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

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.
