Calderón-Zygmund operators with positive kernel

Question

Let $T$ be a Calderón-Zygmund operator. That is, $T$ maps $L^2(\mathbb{R}^d)$ to itself and satisfies the representation formula $$ Tf(x) = \int_{\mathbb{R}^d}K(x,y)f(y)\, dy $$ for all $f \in L^2$ with compact support and $x \notin \operatorname{supp} f$, for some kernel $K$ defined outside the diagonal and satisfying $$ |K(x,y)| \leq C|x-y|^{-d}, $$ $$ \int_{|x-y|>2|y-z|} |K(x,y)-K(x,z)|\, dx \leq C $$ and $$ \int_{|x-y|>2|x-w|} |K(x,y)-K(w,y)|\, dy \leq C. $$

If $\psi$ is any nonnegative $C^\infty_c$ function then $f \mapsto f \ast \psi$ is cleary a Calderón-Zygmund operator with positive kernel.

My question is, do there exist Calderón-Zygmund operators with positive kernels which are not so well-behaved. For example: does there exist a Calderón-Zygmund operator with a positive kernel which is not locally $y$-integrable near the diagonal $\{x=y\}$?

Answer 1

No, the fact that some non-integrable kernels yield bounded operators is due to cancellation. Suppose $K$ is a nonnegative kernel which is nonintegrable with respect to $y$ along the diagonal. Let $f$ be the characteristic function of an open ball $B$. Then for every $x\in B$ we have $$Tf(x)=\int_B K(x,y)\,dy = \infty$$ which is rather far from having $Tf\in L^2$.

Answer 2

A natural positive kernel, that satisfies the Calderón-Zygmund condition, but retains some singularity, could be the (positive) Bergman kernel in the upper half-plane. Namely, for points $z=x+iy$ and $w=u+iv$ in $\mathbb{R}^2_+$ (ie, $y,v>0$), define $$ K(z,w)=\frac1{|z-\overline{w}|^2}=\frac1{(x-u)^2+(y+v)^2}. $$ For each fixed $z$, one can check that $\int_{\mathbb{R}^2_+}K(z,w)\,dw=\infty$. As a consequence, the corresponding operator $T$ is not bounded in $L^1(\mathbb{R}^2_+)$ (just test with $f=\chi_{B_{1/2}(i)}$, so that $Tf(z)\approx \frac1{x^2+(y+1)^2}$ if $|x|\geq1$). On the other hand, it is well known (and not hard to show) that $T$ is bounded in $L^p(\mathbb{R}^2_+)$ for all $1<p<\infty$ (for example, one can use the Schur test). Also, $K(z,w)=K(w,z)$ satisfies the usual Calderón-Zygmund conditions, namely $$ |K(z,w)|=\frac1{|z-\overline w|^2}\leq \frac1{|z-w|^2} $$ and, for $|z-w_0|\geq 2|w-w_0|$ $$ \Big|K(z,w)-K(z,w_0)\Big|\leq \frac{C\,|w-w_0|}{|z-w_0|^3}. $$
This last condition is known to imply the integral condition stated above. In particular, the operator $T$ will also be bounded from $L^1\to L^{1,\infty}$, a fact which cannot be easily proved with elementary techniques (like the Schur test). The kernel, however, is not "singular at the diagonal" as stated in the original question, since $K(z,\cdot)\in L^1(B_R(z))$, for all $z$. In this case, the singularity occurs at infinity.