t)=0;$

  • c) for almost every $t \in [\alpha _2,\beta _2]\setminus G,\ g(t)=0.$

Proof

$(ii)\, \Rightarrow \, (i)$ follows from direct computation.

Suppose that (i) 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,

$$\begin{aligned}\begin{gathered} \int \limits _{\alpha _1}^{\beta _1} f(t)x(t)dt=\int \limits _{\alpha _2}^{\beta _2} g(t)x(t)dt= \int \limits _{\alpha _1}^{\beta _1} f (t)dt=\int \limits _{\alpha _2}^{\beta _2} g (t)dt =\eta , \end{gathered}\end{aligned}$$

$\eta$ is a constant. Now by taking $x(t)=I_{[\alpha _1,\beta _1]\setminus G}(t)$ we get

$$\begin{aligned}\begin{gathered} \int \limits _{\alpha _1}^{\beta _1} f(t)x(t)dt=\int \limits _{\alpha _2}^{\beta _2} g(t)x(t)dt= \int \limits _{[\alpha _1,\beta _1]\setminus G} f (t)dt=\int \limits _{\alpha _2}^{\beta _2} g (t)\cdot 0dt =0. \end{gathered}\end{aligned}$$

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 $S_1,\ldots ,S_n,\ldots$ of the set $[\alpha _1,\beta _1]\setminus G$ such that each set $S_i$, $i=1,2,3,\ldots$ has positive measure. For each $x_i(t)=I_{S_i}(t)$, $i=1,2,3,\ldots$ we have

$$\begin{aligned}\begin{gathered} \int \limits _{\alpha _1}^{\beta _1} f(t)x(t)dt=\int \limits _{\alpha _2}^{\beta _2} g(t)x(t)dt= \int \limits _{S_i} f (t)dt=\int \limits _{\alpha _2}^{\beta _2} g (t)\cdot 0dt =0. \end{gathered}\end{aligned}$$

Thus, $\int \limits _{S_i} f (t)dt=0, \ i=1,2,3,\ldots .$ Since we can choose arbitrary partition with positive measure on each of its elements we have

$$\begin{aligned}\begin{gathered} f(t)=0 \quad \text{ for } \text{ almost } \text{ every } t\in [\alpha _1,\beta _1]\setminus G. \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} \int \limits _G f (t)x(t)dt=\int \limits _G g (t)x(t)dt \Longleftrightarrow \int \limits _G [f(t)-g(t)]x(t)dt=0. \end{gathered}\end{aligned}$$

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

Remark 9.3.1

When operators are given in abstract form, we use the notation $A:L_p(\mathbb {R})\rightarrow L_p(\mathbb {R})$ 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

$$\begin{aligned} (Ax)(t)= \int \limits _{\mathbb {R}} k(t,s)x(s)ds \end{aligned}$$

there are sufficient conditions on kernels $k(\cdot ,\cdot )$ such that operator A is well defined from $L_p(\mathbb {R})$ to $L_p(\mathbb {R})$ and bounded [9, 18]. For instance, [18, Theorem 6.18] states the following: if $1<p<\infty$ and $k:\mathbb {R}\times [\alpha ,\beta ]\rightarrow \mathbb {R}$ is a measurable function, $\alpha ,\beta \in \mathbb {R}$, $\alpha <\beta$, and there is a constant $\lambda >0$ such that

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{s\in [\alpha ,\beta ]}\int \limits _{\mathbb {R}}|k(t,s)| dt\le \lambda , \quad {{\,\mathrm{ess\;sup}\,}}_{t\in \mathbb {R}}\int \limits _{\alpha }^\beta |k(t,s)| ds \le \lambda , \end{aligned}$$

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

9.3.1.1 Representations When A is Integral Operator and B is Multiplication Operator

Proposition 9.3.3

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\alpha }^{\beta }k(t,s) x(s)ds,\quad (Bx)(t)= b(t) x(t), \quad \alpha ,\beta \in \mathbb {R}, \ \alpha <\beta , \end{gathered}\end{aligned}$$

where $k:\mathbb {R}\times [\alpha ,\beta ]\rightarrow \mathbb {R}$ is a measurable function, and $b:\mathbb {R}\rightarrow \mathbb {R}$ is a measurable function. Consider a polynomial defined by $F(z)=\delta _0+\delta _1 z +\cdots +\delta _n z^n$, where $\delta _0,\delta _1,\ldots ,\delta _n$ are real numbers. We set

$$\begin{aligned} k_0(t,s)=k(t,s), \quad \quad k_m(t,s)=\int \limits _{\alpha }^{\beta } k(t,\tau )k_{m-1}(\tau ,s)d\tau ,\quad m\in&\{1,\ldots ,n\} \nonumber \\ F_n(k(t,s))=\sum _{j=1}^{n} \delta _j k_{j-1}(t,s),\quad n\in \{1,2,3,\ldots \}.&\end{aligned}$$

(9.2)

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

$$\begin{aligned} \forall \ x\in L_p(\mathbb {R}): \quad b(t)\delta _0 x(t)+ b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds = \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds. \end{aligned}$$

(9.3)

If $\delta _0=0$, that is, $F(z)=\delta _1 z+\cdots +\delta _n z^n$, then the condition (9.3) reduces to the following: for almost every (t, s) in $\mathbb {R}\times [\alpha ,\beta ]$,

$$\begin{aligned} b(t)F_n(k(t,s))=k(t,s)b(s). \end{aligned}$$

(9.4)

Proof

By applying Fubini Theorem from [1] and iterative kernels from [25], We have

$$\begin{aligned} (A^2x)(t)= & {} \int \limits _{\alpha }^{\beta } k(t,s)(Ax)(s)ds= \int \limits _{\alpha }^{\beta } k(t,s)\left( \int \limits _{\alpha }^{\beta } k(s,\tau )x(\tau )d\tau \right) ds\\= & {} \int \limits _{\alpha }^{\beta } \left( \int \limits _{\alpha }^{\beta } k(t,s)k(s,\tau )ds\right) x(\tau ) d\tau = \int \limits _{\alpha }^{\beta } k_1(t,\tau )x(\tau )d\tau , \end{aligned}$$

where $k_1(t,s)=\int \limits _{\alpha }^{\beta } k(t,\tau )k(\tau ,s)d\tau .$ In the same way,

$$\begin{aligned}\begin{gathered} (A^3x)(t)=\int \limits _{\alpha }^{\beta } k(t,s)(A^2x)(s)ds =\int \limits _{\alpha }^{\beta } k(t,s)\left( \int \limits _{\alpha }^{\beta } k_1(s,\tau )x(\tau )d\tau \right) ds \\ =\int \limits _{\alpha }^{\beta } k_2(t,s)x(s)ds, \end{gathered}\end{aligned}$$

