Case 1

Suppose $d - j > -n$. Then $k := j - d - n < 0$, and $R_f p_{d-j}(x,\xi)$ is homogeneous of degree greater than $-n$, so $h_k(x, z)$ is homogeneous of degree $k$. These terms have no failure of homogeneity.

Before examining the other two cases, we return to the context of functions on $\bR^n \setminus \{0\}$, and look first at $w_0(z) := (2\pi)^n \sF^{-1}(R_f \vert\xi\vert^{-n})$. Since () holds with $u(\xi) = \vert\xi\vert^{-n}$ for $\xi \neq 0$, and since $(2\pi)^n \sF^{-1}(\dl) = 1$, we get

$$t^{-n} w_0(z/t) - t^{-n} w_0(z) = \Om_n t^{-n}\log t \word{for} t > 0,$$

or more simply,

$$w_0(z/t) - w_0(z) = \Om_n \log t.$$ (23)

Notice that $C = w_0(z/\vert z\vert)$ is a constant, because $w_0$ is rotation-invariant. Substituting $t := \vert z\vert$ in () gives

$$w_0(z) = C - \Om_n \log \vert z\vert,$$ (24)

so that $w_0$ ``diverges logarithmically''. We can suppress the constant term if we replace $R_f\vert\xi\vert^{-n}$ by $R_f\vert\xi\vert^{-n} - C\dl$, since we must then subtracting the constant $C$ from the inverse Fourier transform.

For $j = 1,2,\dots$, we define $w_j(z) := (2\pi)^n \sF^{-1}(R_f \vert\xi\vert^{-n-j})$. A similar analysis shows that $w_j(z) = q_j(z) - r_j(z)\log \vert z\vert$, where both $q_j$ and $r_j$ are homogeneous of degree $j > 0$. In this case, $w_j(z)$ remains bounded as $z \to 0$.

We now return to the examination of the terms $h_{j-d-n}$ in the integral kernel $k(x,y)$.