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.
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 $$ \begin{aligned} \langle \alpha \delta_0 + g,v\rangle_{H^1((0,1))} &= \langle \alpha \delta_0 + g,v-v(0)(1-x)\rangle_{H^1((0,1))} + \langle \alpha \delta_0 + g,v(0)(1-x)\rangle_{H^1((0,1))} \\ &= \langle g, v - v(0)(1-x)\rangle_{H^1((0,1))} + \alpha v(0) + \langle g,v(0)(1-x)\rangle_{H^1((0,1))} \\ &= \langle f, v-v(0)(1-x)\rangle_V + \bigl(\langle f ,(1-x) \rangle_V - \langle g,(1-x)\rangle_{H^1((0,1))}\bigr)v(0) + \langle g,v(0)(1-x)\rangle_{H^1((0,1))} \\ &= \langle f,v\rangle_V. \end{aligned} $$