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  pdftitle={Least-Atom Extremizers for Resampling Noise}}

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\title{Least-Atom Extremizers for Resampling Noise\thanks{This work was produced within the Bolzano project (\url{https://bolzano.app}), using AI systems for mathematical research and manuscript preparation. The listed authors contributed to the project and the preparation of this release. The manuscript received only limited human checking.}}
\author{Mat\v{e}j Kripner \qquad V\'{a}clav Rozho\v{n} \qquad Robert \v{S}\'{a}mal\\[0.4em]
Martin \v{S}indel\'{a}\v{r} \qquad Adri\'{a}n Z\'{a}me\v{c}n\'{i}k}
\date{}


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\begin{abstract}
A single least-mass atom determines the extremal behavior of resampling
noise on a finite probability space.  A fixed-moment comparison gives the
sharp generalized Samorodnitsky inequality for every \(1<q<2\) through one
scalar curve, and identifies all nonconstant scalar extremizers.  The same
principle determines every \(L^p\)-to-\(L^q\) resampling norm for
\(1<p<q<\infty\), resolving a conjecture of Cao, Fan, Han, Qiu, and Wang.
The sharp hypercontractive parameter increases strictly with the least atom.
The uniform-binary phase diagram is explicit, and product equality cases
are classified for every nonuniform reference law and for the uniform
binary law.
\end{abstract}

\section{Introduction}

\subsection{Background and the source problem}

Resampling is a basic model of noise on a finite probability space
\((\Omega,\mu)\).  With probability \(\rho\) an observation is kept, and
with probability \(1-\rho\) it is replaced by an independent sample from
\(\mu\).  Its averaging operator is
\[
 T_\rho f=\rho f+(1-\rho)\E_\mu f.
\]
The question is how much a function can remain concentrated after this
averaging.  Two familiar measurements are its logarithmic moments and the
ratio of its output \(L^q\)-norm to its input \(L^p\)-norm.  For example,
a density concentrated on a very rare outcome may retain a large norm
even after substantial resampling.  It is less clear whether spreading
the high values over several outcomes can be more effective.

On a product space, noise and partial observation give two ways to average
a function.  Product resampling acts independently in each coordinate.
Partial observation retains a random set \(S\) of coordinates and averages
over the rest.  The generalized Samorodnitsky inequality compares them:
\begin{equation}\label{eq:product-sam}
 \log\|T_\rho^{[n]}f\|_q
 \le \E_{S\sim\lambda}\log\|\E(f\mid S)\|_q.
\end{equation}
Here \(f\) is nonzero and nonnegative on \(\Omega^n\), the measure is
\(\mu^{\otimes n}\), and each coordinate is retained in \(S\)
independently with probability \(\lambda\).  The sharp inclusion parameter
is the smallest \(\lambda\) for which this holds in every dimension.
Determining it quantifies the information that survives noise in terms of
information available from a partial observation.

Samorodnitsky proved this comparison on the uniform Boolean cube for
\(q>1\) \cite[Thm.~1.1]{Sam19}, and obtained the sharp parameter for
integer \(q\ge2\) \cite[Thm.~1.1]{Sam20}.  Abawonse, H\k{a}z\l{}a, and
O'Donnell extended the sharp result to finite product probability spaces
and real \(q\ge2\) \cite[Thms.~1 and~19]{AHO}.  Their nonconstant scalar
extremizers are normalized indicators of least-mass atoms.  Section~1.4
of the full version explicitly leaves \(1<q<2\) untreated and explains
why this indicator description should not persist throughout that range.
Numbering in \cite{AHO} below refers to its full arXiv version.

\begin{question}[The direction in {\cite[Sec.~1.4]{AHO}}]\label{q:formulated}
For a finite probability space, what is the sharp generalized
Samorodnitsky inequality when \(1<q<2\), and which scalar functions
replace the least-atom indicators from the range \(q\ge2\)?
\end{question}

A related question concerns hypercontractivity: for given \(1<p<q<\infty\),
how much noise is necessary to ensure \(\|T_\rho f\|_q\le\|f\|_p\) for
every \(f\)?  Write \(\sigma_{p,q}(\mu)\) for the largest admissible
\(\rho\), and \(\sigma_{p,q}(m)\) for the two-point law \((m,1-m)\).
Wolff's two-value reduction \cite[Thm.~3.1]{Wolff} gives, in our order of
exponents,
\[
 \inf_{a\in[m,1/2]}\sigma_{p,q}(a)
 \le \sigma_{p,q}(\mu)\le\sigma_{p,q}(m),\qquad
 m=\min_{x\in\Omega}\mu(x).
\]
It leaves a comparison among the two-point masses.  Cao, Fan, Han, Qiu,
and Wang determine the biased two-point constants and prove equality
for conjugate exponents \cite[Thm.~1.6 and Cor.~1.7]{CaoEtAl}.  They pose
the following conjecture immediately after Figure~3 on page~8.

\begin{conjecture}[{\cite[Sec.~1.2, p.~8]{CaoEtAl}}]\label{conj:least}
If \(\mu\) is a finite probability law with least atom \(m\in(0,1/2)\),
then \(\sigma_{p,q}(\mu)=\sigma_{p,q}(m)\) for every
\(1<p<q<\infty\).
\end{conjecture}

\subsection{Contribution}

Both questions are governed by a comparison that keeps the mean and one
convex moment fixed.  Under a monotonicity condition on the ratio of two
second derivatives, the other moment is largest when the high value lies
on a single least atom (Theorem~\ref{thm:moment-principle}).  This identifies
the mass as well as the number of levels.  For a least atom \(a\), the
resulting mean-one family is
\begin{equation}\label{eq:least-density}
 h_s^a=1-s+\frac{s}{m}\1_{\{a\}},\qquad 0\le s\le1.
\end{equation}
It interpolates between a constant and the normalized indicator of \(a\).
The terms \emph{soft} and \emph{hard} refer respectively to \(0<s<1\)
and \(s=1\).  The key comparison moves a high level onto a smaller mass
while preserving the mean and the input moment; the noisy output moment
then increases.  Thus the least atom is selected before any optimization
of the contrast.  The main consequences are as follows.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.3em]
\item For \(1<q<2\), the entire output \(q\)-moment profile at a fixed
input \(q\)-moment is attained by \eqref{eq:least-density}.
Consequently the sharp parameter in \eqref{eq:product-sam} is a maximum
over the single interval \(0\le s\le1\), depending on \(\mu\) only through
\(m\) (Theorems~\ref{thm:finite} and~\ref{thm:sharp}).  Every nonconstant
scalar extremizer belongs to this family.
\item For every \(1<p<q<\infty\) and every \(0\le\rho\le1\), the full
operator norm of \(T_\rho:L^p(\mu)\to L^q(\mu)\) equals that on
\((m,1-m)\) (Theorem~\ref{thm:hyper-norm}).  This proves
Conjecture~\ref{conj:least}, and the sharp noise parameter is strictly
increasing in \(m\) (Corollary~\ref{cor:hyper-monotone}).
\item For every nonuniform \(\mu\), all product equality cases in
\eqref{eq:product-sam} factor into scalar equality cases and constants
(Theorem~\ref{thm:nonuniform-equality}).  For arbitrary \(\mu\), scalar
extremizers are soft near \(q=1\), provided \(m<1/2\), and hard near
\(q=2\).  In the latter regime all product equality cases also factor
(Theorems~\ref{thm:nearone} and~\ref{thm:neartwo}).
\item For the uniform binary law, Theorem~\ref{thm:binary-phase} determines
all scalar phases, with thresholds
\((\sqrt{13}-1)/2\) and \(1+(4\log2-1)^{-1}\), and classifies product
equality throughout \(1<q<2\).
\end{enumerate}
No choice among subset masses remains.  The unresolved scalar question
for general \(m<1/2\) is the maximizing contrast \(s\), including its
uniqueness and transition curve.  Product equality for a uniform alphabet
with at least three symbols is not classified here outside the hard regime.

