Duals of Sobolev Spaces vanishing on parts of the boundary I am revising for a Finite Elements course and have the following question about the definition of $H^{-1}$.
Let $D\subseteq \mathbb{R}^2$ be bounded Lipschitz domain and let $\Gamma_0 \subseteq \partial D$ be a part of positive measure of the boundary where we put Dirichlet conditions.
We define $V_0:=\{v \in H^1(D) : v|_{\Gamma_0} \equiv 0\}$ and $H_0^1(D):=\{v \in H^1(D) : v|_{\partial D}\equiv 0\}$. 
Taking duals obviously yields $(V_0)^\ast \hookrightarrow (H_0^1(D))^\ast$.
My question: How can I construct an example of a function in $H_0^1(D)^\ast$ that does not lie in $V_0^\ast$? 
 A: The counterexample could be intuitively constructed exploiting the fact that any singular part of the functional on boundary will get killed by the $H^1_0(D)$.
The simplest possible example is: choose a singular $g\in H^{-1/2-\epsilon}(\partial D)$ for some $\epsilon >0$
$$
\mathcal{L}(v) := \int_{\partial D} g \cdot \operatorname{tr}(v) dS
$$
(omitting the independent variable here). 
We can check that $\mathcal{L}(\cdot)$ vanishes for any $v\in H^1_0(D)$, but we can choose certain $g$ that $\mathcal{L}(\cdot)$ is not bounded for all $v\in V_0$, because the singularites do not get killed on the other part of the boundary except the Dirichlet part. 
And this is part of the reason we need Neumann data in at least $H^{-1/2}(\Gamma_N)$ where $\Gamma_N$ is the Neumann boundary, if I assume we are interested in the weak formulation in the Hilbertian setting. If $g \in H^{-1/2}(\partial D)$ above, above functional will be bounded for any $H^1(D)$ which can be proved by trace inequality.

A concrete example of above functional: let $D= (0,1)^2$, let $\Gamma_0$ be the top three sides' union, the rest of the boundary, which lies on the $x$-axis, is say $\Gamma_N$, then the singular functional is
$$\mathcal{L}(v):=\int_{[0,1]\times \{0\}} \frac{1}{x^{\alpha}\ln x} v|_{y=0}\, dx, \quad \text{where } \alpha>1.
$$
On the boundary where $y=0$, we can simply let the trace of $v$ be $x$ from $[0,1/2]$, $1-x$ from $[1/2, 1]$, and $0$ on all three other sides, and this trace is continuous on $\partial D$ so it can be extended continuously inside as an $E(v) \in H^1(D)$ function. It can be checked that the integral diverges for this $E(v) \in H^1(D)$. 
