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  \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}%
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    Email: \texttt{s.sra@tum.de}.}}

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\keywords{Kubo--Mori metric, scalar curvature, logarithmic mean, strict concavity, Stieltjes function, Morozova--Chentsov kernel}

\hypersetup{
  pdftitle={A proof of Petz's conjecture on the scalar curvature of the Kubo--Mori metric},
  pdfauthor={Suvrit Sra},
  pdfsubject={Strict concavity of Dittmann's function and Petz's
              conjecture for the Kubo--Mori metric}
}

\makeatletter
\newcommand{\preprintslink}[2]{%
  \thanks{First public release: #2, as v1 in the author's preprint repository,
    \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every
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\begin{document}
\title[A proof of Petz's conjecture]{A Proof of Petz's Conjecture on the Scalar Curvature of the Kubo--Mori Metric}
\aiauthor
\begin{abstract}
Petz (1994) conjectured that the scalar curvature of the Kubo--Mori metric
on invertible density matrices increases as the state becomes more mixed.
Dittmann (2000) reduced the conjecture to the concavity of an explicit
symmetric function \(h\) on \((0,\infty)^3\).  We prove that \(h\) is strictly
concave, and hence Petz's conjecture.  The proof writes \(h=-1/(8M)\) with
\(M\) positive and homogeneous of degree one, and its central idea is to show that
\(t\mapsto 1/M(X+tY)\) is a Stieltjes function for all \(X,Y>0\);
Cauchy--Schwarz and homogeneity then give concavity of \(h\) on
\((0,\infty)^3\).  %
\end{abstract}
\preprintslink{kubo-metric}{9 October 2026}
\maketitle

\section{Introduction}\label{sec:intro}

The Kubo--Mori metric, also called the Bogoliubov--Kubo--Mori metric, is one
of the monotone Riemannian metrics on the manifold of invertible
\(n\times n\) density matrices~\citep{petz1994geometry,petz1996monotone}.
For two such matrices \(\omega_1\) and
\(\omega_2\), call \(\omega_1\) \emph{more mixed} than \(\omega_2\) if, for
each \(k\), the sum of the \(k\) largest eigenvalues of \(\omega_1\) is at
most the corresponding sum for \(\omega_2\).  \citet{petz1994geometry}
conjectured that the scalar curvature of the Kubo--Mori metric increases
under this order: \(\operatorname{Scal}(\omega_1)\ge\operatorname{Scal}(\omega_2)\)
whenever \(\omega_1\) is more mixed than
\(\omega_2\)~\citep[Section~VI]{dittmann2000curvature}.

\citet{dittmann2000curvature} reduced this conjecture to the concavity of
one explicit function.  His curvature calculation leads to a symmetric
function of three positive variables, written here in terms of the
Morozova--Chentsov kernel of the Kubo--Mori metric.  This kernel is the
reciprocal of the logarithmic mean: for \(x,y>0\),
\[
  L(x,y)=\frac{x-y}{\log x-\log y}
        =\int_0^1 x^{1-s}y^s\dd s,
\]
with the usual continuous value \(L(x,x)=x\), and
\begin{equation}\label{eq:c-def}
  c(x,y):=\frac{1}{L(x,y)}
  =\frac{\log x-\log y}{x-y},
  \qquad c(x,x)=\frac1x.
\end{equation}
For pairwise distinct \(x,y,z>0\), define
\begin{align}
 h(x,y,z)
 :={}&
 \frac{(y-z)\,c(x,y)c(x,z)-(x-z)\,c(x,y)c(y,z)+(x-y)\,c(x,z)c(y,z)}
 {6(x-y)(x-z)(y-z)\,c(x,y)c(x,z)c(y,z)}
 \notag\\
 &\quad+
 \frac{-xy(x-y)\,c(x,y)+xz(x-z)\,c(x,z)-yz(y-z)\,c(y,z)}
 {3xyz(x-y)(x-z)(y-z)\,c(x,y)c(x,z)c(y,z)}.
 \label{eq:h-original}
\end{align}
All diagonal and partial-diagonal values are understood by continuous
extension.  Dittmann stated the concavity of this function as a labelled
\emph{Assertion}, writing that he had no formal proof of it but that it was
confirmed by several million numerical tests~\citep[Eq.~(61) and the discussion following it]{dittmann2000curvature}.
He proved that the Assertion implies Petz's
conjecture~\citep[Theorem~3]{dittmann2000curvature}.

The Assertion has remained unproved since 2000, and Petz's conjecture with
it.  Two later works approach the same conjecture.
\citet{andai2003monotonicity} regroups the summands of the scalar curvature and
proves monotonicity for some of them, leaving the remainder supported only by
numerics; \citet{gibilisco2005monotonicity} conjecture that the
Wigner--Yanase--Dyson metrics near the Kubo--Mori metric have monotone scalar
curvature, which would imply Petz's
conjecture~\citep[Conjecture~8.1]{gibilisco2005monotonicity}.  Neither
closes the gap, and neither establishes the Assertion.
Not every monotone metric has monotone scalar curvature:
\citet{andai2003nonmonotone} exhibits monotone metrics for which it fails.

Our main result is the following.

\begin{theorem}[Strict concavity of Dittmann's function]
\label{thm:main}
The function \(h\) defined by \eqref{eq:h-original} extends real-analytically
to \((0,\infty)^3\) and is strictly concave there.  That is, for
all distinct \(X,Y\in(0,\infty)^3\) and every \(0<\lambda<1\),
\[
 h(\lambda X+(1-\lambda)Y)
 >\lambda h(X)+(1-\lambda)h(Y).
\]
\end{theorem}

Theorem~\ref{thm:main} proves Dittmann's Assertion, so his Theorem~3 gives
the following.

\begin{corr}[Petz's conjecture]\label{cor:petz}
Let \(\omega_1\) and \(\omega_2\) be invertible \(n\times n\) density
matrices.  If \(\omega_1\) is more mixed than \(\omega_2\), then the scalar
curvature of the Kubo--Mori metric satisfies
\(\operatorname{Scal}(\omega_1)\ge\operatorname{Scal}(\omega_2)\).
\end{corr}

Corollary~\ref{cor:petz} rests on Dittmann's Theorem~3, whose proof we use as
published and do not re-derive.  Our own work is the proof of
Theorem~\ref{thm:main}.  It passes through a positive, symmetric function \(M\) on
\((0,\infty)^3\), homogeneous of degree one, defined in~\eqref{eq:M-F-def}
below, such that
\begin{equation}\label{eq:h-M-intro}
  h=-\frac{1}{8M}.
\end{equation}
Set
\begin{equation}\label{eq:m-def-intro}
  m(a,b):=M(a,b,1).
\end{equation}
By homogeneity \(M(x,y,z)=z\,m(x/z,y/z)\), so \(M\) is the perspective
of \(m\), and concavity of \(M\) is equivalent to concavity of \(m\).  The
two-variable form is the statement that a Hessian computation would target,
and we record it separately.

\begin{theorem}[Strict concavity in two variables]
\label{thm:m-main}
The function \(m\) in \eqref{eq:m-def-intro} is strictly concave on
\((0,\infty)^2\).
\end{theorem}

The proof has three by-products that do not depend on Petz's conjecture.
The normal form of Section~\ref{sec:normal} writes Dittmann's two-line
expression~\eqref{eq:h-original} as \(h=-1/(8M)\) with \(M\) positive,
symmetric, homogeneous of degree one, and normalized by \(M(x,x,x)=x\).
Theorem~\ref{thm:line-stieltjes} shows that \(1/M\) is a Stieltjes function
along every line with positive base point and positive direction, so
\(M\) is operator monotone along such lines
(Remark~\ref{rmk:operator-monotone}), which is more than the concavity the
proof uses.  Lemma~\ref{lem:positive-line} shows that a function homogeneous
of degree one that is concave along lines with positive direction is concave
on the whole cone, which is why positive directions suffice.

