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\hypersetup{pdftitle={The Kopotun Subpermanent Transform Inequality}, pdfauthor={Yair Lavi}}
\title{The Kopotun Subpermanent Transform Inequality}
\author{Yair Lavi}
\date{11 October 2026}
\begin{document}
\maketitle

\begin{abstract}
We prove the subpermanent transform inequality conjectured by
Kopotun in 1996: for every doubly stochastic
matrix $A$ of order $n\ge2$ and every $2\le k\le n$,

\[
\sigma_k(A)\ge
\sigma_k\!\left(\frac{nJ_n+A}{n+1}\right).
\]

Here $\sigma_k$ is the sum of all subpermanents of order $k$, and $J_n$ has
every entry $1/n$. Equality holds exactly at $A=J_n$.
\end{abstract}

\noindent\textbf{Keywords.} Permanent; subpermanent sum; doubly stochastic matrix; matrix inequality.

\noindent\textbf{2020 Mathematics Subject Classification.} 15A15, 05A20, 15B51.

\section{Introduction and main results}

For a positive integer $r$, write $[r]=\{1,\ldots,r\}$ and let $S_r$ be
the set of permutations of $[r]$. The permanent of an $r\times r$ matrix
$D=(d_{ij})$ is
\[
\operatorname{per}D=\sum_{\pi\in S_r}\prod_{i=1}^r d_{i,\pi(i)}.
\tag{1.1}
\]
If $D$ is an $n\times n$ matrix and $S,T\subseteq[n]$ have the same
cardinality, let $D[S,T]$ be the submatrix with those rows and columns,
listed in increasing order. For $1\le k\le n$, define
\[
\sigma_k(D)=
\sum_{\substack{S,T\subseteq[n]\\ |S|=|T|=k}}
\operatorname{per}D[S,T],
\qquad \sigma_0(D)=1.
\tag{1.2}
\]
Thus $\sigma_n(D)=\operatorname{per}D$.

A real matrix is \textbf{doubly stochastic} if its entries are nonnegative and
every row and column sums to one. Denote the set of these $n\times n$
matrices by $\Omega_n$. Throughout, $n\ge2$, $\mathbf1$ is the all-ones
column vector, and
\[
J_n=\frac1n\mathbf1\mathbf1^{\mathsf T},
\qquad
\tau_n(A)=\frac{nJ_n+A}{n+1}
=J_n+\frac{A-J_n}{n+1}.
\tag{1.3}
\]
The map $\tau_n$ takes $\Omega_n$ into itself, since it is a convex
combination of $J_n$ and $A$.

Kopotun proposed the inequality
$\sigma_k(\tau_n(A))\le\sigma_k(A)$ for every $A\in\Omega_n$ and
$2\le k\le n$ in 1996 \cite[Conjecture~2, p.~167]{ref1}. Since
$\sigma_n=\operatorname{per}$, the case $k=n$ is Wang's permanent
conjecture \cite{ref6}, as Kopotun notes \cite[Remark~1, p.~167]{ref1}:
\[
\operatorname{per}A\ge
\operatorname{per}\!\left(\frac{nJ_n+A}{n+1}\right).
\]
We recently proved this permanent case \cite{ref5}. Kopotun also recorded the
subpermanent inequality for every $2\le k\le4$ and every order $n\ge k$
\cite[Remark~1]{ref1}, with the endpoint cases $(n,k)=(3,3)$ and $(4,4)$ due to
Wang \cite{ref6} and Chang \cite{ref7}, respectively. Kopotun observed \cite[Remark~1]{ref1}
that the transform conjecture follows from the stronger Holens--{\DJ}okovi\'c
conjecture; Wanless later disproved the latter \cite{ref8}. His paper studies
stronger convexity and monotonicity questions as well; the assertion
proved here is the specified transform comparison.

\begin{theorem}
For every integer $n\ge2$, every $A\in\Omega_n$, and
every integer $k$ with $2\le k\le n$,
\[
\sigma_k(A)\ge\sigma_k(\tau_n(A)).
\tag{1.4}
\]
Equality holds if and only if $A=J_n$.

\end{theorem}

We prove the theorem by comparing a quantitative lower bound for
$\sigma_k(A)$ with an upper bound for $\sigma_k(\tau_n(A))$. To state
these bounds, define the Frobenius norm by
$\|X\|_F^2=\sum_{i,j}x_{ij}^2$ for $X=(x_{ij})$, and put
\[
\begin{aligned}
S_{n,k}&=k!\binom nk^2n^{-k}=\sigma_k(J_n),\\
\alpha_{n,k}&=\frac{k(k-1)}{n(n-1)},\qquad
\Delta_{n,k}(A)=\alpha_{n,k}\|A-J_n\|_F^2,\\
C_n&=\frac7{12n^2},\\
b_n&=\frac{1}{n-1}\left[\left(\frac{n}{n+1}\right)^n
\sum_{r=0}^n\frac{1}{r!}-1\right].
\end{aligned}
\tag{1.5}
\]
The expression for $b_n$ comes from evaluating the permanent at the
transform of the identity matrix; see (6.5)--(6.6).
\begin{theorem}
For every $n\ge2$, $A\in\Omega_n$, and $2\le k\le n$,
\[
\frac{\sigma_k(A)}{S_{n,k}}
\ge \exp\!\bigl(C_n\Delta_{n,k}(A)\bigr),
\qquad
\frac{\sigma_k(\tau_n(A))}{S_{n,k}}
\le1+b_n\Delta_{n,k}(A).
\tag{1.6}
\]
Moreover, $0<b_n<C_n$ for every $n\ge2$.

