The question is from Gilbarg and Trudinger Elliptic PDE of second order. Here's the statement:
Let $\Omega$ be a $C^1$ domain, prove the following interpolation inequality: \begin{eqnarray} \|D^{\alpha}u\|_{L^p(\Omega)} \le \epsilon\|u\|_{W^{k,p}(\Omega)} + C_\epsilon\|u\|_{L^p(\Omega)} \end{eqnarray} where $0<|\alpha|<k$, $1 \le p < \infty$, and $C_{\epsilon}$ is a constant independent of u.
Any help, hints, or approaches will be appreciated.
Also, could you explain to me what exactly interpolation means here? Interpolation in what sense? What I mean is that say, for example, the interpolation in $L^p$ and $L^q$, WLOG, say $p < q$. If certain thing holds for p and q, then interpolation would mean that it holds for all $p < x < q$. So what exactly interpolation means for this question?
Interpolation means here: you estimate the norm of some intermediate derivative by the $k$-th and the zeroth derivative, which is in some sense an interpolation between the orders of differentiation.
The correct statement should be: For every $\epsilon>0$ there is $C_\epsilon>0$ such that for all $u$ ...
Now, assume the claim is not valid. Then there is $\epsilon>0$ and a sequence $(u_n)$ such that $$ \|D^\alpha u_n \|_{L^p} \ge \epsilon \|u_n\|_{W^{k,p}} + n \|u_n\|_{L^p}. $$ We can rescale $(u_n)$ such that $\|u_n\|_{W^{k,p}}=1$. By compact embeddings (here we need boundedness of $\Omega$), $u_n$ converges to some $u$ in $W^{k-1,p}(\Omega)$ (after extracting a subsequence).
Dividing the above inequality by $n$ and passing to the limit yields $u=0$. Passing to the limit in $$ \|D^\alpha u_n \|_{L^p} \ge \epsilon \|u_n\|_{W^{k,p}} + n \|u_n\|_{L^p} \ge \epsilon $$ yields the final contradiction.