1
$\begingroup$

This is related to my previous question An inequality involving Sobolev embedding with epsilon. There I wished to get that, for given a nice bounded domain $\Omega$ in $\mathbb{R}^n$, $\forall \epsilon>0$, $\exists C_\epsilon$ s.t. $$ \|u\|_{2^*}\leq \epsilon \|\nabla u\|_2+C_\epsilon \|u\|_2, \quad \forall u\in H^1(\Omega), $$ where $2^*=2n/(n-2)$ is the critical Sobolev exponent. Due to the lack of compact embedding from $H^1$ into $L^{2^*}$. The above inequality is inedd incorrect, see my previous thread. Now, I hope to get a stengthened version of it: given $p\in (2,2(n+2)/n)$ (or $p\in(2,2^*)$ in the worst case), $\forall \epsilon>0$, $\exists C_\epsilon$ s.t. $$ \|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{\frac{p}{2}})+C_\epsilon(1+ \|u\|_2), \quad \forall u\in H^1(\Omega). \tag{MCIS} $$ I tried the example listed in MathOverflow: Counterexample Showing The Rellich-Kondrachov Theorem Is Sharp, which does not give a counterexample. Also, arguing by contradiction seems not to work. Any help is greatly acknowledged.

$\endgroup$
1
$\begingroup$

The $1$ in parentheses is useless; it doesn't scale if you multiply $u$ by a constant. So your inequality can hold for all $u\in H^1$ if and only if $$\|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{\frac{p}{2}})+C_\epsilon \|u\|_2$$ holds. And here the scaling of the variable $u_t = u^{n/2^*}u(tx)$ becomes an issue: namely, $\|u_t\|_{2^*}$ and $\|\nabla u_t\|_2$ are constant, while $\|u_t\|_p\to 0$ for all $p< 2^*$.

So, you don't get this for $p<2^*$. (And for $p\ge 2^*$, the inequality follows just from Hölder's inequality).

$\endgroup$
  • $\begingroup$ Thank you @1999. The inequality is incorrect, and has been answered in [mathoverflow.net/questions/210446/… $\endgroup$ – teh Jul 2 '15 at 1:44
  • $\begingroup$ So you are posting in two places and wait where you get an answer first? Your choice. But this is the last time I answered your question. $\endgroup$ – user147263 Jul 2 '15 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.