where $k_2(t,s)=\int \limits _{\alpha }^{\beta } k(t,\tau )k_1(\tau ,s)d\tau .$ For every $n \ge 1$,

$$ (A^nx)(t)=\int \limits _{\alpha }^{\beta } k_{n-1}(t,s)x(s)ds, $$

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,

$$\begin{aligned} (F(A)x)(t)= & {} \delta _0 x(t)+ \sum \limits _{j=1}^{n} \delta _j (A^j x)(t)=\delta _0 x(t)+\sum \limits _{j=1}^{n} \delta _{j} \int \limits _{\alpha }^{\beta } k_{j-1}(t,s)x(s)ds \\= & {} \delta _0 x(t)+\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds, \end{aligned}$$

where $F_n(k(t,s))=\sum \limits _{j=1}^{n} \delta _j k_{j-1}(t,s),$ for $n=1,2,3,\ldots$. So, we can compute BF(A)x and (AB)x as follows:

$$\begin{aligned} (BF(A)x)(t)&=b(t)(F(A)x)(t)=b(t)\delta _0 x(t)+ b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds, \\ (ABx)(t)&=A(Bx)(t)= \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds. \end{aligned}$$

It follows that $ABx=BF(A)x$ if and only if condition (9.3) holds.

If $\delta _0=0$ then condition (9.3) reduces to the following:

$$\begin{aligned}\begin{gathered} \forall \ x\in L_p(\mathbb {R}): \quad \int \limits _{\alpha }^{\beta } b(t)F_n(k(t,s))x(s)ds = \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds. \end{gathered}\end{aligned}$$

Let $f(t,s)= b(t)F_n(k(t,s))-k(t,s)b(s)$. By applying Lemma 9.3.1 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 [35], the set N has Lebesgue measure zero on the plane $\mathbb {R}^2$.    $\square$

Corollary 9.3.4

For $M_1,M_2\in \mathbb {R}$, $M_1<M_2$ and $1\le p\le \infty$, let $A:L_p([M_1,M_2])\rightarrow L_p([M_1,M_2])$ and $B:L_p([M_1,M_2])\rightarrow L_p([M_1,M_2])$ be nonzero operators defined, for almost all t, by

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\alpha }^{\beta }k(t,s) x(s)ds,\quad (Bx)(t)= b(t) x(t), \quad \alpha ,\beta \in \mathbb {R},\ \alpha <\beta , \end{gathered}\end{aligned}$$

where $[M_1,M_2]\supseteq [\alpha ,\beta ]$, and $k(\cdot ,\cdot ):[M_1,M_2]\times [\alpha ,\beta ]\rightarrow \mathbb {R}$, $b:[M_1,M_2]\rightarrow \mathbb {R}$ are given by

$$\begin{aligned}\begin{gathered} k(t,s)=a_0+a_1t+c_1s,\qquad b(t)=\sum \limits _{j=0}^n b_j t^j, \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} \forall \ x\in L_p([M_1,M_2]): \quad b(t)\delta _0 x(t)+ b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds = \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds, \end{gathered}\end{aligned}$$

where $F_n(k(t,s))$ is given by (9.2).

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 $[M_1,M_2]\times [\alpha ,\beta ]$

$$\begin{aligned} b(t)F_2(k(t,s))=k(t,s)b(s). \end{aligned}$$

(9.5)

Condition (9.5) is equivalent to that $b(\cdot )\equiv b_0 \ne 0$ is a nonzero constant ($b_j=0,$ $j=1,\ldots ,n$) and one of the following cases holds:

• (i) if $\delta _2=0$, $\delta _1=1$, then $a_0,a_1,c_1\in \mathbb {R}$ can be arbitrary;

• (ii) if $\delta _2\not =0$, $\delta _1=1$, $a_1\not =0$, $c_1=0$, then

  $$\begin{aligned} a_0=-\frac{\beta +\alpha }{2}a_1; \end{aligned}$$

• (iii) if $\delta _2\not =0$, $\delta _1=1$, $a_1=0$, $c_1\not =0$, then

  $$\begin{aligned} a_0=-\frac{\beta +\alpha }{2}c_1; \end{aligned}$$

• (iv) if $\delta _2\not =0$, $\delta _1\not =1$, $a_1\not =0$, $c_1=0$, then

  $$\begin{aligned} a_0=\frac{2-2\delta _1-\delta _2(\beta ^2-\alpha ^2)a_1}{2\delta _2(\beta -\alpha )}; \end{aligned}$$

• (v) if $\delta _2\not =0$, $\delta _1\not =1$, $c_1\not =0$, $a_1=0$, then

  $$\begin{aligned} a_0=\frac{2-2\delta _1-\delta _2(\beta ^2-\alpha ^2)c_1}{2\delta _2(\beta -\alpha )}; \end{aligned}$$

• (vi) if $\delta _2\not =0$, $\delta _1\not =1$, $a_1=0$ and $c_1=0$, then

  $$\begin{aligned} a_0=\frac{1-\delta _1}{\delta _2(\beta -\alpha )}. \end{aligned}$$

Proof

Operator A is defined on $L_p[M_1,M_2]$, $1\le p \le \infty$. Therefore, by applying [19, Theorem 3.4.10], we conclude that A is well defined. 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 9.3.3 we just need to check when the condition (9.4) is satisfied for $k(\cdot ,\cdot )$ and $b(\cdot )$. We compute

$$\begin{aligned} \nonumber k_1(t,s)=&\int \limits _{\alpha }^{\beta }k(t,\tau )k(\tau ,s)d\tau =\int \limits _{\alpha }^{\beta }(a_0+a_1t+c_1\tau )(a_0+a_1\tau +c_1s)d\tau \\ \nonumber =&\int \limits _{\alpha }^{\beta } [(a_0^2+a_0a_1t+a_0c_1s+a_1c_1ts)\\ \nonumber&+(a_0a_1+a_0c_1+a_1^2t+c_1^2s)\tau +a_1c_1\tau ^2]d\tau \\ \nonumber =&(\beta -\alpha )(a_0^2+a_0a_1t+a_0c_1s+a_1c_1ts)\\ \nonumber&+\frac{\beta ^2-\alpha ^2}{2}\cdot (a_0a_1+a_0c_1+a_1^2t+c_1^2s)\\&+\frac{\beta ^3-\alpha ^3}{3}a_1c_1 =\nu _0 + \nu _1 t +\nu _2 s +\nu _3 ts, \end{aligned}$$

(9.6)

where

$$\begin{aligned}\begin{gathered} \begin{array}{cclccl} \nu _0&{}=&{}a_0^2(\beta -\alpha )+\frac{\beta ^2-\alpha ^2}{2}a_0(a_1+c_1)+a_1c_1 \frac{\beta ^3-\alpha ^3}{3}, &{} \nu _2&{}=&{}a_0 c_1(\beta -\alpha )+c_1^2 \frac{\beta ^2-\alpha ^2}{2}, \\ \nu _1&{}=&{}a_1^2\frac{\beta ^2-\alpha ^2}{2}+a_1a_0 (\beta -\alpha ), &{} \nu _3&{}=&{}a_1c_1(\beta -\alpha ). \end{array} \end{gathered}\end{aligned}$$

