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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.

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.

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 $\forall \epsilon>0$, $\exists C_\epsilon$ such that $$ \|u\|_{2^*}\leq \epsilon \|\nabla u\|_2+C_\epsilon \|u\|_2, \quad \forall u\in H^1(\Omega). $$ This is due to the lack of a compact embedding from $H^1$ into $L^{2^*}$. The above inequality is indeed not true by the example listed this question. 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), $\forall \epsilon>0$, $\exists C_\epsilon$ 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 is greatly acknowledged.

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.

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 $\forall \epsilon>0$, $\exists C_\epsilon$ such that $$ \|u\|_{2^*}\leq \epsilon \|\nabla u\|_2+C_\epsilon \|u\|_2, \quad \forall u\in H^1(\Omega). $$ This is due to the lack of a compact embedding from $H^1$ into $L^{2^*}$. The above inequality does indeed hold (see this question). 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), $\forall \epsilon>0$, $\exists C_\epsilon$ 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 is greatly acknowledged.

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 $\forall \epsilon>0$, $\exists C_\epsilon$ such that $$ \|u\|_{2^*}\leq \epsilon \|\nabla u\|_2+C_\epsilon \|u\|_2, \quad \forall u\in H^1(\Omega). $$ This is due to the lack of a compact embedding from $H^1$ into $L^{2^*}$. The above inequality is indeed not true by the example listed this question. 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), $\forall \epsilon>0$, $\exists C_\epsilon$ 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 is greatly acknowledged.