Alpha-Beta Log-Determinant Divergences Between Positive Definite Trace Class Operators

Abstract

This work presents a parametrized family of divergences, namely Alpha-Beta Log-Determinant (Log-Det) divergences, between positive definite unitized trace class operators on a Hilbert space. This is a generalization of the Alpha-Beta Log-Determinant divergences between symmetric, positive definite matrices to the infinite-dimensional setting. The family of Alpha-Beta Log-Det divergences is highly general and contains many divergences as special cases, including the recently formulated infinite-dimensional affine-invariant Riemannian distance and the infinite-dimensional Alpha Log-Det divergences between positive definite unitized trace class operators. In particular, it includes a parametrized family of metrics between positive definite trace class operators, with the affine-invariant Riemannian distance and the square root of the symmetric Stein divergence being special cases. For the Alpha-Beta Log-Det divergences between covariance operators on a Reproducing Kernel Hilbert Space (RKHS), we obtain closed form formulas via the corresponding Gram matrices.

Proofs of Main results

1.1 Proofs for the General Alpha-Beta Log-Determinant divergences

In this section, we prove Lemma 1, Proposition 1, and Theorems 7, 8, 9, and 10 .

Proof of Lemma 1

Since any $A \in {\mathcal {L}}({\mathcal {H}})$ commutes with the identity I, we have

$$\begin{aligned} \exp (A+\gamma I) = e^{\gamma }\exp (A) = e^{\gamma }\left( I + \sum _{j=1}^{\infty }\frac{A^j}{j!}\right) = e^{\gamma }I + e^{\gamma } \sum _{j=1}^{\infty }\frac{A^j}{j!} \in \mathrm{Tr}_X({\mathcal {H}}), \end{aligned}$$

since $\sum _{j=1}^{\infty }\frac{A^j}{j!}$ is trace class, with $\left\| \sum _{j=1}^{\infty }\frac{A^j}{j!}\right\| _{\mathrm{tr}} \le \sum _{j=1}^{\infty }\frac{||A||^j_{\mathrm{tr}}}{j!} = \exp (||A||_{\mathrm{tr}}) -1 < \infty .$$\square$

Proof of Proposition 1

For $(A+\gamma I), (B+\mu I) \in \mathrm{PTr}({\mathcal {H}})$, we have $(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2} \in \mathrm{PTr}({\mathcal {H}})$ and the logarithm $\log [(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}] \in \mathrm{Tr}_X({\mathcal {H}})$ is well-defined. By the discussion preceding Proposition 1,

$$\begin{aligned}&\log [(A+\gamma I)(B+\mu I)^{-1}] \\&\quad = \log [(B+\mu I)^{1/2}(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}(B+\mu I)^{-1/2}] \\&\quad = (B+\mu I)^{1/2}\log [(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}] (B+ \mu I)^{-1/2} \\&\quad = (B+\mu I)^{1/2}\log (\varLambda + \frac{\gamma }{\mu }I)(B+ \mu I)^{-1/2} \in \mathrm{Tr}_X({\mathcal {H}}). \end{aligned}$$

For the power function, we have

$$\begin{aligned}&[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha } = \exp (\alpha \log [(A+\gamma I)(B+\mu I)^{-1}]) \\&\quad = \exp [(B+\mu I)^{1/2}\alpha \log (\varLambda + \frac{\gamma }{\mu }I)(B+ \mu I)^{-1/2}] \\&\quad = (B+\mu I)^{1/2}\exp [\alpha \log (\varLambda + \frac{\gamma }{\mu }I)](B+ \mu I)^{-1/2} \\&\quad = (B+\mu I)^{1/2}(\varLambda + \frac{\gamma }{\mu }I)^{\alpha }(B+ \mu I)^{-1/2}. \end{aligned}$$

For the sum of two power functions, we then have

$$\begin{aligned}&\frac{\alpha [(A+\gamma I)(B+\mu I)^{-1}]^{p} + \beta [(A+\gamma I)(B+ \mu I)^{-1}]^{q}}{\alpha + \beta } \\&\quad = (B+\mu I)^{1/2}\left[ \frac{\alpha (\varLambda + \frac{\gamma }{\mu } I )^{p} + \beta (\varLambda + \frac{\gamma }{\mu } I)^{q}}{\alpha + \beta }\right] (B+\mu I)^{-1/2}. \end{aligned}$$

By Lemma 5 in [24], $\mathrm{det_X}[C(A+\gamma I)C^{-1}] = \mathrm{det_X}(A+\gamma I)$ for any invertible $C \in {\mathcal {L}}({\mathcal {H}})$. Thus

$$\begin{aligned}&\mathrm{det_X}\left[ \frac{\alpha [(A+\gamma I)(B+\mu I)^{-1}]^{p} + \beta [(A+\gamma I)(B+ \mu I)^{-1}]^{q}}{\alpha + \beta }\right] \\&\quad =\mathrm{det_X}\left[ \frac{\alpha (\varLambda + \frac{\gamma }{\mu } I )^{p} + \beta (\varLambda + \frac{\gamma }{\mu } I)^{q}}{\alpha + \beta }\right] . \end{aligned}$$

$\square$

Proof of Theorem 7

By definition of the power function, we have

$$\begin{aligned}&\alpha (A+ \gamma I)^p + (1-\alpha )(B+ \mu I)^q = \alpha \exp [p\log (A+\gamma I)] + (1-\alpha )\exp [q\log (B + \mu I)] \\&\quad = \alpha \exp \left[ p\log \left( \frac{A}{\gamma }+I\right) + p(\log \gamma )I\right] + (1-\alpha )\exp \left[ q\log \left( \frac{B}{\mu }+I\right) +q(\log \mu )I\right] \\&\quad = \alpha \gamma ^{p}\left( \frac{A}{\gamma }+I\right) ^p + (1-\alpha )\mu ^q\left( \frac{B}{\mu }+I\right) ^q. \end{aligned}$$

It follows that for $\delta = \frac{\alpha \gamma ^{p}}{\alpha \gamma ^p + (1-\alpha ) \mu ^q}$, $1-\delta =\frac{(1-\alpha ) \mu ^q}{\alpha \gamma ^p + (1-\alpha ) \mu ^q}$, we have

$$\begin{aligned}&\mathrm{det_X}[\alpha (A+\gamma I)^p + (1-\alpha )(B+\mu I)^q] \\&\quad = [\alpha \gamma ^p + (1-\alpha ) \mu ^q] \det \left[ \frac{\alpha \gamma ^{p}}{\alpha \gamma ^p + (1-\alpha ) \mu ^q}\left( \frac{A}{\gamma }+I\right) ^p \right. \\&\qquad \left. + \frac{(1-\alpha )\mu ^q}{\alpha \gamma ^p + (1-\alpha ) \mu ^q}\left( \frac{B}{\mu }+I\right) ^q\right] \\&\quad \ge [\alpha \gamma ^p + (1-\alpha ) \mu ^q]\det \left( \frac{A}{\gamma }+I\right) ^{p\delta } \det \left( \frac{B}{\mu }+I\right) ^{q(1-\delta )} \;\;\;\;\; \text {by Proposition 7 in [24]} \\&\quad \ge \gamma ^{p\alpha }\mu ^{(1-\alpha )q}\det \left( \frac{A}{\gamma }+I\right) ^{p\delta } \det \left( \frac{B}{\mu }+I\right) ^{q(1-\delta )} \\&\quad \;\;\;\;\; \text {by Ky Fan's Inequality applied to }\alpha \gamma ^p + (1-\alpha ) \mu ^q \\&\quad = \gamma ^{p(\alpha - \delta )}\mu ^{-q(\alpha -\delta )}\mathrm{det_X}(A+\gamma I)^{p\delta }\mathrm{det_X}(B+\mu I)^{q(1-\delta )} \\&\quad = \left( \frac{\gamma ^p}{\mu ^q}\right) ^{\alpha -\delta }\mathrm{det_X}(A+\gamma I)^{p\delta }\mathrm{det_X}(B+\mu I)^{q(1-\delta )}. \end{aligned}$$

For $0<\alpha < 1$, equality happens if and only if simultaneously, we have

$$\begin{aligned} \left( \frac{A}{\gamma }+I\right) ^{p} = \left( \frac{B}{\mu }+I\right) ^{q} \;\; \text {and}\;\;\; \gamma ^p = \mu ^q \Longleftrightarrow (A+\gamma I)^p = (B+\mu I)^q. \end{aligned}$$

In particular, for $\gamma = \mu$, the condition $\gamma ^{p} = \mu ^{q}$ becomes

$$\begin{aligned} \gamma ^p = \gamma ^q \Longleftrightarrow \gamma ^{p-q} = 1 \Longleftrightarrow p = q \;\;\; \text {if}\;\;\; \gamma \ne 1. \end{aligned}$$

With the conditions $\gamma = \mu \ne 1$ and $p =q$, we then have $(\frac{A}{\gamma }+I)^{p} = (\frac{B}{\gamma }+I)^{p} \Longleftrightarrow A = B$.$\square$

Proof of Theorem 8

Recall that we write $(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2} = \varLambda + (\gamma /\mu )I \in \mathrm{PTr}({\mathcal {H}}).$ Its inverse, also in $\mathrm{PTr}({\mathcal {H}})$, has the form $(B+\mu I)^{1/2}(A+\gamma I)^{-1}(B+\mu I)^{1/2} = [\varLambda + (\gamma /\mu )I]^{-1} = \frac{\mu }{\gamma }I - (\frac{\mu }{\gamma })^2\varLambda (I + \frac{\mu }{\gamma }\varLambda )^{-1}.$ It follows from Corollary 1 that

$$\begin{aligned}&\mathrm{det_X}\left[ \frac{\alpha [(\varLambda + (\gamma /\mu )I]^{p} + \beta [(\varLambda + (\gamma /\mu )I)^{-1}]^{q}}{\alpha + \beta }\right] \nonumber \\&\quad \ge \left( \frac{(\gamma /\mu )^p}{(\mu /\gamma )^q}\right) ^{\frac{\alpha }{\alpha +\beta } -\delta }\mathrm{det_X}(\varLambda + (\gamma /\mu )I]^{p\delta }\mathrm{det_X}[(\varLambda + (\gamma /\mu )I]^{-q(1-\delta )} \nonumber \\&\quad = \left( \frac{\gamma }{\mu }\right) ^{(p+q)(\frac{\alpha }{\alpha +\beta }-\delta )} \mathrm{det_X}(\varLambda + (\gamma /\mu )I]^{p\delta }\mathrm{det_X}[(\varLambda + (\gamma /\mu )I]^{-q(1-\delta )}, \end{aligned}$$

(129)

where $\delta = \frac{\alpha (\frac{\gamma }{\mu })^{p}}{\alpha (\frac{\gamma }{\mu })^{p} + \beta (\frac{\mu }{\gamma })^{q}} = \frac{\alpha (\frac{\gamma }{\mu })^{p+q}}{\alpha (\frac{\gamma }{\mu })^{p+q} + \beta }$, $1-\delta = \frac{\beta (\frac{\mu }{\gamma })^{q}}{\alpha (\frac{\gamma }{\mu })^{p} + \beta (\frac{\mu }{\gamma })^{q}} = \frac{\beta }{\alpha (\frac{\gamma }{\mu })^{p+q} + \beta }$.

For the two determinants on the right hand side of (129) to cancel each other out, we need

$$\begin{aligned} p\delta = q(1-\delta ) \Longleftrightarrow \alpha p\left( \frac{\gamma }{\mu }\right) ^{p} = \beta q \left( \frac{\mu }{\gamma }\right) ^{q} \Longleftrightarrow \alpha p \left( \frac{\gamma }{\mu }\right) ^{p+q} = \beta q. \end{aligned}$$

Assuming that this condition holds, then by the definition of $D^{(\alpha , \beta )}_{(p,q)}$, (129) gives

$$\begin{aligned}&\left[ \left( \frac{\gamma }{\mu }\right) ^{(p+q)(\delta -\frac{\alpha }{\alpha +\beta })} \mathrm{det_X}\left( \frac{\alpha (\varLambda +\frac{\gamma }{\mu }I)^p + \beta (\varLambda +\frac{\gamma }{\mu }I)^{-q}}{\alpha + \beta }\right) \right] \ge 1 \\&\quad \Longleftrightarrow D^{(\alpha , \beta )}_{(p,q)}[(A+\gamma I), (B+ \mu I)] \ge 0. \end{aligned}$$

In the inequality in (129), the equality sign happens if and only if

$$\begin{aligned}{}[(\varLambda + (\gamma /\mu )I]^{p} = [(\varLambda + (\gamma /\mu )I]^{-q} \Longleftrightarrow [(\varLambda + (\gamma /\mu )I]^{p+q} = I. \end{aligned}$$

If $p+q = 0$, then this is always true, so that $D^{(\alpha , \beta )}_{(p,q)}[(A+\gamma I), (B+\mu I)] = 0$ for all pairs $(A+\gamma I), (B+\mu I) \in \mathrm{PTr}({\mathcal {H}})$, which is not what we want. In fact, with $p+q = 0$, the condition $\alpha p \left( \frac{\gamma }{\mu }\right) ^{p+q} = \beta q$ gives $(\alpha + \beta ) p = 0 \Rightarrow p = 0 \Rightarrow q = 0$.

If $p +q \ne 0$, since $\varLambda + (\gamma /\mu )I > 0$, this happens if and only if

$$\begin{aligned}&\varLambda + (\gamma /\mu )I = I \Longleftrightarrow (B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2} = I \\&\quad \Longleftrightarrow A+\gamma I = B+\mu I \Longleftrightarrow A=B\;\;\text {and}\;\;\;\gamma = \mu . \end{aligned}$$

$\square$

Proof of Theorem 9

Under the condition $p+q = r$, by Theorem 8, we have $\alpha p\left( \frac{\gamma }{\mu }\right) ^{r} = \beta (r-p) \Rightarrow p = \frac{\beta r}{\alpha \left( \frac{\gamma }{\mu }\right) ^{r} + \beta }$ Thus $q = r - p = \frac{r \alpha \left( \frac{\gamma }{\mu }\right) ^{r}}{\alpha \left( \frac{\gamma }{\mu }\right) ^{r} + \beta }$. The equivalence of Eqs. (18) and (19) then follows from Proposition 1. $\square$

Proof of Theorem 10

We have

$$\begin{aligned}&\frac{\alpha (\varLambda + \frac{\gamma }{\mu }I)^p + \beta (\varLambda + \frac{\gamma }{\mu }I)^{-q}}{\alpha + \beta } = \frac{\alpha (\frac{\gamma }{\mu })^p (\frac{\mu }{\gamma }\varLambda + I)^p + \beta (\frac{\gamma }{\mu })^{-q}(\frac{\mu }{\gamma }\varLambda + I)^{-q}}{\alpha + \beta } \\&\quad = \frac{\alpha (\frac{\gamma }{\mu })^p (I+C_1) + \beta (\frac{\gamma }{\mu })^{-q} (I+C_2)}{\alpha + \beta } \\&\quad = \frac{\left[ \alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}\right] I + \left[ \alpha (\frac{\gamma }{\mu })^p C_1 + \beta (\frac{\gamma }{\mu })^{-q} C_2\right] }{\alpha + \beta } \\&\quad = \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\left[ I + \frac{\alpha (\frac{\gamma }{\mu })^p C_1 + \beta (\frac{\gamma }{\mu })^{-q} C_2}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] , \end{aligned}$$

where the operators $C_1$ and $C_2$ are given by $C_1 = \sum _{k=1}^{\infty }\frac{p^k}{k!}\left[ \log \left( \frac{\mu }{\gamma }\varLambda + I\right) \right] ^k \in \mathrm{Tr}({\mathcal {H}})$, $C_2 = \sum _{k=1}^{\infty }\frac{(-1)^kq^k}{k!}\left[ \log \left( \frac{\mu }{\gamma }\varLambda + I\right) \right] ^k \in \mathrm{Tr}({\mathcal {H}})$. By definition of the $\mathrm{det_X}$ function, we then have

$$\begin{aligned}&\log \mathrm{det_X}\left[ \frac{\alpha (\varLambda + \frac{\gamma }{\mu }I)^p + \beta (\varLambda + \frac{\gamma }{\mu }I)^{-q} )}{\alpha + \beta }\right] \\&\quad = \log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) + \log \det \left[ I + \frac{\alpha (\frac{\gamma }{\mu })^p C_1 + \beta (\frac{\gamma }{\mu })^{-q} C_2}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad = \log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) +\log \det \left[ \frac{\alpha (\varLambda +\frac{\gamma }{\mu }I)^p + \beta (\varLambda + \frac{\gamma }{\mu }I)^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] . \end{aligned}$$

This, together with the definition of $D^{(\alpha , \beta )}_{(p,q)}$, gives us the desired expression. $\square$

1.2 Proofs for the affine-invariant Riemannian distance

In this section, we prove Theorem 2 (part 1). In Definition 1, with $\alpha = \beta$, $\delta = \frac{(\frac{\gamma }{\mu })^{r}}{(\frac{\gamma }{\mu })^{r} + 1}$, we have

$$\begin{aligned}&D^{(\alpha , \alpha )}_r[(A+\gamma I), (B+\mu I)] \nonumber \\&\quad = \frac{1}{\alpha ^2}\log \left[ \left( \frac{\gamma }{\mu }\right) ^{r(\delta -\frac{1}{2})} \mathrm{det_X}\left( \frac{(\varLambda +\frac{\gamma }{\mu }I)^{r(1-\delta )} + (\varLambda +\frac{\gamma }{\mu }I)^{-r\delta }}{2}\right) \right] . \end{aligned}$$

(130)

We first need the following results.

Lemma 6

Let $\gamma > 0$. Assume that $r = r(\alpha )$ is smooth, with $r(0) = 0$. Let $\delta = \frac{\gamma ^r}{\gamma ^r+1}$. Then

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{r(\delta - \frac{1}{2})}{\alpha ^2} = \frac{[r{'}(0)]^2}{4}\log \gamma . \end{aligned}$$

(131)

In particular, for $r = 2\alpha$, we have $\lim _{\alpha \rightarrow 0}\frac{r(\delta - \frac{1}{2})}{\alpha ^2} = \log \gamma$.

Lemma 7

Let $\gamma > 0$ be fixed. Let $\lambda > 0$ be fixed. Assume that $r = r(\alpha )$ is smooth, with $r(0) = 0$. Define $\delta = \frac{\gamma ^r}{\gamma ^r+1},\; p = r(1-\delta ), \; q = r\delta$. Then

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{\lambda ^p + \lambda ^{-q}}{2}\right) = \frac{[r{'}(0)]^2}{4}\left[ -(\log \gamma )(\log \lambda ) + \frac{1}{2}(\log \lambda )^2\right] . \end{aligned}$$

(132)

In particular, if $\gamma = \lambda$, then $\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{\gamma ^p + \gamma ^{-q}}{2}\right) = -\frac{[r{'}(0)]^2}{8}(\log \gamma )^2.$ If $\gamma = 1$, then $\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{\lambda ^{r/2} + \lambda ^{-r/2}}{2}\right) = \frac{[r{'}(0)]^2}{8}(\log \lambda )^2.$

Lemma 8

Let $\gamma > 0$ be fixed. Let $\lambda \in {\mathbb {R}}$ be fixed such that $\lambda + \gamma > 0$. Assume that $r = r(\alpha )$ is smooth, with $r(0) = 0$. Define $\delta = \frac{\gamma ^r}{\gamma ^r+1},\; p = r(1-\delta ), \; q = r\delta$. Then

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{(\lambda + \gamma )^p + (\lambda + \gamma )^{-q}}{\gamma ^p + \gamma ^{-q}}\right) = \frac{[r{'}(0)]^2}{8}\left[ \log \left( \frac{\lambda }{\gamma } +1\right) \right] ^2. \end{aligned}$$

(133)

In particular, if $r = r(\alpha ) = 2\alpha$, then $\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{(\lambda + \gamma )^p + (\lambda + \gamma )^{-q}}{\gamma ^p + \gamma ^{-q}}\right) = \frac{1}{2}\left[ \log \left( \frac{\lambda }{\gamma } +1\right) \right] ^2$.

Lemma 9

Let $\gamma > 0$ be fixed. Let $\lambda \in {\mathbb {R}}$ be fixed such that $\lambda + \gamma > 0$. Assume that $r = r(\alpha )$ is smooth, with $r(0) = 0$. Define $\delta = \frac{\gamma ^r}{\gamma ^r+1},\; p = r(1-\delta ), \; q = r\delta$. Then

$$\begin{aligned} \frac{(\lambda + \gamma )^p + (\lambda + \gamma )^{-q}}{\gamma ^p + \gamma ^{-q}} \ge 1, \;\;\; \log \left( \frac{(\lambda + \gamma )^p + (\lambda + \gamma )^{-q}}{\gamma ^p + \gamma ^{-q}}\right) \ge 0. \end{aligned}$$

(134)

Proof of Theorem 2, part 1

For $\alpha = \beta$, we have $\delta = \frac{(\frac{\gamma }{\mu })^r}{(\frac{\gamma }{\mu })^r+1},\; p = r(1-\delta ), \; q = r\delta$. Let $\{\lambda _j\}_{j \in {\mathbb {N}}}$ be the eigenvalues of $\varLambda$. By Theorem 10, we have

$$\begin{aligned}&D^{(\alpha , \alpha )}_r[(A+\gamma I), (B+ \mu I)] = \frac{r(\delta -\frac{1}{2})}{\alpha ^2}\log \left( \frac{\gamma }{\mu }\right) +\frac{1}{\alpha ^2} \log \left( \frac{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}{2}\right) \\&\qquad + \frac{1}{\alpha ^2}\log \det \left[ \frac{(\varLambda + \frac{\gamma }{\mu }I)^p + (\varLambda +\frac{\gamma }{\mu } I)^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad = \frac{r(\delta -\frac{1}{2})}{\alpha ^2}\log \left( \frac{\gamma }{\mu }\right) +\frac{1}{\alpha ^2} \log \left( \frac{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}{2}\right) \\&\qquad + \frac{1}{\alpha ^2}\sum _{j=1}^{\infty }\log \left( \frac{(\lambda _j + \frac{\gamma }{\mu })^p + (\lambda _j +\frac{\gamma }{\mu })^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right) . \end{aligned}$$

By Lemma 6, we have $\lim _{\alpha \rightarrow 0}\frac{r(\delta -\frac{1}{2})}{\alpha ^2}\log \left( \frac{\gamma }{\mu }\right) = \frac{[r{'}(0)]^2}{4}\left[ \log \frac{\gamma }{\mu }\right] ^2$.

