$L^p$-bounding inequality Do we have that$$\|Du\|_{L^{2p}} \le C\|u\|_{L^\infty}^{1\over2} \|D^2u\|_{L^p}^{1\over2}$$for $1 \le p < \infty$ and all $u \in C_c^\infty(U)$? Here, $U$ denotes an open subset of $\mathbb{R}^n$.
 A: Let $u$ be a smooth compactly supported real-valued function defined on $\mathbb R^n$.We have
$$
\int \vert\partial_j u\vert^{2p} dx=\langle\partial_j u,\text{sign}({\partial_j}u )\vert\partial_j u\vert^{2p-1}\rangle=
-\langle u,\partial_j\bigl(\text{sign}({\partial_j}u )\vert{\partial_j}u\vert^{2p-1}\bigr) \rangle,
\tag 1$$
as a distribution bracket of duality. 
$\bullet$ Now a lemma: we consider for $\epsilon_0>0$ and $v$ a $C^1$ real-valued function defined on the real line, the function
$
w=v\vert v\vert^{\epsilon_0}.
$
Then the function $w$ is $C^1$ with 
$
w'=(1+\epsilon_0)\vert v\vert^{\epsilon_0} v'.
$
Proof.
This 
is obvious on the open set $\{v\not=0\}$ and if $v(x_0)=0$, we find easily that 
$w'(x_0)=0$ since
$$
w(x_0+h)-w(x_0)=v(x_0+h)\vert v(x_0+h) \vert^{\epsilon_0}=O(\vert h\vert^{1+\epsilon_0}).
$$
Continuity of $w'$ follows as well from $w'=(1+\epsilon_0)\vert v\vert^{\epsilon_0} v'$ on $\{v\not=0\}$.$\square$
Going back to $(1)$, we apply the lemma to $w=\partial_j u$ and we get, assuming $p>1$,
$$
\partial_j\bigl(\text{sign}({\partial_j}u )\vert{\partial_j}u\vert^{2p-1}\bigr)
=
\partial_j\bigl(({\partial_j}u )\vert{\partial_j}u\vert^{2p-2}\bigr)
=(2p-1)\vert{\partial_j}u\vert^{2p-2}\partial_j^2u.
$$
As a result (1) yields,
$$
\Vert\partial_ju\Vert_{L^{2p}}^{2p}\le(2p-1)
\Vert u\Vert_{L^{\infty}}\int \vert\partial_ju\vert^{2p-2}\vert\partial_j^2u\vert
dx\underbrace{\le}_{\text{Hölder}} (2p-1)
\Vert u\Vert_{L^{\infty}}
\Vert \partial_ju\Vert_{L^{2p}}^{2p-2}
\Vert \partial_j^2u\Vert_{L^{p}},
$$
yielding
$$
\Vert\partial_ju\Vert_{L^{2p}}^{2}\le(2p-1)
\Vert u\Vert_{L^{\infty}}\Vert \partial_j^2u\Vert_{L^{p}}.
$$
Also for $p=1$,
the distribution derivative of $\{\text{sign}({\partial_j}u )\vert{\partial_j}u\vert=\partial _ju\}$ is $\nabla \partial_ju$
so that the estimate holds as well for $p=1$(in fact in that case a direct integration by parts gives the answer) .