\end{theorem}

Friedland proved the lower bound $\sigma_k(A)\ge S_{n,k}$ \cite{ref9}. The first
estimate of Theorem 1.2 is a quantitative strengthening of that bound.

In particular, if
$D_{n,k}(A)=\sigma_k(A)-\sigma_k(\tau_n(A))$ and
$\Delta=\Delta_{n,k}(A)$, then
\[
\begin{aligned}
D_{n,k}(A)
&\ge S_{n,k}\bigl(e^{C_n\Delta}-1-b_n\Delta\bigr)\\
&\ge S_{n,k}(C_n-b_n)\alpha_{n,k}\|A-J_n\|_F^2.
\end{aligned}
\tag{1.7}
\]
All constants are explicit; they are not asserted to be optimal.

The main step in the lower bound is to represent $\sigma_k(A)$ as a
coefficient of a polynomial of degree $n$. Gurvits's coefficient bound
\cite{ref2} then applies. Its one-variable proof also gives a gain when the roots
are unequal. The difficulty is that these roots change when the other
variables change. For $k<n$, the polynomial contains $n-k$ uniform
linear factors; the arithmetic-geometric mean inequality for those
factors controls this movement and preserves enough of the gain. At
$k=n$, a direct one-column estimate supplies the lower bound.

Section 2 gives the coefficient inequality and its scalar refinement.
Sections 3 and 4 prove the lower estimate in (1.6). Section 5 states
the centered subpermanent estimate used for the upper bound and verifies
its hypotheses. Section 6 derives that bound, separates the constants,
and completes the equality argument. The proof is analytic and does not
depend on finite computation.

\section{Coefficients and capacity}

A polynomial $p\in\mathbb R[z_1,\ldots,z_r]$ is \textbf{real stable} if it is
zero or
\[
p(\zeta_1,\ldots,\zeta_r)\ne0
\quad\text{whenever}\quad
\operatorname{Im}\zeta_i>0\quad(1\le i\le r).
\]
It is \textbf{homogeneous of degree $r$} if each monomial with a nonzero
coefficient has total degree $r$. For a homogeneous polynomial of degree
$r$ in $r$ variables with nonnegative coefficients, define its capacity
by
\[
\operatorname{Cap}(p)=
\inf_{z_1,\ldots,z_r>0}\frac{p(z_1,\ldots,z_r)}{z_1\cdots z_r}.
\tag{2.1}
\]
The notation $[z_1\cdots z_r]p$ means the coefficient of that monomial.
For integers $r\ge2$, put
\[
g(r)=\left(\frac{r-1}{r}\right)^{r-1},
\qquad g(1)=1,\qquad \gamma_r=\frac{r!}{r^r}.
\tag{2.2}
\]
We use the following consequence of Gurvits's theorem \cite[Theorem~2.4 and Corollary~2.5]{ref2}. Gurvits uses the condition $\operatorname{Re}z_i>0$;
for a homogeneous polynomial of degree $r$, $p(iz)=i^rp(z)$, so that
convention is equivalent to the upper-half-plane convention used here.
We include the capacity induction and the stability facts it requires.

\begin{proposition}[Gurvits]
If $p$ is real stable, homogeneous of
degree $r$ in $r$ variables, and has nonnegative coefficients, then
\[
[z_1\cdots z_r]p\ge\gamma_r\operatorname{Cap}(p).
\tag{2.3}
\]
\end{proposition}

\begin{proof}
Differentiation and specialization of a variable at a real
number preserve real stability, allowing the zero polynomial. For
specialization, approach the real value from the upper half-plane and
apply Hurwitz's theorem on locally uniform limits of nonvanishing
holomorphic functions.

For differentiation, suppose the degree in the last variable is
$d\ge1$. Its leading coefficient, as a polynomial in the other variables,
does not vanish when those variables have positive imaginary parts:
it is the nonzero locally uniform limit of
$p(\zeta_1,\ldots,\zeta_{r-1},it)/(it)^d$ as $t\to\infty$.
For each such choice of the other variables, the resulting polynomial
in the last variable has degree $d$ and all its roots in the closed
lower half-plane. Gauss--Lucas gives the same root location for its
derivative. Thus the partial derivative is stable. If $d=0$, the
partial derivative is zero.

Let $h$ be a nonzero one-variable polynomial of degree at most $r$,
with nonnegative coefficients and all roots real and nonpositive.
We first establish
\[
h'(0)\ge g(r)\inf_{t>0}\frac{h(t)}t.
\tag{2.4}
\]
When $h(0)>0$, write
$h(t)=h(0)\prod_{i=1}^r(1+c_it)$, with $c_i\ge0$, padding with zeros
if necessary. If $u=\sum_i c_i>0$ and $r\ge2$, the arithmetic-geometric
mean inequality gives
$h(t)\le h(0)(1+ut/r)^r$. Evaluating this upper bound divided by $t$
at $t=r/[u(r-1)]$ gives (2.4), since $h'(0)=h(0)u$.
If $u=0$, the infimum is zero. For $r=1$, the assertion is the identity
for a linear polynomial. When $h(0)=0$, nonnegative coefficients give
$\inf_{t>0}h(t)/t=h'(0)$, so (2.4) again holds.