By Lemma 7, we have $\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left( \frac{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}{2}\right) = -\frac{[r{'}(0)]^2}{8}\left[ \log \frac{\gamma }{\mu }\right] ^2.$

By Lemma 8, we have

$$\begin{aligned}&\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \det \left[ \frac{(\varLambda + \frac{\gamma }{\mu }I)^p + (\varLambda +\frac{\gamma }{\mu } I)^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad = \lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\sum _{j=1}^{\infty }\log \left[ \frac{(\lambda _j + \frac{\gamma }{\mu })^p + (\lambda _j +\frac{\gamma }{\mu })^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad = \sum _{j=1}^{\infty }\lim _{\alpha \rightarrow 0}\frac{1}{\alpha ^2}\log \left[ \frac{(\lambda _j + \frac{\gamma }{\mu })^p + (\lambda _j +\frac{\gamma }{\mu })^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right] \\&\;\;\; \;\;\;\;\;\text {by Lebesgue's Monotone Convergence Theorem, since} \\&\;\;\;\;\;\;\; \; \log \left[ \frac{(\lambda _j + \frac{\gamma }{\mu })^p + (\lambda _j +\frac{\gamma }{\mu })^{-q}}{(\frac{\gamma }{\mu })^p + (\frac{\gamma }{\mu })^{-q}}\right] \ge 0 \; \forall j \in {\mathbb {N}}\;\; \text {by Lemma}~\mathrm{9} \\&\quad = \frac{[r{'}(0)]^2}{8}\sum _{j=1}^{\infty }\left[ \log \left( \lambda _j + \frac{\gamma }{\mu }\right) - \log \left( \frac{\gamma }{\mu }\right) \right] ^2 = \frac{[r{'}(0)]^2}{8}\sum _{j=1}^{\infty }\left[ \log \left( \lambda _j\frac{\mu }{\gamma }+1\right) \right] ^2. \end{aligned}$$

Summing up these three expressions, we obtain

$$\begin{aligned}&\lim _{\alpha \rightarrow 0}D^{(\alpha , \alpha )}_r[(A+\gamma I), (B+\mu I)] = \frac{[r{'}(0)]^2}{8}\left( \left[ \log \frac{\gamma }{\mu }\right] ^2 + \sum _{j=1}^{\infty }\left[ \log \left( \lambda _j\frac{\mu }{\gamma }+1\right) \right] ^2\right) \\&\quad = \frac{[r{'}(0)]^2}{8}\left( \left[ \log \frac{\gamma }{\mu }\right] ^2 + \left\| \log \left( \varLambda \frac{\mu }{\gamma }+ I\right) \right\| ^2_{\mathrm{HS}}\right) = \frac{[r{'}(0)]^2}{8}\left\| \log \left( \varLambda + \frac{\gamma }{\mu }I\right) \right\| ^2_{\mathrm{HS_X}} \\&\quad = \frac{[r{'}(0)]^2}{8}||\log [(B+\mu I)^{-1/2}(A+\gamma I)(B+ \mu I)^{-1/2}]||^2_{\mathrm{HS_X}} \\&\quad =\frac{[r{'}(0)]^2}{8} d_{\mathrm{aiHS}}^2[(A+\gamma I), (B+ \mu I)]. \end{aligned}$$

$\square$

1.3 Proofs for the Alpha Log-Determinant divergences

Proof of Theorem 13

The proof for the cases $\alpha = 0$ and $\alpha =1$ is a special case of the results discussed at the end of Sect. 4.1. Consider now the case $0< \alpha < 1$. We first note that

$$\begin{aligned} d^{1-2\alpha }_{\mathrm{logdet}}[(A+\gamma I), (B+ \mu I)&= \frac{1}{\alpha (1-\alpha )}\log \left[ \frac{\mathrm{det_X}(\alpha (A+\gamma I) + (1-\alpha )(B+\mu I)}{\mathrm{det_X}(A+\gamma I)^q\mathrm{det_X}(B+\mu I)^{1-q}}\right] \\&\quad + \frac{q-\alpha }{\alpha (1-\alpha )}\log \frac{\gamma }{\mu }, \;\;\;\text {where }q = \frac{\alpha \gamma }{\alpha \gamma + (1-\alpha )\mu }. \end{aligned}$$

From Eq. 67 we have

$$\begin{aligned}&D^{(\alpha , 1-\alpha )}_r[(A+\gamma I), (B+\mu I)] \\&\quad = \frac{1}{\alpha (1-\alpha )}\log [(\frac{\gamma }{\mu })^{r(\delta -\alpha )} \mathrm{det_X}({\alpha (\varLambda +\frac{\gamma }{\mu }I)^{r(1-\delta )} + (1-\alpha )(\varLambda +\frac{\gamma }{\mu }I)^{-r\delta }})] \\&\quad = \frac{r(\delta -\alpha )}{\alpha (1-\alpha )}\log (\frac{\gamma }{\mu }) + \frac{1}{\alpha (1-\alpha )}\log \mathrm{det_X}[{\alpha (\varLambda +\frac{\gamma }{\mu }I)^{r(1-\delta )} + (1-\alpha )(\varLambda +\frac{\gamma }{\mu }I)^{-r\delta }}]. \end{aligned}$$

By Proposition 1, we have

$$\begin{aligned}&\mathrm{det_X}[{\alpha (\varLambda +\frac{\gamma }{\mu }I)^{r(1-\delta )} + (1-\alpha )(\varLambda +\frac{\gamma }{\mu }I)^{-r\delta }}] \\&\quad = \mathrm{det_X}\left[ \alpha [(A+\gamma I)(B+\mu I)^{-1}]^{r(1-\delta )} + (1-\alpha )[(A+\gamma I)(B+\mu I)^{-1}]^{-r\delta }\right] \\&\quad = \mathrm{det_X}[(A+\gamma I)(B+\mu I)^{-1}]^{-r\delta }\mathrm{det_X}[\alpha [(A+\gamma I)(B+\mu I)^{-1}]^{r} + (1-\alpha )I]. \end{aligned}$$

In particular, for $r=1$, we have

$$\begin{aligned}&\mathrm{det_X}[\alpha [(A+\gamma I)(B+\mu I)^{-1}] + (1-\alpha )I] \\&\quad = \frac{\mathrm{det_X}[\alpha (A+\gamma I) + (1-\alpha )(B+\mu I)]}{\mathrm{det_X}(B+\mu I)}. \end{aligned}$$

Thus it follows that

$$\begin{aligned}&\mathrm{det_X}[{\alpha (\varLambda +\frac{\gamma }{\mu }I)^{(1-\delta )} + (1-\alpha )(\varLambda +\frac{\gamma }{\mu }I)^{-\delta }}]\\&\quad = \frac{\mathrm{det_X}[\alpha (A+\gamma I) + (1-\alpha )(B+\mu I)]}{\mathrm{det_X}(A+\gamma I)^{\delta }\mathrm{det_X}(B+\mu I)^{1-\delta }}. \end{aligned}$$

For $r=1$, in Eq. 67, $\delta = \delta (r=1) = \frac{\alpha \gamma }{\alpha \gamma + (1-\alpha )\mu }$. Combining all of these expressions and comparing with $d^{1-2\alpha }_{\mathrm{logdet}}$, we obtain the first desired statement. The case $r=-1$ is proved similarly. $\square$

1.4 Proofs for the other limiting cases

In this section, we prove Theorems 11 and 12 . We need the following results.

Lemma 10

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $A \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$ be such that $A+ I > 0$. Then $\forall \alpha \in {\mathbb {R}}$, $(A+I)^{\alpha }$ is well defined and $(A+I)^{\alpha } - I\in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$. Equivalently, let $\{\lambda _k\}_{k\in {\mathbb {N}}}$ be the eigenvalues of A, then

$$\begin{aligned} \mathrm{tr}[(A+I)^{\alpha } - I] = \sum _{k=1}^{\infty }[(\lambda _k + 1)^{\alpha } - 1] \;\;\; \text {has a finite value}. \end{aligned}$$

(135)

Proof of Lemma 10

By Lemma 3 in [24], if $A \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$ and $A+I > 0$, then $\log (A+I) \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$. By definition of the power function, we have

$$\begin{aligned} (A+I)^{\alpha } = \exp [\alpha \log (A+I)] = I + \sum _{j=1}^{\infty }\frac{\alpha ^j}{j!}[\log (A+I)]^j. \end{aligned}$$

Since $\mathrm{Tr}({\mathcal {H}})$ is a Banach algebra under the trace norm, we have

$$\begin{aligned} ||(A+I)^{\alpha }-I||_{\mathrm{tr}}&= \left\| \sum _{j=1}^{\infty }\frac{\alpha ^j}{j!}[\log (A+I)]^j\right\| _{\mathrm{tr}} \le \sum _{j=1}^{\infty }\frac{|\alpha |^j}{j!}||\log (A+I)||^j_{\mathrm{tr}} \\&= \exp (|\alpha |\;||\log (A+I)||_{\mathrm{tr}}) -1 < \infty . \end{aligned}$$

Thus $(A+I)^{\alpha } - I \in \mathrm{Tr}({\mathcal {H}})$. The equivalent statement is then obvious. $\square$

Lemma 11

Let ${\mathcal {H}}$ be a separable Hilbert space. Assume that $(A+\gamma I) \in \mathrm{PTr}({\mathcal {H}})$. Then for any $\alpha \in {\mathbb {R}}$, we have $(A+\gamma I)^{\alpha } - \gamma ^{\alpha }I \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$ and

$$\begin{aligned} \mathrm{tr}[(A+\gamma I)^{\alpha } - \gamma ^{\alpha }I]&= \gamma ^{\alpha }\mathrm{tr}\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] , \end{aligned}$$

(136)

$$\begin{aligned} \mathrm{tr}_X[(A+\gamma I)^{\alpha }]&= \gamma ^{\alpha }\left( 1 + \mathrm{tr}\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] \right) . \end{aligned}$$

(137)

Proof of Lemma 11

By definition of the power function, we have

$$\begin{aligned} (A+\gamma I)^{\alpha }&= \exp [\alpha \log (A+\gamma I)] = \exp \left[ (\alpha \log \gamma )I + \alpha \log \left( \frac{A}{\gamma } + I\right) \right] \\&= \gamma ^{\alpha }\left( \frac{A}{\gamma } +I\right) ^{\alpha } = \gamma ^{\alpha }\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] + \gamma ^{\alpha }I, \end{aligned}$$

where $\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] \in \mathrm{Tr}({\mathcal {H}})$ by Lemma 10. Thus $(A+\gamma I)^{\alpha } - \gamma ^{\alpha }I \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$ and

$$\begin{aligned} \mathrm{tr}[(A+\gamma I)^{\alpha }- \gamma ^{\alpha }I] = \gamma ^{\alpha }\mathrm{tr}\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] , \end{aligned}$$

which is the first identity. For the second identity, by definition of the extended trace

$$\begin{aligned} \mathrm{tr}_X[(A+\gamma I)^{\alpha }] = \mathrm{tr}_X([(A+\gamma I)^{\alpha }- \gamma ^{\alpha }I] + \gamma ^{\alpha }I) = \gamma ^{\alpha }\mathrm{tr}\left[ \left( \frac{A}{\gamma } +I\right) ^{\alpha } - I\right] +\gamma ^{\alpha }. \end{aligned}$$

$\square$

Lemma 12

Let $(A+\gamma I), (B+ \mu I) \in \mathrm{PTr}({\mathcal {H}})$. Let $\varLambda + \frac{\gamma }{\mu }I = (B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}$. Then for any $\alpha \in {\mathbb {R}}$,

$$\begin{aligned} \mathrm{tr}_X[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha }&= \mathrm{tr}_X\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{\alpha }\right] = \mathrm{tr}_X[(B+\mu I)^{-1}(A+\gamma I)]^{\alpha }. \end{aligned}$$

(138)

$$\begin{aligned} \mathrm{det_X}[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha }&= \mathrm{det_X}\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{\alpha }\right] \nonumber \\&= \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]^{\alpha }. \end{aligned}$$

(139)

Proof of Lemma 12

By Proposition 1, we have

$$\begin{aligned}{}[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha }&= (B+\mu I)^{1/2}\left( \varLambda + \frac{\gamma }{\mu }\right) ^{\alpha }(B+\mu I)^{-1/2}. \\ [(B+\mu I)^{-1}(A+\gamma I)]^{\alpha }&= (B+\mu I)^{-1/2}\left( \varLambda + \frac{\gamma }{\mu }\right) ^{\alpha }(B+\mu I)^{1/2}. \end{aligned}$$

By the commutativity of the $\mathrm{tr}_X$ operation (Lemma 4 in [24]), we then have

$$\begin{aligned} \mathrm{tr}_X[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha } = \mathrm{tr}_X\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{\alpha }\right] = \mathrm{tr}_X[(B+\mu I)^{-1}(A+\gamma I)]^{\alpha }. \end{aligned}$$

Similarly, by the product property of the $\mathrm{det_X}$ operation (Proposition 4 in [24]),

$$\begin{aligned} \mathrm{det_X}[(A+\gamma I)(B+\mu I)^{-1}]^{\alpha } {=} \mathrm{det_X}\left[ \left( \varLambda {+} \frac{\gamma }{\mu }\right) ^{\alpha }\right] {=} \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]^{\alpha }. \end{aligned}$$

$\square$

Lemma 13

Assume that $\lambda> 0, \gamma> 0, \alpha > 0$ are fixed. Assume that $r = r(\beta )$ is smooth. Then for $\delta = \frac{\alpha \gamma ^{r}}{\alpha \gamma ^{r} + \beta }, \;\; p = r(1-\delta ), \;\; q = r\delta$, we have

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\log \left( \frac{\alpha \lambda ^p + \beta \lambda ^{-q}}{\alpha + \beta }\right) = \frac{1}{\alpha ^2}\left( (\log \lambda )\frac{r(0)}{\gamma ^{r(0)}} + \lambda ^{-r(0)}-1\right) . \end{aligned}$$

(140)

In particular, for $\lambda = \gamma$, $\lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\log \left( \frac{\alpha \gamma ^p + \beta \gamma ^{-q}}{\alpha + \beta }\right) = \frac{1}{\alpha ^2}\left( [(\log \gamma ){r(0)} + 1]\gamma ^{-r(0)}-1\right) .$

Lemma 14

Assume that $\gamma> 0, \alpha > 0$ are fixed. Assume that $\lambda \in {\mathbb {R}}$ is also fixed, such that $\lambda + \gamma > 0$. Assume that $r = r(\beta )$ is smooth. Then for $\delta = \frac{\alpha \gamma ^{r}}{\alpha \gamma ^{r} + \beta }, \;\; p = r(1-\delta ), \;\; q = r\delta$, we have

$$\begin{aligned}&\lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\log \left( \frac{\alpha (\lambda +\gamma )^p + \beta (\lambda +\gamma )^{-q}}{\alpha \gamma ^p + \beta \gamma ^{-q}}\right) \nonumber \\&\quad = \frac{1}{\alpha ^2}\left[ \log \left( \frac{\lambda }{\gamma }+1\right) \frac{r(0)}{\gamma ^{r(0)}} + (\lambda +\gamma )^{-r(0)} - \gamma ^{-r(0)}\right] . \end{aligned}$$

(141)

Lemma 15

Assume that $\gamma> 0, \alpha > 0$ are fixed. Assume that $\lambda \in {\mathbb {R}}$ is also fixed, such that $\lambda + \gamma > 0$. Assume that $r = r(\beta )$ is smooth. Then for $\delta = \frac{\alpha \gamma ^{r}}{\alpha \gamma ^{r} + \beta }, \;\; p = r(1-\delta ), \;\; q = r\delta$, we have

$$\begin{aligned} \frac{\alpha (\lambda +\gamma )^p + \beta (\lambda +\gamma )^{-q}}{\alpha \gamma ^p + \beta \gamma ^{-q}} \ge 1, \;\; \log \left( \frac{\alpha (\lambda +\gamma )^p + \beta (\lambda +\gamma )^{-q}}{\alpha \gamma ^p + \beta \gamma ^{-q}}\right) \ge 0. \end{aligned}$$

(142)

Lemma 16

Assume that $\gamma> 0, \alpha > 0$ are fixed. Assume that $r = r(\beta )$ is smooth. Then for $\delta = \frac{\alpha \gamma ^{r}}{\alpha \gamma ^{r} + \beta }$,

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{r(\delta -\frac{\alpha }{\alpha +\beta })}{\alpha \beta } = \frac{1}{\alpha ^2}r(0)[-\gamma ^{-r(0)} + 1]. \end{aligned}$$

(143)

Proof of Theorem 11

Let $\{\lambda _j\}_{j=1}^{\infty }$ be the eigenvalues of $\varLambda$. By Theorem 10,

$$\begin{aligned} D^{(\alpha , \beta )}_r[(A+\gamma I), (B+\mu I)]&= \frac{r(\delta -\frac{\alpha }{\alpha +\beta })}{\alpha \beta }\log \left( \frac{\gamma }{\mu }\right) + \frac{1}{\alpha \beta }\log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) \\&\quad + \frac{1}{\alpha \beta }\log \det \left( \frac{\alpha (\varLambda + \frac{\gamma }{\mu }I)^p + \beta (\varLambda + \frac{\gamma }{\mu } I)^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right) \\&= \frac{r(\delta -\frac{\alpha }{\alpha +\beta })}{\alpha \beta }\log \left( \frac{\gamma }{\mu }\right) + \frac{1}{\alpha \beta }\log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) \\&\quad + \frac{1}{\alpha \beta }\sum _{j=1}^{\infty }\log \left( \frac{\alpha (\lambda _j + \frac{\gamma }{\mu })^p + \beta (\lambda _j + \frac{\gamma }{\mu })^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right) , \end{aligned}$$

where $p= p(\beta ) = r(1-\delta ) = \frac{r \beta }{\alpha (\frac{\gamma }{\mu })^{r} + \beta }$, $q= q(\beta ) = r\delta = \frac{r \alpha (\frac{\gamma }{\mu })^{r}}{\alpha (\frac{\gamma }{\mu })^{r} + \beta }$.

