Let $U$ be an open, bounded subset of $R^n.$
Evans' well known PDE book defines the spaces:
-$H_0^1(U)$:= $\{f\in H^1(U): \text{there exists a sequence} \; \phi_n \to f \; \text{in the} \; H^1(U) \;\text{norm, with} \; \phi_n \in C_c(U)\}.$
-$H^{-1}(U)$ is the dual space of $H_0^1(U).$
My question: how come $H_0^1(U)$ is not self-dual?
Indeed, if we consider the inner product $(f,g):= \int f g + \int f' g',$ then it seems to me that all the Hilbert space axioms are satisfied. So the Riesz Representation Theorem would imply that $H_0^1(U)$ is self-dual. But then Evans would have defined the same space twice, which seems strange…
Best Answer
Any Hilbert space is of course self dual. But this is only true if you use the inner product as the pairing. Note that $g\in H^{-1}$ acts on $f\in H^1_0$ (formally) by $g(f)=\int_U gf$, where no derivatives appear. Often the dual space of a function space is considered to act by "a simple integral", which may not be the inner product (if we are not in $L^2$).
Let me try to make this a bit more concrete. Let us write $\langle f,g\rangle=\int_Ufg$ and $(f,g)=\langle f,g\rangle+\langle \nabla f,\nabla g\rangle$. The second one is the inner product on $H^1_0$. Then $H^1_0\ni f\mapsto(f,\cdot)\in (H^1_0)'$ is surjective (as always in a Hilbert space), but $H^1_0\ni f\mapsto\langle f,\cdot\rangle\in (H^1_0)'$ is not.