Representations of Polynomial Covariance Type Commutation Relations by Linear Integral Operators on $L_p$ Over Measure Spaces

Abstract

Representations of polynomial covariance type commutation relations by linear integral operators on $L_p$ over measures spaces are constructed. Conditions for such representations are described in terms of kernels of the corresponding integral operators. Representation by integral operators are studied both for general polynomial covariance commutation relations and for important classes of polynomial covariance commutation relations associated to arbitrary monomials and to affine functions. Examples of integral operators on $L_p$ spaces representing the covariance commutation relations are constructed. Representations of commutation relations by integral operators with special classes of kernels such as separable kernels and convolution kernels are investigated.

1 Introduction

Commutation  relations of the form

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

(4.1)

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 (4.1) are called a representation of this relation in that algebra. Representations of covariance commutation relations (4.1) 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 applications in physics and engineering [4, 5, 16,17,18, 26,27,28, 34, 35, 42].

A description of the structure of representations for the relation (4.1) 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 maps involved in the commutation relations [3, 7,8,9,10,11,12,13, 19,20,21, 29,30,34, 36,37,40, 42,43,55].

In this paper, we construct representations of the covariance commutation relations (4.1) by linear integral operators on Banach spaces $L_p$ over measure spaces. When $B=0$, the relation (4.1) is trivially satisfied for any A. Thus, we focus on construction and properties of nontrivial representations of (4.1). We consider representations by the linear integral operators defined by kernels satisfying different conditions. We derive conditions on such kernel functions so that the corresponding operators satisfy (4.1) for polynomial F when both operators are of linear integral type. Representations of polynomial covariance type commutation relations by linear integral operators on $L_p$ over measure spaces are constructed. Conditions for such representations are described in terms of kernels of the corresponding integral operators. Representation by integral operators are studied both for general polynomial covariance commutation relations and for important classes of polynomial covariance commutation relations associated to arbitrary monomials and to affine functions. Examples of integral operators on $L_p$ spaces representing the covariance commutation relations are constructed. Representations of commutation relations by integral operators with special classes of kernels such as separable kernels and convolution kernels are investigated. In particular, we prove that there are no nonzero one sided convolution linear integral operators representing covariance type commutation relation for monomial $t^m$, where m a nonnegative integer except 1. This paper is organized in four sections. After the introduction, we present in Sect. 4.2 some preliminaries, notations, basic definitions and two useful lemmas. In Sect. 4.3, we present some representations when both operators A and B are linear integral operators acting on the Banach spaces $L_p$. In particular, we consider cases when operators are convolution type and operators with separable kernels.

2 Preliminaries and Notations

In this section we present preliminaries, basic definitions and notations for this article [1, 2, 6, 14, 22,23,24, 41].

Let $\mathbb {R}$ be the set of all real numbers, X be a non-empty space, and $S\subseteq X$. Let $(S,\varSigma , \mu )$ be a $\sigma$-finite measure space, where $\varSigma$ is a $\sigma -$algebra with measurable subsets of S, and S can be covered with at most countably many disjoint sets $E_1,E_2,E_3,\ldots$ such that $E_i\in \varSigma , \, \mu (E_i)<\infty$, $i=1,2,\ldots$ and $\mu$ is a measure. For $1\le p<\infty ,$ we denote by $L_p(S,\mu )$, the set of all classes of equivalent (different on a set of zero measure) measurable functions $f:S\rightarrow \mathbb {R}$ such that $\int \limits _{S} |f(t)|^p d\mu < \infty .$ 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 $L_\infty (S,\mu )$ the set of all classes of equivalent measurable functions $f:S\rightarrow \mathbb {R}$ such that exists $C>0$, $|f(t)|\le C$ almost everywhere. This is a Banach space with norm $\Vert f\Vert _{\infty }=\mathop {{{\,\mathrm{ess\;sup}\,}}}_{t\in S} |f(t)|.$ The support of a function $f:\, X\rightarrow \mathbb {R}$ is $\textrm{supp }\, f = \{t\in X :\, f(t)\not =0\}.$ We will use notation

$$\begin{aligned} Q_{G}(u,v)=\int \limits _{G} u(t)v(t)d\mu \end{aligned}$$

(4.2)

for $G\in \varSigma$ and such functions $u,v:\, G\rightarrow \mathbb {R}$ that integral exists and is finite. The convolution of functions $f:\, \mathbb {R}\rightarrow \mathbb {R}$ and $g:\, \mathbb {R}\rightarrow \mathbb {R}$ is defined by $(f\star g)(t)=\int \limits _{-\infty }^{+\infty } f(\tau )g(t-\tau )d\tau .$

Now we will consider two useful lemmas for integral operators which will be used throughout the article. Lemma 1 is used in the proof of Theorem 1 and Lemma 2 is used in the proof of Theorem 2.

Lemma 1

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $f, g\in L_q(X,\mu )$ for $1\le q\le \infty$ and let $G_1,G_2\in \varSigma$ such that $\mu (G_i)<\infty$, $i=1,2$. Let $G=G_1\cap G_2.$ Then the following statements are equivalent:

• 1. For all $x\in L_p(X,\mu )$, $1\le p \le \infty$ such that $\displaystyle \frac{1}{p}+\frac{1}{q}=1$,

  $$\begin{aligned} Q_{G_1}(f,x)= \int \limits _{G_1} f(t)x(t)d\mu =\int \limits _{G_2} g(t)x(t)d\mu =Q_{G_2}(g,x). \end{aligned}$$

• 2. The following conditions hold:

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

  • (b) for almost every $t \in G_1\setminus G,\ f(t)=0,$

  • (c) for almost every $t \in G_2\setminus G,\ g(t)=0.$

Proof

2 $\Rightarrow$ 1 By additivity of the measure of integration $\mu$ on $\varSigma$,

$$\begin{aligned} \textstyle \int \limits _{G_1} f(t)x(t)d\mu&= \int \limits _{G_1\setminus G} f(t)x(t)d\mu + \int \limits _{G} f(t)x(t)d\mu = \int \limits _{G} f(t)x(t)d\mu \\ \textstyle =&\int \limits _{G} g(t)x(t)d\mu = \int \limits _{G_2\setminus G} g(t)x(t)d\mu + \int \limits _{G} g(t)x(t)d\mu = \int \limits _{G_2} g(t)x(t)d\mu . \end{aligned}$$

1 $\Rightarrow$ 2 For the indicator function $x(t)=I_{H_1}(t)$ of the set $H_1=G_1\cup G_2$,

$$ \textstyle \int \limits _{G_1} f(t)x(t)d\mu =\int \limits _{G_2} g(t)x(t)d\mu = \int \limits _{G_1} f (t)d\mu =\int \limits _{G_2} g (t)d\mu =\eta , $$

where $\eta$ is a constant. Now by taking $x(t)=I_{G_1\setminus G}$ we get

$$ \textstyle \int \limits _{G_1} f(t)x(t)d\mu =\int \limits _{G_2} g(t)x(t)d\mu = \int \limits _{G_1\setminus G} f (t)d\mu =\int \limits _{G_2} g (t)\cdot 0d\mu =0. $$

Then $\int \limits _{G_1\setminus G} f (t)d\mu =0.$ Analogously by taking $x(t)=I_{G_2\setminus G}(t)$ we get $\int \limits _{G_2\setminus G} g (t)d\mu =0$ We claim that $f(t)=0$ for almost every $t\in G_1\setminus G$ and $g(t)=0$ for almost every $t\in G_2\setminus G$. We take a partition $S_1,S_2,\ldots ,S_n,\ldots$ of the set $G_1\setminus G$ 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 $\int \limits _{G_1} f(t)x(t)d\mu =\int \limits _{G_2} g(t)x(t)d\mu = \int \limits _{S_i} f (t)d\mu =\int \limits _{G_2} g (t)\cdot 0d\mu =0.$ Thus, $\int \limits _{S_i} f (t)d\mu =0, \ i=1,2,3,\ldots$ Since we can choose arbitrary partition with positive measure on each of its elements, $f(t)=0 \ \text{ for } \text{ almost } \text{ every } t\in G_1\setminus G.$ Analogously, $g(t)=0 \ \text{ for } \text{ almost } \text{ every } t\in G_2\setminus G.$ Therefore $\eta = \int \limits _{G_1} f (t)d\mu =\int \limits _{G_2} g (t)d\mu =\int \limits _G f (t)d\mu =\int \limits _G g (t)d\mu .$ Then, for all function $x\in L_p(X,\mu )$ we have $\int \limits _G f (t)x(t)d\mu =\int \limits _G g (t)x(t)d\mu \Leftrightarrow \int \limits _G [f(t)-g(t)]x(t)d\mu =0.$ This implies that $f(t)=g(t)$ for almost every $t\in G$. $\square$

Let n be a positive integer, $(\mathbb {R}^n,\varSigma ,\mu )$ be the standard Lebesgue measure space and $\varOmega \in \varSigma$. We denote by $C(\varOmega )$ the set of all continuous functions $f:\varOmega \rightarrow \mathbb {R}$. This is a Banach space with norm $\Vert f\Vert =\max _{t\in \varOmega } |f(t)|.$ We denote by $C_c(\mathbb {R}^n)$ the set of all continuous functions with compact support.

The following statement is similar to Lemma 1 under conditions: $X=\mathbb {R}^n$ and sets $G_1$, $G_2$ can have infinite measure.

Lemma 2

Let $(\mathbb {R}^n,\varSigma , \mu )$ be the standard Lebesgue measure space and $f, g\in L_q(\mathbb {R}^n,\mu )$ for $1< q < \infty$, $G_1\in \varSigma$ and $G_2\in \varSigma$. Let $G=G_1\cap G_2.$ Then the following statements are equivalent:

• 1. For all $x\in L_p(\mathbb {R}^n,\mu )$, where $1 \le p < \infty$ such that $\displaystyle \frac{1}{p}+\frac{1}{q}=1$,

  $$\begin{aligned}\begin{gathered} Q_{G_1}(f,x)=\int \limits _{G_1} f(t)x(t)d\mu =\int \limits _{G_2} g(t)x(t)d\mu =Q_{G_2}(g,x). \end{gathered}\end{aligned}$$

• 2. The following conditions hold:

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

  • (b) for almost every $t \in G_1\setminus G$ $f(t)=0,$

  • (c) for almost every $t \in G_2\setminus G$ $g(t)=0.$

Proof

2 $\,\Rightarrow \,$ 1 This follows by direct computation as in the proof of Lemma 1.

1 $\,\Rightarrow \,$ 2 Suppose that 2 is true. If $G_1\in \varSigma$ and $G_2\in \varSigma$ have finite measure then it follows from Lemma 1. Suppose that either $G_1$ has infinite measure or $G_2$ has infinite measure. For any $\alpha >0$ and $\varOmega _\alpha =[-\alpha ,\alpha ]^n \subset \mathbb {R}^n$, the set $V_\alpha =\{x\in C_c(\mathbb {R}^n):\, x(t)=0,\,\forall t\in \mathbb {R}^n \setminus \varOmega _\alpha \}$ is a subspace of $C_c(\mathbb {R}^n)$. Since condition 1 is satisfied for any $x\in V_\alpha$, and any $x\in V_\alpha$ vanishes outside the set $\varOmega _\alpha$, with finite measure, we have from Lemma 1:

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

• (b) for almost every $t \in (G_1\cap \varOmega _\alpha )\setminus G$, $f(t)=0$;

• (c) for almost every $t \in (G_2\cap \varOmega _\alpha )\setminus G$, $g(t)=0$.

These conclusions are true for any fixed $\alpha >0$, and so for the corresponding $\varOmega _\alpha$, $V_\alpha$. Since $f,g\in L_p(\mathbb {R}^n,\mu )\,1<p<\infty$ then there exist compact sets $K_m$ such that

$$ \lim \limits _{m\rightarrow +\infty }\mu (\{t\in \mathbb {R}^n\setminus K_m:\, f(t)>0\})=\lim _{m\rightarrow +\infty }\mu (\{t\in \mathbb {R}^n\setminus K_m:\, g(t)>0\})=0. $$

Hence condition 1 holds for all $x\in C_c(\mathbb {R}^n)$ if and only if condition 2 holds. The conclusion follows from [6, Theorem 4.3 and Theorem 4.12], that is, the set $C_c(\mathbb {R}^n)$ is dense in $L_p(\mathbb {R}^n,\mu )$, for $1< p<\infty$. $\square$

3 Representations by Linear Integral Operators

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. In this section we consider representations of the covariance type commutation relation (4.1) when both A and B are linear integral operators acting from the Banach space $L_p(X,\mu )$ to itself for a fixed p such that $\ 1\le p\le \infty$ defined as follows:

$$ (Ax)(t)= \int \limits _{S_A} k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{S_B}{k}_B(t,s)x(s)d\mu _s, $$

almost everywhere, where the index in $\mu _s$ indicates the variable of integration, $S_A,S_B\in \varSigma$, $\mu (S_A)<\infty$, $\mu (S_B)<\infty$, $k_A(t,s):X\times S_A\rightarrow \mathbb {R}$, ${k}_B(t,s):\, X\times S_B\rightarrow \mathbb {R}$ are measurable functions satisfying conditions bellow. For $1<p<\infty$ we have from [15] that the operators $A:\, L_p(X,\mu )\rightarrow L_p(X,\mu )$ and $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$ are well-defined if kernels satisfy the following conditions

$$\begin{aligned} \int \limits _{X}\left( \ \int \limits _{S_A}|k_A(t,s)|^q d\mu _s\right) ^{p/q}d\mu _t<\infty ,\ \int \limits _{X}\left( \ \int \limits _{S_B}|{k}_B(t,s)|^q d\mu _s\right) ^{p/q}d\mu _t <\infty , \end{aligned}$$