For $\alpha > 0$ fixed, as functions of $\beta$, we have $\lim _{\beta \rightarrow 0} p(\beta ) = 0, \lim _{\beta \rightarrow 0} q(\beta ) = r(0)$. For simplicity, in the following, we replace $\frac{\gamma }{\mu }$ by $\gamma$. By Lemma 13,

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta } \log \left( \frac{\alpha \gamma ^p + \beta \gamma ^{-q}}{\alpha + \beta }\right) =\frac{1}{\alpha ^2}\left( [(\log \gamma ){r(0)} + 1]\gamma ^{-r(0)}-1\right) . \end{aligned}$$

By Lemma 14,

$$\begin{aligned}&\lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\log \left( \frac{\alpha (\lambda _j+\gamma )^p + \beta (\lambda _j+\gamma )^{-q}}{\alpha \gamma ^p + \beta \gamma ^{-q}}\right) \\&\quad = \frac{1}{\alpha ^2}\left[ \log \left( \frac{\lambda _j}{\gamma }+1\right) \frac{r(0)}{\gamma ^{r(0)}} + (\lambda _j+\gamma )^{-r(0)} - \gamma ^{-r(0)}\right] . \end{aligned}$$

By Lemma 15, we have $\log \left( \frac{\alpha (\lambda _j+\gamma )^p + \beta (\lambda _j+\gamma )^{-q}}{\alpha \gamma ^p + \beta \gamma ^{-q}}\right) \ge 0$$\forall j \in {\mathbb {N}}$, so that by Lebesgue’s Monotone Convergence Theorem, we obtain

$$\begin{aligned}&\lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\sum _{j=1}^{\infty }\log \left( \frac{\alpha (\lambda _j + {\gamma })^p + \beta (\lambda _j + {\gamma })^{-q}}{\alpha {\gamma }^p + \beta {\gamma }^{-q}}\right) \\&\quad = \sum _{j=1}^{\infty }\lim _{\beta \rightarrow 0}\frac{1}{\alpha \beta }\log \left( \frac{\alpha (\lambda _j + {\gamma })^p + \beta (\lambda _j + \gamma )^{-q}}{\alpha {\gamma }^p + \beta {\gamma }^{-q}}\right) \\&\quad = \frac{1}{\alpha ^2}\sum _{j=1}^{\infty }\left[ \log \left( \frac{\lambda _j}{\gamma }+1\right) \frac{r(0)}{\gamma ^{r(0)}} + (\lambda _j+\gamma )^{-r(0)} - \gamma ^{-r(0)}\right] . \end{aligned}$$

By Lemma 16, $\log (\gamma )\lim _{\beta \rightarrow 0}\frac{r(\delta -\frac{\alpha }{\alpha +\beta })}{\alpha \beta } = \frac{1}{\alpha ^2}r(0)[-\gamma ^{-r(0)} + 1]\log (\gamma ).$ Combining all three expressions, we obtain the desired limit as the sum

$$\begin{aligned}&\frac{1}{\alpha ^2}[\gamma ^{-r(0)}+r(0)\log (\gamma ) -1] \nonumber \\&\qquad + \frac{1}{\alpha ^2}\left\{ \frac{r(0)}{\gamma ^{r(0)}}\sum _{j=1}^{\infty } \log \left( \frac{\lambda _j}{\gamma }+1\right) +\sum _{j=1}^{\infty }\left[ \frac{1}{\left( \lambda _j+\gamma \right) ^{r(0)}} - \frac{1}{\gamma ^{r(0)}}\right] \right\} . \end{aligned}$$

(144)

By Lemmas 10 and 11 , we have

$$\begin{aligned}&\sum _{j=1}^{\infty }\left[ \frac{1}{(\lambda _j+\gamma )^{r(0)}} - \frac{1}{\gamma ^{r(0)}}\right] = \gamma ^{-r(0)}\sum _{j=1}^{\infty }\left[ \left( \frac{\lambda _j}{\gamma } + 1\right) ^{-r(0)} -1\right] \\&\quad = \gamma ^{-r(0)}\mathrm{tr}\left[ \left( \frac{\varLambda }{\gamma } + I\right) ^{-r(0)} - I\right] = \mathrm{tr}[(\varLambda + \gamma I)^{-r(0)} - \gamma ^{-r(0)}I]. \end{aligned}$$

Thus it follows that

$$\begin{aligned}&\gamma ^{-r(0)} - 1 + \sum _{j=1}^{\infty }\left[ \frac{1}{\left( \lambda _j+\gamma \right) ^{r(0)}} - \frac{1}{\gamma ^{r(0)}}\right] \\&\quad = \gamma ^{-r(0)} - 1 + \mathrm{tr}[(\varLambda + \gamma I)^{-r(0)} - \gamma ^{-r(0)}I] = \mathrm{tr}_X[(\varLambda + \gamma I)^{-r(0)} - I]. \end{aligned}$$

Furthermore,

$$\begin{aligned}&\frac{r(0)}{\gamma ^{r(0)}}\sum _{j=1}^{\infty } \log \left( \frac{\lambda _j}{\gamma }+1\right) = r(0)\gamma ^{-r(0)}\log \det \left( \frac{\varLambda }{\gamma }+I\right) \\&\quad = r(0)\gamma ^{-r(0)}\log \mathrm{det_X}(\varLambda + \gamma I) - r(0)\gamma ^{-r(0)}\log \gamma \\&\quad = - \gamma ^{-r(0)}\log \mathrm{det_X}(\varLambda + \gamma I)^{-r(0)} - r(0)\gamma ^{-r(0)}\log \gamma . \end{aligned}$$

Plugging the last two expressions into (144), we obtain the desired limit as

$$\begin{aligned}&\frac{1}{\alpha ^2}\left\{ r(0)(1-\gamma ^{-r(0)})\log \gamma \right\} \nonumber \\&\quad + \frac{1}{\alpha ^2}\left\{ \mathrm{tr}_X[(\varLambda + \gamma I)^{-r(0)} - I]- \gamma ^{-r(0)}\log \mathrm{det_X}(\varLambda + \gamma I)^{-r(0)} \right\} . \end{aligned}$$

(145)

We now replace $\gamma$ by $\frac{\gamma }{\mu }$. We have by Lemma 12,

$$\begin{aligned} \mathrm{tr}_X\left[ \left( \varLambda + \frac{\gamma }{\mu } I\right) ^{-r(0)}\right]&= \mathrm{tr}_X[(B+\mu I)^{-1}(A+\gamma I)]^{-r(0)} \\&= \mathrm{tr}_X[(A+\gamma I)^{-1}(B+ \mu I)]^{r(0)}, \\ \mathrm{det_X}\left( \varLambda + \frac{\gamma }{\mu }I\right) ^{-r(0)}&= \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]^{-r(0)} \\&= \mathrm{det_X}\left[ (A+\gamma I)^{-1}(B+\mu I)\right] ^{r(0)}. \end{aligned}$$

Then (145) becomes

$$\begin{aligned}&\frac{r(0)}{\alpha ^2}[(\frac{\mu }{\gamma })^{r(0)} -1]\log \frac{\mu }{\gamma } + \frac{1}{\alpha ^2}\mathrm{tr}_X([(A+\gamma I)^{-1}(B+ \mu I)]^{r(0)} -I) \\&- \frac{1}{\alpha ^2}(\frac{\mu }{\gamma })^{r(0)}\log \mathrm{det_X}[(A+\gamma I)^{-1}(B+ \mu I)]^{r(0)}. \end{aligned}$$

$\square$

Proof of Theorem 12

The dual symmetry in Theorem 14 gives

$$\begin{aligned} \lim _{\alpha \rightarrow 0}D^{(\alpha , \beta )}_r[(A+\gamma I), (B+\mu I)] = \lim _{\alpha \rightarrow 0}D^{(\beta , \alpha )}_r[(B+ \mu I), (A+\gamma I)]. \end{aligned}$$

The limit on the right hand side then follows from Theorem 11. $\square$

1.5 Proofs of the properties of the Alpha-Beta Log-Determinant divergences

In this section, we prove Theorems 14, 15, 16, 17, 18, and 19 . For the case $\alpha = \beta = 0$, we have $D^{(0,0)}_0[(A+\gamma I), (B+\mu I)] = \frac{1}{2}d_{\mathrm{aiHS}}^2[(A+\gamma I), (B+\mu I)]$, with $d_{\mathrm{aiHS}}$ being the affine-invariant Riemannian distance on $\mathrm{PTr}({\mathcal {H}})$. Thus these properties are either automatic or straightforward to verify. We thus focus on the three cases $(\alpha> 0, \beta > 0)$, $(\alpha > 0, \beta = 0)$, and $(\alpha = 0, \beta > 0)$.

Proof of Theorem 14

The cases $(\alpha > 0, \beta = 0)$ and $(\alpha =0, \beta > 0)$ follow immediately from Eqs. (20) and (21). Consider now the case $\alpha> 0, \beta > 0$. Write $\delta = \delta (\alpha , \beta )$ to emphasize its dependence on $\alpha$ and $\beta$, we have $\delta (\alpha , \beta ) = \frac{\alpha \gamma ^r}{\alpha \gamma ^r + \beta \mu ^r}$ in $D^{(\alpha ,\beta )}_r[(A+\gamma I), (B+ \mu I)]$. Then for $D^{(\beta , \alpha )}_r[(B+\mu I), (A+\gamma I)]$, $\delta (\beta , \alpha ) = \frac{\beta \mu ^r}{\alpha \gamma ^r + \beta \mu ^r} = 1 -\delta (\alpha , \beta ), \;\;\; 1-\delta (\beta , \alpha ) = \delta (\alpha , \beta ),$$\delta (\beta , \alpha ) - \frac{\beta }{\alpha + \beta } = 1 -\delta (\alpha , \beta ) - \frac{\beta }{\alpha + \beta } = -\left( \delta (\alpha , \beta ) - \frac{\alpha }{\alpha + \beta }\right) .$ By Definition 1, we have

$$\begin{aligned}&D^{(\beta , \alpha )}_r[(B+ \mu I),(A+\gamma I)] = \frac{1}{\alpha \beta }\log (\frac{\mu }{\gamma })^{r(\delta (\beta , \alpha )-\frac{\beta }{\alpha +\beta })} \\&\qquad + \frac{1}{\alpha \beta }\log \mathrm{det_X}\\&\quad \left( \frac{\beta [(B+\mu I)(A+\gamma I)^{-1}]^{r(1-\delta (\beta , \alpha ))} + \alpha [(B+\mu I)(A+\gamma I)^{-1}]^{-r\delta (\beta , \alpha )}}{\alpha + \beta }\right) \\&\quad = \frac{1}{\alpha \beta }\log (\frac{\gamma }{\mu })^{r(\delta (\alpha , \beta )-\frac{\alpha }{\alpha +\beta })} \\&\qquad + \frac{1}{\alpha \beta }\log \mathrm{det_X}\\&\quad \left( \frac{\beta [(A+\gamma I)(B+\mu I)^{-1}]^{-r\delta (\alpha , \beta )} + \alpha [(A+\gamma I)(B+\mu I)^{-1}]^{r(1-\delta (\alpha , \beta ))}}{\alpha + \beta }\right) \\&\quad = D^{(\alpha , \beta )}_r[(A+ \gamma I),(B+\mu I)]. \end{aligned}$$

$\square$

Proof of Theorem 15

We prove Eq. (77) - the proof for Eq. (78) is similar. We write $(A+ \gamma I)^{-1} = \frac{1}{\gamma }I - \frac{A}{\gamma }(A+\gamma I)^{-1}, (B+\mu I)^{-1} = \frac{1}{\mu }I - \frac{B}{\mu }(B+\mu I)^{-1}, (B+\mu I)^{1/2}(A+\gamma I)^{-1}(B+\mu I)^{1/2} = [(B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}]^{-1}$. Consider the case $\alpha> 0, \beta > 0$. By Definition 1,

$$\begin{aligned}&D^{(\alpha , \beta )}_r[(A+\gamma I)^{-1}, (B+\mu I)^{-1}] \\= & {} \frac{1}{\alpha \beta }\log \left( \frac{1/\gamma }{1/\mu }\right) ^{r(\delta _2-\frac{\alpha }{\alpha +\beta })} + \frac{1}{\alpha \beta }\log \mathrm{det_X}\left( \frac{\alpha (\varLambda + \frac{\gamma }{\mu }I)^{-r(1-\delta _2)} + \beta (\varLambda + \frac{\gamma }{\mu })^{r\delta _2}}{\alpha + \beta }\right) \end{aligned}$$

where $\delta _2 = \frac{\alpha (1/\gamma )^r}{\alpha (1/\gamma )^r + \beta (1/\mu )^r} = \frac{\alpha \mu ^r}{\alpha \mu ^r + \beta \gamma ^r} = \delta (-r)$. Thus

$$\begin{aligned} D^{(\alpha , \beta )}_r[(A+\gamma I)^{-1}, (B+\mu I)^{-1}] = D^{(\alpha , \beta )}_{-r}[(A+\gamma I), (B+\mu I)]. \end{aligned}$$

Consider the case $\alpha = 0, \beta > 0$ (the case $\alpha > 0, \beta = 0$ then follows by dual symmetry). We have

$$\begin{aligned}&D^{(0, \beta )}_r[(A+\gamma I)^{-1}, (B+\mu I)^{-1}] = \frac{r}{\beta ^2}[(\frac{1/\gamma }{1/\mu })^{r} -1]\log \frac{1/\gamma }{1/\mu } \\&\qquad +\frac{1}{\beta ^2}\mathrm{tr}_X([(B+\mu I)(A+ \gamma I)^{-1}]^{r} -I) - \frac{1}{\beta ^2}(\frac{1/\gamma }{1/\mu })^{r}\log \mathrm{det_X}[(B+\mu I)(A+ \gamma I)^{-1}]^{r} \\&\quad = -\frac{r}{\beta ^2}[(\frac{\gamma }{\mu })^{-r} -1]\log \frac{\gamma }{\mu } +\frac{1}{\beta ^2}\mathrm{tr}_X([(A+ \gamma I)(B+\mu I)^{-1}]^{-r} -I) \\&\qquad - \frac{1}{\beta ^2}(\frac{\gamma }{\mu })^{-r}\log \mathrm{det_X}[(A+ \gamma I)(B+\mu I)^{-1}]^{-r}. \end{aligned}$$

By Lemma 12, we have

$$\begin{aligned}&\mathrm{tr}_X[(A+\gamma I)(B+\mu I)^{-1}]^{-r} = \mathrm{tr}_X\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{-r}\right] = \mathrm{tr}_X[(B+\mu I)^{-1}(A+\gamma I)]^{-r}, \\&\mathrm{det_X}[(A+\gamma I)(B+\mu I)^{-1}]^{-r} = \mathrm{det_X}\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{-r}\right] = \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]^{-r}. \end{aligned}$$

Thus it follows that

$$\begin{aligned}&D^{(0, \beta )}_r[(A+\gamma I)^{-1}, (B+\mu I)^{-1}] \\&\quad = -\frac{r}{\beta ^2}[(\frac{\gamma }{\mu })^{-r} -1]\log \frac{\gamma }{\mu } +\frac{1}{\beta ^2}\mathrm{tr}_X([(B+\mu I)^{-1}(A+\gamma I)]^{-r} -I) \\&\qquad - \frac{1}{\beta ^2}\left( \frac{\gamma }{\mu }\right) ^{-r}\log \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]^{-r} \\&\quad = D^{(0,\beta )}_{-r}[(A+\gamma I), (B+\mu I)]. \end{aligned}$$

$\square$

Proof of Theorem 16 - Affine-invariance

For $(A + \gamma I) \in \mathrm{PTr}({\mathcal {H}})$ and $(C+\nu I) \in \mathrm{Tr}_X({\mathcal {H}})$, $\nu \ne 0$,

$$\begin{aligned}&(C+\nu I)(A+\gamma I)(C+\nu I)^{*} \\&\quad = CAC^{*} + \nu (CA+AC^{*}) + \nu ^2A + \gamma CC^{*} + \gamma \nu (C + C^{*}) + \gamma \nu ^2 I \in \mathrm{Tr}_X({\mathcal {H}}). \end{aligned}$$

Since $(C+\nu I)$ is invertible, $(C+\nu I)(A+\gamma I)(C+\nu I)^{*}$ is also invertible, with inverse $[(C+\nu I)^{*}]^{-1}(A+\gamma I)^{-1}(C+\nu I)^{-1}$. Furthermore, $\forall x \in {\mathcal {H}}$,

$$\begin{aligned} \langle x, (C+\nu I)(A+\gamma I)(C+\nu I)^{*}x\rangle&= \langle (C+\nu I)^{*}x, (A+\gamma I)(C+ \nu I)^{*}x\rangle \\&\ge M_A||(C+\nu I)^{*}x||^2 \ge 0, \end{aligned}$$

with equality if and only if $(C+\nu I)^{*}x = 0 \Longleftrightarrow x = 0$. Thus $(C+\nu I)(A+\gamma I)(C+\nu I)^{*}$ is strictly positive. Together with its invertibility, this shows that this is a positive definite operator. Hence $(C+\nu I)(A+\gamma I)(C+\nu I)^{*} \in \mathrm{PTr}({\mathcal {H}})$. For $(A+\gamma I), (B + \mu I) \in \mathrm{PTr}({\mathcal {H}})$, we then have

$$\begin{aligned}&[(C+\nu I)(A+\gamma I)(C+\nu I)^{*}][(C+\nu I)(B+\mu I)(C+\nu I)^{*}]^{-1} \\&\quad = (C+\nu I)[(A+\gamma I)(B+\mu I)^{-1}](C+\nu I)^{-1}. \end{aligned}$$

Then for any $p \in {\mathbb {R}}$, we have

$$\begin{aligned}&([(C+\nu I)(A+\gamma I)(C+\nu I)^{*}][(C+\nu I)(B+\mu I)(C+\nu I)^{*}]^{-1})^p \\&\quad = (C+\nu I)[(A+\gamma I)(B+\mu I)^{-1}]^p(C+\nu I)^{-1}. \end{aligned}$$

Thus for any $a, b > 0$ and any $p, q \in {\mathbb {R}}$.

$$\begin{aligned}&a([(C+\nu I)(A+\gamma I)(C+\nu I)^{*}][(C+\nu I)(B+\mu I)(C+\nu I)^{*}]^{-1})^p \\&\quad +b ([(C+\nu I)(A+\gamma I)(C+\nu I)^{*}][(C+\nu I)(B+\mu I)(C+\nu I)^{*}]^{-1})^q \\&= (C+\nu I)(a[(A+\gamma I)(B+\mu I)^{-1}]^p + b [(A+\gamma I)(B+\mu I)^{-1}]^q)(C+\nu I)^{-1}. \end{aligned}$$

By the definition of $D^{(\alpha , \beta )}_r$ and the following invariances of the extended Fredholm determinant $\mathrm{det_X}$ as well as of the extended trace operation $\mathrm{tr}_X$, namely,

$$\begin{aligned} \mathrm{det_X}[C(A+\gamma I)C^{-1}] = \mathrm{det_X}[(A+\gamma I)], \;\;\; \mathrm{tr}_X[C(A+\gamma I)C^{-1}] = \mathrm{tr}_X[(A+\gamma I)], \end{aligned}$$

for $A + \gamma I \in \mathrm{Tr}_X({\mathcal {H}})$, $\gamma \ne 0$, and $C \in {\mathcal {L}}({\mathcal {H}})$ invertible (Lemma 5 in [24]), we obtain

$$\begin{aligned}&D^{(\alpha , \beta )}_r[(C+\nu I)(A+\gamma I)(C+\nu I)^{*}, (C+\nu I)(B+ \mu I)(C+\nu I)^{*}] \nonumber \\&=D^{(\alpha , \beta )}_r[(A+\gamma I), (B+ \mu I)]. \end{aligned}$$

$\square$

Proof of Theorem 17 - Invariance under unitary transformations

The proof of this theorem is similar to that of the proof for Theorem 16 , using the fact that $C^{*} = C^{-1}$ and the properties

$$\begin{aligned} \mathrm{det_X}[C(A+\gamma I)C^{-1}] = \mathrm{det_X}[(A+\gamma I)], \;\;\; \mathrm{tr}_X[C(A+\gamma I)C^{-1}] = \mathrm{tr}_X[(A+\gamma I)], \end{aligned}$$

of the operations $\mathrm{det_X}$ and $\mathrm{tr}_X$.$\square$

Proof of Theorem 18

For the case $\alpha > 0$, $\beta > 0$, this follows immediately from Definition 1. For the case $\alpha > 0, \beta = 0$ (the case $\alpha =0, \beta > 0$ is entirely similar), by Definition 2 and Lemma 12, we have

$$\begin{aligned}&D^{(\alpha , 0)}_r[(A+\gamma I), (B+\mu I)] = \frac{r}{\alpha ^2}\left[ (\frac{\mu }{\gamma })^{r} -1\right] \log \left( \frac{\mu }{\gamma }\right) \\&\qquad +\frac{1}{\alpha ^2}\mathrm{tr}_X\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) ^{-r} -I\right] - \frac{1}{\alpha ^2}\left( \frac{\mu }{\gamma }\right) ^{r}\log \mathrm{det_X}\left( \varLambda + \frac{\gamma }{\mu }\right) ^{-r} \\&\quad = D^{(\alpha , 0)}_r\left[ \left( \varLambda + \frac{\gamma }{\mu }\right) , I\right] . \end{aligned}$$

$\square$

Proof of Theorem 19

We first note that $(\varLambda + \frac{\gamma }{\mu } I)^{\omega } = (\frac{\gamma }{\mu })^{\omega }(\frac{\mu }{\gamma }\varLambda + I)^{\omega }$. For $\alpha> 0, \beta > 0$, this follows immediately from Definition 1. For $\alpha > 0, \beta = 0$, by Definition 2 and Lemma 12,

$$\begin{aligned}&D^{(\omega \alpha , 0)}_{\omega r}[(A+\gamma I), (B+\mu I)] = \frac{r}{\omega ^2\alpha ^2}[(\frac{\mu }{\gamma })^{\omega r} -1]\log (\frac{\mu }{\gamma })^{\omega } \\&\quad +\frac{1}{\omega ^2\alpha ^2}\mathrm{tr}_X[(\varLambda + \frac{\gamma }{\mu })^{-\omega r} -I] - \frac{1}{\omega ^2\alpha ^2}(\frac{\mu }{\gamma })^{\omega r}\log \mathrm{det_X}(\varLambda + \frac{\gamma }{\mu })^{-\omega r} \\&= \frac{1}{\omega ^2}D^{(\alpha , 0)}_r[(\varLambda + \frac{\gamma }{\mu })^{\omega }, I]. \;\;\; \text {The case }\alpha =0, \beta > 0 \text { is entirely similar.} \end{aligned}$$

$\square$

1.6 Proofs of Theorems 1, 2, and 3

We are now ready to provide the proofs for Theorems 1, 2, and 3 . We first need the following result.

Lemma 17

• (i ) Let $r \ne 0$ be fixed. The function $f(x) = x^r-1 - r\log (x)$ for $x > 0$ has a unique global minimum $f_{\min } = f(1) = 0$. In other words, $f(x) \ge 0$$\forall x > 0$, with equality if and only if $x = 1$.

• (ii) Let $\nu > 0, r \ne 0$ be fixed. For $r \ne 0$, the function $g(x) = (\frac{x}{\nu }+1)^r - 1 - r \log (\frac{x}{\nu }+1)$ for $x > - \nu$ has a unique global minimum $g_{\min } = g(0) = 0$. In other words, $g(x) \ge 0$$\forall x > -\nu$, with equality if and only if $x = 0$.