Fix positive $z_1,\ldots,z_{r-1}$ and apply (2.4) to
$h(t)=p(z_1,\ldots,z_{r-1},t)$. Real specialization makes this slice
stable; its real coefficients then force all roots to be real, and
nonnegative coefficients exclude positive roots. For
\[
q(z_1,\ldots,z_{r-1})
=\left.\frac{\partial p}{\partial z_r}\right|_{z_r=0},
\]
the definition of capacity gives
$q(z)\ge g(r)\operatorname{Cap}(p)\prod_{i<r}z_i$, and hence
$\operatorname{Cap}(q)\ge g(r)\operatorname{Cap}(p)$.
The closure facts permit iteration until the coefficient is reached.
The product of the factors is
$\prod_{s=2}^r g(s)=\gamma_r$. If a zero polynomial occurs, the preceding
capacity bound forces zero capacity, and the conclusion remains valid.
This proves (2.3).
\end{proof}

The next lemma refines the scalar step. For a vector $v=(v_i)$, write
$\|v\|_2^2=\sum_i v_i^2$.

\begin{lemma}
Let $h(t)$ be a polynomial of degree $n\ge2$ with positive
coefficients and only real negative roots. The function $h(t)/t$ has a unique minimizer
$t_*>0$. Write
\[
h(t)=h(0)\prod_{i=1}^n(1+t/\lambda_i),
\qquad
p_i=\frac{t_*}{t_*+\lambda_i},
\qquad
d=\sum_i p_i^2-\frac1n,
\tag{2.5}
\]
where $\lambda_i>0$. Then $\sum_i p_i=1$, $d\ge0$, and
\[
h'(0)\ge g(n)\frac{h(t_*)}{t_*}
\exp\!\left[\left(\frac12+\frac1{3n}\right)d+\frac{d^2}{6}\right].
\tag{2.6}
\]
\end{lemma}

\begin{proof}
The derivative of $\log(h(t)/t)$ with respect to $\log t$ is
$\sum_i t/(t+\lambda_i)-1$. It is strictly increasing from $-1$ to
$n-1$, and therefore has exactly one zero. This proves the minimizer
claim and $\sum_i p_i=1$.

Put $C_h=h(t_*)/t_*$. Factorization and weighted arithmetic-geometric
mean, with weights $p_i$, give
\[
\frac{h'(0)}{C_h}
=\sum_i p_i\prod_{\ell\ne i}(1-p_\ell)
\ge\prod_i(1-p_i)^{1-p_i}.
\tag{2.7}
\]
For $0\le u<1$, define $f(u)=(1-u)\log(1-u)$, with the continuous
value $f(1)=0$. Its expansion is
\[
f(u)=-u+\sum_{\ell=2}^{\infty}\frac{u^\ell}{\ell(\ell-1)}.
\tag{2.8}
\]
For each $\ell\ge2$, convexity gives
$\sum_i p_i^\ell\ge n^{1-\ell}$. In addition, Cauchy--Schwarz gives
$\sum_i p_i^3\ge(\sum_i p_i^2)^2$. Thus
\[
\begin{aligned}
\sum_i f(p_i)-nf(1/n)
&\ge\frac d2+\frac16\left(\sum_i p_i^3-\frac1{n^2}\right)\\
&\ge\left(\frac12+\frac1{3n}\right)d+\frac{d^2}{6}.
\end{aligned}
\tag{2.9}
\]
Since $nf(1/n)=\log g(n)$, (2.7)--(2.9) prove (2.6).
\end{proof}

\section{A polynomial for subpermanent sums}

We initially assume $A=(a_{ij})\in\Omega_n$ has strictly positive
entries. For $y=(y_1,\ldots,y_n)$, define the elementary symmetric
polynomial
\[
e_k(y)=\sum_{\substack{S\subseteq[n]\\ |S|=k}}\prod_{i\in S}y_i.
\]
Put $m=n-k$, $X=A-J_n$, and define
\[
P(z)=\frac{e_k(Az)}{\binom nk}
\left(\frac{z_1+\cdots+z_n}{n}\right)^m.
\tag{3.1}
\]
Friedland and Gurvits used the degree-$n$ polynomial
$e_k(Az)((z_1+\cdots+z_n)/n)^{n-k}$ to derive the subpermanent lower
bound \cite[Proposition~2.5, proof of Theorem~3.1, and Corollary~3.2]{ref10}.
Our $P$ is their polynomial normalized by $\binom nk$.
The additional $m$ linear factors raise the degree from $k$ to $n$
without changing the normalization at $\mathbf1$. They will also control
the loss incurred when a slice is optimized.

\subsection{Stability, capacity, and normalization}

The polynomial $P$ is homogeneous of degree $n$ and has a positive
coefficient for every monomial of that degree. To see stability, consider
$\prod_{i=1}^n(y+(Az)_i)$. If $y$ and all $z_j$ have positive
imaginary parts, so does every linear factor. Differentiating $n-k$
times in $y$ and then setting $y=0$ gives $(n-k)!\,e_k(Az)$.
Proposition 2.1's closure facts therefore show that $e_k(Az)$ is stable.
The same differentiation argument with independent arguments shows
that $e_k$ itself is stable. The uniform linear factor in (3.1) is
stable too.