(4.3)

where $1< q <\infty$ is such that $\frac{1}{p}+\frac{1}{q}=1$. For $p=1$, operators $A:\, L_1(X,\mu )\rightarrow L_1(X,\mu )$ and $B: \, L_1(X,\mu )\rightarrow L_1(X,\mu )$ are well-defined if kernels satisfy the following conditions

$$\begin{aligned} \int \limits _{X} \mathop {{{\,\mathrm{ess\;sup}\,}}}_{s\in S_A } |k_A(t,s)| d\mu _t<\infty ,\qquad \int \limits _{X} \mathop {{{\,\mathrm{ess\;sup}\,}}}_{s\in S_B } |{k}_B(t,s)| d\mu _t <\infty . \end{aligned}$$

(4.4)

For $p=\infty$, operators $A:\, L_\infty (X,\mu )\rightarrow L_\infty (X,\mu )$ and $B: \, L_\infty (X,\mu )\rightarrow L_\infty (X,\mu )$ are well-defined if kernels satisfy the following conditions

$$\begin{aligned} \mathop {{{\,\mathrm{ess\;sup}\,}}}_{t\in X } \left( \ \int \limits _{S_A} |k_A(t,s)| d\mu _s\right)<\infty ,\qquad \mathop {{{\,\mathrm{ess\;sup}\,}}}_{t\in X } \left( \ \int \limits _{S_B} |{k}_B(t,s)| d\mu _s\right) <\infty . \end{aligned}$$

(4.5)

Theorem 1

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:\, L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined as follows

$$ (Ax)(t)= \int \limits _{G_A} k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s, $$

almost everywhere, where the index in $\mu _s$ indicates the variable of integration, $G_A,\, G_B \in \varSigma$, $\mu (G_A)<\infty$, $\mu (G_B)<\infty$, $k_A(t,s):\mathbb {R}\times S_A\rightarrow \mathbb {R}$, ${k}_B(t,s):\mathbb {R}\times S_B\rightarrow \mathbb {R}$ are measurable functions satisfying either relation (4.3) or (4.4) or (4.5), respectively. Consider a polynomial $F:\mathbb {R}\rightarrow \mathbb {R}$ defined by $F(z)=\sum \limits _{j=0}^{n} \delta _j z^j$, where $\delta _j \in \mathbb {R}$, $j=0,1,2,\ldots ,n$. Set $G=G_A\cap G_B$, and

$$\begin{aligned}\begin{gathered} k_{0,A}(t,s)=k_A(t,s), \ k_{m,A}(t,s)=\int \limits _{G_A} k_A(t,\tau )k_{m-1,A}(\tau ,s)d{\mu _\tau },\ m=1,2,3,\ldots ,n \\ F_n(k_A(t,s))=\sum _{j=1}^{n} \delta _j k_{j-1}(t,s),\quad n=1,2,3,\ldots \end{gathered}\end{aligned}$$

Then $AB=BF(A)$ if and only if the following conditions are fulfilled :

• 1. for almost every $(t,\tau )\in X\times G$,

  $$ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s-\delta _0\tilde{k}(t,\tau ) = \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s, $$

  where $\mu _s$ indicates that integration is taken with respect to the variable s;

• 2. for almost every $(t,\tau )\in X\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s=\delta _0 {k}_B(t,\tau );$

• 3. for almost every $(t,\tau )\in X\times (G_A\setminus G)$, $\int \limits _{G_B} {k}_B(t,s) F_n(k(s,\tau ))d\mu _s=0.$

Proof

By applying Fubini theorem from [1] and iterative kernels from [25] we have

$$\begin{aligned}&(A^2x)(t) \textstyle =\int \limits _{G_A} k_A(t,s)(Ax)(s)d\mu _s= \int \limits _{G_A} k_A(t,s)\left( \ \int \limits _{G_A} k_A(s,\tau )x(\tau )d{\mu _\tau }\right) d\mu _s \\&\textstyle =\int \limits _{G_A} \big (\ \int \limits _{G_A} k_A(t,s)k(s,\tau )d\mu _s\big )x(\tau ) d{\mu _\tau } = \int \limits _{G_A} k_{1,A}(t,\tau )x(\tau )d{\mu _\tau }, \\&k_{1,A}(t,s) \textstyle = \int \limits _{G_A} k_A(t,\tau )k_A(\tau ,s)d{\mu _\tau }; \\&(A^3x)(t) \textstyle =\int \limits _{G_A} k_A(t,s)(A^2x)(s)d\mu _s =\int \limits _{G_A} k_A(t,s)\big (\ \int \limits _{G_A} k_{1,A}(s,\tau )x(\tau )d\mu _{\tau }\big )d\mu _s \\&\textstyle =\int \limits _{G_A} \big (\ \int \limits _{G_A} k_A(t,s)k_1(s,\tau )d\mu _s\big )x(\tau )d\mu _{\tau }=\int \limits _{G_A} k_{2,A}(t,\tau )x(\tau )d{\mu _\tau }, \\&k_{2,A}(t,s) \textstyle = \int \limits _{G_A} k_A(t,\tau )k_{1,A}(\tau ,s)d{\mu _\tau }; \\&(A^nx)(t)\textstyle = \int \limits _{G_A} k_{{n-1},A}(t,s)x(s)d\mu _s,\quad n\ge 1, \\&k_{m,A}(t,s)\textstyle = \int \limits _{G_A} k_A(t,\tau )k_{m-1,A}(\tau ,s)d{\mu _\tau },\ m=1,2,3,\ldots ,n, \\&k_{0,A}(t,s) = k_A(t,s); \\&(F(A)x)(t) \textstyle =\delta _0 x(t)+ \sum \limits _{j=1}^{n} \delta _j (A^j x)(t) \\&\textstyle =\delta _0 x(t)+\sum \limits _{j=1}^{n} \delta _{j} \int \limits _{G_A} k_{j-1,A}(t,s)x(s)d\mu _s \\&\textstyle =\delta _0 x(t)+\int \limits _{G_A} F_n(k_A(t,s))x(s)d\mu _s, \\&F_n(k_A(t,s)) \textstyle = \sum \limits _{j=1}^{n} \delta _j k_{j-1,A}(t,s),\quad n=1,2,3,\ldots ; \\&(BF(A)x)(t) \textstyle =\int \limits _{G_B} {k}_B(t,s)(F(A)x)(s)d\mu _s \\&\textstyle =\int \limits _{G_B} {k}_B(t,s) \left( \delta _0 x(s)+\int \limits _{G_A} F_n(k_A(s,\tau )x(\tau )d{\mu _\tau }) \right) d\mu _s \\&\textstyle = \delta _0\int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s+ \int \limits _{G_A} \left( \ \int \limits _{G_B} {k}_B(t,s)F_n(k_A(s,\tau ))d\mu _s\right) x(\tau ) d\mu _{\tau } \\&\textstyle = \delta _0\int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s+ \int \limits _{G_A} k_{BFA}(t,\tau )x(\tau )d\mu _\tau , \\&k_{BFA}(t,\tau ) \textstyle =\int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s; \\&(ABx)(t) \textstyle =\int \limits _{G_B} k_A(t,s)(Bx)(s)d\mu _s =\int \limits _{G_A} k_A(t,s)\left( \ \int \limits _{G_B} {k}_B(s,\tau )x(\tau )d\mu _\tau \right) d\mu _s \\&\textstyle =\int \limits _{G_B} \left( \ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s\right) x(\tau )d\mu _\tau = \int \limits _{G_B} k_{AB}(t,\tau ) x(\tau )d\mu _\tau , \\&k_{AB}(t,\tau ) \textstyle =\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s. \end{aligned}$$

Therefore, $(ABx)(t)=(BF(A)x)(t)$ for all $x\in L_p(X,\mu )$ if and only if

$$ \textstyle \int \limits _{G_B} [ k_{AB}(t,\tau )-\delta _0 {k}_B(t,\tau )] x(\tau )d\mu _\tau =\int \limits _{G_A} k_{BFA}(t,\tau )x(\tau )d\mu _\tau . $$

By applying Lemma 1 we have $AB=BF(A)$ if and only if

• 1. for almost every $(t,\tau )\in X\times G$,

  $$ \textstyle \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s-\delta _0 {k}_B(t,\tau ) = \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s; $$

• 2. for almost every $(t,\tau )\in X\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_A(s,\tau )d\mu _s=\delta _0\tilde{k}(t,\tau );$

• 3. for almost every $(t,\tau )\in X\times (G_A\setminus G)$, $\int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s=0.$$\square$

Remark 1

In Theorem 1 when $G_A=G_B=G$ conditions 2 and 3 are taken on set of measure zero so we can ignore them. Thus, we only remain with condition 1. When $G_A\not =G_B$, then we need to check conditions 2 and 3 outside the intersection $G=G_A\cap G_B$. Moreover, condition 3 that for almost every $(t,\tau )\in X\times (G_A\setminus G)$,

$$\begin{aligned} \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s=0, \end{aligned}$$

(4.6)

does not imply $\displaystyle B\left( \sum _{k=1}^{n} \delta _k A^k\right) =0$ because its kernel has to satisfy (4.6) only on the set $X\times (G_A\setminus G)$ and not on the whole set of definition. On the other hand, the same kernel has to satisfy condition 1, which is, for almost every $(t,\tau )\in X\times G$,

$$ \int \limits _{G_A} k(t,s)\tilde{k}(s,\tau )d\mu _s-\delta _0\tilde{k}(t,\tau ) = \int \limits _{G_B} \tilde{k}(t,s) F_n(k(s,\tau ))d\mu _s. $$

Note that Theorem 1 does not imply $\displaystyle \sum _{k=1}^n \delta _k A^k=0$. In fact, $\displaystyle \sum _{k=1}^n \delta _k A^k=0$ implies $\displaystyle B\left( \sum _{k=1}^{n} \delta _k A^k\right) =0$ but as mentioned above it can be non zero in general.

Example 1

Let $(\mathbb {R}, \varSigma ,\mu )$ be the standard Lebesgue measure space. Consider integral operators acting on $L_p(\mathbb {R},\mu )$ for $1<p<\infty$. Let $A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $1< p<\infty$ defined as follows

$$ \textstyle (Ax)(t)= \int \limits _0^\pi k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _0^\pi {k}_B(t,s)x(s)d\mu _s, $$

almost everywhere, where the index in $\mu$ indicates the variable of integration,

$$\begin{array}{lll} k_A(t,s)&{}=&{} I_{[\alpha ,\beta ]}(t)\frac{2}{\pi }(\cos t \cos s+\sin t\sin s+\cos t\sin s), \\ k_B(t,s)&{}=&{} I_{[\alpha ,\beta ]}(t)\frac{2}{\pi }(\cos t \cos s+2\sin t\sin s), \end{array} $$

almost everywhere $(t,s)\in \mathbb {R}\times [0,\pi ]$, $\alpha ,\,\beta$ are real constants such that $\alpha \le 0$, $\beta \ge \pi$ and $I_{E}(t)$ is the indicator function of the set E. These operators are well defined, since the kernels satisfy (4.3). In fact,

$$\begin{aligned}&\textstyle \int \limits _{\mathbb {R}}\big (\ \int \limits _{0}^{\pi }|k_A(t,s)|^q d\mu _s\big )^{\frac{p}{q}} d\mu _t \\&\quad = \int \limits _{\alpha }^\beta \big (\ \int \limits _{0}^{\pi }\big |\frac{2}{\pi }(\cos t \cos s + \sin t\sin s+\cos t\sin s)\big |^q d\mu _s\big )^{\frac{p}{q}}d\mu _t\\&\quad \textstyle \le \int \limits _{\alpha }^\beta \frac{6^p}{\pi ^{p-1}}dt=\frac{6^p(\beta -\alpha )}{\pi ^{p-1}}<\infty , \\&\textstyle \int \limits _{\mathbb {R}}\big (\ \int \limits _{0}^{\pi }|{k}_B(t,s)|^q d\mu _s\big )^{p/q}d\mu _t \\&\quad \textstyle =\int \limits _{\alpha }^\beta \big (\ \int \limits _{0}^{\pi }\big |\frac{2}{\pi }(\cos t \cos s+2\sin t\sin s)\big |^q d\mu _s\big )^{p/q}d\mu _t \le \int \limits _{\alpha }^\beta \frac{6^p}{\pi ^{p-1}}dt=\frac{6^p(\beta -\alpha )}{\pi ^{p-1}} <\infty , \end{aligned}$$

where $q\ge 1$ such that $\displaystyle \frac{1}{p}+\frac{1}{q}=1$. In the estimations above we used the inequalities:

$$\begin{aligned} |2(\cos t \cos s+\sin t\sin s+\cos t\sin s)|^q\le & {} 2^q\cdot 3^q=6^q, \\ |2(\cos t \cos s+2\sin t\sin s)|^q\le & {} 2^q\cdot 3^q=6^q, \ 1<q<\infty . \end{aligned}$$

Note that in this case conditions 1, 2 and 3 of Theorem 1 reduce just to condition 1 because the sets $G_A=G_B=[0,\pi ]$, and so $G=[0,\pi ]$, $G_A\setminus G=G_B\setminus G=\emptyset$. Therefore, according to Remark 1 conditions 2 and 3 are taken on a set of measure zero.

