I'm currently working in a setting where Sobolev spaces of the form $H^2(I, \mathbb{R}^n)$ appear, where $I \subset \mathbb{R}$ is a closed interval. We construct such a set by considering the space $C^{\infty}(I, \mathbb{R}^n)$ of smooth functions from $I$ to $\mathbb{R}^n$ and then completing it with respect to the sobolev norm $\| \cdot\|_{H^2}$. We find that $H^2(I, \mathbb{R}^n)$ is, in fact, a Hilbert space, and it is a subset of $C^1(I, \mathbb{R}^n)$.
I'm wondering what happens if instead we complete the space $C^{\infty}_p(I, \mathbb{R}^n)$ of piecewise-smooth functions with respect to the $H^2$ norm (let's call this completion $H^2_p(I, \mathbb{R}^n)$). It's clear that $H^2_p(I, \mathbb{R}^n)$ is a Banach space by construction, and, as far as I can tell, the inner product on $H^2$ is also an inner product on $H^2_p$, making it a Hilbert space. I suspect that the inclusion $H^2_p(I, \mathbb{R}^n) \subset C^1_p(I, \mathbb{R}^n)$ will hold as well. Is this correct?
Edit: I will say that $f \in C^{\infty}_p(I, \mathbb{R}^n)$ if and only if there exists a finite collection of open intervals $\{I_i\}$ such that the corresponding collection of closed intervals $\{\bar{I_i}\}$ is a partition of $I$, and the restrictions $f \vert_{I_i}$ are smooth.