Case 3

Suppose $\la = -n$, and that $\int_{\bS^{n-1}} v(\om) \sg = 0$.

We define a distribution $\Ppart u$ by the following trick. Let $f\: [0,\infty) \to \bR$ be a cutoff function, such that:

$$f(t) := \begin{cases} 1 &\word{if} 0 \leq t \leq \half, \\ 0 &\word{if} t \geq 1, \end{cases}$$

and $f$ decreases smoothly from 1 to 0 on $[\half, 1]$. Replace the test function $\phi$ by $\phi(\xi) - \phi(0) f(\vert\xi\vert)$, and put

$$\dst{\Ppart u}{\phi} := \int_{\bR^n} u(\xi)\bigl(\phi(\xi) - \phi(0) f(\vert\xi\vert)\bigr) d^n\xi.$$ (21)

If $g(t)$ is another cutoff function with the same properties, the right hand side of this formula changes by

$$\int_{\bR^n} u(\xi) \phi(0) \bigl(f(r) - g(r)\bigr) d^n\xi... ... \sg \int_{1/2}^1 \bigl(f(r) - g(r)\bigr) \frac{dr}{r} = 0,$$

since $u(\xi) d^n\xi = r^{-n} v(\om) \sg r^{n-1} dr = v(\om) \sg dr/r$ by homogeneity. Thus $\dst{\Ppart u}{\phi}$ is independent of the cutoff chosen. Indeed, since

$$\int_{\vert\xi\vert>\eps} u(\xi) f(\vert\xi\vert) d^n\xi = \int_\eps^1 f(r) \frac{dr}{r} \int_{\bS^{n-1}} v(\om) \sg = 0,$$

for any $\eps > 0$, we get another formula for $\Ppart u$:

$$\dst{\Ppart u}{\phi} = \lim_{\eps\downarrow 0} \int_{\vert\xi\vert>\eps} u(\xi) \phi(\xi) d\xi.$$

Therefore, $\Ppart u$ is just the ``Cauchy principal part'' of $u$ at $\xi = 0$.