Invertibility of a semi-positive perturbation of identity operator Let $X$ be a Hilbert space and $A$ be (not necessarily bounded) self-adjoint and semi-positive definite operator, i.e., $(x,Ax)\ge 0$ for all $x\in X$. Then does $I+A$ have an inverse? That is, is $(I+A)^{-1}$ well defined? Also, is requiring $A$ to be bounded helpful?

I have seen the same question asked in this comment but without any answer. Personally, I don't know how to approach this question. Could you give me some hints or suggest some reference? 
 A: As I mentioned in a comment, if $(Ax,x) \ge 0$ for all $x\in\mathcal{D}(A)$, then
$$
           \|x\|^2 \le ((A+I)x,x) \le \|(A+I)x\|\|x\|,\\
                   \|x\| \le \|(A+I)x\|,\;\; x\in\mathcal{D(A)}.
$$
The range of $A+I$ is dense because
$$
            \overline{\mathcal{R}(A+I)} = \mathcal{N}(A+I)^{\perp} = \{0\}^{\perp}= \mathcal{H}.
$$
The derived inequality $\|x\| \le \|(A+I)x\|$ shows that the range of $A+I$ is closed. Indeed, if $y\in \mathcal{H}$, then there exists $\{ x_n \} \subset \mathcal{D}(A)$ such that $\{y_n=(A+I)x_n\}$ converges to $y$. By the derived inequality,
$$
         \|x_n-x_m\| \le \|(A+I)(x_n-x_m)\| = \|y_n-y_m\|,
$$
from which it follows that $\{ x_n \}$ is a Cauchy sequence and, hence, converges to some $x\in\mathcal{H}$. Every selfadjoint $A$ is closed. Therefore, because $x_n, Ax_n$ converge to $x, y-x$, then $x\in\mathcal{D}(A)$ and $Ax=y-x$ or $(A+I)x=y$. So the range of $A+I$ is closed; because the range of $A+I$ is dense, then $A+I$ is surjective. So $(A+I)^{-1}$ is defined on all of $\mathcal{H}$, and this inverse is bounded because
$$
           \|(A+I)^{-1}y\| \le \|(A+I)(A+I)^{-1}y\|=\|y\|,\;\; y \in\mathcal{H}.
$$
Note: The argument above works for unbounded selfadjoint $A : \mathcal{D}(A)\subset \mathcal{H} \rightarrow \mathcal{H}$. However, a selfadjoint operator defined on all of $\mathcal{H}$ is necessarily bounded, because it is closed.
A: I assume that $A$ is linear and the ground field is $\Bbb C$, then $A$ is self-adjoint. 
A must be bounded: Let $x_n \rightarrow x, Ax_n \rightarrow y $,  then for any z $\in H$ we have $<y,z>=\lim<Ax_n,z>=\lim<x_n,Az>=<x,Az>=<Ax,z>$ hence $y=Ax$. By closed graph theorem, A is bounded.
Now $\sigma(A) \subseteq [0,+\infty]$, hence I+A is invertible.