\subsection{Idea of the proof}\label{sec:idea}
By~\eqref{eq:h-M-intro} and homogeneity, concavity of \(h\) follows from
concavity of \(m\), so the direct route is to show that the \(2\times2\)
Hessian of \(m\) is negative semidefinite at every point of
\((0,\infty)^2\).  Its entries are second derivatives of a rational
expression in logarithmic divided differences.  The route we take replaces
this second-order condition by a sign condition that involves no
derivatives.  For \(X,Y>0\), consider the restriction
\(t\mapsto F(X+tY)=1/M(X+tY)\) of \(F\) to a line with positive base
point and positive direction.  The resolvent integrals behind the normal
form continue this function analytically to the complex plane cut along a
compact interval of the negative axis, and we show that it is a Stieltjes
function, the integral of \(1/(t-s)\) against a finite positive measure on
that interval (Theorem~\ref{thm:line-stieltjes}).  The Stieltjes property
gives concavity for free: a single Cauchy--Schwarz inequality shows that the
reciprocal of a Stieltjes function is concave, so \(M\) is concave along
each of these lines, and the same property makes each of these restrictions
of \(M\) operator monotone (Remark~\ref{rmk:operator-monotone}).  Degree-one homogeneity then shows that positive
directions already control every chord of the cone: concavity along lines
with positive direction gives superadditivity of \(M\), hence concavity, and
excluding point masses gives the strict statements.

The Stieltjes property is where the work lies.  We continue
\(t\mapsto F(X+tY)\) analytically to the plane cut along a compact interval
of the negative axis, and a keyhole contour represents it through its
boundary values on the cut.  The representing measure is positive when the
upper boundary value has nonpositive imaginary part.  This exchange is the
point of the detour: concavity, a second-order condition at every interior
point, becomes a sign condition on the boundary values alone.  On the cut, the three
coordinates of \(X+sY\) are real with mixed signs, and symmetry,
homogeneity and Schwarz reflection reduce them to \((-1,p,q)\) with
\(p,q>0\).  There the logarithms are explicit, since the logarithm of a
negative number is \(\log|x|\pm\ii\pi\), and the sign condition becomes
the positivity of the function \(N(p,q)\) of Lemma~\ref{lem:boundary-N}, which we prove in logarithmic coordinates by elementary
estimates for hyperbolic functions.  No computer calculation is part of the
proof, and the only numerical facts it uses are \(\pi>3\), \(\pi^2<10\) and
\(e>8/3\).

The argument uses two properties of the
logarithmic mean: its resolvent integral, which gives \(h\) the nonsingular
form of Section~\ref{sec:normal}, and the explicit values of the principal
logarithm on the negative axis, which turn the sign condition of
Section~\ref{sec:boundary} into an explicit inequality.  It is therefore
specific to the Kubo--Mori metric, and it does not say which other monotone
metrics have monotone scalar curvature; the conjecture of
\citet{gibilisco2005monotonicity}, which places the Kubo--Mori metric at the
end of a one-parameter family of such metrics, is not addressed here.

Section~\ref{sec:normal} derives the normal form~\eqref{eq:h-M-intro} and
reduces both theorems to the Stieltjes property, Section~\ref{sec:proof}
proves it, and the elementary estimates behind the boundary sign are in
Appendix~\ref{app:core}.

\section{The normal form and the reduction to lines}\label{sec:normal}

This section writes \(h\) as \(-1/(8M)\) for an explicit positive function
\(M\), removing the apparent singularities of~\eqref{eq:h-original}, and
then reduces the concavity of \(M\) to a Stieltjes property of \(1/M\)
along lines, Theorem~\ref{thm:line-stieltjes}.

\subsection{Resolvent integrals}

For \(x,y,z>0\), set
\begin{equation}\label{eq:TU-def}
  T(x,y,z):=\int_0^\infty
  \frac{\dd t}{(t+x)(t+y)(t+z)},
  \qquad
  U(x,y,z):=\int_0^\infty
  \frac{t\,\dd t}{(t+x)(t+y)(t+z)}.
\end{equation}
We shall usually abbreviate
\begin{equation}\label{eq:ABC-def}
  A:=c(x,y),\qquad B:=c(x,z),\qquad C:=c(y,z).
\end{equation}

\begin{lemma}\label{lem:basic-integrals}
For \(x,y,z>0\),
\begin{equation}\label{eq:c-resolvent}
 c(x,y)=\int_0^\infty\frac{\dd t}{(t+x)(t+y)},
\end{equation}
and
\begin{equation}\label{eq:AUT}
 A=U+zT,\qquad B=U+yT,\qquad C=U+xT.
\end{equation}
\end{lemma}

\begin{proof}
The identity \eqref{eq:c-resolvent} follows from partial fractions.  Then
\[
 A=\int_0^\infty
 \frac{t+z}{(t+x)(t+y)(t+z)}\dd t=U+zT,
\]
and the other two identities are cyclic variants.
\end{proof}

\subsection{Exact reduction of \texorpdfstring{\(h\)}{h}}

Define
\begin{equation}\label{eq:Q-def}
 Q(x,y,z):=\frac{2U(x,y,z)}{xyz}-T(x,y,z)^2.
\end{equation}

\begin{prop}[Nonsingular normal form]\label{prop:h-normal}
For all \(x,y,z>0\),
\begin{equation}\label{eq:h-normal}
  h(x,y,z)=-\frac{Q(x,y,z)}{6A B C}.
\end{equation}
Consequently, \(h\) has a real-analytic extension through every partial
diagonal.
\end{prop}

\begin{proof}
It suffices first to work with pairwise distinct variables.  In the
abbreviations \eqref{eq:ABC-def}, the first fraction in
\eqref{eq:h-original} reads
\begin{equation}\label{eq:first-frac-ABC}
 \frac{(y-z)AB-(x-z)AC+(x-y)BC}{6(x-y)(x-z)(y-z)ABC}.
\end{equation}
By \eqref{eq:AUT},
\[
 B=A+(y-z)T,\qquad C=A+(x-z)T.
\]
A direct expansion gives
\begin{equation}\label{eq:quadratic-cancellation}
 (y-z)AB-(x-z)AC+(x-y)BC=(x-y)(x-z)(y-z)T^2.
\end{equation}
Hence \eqref{eq:first-frac-ABC} equals \(T^2/(6ABC)\).

The numerator of the second fraction is
\[
 -xy(x-y)A+xz(x-z)B-yz(y-z)C.
\]
Substituting \eqref{eq:AUT} and simplifying yields
\begin{equation}\label{eq:linear-cancellation}
 -xy(x-y)A+xz(x-z)B-yz(y-z)C=-(x-y)(x-z)(y-z)U.
\end{equation}
Thus the second fraction equals \(-U/(3xyzABC)\), proving
\eqref{eq:h-normal}.  The integral expressions in
Lemma~\ref{lem:basic-integrals} are real-analytic on the positive cone, including
coincident variables, and \(ABC>0\); therefore the extension is analytic.
\end{proof}

\subsection{Positivity of \texorpdfstring{\(Q\)}{Q}}

Let
\[
 \Delta_2:=\{\lambda\in[0,\infty)^3:\lambda_1+\lambda_2+\lambda_3=1\}
\]
with ordinary two-dimensional Lebesgue measure, and write
\[
 s_\lambda:=\lambda_1x+\lambda_2y+\lambda_3z,
 \qquad
 I_k:=\int_{\Delta_2}s_\lambda^{-k}\dd\lambda.
\]

\begin{lemma}[Simplex representation]\label{lem:simplex}
One has
\begin{equation}\label{eq:simplex-TU}
 U=I_1,\qquad T=I_2,\qquad I_3=\frac{1}{2xyz}.
\end{equation}
Consequently,
\begin{equation}\label{eq:Q-moments}
 Q=4I_1I_3-I_2^2>0.
\end{equation}
\end{lemma}

\begin{proof}
The elementary Feynman-parameter identity
\begin{equation}\label{eq:feynman}
 \frac{1}{abc}
 =2\int_{\Delta_2}
   \frac{\dd\lambda}{(\lambda_1a+\lambda_2b+\lambda_3c)^3}
\end{equation}
holds for \(a,b,c>0\).  Applying it to
\(a=t+x,b=t+y,c=t+z\), followed by integration in \(t\), gives
\[
 T=2\int_{\Delta_2}\int_0^\infty(t+s_\lambda)^{-3}\dd t\dd\lambda
   =I_2,
\]
whereas
\[
 U=2\int_{\Delta_2}\int_0^\infty t(t+s_\lambda)^{-3}\dd t\dd\lambda
   =I_1.
\]
Putting \(t=0\) in \eqref{eq:feynman} gives
\(1/(xyz)=2I_3\).  Finally, Cauchy--Schwarz gives
\[
 I_2^2
 =\left(\int_{\Delta_2}s_\lambda^{-1/2}s_\lambda^{-3/2}\dd\lambda\right)^2
 \le I_1I_3.
\]
Thus
\(Q=4I_1I_3-I_2^2\ge3I_1I_3>0\); equivalently one may use
\(Q\ge3I_2^2>0\).
\end{proof}

\begin{rmk}
Formula \eqref{eq:feynman} can be verified directly from
\[
 \frac1{abc}=\int_{[0,\infty)^3}e^{-au-bv-cw}\dd u\dd v\dd w
\]
by the change of variables
\((u,v,w)=r(\lambda_1,\lambda_2,\lambda_3)\).
\end{rmk}

