Let $a, b \in \mathbb{R}$, such that $a < b$.
How do we prove that if $\;v \in H^1_0(a,b),\,$ then $\,v(a)= v(b)= 0$?
By definition, $H^1_0(a,b)$ is the completion of smooth compactly supported functions on $(a,b)$ with respect to the $H^1$ norm.
Thanks for your help.
in one dimension the Sobolev embedding theorems imply that $H^{1,2}(a,b)$ is continuously embedded into $C^0$ (even $C^{0,\alpha}$ for appropriately chosen $\alpha$). This implies in particular that approximation in the $H^{1,2}$ norm implies uniform convergence hence -- of course -- pointwise convergence.
Edit (responding to questions in comments): it is actually easy to see in the case you are looking at: suppose $f_n$ converges in $H^1$ to $f$, where the $f_n$ are smooth with compact support in $]a,b[$. This, first of all, implies that the $f_n$ can be extended continuously to $[a,b]$ by setting $f_n(a)=f_n(b) =0$.
Then (by the fundamental theorem of calculus) $|f_n(x)-f_m(x)| $ equals $$|f_n(x)-f_m(x) - (f_n(a)-f_m(a))| =| \int\limits_a^x (f_n-f_m)^\prime(t)dt| \le (b-a) ||f_n-f_m||_{H^1} $$ So you get uniform bounds on the difference of the functions. Hence the limit $f$ is continous and allows continuous extension to the closed interval $[a,b]$ where it has to attain the same value as all the $f_n$, namely $0$.