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I need some help to understand the following statement.

Let $A$ be an operator defined as follows: $Av = -\Delta v - \nabla \text{div} u$

It is known that the operator $A$ is positive self-adjoint operator from $H^2(\Omega) \bigcap H^1_0(\Omega)$ onto $L^2(\Omega)$

Then, the author considers $A^{-1}u$ for $u\in H$, where $H$ is the dual space of the $H^2(\Omega) \bigcap H^1_0(\Omega)$.

But, this makes no sense to me. Doesn't $A^{-1}$ map from $L^2$ into $H^2(\Omega)\bigcap H^1_0(\Omega)$?

So, the domain of $A^{-1}$ is $L^2$, but using $L^2$ as a pivot space, I know that $L^2\subset H$, so how can one apply $A^{-1}$ to any member of $H$?

Am I missing something?

mononono
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the operator $A$ is positive self-adjoint operator from $H^2(\Omega) \bigcap H^1_0(\Omega)$ onto $L^2(\Omega)$

This is confusing because by definition, a self-adjoint operator maps a Hilbert space (or its dense subspace) to the same Hilbert space. I'd say that $A$ is "formally self-adjoint" meaning that $\int (Af)g = \int f(Ag)$ for test functions $f,g$.

The adjoint of $A$ maps the dual of $L^2(\Omega)$ to the dual of $H^2(\Omega) \bigcap H^1_0(\Omega)$. Hence, the inverse of this adjoint maps the dual of $H^2(\Omega) \bigcap H^1_0(\Omega)$ to the dual of $L^2(\Omega)$. Perhaps the author decided to be sloppy and use notation $A^{-1}$ for the inverse of the adjoint of $A$ (after all, $A$ is kind of self-adjoint).

  • So, I suppose that the reason for your saying "formally self-adjoint" is because $H^2(\Omega)\bigcap H^1_0(\Omega)$ is not dense in $L^2(\Omega)$? – mononono Feb 13 '16 at 18:48
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    No, that's not the reason. And it's dense. To be self-adjoint, $A$ must be equal to its adjoint. Equality means, in particular, having the same domain. –  Feb 13 '16 at 18:53
  • I get it. And is it legitimate to say that the adjoint of $A$ in fact maps $L^2(\Omega)$ into $H^2(\Omega)\bigcap H^1_0(\Omega) \subset L^2(\Omega)$ since these spaces are hilbert spaces, hence self dual? – mononono Feb 13 '16 at 18:56
  • "Self-dual" is more sloppiness, often leading to more confusion. A Hilbert space is isomorphic to its dual. This isomorphism may or may not be useful. There are situations where you do not want to use this isomorphism, and dealing with Sobolev spaces is such a situation. it is misguided to too-automatically identify a Hilbert space and its dual. –  Feb 13 '16 at 19:07