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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

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\mu2012-08-19
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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\mu2012-08-19
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    Suppose for a moment that $R_0<1$. Then $n-2\leq \mu $r \leq R_0$ 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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The second part goes into the first part, if n is shiftet up, to the right or whatever, by two. The fixed value is $\mu$, not $n$.So for the case $n-2 ≤ \mu < n$, use $n' = n+2$, then $\mu < n = n+2-2 = n'-2$. Get
$$ \int_{B_r(x_0)} |u|^2 \leq C(n',\lambda,\mu,\Omega,\Omega') \left\{ M+\int_\Omega u^2 \right\}r^{\lambda} $$ where $\lambda =\mu+2$ and use $\left(\frac{R_0}{r}\right)^{\lambda} ≤ \left(\frac{R_0}{r}\right)^{\mu + 2}$ or $\quad$ $r^{\mu+2} ≤ R_0^{\mu + 2 - \lambda} r^\lambda$ because $\lambda \mapsto x^\lambda $ is non-increasing for $x \leq 1$.

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    Sorry, but did not understand.2012-08-28
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    If $\mu \in [0, n)$ then also $\mu \in [0, n'=n+2)$2012-08-28