We now introduce the functions that drive the proof.

\begin{defn}\label{def:MF}
Set
\begin{equation}\label{eq:M-F-def}
 M(x,y,z):=\frac{3ABC}{4Q},
 \qquad
 F(x,y,z):=\frac1{M(x,y,z)}=\frac{4Q}{3ABC}.
\end{equation}
\end{defn}

\begin{prop}\label{prop:basic-MF}
The functions \(M\) and \(F\) are positive and symmetric on
\((0,\infty)^3\).  They are homogeneous of degrees \(1\) and \(-1\),
respectively, and
\begin{equation}\label{eq:h-M}
 h=-\frac1{8M}=-\frac{F}{8}.
\end{equation}
Moreover,
\begin{equation}\label{eq:diagonal}
 M(x,x,x)=x,
 \qquad
 h(x,x,x)=-\frac1{8x}.
\end{equation}
\end{prop}

\begin{proof}
Positivity follows from Lemma~\ref{lem:simplex}.  Symmetry follows either from
\eqref{eq:h-original} or from the integral expressions.  Under the scaling
\((x,y,z)\mapsto(r x,r y,r z)\), one has
\[
 A,B,C\mapsto r^{-1}A,r^{-1}B,r^{-1}C,
 \quad U\mapsto r^{-1}U,
 \quad T\mapsto r^{-2}T,
 \quad Q\mapsto r^{-4}Q,
\]
which proves the homogeneity statements.  Equation \eqref{eq:h-M} follows
from \eqref{eq:h-normal}.  At the diagonal,
\[
 A=B=C=\frac1x,
 \qquad U=\frac1{2x},
 \qquad T=\frac1{2x^2},
 \qquad Q=\frac{3}{4x^4},
\]
which gives \eqref{eq:diagonal}.
\end{proof}

\subsection{Reduction to a Stieltjes property along lines}\label{sec:lines}

We now reduce Theorems~\ref{thm:main} and~\ref{thm:m-main} to one
statement about the restrictions of \(F\) to lines,
Theorem~\ref{thm:line-stieltjes}.  Two elementary lemmas carry the
reduction: the reciprocal of a Stieltjes function is concave
(Lemma~\ref{lem:reciprocal-stieltjes}), and a function homogeneous of degree
one that is concave along every line with positive direction is concave
(Lemma~\ref{lem:positive-line}).  Both are proved where they are stated.
Theorem~\ref{thm:line-stieltjes} itself is proved in
Section~\ref{sec:proof}.

\begin{defn}
A function \(f:(0,\infty)\to(0,\infty)\) will be called a compactly
supported Stieltjes function if
\begin{equation}\label{eq:stieltjes-def}
 f(t)=\int_I\frac{\dd\mu(s)}{t-s}
\end{equation}
for a finite positive measure \(\mu\) supported on a compact interval
\(I\subset(-\infty,0)\).
\end{defn}

\citet{schilling2012bernstein} is the general reference for Stieltjes and
Bernstein functions.

\begin{lemma}[Reciprocal Stieltjes concavity]\label{lem:reciprocal-stieltjes}
Let \(f\) have the representation \eqref{eq:stieltjes-def}, and set
\(g=1/f\).  Then \(g''\le0\) on \((0,\infty)\).  If \(\mu\) is not a point
mass, then \(g''<0\).
\end{lemma}

\begin{proof}
Differentiating under the integral sign gives
\[
 -f'(t)=\int_I(t-s)^{-2}\dd\mu(s),
 \qquad
 \frac12f''(t)=\int_I(t-s)^{-3}\dd\mu(s).
\]
Therefore
\begin{equation}\label{eq:g-second}
 g''(t)=\frac{2f'(t)^2-f(t)f''(t)}{f(t)^3}.
\end{equation}
Cauchy--Schwarz yields
\[
 \left(\int_I(t-s)^{-2}\dd\mu(s)\right)^2
 \le
 \left(\int_I(t-s)^{-1}\dd\mu(s)\right)
 \left(\int_I(t-s)^{-3}\dd\mu(s)\right),
\]
which is exactly \(2f'^2\le ff''\).  Equality holds only when
\((t-s)^{-1/2}\) and \((t-s)^{-3/2}\) are proportional
\(\mu\)-almost everywhere, that is, only when \(s\) is constant on the
support of \(\mu\).
\end{proof}

The next elementary lemma explains why it is enough to control lines with a
positive direction.

\begin{lemma}[Concavity from positive directions]
\label{lem:positive-line}
Let \(G:(0,\infty)^n\to(0,\infty)\) be continuous and homogeneous of degree
one.  Suppose that, for every \(X,Y>0\), the map
\[
 t\longmapsto G(X+tY),\qquad t\ge0,
\]
is concave.  Then \(G\) is concave on \((0,\infty)^n\).

If, in addition, \(G\) is continuously differentiable and every such line
restriction is strictly concave whenever \(X\) and \(Y\) are not
proportional, then
\begin{equation}\label{eq:strict-superadd}
 G(X+Y)>G(X)+G(Y)
\end{equation}
for nonproportional \(X,Y>0\), and equality holds in superadditivity exactly
on common rays.
\end{lemma}

\begin{proof}
Fix \(X,Y>0\) and write \(g(t)=G(X+tY)\).  Concavity gives, for \(t>1\),
\[
 g(1)-g(0)\ge\frac{g(t)-g(1)}{t-1}.
\]
By homogeneity and continuity,
\[
 \lim_{t\to\infty}\frac{g(t)-g(1)}{t-1}=G(Y).
\]
Hence
\begin{equation}\label{eq:superadd}
 G(X+Y)\ge G(X)+G(Y).
\end{equation}
Homogeneity and \eqref{eq:superadd} imply, for \(0<\lambda<1\),
\[
 G(\lambda X+(1-\lambda)Y)
 \ge \lambda G(X)+(1-\lambda)G(Y),
\]
so \(G\) is concave.

For strictness, assume \(X,Y\) are not proportional.  Since \(g\) is
differentiable and strictly concave, its derivative is strictly decreasing.
Homogeneity gives
\[
 g(t)=tG\left(Y+\frac Xt\right),
 \qquad
 g'(t)=G\left(Y+\frac Xt\right)
       -\frac1t\dd G_{Y+X/t}(X),
\]
so \(g'(t)\to G(Y)\) as \(t\to\infty\).  Thus \(g'(t)>G(Y)\) for finite
\(t\), and
\[
 G(X+Y)-G(X)=\int_0^1g'(t)\dd t>G(Y).
\]
If \(X\) and \(Y\) are proportional, equality in \eqref{eq:superadd} follows
from homogeneity.
\end{proof}

The analytic heart of the paper is the following theorem.

\begin{theorem}[Stieltjes property along lines]
\label{thm:line-stieltjes}
For every \(X,Y\in(0,\infty)^3\), there is a finite positive measure
\(\mu_{X,Y}\), supported on a compact interval in \(( -\infty,0)\), such
that
\begin{equation}\label{eq:line-stieltjes}
 F(X+tY)=\int\frac{\dd\mu_{X,Y}(s)}{t-s},
 \qquad t\ge0.
\end{equation}
If \(X\) and \(Y\) are not proportional, the measure is not a point mass.
\end{theorem}

Assuming this theorem, the main concavity statements are immediate.

\begin{prop}\label{prop:main-from-stieltjes}
Theorem~\ref{thm:line-stieltjes} implies Theorems~\ref{thm:m-main} and~\ref{thm:main}.
\end{prop}

\begin{proof}
By Lemma~\ref{lem:reciprocal-stieltjes},
\[
 t\longmapsto M(X+tY)=\frac1{F(X+tY)}
\]
is concave, and is strictly concave unless \(X,Y\) are proportional.
By Proposition~\ref{prop:h-normal} and \eqref{eq:h-M}, \(M=-1/(8h)\) is
real-analytic, so Lemma~\ref{lem:positive-line} applies and makes \(M\)
globally concave and strictly superadditive off common rays.

