Let $\Omega$ be a $C^1$ domain, for any $\epsilon>0, 0<|\alpha|<k$, there exists a $C_\epsilon$ such that $$||D^\alpha u||_{L^p}\leq \epsilon||u||_{W^{k,p}}+C_\epsilon||u||_{L^p}$$
for all $u\in W^{k,p}$.
I was trying to prove it by contradiction.
$\forall n \in \mathbb{N}, \exists u_n \in W^{k,p}$ such that $||D^\alpha u_n||_{L^p}>\epsilon||u_n||_{W^{k,p}}+n||u_n||_{L^p}$.
Anyone could give me a hint about using Sobolev compactness embedding?
I was trying to use the $W^{k,p}\subset \subset L^p$. However, I do not how to show that the sequence is bounded in $sup$.
By taking normalisation, $v_n=\frac{u_n}{||u_n||_{W^{k,p}}}$, we would have
$$\frac{1}{||u_n||_{W^{k,p}}}||D^\alpha u_n||_{L^p}>\epsilon+\frac{n}{||u_n||_{W^{k,p}}}||u_n||_{L^p}$$
Best Answer
It turns out that the embedding $W^{k,p} \subset \subset L^p$ being compact is not good enough. We need something slightly stronger : for $0<|\alpha| < k$ the embedding $W^{k,p} \subset \subset W^{|\alpha|,p}$ is compact as well.
Once you have (following normalization) $\|u_n\|_{W^{k,p}} = 1$ for all $n$, the sequence $u_n$ is bounded in $W^{k,p}$ (because $\sup_n \|u_n\|_{W^{k,p}} =1$ as we normalized each of them), therefore by compact inclusion there is a convergent subsequence $u_{n_j} \xrightarrow{W^{|\alpha|,p}} u$.
Now, write down the contradictory statement : $$ \|D^{\alpha} u_{n_j}\|_{L^p} > \epsilon\|u_{n_j}\|_{W^{k,p}} + n \|u_{n_j}\|_{L^p} = \epsilon + n \|u_{n_j}\|_{L^p} \tag{*} $$
Note that since $1 = \|u_{n_j}\|_{W^{k,p}} \geq \|D^{\alpha}u_{n_j}\|_{L^p}$, the LHS is bounded by $1$. From here, the right hand side must be bounded by $1$. But for that to occur, we must have $\|u\|_{L^p} = 0$, since if not , then the RHS will look like $\epsilon + n \|u\|_{L^p}$ for large $n$, which is not bounded unless $\|u\|_{L^p} = 0$. Hence $u=0$.
But we also have $D^{\alpha} u_{n_j} \xrightarrow{L^p} D^{\alpha} u = 0$. However, if we take limits in $*$ we get $\|D^{\alpha}u\|_{L^p} > \epsilon$. This completes the contradiction.