Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $1,1/2,1/3,\ldots$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.