Extending $1/|x|$ to a homogeneous distribution

Question

On $\mathbb{R} \setminus \{0\}$, $1/|x|$ is a homogeneous function of degree $-1$. Is there a way to extend it to a homogeneous distribution on all of $\mathbb{R}$?

The standard way I've found of extending $1/|x|$ to a distribution on all of $\mathbb{R}$ is (see e.g. here, here, Chapter 1 Section 4.3 of Gel’fand and Shilov, "Generalized functions. Vol. I: Properties and operations", or Exercise 13 of this) is by writing $\mathcal{P} (1/|x|) = (\mathrm{sgn}(x)\ln|x|)'$ as a distributional derivative, so its action on any test function $\phi(x)$ is

$$ \begin{align} \left\langle \mathcal{P}\frac{1}{|x|}, \phi(x)\right\rangle &= -\int_{-\infty}^\infty \phi'(x) \, \mathrm{sgn}(x) \ln|x| \, dx \\ &= \int_{|x| \leq r} \frac{\phi(x)-\phi(0)}{|x|} \, dx + \int_{|x|>r} \frac{\phi(x)}{|x|} \, dx + 2 \phi(0) \ln r, \end{align} $$ where $r > 0$ is arbitrary. (It looks like folks often set $r = 1$ to get rid of the final term in the second line.)

But $\mathcal{P}(1/|x|$) is not homogeneous. In general, a homogeneous function $f(x)$ of degree $\lambda$ has the property that $f(ax) = a^\lambda f(x)$ for any $a> 0$. The analogous property for distributions is $\langle f(x), \phi(x/a) \rangle = a^{\lambda+1} \langle f(x), \phi(x)\rangle$ for any test function $\phi$. But for $\mathcal{P}(1/|x|)$, we have

$$ \begin{align} \left\langle \mathcal{P}\frac{1}{|x|}, \phi\left(\frac{x}{a}\right)\right\rangle &= -\int_{-\infty}^\infty \frac{1}{a}\phi'\left(\frac{x}{a}\right) \, \mathrm{sgn}(x) \ln|x| \, dx \\ &= -\int_{-\infty}^\infty \phi'(x) \, \mathrm{sgn}(x) \ln|a x| \, dx \\ &= -\int_{-\infty}^\infty \phi'(x) \, \mathrm{sgn}(x) \ln|x| \, dx - \ln a \int_{-\infty}^\infty \phi'(x) \, \mathrm{sgn}(x) \, dx \\ &= \left\langle \mathcal{P}\frac{1}{|x|}, \phi(x)\right\rangle + 2\ln a \, \phi(0), \end{align} $$

and hence

$$ \mathcal{P} \frac{1}{|ax|} = \frac{1}{a} \mathcal{P}\frac{1}{|x|} + \frac{2 \ln a}{a} \, \delta(x). $$

Based on Chapter 1, Section 4.1 of Gel’fand and Shilov, this result suggests $\mathcal{P}(1/|x|)$ would be deemed an associated homogeneous distribution of degree $-1$.

So my question of whether $1/|x|$ can be extended to a homogeneous distribution amounts to asking whether or not an associated homogeneous distribution of degree $-1$ can be modified to turn it into a homogeneous distribution (of the same degree). I'm sure the answer is likely somewhere in Gel’fand and Shilov, but I couldn't find anything definitive.

Answer 1

No, there is no homogeneous distribution on $\mathbb{R}$ that agrees with $1/|x|$ on $\mathbb{R} \setminus \{0\}$.

You've already found one extension $f(x) = \mathcal{P} (1/|x|)$ and showed that it's not homogeneous. For any other extension $g$, the difference between $g$ and $f$ is only at the origin, i.e. $g - f$ must be a linear combination of the Dirac delta and its derivatives, $\delta, \delta', \delta'', ....$ And $f$ plus any such extra terms also isn't homogeneous. (If we explicitly check, we still get the term involving $\ln a$, just like for $\mathcal{P} (1/|ax|)$.)