Case 5

$u(\xi) = \vert\xi\vert^{-n-j}$ for $j=1,2,3,\dots$.

Any cutoff function $f$ gives a regularization by ``Taylor subtraction'', as follows:

$$\dst{\Tilde{R}_f u}{\phi} := \int_{\bR^n} \vert\xi\vert^{-n-... ...al!} D^\al\phi(0) \xi^\al f(\vert\xi\vert) \biggr) d^n\xi.$$

Again one finds that $\Tilde{R}_f u$ is not homogeneous, by a straightforward calculation along the lines of the previous Lemma. This can be simplified a little by the following observation [GVF]. One can find constants $c_\al$ for $\vert\al\vert\leq j$, such that the modified regularization $R_f u := \Tilde{R}_f u - \sum_{\vert\al\vert<j} c_\al D^\al\dl$ has a ``failure of homogeneity'' of the form

$$(R_f u)_t - t^{-n-j} R_f u = t^{-n-j} \log t \biggl( \sum_{\vert\al\vert=j} c_\al D^\al\dl \biggr).$$

That completes our study of the extensions of homogeneous functions to distributions on $\bR^n$. We need a remark about their Fourier transforms. Recall that the Fourier transformation $\sF$ preserves the Schwartz space $\sS(\bR^n)$, and by duality it also preserves $\sS'(\bR^n)$. If $u$ is a $\la$-homogeneous function on $\bR^n \setminus \{0\}$, its Fourier transform is $\sF u(\xi) := \int_{\bR^n} e^{-ix\xi} u(x) d^nx$, thus

$$(\sF u)_t(\xi) = \int_{\bR^n} e^{itx\xi} u(x) d^nx = t^{... ..._{\bR^n} e^{-iy\xi} u(y/t) d^ny = t^{-n-\la} \sF u(\xi).$$

It follows that $\sF$, and also the inverse transformation $\sF^{-1}$, take homogeneous functions (or distributions) of degree $\la$ to homogeneous functions (or distributions) of degree $(-n-\la)$.