Distributional Laplacian of logarithm and the Dirac delta distribution

Question

Consider $S(\mathbb{R}^2)$ the space of rapidly decreasing functions.

Consider $F(x) = \displaystyle\frac{1}{2 \pi} \ln|x| , x \in \mathbb{R}^2 - \{ 0 \}$.

I want to prove this : $\Delta F = \delta $ in $S^{'}(\mathbb{R}^2).$

My try: For every $\varphi \in S({\mathbb{R}}^2)$ we have

$$\langle \Delta F, \varphi\rangle = \langle F , \Delta \varphi\rangle = \displaystyle\int_{\mathbb{R}^2} F(x) \Delta\varphi(x) \ dx = \displaystyle\lim_{r \rightarrow + \infty} \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{\epsilon \leq |x| \leq R} F(x) \Delta \varphi(x) \ dx $$

I don't know how to prove this :$\displaystyle\lim_{r \rightarrow + \infty} \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{\epsilon \leq |x| \leq R} F(x) \Delta\varphi(x) \ dx = \varphi (0) = \delta(\varphi)$

I don't know what to do from here.

Answer 1

(Indeed, $F$ is not in $\mathscr{S}$; it is any rate more convenient to consider $\mathscr{D}=C^\infty_0$.) Let $\varepsilon>0$ and $\Omega_\varepsilon:=\mathbf{R}^2-B_\varepsilon(0)$; now $$\left\langle F,\Delta\phi\right\rangle=\lim_{\varepsilon\downarrow 0}\int_{\Omega_\varepsilon} F\Delta\phi \ dx,$$ and by Green's identity $$\int_{\Omega_\varepsilon} F\Delta\phi \ dx=\int_{\Omega_\varepsilon} \phi\Delta F \ dx+\int_{\partial\Omega_\varepsilon} F\frac{\partial\phi}{\partial n}ds-\int_{\partial\Omega_\varepsilon}\phi\frac{\partial F}{\partial n} \ ds.$$ In the RHS the first integral vanishes as $\Delta F=0$ on $\mathbf{R}^2-\left\lbrace 0\right\rbrace$; the second integral can be estimated in absolute value to be at most $$C\int_{\partial\Omega_\varepsilon} Fds=\frac{C}{2\pi}2\pi\varepsilon\log \varepsilon$$ which tends to $0$ as $\varepsilon\downarrow 0$ ($C$ is an upper bound for $\partial\phi/\partial n$). Now $\partial F/\partial n=-1/2\pi|x|$ (the normal to $\partial\Omega_\varepsilon$ is $-x/|x|$) so that we are left with $$\left\langle F,\Delta\phi\right\rangle=\lim_{\varepsilon\downarrow 0}\frac{1}{2\pi\varepsilon}\int_{\partial\Omega_\varepsilon}\phi\ ds=\phi(0),$$ by a standard argument using, say, polar coordinates.

Answer 2

F(x) appears to be an increasing function! Nonetheless, I think the proposition is true (but with a minus sign), and it can easily be proved using Gauss's theorem: http://en.wikipedia.org/wiki/Divergence_theorem .

Take the gradient of F to obtain a vector field. The Laplacian of F is the divergence of this vector field. The field points everywhere radially outwards, and has magnitude 1/(2*pi*|x|). The flux of the field across a sphere (circle) of radius r is therefore 1. The integral of the divergence of the field inside the sphere is therefore 1. Since this is true for all r, the divergence must be confined to the origin.