# Inverse of bounded self adjoint operator on HS is self adjoint?

Let $A=A^{*}$ be a bounded self adjoint operator on a Hilbert space $\mathcal{H}$ with Range Ran$(A) = D$ dense in $\mathcal{H}$. $A$ is injective, since Ran$(B) \perp ker(B^{*}) = ker(B)$.

So $A^{-1}: D \to \mathcal{H}$ is a densely defined operator and has adjoint $(A^{-1})^{*}$, defined on $D^{*}$. My intuition tells me that the inverse should be self adjoint and I know the proof for the case that $A$ is bijective:

$\langle (A^{-1})^{*} A^{*} \varphi, \psi \rangle = \langle A^{*} \varphi, A^{-1} \psi \rangle = \langle \varphi, A(A^{*})^{-1} \psi \rangle = \langle \varphi, \psi \rangle$.

In my case this is only true for $\psi \in D$. In the same way one gets $A^{*}(A^{-1})^{*} =$ Id.

So I have $(A^{*})^{-1} = (A^{-1})^{*}$ on $D$. But how can I show know that $D=D^{*}$?

Any ideas?