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Let $\Omega\in\mathbb{R}^n$ be a bounded open set with smooth boundary. How to prove the invertibility of $$- \triangle:H^2_0(\Omega) \to L²(\Omega) $$

The injectivity is easy. But how to prove surjectivity without the use of weak notion of solution (when the domain becomes $H^1_0(\Omega)$ and this can be easily found in books)?

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Are you sure it is $H_0^2$? – Tomás Jan 21 '13 at 14:20
yes. @Tomás. This article has a citation to the result: – ningbo polymer Jan 21 '13 at 14:38
I think there is an mistake there. You have to change $H_0^2$ by $H_0^1$. If not you are saying that every solution of the Dirichlet problem has null derivative in the boundary of $\Omega$. This is not true as you can easily see contructing one dimensional examples. – Tomás Jan 21 '13 at 14:50
@Tomas That would not fix things, because the inverse operator (convolution with $|x|^{2-n}$) converts $L^2$ functions into $H^2$ functions. – user53153 Jan 21 '13 at 17:04
@5PM, so is it better to consider $H_0^1\cap H^2$? I think in this case we have a homeomorphism $L^2\rightarrow H_0^1\cap H^2$. – Tomás Jan 21 '13 at 17:27
up vote 4 down vote accepted

The operator $$- \triangle:H^1_0(\Omega) \to L²(\Omega) $$ is "surjective" in the weak sense (as a direct use of Riesz representation theorem).

Then you can use regularity theorems to prove that, in fact, this weak solution is in $H^2(\Omega)$. A good reference to a result like this is the Brezis book "functional analysis, Sobolev spaces and PDE". See theorem 9.25 and note that a $C^2$ domain $\Omega$ is required. So your result is valid from $H^1_0(\Omega)\cap H^2(\Omega) $ to $L²(\Omega) $.

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