Then, we have

$$\begin{aligned}&b(t)F_2(k(t,s))=b(t)[\delta _1 k(t,s)+\delta _2 k_1(t,s) ]=(a_0 \delta _1 +\delta _2 \nu _0)\sum _{j=0}^n b_j t^j \\&+(a_1\delta _1+\delta _2 \nu _1) \sum _{j=0}^n b_j t^{j+1}+ ( c_1\delta _1+\delta _2 \nu _2) \sum _{j=0}^n b_j t^js + \nu _3 \delta _2 \sum _{j=0}^n b_j t^{j+1}s \\&\displaystyle = (\delta _1a_0+\delta _2\nu _0)b_0+(c_1\delta _1+\nu _2\delta _2)b_0s+ \sum _{j=1}^{n}[(\delta _1a_0+\delta _2\nu _0)b_j+(\delta _1a_1+\delta _2\nu _1)b_{j-1}]t^j \\&+ \sum _{j=1}^{n}[(c_1\delta _1+\nu _2\delta _2)b_j+\nu _3\delta _2b_{j-1}]t^js+(\delta _1 a_1+\delta _2\nu _1)b_nt^{n+1}+\nu _3\delta _2 b_n t^{n+1}s\\&k(t,s)b(s)=a_0\sum _{j=0}^n b_j s^j+a_1 \sum _{j=0}^n b_j s^j t +c_1 \sum _{j=0}^n b_j s^{j+1}=a_0b_0+a_1b_0t \\&+\sum _{j=1}^{n} (a_0b_j+c_1b_{j-1})s^j +\sum _{j=1}^{n}a_1b_js^j t + c_1b_ns^{n+1}. \end{aligned}$$

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

$$\begin{aligned} a_0b_0= & {} (a_0\delta _1 + \delta _2 \nu _0 )b_0 \nonumber \\ \nonumber a_1b_0= & {} (a_0\delta _1+\delta _2 \nu _0 )b_1+(a_1\delta _1 +\delta _2 \nu _1) b_0\end{aligned}$$

(9.7)

$$\begin{aligned} a_0b_1+c_1b_0= & {} (c_1\delta _1+\delta _2 \nu _2) b_0\end{aligned}$$

(9.8)

$$\begin{aligned} a_1b_1= & {} (c_1\delta _1 + \delta _2 \nu _2) b_1+ \delta _2 \nu _3 b_0 \end{aligned}$$

(9.9)

$$\begin{aligned} 0= & {} a_0b_j +c_1 b_{j-1},\quad 2\le j\le n \\ \nonumber 0= & {} (a_0\delta _1+\delta _2 \nu _0) b_j+(a_1\delta _1 +\delta _2 \nu _1) b_{j-1},\quad 2\le j\le n \end{aligned}$$

(9.10)

$$\begin{aligned} a_1b_j= & {} 0,\quad 2\le j \le n\\ \nonumber 0= & {} c_1\delta _1b_j+\delta _2 \nu _3 b_{j-1}+\delta _2 \nu _2 b_j\quad 2\le j\le n\\ \nonumber 0= & {} a_1\delta _1 b_n+\delta _2 \nu _1 b_n,\quad \text{ if } n\ge 1 \\ c_1b_{n}= & {} 0,\quad \text{ if } n\ge 1 \\ \nonumber 0= & {} \delta _2 \nu _3 b_n,\quad \text{ if } n\ge 1. \end{aligned}$$

(9.11)

Suppose that $n\ge 1$. We proceed by induction to prove that $b_j=0$, for all $j=1,2,\ldots ,n$. For $i=0$, we suppose that $b_n=b_{n-i}\not =0$. Then from (9.10) we have $a_1b_n=0$ and thus $a_1=0$. From Eq. (9.11) we have $c_1b_n=0$ and thus $c_1=0$. From  (9.9) we have $0= a_0b_n +c_1 b_{n-1}=a_0b_n$ 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 $1<m\le n-2$ and suppose that $b_{n-i}=0$ for all $i=1,2,\ldots ,m-1$. Let us show that then $b_{n-m}=0$. If $b_{n-m}\not =0$, then from  (9.10) we have $a_1b_{n-m}=0$ which implies $a_1=0$. From  (9.9) and for $j=n-m+1$ by induction assumption $a_0b_{n-m+1}+c_1b_{n-m}=c_1b_{n-m}=0$ which implies $c_1=0$. Therefore from (9.9) and for $j=n-m$ we have $a_0b_{n-m}=0$ 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 (9.9) gives $c_1b_{n-m}=c_1b_1=0$ when $j=n-m+1=2$. Then $c_1=0$ and by (9.8), since $\nu _2=\nu _3=0$ we get $a_1b_1=0$ which yields $a_1=0$. Therefore, (9.7) gives $a_0b_1=0$ 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 $B\ne 0$ and $B=b_0 I$ (multiple of identity operator), $b_0\ne 0$ and the commutation relation is equivalent to $F(A)=A$. By (9.4) we have $F_2(k(t,s))=k(t,s)$ which can be written as follows

$$\begin{aligned} \delta _1 k(t,s)+\delta _2 k_1(t,s)=k(t,s), \end{aligned}$$

(9.12)

where $k(t,s)=a_0+a_1t+c_1s$ and $k_1(t,s)=\nu _0+\nu _1 t+\nu _2 s+\nu _3 ts$,

$$\begin{aligned}\begin{gathered} \begin{array}{cclccl} \nu _0&{}=&{}a_0^2(\beta -\alpha )+\frac{\beta ^2-\alpha ^2}{2}a_0(a_1+c_1)+a_1c_1 \frac{\beta ^3-\alpha ^3}{3}, &{} \nu _2&{}=&{}a_0 c_1(\beta -\alpha )+c_1^2 \frac{\beta ^2-\alpha ^2}{2}, \\ \nu _1&{}=&{}a_1^2\frac{\beta ^2-\alpha ^2}{2}+a_1a_0 (\beta -\alpha ), &{} \nu _3&{}=&{}a_1c_1(\beta -\alpha ). \end{array} \end{gathered}\end{aligned}$$

If $\delta _2=0$, then (9.12) 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 (9.12) is satisfied for any $a_0,a_1,c_1\in \mathbb {R}$.

