Characteristics of dual of $V=\{u\in H^1((0,1))\mid u(1)=0\}$ Let $\Omega \subset \mathbb{R}^n$ be an open domain. I know that the elements in $H^{-1}(\Omega)$ denoted the dual of $H_0^1(\Omega)$ have representation as follows,
$$f\in H^{-1}(\Omega) \Leftrightarrow f=f_0+\sum_{i=1}^n\partial_{x_i}f_i,~~~f_0,f_i\in L^2(\Omega)$$
where $\partial$ is derivative of distribution. However, I’m confused whether functionals in the dual of $V$ denoted in the title have similar representation?
Any help will be appreciated.
 A: What happens is that in addition to the elements from $H^{-1}$, you can also get functionals/Dirac measures of the type
$\alpha\delta_0$ (with $\alpha\in\Bbb R$), where $\delta_0$ is defined via $\langle \delta_0 , v \rangle = v(0)$.
First, one can check that $\delta_0$ is really a functional over $V$
(using the embedding oh $H^1$ into the continuous functions)
and it is also not representable using $L^2$-functions or using
distributional derivatives of $L^2((0,1))$.
The slightly more difficult thing is to show that all functionals $f$ in the dual of $V$
can be represented as a sum of a $H^{-1}((0,1))$-functional and $\alpha\delta_0$ for some $\alpha\in\Bbb R$.
Here, one can use
$$
\langle f , v \rangle_V
= \langle f, v - v(0)(1-x) \rangle_V
+ \langle f,  v(0)(1-x) \rangle_V
\qquad\forall v\in V.
$$
Since $v-v(0)(1-x) \in H_0^1((0,1))$ for all $v\in V$,
we can find $g\in H^{-1}((0,1))$ such that
$\langle f,  v-v(0)(1-x) \rangle_V =\langle g, v- v(0)(1-x) \rangle_{H_0^1((0,1))}$.
Since $g\in H^{-1}((0,1))$, we can use a representation of $g$:
$g= g_0+\partial_x g_1'$, where $g_0,g_1\in L^2((0,1))$.
With this representation, we can also interpret $g$ as a functional on $H^1((0,1))$
or $V$.
For the second term, we set
$\alpha := \langle f, 1-x\rangle_V-\langle g,1-x\rangle_{H^1((0,1))}$.
We claim now that $f=\alpha \delta_0 + g_0 + \partial g_1$.
Indeed, for $v\in V$ we have
$$
\begin{aligned}
 \langle \alpha \delta_0 + g,v\rangle_{H^1((0,1))}
 &= \langle \alpha \delta_0 + g,v-v(0)(1-x)\rangle_{H^1((0,1))}
 + \langle \alpha \delta_0 + g,v(0)(1-x)\rangle_{H^1((0,1))}
\\
&= \langle g, v - v(0)(1-x)\rangle_{H^1((0,1))}
+ \alpha v(0) + \langle g,v(0)(1-x)\rangle_{H^1((0,1))}
\\
&= \langle f, v-v(0)(1-x)\rangle_V
+ \bigl(\langle f ,(1-x) \rangle_V - \langle g,(1-x)\rangle_{H^1((0,1))}\bigr)v(0)
+ \langle g,v(0)(1-x)\rangle_{H^1((0,1))}
\\
&= \langle f,v\rangle_V.
\end{aligned}
$$
