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Let $U$ be an open, bounded subset of $\mathbb{R}^n.$

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...

Stephen
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user142700
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1 Answers1

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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.

Joonas Ilmavirta
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  • If I have understood you correctly, you say the following: given an arbitrary element $\phi$ of $H_0^1(U),$ this functional can be realized either as an element of $H_0^1(U)$ via the inner product, or as an element of $H^{-1}(U)$ via direct integration (or, in this case, some related operation). Is this right? The point is that the spaces are different because they act on $H_0^1(U)$ in different ways. – user142700 Aug 13 '14 at 15:14
  • @user142700 Any $\phi\in H^1_0$ can be realized as a functional $H^1_0\to\mathbb R$ in two natural ways: by the inner product and by direct integration. People tend to be sloppy with these two and the exact meaning of dual spaces, so be careful. – Joonas Ilmavirta Aug 13 '14 at 15:20