Finite modular energy and entropy balance

For $\mathrm{\mid }x\mathrm{\mid }<1$, the power series $P(x):={\sum }_{n=1}^{\mathrm{\infty }}{x}^n\mathrm{/}n$ converges absolutely because it is termwise bounded in absolute value by the geometric series in [F5]. Its radius is therefore at least $1$, and [F6] gives $P^{\mathrm{\prime }}(x)={\sum }_{n=1}^{\mathrm{\infty }}{x}^{n-1}=1\mathrm{/}(1-x)$. The difference quotient for $\log$ in [F4], with increment $-h$, gives $\frac{d}{dx}[-\log (1-x)]=1\mathrm{/}(1-x)$. Both functions are differentiable on $(-1,1)$ and hence continuous there by [F8]: at a differentiability point, the difference quotient is bounded near the point, so multiplying it by the increment shows that the function increment tends to zero. Since both functions vanish at $0$ by [F3], [F7] applied to their difference yields ${\sum }_{n=1}^{\mathrm{\infty }}{x}^n\mathrm{/}n=-\log (1-x)$ for every $\mathrm{\mid }x\mathrm{\mid }<1$.

Write $d:=\dim H$. For any endomorphism $T$, let $M:={\max }_{i,j}\mathrm{\mid }{T}_{ij}\mathrm{\mid }$ in the fixed orthonormal basis. For $x={\sum }_j{x}_j{e}_j$, Cauchy–Schwarz [F12] applied in these coordinates to $(T_{ij}{)}_j$ and $(\overline{x_j}{)}_j$ gives $\mathrm{\mid }(Tx{)}_i\mathrm{\mid }=\mathrm{\mid }{\sum }_j{T}_{ij}{x}_j\mathrm{\mid }\le ({\sum }_j\mathrm{\mid }{T}_{ij}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}({\sum }_j\mathrm{\mid }{x}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\le \sqrt{d}M\mathrm{\parallel }x\mathrm{\parallel }$ for each $i$, hence $\mathrm{\parallel }Tx\mathrm{\parallel }\le dM\mathrm{\parallel }x\mathrm{\parallel }$. Thus $T$ is bounded, so the operator norm [F9] is defined and finite, with $\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le dM$. For any endomorphisms $S,T$, [F9] then gives $\mathrm{\parallel }STx\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }x\mathrm{\parallel }$ for every vector $x$. Taking the supremum over unit $x$ proves $\mathrm{\parallel }ST{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$.

Put $A_0:=A(t_0)$ and let ${\lambda }_{\max }$ and ${\lambda }_{\min }$ be its largest and smallest eigenvalues. They satisfy $0<{\lambda }_{\min }\le {\lambda }_{\max }$ by [F1] and the hypothesis. Set $c:=2{\lambda }_{\max }$, $B:=I_H-A_0\mathrm{/}c$, and, for $t\in U$, $B_t:=I_H-A(t)\mathrm{/}c$. The eigenvalues of $B$ lie in $[1\mathrm{/}2,1-{\lambda }_{\min }\mathrm{/}c]$. In an orthonormal eigenbasis of $B$, writing its eigenvalues as $b_j$, [F9] gives $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}={\max }_j\mathrm{\mid }{b}_j\mathrm{\mid }$: for $x={\sum }_j{x}_j{e}_j$, $\mathrm{\parallel }Bx{\mathrm{\parallel }}^2={\sum }_j\mathrm{\mid }{b}_j{\mathrm{\mid }}^2\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le ({\max }_j\mathrm{\mid }{b}_j\mathrm{\mid }{)}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$, with equality on an eigenvector for a largest $\mathrm{\mid }{b}_j\mathrm{\mid }$. Hence $q:=\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1$ by [F1]. Choose $r$ with $q<r<1$. By [F8], every entry of $A(t)-A_0$ tends to zero as $t\to t_0$. Since there are finitely many entries, $M(t):={\max }_{i,j}\mathrm{\mid }(A(t)-A_0{)}_{ij}\mathrm{\mid }\to 0$, and the matrix bound in step 1.2 gives $\mathrm{\parallel }A(t)-A_0{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\to 0$, so choose a neighborhood $V\subseteq U$ of $t_0$ where $\mathrm{\parallel }A(t)-A_0{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}c<(r-q)\mathrm{/}2$. For every unit vector $x$, the inner-product norm triangle inequality [F11] and [F9] give $\mathrm{\parallel }{B}_{t}x\mathrm{\parallel }\le \mathrm{\parallel }Bx\mathrm{\parallel }+\mathrm{\parallel }(A(t)-A_0)x\mathrm{\parallel }\mathrm{/}c\le q+\mathrm{\parallel }A(t)-A_0{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}c<r$ for $t\in V$; taking the supremum proves $\mathrm{\parallel }{B}_t{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<r$. Since $B_t$ is self-adjoint, each eigenvalue $\beta$ satisfies $\mathrm{\mid }\beta \mathrm{\mid }\le \mathrm{\parallel }{B}_t{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<r<1$. For $t\in V$, the scalar identity in step 1.1 applied to these eigenvalues, the product law [F10], and spectral calculus [F1] give $\log A(t)=(\log c)I_H-{\sum }_{n=1}^{\mathrm{\infty }}{B}_t^n\mathrm{/}n$, with convergence in operator norm because repeated application of step 1.2 gives $\mathrm{\parallel }{B}_t^n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }{B}_t{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^n<r^n$ and $\sum r^n\mathrm{/}n\le \sum r^n<\mathrm{\infty }$.