Proof of Theorem 1 - Positivity

For the case $\alpha> 0, \beta > 0$, this is a special case of Theorem 8, with $p +q = r$. Consider now the case $\alpha = 0, \beta > 0$ (the case $\alpha > 0, \beta =0$ then follows by dual symmetry). It suffices to consider $D^{(0,1)}_r$. Recall that $\varLambda + \nu I = (B+\mu I)^{-1/2}(A+\gamma I)(B+ \mu I)^{-1/2}$, where $\nu = \frac{\gamma }{\mu }$. Then, since $\mathrm{det_X}[(B+\mu I)^{-1/2}(A+\gamma I)(B+ \mu I)^{-1/2}] = \mathrm{det_X}[(B+\mu I)^{-1}(A+\gamma I)]$ and $\mathrm{tr}_X[(B + \mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}] = \mathrm{tr}_X[(B+\mu I)^{-1}(A + \gamma I)]$, we have

$$\begin{aligned} D^{(0,1)}_r[(A+\gamma I), (B+ \mu I)]&= {r}(\nu ^{r} -1)\log \nu +\mathrm{tr}_X[(\varLambda + \nu I)^{r} -I] \\&\quad - \nu ^{r}\log \mathrm{det_X}(\varLambda + \nu I)^{r}. \end{aligned}$$

By Lemma 11, $\mathrm{tr}_X[(\varLambda + \nu I)^{r} -I] = \nu ^r - 1 +\nu ^r\mathrm{tr}\left[ \left( \frac{\varLambda }{\nu } + I\right) ^r - I\right] .$ Also

$$\begin{aligned} \log \mathrm{det_X}(\varLambda + \nu I)^{r} = \log \left[ \nu ^r\det \left( \frac{\varLambda }{\nu } + I\right) ^r\right] = r\log \det \left( \frac{\varLambda }{\nu } +I\right) + r\log \nu . \end{aligned}$$

Thus we have

$$\begin{aligned} \begin{aligned}&D^{(0,1)}_r[(A+\gamma I), (B+ \mu I)] \\&\quad = \nu ^r - 1 - r\log \nu + \nu ^r\left( \mathrm {tr}\left[ \left( \frac{\varLambda }{\nu } + I\right) ^r - I\right] \right. \left. -\, r\log \det \left( \frac{\varLambda }{\nu } +I\right) \right) \\&\quad = \nu ^r - 1 - r\log \nu + \nu ^r\left[ \sum _{k=1}^{\infty }\left( \frac{\lambda _k}{\nu }+1\right) ^r - 1 - r\log \left( \frac{\lambda _k}{\nu } + 1\right) \right] . \end{aligned} \end{aligned}$$

By the first part of Lemma 17, we have for all $\nu > 0$, $\nu ^r - 1 - r\log \nu \ge 0$, with equality if and only if $\nu =1$. By the second part of Lemma 17, we have $\forall k \in {\mathbb {N}}$, $\left( \frac{\lambda _k}{\nu }+1\right) ^r - 1 - r\log \left( \frac{\lambda _k}{\nu } + 1\right) \ge 0$, with equality if and only $\lambda _k = 0$. Combining these two inequalities, we obtain

$$\begin{aligned}&D^{(0,1)}_r[(A+\gamma I), (B+ \mu I)] \ge 0, \end{aligned}$$

with equality if and only if $\nu = \frac{\gamma }{\mu } = 1$ and $\lambda _k = 0 \forall k \in {\mathbb {N}}\Longleftrightarrow \varLambda = I$, that is if and only $(B+\mu I)^{-1/2}(A + \gamma I)(B+ \mu I)^{-1/2} = I \Longleftrightarrow A+\gamma I = B+\mu I \Longleftrightarrow A = B$ and $\gamma = \mu$. $\square$

Proof of Theorem 2 - Special cases I

The first statement of the theorem was proved in Sect. A.2. The second statement is the content of Theorem 13. $\square$

Proof of Theorem 3 - Special cases II

This theorem follows from Theorem 2

as well as the symmetry of $D^{(\alpha , \alpha )}_r$ as proved in Theorem 14. $\square$

1.7 Proofs for the Divergences between RKHS covariance operators

In this section, we prove Theorems 26, 27, 28, and 29 . We first need the following preliminary results.

Lemma 18

Let ${\mathcal {H}}_1, {\mathcal {H}}_2$ be separable Hilbert spaces. Let $A:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_2,B:{\mathcal {H}}_2 \rightarrow {\mathcal {H}}_1$ be compact operators such that $AB: {\mathcal {H}}_2 \rightarrow {\mathcal {H}}_2, BA:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_1$ are trace class. Let $\alpha , \beta > 0$ be fixed. Then

$$\begin{aligned}&\det \left[ \frac{\alpha (AB + I_{{\mathcal {H}}_2})^p + \beta (AB + I_{{\mathcal {H}}_2})^q}{\alpha + \beta }\right] \nonumber \\&\quad = \det \left[ \frac{\alpha (BA + I_{{\mathcal {H}}_1})^{p} + \beta (BA + I_{{\mathcal {H}}_1})^q}{\alpha + \beta }\right] \forall p,q \in {\mathbb {R}}. \end{aligned}$$

(146)

Proof

Since the nonzero eigenvalues of $AB:{\mathcal {H}}_2 \rightarrow {\mathcal {H}}_2$ and $BA:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_1$ are the same, we have

$$\begin{aligned} \det [(AB + I_{{\mathcal {H}}_2})^p] = \det [(BA+I_{{\mathcal {H}}_1})^p], \;\;\;\forall p \in {\mathbb {R}}. \end{aligned}$$

For any $p,q \in {\mathbb {R}}$,

$$\begin{aligned}&\det \left[ \frac{\alpha (AB + I_{{\mathcal {H}}_2})^p + \beta (AB + I_{{\mathcal {H}}_2})^q}{\alpha + \beta }\right] = \det \left[ \frac{\alpha (BA + I_{{\mathcal {H}}_1})^{p} + \beta (BA + I_{{\mathcal {H}}_1})^q}{\alpha + \beta }\right] . \end{aligned}$$

In the above equality, we have used the fact that a zero eigenvalue of AB and BA corresponds to an eigenvalue equal to 1 for $\frac{\alpha (AB + I_{{\mathcal {H}}_2})^{p} + \beta (AB + I_{{\mathcal {H}}_2})^q}{\alpha + \beta }:{\mathcal {H}}_2 \rightarrow {\mathcal {H}}_2$ and $\frac{\alpha (BA + I_{{\mathcal {H}}_1})^{p} + \beta (BA + I_{{\mathcal {H}}_1})^q}{\alpha + \beta }:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_1$, respectively, which does not change the determinant. $\square$

Lemma 19

Let ${\mathcal {H}}_1, {\mathcal {H}}_2$ be separable Hilbert spaces. Let $A,B:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_2$ be compact operators such that $AA^{*}: {\mathcal {H}}_2 \rightarrow {\mathcal {H}}_2,BB^{*}:{\mathcal {H}}_2 \rightarrow {\mathcal {H}}_2$ are trace class. Let $\alpha , \beta > 0$ be fixed. For any $p, q \in {\mathbb {R}}$,

$$\begin{aligned}&\det \left[ \frac{\alpha [(AA^{*}+I_{{\mathcal {H}}_2})(BB^{*} + I_{{\mathcal {H}}_2})^{-1}]^p + \beta [(AA^{*}+I_{{\mathcal {H}}_2})(BB^{*} + I_{{\mathcal {H}}_2})^{-1}]^q}{\alpha + \beta }\right] \nonumber \\&=\det \left[ \frac{\alpha (C + I_{{\mathcal {H}}_1}\otimes I_3)^{p}+ \beta (C + I_{{\mathcal {H}}_1} \otimes I_3)^q}{\alpha + \beta }\right] , \end{aligned}$$

(147)

$$\begin{aligned}&\text {where}\;\;C = \begin{pmatrix} A^{*}A \;\; -A^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -A^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \\ B^{*}A \;\; -B^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -B^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \\ B^{*}A \;\; -B^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -B^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \end{pmatrix}. \end{aligned}$$

(148)

Proof of Lemma 19

We make use of the following notation. Let $A,B,C: {\mathcal {H}}_1 \rightarrow {\mathcal {H}}_2$ be three bounded linear operators. Consider the operator $(A \; B \; C): {\mathcal {H}}_1^3 \rightarrow {\mathcal {H}}_2$, with $(A \; B \; C)^{*} = \begin{pmatrix} A^{*}\\ B^{*}\\ C^{*} \end{pmatrix}: {\mathcal {H}}_2 \rightarrow {\mathcal {H}}_1^3$. Here ${\mathcal {H}}_1^3 = {\mathcal {H}}_1 \oplus {\mathcal {H}}_1 \oplus {\mathcal {H}}_1$ denotes the direct sum of ${\mathcal {H}}_1$ with itself, that is

$$\begin{aligned} {\mathcal {H}}_1^3 = {\mathcal {H}}_1 \oplus {\mathcal {H}}_1 \oplus {\mathcal {H}}_1 = \{(v_1, v_2, v_3) \; : \; v_1, v_2, v_3 \in {\mathcal {H}}_1\}, \end{aligned}$$

equipped with the inner product

$$\begin{aligned} \langle (v_1, v_2, v_3) , (w_1, w_2, w_3)\rangle _{{\mathcal {H}}_1^3} = \langle v_1, w_1\rangle _{{\mathcal {H}}_1} + \langle v_2, w_2\rangle _{{\mathcal {H}}_1} + \langle v_3, w_3\rangle _{{\mathcal {H}}_1}. \end{aligned}$$

By the Sherman–Morrison–Woodbury formula, $(BB^{*}+I_{{\mathcal {H}}_2})^{-1} = I_{{\mathcal {H}}_2} - B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1}B^{*}.$ Thus

$$\begin{aligned} (AA^{*} + I_{{\mathcal {H}}_2})(BB^{*} + I_{{\mathcal {H}}_2})^{-1}&= I_{{\mathcal {H}}_2} + AA^{*} - B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1}B^{*} \\&\quad - AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1}B^{*} = I_{{\mathcal {H}}_2} + C_1C_2. \end{aligned}$$

Here the operators $C_1, C_2$ are defined as follows.

$$\begin{aligned} C_1&= [A \;\; -B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1}]:{\mathcal {H}}_1^3 \rightarrow {\mathcal {H}}_2, \\ C_2&= \begin{pmatrix} A^{*} \\ B^{*} \\ B^{*} \end{pmatrix}: {\mathcal {H}}_2 \rightarrow {\mathcal {H}}_1^3. \end{aligned}$$

The operator $C_2C_1: {\mathcal {H}}_1^3 \rightarrow {\mathcal {H}}_1^3$ is given by

$$\begin{aligned} C_2C_1 = \begin{pmatrix} A^{*}A \;\; -A^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -A^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \\ B^{*}A \;\; -B^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -B^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \\ B^{*}A \;\; -B^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \;\; -B^{*}AA^{*}B(I_{{\mathcal {H}}_1} + B^{*}B)^{-1} \end{pmatrix} = C. \end{aligned}$$

It follows from Lemma 18 that

$$\begin{aligned}&\det \left[ \frac{\alpha [(AA^{*}+I_{{\mathcal {H}}_2})(BB^{*} + I_{{\mathcal {H}}_2})^{-1}]^p + \beta [(AA^{*}+I_{{\mathcal {H}}_2})(BB^{*} + I_{{\mathcal {H}}_2})^{-1}]^q}{\alpha + \beta }\right] \\&\quad =\det \left[ \frac{\alpha (I_{{\mathcal {H}}_2} + C_1C_2)^p + \beta (I_{{\mathcal {H}}_2} + C_1C_2)^q}{\alpha + \beta }\right] \\&\quad = \det \left[ \frac{\alpha (C_2C_1 + I_{{\mathcal {H}}_1}\otimes I_3)^{p}+ \beta (C_2C_1 + I_{{\mathcal {H}}_1} \otimes I_3)^q}{\alpha + \beta }\right] . \end{aligned}$$

$\square$

Proof of Theorem 26

Let $\varLambda +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2} = (BB^{*} + \mu I_{{\mathcal {H}}_2})^{-1/2}(AA^{*}+\gamma I_{{\mathcal {H}}_2})(BB^{*}+ \mu I_{{\mathcal {H}}_2})^{-1/2}, Z + \frac{\gamma }{\mu } I_{{\mathcal {H}}_2} = (AA^{*}+\gamma I_{{\mathcal {H}}_2})(BB^{*}+ \mu I_{{\mathcal {H}}_2})^{-1}$. By Theorem 10,

$$\begin{aligned}&D^{(\alpha , \beta )}_{r}[(AA^{*}+\gamma I_{{\mathcal {H}}_2}), (BB^{*}+\mu I_{{\mathcal {H}}_2})] \\&\quad = \frac{r(\delta -\frac{\alpha }{\alpha +\beta })}{\alpha \beta }\left( \log \frac{\gamma }{\mu }\right) +\frac{1}{\alpha \beta }\log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) \nonumber \\&\qquad +\frac{1}{\alpha \beta }\log \det \left[ \frac{\alpha (\varLambda +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^p + \beta (\varLambda + \frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] , \\&\qquad \;\;\text {with }p = r(1-\delta )\text { and }q = r\delta . \end{aligned}$$

By Lemma 19, with $\frac{\mu }{\gamma }Z + I_{{\mathcal {H}}_2} = (\frac{AA^{*}}{\gamma } + I_{{\mathcal {H}}_2})(\frac{BB^{*}}{\mu } + I_{{\mathcal {H}}_2})^{-1}$, the determinant in the last term is

$$\begin{aligned}&\det \left[ \frac{\alpha (\varLambda +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^p + \beta (\varLambda + \frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad = \det \left[ \frac{\alpha (\frac{\gamma }{\mu })^p(\frac{\mu }{\gamma }\varLambda + I_{{\mathcal {H}}_2})^p + \beta (\frac{\gamma }{\mu })^{-q}(\frac{\mu }{\gamma }\varLambda + I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad =\det \left[ \frac{\alpha (\frac{\gamma }{\mu })^p(\frac{\mu }{\gamma }Z + I_{{\mathcal {H}}_2})^p + \beta (\frac{\gamma }{\mu })^{-q}(\frac{\mu }{\gamma }Z + I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] \\&\quad =\det \left[ \frac{ \alpha (\frac{\gamma }{\mu })^p(C + I_{{\mathcal {H}}_1}\otimes I_3)^{p}+ \beta (\frac{\gamma }{\mu })^{-q} (C + I_{{\mathcal {H}}_1} \otimes I_3)^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] , \;\;\text {where} \\&C = \begin{pmatrix} \frac{A^{*}A}{\gamma } \;\; -\frac{A^{*}B}{\sqrt{\gamma \mu }}(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \;\; -\frac{A^{*}AA^{*}B}{\gamma \sqrt{\gamma \mu }}(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \\ \frac{B^{*}A}{\sqrt{\gamma \mu }} \;\; -\frac{B^{*}B}{\mu }(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \;\; -\frac{B^{*}AA^{*}B}{\gamma \mu }(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \\ \frac{B^{*}A}{\sqrt{\gamma \mu }} \;\; -\frac{B^{*}B}{\mu }(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \;\; -\frac{B^{*}AA^{*}B}{\gamma \mu }(I_{{\mathcal {H}}_1} + \frac{B^{*}B}{\mu })^{-1} \end{pmatrix}, \end{aligned}$$

which is obtained by replacing $AA^{*}$ and $BB^{*}$ in Lemma 19 with $\frac{AA^{*}}{\gamma }$ and $\frac{BB^{*}}{\mu }$, respectively. $\square$

Proof of Theorem 27

Let $Z + \frac{\gamma }{\mu } I_{{\mathcal {H}}_2} = (AA^{*}+\gamma I_{{\mathcal {H}}_2})(BB^{*}+ \mu I_{{\mathcal {H}}_2})^{-1}$. By Eq. (29), when $\dim ({\mathcal {H}}_2) < \infty$,

$$\begin{aligned}&D^{(\alpha , \beta )}_{r}[(AA^{*}+\gamma I_{{\mathcal {H}}_2}), (BB^{*}+\mu I_{{\mathcal {H}}_2})] \\&\quad = \frac{1}{\alpha \beta }\log \det \left[ \frac{\alpha (Z +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^p + \beta (Z + \frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^{-q}}{\alpha + \beta }\right] \\&\quad =\frac{1}{\alpha \beta }\left[ \log \left( \frac{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}{\alpha + \beta }\right) \right] \dim ({\mathcal {H}}_2) \\&\qquad +\frac{1}{\alpha \beta }\log \det \left[ \frac{\alpha (Z +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^p + \beta (Z + \frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] . \end{aligned}$$

As in the proof of Theorem 26, the determinant in last term in the above expression is

$$\begin{aligned}&\det \left[ \frac{\alpha (\varLambda +\frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^p + \beta (\varLambda + \frac{\gamma }{\mu }I_{{\mathcal {H}}_2})^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] \\&= \det \left[ \frac{ \alpha (\frac{\gamma }{\mu })^p(C + I_{{\mathcal {H}}_1}\otimes I_3)^{p}+ \beta (\frac{\gamma }{\mu })^{-q} (C + I_{{\mathcal {H}}_1} \otimes I_3)^{-q}}{\alpha (\frac{\gamma }{\mu })^p + \beta (\frac{\gamma }{\mu })^{-q}}\right] . \end{aligned}$$

This gives us the final expression. $\square$

Proof of Theorems 28 and 29

Theorem 28 follows from Theorem 26 by considering the linear operators $A = \frac{1}{\sqrt{m}}\varPhi ({\mathbf {x}})J_m: {\mathbb {R}}^m \rightarrow {\mathcal {H}}_K, \;\;\; B = \frac{1}{\sqrt{m}}\varPhi ({\mathbf {y}})J_m:{\mathbb {R}}^m \rightarrow {\mathcal {H}}_K$. The proof of Theorem 29 is similar to that of Theorem  28, except that we invoke Theorem 27.$\square$

1.8 Proofs for the Metric properties

We now prove Theorems 20, 22, 23, which lead to the proofs of Theorems 4 and 21 . We first prove Theorems 4 and 21 for the case $\alpha =1/2$, which corresponds to the infinite-dimensional symmetric Stein divergence, and then for the general case $\alpha > 0$. The former case utilizes Theorem 31 and the latter case utilizes Theorem 33 and the case $\alpha = 1/2$. Both Theorems 31 and 33 are of interest in their own right.

The case of the infinite-dimensional symmetric Stein divergence. Consider the first case $\alpha =\frac{1}{2}$.

Lemma 20

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $A, B,C:{\mathcal {H}}\rightarrow {\mathcal {H}}$ be self-adjoint finite-rank operators, such that $A+I > 0$, $B + I > 0$, $C+I > 0$. Then

$$\begin{aligned} \sqrt{\log \frac{\det (\frac{A+B}{2} + I)}{\sqrt{\det (A+I)\det (B+I)}}}&\le \sqrt{\log \frac{\det (\frac{A+C}{2} + I)}{\sqrt{\det (A+I)\det (C+I)}}} \nonumber \\&+ \sqrt{\log \frac{\det (\frac{C+B}{2} + I)}{\sqrt{\det (C+I)\det (B+I)}}}. \end{aligned}$$

(149)

Proof

Since A, B, C are finite-rank operators, there exists a finite-dimensional subspace ${\mathcal {H}}_n \subset {\mathcal {H}}$, with $\dim ({\mathcal {H}}_n) = n$ for some $n \in {\mathbb {N}}$, such that $\mathrm{range}(A) \subset {\mathcal {H}}_n, \mathrm{range}(B) \subset {\mathcal {H}}_n$, $\mathrm{range}(C) \subset {\mathcal {H}}_n$. Let

$$\begin{aligned} A_n = A \big \vert _{{\mathcal {H}}_n}: {\mathcal {H}}_n \rightarrow {\mathcal {H}}_n,\;\; B_n = B\big \vert _{{\mathcal {H}}_n}: {\mathcal {H}}_n \rightarrow {\mathcal {H}}_n, \;\; C_n = C\big \vert _{{\mathcal {H}}_n}: {\mathcal {H}}_n \rightarrow {\mathcal {H}}_n. \end{aligned}$$

Then $A_n,B_n, C_n$ are linear operators on the finite-dimensional space ${\mathcal {H}}_n$ and thus are represented by $n \times n$ matrices, which we denote by the same symbols. We have $(A+B)_n = (A+B) \big \vert _{{\mathcal {H}}_n} = A \big \vert _{{\mathcal {H}}_n} + B\big \vert _{{\mathcal {H}}_n} = A_n + B_n, (A+C)_n = A_n + C_n, (C+B)_n = B_n + C_n.$ Applying the finite-dimensional result in [37], we then obtain

$$\begin{aligned} \sqrt{\log \frac{\det (\frac{A_n+B_n}{2} + I_n)}{\sqrt{\det (A_n+I_n)\det (B_n+I_n)}}}&\le \sqrt{\log \frac{\det (\frac{A_n+C_n}{2} + I_n)}{\sqrt{\det (A_n+I_n)\det (C_n+I_n)}}} \nonumber \\&+ \sqrt{\log \frac{\det (\frac{C_n+B_n}{2} + I_n)}{\sqrt{\det (C_n+I_n)\det (B_n+I_n)}}}. \end{aligned}$$

Since the non-zero eigenvalues of A and $A_n$ are the same, we have $\det (A+I) = \det (A_n+I_n)$ and the same holds true for the other operators. This, together with the last expression, gives us the final result. $\square$

Proof of Theorem 23 - Triangle inequality for square root of symmetric Stein divergence

Let $\{A_n\}_{n\in {\mathbb {N}}}$, $\{B_n\}_{n\in {\mathbb {N}}}$, $\{C_n\}_{n\in {\mathbb {N}}}$ be sequences of finite-rank operators with $||A_n - A||_{\mathrm{tr}} \rightarrow 0,\;\;||B_n - B||_{\mathrm{tr}} \rightarrow 0,\;\; ||C_n - C||_{\mathrm{tr}} \rightarrow 0, \;\;\hbox { as}\ n \rightarrow \infty .$ By Lemma 20, we have

$$\begin{aligned} \sqrt{\log \frac{\det (\frac{A_n+B_n}{2} + I)}{\sqrt{\det (A_n+I)\det (B_n+I)}}}&\le \sqrt{\log \frac{\det (\frac{A_n+C_n}{2} + I)}{\sqrt{\det (A_n+I)\det (C_n+I)}}} \nonumber \\&+ \sqrt{\log \frac{\det (\frac{C_n+B_n}{2} + I)}{\sqrt{\det (C_n+I)\det (B_n+I)}}}. \end{aligned}$$

By Theorem 3.5 in [36], as $n \rightarrow \infty$, we have

$$\begin{aligned}&\det \left( A_n + I\right) \rightarrow \det (A+I), \;\; \det (B_n + I) \rightarrow \det (B+I), \\&\det \left( \frac{A_n+B_n}{2} + I\right) \rightarrow \det (\frac{A+B}{2} + I),\; \hbox { and similarly for operators involving}\ C_n. \end{aligned}$$

Taking the limit $n \rightarrow \infty$ in the above triangle inequality for $(A_n+I), (B_n+I)$, and $(C_n+I)$ gives the final triangle inequality for $(A+I), (B+I)$, and $(C+I)$. $\square$

The following is the specialization of Theorem 4 when $\alpha = 1/2$.

