I'm studying a paper, which I'll include a small piece here. And I'm struggling to calculate 

$$C_n\|u_{m,n}\|^{\left(\frac{2*}{2}\right)^k\frac{2*-q}{(r_k)^k}}_{L^{2*}(\Omega)}$$
Where $\Omega$ is an open, bounded subset of $\mathbb{R}^N$.  We also are considering values of $m$ and $n$ such that $m>\overline{t}$, and $n>\max\{\overline{t}, \overline{n}\}$, given that $\overline{n}>0$, $\overline{t}=t_0+1$ with $t_0>0$. We also are considering $\gamma>1$, $2^{*}=\frac{2N}{N-2}> \gamma +2$ and $p$, $q$ are such that
\begin{align*}
\gamma+2&<p<\frac{2^*}{2}(\gamma +2)\\
\gamma+2&< q< \min\{p,2^*\}
\end{align*}
Remark: $\mathcal{S}^2$ is a constant of Sobolev, I mean appears after I used one of Sobolev embeddings theorem. Here $\alpha$ is a positive constant.


I tried to use Moser iteration. However, I do not understand where I'm making mistakes. I found the expression in \eqref{5} which is clearly different. Would you check if the expression in \eqref{6} is right?

I bet there is a very big problem, because since the basis $\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2 (r_{k-1}-q+1)}\right)$ goes to infinity (by \eqref{4}), how can the whole expression obtained in \eqref{5} be smaller or equal to $\tilde{C}_n$?
[![partial of page 165][1]][1]


I've included the original paper in the link of google drive below. This part is on page 165 of the following paper:

David Arcoya, Lucio Boccardo, Luigi Orsina, "[Critical points for functionals with quasilinear singular Euler-Lagrange equations.][2]", Calculus of Variations and Partial Differential Equations 47, No. 1-2, 159-180 (2013), [MR3044135](https://mathscinet.ams.org/mathscinet-getitem?mr=MR3044135), [Zbl 1266.35102](https://www.zbmath.org/?q=an%3A1266.35102).


**MY ATTEMPT**
>Let's see the iteration

>\begin{equation}
|u_{m,n}|_{L^{\frac{r-q+2}{2}2^*}(\Omega)}\leq \left(\frac{(r-q+2)^2}{4\alpha S^2 (r-q+1)}\right)^{\frac{1}{r-q+2}}n^{\frac{p-q}{r-q+2}}|u_{m,n}|_{L^r(\Omega)}^{\frac{r}{r-q+2}}\label{1}\tag{1}
\end{equation}

>Defining $r_0=2^*$ and $r_k=\left(\frac{2^*}{2}r_{k-1}+\frac{2^*}{(2-q)}\right)$. From \eqref{1} one has that

