Problem concerning $H^{-1}$ space

Question

Let $I=(0,a)\subseteq\mathbb{R}$ be an interval. I need to show that $l\in H^{-1}(I)$ iff $\exists u\in L^{2}(I)$ such that $l=u'$. I know that $H^{1}(I)$ stands for set of functions from $L^{2}(I)$ whose weak derivatives are also from $L^{2}(I)$, but I have problem understanding what are the elements of $H^{-1}(I)$, dual space of $H^{1}(I)$. Can somebody help me?

Answer 1

Since $l \in H^{-1}(I)= (H^1_0(I))'$ then, fom Riesz representation Theorem, there is a unique $u\in H^1_0(I)$ such that $\forall \phi \in H^1_0(I) $ in particular, for every $\phi \in C^\infty_0(I) $

$$\langle l,\phi\rangle = (\phi, u)_{H^1_0(I)} \\= \int_I \phi u ~dx + \int_I \phi' u' ~dx \\= \int_I \phi u ~dx - \int_I \phi u'' ~dx \\= \int_I \phi (u- u'') ~dx $$

Where the second derivation of $u$ in the sense of distributions given by $u''$ is a distribution not necessary a function. Therefor, for every $\phi \in C^\infty_0(I) $ we have $$\langle l,\phi\rangle = \int_I \phi (u- u'') ~dx $$ hence we conclude that, $l= u-u''$ in the sense of distribution with $u\in L^2(I)$.

The converse is easy and left to the OPs