let $\Omega = (0,1)$ and $H^1 (\Omega)$ be the Sobolev space defined as follows :
$$ H^1 (\Omega) = \{v \in L^2(\Omega )\; ;\; v' \in L^2(\Omega )\}$$
where $v'$ is the weak derivative of $v$. meaning :
$v' \in L^2(\Omega ) $ and $\int_0^1 v(x) \phi'(x)dx = - \int_0^1v'(x)\phi(x)dx$
for every continuously differentiable function $\phi$ of compact support
then in some lecture notes a friend gave me they prove that the set $F = \{v \in H^1 (\Omega), \; v(1) = 0\}$ is a Hilbert sub-space of $H^1 (\Omega)$ by showing that it's the kernel of a continuous map but that's not the thing that's bothering me.
something I couldn't find an explanation for : is how $v(1)$ is guaranteed to exist ?
as far as I know discontinuous functions could be elements of $L^2(\Omega )$, not sure concerning $H^1 (\Omega)$ as it's smaller than $L^2(\Omega )$.
if anybody could explain why $v(1)$ is assured to exist, it'd be great. thanks !
In fact the functions in $H^1$ are continuous. A very sloppy argument to that effect, that can be cleaned up:
If $t>0$ then $$f(x+t)-f(x)=\int_x^{x+t}f'(s)\,ds,$$hence
$$|f(x+t)-f(x)\le||f'||_2\left(\int_x^{x+t}1^2\right)^{1/2}=t^{1/2}||f'||_2.$$
(The application of the fundamental theorem of calculus is the sloppy part. You could clean this up by applying more or less the same argument to $f*\phi_n$, where $(\phi_n)$ is a smooth approximate identity. In fact $\phi_n*f$ is smooth and $(\phi_n*f)'=\phi_n*f'$, so $||(\phi_n*f)'||_2\le||f'|||_2$. So the argument above shows that $(\phi_n*f)$ is equicontinuous; now $\phi_n*f\to f$ almost everywhere and some subsequence converges uniformly, so $f$ is continuous.)