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Let

  • $\Omega\subseteq\mathbb{R}^n$ be a domain with a smooth boundary
  • $H:=W_0^{1,2}(\Omega)$ be the Sobolev space
  • $p>1$ such that $$p<\begin{cases}\infty&\text{, if }n=2\\\frac{n+2}{n-2}&\text{, if }n> 2\end{cases}$$

How can we prove, that $$S:=\left\{u\in H:\int_\Omega|u|^{p+1}\;d\lambda^n=1\right\}$$ is a well-defined and closed set? Moreover, why is there a non-negative $v\in S$, such that $$\int_\Omega|\nabla v|^2\;d\lambda^n\le \int_\Omega|\nabla u|^2\;d\lambda^n\;\;\;\text{for all }u\in S\;?$$ I suppose we need to use the smoothness of the boundary $\partial\Omega$, but I absolutely don't get how I need to process in detail.

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  • $\begingroup$ By the continuity and compactness of the Sobolev embedding. $\endgroup$
    – user72012
    Commented Jul 2, 2015 at 22:18
  • $\begingroup$ Sorry, what do you mean by $\lambda^n$ here? It is not a standard notation for integration. $\endgroup$
    – spatially
    Commented Jul 3, 2015 at 12:18

1 Answer 1

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I think here you probability need $\Omega$ to be bounded as well. (however since $u\in W_0^{1,2}(\Omega)$, you do not need smooth boundary condition) And I will assume by $d\lambda^n$ you just mean $x\in \mathbb R^n$, the standard integration notation.

For your first question, since $u\in W_0^{1,2}(\Omega)$, by Sobolev embedding you have $u\in L^{p^*}$ where $p^*\leq \frac{2n}{n-2}$ and clearly $p^*> p+1$. So at least $S$ is well-defined. To show closeness, you again use the fact that $p^*> p+1$ and call compact embedding. (suppose a sequence convergence in $S$ and there is a subsequence convergence to...)

For your second question, it is known as minimizing problem. It is a big topic in calculus of variation. I would suggest you to read Evans PDE book chapter 8 for more details. I will give you a start on second question and the rest would be the compact embedding again.

Notice that the number $m\geq 0$ defined as $$ m:=\inf_{u\in S} \left\{\int_\Omega |\nabla u|^2dx\right\} $$ exists. So you could have a sequence $(u_n)\subset S$ such that $\|\nabla u_n\|_{L^2}\to m$. Notice that you actually have the norm of $\nabla u_n$ is bounded and hence by compact embedding...

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  • $\begingroup$ To prove closedness of $S$ boundedness of $\Omega$ and compact embedding is not needed (only continuity of the embedding). $\endgroup$
    – daw
    Commented Jul 3, 2015 at 14:45

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