If $\delta _2\not =0$ and $\delta _1=1$ then (9.12) becomes $k_1(\cdot ,\cdot )=0$, that is, $\nu _0=\nu _1=\nu _2=\nu _3=0$, where

$$\begin{aligned}\begin{gathered} \begin{array}{cclccl} \nu _0&{}=&{}a_0^2(\beta -\alpha )+\frac{\beta ^2-\alpha ^2}{2}a_0(a_1+c_1)+a_1c_1 \frac{\beta ^3-\alpha ^3}{3}, &{} \nu _2&{}=&{}a_0 c_1(\beta -\alpha )+c_1^2 \frac{\beta ^2-\alpha ^2}{2}, \\ \nu _1&{}=&{}a_1^2\frac{\beta ^2-\alpha ^2}{2}+a_1a_0 (\beta -\alpha ), &{} \nu _3&{}=&{}a_1c_1(\beta -\alpha ). \end{array} \end{gathered}\end{aligned}$$

Since $\alpha <\beta$, $a_1c_1(\beta -\alpha )=0$ is equivalent to either $a_1=0$ or $c_1=0$. If $a_1\ne 0$, $c_1=0$, then

$$\begin{aligned}\begin{gathered} \left\{ \begin{array}{c} \nu _0=0 \\ \nu _1=0\\ \nu _2=0 \\ \nu _3=0 \end{array}\right. \Leftrightarrow \left\{ \begin{array}{c} (\beta -\alpha )a_0^2+\frac{\beta ^2-\alpha ^2}{2}a_0a_1=0 \\ (\beta -\alpha )a_1a_0+\frac{\beta ^2-\alpha ^2}{2}a^2_1=0 \end{array}\right. \Leftrightarrow a_0+\frac{\beta +\alpha }{2}a_1=0, \end{gathered}\end{aligned}$$

which is equivalent to $a_0=-\frac{\beta +\alpha }{2}a_1$. If $a_1=0$, $c_1\ne 0$, then

$$\begin{aligned}\begin{gathered} \left\{ \begin{array}{c} \nu _0=0 \\ \nu _1=0\\ \nu _2=0 \\ \nu _3=0 \end{array}\right. \Leftrightarrow \left\{ \begin{array}{c} (\beta -\alpha )a_0^2+\frac{\beta ^2-\alpha ^2}{2}a_0c_1=0 \\ (\beta -\alpha )c_1a_0+\frac{\beta ^2-\alpha ^2}{2}c^2_1=0 \end{array}\right. \Leftrightarrow a_0+\frac{\beta +\alpha }{2}c_1=0, \end{gathered}\end{aligned}$$

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 (9.12) is equivalent to:

$$\begin{aligned} \left\{ \begin{array}{cl} a_0&{}= \delta _1a_0+ \delta _2 a_0^2(\beta -\alpha )+\delta _2\frac{\beta ^2-\alpha ^2}{2}a_0(a_1+c_1)+\delta _2a_1c_1 \frac{\beta ^3-\alpha ^3}{3} \\ a_1&{}= \delta _1a_1+\delta _2 a_1^2\frac{\beta ^2-\alpha ^2}{2}+\delta _2 a_1a_0 (\beta -\alpha )\\ c_1&{}= \delta _1 c_1+\delta _2 a_0 c_1(\beta -\alpha )+\delta _2 c_1^2 \frac{\beta ^2-\alpha ^2}{2} \\ 0&{}= \delta _2 a_1c_1(\beta -\alpha ). \end{array}\right. \end{aligned}$$

(9.13)

Since $\alpha <\beta$ and $\delta _2\not =0$, equation $\delta _2 a_1c_1(\beta -\alpha )=0$ 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 (9.13) becomes

$$\begin{aligned} a_0= & {} \delta _1a_0+ \delta _2 a_0^2(\beta -\alpha )+\delta _2\frac{\beta ^2-\alpha ^2}{2}a_0a_1\\ a_1= & {} \delta _1a_1+\delta _2 a_1^2\frac{\beta ^2-\alpha ^2}{2}+\delta _2 a_1a_0 (\beta -\alpha ) \end{aligned}$$

which is equivalent to $1=\delta _1+\delta _2(\beta -\alpha )a_0+\delta _2 \frac{\beta ^2-\alpha ^2}{2}a_1$. Then,

$$\begin{aligned} a_0=\frac{2-2\delta _1-\delta _2(\beta ^2-\alpha ^2)a_1}{2\delta _2(\beta -\alpha )}. \end{aligned}$$

If $\delta _2\not =0$, $\delta _1\not =1$, $a_1=0$ and $c_1\ne 0$, then (9.13) becomes

$$\begin{aligned} a_0= & {} \delta _1a_0+ \delta _2 a_0^2(\beta -\alpha )+\delta _2\frac{\beta ^2-\alpha ^2}{2}a_0c_1\\ c_1= & {} \delta _1c_1+\delta _2 c_1^2\frac{\beta ^2-\alpha ^2}{2}+\delta _2 c_1a_0 (\beta -\alpha ) \end{aligned}$$

which is equivalent to $1=\delta _1+\delta _2(\beta -\alpha )a_0+\delta _2 \frac{\beta ^2-\alpha ^2}{2}c_1$. Then,

$$\begin{aligned} a_0=\frac{2-2\delta _1-\delta _2(\beta ^2-\alpha ^2)c_1}{2\delta _2(\beta -\alpha )}. \end{aligned}$$

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 (9.13) becomes

$$\begin{aligned}{} & {} a_0=\delta _1a_0+ \delta _2 a_0^2(\beta -\alpha ) \end{aligned}$$

which is equivalent to $a_0=\frac{1-\delta _1}{\delta _2(\beta -\alpha )}.$    $\square$

Remark 9.3.2

The integral operator given by $(Ax)(t)=\int \limits _{\alpha _1}^{\beta _1} k(t,s)x(s)ds$ for almost all t, where $k:[\alpha _1,\beta _1]\times [\alpha _1,\beta _1]\rightarrow \mathbb {R}$ is a measurable function that satisfies

$$\begin{aligned} \int \limits _{\alpha _1}^{\beta _1} \left( \int \limits _{\alpha _1}^{\beta _1} |k(t,s)|^qds\right) ^\frac{p}{q}dt <\infty , \end{aligned}$$

by [19, Theorem 3.4.10] is well defined from $L_p{[\alpha _1,\beta _1]}$ to $L_p{[\alpha _1 ,\beta _1]}$, $1<p<\infty$ and bounded.

