# Regularizations of general singular integral operators

### Constanze Liaw

Baylor University, Waco, USA### Sergei Treil

Brown University, Providence, USA

## Abstract

In the theory of singular integral operators significant effort is often required to rigorously define such an operator. This is due to the fact that the kernels of such operators are not locally integrable on the diagonal $s=t$, so the integral formally defining the operator $T$ or its bilinear form $\langle Tf, g \rangle$ is not well defined (the integrand in not in $L^1$) even for “nice” $f$ and $g$. However, since the kernel only has singularities on the “diagonal” $s=t$, the bilinear form $\langle Tf, g \rangle$ is well defined, say, for bounded compactly supported functions with separated supports.

One of the standard ways to interpret the boundedness of a singular integral operators is to consider the *regularized kernel*

where the cut-off function $m$ is $0$ in a neighborhood of the origin, so the integral operators $T_\varepsilon$ with kernel $K_\varepsilon$ are well defined (at least on a dense set). Then instead of asking about the boundedness of the operator $T$, which is not well defined, one can ask about uniform boundedness (in $\varepsilon$) of the regularized operators $T_\varepsilon$.

For the standard regularizations one usually considers *truncated* operators $T_\varepsilon$ with $m(s) =\mathbf{1}_{[1, \infty)} (|s|)$, although smooth cut-off functions were also considered in the literature.

The main result of the paper is that for a wide class of singular integral operators (including the classical Calderón–Zygmund operators in nonhomogeneous two weight settings), the so called restricted $L^p$ boundedness, i.e., the uniform estimate

for bounded compactly supported $f$ and $g$ with separated supports implies the uniform $L^p$-boundedness of regularized operators $T_\varepsilon$ for any reasonable choice of smooth cut-off function $m$. For example, any $m \in C^\infty(\mathbb{R}^N)$}, $m\equiv 0$ in a neighborhood of $0$, and such that $1-m$ is compactly supported would work.

If the kernel $K$ satisfies some additional assumptions (which are satisfied for classical singular integral operators like the Hilbert transform, Cauchy transform, Ahlfors–Beurling transform, and generalized Riesz transforms), then the restricted $L^p$ boundedness also implies the uniform $L^p$ boundedness of the classical truncated operators $T_\varepsilon$ ($m(s) =\mathbf{1}_{[1, \infty)} (|s|)$).

## Cite this article

Constanze Liaw, Sergei Treil, Regularizations of general singular integral operators. Rev. Mat. Iberoam. 29 (2013), no. 1, pp. 53–74

DOI 10.4171/RMI/712