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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 $\mu_0$. – vesszabo Sep 16 '12 at 17:08
@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. – Ewan Delanoy Sep 17 '12 at 5:05

The first line says two things :

(1) We may assume $\beta < \alpha$ in (2.6)


(2) In this case, we may further assume that $\sigma=\beta$.

Note that (2.6) still holds if we replace $\alpha$ by a larger value $\alpha’ > \alpha$ , because

$$ \Bigl(\dfrac{r}{R}\Bigr)^\alpha \le \Bigl(\dfrac{r}{R}\Bigr)^{\alpha’} $$

So we may enlarge $\alpha$ as we please, and in particular we can make it $> \beta$. This justifies (1).

(2.7) can be written as $\phi (r) \leq F(C_2,C_3,\sigma,r,R)$ where $F$ is the complicated function defined by

$$ F(C_2,C_3,\sigma,r,R)=C_3\Bigl(\dfrac{r}{R}\Bigr)^\sigma \left[\phi(R) + C_2R^\sigma \right] $$

Suppose that we have shown the estimate (2.7) when $\sigma=\beta$, so that $\phi (r) \leq F(C_2,C_3,\beta,r,R)$ for $r \leq R \leq R_0$. Now let

$$ C’_3={\sf max}\Bigg(C_3,C_3R^{\beta-\sigma}\Bigg) $$

This is a positive constant, and we have for $r \leq R$,

$$ C_3\Bigl(\dfrac{r}{R}\Bigr)^\sigma \leq C_3 \leq C’_3 \ \ {\text{and}} \ \ C_3r^{\beta-\sigma} \leq C_3R^{\beta-\sigma} \leq C’_3 $$

and hence,

$$ C_3\Bigl(\dfrac{r}{R}\Bigr)^\beta \phi (R) \leq C’_3\Bigl(\dfrac{r}{R}\Bigr)^\sigma \phi (R) \ \ {\text{and}} \ \ C_2C_3 r^{\beta} \le C_2C’_3 r^{\sigma} $$

Adding up, we deduce

$$ \phi (r) \le F(C_2,C_3,\beta,r,R) \le F(C_2,C’_3,\sigma,r,R) $$

This justifies (2).

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First, thank you for your answer. But I need to say the following: in the first part I know that \begin{equation} C_1\Bigr[ \Bigr(\dfrac{r}{R}\Bigl)^\alpha + \mu \Bigl] ]+C_2 R^\beta \le C_1\Bigr[ \Bigr(\dfrac{r}{R}\Bigl)^{\alpha´} + \mu \Bigl] +C_2 R^\beta \end{equation} What I guess strange is that, if I do this, then I always will get $\min \{\alpha, \beta\} = \beta$. – user29999 Sep 17 '12 at 19:37
In the second part. I'm not convinced that \begin{equation} C_3\Bigl(\dfrac{r}{R}\Bigr)^\beta \phi (R) \leq C’_3\Bigl(\dfrac{r}{R}\Bigr)^\sigma \phi(R) \end{equation} because this is equivalent to $C_3r^{\beta-\sigma}\dfrac{1}{R^\beta} \le C`_3\dfrac{1}{R^\sigma}$. I know that $C_3r^{\beta-\sigma} \le C’_3 $ but $\dfrac{1}{R^\beta}$ can be greater than $\dfrac{1}{R^\sigma}$ for $R, \beta, \sigma $ convenient. I will be waiting some commentary. – user29999 Sep 17 '12 at 19:39

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