Lagrange multipliers and a one-sided boundary variation reduce a
constrained maximizer to two levels.  The comparison of their masses has
a strictly negative covariance as its derivative.  This supplies the
least-atom conclusion without requiring a phase analysis.  For hypercontractivity it applies
directly when the input exponent is at most two; self-adjoint duality
covers the remaining exponents.

\subsection{Related work and scope}

The sharp Samorodnitsky parameter is the \emph{fixed-reference} R\'enyi
contraction coefficient of the resampling channel.  This interpretation
already appears for \(q\ge2\) in \cite[Thm.~3 and Rem.~20]{AHO}, whose
Theorem~12 gives the scalar-to-product passage.  Vandenbroucque, Esposito,
and Gastpar extend that passage to arbitrary product channels
\cite[Thm.~10(a)]{VEGfull26}.  The tensorization proof below is included
to make the equality discussion self-contained.  The scalar contribution
is the fixed-moment comparison and its least-atom equality statement for
\(1<q<2\).

At the KL endpoint, Gu and Polyanskiy already give the sharp coefficient
and a one-variable description for the uniform Potts channel
\cite[Thm.~4 and Props.~17--18]{GuPolyanskiy}.  Positivity of KL extremizers
for strictly positive reversible kernels follows more generally from
\cite[Lem.~31]{CCGP}, and continuity in the R\'enyi order at one is known
for arbitrary finite channels \cite[Cor.~1]{VEGfull26}.  The least-atom
comparison extends the scalar entropy formula to nonuniform reference laws.

For hypercontractivity, the two-value method and the reduction to a range
of two-point masses are due to Wolff \cite[Thm.~3.1]{Wolff}.  The extra
comparison here selects its smallest mass and identifies the norm even
outside the hypercontractive region.  The operator is precisely
\(\E_\mu+\rho(I-\E_\mu)\); the result makes no assertion about a different
Fourier multiplier on a general cyclic group.  The two-point constant
\(\sigma_{p,q}(1/2)=\sqrt{(p-1)/(q-1)}\) is classical; see
\cite[Eq.~(2.1)]{Wolff}.

A fixed reference law must be distinguished from optimization over both
laws in a divergence ratio.  The binary-input reductions of
Ordentlich and Polyanskiy \cite[Thms.~1--2]{OP22}, and their R\'enyi
counterparts \cite[Thm.~1]{VEG26} and \cite[Thm.~6]{VEGfull26}, concern
the latter problem.  A two-valued density relative to a prescribed \(\mu\)
can still have full support on an arbitrarily large alphabet.  General
contraction theory is surveyed in \cite{PW}; related R\'enyi bounds appear
in \cite{JEG,VEG26,VEGfull26}, and constrained-input divergence bounds
motivated by privacy amplification in \cite{GrosseEtAl}.

The threshold \((\sqrt{13}-1)/2\) already occurs in the binary
R\'enyi analysis of Song et al.\ \cite[Lem.~2]{SongEtAl}.  The Riccati
representation and negative second derivative used below also appear
in trureturing's paper-level analysis of Cachin R\'enyi information
combining \cite{TrureTuring}.  These ingredients are credited there;
the additional third-derivative sign yields a one-valley property and
the complete noisy-ratio phase diagram.  No formal verification claim
about that external development is needed for the proof given here.

\section{Notation and the scalar inequality}

Write \(\mu^*=\min_{x\in\Omega}\mu(x)\).  All expectations and
\(L_q\)-norms without a displayed measure are taken with respect to \(\mu\),
or to the appropriate product measure.  For a function
\(h:\Omega\to\mathbb R\), set
\begin{equation}\label{eq:one-noise}
 (T_\rho h)(x)=\rho h(x)+(1-\rho)\E h.
\end{equation}
The product operator \(T_\rho^{[n]}\) is the tensor product of these
one-coordinate operators.  Equivalently, given \(x\in\Omega^n\), it retains
each coordinate independently with probability \(\rho\) and otherwise
resamples that coordinate from \(\mu\), then averages the value of \(f\) at
the resulting point.

For \(S\subseteq[n]\), \(\E(f\mid S)\) denotes the function of the
coordinates in \(S\) obtained by integrating all coordinates outside \(S\).
We regard it as a function on \(\Omega^n\), constant in the integrated
coordinates; this convention leaves its \(L_q\)-norm unchanged.  The notation
\(S\sim\lambda\) means that each coordinate belongs to \(S\) independently
with probability \(\lambda\).

Theorem~1 of Abawonse, H\k{a}z\l{}a, and O'Donnell~\cite{AHO} treats
\(q\in[2,\infty]\).  For \(n=1\) it proves
\begin{equation}\label{eq:source-scalar}
 \|T_\rho h\|_q\le \|h\|_1^{1-\lambda}\|h\|_q^\lambda
\end{equation}
with a sharp constant depending only on \(\mu^*\); its nonconstant equality
cases are precisely the indicators of atoms of mass \(\mu^*\).  Theorem~12
of that paper supplies the induction from the scalar inequality to products,
and Theorems~15 and~19 contain the precise product and scalar formulations,
respectively.  The fixed-reference R\'enyi formulation is made explicit in
Theorem~3 and Remark~20.
The first item of \cite[Sec.~1.4]{AHO} isolates \(1<q<2\) as an untreated
direction.  We retain the same operator, positivity assumption, and
independent-subset formulation.  For the variational characterization and
tensorization, the standing regime is
\begin{equation}\label{eq:regime}
 2\le |\Omega|<\infty,\quad \mu(x)>0\ \text{for all }x,\qquad
 1<q<2,\qquad 0<\rho<1.
\end{equation}
Hypercontractive norms in Section~\ref{sec:hyper} use separate exponents
\(1<p<q<\infty\), explicitly replacing the restriction \(q<2\).

