If $u \in H^{1}(I)$ and $\int_{I}|u^{\prime}(s)|^{2}ds =0$, then $u$ is constant in $\overline{I}$, with $|I| < \infty$?
I know that, since $u \in H^{1}(I)$, then there exists a function $v \in C(\overline{I})$ such that $$ u = v, \quad \text{a.e.} \quad \text{on} \quad I, $$ and $$ v(x) - v(y) = \int_{x}^{y}v^{\prime}(s)ds, \quad \forall x,y \in \overline{I} $$ See Brezis, Theorem 8.2.
Now, since $\int_{I}|u^{\prime}|^{2}(s)ds = 0$, then $u^{\prime} = 0$ a.e. in I. Hence, $v^{\prime} = 0$, where $u = v$ a.e. in $I$. Therefore, we have $$ v(x) - v(y) = \int_{x}^{y}v^{\prime}(s)ds = 0, \quad \forall x,y \in \overline{I}, $$ this is, $$ v(x) = v(y) , \quad \forall x,y \in \overline{I} $$ Can I say that $u$ is constant in $\overline{I}$?