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 $\mathbb{R}^n$.



$L^p$-bounding inequality
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) .


