Distributional derivative of $\log(|x|)$ in higher dimensions

Question

I was attempting to compute the distributional Laplacian of the function $|x|^{2-n}$ in $\mathbb{R}^n$ for $n \geq 3$.

I have reduced it down to showing that, setting $r = |x|$, the identity $$\partial_r \log r = C(n)r^{n-1}\delta$$ holds in the distributional sense, where $C(n)$ is some dimensional constant and $\delta$ is the Dirac delta at the origin.

I was unable to find this computation anywhere before for this higher dimensional case. The one dimensional case is almost tautologically the principal value distribution of $1/x$, which is certainly not a Dirac delta. Therefore, I'd at least like some idea of why it would become a Dirac delta in higher dimensions, if not a full computation.

Answer 1

Proposition: If $n\geq 3$, then $\Delta\left(|x|^{2-n}\right)=C_n\delta$ on $\mathbb{R}^n$, where $C_n=-(n-2)A(S^{n-1}).$

I'll give a sketched proof, and I can fill in details tomorrow morning, if need be.

Proof: Let $u\in C_0^\infty(\mathbb{R}^n)$, and set $\Omega_\epsilon=\mathbb{R}^n\setminus B_\epsilon(0).$ Then, \begin{align*}\langle \Delta u,|x|^{2-n}\rangle &=\lim_{\epsilon\rightarrow 0}\int\limits_{\Omega_\epsilon}\Delta u\cdot |x|^{2-n} dx\\ &=\lim_{\epsilon\rightarrow 0}\int\limits_{\Omega_\epsilon}(\Delta u|x|^{2-n}-u\Delta |x|^{2-n})dx,\end{align*} using that $\Delta|x|^{2-n}=0$ for $x\neq 0.$ Applying Green's formula, we get that the above equals $$-\lim_{\epsilon\rightarrow 0}\int\limits_{\partial B_\epsilon(0)}(\epsilon ^{2-n}\partial_r u-(2-n)\epsilon^{1-n})dS.$$ Since $A(\partial B_\epsilon(0))=\epsilon^{n-1} A(S^{n-1}),$ the above limit is $-(n-2)u(0) A(S^{n-1})$, as desired.

Answer 2

There is a correct proof already, so some remarks on your attempt.

• the correct formula cannot be $r^{n-1}\delta$ as this can only be interpreted as the zero distribution. Also multiplying by the nonsmooth function $r^{n-1}$ is a little suspect.
• The correct laplacian formula for smooth radial functions is $$ \Delta f=\frac{1}{r^{n-1}} \frac{d}{d r}\left(r^{n-1} f'\right) $$ and you have $f=r^{2-n}$. Correctly computing this should give you $f'=r^{1-n}$ and therefore $\Delta f = 0$. The reason this doesn't give you the $\delta$ is because this formula only holds away from the singularity $x=0$. For test functions whose support contain 0, you should perform the careful limitting procedure in the other answer.