For $z>0$, arithmetic-geometric mean over the $\binom nk$ subset
products gives
\[
\frac{e_k(Az)}{\binom nk}
\ge\left(\prod_i(Az)_i\right)^{k/n}
\ge\left(\prod_j z_j\right)^{k/n}.
\tag{3.2}
\]
In the second inequality, apply weighted arithmetic-geometric mean to
each row, then use the column sums. Applying ordinary
arithmetic-geometric mean to the uniform factor proves
$P(z)\ge\prod_jz_j$. Write $\nabla P$ for the vector of first partial
derivatives, and $\nabla^2\log P$ for the Hessian, the matrix of second
partial derivatives of $\log P$. Since $P(\mathbf1)=1$, we have
\[
\operatorname{Cap}(P)=1,\qquad
P(\mathbf1)=1,\qquad \nabla P(\mathbf1)=\mathbf1.
\tag{3.3}
\]
For the gradient identity, each derivative of $e_k$ at $\mathbf1$ is
$\binom{n-1}{k-1}$; composition with $A$ uses its column sums, giving
$k/n$ in each coordinate. The uniform factor contributes $m/n$.

Expanding the coefficient of $z_1\cdots z_n$ gives
\[
[z_1\cdots z_n]P
=\frac{m!}{n^m\binom nk}\sigma_k(A)
=\gamma_n\frac{\sigma_k(A)}{S_{n,k}}.
\tag{3.4}
\]
Indeed, $e_k(Az)$ contributes each permanent with $k$ distinct columns,
and the other $m$ distinct columns must each occur once in the uniform
factor. Their coefficient is $m!/n^m$.

Write $I_n$ for the $n\times n$ identity matrix. We also need the
second derivatives. The Hessian of $\log e_k$ at
$\mathbf1$ is
\[
-\alpha_{n,k}I_n+
\left(\alpha_{n,k}-\frac{k^2}{n^2}\right)
\mathbf1\mathbf1^{\mathsf T},
\]
because its first derivatives, divided by $e_k(\mathbf1)$, equal
$k/n$, its pure second derivatives vanish, and its mixed second
derivatives, divided by $e_k(\mathbf1)$, equal $\alpha_{n,k}$.
Using $A^{\mathsf T}A=J_n+X^{\mathsf T}X$ and including the uniform
factor yields
\[
\nabla^2\log P(\mathbf1)
=-J_n-\alpha_{n,k}X^{\mathsf T}X.
\tag{3.5}
\]
\subsection{Optimized slices}

Fix a column index $j$ and positive coordinates
$x=(x_h)_{h\ne j}$. Let
\[
h_x(t)=P(z),\qquad z_j=t,\quad z_h=x_h\ (h\ne j).
\]
The slice has degree $n$, positive coefficients, and only real negative
roots. Write these roots as $-\lambda_i(x)$, and let $t_*(x)$ be its
minimizer from Lemma 2.2. Define
\[
\begin{aligned}
p_i(x)&=\frac{t_*(x)}{t_*(x)+\lambda_i(x)},\\
d_j(x)&=\sum_i p_i(x)^2-\frac1n,\\
F_j(x)&=\log\frac{h_x(t_*(x))}
{t_*(x)\prod_{h\ne j}x_h},\\
M_j&=\inf_{x>0}\left(F_j(x)+\frac12d_j(x)\right).
\end{aligned}
\tag{3.6}
\]
The vector $p(x)$ is a probability vector, meaning that its coordinates
are nonnegative and sum to one. Also $F_j(x)\ge0$ by (3.3). When every
$x_h=1$, (3.3) makes $t_*=1$. Write $p^0=p(\mathbf1)$ and
$d_0=d_j(\mathbf1)$, suppressing their dependence on the fixed column
$j$. Differentiating the logarithm of the factorization
of $h_x$ and using (3.5) gives
\[
d_j(\mathbf1)=\|p^0-\mathbf1/n\|_2^2
=\alpha_{n,k}\|X_{:j}\|_2^2,\qquad
\sum_jd_j(\mathbf1)=\Delta_{n,k}(A).
\tag{3.7}
\]
The notation $X_{:j}$ denotes column $j$. Any probability vector
satisfies $\sum_i(p_i^0)^2\le1$, so $0\le d_0\le1-1/n$.

For $q_j=\left.\partial P/\partial z_j\right|_{z_j=0}$, Lemma 2.2
and (3.6) imply, pointwise in $x$,
\[
q_j(x)\ge g(n)\left(\prod_{h\ne j}x_h\right)
\exp\!\left(F_j(x)+\frac12d_j(x)\right).
\]
Thus $\operatorname{Cap}(q_j)\ge g(n)e^{M_j}$. The polynomial $q_j$
is stable, homogeneous of degree $n-1$ in $n-1$ variables, and has
nonnegative coefficients. Proposition 2.1 applies, and
$\gamma_{n-1}g(n)=\gamma_n$. Its coefficient is the one in (3.4).
Consequently
\[
\frac{\sigma_k(A)}{S_{n,k}}\ge
\exp\!\left(\max_{1\le j\le n}M_j\right).
\tag{3.8}
\]
Our task is now to bound the infimum $M_j$ in terms of the value $d_0$
at $\mathbf1$. Evaluating (3.6) at $\mathbf1$ would give an upper bound
on $M_j$; a lower bound requires control at every positive $x$.

