I'm stuck on the following problem:
Assume that $U$ is bounded and there exists a smooth vector field $\alpha$ such that $\alpha\cdot\nu\ge 1$ along $\partial U$, where $\nu$ as usual denotes the outward unit normal. Assume $1\le p<\infty$.
Apply the Gauss-Green theorem to $$\int_{\partial U} |u|^p dS\le C\int_U |Du|^p+|u|^p dx$$ for all $u\in C^1(\bar{U})$.
I have two issues with this:
First, how can we apply the Gauss-Green theorem in the case $p=1$, since $|u|$ is not smooth? I'm thinking maybe a mollification argument?
In the case $1<p<\infty$, when we go to compute: $$\int_{\partial U}|u|^p dS \le \int_{\partial U}(|u|^p\alpha)\cdot\nu dS=\int_U \text{div} (|u|^p\alpha)dx=\int_U(D|u|^p)\cdot \alpha+|u|^p\text{div}(\alpha)dx$$ The second term can be bounded by $C\int_U |u|^pdx$. As for the first, we have that $|D|u|^p|=p|u|^{p-1}|Du|$, and I don't see how to bound this in terms of $|Du|^p$ (Hölder's inequality introduces a product of integrals which does not seem very tractable).
Any hints?
In the $p=1$ case you could try proving the result for $u_\epsilon(x) = \sqrt{u^2 + \epsilon^2}$ and taking a limit.
In case $p > 1$ what you need is Young's inequality: if $\frac 1p + \frac 1q = 1$ then for $a,b \ge 0$ you have $$ab \le \frac{a^p}{p} + \frac{b^q}{q}.$$ In particular with $q = p/(p-1)$ you get $$|u|^{p-1} |Du| \le \frac{|Du|^p}{p} + \frac{(|u|^{p-1})^{p/(p-1)}}{p/(p-1)} = (p-1) \frac{|u|^p}{p} + \frac{|Du|^p}{p}.$$