## Hardy inequality in dimesion 2

Suppose ${u}$ is a smooth function defined in ${B^c=\{|x|>1\}}$ in the plane ${\mathbb{R}^2}$, assume ${u=0}$ on ${\partial B^c}$ and also has compact support, then

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx\leq 4\int_{|x|>1}|\nabla u|^2dx$

There is a way presented by my advisor to explain why the strange function ${|x|^2\ln^2|x|}$ pops up here. First we transform LHS to polar coordinates

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx=\int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{r^2\ln^2 r}rdrd\theta$

which leads us to start with a very general ${f(r)}$

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta$

Suppose ${F'(r)=r/f(r)}$, then

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta=\int_{0}^{2\pi}\int_{1}^\infty u(r,\theta)dF(r)d\theta$

$\displaystyle =\int_{0}^{2\pi}\left(u(r,\theta)F(r)|_1^\infty-\int_{1}^\infty F(r)\partial_r u(r,\theta)dr\right)d\theta$

$\displaystyle =-\int_{0}^{2\pi}\int_{1}^\infty F(r)\partial_r u(r,\theta)drd\theta=-\int_{|x|>1} \frac{F(r)\partial_r u(x)}{r}dx$

$\displaystyle \leq 2\left(\int_{|x|>1}\frac{u(x)^2}{f(r)}dx\right)^{1/2}\left(\int_{|x|>1}\frac{f(r)F^2(r)}{r^2}|\nabla u|^2dx\right)^{1/2}$

So we only need to find ${f(r)}$ and ${F(r)}$ such that

$\displaystyle \frac{f(r)F^2(r)}{r^2}\leq C$

Actually one can solve the ODE

$\displaystyle \frac{f(r)F^2(r)}{r^2}=1, \quad F'(r)=\frac{r}{f(r)}$

to get ${F(r)=\frac{-1}{\ln r}}$, ${f(r)=r^2\ln^2 r}$. Plugging in this function back to the above proof gives you the desired inequality.