Consider the polynomial $F(t)=t^2$, $t\in \mathbb {R}$. These operators satisfy $AB=BF(A)$. In fact, by applying Theorem 1 we have $\delta _0=\delta _1=0$, $\delta _2=1$, $n=2$,

$$\begin{aligned}&k_{AB}(t,\tau ) \textstyle = \int \limits _{0}^{\pi } k_A(t,s){k}_B(s,\tau )d\mu _s\\&\quad \textstyle = \frac{4}{\pi ^2} \int \limits _{0}^{\pi } I_{[\alpha ,\beta ]}(t)(\cos (t)\cos (s)+\sin (t)\sin (s) + \cos (t)\sin (s))\cdot \\&\textstyle \ I_{[\alpha ,\beta ]}(s)(\cos (s)\cos (\tau )+2\sin s\sin \tau )d\mu _s\\&\quad \textstyle = \frac{4}{\pi }I_{[\alpha ,\beta ]}(t)\left( \frac{\cos t\cos \tau }{2}+\cos t\sin \tau +\sin t \sin \tau \right) \\&\quad \textstyle = \frac{2}{\pi }I_{[\alpha ,\beta ]}(t)({\cos t\cos \tau }+2\cos t\sin \tau +2\sin t \sin \tau ), \end{aligned}$$

for almost every $(t,\tau )\in \mathbb {R}\times [0,\pi ]$. Moreover,

$$\begin{aligned}&\textstyle F_2(k_A(t,s))=k_{1,A}(t,s) = \int \limits _{0}^{\pi } k_A(t,\tau )k_A(\tau ,s)d\mu _\tau =\\&\quad \textstyle = \frac{4}{\pi ^2}\int \limits _{0}^{\pi } I_{[\alpha ,\beta ]}(t)(\cos t \cos \tau +\sin t\sin \tau +\cos t\sin \tau )\cdot \\&\textstyle I_{[\alpha ,\beta ]}(\tau )(\cos \tau \cos s+\sin \tau \sin s+\cos \tau \sin s)d\mu _\tau \\&\quad \textstyle = \frac{4}{\pi } I_{[\alpha ,\beta ]}(t)\left( \frac{\cos t\cos s}{2}+\cos t\sin s+\frac{\sin t\sin s}{2}\right) \\&\quad \textstyle = \frac{2}{\pi } I_{[\alpha ,\beta ]}(t)({\cos t\cos s}+2\cos t\sin s+{\sin t\sin s}), \end{aligned}$$

for almost every $(t,s)\in \mathbb {R}\times [0,\pi ]$. Therefore,

$$\begin{aligned}&\textstyle k_{BFA}(t,\tau )=\int \limits _{0}^{\pi } {k}_B(t,s)F_2(k_A(s,\tau ))d\mu _s\\&\quad \textstyle = \frac{4}{\pi ^2} \int \limits _{0}^{\pi } I_{[\alpha ,\beta ]}(t)(\cos (t)\cos (s)+2\sin (t)\sin (s))\\& \textstyle \cdot \textstyle I_{[\alpha ,\beta ]}(s)({\cos s\cos \tau }+2\cos s\sin \tau +{\sin s\sin \tau })d\mu _s\\&\quad \textstyle = \frac{4}{\pi } I_{[\alpha ,\beta ]}(t)\left( \frac{\cos t\cos \tau }{2}+\cos t\sin \tau +\sin t \sin \tau \right) \\&\quad \textstyle = \frac{2}{\pi } I_{[\alpha ,\beta ]}(t)({\cos t\cos \tau }+2\cos t\sin \tau +2\sin t \sin \tau ), \end{aligned}$$

for almost every $(t,\tau )\in \mathbb {R}\times [0,\pi ]$, which coincides with the kernel $k_{AB}$. Thus, conditions of Theorem 1 are fulfilled and so $AB=BA^2$. Moreover, $BA^2\not =0$ as mentioned in Remark 1, in fact

$$ \textstyle (BA^2x)(t)=\frac{2}{\pi }I_{[\alpha ,\beta ]}(t)\int \limits _{0}^{\pi /2} ({\cos t\cos \tau }+2\cos t\sin \tau +2\sin t \sin \tau )x(\tau )d\mu _\tau $$

almost everywhere.

The following corollary is a special case of Theorem 1 for the important class of covariance commutation relations, associated to affine (degree 1) polynomials F.

Corollary 1

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined as follows

$$ (Ax)(t)= \int \limits _{G_A} k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s, $$

where the index in $\mu _s$ indicates variable of integration, $G_A,\ G_B\in \varSigma$, $\mu (G_A)<\infty$, $\mu (G_B)<\infty$, $k_A(t,s):X\times G_A\rightarrow \mathbb {R}$, ${k}_B(t,s):X\times G_B\rightarrow \mathbb {R}$ are measurable functions satisfying either relation (4.3) or (4.4) or (4.5). Let $F:\mathbb {R}\rightarrow \mathbb {R}$ be a polynomial of degree at most 1 given by $F(z)=\delta _0+\delta _1 z$, where $\delta _0,\ \delta _1 \in \mathbb {R}$. We set $G=G_A\cap G_B.$

Then $AB-\delta _1 BA=\delta _0 B$ if and only if the following conditions are fulfilled

• 1. for almost every $(t,\tau )\in X\times G$,

  $$ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s-\delta _0 {k}_B(t,\tau ) = \delta _1\int \limits _{G_B} {k}_B(t,s) k_A(s,\tau )d\mu _s. $$

• 2. for almost every $(t,\tau )\in X\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s=\delta _0 {k}_B(t,\tau ).$

• 3. for almost every $(t,\tau )\in X\times (G_A\setminus G)$, $\delta _1\int \limits _{G_B} k_B(t,s) k_A(s,\tau )d\mu _s=0.$

The following corollary of Theorem 1 is concerned with representations by integral operators of another important family of covariance commutation relations associated to monomials F.

Corollary 2

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined as follows

$$ (Ax)(t)= \int \limits _{G_A} k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s, $$

where the index in $\mu _s$ indicates variable of integration, $G_A,\ G_B\in \varSigma$, $\mu (G_A)<\infty$, $\mu (G_B)<\infty$, $k_A(t,s):X\times G_A\rightarrow \mathbb {R}$, ${k}_B(t,s):X\times G_B\rightarrow \mathbb {R}$ are measurable functions satisfying either relation (4.3) or (4.4) or (4.5). Let $F:\mathbb {R}\rightarrow \mathbb {R}$ be a monomial defined by $F(z)=\delta z^d$, where $\delta \not =0$ is a real number and d is a positive integer. Let $G=G_A\cap G_B$ and

$$\begin{aligned}\begin{gathered} k_{0,A}(t,s)=k_A(t,s), \ k_{m,A}(t,s)=\int \limits _{G_A} k_A(t,\tau )k_{m-1,A}(\tau ,s)d\mu _\tau ,\ m=1,2,3,\ldots ,d. \end{gathered}\end{aligned}$$

Then $AB=\delta BA^d$ if and only if the following conditions are fulfilled

• 1. for almost every $(t,\tau )\in X\times G$,

  $$ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s = \delta \int \limits _{G_B} {k}_B(t,s) k_{d-1,A}(s,\tau )d\mu _s. $$

• 2. for almost every $(t,\tau )\in X\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s=0.$

• 3. for almost every $(t,\tau )\in X\times (G_A\setminus G)$, $\int \limits _{G_B} {k}_B(t,s) k_{d-1,A}(s,\tau )d\mu _s=0.$

Remark 2

Example 1 describes a specific case for Corollary 2 when $G_A=G_B=[0,\pi ]$, $\delta =1$, $d=2$.

Consider now the case when $X=\mathbb {R}^l$ and $\mu$ is the Lebesgue measure. In the following theorem we allow the sets $G_A$ and $G_B$ to have infinite measure.

Theorem 2

Let $(\mathbb {R}^l,\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p(\mathbb {R}^l,\mu )\rightarrow L_p(\mathbb {R}^l,\mu ),\ B:L_p(\mathbb {R}^l,\mu )\rightarrow L_p(\mathbb {R}^l,\mu ),\ 1< p<\infty \end{aligned}$$

be nonzero operators defined by

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{G_A} k_A(t,s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s, \end{gathered}\end{aligned}$$

where the index in $\mu$ indicates the variable of integration, $G_A\in \varSigma$ and $G_B\in \varSigma$, and kernels $k_A(t,s):\mathbb {R}^l\times G_A\rightarrow \mathbb {R}$, ${k}_B(t,s):\mathbb {R}^l\times G_B\rightarrow \mathbb {R}$ are measurable functions satisfying either relation (4.3) or (4.4). Consider a polynomial $F:\mathbb {R}\rightarrow \mathbb {R}$ defined by $F(z)=\sum \limits _{j=0}^{n} \delta _j z^j$, where $\delta _j \in \mathbb {R}$, $j=0,1,2,\ldots ,n$. Let $G=G_A\cap G_B$ and

$$\begin{aligned}\begin{gathered} k_{A,0}(t,s)=k_A(t,s), \ k_{A,m}(t,s)=\int \limits _{G_A} k_A(t,\tau )k_{A,{m-1}}(\tau ,s)d\mu _\tau ,\ m=1,2,3,\ldots ,n \\ F_m(k_A(t,s))=\sum _{j=1}^{m} \delta _j k_{A,{j-1}}(t,s),\quad m=1,2,3,\ldots ,n. \end{gathered}\end{aligned}$$

Then $AB=BF(A)$ if and only if the following conditions are fulfilled:

• 1. for almost every $(t,\tau )\in \mathbb {R}^n\times G$,

  $$ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s-\delta _0{k}_B(t,\tau ) = \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s. $$

• 2. for almost every $(t,\tau )\in \mathbb {R}^n\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s=\delta _0{k}_B(t,\tau ).$

• 3. for almost every $(t,\tau )\in \mathbb {R}^n\times (G_A\setminus G)$, $\int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s=0.$

Proof

By applying Fubini theorem from [1] and iterative kernels from [25] we have

$$\begin{aligned} (A^2x)(t)&\textstyle = \int \limits _{G_A} k_A(t,s)(Ax)(s)d\mu _s= \int \limits _{G_A} k_A(t,s)\big (\ \int \limits _{G_A} k_A(s,\tau )x(\tau )d\mu _\tau \big )d\mu _s\\&\textstyle =\int \limits _{G_A} \big (\ \int \limits _{G_A} k_A(t,s)k_A(s,\tau )ds\big )x(\tau ) d\tau = \int \limits _{G_A} k_{1,A}(t,\tau )x(\tau )d\mu _\tau , \\ k_{1,A}(t,s)&\textstyle = \int \limits _{G_A} k_A(t,\tau )k_A(\tau ,s)d\mu _\tau ; \\ (A^3x)(t)&\textstyle =\int \limits _{G_A} k_A(t,s)(A^2x)(s)d\mu _s= \int \limits _{G_A} k_A(t,s)\big (\ \int \limits _{G_A} k_{1,A}(s,\tau )x(\tau )d\mu _\tau \big )d\mu _s\\&\textstyle = \int \limits _{G_A} \big (\ \int \limits _{G_A} k_A(t,s)k_{1,A}(s,\tau )d\mu _s\big )x(\tau )d\mu _\tau =\int \limits _{G_A} k_{2,A}(t,\tau )x(\tau )d\mu _\tau , \\ k_{2,A}(t,s)&\textstyle =\int \limits _{G_A} k_A(t,\tau )k_{1,A}(\tau ,s)d\mu _{\tau }; \\ (A^nx)(t)&\textstyle =\int \limits _{G_A} k_{{n-1},A}(t,s)x(s)d\mu _s,\quad n\ge 1\\ k_{m,A}(t,s)&\textstyle =\int \limits _{G_A} k_A(t,\tau )k_{{m-1},A}(\tau ,s)d\mu _\tau ,\ m=1,2,3,\ldots ,n,\ k_{0,A}(t,s)=k_A(t,s); \\ (F(A)x)(t)&\textstyle =\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 _{G_A} k_{j-1,A}(t,s)x(s)d\mu _s \\&\textstyle =\delta _0 x(t)+\int \limits _{G_A} F_n(k_A(t,s))x(s)d\mu _s, \\ F_n(k_A(t,s))&\textstyle =\sum \limits _{j=1}^{n} \delta _j k_{{j-1},A}(t,s),\ n=1,2,3,\ldots ; \\ (BF(A)x)(t)&\textstyle =\int \limits _{G_B} {k}_B(t,s)(F(A)x)(s)d\mu _s= \\&\textstyle =\int \limits _{G_B} {k}_B(t,s) \big (\delta _0 x(s)+\int \limits _{G_A} F_n(k_A(s,\tau )x(\tau )d\mu _\tau ) \big )d\mu _s \\&\textstyle = \delta _0\int \limits _{G_2} {k}_B(t,s)x(s)d\mu _s+ \int \limits _{G_A} \big (\ \int \limits _{G_B} k_B(t,s)F_n(k_B(s,\tau ))d\mu _s\big )x(\tau ) d\mu _\tau =\\&\textstyle = \delta _0\int \limits _{G_B} {k}_B(t,s)x(s)d\mu _s+ \int \limits _{G_A} k_{BF}(t,\tau )x(\tau )d\mu _\tau \\ k_{BF}(t,\tau )&\textstyle =\int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s; \\ (ABx)(t)&\textstyle =\int \limits _{G_B} k_A(t,s)(Bx)(s)d\mu _s= \int \limits _{G_A} k_A(t,s)\big (\ \int \limits _{G_B} {k}_B(s,\tau )x(\tau )d\mu _\tau \big )d\mu _s \\&\textstyle =\int \limits _{G_B} \big (\ \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s\big )x(\tau )d\mu _\tau = \int \limits _{G_B} k_{AB}(t,\tau ) x(\tau )d\mu _\tau , \\ k_{AB}(t,\tau )&\textstyle =\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s. \end{aligned}$$

Thus for all $x\in L_p(\mathbb {R}^l,\mu )$, $1< p<\infty$ we have $(ABx)(t)=(BF(A)x)(t)$ almost everywhere if and only if $\int \limits _{G_B} [ k_{AB}(t,\tau )-\delta _0 \tilde{k}(t,\tau )] x(\tau )d\mu _\tau =\int \limits _{G_A} k_{BF}(t,\tau )x(\tau )d\mu _\tau$ almost everywhere. By Lemma 2 we have $AB=BF(A)$ if and only if

• 1. for almost every $(t,\tau )\in \mathbb {R}\times G$,

  $$ \textstyle \int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s-\delta _0{k}_B(t,\tau ) = \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s; $$

• 2. for almost every $(t,\tau )\in \mathbb {R}\times (G_B\setminus G)$, $\int \limits _{G_A} k_A(t,s){k}_B(s,\tau )d\mu _s=\delta _0{k}_B(t,\tau ).$

• 3. for almost every $(t,\tau )\in \mathbb {R}\times (G_A\setminus G)$, $\int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s=0.$$\square$

Remark 3

Similar to Remark 1, in Theorem 2 when $G_A=G_B=G$ conditions 2 and 3 are taken on set of measure zero so we can ignore them. Thus, we only remain with condition 1. When $G_A\not =G_B$ we need to check also conditions 2 and 3 outside the intersection $G=G_A\cap G_B$. Moreover condition 3, which is, for almost every $(t,\tau )\in \mathbb {R}^n\times (G_A\setminus G)$,

$$\begin{aligned} \int \limits _{G_B} {k}_B(t,s) F_n(k_A(s,\tau ))d\mu _s=0. \end{aligned}$$

(4.7)

does not imply $\displaystyle B\left( \sum _{k=1}^{n} \delta _k A^k\right) =0$ because its kernel has to satisfy (4.7) only on the set $\mathbb {R}^n\times (G_A\setminus G)$ and not on the whole set of definition. On the other hand, the same kernel has to satisfy condition 2, that for almost every $(t,\tau )\in \mathbb {R}^n\times G$,

$$ \int \limits _{G_A} k(t,s)\tilde{k}(s,\tau )d\mu _s-\delta _0\tilde{k}(t,\tau ) = \int \limits _{G_B} \tilde{k}(t,s) F_n(k(s,\tau ))d\mu _s. $$

Note that Theorem 2 does not imply $\displaystyle \sum _{k=1}^n \delta _k A^k=0$. In fact, $\displaystyle \sum _{k=1}^n \delta _k A^k=0$ implies $\displaystyle B\left( \sum _{k=1}^{n} \delta _k A^k\right) =0$ but as mentioned above it can be non zero in general.

