I was reading a bit and came across this:
It's kind of confusing. I recall hearing that functions in $H^1_{0}(\Omega)$ were just functions in $H^1(\Omega)$ (the function and its derivatives are square integrable) with value of 0 on the boundary. But apparently this is just something that can be shown to equivalent to, but not the actual definition. I'm curious about how one would show this but I have more basic questions to ask.
So I know that the closure of a set is what we get when we throw in that set's limit points. For example the closure of $(0,1)$ is $[0,1]$ But it looks like I'm having trouble making the transition to when the elements of the set are functions. I'm having difficulty understanding what the closure of $C^{\infty}_{0}(\Omega)$ would look like.
I may have been thinking that C-infinity functions (with compact support) were a subset of $H^1_{0}(\Omega)$ functions. But that doesn't really make sense because $H^1_{0}(\Omega)$ functions are defined by integrals and so don't even need to be continuous, let alone smooth. I think the correct interpretation is that functions in $H^1_{0}(\Omega)$ can be "approximated" by smooth functions with compact support(BTW why does the definition of support of a function involves taking closure?)
So a few more questions.
the u is the limit in H1 bit is a bit confusing. What does that mean exactly?
If we drop the "0" can we still make the same definition and have it make sense? i.e can we say that $H^1(\Omega)$ is the closure of $C^{\infty}(\Omega)$? I would imagine so since it seems that the "0" is only needed to take into account the fact that we have homogeneous Dirichlet boundary conditions.
Why do we need the boundary to be sufficiently smooth? what happens if it isn't?
Can someone provide an actual example of an $H^{1}_{0}(\Omega)$ function actually being approximated by smooth functions?