Theorem 30

(Metric property - square root of symmetric Stein divergence) Let $\gamma > 0, \gamma \in {\mathbb {R}}$ be fixed. The square root $\sqrt{D^{(1/2,1/2)}_1[(A+\gamma I), (B+\gamma I)]}$ of the infinite-dimensional symmetric Stein divergence is a metric on $\mathrm{PTr}({\mathcal {H}})(\gamma )$.

Proof of Theorem 30

The positivity and symmetry of $D^{(1/2,1/2)}_1[(A+\gamma I), (B+\gamma I)]$ are shown in Theorems 1 and 14, respectively. It remains for us to show the triangle inequality, namely

$$\begin{aligned} \sqrt{D^{(1/2,1/2)}_1[(A+\gamma I), (B+\gamma I)]}&\le \sqrt{D^{(1/2,1/2)}_1[(A+\gamma I), (C+\gamma I)]} \\&+\sqrt{D^{(1/2,1/2)}_1[(C+\gamma I), (B+\gamma I)]}, \end{aligned}$$

for any three operators $(A+\gamma I), (B+\gamma I), (C+\gamma I) \in \mathrm{PTr}({\mathcal {H}})$. We have

$$\begin{aligned} D^{(1/2,1/2)}_1[(A+\gamma I), (B+\gamma I)]&= 4 \log \left[ \frac{\mathrm{det_X}(\frac{(A+\gamma I) + (B+\gamma I)}{2})}{\mathrm{det_X}(A+\gamma I)^{1/2}\mathrm{det_X}(B+\gamma I)^{1/2}}\right] \\&= 4\log \left[ \frac{\det (\frac{A+B}{2\gamma }+I)}{\det (\frac{A}{\gamma } + I)^{1/2}\det (\frac{B}{\gamma }+I)^{1/2}}\right] . \end{aligned}$$

Thus the triangle inequality for $D^{(1/2,1/2)}_1[(A+\gamma I), (B+\gamma I)]$ follows that stated in Theorem 23. $\square$

Lemma 21

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $A, B \in \mathrm{Sym}({\mathcal {H}})$ be finite-rank operators, with maximum rank $n \in {\mathbb {N}}$, such that $A+I > 0$, $B + I > 0$. Then

$$\begin{aligned} \prod _{j=1}^n\left[ \frac{\lambda _j(A) + \lambda _j(B)}{2} + 1\right]&\le \det \left( \frac{A+B}{2} + I\right) . \end{aligned}$$

(150)

Proof of Lemma 21

Since A, B are both finite-rank operators, there exists a finite-dimensional subspace ${\mathcal {H}}_n \subset {\mathcal {H}}$, with $\dim ({\mathcal {H}}_n) = n$, such that $\mathrm{range}(A) \subset {\mathcal {H}}_n$, $\mathrm{range}(B) \subset {\mathcal {H}}_n$. Let $A_n = A \big \vert _{{\mathcal {H}}_n}: {\mathcal {H}}_n \rightarrow {\mathcal {H}}_n,\;\; B_n = B\big \vert _{{\mathcal {H}}_n}: {\mathcal {H}}_n \rightarrow {\mathcal {H}}_n$. Then $A_n,B_n$ are linear operators on the finite-dimensional space ${\mathcal {H}}_n$ and thus are represented by $n \times n$ matrices, which we denote by the same symbols. Furthermore, $(A+B)_n = (A+B) \big \vert _{{\mathcal {H}}_n} = A \big \vert _{{\mathcal {H}}_n} + B\big \vert _{{\mathcal {H}}_n} = A_n + B_n.$ Thus we can apply the following inequality for finite SPD matrices ( [4])

$$\begin{aligned} \prod _{j=1}^n\left[ \frac{\lambda _j(A_n) + \lambda _j(B_n)}{2} + 1\right]&= \prod _{j=1}^n\left[ \frac{\lambda _j(A_n+I_n) + \lambda _j(B_n+I_n)}{2}\right] \\&\quad \le \det \left( \frac{A_n+B_n}{2} + I_n\right) . \end{aligned}$$

We note that the non-zero eigenvalues of $A_n, B_n$ are the same as those of A, B, respectively, with the maximum number being n, and $\det (\frac{A+B}{2}+I) = \det (\frac{A_n+B_n}{2} + I_n)$. Together with the previous inequality, this gives us the final result. $\square$

Theorem 31

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $A, B:{\mathcal {H}}\rightarrow {\mathcal {H}}$ be self-adjoint trace class operators, such that $A+I > 0$, $B + I > 0$. Then

$$\begin{aligned} \prod _{j=1}^{\infty }\left[ \frac{\lambda _j(A) + \lambda _j(B)}{2} + 1\right]&\le \det \left( \frac{A+B}{2} + I\right) . \end{aligned}$$

(151)

Proof of Theorem 31

Let $A = \sum _{j=1}^{\infty }\lambda _j(A)\phi _j \otimes \phi _j$ denote the spectral decomposition for A. For each $n \in {\mathbb {N}}$, define $A_n = \sum _{j=1}^n\lambda _j(A)\phi _j \otimes \phi _j$. Then $A_n$ is a finite-rank operator with the eigenvalues being the first n eigenvalues of A and $\lim _{n \rightarrow \infty }||A_n - A||_{\mathrm{tr}} = 0$. In the same way, we construct a sequence of finite-rank operators $B_n$ with $\lim _{n\rightarrow \infty }||B_n - B||_{\mathrm{tr}} = 0$, so that $\lim _{n \rightarrow \infty } ||(A_n+B_n) -(A+B)||_{\mathrm{tr}} = 0.$ By Theorem 3.5 in [36], as $n \rightarrow \infty$, we then have

$$\begin{aligned} \lim _{n \rightarrow \infty }\det \left( \frac{A_n+B_n}{2}+I\right) = \det \left( \frac{A+B}{2}+I\right) . \end{aligned}$$

Applying Lemma 21 to $A_n$ and $B_n$, we have

$$\begin{aligned} \prod _{j=1}^{n}\left[ \frac{\lambda _j(A_n) + \lambda _j(B_n)}{2} + 1\right]&\le \det \left( \frac{A_n+B_n}{2} + I\right) . \end{aligned}$$

(152)

The final result is then obtained by taking the limit as $n \rightarrow \infty$, noting that the eigenvalues of $A_n$, $B_n$, are precisely the first n eigenvalues of A, B, respectively. $\square$

The following is the specialization of Theorem 21 when $\alpha = 1/2$.

Theorem 32

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $A,B \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$ be such that $A+I> 0,B + I > 0$. Let $\mathrm{Eig}(A), \mathrm{Eig}(B): \ell ^2 \rightarrow \ell ^2$ be diagonal operators with the diagonals consisting of the eigenvalues of A and B, respectively, in decreasing order. Then

$$\begin{aligned} D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(A) + I), (\mathrm{Eig}(B) + I)] \le D^{(1/2,1/2)}_1[(A+I), (B+I)]. \end{aligned}$$

(153)

Proof of Theorem 32

By definition, we have

$$\begin{aligned}&D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(A) + I), (\mathrm{Eig}(B) + I)]\\&\quad = 4\log \left[ \frac{\det (\frac{\mathrm{Eig}(A) + \mathrm{Eig}(B)}{2}+I)}{\sqrt{\det (\mathrm{Eig}(A)+I)\det (\mathrm{Eig}(B)+I)}}\right] \\&\quad = 4 \log \left[ \frac{\prod _{j=1}^{\infty }\left[ \frac{\lambda _j(A) + \lambda _j(B)}{2} + 1\right] }{\sqrt{\det (A+I)\det (B+I)}}\right] \\&\quad \le 4\log \left[ \frac{\det (\frac{A + B}{2}+I)}{\sqrt{\det (A+I)\det (B+I)}}\right] \;\;\;\text {by Theorem~}\mathrm{31} \\&\quad = D^{(1/2,1/2)}_{1}[(A + I), (B + I)]. \end{aligned}$$

$\square$

The general case. We now consider the general case $\alpha > 0$. In the following, let ${\mathscr {C}}_p({\mathcal {H}})$ denote the set of pth Schatten class operators on ${\mathcal {H}}$, under the norm $||\;||_p$, which is defined by

$$\begin{aligned} ||A||_p = (\mathrm{tr}|A|^p)^{1/p}, \;\;\; 1 \le p \le \infty , \end{aligned}$$

(154)

with ${\mathscr {C}}_1({\mathcal {H}})$ being the space of trace class operators $\mathrm{Tr}({\mathcal {H}})$, ${\mathscr {C}}_2({\mathcal {H}})$ being the space of Hilbert–Schmidt operators $\mathrm{HS}({\mathcal {H}})$, and ${\mathscr {C}}_{\infty }({\mathcal {H}})$ being the set of compact operators under the operator norm $||\;||$.

Theorem 33

Let $r \in {\mathbb {R}}$ be fixed but arbitrary. Assume that $1 \le p \le \infty$. Let $A,\{A_n\}_{n \in {\mathbb {N}}} \in \mathrm{Sym}({\mathcal {H}}) \cap {\mathscr {C}}_p({\mathcal {H}})$, be such that $I+A > 0$, $I+A_n > 0$$\forall n \in {\mathbb {N}}$. Assume that $\lim _{n \rightarrow \infty }||A_n - A||_{p} = 0$. Then

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+A_n)^r - (I+A)^r||_{p} = 0. \end{aligned}$$

(155)

Proof of Theorem 33

(i) The case $r = 0$ is trivial. We first prove that

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+A_n)^r - (I+A)^r||_{p} = 0, \;\; 0 < r \le 1. \end{aligned}$$

(156)

We make use of the following result from [18] (Corollary 3.2), which states that for any two positive operators A, B on ${\mathcal {H}}$ such that $A \ge c > 0$, $B \ge c > 0$, and any operator X on ${\mathcal {H}}$,

$$\begin{aligned} ||A^rX - XB^r||_p \le rc^{r-1}||AX-XB||_p, \;\;\; 0 < r \le 1. \end{aligned}$$

(157)

By the assumption $I+A > 0$, there exists $M_A > 0$ such that $\langle x, (I+A)x\rangle \ge M_A||x||^2 \;\; \; \forall x \in {\mathcal {H}}.$ By the assumption $\lim _{n \rightarrow \infty }||A_n -A||_{p} = 0$, for any $\epsilon$ satisfying $0< \epsilon < M_A$, there exists $N = N(\epsilon ) \in {\mathbb {N}}$ such that $||A_n - A||_{p} < \epsilon$$\forall n \ge N$. Then $\forall x \in {\mathcal {H}}$,

$$\begin{aligned} |\langle x, (A_n -A)x\rangle | \le ||A_n - A||\;||x||^2 \le ||A_n-A||_{p}||x||^2 \le \epsilon ||x||^2. \end{aligned}$$

It thus follows that $\forall x \in {\mathcal {H}}$,

$$\begin{aligned} \langle x, (I+A_n)x\rangle = \langle x, (I+A)x\rangle + \langle x, (A_n-A)x\rangle \ge (M_A -\epsilon )||x||^2. \end{aligned}$$

Thus we have $I+A \ge M_A > 0$, $I+A_n \ge M_A-\epsilon > 0$$\forall n \ge N = N(\epsilon )$. Then, applying Eq. (157),

$$\begin{aligned} ||(I+A_n)^r - (I+A)^r||_{p}&\le r (M_A - \epsilon )^{r-1}||A_n - A||_{p} \;\;\;\forall n \ge N, \end{aligned}$$

which implies $\lim _{n \rightarrow \infty }||(I+A_n)^r - (I+A)^r||_{p} = 0$.

(ii) For $r > 1$, we proceed by induction as follows. We have

$$\begin{aligned}&||(I+A_n)^r - (I+A)^r||_p \\&\quad \le ||(I+A_n)^r - (I+A_n)(I+A)^{r-1}||_p + ||(I+A_n)(I+A)^{r-1} - (I+A)^r||_p \\&\quad \le ||I+A_n||\;||(I+A_n)^{r-1} - (I+A)^{r-1}||_p + ||A_n - A||_p||(I+A)^{r-1}||. \end{aligned}$$

Thus this case follows from the case $0 \le r \le 1$ by induction.

(iii) For the case $r < 0$, we first prove that

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+A_n)^{-1} - (I+A)^{-1}||_p = 0. \end{aligned}$$

(158)

This follows from the fact that $\forall n \ge N = N(\epsilon )$,

$$\begin{aligned}&||(I+A_n)^{-1} - (I+A)^{-1}||_{p} = ||(I+A_n)^{-1}[(I+A_n) - (I+A)](I+A)^{-1}||_{p} \\&\quad \le ||(I+A_n)^{-1}||\;||A_n - A||_{p}||(I+A)^{-1}|| \le \frac{1}{M_A(M_A-\epsilon )}||A_n -A||_{p}. \end{aligned}$$

(iii) We next prove that

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+A_n)^{-r} - (I+A)^{-r}||_p = 0, \;\; 0 < r \le 1. \end{aligned}$$

(159)

We have $(I+A)^{-1} \ge \frac{1}{\max \{(1+\lambda _k(A)): k \in {\mathbb {N}}\}} = \frac{1}{||I+A||}> 0$. From the limit $\lim _{n \rightarrow \infty }||A_n-A|| = 0$, it follows that for any $\epsilon$ satisfying $0< \epsilon < ||I+A||$, there exists $M = M(\epsilon ) \in {\mathbb {N}}$ such that $\forall n \ge M$,

$$\begin{aligned} ||I+A|| - \epsilon \le ||I+A_n|| \le ||I+A|| + \epsilon . \end{aligned}$$

It follows that $\forall n \ge M$, $(I+A_n)^{-1} \ge \frac{1}{\max \{(1+\lambda _k(A_n)): k \in {\mathbb {N}}\}} = \frac{1}{||I+A_n||} \ge \frac{1}{||I+A||+\epsilon }.$ Hence invoking Eq. (157) again, we obtain $\forall n \ge M$

$$\begin{aligned} ||(I+A_n)^{-r} - (I+A)^{-r}||_{p} \le r(||I+A||+\epsilon )^{1-r}||(I+A_n)^{-1} - (I+A)^{-1}||_p, \end{aligned}$$

which implies that $\lim _{n \rightarrow \infty }||(I+A_n)^{-r} - (I+A)^{-r}||_{p} = 0$ by the previous limit, when $r = 1$.

(iv) By an induction argument as in step (ii), we then obtain the final part of the theorem, namely

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+A_n)^{-r} - (I+A)^{-r}||_p = 0, \;\; \forall r > 1. \end{aligned}$$

Lemma 22

Let ${\mathcal {H}}$ be a separable Hilbert space. Assume that $\{A_n\}_{n\in {\mathbb {N}}}$, A are self-adjoint trace class operators on ${\mathcal {H}}$ such that $(I+A) > 0$, $(I+A_n) > 0$$\forall n \in {\mathbb {N}}$. Assume that $||A_n - A||_{\mathrm{tr}} = 0$ as $n \rightarrow \infty$. Then $A_n(I+A_n)^{-1}$ and $A(I+A)^{-1}$ are trace class operators and

$$\begin{aligned} \lim _{n \rightarrow \infty }||A_n(I+A_n)^{-1} - A(I+A)^{-1}||_{\mathrm{tr}} = 0. \end{aligned}$$

(160)

Lemma 23

Let ${\mathcal {H}}$ be a separable Hilbert space. Let $\{A_n\}_{n \in {\mathbb {N}}}$, A, $\{B_n\}_{n \in {\mathbb {N}}}$, B, in $\mathrm{Tr}({\mathcal {H}})$ be self-adjoint operators with $\lim _{n \rightarrow \infty }||A_n -A||_{\mathrm{tr}} = 0$, $\lim _{n \rightarrow \infty }||B_n - B||_{\mathrm{tr}} = 0$. Assume that $I+A> 0, I+B> 0, I+A_n> 0, I+ B_n > 0$$\forall n \in {\mathbb {N}}$. Then $(I+B_n)^{-1/2}(I+A_n)(I+B_n)^{-1/2} - I$ and $(I+B)^{-1/2}(I+A)(I+B)^{-1/2} - I$ are self-adjoint, trace class operators on ${\mathcal {H}}$ and

$$\begin{aligned}&\lim _{n \rightarrow \infty }||(I+B_n)^{-1/2}(I+A_n)(I+B_n)^{-1/2} - (I+B)^{-1/2}(I+A)(I+B)^{-1/2}||_{\mathrm{tr}}\\&\quad = 0. \end{aligned}$$

Proof of Theorem 20 - Convergence in trace norm

Let $I+\varLambda = (I+B)^{-1/2}(I+A)(I+B)^{-1/2}$ and $I+\varLambda _n = (I+B_n)^{-1/2}(I+A_n)(I+B_n)^{-1/2}$, with $\varLambda , \varLambda _n \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}({\mathcal {H}})$.

By Lemma 23, we have $\lim _{n\rightarrow \infty }||\varLambda _n -\varLambda ||_{\mathrm{tr}} = 0$. Thus by Theorem 33,

$$\begin{aligned} \lim _{n \rightarrow \infty }||(I+\varLambda _n)^{\alpha } - (I+\varLambda )^{\alpha }||_{\mathrm{tr}} = 0 \;\;\forall \alpha \in {\mathbb {R}}. \end{aligned}$$

From Eq.(130), we have

$$\begin{aligned}&D^{(\alpha , \alpha )}_{2\alpha }[(I+A_n), (I+B_n)] = \frac{1}{\alpha ^2}\log \det \left[ \frac{(I+\varLambda _n)^{\alpha } + (I+\varLambda _n)^{-\alpha }}{2}\right] . \end{aligned}$$

Taking limit as $n \rightarrow \infty$ and applying the continuity of the Fredholm determinant in the trace norm (e.g. Theorem 3.5 in [36]), we obtain

$$\begin{aligned}&\lim _{n \rightarrow \infty }D^{(\alpha , \alpha )}_{2\alpha }[(I+A_n), (I+B_n)] \\&\quad = \frac{1}{\alpha ^2}\log \det \left[ \frac{(I+\varLambda )^{\alpha } + (I+\varLambda )^{-\alpha }}{2}\right] = D^{(\alpha , \alpha )}_{2\alpha }[(I+A), (I+B)]. \end{aligned}$$

$\square$

Proof of Theorem 21 - Diagonalization

Consider first the case $\alpha > 0$. As in the proof of Theorem 22, it suffices to prove for the case $\gamma = 1$. Let $A = \sum _{j=1}^{\infty }\lambda _j(A)\phi _j \otimes \phi _j$ be the spectral decomposition for A. For each $n \in {\mathbb {N}}$, define $A_n = \sum _{j=1}^n\lambda _j(A)\phi _j \otimes \phi _j$. Then $A_n$ is a finite-rank operator with the eigenvalues being the first n eigenvalues of A and $\lim _{n \rightarrow \infty }||A_n - A||_{\mathrm{tr}} = 0$. In the same way, we construct a sequence of finite-rank operators $B_n$ with $\lim _{n\rightarrow \infty }||B_n - B||_{\mathrm{tr}} = 0$. By construction,

$$\begin{aligned} \lim _{n \rightarrow \infty }||\mathrm{Eig}(A_n) - \mathrm{Eig}(A)||_{\mathrm{tr}} = 0, \;\; \lim _{n \rightarrow \infty }||\mathrm{Eig}(B_n) - \mathrm{Eig}(B)||_{\mathrm{tr}} = 0. \end{aligned}$$

Thus by Theorem 20, we have

$$\begin{aligned}&\lim _{n \rightarrow \infty }D^{(\alpha ,\alpha )}_{2\alpha }[(\mathrm{Eig}(A_n) + I), (\mathrm{Eig}(B_n) + I)] = D^{(\alpha ,\alpha )}_{2\alpha }[(\mathrm{Eig}(A) + I), (\mathrm{Eig}(B) + I)], \\&\lim _{n \rightarrow \infty }D^{(\alpha ,\alpha )}_{2\alpha }[(A_n + I), (B_n + I)] = D^{(\alpha ,\alpha )}_{2\alpha }[(A + I), (B + I)]. \end{aligned}$$

Since $A_n, B_n$ can be identified with finite-dimensional matrices, as in the proof of Lemma 20, we can apply the corresponding finite-dimensional result in [9] to obtain

$$\begin{aligned} D^{(\alpha ,\alpha )}_{2\alpha }[(\mathrm{Eig}(A_n) + I), (\mathrm{Eig}(B_n) + I)] \le D^{(\alpha ,\alpha )}_{2\alpha }[(A_n + I), (B_n + I)]. \end{aligned}$$

Thus taking limits as $n \rightarrow \infty$ gives

$$\begin{aligned} D^{(\alpha ,\alpha )}_{2\alpha }[(\mathrm{Eig}(A) + I), (\mathrm{Eig}(B) + I)] \le D^{(\alpha ,\alpha )}_{2\alpha }[(A + I), (B + I)]. \end{aligned}$$

Letting $\alpha \rightarrow 0$ on both sides of the above expression, we also obtain the result for the case $\alpha = 0$. $\square$

Proof of Theorem 22 - Triangle inequality

For a fixed $\gamma > 0$, we have

$$\begin{aligned}&D^{(\alpha , \alpha )}_{2\alpha }[(A + \gamma I), (B+ \gamma I)] \\&\quad = \frac{1}{\alpha ^2}\log \mathrm{det_X}\left( \frac{[(A+\gamma I)(B+\gamma I)^{-1}]^{\alpha } + (A+\gamma I)(B+\gamma I)^{-1}]^{-\alpha }}{2}\right) \\&\quad = \frac{1}{\alpha ^2}\mathrm{logdet}\left( \frac{[(\frac{A}{\gamma }+ I)(\frac{B}{\gamma }+ I)^{-1}]^{\alpha } + (\frac{A}{\gamma }+ I)(\frac{B}{\gamma }+I)^{-1}]^{-\alpha }}{2}\right) , \end{aligned}$$

which reduces to the case $\gamma = 1$. Thus it suffices to prove for $\gamma = 1$.

