# Inverse of trace class operator restricted to it's range

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.

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Are you sure they are claiming that $B^{-1}$ maps the range into itself? Because that is certainly wrong, for any example where $H$ is infinite-dimensional. – Lukas Geyer Nov 21 '12 at 0:06
Yeah seems wrong to me too, but the above is a direct quote from the paper. – Nick Alger Nov 21 '12 at 1:01

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 $1,1/2,1/3,\ldots$
Hmm, ok. However in the next paragraph they define a scalar product $\langle .,. \rangle:\mathcal{R}(B) \times \mathcal{R}(B) \rightarrow \mathbb{R}$ via $\langle \xi, \eta \rangle := (\xi,B^{-1} \eta)$, and then go on to form a new space $\mathcal{R}(B) \subset H_\mu \subset H$ as the completion of $\mathcal{R}(B)$ under $\langle .,. \rangle$. If $B^{-1}$ is unbounded and we're really working in a dense subspace, then this seems like nonsense! What am I missing here? – Nick Alger Nov 21 '12 at 3:51
$B^{-1}$ is definitely unbounded. As the authors of your paper say, if $B^{-1}$ is bounded, then $I=BB^{-1}$ is trace-class, and then $H$ is finite-dimensional. Not sure what looks like nonsense to you. – Martin Argerami Nov 21 '12 at 3:53
Yeah that makes sense, but then what does it mean to form the completion of $\mathcal{R}(B)$ under $(\cdot,B^{-1} \cdot)$ if $B^{-1}$ is unbounded? – Nick Alger Nov 21 '12 at 3:55
Ok. There exists a dense subpace $D \subset \mathcal{R}(B)$ on which $B^{-1}$ is defined. This subspace is not closed under the standard topology of $H$. However, on $D$, viewed as a set, we can define the scalar product $(\cdot,B^{-1} \cdot)$, and then complete $D$ with respect to this new scalar product to form a new space $H_\mu$. Now the question becomes, how do we know that $\mathcal{R}(B) \subset H_\mu$? – Nick Alger Nov 21 '12 at 4:05