\subsection{Movement of the roots}

Assume $k<n$, so $m\ge1$ and $n\ge3$. For the fixed column $j$, put
\[
a_i=a_{ij},\qquad
b_i=\sum_{h\ne j}a_{ih}x_h,\qquad b_i^0=1-a_i,
\]
and
\[
u=\min_{h\ne j}x_h,\qquad
v=\max_{h\ne j}x_h,\qquad R=\log(v/u).
\tag{3.9}
\]
Then $ub^0\le b\le vb^0$ coordinatewise.

\begin{lemma}
Let $a,b\in(0,\infty)^n$. Viewed as a polynomial in $t$,
$e_k(b+at)$ has $k$ real roots, counted with multiplicity. List them in
nondecreasing order. Each listed root is nonincreasing in every coordinate
of $b$, and multiplying $b$ by $c>0$ multiplies every root by $c$.

\end{lemma}

\begin{proof}
Let $\mathbf e_i$ denote the $i$-th standard basis vector.
Fix $b$ and increase one coordinate $b_i$ by $s$. Since
$e_k$ has degree at most one in each coordinate, the resulting polynomial
has the form $H(t)+sL(t)$, with $\deg H=k$ and $\deg L\le k-1$.
It is stable in the two variables $t,s$, because
$e_k(b+at+s\mathbf e_i)$ is stable: for positive imaginary parts of $t,s$,
every argument of $e_k$ has positive imaginary part.

For $\operatorname{Im}t>0$, $H(t)\ne0$. Because $H$ has real coefficients,
its nonreal zeros occur in conjugate pairs. Since $H$ has no zero in the
open upper half-plane, all its zeros are real. If $L(t)\ne0$, the root
$s=-H(t)/L(t)$ cannot lie in the upper half-plane. It follows that
$\operatorname{Im}(L(t)/H(t))\le0$, also when $L(t)=0$.
The real rational function $L/H$ tends to zero at infinity. After
cancellation, its poles are real and simple, and its residues are
nonnegative. To verify this last assertion directly, expand at a real
pole: a pole of order at least two changes the sign of the imaginary
part on small upper semicircles, whereas a simple pole
$\rho/(t-r)$ has nonpositive imaginary part there precisely when
$\rho\ge0$.

Let the remaining poles be $r_\ell$, with residues $\rho_\ell>0$,
and put $w_\ell=\sqrt{\rho_\ell}$. The determinant identity
\[
\det\!\left(tI-\operatorname{diag}(r_\ell)+sww^{\mathsf T}\right)
=\prod_\ell(t-r_\ell)
\left(1+s\sum_\ell\frac{\rho_\ell}{t-r_\ell}\right)
\tag{3.10}
\]
shows that the moving roots are the eigenvalues of
$\operatorname{diag}(r_\ell)-sww^{\mathsf T}$. The cancelled factors
give constant roots. The variational characterization of ordered
eigenvalues shows that subtracting $sww^{\mathsf T}$ makes every
ordered eigenvalue nonincreasing for $s\ge0$, since
$y^{\mathsf T}ww^{\mathsf T}y=(w^{\mathsf T}y)^2\ge0$.
Adding the constant roots preserves this ordering conclusion, including
multiple roots. Finally,
$e_k(cb+at)=c^k e_k(b+a(t/c))$ proves the scaling assertion.
\end{proof}

Pair the $k$ roots of the $e_k$ factor with their ordered counterparts
at $x=\mathbf1$. Lemma 3.1 and the coordinate bounds on $b$ give
\[
u\lambda_i^0\le\lambda_i(x)\le v\lambda_i^0.
\tag{3.11}
\]
Each of the $m$ uniform factors has root magnitude
$\sum_{h\ne j}x_h$, compared with $n-1$ at $\mathbf1$, so (3.11)
also holds for those roots. We pair these factors separately; no
ordering of the combined list is needed.

Abbreviate $p=p(x)$, $d=d_j(x)$, $F=F_j(x)$ and $t=t_*(x)$.
The quantities
\[
\delta_i=\log\frac{p_i^0/(1-p_i^0)}{p_i/(1-p_i)}
=\log\frac{\lambda_i(x)}{t\lambda_i^0}
\]
lie in an interval of length $R$. Since $p$ and $p^0$ both sum to one,
$\min_i\delta_i\le0\le\max_i\delta_i$. The identity
$p_i^0=e^{\delta_i}p_i/(1-p_i+e^{\delta_i}p_i)$ shows that
$\log(p_i^0/p_i)$ lies between $0$ and $\delta_i$, regardless of its
sign. Thus, with $r_i=p_i^0/p_i$,
\[
\frac{\max_i r_i}{\min_i r_i}\le e^R,\qquad
\sum_i p_i r_i=1.
\tag{3.12}
\]
For completeness, if $\ell=\min_i r_i$, $w=\max_i r_i$, and
$q=w/\ell$, the inequality $(r_i-\ell)(w-r_i)\ge0$ yields
\[
\sum_i p_i(r_i-1)^2\le(w-1)(1-\ell)
\le\frac{(q-1)^2}{4q}\le\sinh^2(R/2).
\tag{3.13}
\]
The middle inequality maximizes
$(q\ell-1)(1-\ell)$ over $\ell$; the last uses $q\le e^R$.
Here $\sinh t=(e^t-e^{-t})/2$. Let $p_{\max}=\max_i p_i$.
It follows that
\[
\|p^0-p\|_2^2
\le p_{\max}\sinh^2(R/2),\qquad
p_{\max}\le\frac1n+\sqrt{(1-1/n)d}.
\tag{3.14}
\]
For the second bound, if the largest deviation from $1/n$ is $a$,
the remaining deviations sum to $-a$ and their squares sum to at
least $a^2/(n-1)$.

