# Nonintegrable inverse powers as distributions

I am working through Lieb/Loss's "Analysis", and have been stuck on one of the problems for a while;

Suppose we are on $\mathbb{R}^n$ and define $f(x) = |x|^{-n}$. This is not a locally integrable function. However if $\phi \in C_c^{\infty}(\mathbb{R}^n)$ is a function vanishing at the origin, we can still define the action of $f$ on $\phi$ as a distribution: \begin{align*} T_f (\phi) = \int_{\mathbb{R}^n} \frac{\phi(x)}{|x|^n} dx \end{align*} Which is always well defined when $\phi$ vanishes at the origin, as can be seen by converting to polar coordinates for example. However Lieb states that there are many actual distributions $T$ that agree with $T_f$ on (test) functions that vanish at the origin, and he wants the reader to find all of them.

I have tried a couple of things so far. In one-dimension, I noticed (i'm pretty sure at least) that $T_f$ agrees with the derivative of the distribution given by $-\ln(1/|x|)$. However i'm really stuck as to how to classify all such distributions, or how to generalize to $n$ dimensions.

Thanks for the help in advance.

• This question is more suitable at MathStackExchange. Anyways, there is a characterization of distributions supported at a point. – timur Apr 15 '13 at 3:06
• I am not so sure this is below MO level, although I admit I don't know the book of L and L so don't know if they are guiding the reader to a solution of this question. – Yemon Choi Apr 15 '13 at 3:50
• They don't say anything that wasn't mentioned above, save for a hint about using the following theorem: If $T$ is zero on the nullspaces of the distributions $S_1 \dots S_m$ then $T$ is a linear combination of the $S_i$. I am familiar with the characterization of point supported distributions as linear combinations of the delta's derivatives, but those that agree with $T_f$ are not point supported, i can't see how to apply that. – Mesoscopic_P Apr 15 '13 at 4:57
• @Mesoscopic_P, you are almost there. If $T_f$ and $T'_f$ are two candidates, what is the support of $T_f-T'_f$? – Igor Khavkine Apr 15 '13 at 7:38
• Oh, gotcha. So Using Shanlin's answer from below, every such $T$ is just a particular extension plus a linear combination of $\delta$ and its derivatives, since the support of the difference is just the origin. But any derivative of the delta other than zeroth order clearly doesn't work, so we end up with just a multiple of $\delta$ in the linear combination. Thanks! – Mesoscopic_P Apr 15 '13 at 8:06

The related topic here is the homogeneous distribution on $\mathbb{R}^n\0$ and its extension to $\mathbb{R}^n$. In your case $T_{f}$ is a homogeneous distribution on $\mathbb{R}^n\0$ of degreee $-n$. And it's always possible to extend it to a distribution on $\mathbb{R}^n$,which may not be homogeneous any more. In fact,you can define the extension $\dot{T_{f}}$ as follows $$\dot{T_{f}}(\phi)=T_{f}(\psi R_{-n}\phi), \quad \phi\in C_{0}^{\infty}({\mathbb{R}^n})\\ R_{-n}\phi=<t_{+}^{-1},\phi(tx)>,\quad x\neq 0$$ where $\psi$ is a fixed function in $C_{0}^{\infty}({\mathbb{R}^n}\0)$ satisfies that $\int_{0}^{+\infty}\frac{\psi(tx)}{t}dt=1$.
Now one can see that when restricting test functions in $C_{0}^{\infty}({\mathbb{R}^n}\0)$, then $\dot{T_{f}}=T_{f}$, and the extension depends on the choice of $\psi$. However, all such extension has the above form. And in one dimensin, it's particularly clear. You can find them in chaper $3$ of Hormander's book "The Analysis of Linear Parital Differential Operators"