# Is the following operator invertible?

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?

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 $$(I+A^*A)^{-1}$$.
• Almost all clear: but why the inverse of $I+AA^*$ is continuous? Oct 25 '21 at 14:37
• because $\lVert x \rVert \leq \lVert (I +AA^\star) x \rVert$ Oct 26 '21 at 13:59