Let $\Omega\subset \mathbb{R}^n$ be a connected open set containing $0$, $u\in H_\mathrm{loc}^{1}(\Omega)$, $A(x)\leq C|x|^{-1+\epsilon}$, where $\epsilon$ is small, and we also have $$ \|\nabla u\|_{L^2(|x|\leq R)}\leq C_{N}R^N,~~R\to 0, $$ for all $N>0$ , then do we have $A(x)\nabla u\in L_\mathrm{loc}^{2}(\Omega) $?
how to show that $A(x)\nabla u\in L_\mathrm{loc}^{2}(\Omega) $ for $u\in H_\mathrm{loc}^{1}(\Omega)$
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What is $\varepsilon$? Also, $C$ depends on $N$? Is $\Omega$ subset of $\mathbb R^n$? – detnvvp Oct 17 '13 at 03:31
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sorry to make you confused, I add some details – Tomas Oct 17 '13 at 03:39
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The answer is yes.
The only problem can be caused around zero. There, if $n\in\mathbb N$ and $B_n$ is the ball with center $0$ and radius $\frac{1}{n}$, you have that $$\int_{B_n\setminus B_{n+1}}|A(x)|^2|\nabla u(x)|^2\,dx\leq$$$$C\int_{B_n\setminus B_{n+1}}|x|^{-2+2\varepsilon}|\nabla u(x)|^2\,dx\leq C\int_{B_n\setminus B_{n+1}}\frac{1}{(n+1)^{-2+2\varepsilon}}|\nabla u(x)|^2dx=$$$$C(n+1)^{2-2\varepsilon}\int_{B_n}|\nabla u(x)|^2\,dx\leq CC_N(n+1)^{2-2\varepsilon}\frac{1}{n^N}.$$ So, by choosing $N=4-2\varepsilon$, you get a convergent series, which shows that $A|\nabla u|$ is square integrable around $0$.
detnvvp
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