Let

$$\begin{aligned} \varLambda _1+ I&= (B+I)^{-1/2}(A+I)(B+I)^{-1/2}> 0, \\ \varLambda _2 + I&= (B+I)^{-1/2}(C+I)(B+I)^{-1/2} > 0. \end{aligned}$$

Let $\mathrm{Eig}(\varLambda _1), \mathrm{Eig}(\varLambda _2):\ell ^2 \rightarrow \ell ^2$ denote diagonal operators with the diagonals consisting of the eigenvalues of $\varLambda _1, \varLambda _2$, respectively, in decreasing order. We have by definition

$$\begin{aligned} D^{(1/2,1/2)}_{1}[(A+I),(B+I)]&= 4\log \det \left[ \frac{(\varLambda _1 + I)^{1/2} + (\varLambda _1 + I)^{-1/2}}{2}\right] \\&= D^{(1/2,1/2)}_{1}[(\varLambda _1 + I),I] \\&= D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1) + I),I]. \end{aligned}$$

Then for any $\alpha > 0$, we have

$$\begin{aligned}&D^{(\alpha , \alpha )}_{2\alpha }[(A+ I), (B+ I)] = \frac{1}{\alpha ^2}\log \det \left[ \frac{(\varLambda _1 + I)^{\alpha } + (\varLambda _1 +I)^{-\alpha }}{2}\right] \nonumber \\&= \frac{1}{4\alpha ^2}D^{(1/2,1/2)}_{1}[(\varLambda _1+I)^{2\alpha }, I] = \frac{1}{4\alpha ^2}D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1)+I)^{2\alpha }, I]. \end{aligned}$$

(161)

Similarly,

$$\begin{aligned} D^{(\alpha , \alpha )}_{2\alpha }[(C+ I), (B+ I)]&= \frac{1}{4\alpha ^2}D^{(1/2,1/2)}_{1}[(\varLambda _2+I)^{2\alpha }, I] \nonumber \\&= \frac{1}{4\alpha ^2}D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _2)+I)^{2\alpha }, I]. \end{aligned}$$

(162)

Furthermore, by the Affine Invariance property (Theorem 16), we have

$$\begin{aligned} D^{(\alpha , \alpha )}_{2\alpha }[(A+ I), (C+ I)]&= D^{(\alpha , \alpha )}_{2\alpha }[(\varLambda _1 + I), (\varLambda _2 + I)]. \end{aligned}$$

(163)

By the triangle inequality in Theorem 23,

$$\begin{aligned} \sqrt{D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1) + I)^{2\alpha }, I]}&\le \sqrt{D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1) + I)^{2\alpha }, (\mathrm{Eig}(\varLambda _2) + I)^{2\alpha }]} \nonumber \\&+ \sqrt{D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _2) + I)^{2\alpha }, I]}. \end{aligned}$$

(164)

The desired triangle inequality for $\sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(A+I), (B+I)]}$ requires that for all $(A+I), (B+I), (C+I) \in \mathrm{PTr}({\mathcal {H}})$,

$$\begin{aligned} \sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(A+I), (B+I)]}&\le \sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(A+I), (C+I)]} + \sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(C+I), (B+I)]}. \end{aligned}$$

(165)

By Eqs. (161,162,163,164), the triangle inequality in Eq. (165) is satisfied if

$$\begin{aligned} D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1) + I)^{2\alpha }, (\mathrm{Eig}(\varLambda _2) + I)^{2\alpha }] \le 4\alpha ^2 D^{(\alpha ,\alpha )}_{2\alpha }[(A + I), (C + I)]. \end{aligned}$$

(166)

Since $\mathrm{Eig}(\varLambda _1)$ and $\mathrm{Eig}(\varLambda _2)$ are diagonal operators, they commute and thus

$$\begin{aligned}&D^{(1/2,1/2)}_{1}[(\mathrm{Eig}(\varLambda _1) + I)^{2\alpha }, (\mathrm{Eig}(\varLambda _2) + I)^{2\alpha }] \\&\quad =4 \log \left[ \frac{\det \left[ \frac{(\mathrm{Eig}(\varLambda _1) + I)^{2\alpha } + (\mathrm{Eig}(\varLambda _2) + I)^{2\alpha }}{2}\right] }{\det (\mathrm{Eig}(\varLambda _1) +I)^{\alpha }\det (\mathrm{Eig}(\varLambda _2) + I)^{\alpha }}\right] \\&\quad = 4\log \det \left[ \frac{[(\mathrm{Eig}(\varLambda _1) + I)(\mathrm{Eig}(\varLambda _2) + I)^{-1}]^{\alpha } + (\mathrm{Eig}(\varLambda _1) + I)(\mathrm{Eig}(\varLambda _2) + I)^{-1}]^{-\alpha }}{2}\right] \\&\quad = 4\alpha ^2D^{(\alpha ,\alpha )}_{2\alpha }[(\mathrm{Eig}(\varLambda _1) + I), (\mathrm{Eig}(\varLambda _2) + I)] \\&\quad \le 4\alpha ^2 D^{(\alpha ,\alpha )}_{2\alpha }[(\varLambda _1 + I), (\varLambda _2 + I)] \;\;\;\text {by Theorem}~\mathrm{21} \\&\quad = 4\alpha ^2 D^{(\alpha ,\alpha )}_{2\alpha }[(A + I), (C + I)]. \end{aligned}$$

This is precisely the desired inequality stated in Eq. (166). $\square$

Proof of Theorem 4 - Metric property

The case $\alpha = 0$ corresponds to the affine-invariant Riemannian distance on the Hilbert manifold $\varSigma ({\mathcal {H}})$ [20], which is still a metric when restricted to $\mathrm{PTr}({\mathcal {H}})$. Consider the case $\alpha > 0$. The positivity and symmetry of the divergence $D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+\gamma I)]$ are from Theorems 1 and 14 , respectively. The triangle inequality for the square root $\sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+\gamma I)]}$ is from Theorem 22. Thus $\sqrt{D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+\gamma I)]}$ is a metric on $\mathrm{PTr}({\mathcal {H}})(\gamma )$. $\square$

To prove Theorem 5, we first need the following technical result.

Lemma 24

Let $\lambda > 0, \lambda \ne 1$ be fixed. Consider the function

$$\begin{aligned} f(x) = \left\{ \begin{array}{ll} \frac{1}{x^2}\log \left( \frac{\lambda ^{x} + \lambda ^{-x}}{2}\right) , &{} x > 0. \\ \frac{1}{2}(\log \lambda )^2, &{} x = 0. \end{array} \right. \end{aligned}$$

(167)

Then f is continuous on $[0, \infty )$ and strictly decreasing on $(0, \infty )$, with $f_{\max } = f(0) = \frac{1}{2}(\log \lambda )^2$. Furthermore, $\lim _{x \rightarrow \infty }f(x) = 0$.

Proof of Theorem 5 - Monotonicity

Let $\{\lambda _j\}_{j=1}^{\infty }$ be the eigenvalues of the trace class operator $\varLambda$, where $\varLambda + \frac{\gamma }{\mu }I = (B+\mu I)^{-1/2}(A+\gamma I)(B+\mu I)^{-1/2}$. For $\gamma = \mu$, we have $\delta = \frac{\alpha }{\alpha + \beta }$, $p = r(1-\delta ) = \frac{r\beta }{\alpha + \beta }$, $q = \frac{r\alpha }{\alpha + \beta }$. If furthermore $\alpha = \beta$ and $r = 2\alpha$, then $\delta = \frac{1}{2}$, $p = \alpha$, $q = \alpha$. By Theorem 10, for $\alpha > 0$,

$$\begin{aligned} D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+ \gamma I)]&= \frac{1}{\alpha ^2} \log \det \left[ \frac{(\varLambda + I)^{\alpha } + (\varLambda + I)^{-\alpha }}{2}\right] \\&= \frac{1}{\alpha ^2}\sum _{j=1}^{\infty }\log \left( \frac{(\lambda _j+1)^{\alpha } + (\lambda _j + 1)^{-\alpha }}{2}\right) . \end{aligned}$$

We note that since $\varLambda +I > 0$, we have $\lambda _j +1> 0$$\forall j \in {\mathbb {N}}$. Furthermore, if $A \ne B$, then there exists at least one index $j \in {\mathbb {N}}$ for which $\lambda _j \ne 0$, so that $\lambda _j + 1 > 0$, $\lambda _j +1 \ne 1$, since otherwise $\varLambda = 0$ and $A+\gamma I = B+ \gamma I \Longleftrightarrow A = B$. The terms with $\lambda _j = 0$ simply vanish and play no contribution in the infinite series expansion above for $D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+ \gamma I)]$.

By Lemma 24, as a function of the parameter $\alpha$, $D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+ \gamma I)]$ is strictly decreasing on $(0, \infty )$, with maximum value achieved at $\alpha = 0$.

By Lemma 24 and Lebesgue’s Monotone Convergence Theorem, we have

$$\begin{aligned}&\lim _{\alpha \rightarrow \infty }D^{(\alpha , \alpha )}_{2\alpha }[(A+\gamma I), (B+ \gamma I)] = \lim _{\alpha \rightarrow \infty }\frac{1}{\alpha ^2}\sum _{j=1}^{\infty }\log \left( \frac{(\lambda _j+1)^{\alpha } + (\lambda _j + 1)^{-\alpha }}{2}\right) \\&\quad = \sum _{j=1}^{\infty }\lim _{\alpha \rightarrow \infty }\log \left( \frac{(\lambda _j+1)^{\alpha } + (\lambda _j + 1)^{-\alpha }}{2}\right) = 0. \end{aligned}$$

$\square$

1.9 Proofs for the Derivatives

Proof of Lemma 2

This follows from the linearity of the $\mathrm{tr}_X$ operation, which gives $\mathrm{tr}_X[A_0+\gamma _0I + t(A+\gamma I)] - \mathrm{tr}_X(A_0+\gamma _0I) - t\mathrm{tr}_X(A+\gamma I) = 0$. $\square$

Proof of Proposition 2

For $\gamma = 0$, this reduces to the Plemelj-Smithies formula. Assume that $\gamma \ne 0$. By definition of the extended Fredholm determinant $\mathrm{det_X}$, for $t \ne -1/\gamma$, we have

$$\begin{aligned}&\mathrm{det_X}[I+ t(A+\gamma I)] = \mathrm{det_X}[(1+\gamma t)I + tA] = (1+\gamma t)\det \left( \frac{tA}{1+\gamma t} + I\right) \\&\quad = (1+\gamma t)\left[ 1 +\sum _{k=1}^{\infty }\frac{t^k}{k!}\frac{\alpha _k(A)}{(1+\gamma t)^k}\right] = (1+\gamma t) + t\mathrm{tr}(A) +\sum _{k=2}^{\infty }\frac{t^k}{k!}\frac{\alpha _k(A)}{(1+\gamma t)^{k-1}} \\&\quad = 1 +t\mathrm{tr}_X(A+\gamma I) + \sum _{k=2}^{\infty }\frac{t^k}{k!}\frac{\alpha _k(A)}{(1+\gamma t)^{k-1}} \;\; \text {by the Plemelj-Smithies formula}. \end{aligned}$$

Let $0< \epsilon < 1$ be fixed. For t sufficiently close to zero such that $1+ \gamma t > \epsilon$, we have

$$\begin{aligned}&\left| \frac{\mathrm{det_X}[I+t(A+\gamma I)] - [1 + t\mathrm{tr}_X(A+\gamma I)]}{t}\right| = \left| \sum _{k=2}^{\infty }\frac{t^{k-1}}{k!}\frac{\alpha _k(A)}{(1+\gamma t)^{k-1}}\right| \\&\quad \le \sum _{k=1}^{\infty }\frac{|t|^k}{(k+1)!}\frac{e^{k+1}||A||_{\mathrm{tr}}^{k+1}}{\epsilon ^k} = \frac{|t|e^2||A||^2_{\mathrm{tr}}}{\epsilon }\left[ \frac{1}{2!} + \frac{|t|e||A||_{\mathrm{tr}}}{3!\epsilon } + \cdots \right] \;\;\text {by Eq.~({95})} \\&\quad \le \frac{|t|e^2||A||^2_{\mathrm{tr}}}{\epsilon }\exp (|t|e||A||_{\mathrm{tr}}/\epsilon ), \;\;\;\text {from which it follows that} \\&\mathrm{det_X}[I+ t(A+\gamma I)] = 1 +t\mathrm{tr}_X(A+\gamma I) + O(t^2), \;\;\; \text {and consequently}, \\&\lim _{t \rightarrow 0}\frac{\mathrm{det_X}[I+ t(A+\gamma I)] - [1 +t\mathrm{tr}_X(A+\gamma I)]}{t} = 0. \end{aligned}$$

$\square$

Proof of Lemma 3

By the product rule of $\mathrm{det_X}$,

$$\begin{aligned}&\mathrm{det_X}(A_0 +\gamma _0 I + t(A+\gamma I)) - \mathrm{det_X}(A_0 + \gamma _0I) \\&\quad -t \mathrm{det_X}(A_0+\gamma _0I)\mathrm{tr}_X[(A_0+\gamma _0I)^{-1}(A+\gamma I))] \\&\quad = \mathrm{det_X}(A_0+\gamma _0I) \\&\quad \times [\mathrm{det_X}(I + t(A_0+\gamma _0I)^{-1}(A+\gamma I)] - 1 - t \mathrm{tr}_X[(A_0+\gamma _0I)^{-1}(A+\gamma I))]. \end{aligned}$$

By Proposition 2, we have

$$\begin{aligned} \lim _{t \rightarrow 0}\frac{\mathrm{det_X}(I + t(A_0+\gamma _0I)^{-1}(A+\gamma I)] - 1 - t \mathrm{tr}_X[(A_0+\gamma _0I)^{-1}(A+\gamma I))}{t} = 0, \end{aligned}$$

from which we have the desired formula. $\square$

Proof of Lemma 4

For the function $\log :{\mathbb {R}}^{+}\rightarrow {\mathbb {R}}$, $D\log (x_0)(x) = \frac{x}{x_0}$. By the chain rule, we have

$$\begin{aligned}&D\log \mathrm{det_X}(A_0+\gamma _0 I)(A+\gamma I) = D(\log \circ \mathrm{det_X})(A_0 + \gamma _0I)(A+\gamma I) \\&\quad = [ D\log (\mathrm{det_X}(A_0+\gamma _0I))\circ D\mathrm{det_X}(A_0 + \gamma _0I)](A+\gamma I) \\&\quad = \mathrm{tr}_X[(A_0+\gamma _0I)^{-1}(A+\gamma I)], \end{aligned}$$

by applying the formula for the derivative of $D\mathrm{det_X}(A_0+\gamma _0)(A+\gamma I)$ from Lemma 3. $\square$

We now prove Proposition 3. Recall the chain rule for differentiation on Banach spaces. Let V, W, U be Banach spaces. Let $\varOmega \subset V$ be open and assume that $f: \varOmega \rightarrow W$ is differentiable at $x_0 \in \varOmega$. Let $\varSigma \subset W$ be open and assume that $g:\varSigma \rightarrow U$ is differentiable at $y_0 = f(x_0) \in \varSigma$. Then $g\circ f$ is defined in an open neighborhood of $x_0$ and is differentiable at $x_0$, with

$$\begin{aligned} D(g\circ f)(x_0) = Dg(y_0) \circ Df(x_0). \end{aligned}$$

(168)

Let ${\mathcal {A}}$ be a Banach algebra with identity I. Consider the mapping $f: {\mathcal {A}}\rightarrow {\mathcal {A}}$ defined by $f(A) = A^k$, with $k \in {\mathbb {N}}$. Its derivative $Df(A_0)$ at $A_0 \in {\mathcal {A}}$ is given by

$$\begin{aligned}&Df(A_0)A = A_0^{k-1}A + A_0^{k-2}AA_0 + \cdots + AA_0^{k-1}, \;\; A \in {\mathcal {A}}. \end{aligned}$$

(169)

For the exponential mapping $\exp (A) = \sum _{k=0}^{\infty }\frac{A^k}{k!}$, we then have

$$\begin{aligned}&D\exp (A_0)(A) = \sum _{k=1}^{\infty }\frac{1}{k!}(A_0^{k-1}A + A_0^{k-2}AA_0 + \cdots + AA_0^{k-1}). \end{aligned}$$

(170)

In particular, for $A_0 = I$ and $A_0 = 0$, respectively,

$$\begin{aligned}&D\exp (I)A = A + \frac{2A}{2!} + \cdots + \frac{kA}{k!} + \cdots = A\sum _{k=0}^{\infty }\frac{1}{k!} = \exp (1)A = eA, \end{aligned}$$

(171)

$$\begin{aligned}&D\exp (0)A = A \Longleftrightarrow D\exp (0) = \mathrm{id}_{{\mathcal {A}}}, \; \; \text {where }\mathrm{id}_{{\mathcal {A}}}\text { is the identity operator on }{\mathcal {A}}. \end{aligned}$$

(172)

Let $\mathrm{GL}({\mathcal {A}})$ denote the group of invertible elements in ${\mathcal {A}}$. For the function $f:\mathrm{GL}({\mathcal {A}}) \rightarrow \mathrm{GL}({\mathcal {A}})$ defined by $f(A) = A^{-1}$, its Fréchet derivative $Df(A_0)$ at $A_0 \in \mathrm{GL}({\mathcal {A}})$ is given by

$$\begin{aligned}&Df(A_0)(A) = - A_0^{-1}AA_0^{-1}, \;\; A \in {\mathcal {A}}. \end{aligned}$$

(173)

Consider now the case ${\mathcal {A}}= {\mathcal {L}}({\mathcal {H}})$. Consider the sets

$$\begin{aligned} \mathrm{K}_X({\mathcal {H}}) = \{A + \gamma I \; : A \in {\mathcal {L}}({\mathcal {H}}), \; A \text { compact }, \gamma \in {\mathbb {R}}\}, \end{aligned}$$

(174)

$$\begin{aligned} \mathrm{PK}({\mathcal {H}}) = \{A + \gamma I > 0 \; : A \in {\mathcal {L}}({\mathcal {H}}), \; A \text { compact },\; A = A^{*}, \gamma \in {\mathbb {R}}\}, \end{aligned}$$

(175)

of unitized compact and positive definite unitized compact operators on ${\mathcal {H}}$, respectively. Then the inverse logarithm $\log$ of the exponential map $\exp$$\log :\mathrm{PK}({\mathcal {H}}) \rightarrow \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}})$ is well-defined. Let us find $D\log (A_0)$ for $A_0 \in \mathrm{PK}({\mathcal {H}})$. We first have the following result.

Lemma 25

The map $\exp :\mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}}) \rightarrow \mathrm{PK}({\mathcal {H}})$ and its inverse $\log :\mathrm{PK}({\mathcal {H}}) \rightarrow \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}})$ are bijections.

