Is the Cauchy principal value "invariant" under change of variables?

Question

Let $f \in C^{\gamma}_c(\mathbb{R}) $. 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. 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 \\ &=& \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}

Answer 1

It should not hold as stated, though integrating over $\lvert G(v) - G(w) \rvert > \delta$ in the right integral will work. It is instructive to try to prove it to see how to construct a counterexample.

Define the truncated kernel $\tilde{K}(x) = K(x) \chi_{[\lvert{x}\rvert \geq 1]}$. We know $K$ is of the form $\Omega(x/\lvert x \rvert) \lvert x \rvert^{-n}$ for $\Omega : S^{n-1} \to \mathbb{R}$ smooth with mean value $0$. For any function $f$, define $f_\varepsilon(x) = \varepsilon^{-n}f(x/\varepsilon)$. We may apply standard change of variables: $$ \int \tilde{K}_\varepsilon(x-y) f(y) dy = \int \tilde{K}_\varepsilon(G(v) - G(w)) f(G(w)) \lvert \det \nabla G(w) \rvert dw. $$ Note $\tilde{K}_\varepsilon(x) = K(x) \chi_{[\lvert x \rvert > \varepsilon]}$. Now the problem can be seen: the sets $\{ \lvert G(v) - G(w) \rvert > \varepsilon \}$ are symmetric w.r.t the kernel arguments in the right integral, but the set $\{ \lvert v - w \rvert > \varepsilon \}$ is not. Since singular integrals rely on cancellation, they are sensitive to this difference, and we can use this to find a counterexample.

I now construct a counterexample. There is some computational difficulty here, so be cautious taking my word for the following. Set $n = 2$ for simplicity. Let $G(x_1,x_2)= (x_1, 2x_2)$. Let $\Omega$ be a smooth function on the unit circle that approximates $$ H(e^{i\theta}) = \begin{cases} 2 \quad &\text{if } \theta \in (-\pi/4,\pi/4) \\ -2 \quad &\text{if } \theta \in (\pi/4,3\pi/4) \\ -1 \quad &\text{if } \theta \in (3\pi/4,5\pi/4) \\ 1 \quad &\text{if } \theta \in (5\pi/4,7\pi/4). \end{cases} $$ Let $K$ be of the form specified with $\Omega$. Let $v = x = (0,0)$. Let $f$ be $1$ on the unit ball. Then $$ \int_{[1 > \lvert w \rvert > \varepsilon]} K(w_1,2w_2)dw $$ will diverge to $-\infty$ as $\varepsilon \to 0$. This is because the kernel $K(\cdot,2\cdot)$ does not have average value $0$ over spheres, so of course the associated principal value integral can not converge. Switching to polar coordinates makes this particularly clear.

Please let me know if anything should be clarified.

Answer 2

No, it's not invariant. The Cauchy principal value is crucially depended on the notion of the unit sphere and that you get closer to the singularity in all directions uniformly. If your diffeomorphism weighs different directions differently, the contributions to the singularity from different directions no longer cancel each other out.

Easiest example: $n=1$ and $G(x)=\begin{cases} -x^2 & x\leq 0 \\ 2x^2 & x\geq 0\end{cases}$. Note that $G'(x)=\begin{cases} -2x & x\leq 0 \\ 4x & x\geq 0\end{cases}$ so it really weighs the positive and negative direction differently). Let $K(x)=\frac{1}{x}$ and $f(x)=\chi_{[-1,+1]}$ or something easy like that (yeah, I know you want something continuous, but that obviously doesn't help. If the claim were true for continuous $f$, it would also be true for some $L^1$-space by completion)