Proposition 1

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. Let $A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $1< p<\infty$ be nonzero operators defined as follows

$$\begin{aligned}\begin{gathered} (Ax)(t)= \int \limits _{\mathbb {R}} \tilde{k}_A(t-s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{\mathbb {R}} \tilde{k}_B(t-s)x(s)d\mu _s, \end{gathered}\end{aligned}$$

where the index in $\mu$ indicates the variable of integration, kernels $\tilde{k}_A(\cdot )\in L_1(\mathbb {R},\mu )$, $\tilde{k}_B(\cdot )\in L_1(\mathbb {R},\mu )$, that is,

$$\begin{aligned}\begin{gathered} \int \limits _{\mathbb {R}}|\tilde{k}_A(t)|d\mu _t<\infty ,\qquad \int \limits _{\mathbb {R}}|\tilde{k}_B(t)|d\mu _t <\infty . \end{gathered}\end{aligned}$$

Consider a polynomial $F:\mathbb {R}\rightarrow \mathbb {R}$, $F(z)=\sum \limits _{j=0}^{n} \delta _j z^j$, where $\delta _j \in \mathbb {R}$, $j=0,1,2,\ldots ,n$. Then $AB=BF(A)$ if and only if for almost every $t\in \mathbb {R}$,

$$\begin{aligned} \tilde{k}_B \star \left( \tilde{k}_A-\delta _0 -\sum _{j=1}^n \delta _j(\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}})\right) (t)=0. \end{aligned}$$

(4.8)

In particular, if $\delta _0=0$, that is, $F(z)=\delta _1 t+\delta _2 z^2+\ldots +\delta _n z^n$, then $AB=BF(A)$ if and only if the set $\textrm{supp }\, K_B \, \cap \, \textrm{supp }\, \left( K_A -\sum _{j=1}^n \delta _j K_A^j\right)$ has measure zero in $\mathbb {R}$, where

$$\begin{aligned}\begin{gathered} K_B(s)=\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B(t)d\mu _t,\quad K_A(s)=\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_A(t)d\mu _t. \end{gathered}\end{aligned}$$

Proof

Operators A and B are well defined by Young theorem ([6], Theorem 4.15). By Fubbini theorem for composition of operators A, B and $A^n$, similarly to the proof of Theorem 2 when $k_A(t,s)=\tilde{k}_A(t-s)$, $k_B(t,s)=\tilde{k}_B(t-s)$ and $G_A=G_B=\mathbb {R}$ we get from Lemma 2 that $AB=BF(A)$ if and only if for almost every $(t,s)\in \mathbb {R}^2$,

$$\begin{aligned} \int \limits _{\mathbb {R}} \tilde{k}_A(t-\tau )\tilde{k}_B(\tau -s)d\mu _\tau -\delta _0 \tilde{k}_B(t-s) = \int \limits _{\mathbb {R}} \tilde{k}_B(t-\tau ) F_n({k}_A(\tau ,s))d\mu _\tau , \end{aligned}$$

(4.9)

where

$$\begin{aligned}\begin{gathered} \tilde{k}_{0, A}(t,s)=\tilde{k}_A(t-s), \ \tilde{k}_{m,A}(t,s)=\int \limits _{\mathbb {R}} \tilde{k}_A(t-\tau ) {k}_{{m-1},A}(\tau ,s)d\mu _\tau ,\quad m=1,2,3,\ldots ,n \\ F_m(\tilde{k}_A(t,s))=\sum _{j=1}^{m} \delta _j \tilde{k}_{{j-1},A}(t,s),\quad m=1,2,3,\ldots ,n. \end{gathered}\end{aligned}$$

Computing $\tilde{k}_{m,A}(t,s)$ we have for $m=1$,

$$ \tilde{k}_{1,A}(t,s)= \int \limits _{\mathbb {R}}\tilde{k}_A(t-\tau )\tilde{k}_A( \tau -s)d\mu _\tau =\int \limits _{\mathbb {R}}\tilde{k}_A(t-s-\nu )\tilde{k}_A(\nu )d\mu _\nu =(\tilde{k}_A \star \tilde{k}_A)(t-s), $$

for $m=2$,

$$\begin{aligned} \tilde{k}_{2,A}(t,s)= & {} \int \limits _{\mathbb {R}}\tilde{k}_A(t-\tau )(\tilde{k}_A \star \tilde{k}_A(\tau -s)d\mu _\tau =\\= & {} \int \limits _{\mathbb {R}}\tilde{k}_A(t-s-\nu )(\tilde{k}_A \star \tilde{k}_A(\nu )d\mu _\nu =(\tilde{k}_A\star \tilde{k}_A\star \tilde{k}_A)(t-s). \end{aligned}$$

and for all $2\le m\le n$, $\tilde{k}_{m-1,A}(t,s)=(\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{m\ \text {times}})(t-s).$ Thus, for all $1\le m\le n$

$$\begin{aligned}&F_m(\tilde{k}_A(t,s))=\sum _{j=1}^m \delta _j (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}})(t-s), \\&\int \limits _{\mathbb {R}} \tilde{k}_B(t-s) F_n(\tilde{k}_A(s,\tau ))d\mu _s= \int \limits _{\mathbb {R}}\tilde{k}_{B}(t-\tau )\cdot \sum _{j=1}^n \delta _j(\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}})(\tau -s)d\mu _\tau \\&= \int \limits _{\mathbb {R}} \sum _{j=1}^n \delta _j\tilde{k}_{B}(t-s-\nu )\cdot (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}})(\nu )d\mu _\nu \\&= \sum _{j=1}^n \delta _j\tilde{k}_B\star (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}})(t-s). \end{aligned}$$

Therefore, for almost every pairs $(t,s)\in \mathbb {R}^2$, the equality (4.9) is equivalent to

$$ (\tilde{k}_A\star \tilde{k}_B)(t-s)= \delta _0 \tilde{k}_B(t-s)+\sum _{j=1}^n \delta _j\tilde{k}_B\star \left( \underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{j\ \text {times}}\right) (t-s) $$

which is equivalent to (4.8). If $\delta _0=0$, then by applying the two-sided Laplace transform we get that (4.8) is equivalent to

$$ \int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B\star \left( \tilde{k}_A -\sum _{j=1}^n \delta _j \tilde{k}_A^{\star \,j}\right) (t)d\mu _t=0, $$

which is equivalent to

$$\begin{aligned}&\textstyle K_B(s)\cdot (K_A(s)-\sum _{j=1}^n \delta K_A^j (s))=0, \\&\text {where} \ \textstyle K_B(s)=\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B(t)d\mu _t,\ K_A(s) =\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_A(t)d\mu _t. \nonumber \end{aligned}$$

(4.10)

Equation (4.10) is equivalent to the set $\textrm{supp }\, K_B \, \cap \, \textrm{supp }\, \left( K_A -\sum _{j=1}^n \delta _j K_A^j\right)$ to have measure zero in $\mathbb {R}$. $\square$

Proposition 2

Let $(\mathbb {R},\Sigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ 1< p<\infty \end{aligned}$$

be non-zero operators defined as follows

$$\begin{aligned} (Ax)(t)= \int \limits _{\mathbb {R}} \tilde{k}_A(t-s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{\mathbb {R}} \tilde{k}_B(t-s)x(s)d\mu _s, \end{aligned}$$

(4.11)

where $\tilde{k}_A(\cdot )\in L_1(\mathbb {R},\mu )$, $\tilde{k}_B(\cdot )\in L_1(\mathbb {R},\mu )$, that is,

$$\begin{aligned} \int \limits _{\mathbb {R}}|\tilde{k}_A(t)|d\mu _t<\infty ,\quad \int \limits _{\mathbb {R}}|\tilde{k}_B(t)|d\mu _t <\infty . \end{aligned}$$

(4.12)

and the index in $\mu$ indicates the variable of integration. Suppose that

$$ \int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_A(t)d\mu _t={K}_A(s),\quad \int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B(t)d\mu _t={K}_B(s) $$

exist and the domain of $K_A(\cdot )$ is equal to the domain of $K_B(\cdot )$ with exception of a set of measure zero. Then, $AB=\delta BA^n$, for a fixed $n\in \mathbb {Z},\ n\ge 2$ and $\delta \in \mathbb {R}\setminus \{0\}$ if and only if $(\tilde{k}_A \star \tilde{k}_B)(t)=0$ almost everywhere.

Proof

Operators A and B are well defined by Young theorem ([6], Theorem 4.15). Let $n\ge 1$. By Fubbini Theorem for composition of operators A, B and $A^n$, similarly to the proof of Theorem 2 when $k_A(t,s)=\tilde{k}_A(t-s)$, $k_B(t,s)=\tilde{k}_B(t-s)$, $G_A=G_B=\mathbb {R}$, we get from Lemma 2 that $AB=\delta BA^n$ if and only if, for almost every $(t,s)\in \mathbb {R}^2$,

$$\begin{aligned}&\int \limits _{\mathbb {R}} \tilde{k}_A(t-\tau )\tilde{k}_B(\tau -s)d\mu _\tau = \int \limits _{\mathbb {R}} \delta \tilde{k}_B(t-\tau ) \tilde{k}_{n-1,A}(\tau ,s)d\mu _\tau , \\&\tilde{k}_{0,A}(t,s)=\tilde{k}_A(t-s), \ \tilde{k}_{n,A}(t,s)=\int \limits _{\mathbb {R}} \tilde{k}_A(t-\tau ) \tilde{k}_{{n-1},A}(\tau ,s)d\mu _\tau ,\ n\ge 1. \nonumber \end{aligned}$$

(4.13)

Computing $\tilde{k}_{n,A}(t,s)$, we get for $n=1$,

$$ \tilde{k}_{1,A}(t,s)= \int \limits _{\mathbb {R}}\tilde{k}_A(t-\tau )\tilde{k}_A( \tau -s)d\mu _\tau =\int \limits _{\mathbb {R}}\tilde{k}_A(t-s-\nu )\tilde{k}_A(\nu )d\mu _\nu =(\tilde{k}_A \star \tilde{k}_A)(t-s), $$

for $n=2$,

$$\begin{aligned} \tilde{k}_{2,A}(t,s)&=\int \limits _{\mathbb {R}}\tilde{k}_A(t-\tau )(\tilde{k}_A \star \tilde{k}_A(\tau -s)d\mu _\tau =\\&= \int \limits _{\mathbb {R}}\tilde{k}_A(t-s-\nu )(\tilde{k}_A \star \tilde{k}_A(\nu )d\mu _\nu =(\tilde{k}_A\star \tilde{k}_A\star \tilde{k}_A)(t-s). \end{aligned}$$

and for all $n\ge 2$,

$$\begin{aligned}&\tilde{k}_{n-1,A}(t,s)=(\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(t-s), \\&\int \limits _{\mathbb {R}} \tilde{k}_B(t-s) (\tilde{k}_{n-1,A}(s,\tau ))d\mu _s = \int \limits _{\mathbb {R}}\tilde{k}_{B}(t-\tau )\cdot (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(\tau -s)d\mu _\tau \nonumber \\&\quad = \int \limits _{\mathbb {R}} \tilde{k}_{B}(t-s-\nu )\cdot (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(\nu )d\mu _\nu . \nonumber \end{aligned}$$

(4.14)

Therefore, for almost all pairs $(t,s)\in \mathbb {R}^2$, the equality (4.13) is equivalent to

$$ (\tilde{k}_A\star \tilde{k}_B)(t-s)= \delta \tilde{k}_B\star (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(t-s), $$

which is equivalent to

$$\begin{aligned} (\tilde{k}_A\star \tilde{k}_B)(t)= \delta \tilde{k}_B\star (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(t) \end{aligned}$$

(4.15)

almost everywhere. By applying the two-sided Laplace transform in both cases $n\ge 2$ we get that (4.15) is equivalent to $\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B\star \left( \tilde{k}_A-\delta \tilde{k}_A^{\star \,n}\right) (t)d\mu _t=0$ almost everywhere, which can be written as follows

$$\begin{aligned} K_B(s)\cdot \left( K_A(s)-\delta K_A^n (s)\right) =0, \ n\ge 2, \end{aligned}$$

(4.16)

almost everywhere, $K_B(s)=\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_B(t)d\mu _t,\ K_A(s)=\int \limits _{-\infty }^{\infty } \exp ({-st})\tilde{k}_A(t)d\mu _t.$ Equation (4.16) is equivalent to the set $\textrm{supp }\, K_B \, \cap \, \textrm{supp }\, \left( K_A - \delta K_A^n\right) ,\ n\ge 2,$ to have measure zero in $\mathbb {R}$, that is, $K_B(\cdot )\cdot I_{(\textrm{supp }(K_A-\delta K^n_A))}(\cdot )=0$ almost everywhere and $(K_A(\cdot )-\delta K^n_A(\cdot ))\cdot I_{(\textrm{supp }\,{K_B})}(\cdot )=0$ almost everywhere, where $I_E(\cdot )$ is the indicator function of the set E. If $\textrm{supp }(K_A-\delta K^n_A)=\mathbb {R}$ then $\textrm{supp }\,{K_B}$ has measure zero, that is, $B=0$. Similarly, if $\textrm{supp }{K_B} =\mathbb {R}$ then $A=0$. Suppose that $\textrm{supp }\,{K_B}\not =\mathbb {R}$ and has positive measure. If $(K_A(\cdot )-\delta K^n_A(\cdot ))\cdot I_{(\textrm{supp }\,{K_B})}(\cdot )=0$ almost everywhere, then $K_A(s)-\delta K^n_A(s)=0$ for almost every $s\in \textrm{supp }\,{K_B}$. Let $p(z)=z-\delta z^n$. Suppose that p(z) has $m>0$ roots $z_i$, $i=1,2,\ldots , m$, $m\le n$, $n\ge 2$. We consider the following cases:

• If $n> 1$ and p(z) has $m\ge 2$ roots $z_i$, $i=1,2,\ldots , m$, $m\le n$, then $\displaystyle \tilde{k}_A(t)= \sum _{i=1}^m z_i \Delta (t-z_i)$, where $\Delta (t-t_0)$, $t,t_0\in \mathbb {R}$, is the Dirac function defined as follows $\Delta (t-t_0)=\left\{ \begin{array}{cc} 0, &{} t\not =t_0 \\ \infty , &{} t=t_0 \end{array} \right. .$ In this case $K_A(s)-\delta K_A^n (s)=0$ for almost every s in $\textrm{supp }\,{K_B}$. But this implies $K_A(s)=0$ for almost every s in $\textrm{supp }\,{K_B}$ since the Dirac function $\Delta (\cdot )$ is equivalent to zero function.

