Differences between $\Delta: H_0^1(\Omega)\to H^{1}(\Omega)$ and $\Delta: H^2(\Omega)\cap H_0^1(\Omega)\to L^2(\Omega)$
$H^{1}=(H^1_0)^*$, so $−\Delta:H^1_0(Ω) \rightarrow H^{−1}(Ω)$ means $\Delta u \in H^{−1}(Ω)$ for any $u \in H^1_0(Ω)$, and that is obviously right. Because $\Delta u \in H^{−1}(Ω)$ means $(\Delta u,v)$ is a scalar, for any $v \in H^1_0(Ω)$. this also satisfies the definition of weak solution.
Therefore, $\Delta u \in L^2(\Omega)$, you mean this function is in $L^2$, while, if you say $\Delta u \in H^{1}(Ω)$, you mean $\Delta u$ is operator on $H^1_0(Ω)$. These are both correct, I believe if you know something about distribution, this would be easy to understand.
I don't know if I catch your point.
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Updated on August 01, 2022Comments

ningbo polymer 10 months
I'll try to explain what I want to know: Let $\Omega\subset\mathbb{R}^n$ be a bounded domain.
When we look to $\Delta: H^2(\Omega)\cap H_0^1(\Omega)\to L^2(\Omega)$, the meaning of $\Delta$ is precisely "the sum of second partial derivatives with Dirichlet boundary conditions" since the partial (weak) derivatives of a $ H^2(\Omega)$ exist and lie on $L^2(\Omega)$. So when we say "let's solve $\Delta u=f$ !" this equation means what it means. To solve this we can define the weak problem "find an $u\in H_0^1(\Omega)$ such that $\langle \nabla u, \nabla v\rangle_0=\langle f,v\rangle_0$ for all $v\in H_0^1(\Omega)$", solve it applying Riesz Representation Theorem and then use regularity theory to show that $u\in H^2(\Omega)$.
It's not the case with $\Delta: H_0^1(\Omega)\to H^{1}(\Omega)$. What's the meaning of $\Delta u$ for $u\in H_0^1(\Omega)$?

user127096 about 9 years$\Delta u$ is a bounded linear functional on $H_0^1$, defined by $v\mapsto \int_\Omega \nabla u\cdot \nabla v$.


ningbo polymer about 9 yearsThanks for you answer, chuck. But what do you mean with $(\Delta u,v)$? Is this the pairing $H^{1}$,$H_0^1$? If yes, you are saying $\Delta$ is simply the Riesz duality map, right? (I think I'll edit the question to include more details)

chuck about 9 years@ningbo polymer $(\Delta u, v)$ is the pair $(H^{1},H^1_0)$ because you can view $\Delta u$ (actually, you should write $(\Delta u,.)$, but just like we do in Representation theory, we can regard it as $\Delta u$) as all the continuous linear functionals on $H^1_0$, which means $\Delta u$ should be in $H^1_0$. Then, $\Delta$ maps u which is in $H^1_0$ into $H^{1}$