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I am studying the paper On some nonlinear elliptic problems for $p$-Laplacian in $\mathbb{R}^n$ by Abdelouahed El Khalil and Said El Manouni Mohammed Ouanan. I have a problem understanding one step of the proof needed to get a $ L^\infty$ estimate for the solution $u$. The step goes like this:

$$ \int f(x)u^{a+kp+1} dx \leq \left(\int (f(x))^w\right)^{\frac{1}{w}} \left(\int u^q\right)^{\frac{q}{a}} \left(\int (u)^{(k+1)t}\right)^{\frac{p}{t}}\,dx. $$

where $w:=p^*/(p^* -(a+1))$ , $t:=p^* q/(a(q-p^*) +q)$ and $q > p^*$ fixed, here $p^*$ is the Sobolev exponent of $p$. $f$ is in $ L^\infty \cap L^w $ and $u$ is in $L^s$ for all $p^* \leq s < \infty $. Also $1/w +1/t +a/q =1$ and $0 < a < p^*-1$.

I think it has to be with the Hölder inequality but I can not figure out how. Is there a mistake in this? Here is a link to the paper.

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    $\begingroup$ A simple way to check for obvious errors is to check that the inequality is invariant under rescaling of each function appearing in it, as well as rescaling space. And why is there a $dx$ at the end of the right side? $\endgroup$ – Deane Yang Mar 30 '16 at 4:50
  • $\begingroup$ @DeaneYang Thank you for your advice. I tried it (rescaled f(x) and u(x) by 1/2) and it does not check out. So there must be a mistake in this inequality. Since the last term on the right side is extensively used in the paper after this point, it would be crucial to make the inequality work with the last term staying the same (my guess is by changing up the second term on the left). Anyway, i can not figure it out by myself, maybe you could take another try at it. By the way the dx at the end is a typo, it is my first question here... $\endgroup$ – Julian Mar 30 '16 at 20:31
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Some typing mistakes are detected, especially in these two lines. First, it should be mentioned that at this step, we have to consider the case where $p-1< \alpha.$ Take $t=\frac{p^*qp}{\alpha(q-p^*)+q+(p-1)p^*}.$ One can easily check that $\frac{t}{p}>1$ and $t<p^*.$ Then write in the integral $u^{\alpha+kp+1}$ as $u^{\alpha+1-p+(k+1)p}$ and use H\"older inequality with $w, \frac{q}{\alpha+1-p}$ and $\frac{t}{p}$ as exponents in the right side of the inequality. We get $\int f(x)u^{\alpha+kp+1}dx=\int f(x)u^{\alpha+1-p}u^{(k+1)p}dx\leq \left(\int (f(x))^wdx\right)^{\frac{1}{w}} \left(\int u^qdx\right)^{\frac{\alpha+1-p}{q}}\left(\int u^{(k+1)t}dx\right)^{\frac{p}{t}},$ with $\frac{1}{w}+\frac{\alpha+1-p}{q}+\frac{p}{t}=1.$

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