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.
Note that by choosing an appropriate basis we can assume $B$ diagonal, with $B_{kk}\geq0$ for all $k$. It is then clear that $\mathcal R(B)$ is spanned by the eigenvectors corresponding to all the nonzero eigenvalues. So, as an operator $\mathcal R(B)\to\mathcal R(B)$, $B$ has dense range. Its inverse $B^{-1}$ is then densely defined and selfadjoint (and, most likely, unbounded).
It is easier to believe this by considering a diagonal $B$ with diagonal $1,1/2,1/3,\ldots$