Let $I$ be an open interval in $\mathbb{R}$.
We define $H= \{v \in H^1(I) \ | \ v(0)=0 \} \subset H^1(I)$ with the scalar product of the Sobolev space $H^1(I)$, i.e. $(u,v)=(u,v)_{L^2(I)}+(u',v')_{L^2(I)}$.
I want to show that $H$ is a Hilbert space. We already know that $(u,v)$ is a scalar product in $H$. We have to show that $H$ is complete i.e. $H$ is closed on $H^1(I)$.
We pick a convergent sequence $(v_n) \subset H$ such that $v_n \to v$. We want to show that $v \in H$. As $(v_n)$ converges in the norm $||\cdot ||_{H^1(I)}$ and $$||v_n||_{L^{\infty}(I)} \leq ||v_n||_{H^1(I)}$$ (Sobolev embedding) then $v_n \to v$ uniformly. Then, as $v_n(0)=0$, $v(0)=0$.
In one dimension $v$ has a continuous representative. Then $v(0)=0$. Otherwise $v(0) $ might be non defined.