0

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 $L^{p^{\prime}}(I)$?

Kawai Suta
  • 65
  • 1
    A few comments: I think you mean $L^2$ in place of $L^{p'}$, and I think it should be $f_1 u'$ in the second integral in the last display. For the second point, notice that the dual of $H^1_a(I)$ can't be the same (as distributions) as $H^{-1}(I)$; for example the delta mass $\delta_b$ is just the zero functional in $H^{1}_0$, but it's non-trivial in $H_a^1(I)$. – Jose27 Sep 13 '23 at 17:52
  • @Jose27 So the dual of $H_{a}^{1}$ is the same dual as $H^{1}(I)$? One question, is the dual of $H^{1}(I)$ the same as the dual of $L^{2}(I)$? – Kawai Suta Sep 13 '23 at 18:07
  • 1
    No to both questions: Think of $\delta_a$ for both of them. What's true in the second one is, as you noted, that the dual of $H^1(I)$ can be identified with a quotient of $L^2(I)\times L^2(I)$. – Jose27 Sep 13 '23 at 23:28
  • @Jose27 thank you very much. – Kawai Suta Sep 14 '23 at 00:42

0 Answers0