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Lemma:Let $\phi(s)$ be a non-negative and non-decreasing function. Suposse that \begin{equation} \phi(r) \le C_1 \left[\Bigl(\dfrac{r}{R}\Bigr)^\alpha + \mu \right]\phi(R) + C_2 R^\beta] \end{equation} for all $r\le R \le R_0$, with $C_1,\alpha,\beta$ positive constants. Then, for any $\sigma<\min\{\alpha,\beta\}$ there exists a constant $\mu_0 = \mu_0(C_1,\alpha,\beta,\sigma)$ such that if $\mu<\mu_0$, then for for all $r\le R\le R_0$ we have \begin{equation} \phi(r)\le C_3\Bigl(\dfrac{r}{R}\Bigr)^\sigma \left[\phi(R) + C_2R^\sigma \right] \end{equation} where $C_3=C_3(C_1,\sigma-\min\{\alpha,\beta\})$ is a positive constant. In turn, \begin{equation} \phi(r)\le C_4r^\sigma, \end{equation} where $C_4=C_4(C_2,C_3,R_0,\phi,\sigma)$ is a positive constant.

proof: We can assume $\beta < \alpha$ and, in this case, it suffices to show the estimate for $\sigma = \beta. \cdots$

Ask I'd like to prove the assertion in the first line of the proof above. The lemma can be found here on page 9 and a similar lemma can be found in article,book in the end on page 10.

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    There is no $\mu$ below the definition of $\mu_0$.2012-09-16
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    @vesszabo : this is not a problem as $\mu$ is already defined before $\mu_0$. In fact the lemma is later used in the paper with $\mu=0$, so that $\mu < \mu_0$ will hold.2012-09-17

1 Answers 1