Let $D\subset D'\subset\mathbb{R}^n$ be two open, bounded sets, with $\bar{D}\subset D'$, and let $u$ be a smooth function supported in $D$.
Then $u$ can be thought of as functional acting on $H^1_0(D)$. Given any $v\in H^1_0(D)$,
$$(u,v) = \int u v\,dx.$$
So $u$ belongs to the space $H^{-1}(D)$, the dual of $H^1(D)$, and has norm
$$||u||_{H^{-1}(D)} := \sup_{v\in H^1(D)}\frac{\int uv\,dx}{||v||_{H^1_0(D)}}.$$
Of course $u$ is also a functional on $H^1_0(D')$, defined as above, and so it has a norm in $H^{-1}(D')$.
Are these two norms comparable? Clearly $||u||_{H^{-1}(D)}\leq ||u||_{H^{-1}(D')}$, but I'm struggling to determine whether the reverse inequality holds (up to a constant).
The support of $u$ is a compact set contained in $D$. Now let me take a smooth cut-off function $\phi\in C_c^\infty(\mathbb R^n)$ with $\phi\ge0$, $\phi=1$ on the support of $u$, $\phi=0$ on the complement of $D$.
Let $v\in H^1_0(D')$. Then $\phi v \in H^1_0(D)$. In addition, $$\begin{split} \int u v dx =& \int u v \phi dx \le \|u\|_{H^{-1}(D)} \|v \phi\|_{H^1(D)}\\ &\le \|u\|_{H^{-1}(D)} ( \|v\|_{L^2(D)} + \|v\|_{H^1(D)} + \|v\|_{L^2(D)} \|\nabla \phi\|_{L^\infty(D)} \le C \|u\|_{H^{-1}(D)} \|v\|_{H^1(D)}. \end{split} $$ The constant $C$ depends on the distance between the support of $u$ and the complement of $D$.