Completion of $C^{\infty,2}_0$ is a subset of $L^2(\mathbb{R}^n)$? this question comes from quantum mechanics in physics. Let $C^{\infty}_0$ be the space of smooth functions with compact support. Then define $C^{\infty,2}_0 := C^{\infty}_0 \oplus C^{\infty}_0 $. and define an inner product $(\cdot , \cdot)_0$ by
$$
\begin{equation}
(f,g)_0 = \sum^n_{i=1} (D_i f_1, D_i g_1) + M^2 (f_1,g_1) + (f_2, g_2),
\end{equation}
$$
where $f,g \in C^{\infty,2}_0 $, $f_i$ are the components of $f$, $M^2$ is a positive constant and $(\cdot, \cdot)$ denotes the $L^2$ scalar product on $\mathbb{R}^n$.
My question is if the completion of $C^{\infty,2}_0$ with respect to the induced norm is a subset of $L^2(\mathbb{R}^n)^2$.
 A: $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
I think I found an answer to this:
Following the discourse in [1] gives us a simple criterion if the above claim is true. Note that we have 
\begin{equation}
    \norm{f}_{L^2} \leq \norm{f}_0, \quad \forall f \in C^{\infty,2}_0.
    \label{comparisonnorms}
\end{equation}
To prove this, we change to a more general setting and let $L$ be a linear space with two norms $\norm{\cdot}_{E^1}$ and $\norm{\cdot}_{E^2}$ such that $\norm{f}_{E_1} \leq \norm{f}_{E_2}$ where $E_1$ and $E_2$ are the corresponding completions. Assume that $(f_n)_{n \in \mathbb{N}}$ with $f_n \in L$ is a Cauchy sequence with respect to the norm $\norm{\cdot}_{E_2}$. Then by virtue of of the above realtion it is also a Cauchy sequence in the norm $\norm{\cdot}_{E_1}$. Let $(f_n)_{n \in \mathbb{N}} \in f_{E_2} \in E_2 $ and $(f_n)_{n \in \mathbb{N}} \in f_{E_1} \in E_1 $ and associate the vector $f_{E_2}$ with $f_{E_1}$. This mapping is indeed well-defined since if, $(g_n)_{n \in \mathbb{N}} \in f_{E_2}$. i.e. $(g_n)_{n \in \mathbb{N}} \sim (f_n)_{n \in \mathbb{N}}$ with respect to $\norm{\cdot}_{E_2}$, then the same equivalence relation can be written for $\norm{\cdot}_{E_1}$. We denote the constructed map by $Q: E_2 \to E_1$ with $ f_{E_2} \mapsto Q f_{E_2} = f_{E_1}$ and the following theorem gives us a precise statement when $E_2 \subset E_1$. 
Theorem Let $L$ be a linear space with two norms $\norm{\cdot}_{E^1}$ and $\norm{\cdot}_{E^2}$ comparable in the sense of the above realtion, let $E_1$ and $E_2$ be the corresponding completions of $L$ and let $Q$ be the operator introduced above. Then
\begin{equation}
    E_2 /ker(Q) \subset E_1, \quad \ \norm{f}_{E_1} \leq \norm{f}_{E_2 / ker(Q) }, \quad ( f \in E_2 / ker(Q) ).
    \label{7.3}
\end{equation}
If $ker(Q)=\{ 0 \}$, then $ E_2 \subset E_1$.
Proof See [1] Theorem 7.1.
In order to apply  the theorem we need to show that $ker(Q) = \{ 0 \}$ and this can be verified by the following simple condition.
Proposition The kernel $ker(Q)=\{ 0 \}$ if and only if any Cauchy sequence $(f_n)_{n \in \mathbb{N}}$ in the norm $\norm{\cdot}_0$ which approaches to zero in the norm $\norm{\cdot}_{L^2}$ approaches to zero in $\norm{\cdot}_0$.
Proof See [1] Theorem 7.2.
Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence on $C^{\infty,2}_0 $ with respect to the norm $\norm{\cdot}_0$ which converges to zero in $\norm{\cdot}_{L^2}$. This implies that the part of the induced norm $\norm{\cdot}_0$ without the derivatives converges to zero. Now assume $M \geq 1$, let $\Omega'$ the support of $\nabla f_1$ and consider
\begin{equation}
    ( \nabla f_{1,n} , \nabla f_{1,n} ) = \int_{\Omega'} (\nabla f_{1,n})^2 = - \int_{\Omega} f_{1,n} \Delta f_{1,n} \leq C \int_{\Omega} |f_{1,n}| .
    \label{29}
\end{equation}
Here  we introduced a shorthanded notation where $\Omega$ denotes the smallest support of the considered functions and $C \sim \max(\max_{\Omega} ( \Delta f_{1,n} ),\min_{\Omega} ( \Delta f_{1,n} ))$ since continuous functions attend their maximum/minimum on closed sets. Applying the Hölder inequality gives us
\begin{equation}
    \norm{f_{n,1}}_{L^1} = \norm{f_{n,1} \cdot 1_{supp(f_n,1)}}_{L^1} \leq \norm{f_{n,1}}_{L^2} \cdot \norm{1_{supp(f_{n,1})}}_{L^2}.
\end{equation}
Hence $\norm{1_{supp(f_{n,1})}}_{L^2}$ is bounded by compactness. Since $\norm{f_{n,1}}_{L^2}$ is approaching to $0$, $\norm{f_{n,1}}_{L^1}$ vanishes and this implies $ker(Q)=\{ 0 \}$. This finally shows that $\overline{C^{\infty,2}_0}$ is contained in $L^2(\mathbb{R}^m)^2$.

[1] Berezanskiĭ, I͡Uriĭ Makarovich and Sheftel, Zinovij G. and Us, Georgij F. “Functional Analysis Vol.II”,Birkhäuser Verlag,1996