\section{The lower bound}

The root comparison in Section 3 controls dispersion in terms of the
range $R$ of the other coordinates. The uniform factors now bound that
range by the capacity cost $F$. Combining the two estimates gives a
bound that holds before taking the infimum in (3.6).

\subsection{The cost of changing the other coordinates}

Continue with $2\le k<n$ and the notation of Section 3. Define
\[
G(z)=n\log\left(\frac{\sum_i z_i}{n}\right)-\sum_i\log z_i.
\]
Keeping the uniform factor in (3.2), rather than replacing it by its
geometric mean, gives
\[
\frac{P(z)}{\prod_i z_i}\ge
\left[\frac{(\sum_i z_i/n)^n}{\prod_i z_i}\right]^{m/n}.
\]
At $z_j=t_*(x)$ this says
\[
F\ge\frac mn G(z).
\tag{4.1}
\]
We claim that $G(z)\ge2\log\cosh(R/2)$, where
$\cosh t=(e^t+e^{-t})/2$. Choose two of the coordinates $x_h$ having
ratio $r=v/u$ and scale all coordinates so that these two have
arithmetic mean one. Their product is $4r/(1+r)^2$. Let $w>0$ be the
arithmetic mean of the remaining $n-2$ coordinates of the scaled $z$,
including the coordinate indexed by $j$ (which was $t_*(x)$ before
scaling). Their product is at most $w^{n-2}$. Weighted
arithmetic-geometric mean gives
\[
\left(\frac{2+(n-2)w}{n}\right)^n\ge w^{n-2}.
\]
Since $G$ is unchanged by a common scaling, we obtain
\[
G(z)\ge\log\frac{(1+r)^2}{4r}
=2\log\cosh(R/2).
\tag{4.2}
\]
This argument also covers $R=0$, by choosing any two of the other
coordinates. Consequently
$\sinh^2(R/2)\le e^{nF/m}-1$. The triangle inequality and (3.14)
now yield
\[
\sqrt{d_0}\le\sqrt d+
\sqrt{p_{\max}\bigl(e^{nF/m}-1\bigr)}.
\tag{4.3}
\]
The estimate allows $d$ to vanish away from $\mathbf1$; the cost $F$
then accounts for the displacement.

\subsection{A pointwise capacity bound}

\begin{proposition}
For every positive $A\in\Omega_n$, every
$2\le k<n$, every column $j$, and every $x>0$,
\[
F_j(x)+\frac12d_j(x)\ge\frac7{12n}d_j(\mathbf1).
\tag{4.4}
\]
\end{proposition}

\begin{proof}
If $d_0=0$, the conclusion follows from $F,d\ge0$.
Suppose $d_0>0$ and, seeking a contradiction, assume
$F+d/2\le Bd_0/n$, where $B=7/12$. Set
\[
s=\sqrt{d/d_0},\qquad \eta=F/d_0.
\]
Then
\[
\eta+\frac{s^2}{2}\le\frac Bn,\qquad
0\le s\le\sqrt{\frac7{6n}}<1,\qquad
0\le\frac{nF}{m}\le B.
\]
For $0\le y\le7/12$, the exponential series gives
\[
e^y-1\le y+\frac{y^2}{2(1-y/3)}\le\frac{11}{8}y.
\tag{4.5}
\]
Indeed, $\ell!\ge2\cdot3^{\ell-2}$ for $\ell\ge2$, and the middle
expression divided by $y$ is at most $79/58<11/8$.
By (3.14) and $d_0\le1-1/n$,
$np_{\max}\le1+(n-1)s$. Since $m\ge1$, (4.5) and the preceding
bounds give
\[
\begin{aligned}
\frac{p_{\max}(e^{nF/m}-1)}{d_0}
&\le\frac{11}{8}(np_{\max})\eta\\
&\le\frac{11}{8}(1+(n-1)s)
\left(\frac Bn-\frac{s^2}{2}\right).
\end{aligned}
\]
Dividing (4.3) by $\sqrt{d_0}$ and using this bound gives
\[
1\le s+
\sqrt{\frac{11}{8}\bigl(1+(n-1)s\bigr)
\left(\frac7{12n}-\frac{s^2}{2}\right)}.
\tag{4.6}
\]
For fixed $0\le s<1$, the expression under the square root, before
its factor $11/8$, equals
\[
Bs+\frac{B(1-s)}n-\frac{s^2(1-s)}2-\frac{ns^3}{2}.
\]
It decreases as $n$ increases and hence is at most its value at $n=3$.
However,
\[
\begin{aligned}
\Phi(s)
&=(1-s)^2-\frac{11}{8}(1+2s)
\left(\frac7{36}-\frac{s^2}{2}\right)\\
&=\frac{211}{288}-\frac{365}{144}s
+\frac{27}{16}s^2+\frac{11}{8}s^3>0
\qquad(s\ge0).
\end{aligned}
\tag{4.7}
\]
To check positivity entirely with rational constants, put $w=s-1/2$.
Since $w\ge-1/2$,
\[
\begin{aligned}
\Phi(s)
&=\frac{17}{288}+\frac{53}{288}w
+\left(\frac{15}{4}+\frac{11}{8}w\right)w^2\\
&\ge\frac{17}{288}+\frac{53}{288}w+\frac{49}{16}w^2\\
&\ge\frac{57167}{1016064}>0.
\end{aligned}
\tag{4.8}
\]
The last inequality minimizes the quadratic. In the range of $s$
under consideration, $1-s>0$ and the square root in (4.6) is defined.
Equations (4.7)--(4.8) make that square root strictly less than $1-s$,
contradicting (4.6). This proves (4.4).
\end{proof}

