Let $2\le p<\infty$, $U\subset \mathbb R^n$ being bounded with $C^1$ boundary, and $u\in W^{2,p}(U)\cup W^{1,p}_0(U)$. (I guess this is a typo in my textbook and it should be $u\in W^{2,p}(U)\cap W^{1,p}_0(U)$.)
The inequality to be proved is $$\int_U|Du|^pdx\le C \left(\int_U|u|^pdx\right)^{1\over2} \left(\int_U|D^2u|^pdx\right)^{1\over2}.$$
I was told to use integration by parts to prove this inequality. But there is only one term on the left-hand side, and there is no "parts" to integrate. I am really confused now.
Thanks for help.
In case $p=2$ you have $$\int_U |Du|^2 \, dx = \sum_{k=1}^n \int_U u_{x_k}^2 \, dx.$$ You can integrate by parts to obtain $$\int_U u_{x_k}^2 \, dx = \int_U u_{x_k} u_{x_k} \, dx = - \int_U u \, u_{x_k x_k} \, dx$$ and apply Cauchy-Schwarz to get $$ - \int_U u \, u_{x_k x_k} \, dx \le \left( \int_U |u|^2 \, dx \right)^{1/2} \left( \int_U |u_{x_k x_k}|^2 \, dx \right)^{1/2} \le \left( \int_U |u|^2 \, dx \right)^{1/2} \left( \int_U |D^2 u|^2 \, dx \right)^{1/2}.$$
After you return to the original sum the inequality holds with $C=n$.
If $p > 2$ the details are messier so I will just get it started. Note that $$|Du|^p = |Du|^{p-2} |Du|^2 = \sum_{k=1}^n |Du|^{p-2} u_{x_k}^2= \sum_{k=1}^n |Du|^{p-2} u_{x_k}u_{x_k}.$$ Integration by parts gives you $$ \int_U |Du|^{p-2} u_{x_k}u_{x_k} \, dx = - \int_U u \left(|Du|^{p-2} u_{x_k} \right)_{x_k} \, dx.$$ You need to estimate the last term.