Consider $I = (a,b)$ a limited interval. I have a doubt regarding the dual space of $H^{1}(I)$ according to some conditions about the values on the boundary of $(a,b)$. For example:
1- If $f \in H^{1}(I)$ such that $f(a)=f(b) = 0$, then
$$f \in H^{1}_{0}(I),$$
and I know the dual of $H^{1}_{0}(I)$ is $H^{-1}(I)$.
Now I start to have doubts when I think about 2 - If $f \in H^{1}(I)$ such that $f(a)=0$ and $f(b) \neq 0$, for example. Who would be the dual of the set $H_{a}^{1}(I) = \lbrace f \in H^{1}(I): f(a) = 0 \rbrace$ ?
I think it's $H^{-1}$, because
$\parallel f \parallel_{H_{a}^{1}(I)} \leq C \parallel f_{x}\parallel_{L^{2}(I)}$ and I is bounded. Am I right?
Now, if $f \in H^{1}(I)$ with $f(a) \neq 0$ and $f(b) \neq 0$, who is the dual of $H^{1}(I )$? I know that if $F$ is in the dua of $H^{1}(I)$, then there are $g_{0}, g_{1} \in L^{p^{\prime}}(I)$ such that $$ \langle F , u \rangle = \int_{I} f_{0}udx + \int_{I}f_{1}u^{\prime} dx, \quad u \in H^{1}(I) $$ Is the dual of $H^{1}(I)$ with limited $I$ the $L^{p^{\prime}}(I)$?