\section{A fixed-moment least-atom principle}\label{sec:moment}

The following comparison is formulated independently of noise.  Its
strictness will also identify the extremizers in the applications.

\begin{theorem}[Least-atom moment comparison]\label{thm:moment-principle}
Let \(\mu\) be fully supported on a finite set with at least two points,
and let \(m=\min_x\mu(x)\).  Suppose \(f,g\in C^1([0,\infty))\cap
C^2((0,\infty))\), \(f''>0\), and \(g''/f''\) is strictly increasing.
For every \(h\ge0\) with \(\E_\mu h=1\), there is a unique
\(s\in[0,1]\) such that
\begin{equation}\label{eq:fixed-f-moment}
 \E_\mu f(h)=m f\left(1+\frac{1-m}{m}s\right)+(1-m)f(1-s).
\end{equation}
For this \(s\),
\begin{equation}\label{eq:moment-comparison}
 \E_\mu g(h)\le
 m g\left(1+\frac{1-m}{m}s\right)+(1-m)g(1-s).
\end{equation}
If \(s>0\), equality holds exactly for \(h=h_s^a\) with \(a\) a least
atom, as in \eqref{eq:least-density}.  If \(s=0\), then \(h=1\).
\end{theorem}

\begin{proof}
Write \(M=\E f(h)\).  On the compact mean-one simplex the strictly convex
functional \(\E f(h)\) is minimized only by \(h=1\) and maximized at a
vertex.  The vertex supported on an atom of mass \(a\) has value
\[
 J(a)=a f(1/a)+(1-a)f(0).
\]
Strict convexity gives
\(J'(a)=f(1/a)-f(0)-f'(1/a)/a<0\).  Hence
\(f(1)\le M\le J(m)\).  The right-hand side of
\eqref{eq:fixed-f-moment} has these endpoint values, and its derivative
for \(s>0\) is
\[
 (1-m)\left[f'\left(1+\frac{1-m}{m}s\right)-f'(1-s)\right]>0.
\]
This proves existence and uniqueness of \(s\).  Strict convexity also
shows that the upper endpoint is attained only at vertices of mass \(m\).
It remains to consider \(f(1)<M<J(m)\).

First compare two-level variables without restricting their masses to
subset masses of \(\mu\).  Let the high value \(x>1\) have mass \(a\),
and the low value \(0\le y<1\) have mass \(1-a\), with
\[
 ax+(1-a)y=1,\qquad af(x)+(1-a)f(y)=M.
\]
For a fixed \(a\) this pair is unique whenever it exists.  It exists
exactly when \(M\le J(a)\).  Thus a feasible pair at \(a\) can be continued
to every smaller positive mass, in particular down to \(m\) when
\(a\ge m\).  For \(y>0\), put
\(\Delta=x-y\), \(F=f(x)-f(y)\), and \(D=f'(x)-f'(y)>0\).
Differentiation of the two constraints gives
\begin{equation}\label{eq:two-level-derivatives}
 x'=-\frac{F-f'(y)\Delta}{aD},\qquad
 y'=-\frac{f'(x)\Delta-F}{(1-a)D}.
\end{equation}
For \(N(a)=a g(x)+(1-a)g(y)\), substitution yields
\begin{align*}
 N'(a)
 &=g(x)-g(y)-g'(y)\Delta
   -\frac{g'(x)-g'(y)}{D}\bigl(F-f'(y)\Delta\bigr)\\
 &=D\,\operatorname{Cov}\left(x-U,\frac{g''(U)}{f''(U)}\right)<0,
\end{align*}
where \(U\) has density \(f''(u)/D\) on \((y,x)\).  The second identity
uses
\(F-f'(y)\Delta=\int_y^x(x-u)f''(u)\,\dd u\) and its analogue for
\(g\).  For independent copies \(U,V\), twice the covariance is
\[
 \E\left[(V-U)\left(\frac{g''(U)}{f''(U)}
                         -\frac{g''(V)}{f''(V)}\right)\right]<0.
\]
The strict sign follows because the ratio increases strictly and the
density is positive.  A pair with \(y=0\) is an endpoint of this
comparison path; continuity of \(f,g\) and their first derivatives
extends the strict comparison to that endpoint.  In particular,
decreasing the high-level mass strictly increases \(N\).

Now maximize \(\E g(h)\) subject to the prescribed mean and \(f\)-moment.
Compactness gives a maximizer.  If all its positive coordinates are equal,
it is a normalized subset indicator, and the two-level comparison just
proved applies.  Otherwise two positive coordinates are distinct.  The
gradients of the two constraints are then independent on the positive
support.  Lagrange multipliers give constants \(\lambda,\beta\) with
\[
 \psi(h_i)=\beta\quad(h_i>0),\qquad
 \psi(t)=g'(t)-\lambda f'(t).
\]
On \((0,\infty)\),
\(\psi'(t)=f''(t)(g''(t)/f''(t)-\lambda)\).  Its sign can change at most
once, from negative to positive.  Thus a horizontal line meets \(\psi\)
in at most two positive points.  Since two such points occur, \(\psi\)
has a strictly decreasing branch followed by a strictly increasing branch,
and the level \(\beta\) is strictly below \(\psi(0)\).

There can be no zero coordinate.  Indeed, activate a zero coordinate
\(i\) to a small positive value \(\varepsilon\), and adjust two distinct
positive coordinates to keep both constraints fixed.  The implicit
function theorem applies because their constraint Jacobian has nonzero
determinant.  The one-sided objective derivative is
\(\mu_i(\psi(0)-\beta)\); maximality requires it to be nonpositive,
contradicting \(\beta<\psi(0)\).  The maximizer is therefore positive and
two-valued.

In either case the mass \(a\) of its high level is a sum of atom masses,
so \(a\ge m\).  If \(a>m\), moving to the two-level profile of mass \(m\)
with the same constraints strictly increases the objective.  This is a
feasible function on \(\Omega\), supported at its high level on any least
atom, and contradicts maximality.  Hence \(a=m\), which consists of one
atom.  The constraint determines its contrast to be \(s\).  This proves
both the inequality and its equality cases.
\end{proof}

\section{The one-coordinate coefficient}

\begin{definition}[The scalar ratio]\label{def:ratio}
Let
\[
\C=\{h:\Omega\to[0,\infty):\E h=1\}.
\]
For nonconstant \(h\in\C\), put
\[
A(h)=\E h^q,\qquad
B(h)=\E(1-\rho+\rho h)^q,\qquad
R_q(h)=\frac{\log B(h)}{\log A(h)}.
\]
At the constant density \(1\), define \(R_q(1)=\rho^2\), the continuous
quadratic perturbation value.  Finally set
\[
\lambda_*(q,\mu,\rho)=\max_{h\in\C}R_q(h).
\]
\end{definition}

\begin{lemma}[Existence and nonlocality]\label{lem:exist}
For \(1<q<2\) and \(0<\rho<1\), the maximum in Definition~\ref{def:ratio}
exists and satisfies \(0<\lambda_*<1\).  If \(|\Omega|>2\), then
\(\lambda_*>\rho^2\).  More generally, the latter conclusion holds whenever
\(\mu\) has a subset of mass strictly below \(1/2\).
\end{lemma}

\begin{proof}
Since \(\mu\) is fully supported, \(0\le h(x)\le1/\mu(x)\) for every
\(h\in\C\).  Hence \(\C\) is compact.  If \(h=1+u\), \(\E u=0\), and
\(\|u\|_\infty\to0\), Taylor expansion gives
\[
\log\E(1+u)^q=\frac{q(q-1)}2\E u^2+O(\|u\|_\infty\E u^2)
\]
and the same formula with \(u\) replaced by \(\rho u\) for the numerator.
Thus \(R_q(h)\to\rho^2\) as \(h\to1\), so the definition at \(1\) makes
\(R_q\) continuous on \(\C\).  Compactness gives existence.

