Let $\alpha(x) : \mathbb{R} \to (0,\infty)$ have bounded variation (BV) and suppose $\inf\_{\mathbb{R}} \alpha > 0$. Consider the second order differential operator
$$H : =-\partial\_x (\alpha(x) \partial\_x) : L^2(\mathbb{R}) \to L^2(\mathbb{R}).$$
It's not too hard to show that  

$$ \{u \in L^2(\mathbb{R}) : u, u' \in L^2(\mathbb{R}) \cap L^\infty(\mathbb{R}) \text{ and } Hu \in L^2(\mathbb{R}) \} \\
= \{u \in L^2(\mathbb{R}) : u, \, \alpha u' \text{ are locally absolutely continuous and } Hu \in L^2(\mathbb{R})\},
$$
and that $H$ is self-adjoint with respect to this domain $\mathcal{D}(H)$, which is a dense subset of $L^2(\mathbb{R})$.



> 
> I would like to know whether the Sobolev space $H^1(\mathbb{R})$ is contained in the domain of the square root operator $H^{1/2}$ (defined via the spectral theorem).
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Recall the well-known fact that $\mathcal{D}(H^{1/2})$ is equal to the form domain associated to $H$ (i.e., the completion of $\mathcal{D}(H)$ with respect to the norm $\| u\|^2\_{+1} := \langle Hu,u \rangle\_{L^2} + \langle u,u \rangle\_{L^2} \approx \|u\|^2\_{H^1}$.)


If $\alpha$ has some more regularity, for instance if $\alpha$ is $W^{1, \infty}$ instead of merely having BV, then it is immediate that $C\_0^\infty \subseteq \mathcal{D}(H) \subseteq \mathcal{D}(H^{1/2})$. So in that case, for any $u \in H^1(\mathbb{R})$, if we approximate $u$ in $H^1(\mathbb{R})$ by a sequence $u\_j \in C^\infty\_0(\mathbb{R})$, these $u\_j$ are also Cauchy with respect to $\| \cdot \|\_{+1}$, and hence $u \in \mathcal{D}(H^{1/2})$.


We don't necessarily have the inclusion $C\_0^\infty \subseteq \mathcal{D}(H)$ in the BV case. But can we come up with a different subspace of functions contained in $\mathcal{D}(H)$, that has a sequence converging to $u$ in $L^2(\mathbb{R})$ and Cauchy with respect to $\| \cdot \|\_{+1}$? This is where I am stuck.



Is the Sobolev space $H^1(\mathbb{R})$ contained in the domain of $(-\partial_x \alpha(x) \partial_x)^{1/2}$?
We can actually do this directly and my comment above is not that relevant. Let $u\in H^1$ and also assume that $u$ is compactly supported, so $\int u'=0$. Approximate $\alpha u'$ in $L^2$ by $v\_n\in C\_0^{\infty}$. Here we can also insist that $\int v\_n/\alpha=0$. Then also $v\_n/\alpha\to u'$ in $L^2$, so $u\_n(x)=\int\_{-\infty}^x v\_n/\alpha\, dt\to u$ locally uniformly and thus $u\_n\to u$ in $H^1$.


Clearly, $u\_n\in D(H)\subseteq D(H^{1/2})$, and as you already explained yourself, this implies that $H^1\subseteq D(H^{1/2})$.


