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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 $\mathbb{R}^n$.

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    $\begingroup$ Crossposted on MSE. $\endgroup$
    – Silvia Ghinassi
    Commented Mar 8, 2016 at 14:34
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    $\begingroup$ Very roughly: $$ \int |Du|^{2p} = \int |Du|^{2p-2} Du\cdot Du = - \int D( |Du|^{2p-2} Du) u$$. The last term is $$ \lesssim \int |D^2 u||Du|^{2p-2} |u|$$ and use Holder with $\frac{1}{\infty} + \frac{1}{p} + \frac{p-1}{p} = 1$. $\endgroup$
    – Willie Wong
    Commented Mar 8, 2016 at 14:38
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    $\begingroup$ This is a special case of the Gagliardo-Nirenberg inequalities, and, as Willie shows, they can all be proved using integration by parts and the Holder inequality. $\endgroup$
    – Deane Yang
    Commented Mar 8, 2016 at 18:49

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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) .

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