Riesz transform

Remarks

With $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} \, dx$, the Fourier transform of $R_j f$ is $$\widehat{R_j f}(\xi) = -i \frac{\xi_j}{\abs{\xi}} \hat f(\xi).$$ So $\sum_{j=1}^n R_j^2 = -I$ on $L^2(\R^n)$, and each $R_j$ is bounded on $L^2$ with norm $1$. Each $R_j$ extends to a bounded operator on $L^p(\R^n)$ for $1 < p < \infty$.

Example: for $n = 1$, $c_1 = \Gamma(1)/\pi = 1/\pi$ and the kernel is $\frac{1}{\pi}\frac{y}{\abs{y}^2} = \frac{1}{\pi y}$. So $R_1$ is the Hilbert transform, with multiplier $-i\sgn\xi$.

Non-example: the kernel $\abs{y}^{-n}$, with the same homogeneity but without the odd factor $y_j/\abs{y}$, does not define a principal value operator. It has no cancellation, and $\int_{\varepsilon < \abs{y} < 1} \abs{y}^{-n} \, dy$ grows like $\log(1/\varepsilon)$.

The kernel $c_n y_j/\abs{y}^{n+1}$ is odd, homogeneous of degree $-n$, and smooth away from $0$, which makes $R_j$ a singular integral operator.