# On a Weighted Singular Integral Operator with Shifts and Slowly Oscillating Data

Research paper by Alexei Yu. Karlovich, Yuri I. Karlovich; Amarino B. Lebre

Indexed on: 11 Aug '16Published on: 01 Aug '16Published in: Complex Analysis and Operator Theory

#### Abstract

Let $$\alpha ,\beta$$ be orientation-preserving diffeomorphism (shifts) of $$\mathbb {R}_+=(0,\infty )$$ onto itself with the only fixed points $$0$$ and $$\infty$$ and $$U_\alpha ,U_\beta$$ be the isometric shift operators on $$L^p(\mathbb {R}_+)$$ given by $$U_\alpha f=(\alpha ')^{1/p}(f\circ \alpha )$$ , $$U_\beta f=(\beta ')^{1/p}(f\circ \beta )$$ , and $$P_2^\pm =(I\pm S_2)/2$$ where \begin{aligned} (S_2 f)(t):=\frac{1}{\pi i}\int \limits _0^\infty \left( \frac{t}{\tau }\right) ^{1/2-1/p}\frac{f(\tau )}{\tau -t}\,d\tau , \quad t\in \mathbb {R}_+, \end{aligned} is the weighted Cauchy singular integral operator. We prove that if $$\alpha ',\beta '$$ and $$c,d$$ are continuous on $$\mathbb {R}_+$$ and slowly oscillating at $$0$$ and $$\infty$$ , and \begin{aligned} \limsup _{t\rightarrow s} c(t) <1, \quad \limsup _{t\rightarrow s} d(t) <1, \quad s\in \{0,\infty \}, \end{aligned} then the operator $$(I-cU_\alpha )P_2^++(I-dU_\beta )P_2^-$$ is Fredholm on $$L^p(\mathbb {R}_+)$$ and its index is equal to zero. Moreover, its regularizers are described. Let $$\alpha ,\beta$$ be orientation-preserving diffeomorphism (shifts) of $$\mathbb {R}_+=(0,\infty )$$ onto itself with the only fixed points $$0$$ and $$\infty$$ and $$U_\alpha ,U_\beta$$ be the isometric shift operators on $$L^p(\mathbb {R}_+)$$ given by $$U_\alpha f=(\alpha ')^{1/p}(f\circ \alpha )$$ , $$U_\beta f=(\beta ')^{1/p}(f\circ \beta )$$ , and $$P_2^\pm =(I\pm S_2)/2$$ where \begin{aligned} (S_2 f)(t):=\frac{1}{\pi i}\int \limits _0^\infty \left( \frac{t}{\tau }\right) ^{1/2-1/p}\frac{f(\tau )}{\tau -t}\,d\tau , \quad t\in \mathbb {R}_+, \end{aligned} is the weighted Cauchy singular integral operator. We prove that if $$\alpha ',\beta '$$ and $$c,d$$ are continuous on $$\mathbb {R}_+$$ and slowly oscillating at $$0$$ and $$\infty$$ , and \begin{aligned} \limsup _{t\rightarrow s} c(t) <1, \quad \limsup _{t\rightarrow s} d(t) <1, \quad s\in \{0,\infty \}, \end{aligned} then the operator $$(I-cU_\alpha )P_2^++(I-dU_\beta )P_2^-$$ is Fredholm on $$L^p(\mathbb {R}_+)$$ and its index is equal to zero. Moreover, its regularizers are described. $$\alpha ,\beta$$ $$\alpha ,\beta$$ $$\mathbb {R}_+=(0,\infty )$$ $$\mathbb {R}_+=(0,\infty )$$ $$0$$ $$0$$ $$\infty$$ $$\infty$$ $$U_\alpha ,U_\beta$$ $$U_\alpha ,U_\beta$$ $$L^p(\mathbb {R}_+)$$ $$L^p(\mathbb {R}_+)$$ $$U_\alpha f=(\alpha ')^{1/p}(f\circ \alpha )$$ $$U_\alpha f=(\alpha ')^{1/p}(f\circ \alpha )$$ $$U_\beta f=(\beta ')^{1/p}(f\circ \beta )$$ $$U_\beta f=(\beta ')^{1/p}(f\circ \beta )$$ $$P_2^\pm =(I\pm S_2)/2$$ $$P_2^\pm =(I\pm S_2)/2$$ \begin{aligned} (S_2 f)(t):=\frac{1}{\pi i}\int \limits _0^\infty \left( \frac{t}{\tau }\right) ^{1/2-1/p}\frac{f(\tau )}{\tau -t}\,d\tau , \quad t\in \mathbb {R}_+, \end{aligned} \begin{aligned} (S_2 f)(t):=\frac{1}{\pi i}\int \limits _0^\infty \left( \frac{t}{\tau }\right) ^{1/2-1/p}\frac{f(\tau )}{\tau -t}\,d\tau , \quad t\in \mathbb {R}_+, \end{aligned} $$\alpha ',\beta '$$ $$\alpha ',\beta '$$ $$c,d$$ $$c,d$$ $$\mathbb {R}_+$$ $$\mathbb {R}_+$$ $$0$$ $$0$$ $$\infty$$ $$\infty$$ \begin{aligned} \limsup _{t\rightarrow s} c(t) <1, \quad \limsup _{t\rightarrow s} d(t) <1, \quad s\in \{0,\infty \}, \end{aligned} \begin{aligned} \limsup _{t\rightarrow s} c(t) <1, \quad \limsup _{t\rightarrow s} d(t) <1, \quad s\in \{0,\infty \}, \end{aligned} $$(I-cU_\alpha )P_2^++(I-dU_\beta )P_2^-$$ $$(I-cU_\alpha )P_2^++(I-dU_\beta )P_2^-$$ $$L^p(\mathbb {R}_+)$$ $$L^p(\mathbb {R}_+)$$