A rigorous way to define the singular integral

Suppose K(x) satisfies the following conditions

(1)|K(x)|\leq \frac{C}{|x|^n}

(2)|\nabla K(x)|\leq frac{C}{|x|^{n+1}}

(3)\displaystyle \int_{r_1<|x|<r_2}K(x)dx=0

Then we can define the singular integral operator

\displaystyle Tf=\int_{\mathbb{R}^n}K(x-y)f(y)dy

On the face of it, this integrand may not be Lebesgue integrable. There is a way to interpret this integral

\textbf{Thm: } Suppose (1),(2),(3) are satisfied, define

\displaystyle T_{\epsilon,R}f(x)=\int_{\epsilon<|x-y|<R}K(x-y)f(y)dy

Then ||Tf||_2\leq C||f||_2 with C independent of \epsilon,R,f. Moreover, one can prove T_{\epsilon,R}f is a cauchy sequence, then there exist g\in L^2(\mathbb{R}^n), such that

\displaystyle T_{\epsilon,R}f\to g in L^2

From this, we define Tf=g.

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