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.

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