• If $n > 1$ and p(z) has only one real root, which is $z=0$, then $\textrm{supp }\, \left( K_A - \delta K_A^n\right) = \textrm{supp }\, K_A$ for all s in $\textrm{supp }\,{K_B}$. This implies that Equality (4.16) is satisfied if and only if $K_A(\cdot )=0$ almost everywhere in $\textrm{supp }\,{K_B}$.

In both cases we conclude that $K_A(\cdot )=0$ almost everywhere in $\textrm{supp }\,{K_B}$. Outside of $\textrm{supp }\,{K_B}$, the function $K_A(\cdot )$ can be nonzero. This implies that Equality (4.16) is equivalent to $K_A(s)K_B(s)=0$ almost everywhere. This is equivalent to

$(\tilde{k}_A \star \tilde{k}_B)(t)=0$ almost everywhere.

Remark 4

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. The operators A and B defined in (4.11) as $A:\, L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$ and $B:\, L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $1< p<\infty$, $(Ax)(t)= \int \limits _{\mathbb {R}} \tilde{k}_A(t-s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{\mathbb {R}} \tilde{k}_B(t-s)x(s)d\mu _s,$ almost everywhere, with $\tilde{k}_A(\cdot )\in L_1(\mathbb {R},\mu )$, $\tilde{k}_B(\cdot )\in L_1(\mathbb {R},\mu )$, commute, that is $AB=BA$. In fact, by applying Fubbini theorem for composition of A, B and Lemma 2,

$$\begin{aligned} AB=BA&\Leftrightarrow \int \limits _{\mathbb {R}} \tilde{k}_A(t-s) \tilde{k}_B(s-\tau )ds =\int \limits _{\mathbb {R}} \tilde{k}_B(t-s) \tilde{k}_A(s-\tau )d\mu _s \\&\Leftrightarrow (\tilde{k}_A\star \tilde{k}_B)(t-\tau )=(\tilde{k}_B\star \tilde{k}_A)(t-\tau )\ \text {for almost every }(t,\tau )\in \mathbb {R}^2, \end{aligned}$$

which holds true by the commutativity property of convolution.

Remark 5

If operators A and B commute then they satisfy, simultaneously, the following relations $AB=BF(A)$, $BA=F(A)B$ and $B(A-F(A))=0$. In fact, if A and B commute, then $AB=BA$, $BF(A)=F(A)B$, and thus $AB=BF(A)$ is equivalent to $BA=F(A)B$, which can be then written also as $B(A-F(A))=0$.

Proposition 3

Let $([0,\infty ),\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p([0,\infty ),\mu )\rightarrow L_p([0,\infty ),\mu ),\ B:L_p([0,\infty ),\mu )\rightarrow L_p([0,\infty ),\mu ), 1< p<\infty \end{aligned}$$

be non-zero operators defined by

$$\begin{aligned}&\begin{array}{lll} (Ax)(t) &{}=&{} \int \limits _{0}^\infty \tilde{k}_A(t-s)I_{[0,\infty )}(t-s)x(s)d\mu _s, \\ (Bx)(t) &{}=&{} \int \limits _{0}^\infty \tilde{k}_B(t-s)I_{[0,\infty )}(t-s)x(s)d\mu _s, \end{array} \\&\tilde{k}_A(\cdot )\in L_1([0,\infty ),\mu ),\ \tilde{k}_B(\cdot )\in L_1([0,\infty ),\mu ) \nonumber \\&\text {that is,}\ \textstyle \int \limits _{0}^\infty |\tilde{k}_A(t)|d\mu _t<\infty ,\ \int \limits _{0}^\infty |\tilde{k}_B(t)|d\mu _t <\infty , \nonumber \end{aligned}$$

(4.17)

where $I_{E}(\cdot )$ is the indicator function of the set E and the index in $\mu$ is the variable of integration. Then, there are no non-zero operators A and B satisfying $AB=\delta BA^n$ for a fixed $n\in \mathscr {Z},\ n\ge 2$, $\delta \in \mathbb {R}\setminus \{0\}$.

Proof

Operators A and B are well defined by Young’s theorem ([6], Theorem 4.15). Let $n\ge 1$. By applying Fubbini theorem for composition of operators A, B and $A^n$, similarly to the proof of Theorem 2 when $k_A(t,s)=\tilde{k}_A(t-s)I_{[0,\infty )}$, $k_B(t,s)=\tilde{k}_B(t-s)I_{[0,\infty )}(t-s)$ and $G_1=G_2=[0,\infty )$, we get from Lemma 2 that $AB=\delta BA^n$ if and only if for almost every $(t,s)\in \mathbb {R}^2$,

$$\begin{aligned}&\begin{array}{l} \int \limits _{0}^\infty \tilde{k}_A(t-\tau )I_{[0,\infty )}(t-\tau )\tilde{k}_B(\tau -s)I_{[0,\infty )}(\tau -s)d\mu _\tau \\ \quad \quad = \int \limits _{0}^\infty \tilde{k}_B(t-\tau ) (\tilde{k}_{n-1,A}(\tau ,s))d\mu _\tau , \end{array} \\&\textstyle \tilde{k}_{0,A}(t,s)=\tilde{k}_A(t-s)I_{[0,\infty )}(t-s), \nonumber \\&\textstyle \tilde{k}_{n,A}(t,s)=\int \limits _{0}^\infty \tilde{k}_A(t-\tau )I_{[0,\infty )}(t-\tau ) {k}_{n-1,A}(\tau ,s) d\mu _\tau ,\quad n\ge 1. \nonumber \end{aligned}$$

(4.18)

Computing $\tilde{k}_{n-1,A}(t,s)$ for $n\ge 1$, using (4.14), yields

$$\begin{aligned} \tilde{k}_{n-1,A}(t,s)= & {} \textstyle (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(t-s)I_{[0,\infty )}(t-s)=\\= & {} \textstyle \int \limits _{s}^{t} \tilde{k}_A(t-\tau )\cdot (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n-1\ \text {times}})(\tau -s)d\mu _\tau =\\= & {} \textstyle \int \limits _{0}^{t-s} \tilde{k}_A(t-s-\nu )\cdot (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n-1\ \text {times}})(\nu )d\mu _\nu \\= & {} (\underbrace{\tilde{k}_A\star \tilde{k}_A\star \ldots \star \tilde{k}_A}_{n\ \text {times}})(t-s). \end{aligned}$$

Therefore, from (4.18) we have for $n\ge 2$,

$$ \textstyle \int \limits _{0}^{t-s} \tilde{k}_A(t-s-\tau ) \tilde{k}_B(\tau )d\mu _\tau =\int \limits _{0}^{t-s} \tilde{k}_B(t-s-\tau )\delta (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}})(\tau )d\mu _\tau $$

which we can write as follows

$$\begin{aligned} \textstyle (\tilde{k}_A{\star }\tilde{k}_B)(t-s)= \delta (\tilde{k}_B\star (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}}))(t-s). \end{aligned}$$

(4.19)

By commutativity, linearity of convolution and the Titchmarsh convolution theorem, (4.19) is equivalent to either

$$\begin{aligned} \tilde{k}_B( t-s)=0\ \text {or}\ \delta (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}})(t-s)=\tilde{k}_A(t-s) \end{aligned}$$

for almost every $(t,s)\in \mathbb {R}^2$ such that $t\ge 0$, $0\le s\le t$. This is equivalent to either

$$\begin{aligned}\begin{gathered} \tilde{k}_B( t)=0\, \text{ or } \, \delta (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}})(t)=\tilde{k}_A(t) \end{gathered}\end{aligned}$$

almost everywhere, $n\ge 2$. Suppose that $\tilde{k}_B( t)\not =0$ for almost every t in a set of positive measure. Then $\delta (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}})(t)=\tilde{k}_A(t)$ for almost every $t\in [0,\infty )$. By applying the one sided Laplace transform $K_A(s)=\int \limits _0^\infty \tilde{k}_A(t)\exp (-ts)dt$, which exists for certain $s>0$ since $\exp (-st)\in L_p([0,\infty ),\mu )$, $1< p<\infty$, we have for $n\ge 2$ $\delta (\underbrace{\tilde{k}_A\star \tilde{k}_A{\star } \ldots {\star }\tilde{k}_A}_{n\ \text {times}})(t)=\tilde{k}_A(t) \Longleftrightarrow \delta K^n_A(s)=K_A(s).$ Let $p(z)=z-\delta z^n$ and suppose that p(z) has $m>0$ roots $z_i$, $i=1,2,\ldots , m$, $m\le n$, $n>1$. We consider the following cases:

• If $n> 1$ and p(z) has $m\ge 2$ roots, then $\tilde{k}_A(t)= \sum \limits _{i=1}^{m} z_i\varDelta (t-z_i)$, In this case $K_A(s)-\delta K_A^n (s)=0$ for all s in the domain of ${K}_A(\cdot )$. But this implies $A=0$ since the Dirac function $\varDelta (\cdot )$ is equivalent to zero function.

• If $n> 1$ and p(z) has only one real root, which is $z=0$, then $K_A(s)-\delta K^n_A(s)=0$ implies $A=0$. $\square$

Remark 6

Let $([0,\infty ),\varSigma ,\mu )$ be the standard Lebesgue measure space. The operators $A:\, L_p([0,\infty ),\mu )\rightarrow L_p([0,\infty ),\mu )$, $B:\, L_p([0,\infty ),\mu )\rightarrow L_p([0,\infty ),\mu )$, $1\le p<\infty$, defined in (4.17) as

$$ \textstyle (Ax)(t)= \int \limits _{0}^\infty \tilde{k}_A(t-s)\cdot I_{[0,\infty )}(t-s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{0}^\infty \tilde{k}_B(t-s)\cdot I_{[0,\infty )}x(s)d\mu _s, $$

almost everywhere, with $\tilde{k}_A(\cdot )\in L_1([0,\infty ),\mu )$, $\tilde{k}_B(\cdot )\in L_1([0,\infty ),\mu )$ (where $I_E(\cdot )$ denotes the indicator function of the set E, and the index in $\mu$ indicates the variable of integration) commute, $AB=BA$. In fact, by applying Fubbini theorem for composition of operators A, B and Lemma 2 we have $AB=BA$ if and only if

$$\begin{aligned} \nonumber&\begin{array}{l} \int \limits _{0}^\infty \tilde{k}_A(t-s)\cdot I_{[0,\infty )}(t-s) \tilde{k}_B(s-\tau )\cdot I_{[0,\infty )}(s-\tau )d\mu _s \\ \quad =\int \limits _{0}^\infty \tilde{k}_B(t-s)\cdot I_{[0,\infty )}(t-s) \tilde{k}_A(s-\tau )\cdot I_{[0,\infty )}(s-\tau ) d\mu _s \end{array} \Leftrightarrow \\&\textstyle \int \limits _{\tau }^t \tilde{k}_A(t-s)\tilde{k}_B(s-\tau )d\mu _s =\int \limits _{\tau }^t \tilde{k}_B(t-s)\cdot \tilde{k}_A(s-\tau )d\mu _s \,\Leftrightarrow \nonumber \\&\textstyle \int \limits _{0}^{t-\tau } \tilde{k}_A(t-\tau -\nu )\tilde{k}_B(\nu )d\mu _\nu =\int \limits _{0}^{t-\tau } \tilde{k}_B(t-\tau -\nu )\cdot \tilde{k}_A(\nu )d\mu _\nu , \end{aligned}$$

(4.20)

for almost every $(t,\tau )\in \mathbb {R}^2$. By changing variable $\xi =t-\tau -\nu$ on the right hand side of (4.20) we get

$$\begin{aligned} \textstyle \int \limits _{0}^{t-\tau } \tilde{k}_B(t-\tau -\nu )\cdot \tilde{k}_A(\nu )d\mu _\nu= & {} - \textstyle \int \limits _{t-\tau }^{0} \tilde{k}_B(\xi )\cdot \tilde{k}_A(t-\tau -\xi )d\mu _\xi \\ {}= & {} \textstyle \int \limits _{0}^{t-\tau } \tilde{k}_A(t-\tau -\xi )\cdot \tilde{k}_B(\xi )d\mu _\xi \end{aligned}$$

which proves (4.20). This completes the proof.