For nonconstant \(h\), Jensen gives \(A(h)>1\).  Since \(\rho>0\), the
function \(1-\rho+\rho h\) is also nonconstant and has mean one, so strict
Jensen gives \(B(h)>1\).  Also
\[
1-\rho+\rho h=(1-\rho)\cdot 1+\rho h,
\]
and strict convexity of \(x^q\) gives
\[
B(h)<(1-\rho)+\rho A(h)<A(h).
\]
Thus \(0<R_q(h)<1\), and \(R_q(1)=\rho^2<1\).

Assume that \(A\subset\Omega\) has mass \(a=\mu(A)<1/2\), and put
\[
g=\1_A-\frac{a}{1-a}\1_{A^c}.
\]
Then \(\E g=0\), \(\E g^2=a/(1-a)\), and
\(\E g^3=a(1-2a)/(1-a)^2>0\).  For small \(t>0\), \(h_t=1+tg\) lies in
\(\C\).  Writing \(L(t)=\log\E(1+tg)^q\), expansion gives
\[
L(t)=\frac{q(q-1)}2t^2\E g^2
+\frac{q(q-1)(q-2)}6t^3\E g^3+O(t^4).
\]
Therefore
\[
R_q(h_t)
=\rho^2\left(1+\frac{q-2}{3}\frac{\E g^3}{\E g^2}(\rho-1)t+O(t^2)\right).
\]
Since \(q-2<0\), \(\rho-1<0\), and \(\E g^3>0\), the coefficient of \(t\)
is positive.  Thus \(R_q(h_t)>\rho^2\) for all sufficiently small positive
\(t\).
\end{proof}

\begin{theorem}[Sharp moment profile and scalar parameter]\label{thm:finite}
Let \(m=\min_x\mu(x)\), \(1<q<2\), and \(0<\rho<1\).  Define
\begin{equation}\label{eq:moment-curve}
 M_{m,q}(s)=m\left(1+\frac{1-m}{m}s\right)^q+(1-m)(1-s)^q,
 \qquad 0\le s\le1.
\end{equation}
If \(\E h=1\) and \(\E h^q=M_{m,q}(s)\), then
\begin{equation}\label{eq:sharp-profile}
 \E(T_\rho h)^q\le M_{m,q}(\rho s).
\end{equation}
For \(s>0\), equality holds exactly when \(h=h_s^a\) for a least atom
\(a\).  In particular,
\begin{equation}\label{eq:least-ratio}
 \lambda_*(q,\mu,\rho)=
 \max_{0\le s\le1}
 \frac{\log M_{m,q}(\rho s)}{\log M_{m,q}(s)},
\end{equation}
where the ratio at zero is \(\rho^2\).
\end{theorem}

\begin{proof}
Apply Theorem~\ref{thm:moment-principle} with
\(f(t)=t^q\) and \(g(t)=(1-\rho+\rho t)^q\).  Its decisive hypothesis is
\[
 \frac{g''(t)}{f''(t)}
 =\rho^2\left(\frac{t}{1-\rho+\rho t}\right)^{2-q},
\]
which is strictly increasing for \(t>0\).  The function \(h_s^a\) is taken
by \(T_\rho\) to \(h_{\rho s}^a\), giving \eqref{eq:sharp-profile} and
its equality statement.  Every mean-one density has a unique input
moment parameter \(s\), while every \(s\in[0,1]\) is attained by
\(h_s^a\).  Dividing logarithms proves \eqref{eq:least-ratio}, with the
value at zero supplied by Lemma~\ref{lem:exist}.
\end{proof}

\begin{corollary}[Scalar extremizers]\label{prop:twolevel}
Every nonconstant scalar equality case, normalized to mean one, is
\(h_s^a\), where \(a\) is a least atom and \(s>0\) maximizes
\eqref{eq:least-ratio}.  Thus its high value occurs on one least atom and
its low value is common to all other atoms.  Constants are always scalar
equality cases, whether or not the ratio at zero is maximal.
\end{corollary}

\begin{proof}
At a fixed nonconstant input moment the denominator of the scalar ratio
is fixed and positive.  A scalar equality case must therefore attain the
sharp output moment in \eqref{eq:sharp-profile}.  Conversely each maximizing
profile attains the ratio.  The constant equality follows directly from
\(T_\rho1=1\).
\end{proof}

\section{Sharp product tensorization}

The scalar-to-product passage itself is already contained in the general
induction theorem of Abawonse, H\k{a}z\l{}a, and O'Donnell
\cite[Theorem~12]{AHO}.  In their notation, take the target exponent
\(Q\in(1,2)\), set \(r=0\) and \(p=q=s=Q\), use deterministic resampling
parameters \(\rho_i=\rho\) on the left, and independent Bernoulli inclusion
variables on the right.  The direct H\"older proof below is retained for
self-containedness; the new input is the exact scalar coefficient and its
extremizers.

For probability measures \(\nu\ll\mu\), the order-\(q\) R\'enyi divergence is
\begin{equation}\label{eq:renyi}
D_q(\nu\|\mu)=\frac{1}{q-1}
\log\E_\mu\!\left(\frac{\dd\nu}{\dd\mu}\right)^q.
\end{equation}
For a channel \(K\) and a fixed reference input law \(\pi\), define the
\emph{fixed-reference} coefficient
\begin{equation}\label{eq:fixed-reference}
 \eta_q(\pi,K)=
 \sup_{\substack{\nu\ne\pi\\0<D_q(\nu\|\pi)<\infty}}
 \frac{D_q(\nu K\|\pi K)}{D_q(\nu\|\pi)}.
\end{equation}
The channel-level coefficient
\[
 \eta_q(K)=\sup_\pi\eta_q(\pi,K)
\]
also optimizes the reference law.  Throughout this paper the reference law
remains \(\mu\).  General properties and bounds for these two coefficients
are discussed in \cite{JEG,VEG26,VEGfull26}; their distinction is explained
in the introduction.  For the kernel below, stationarity gives
\(\mu K=\mu\), and the density of \(\nu K\) is \(T_\rho h\) when
\(h=\dd\nu/\dd\mu\).

\begin{theorem}[Sharp product inequality]\label{thm:sharp}
Let \(1<q<2\), \(0<\rho<1\), and \(\lambda\in[0,1]\).  The product
inequality
\[
\log\|T_\rho^{[n]} f\|_q
\le
\E_{S\sim\lambda}\log\|\E(f\mid S)\|_q
\]
holds for every \(n\ge1\) and every nonzero nonnegative
\(f:\Omega^n\to\R\) if and only if
\[
\lambda\ge\lambda_*(q,\mu,\rho),
\]
where \(\lambda_*\) is given by Theorem~\ref{thm:finite}.  Equivalently,
\(\lambda_*=\eta_q(\mu,K)\) is the fixed-reference order-\(q\) R\'enyi
contraction coefficient of the one-coordinate channel
\[
K(x,y)=\rho\1_{x=y}+(1-\rho)\mu(y).
\]
\end{theorem}

\begin{proof}
The one-coordinate inequality is
\[
\log\|T_\rho h\|_q
\le
\lambda\log\|h\|_q+(1-\lambda)\log\E h
\]
for all nonzero \(h\ge0\).  By homogeneity, with \(u=h/\E h\), this is
equivalent to
\[
\log B(u)\le\lambda\log A(u)
\]
for all \(u\in\C\).  Hence it holds if and only if
\(\lambda\ge\lambda_*\).  If \(\dd\nu/\dd\mu=h\), then \(\mu K=\mu\),
\(\dd(\nu K)/\dd\mu=1-\rho+\rho h\), and the same quotient is
\[
\frac{\Dq(\nu K\|\mu)}{\Dq(\nu\|\mu)}.
\]

