Find extension for distribution defined on $C_c^\infty(\mathbb R^n \setminus \{0\})$

Question

Let $\mathcal D(\mathbb R^n \setminus \{0\}) := C_c^\infty(\mathbb R^n \setminus \{0\}), f(x) := \frac{1}{\lvert x \vert^n}$ and consider the function $$T_f : \mathcal D(\mathbb R^n \setminus \{0\}) \to \mathbb C; \quad \phi \mapsto \int_{\mathbb R^n} \frac{\phi(x)}{\lvert x \rvert^n} \, dx.$$ Find an explicit distribution $T\in \mathcal D'(\mathbb R^n)$ such that $T\restriction_{\mathcal D(\mathbb R^n \setminus \{0\})} = T_f$.

I first tried to do it when $n=1$ for simplicity. But even then I had no good idea how to start. My idea was to use integration by parts or Taylor to "weaken the singularity" and make the integral well-defined for all test functions, not necessarily vanishing at $0$, but I could not quite get that to work. I believe the solution should be something involing a delta distribution at $0$. Any help appreciated!

Answer 1

Try the distribution $T$ defined by \begin{align*} T(\phi)=\lim_{N\rightarrow\infty}\int_{1/N\leq|x|\leq 1}\dfrac{\phi(x)-\phi(0)}{|x|^{n}}dx+\int_{|x|\geq 1}\dfrac{\phi(x)}{|x|^{n}}dx. \end{align*}

Answer 2

Define $f : \mathbb{R}^n \to \mathbb{R}$ by $f(0)=0$ and for $x \neq 0$, $$ f(x) = \begin{cases} \frac{1}{2-n} |x|^{2-n} \ln|x| & (n\neq 2) \\ \frac{1}{2}(\ln |x|)^2 & (n=2) \end{cases} $$

Then $f \in L^1_{\text{loc}}(\mathbb{R}^n)$ and therefore defines a distribution $u$ on $\mathbb{R}^n$. Also, $\Delta f(x) = |x|^{-n}$ for $x \neq 0$.

Thus we can take $T = \Delta u$.