In the following theorem we consider a special case of operators in Theorem 1 when the kernels have the separated variables.

Theorem 3

Let $(X,\varSigma ,\mu )$ be $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined as follows

$$\begin{aligned} (Ax)(t)= \int \limits _{G_A} a(t)b(s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} c(t)e(s)x(s)d\mu _s, \end{aligned}$$

(4.21)

almost everywhere, where the index in $\mu _s$ indicates the variable of integration, $G_A\in \varSigma$ and $G_B\in \varSigma$ with finite measure, $a,c\in L_p(X,\mu )$, $b\in L_q(G_A,\mu )$, $e\in L_q(G_B,\mu )$, $1\le q\le \infty$, $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F:\mathbb {R}\rightarrow \mathbb {R}$ defined by $F(z)=\sum _{j=0}^{n} \delta _j z^j$, where $\delta _j \in \mathbb {R}$ $j=0,1,2,\ldots ,n$. let $G=G_A\cap G_B,$ and

$$ k_1=\sum _{j=1}^{n} \delta _j Q_{G_A} (a,b)^{j-1}Q_{G_B} (a,e), \quad k_2=Q_{G_B} (b,c), $$

where $Q_{\varLambda }(u,v)$, $\varLambda \in \varSigma$, is defined by (4.2). Then $AB=BF(A)$ if and only if the following conditions are fulfilled:

• 1.

  • (a) for almost every $(t,s)\in \textrm{supp }\, c\times [(\textrm{supp }\, e)\cap G]$, we have;

    • (i) if $k_2\not =0$ then $b(s)k_1= \lambda e(s)$ and $a(t)=\frac{(\delta _0+\lambda )c(t)}{k_2}$ for some real scalar $\lambda$,

    • (ii) if $k_2=0$ then $k_1b(s)=-\delta _0e(s)$.

  • (b) If $t\not \in \textrm{supp }\, c$ then either $k_2=0$ or $a(t)=0$ for almost all $t\not \in \textrm{supp }\, c$.

  • (c) If $s\in G\setminus \textrm{supp }\, e$ then either $k_1=0$ or $b(s)=0$ for almost all $s\in G\setminus \textrm{supp }\, e$.

• 2. $k_2 a(t)-\delta _0 c(t)=0$ for almost every $t\in X$ or $e(s)=0$ for almost every $s\in G_B\setminus G$.

• 3. $k_1=0$ or $b(s)=0$ for almost every $s\in G_A\setminus G$.

Proof

We observe that since $a,c\in L_p(X,\mu ),\ 1\le p\le \infty$, $b\in L_q(G_A,\mu )$, $e\in L_q(G_B,\mu )$, where $1\le q\le \infty$, with $\frac{1}{p}+\frac{1}{q}=1,$ then either condition (4.3) or (4.4) or (4.5) is satisfied and therefore the operators A and B are well-defined. By direct calculation, we have

$$\begin{aligned} (A^2x)(t)= & {} \int \limits _{G_A} a(t)b(s)(Ax)(s)d\mu _s=\int \limits _{G_A} a(t)b(s)a(s)d\mu _s\int \limits _{G_A} b(\tau _1)x(\tau _1)d\mu _{\tau _1}\\= & {} Q_{G_A}(a,b)(Ax)(t), \\ \\ (A^3x)(t)= & {} A(A^2x)(t)=Q_{G_A}(a,b)(A^2x)(t)=Q_{G_A}(a,b)^2(Ax)(t) \end{aligned}$$

almost everywhere. We suppose that

$$\begin{aligned} (A^{m}x)(t)=Q_{G_A}(a,b)^{m-1} (Ax)(t),\quad m=1,2,\ldots \end{aligned}$$

almost everywhere. Then

$$\begin{aligned} (A^{m+1}x)(t)= & {} A(A^{m}x)(t)=Q_{G_A}(a,b)^{m-1} (A^2x)(t)=Q_{G_A}(a,b)^{m} (Ax)(t) \end{aligned}$$

almost everywhere. Then, we compute

$$\begin{aligned}&\begin{array}{lll} (ABx)(t)&{}=&{}\displaystyle \int \limits _{G_A} a(t)b(s)c(s)d\mu _s\int \limits _{G_B} e(\tau _1) x(\tau _1)d\mu _{\tau _1} \\ &{}=&{} k_2 \displaystyle \int \limits _{G_B} a(t) e(\tau _1) x(\tau _1)d\mu _{\tau _1}, \end{array}\end{aligned}$$

(4.22)

$$\begin{aligned}&(F(A)x)(t)=\delta _0 x(t)+a(t)\sum _{j=1}^{n} \delta _j \left( Q_{G_A}(a,b) \right) ^{j-1}\int \limits _{G_A} b(\tau ) x(\tau ) d\mu _\tau , \nonumber \\&(BF(A)x)(t)=\delta _0 c(t)\int \limits _{G_B} e(\tau _1)x(\tau _1)d\mu _{\tau _1}\end{aligned}$$

(4.23)

$$\begin{aligned}&+ c(t)\sum _{j=1}^{n} \delta _j \left( Q_{G_A}(a,b) \right) ^{j-1}\int \limits _{G_B} e(\tau )a(\tau )d\mu _\tau \int \limits _{G_B} b(\tau _1) x(\tau _1) d\mu _{\tau _1} \nonumber \\&= \delta _0 c(t)\int \limits _{G_B} e(\tau _1) x(\tau _1)d\mu _{\tau _1}+c(t)k_1\int \limits _{G_A} b(\tau _1) x(\tau _1)d\mu _{\tau _1}. \end{aligned}$$

(4.24)

Thus, $(ABx)(t)=(BF(A)x)(t)$ for all $x\in L_p( X,\mu )$ if and only if

$$ \int \limits _{G_B} [k_2a(t)-\delta _0 c(t)]e(s)x(s)d\mu _s= \int \limits _{G_A} k_1 c(t)b(s)x(s)d\mu _s. $$

Then by Lemma 1, $AB=BF(A)$ if and only if

• 1. for almost every $(t,s)\in X\times G$,

  $$\begin{aligned}{}[k_2a(t)-\delta _0 c(t)]e(s)=k_1 c(t)b(s); \end{aligned}$$

• 2. $k_2a(t)-\delta _0 c(t)=0$ for almost every $t\in X$ or $e(s)=0$ for almost every $s\in G_B\setminus G$;

• 3. $k_1=0$ or $c(t)=0$ for almost every $t\in X$ or $b(s)=0$ for almost every $s\in G_A\setminus G$.

We can rewrite the first condition as follows:

• (a) Suppose $(t,s)\in \textrm{supp }\, c\times [(\textrm{supp }\, e)\cap G]$.

  • (i) If $k_2\ne 0,$ then $k_1\frac{b(s)}{e(s)}=k_2\frac{a(t)}{c(t)}-\delta _0=\lambda$ for some real scalar $\lambda$. From this, it follows that $k_1 b(s)=e(s)\lambda$ and $a(t)=\frac{\delta _0+\lambda }{k_2} c(t)$.

  • (ii) If $k_2=0$ then $-\delta _0 c(t)e(s)=k_1 c(t)b(s)$ from which we get that $k_1b(s)=-\delta _0e(s)$.

• (b) If $t\not \in \textrm{supp }\, c$ then $k_2a(t)e(s)=0$ from which we get that either $k_2=0$ or $a(t)=0$ for almost all $t\not \in \textrm{supp }\, c$ or $e(s)=0$ almost everywhere (this implies $B=0$).

• (c) If $s\in G\setminus \textrm{supp }\, e,$ then $k_1c(t)b(s)=0$ which implies that either $k_1=0$ or $b(s)=0$ for almost all $s\in G\setminus \textrm{supp }\, e$, or $c(t)=0$ almost everywhere (this implies that $B=0$). $\square$

Remark 7

Observe that operators A and B as defined in (4.21) take the form $(Ax)(t)=a(t)\phi (x)$ and $(Bx)(t)=c(t)\psi (x)$ for some functions $a,c\in L_p(X,\mu )$, $1\le p \le \infty$ and linear functionals $\phi ,\psi : X\rightarrow \mathbb {R}.$ In this case $AB=BF(A)$ if and only in $\phi (\psi (x)c(t))a(t)=\psi (F(\phi (x)a(t)))c(t)$ in $L_p(X,\mu )$, $1\le p\le \infty$.

Corollary 3

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators such that

$$\begin{aligned}&(Ax)(t)= \int \limits _{G} a(t)b(s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G} c(t)e(s)x(s)d\mu _s, \\&\quad \quad \quad {\text {(the index in }}\mu _s\text { indicates the variable of integration)} \end{aligned}$$

almost everywhere, $G\in \varSigma$ is a set with finite measure, $a, c \in L_p(X,\mu )$, $b, e\in L_q(G,\mu )$, $1\le q\le \infty$, $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\ldots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,\ldots ,n$. Set

$$ k_1=\sum _{j=1}^{n} \delta _j Q_{G}(a,b)^{j-1}Q_{G}(a,e), \quad k_2=Q_{G}(b,c), $$

Then $AB=BF(A)$ if and only if the following is true

• 1. for almost every $(t,s)\in \textrm{supp }\, c\times \textrm{supp }\, e$, we have

  • a) If $k_2\ne 0,$ then $k_1 b(s)=e(s)\lambda$ and $a(t)=\frac{\delta _0+\lambda }{k_2} c(t)$ for some $\lambda \in \mathbb {R}$.

  • b) If $k_2=0$ then $k_1b(s)=-\delta _0e(s)$;

• 2. If $t\not \in \textrm{supp }\, c$ then either $k_2=0$ or $a(t)=0$ for almost all $t\not \in \textrm{supp }\, c.$

• 3. If $s\in G \setminus \textrm{supp }\, e,$ then either $k_1=0$ or $b(s)=0$ for almost all $s\in G\setminus \textrm{supp }\, e$

Proof

This follows immediately from Theorem 3 as $G_A=G_B=G$. $\square$

Remark 8

From Theorem 3 and Corollary 3 we observe that if $k_1,k_2 \ne 0$, then given operator B as defined by (4.21), we can obtain the kernel of operator A using relations, $a(t)=\frac{\delta _0+\lambda }{k_2}c(t)$ and $b(s)=\frac{\lambda }{k_1}e(s)$ for some $\lambda \in \mathbb {R}$. In the next two propositions we state necessary and sufficient conditions for the choice of $\lambda$.

Proposition 4

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let

$$\begin{aligned} A:L_p(X,\mu )\rightarrow L_p(X,\mu ),\ B:L_p(X,\mu )\rightarrow L_p(X,\mu ),\ 1\le p\le \infty \end{aligned}$$

be nonzero operators such that

$$\begin{aligned}&(Ax)(t)= \int \limits _{G} a(t)b(s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{G} c(t)e(s)x(s)d\mu _s, \\&\quad \quad \quad {\text {(the index in }}\mu _s{\text { indicates the variable of integration)}} \end{aligned}$$

almost everywhere, $G\in \varSigma$ is a set with finite measure, and $a, c \in L_p(X,\mu )$, $b, e\in L_q(G,\mu )$, $1\le q\le \infty$, $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\cdots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,3,\ldots ,n$. Set

$$ k_1=\sum _{j=1}^{n} \delta _j Q_{G}(a,b)^{j-1} Q_{G}(a,e), \quad k_2=Q_{G}(b,c). $$

Suppose that $AB=BF(A).$ If $k_2\not =0$ and $k_1\not =0$ in condition 1(a) in Corollary 3, then the corresponding nonzero $\lambda$ satisfy

$$\begin{aligned} F(\lambda +\delta _0)=\lambda +\delta _0. \end{aligned}$$

(4.25)

Proof

By definition $k_1=\sum _{j=1}^{n} \delta _j Q_{G}(a,b)^{j-1}Q_{G}(a,e), \quad k_2=Q_{G}(b,c).$ If $k_1\not =0$, $k_2\not =0$, by condition 1(a) in Corollary 3 we have $a(t)=\frac{\lambda +\delta _0}{k_2}c(t),\quad b(s)=\frac{\lambda }{k_1}e(s)$ almost everywhere. If $\lambda \not =0$ then we replace $k_2=Q_{G}(b,c)=Q_{G}(\frac{\lambda }{k_1} e, c)$ in the following equality $k_1=\sum _{j=1}^{n} \delta _j Q_{G}\left( \frac{\lambda +\delta _0}{k_2}c,\frac{\lambda }{k_1}e \right) ^{j-1}Q_{G}\left( \frac{\lambda +\delta _0}{k_2}c,e\right) .$ Then, by using the bilinearity of $Q_{G}(\cdot ,\cdot )$ and after simplification, this is equivalent to $\lambda =\sum _{j=1}^{n}\delta _j(\lambda +\delta _0)^j.$ By adding $\delta _0$ on both sides we can write this as (4.25). $\square$

Proposition 5

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let

$$\begin{aligned} A:L_p(X,\mu )\rightarrow L_p(X,\mu ),\ B:L_p(X,\mu )\rightarrow L_p(X,\mu ),\ 1\le p\le \infty \end{aligned}$$

be nonzero operators such that

$$\begin{aligned}&(Ax)(t)= \int \limits _{G} a(t)b(s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G} c(t)e(s)x(s)d\mu _s, \\&\quad \quad \quad \text {(the index in }\mu _s\text { indicates the variable of integration)} \end{aligned}$$

almost everywhere, $G\in \varSigma$ is a set with finite measure, $a, c \in L_p(X,\mu )$, $b, e\in L_q(G,\mu )$, $1\le q\le \infty$, $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\ldots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,3,\ldots ,n$. Suppose that for almost every $(t,s)\in \textrm{supp }\, c\times \textrm{supp }\, e$, we have

$$\begin{aligned} a(t)=\frac{\lambda +\delta _0}{k_2}c(t),\quad b(s)=\frac{\lambda }{k_1}e(s) \end{aligned}$$

for nonzero constants $\lambda ,$ $k_1$ and $k_2$. If $F(\lambda +\delta _0)=\lambda +\delta _0$ and $k_2=\frac{\lambda }{k_1}Q_{G}(e,c)$, then

