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.
Best Answer
(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.