For $t$ sufficiently near $t_0$, put $E:=A(t)-A_0$, $Z:=-E\mathrm{/}c$, and $z:=\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Then $B_t=B+Z$. The supremum definition gives $z=\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}c$, so $z\to 0$ as $t\to t_0$; shrink the neighborhood so that $q+z\le r$. Distributivity writes $(B+Z{)}^n=B^n+S_n+R_n$, where $S_n:={\sum }_{j=0}^{n-1}{B}^{j}Z{B}^{n-1-j}$ contains the terms with exactly one $Z$, and $R_n$ contains all terms with at least two $Z$ factors. The submultiplicative bound in step 1.2 gives $\mathrm{\parallel }{S}_n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le n{q}^{n-1}z$. For $R_n$, each word with exactly $k$ factors $Z$ has norm at most $q^{n-k}{z}^k$, and there are $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ such words, so $\mathrm{\parallel }{R}_n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\sum }_{k=2}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)q^{n-k}{z}^k\le \left(\genfrac{}{}{0px}{}{n}{2}\right)z^2(q+z{)}^{n-2}\le \left(\genfrac{}{}{0px}{}{n}{2}\right)z^2{r}^{n-2}$. The middle bound follows from $\left(\genfrac{}{}{0px}{}{n}{k}\right)\le \left(\genfrac{}{}{0px}{}{n}{2}\right)\left(\genfrac{}{}{0px}{}{n-2}{k-2}\right)$ for $k\ge 2$.

Substituting the series of step 2.1 into $G(A):=\mathrm{tr}(A\log A)$, and using step 3.1, yields $G(A_0+E)-G(A_0)=(\log c)\mathrm{tr}E-{\sum }_{n=1}^{\mathrm{\infty }}\frac{\mathrm{tr}(E{B}^n)+\mathrm{tr}(A_0{S}_n)}{n}+R(E),$ where $\mathrm{\mid }R(E)\mathrm{\mid }\le C\mathrm{\parallel }E{\mathrm{\parallel }}^2$ for a constant $C$ independent of sufficiently small $E$. Indeed, the omitted terms are $\mathrm{tr}(A_0{R}_n)$, $\mathrm{tr}(E{S}_n)$, and $\mathrm{tr}(E{R}_n)$; for each orthonormal basis vector $e_i$, the $i$th coordinate of $T{e}_i$ is $⟨T{e}_i,e_i⟩$ and [F12] gives its magnitude at most $\mathrm{\parallel }T{e}_i\mathrm{\parallel }\mathrm{\parallel }{e}_i\mathrm{\parallel }=\mathrm{\parallel }T{e}_i\mathrm{\parallel }\le \mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ by [F9], since $\mathrm{\parallel }{e}_i\mathrm{\parallel }=1$, so [F2] gives $\mathrm{\mid }\mathrm{tr}T\mathrm{\mid }\le d\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ with $d=\dim H$, and the bounds in step 3.1 make their series after division by $n$ bounded by constants times $z^2{\sum }_{n\ge 2}n{r}^{n-2}$, $z^2{\sum }_{n\ge 1}{q}^{n-1}$, and $z^3{\sum }_{n\ge 2}n{r}^{n-2}$, respectively. Each geometric derivative sum is finite by [F5]–[F6], and step 3.1 gives $z=\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}c$.