Proof

For $A \in \mathrm{Sym}({\mathcal {H}})$, A compact, let $\{\lambda _j\}_{j=1}^{\infty }$ be its eigenvalues, then $\lim _{j \rightarrow \infty }\lambda _j = 0$. For $A+\gamma I \in \mathrm{PK}({\mathcal {H}})$, $\log (\frac{A}{\gamma }+I)$ is compact, with eigenvalues $\{\log (\frac{\lambda _j}{\gamma } + 1)\}_{j=1}^{\infty } \rightarrow 0$ as $j \rightarrow \infty$. Hence $\log (A+\gamma I) = \log (\frac{A}{\gamma } +I) + (\log \gamma )I \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}}).$ Conversely, for $A+\gamma I \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}})$, we have $\exp (A+\gamma I) = e^{\gamma }\exp (A) = e^{\gamma }I + e^{\gamma }[\exp (A) - I] \in \mathrm{PK}({\mathcal {H}}),$ since the operator $(\exp (A) - I)$, with eigenvalues $\{e^{\lambda _j} - 1\}_{j=1}^{\infty } \rightarrow 0$ as $j \rightarrow \infty$, is compact. $\square$

From the relation $\exp (\log (A_0)) = A_0$$\forall A_0 \in \mathrm{PK}({\mathcal {H}})$ and the chain rule, it follows that

$$\begin{aligned}{}[D\exp (\log (A_0)) \circ D\log (A_0)]A = A \;\forall A_0 \in \mathrm{PK}({\mathcal {H}}), A \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}}). \end{aligned}$$

Similarly, from the relation $\log (\exp (B_0)) = B_0$$\forall B_0 \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}})$ and the chain rule, we have

$$\begin{aligned}{}[D\log (\exp (B_0)) \circ D\exp (B_0)]A = A \;\;\forall A,B_0 \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}}). \end{aligned}$$

For each $A_0 \in \mathrm{PK}({\mathcal {H}})$, we have $A_0 = \exp (B_0)$ where $B_0 =\log (A_0) \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}})$, thus

$$\begin{aligned}{}[D\log (A_0) \circ D\exp (\log (A_0))]A = A \;\forall A_0 \in \mathrm{PK}({\mathcal {H}}), A \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{K}_X({\mathcal {H}}). \end{aligned}$$

Hence it follows that $\forall A_0 \in \mathrm{PK}({\mathcal {H}})$,

$$\begin{aligned}{}[D\exp (\log (A_0)) \circ D\log (A_0)] = [D\log (A_0) \circ D\exp (\log (A_0))] = \mathrm{id}_{{\mathcal {A}}}. \end{aligned}$$

Thus $D\log (A_0)$ and $D\exp (\log (A_0))$ are invertible operators in ${\mathcal {L}}({\mathcal {A}})$ and

$$\begin{aligned} D\log (A_0) = [D\exp (\log (A_0))]^{-1}, \;\;\;\text {with} \;\;\; D\log (I) = [D\exp (0)]^{-1} = \mathrm{id}_{{\mathcal {A}}}. \end{aligned}$$

(176)

We first consider the Fréchet derivative of $f(t) = (I+tA)^r$, $r \in {\mathbb {R}}$, for $A \in \mathrm{Sym}({\mathcal {H}})$.

Lemma 26

Assume that $A,B \in \mathrm{Sym}({\mathcal {H}})$ are compact, $A \ne 0$. Let $r \in {\mathbb {R}}$ be fixed. Consider the function $f:(-1/||A||, 1/||A||) \rightarrow \mathrm{Sym}({\mathcal {H}})$ defined by $f(t) = (I+ tA)^r + B$. Then

$$\begin{aligned} Df(0)(t) = rAt, \;\; \; t \in {\mathbb {R}}. \end{aligned}$$

(177)

Proof

Since $t \in (-1/||A||, 1/||A||)$, $I+tA > 0$ and thus $(I+tA)^r$ is well-defined $\forall r \in {\mathbb {R}}$. The derivative of f does not depend on the constant term B so we can set $B = 0$. Write $f(t) = (I+tA)^r = \exp (r\log (I+tA))$. By the chain rule, we have

$$\begin{aligned} Df(0)(t) = [D\exp (0) \circ rD\log (I)](At) = rtD\exp (0)(D\log (I)(A)) = rAt. \end{aligned}$$

$\square$

Lemma 27

Assume that $A \in \mathrm{Sym}({\mathcal {H}})$ is compact, $A \ne 0$. Let $r,s,c \in {\mathbb {R}}, c\ge 0$ be fixed. Consider the function $f:(-1/||A||, 1/||A||) \rightarrow \mathrm{Sym}({\mathcal {H}})$ defined by $f(t) = [(I+tA)^r + cI]^{-1}(I+tA)^{s}$. Then

$$\begin{aligned} Df(0)(t) = -\frac{r- (1+c)s}{(1+c)^2}At, \;\;\; \forall t \in {\mathbb {R}}. \end{aligned}$$

(178)

Proof

Since $t \in (-1/||A||, 1/||A||)$, $I+tA > 0$ and thus $(I+tA)^r, (I+tA)^s$ are well-defined $\forall r,s \in {\mathbb {R}}$. Write $f(t) = [(I + tA)^{r-s} + c(I+tA)^{-s}]^{-1} = [g(t)]^{-1}$, where $g(t) = (I + tA)^{r-s} + c(I+tA)^{-s}$, where $g(0) = (1+c)I$. By Lemma 26,

$$\begin{aligned} Dg(0)(t) = (r-s)At -csAt = [r - (1+c)s]At. \end{aligned}$$

By Eq. (173), we then have $Df(0)(t) = -g(0)^{-1}[Dg(0)(t)]g(0)^{-1} = -\frac{ r- (1+c)s}{(1+c)^2}At.$$\square$

Lemma 28

Let ${\mathcal {A}}$ be a Banach algebra with identity I. Let $A,B \in {\mathcal {A}}$ be fixed and $k \in {\mathbb {N}}$. Consider the function $f:{\mathbb {R}}\rightarrow {\mathcal {A}}$ defined by $f(t) = (A+tB)^k$. Then

$$\begin{aligned} Df(0)(t) = t(A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}). \end{aligned}$$

(179)

Proof

Write $f(t) = [g(t)]^k = h(g(t))$, where $g(t) = (A+tB)$, with $g(0) = A$ and $Dg(0)(t) = Bt$, and $h(C) = C^k$. By the chain rule and Eq. (169), we have

$$\begin{aligned} Df(0)(t)&= [Dh(g(0)) \circ Dg(0)](t) = Dh(A)(Dg(0)(t)) = Dh(A)(Bt) \\&= t(A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}). \end{aligned}$$

$\square$

Lemma 29

Let V be a Banach space. Let $\varOmega \subset {\mathbb {R}}$ be an open subset. Consider $F(t) = g(t)H(t)$, where $H:\varOmega \rightarrow V$ and $g:\varOmega \rightarrow {\mathbb {R}}$ are both differentiable on $\varOmega$. Then F is differentiable on $\varOmega$ and

$$\begin{aligned} DF(x_0)(1) = g(x_0)DH(x_0)(1) + g{'}(x_0)H(x_0), \;\;\forall x_0 \in \varOmega . \end{aligned}$$

(180)

Lemma 30

Let ${\mathcal {A}}$ be a Banach algebra with identity I. Let $A,B \in {\mathcal {A}}$ be fixed, $\gamma \in {\mathbb {R}},\gamma \ne 0$, $\mu \in {\mathbb {R}}$, and $k \in {\mathbb {N}}$. Consider the function $f:{\mathbb {R}}\rightarrow {\mathcal {A}}$ defined by $f(t) = \frac{(A+tB)^k}{(\gamma + t\mu )^k}$. Then

$$\begin{aligned} Df(0)(1)=\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} - \frac{k\mu }{\gamma ^{k+1}}A^k. \end{aligned}$$

(181)

Proof

For $\mu = 0$, f is well-defined $\forall t \in {\mathbb {R}}$. For $\mu \ne 0$, f is well-defined on $(-\infty , -\frac{\gamma }{\mu }) \cup (-\frac{\gamma }{\mu }, \infty )$. Let $g(t) = (A+tB)^k$, then $g(0) = A^k$ and $Dg(0)(1) = A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}$ by Lemma 28. Let $h(t) = (\gamma + t \mu )^{-k}$, with $h(0) = \gamma ^{-k}$, $h{'}(0) = -k \gamma ^{-k-1}\mu$. By Lemma 29,

$$\begin{aligned} Df(0)(1)&= h(0)Dg(0)(1) + h{'}(0)g(0) \\&= \frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} - \frac{k\mu }{\gamma ^{k+1}}A^k. \end{aligned}$$

$\square$

We now consider the function $f(t) = [(A+\gamma I) + t(B+\mu I)]^r$, $r \in {\mathbb {R}}$, for $(A+\gamma I) \in \mathrm{PTr}({\mathcal {H}})$ and $(B+\mu I)\in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}_X({\mathcal {H}})$. We recall the binomial series expansion in a unital Banach algebra ${\mathcal {A}}$. Let $A \in {\mathcal {A}}$ be such that $||A|| < 1$. Then the following binomial series converges absolutely

$$\begin{aligned} (I+A)^r = I + rA + \frac{r(r-1)}{2!}A^2 + \frac{r(r-1)(r-2)}{3!}A^3 + \cdots . \end{aligned}$$

(182)

Proof of Proposition 3

We consider three cases: $r= k$, $r=-k$ ($k \in {\mathbb {N}}$), and $r \in {\mathbb {R}}$ in general. While the first two cases are special cases of the last one, they do not require the assumption $||A||_{\mathrm{tr}} < \gamma$ and result in considerably simpler expressions for the Fréchet derivative.

(a) Consider first the function f of the form $f(t) = (A+\gamma I + t(B+\mu I))^k$, $k \in {\mathbb {N}}$, which is well-defined $\forall t \in {\mathbb {R}}$. Write $f(t) = h(g(t)))^k$, where $h(A) = A^k$ and $g(t) = A+\gamma I + t(B+\mu I)$. By the chain rule and Eq. (169), we have

$$\begin{aligned}&Df(0)(1) = [Dh(g(0))\circ Dg(0)](1) = Dh(g(0))(Dg(0)(1))\\&\quad = Dh(A+\gamma I)(B+\mu I) \\&\quad = (A+\gamma I)^{k-1}(B+\mu I) + (A+\gamma I)^{k-2}(B+\mu I)(A+\gamma I)\\&\qquad + \cdots + (B+\mu I)(A+\gamma I)^{k-1} \\&\quad = k\mu (A+\gamma I)^{k-1} + \sum _{j=0}^{k-1}(A+\gamma I)^{k-1-j}B(A+\gamma I)^j. \end{aligned}$$

(b) Consider next the case $r = -k$, $k \in {\mathbb {N}}$. Write $f(t) = h(g(t))$, where $g(t) = (A+\gamma I + t(B+\mu I))^k$ and $h(A) = A^{-1}$. Let t be sufficiently close to zero such that $A+\gamma I + t(B+\mu I) > 0$, then g(t) is invertible and f(t) is well-defined. By the chain rule and Eq. (173), we have

$$\begin{aligned}&Df(0)(1) = Dh(g(0))(Dg(0)(1)) = -g(0)^{-1}Dg(0)(1)g(0)^{-1} \\&= - (A+\gamma I)^{-k}[k\mu (A+\gamma I)^{k-1} + \sum _{j=0}^{k-1}(A+\gamma I)^{k-1-j}B(A+\gamma I)^j](A+\gamma I)^{-k}. \end{aligned}$$

(c) Consider now the general case $r \in {\mathbb {R}}$. Let t be sufficiently close to zero such that $A+\gamma I + t(B+\mu I) > 0$. Then $[A+\gamma I + t(B+\mu I)]^r$ is well-defined for any $r \in {\mathbb {R}}$. We write

$$\begin{aligned}&[A +\gamma I + t(B+\mu I)]^r = [(A+tB) + (\gamma + t\mu ) I]^r = (\gamma + t\mu )^r\left[ I + \frac{A +tB}{\gamma + t\mu }\right] ^r \\&\quad = h(t)g(t), \;\;\;\text {where} \;\;\; h(t) =(\gamma + t\mu )^r, \;\; g(t) = \left[ I + \frac{A +tB}{\gamma + t\mu }\right] ^r. \end{aligned}$$

Let $\epsilon = \gamma - ||A||_{\mathrm{tr}} > 0$ by the assumption that $||A||_{\mathrm{tr}} < \gamma$. Then

$$\begin{aligned} ||A+tB||_{\mathrm{tr}} \le ||A||_{\mathrm{tr}} + |t|\;||B||_{\mathrm{tr}} = \gamma - \epsilon + |t|\;||B||_{\mathrm{tr}} < \gamma + t\mu \end{aligned}$$

for all t satisfying $|t|\;||B|| - t\mu < \epsilon$. Let $\varOmega \in {\mathbb {R}}$ be an open set such that $0 \in \varOmega$ and

$$\begin{aligned} \{A+ \gamma I + t(B+ \mu I) > 0, \;\; |t|\;||B||_{\mathrm{tr}} - t\mu < \epsilon \;\;\;\forall t \in \varOmega \}. \end{aligned}$$

Then f(t) is well-defined on $\varOmega$ and furthermore $||\frac{A+tB}{\gamma + t\mu }||_{\mathrm{tr}} < 1$$\forall t \in \varOmega$. Thus g(t) admits the following absolutely convergent binomial series expansion on $\varOmega$

$$\begin{aligned} g(t) =\left[ I + \frac{A +tB}{\gamma + t\mu }\right] ^r = I + \sum _{k=1}^{\infty }\frac{r(r-1)\ldots (r-k+1)}{k!}\frac{(A+tB)^k}{(\gamma + t\mu )^k}. \end{aligned}$$

Since each term in the series is differentiable on $\varOmega$, it follows that g and hence f are both differentiable on $\varOmega$. Let $g_k(t) = \frac{(A+tB)^k}{(\gamma + t\mu )^k}$, then by Lemma 30,

$$\begin{aligned} Dg_k(0)(1) = \frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} - \frac{k\mu }{\gamma ^{k+1}}A^k \in \mathrm{Tr}({\mathcal {H}}). \end{aligned}$$

It follows that

$$\begin{aligned} Dg(0)(1)&= \sum _{k=1}^{\infty }\frac{r(r-1) \ldots (r-k+1)}{k!}Dg_k(0)(1) \\&= \sum _{k=1}^{\infty }\frac{(r)_k}{k!}\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} - \mu \frac{A}{\gamma ^2}\sum _{k=1}^{\infty }\frac{(r)_k}{(k-1)!}\frac{A^{k-1}}{\gamma ^{k-1}} \\&\quad = \sum _{k=1}^{\infty }\frac{(r)_k}{k!}\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} -r \mu \frac{A}{\gamma ^2}\left( I+\frac{A}{\gamma }\right) ^{r-1}. \end{aligned}$$

By Lemma 29, we then have

$$\begin{aligned}&Df(0)(1) = h(0)Dg(0)(1) + h{'}(0)g(0) \\&\quad = \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_k}{k!}\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} \\&\qquad - r\gamma ^{r-1}\mu \frac{A}{\gamma }\left( I+\frac{A}{\gamma }\right) ^{r-1} + {r\gamma ^{r-1}\mu }\left( I +\frac{A}{\gamma }\right) ^r \\&\quad = \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_k}{k!}\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} + r\gamma ^{r-1}\mu \left( I+\frac{A}{\gamma }\right) ^{r-1} \\&\quad = \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_k}{k!}\frac{A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1}}{\gamma ^k} + r\mu (A+\gamma I)^{r-1}. \end{aligned}$$

$\square$

Setting $A = 0$ and $\gamma = 1$ in Proposition 3, we obtain the following result.

Corollary 3

Let $(B+ \mu I) \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}_X({\mathcal {H}}), r \in {\mathbb {R}}$ be fixed. Then there exists an open set $\varOmega \in {\mathbb {R}}$ containing 0 such that the function $f(t) = [I+ t(B+ \mu I)]^r$ is differentiable on $\varOmega$. Furthermore,

$$\begin{aligned} Df(0)(1) = r(B+\mu I). \end{aligned}$$

(183)

Corollary 4

Assume that $(B+ \mu I) \in \mathrm{Sym}({\mathcal {H}}) \cap \mathrm{Tr}_X({\mathcal {H}})$ is fixed. Let $r \in {\mathbb {R}}$ be fixed. Then for any $(C+\nu I ) \in \mathrm{Tr}_X({\mathcal {H}})$,

$$\begin{aligned} \frac{d}{dt}\mathrm{tr}_X[(I+ t(B+ \mu I))^r(C+\nu I)]\big \vert _{t=0} = r\mathrm{tr}_X[(B+ \mu I)(C+\nu I)]. \end{aligned}$$

(184)

Proof

Since $(C+\nu I)$ is a constant, for $g(t) = (I+ t(B+\mu I))^r(C+\nu I)$, from Corollary 3, we have $Dg(0)(1) = r(B+\mu I)(C+\nu I)$. Thus by the chain rule and Lemma 2, for the function $f(t) = \mathrm{tr}_X[(I+ t(B+ \mu I))^r(C+\nu I)]$,

$$\begin{aligned} Df(0)(1)&= [D\mathrm{tr}_X(g(0)) \circ Dg(0)](1) = D\mathrm{tr}_X(C+\nu I)(r(B+\mu I)(C+\nu I) \\&= r\mathrm{tr}_X[(B+ \mu I)(C+\nu I)]. \end{aligned}$$

Since $f{'}(0) = Df(0)(1)$, this gives us the desired result. $\square$

Proof of Proposition 4

Write $f(t) = \mathrm{tr}_X[(A+\gamma I+t(B+\mu I))^r] = \mathrm{tr}_X[g(t)]$, where $g(t) = (A+\gamma I +t(B+\mu I))^r$. By the chain rule and Lemma 2, we have

$$\begin{aligned} Df(0)(t) = [D\mathrm{tr}_X((A+\gamma I)^r) \circ Dg(0)](t) = \mathrm{tr}_X[Dg(0)(t)]. \end{aligned}$$

By Proposition 3, we then have

$$\begin{aligned} Df(0)(1)&= \mathrm{tr}_X\left[ \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_k(A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1})}{k!\gamma ^k}\right] \nonumber \\&\quad + r\mu \mathrm{tr}_X[(A+\gamma I)^{r-1}], \;\;\text {where}\;\; (r)_k = r(r-1)\ldots (r-k+1) \\&= \mathrm{tr}_X\left[ \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_kA^{k-1}B}{(k-1)!\gamma ^k}\right] + r\mu \mathrm{tr}_X[(A+\gamma I)^{r-1}] \\&= r\mathrm{tr}_X\left[ \gamma ^{r-1}\sum _{k=1}^{\infty }\frac{(r-1)_{k-1}A^{k-1}B}{(k-1)!\gamma ^{k-1}}\right] + r\mu \mathrm{tr}_X[(A+\gamma I)^{r-1}] \\&= r\mathrm{tr}_X\left[ \gamma ^{r-1}\left( I+\frac{A}{\gamma }\right) ^{r-1}B\right] + r\mu \mathrm{tr}_X[(A+\gamma I)^{r-1}]\\&= r\mathrm{tr}_X[(A+\gamma I)^{r-1}(B+\mu I)]. \end{aligned}$$

$\square$

Proof of Proposition 5

Write $f(t) = \log \mathrm{det_X}[(A+\gamma I+t(B+\mu I))^r + cI] = \log \mathrm{det_X}[g(t)]$, where $g(t) = (A+\gamma I +t(B+\mu I))^r + cI$. By the chain rule and Lemma 4, we have

$$\begin{aligned} Df(0)(t)&= [D\log \mathrm{det_X}((A+\gamma I)^r +cI) \circ Dg(0)](t)\\&= \mathrm{tr}_X[((A+\gamma I)^r +cI)^{-1}Dg(0)(t)]. \end{aligned}$$

By Proposition 3, noting that the Fréchet derivatives of $(A+\gamma I +t(B+\mu I))^r + cI$ and $(A+\gamma I + t(B+\mu I))^r$ are the same, we have

$$\begin{aligned} Df(0)(1)&= \mathrm{tr}_X\left[ ((A+\gamma I)^r+cI)^{-1} \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_k(A^{k-1}B + A^{k-2}BA + \cdots + BA^{k-1})}{k!\gamma ^k}\right] \nonumber \\&\quad + r\mu \mathrm{tr}_X[((A+\gamma I)^r + cI)^{-1}(A+\gamma I)^{r-1}], \;\;\text {where}\;\; (r)_k = r(r-1)\ldots (r-k+1) \\&= \mathrm{tr}_X\left[ ((A+\gamma I)^r+cI)^{-1} \gamma ^r\sum _{k=1}^{\infty }\frac{(r)_kA^{k-1}B}{(k-1)!\gamma ^k}\right] \\&\quad + r\mu \mathrm{tr}_X[((A+\gamma I)^r + cI)^{-1}(A+\gamma I)^{r-1}] \\&= r\mathrm{tr}_X\left[ ((A+\gamma I)^r+cI)^{-1} \gamma ^{r-1}\sum _{k=1}^{\infty }\frac{(r-1)_{k-1}A^{k-1}B}{(k-1)!\gamma ^{k-1}}\right] \\&\quad + r\mu \mathrm{tr}_X[((A+\gamma I)^r + cI)^{-1}(A+\gamma I)^{r-1}] \\&= r\mathrm{tr}_X\left[ ((A+\gamma I)^r+cI)^{-1} \gamma ^{r-1}\left( I+\frac{A}{\gamma }\right) ^{r-1}B\right] \\&\quad + r\mu \mathrm{tr}_X[((A+\gamma I)^r + cI)^{-1}(A+\gamma I)^{r-1}] \\&= r\mathrm{tr}_X[((A+\gamma I)^r + cI)^{-1}(A+\gamma I)^{r-1}(B+\mu I)]. \end{aligned}$$

