# About a counterexample of an inequality？

I have known how to use the compactive argument to prove the inequality (1), i.e. $1\leqslant p<n$, $\Omega\subset R^n$ is a bounded domain,$\forall \varepsilon>0$, there is $C_\varepsilon>0$, such that $\forall u\in W^{1,p}(\Omega)$, we have $$\|u\|_{L^p(\Omega)}\leqslant\varepsilon\|\nabla u\|_{L^p(\Omega)}+C_\varepsilon\|u\|_{L^1(\Omega)} \tag{1}$$

Now,I want to find out a counterexample to demonstrate that $$\|u\|_{L^{p*}(\Omega)}\leqslant\varepsilon\|\nabla u\|_{L^p(\Omega)}+C_\varepsilon\|u\|_{L^1(\Omega)} \tag{2}$$

namely if we replace $\|u\|_{L^p(\Omega)}$ in $(1)$ with $\|u\|_{L^{p*}(\Omega)}$ to get $(2)$, then $(2)$ is not true in general, where $\frac{1}{p*}=\frac{1}{p}-\frac{1}{n}$.

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You can do it with scaling. Pick the unit ball for $\Omega$, and let $u$ be any nonzero smooth function with support contained in $\Omega$. Now for each $\lambda<1$, let $u_\lambda(x)=u(x/\lambda)$. If (2) holdes for $u_\lambda$, we find $$\lambda^{n/p^*}\lVert u\rVert_{L^{p^*}}\le\epsilon\lambda^{n/p-1}\lVert\nabla u\rVert_{L^p}+\lambda^nC_\epsilon\lVert u\rVert_{L^1}.$$ Now the first two exponents of $\lambda$ are the same, so this becomes $$\lVert u\rVert_{L^{p^*}}\le\epsilon\lVert\nabla u\rVert_{L^p}+\lambda^{p^*}C_\epsilon\lVert u\rVert_{L^1}.$$ Finally, pick $\epsilon$ small enough so that letting $\lambda\to0$ yields a contradiction.
There was a misprint in my answer: $L^p$ on the LHS of the two inequalities should have been $L^{p*}$. Oddly, this edit was suggested by an anonymous user and rejected by two reviewers. I found no way to override the two, other than to do the same edit myself. – Harald Hanche-Olsen Jan 18 '13 at 13:32