>\begin{align*}
&|u_{m,n}|_{L^{r_k}(\Omega)}=|u_{m,n}|_{L^{\frac{(r_{k-1}-q+2)}{2}2^*}(\Omega)}\label{2}\tag{2}\\
&\leq\underbrace{\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2 (r_{k-1}-q+1)}\right)^{\frac{1}{r_{k-1}-q+2}}n^{\frac{p-q}{r_{k-1}-q+2}}}_{C_1}|u_{m,n}|_{L^{r_{k-1}}(\Omega)}^{\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}}\\
&=C_1|u_{m,n}|_{L^{r_{k-1}}(\Omega)}^{\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}}\\
&\leq C_1\left\{\left(\frac{(r_{k-2}-q+2)^2}{4\alpha S^2 (r_{k-2}-q+1)}\right)^{\frac{1}{r_{k-2}-q+2}}n^{\frac{p-q}{r_{k-2}-q+2}}|u_{m,n}|_{L^{r_{k-2}}(\Omega)}^{\left(\frac{2^*}{2}\right)\frac{r_{k-2}}{r_{k-1}}}\right\}^{\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}}\\
&\leq C_1\underbrace{\left(\frac{(r_{k-2}-q+2)^2}{4\alpha S^2 (r_{k-2}-q+1)}\right)^{\frac{1}{r_{k-2}-q+2}\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}}n^{\frac{p-q}{r_{k-2}-q+2}\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}}}_{C_2}|u_{m,n}|_{L^{r_{k-2}}(\Omega)}^{\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_{k-1}}\frac{r_{k-1}}{r_k}}\\
&=C_1C_2|u_{m,n}|_{L^{r_{k-2}}(\Omega)}^{\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}}\\
&\leq C_1C_2\left\{\left(\frac{(r_{k-3}-q+2)^2}{4\alpha S^2(r_{k-3}-q+2)}\right)^{\frac{1}{r_{k-3}-q+2}}n^{\frac{p-q}{r_{k-3}-q+2}}|u_{m,n}|_{L^{r_{k-3}}(\Omega)}^{\frac{2^*}{2}\frac{r_{k-3}}{r_{k-2}}}\right\}^{\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}}\\
&= C_1C_2\underbrace{\left(\frac{(r_{k-3}-q+2)^2}{4\alpha S^2(r_{k-3}-q+2)}\right)^{\frac{1}{r_{k-3}-q+2}\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}}n^{\frac{p-q}{r_{k-3}-q+2}\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}}}_{C_3}|u_{m,n}|_{L^{r_{k-3}}(\Omega)}^{\left(\frac{2^*}{2}\right)^3\frac{r_{k-3}}{r_{k-2}}\frac{r_{k-2}}{r_k}}\\
&= C_1C_2C_3|u_{m,n}|_{L^{r_{k-3}}(\Omega)}^{\left(\frac{2^*}{2}\right)^3\frac{r_{k-3}}{r_k}}\\
&\leq \cdots \leq\\
&\leq C_1C_2C_3\cdots C_{k-1}\underbrace{\left(\frac{(r_{0}-q+2)^2}{4\alpha S^2(r_{0}-q+2)}\right)^{\frac{1}{r_{0}-q+2}\left(\frac{2^*}{2}\right)^{k-1}\frac{r_1}{r_k}}n^{\frac{p-q}{r_{0}-q+2}\left(\frac{2^*}{2}\right)^{k-1}\frac{r_1}{r_k}}}_{C_k}|u_{m,n}|_{{L^{2^*}}(\Omega)}^{\left(\frac{2^*}{2}\right)^{k}\frac{r_0}{r_k}}
\end{align*}

>Now rewriting the first powers in $C_1, C_2,\ldots, C_k$, it follows that in

    

> 1. item $C_1)$ \begin{align*}
        \frac{1}{r_{k-1}-q+2}&=\frac{\left(\frac{2^*}{2}\right)}{\left(\frac{2^*}{2}\right)(r_{k-1}-q+2)}\\
        &=\left(\frac{2^*}{2}\right)\frac{1}{r_k}
    \end{align*}
    
> 2. item $C_2)$ \begin{align*}
        \frac{1}{r_{k-2}-q+2}\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}&=\frac{\left(\frac{2^*}{2}\right)}{\left(\frac{2^*}{2}\right)(r_{k-2}-q+2)}\left(\frac{2^*}{2}\right)\frac{r_{k-1}}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^2\frac{1}{r_{k-1}}\frac{r_{k-1}}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^2\frac{1}{r_k}\\
        \end{align*}
        
> 3. item $C_3)$ \begin{align*}
        \frac{1}{r_{k-3}-q+2}\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}&=\frac{\left(\frac{2^*}{2}\right)}{\left(\frac{2^*}{2}\right)(r_{k-3}-q+2)}\left(\frac{2^*}{2}\right)^2\frac{r_{k-2}}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^3\frac{1}{r_{k-2}}\frac{r_{k-2}}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^3\frac{1}{r_k}
        \end{align*}
        $\vdots$
        
