Two density results, $C_{0}^{\infty}$ is dense in $L^{2}$

Question

I really lack analysis background but I'm in a situation with PDE's (Finite elements specifically) where I'm trying to prove a couple intermediate results.

1. Prove $C_{0}^{\infty}(\Omega)$ dense in $L^{2}(\Omega)$ ($\Omega$ is some domain): The hint for the problem is to show $(C_{0}^{\infty}(\Omega))^{\bot}=\{0\}$, which I guess I simply did by invoking a previous exercise whose result is: for open set $\Omega$ and $f \in L^{1}_{\text{loc}}(\Omega)$ such that $\int_{\Omega}f \phi = 0, ~\forall \phi \in C_{0}^{\infty}(\Omega)$, then $ f=0, ~ a.e.~~~~ $ So now what?

2. For domain $\Omega, \text{such that } \partial \Omega=\Gamma_{1} \cup \Gamma_{2}$, show that if $\int_{\Gamma_{2}} v \frac{\partial u}{\partial n} = 0~~ \forall v \in V=\{v \in H^{1}(\Omega): v\big|_{\Gamma_{1}}=0\}$, then $\frac{\partial u}{\partial n}=0,~ a.e.~~$ I'm told I need to prove that $V \big|_{\Gamma_{2}} $ is dense in $L^2(\Gamma_{2})$. In general I don't really know how to come up with density arguments (I understand the very general premise only) and certainly here don't understand what this restriction would look like in the proof. Thanks.

Answer 1

For the first part: $L^2(\Omega)\subset L^1_{loc}(\Omega)$, hence indeed you can apply your previous exercise. Then again, what is $(C^{\infty}_0)^\bot$?

It is merely the set $\{f\in L^2(\Omega) : \, \forall \phi\in C^{\infty}_0\int_\Omega f \phi =0\}$, which results in $(C^{\infty}_0)^\bot=\{0\}$.

Finally, the result "orthogonal complement of the subspace is zero, hence this subspace is dense" should be in your textbook.

For the second part, you have the inclusion $$L^2(\Gamma_2)\subset \left\{v\big|_{\Gamma_2}:v\in H^1(\Omega)\right\}$$ i.e. the restrictions of functions in $H^1(\Omega)$ on $\Gamma_2$ cover the whole $L^2(\Gamma_2)$. This result should also be in most books on functional analisys and Sobolev spaces. This inclusion implies that $\frac{\partial u}{\partial n}$ restricted to $\Gamma_2$ is orthogonal to all functions in $L^2(\Gamma_2)$, hence by the already mentioned density result we have that $\frac{\partial u}{\partial n}=0$ on $\Gamma_2$.

Another approach would be to say that $C^\infty(\Omega)\subset H^1(\Omega)$, and then $$C^\infty(\Omega)\cap \{v\big|_{\Gamma_1}=0\}\subset V.$$ Then again, $$C^\infty(\Gamma)\subset C^\infty(\Omega)\big|_{\Gamma },$$ hence $$\big(C^\infty(\Omega)\cap \{v\big|_{\Gamma_1}=0\}\big)\big|_{\Gamma_2}\supset C^\infty(\Gamma_2), $$ and the last space is dense in $L^2(\Gamma_2)$.