For distinct \((a,b),(a',b')\in(0,\infty)^2\), set
\[
 X=\lambda(a,b,1),\qquad Y=(1-\lambda)(a',b',1).
\]
These vectors are nonproportional.  Strict superadditivity gives
\[
 m(\lambda a+(1-\lambda)a',\lambda b+(1-\lambda)b')
 >\lambda m(a,b)+(1-\lambda)m(a',b'),
\]
proving Theorem~\ref{thm:m-main}.

Finally, \(h=\phi\circ M\), where \(\phi(s)=-1/(8s)\) is increasing and
strictly concave.  If two points are not proportional, strictness already
comes from \(M\).  On a common ray, \(M\) is affine but takes distinct
endpoint values, so strictness comes from \(\phi\).  This proves
Theorem~\ref{thm:main}.
\end{proof}

\begin{rmk}\label{rmk:operator-monotone}
Theorem~\ref{thm:line-stieltjes} gives more than concavity.  Fix
\(X,Y>0\) and let \(\beta:=\min_jx_j/y_j\), so that \(X+tY>0\) exactly
for \(t>-\beta\).  The proof in Section~\ref{sec:stieltjes} places
\(\mu=\mu_{X,Y}\) in the interval \(I\) of~\eqref{eq:cut-I}, whose right
endpoint is \(-\beta\), and represents the analytic continuation \(\Fc\)
of \(t\mapsto F(X+tY)\) to \(\C\setminus I\), defined
in~\eqref{eq:calF-def}, as
\(\Fc(\zeta)=\int(\zeta-s)^{-1}\dd\mu(s)\)~\eqref{eq:measure-representation}.
For \(\imag\zeta>0\),
\[
 \imag\Fc(\zeta)
 =-\imag\zeta\int\frac{\dd\mu(s)}{|\zeta-s|^2}<0,
\]
so \(1/\Fc\) is analytic in the upper half-plane with positive imaginary
part.  It is also analytic across \((-\beta,\infty)\), where it is real and
equals \(M(X+tY)\).  By Löwner's
theorem~\citep[Chapter~V]{bhatia1997matrix}, \(t\mapsto M(X+tY)\) is
therefore operator monotone on \((-\beta,\infty)\).  Operator monotone
functions are also the functions that classify monotone
metrics~\citep{petz1996monotone}.  Here they appear in a different role,
along lines in the space of eigenvalue triples.  The proof of
Theorem~\ref{thm:main} uses only the concavity.
\end{rmk}

\section{The Stieltjes property along lines}\label{sec:proof}

This section proves Theorem~\ref{thm:line-stieltjes}.  The theorem asks
for a Stieltjes representation rather than for concavity directly because
the representation turns a second-order condition into a boundary sign.
Concavity of \(M\) along a line is a condition on second derivatives, and
the second derivatives of \(F=4Q/(3ABC)\) are unwieldy.  If instead the
continuation \(\Fc\) of \(t\mapsto F(X+tY)\), defined
in~\eqref{eq:calF-def}, is analytic off a compact interval \(I\) of the
negative axis, decays at infinity, and has nonpositive imaginary part on the
upper edge of \(I\), then Cauchy's formula on a keyhole contour writes it
as the integral of \(1/(\zeta-s)\) against the positive measure
\(-\pi^{-1}\imag\Fc(s+\ii0)\dd s\), and
Lemma~\ref{lem:reciprocal-stieltjes} converts this representation into
concavity along the line.  The analytic continuation comes from the
resolvent integrals of Section~\ref{sec:normal}, and on the cut the
logarithms in \(A\), \(B\), \(C\) take explicit values, so the sign of
the boundary value is a question about one explicit real function.

Fix \(X,Y>0\).  The proof has three steps, one per subsection, in this
order.
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2em]
\item \emph{Continuation.}  The function \(t\mapsto F(X+tY)\) continues
      analytically, as the function \(\Fc\) of~\eqref{eq:calF-def}, to the
      complement of a compact interval \(I\) of the negative real axis, and
      decays like \(F(Y)/\zeta\) at infinity (Lemma~\ref{lem:analyticity}).
\item \emph{Boundary sign.}  Away from the branch points, its upper boundary
      value on \(I\) has negative imaginary part.  Symmetry, homogeneity
      and Schwarz reflection reduce this to boundary triples \((-1,p,q)\), and there to the
      positivity of an explicit function of two real variables
      (Section~\ref{sec:boundary});
      the elementary estimates that prove this positivity are in
      Appendix~\ref{app:core}.
\item \emph{Reconstruction.}  A keyhole contour recovers the function from
      its boundary values; the representing measure has density
      \(-\pi^{-1}\imag\Fc(s+\ii0)\), which is positive by step~(2).
      Coincident branch points follow by approximation
      (Section~\ref{sec:stieltjes}).
\end{enumerate}

\subsection{Analytic continuation along a line}\label{sec:continuation}

To prove Theorem~\ref{thm:line-stieltjes} we continue \(t\mapsto F(X+tY)\)
from \(t\ge0\) to the complex plane cut along an interval of the negative
axis.  Fix
\[
 X=(x_1,x_2,x_3),\qquad Y=(y_1,y_2,y_3)\in(0,\infty)^3,
\]
and define
\begin{equation}\label{eq:Z-line}
 Z_j(\zeta):=x_j+\zeta y_j,
 \qquad
 \alpha_j:=\frac{x_j}{y_j}>0.
\end{equation}
Let
\begin{equation}\label{eq:cut-I}
 I:=[-\max_j\alpha_j,-\min_j\alpha_j]\subset(-\infty,0),
\end{equation}
and set
\begin{equation}\label{eq:calF-def}
 \Fc(\zeta):=F(Z_1(\zeta),Z_2(\zeta),Z_3(\zeta)).
\end{equation}

The integral formulas admit analytic continuation to the slit plane.  For
complex arguments, use
\begin{equation}\label{eq:c-segment-complex}
 c(z_i,z_j)=\int_0^1
 \frac{\dd\lambda}{(1-\lambda)z_i+\lambda z_j},
\end{equation}
which agrees with the divided difference of the principal logarithm wherever
the latter expression is directly available.  Similarly,
\begin{equation}\label{eq:UT-simplex-complex}
 U(z_1,z_2,z_3)=\int_{\Delta_2}
 \frac{\dd\lambda}{\lambda_1z_1+\lambda_2z_2+\lambda_3z_3},
\end{equation}
\begin{equation}\label{eq:T-simplex-complex}
 T(z_1,z_2,z_3)=\int_{\Delta_2}
 \frac{\dd\lambda}{(\lambda_1z_1+\lambda_2z_2+\lambda_3z_3)^2}.
\end{equation}

\begin{lemma}[Slit-plane analyticity]\label{lem:analyticity}
The function \(\Fc\) is analytic on \(\C\setminus I\), satisfies
\begin{equation}\label{eq:schwarz}
 \Fc(\overline\zeta)=\overline{\Fc(\zeta)},
\end{equation}
and has the expansion
\begin{equation}\label{eq:infinity-asymptotic}
 \Fc(\zeta)=\frac{F(Y)}{\zeta}+O(|\zeta|^{-2})
 \qquad (\zeta\to\infty).
\end{equation}
\end{lemma}

\begin{proof}
Every possible zero of a denominator in
\eqref{eq:c-segment-complex}--\eqref{eq:T-simplex-complex} has the form
\[
 -\frac{\sum_j\lambda_jx_j}{\sum_j\lambda_jy_j},
\]
which lies in \(I\).  Hence \(A,B,C,U,T\) are analytic on \(\C\setminus I\).
The factor \(1/(Z_1Z_2Z_3)\) in \(Q\) is analytic there as well, because its
poles \(-\alpha_j\) lie in \(I\).  Therefore \(Q\) is analytic on
\(\C\setminus I\).

We must also check that the factors \(c(Z_i,Z_j)\) do not vanish off the cut.
If \(\imag\zeta>0\), then every denominator in
\eqref{eq:c-segment-complex} has positive imaginary part, so its reciprocal
has strictly negative imaginary part.  Thus
\[
 \imag c(Z_i(\zeta),Z_j(\zeta))<0.
\]
The lower-half-plane assertion follows by conjugation.  On either real
component of \(\reals\setminus I\), all affine denominators have the same
nonzero sign, so the integral is again nonzero.  This proves analyticity of
\(\Fc\).  Equation \eqref{eq:schwarz} is immediate.

