Can someone tell me whether $|D^2 u|$ is equivalent of writing $\frac{\nabla u}{|\nabla u|}\, D^2 u$? This relates to the post Estimate $L^{2p}$ norm of the gradient by the supremum of the function and $L^p$ norm of the Hessian
I wasn't sure why $$\int_{U}|Du|^{2p}\,dx\le C\int_{U} u|Du|^{2p-2}|D^2 u| dx$$ But to answer this very basic question it is not necessary to read in whole.
No, $|D^2u|$ is not the same as $\frac{\nabla u}{|\nabla u|}\, D^2 u$. One of them is a number and the other is a vector. One involves first order derivative and the other doesn't.
The argument you read is flawed: the gradient is not the sum of derivatives, nor can it be estimated by this sum. Instead, use the fact that the square of the gradient is the sum of squares of derivatives. So (using $i$ subscript for $x_i$ derivative),
$$\int_{U}|Du|^{2p}\,dx = \sum_i \int_{U} u_i u_i |Du|^{2p-2} dx = - \sum_i \int_{U} u ( u_i |Du|^{2p-2})_{i} dx $$ By the product rule,
$$ |( u_i |Du|^{2p-2})_{i}| \le |u_{ii}||Du|^{2p-2}+ (2p-2) |u_i||D^2u| |Du|^{2p-3} \le (2p-1)|D^2u||Du|^{2p-2} $$ This is how we end up with $$\int_{U}|Du|^{2p}\,dx \le C \int_U |u| |D^2u||Du|^{2p-2}$$ As Jose27 said, an application of Hölder's inequality yields $$ \int_U |u| |D^2u||Du|^{2p-2} \le \left(\int_U (|u| |D^2u|)^p \right)^{1/p} \left(\int_U|Du|^{2p } \right)^{1-\frac1p}$$ Moving the latter factor to the left yields $$ \int_U|Du|^{2p } \le C \int_U |u|^p |D^2u|^p $$