How to prove that if $f \in H^{1,2}(\mathbb{R}^n)$ and $\nabla f \in H^{1,2}(\mathbb{R}^n)$, then $f \in H^{2,2}(\mathbb{R}^n)$?

Question

I'm trying to learn about Sobolev spaces as completions and I was pondering the following question. Let $\Omega \subseteq \mathbb{R}^n$ be a domain. For $k \in \mathbb{N}$ and $1 \le p < \infty$, define the space $H^{k,p}(\Omega)$ to be the completion of the subset of $C^{\infty}(\Omega)$ consisting of functions with finite Sobolev norm. The completion is taken w.r.t. the Sobolev norm. Now suppose for concreteness that $f \in H^{1,2}(\Omega)$ and $\nabla f \in H^{1,2}(\Omega)$. Then it is reasonable to think that $f \in H^{2,2}(\Omega)$ to begin with. I think there should be an elementary proof of this without resorting to the fact that $H = W$ i.e. the completion definition of Sobolev spaces agrees with the definition using weak/distributional derivatives. However, I have not been able to show this. Does anybody know an elementary way to show that this indeed holds? You can take $\Omega$ to be bounded, for example the open unit ball $B := B(0,1) \subseteq \mathbb{R}^n$ if that makes any difference. Thanks for help!