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I am reading Le Dret's book on Nonlinear Elliptic Partial Differential Equations.

On chapter 7 (page 209) I am trying to prove that the functional $J(u) = \frac{1}{2}\int \lVert \nabla u\rVert^2 - \int G(u)$, that arise from the study of the boundary-value problem $-\Delta u = G'(u)\doteq g(u)$ in $H_0^1(\Omega)$, satisfies the Palais Smale condition if $g$ has some growth property:enter image description here

My issue is the following: Here the author says to conclude that $u_n$ is bounded just like on a previous propositionenter image description here

If you go to the previous proposition this is what he is referring to: enter image description here enter image description here

My problem is: On the proposition 7.3 we have the equality $DJ(u_n)u_n = (p+1)J(u_n) - \frac{p-1}{2}\int \lVert \nabla u_n\rVert^2$ and then you may use the norm inequality for $DJ(u_n)$. But in the lemma 7.5, the one I'm trying to prove, we only have that $DJ(u_n)u_n\leq C m(\Omega)+\theta J(u_n) +(1-\frac{\theta}{2})\int \lVert \nabla u_n\rVert^2 $. I am not sure how to conclude that $u_n$ is bounded from this.

Could someone help me in this passage?

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Since $\theta>2$, one gets from the upper bound of $DJ(u_n)u_n$ $$ (\frac\theta2-1) \|\nabla u\|_{L^2}^2 \le Cm(\Omega) + \theta J(u_n) - DJ(u_n)u_n. $$ Using the estimate as in Prop 7.3 $$ (\frac\theta2-1) \|\nabla u\|_{L^2}^2 \le Cm(\Omega) + \theta J(u_n) +c\|DJ(u_n)\|_{H^{-1}}^2 +\frac12(\frac\theta2-1) \|\nabla u\|_{L^2}^2 $$ with $c = \frac12 (\frac\theta2-1)^{-1}$.

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  • $\begingroup$ Thanks for the tip, I think I get most of it now. My only problem is: why does he references prop 7. 3? Didn't you use only something along the lines of $\lvert ab\rvert\leq \frac{1}{2k} a^2 + \frac{k}{2} b^2$ ? How is this any similar as 7.3? He is referring to 7.3 only for the norm of a linear functional inequality ? it seems odd to me $\endgroup$ Oct 22 '20 at 14:14
  • $\begingroup$ Just realised that my question was really dumb, I didn't realize that the $\int \lVert \nabla u_n\rVert^2$ had a negative sign, thank you though $\endgroup$ Oct 22 '20 at 18:15
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The solution is actually very simple:/, I did not realize that the term $\int \lVert \nabla u_n\rVert^2$ had a negative sign on the inequality that I have on lemma 7.5.

Since it does, you can multiply everything by -1, and get the inequality that helps you to conclude just as 7.3.

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