Let $f \in C^{\gamma}_c(\mathbb{R}^n) $. Let $K:\mathbb{R}^n \backslash \{\vec{0}\} \rightarrow \mathbb{R}^n$ be a singular integral kernel with the following properties:

1) K smooth everywhere except at $\vec{0}$

2) K homogeneous of degree $-n$, in particular $\|K(x)\| \leq \frac{c}{\|x \|^{n}}$

3) K has mean value zero on the unit sphere, ie $\int_{\|x\|=1}K(x)dS=0$

I was wondering if the Cauchy principal value of the convolution of $K$ with $f$ is "invariant" under a change of variables in the following sense. That is, for a $C^1$ diffeomorphism $G: \mathbb{R}^n \rightarrow \mathbb{R}^n$, denoting $y=G(w)$ and $x=G(v)$, do we have:

\begin{eqnarray} \text{P.V.} \int_{\mathbb{R}^n} K(x-y)f(y)dy &\equiv& \lim_{\delta \searrow 0} \int_{\|x-y\|> \delta} K(x-y)f(y)dy \end{eqnarray} \begin{eqnarray} &=& \lim_{\delta \searrow 0} \int_{\|v-w\|> \delta} K \left(x-G(w) \right)f \left( G(w) \right) \left|\det \nabla G(w) \right| dw \quad \text{?} \end{eqnarray}


1 Answer 1


Changing variables, we have $$(pv(K)\ast f)(x)=\lim_{\delta\searrow 0} \int_{\vert x-y\vert>\delta}K(x-y) f(y) dy=\lim_{\delta\searrow 0} \int_{\vert x-G(w)\vert>\delta}K(x-G(w)) f(G(w))\vert \nabla G(w)\vert dw, $$ so that $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert G(\nu)-G(w)\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw. $$ Now, if $G$ is globally Lipschitz continuous, i.e. $\nabla G\in L^\infty$, we have $$ \{w,\vert G(\nu)-G(w)\vert>\delta\}\subset\{w,\Vert\nabla G\Vert_{L\infty}\vert\nu-w\vert>\delta\}. $$ Since $\nabla G\circ \nabla G^{-1}=Id$, we have $ \Vert\nabla G\Vert\Vert \nabla G^{-1}\Vert\ge 1. $ As a result, if $G$ is globally bi-Lipschitz continuous (i.e. $G,G^{-1}$ are both globally Lipschitz continuous), we find $\alpha\ge 1$ such that $$ \{w,\vert w-v\vert\le \delta/\ \alpha\}\subset \{w,\vert G(w)-G(v)\vert\le \delta\}\subset \{w,\vert w-v\vert\le \alpha\delta\}.\tag E $$ We may now consider what is now a general singular integral, and not only a Fourier multiplier, that is the operator with kernel $$ k(v,w)=K(G(v)-G(w))\vert\nabla G(w)\vert. $$ We have indeed the following estimates $$ \vert k(v,w)\vert\lesssim\vert v-w\vert^{-n},\quad\vert\partial_vk(v,w)\vert+\vert\partial_wk(v,w)\vert\lesssim\vert v-w\vert^{-n-1} \tag {CZ}$$ $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert G(\nu)-G(w)\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw. $$ Using the embeddings (E), we see that that $\{w,\vert G(w)-G(v)\vert> \delta\}$ is the union of a set $\{w,\vert w-v\vert> \alpha\delta\}$ with a set where $\vert G(w)-G(v)\vert\sim \delta\sim \vert v-w\vert$ with volume $\delta^{n}$ on which $K$ is of size $\delta^{-n}$. From the Lebesgue differentiation theorem, this part of the integral converges as well. $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert\nu-w\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw+\Omega f, $$ where $\Omega$ is $L^p$ bounded for $p\in(1,+\infty)$. So when you change variables in a singular integral appearing as a Fourier multiplier or a convolution, you get a more general type of operator, a Calder\'on-Zygmund type of operator with a kernel satisfying (CZ).

  • $\begingroup$ G is indeed globally bi-Lipschitz. But unfortunately I don't understand the last step. $\endgroup$
    – Jim Beech
    May 19, 2013 at 20:42
  • $\begingroup$ With the bi-Lipschitz continuity hypothesis, for every $\delta>0$, you get $\epsilon,\sigma >0$ such that $$ \\{\vert G(v)-G(w)\vert\le \sigma\\}\subset\\{\vert v-w\vert\le \delta\\}\subset\\{\vert G(v)-G(w)\vert\le \epsilon\\} $$ implying that last step. $\endgroup$
    – Bazin
    May 20, 2013 at 9:58
  • $\begingroup$ Wouldn't we then be assuming \begin{equation} \lim_{\delta \searrow 0} \int_{\|G(v)-G(w) \|>\delta} \|K(x-y)f(y) \|dy \end{equation} exists in order to set up the inequality? $\endgroup$
    – Jim Beech
    May 20, 2013 at 14:34
  • $\begingroup$ You are right, I have changed my answer and added some explanations on Calderon-Zygmund operators. $\endgroup$
    – Bazin
    May 20, 2013 at 21:28
  • $\begingroup$ One last question. Is $\Omega$ an operator standing for \begin{equation} \lim_{\delta \searrow 0} \int_{\{|G(v)-G(w)|>\delta\}\cap\{|v -w|< \alpha \delta\}} K(G(v)-G(w))f(G(w))|\det \nabla G(w)|dw \end{equation} $\endgroup$
    – Jim Beech
    May 20, 2013 at 23:21

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