• 1. $A=\frac{\lambda +\delta _0}{Q_{G}(e,c)}B,$

• 2. for all $x\in L_p(X,\mu )$ and almost all $t\in \textrm{supp }\, c$, $(ABx)(t)=(BF(A)x)(t).$

Proof

We have, almost everywhere,

$$\begin{aligned}&(Ax)(t)=\int \limits _{G} a(t)b(s)x(s)d\mu _s=\frac{(\lambda +\delta _0)\lambda }{k_1k_2}\int \limits _{G} c(t)e(s)x(s)d\mu _s=\frac{(\lambda +\delta _0)\lambda }{k_1k_2}(Bx)(t) \\&(ABx)(t)=\frac{(\lambda +\delta _0)\lambda }{k_1k_2}(B^2x)(t)=\frac{(\lambda +\delta _0)\lambda }{k_1k_2}Q_{G}(c,e)(Bx)(t), \\&(A^2x)(t)=\left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^2(B^2x)(t)= \left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^2Q_{G}(e,c)(Bx)(t). \end{aligned}$$

Similarly, for $m \ge 2$, almost everywhere

$$\begin{aligned}&(A^mx)(t)=\left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^m Q_{G}(c,e)^{m-1}(Bx)(t) \\&(F(A)x)(t)=\delta _0 (Bx)(t)+\sum _{j=1}^{n}\delta _j \left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^j Q_{G}(c,e)^{j-1}(Bx)(t). \end{aligned}$$

Therefore, almost everywhere,

$$\begin{aligned} (BF(A)x)(t)= & {} \delta _0 (B^2x)(t)+\sum _{j=1}^{n}\delta _j \left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^j Q_{G}(c,e)^{j-1}(B^2x)(t)\\= & {} \delta _0 Q_{G}(c,e) (Bx)(t)+\sum _{j=1}^{n}\delta _j \left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\right) ^j Q_{G}(c,e)^{j}(Bx)(t) \\= & {} F\left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}Q_{G}(c,e)\right) (Bx)(t), \end{aligned}$$

Hence, $(ABx)(t)=(BF(A)x)(t),$ for all $x\in L_p(X,\mu )$ and almost all $t\in \textrm{supp }\, c$ if and only if

$$\begin{aligned} \frac{(\lambda +\delta _0)\lambda }{k_1k_2}Q_{G}(c,e)= F\left( \frac{(\lambda +\delta _0)\lambda }{k_1k_2}Q_{G}(c,e)\right) \end{aligned}$$

(4.26)

for almost every $t\in \textrm{supp }\, c$. If $k_2=\frac{\lambda }{k_1}Q_{G}(c,e)$ and $\lambda$ satisfies (4.25), then (4.26) holds. $\square$

Corollary 4

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1< p<\infty$ be nonzero operators such that

$$\begin{aligned}&(Ax)(t)= \int \limits _{G} a(t)b(s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G} c(t)e(s)x(s)d\mu _s, \\&\quad \quad \quad \text {(the index in }\mu _s\text { indicates the variable of integration)} \end{aligned}$$

almost everywhere, $G\in \varSigma$ is a set with finite measure, $a, c \in L_p(X,\mu )$, $b, e\in L_q(G,\mu )$, $1<q<\infty$, $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\delta _2 z^2$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2$. Suppose that for almost every $(t,s)\in \textrm{supp }\, c\times \textrm{supp }\, e$, $a(t)=\frac{\lambda +\delta _0}{k_2}c(t),\ b(s)=\frac{\lambda }{k_1}e(s)$ for nonzero constants $\lambda ,$ $k_1$ and $k_2$. If $k_2=\frac{\lambda }{k_1}Q_{G}(e,c)$, then $(ABx)(t)=(BF(A)x)(t),$ for all $x\in L_p(G,\mu )$ and almost all $t\in \textrm{supp }\, c$ if either $\delta _0\delta _2<0$, or $\delta _0\delta _2\ge 0$ and either $\delta _1\ge 1+2\sqrt{\delta _0\delta _2}$ or $\delta _1\le 1-2\sqrt{\delta _0\delta _2}.$

Proof

From Propositions 4 and 5 we have that $AB=BF(A)$ if $F(\lambda +\delta _0)=\lambda +\delta _0.$ This is equivalent to

$$\begin{aligned} \delta _2\lambda ^2+(2\delta _0\delta _2+\delta _1-1)\lambda +\delta _2\delta _0^2+\delta _1\delta _0=0 \end{aligned}$$

(4.27)

Equation (4.27) has real solutions if and only if $(\delta _1-1)^2-4\delta _0\delta _2\geqslant 0.$ This is equivalent to either $\delta _0\delta _2<0$, or $\delta _0\delta _2\ge 0$ and either $\delta _1\ge 1+2\sqrt{\delta _0\delta _2}$ or $\delta _1\le 1-2\sqrt{\delta _0\delta _2},$ which completes the proof. $\square$

Example 2

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ 1< p<\infty \end{aligned}$$

be nonzero operators defined as follows

$$ (Ax)(t)= \int \limits _0^1 a(t)b(s)x(s)ds,\quad (Bx)(t)= \int \limits _0^1 c(t)e(s)x(s)ds, $$

where $a \in L_p(\mathbb {R},\mu )$, $b\in L_q([0,1],\mu )$, $1<q<\infty$, $\frac{1}{p}+\frac{1}{q}=1,$ and $c(t)=tI_{[0,1]}(t)$, $e(s)=s+1$. Consider the polynomial $F(z)=z^2+z-1$ and suppose that for almost every $(t,s)\in \textrm{supp }\, c\times \textrm{supp }\, e$, $a(t)=\frac{\lambda +\delta _0}{k_2}c(t),\ b(s)=\lambda e(s)$ for nonzero constants $\lambda$ and $k_2=\lambda Q_{[0,1]}(e,c)=\frac{5}{6}\lambda$. From Propositions 4 and 5 we have that $AB=BF(A)$ if $F(\lambda -1)=\lambda -1,$ or $\lambda ^2-2\lambda =0.$ Therefore, we take $\lambda =2.$ Then, $A=\frac{\lambda +\delta _0}{Q_{[0,1]}(e,c)}B=\frac{6}{5}B.$ Hence, $A^2=\left( \frac{6}{5}B\right) \left( \frac{6}{5}B\right) =\left( \frac{6}{5}\right) ^2B^2.$ But

$$(B^2x)(t)=\int \limits _0^1tI_{[0,1]}(t)(s+1)\int \limits _0^1sI_{[0,1]}(s)(\tau +1)x(\tau )d\tau =\frac{5}{6}(Bx)(t).$$

Therefore, $A^2=(\frac{6}{5})^2B^2=\frac{6}{5}B=A.$ Thus, $F(A)=A^2+A-I=2A-I=\frac{12}{5}B-I$ and $BF(A)=B\left( \frac{12}{5}B-I\right) =\frac{12}{5}B^2-B=\frac{12}{5}\cdot \frac{5}{6}B-B=B.$ Finally, $AB=\frac{6}{5}B^2=\frac{6}{5}\cdot \frac{5}{6}B=B=BF(A).$

Remark 9

Example 2 is a case when operator $\displaystyle B\left( \sum _{k=1}^{1}\delta _k A^k\right) \not =0$ as mentioned in Remark 1 and Remark 3. In this case we have $G_A=G_B=G=[0,1]$, operators $A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu )$, $1< p<\infty$ are defined as follows $(Ax)(t)=\frac{6}{5}(Bx)(t)$, $(Bx)(t)=\int \limits _0^1 tI_{[0,1]}(t)(s+1)x(s)ds$ almost everywhere, the polynomial is $F(z)=-1+z+z^2$ with coefficients $\delta _0=-1$, $\delta _1=\delta _2=1$. We have that Condition 2 and 3 are satisfied because they are taken on the set $\mathbb {R}\times \emptyset =\emptyset$ which has measure zero in $\mathbb {R}\times [0,1]$. Condition 1 is satisfied as showed in Example 2. Moreover, $B(A+A^2)=2BA=2\cdot \frac{6}{5}B^2=2B=2\int \limits _0^1 tI_{[0,1]}(t)(s+1)x(s)ds\not =0,$ and $\displaystyle A+A^2=2A=\frac{12}{5}B=\int \limits _0^1 tI_{[0,1]}(t)(s+1)x(s)ds\not =0$.

Remark 10

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. From Proposition 5 we have that if $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ are nonzero operators defined as follows $(Ax)(t)= \int \limits _{G} a(t)b(s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{G} c(t)e(s)x(s)d\mu _s,$ almost everywhere, $G\in \varSigma$ is a set with finite measure, $a, c \in L_p(X,\mu )$, $b, e\in L_q(G,\mu )$, $1\le q\le \infty$, $\frac{1}{p}+\frac{1}{q}=1$ and $F(z)=\delta _0+\delta _1z+\ldots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,3,\ldots ,n$. If we suppose that for almost every $(t,s)\in \textrm{supp }\, c\times \textrm{supp }\, e$, $a(t)=\frac{\lambda +\delta _0}{k_2}c(t)$ and $b(s)=\frac{\lambda }{k_1}e(s)$ for some nonzero constants $\lambda$, $k_1$ and $k_2$ and if $F(\lambda +\delta _0)=\lambda +\delta _0$ and $k_2=\frac{\lambda }{k_1}Q_{G}(e,c)$, then $A=\frac{\lambda +\delta _0}{Q_{G}(e,c)}B$ and $AB=BF(A).$ Now suppose that $A=\omega B$ for some $\omega \in \mathbb {R}$, then $AB=BF(A)$ if and only if

$$\begin{aligned} F(\omega Q_{G}(c,e))=\omega Q_{G}(c,e). \end{aligned}$$

(4.28)

This relation is the same as Equation (4.26) with $\omega =\frac{(\lambda +\delta _0)\lambda }{k_1k_2}.$

Corollary 5

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined by $(Ax)(t)= \int \limits _{G_A} a(t)b(s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{G_B} c(t)e(s)x(s)d\mu _s,$ almost everywhere, where $G_A\in \varSigma$, $G_B\in \varSigma$ are sets with finite measure, $a, c \in L_p(X,\mu )$, $b\in L_q(G_A,\mu )$, $e\in L_q(G_B,\mu )$, $1\le q\le \infty$ and $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\ldots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,3,\ldots ,n$. Let $G=G_A\cap G_B$, and

$$ k_1= \sum _{j=1}^{n}\delta _j Q_{G_A} (a,b)^{j-1}Q_{G_B} (a,e), \ k_2=Q_{G_B} (b,c). $$

Then,

• 1. if $k_1\not =0$, $k_2\not =0$, then $AB=BF(A)$ if and only if $A=\omega B$, for some constant $\omega$ which satisfies (4.28);

• 2. if $k_2=0$ then $AB=0$ and, $AB=BF(A)$ if and only if $BF(A)=0$. Moreover,

  • (a) if $k_1\not =0$ then $BF(A)=0$ if and only if $b(s)=-\frac{\delta _0}{k_1}e(s)I_{G}(s)$ almost everywhere;

  • (b) if $k_1=0$ then $AB=BF(A)$ if $\delta _0=0$, that is, $F(t)=\sum _{j=1}^{n}\delta _j t^{j}$;

• 3. if $k_2\not =0$ and $k_1=0$ then $AB=BF(A)$ if and only if $AB=\delta _0 B$, that is

  $$(Ax)(t)=\frac{\delta _0}{k_2}\int \limits _{G_A}c(t)b(s)x(s)d\mu _s.$$

Proof

• 1. By applying Theorem 3 if $k_1\not =0$ and $k_2\not =0$ we have $AB=BF(A)$ if and only if the following is true:

  • for almost every $t\in \textrm{supp }\, c$ $a(t)=\frac{\delta _0+\lambda }{k_2}c(t)$ and $b(s)=\frac{\lambda }{k_2}e(s)$ for almost every $s\in G \cap \textrm{supp }\, e$ and nonzero constant $\lambda$ satisfying (4.26);

  • $e(s)=0$ for almost every $s\in G_B\setminus G$;

  • $b(s)=0$ for almost every $s\in G_A\setminus G$;

  From which we have,

  $$\begin{aligned} (Ax)(t)= & {} \int \limits _G a(t)b(s)x(s)d\mu _s+\int \limits _{G_A\setminus G} a(t)b(s)x(s)d\mu _s=\\= & {} \frac{(\lambda +\delta _0)\lambda }{k_1k_2}\int \limits _G c(t)e(s)x(s)d\mu _s =\frac{(\lambda +\delta _0)\lambda }{k_1k_2} (Bx)(t) \end{aligned}$$

  almost everywhere. If $\lambda =0$ then $A=0$.

• 2. If $k_2=0$ then from (4.22) we have $AB=0$ and, hence $AB=BF(A)$ if and only if $BF(A)=0$. Moreover, by applying Theorem 3 we have

  • (a) if $k_1\not =0$ then $AB=BF(A)$ if and only if for almost every $s\in \textrm{supp }\, e\cap G$, $b(s)=-\frac{\delta _0}{k_1}e(s),$ $b(s)=0$ for almost every $s\in G\setminus \textrm{supp }\, e$, $e(s)=0$ for almost every $s\in G_B\setminus G$ and $b(s)=0$ for almost every $s\in G_A\setminus G$. Therefore, almost everywhere, $b(s)=-\frac{\delta _0}{k_1}e(s)I_G(s).$

  • (b) if $k_1=0$ and $\delta _0=0$, then $AB=BF(A)$.