If \(\lambda=1\), the asserted product inequality is simply the
\(L_q\)-contraction \(\|T_\rho^{[n]}f\|_q\le\|f\|_q\).  We may therefore
assume \(\lambda<1\).  It remains to prove that the one-coordinate inequality
tensorizes.  Let
\((Y,\nu)\) be another finite probability space and let
\(F:\Omega\times Y\to[0,\infty)\).  Write
\[
m(y)=\E_xF(x,y).
\]
For \(m(y)>0\), put \(h_y(x)=F(x,y)/m(y)\), and ignore zero fibers.  Then
\(\E_xh_y=1\) and
\[
T_\rho^{(x)}F(x,y)=m(y)(1-\rho+\rho h_y(x)).
\]
If \(A_y=\E_xh_y^q\), the scalar inequality gives
\[
\|T_\rho^{(x)}F\|_q^q
\le
\E_y m(y)^q A_y^\lambda .
\]
H\"older's inequality, in the form
\[
m^qA_y^\lambda=(m^qA_y)^\lambda(m^q)^{1-\lambda},
\]
gives
\[
\E_y m(y)^qA_y^\lambda
\le
\left(\E_y m(y)^qA_y\right)^\lambda
\left(\E_y m(y)^q\right)^{1-\lambda}.
\]
But \(\E_y m(y)^qA_y=\|F\|_q^q\) and
\(\E_y m(y)^q=\|\E_xF\|_q^q\).  Taking logarithms and dividing by \(q\),
\[
\log\|T_\rho^{(x)}F\|_q
\le
\lambda\log\|F\|_q+(1-\lambda)\log\|\E_xF\|_q.
\]

The induction uses the following auxiliary-space statement.  For every finite
probability space \((Y,\nu)\) and every nonnegative
\(F:\Omega^n\times Y\to\R\),
\begin{equation}\label{eq:auxiliary-induction}
 \log\|(T_\rho^{[n]}\otimes I_Y)F\|_{L_q(\mu^n\otimes\nu)}
 \le
 \E_{S\sim\lambda}
 \log\|\E(F\mid S,Y)\|_{L_q(\mu^n\otimes\nu)},
\end{equation}
where \(\E(F\mid S,Y)\) integrates the \(\Omega\)-coordinates outside
\(S\) and retains the auxiliary variable.  The case \(n=0\) is equality,
and the one-coordinate estimate above proves the induction step with all
unprocessed coordinates and \(Y\) carried as the auxiliary space.

Indeed, if the last coordinate is handled after the first \(n-1\), the
induction gives
\[
\log\|T_\rho^{[n]}f\|_q
\le
\lambda\E_R\log\|\E(f\mid R\cup\{n\})\|_q
+(1-\lambda)\E_R\log\|\E(f\mid R)\|_q,
\]
where \(R\subseteq[n-1]\) includes each coordinate independently with
probability \(\lambda\).  This is exactly the expectation over
\(S\sim\lambda\).  Necessity follows from \(n=1\).
\end{proof}

\begin{lemma}[Equality in one tensorization step]\label{lem:fiber-equality}
Let \(\lambda=\lambda_*(q,\mu,\rho)\), let \((Y,\nu)\) be finite and fully
supported, and let
\(F:\Omega\times Y\to[0,\infty)\) be nonzero.  Put
\(m(y)=\E_xF(x,y)\), and for \(m(y)>0\) put
\[
h_y(x)=\frac{F(x,y)}{m(y)},\qquad A_y=\E_xh_y^q.
\]
Equality holds in
\begin{equation}\label{eq:one-step}
\log\|T_\rho^{(x)}F\|_q
=\lambda\log\|F\|_q+(1-\lambda)\log\|\E_xF\|_q
\end{equation}
if and only if both of the following conditions hold:
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em]
\item every positive normalized fiber \(h_y\) is a scalar equality case;
\item \(A_y\) is constant on \(\{y:m(y)>0\}\).
\end{enumerate}
If one positive fiber is constant, then all positive fibers are constant.
\end{lemma}

\begin{proof}
Write \(B_y=\E_x(1-\rho+\rho h_y)^q\).  The two inequalities in the
one-step proof are
\begin{align}
\E_y m(y)^qB_y
&\le \E_y m(y)^qA_y^\lambda,
\label{eq:fiber-scalar}\\
\E_y m(y)^qA_y^\lambda
&\le
\bigl(\E_y m(y)^qA_y\bigr)^\lambda
\bigl(\E_y m(y)^q\bigr)^{1-\lambda}.
\label{eq:fiber-holder}
\end{align}
All summands in the difference in \eqref{eq:fiber-scalar} are nonnegative,
and \(m(y)^q>0\) precisely on the positive fibers.  Equality therefore holds
there exactly when \(B_y=A_y^\lambda\) for every positive fiber, which is
condition~(a).

For \eqref{eq:fiber-holder}, H\"older is applied to
\(U(y)=m(y)^qA_y\) and \(V(y)=m(y)^q\) with exponents
\(1/\lambda\) and \(1/(1-\lambda)\).  Since \(0<\lambda<1\), its equality
condition is that \(U\) and \(V\) be proportional on their common positive
support.  This is exactly the constancy of \(A_y\), condition~(b).  Both
intermediate inequalities must be equalities if and only if the final
one-step estimate is an equality.

Finally, a constant mean-one fiber has \(A_y=1\).  Condition~(b) then gives
\(A_y=1\) on every positive fiber; strict convexity of \(r\mapsto r^q\) and
\(\E h_y=1\) force every such \(h_y\) to be constant.
\end{proof}

\begin{theorem}[Product equality for nonuniform laws]\label{thm:nonuniform-equality}
Suppose \(\mu\) is nonuniform.  For every \(1<q<2\) and \(0<\rho<1\),
the nonzero equality cases in the sharp product inequality are precisely
\[
 f(x_1,\ldots,x_n)=c\prod_{i=1}^n h_i(x_i),\qquad c>0,
\]
where each \(h_i\) is either \(1\) or a mean-one scalar extremizer
from Corollary~\ref{prop:twolevel}.  Uniqueness of the maximizing contrast
in \eqref{eq:least-ratio} is not required.
\end{theorem}

\begin{proof}
Fix one coordinate \(x\), write the others as \(y\), and let \(U\) be
product noise on \(y\).  Two orders of tensorization show that equality
forces the one-step equality of Lemma~\ref{lem:fiber-equality} both for
\(F=f\) and for \(F=Uf\).  To see the first assertion, apply
\eqref{eq:auxiliary-induction} in the \(y\)-coordinates to \(T_\rho^{(x)}f\),
with \(x\) as auxiliary space, and then apply the \(x\)-step to every
conditional expectation in the resulting average.  The term retaining
all \(y\)-coordinates has weight \(\lambda^{n-1}>0\) and contains the
\(x\)-step on raw \(f\).  For the second assertion, apply the \(x\)-step
first to \(Uf\), and then tensorize the two resulting terms in \(y\).
Both chains end at the same sharp product bound.  Every gap is
nonnegative and \(0<\lambda<1\), so equality forces the stated steps.