Remark 9.3.3

If in the Corollary 9.3.4 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 $k(t,s)=a_0+a_1t+c_1s$ (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 (9.6). Then we have

$$\begin{aligned}&b(t)F_2(k(t,s))=b(t)[\delta _1 k(t,s)+\delta _2 k_1(t,s) ]=(a_0 \delta _1 +\delta _2 \nu _0)\sum _{j=1}^n b_j t^{-j} \\&+(a_1\delta _1+\delta _2 \nu _1) \sum _{j=1}^n b_j t^{-j+1}+ ( c_1\delta _1+\delta _2 \nu _2) \sum _{j=1}^n b_j t^{-j}s + \nu _3 \delta _2 \sum _{j=1}^n b_j t^{-j+1}s\\&= (a_1\delta _1+\delta _2\nu _1)b_1+\nu _3\delta _2b_1s+\sum _{j=1}^{n-1} [(a_0 \delta _1 +\delta _2 \nu _0)b_j +(a_1\delta _1+\delta _2 \nu _1)b_{j+1}]t^{-j}\\&\displaystyle +(a_0 \delta _1 +\delta _2 \nu _0)b_{n}t^{-n}+ \sum _{j=1}^{n-1} [( c_1\delta _1+\delta _2 \nu _2)b_j +\nu _3 \delta _2b_{j+1}]t^{-j}s+( c_1\delta _1+\delta _2 \nu _2)b_nt^{-n}s \\[0,2cm]&k(t,s)b(s)=\,a_0\sum \limits _{j=1}^n b_j s^{-j}+a_1 \sum \limits _{j=1}^n b_j s^{-j} t +c_1 \sum \limits _{j=1}^n b_j s^{-j+1}\\ =&\,c_1b_1+\sum \limits _{j=1}^{n-1} (a_0b_j +c_1b_{j+1})s^{-j}+\sum \limits _{j=1}^{n} a_1b_j s^{-j} t+a_0b_ns^{-n}. \end{aligned}$$

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

$$\begin{aligned} c_1b_1= & {} a_1\delta _1 b_1 + \delta _2 \nu _1 b_1, \nonumber \\ \nonumber 0= & {} \delta _2 \nu _3 b_1,\\ \nonumber 0= & {} (a_0\delta _1+\delta _2\nu _0) b_{j}+(\delta _1 a_1 +\delta _2 \nu _1) b_{j+1},\quad 1\le j\le n-1,\end{aligned}$$

(9.14)

$$\begin{aligned} a_0 b_j+c_1 b_{j+1}= & {} 0,\quad 1\le j \le n-1, \\ \nonumber 0= & {} c_1\delta _1b_j+\delta _2\nu _2 b_{j}+\delta _2 \nu _3 b_{j+1},\quad 1\le j\le n-1,\end{aligned}$$

(9.15)

$$\begin{aligned} a_1 b_j= & {} 0, \quad 1\le j\le n, \\ \nonumber 0= & {} a_0\delta _1b_n+\delta _2 \nu _0 b_n,\\ 0= & {} a_0b_n, \\ \nonumber 0= & {} c_1\delta _1 b_n + \delta _2 \nu _3 b_n. \end{aligned}$$

(9.16)

Since $b_n\not =0$ then from (9.16) we have $a_0b_n=0$ and thus $a_0=0$. From (9.14) for $j=n-1$ we get $c_1b_n=0$ and thus $c_1=0$. Finally from (9.15) we have $0= a_1b_j$ 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 9.3.5

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\alpha }^{\beta }k(t,s) x(s)ds,\quad (Bx)(t)= b(t) x(t), \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta , \end{gathered}\end{aligned}$$

where $k(t,s):\mathbb {R}\times [\alpha ,\beta ]\rightarrow \mathbb {R}$ is a measurable function, and $b\in L_\infty (\mathbb {R})$ is a nonzero function such that the set ${{\,\mathrm{\mathrm supp\,}\,}}b(t)\cap [\alpha ,\beta ]$ has measure zero.

Consider a polynomial defined by $F(z)=\delta _0+\delta _1 z +\cdots +\delta _n z^n$, where $\delta _0,\ldots ,\delta _n$ are real numbers. We set

$$\begin{aligned}\begin{gathered} \nonumber k_0(t,s)=k(t,s), \quad \quad k_m(t,s)=\int \limits _{\alpha }^{\beta } k(t,\tau )k_{m-1}(\tau ,s)d\tau ,\quad m=1,\ldots ,n, \\ F_n(k(t,s))=\sum _{j=1}^{n} \delta _j k_{j-1}(t,s),\quad n=1,2,3,\ldots \end{gathered}\end{aligned}$$

Then $AB=BF(A)$ if and only if $\delta _0=0$ and the set

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}b(t)\,\cap \, {{\,\mathrm{\mathrm supp\,}\,}}F_n(k(t,s)) \end{gathered}\end{aligned}$$

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

Proof

Suppose that the set ${{\,\mathrm{\mathrm supp\,}\,}}b\cap [\alpha ,\beta ]$ has measure zero. By Proposition 9.3.3 we have $AB=BF(A)$ if and only if condition (9.3) holds, that is,

$$\begin{aligned}\begin{gathered} \forall \ x\in L_p(\mathbb {R}): \quad b(t)\delta _0 x(t)+ b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds = \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds, \end{gathered}\end{aligned}$$

almost everywhere. By taking $x(\cdot )=I_{[M_1,M_2]}(\cdot )b(\cdot )$, where $M_1,M_2\in \mathbb {R}$, $M_1<M_2$, $[M_1,M_2]\supset [\alpha ,\beta ]$, $\mu ([M_1,M_2]\setminus [\alpha ,\beta ])>0$, $I_E(\cdot )$ is the indicator function of the set E, the condition (9.3) reduces to

$$\begin{aligned} I_{[M_1,M_2]}(\cdot ) b^2(\cdot )\delta _0 =0. \end{aligned}$$

Since b has support with positive measure (otherwise $B\equiv 0$), then $\delta _0=0$. By using this, condition (9.3) reduces to the following

$$\begin{aligned}\begin{gathered} \forall \ x\in L_p(\mathbb {R}): \quad b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds = \int \limits _{\alpha }^{\beta } k(t,s)b(s) x(s)ds. \end{gathered}\end{aligned}$$

By hypothesis the right hand side is equal zero. Then condition (9.3) reduces to

$$\begin{aligned}\begin{gathered} \forall \ x\in L_p(\mathbb {R}): \quad b(t)\int \limits _{\alpha }^{\beta } F_n(k(t,s))x(s)ds =0. \end{gathered}\end{aligned}$$

This is equivalent to

$$\begin{aligned} b(t)F_n(k(t,s))=0 \quad \text{ for } \text{ almost } \text{ every } \ s\in [\alpha ,\beta ]. \end{aligned}$$

(9.17)

By applying a similar argument as in the proof of Proposition 9.3.3 we conclude that condition (9.17) is equivalent to that the set

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}b(t)\,\cap \, {{\,\mathrm{\mathrm supp\,}\,}}F_n(k(t,s)) \end{gathered}\end{aligned}$$

