# An inequality with critical Sobolev exponent

Let $\Omega\subset \mathbb{R}^n, n\geq 3$ be a nice bounded domain and $2^*=2n/(n-2)$ 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.

• 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...) Commented Jun 30, 2015 at 7:48
• @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!
– teh
Commented Jun 30, 2015 at 8:26

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.

• 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.
– teh
Commented Jun 30, 2015 at 12:08