$\square$

Proof of Theorem 25

(The case$\alpha> 0, \beta > 0$) Consider the case $\alpha> 0, \beta > 0$. Let $B = A + sA_1 + tA_2$ and $\mu = \gamma +s\gamma _1 + t\gamma _2$. By definition of $D^{(\alpha , \beta )}_r$ and using the product and inverse rules of $\mathrm{det_X}$,

$$\begin{aligned}&D^{(\alpha , \beta )}_r[(A+\gamma I), (B+\mu I)] \nonumber \\&\quad = \frac{{r(\delta -\frac{\alpha }{\alpha +\beta })}}{\alpha \beta }\log \left( \frac{\gamma }{\mu }\right) + \frac{1}{\alpha \beta }\log \mathrm{det_X}\left( \frac{\alpha (Z + \frac{\gamma }{\mu }I)^{r(1-\delta )} + \beta (Z + \frac{\gamma }{\mu }I)^{-r\delta }}{\alpha + \beta }\right) \nonumber \\&\quad = \frac{{r(\delta -\frac{\alpha }{\alpha +\beta })}}{\alpha \beta }\log \left( \frac{\gamma }{\mu }\right) + \frac{1}{\alpha \beta }\log \mathrm{det_X}\left( \frac{\alpha (Z + \frac{\gamma }{\mu }I)^{r} + \beta I}{\alpha + \beta }\right) -\frac{r\delta }{\alpha \beta }\log \mathrm{det_X}(Z+\frac{\gamma }{\mu } I). \end{aligned}$$

(185)

For $\mu = \gamma + s\gamma _1 + t \gamma _2$, $\mu (t=0) = \gamma + s \gamma _1$, $\mu (s=0,t=0) = \gamma$, $\frac{\partial \mu }{\partial s} = \gamma _1$, $\frac{\partial \mu }{\partial t} = \gamma _2$. With $\delta = \frac{\alpha \gamma ^r}{\alpha \gamma ^r + \beta \mu ^r}$, we have $\delta (t=0) = \frac{\alpha \gamma ^r}{\alpha \gamma ^r + \beta (\gamma + s\gamma _1)^r}$ and $\frac{\partial \delta }{\partial t} = -\frac{(\alpha \gamma ^r)(r\beta \mu ^{r-1}\gamma _2)}{(\alpha \gamma ^r + \beta \mu ^r)^2}$, $\frac{\partial \delta }{\partial t}\big \vert _{t=0} = -\frac{(\alpha \gamma ^r)(r\beta (\gamma + s\gamma _1)^{r-1}\gamma _2)}{(\alpha \gamma ^r + \beta (\gamma + s\gamma _1)^r)^2}$, $\frac{\partial }{\partial t}\log \frac{\gamma }{\mu }\big \vert _{t=0} = -\frac{\gamma _2}{\gamma + s\gamma _1}.$

(a) Consider the first term in Eq. (185), we have

$$\begin{aligned} \frac{\partial }{\partial t}\left[ \left( \delta - \frac{\alpha }{\alpha +\beta }\right) \log \frac{\gamma }{\mu } \right] \big \vert _{t=0}&= -\frac{(\alpha \gamma ^r)(r\beta (\gamma + s\gamma _1)^{r-1}\gamma _2)}{(\alpha \gamma ^r + \beta (\gamma + s\gamma _1)^r)^2}\log \frac{\gamma }{\gamma + s\gamma _1} \\&\quad -\left[ \frac{\alpha \gamma ^r}{\alpha \gamma ^r + \beta (\gamma + s\gamma _1)^r}-\frac{\alpha }{\alpha + \beta }\right] \frac{\gamma _2}{\gamma + s\gamma _1}. \end{aligned}$$

Differentiating with respect to s then gives

$$\begin{aligned}&\frac{\partial ^2 }{\partial s\partial t}\left[ \left( \delta - \frac{\alpha }{\alpha +\beta }\right) \log \frac{\gamma }{\mu } \right] \big \vert _{s =0, t=0} = \frac{2r\alpha \beta }{(\alpha +\beta )^2}\frac{\gamma _1\gamma _2}{\gamma ^2}. \end{aligned}$$

(186)

(b) Consider the second term in Eq. (185). With $\nu = \frac{\gamma }{\mu }$, we write

$$\begin{aligned}&\log \mathrm{det_X}\left( \frac{\alpha (Z + \nu I)^r + \beta I}{\alpha + \beta }\right) = \log \frac{\alpha }{\alpha +\beta } + \log \mathrm{det_X}\left( {(Z + \nu I)^{r} + \frac{\beta }{\alpha } I}\right) \end{aligned}$$

With $B+\mu I = (A+\gamma I) + s(A_1 + \gamma _1I) + t(A_2 + \gamma _2I)$, we have

$$\begin{aligned}&[(A+\gamma I)^{-1/2}(B+\mu I)(A+\gamma I)^{-1/2}]^r = [I+ s(Z_1 + \nu _1I) + t(Z_2 + \nu _2I)]^r, \end{aligned}$$

where $(Z_1 + \nu _1 I) = (A+\gamma I)^{-1/2}(A_1+\gamma _1I)(A+\gamma I)^{-1/2}$, $(Z_2 + \nu _2I) = (A+\gamma I)^{-1/2}(A_2+\gamma _2I)(A+\gamma I)^{-1/2}$, $\nu _1 = \frac{\gamma _1}{\gamma }$, $\nu _2 = \frac{\gamma _2}{\gamma }$, $Z_1, Z_2 \in \mathrm{Tr}({\mathcal {H}})$.

By definition, $Z +\nu I = (A+\gamma I)(B+ \mu I)^{-1}$, with $\nu = \frac{\gamma }{\mu }$, so that

$$\begin{aligned}&\log \mathrm{det_X}\left( {(Z + \nu I)^{r} + \frac{\beta }{\alpha } I}\right) = \log \mathrm{det_X}\left( [(A+\gamma I)(B+\mu I)^{-1}]^{r} + \frac{\beta }{\alpha } I\right) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \log \mathrm{det_X}\left( [(B+\mu I)(A+\gamma I)^{-1}]^{-r} + \frac{\beta }{\alpha } I\right) \\&\quad = \log \mathrm{det_X}\left( [(A+\gamma I)^{-1/2}(B+\mu I)(A+\gamma I)^{-1/2}]^{-r} + \frac{\beta }{\alpha } I\right) \\&\quad = \log \mathrm{det_X}[(I + s(Z_1 + \nu _1I) + t(Z_2 + \nu _2I))^{-r} + \frac{\beta }{\alpha } I]. \end{aligned}$$

By Proposition 5, for s sufficiently close to zero so that $|s|\;||Z_1||_{\mathrm{tr}} < 1 + s\nu _1$,

$$\begin{aligned}&\frac{\partial }{\partial t}\log \mathrm{det_X}\left( {(Z + \nu I)^{r} + \frac{\beta }{\alpha } I}\right) \big \vert _{t=0} \\&\quad = -r\mathrm{tr}_X\left[ \left( (I + s(Z_1 + \nu _1I))^{-r} + \frac{\beta }{\alpha } I\right) ^{-1}(I + s(Z_1 +\nu _1 I))^{-r-1}(Z_2 + \nu _2I)\right] . \end{aligned}$$

By Lemma 2, Lemma 27, and the chain rule, differentiating the last expression with respect to s gives

$$\begin{aligned}&\frac{\partial ^2}{\partial s\partial t}\log \mathrm{det_X}\left( {(Z + \nu I)^{r} + \frac{\beta }{\alpha } I}\right) \big \vert _{s=0, t=0} \nonumber \\&= \frac{r[1+(1+r)\frac{\beta }{\alpha }]}{(1+\frac{\beta }{\alpha })^2} \mathrm{tr}_X[(Z_1 +\nu _1I)(Z_2 + \nu _2I)] \nonumber \\&= \frac{r\alpha (\alpha + (1+r)\beta )}{(\alpha + \beta )^2} \mathrm{tr}_X[(Z_1 + \nu _1I)(Z_2 + \nu _2I)] \nonumber \\&= \frac{r\alpha (\alpha + (1+r)\beta )}{(\alpha + \beta )^2} \mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 +\gamma _1 I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)]. \end{aligned}$$

(187)

(c) Consider the third term in Eq. (185). By definition, $Z +\nu I = (A+\gamma I)(B+ \mu I)^{-1}$, with $\nu = \frac{\gamma }{\mu }$, so that with $B+\mu I = (A+\gamma I) + s(A_1 + \gamma _1I) + t(A_2 + \gamma _2I)$, we have

$$\begin{aligned} \log \mathrm{det_X}(Z+\nu I)&= \log \mathrm{det_X}(A+\gamma I) - \log \mathrm{det_X}(B + \mu I) \\&= \log \mathrm{det_X}(A+\gamma I) \\&\quad -\log \mathrm{det_X}[A+\gamma I + s(A_1+\gamma _1 I) +t(A_2 + \gamma _2I)]. \end{aligned}$$

By Proposition 5, differentiating with respect to t gives

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ \delta \log \mathrm{det_X}(Z + \nu I)\right] \big \vert _{t=0} = \\&\quad -\frac{(\alpha \gamma ^r)(r\beta (\gamma {+} s\gamma _1)^{r-1}\gamma _2)}{(\alpha \gamma ^r + \beta (\gamma {+} s\gamma _1)^r)^2} [\log \mathrm{det_X}(A{+}\gamma I) - \log \mathrm{det_X}(A+\gamma I + s(A_1 + \gamma _1I))] \\&\quad - \frac{\alpha \gamma ^r}{\alpha \gamma ^r + \beta (\gamma + s\gamma _1)^r}\mathrm{tr}_X[(A+\gamma I + s(A_1 + \gamma _1I))^{-1}(A_2 +\gamma _2I)]. \end{aligned}$$

By Proposition 5, $\frac{d}{ds}\log \mathrm{det_X}(A+\gamma I + s(A_1 + \gamma _1I))\big \vert _{s=0} = \mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)].$ Also

$$\begin{aligned}&\frac{d}{ds}\mathrm{tr}_X[(A+\gamma I + s(A_1 + \gamma _1I))^{-1}(A_2 +\gamma _2I)] \big \vert _{s=0} \\&\quad = \frac{d}{ds}\mathrm{tr}_X[(I+ s(A+\gamma I)^{-1}(A_1 + \gamma _1I))^{-1}(A+\gamma I)^{-1}(A_2 +\gamma _2I)]\big \vert _{s=0} \\&\quad = - \mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \;\text {by Corollary}~\mathrm{4} . \end{aligned}$$

Combining the last three expressions, we obtain

$$\begin{aligned}&\frac{\partial ^2}{\partial s \partial t}\left[ \delta \log \mathrm{det_X}(Z + \nu I)\right] \big \vert _{s=0, t=0} = \frac{r\alpha \beta \gamma _2}{(\alpha + \beta )^2\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)] \nonumber \\&\quad + \frac{r\alpha \beta \gamma _1}{(\alpha + \beta )^2\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \nonumber \\&\quad +\frac{\alpha }{\alpha + \beta }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)]. \end{aligned}$$

(188)

Combining Eqs. (186), (187), (188), we obtain

$$\begin{aligned}&\frac{\partial ^2}{\partial s\partial t}D^{(\alpha , \beta )}_r[(A+\gamma I), (B+ \mu I)] \big \vert _{s=0, t=0} = \frac{r^2}{(\alpha +\beta )^2}\frac{2\gamma _1\gamma _2}{\gamma ^2} \\&\quad - \frac{r^2}{(\alpha +\beta )^2}\frac{\gamma _2}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)] \\&\quad - \frac{r^2}{(\alpha +\beta )^2}\frac{\gamma _1}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \\&\quad + \frac{r^2}{(\alpha +\beta )^2}\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)]. \end{aligned}$$

$\square$

Proof of of Theorem 25 - Limiting cases

We now prove Theorem 25 for the case $\alpha > 0, \beta = 0$ (the case $\alpha =0, \beta > 0$ is then obtained by dual symmetry). It suffices to carry out the proof for $\alpha = 1$, where

$$\begin{aligned}&D^{(1, 0)}_r[(A+\gamma I), (B+\mu I)] = r[(\frac{\mu }{\gamma })^{r} -1]\log \frac{\mu }{\gamma } \nonumber \\&\quad +\mathrm{tr}_X([(A+\gamma I)^{-1}(B+ \mu I)]^{r} -I) - r(\frac{\mu }{\gamma })^{r}\log \mathrm{det_X}[(A+\gamma I)^{-1}(B+ \mu I)]. \end{aligned}$$

(189)

Let $B = A + sA_1 + tA_2, \mu = \gamma + s \gamma _1 + t \gamma _2$, with $\mu (t=0) = \gamma + s\gamma _1$, $\mu (s=0, t=0) = \gamma$, $\frac{\partial \mu }{\partial s} = \gamma _1$, $\frac{\partial \mu }{\partial t} = \gamma _2$. By definition, $\frac{\partial }{\partial t}\left( \frac{\mu }{\gamma }\right) ^r = r\frac{\gamma _2}{\gamma }\left( \frac{\mu }{\gamma }\right) ^{r-1}, \frac{\partial }{\partial t}\left( \frac{\mu }{\gamma }\right) ^r \big \vert _{t=0} = r\frac{\gamma _2}{\gamma }\left( \frac{\gamma + s\gamma _1}{\gamma }\right) ^{r-1}.$

(a) Consider the first term, we have $\frac{\partial }{\partial t}\left[ \left( \frac{\mu }{\gamma }\right) ^r-1\right] \log \frac{\mu }{\gamma } \big \vert _{t=0} = r\frac{\gamma _2}{\gamma }\left( \frac{\gamma + s\gamma _1}{\gamma }\right) ^{r-1}\log \frac{\gamma + s\gamma _1}{\gamma } +\left[ \left( \frac{\gamma +s\gamma _1}{\gamma }\right) ^r-1\right] \frac{\gamma _2}{\gamma + s\gamma _1}.$ Differentiating with respect to s then gives

$$\begin{aligned}&\frac{\partial ^2}{\partial s \partial t}\left[ \left( \frac{\mu }{\gamma }\right) ^r-1\right] \log \frac{\mu }{\gamma } \big \vert _{s=0, t=0} = 2r\frac{\gamma _1\gamma _2}{\gamma ^2}. \end{aligned}$$

(190)

(b) Consider the second term. For $B = A+sA_1 + tA_2$ and $\mu = \gamma + s\gamma _1 + t\gamma _2$, by the definition of the power function, for any $r \in {\mathbb {R}}$,

$$\begin{aligned}&[(A+\gamma I)^{-1/2}(B+\mu I)(A+\gamma I)^{-1/2}]^r = [I+ s(Z_1 + \nu _1I) + t(Z_2 + \nu _2I)]^r, \end{aligned}$$

where $(Z_1 + \nu _1 I) = (A+\gamma I)^{-1/2}(A_1+\gamma _1I)(A+\gamma I)^{-1/2}$, $(Z_2 + \nu _2I) = (A+\gamma I)^{-1/2}(A_2+\gamma _2I)(A+\gamma I)^{-1/2}$, $\nu _1 = \frac{\gamma _1}{\gamma }$, $\nu _2 = \frac{\gamma _2}{\gamma }$, $Z_1, Z_2 \in \mathrm{Tr}({\mathcal {H}})$.

For s sufficiently close to zero so that $|s|\;||Z_1||_{\mathrm{tr}} < 1 + s\nu _1$, by Proposition 4,

$$\begin{aligned}&\frac{\partial }{\partial t}\mathrm{tr}_X[(A+\gamma I)^{-1}(B+\mu I)]^r \big \vert _{t=0}\\&\quad = \frac{\partial }{\partial t}\mathrm{tr}_X[(A+\gamma I)^{-1/2}(B+\mu I)(A+\gamma I)^{-1/2}]^r \big \vert _{t=0} \\&\quad = r\mathrm{tr}_X[(I+ s(Z_1 + \nu _1I))^{r-1}(Z_2 + \nu _2I)]. \end{aligned}$$

By Corollary 4, differentiating with respect to s, we obtain

$$\begin{aligned}&\frac{\partial ^2 }{\partial s\partial t}\mathrm{tr}_X[(A+\gamma I)^{-1}(B+\mu I)]^r \big \vert _{s=0, t=0}\nonumber \\&\quad = r(r-1)\mathrm{tr}_X[(Z_1+\nu _1 I)(Z_2 + \nu _2I)] \nonumber \\&\quad = r(r-1)\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1+\gamma _1I)(A+\gamma I)^{-1}(A_2+\gamma _2I)]. \end{aligned}$$

(191)

(c) Consider the third term. For $B = A+sA_1 + tA_2$ and $\mu = \gamma + s\gamma _1 + t\gamma _2$, we have by the product and inverse rules of $\mathrm{det_X}$ that

$$\begin{aligned} \log \mathrm{det_X}[(A+\gamma I)^{-1}(B+\mu I)]&= -\log \mathrm{det_X}(A+\gamma I) \\&\quad + \log \mathrm{det_X}[A+ \gamma I + s(A_1 {+} \gamma _1I) + t(A_2 {+} \gamma _2I)]. \end{aligned}$$

It follows from Proposition 5 and the above calculations that

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ r\left( \frac{\mu }{\gamma }\right) ^{r}\log \mathrm{det_X}[(A+\gamma I)^{-1}(B+ \mu I)]\right] \big \vert _{t=0} \\&\quad = -r^2\frac{\gamma _2}{\gamma }(\frac{\gamma + s\gamma _1}{\gamma })^{r-1}\log \mathrm{det_X}(A+\gamma I) \\&\qquad + r^2\frac{\gamma _2}{\gamma }(\frac{\gamma + s\gamma _1}{\gamma })^{r-1}\log \mathrm{det_X}(A+\gamma I + s(A_1+\gamma _1 I)] \\&\qquad + r(\frac{\gamma + s\gamma _1}{\gamma })^{r}\mathrm{tr}_X[(A+\gamma I +s(A_1 + \gamma _1I))^{-1}(A_2 +\gamma _2I)]. \end{aligned}$$

By Proposition 5, $\frac{d}{ds}\log \mathrm{det_X}[(A+ \gamma I + s(A_1 + \gamma _1I)]\big \vert _{s=0} = \mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)].$ Also

$$\begin{aligned}&\frac{d}{ds}\mathrm{tr}_X[(A+\gamma I +s(A_1 + \gamma _1I))^{-1}(A_2 +\gamma _2I)] \\&\quad = \frac{d}{ds}\mathrm{tr}_X[(I +s(A+\gamma I)^{-1}(A_1 + \gamma _1I))^{-1}(A+\gamma I)^{-1}(A_2 +\gamma _2I)] \\&\quad = -\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \;\;\text {by Corollary~4} . \end{aligned}$$

Combining the last three expressions, we obtain

$$\begin{aligned}&\frac{\partial ^2}{\partial s\partial t}\left[ r\left( \frac{\mu }{\gamma }\right) ^{r}\log \mathrm{det_X}[(A+\gamma I)^{-1}(B+ \mu I)]\right] \big \vert _{s=0,t=0} \nonumber \\&\quad = r^2\frac{\gamma _2}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)] + r^2\frac{\gamma _1}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \nonumber \\&\qquad - r\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1+\gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)]. \end{aligned}$$

(192)

Combining Eqs. (190), (191), and (192), we obtain

$$\begin{aligned}&\frac{\partial ^2}{\partial s \partial t}D^{(1,0)}_r[(A+\gamma I), (B+\mu I)] \big \vert _{s=0,t=0} \\&\quad = 2r^2\frac{\gamma _1\gamma _2}{\gamma ^2} + r^2\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1+\gamma _1I)(A+\gamma I)^{-1}(A_2 + \gamma _2I)] \\&\qquad - r^2\frac{\gamma _2}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_1 + \gamma _1I)] - r^2\frac{\gamma _1}{\gamma }\mathrm{tr}_X[(A+\gamma I)^{-1}(A_2 + \gamma _2I)]. \end{aligned}$$

$\square$

Proof of Lemma 5

By definition of the extended trace $\mathrm{tr}_X$ and the extended inner product $\langle \;, \;\rangle _{\mathrm{HS_X}}$,

$$\begin{aligned}&\mathrm{tr}_X[(A+\gamma I)^{*}(B+\mu I)] = \mathrm{tr}_X[A^{*}B + \mu A^{*} + \gamma B + \gamma \mu I] \\&\quad = \mathrm{tr}(A^{*}B) + \mu \mathrm{tr}(A^{*}) + \gamma \mathrm{tr}(B) + \gamma \mu \\&\quad = [\mathrm{tr}(A^{*}B) + \gamma \mu ] + \mu \mathrm{tr}_X(A+\gamma I)^{*} + \gamma \mathrm{tr}_X(B+\mu I) - 2\gamma \mu \\&\quad = \langle A+\gamma I, B + \mu I\rangle _{\mathrm{HS_X}} + \mu \mathrm{tr}_X(A+\gamma I)^{*} + \gamma \mathrm{tr}_X(B+\mu I) - 2\gamma \mu . \end{aligned}$$

$\square$

Proof of Corollary 2

This follows from Theorem 25, Lemma 5, and the property

$$\begin{aligned} \mathrm{tr}_X[(A+\gamma I)^{-1/2}(B+\mu I)(A+\gamma I)^{-1/2} = \mathrm{tr}_X[(A+\gamma I)^{-1}(B +\mu I)] \end{aligned}$$

for any pair $(A+\gamma I) \in \mathrm{PTr}({\mathcal {H}})$ and $(B+\mu I) \in \mathrm{Tr}_X({\mathcal {H}})$. $\square$