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Let $\mathcal{S}$ be the space of Schwartz test functions on $\mathbb{R}^N$ and let $\mathcal{S'}$ be the space of tempered distributions on $\mathbb{R}^N$. Let $\mathcal{O}$ be the space of unitary linear transformations of $\mathbb{R}^N$. We say that $f\in\mathcal{S'}$ is radial if $$\forall O\in\mathcal{O}, \forall \varphi\in\mathcal{S}, f(\varphi\circ O) = f(\varphi).$$ We say that $f\in\mathcal{S'}$ is positive homogeneous of degree $\alpha\in\mathbb{R}$ if $$\forall\lambda>0, \forall \varphi\in\mathcal{S}, f\left(x\mapsto\frac{1}{\lambda^N} \varphi\left(\frac{x}{\lambda}\right)\right) = \lambda^{\alpha}f(\varphi).$$

Now, for $\alpha\in(-N,0)$, the function $$f_\alpha :x\mapsto |x|^\alpha$$ is locally integrable of moderate growth and represents, by integral pairing, a radial, positive homogeneous of degree $\alpha$, element of $\mathcal{S}'$.

I proved that every element in $\mathcal{S}'$ that is radial, positive homogeneous of degree $\alpha\in(-N,0)$, and representable by integral pairing by a locally integrable function of moderate growth is proportional to $f_\alpha$. What if we remove the hypothesis of representability? I.e.

For $\alpha\in(-N,0)$, are all the radial, positive homogeneous of degree $\alpha$ elements in $\mathcal{S}'$ proportional to $f_\alpha$?

Bob
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1 Answers1

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The answer is yes. More generally, for any degree of homogeneity $\alpha\in\mathbb{R}$, there is up to constant multiples a unique radial element of $S'(\mathbb{R}^d)$ with that degree of homogeneity. You can find a detailed proof in my former student's thesis https://libraetd.lib.virginia.edu/public_view/2r36tx772

In particular you are missing Prop 2.5 (Euler's equation in the sense of distributions) and Cor 2.2 in that thesis (solving the 1d equation $f'=0$).

Abdelmalek Abdesselam
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