Let $\Omega\subset \mathbb{R}^n, n\geq 3$ be a nice bounded domain and $2^*=2n/(n2)$ the critical Sobolev exponent. One may expect that $\forall \epsilon>0$, $\exists C_\epsilon<\infty$ such that $$ \u\_{2^*}\leq \epsilon \\nabla u\_2+C_\epsilon \u\_2, \quad \forall u\in H^1(\Omega). $$ Due to the lack of compact embedding from $H^1$ into $L^{2^*}$, the above inequality is indeed not true by the example listed this question. Now, I wish to make it right by formulating it in a strengthened version as follows: given $p\in (2,2(n+2)/n)$ (or $p\in(2,2^*)$ in the worst case), $\forall \epsilon>0$, $\exists C_\epsilon<\infty$ such that $$ \u\_{2^*}\leq \epsilon (\\nabla u\_2+\u\_p^{p/2})+C_\epsilon(1+ \u\_2), \quad \forall u\in H^1(\Omega). \tag{MCIS} $$ I tried the example listed here, which does not give a counterexample. Also, arguing by contradiction seems not to work. Any help to prove or disprove (MCIS) is greatly acknowledged.

$\begingroup$ For the example Pietro gave, doesn't $\v_\epsilon\_p$ for $p < 2^*$ (where $v_\epsilon = \epsilon^{n/2^*} u(x / \epsilon)$) go to zero for $\epsilon \to 0$? How is adding $\u\_p^{p/2}$ supposed to help? (With the $\epsilon$ in front it can only help if it diverges...) $\endgroup$– Willie WongCommented Jun 30, 2015 at 7:48

$\begingroup$ @Wille Wong, yeah, $\v_\epsilon\_p\rightarrow 0$ as $\epsilon\rightarrow 0$ for $p<2^*$. The large constant term $C_\epsilon$ may help in this case. While, in the otherwise case, $\u\_p$ may contribute! $\endgroup$– tehCommented Jun 30, 2015 at 8:26
1 Answer
Start with Pietro's example in the linked question. Fix $u$. Define $$ v_{\delta,M} = M\delta^{n/2^*} u(x/\delta) $$ We know that $\\nabla v_{\delta,M}\_2 = M\\nabla u\_2$ and $\v_{\delta,M}\_{2^*} = M\u\_{2^*}$. Choose $\epsilon < \frac12 (\u\_{2^*} / \\nabla u\_2 )$.
We also know that for $M$ fixed, as $\delta \to 0$ we have that $\u\_p \to 0$ in the range of $p$ you allow.
Now, let $\Lambda > 0$ be arbitrary. We show that for every fixed $\Lambda$ there exists $\delta$ and $M$ such that $$ \v_{\delta,M} \_{2^*} > \epsilon (\\nabla v_{\delta,M}\_2 + \v_{\delta,M}\_p^{p/2} ) + \Lambda (1 + \v_{\delta,M}\_2) $$ which will disprove your desired inequality.
By our choice of small $\epsilon$, it suffices to prove $$ \frac12 \v_{\delta,M} \_{2^*} = \frac{M}{2} \u\_{2^*} > \epsilon \v_{\delta,M}\_{p}^{p/2} + \Lambda + \Lambda \v_{\delta,M}\_2 = \epsilon M \v_{\delta,1}\_p^{p/2} + \Lambda M \v_{\delta,1}\_2 + \Lambda $$
First choose $M$ sufficiently large that $$ \frac{M}{4} \u\_{2^*} > \Lambda $$ then choose $\delta$ sufficiently small so that the first two terms on the right hand side of the desired inequality is negligible.

$\begingroup$ thank you for your cute observation! This once again shows the compact embedding is vital to have $\epsilon$ in front of the highest order term. $\endgroup$– tehCommented Jun 30, 2015 at 12:08