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I'm a bit confused, why do people not define $H^1(\Omega)^*$? Instead they only say that $H^{-1}(\Omega)$ is the dual of $H^1_0(\Omega).$

$H^1(\Omega)$ is a Hilbert space so it has a well-defined dual space. Can someone explain the issue with this?

matt.w
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2 Answers2

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I don't quite agree with Dirk. The dual of $H^1$ sure is well-defined, it is just that it is hard to be characterized like $(H^1_0)^* = H^{-1}$, which is people usually say.

The isomorphism between $H^1_0$ and $H^{-1}$ can interpreted in boundary value problem sense: The Dirichlet Laplacian can be defined as $$ -\Delta^{\mathrm{Dir}}: H^1_0(\Omega)\to H^{-1}(\Omega) $$ in the sense that $$ \langle-\Delta^{\mathrm{Dir}} u,v \rangle = \int_{\Omega} \nabla u\cdot \nabla v, \quad \text{for } \forall u,v\in H^1_0(\Omega), $$ which is to say, by Riesz's representation, that $$\forall f\in H^{-1}(\Omega), \quad f = -\Delta^{\mathrm{Dir}} u \quad\text{for some } \; u\in H^1_0(\Omega).\tag{1}$$

Mimicking above, we can define Neumann Laplacian: $$-\Delta^{\mathrm{Neu}}: H^1(\Omega) \to H^1(\Omega)^*,$$ in the sense that $$ \langle-\Delta^{\mathrm{Neu}} u,v \rangle = \int_{\Omega} \nabla u\cdot \nabla v, \quad \text{for } \forall u,v\in H^1 (\Omega), $$ by the compatibility condition of a Neumann boundary value problem, the Neumann Laplacian $-\Delta^{\mathrm{Neu}}$ is rather a bijection from $H^1(\Omega)/\mathbb{R} \to (H^1(\Omega)/\mathbb{R})^*$.

Only if we use the full $H^1$-inner product to define an operator that maps an $H^1$-function to its dual: $$ \langle\mathcal{A} u,v \rangle = \int_{\Omega} (\nabla u\cdot \nabla v+u\,v),\quad \text{for } \forall u,v\in H^1(\Omega), $$ where $\mathcal{A}: H^1(\Omega) \to H^1(\Omega)^*$. However we don't have a (1) kind like statement because the unspecified boundary value.

Dreamer
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Shuhao Cao
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  • Can you elaborate on the last line more? I am trying to use Faedo-Galerkin on a parabolic equation with Neumann boundary conditions. Does that mean I can not assume $\partial_t u \in H^1(\Omega)^*$? Because in the case of the Dirichlet boundary condition, we use (1) for this purpose. – Tarek Acila Jul 02 '23 at 07:33
  • @TarekAcila It is safe to assume that. For Dirichlet BC, modulo the appropriate function (with its harmonic extension or other proper extension to the interior), you have zero boundary condition. For Neumann or Robin, you just assume the function you like in $H^1$ without specifying a pointwise-like boundary value (I am using -like because pointwise values are not defined, but you get the idea). In your exam, once you can assume your solution live in the correct Bochner space (without a boundary value). – Shuhao Cao Jul 06 '23 at 16:15
  • Sorry to comment on an old post but I don't fully understand why the same argument used for $\Delta^{\mathrm{Dir}}$ doesn't apply to $\Delta^{\mathrm{Neu}}$, why do you need to quotient by $\mathbb{R}$? – CBBAM Oct 09 '24 at 20:26
  • @CBBAM Because pure Neumann problem needs to quotient a constant to get well-posedness (otherwise adding a constant to the solution makes another solution, due to how the BC is formulated without the trace of $u$ itself). – Shuhao Cao Oct 10 '24 at 14:26
  • @ShuhaoCao Thanks. So the whole reason $H^{-1}$ isn't the dual of $H^1$ is because, in general, the values at the boundary are not specified? I don't see why this prevents us from identifying the two using the full $H^1$ inner product as in the last part of your answer. – CBBAM Oct 10 '24 at 23:47
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As far as I remember, one usually defines $H^{-1}(\Omega)$ to be the dual space of $H^1(\Omega)$. The reason for that is that one usually does not identify $H^1(\Omega)^*$ with $H^1(\Omega)$ (which would be possible) but instead works with a different representation. E.g. one works with the $L^2$-inner product as dual pairing between $H^{-1}(\Omega)$ and $H^1(\Omega)$ (in case the element in $H^{-1}(\Omega)$ is an $L^2$-function).

Dirk
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  • So there is nothing wrong in me making use of $H^1 \subset L^2 \subset (H^1)^$ as Hilbert triple, so if $f \in L^2$, the dual pairing $\langle f, u \rangle_{(H^1)^, H^1} = (f,u)_{L^2}$ right? As your answer says this is true for $H^1_0$ and its dual but I see no reason why it can't work for $H^1$. – matt.w May 30 '13 at 10:35