Finally, degree \(-1\) homogeneity gives
\[
 \Fc(\zeta)
 =\frac1\zeta F\left(Y+\frac{X}{\zeta}\right),
\]
and smoothness of \(F\) near \(Y\) yields
\eqref{eq:infinity-asymptotic}.
\end{proof}

The remaining task is to prove
\begin{equation}\label{eq:boundary-target}
 \imag\Fc(s+\ii0)\le0,
 \qquad s\in I.
\end{equation}
Section~\ref{sec:boundary} establishes this sign
(Lemma~\ref{lem:all-boundary-sign}).

\subsection{The boundary sign}\label{sec:boundary}

We now prove the sign condition~\eqref{eq:boundary-target}.
At a non-branch point of the cut, the real limiting triple
\((Z_1(s),Z_2(s),Z_3(s))\) has mixed signs.  The case with two negative
coordinates reduces to the case with one by reflection
(Lemma~\ref{lem:all-boundary-sign}).  By permutation symmetry and positive
homogeneity, the case with exactly one negative coordinate reduces to
\begin{equation}\label{eq:normalized-boundary}
 (-1,p,q),\qquad p>q>0.
\end{equation}
The case \(p=q\) will be obtained by a limit.  We write \(F_+(Z)\) and
\(F_-(Z)\) for the upper and lower boundary values of \(F\) at a real
triple \(Z\) with mixed signs, so that
\(\Fc(s\pm\ii0)=F_\pm(Z_1(s),Z_2(s),Z_3(s))\).

Put
\begin{equation}\label{eq:uvd}
 u:=\log p,
 \qquad
 v:=\log q,
 \qquad
 \delta:=u-v>0,
 \qquad
 d:=p-q>0.
\end{equation}
For the upper boundary value, the principal logarithm gives, in the
abbreviations \eqref{eq:ABC-def},
\begin{equation}\label{eq:boundary-ABC}
 A=c(-1,p)=\frac{u-\ii\pi}{1+p},
 \qquad
 B=c(-1,q)=\frac{v-\ii\pi}{1+q},
 \qquad
 C=c(p,q)=\frac{\delta}{d}>0.
\end{equation}
The relations \eqref{eq:AUT} become
\begin{equation}\label{eq:boundary-TU}
 T=\frac{B-A}{d},
 \qquad
 U=\frac{pA-qB}{d}.
\end{equation}
Since \(xyz=-pq\),
\begin{equation}\label{eq:Q-over-AB}
 \frac{Q}{AB}
 =-\frac{2}{pqd}\left(\frac pB-\frac qA\right)
  -\frac1{d^2}\left(\frac BA+\frac AB-2\right).
\end{equation}

The required sign is encoded by one real expression.

\begin{lemma}[Boundary numerator]\label{lem:boundary-N}
Define
\begin{align}
 N(p,q):={}&(1+q)^2(u^2+\pi^2)
 \left(\delta+\frac{2d(1+p)}{q}\right)
 \notag\\
 &-(1+p)^2(v^2+\pi^2)
 \left(\delta+\frac{2d(1+q)}{p}\right).
 \label{eq:N-def}
\end{align}
Then
\begin{equation}\label{eq:ImQAB}
 \frac1\pi\imag\frac{Q}{AB}
 =-
 \frac{N(p,q)}
 {(u^2+\pi^2)(v^2+\pi^2)(1+p)(1+q)d^2}.
\end{equation}
Consequently,
\begin{equation}\label{eq:sign-equivalence}
 N(p,q)>0
 \quad\Longrightarrow\quad
 \imag F_+(-1,p,q)<0.
\end{equation}
\end{lemma}

\begin{proof}
From \eqref{eq:boundary-ABC},
\[
 \imag\frac1A=\frac{\pi(1+p)}{u^2+\pi^2},
 \qquad
 \imag\frac1B=\frac{\pi(1+q)}{v^2+\pi^2},
\]
while
\[
 \imag\frac BA
 =-\frac{\pi\delta(1+p)}{(1+q)(u^2+\pi^2)},
 \qquad
 \imag\frac AB
 =\frac{\pi\delta(1+q)}{(1+p)(v^2+\pi^2)}.
\]
Substitution in \eqref{eq:Q-over-AB} gives
\eqref{eq:ImQAB}.  Since
\[
 F_+(-1,p,q)=\frac{4}{3C}\frac{Q}{AB}
\]
and \(C>0\), the sign implication follows.
\end{proof}

We now convert \(N\) into hyperbolic coordinates.  Write
\begin{equation}\label{eq:hyperbolic-coordinates}
 p=we^\rho,
 \qquad
 q=we^{-\rho},
 \qquad
 w=e^\sigma,
 \qquad
 \rho>0.
\end{equation}
Then
\[
 u=\sigma+\rho,
 \qquad
 v=\sigma-\rho.
\]
Set
\begin{equation}\label{eq:SC-E}
 S:=\sinh\rho,
 \qquad
 \gamma:=\cosh\rho,
 \qquad
 E:=1+2w\gamma+w^2,
\end{equation}
\begin{equation}\label{eq:Dcal-def}
 \Dc:=ES-\rho w(1+w\gamma),
\end{equation}
and
\begin{equation}\label{eq:Bcal-def}
 \Bc:=
 \rho\bigl(1+2w\gamma+w^2\cosh(2\rho)\bigr)
 +2SE(w+\gamma).
\end{equation}

\begin{lemma}[Hyperbolic normal form]\label{lem:N-hyperbolic}
With the notation above,
\begin{equation}\label{eq:N-hyperbolic}
 \boxed{
 N=8\left[
 S\Dc(\sigma^2+\rho^2+\pi^2)
 +\rho\sigma\Bc
 \right].}
\end{equation}
\end{lemma}

\begin{proof}
Insert \eqref{eq:hyperbolic-coordinates} into \eqref{eq:N-def}, use
\(d=2wS\) and \(\delta=2\rho\), and collect the coefficients of
\(\sigma^2,\sigma\), and \(\rho^2+\pi^2\).  The resulting identity is
\eqref{eq:N-hyperbolic}.  No inequality is used in this step.
\end{proof}

The proof of the boundary sign is therefore reduced to the following
real-variable proposition.

\begin{prop}[Core inequality]\label{prop:core-N}
For every \(\rho>0\) and every \(\sigma\in\reals\), the right-hand side of
\eqref{eq:N-hyperbolic} is strictly positive.
\end{prop}

Proposition~\ref{prop:core-N} is proved in Appendix~\ref{app:core}.  The
proof splits according to the sign of \(\sigma\), because the second term
in~\eqref{eq:N-hyperbolic} changes sign with \(\sigma\).  For \(\sigma\le0\)
the factor \(\Dc\) is positive, and dividing \(N\) by \(8S\Dc\) leaves a
quadratic in \(\tau=-\sigma\) whose linear coefficient is bounded by its
value at \(w=0\) plus \(16w\).  For \(\sigma>0\) only the case \(\Dc<0\)
needs an argument; that case forces \(\tau\) beyond the threshold
\(\rho-\log\rho\), where a lower bound for \(N\) that increases in \(\tau\)
is already positive.  The case \(p=q\), that is \(\rho\downarrow0\), is
decided by the limit of \(N/(8\rho^2)\), computed in
Appendix~\ref{app:coalescent}.  Together these give the following.

\begin{theorem}[Boundary sign with one negative coordinate]
\label{thm:one-negative-boundary}
For every \(p,q>0\), the upper boundary value satisfies
\begin{equation}\label{eq:one-negative-sign}
 \imag F_+(-1,p,q)<0.
\end{equation}
\end{theorem}

\begin{lemma}\label{lem:all-boundary-sign}
For every interior point \(s\in I\) that is not a branch point,
\begin{equation}\label{eq:all-boundary-sign}
 \imag\Fc(s+\ii0)<0.
\end{equation}
\end{lemma}

\begin{proof}
If exactly one limiting coordinate is negative, positive scaling and
permutation symmetry reduce the boundary value to
Theorem~\ref{thm:one-negative-boundary}.

