I'm trying to prove the Friedrich's inequality in $\mathbb{R}^n$, but I first want to prove it in $\mathbb{R}$ because using iterated integrals I can then use the same idea in $\mathbb{R}^n$. So let $\Omega$ be an open bounded subset of $\mathbb{R}$.
As $\Omega$ is bounded and $\mathbb{R}$ is well ordered it follows that there exists a least $a\in \mathbb{R}^+$ such that $\Omega \subseteq [-a,a]$. We consider first $v \in C^\infty_0(\Omega)$. Without loss of generality then we can assume that $\Omega$ is connected. Now fix an arbitrary $x\in \Omega$, because $v$ can be taken to be $0$ outside of $\Omega$. By the Cauchy-Schwarz inequality and the fundamental theorem of calculus it follows that \begin{align*} v(x)^2 & =\left (\int_{-a}^xv'(t)dt\right) ^2 \\ & \leq \int_{-a}^{x}(v'(t))^2dt\int_{-a}^{x}dt \\ & \leq \int_{-a}^{x}(v'(t))^2dt\int_{-a}^{a}dt \\ & \leq 2a\int_{\Omega}(v'(x))^2dx. \end{align*} As this is true for any $x\in \Omega$ we can integrate over $\Omega$ on both sides \begin{align*} \int_{\Omega}v(x)^2dx & \leq \int_{\Omega}2a \int_{\Omega}(v'(x'))^2dx'dx \\ & \leq 2a\int_{-a}^a \int_{\Omega}(v'(x'))^2dx'dx\\ & = 4a\int_{\Omega}(v'(x))^2dx. \end{align*} This proves the result for all $v \in C^\infty_0(\Omega)$. For any function $u\in H_0^1(\Omega)$ there exists a sequence of functions $\varphi_n$ in $C^\infty_0$ converging to $u$ in $H^1(\Omega)$. It follows from above that $$ \int_{\Omega}\varphi_n(x)^2dx\leq 4a\int_{\Omega}(\varphi_n'(x))^2dx$$ for all $n\in \mathbb{N}$. As the integral and $t\to t^2$ are continuous functions, and because $\varphi_n \to u$ and $\partial x \varphi_n \to \partial x u$ in $L^2$ it follows that $$\int_{\Omega}u(x)^2dx\leq 4a\int_{\Omega}(u'(x))^2dx.$$
I believe my proof is essentially correct, if not please let me know, but my main issue is in the extension from $C_0^\infty$ to $H_0^1$ using a limiting argument. Am I taking my limits in the right spaces? Am I looking for convergence in $L^2$ or $H^1$ and why? and how can I be more rigorous about it? Any help would be appreciated.