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

Corollary 9.3.6

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\alpha }^{\beta }a(t)c(s) x(s)ds,\quad (Bx)(t)= b(t) x(t), \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta , \end{gathered}\end{aligned}$$

where $a:\mathbb {R}\rightarrow \mathbb {R}$, $c:[\alpha ,\beta ]\rightarrow \mathbb {R}$, $b:\mathbb {R}\rightarrow \mathbb {R}$ are measurable functions. Consider a polynomial defined by $F(z)=\delta _1 z +\delta _2 z^2+\cdots +\delta _n z^n$, where $\delta _1,\ldots ,\delta _n$ are real constants. We set $\mu =\int \limits _{\alpha }^{\beta } a(s)c(s)ds.$ Then, we have $AB=BF(A)$ if and only if the set

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)c(s)]\cap {{\,\mathrm{\mathrm supp\,}\,}}\left[ b(t) \sum \limits _{j=1}^{n} \delta _j \mu ^{j-1}-b(s)\right] , \end{gathered}\end{aligned}$$

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

Proof

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

$$\begin{aligned}\begin{gathered} k_0(t,s)=k(t,s)=a(t)c(s),\quad \\ k_m(t,s)=\int \limits _{\alpha }^{\beta } k(t,\tau )k_{m-1}(\tau ,s)d\tau =a(t)c(s)\left( \int \limits _{\alpha }^{\beta } a(s)c(s)ds\right) ^{m}, \ m=1,\ldots ,n\\ F_n(k(t,s))=\sum _{j=1}^{n} \delta _j k_{j-1}(t,s)=\sum _{j=1}^{n} \delta _j a(t)c(s)\left( \int \limits _{\alpha }^{\beta } a(s)c(s)ds\right) ^{j-1} \quad n=1,2,3,\ldots \end{gathered}\end{aligned}$$

By applying Proposition 9.3.3 we have $AB=BF(A)$ if and only if

$$\begin{aligned}\begin{gathered} b(t) \sum \limits _{j=1}^{n} \delta _j a(t)c(s)\left( \int \limits _{\alpha }^{\beta } a(s)c(s)ds\right) ^{j-1}=a(t)c(s)b(s) \ \Longleftrightarrow \\ a(t)c(s)\left[ b(t) \sum \limits _{j=1}^{n} \delta _j \left( \int \limits _{\alpha }^{\beta } a(s)c(s)ds\right) ^{j-1}-b(s)\right] =0 \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)c(s)]\cap {{\,\mathrm{\mathrm supp\,}\,}}\left[ b(t) \sum \limits _{j=1}^{n} \delta _j \left( \int \limits _{\alpha }^{\beta } a(s)c(s)ds\right) ^{j-1}-b(s)\right] \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)c(s)]\cap {{\,\mathrm{\mathrm supp\,}\,}}\left[ b(t) \sum \limits _{j=1}^{n} \delta _j \mu ^{j-1}-b(s)\right] , \end{gathered}\end{aligned}$$

where $\displaystyle \mu =\int \limits _{\alpha }^{\beta } a(s)c(s)ds$.    $\square$

Example 9.3.7

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{0}^{2} a(t)c(s) x(s)ds,\quad (Bx)(t)= b(t)x(t), \end{gathered}\end{aligned}$$

where $a(t)=I_{[0,1]}(t) (1+t^2)$, $c(s)=1$, $b(t)=I_{[1,2]}(t) t^2$. Since kernel has compact support, we can apply [19, 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 $\Vert B\Vert _{L_p} \le 4$. Hence operator B is well defined and bounded. Consider a polynomial defined by $F(z)=\delta _1 z +\cdots +\delta _n z^n$, where $\delta _1,\ldots ,\delta _n$ are real constants. Then, the above operators does not satisfy the relation $AB=BF(A).$ In fact for $\lambda \ne 0$, by applying Corollary 9.3.6 and setting $\lambda =\sum \limits _{j=1}^{n} \delta _j (\beta -\alpha )^{j-1},$ we have

$$ {{\,\mathrm{\mathrm supp\,}\,}}\, \{b(t)\lambda -b(s) \}=\left( {\mathbb {R} \times [1,2] \cup [1,2] \times [0,1])\setminus W,} \right. $$

where $W=\{(t,s)\in [1,2]\times [1,2]:\ b(t)\lambda -b(s)=0 \}$ is a set of measure zero in the plane. Moreover, ${{\,\mathrm{\mathrm supp\,}\,}}a(t)c(s)=[0,1]\times [0,2].$ The set

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)c(s)]\cap {{\,\mathrm{\mathrm supp\,}\,}}\left[ b(t) \lambda -b(s)\right] , \end{gathered}\end{aligned}$$

has positive measure in $\mathbb {R}\times [0,2]$.

Example 9.3.8

Let $A:L_p(\mathbb {R})\rightarrow L_p(\mathbb {R})$, $B:L_p(\mathbb {R})\rightarrow L_p(\mathbb {R})$, $1<p<\infty$ 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 [19, 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 $\Vert B\Vert _{L_p} \le 4$. Hence operator B is well defined and bounded. Consider a polynomial defined by $F(z)=\delta _1 z +\cdots +\delta _n z^n$, where $\delta _1,\ldots ,\delta _n$ are real constants. 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 9.3.6 we have

$$\begin{aligned}\begin{gathered} \mu =\int \limits _{0}^2 a(s)c(s)ds=1. \end{gathered}\end{aligned}$$

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

Example 9.3.9

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{0}^{2} a(t)c(s) x(s)ds,\quad (Bx)(t)= b(t)x(t), \end{gathered}\end{aligned}$$

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 $\Vert B\Vert _{L_p}\le \Vert b\Vert _{L_\infty }$ 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 $d\ge 2$. Then, the above operators satisfy the relation $AB=\delta BA^d.$ In fact, by applying Corollary 9.3.6 we have $\mu =\int \limits _{0}^2 a(s)c(s)ds=0.$ Hence, ${{\,\mathrm{\mathrm supp\,}\,}}\{b(t)\cdot 0-b(s) \}=\mathbb {R}\times [1,2].$ Moreover, ${{\,\mathrm{\mathrm supp\,}\,}}a(t)c(s)=[0,2]\times [0,1].$ The set ${{\,\mathrm{\mathrm supp\,}\,}}[a(t)c(s)]\cap {{\,\mathrm{\mathrm supp\,}\,}}\left[ -b(s)\right] ,$ has measure zero in $\mathbb {R}\times [0,2]$.