Suppose instead that two coordinates are negative, and write
\(Z:=(Z_1(s),Z_2(s),Z_3(s))\).  Multiplication of the upper boundary triple
by \(-1\) produces the lower boundary value at \(-Z\), which has exactly
one negative coordinate.  The integral definitions
\eqref{eq:c-segment-complex}--\eqref{eq:T-simplex-complex} show directly
that the analytic continuation of \(F\) remains homogeneous of degree
\(-1\), even for the scalar \(-1\).  Therefore
\[
 F_+(Z)=-F_-(-Z).
\]
By Schwarz reflection,
\(F_-(-Z)=\overline{F_+(-Z)}\).  Theorem~\ref{thm:one-negative-boundary}
then gives \(\imag F_+(Z)<0\).
\end{proof}

\subsection{The Stieltjes representation}\label{sec:stieltjes}

We now return to the restriction \(\Fc\) of \(F\) to the line
\(X+\zeta Y\), defined in~\eqref{eq:calF-def}, and complete the proof of
Theorem~\ref{thm:line-stieltjes}.

\emph{\textbf{Behavior at branch points.}}
Assume first that the three numbers \(\alpha_j\) in
\eqref{eq:Z-line} are distinct.  Near a branch point
\(\zeta_0=-\alpha_j\), precisely one coordinate, say \(Z_1\), tends to
zero while the other two tend to nonzero limits.  If
\(\xi=Z_1(\zeta)\), then
\begin{equation}\label{eq:c-zero-asymptotic}
 c(\xi,r)=-\frac{\Log\xi}{r}+O(1)
 \qquad (\xi\to0),
\end{equation}
for every fixed \(r\ne0\) in a slit neighborhood.  Consequently,
\begin{equation}\label{eq:branch-asymptotic}
 \Fc(\zeta)
 =O\left(
 \frac{1}{(\zeta-\zeta_0)\Log^2(\zeta-\zeta_0)}
 \right).
\end{equation}
Indeed, the two \(c\)-factors involving \(\xi\) each grow logarithmically,
the third tends to a nonzero limit, \(T=O(\Log\xi)\), and
\(U\) stays bounded; the term \(2U/(Z_1Z_2Z_3)\) is therefore
\(O(1/\xi)\) and dominates the possible logarithmic-square growth of
\(T^2\).  In particular, the integral of \(\Fc\) over a circle of radius
\(\eps\) around \(\zeta_0\) is
\begin{equation}\label{eq:small-circle}
 O\left(\frac1{|\log\eps|^2}\right)\longrightarrow0.
\end{equation}

\emph{\textbf{Keyhole reconstruction.}}
\begin{prop}[Cauchy representation for distinct branch points]
\label{prop:keyhole}
If the \(\alpha_j\) are pairwise distinct, then
\begin{equation}\label{eq:measure-representation}
 \Fc(\zeta)=\int_I\frac{\dd\mu(s)}{\zeta-s},
 \qquad \zeta\in\C\setminus I,
\end{equation}
where
\begin{equation}\label{eq:density}
 \dd\mu(s)=-\frac1\pi\imag\Fc(s+\ii0)\dd s.
\end{equation}
The measure is finite, positive, and has total mass
\begin{equation}\label{eq:mass}
 \mu(I)=F(Y).
\end{equation}
\end{prop}

\begin{proof}
Apply Cauchy's formula to a large circle with a thin keyhole removed around
\(I\), and remove small circles around the three branch points.  The large
circle contributes zero in the limit by \eqref{eq:infinity-asymptotic}; the
small circles contribute zero by \eqref{eq:small-circle}.  The two lips of
the cut give
\[
 \Fc(\zeta)
 =\frac{1}{2\pi\ii}\int_I
 \frac{\Fc(s+\ii0)-\Fc(s-\ii0)}{s-\zeta}\dd s.
\]
Using Schwarz reflection and rearranging yields
\eqref{eq:measure-representation}--\eqref{eq:density}.  Positivity follows
from Lemma~\ref{lem:all-boundary-sign}.  Finally, comparing the coefficient of
\(1/\zeta\) at infinity in \eqref{eq:measure-representation} with
\eqref{eq:infinity-asymptotic} gives \eqref{eq:mass}.
\end{proof}

\emph{\textbf{Coincident branch points.}}
If two or three \(\alpha_j\) coincide, perturb \(X,Y\) slightly so that the
ratios become distinct while remaining positive.  The resulting measures
are supported in a common compact negative interval and have masses tending
to \(F(Y)\), hence uniformly bounded.  By weak compactness, a subsequence
converges to a finite positive measure \(\mu\).  The corresponding Stieltjes
transforms converge locally uniformly off the interval, while the analytic
functions \(F(X^{(n)}+\zeta Y^{(n)})\) converge locally uniformly to
\(F(X+\zeta Y)\).  Thus \eqref{eq:measure-representation} passes to the
limit.  Uniqueness of the Stieltjes transform makes the limiting measure
independent of the subsequence.

\emph{\textbf{Point masses.}}
If all three ratios coincide, then \(X=\kappa Y\) for some \(\kappa>0\), and
homogeneity gives the explicit point-mass representation
\begin{equation}\label{eq:proportional-atom}
 F(X+tY)=\frac{F(Y)}{t+\kappa}.
\end{equation}
If the ratios are not all equal, the cut contains a nondegenerate mixed-sign
interval.  On every compact subinterval avoiding the branch points, the
Stieltjes inversion formula identifies the absolutely continuous part as
\[
 \frac{\dd\mu}{\dd s}(s)
 =-\frac1\pi\imag\Fc(s+\ii0).
\]
By Lemma~\ref{lem:all-boundary-sign}, this density is strictly positive there;
therefore the representing measure is not a point mass.

\emph{\textbf{Conclusion.}}  This completes the proof of
Theorem~\ref{thm:line-stieltjes}, and hence, by
Proposition~\ref{prop:main-from-stieltjes}, the proofs of
Theorems~\ref{thm:m-main} and~\ref{thm:main}.

\begin{aiusage}  %
The results in this paper were developed with an AI assistant (OpenAI's
GPT~Pro). None of the claims were obtained zero-shot: they emerged through
sustained back-and-forth discussion, including failed attempts,
counterexamples, audits, and repairs.
\end{aiusage}

\aiacknowledgements

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\bibitem[Andai(2003{\natexlab{a}})]{andai2003monotonicity}
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Attila Andai.
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\newblock \emph{Matrix Analysis}.
\newblock Graduate Texts in Mathematics 169. Springer, New York, 1997.
\newblock \doi{10.1007/978-1-4612-0653-8}.

\bibitem[Dittmann(2000)]{dittmann2000curvature}
Jochen Dittmann.
\newblock On the curvature of monotone metrics and a conjecture concerning the
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\bibitem[Gibilisco and Isola(2005)]{gibilisco2005monotonicity}
Paolo Gibilisco and Tommaso Isola.
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\end{thebibliography}

\appendix

\section{Proof of the core inequality}\label{app:core}

This appendix proves Proposition~\ref{prop:core-N} and treats the case
\(p=q\).  We split according to the sign of \(\sigma\).  The two halves
require different estimates because the second term in
\eqref{eq:N-hyperbolic} changes sign.

\subsection{The half-plane \texorpdfstring{\(\sigma\le0\)}{sigma <= 0}}

The case \(\sigma=0\) is immediate: then \(w=1\), and
\begin{equation}\label{eq:D-at-one}
 \Dc=(\gamma+1)(2S-\rho)>0.
\end{equation}
Assume henceforth
\[
 \sigma=-\tau,
 \qquad
 \tau>0,
 \qquad
 w=e^{-\tau}\in(0,1).
\]

\begin{lemma}\label{lem:D-positive-negative-sigma}
For \(0\le w\le1\), one has \(\Dc>0\).
\end{lemma}

\begin{proof}
Expanding \eqref{eq:Dcal-def} as a quadratic in \(w\) gives
\begin{equation}\label{eq:D-quadratic}
 \Dc
 =S+(2\gamma S-\rho)w+(S-\rho \gamma)w^2.
\end{equation}
Since \(\rho \gamma-S>0\), this quadratic is concave.  Its endpoint values
are
\[
 \Dc(0)=S>0,
 \qquad
 \Dc(1)=(\gamma+1)(2S-\rho)>0.
\]
A concave function lies above its endpoint chord, proving the claim.
\end{proof}