> 4. item $C_k)$ \begin{align*}
        \frac{1}{{r_0}-q+2}\left(\frac{2^*}{2}\right)^{k-1}\frac{r_1}{r_k}&=\frac{\left(\frac{2^*}{2}\right)}{\left(\frac{2^*}{2}\right)(r_0-q+2)}\left(\frac{2^*}{2}\right)^{k-1}\frac{r_1}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^k\frac{1}{r_1}\frac{r_1}{r_k}\\
        &=\left(\frac{2^*}{2}\right)^k\frac{1}{r_k}
        \end{align*}

>Replacing properly the expressions above in every power of $C_1, C_2,\ldots, C_k$ it follows that
\begin{align*}
    C_1&=\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2(r_{k-1}-q+1)}\right)^{\left(\frac{2^*}{2}\right)\frac{1}{r_k}}n^{(p-q)\left(\frac{2^*}{2}\right)\frac{1}{r_k}}\\
    C_2&=\left(\frac{(r_{k-2}-q+2)^2}{4\alpha S^2(r_{k-2}-q+1)}\right)^{\left(\frac{2^*}{2}\right)^2\frac{1}{r_k}}n^{(p-q)\left(\frac{2^*}{2}\right)^2\frac{1}{r_k}}\\
    &\vdots\\
 C_k&=\left(\frac{(r_0-q+2)^2}{4\alpha S^2(r_0-q+1)}\right)^{\left(\frac{2^*}{2}\right)^k\frac{1}{r_k}}n^{(p-q)\left(\frac{2^*}{2}\right)^k\frac{1}{r_k}}\\
\end{align*}
Rewriting the whole expression in \eqref{2} replacing the exponents as above, one has that
\begin{align*}
   |u_{m,n}|_{L^{r_k}}&\leq \left[\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2(r_{k-1}-q+1)}\right)n^{(p-q)}\right]^{\left(\frac{2^*}{2}\right)\frac{1}{r_k}}\\
    &\cdot \left[\left(\frac{(r_{k-2}-q+2)^2}{4\alpha S^2(r_{k-2}-q+1)}\right)n^{(p-q)}\right]^{\left(\frac{2^*}{2}\right)^2\frac{1}{r_k}}\\
    &\vdots\\
    &\cdot\left[\left(\frac{(r_0-q+2)^2}{4\alpha S^2(r_0-q+1)}\right)n^{(p-q)}\right]^{\left(\frac{2^*}{2}\right)^k\frac{1}{r_k}}\cdot |u_{m,n}|_{L^{2^*}(\Omega)}^{\left(\frac{2^*}{2}\right)^k\frac{r_0}{r_k}}\label{3}\tag{3}\\
\end{align*}
>Considering $r_0=2^*$. Since $\frac{2^*}{2}>1$, let's prove that $r_k$ is an increasing sequence which diverges to infinity.

>Indeed, note that if
> 1. item $(k=1)$
    \begin{align*}
        r_1&=\frac{2^*}{2}(r_0-q+2)\\
        &=\frac{2}{2^*}=r_0+(2-q)\\
    \end{align*}
    Subtracting $\frac{2}{2^*}r_0$ in both sides, it follows that
    \begin{align*}
        \frac{2}{2^*}(r_1-r_0)&=r_0-\frac{2}{2^*}r_0+(2-q)\\
        &=2^*-\frac{2}{2^*}2^*+2-q \\
        &=2^*-q>0
\end{align*}
Thus $r_1>r_0$.
> 2. item $(k=2)$ , similarly
    \begin{align*}
       \frac{2}{2^*}(r_2-r_1)&=r_1-\frac{2}{2^*}r_1+(2-q)\\
        &=\frac{2^*}{2}(r_0-q+2)-r_0+q-2-q+2\\
        &=\frac{(2^*)^2-2^*q+22^*-22^*}{2}\\
        &=\frac{2^*}{2}(2^*-q)>0
\end{align*}
>Therefore, $r_2>r_1$.