• 3. By applying Theorem 3 and if $k_2\not =0$, $k_1=0$ we have $AB=BF(A)$ if and only if for almost every $t\in \textrm{supp }\, c$, $a(t)=\frac{\delta _0+\lambda }{k_2}c(t)$ and $\lambda e(s)=0$ for almost every $s\in G \cap \textrm{supp }\, e$, from which we get $\lambda =0$. Therefore, $a(t)=\frac{\delta _0}{k_2}c(t)$ almost everywhere. So we can write $(Ax)(t)=\int \limits _{G_A} a(t)b(s)x(s)d\mu _s=\frac{\delta _0}{k_2}\int \limits _{G_A} c(t)b(s)x(s)d\mu _s$ almost everywhere. Hence, almost everywhere,

  $$\begin{aligned} (ABx)(t)= & {} \int \limits _{G_A} c(t)b(s)\left( \int \limits _{G_B} e(\tau )x(\tau )d\mu _\tau \right) d\mu _s \\= & {} \frac{\delta _0}{k_2} c(t)\int \limits _{G_A} c(s)b(s)d\mu _s\int \limits _{G_B} e(\tau )x(\tau )d\mu _\tau \\= & {} \frac{\delta _0}{k_2}Q_{G}(b,c)\int \limits _{G_B}c(t) e(\tau )x(\tau )d\mu _\tau =\delta _0(Bx)(t) \end{aligned}$$

  On the other hand, from (4.24) follows that $BF(A)=\delta _0B$ if $k_1=0$. $\square$

Example 3

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ),\ 1< p<\infty \end{aligned}$$

be nonzero operators $(Ax)(t)= \int \limits _{0}^{1} a(t)b(s)x(s)ds,\ (Bx)(t)= \int \limits _{0}^{1} c(t)e(s)x(s)ds,$ where $a(t)=t^2I_{[0,1]}(t),$ $b(s)=s^3$, $c(t)=-6t^2I_{[0,1]}(t)$ and $e(s)=s$. Consider a polynomial $F(t)=\delta _0+\delta _1t+\delta _2 t^2$, where $t\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2$. We have $k_2= Q_{[0,1]}(b,c)=\int \limits _{0}^{1} b(s)c(s)ds=\int \limits _{0}^{1} -6s^3s^2ds=-1.$ If

$$\begin{aligned} k_1=\delta _1 Q_{[0,1]}(a,e)+\delta _2 Q_{[0,1]}(a,b) Q_{[0,1]}(a,e)=0, \end{aligned}$$

then choose $\delta _i$, $i=1,2$ such that $0=\delta _1+\delta _2 Q_{[0,1]}(a,b)=\delta _1-\frac{1}{6}\delta _2 Q_{[0,1]}(c,b)=\delta _1+\frac{1}{6}\delta _2$. Thus $\delta _2=-6\delta _1$ and $\frac{\delta _0}{k_2}=-\frac{1}{6}$ from which we get $\delta _0=\frac{1}{6}$. Hence, $F(t)=-6\delta _1 t^2+\delta _1 t+\frac{1}{6}$. We have almost everywhere

$$ (Ax)(t)=\int \limits _{0}^{1} t^2I_{[0,1]}(t)s^3 x(s)ds,\ (Bx)(t)=-6\int \limits _{0}^{1} t^2I_{[0,1]}(t)s x(s)ds, $$

and thus almost everywhere

$$\begin{aligned}\begin{gathered} (ABx)(t)=\int \limits _{0}^{1} t^2I_{[0,1]}(t)s^3 \left( -6\int \limits _{0}^{1} s^2I_{[0,1]}(s)\tau x(\tau )d\tau \right) ds=\frac{1}{6}(Bx)(t),\\ (A^2x)(t)=\int \limits _{0}^{1} t^2I_{[0,1]}(t)s^3 \left( \ \int \limits _{0}^{1} s^2I_{[0,1]}(s)\tau ^3x(\tau )d\tau \right) ds=\frac{1}{6}(Ax)(t). \end{gathered}\end{aligned}$$

Finally, we have

$$ BF(A)=B\left( -6\delta _1 A^2+\delta _1 A+\frac{1}{6}I\right) =-\delta _1 BA+\delta _1 BA+\frac{1}{6}B=\frac{1}{6}B=AB. $$

Example 4

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ), \ B:L_p(\mathbb {R},\mu )\rightarrow L_p(\mathbb {R},\mu ), \ 1< p<\infty \end{aligned}$$

be nonzero operators $(Ax)(t)= \int \limits _{\alpha }^{\beta } a(t)b(s)x(s)ds,\ (Bx)(t)= \int \limits _{\alpha }^{\beta } c(t)e(s)x(s)ds,$ where $\alpha ,\beta \in \mathbb {R}$, $-\infty<\alpha \le \beta <\infty$, $a,c\in L_p(\mathbb {R},\mu )$, $b,e\in L_q([\alpha ,\beta ],\mu )$ where $1<q<\infty$ such that $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(t)=\delta _0+\delta _1t+\delta _2 t^2$, where $t\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2$. We set

$$ \textstyle k_2= Q_{[\alpha ,\beta ]}(b,c)=\int \limits _{\alpha }^{\beta } b(s)c(s)ds, \ k_1= \delta _1 Q_{[\alpha ,\beta ]}(a,e)+\delta _2 Q_{[\alpha ,\beta ]}(a,b) Q_{[\alpha ,\beta ]}(a,e). $$

If $k_2\not =0$ and $k_1=0$ then we choose either $Q_{[\alpha ,\beta ]}(a,e)=0$ or $\delta _i$, $i=1,2$ such that $\delta _1+\delta _2 Q_{[\alpha ,\beta ]}(a,b)=0$. Thus from Corollary 5 we have $a(t)=\frac{\delta _0}{k_2}c(t)$ almost everywhere. Thus $k_1=0$ implies that either $Q_{[\alpha ,\beta ]}(a,e)=0\ \text{ or }\ \delta _1+\frac{\delta _0}{k_2}\delta _2 k_2=0.$ We choose coefficients $\delta _j$, $j=0,1,2$ such that $\delta _1=-\delta _0\delta _2$, and hence $F(t)=\delta _2 t^2-\delta _0\delta _2 t+\delta _0$. Then, the operators

$$ (Ax)(t)=\frac{\delta _0}{k_2}\int \limits _{\alpha }^{\beta } c(t)b(s) x(s)ds,\ (Bx)(t)=\int \limits _{\alpha }^{\beta } c(t)e(s) x(s)ds $$

almost everywhere, satisfy the relation

$$\begin{aligned} AB=\delta _2 BA^2-\delta _0\delta _2 BA+\delta _0B. \end{aligned}$$

(4.29)

In fact,

$$\begin{aligned}\begin{gathered} (ABx)(t)=\frac{\delta _0}{k_2}\int \limits _{\alpha }^{\beta } c(t)b(s) \left( \ \int \limits _{\alpha }^{\beta } c(s)e(\tau ) x(\tau )d\tau \right) ds=\delta _0(Bx)(t),\\ (A^2x)(t)=\frac{\delta _0}{k_2}\int \limits _{\alpha }^{\beta } c(t)b(s) \left( \frac{\delta _0}{k_2}\int \limits _{\alpha }^{\beta } c(s)b(\tau )x(\tau )d\tau \right) ds=\delta _0(Ax)(t), \end{gathered}\end{aligned}$$

almost everywhere. Finally, we have $BF(A)=B\left( \delta _2 A^2-\delta _2 \delta _0 A+\delta _0I\right) =\delta _2\delta _0 BA-\delta _2\delta _0 BA+\delta _0 B=\delta _0B=AB.$ In particular, if $\alpha =0$, $\beta =1$, $b(s)=s$ and $c(t)=t^2I_{[0,1]}(t)$, $e(s)=s^3$ we have $k_2=Q_{[0,1]}(b,c)=\frac{1}{4}$. Hence the operators

$$\begin{aligned} (Ax)(t)=4\delta _0\int \limits _{0}^{1} t^2I_{[0,1]}(t)sx(s)ds,\ (Bx)(t)=\int \limits _{0}^{1} t^2I_{[0,1]}(t)s^3x(s)ds \end{aligned}$$

(4.30)

satisfy the Relation (4.29). In particular, if $\delta _2=1$ and $\delta _0=-1$, that is, $F(t)=t^2+t-1$ then the corresponding operators in (4.30) satisfy $AB=BA^2+BA-B.$

Corollary 6

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined by

$$ (Ax)(t)= \int \limits _{G_A} a(t)b(s)x(s)d\mu _s,\ (Bx)(t)= \int \limits _{G_B} c(t)e(s)x(s)d\mu _s, $$

almost everywhere, $G_A\in \varSigma$, $G_B\in \varSigma$ are sets with finite measure, $a, c \in L_p(X,\mu )$, $b\in L_q(G_A,\mu )$, $e\in L_q(G_B,\mu )$, $1\le q\le \infty$ and $\frac{1}{p}+\frac{1}{q}=1$. Consider a polynomial $F(z)=\delta _0+\delta _1z+\ldots +\delta _n z^n$, where $z\in \mathbb {R}$, $\delta _j\in \mathbb {R}$, $j=0,1,2,3,\ldots ,n$. Let $G=G_A\cap G_B$ and $k_1= \sum _{j=1}^{n}\delta _j Q_{G_A} (a,b)^{j-1} Q_{G_B} (a,e), \ k_2=Q_{G_B} (b,c).$ If $k_2\not =0$ and $Q_{G_B} (a,e)=0$, then $AB=BF(A)$ if and only if $AB=\delta _0 B$, that is $a(t)=\frac{\delta _0}{k_2}c(t),$ almost everywhere.

Proof

This follows by Corollary 5 since $k_2\not =0$ and $k_1=0$. $\square$

Corollary 7

Let $(X,\varSigma ,\mu )$ be a $\sigma$-finite measure space. Let $A:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $B:L_p(X,\mu )\rightarrow L_p(X,\mu )$, $1\le p\le \infty$ be nonzero operators defined by

$$ (Ax)(t)= \int \limits _{G_A} a(t)b(s)x(s)d\mu _s,\quad (Bx)(t)= \int \limits _{G_B} c(t)e(s)x(s)d\mu _s, $$

almost everywhere, $G_A\in \varSigma$, $G_B\in \varSigma$ are sets with finite measure, $a, c \in L_p(X,\mu )$, $b\in L_q(G_A,\mu )$, $e\in L_q(G_B,\mu )$, $1\le q\le \infty$ and $\frac{1}{p}+\frac{1}{q}=1$. Consider a monomial $F(z)=\delta z^d$, where $z\in \mathbb {R}$, d is a positive integer and $\delta \not =0$ is a real number. Let $G=G_A\cap G_B$ and $k_1= \delta Q_{G_A} (a,b)^{d-1} Q_{G_B} (a,e), \ k_2=Q_{G_B} (b,c).$ Then $AB=\delta BA^d$ if and only the following conditions are fulfilled:

• 1.

  • (a) for almost every $(t,s)\in \textrm{supp }\, c\times [(\textrm{supp }\, e)\cap G]$ we have the following:

    • (i) If $k_2\ne 0,$ then $k_1 b(s)=e(s)\lambda$ and $a(t)=\frac{\lambda }{k_2} c(t)$ for some $\lambda \in \mathbb {R}$.

    • (ii) If $k_2=0$ then either $k_1=0$ or $b(s)=0$ for almost all $s\in \textrm{supp }\, e\cap G$.

  • (b) If $t\not \in \textrm{supp }\, c$ then either $k_2=0$ or $a(t)=0$ for almost all $t\not \in \textrm{supp }\, c$.

  • (c) If $s\in G\setminus \textrm{supp }\, e$ then either $k_1=0$ or $b(s)=0$ for almost all $s\in G\setminus \textrm{supp }\, e.$

• 2. $k_2 =0$ , or $e(s)=0$ for almost every $s\in G_B\setminus G$.

• 3. $k_1=0$ or $b(s)=0$ for almost every $s\in G_A\setminus G$.

Proof

This follows from Theorem 3 and the fact that $\delta _0=0$ in this case. $\square$

Example 5

Let $(\mathbb {R},\varSigma ,\mu )$ be the standard Lebesgue measure space. Let

$$\begin{aligned} A:L_2([\alpha ,\beta ],\mu )\rightarrow L_2([\alpha ,\beta ],\mu ),\ B:L_2([\alpha ,\beta ],\mu )\rightarrow L_2([\alpha ,\beta ],\mu ) \end{aligned}$$

be defined by $(Ax)(t)= \int \limits _{\alpha }^{\beta } a(t)b(s)x(s)ds,\ (Bx)(t)= \int \limits _{\alpha }^{\beta } c(t)e(s)x(s)ds,$ where $\alpha ,\beta$ are real numbers with $\alpha <\beta$, $a, b, c, e \in L_2([\alpha ,\beta ],\mu )$, such that $a\perp b$ and $b \perp c$, that is, $\int \limits _{\alpha }^{\beta } a(t)b(t)dt=\int \limits _{\alpha }^{\beta } b(t)c(t)dt=0.$ Then the above operators satisfy $AB=\delta BA^d$, $d=2,3,\ldots$. In fact, by using Corollary 7 and putting

$$\begin{aligned}\begin{gathered} F(t)=\delta t^d,\quad d=2,3,\ldots \\ k_1= Q_{[\alpha ,\beta ]}(a,b)^{d-1}Q_{[\alpha ,\beta ]}(a,e), \quad k_2=Q_{[\alpha ,\beta ]}(b,c), \end{gathered}\end{aligned}$$

we get $k_1=k_2=0$. So we have all conditions in Corollary 7 satisfied. In particular, if $a(t)=\left( \frac{5}{3}t^3-\frac{3}{2}t\right) I_{[-1,1]}(t)$, $b(s)=\frac{3}{2}s^2-\frac{1}{2}$ and $c(t)=tI_{[-1,1]}(t)$, then the operators

$$\begin{aligned}\begin{gathered} (Ax)(t)=\int \limits _{-1}^1 \left( \frac{5}{3}t^3-\frac{3}{2}t\right) I_{[-1,1]}(t)\left( \frac{3}{2}s^2-\frac{1}{2}\right) x(s)ds, \\ (Bx)(t)=\int \limits _{-1}^1 tI_{[-1,1]}(t) e(s) x(s)ds \end{gathered}\end{aligned}$$

satisfy the relation $AB=BA^d$, $d=2,3,\ldots$. In fact $a,\, b,\, c$ are pairwise orthogonal in $[-1,1]$.