Put \(b(y)=\E_x f(x,y)\).  If one positive normalized raw fiber is
constant, Lemma~\ref{lem:fiber-equality} makes them all constant, and
\(f(x,y)=b(y)\).  Otherwise all positive raw fibers are nonconstant scalar
extremizers with a common \(q\)-moment.  Since \(M_{m,q}\) is strictly
increasing on \([0,1]\), they have a common contrast \(s>0\).  Thus
\[
 f(x,y)=b(y)\left(1-s+\frac{s}{m}\1_{\{x=a(y)\}}\right),
 \qquad b(y)>0,
\]
where each \(a(y)\) is a least atom.  A nonuniform law has an atom \(x_0\)
with \(\mu(x_0)>m\).  At that atom,
\[
 f(x_0,y)=(1-s)b(y),\qquad
 (Uf)(x_0,y)=(1-s)(Ub)(y).
\]
The kernel of \(U\) is strictly positive and \(b\not\equiv0\), so
\((Ub)(y)>0\).  Every normalized \(Uf\) fiber is a scalar equality case.
Its value at \(x_0\) is \(1-s<1\), excluding a constant fiber; as
\(x_0\) cannot be its peak, this value also forces its contrast to be
exactly \(s\).

Each normalized \(Uf\) fiber is a convex combination of the nonzero raw
normalized fibers, with strictly positive weights
\[
 \frac{U(y,y')b(y')}{(Ub)(y)}.
\]
A mixture of the vectors \(1-s+(s/m)\1_{\{a(y')\}}\) can itself equal
such a vector with the same \(s>0\) only when every peak label \(a(y')\)
is the same.  Indeed, after subtracting \(1-s\) and multiplying by
\(m/s\), a convex combination of point indicators is a point indicator
only in that case.  Hence \(f(x,y)=h_s^a(x)b(y)\).

Both sides of the product inequality add under products.  The factor
\(h_s^a\) is a scalar equality case, so equality for \(f\) is equivalent
to equality for \(b\) in one fewer coordinate.  Induction proves necessity,
and the same factorization verifies the converse.
\end{proof}

\section{Hypercontractive norms}\label{sec:hyper}

This section allows arbitrary \(1<p<q<\infty\).  The operator remains
\(T_\rho=\rho I+(1-\rho)\E_\mu\), and
\(\nu_m=(m,1-m)\) denotes the two-point probability law.

\begin{theorem}[Exact least-atom norm identity]\label{thm:hyper-norm}
For every finite full-support law \(\mu\) with at least two atoms,
\(m=\min_x\mu(x)\), \(1<p<q<\infty\), and \(0\le\rho\le1\),
\begin{equation}\label{eq:hyper-norm}
 \|T_\rho\|_{L^p(\mu)\to L^q(\mu)}
 =\|T_\rho\|_{L^p(\nu_m)\to L^q(\nu_m)}.
\end{equation}
For \(0<\rho<1\), every nonconstant nonnegative norm extremizer,
normalized to mean one, is \(h_s^a\) for a least atom \(a\) and some
\(0<s<1\).  In particular, with \(M_{m,r}\) defined by
\eqref{eq:moment-curve} for any \(r>1\),
\begin{equation}\label{eq:hyper-scalar}
 \|T_\rho\|_{p\to q}
 =\max_{0\le s\le1}
   \frac{M_{m,q}(\rho s)^{1/q}}{M_{m,p}(s)^{1/p}}.
\end{equation}
The identity holds for real or complex operator norms.
\end{theorem}

\begin{proof}
Assume first \(0<\rho<1\) and \(p\le2\).  Positivity gives
\(|T_\rho f|\le T_\rho|f|\), so nonnegative functions suffice even for
the complex norm.  Normalize a nonzero such function to \(\E f=1\).
In Theorem~\ref{thm:moment-principle}, use the functions
\(F(t)=t^p\) and \(G(t)=(1-\rho+\rho t)^q\).  Their derivative ratio is
positive, and
\begin{align*}
 \frac{G''(t)}{F''(t)}
 &=\frac{q(q-1)\rho^2}{p(p-1)}
       (1-\rho+\rho t)^{q-2}t^{2-p},\\
 \frac{\dd}{\dd t}\log\frac{G''(t)}{F''(t)}
 &=\frac{(q-p)\rho t+(2-p)(1-\rho)}{t(1-\rho+\rho t)}>0.
\end{align*}
The theorem replaces \(f\) by \(h_s^a\) with the same \(p\)-moment
and no smaller noisy \(q\)-moment.  Its norm ratio is that of a function
on \(\nu_m\), proving the upper bound in \eqref{eq:hyper-norm}.
Conversely every function on \(\nu_m\) embeds into \(\Omega\) by using
one least atom and its complement; this embedding preserves the two
norms and commutes with \(T_\rho\).  It gives the reverse bound.
Strictness in the moment comparison also gives the asserted two-level
shape of every nonconstant nonnegative extremizer in this range.

If \(p>2\), let \(p'=p/(p-1)\) and \(q'=q/(q-1)\).
Self-adjointness for the \(\mu\) pairing and finite-dimensional duality give
\[
 \|T_\rho\|_{p\to q}=\|T_\rho\|_{q'\to p'}.
\]
Here \(1<q'<p'<2\), so the case just proved applies, first on \(\mu\)
and then on \(\nu_m\), proving \eqref{eq:hyper-norm}.  To transfer the
extremizer statement, normalize \(f\ge0\) by \(\|f\|_p=1\) and take
\(g=(T_\rho f)^{q-1}/\|T_\rho f\|_q^{q-1}\).  Then \(\|g\|_{q'}=1\)
and
\[
 \|T_\rho\|_{p\to q}
 =\langle T_\rho f,g\rangle
 =\langle f,T_\rho g\rangle
 \le\|T_\rho g\|_{p'}
 \le\|T_\rho\|_{q'\to p'}.
\]
Every inequality is an equality.  Thus \(g\) is a dual extremizer and
H\"older equality gives \(f\) proportional to
\((T_\rho g)^{p'-1}\).  If \(f\) is nonconstant, then \(g\) is
nonconstant, since \(T_\rho\) is invertible.  The dual result makes \(g\),
and hence \(f\), two-valued with its high level on one least atom.

Every nonconstant nonnegative norm extremizer is strictly positive.
If \(f_i=0\), replacing it by \(\varepsilon>0\) changes
\(\|f\|_p^p\) by \(\mu_i\varepsilon^p=o(\varepsilon)\).  The strictly
positive kernel gives \(T_\rho f>0\), and the corresponding change in
\(\|T_\rho f\|_q^q\) has strictly positive first derivative.  This would
increase the norm ratio, a contradiction.  Therefore its contrast is
less than one.  Constants and the preceding extremizer classification
prove \eqref{eq:hyper-scalar}.  Finally, all norms in question are
continuous in \(\rho\), which includes \(\rho=0,1\).
\end{proof}

\begin{corollary}[Sharp boundary and its dependence on the least atom]
\label{cor:hyper-monotone}
For every \(1<p<q<\infty\),
\[
 \sigma_{p,q}(\mu)=\sigma_{p,q}(m).
\]
The function \(m\mapsto\sigma_{p,q}(m)\) is strictly increasing on
\((0,1/2]\).  Thus Conjecture~\ref{conj:least} holds.
\end{corollary}