Taking the infimum in (4.4) gives
$M_j\ge7d_j(\mathbf1)/(12n)$. By (3.7), some column satisfies
$d_j(\mathbf1)\ge\Delta_{n,k}(A)/n$. Therefore (3.8) proves
\[
\frac{\sigma_k(A)}{S_{n,k}}
\ge\exp\!\left(\frac7{12n^2}\Delta_{n,k}(A)\right)
\qquad(2\le k<n).
\tag{4.9}
\]
\subsection{Permanent order and the boundary}

When $k=n$, the uniform factor is absent. In this case
$P(z)=\prod_i(Az)_i$ is a product of row forms, and a direct
one-column argument is available. This estimate is implicit in the
proof of Laurent and Schrijver \cite[proof of Corollary~1d, equation~(39)]{ref3}.

Fix $j$, put $a_i=a_{ij}$ and
$L_i(x)=\sum_{h\ne j}a_{ih}x_h$. Differentiating at $z_j=0$ gives
\[
q_j(x)=\sum_i a_i\prod_{\ell\ne i}L_\ell(x)
\ge\prod_i L_i(x)^{1-a_i}.
\]
For each row, its remaining sum is $1-a_i$, so a second weighted
arithmetic-geometric mean inequality gives
\[
L_i(x)^{1-a_i}\ge
(1-a_i)^{1-a_i}\prod_{h\ne j}x_h^{a_{ih}}.
\]
Multiplying and using the remaining column sums proves
$\operatorname{Cap}(q_j)\ge\prod_i(1-a_i)^{1-a_i}$.
Proposition 2.1 and $\gamma_{n-1}g(n)=\gamma_n$ imply
\[
\log\frac{\operatorname{per}A}{\gamma_n}
\ge\sum_i f(a_i)-nf(1/n)
\ge\frac12\sum_i(a_i-1/n)^2.
\tag{4.10}
\]
The final inequality follows from (2.8), since $\sum_i a_i=1$ and
every power-sum difference of degree at least two is nonnegative;
the quadratic difference is $\sum_i(a_i-1/n)^2$.
Choosing a column with squared deviation at least
$\|A-J_n\|_F^2/n$ gives
\[
\frac{\operatorname{per}A}{\gamma_n}
\ge\exp\!\left(\frac1{2n}\|A-J_n\|_F^2\right).
\tag{4.11}
\]
Here $1/(2n)\ge7/(12n^2)$ for $n\ge2$, while
$S_{n,n}=\gamma_n$ and $\alpha_{n,n}=1$. Thus (4.11) proves the
lower estimate in Theorem 1.2 at $k=n$.

For an arbitrary nonnegative $A\in\Omega_n$, apply the lower estimate
to $A_\varepsilon=(1-\varepsilon)A+\varepsilon J_n$,
$0<\varepsilon<1$. Its entries are positive, and
$A_\varepsilon-J_n=(1-\varepsilon)(A-J_n)$. Letting
$\varepsilon\downarrow0$ proves the same non-strict estimate for $A$
by continuity of the subpermanent sums and the norm. The strict
transform comparison will follow from the positive coefficient in
(1.7), so it requires no passage of a strict inequality through a limit.

\section{A tensor estimate for centered subpermanents}

A real $n\times n$ matrix has \textbf{zero margins} if every row and column
sum is zero. Write
$\|X\|_{2\to2}=\sup_{v\ne0}\|Xv\|_2/\|v\|_2$,
$Q=I_n-J_n$, and $q_{n,r}=\sigma_r(Q)$.
The centered subpermanent estimate is \cite[Lemma~3.1]{ref4}, and
$q_{n,r}\ge0$ follows from \cite[equation~(3.5)]{ref4}.

\begin{lemma}
For $n\ge2$, if $X$ has zero margins and
$\|X\|_{2\to2}\le1$, then for $2\le r\le n$,
\[
q_{n,r}\ge0,\qquad
|\sigma_r(X)|\le\frac{q_{n,r}}{n-1}\|X\|_F^2.
\tag{5.1}
\]
\end{lemma}

For $A\in\Omega_n$ and $v\in\mathbb R^n$, weighted
Cauchy--Schwarz in each row gives
\[
\sum_i\left(\sum_j a_{ij}v_j\right)^2
\le\sum_{i,j}a_{ij}v_j^2=\sum_jv_j^2.
\tag{5.2}
\]
Therefore $\|A\|_{2\to2}\le1$. The row and column sums give
$A-J_n=QAQ$, so $X=A-J_n$ has zero margins and
$\|X\|_{2\to2}\le1$. Lemma 5.1 applies to every such $A$, including
those with zero entries.

\section{The upper bound and completion of the proof}

\subsection{Expanding the transform}

For $0\le r\le k$, put
\[
c_{n,k,r}=(k-r)!\binom{n-r}{k-r}^{\!2}n^{-(k-r)}.
\]
For every real $t$ and every $n\times n$ matrix $X$,
\[
\sigma_k(J_n+tX)=
\sum_{r=0}^k c_{n,k,r}t^r\sigma_r(X).
\tag{6.1}
\]
This identity is also a specialization of Kopotun's Lemma B
\cite[equation~(26)]{ref1}; we include the counting argument for completeness.
To verify the multiplicity, choose the $r$ entries from $X$ in a
partial matching, meaning a collection of positions with distinct row
indices and distinct column indices. Its rows and columns can be extended to $k$ rows
and columns in $\binom{n-r}{k-r}^2$ ways, and the remaining
$k-r$ rows can be matched in $(k-r)!$ ways. Each entry supplied by
$J_n$ contributes $1/n$.