Define
\begin{equation}\label{eq:R-def}
 R(\rho,w):=\frac{\rho\Bc}{S\Dc},
 \qquad
 R_0(\rho):=R(\rho,0)
 =2\rho\coth\rho+\rho^2\operatorname{csch}^2\rho.
\end{equation}
Then \eqref{eq:N-hyperbolic} becomes
\begin{equation}\label{eq:N-negative-reduced}
 \frac{N}{8S\Dc}
 =\tau^2+\rho^2+\pi^2-\tau R(\rho,w).
\end{equation}

\begin{lemma}[Comparison with the value at \(w=0\)]
\label{lem:R-comparison}
For \(\rho>0\) and \(0\le w\le1\),
\begin{equation}\label{eq:R-comparison}
 R(\rho,w)\le R_0(\rho)+16w.
\end{equation}
\end{lemma}

\begin{proof}
Set
\begin{equation}\label{eq:Delta-def}
 \Delta
 :=S\bigl[(R_0+16w)S\Dc-\rho\Bc\bigr].
\end{equation}
It is enough to prove \(\Delta\ge0\).  Put
\[
 \theta:=\frac{\rho}{S}\in(0,1).
\]
A direct expansion in the Bernstein basis on \([0,1]\) gives
\begin{equation}\label{eq:Delta-Bernstein}
 \Delta
 =w\left[
 \beta_0(1-w)^2+2\beta_1w(1-w)+\beta_2w^2
 \right],
\end{equation}
where
\begin{equation}\label{eq:beta0}
 \frac{\beta_0}{S^3}
 =16-2\theta-2\gamma\theta^2-\theta^3,
\end{equation}
\begin{align}
 \frac{2\beta_1}{S^3}
 ={}&32(\gamma+1)-4\gamma\theta-4\gamma\theta^2
     -\gamma\theta^3
 \notag\\
 &-4\rho^2-20\theta-2\theta^2-2\theta^3,
 \label{eq:beta1}
\end{align}
and
\begin{equation}\label{eq:beta2}
 \frac{\beta_2}{S^3}
 =(\gamma+1)(32-20\theta-2\theta^2-\theta^3)-4\rho^2.
\end{equation}

We use only
\begin{equation}\label{eq:elementary-hyperbolic-bounds}
 \rho^2\le2(\gamma-1),
 \qquad
 \gamma\theta^2
 =\frac{\gamma\rho^2}{S^2}
 \le\frac{2\gamma}{\gamma+1}\le2.
\end{equation}
The first inequality is the quadratic Taylor lower bound for \(\cosh\rho\).
Equations \eqref{eq:beta0} and \eqref{eq:elementary-hyperbolic-bounds} give
\[
 \beta_0\ge9S^3.
\]
For \(\beta_1\), use
\(\theta\le1\),
\(\gamma\theta^3\le \gamma\theta^2\le2\), and
\(4\rho^2\le8(\gamma-1)\); then
\[
 2\beta_1\ge(20\gamma+6)S^3>0.
\]
Finally,
\[
 32-20\theta-2\theta^2-\theta^3\ge9,
\]
so
\[
 \beta_2
 \ge S^3\{9(\gamma+1)-8(\gamma-1)\}
 =(\gamma+17)S^3>0.
\]
All Bernstein coefficients in \eqref{eq:Delta-Bernstein} are positive;
therefore \(\Delta\ge0\), proving \eqref{eq:R-comparison}.
\end{proof}

\begin{lemma}[A bound for \(R_0\)]\label{lem:R0-bound}
For every \(\rho>0\),
\begin{equation}\label{eq:R0-bound}
 R_0(\rho)^2\le4\rho^2+9.
\end{equation}
\end{lemma}

\begin{proof}
Let \(\theta=\rho/S\), as in the proof of Lemma~\ref{lem:R-comparison}.
Since \(\theta\gamma=\sqrt{\rho^2+\theta^2}\),
\begin{equation}\label{eq:R0-a}
 R_0=2\sqrt{\rho^2+\theta^2}+\theta^2.
\end{equation}
The estimate
\(
 \sinh\rho\ge\rho+\rho^3/6
\)
implies
\begin{equation}\label{eq:a-bound}
 \theta^2\le\frac{3}{\rho^2+3}.
\end{equation}
Indeed,
\((1+\rho^2/6)^{-2}\le3/(\rho^2+3)\).
The right-hand side of \eqref{eq:R0-a} is increasing in \(\theta^2\).  It is
therefore enough, with
\[
 t=\rho^2,
 \qquad
 \nu=\frac3{t+3},
\]
to prove
\begin{equation}\label{eq:sqrt-ineq}
 2\sqrt{t+\nu}+\nu\le\sqrt{4t+9}.
\end{equation}
The quantities on the right after each rearrangement below are nonnegative.
Squaring twice reduces \eqref{eq:sqrt-ineq} to
\[
 (9+\nu^2-4\nu)^2-4\nu^2(4t+9)\ge0.
\]
After substituting \(\nu=3/(t+3)\), the left-hand side is
\begin{equation}\label{eq:R0-polynomial}
 \frac{9t(t+2)(9t^2+50t+72)}{(t+3)^4}\ge0.
\end{equation}
This proves \eqref{eq:R0-bound}.
\end{proof}

\begin{proof}[\textbf{Completion for \(\sigma<0\)}]
By Lemma~\ref{lem:R-comparison},
\begin{align*}
 \tau^2+\rho^2+\pi^2-\tau R
 &\ge
 \tau^2+\rho^2+\pi^2-\tau R_0-16\tau e^{-\tau}
 \\
 &=\left(\tau-\frac{R_0}{2}\right)^2
   +\rho^2+\pi^2-\frac{R_0^2}{4}-16\tau e^{-\tau}.
\end{align*}
Using Lemma~\ref{lem:R0-bound} and
\(\tau e^{-\tau}\le1/e\),
\begin{equation}\label{eq:negative-final-bound}
 \tau^2+\rho^2+\pi^2-\tau R
 \ge\pi^2-\frac94-\frac{16}{e}>0.
\end{equation}
For example, \(\pi>3\) and \(e>8/3\) make the final lower bound exceed
\(3/4\).  Equations \eqref{eq:N-negative-reduced} and
Lemma~\ref{lem:D-positive-negative-sigma} now give \(N>0\).
\end{proof}

\subsection{The half-plane \texorpdfstring{\(\sigma>0\)}{sigma > 0}}

Write
\[
 \sigma=\tau>0,
 \qquad
 w=e^\tau>1.
\]
Set
\begin{equation}\label{eq:a-d-rho}
 a_\rho:=\rho \gamma-S>0,
 \qquad
 d_\rho:=2\gamma S-\rho>0.
\end{equation}
Then
\begin{equation}\label{eq:D-positive-case-expansion}
 \Dc=S+d_\rho w-a_\rho w^2.
\end{equation}
Also,
\begin{equation}\label{eq:B-positive-expansion}
 \Bc=b_0+b_1w+b_2w^2+2Sw^3,
\end{equation}
where
\begin{equation}\label{eq:b-coeffs}
 b_0=\rho+2\gamma S>0,
\end{equation}
\begin{equation}\label{eq:b1}
 b_1=4\gamma^2S+2\gamma\rho+2S>0,
\end{equation}
and
\begin{equation}\label{eq:b2}
 b_2=\rho\cosh(2\rho)+6\gamma S>0.
\end{equation}
If \(\Dc\ge0\), both terms in \eqref{eq:N-hyperbolic} are positive and
there is nothing to prove.  Assume from now on that
\begin{equation}\label{eq:D-negative-assumption}
 \Dc<0.
\end{equation}
Discarding the positive \(b_0,b_1\) terms in \(\Bc\), and the positive
\(S+d_\rho w\) part of \(\Dc\), gives
\begin{equation}\label{eq:N-lower-P}
 \frac N8>w^2P_\rho(\tau),
\end{equation}
where
\begin{equation}\label{eq:P-def}
 P_\rho(t)
 :=\rho t\bigl(b_2+2Se^t\bigr)
   -Sa_\rho(t^2+\rho^2+\pi^2).
\end{equation}
We shall show \(P_\rho(\tau)>0\).

