I have been reading the paper - "[Introduction to Quantum Fisher Information](http://arxiv.org/abs/1008.2417)".
In section 1.2 the author talks about the linear map $\mathbb{J}\_D$, which he defines as follows:


Let $D \in M\_n$ be a positive invertible matrix. The linear mapping $\mathbb{J}^f\_D:M\_n \to M\_n$ is defined by the formula


$\mathbb{J}\_D^f=f(\mathbb{L}\_D\mathbb{R}^{-1}\_D)\mathbb{R}\_D$


where $f:\mathbb{R}^+\to\mathbb{R}^+$,


$\mathbb{L}\_D(X)=DX$ and $\mathbb{R}\_D(X)=XD$.


Also, the author points out that the inverse mapping of $\mathbb{J}\_D^f$ is the given by


$\frac{1}{f}(\mathbb{L}\_D\mathbb{R}^{-1}\_D)\mathbb{R}\_D^{-1}$


Then he gives the following example, of which I am able to get the first part but not the inverse map. 


*Example 1*: If $f(x)=(x+1)/2$, then


$\mathbb{J}\_DB=\frac{1}{2}(DB+BD)$.


$\mathbb{J}\_DB=\int\_0^\infty \text{exp}(-tD/2)B\text{exp}(-tD/2)dt$


Can you tell me how does he get the required formula for the inverse map? Also, I would like to know what is the meaning of the map $\mathbb{J}\_D$? What does the integral of the linear map represent? Can you also provide me some reference book where I could look for these topics?



Linear Mapping and integration
One way to think of the operation ${J}\_D^f$ is as follows: $M\_n$ has the structure of a Hilbert space by the inner product $\langle X, Y \rangle = {\rm Tr}(X^\* Y) = {\rm Tr}(Y X^\*)$. Then the transformations $L\_D R\_{D}^{-1}$ and ${R}\_D$ are positive operators on this Hilbert space. Since they commute, you may take the functional calculus $f(s) \otimes g(t) \mapsto f(L\_D R\_{D}^{-1}) g({R}\_D)$ from the space of functions on ${\rm Sp}(L\_D R\_D^{-1}) \times {\rm Sp}(R\_D)$ into the space of bounded operators on $M\_n$ (since the spectrum ${\rm Sp}(L\_D{R}\_{D^{-1}})$ of $L\_D$ is a finite subset of ${R}$, there is no need to worry about the regularity of $f$. Ditto for $g$). Then ${J}\_D^f$ is simply the image of $f(s) \otimes t$. Since the functional calculus is an algebra homomorphism, the inverse of ${J}\_D^f$ is represented by $(1/f(s)) \otimes (1/t)$, which equals $\frac{1}{f}(L\_D{R}\_D^{-1}){R}\_D^{-1}$. When $E \subset M\_n$ is the joint eigenspace for $L\_D {R}\_{D^{-1}}$ and ${R}\_D$ with the corresponding eigenvalues $a b^{-1}$ and $b$ (i.e. $L\_D$ has eigenvalue $a$ on this subspace), ${J}\_D^f$ acts by $f(a/b)b$ on $E$.


Any book containing the spectral theory of self adjoint operators on Hilbert spaces will do, like Pedersen's Analysis Now (GTM 118).


About the computation of $\mathbb{J}\_D$ for the case of $f(t) = \frac{t+1}{2}$, the integral formula follows from the identity $\int\_0^\infty exp(-t a / 2) exp(-t b / 2) = \frac{2}{a+b} = \frac{2}{ab^{-1}+1} \frac{1}{b}$.


Edit: I made a stupid mistake in the first version, and this is the corrected version. Sorry for the change of notation from $\mathbb{L}\_D$ to $L\_D$, etc. I somehow couldn't make it to work.


