I have a kind of vague question about the property of Calderon-Zygmund kernel. If given a k dimensional Calderon-Zygmund kernel $K$, can we say immediately that it is Lipschitz continuous except at the origin, $K(rx)=r^{-k}K(x)$ for all $r>0$ and $x\neq0$, $\int_{S^{k-1}}Kd\sigma=0$ where $\sigma$ is the surface measure of the unit sphere? Is there any reference for this? Or we actually need some more assumptions to make this claim? Thanks!
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Given those two properties, no, unless of course $k=1$. Each kernel is determined by its value on the unit sphere, and conversely any mean-value zero function on the unit sphere defines such a $K$. So just take a non-Lipschitz continuous mean value $0$ function on the unit sphere. However, for nice boundedness properties of $K$ as a convolution operator, we need some regularity assumption. I've seen $K \in C^1$ used, or an averaged form of regularity. – Eric Thoma Aug 17 '16 at 04:46