\begin{lemma}[The negative-\(\Dc\) threshold]\label{lem:D-threshold}
Assumption \eqref{eq:D-negative-assumption} implies
\begin{equation}\label{eq:tau-threshold}
 \tau>t_*:=\rho-\log\rho.
\end{equation}
Moreover, \(t_*\ge1\).
\end{lemma}

\begin{proof}
From \eqref{eq:D-positive-case-expansion} and \(\Dc<0\),
\[
 a_\rho w^2>S+d_\rho w>d_\rho w,
\]
so \(w>d_\rho/a_\rho\).  We claim
\begin{equation}\label{eq:d-over-a}
 \frac{d_\rho}{a_\rho}\ge\frac{e^\rho}{\rho}.
\end{equation}
Indeed,
\begin{align}
 \rho d_\rho-e^\rho a_\rho
 =\frac12\bigl[
 e^{2\rho}-1-\rho(1+e^{-2\rho})-2\rho^2
 \bigr]\ge0,
 \label{eq:d-over-a-proof}
\end{align}
where we used
\(e^{2\rho}\ge1+2\rho+2\rho^2\) and
\(1+e^{-2\rho}\le2\).  Thus
\[
 e^\tau=w>\frac{e^\rho}{\rho},
\]
which is \eqref{eq:tau-threshold}.  Finally,
\(\rho-\log\rho\ge1\) because this function has its global minimum at
\(\rho=1\).
\end{proof}

\begin{lemma}[Monotonicity of \(P_\rho\)]\label{lem:P-increasing}
For every \(\rho>0\), the function \(P_\rho(t)\) is strictly increasing on
\((0,\infty)\).
\end{lemma}

\begin{proof}
First,
\begin{equation}\label{eq:b2-basic}
 b_2>2Sa_\rho.
\end{equation}
Indeed, direct simplification gives
\begin{equation}\label{eq:b2-basic-exact}
 b_2-2Sa_\rho
 =2e^{2\rho}-1+(\rho-1)e^{-2\rho}>0.
\end{equation}
For \(\rho\ge1\) positivity is immediate.  If \(0<\rho<1\), then
\(2e^{2\rho}-1>1\) and
\((\rho-1)e^{-2\rho}>\rho-1\), so the sum exceeds \(\rho\).

Differentiating \eqref{eq:P-def},
\begin{equation}\label{eq:P-prime}
 P_\rho'(t)
 =\rho b_2+2\rho Se^t(t+1)-2Sa_\rho t.
\end{equation}
If \(t\le\rho\), \eqref{eq:b2-basic} gives
\[
 P_\rho'(t)
 >2Sa_\rho(\rho-t)+2\rho Se^t(t+1)>0.
\]
If \(t>\rho\), then
\[
 a_\rho=\rho \gamma-S<\rho e^\rho<\rho e^t,
\]
so
\[
 \rho e^t(t+1)-a_\rho t
 =t(\rho e^t-a_\rho)+\rho e^t>0.
\]
Equation \eqref{eq:P-prime} is again positive.
\end{proof}

We need a somewhat stronger lower bound for \(b_2\).

\begin{lemma}\label{lem:b2-strong}
For every \(\rho>0\),
\begin{equation}\label{eq:b2-strong}
 b_2>\left(2+\frac8\rho\right)Sa_\rho.
\end{equation}
\end{lemma}

\begin{proof}
An exact calculation gives
\begin{align}
 \rho\left[
 b_2-\left(2+\frac8\rho\right)Sa_\rho
 \right]
 ={}&(\rho^2+\rho)e^{-2\rho}-\rho
 \notag\\
 &+4(\cosh(2\rho)-1).
 \label{eq:b2-strong-exact}
\end{align}
If \(0<\rho\le3\), use
\(e^{-2\rho}\ge1-2\rho\) and
\(\cosh(2\rho)-1\ge2\rho^2\).  The right-hand side of
\eqref{eq:b2-strong-exact} is then at least
\[
 7\rho^2-2\rho^3=\rho^2(7-2\rho)>0.
\]
If \(\rho\ge3\), discard the first positive term and use
\[
 4(\cosh(2\rho)-1)-\rho\ge8\rho^2-\rho>0.
\]
\end{proof}

\begin{lemma}[Positivity at the threshold]\label{lem:P-threshold-positive}
For \(t_*=\rho-\log\rho\), one has
\begin{equation}\label{eq:P-threshold-positive}
 P_\rho(t_*)>0.
\end{equation}
\end{lemma}

\begin{proof}
Since \(e^{t_*}=e^\rho/\rho\), Lemma~\ref{lem:b2-strong} and
\eqref{eq:P-def} give
\begin{align}
 \frac{P_\rho(t_*)}{Sa_\rho}
 >{}&2\rho t_*+8t_*
 +\frac{2t_*e^\rho}{a_\rho}
 -t_*^2-\rho^2-\pi^2
 \notag\\
 ={}&8t_*+\frac{2t_*e^\rho}{a_\rho}
 -(\log\rho)^2-\pi^2.
 \label{eq:P-threshold-step1}
\end{align}
Furthermore,
\begin{equation}\label{eq:a-upper}
 a_\rho<\frac{\rho e^\rho}{2},
\end{equation}
because
\[
 \rho e^\rho-2a_\rho
 =e^\rho-(\rho+1)e^{-\rho}>0.
\]
Thus
\begin{equation}\label{eq:P-lower-B}
 \frac{P_\rho(t_*)}{Sa_\rho}>\Psi(\rho),
\end{equation}
where
\begin{equation}\label{eq:B-rho-def}
 \Psi(\rho)
 :=8(\rho-\log\rho)
   +\frac{4(\rho-\log\rho)}{\rho}
   -(\log\rho)^2-\pi^2.
\end{equation}

We prove \(\Psi(\rho)>0\) in two ranges.  If \(0<\rho\le1\), write
\(\rho=e^{-s}\), \(s\ge0\).  Then
\[
 \Psi(\rho)=8e^{-s}+8s+4+4se^s-s^2-\pi^2.
\]
Its derivative with respect to \(s\) is
\[
 8(1-e^{-s})+4e^s(1+s)-2s>0.
\]
Therefore
\begin{equation}\label{eq:B-small-rho}
 \Psi(\rho)\ge \Psi(1)=12-\pi^2>0.
\end{equation}

If \(\rho\ge1\), write \(\rho=e^s\), \(s\ge0\).  Then
\[
 \Psi(\rho)=8e^s-8s+4-4se^{-s}-s^2-\pi^2.
\]
Using
\[
 e^s\ge1+s+\frac{s^2}{2}+\frac{s^3}{6},
 \qquad
 e^{-s}\le1,
\]
we obtain
\begin{align}
 \Psi(\rho)
 &\ge12-4s+3s^2+\frac43s^3-\pi^2
 \notag\\
 &=\frac{131}{12}
   +\frac{(2s-1)^2(4s+13)}{12}
   -\pi^2>0,
 \label{eq:B-large-rho}
\end{align}
where the last inequality uses \(\pi^2<10\).  This proves
\eqref{eq:P-threshold-positive}.
\end{proof}

\begin{proof}[\textbf{Completion for \(\sigma>0\)}]
Under \(\Dc<0\), Lemma~\ref{lem:D-threshold} gives \(\tau>t_*\).
By Lemmas~\ref{lem:P-increasing} and~\ref{lem:P-threshold-positive},
\[
 P_\rho(\tau)>P_\rho(t_*)>0.
\]
Equation \eqref{eq:N-lower-P} implies \(N>0\).  Together with the immediate
case \(\Dc\ge0\), this completes the proof for \(\sigma>0\).
\end{proof}

\subsection{The case \texorpdfstring{\(p=q\)}{p = q}}\label{app:coalescent}

The preceding argument assumed \(\rho>0\), equivalently \(p>q\).  At
\(p=q=w\), note that \(d=2w\sinh\rho\sim2w\rho\).  Hence
\eqref{eq:ImQAB}, after canceling the common quadratic factor,
shows that the continuous boundary sign is determined by the exact limit
\begin{equation}\label{eq:rho-zero-limit}
 \lim_{\rho\downarrow0}\frac{N}{8\rho^2}
 =(1+w)\left[
 \sigma^2+\pi^2+\sigma(1+w)(3+2w)
 \right].
\end{equation}
For \(\sigma\ge0\), this is plainly positive.  If
\(\sigma=-\tau<0\), then \(w=e^{-\tau}\), and the bracket is
\[
 \tau^2+\pi^2-\tau(1+e^{-\tau})(3+2e^{-\tau}).
\]
Since
\[
 (1+e^{-\tau})(3+2e^{-\tau})
 =3+5e^{-\tau}+2e^{-2\tau},
\]
we get
\begin{align*}
 &\tau^2+\pi^2-\tau(1+e^{-\tau})(3+2e^{-\tau})
 \\
 &\qquad\ge
 \left(\tau-\frac32\right)^2
 +\pi^2-\frac94-\frac7e>0.
\end{align*}
Thus the upper boundary sign remains strict when the two positive coordinates
coincide.

This proves Proposition~\ref{prop:core-N} and the case \(p=q\) of
Theorem~\ref{thm:one-negative-boundary}.

\end{document}