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I am reading the following paper: Takáč, Peter On the Fredholm alternative for the p-Laplacian at the first eigenvalue. Indiana Univ. Math. J. 51 (2002), no. 1, 187–237.

I need help to understand the following argument (page 193 section 2.1):

$\Omega\subset\mathbb{R}^N$ is a bounded regular domain, $p\in (2,\infty)$. Let $\phi_1$ be the first eigenfunction associated with the problem $-\Delta_p u=f$ and $u\in W_0^{1,p}(\Omega)$, i.e. $$\int|\nabla\phi_1|^{p-2}\nabla\phi_1\nabla v=\lambda_1\int|\phi_1|^{p-2}\phi_1v,\ \forall\ v\in W_0^{1,p}$$

where $\lambda_1>0$ os the first eigenvalue. We can assume that $\phi_1\in C^1(\overline{\Omega})$, $\phi_1>0$ in $\Omega$ and $\frac{\partial\phi_1}{\partial\eta}<0$ in $\partial\Omega$, where $\frac{\partial\phi_1}{\partial\eta}$ represents derivative in the normal direction.

Define in $W_0^{1,p}$ the semi-norm $$\|u\|_{\phi_1}=\Big(\int|\nabla\phi_1|^{p-2}|\nabla u|^2\Big)^{\frac{1}{2}}$$

The author says that in fact, $\|\cdot\|_{\phi_1}$ is an norm because of the following argument: if $v\in W_0^{1,p}$, then

\begin{eqnarray} \lambda_1\int\phi_1^{p-1}v^2 &=& \int|\nabla\phi_1|^{p-2}\nabla\phi_1\nabla(v^2) \nonumber \\ &\leq& 2\int|\nabla\phi_1|^{p-1}|\nabla v||v| \nonumber \\ &\le& 2\|v\|_{\phi_1}\Big(\int|\nabla\phi_1|^pv^2\Big)^{\frac{1}{2}} \end{eqnarray}

I can understand the last two inequalities, and I can use it to prove that $\|\cdot\|_{\phi_1}$ is a norm. The problem is the first equality. To use the characterization of the eigenvalue, we have to take $u\in W_0^{1,p}$. Why does $v^2\in W_0^{1,p}$? I think that he is using this fact, is this true?

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  • $\begingroup$ When referring to a 51-page paper, it's not a bad idea to give the page on which the argument is presented. $\endgroup$ – user53153 Feb 18 '13 at 18:23
  • $\begingroup$ Sorry Pavel, I forgot this detail. Take a look in page 193 section 2.1. $\endgroup$ – Tomás Feb 18 '13 at 18:27
  • $\begingroup$ OK, I see that the inequality (4.7) assumes that $v$ is in the domain of some quadratic form, not just an arbitrary $W^{1,p}_0$ function. In general, $v\in W^{1,p}_0$ does not imply $v^2\in W_0^{1,p}$. For example, consider $v(x)=|x|^{-\alpha}-1$ on the unit ball. This function belongs to $W^{1,p}_0$ as long as $p(\alpha+1)<n$. Squaring $v$ will take it out of the Sobolev space if $p(2\alpha+1)\ge n$. $\endgroup$ – user53153 Feb 18 '13 at 18:31
  • $\begingroup$ But the domain of the quadratic form contains $W_0^{1,p}$. $\endgroup$ – Tomás Feb 18 '13 at 18:34
  • $\begingroup$ I see. Then I guess that the validity of the formula $$\int|\nabla\phi_1|^{p-2}\nabla\phi_1\nabla v=\lambda_1\int|\phi_1|^{p-2}\phi_1v$$ extends to $v\in \mathcal D_{\varphi_1}$ by continuity. $\endgroup$ – user53153 Feb 18 '13 at 19:27
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To prove that $\Vert u\Vert_{\phi_1} = \left(\int |\nabla \phi_1|^{p-2}|\nabla u |^2\right)^{\frac 12}$ is a norm on $W_0^{1,p}$ rather than a semi-norm, we have to show that $\Vert u \Vert_{\phi_1} = 0$ implies $u=0$.

Assume that $u\in W^{1,p}_0$ satisfies $\Vert u\Vert_{\phi_1}=0$. Define $u_M := \min(|u|, M)$ for $M>0$. Then $\Vert u_M\Vert_{\phi_1}=0$ and $u_M \in W^{1,p}_0\cap L^\infty$. In particular, the latter implies that $u_M^2\in W^{1,p}_0$ (using $|u_M|^p \le |u|^p$ and $ |\nabla u_M^2|^p = 2^p|u_M|^p|\nabla u_M|^p\le 2^pM^p |\nabla u|^p$) The argument at the end of the original post (and our assumption on $u$) now shows that $$\lambda_1 \int \phi_1^{p-1} u_M^2 \le2 \Vert u_M \Vert_{\phi_1} \left(\int |\nabla \phi_1|^p u_M^2\right)^{\frac 12} = 0.$$ This holds for any $M> 0$. Since $u_M^2\nearrow u^2$ as $M\to +\infty$, the dominated convergence theorem implies that $\lambda_1 \int \phi_1^{p-1} u^2 \le 0$. It follows that $u=0$, so $\Vert \cdot \Vert_{\phi_1}$ is a norm.

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  • $\begingroup$ I think it is right. Let me think a little bit more about it, then I accept it. Thank you. $\endgroup$ – Tomás Jul 22 '13 at 22:23

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