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Let $H$ Hilbert and $A\colon D(A) \subset H \to H$ close (but unbounded) linear operator with dense domain. Assume moreover that it is maximally dissipative, i.e. $\langle Ax,x\rangle \leq 0$ for all $x \in D(A)$ and $Range(A-\lambda_0I)=H$ for some $\lambda_0 >0$.
Denote by $A^*$ the adjoint of $A$.

Then in [Fabbri, Gozzi, Swiech. "Stochastic optimal control in infinite dimension." Probability and Stochastic Modelling. Springer (2017).] p.175 it is claimed that the linear operator $$B=(I+AA^*)^{-1/2}$$ is well defined on all $H$ as a linear bounded operator.

I want to show it: if I prove that $I+AA^*$ is a strictly positive operator and moreover that it is also surjective, then I can take the inverse $(I+AA^*)^{-1}$ and define $B$ which is well defined at this point. Then I will work the details for the boundedness.

In order to prove $I+AA^*$ is a strictly positive operator it I should prove that $AA^*$ is a positive semidefinite operator (which I think should hold but don't know how to do it), how do you show it? How do you show $I+AA^*$ is surjective?

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If $A : \mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ is closed and densely-defined on a Hilbert space $\mathcal{H}$, then the graph $\mathcal{G}(A)\subseteq \mathcal{H}\times\mathcal{H}$ is a closed linear subspace of $\mathcal{H}\times\mathcal{H}$, which gives $$ \mathcal{G}(A)\oplus\mathcal{G}(A)^{\perp}=\mathcal{H}\times\mathcal{H} \;\;\; (\dagger) $$ where $\mathcal{G}(A)^{\perp}$ is the orthogonal complement of $\mathcal{G}(A)$ in $\mathcal{H}\times\mathcal{H}$. Note that $(x,y)\in\mathcal{G}(A)^{\perp}$ iff $$ \langle z,x\rangle+\langle Az,y\rangle=0,\;\;\; z\in\mathcal{D}(A). $$ This condition is restated as $\langle Az,y\rangle = \langle z,-x\rangle$ holding for all $z\in\mathcal{D}(A)$, which is equivalent to having $y\in\mathcal{D}(A^*)$ with $A^*y=-x$. As a consequence, every $(a,b)\in\mathcal{H}\times\mathcal{H}$ may be uniquely written as the following some unique $x\in\mathcal{D}(A)$, $y\in\mathcal{D}(A^{*})$. $$ (x,Ax)+(-A^*y,y)=(a,b) $$ Setting $b=0$ gives $x-A^*y=a$ and $Ax+y=0$, or $x+A^*Ax=a$ for some $x\in\mathcal{D}(A^*A)$. That is enough to prove that $I+A^*A$ is surjective and symmetric. $I+A^*A$ is also injective because $$ \langle (I+A^*A)x,x\rangle=\|x\|^2+\|Ax\|^2,\;\;\; x\in\mathcal{D}(A^*A). $$ From this it follows that $I+A^*A$ self-adjoint, with a continuous inverse. So $(I+A^*A)^{-1/2}$ can be defined as the self-adjoint square root of the bounded self-adjoint operator $(I+A^*A)^{-1}$.

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  • $\begingroup$ Almost all clear: but why the inverse of $I+AA^*$ is continuous? $\endgroup$
    – carlos85
    Oct 25 '21 at 14:37
  • $\begingroup$ because $ \lVert x \rVert \leq \lVert (I +AA^\star) x \rVert $ $\endgroup$ Oct 26 '21 at 13:59

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