The linear matrix-log term read from the series in steps 2.1 and 3.1 is $L(E):=\frac{1}{c}{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n}{\sum }_{j=0}^{n-1}{B}^{j}E{B}^{n-1-j}\mathrm{.}$ Its series converges in operator norm because the $n$th summand has norm at most $q^{n-1}\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}c$. The remainder in the logarithm series is bounded by ${\sum }_{n=2}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{n}{2}\right)z^2{r}^{n-2}\mathrm{/}n=(z^2\mathrm{/}2){\sum }_{n=2}^{\mathrm{\infty }}(n-1)r^{n-2}=O(\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2)$, so $L(E)$ is the full first-order term of $\log (A_0+E)$. In an orthonormal eigenbasis of $A_0$, write $A_0{e}_i={\lambda }_i{e}_i$ and $b_i:=1-{\lambda }_i\mathrm{/}c$. Then $L(E{)}_{ik}=\frac{E_{ik}}{c}{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n}{\sum }_{j=0}^{n-1}{b}_i^j{b}_k^{n-1-j}\mathrm{.}$ If $b_i\mathrm{\ne }{b}_k$, the inner sum is $(b_i^n-b_k^n)\mathrm{/}(b_i-b_k)$, so its coefficient is $\frac{1}{c}\frac{{\sum }_{n=1}^{\mathrm{\infty }}(b_i^n-b_k^n)\mathrm{/}n}{b_i-b_k}=\frac{\log (1-b_k)-\log (1-b_i)}{c(b_i-b_k)}=\frac{\log {\lambda }_i-\log {\lambda }_k}{{\lambda }_i-{\lambda }_k},$ where step 1.1 evaluates the series and [F10] gives $\log {\lambda }_j=\log c+\log (1-b_j)$, while ${\lambda }_i-{\lambda }_k=c(b_k-b_i)$. If $b_i=b_k$, the coefficient is $c^{-1}{\sum }_{n=1}^{\mathrm{\infty }}{b}_i^{n-1}=1\mathrm{/}(c(1-b_i))=1\mathrm{/}{\lambda }_i={\log }^{\mathrm{\prime }}({\lambda }_i)$ by [F4] and [F5]. Defining ${\log }^{[1]}(x,y):=(\log x-\log y)\mathrm{/}(x-y)$ for $x\mathrm{\ne }y$ and ${\log }^{[1]}(x,x):=1\mathrm{/}x$, we have $L(E{)}_{ik}={\log }^{[1]}({\lambda }_i,{\lambda }_k)E_{ik}$: this is the divided-difference derivative of the matrix logarithm.

In a fixed basis, [F2] gives $\mathrm{tr}(XY)={\sum }_{i,j}{X}_{ij}{Y}_{ji}={\sum }_{i,j}{Y}_{ji}{X}_{ij}=\mathrm{tr}(YX)$, so trace is cyclic. Since $A_0$ commutes with $B$ and $Z=-E\mathrm{/}c$, cycling each term gives $\mathrm{tr}(A_0{S}_n)=-(n\mathrm{/}c)\mathrm{tr}(A_0{B}^{n-1}E)$. Thus the $A_0{S}_n$ terms in step 4.1 sum to $\mathrm{tr}(A_{0}L(E))$. In the eigenbasis used in step 4.2, $L(E{)}_{ii}=E_{ii}\mathrm{/}{\lambda }_i$, so $\mathrm{tr}(A_{0}L(E))={\sum }_i{\lambda }_{i}L(E{)}_{ii}=\mathrm{tr}E$. The remaining linear terms in step 4.1 are $\mathrm{tr}(E\log A_0)$ by the spectral series in step 2.1. Therefore $G(A_0+E)-G(A_0)=\mathrm{tr}(E(\log A_0+I_H))+O(\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2)$.

By [F8], for $h\to 0$ the path increment is $E=A(t_0+h)-A_0=h{A}^{\mathrm{\prime }}(t_0)+o(h)$ entrywise. The finite-coordinate bound in step 2.1 gives $\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=O(\mathrm{\mid }h\mathrm{\mid })$. Dividing the expansion in step 5.1 by $h$ makes its $O(\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2)$ remainder tend to $0$ and its linear term tend to $\mathrm{tr}(A^{\mathrm{\prime }}(t_0)(\log A(t_0)+I_H))$. This is the stated derivative.

