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Let us consider the Sobolev inequality $||u||_{L^p} \le C||u||_{H^1}$ for $2 <p< 2^*$, where the constant $C$ depends on $p$ and the domain. My question is, how can one see that the optimal Sobolev constant is attained at a positive function $f>0$?

By taking the absolute value, it is easy to see that we may take $f\ge 0$. Also when the domain is bounded, we may apply maximal principle. How about when the domain is not bounded?

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Symmetrization and ODE analysis. – Daniel Spector Apr 1 '13 at 12:23
Is it not possible to apply the maximum principle to a sequence of compact domains that exhaust the closure of the original domain? – Deane Yang Apr 1 '13 at 18:00
Deane Yang: your idea is natural, but doing it in a precise way is a delicate issue when the boundary of the domain is rough. – Delio Mugnolo Oct 30 '13 at 9:10

Here is a variational argument to prove that the maximizers do not change sign.

If $f \in H^1 (\mathbb{R}^N)$ be a maximizer, $u$ can be written as $$ f = f_+ - f_-, $$ with $f_+ \ne 0$ and $f_- \ne 0$. Moreover $$ \Vert f \Vert_{H^1}^2 = \Vert f_+ \Vert_{H^1}^2 + \Vert f_- \Vert_{H^1}^2 $$ and $$ \Vert f \Vert_{L^p}^p= \Vert f_+ \Vert_{L^p}^p + \Vert f_- \Vert_{L^p}^p. $$

We have thus
$$ \Vert f \Vert_{L^p}^2 = \frac{\Vert f_+\Vert_{L^p}^p}{\Vert f\Vert_{L^p}^p} \Vert f\Vert_{L^p}^2 + \frac{\Vert f_-\Vert_{L^p}^p}{\Vert f\Vert_{L^p}^p} \Vert f\Vert_{L^p}^2. $$ By strict concavity of the map $t \in \mathbb{R} + \mapsto \vert t \vert^{\frac{2}{p}}$ and the optimality of (f), $$ \Vert f \Vert_{L^p}^2 <\Vert f_+ \Vert_{L^p}^2 + \Vert f_- \Vert_{L^p}^2 \le \frac{\Vert f \Vert_{L^p}^2}{\Vert f \Vert_{H^1}^2} \bigl(\Vert f_+ \Vert_{H^1}^2 + \Vert f_- \Vert_{H^1}^2 \bigr) = \Vert f \Vert_{L^p}^p, $$ which is a contradiction.

Essentially, the argument says that if $f$ changes sign then taking either $f_+$ or $f_-$ increases the quotient.

The argument extends to maximizers for embedding of $W^{1, q}$ with $1 \le q < \infty$.

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that's a nice argument. – Delio Mugnolo Oct 30 '13 at 9:12
@ user36236. How do you conclude that $\Vert f\Vert_{H^1}^2=\Vert f_+\Vert_{H^1}^2+\Vert f_- \Vert_{H^1}^2$ hold? I have seen this kind of thing also elsewhere but no proof. – TaQ Nov 3 '13 at 12:14
@TaQ: $f_+$ and $f_-$ have disjoint supports. – Pietro Majer Nov 3 '13 at 12:26
@ Pietro Majer. Usually the support is the closure of the set where the function is nonzero. If you have for example the identity $f:\mathbb R\to\mathbb R$, then the support of $f_+$ is $[0,+\infty[$ , and that of $f_-$ is $]-\infty,0]$ . Their intersection is $\{0\}$ . So they do not have disjoint supports. – TaQ Nov 3 '13 at 12:42
I also have a feeling (but not an explicit example) that one can construct an example where ${\rm supp\kern.5mm}f_+\cap{\rm supp\kern.5mm}f_-$ has positive measure. – TaQ Nov 3 '13 at 12:50

For bounded domains it follows from the Rellich-Kondrashov compactness theorem $H^1\subset\subset L^p$.

If you denote by $$ S = \inf \frac{\|u\|{H^1}}{\|u\|_{L^p}} $$

and take a sequence $\{u_n\}$ such that $\|u_n\|_{H^1}\to S$ and $\|u_n\|_{L^p} = 1$

then there exists a subsequence $\{u_{n_k}\}$ and a function $u$ such that $u_{n_k}\rightharpoonup u$ weakly in $H^1$ and so

$$ \|u\|_{H^1}\le \liminf \|u_{n_k}\|_{H^1} = S $$

Now, you apply R-K Theorem to conclude that $u_{n_k}\to u$ strongly in $L^p$ and so $\|u\|_{L^p}=1$.

By the maximum principle, as you said, it follows that the function $u$ must be positive inside the domain.

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