Singular integral

The archetypal singular integral operator is the Hilbert transform H. It is given by convolution against the kernel K(x)=1/(\pi x) for x in \mathbb R. More precisely, : H(f)(x) = \frac{1}{\pi}\lim_{\varepsilon \to 0} \int_{|x-y|>\varepsilon} \frac{1}{x-y}f(y) \, dy. The most straightforward higher dimension analogues of these are the Riesz transforms, which replace K(x)=1/x with : K_i(x) = \frac{x_i}{|x|^{n+1}} where i = 1, ..., n and x_i is the i-th component of x in \mathbb R^n. All of these operators are bounded on L^p and satisfy weak-type (1,1) estimates. ==Singular integrals of convolution type==

A singular integral of convolution type is an operator T defined by convolution with a kernel K that is locally integrable on R^n\setminus \{0\}, in the sense that {{NumBlk|:|T(f)(x) = \lim_{\varepsilon \to 0} \int_ |K(x-y) - K(x)| \, dx \leq C. Then it can be shown that T is bounded on L^p(\mathbb R^n) and satisfies a weak-type (1,1) estimate. Property 1. is needed to ensure that convolution () with the tempered distribution p.v. K given by the principal value integral :\operatorname{p.v.}\,\, K[\phi] = \lim_{\epsilon\to 0^+} \int_{|x|>\epsilon}\phi(x)K(x)\,dx is a well-defined Fourier multiplier on L^2. Neither of the properties 1. or 2. is necessarily easy to verify, and a variety of sufficient conditions exist. Typically in applications, one also has a cancellation condition : \int_{R_1 0 which is quite easy to check. It is automatic, for instance, if K is an odd function. If, in addition, one assumes 2. and the following size condition : \sup_{R>0} \int_{R then it can be shown that 1. follows. The smoothness condition 2. is also often difficult to check in principle, the following sufficient condition of a kernel K can be used: • K\in C^1(\mathbf{R}^n\setminus\{0\}) • |\nabla K(x)|\le\frac{C}{|x|^{n+1}} Observe that these conditions are satisfied for the Hilbert and Riesz transforms, so this result is an extension of those result. ==Singular integrals of non-convolution type==

These are even more general operators. However, since our assumptions are so weak, it is not necessarily the case that these operators are bounded on L^p. Calderón–Zygmund kernels A function K: \mathbb R^n \times \mathbb R^n \to \mathbb R is said to be a Calderón–Zygmund kernel if it satisfies the following conditions for some constants C>0 and \delta > 0. M_{b_2}TM_{b_1} is weakly bounded; T(b_1) is in BMO; T^t(b_2), is in BMO, where T^t is the transpose operator of T. ==See also==