\begin{proof}
The identity follows immediately from \eqref{eq:hyper-norm}.  For strict
monotonicity, duality reduces the proof to \(p\le2\).  Fix
\(0<m<1/2\), put \(Z=m^{-1}\1_{\{a\}}-1\) on \(\nu_m\), and set
\(\rho_0=\sqrt{(p-1)/(q-1)}\).  Then \(\E Z=0\) and \(\E Z^3>0\).
Taylor expansion at the constant function gives
\begin{align}\label{eq:hyper-expansion}
 \log\|1+\rho t Z\|_q-\log\|1+tZ\|_p
 &=\frac{(q-1)\rho^2-(p-1)}2\E Z^2\,t^2\notag\\
 &\quad+\frac{(q-1)(q-2)\rho^3-(p-1)(p-2)}6
           \E Z^3\,t^3+O(t^4).
\end{align}
At \(\rho=\rho_0\), the quadratic term vanishes, while the cubic bracket is
\[
 (p-1)\bigl[\rho_0(q-2)-(p-2)\bigr]>0.
\]
For \(q\ge2\) its positivity follows from \(p\le2\); for \(q<2\),
\(\rho_0(q-2)>q-2>p-2\).  A small positive \(t\) therefore violates
hypercontractivity at \(\rho_0\), and continuity gives
\(\sigma_{p,q}(m)<\rho_0\).

There is a nonconstant critical extremizer at
\(\sigma=\sigma_{p,q}(m)>0\).  Here are the required compactness details.
Normalize nonnegative inputs to mean one, forming a compact interval.
For sufficiently small \(\rho\), the strictly negative quadratic term
in \eqref{eq:hyper-expansion} handles a neighborhood of the constant,
uniformly in the two signs of \(t\).  On the compact complement,
\(\|T_0f\|_q=1<\|f\|_p\), and continuity preserves this strict inequality
for small \(\rho\).  Hence \(\sigma>0\).  The admissible noise parameters
form a closed interval: if \(\rho'<\rho\), then
\(T_{\rho'}=T_{\rho'/\rho}T_\rho\) and the first factor is an
\(L^q\)-contraction.  Choose violating maximizers at \(\rho_j\downarrow
\sigma\).  A subsequence converges to an input attaining norm ratio one
at \(\sigma\).  Since \(\sigma<\rho_0\), the quadratic term is uniformly
negative near the constant and near \(\sigma\); the violating maximizers
cannot approach the constant.  Their limit is therefore nonconstant.

Now take \(m_1<m_2<1/2\) and a nonconstant critical extremizer for
\(m_2\).  By Theorem~\ref{thm:hyper-norm} its high level has mass \(m_2\).
Decrease that mass to \(m_1\) while preserving its mean and \(p\)-moment.
The strict comparison in the proof of
Theorem~\ref{thm:moment-principle}, with \(F(t)=t^p\) and
\(G(t)=(1-\sigma+\sigma t)^q\), increases the noisy \(q\)-moment.
Thus the norm on \(\nu_{m_1}\) already exceeds one at
\(\rho=\sigma_{p,q}(m_2)\); by continuity it does so slightly earlier.
This proves \(\sigma_{p,q}(m_1)<\sigma_{p,q}(m_2)\).  At \(m_2=1/2\),
the classical value is \(\rho_0\) \cite[Eq.~(2.1)]{Wolff}, and the strict
inequality for biased laws was proved above.  Duality completes the proof.
\end{proof}

\section{Endpoint regimes}

\subsection{\texorpdfstring{The \(q\downarrow1\) soft-dictator regime}{The q to 1 soft-dictator regime}}

For a mean-one density \(h\), define
\[
\Dq(h)=\frac{1}{q-1}\log\E h^q,\qquad q>1,
\]
and
\[
D_1(h)=\Ent(h)=\E h\log h,
\]
with \(0\log0=0\).  Let \(Th=1-\rho+\rho h\) and set
\[
\eta_1=\max_{h\in\C}\frac{\Ent(Th)}{\Ent(h)},
\]
with the constant density assigned the limiting value \(\rho^2\).
This is the ordinary KL strong data-processing coefficient for the
resampling kernel; see Polyanskiy and Wu~\cite{PW} for general background on
KL contraction coefficients.  For uniform \(\mu\), this kernel is the Potts
semigroup, and Gu and Polyanskiy~\cite[Thm.~4 and Props.~17--18]{GuPolyanskiy}
already obtained an exact one-variable formula and the corresponding
one-coordinate entropy extremizers.  Proposition~\ref{prop:kl-finite}
extends that endpoint description to general fully supported laws having a
subset of mass below \(1/2\), in particular to every nonbinary law.
Continuity of the coefficient itself at \(q=1\) follows from the more
general result \cite[Cor.~1]{VEGfull26}.  Positivity of KL maximizers for
this strictly positive reversible kernel is also a special case of
\cite[Lem.~31]{CCGP}.  The direct arguments below additionally control the
ratios uniformly and use the two-level classification to obtain persistence
of interior extremizers in Theorem~\ref{thm:nearone}.

\begin{lemma}[Uniform KL limit]\label{lem:kl-limit}
As \(q\downarrow1\), \(R_q\to R_1\) uniformly on \(\C\), where
\[
R_1(h)=\frac{\Ent(Th)}{\Ent(h)}.
\]
Consequently \(\lambda_*(q,\mu,\rho)\to\eta_1\), and every subsequential
limit of \(R_q\)-maximizers is a KL maximizer.
\end{lemma}

\begin{proof}
Let \(M=1/\mu^*\).  Uniformly for \(x\in[0,M]\),
\[
x^{1+\varepsilon}=x+\varepsilon x\log x+O(\varepsilon^2),
\]
so \(D_{1+\varepsilon}(h)\to\Ent(h)\) uniformly on \(\C\); the same holds
for \(Th\).

Ratios near constants require a separate argument.  Fix \(q_+\in(1,2)\).
If \(h=1+u\), \(\E u=0\), and \(\|u\|_\infty\le s\), Taylor's formula with
a remainder bounded on the finite coordinate range gives, uniformly for
\(q\in(1,q_+]\),
\[
D_q(1+u)=\frac q2\E u^2+O(s\E u^2),
\qquad
D_q(1+\rho u)=\frac q2\rho^2\E u^2+O(s\E u^2).
\]
The same expansion at \(q=1\) follows from
\((1+x)\log(1+x)=x+x^2/2+O(|x|^3)\).  Thus \(R_q(1+u)=\rho^2+O(s)\)
and \(R_1(1+u)=\rho^2+O(s)\), with constants independent of \(q\).