Since $\sigma$ has only positive eigenvalues, $\mathrm{supp}(\sigma )=H$ and hence $\mathrm{supp}(\rho )\subseteq \mathrm{supp}(\sigma )$. By [F1], $D(\rho \mathrm{\parallel }\sigma )$ is finite and equals $\mathrm{tr}(\rho \log \rho )-\mathrm{tr}(\rho \log \sigma )=-s(\rho )+\mathrm{tr}(\rho K_{\sigma })$.

By [F2] and $K_{\sigma }=-\log \sigma$, $s(\sigma )=\mathrm{tr}(\sigma K_{\sigma })$. Linearity of trace [F3] then gives ${\mathrm{\Delta }}_K(\rho ,\sigma )-(s(\rho )-s(\sigma ))=\mathrm{tr}(\rho K_{\sigma })-\mathrm{tr}(\sigma K_{\sigma })-s(\rho )+s(\sigma )=\mathrm{tr}(\rho K_{\sigma })-s(\rho )=D(\rho \mathrm{\parallel }\sigma )$ by step 1.1. This proves the finite modular-energy identity.

With the finite identity established in step 2.1, apply [F5] for the separate partial-trace claim to the nonnegative endomorphisms $\rho ,\sigma$ on $H_A\otimes H_B$, with the partial trace defined in [F4], to obtain the displayed inequality.

Let $m>0$ be the smallest eigenvalue of $\sigma$ from [F3]. Since $\rho$ is entrywise continuous at $0$, its finite matrix difference $T(\epsilon ):=\rho (\epsilon )-\sigma$ tends to zero in operator norm: in a fixed orthonormal basis, if every entry of $T$ has magnitude at most $M$, then for a unit vector $x$ ${\sum }_j\mathrm{\mid }{x}_j{\mathrm{\mid }}^2=1$ makes each coordinate of $x$ have magnitude at most $1$, and each coordinate of $Tx$ has magnitude at most $dM$, and therefore $\mathrm{\parallel }Tx\mathrm{\parallel }\le d^{3\mathrm{/}2}M$, where $d=\dim H$. Choose $\delta >0$ so that $\mathrm{\parallel }T(\epsilon ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<m\mathrm{/}2$ for $\mathrm{\mid }\epsilon \mathrm{\mid }<\delta$. For each $v\in H$, $⟨\sigma v,v⟩\ge m\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by spectral decomposition, while [F5] and [F6] give $\mathrm{\mid }⟨T(\epsilon )v,v⟩\mathrm{\mid }\le \mathrm{\parallel }T(\epsilon ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }v{\mathrm{\parallel }}^2$. Hence $⟨\rho (\epsilon )v,v⟩>(m\mathrm{/}2)\mathrm{\parallel }v{\mathrm{\parallel }}^2$ for nonzero $v$. Thus $\rho (\epsilon )$ is positive definite near $0$ and [F1] applies.

Apply [F1] to $\rho$ at $0$: ${\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_{\epsilon =0}=-\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)(\log \sigma +I))$. Because $\mathrm{tr}(\rho (\epsilon ))=1$ on the whole interval, in the fixed basis [F4] gives $\mathrm{tr}({\rho }^{\mathrm{\prime }}(0))={\sum }_i{\rho }_{ii}^{\mathrm{\prime }}(0)={\lim }_{h\to 0}{\sum }_i({\rho }_{ii}(h)-{\rho }_{ii}(0))\mathrm{/}h={\lim }_{h\to 0}(1-1)\mathrm{/}h=0$; the finite sum commutes with this limit. Therefore ${\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_0=-\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)\log \sigma )=\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)K_{\sigma })$, as claimed.

The identity [F2] holds for every $\rho (\epsilon )$ in the path. Its right-hand side is differentiable at $0$ by linearity of trace and step 2.1, so differentiating gives ${\frac{d}{d\epsilon }D(\rho (\epsilon )\mathrm{\parallel }\sigma )\mid }_0=\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)K_{\sigma })-{\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_0=0$. The additional twice continuously differentiable hypothesis ensures the differentiability required here.