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Lemma 3.3 Suppose $u \in H^{1}(\Omega)$ satisfies $$\int_{B_r(x_0)}|Du|^2 \le M r^{^\mu} \quad \mbox{for any} \ \ B_r(x_0) \subset \Omega,$$ for some $\mu \in [0,n)$. Then, for any $\Omega' \Subset \Omega$ there holds for any $B_r(x_0) \subset \Omega$ with $x_0 \in \Omega'$ $$\int_{B_r(x_0)} |u|^2 \le C(n,\lambda,\mu,\Omega,\Omega')\, \left \{M+\int_{\Omega}u^2\right \} r^{\lambda}$$ where $\lambda = \mu +2$ if $\mu < n-2$ and $\lambda$ is any number in $[0,n)$ if $n-2\le\mu.

Proof. Denote $R_0= \mbox{dist}(x,\partial \Omega)$. For any $x_0 \in \Omega'$ and $0, the Poincaré inequality yields $$\int_{B_r(x_0)}|u- u_{x_0,r}|\le cr^2 \int_{B_r(x_0)}|du|^2 \le c(n)Mr^{\mu +2}.$$ And this implies that $$\int_{B_r(x_0)}|u- u_{x_0,r}| \le c(n)Mr^{\lambda}.$$ where $\lambda$ as in the lemma 3.3.

Then, following the proof, I understand the case $\lambda = \mu +2$ if $\mu < n-2$. But I don't understand how to obtain the case $\lambda$ is any number in $[0,n)$ if $n-2\le\mu?

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    You don't understand why the inequality after "implies that" is true in the case $n-2\leq\mu? Or are you talking about another part of the proof that is not displayed here?2012-08-19
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    I don't know if $$\int_{B_r(x_0)}|u- u_{x_0,r}|\le cr^2 \int_{B_r(x_0)}|du|^2 \le c(n)Mr^{\mu +2}.$$ implies that $$\int_{B_r(x_0)}|u- u_{x_0,r}| \le c(n)Mr^{\lambda}.$$ where $\lambda$ is any number in $[0,n)$ if $n-2\le\mu. Seems that the author proof the two cases in the same way after this. He did not write another part.2012-08-19
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    Suppose for a moment that $R_0<1$. Then $n-2\leq \mu means $n \leq \mu+2 , and thus $\lambda < \mu+2$. But $r^{\mu+2} < r^\lambda$ for $0 \leq r <1$. Can you treat now the case $r \leq R_0$ for any $R_0$?2012-08-19
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    I had thought of that. No, I can not.2012-08-19
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    Of course, it will imply, because $\lambda\leq\mu+2$ in either case.2012-08-19
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    But if $\lambda \le \mu +2$ and $r\ge 1$ we have $r^{\mu+2} \not\leq r^\lambda$.2012-08-19
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    Well, that can be absorbed into the constant, since $r\leq R_0$.2012-08-21

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