>Proceeding in this way indefinitelyt, one can find that $r_k$ is an increasing sequence, that is $r_0<r_1<r_2<...<r_k<...$ and its limit is
$$\lim_{k\to \infty}r_k=+\infty.$$
>One claims that
$$\left(\frac{(r_0-q+2)^2}{4\alpha S^2(r_0-q+1)}\right)<\left(\frac{(r_1-q+2)^2}{4\alpha S^2(r_1-q+1)}\right)<\dots<\left(\frac{(r_k-q+2)^2}{4\alpha S^2(r_k-q+1)}\right)<\dots\label{4}\tag{4}$$
>Indeed, taking $t=r_0, r_1, r_2, ..., r_k,...$ and defining
$$f(t)=\frac{(t-q+2)^2}{4\alpha S^2 (t-q+1)}$$
one has that
\begin{align*}
    f'(t)&=\frac{2(t-q+2)4\alpha S^2(t-q+1)-(t-q+2)^24\alpha S^2}{4\alpha S^2(t-q+1)^2}\\
    &=\frac{(t-q+2)[2t-2q+2-t+q-2]}{(t-q+1)^2}
\end{align*}
>In order to show that $f$ is an increasing function one only needs to get that $(t-q+2)(t-q)>0$ and this is true, first note that if $t=r_0$ then $(2^*-q+2)(2^*-q)>0$ once $2^*-q>0$. Now, by \eqref{4}, replacing every single value for $t$ one has that $f'(t)>0$ which implies that $f$ is an increasing function. Hence one has proved the claim.
>Now turning back to \eqref{3} replacing the result in \eqref{4}, one can rewrite \eqref{3} as
$$|u_{m,n}|_{L^{r_k}(\Omega)}\leq \left[\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2(r_{k-1}-q+1)}\right)n^{(p-q)}\right]^{\frac{1}{r_k}\displaystyle\sum_{i=1}^{k}\left(\frac{2^*}{2}\right)^i}|u_{m,n}|_{L^{2^*}(\Omega)}^{\left(\frac{2^*}{2}\right)^k\frac{r_0}{r_k}}.\label{5}\tag{5}$$
On the other hand, notice that, taking $A=\left(\frac{2^*}{2}\right)(2-q)$ one can rewrite $r_k$ as
\begin{align*}
    r_k&=r_{k-1}\left(\frac{2^*}{2}\right)+\left(\frac{2^*}{2}\right)(2-q)\\
    &=r_{k-1}\left(\frac{2^*}{2}\right)+A\\
    &=\left(r_{k-2}\left(\frac{2^*}{2}+A\right)\right)\left(\frac{2^*}{2}\right)+A\\
    &=\left(\frac{2^*}{2}\right)^2r_{k-2}+\left(1+\frac{2^*}{2}\right)A\\
    &=\left(\frac{2^*}{2}\right)^2\left[r_{k-3}\left(\frac{2^*}{2}+A\right)\right]+\left(1+\frac{2^*}{2}\right)A\\
    &=\left(\frac{2^*}{2}\right)^3r_{k-3}+\left(1+\left(\frac{2^*}{2}\right)+\left(\frac{2^*}{2}\right)^2\right)A\\
    &\vdots\\
    &=\left(\frac{2^*}{2}\right)^kr_0+A+A\left(\frac{2^*}{2}\right)+A\left(\frac{2^*}{2}\right)^2+...+A\left(\frac{2^*}{2}\right)^{k-1}
\end{align*}
>what is the same as $r_k=\left(\frac{2^*}{2}\right)^kr_0+A\displaystyle\sum_{i=0}^{k-1}\left(\frac{2^*}{2}\right)^i$, this is
$$r_k=\left(\frac{2^*}{2}\right)^kr_0+(2-q)\left(\frac{2^*}{2}\right)\left[\frac{1-\left(\frac{2^*}{2}\right)^{k-1}}{1-\left(\frac{2^*}{2}\right)}\right].\label{6}\tag{6}$$


**I appreciate if someone can help me!
Thanks in advance.**


  [1]: https://i.sstatic.net/bTfjO.png  
 [2]:https://drive.google.com/file/d/1OngvGeX5510SpBxmVdaf57dFDEuprBuy/view?usp=sharing