It's well-know that if $\Omega \subset \mathbb{R}^N$ is a smooth bounded domain, then the Sobolev Space $W^{1,2}(\Omega) \hookrightarrow L^p(\Omega)$ for $p \in [1,2^* ]$, where $2^*= \dfrac{2N}{N-2}$. With this, there exists a positive constant $C_1$ in a such way that
$$ \bigg(\int_{\Omega} |u|^p dx\bigg)^{1/p} \le C_1 ||u||_{W^{1,2}(\Omega)}, \quad \forall u \in W^{1,2}(\Omega) $$
Moreover, we have also the trace embedding $W^{1,2}(\Omega) \hookrightarrow L^q(\partial \Omega)$, for $q \in [1,2_* ]$, where $2_* = \dfrac{2(N-1)}{N-2}$. With this, there exists a positive constant $C_2$ such that
$$ \bigg(\int_{\partial \Omega} |u|^q d\sigma\bigg)^{1/q} \le C_2 ||u||_{W^{1,2}(\Omega)}, \quad \forall u \in W^{1,2}(\Omega) $$
Inspired in this informations, my question is: Let $p\ge 1$ and $u \in L^p(\Omega) \cap L^p(\partial \Omega)$. There exists a inequality of type:
$$\int_{\partial \Omega} |u|^p d\sigma\le C \int_{\Omega} |u|^p dx \quad \mbox{?}$$
PS: I think that is the only possibility of this comparison, once know that If $u$ has compact support, then $||u||_{L^p(\partial \Omega)} =0$.
I would need to go back through the proof of the boundedness of the trace operator for Sobolev spaces to see exactly what conditions on the spaces are needed for this to be true, but as stated, it is false.
Let $\Omega = [0,1]$ and consider the sequence of functions $u_n(x) = \begin{cases} n^{1/p},& x\in[0,1/n] \\ 0,&\text{else}\end{cases}$.
Then $\|u_n\|_{L^p(\Omega)} = 1$ and $\|u_n\|_{L^p(\partial\Omega)} = n^{1/p}\to\infty$, so no such $C$ can exist.
You can also divide the entire problem through by $n^{1/p}$ to obtain a sequence where the boundary norm is constant and the domain norm goes to zero.