## 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|

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|

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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