Kernel of a double Riesz transform in three dimensions

Statement

In three dimensions, the double Riesz transform satisfies, for a test function $f$ ,

R_iR_jf(x)=\operatorname{p.v.}\int_{\mathbb R^3} \frac{3z_iz_j-\delta_{ij}|z|^2}{4\pi|z|^5}f(x-z)\,dz -\frac{\delta_{ij}}3 f(x).

Here $\delta_{ij}$ is the Kronecker delta. In particular its off-diagonal kernel obeys $|K_{ij}(z)|\le C|z|^{-3}$ .

Distributional Hessian

Differentiate $\Gamma(z)=1/(4\pi|z|)$ twice outside zero. The displayed kernel results. Its angular mean is zero because the spherical average of $z_iz_j/|z|^2$ is $\delta_{ij}/3$ , making the principal value converge for smooth inputs. Integration by parts across a small sphere contributes $-\delta_{ij}\delta_0/3$ to $\partial_i\partial_j\Gamma$ . Fourier transformation yields the symbol $-\xi_i\xi_j/|\xi|^2$ , proving the formula. Summing $i=j$ gives $\sum_iR_i^2=-I$ , which also checks the local term's sign.

Operator bounds

The off-diagonal kernel and its first derivatives have the Calderón–Zygmund bounds. The Fourier symbol has absolute value at most one, giving $L^2$ boundedness. The Calderón–Zygmund theorem therefore applies for $1<p<\infty$ .