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Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open
  • $H^{-1}(\Lambda):=H_0^1(\Lambda)'$

Let $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H_0^1(\Lambda)$$ and note that (since $\Lambda$ is bounded) the norm $\left\|\;\cdot\;\right\|_{\mathfrak a}$ induced by $\mathfrak a$ is equivalent to $\left\|\;\cdot\;\right\|_{H^1(\Lambda)}$ on $H_0^1(\Lambda)$.

How can we show that $$-\Delta:H_0^1(\Lambda)\to H^{-1}(\Lambda)\;,\;\;\;v\mapsto\mathfrak a(\;\cdot\;,v\rangle$$ is an isomorphism between $H_0^1(\Lambda)$ and $H^{-1}(\Lambda)$?

Clearly, if $v\in H_0^1(\Lambda)$ with $$-\Delta v=0\;,$$ then $$0=(-\Delta v)v=\mathfrak a(v,v)$$ and hence $$v=0\;,$$ i.e. $-\Delta$ is injective. On the other hand, let $\eta\in H^{-1}(\Lambda)$. Since $H_0^1(\Lambda)$ equipped with $\mathfrak a$ is a $\mathbb R$-Hilbert space, we obtain the existence of a unique $v\in H_0^1(\Lambda)$ with $$\eta(u)=\mathfrak a(u,v)=(-\Delta v)u\;\;\;\text{for all }u\in H_0^1(\Lambda)\;,$$ i.e. $$\eta=-\Delta v\tag1$$ by Riesz' representation theorem.

So, we're done, aren't we? The crucial point in the proof of the surjectivity above is that our intention is to show that for each $\eta\in\left(H_0^1(\Lambda),\left\|\;\cdot\;\right\|_{H^1(\Lambda)}\right)'$, there is a $v\in\left(H_0^1(\Lambda),\left\|\;\cdot\;\right\|_{H^1(\Lambda)}\right)$ with $(1)$. What we've used above is that, by the equivalence of $\left\|\;\cdot\;\right\|_{\mathfrak a}$ and $\left\|\;\cdot\;\right\|_{H^1(\Lambda)}$ on $H_0^1(\Lambda)$, $\eta\in\left(H_0^1(\Lambda),\left\|\;\cdot\;\right\|_{\mathfrak a}\right)'$ too.

Question 1: This should be correct or did I made any mistake?

Question 2: I've read that $-\Delta$ is the canonical isomorphism between $H_0^1(\Lambda)$ and $H^{-1}(\Lambda)$. Moreover, $-\Delta$ is said to be the dualization between $H_0^1(\Lambda)$ and and $H^{-1}(\Lambda)$ with respect to $\left\|\;\cdot\;\right\|_{\mathfrak a}$. What exactly is meant by these two statements?

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    1) is a PDE question; one needs to know that the Poisson equation with the source term in $H^{-1}$ has a solution in $H^1_0$. By (2), people mean that $H^{-1}$ is often defined as the dual of $H_0^1(\Lambda)$ (it's certainly __not__ defined to be the same space as $H_0^1(\Lambda)$), and the Laplacian sends each element to the linear functional that it induces on the space. It gives a concrete realization of the map that we see in the Riesz representation theorem. – 2017-01-22
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    @zaq I don't understand your comment to question 1. Is there anything wrong with my proof? If so, please tell me where my mistake is. – 2017-01-22
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    @zaq Oh, and I didn't intend to define $H^{-1}(\Lambda)$ to be $H_0^1(\Lambda)$ but $H_0^1(\Lambda)'$. I've just missed the prime. – 2017-01-23

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