A paper I'm reading constructs the Cameron-Martin space in a way different than I'm used to, and in the process they gloss over a functional analysis result about the existence of an inverse. It should be simple but I'm having trouble proving it.
Let $H$ be a separable Hilbert space, and $B:H \rightarrow H$ a positive trace class (covariance) operator. The sentence I can't prove is the following on page 3:
Consider the covariance operator $B$ restricted to the range of $B$, i.e., $B:\mathcal{R}(B) \rightarrow \mathcal{R}(B),~~~~~~~~~(15)$ then the (self-adjoint) inverse $B^{-1}$ exists on this subspace since $B>0$, and hence $B^{-1}>0$ on $\mathcal{R}(B)$.
This seems somewhat counterintuitive - how do we know that there isn't an element outside $\mathcal{R}(B)$ that gets mapped into $\mathcal{R}(B)$, thereby preventing invertibility?
My first thought was to use the existence of a square root operator $B^{\frac{1}{2}}$ and just unwrap the definitions of inverse functions, positivity of operators, etc. However, I was unsuccessful in this endeavor, and I think a more sophisticated approach may be required.