Set $\rho=1/(n+1)$ and $X=A-J_n$. Since $\sigma_1(X)=0$ and
$c_{n,k,0}=S_{n,k}$, Lemma 5.1 gives
\[
\sigma_k(\tau_n(A))
\le S_{n,k}+U_{n,k}\|X\|_F^2,
\qquad
U_{n,k}=\frac1{n-1}\sum_{r=2}^k
c_{n,k,r}\rho^r q_{n,r}.
\tag{6.2}
\]
All summands defining $U_{n,k}$ are nonnegative. Define the falling
factorial by $(a)_r=a(a-1)\cdots(a-r+1)$ for $r\ge1$, and
$(a)_0=1$. Direct cancellation gives
\[
\frac{c_{n,k,r}}{S_{n,k}}
=\frac{n^r(k)_r}{(n)_r^2}.
\tag{6.3}
\]
For each $r\ge2$, division by
$\alpha_{n,k}=(k)_2/(n)_2$ leaves a positive factor independent of
$k$ times $(k-2)_{r-2}$. This is nondecreasing for integers
$k\ge r$. Increasing $k$ also introduces only nonnegative summands
in (6.2). Consequently
\[
\frac{U_{n,k}}{S_{n,k}\alpha_{n,k}}
\le\frac{U_{n,n}}{S_{n,n}}.
\tag{6.4}
\]
At $k=n$, (6.1) with $X=Q$ identifies the right-hand side as
\[
\frac1{n-1}\left(
\frac{\operatorname{per}(J_n+\rho Q)}{\gamma_n}-1\right).
\tag{6.5}
\]
Now $J_n+\rho Q=(I_n+\mathbf1\mathbf1^{\mathsf T})/(n+1)$.
Expanding the permanent according to the identity entries selected
gives
\[
\operatorname{per}(I_n+\mathbf1\mathbf1^{\mathsf T})
=\sum_{r=0}^n\binom nr(n-r)!
=n!\sum_{r=0}^n\frac1{r!}.
\tag{6.6}
\]
Thus (6.5) equals $b_n$. Combining (6.2)--(6.6) proves the upper
estimate in Theorem 1.2. It also shows $b_n>0$: the $r=2$ summand of
$U_{n,n}$ in (6.2) is positive, since $q_{n,2}=(n-1)/2$. This last identity
follows from
$2\sigma_2(X)=\sum_{i\ne j,\ h\ne\ell}x_{ih}x_{j\ell}
=\sum_{i,h}x_{ih}^2$ for any zero-margin $X$: inclusion-exclusion
eliminates the total sum and the terms with equal rows or equal columns.
Apply this to $Q$, whose squared Frobenius norm is $n-1$.

\subsection{Separating the constants}

It remains to prove $n^2b_n<7/12$. Direct evaluation for $n=2,3,4$
gives, respectively,
\[
n^2b_n=\frac49,\qquad \frac9{16},\qquad \frac{656}{1125},
\tag{6.7}
\]
each strictly less than $7/12$.

For $n\ge5$, the alternating expansion of $\log(1+1/n)$ and
$\sum_{r=0}^n1/r!<e$ give
\[
\left(\frac n{n+1}\right)^n\sum_{r=0}^n\frac1{r!}
<e^{y_n},\qquad
y_n=\frac1{2n}-\frac1{3n^2}+\frac1{4n^3}.
\tag{6.8}
\]
Indeed, $1-n\log(1+1/n)<y_n$. Also $0<y_n<1$.
Since $r!\ge2^{r-1}$ for $r\ge1$, summing the exponential series
gives
\[
e^y-1\le\frac{y}{1-y/2}\qquad(0\le y<2).
\tag{6.9}
\]
It suffices to show
$y_n/(1-y_n/2)<7(n-1)/(12n^2)$. Clearing positive denominators,
this is equivalent to
\[
24n^4-114n^3-2n^2-49n+21>0.
\tag{6.10}
\]
Writing $n=5+t$, the polynomial in (6.10) is
\[
476+3381t+1888t^2+366t^3+24t^4,
\]
which is positive for $t\ge0$. Equations (6.7)--(6.10) prove
$b_n<C_n$ for every $n\ge2$. Together with Section 4 and the upper
bound, this completes Theorem 1.2.

\subsection{Equality and the quantitative gap}

\begin{proof}[Proof of Theorem~1.1]
Subtract the estimates of Theorem 1.2 and
use $e^u\ge1+u$ for $u\ge0$ to obtain (1.7).
Since $S_{n,k}>0$, $\alpha_{n,k}>0$, and $C_n-b_n>0$, that
inequality is strict whenever $A\ne J_n$. At $A=J_n$ the transform
fixes $A$, giving equality. Thus Theorem 1.1 holds in every stated
order, including all support boundaries.
\end{proof}

\clearpage
\section*{Acknowledgments}

The proof of this conjecture was carried out by GPT-5.6-sol, GPT-6 Astra,
Claude Fable 5, and Claude Fable 5.1, under the guidance of the author. The author has reviewed the
resulting proof arguments. Responsibility for the final text rests with the
author.

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\end{document}