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Let $\Omega$ be a nice bounded domain in $\mathbb{R}^n$ and let $2^*=2n/(n-2)$ for $n>2$ or anything finite for $n\leq2$, i.e., the Sobolev exponent. Then, for any $p\leq 2^*$, one has $$ \|u\|_p\leq C\| u\|_{H^1}\leq C\|\nabla u\|_2+C\|u\|_2. $$ Furthermore, if $p<2^*$, then we have a strong inequality with epsilon: for any $\epsilon>0$, there exists $C_\epsilon>0$ such that $$ \|u\|_p\leq \epsilon\|\nabla u\|_2+C_\epsilon\|u\|_2, \quad \quad \forall u\in H^1(\Omega). $$ See "Why is this estimate using a compact embedding in a sobolev space true?" Now, I wish to show that the above inequality continues to hold for $2^*$: for any $\epsilon>0$, there exists $C_\epsilon>0$ such that $$ \|u\|_{2^*}\leq \epsilon\|\nabla u\|_2+C_\epsilon\|u\|_2, \quad \quad \forall u\in H^1(\Omega). \tag{C-S-I} $$ Note the continuity of $L^p$ norm: $\|u\|_p\rightarrow \|u\|_{2^*}$ as $p\rightarrow 2^*$ and the interpolation inequalities bwteen Sobolev spaces.I guess (C-S-I) is true. While, I do not know how to prove or disprove (C-S-I) (mainly one case $n>2$). I tried to argue it by contradiction, but not succeeded. Any help is greatly appreciated!

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  • $\begingroup$ I think this can really help you: mathoverflow.net/questions/81034/…. $\endgroup$
    – Micael
    Jun 28, 2015 at 6:00
  • $\begingroup$ Thank you Micael! Due to the lacking of compact embedding, (C-I-S) may be disproved via the example listed there. $\endgroup$
    – teh
    Jun 29, 2015 at 3:42
  • $\begingroup$ @1999. Yeah, you modified the example in Micael's link. Due to the lack of compact embedding from $H^1$ into $L^{2^*}$, (C-I-S) is indeed incorrect. I thus strengthened it to a new version, see math.stackexchange.com/questions/1342920/… Also, since $\|u_t\|_2=\|u\|_2$, why $\|u_t\|_2\rightarrow 0$ as $t\rightarrow \infty$ if $u\not\equiv 0$. $\endgroup$
    – teh
    Jun 30, 2015 at 5:00

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For completeness, here is the counterexample for $p=2^*$. Let $u$ be any nice compactly supported function. For $t>0$, define $u_t(x) = t^{n/p} u(tx)$. This function has the same $L^2$ norm as $u$. Also, $\nabla u_t(x) = t^{1+n/p} \nabla u(tx)$ which implies $$\|\nabla u\|_2^2 = t^{2(1+n/p) - n}\|\nabla u\|^2=\|\nabla u\|^2$$ because $n/p = n(n-2)/(2n) = n/2-1$.

On the other hand, $\|u_t\|_{2}\to 0$ as $t\to\infty$. Fix $\epsilon<\|u\|_p/\|\nabla u\|_2$ and observe that the ratio $$\frac{\|u_t\|_p-\epsilon\|\nabla u_t\|_2}{\|u_t\|_2}$$ tends to $+\infty$ as $t\to\infty$.

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