Let $D\subseteq$ be an open, bounded Lipshitz domain, $1\le p<\infty$, $u \in W^{2,p}(D)$ and $\delta>0$. Show that there exists $C=C(\delta)>0$ such that
\begin{align} \|Du\|_{L^p(D)} \le \delta\|D^2u\|_{L^p(D)}+C\|u\|_{L^p(D)}. \end{align}
The following proof is adapted from Troianiello's book Elliptic differential equations and obstacle problems:
Assume to the contrary that there exist
$\delta_0>0$ and $u_k\in W^{2,p}(D), k\in \mathbb{N}$ s.t.
$\|u_k\|_{W^{2,p}(D)}=1$ and
\begin{align}
\|Du_k\|_{L^p(D)}
>\varepsilon_0 \|D^2u_k\|_{L^p(D)}
+k\|u_k\|_{L^p(D)}.
\end{align}
Therefore,
$$
\|u_k\|_{L^p(D)}
<
\frac{1}{k}
\|Du_k\|_{L^p(D)}
<
\frac{1}{k}
\|u_k\|_{W^{2,p}(D)}
=\frac{1}{k},
$$
and hence, $\lim\limits_{k\to \infty} \|u_k\|_{L^p(D)}=0$.
Since $\|u_k\|_{W^{2,p}(D)}=1$, by virtue of the Rellich-Kondrachov Theorem
($W^{2,p}(D) \subseteq W^{1,p}(D)$ compactly),
there exists $\{u_{k_l}\}_{l=1}^\infty \subset \{u_k\}_{k=1}^\infty$ and $u\in W^{1,p}(D)$
such that
\begin{align}
\lim\limits_{l\to \infty}
\|u_{k_l}-u\|_{W^{1,p}(D)}=0.
\end{align}
Combining this with $\lim\limits_{l\to \infty}\|u_{k_l}\|_{L^p(D)}=0$ and
$$
\|u\|_{L^p(D)}
\le
\|u_{k_l}-u\|_{L^p(D)}
+
\|u_{k_l}\|_{L^p(D)}
\le
\|u_{k_l}-u\|_{W^{1,p}(D)}
+\|u_{k_l}\|_{L^p(D)},
$$
we have $\|u\|_{L^p(D)}=0$, so $u=0$.
Therefore,
$$
\|Du_{k_l}\|_{L^p(D)}
=
\|Du_{k_l}-Du\|_{L^p(D)}
<
\|u_{k_l}-u\|_{W^{1,p}(D)}
\to 0
$$
as $l\to \infty$. Consequently,
$$
\|D^2u_{k_l}\|_{L^p(D)}
<
\frac{1}{\delta_0}
\|Du_{k_l}\|_{L^p(D)} \to 0
$$
as $l\to \infty$.
Since $$\|u_{k_l}\|_{W^{2,p}(D)} \sim
\|u_{k_l}\|_{L^p(D)}
+
\|Du_{k_l}\|_{L^p(D)}
+\|D^2u_{k_l}\|_{L^p(D)},$$ we have
$\lim\limits_{l\to \infty} \|u_{k_l}\|_{W^{2,p}(D)}=0$, contradicting
$\|u_{k_l}\|_{W^{2,p}(D)}=1$.
I have two questions:
How do we guarantee that the constant $C$ depend on $\delta>0$ only and independent of domain $D$?
Is there a direct proof of this inequality (not using contradiction)?
Thank you very much.