Properties of the trace operator

Question

I am using the book of Evans and try to understand the trace operator. I have the following question:

Let $u \in H^1(U)$ with $Du = 0$ and $Tu = 0$. Then (because of $Du = 0$) I know that $u$ must be constant on each connected component of $U$. But if $u \in H^1(U) \cap C(\overline{U})$, we have $Tu = u \vert_{\partial U}$. Can I use this (or something else) to conclude that $u = 0$?

Answer 1

Since $Du=0$, it follows $u=$ constant on each connected component, hence $u$ is continuous on each component, thus the trace operator yields the values of $u$ on the boundary, hence $u=0$.

In pde analysis, one often works with domains $U$ that are open and connected sets. So you do not need to worry too much about connected components.

Answer 2

If $Tu=0$ then $u \in H^1_0(U)$ and you have the Poincare inequality

$$\|u\|_{L^2(U)} \leq C\|Du\|_{L^2(U)},$$

for any open and bounded $U$ (not necessarily connected). So if $Du=0$ and $Tu=0$ then $u=0$.