Let $f \in H^m(\mathbb{R}^n)$, then we have for $|\alpha| \le m$ that
$$\langle \partial^{\alpha}f, \phi \rangle = (-1)^{|\alpha|} \langle f , \partial^{\alpha} \phi \rangle$$ for all test functions $\phi$,
now let's take a sequence of $C^{\infty}_C$ cut-off functions $\psi_n$ such that $\psi_n \rightarrow 1$ for $n \rightarrow \infty$ and $\partial^{\alpha} \psi_n \rightarrow 0$ for any $|\alpha| \le m$, then we have for any $g \in H^m(\mathbb{R}^n) \cap C^{\infty}$
$\langle \partial^{\alpha}f,\psi_n g\rangle = (-1)^{|\alpha|} \langle f, \partial^{\alpha} (\psi_n g) \rangle = \sum_{\beta \le \alpha} \binom{ \alpha}{\beta}\langle f, \partial^{\beta} \psi_n \partial^{\alpha-\beta}g\rangle$. Now,if I see this correctly, then the dominated convergence theorem implies (cause all derivatives of $\psi_n$ go to zero) that
$$\langle \partial^{\alpha}f,g \rangle = (-1)^{|\alpha|} \langle f, \partial^{\alpha} g\rangle. $$
Now my question is: This is REALLY strange, cause this would imply that instead of testing with $C^{\infty}_C$ functions, if something is weakly differentiable, I could also use the much larger class of $C^{\infty} \cap H^m$ functions. Can this be correct?
So my question is: Is this proof alright?