\(L^p\) boundedness of a class of singular integral operators with rough kernels (Q2776894)

Define the singular integral operator \(T_{\Phi}\) by NEWLINE\[NEWLINE T_{\Phi} = \text{p.v.} \int_{\mathbb{R}^n} f(x - \Phi(|y |) y') \frac{\Omega (y')}{ |y |^n} h( |y |) dy NEWLINE\]NEWLINE where \(\Omega\) is a homogeneous function of degree zero on \(\mathbb{R}^n\) and satisfies the cancellation condition \(\int_{S^{n-1}} \Omega (y') dy' =0\). NEWLINENEWLINENEWLINE\textit{J. Namazi} [Proc. Am. Math. Soc. 96, 421-424 (1986; Zbl 0585.42017)], showed that if \(\Phi = t, h \in L^{\infty}(\mathbb{R}^{+})\) and \(\Omega \in L^q(S^{n-1})\) for some \(1< q <\infty\), then \(T_{\Phi}\) is a bounded operator on \(L^p(\mathbb{R}^n)\) where \(1<p < \infty\). \textit{D. Fan} and \textit{Y. Pan} [Am. J. Math. 119, No. 4, 799-839 (1997; Zbl 0899.42002)], considered the case when \(\Omega \in H^1(S^{n-1})\). NEWLINENEWLINENEWLINEThe main result of this paper is the following. Suppose that \(\Phi\) satisfies certain growth conditions. If \(h \in L^{\infty}(\mathbb{R}^{+})\) and \(\Omega\) belongs to some block space \(B^{0,0}_q\), then \(T_{\Phi}\) is a bounded operator on \(L^p(\mathbb{R}^n)\) where \(1<p < \infty\). NEWLINENEWLINENEWLINEThe authors also consider the maximal singular integral and the oscillatory singular integral.