An inequality involving critical Sobolev exponent

Question

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 indeed incorrect, see my previous thread. Now, I hope to get a strengthened 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.

Answer 1

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).