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 (5) which is clearly different. Would you check if the expression in (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 (4)), how can the whole expression obtained in (5) be smaller or equal to $\tilde{C}_n$?
I've included the original paper in the link of google drive below. This part it is on page 165 of the paper.
David Arcoya, Lucio Boccardo, Luigi Orsina, "Critical points for functionals with quasilinear singular Euler-Lagrange equations.", Calculus of Variations and Partial Differential Equations 47, No. 1-2, 159-180 (2013), MR3044135, Zbl 1266.35102.
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}}\tag1 \end{equation}
Defining $r_0=2^*$ and $r_k=\left(\frac{2^*}{2}r_{k-1}+\frac{2^*}{(2-q)}\right)$. From (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)}\tag2\\ &\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 $C_1, C_2, ..., C_k$, it follows that in
- 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*}
- 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*}
- 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*} $\vdots$
- 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 $C_1, C_2, ..., C_k$ 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 (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}}\tag3\\ \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
- 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 $r_1>r_0$.
- 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 $$\left(\frac{(r_0-q+2)^2}{4\alpha S^2(r_0-q+1)}\right)<\left(\frac{(r_1-q+2)^2}{4\alpha S^2(r_1-q+1)}\right)<...<\left(\frac{(r_k-q+2)^2}{4\alpha S^2(r_k-q+1)}\right)<...\tag4$$ 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 (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 (3) replacing the result in (4), one can rewrite (3) as $$|u_{m,n}|_{L^{r_k}(\Omega)}\leq \left[\left(\frac{(r_{k-1}-q+2)^2}{4\alpha S^2(r_{k-1}-q+1)}\right)n^{(p-q)}\right]^{\frac{1}{r_k}\displaystyle\sum_{i=1}^{k}\left(\frac{2^*}{2}\right)^i}|u_{m,n}|_{L^{2^*}(\Omega)}^{\left(\frac{2^*}{2}\right)^k\frac{r_0}{r_k}}.\tag5$$ 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 $$r_k=\left(\frac{2^*}{2}\right)^kr_0+(2-q)\left(\frac{2^*}{2}\right)\left[\frac{1-\left(\frac{2^*}{2}\right)^{k-1}}{1-\left(\frac{2^*}{2}\right)}\right].\tag6$$
I really appreciate if someone can help me! Thanks in advance.