How do I interpret the completion of $C_{c}^{\infty}$

Let $X$ be the vector space completion of $C_{c}^{\infty}(\mathbb{R}^n)$ under the norm $$ \|u\|_{X}=\left(\int_{\mathbb{R}^{n}}|Du|^{2}dx\right)^{\frac{1}{2}} $$ I'm trying to show that the Sobolev space $H^{1}(\mathbb{R}^{n})$ is a proper subset of $X$. However I'm not even sure what the completion of $C_{c}^{\infty}(\mathbb{R}^n)$ looks like. I only know the fact that it is the closure of its image under the canonical embedding $*:C_{c}^{\infty}(\mathbb{R}^n) \to (C_{c}^{\infty}(\mathbb{R}^n))^{**}$.

Solution 1:

A partial answer. Let $n>2$. Then, for any $f∈ C^\infty_c$, the following inequality (Sobolev's embeddings) holds $$ \|f\|_{L^p} ≤ C_{n} \|∇f\|_{L^2} $$ where $p=\frac{2\,n}{n-2}$ and some constant $C_n$ depending on the dimension. Take $f ∈ X$ and for any $k\in\Bbb N$, take $f_k ∈ C^\infty_c$ such that $f_k \to f$ in $X$ (and so in particular in the sense of distributions). Then $S = \sup_k \|∇f_k\|_{L^2} < ∞$ and so $f_k$ is uniformly bounded in $L^p$. In particular, up to a subsequence, it converges weakly in $L^p$ to a function $g∈L^p$, and so it also converges in the sense of distributions. Hence $f=g∈L^p$. This proves that when $n>2$ $$ X ⊆ L^p. $$