Example 9.3.10

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

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\alpha }^{\beta } I_{[\alpha ,\beta ]}(t) x(s)ds,\quad (Bx)(t)= I_{[\alpha ,\beta ]}(t) x(t), \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta . \end{gathered}\end{aligned}$$

Since kernel has compact support, we can apply [19, Theorem 3.4.10] and, we conclude that operator A is well defined and bounded. Since $\Vert B\Vert _{L_p}\le 1$ then operator B is well defined and bounded. Consider a polynomial defined by $F(z)=\delta _1 z +\cdots +\delta _n z^n$, where $\delta _1,\ldots ,\delta _n$ are constants. 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

$$\begin{aligned}\begin{gathered} \lambda =\sum \limits _{j=1}^{n} \delta _j \left( \int \limits _{\alpha }^{\beta }a(s)c(s)ds\right) ^{j-1}=\sum \limits _{j=1}^{n} \delta _j (\beta -\alpha )^{j-1}, \end{gathered}\end{aligned}$$

then from Corollary 9.3.6 we have the following:

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

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

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

• If $\lambda =1$,

  $$\begin{aligned} {{\,\mathrm{\mathrm supp\,}\,}}[b(t)-b(s)]= & {} \left\{ (t,s)\in \mathbb {R}\times [\alpha ,\beta ]:\ I_{[\alpha ,\beta ]}(t)\not =1\right\} \\= & {} (\mathbb {R}\setminus [\alpha ,\beta ])\times [\alpha ,\beta ]. \end{aligned}$$

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

• If $\lambda =0$,

  $$\begin{aligned} {{\,\mathrm{\mathrm supp\,}\,}}[\lambda b(t)-b(s)]= & {} {{\,\mathrm{\mathrm supp\,}\,}}b(s)=\left\{ (t,s)\in \mathbb {R}^2:\ I_{[\alpha ,\beta ]}(s)\not =0\right\} \\= & {} \left\{ (t,s)\in \mathbb {R}^2:\ \alpha \le s\le \beta \right\} . \end{aligned}$$

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

The conditions in the Corollary 9.3.6 are fulfilled only in the second case, that is, when $\lambda =1$.

9.3.1.2 Representations When A is Multiplication Operator and B is Integral Operator

Proposition 9.3.11

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

$$\begin{aligned}\begin{gathered} (Ax)(t)=a(t) x(t),\quad (Bx)(t)=\int \limits _{\alpha }^{\beta }k(t,s) x(s)ds, \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta , \end{gathered}\end{aligned}$$

where $a:\mathbb {R}\rightarrow \mathbb {R}$, $k: \mathbb {R}\times [\alpha ,\beta ]\rightarrow \mathbb {R}$ are measurable functions. Consider a polynomial defined by $F(z)=\delta _0+\delta _1 z +\cdots +\delta _n z^n$, where $\delta _0, \delta _1,\ldots ,\delta _n$ are constants. Then

$$\begin{aligned}\begin{gathered} AB=BF(A) \end{gathered}\end{aligned}$$

if and only if the set

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}k(t,s) \end{gathered}\end{aligned}$$

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

Proof

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

$$\begin{aligned}\begin{gathered} (ABx)(t)= \int \limits _{\alpha }^{\beta } a(t) k(t,s) x(s)ds \\ (A^nx)(t)=[a(t)]^nx(t) \\ (F(A)x)(t)= \sum _{i=0}^{n} \delta _i (A^ix)(t)=\left( \sum _{i=0}^{n} \delta _i[a(t)]^i \right) x(t)=F(a(t))x(t)\\ (BF(A)x)(t)=\int \limits _{\alpha }^{\beta }k(t,s))F(a(s)) x(s)ds. \end{gathered}\end{aligned}$$

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

$$\begin{aligned} \int \limits _{\alpha }^{\beta }a(t) k(t,s) x(s)ds=\int \limits _{\alpha }^{\beta } k(t,s)F(a(s)) x(s)ds. \end{aligned}$$

(9.18)

almost everywhere. By using Lemma 9.3.1 and by applying the same argument as in the final steps on the proof of Proposition 9.3.3, the condition (9.18) is equivalent to

$$\begin{aligned}\begin{gathered} a(t)k(t,s)=k(t,s)F[a(s)] \Longleftrightarrow k(t,s)[a(t)-F(a(s))]=0 \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}k(t,s) \end{gathered}\end{aligned}$$

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

Example 9.3.12

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

$$\begin{aligned}\begin{gathered} (Ax)(t)=I_{[\alpha ,\beta ]}(t) x(t),\quad (Bx)(t)=\int \limits _{\alpha }^{\beta } I_{[\alpha ,\beta ]^2}(t,s)x(s)ds, \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta \end{gathered}\end{aligned}$$

By using properties of norm and [19, 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

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]=\left\{ (t,s)\in \mathbb {R}\times [\alpha ,\beta ]:\ I_{[\alpha ,\beta ]}(t)\not =1\right\} =(\mathbb {R}\setminus [\alpha ,\beta ])\times [\alpha ,\beta ],\\ \\ {{\,\mathrm{\mathrm supp\,}\,}}k(t,s)=[\alpha ,\beta ]\times [\alpha ,\beta ]. \end{gathered}\end{aligned}$$

The set ${{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}[k(t,s)]$ has measure zero in $\mathbb {R}\times [\alpha ,\beta ]$. So the result follows from Proposition 9.3.11.

Example 9.3.13

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

$$\begin{aligned} (Ax)(t)=[\gamma _1I_{[0,1/2)}(t)-\gamma _2I_{[1/2,1]}(t)] x(t) ,\quad (Bx)(t)=\int \limits _{0}^{1} k(t,s) x(s)ds \end{aligned}$$

$k:\mathbb {R}\times [0,1]\rightarrow \mathbb {R}$ 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

$$\begin{aligned}\begin{gathered} |\delta _0|+|\delta _1|+|\gamma _1|+|\gamma _2|\not =0. \end{gathered}\end{aligned}$$

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

• (i) $\delta _0=-\delta _1\gamma _1$ and ${{\,\mathrm{\mathrm supp\,}\,}}\ k(t,s)\subseteq (\mathbb {R}\setminus [0,1])\times [0,1/2]$;

• (ii) $\delta _0=\delta _1\gamma _2$ and ${{\,\mathrm{\mathrm supp\,}\,}}\ k(t,s)\subseteq (\mathbb {R}\setminus [0,1])\times [1/2,1]$;

• (iii) $\delta _0+\delta _1 \gamma _1-\gamma _1=0$ and ${{\,\mathrm{\mathrm supp\,}\,}}k(t,s)\subseteq [0,1/2]\times [0,1/2]$;

• (iv) $\delta _0+\delta _1 \gamma _1+\gamma _2=0$ and ${{\,\mathrm{\mathrm supp\,}\,}}k(t,s)\subseteq [1/2,1]\times [0,1/2]$;

• (v) $\delta _0-\delta _1 \gamma _2-\gamma _1=0$ and ${{\,\mathrm{\mathrm supp\,}\,}}k(t,s)\subseteq [0,1/2]\times [1/2,1]$;

• (vi) $\delta _0-\delta _1 \gamma _2+\gamma _2=0$ and ${{\,\mathrm{\mathrm supp\,}\,}}k(t,s)\subseteq [1/2,1]\times [1/2,1]$,

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

$$\begin{aligned}\begin{gathered} [a(t)-F(a(s))]=\left\{ \begin{array}{cccc} 0, &{} \text{ if } \delta _0=-\delta _1\gamma _1, &{} t\not \in [0,1], &{} s\in [0,1/2) \\ 0, &{} \text{ if } \delta _0=\delta _1\gamma _2, &{} t\not \in [0,1], &{} s\in [1/2,1] \\ 0, &{} \text{ if } \delta _0+\delta _1\gamma _1-\gamma _1=0, &{} t\in [0,1/2), &{} s\in [0,1/2) \\ 0, &{} \text{ if } \delta _0+\delta _1\gamma _1+\gamma _2=0, &{} t\in [1/2,1), &{} s\in [0,1/2] \\ 0, &{} \text{ if } \delta _0-\delta _1\gamma _2-\gamma _1=0, &{} t\in [0,1/2], &{} s\in [1/2,1] \\ 0, &{} \text{ if } \delta _0-\delta _1\gamma _2+\gamma _2=0, &{} t\in [1/2,1], &{} s\in [1/2,1] \\ \gamma _3, &{} \text{ otherwise } &{} &{} \\ \end{array}\right. \end{gathered}\end{aligned}$$

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:

• (i) $k(t,s)=I_{[2,3]\times [0,1/2]}(t,s)$ if $\delta _0=-\delta _1\gamma _1$;

• (ii) $k(t,s)=I_{[2,3]\times [1/2,1]}(t,s)$ if $\delta _0=\delta _1\gamma _2$;

• (iii) $k(t,s)=I_{[0,1/3]\times [1/3,1/2]}(t,s)$ if $\delta _0+\delta _1\gamma _1-\gamma _1=0$;

• (iv) $k(t,s)=I_{[2/3,1/2]\times [0,1/2]}(t,s)$ if $\delta _0+\delta _1\gamma _1+\gamma _2=0$;

• (v) $k(t,s)=I_{[0,1/3]\times [2/3,1]}(t,s)$ if $\delta _0-\delta _1\gamma _2-\gamma _1=0$;

• (vi) $k(t,s)=I_{[2/3,1]\times [2/3,1]}(t,s)$ if $\delta _0-\delta _1\gamma _2+\gamma _2$.

According to the definition, in all above cases the set

$$\begin{aligned} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}\ [k(t,s)] \end{aligned}$$

has measure zero in $\mathbb {R}\times [0,1]$. So the result follows from Proposition 9.3.11.

Corollary 9.3.14

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

$$\begin{aligned}\begin{gathered} (Ax)(t)=a(t) x(t),\quad (Bx)(t)=\int \limits _{\alpha }^{\beta }b(t)c(s) x(s)ds, \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta , \end{gathered}\end{aligned}$$

where $a:\mathbb {R}\rightarrow \mathbb {R}$, $b:\mathbb {R}\rightarrow \mathbb {R}$, $c:[\alpha ,\beta ]\rightarrow \mathbb {R}$ are measurable functions. For a polynomial defined by $F(z)=\delta _0+\delta _1 z +\cdots +\delta _n z^n$, where $\delta _0, \delta _1,\ldots ,\delta _n$ are real constants, we have

$$\begin{aligned}\begin{gathered} AB=BF(A) \end{gathered}\end{aligned}$$

if and only if the set

$$\begin{aligned} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}[b(t)c(s)] \end{aligned}$$

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

Proof

This follows by Proposition 9.3.11.    $\square$

Example 9.3.15

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

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

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 $\Vert A\Vert _{L_p}\le \Vert a\Vert _{L_\infty }$. Therefore, A is well defined and bounded. Since kernel has compact support in $\mathbb {R}\times [\alpha ,\beta ]$, we can apply [19, 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

$$\begin{aligned} F(a(s))-a(t)= & {} -\delta _1+ \delta _1 I_{[\alpha ,\beta ]}(s)-I_{[\alpha ,\beta ]}(t)= -I_{[\alpha ,\beta ]}(t). \end{aligned}$$

Therefore, we have

$$\begin{aligned}\begin{gathered} {{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]=[\alpha ,\beta ]\times [\alpha ,\beta ],\\ {{\,\mathrm{\mathrm supp\,}\,}}b(t)c(s)={{\,\mathrm{\mathrm supp\,}\,}}\ I_{[\alpha -2,\alpha -1]}(t)I_{[\alpha ,\beta ]}(s)=[\alpha -2,\alpha -1]\times [\alpha ,\beta ]. \end{gathered}\end{aligned}$$

The set ${{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}[I_{[\alpha -2,\alpha -1]}(t)I_{[\alpha ,\beta ]}(s)]$ has measure zero. So the result follows from Corollary 9.3.14.

Example 9.3.16

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

$$\begin{aligned}\begin{gathered} (Ax)(t)=a(t) x(t),\quad (Bx)(t)=\int \limits _{\alpha }^{\beta } b(t)c(s) x(s)ds, \quad \alpha ,\beta \in \mathbb {R}, \alpha <\beta , \end{gathered}\end{aligned}$$

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

$$\begin{aligned}\begin{gathered} AB-\delta _1 BA=\delta _0 B. \end{gathered}\end{aligned}$$

In fact, A is well defined, bounded since $a\in L_\infty$ and this implies $\Vert A\Vert _{L_p}\le \Vert a\Vert _{L_\infty }$. 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 [19, Theorem 3.4.10]. If $\delta _0=\gamma _0-\delta _1 \gamma _0$ then we have

$$\begin{aligned} F(a(s))-a(t)= & {} \delta _0+\gamma _0\delta _1+ \delta _1 I_{\left[ \alpha ,\frac{\alpha +\beta }{2}\right] }(s)s^2-\gamma _0-I_{\left[ \alpha ,\frac{\alpha +\beta }{2}\right] }(t)t^2\\= & {} \delta _1 I_{\left[ \alpha , \frac{\alpha +\beta }{2}\right] }(s)s^2-I_{\left[ \alpha ,\frac{\alpha +\beta }{2}\right] }(t)t^2. \end{aligned}$$

Then we have

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

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

$$\begin{aligned} {{\,\mathrm{\mathrm supp\,}\,}}\ b(t)c(s)= & {} {{\,\mathrm{\mathrm supp\,}\,}}\ (1+t^2)I_{[\beta +1,\beta +2]}(t) I_{\left[ \frac{\alpha +\beta }{2},\beta \right] }(s)(1+s^4)\\= & {} [\beta +1,\beta +2]\times \left[ \frac{\alpha +\beta }{2},\beta \right] . \end{aligned}$$

The set ${{\,\mathrm{\mathrm supp\,}\,}}[a(t)-F(a(s))]\cap {{\,\mathrm{\mathrm supp\,}\,}}[b(t)c(s)]$ has measure zero. So the result follows from Corollary 9.3.14.