For the complementary region, put
\[
 \mathcal K_s=\{h\in\C:\|h-1\|_\infty\ge s\}.
\]
It is compact and contains no constant density, so
\(d_s=\min_{h\in\mathcal K_s}\Ent(h)>0\).  Uniform convergence of
\(D_q(h)\) and \(D_q(Th)\) to their KL limits implies that, for \(q\) close
enough to \(1\), \(D_q(h)\ge d_s/2\) throughout \(\mathcal K_s\).  Division
is therefore uniform there, and
\[
 \sup_{h\in\mathcal K_s}|R_q(h)-R_1(h)|\longrightarrow0.
\]
Given an error tolerance, first choose \(s\) so that the two quadratic-region
errors are smaller than half that tolerance, and then choose \(q\) close
enough to \(1\) for the compact-complement error.  This proves uniform
convergence on all of \(\C\).
Uniform convergence on the compact set gives convergence of maxima and the
usual compactness statement for maximizers.
\end{proof}

\begin{lemma}[KL maximizers are soft]\label{lem:kl-soft}
Assume that \(\mu\) has a subset of mass strictly below \(1/2\).  Then
\(\eta_1>\rho^2\), and every nonconstant KL
maximizer is strictly positive and takes exactly two values.
\end{lemma}

\begin{proof}
The perturbation from Lemma~\ref{lem:exist}, with entropy in place of
\(\log L_q\), gives \(\eta_1>\rho^2\).  Indeed,
\[
\Ent(1+tg)=\frac{t^2}{2}\E g^2-\frac{t^3}{6}\E g^3+O(t^4)
\]
and the numerator has \(t\) replaced by \(\rho t\), so the quotient is
\[
\rho^2\left(1+\frac{(1-\rho)\E g^3}{3\E g^2}t+O(t^2)\right)>\rho^2
\]
for small positive \(t\).

Let \(h\) maximize the KL ratio and put \(\lambda=\eta_1\).  Then
\[
\Phi(g)=\Ent(Tg)-\lambda\Ent(g)
\]
is at most \(0\) on \(\C\) and equals \(0\) at \(h\).  Suppose \(h_i=0\).
Choose \(j\) with \(h_j>0\).  For sufficiently small \(s>0\), define the
weighted mass transfer
\[
 h_s(i)=\frac{s}{\mu_i},\qquad
 h_s(j)=h_j-\frac{s}{\mu_j},\qquad
 h_s(k)=h_k\quad(k\notin\{i,j\}).
\]
Then \(h_s\ge0\) and \(\E h_s=\E h=1\).  At the newly positive coordinate,
\[
 \mu_i h_s(i)\log h_s(i)=s\log s-s\log\mu_i,
\]
while the change at coordinate \(j\) is \(O(s)\).  Consequently
\(\Ent(h_s)-\Ent(h)=s\log s+O(s)\).  Since \(Th\ge1-\rho>0\) coordinatewise
and \(T(h_s-h)=O(s)\), differentiability of \(r\mapsto r\log r\) on this
positive range gives \(\Ent(Th_s)-\Ent(Th)=O(s)\).  Hence
\[
\Phi(h_s)-\Phi(h)=-\lambda s\log s+O(s)>0
\]
for small \(s>0\), a contradiction.  Thus \(h>0\).

The Lagrange equation on the relative interior says that
\[
\rho(\log(1-\rho+\rho r)+1)-\lambda(\log r+1)=c
\]
for each value \(r=h(x)\).  The derivative has sign determined by
\[
\frac{\rho^2r}{1-\rho+\rho r}-\lambda,
\]
which is a strictly increasing function of \(r\).  Therefore the left side
first decreases and then possibly increases, so a horizontal line has at most
two positive intersections.  Since \(h\) is not constant, it has exactly two
positive values.
\end{proof}

\begin{proposition}[Least-atom KL formula]\label{prop:kl-finite}
For \(m=\min_x\mu(x)\), define
\[
 K_m(s)=m\left(1+\frac{1-m}{m}s\right)
             \log\left(1+\frac{1-m}{m}s\right)
       +(1-m)(1-s)\log(1-s),\qquad 0\le s\le1.
\]
Then
\begin{equation}\label{eq:kl-finite}
 \eta_1=\max_{0\le s\le1}\frac{K_m(\rho s)}{K_m(s)},
\end{equation}
with value \(\rho^2\) at zero.  If \(m<1/2\), every nonconstant KL
extremizer is \(h_s^a\) for a least atom \(a\) and some \(0<s<1\).
\end{proposition}

\begin{proof}
For a fixed density \(h\), choose \(s_q\) such that
\(M_{m,q}(s_q)=\E h^q\).  Uniformly on \([0,1]\),
\[
 \frac{M_{m,q}(s)-1}{q-1}\longrightarrow K_m(s).
\]
The function \(K_m\) is strictly increasing from \(0\) to \(\log(1/m)\).
Compactness and the preceding limit therefore imply that \(s_q\to s\),
where \(K_m(s)=\Ent(h)\).  Subtracting one from
\eqref{eq:sharp-profile}, dividing by \(q-1\), and taking the limit yields
\(\Ent(T_\rho h)\le K_m(\rho s)\).  Every \(s\) is attained by
\(h_s^a\), proving \eqref{eq:kl-finite}.

For \(m<1/2\), Lemma~\ref{lem:kl-soft} makes every nonconstant KL
extremizer positive and two-valued.  Apply the fixed-moment comparison
from the proof of Theorem~\ref{thm:moment-principle} with
\(f(t)=t\log t\) and
\(g(t)=(1-\rho+\rho t)\log(1-\rho+\rho t)\).  The high-mass comparison
path stays strictly positive when its starting low value is positive,
so differentiability at zero is not needed.  On that path,
\[
 \frac{g''(t)}{f''(t)}=\frac{\rho^2t}{1-\rho+\rho t}
\]
is strictly increasing.  Reducing any high-level mass larger than \(m\)
would strictly improve its output entropy at fixed input entropy.
Thus its high level has mass \(m\), and positivity gives \(s<1\).
\end{proof}

\begin{theorem}[Uniform soft dictators near \(q=1\)]\label{thm:nearone}
If \(\mu\) has a subset of mass strictly below \(1/2\), then there are
\(\delta>0\) and \(q_1>1\) such that, for
every \(q\in(1,q_1)\), every one-coordinate maximizer \(h_q\) of \(R_q\)
satisfies
\[
h_q(x)\ge\delta\qquad\text{for all }x\in\Omega.
\]
Consequently every nonconstant one-coordinate maximizer in this range is a
strictly positive two-level soft dictator.
\end{theorem}

\begin{proof}
Let \(\mathcal K_1\) be the compact set of KL maximizers.  By
Lemma~\ref{lem:kl-soft}, every element of \(\mathcal K_1\) is strictly
positive, so
\[
\delta_0=\min_{h\in\mathcal K_1}\min_{x\in\Omega}h(x)>0.
\]
If a sequence \(q_j\downarrow1\) had maximizers staying at sup-norm distance
at least \(\delta_0/2\) from \(\mathcal K_1\), compactness would give a
subsequence converging to a KL maximizer by Lemma~\ref{lem:kl-limit}, a
contradiction.  Thus all maximizers for \(q\) close enough to \(1\) lie within
\(\delta_0/2\) of \(\mathcal K_1\), and are bounded below by
\(\delta_0/2\).  Corollary~\ref{prop:twolevel} then rules out subset
indicators and gives the two-level soft form.
\end{proof}

\subsection{\texorpdfstring{The \(q\uparrow2\) least-atom regime}{The q to 2 least-atom regime}}

\begin{theorem}[Least-atom dictators near \(q=2\)]\label{thm:neartwo}
Let \(m=\mu^*\).  There is \(q_0=q_0(\mu,\rho)\in(1,2)\) such that, for every
\(q\in(q_0,2)\),
\[
\lambda_*(q,\mu,\rho)=
\frac{
\log\left[m(1-\rho+\rho/m)^q+(1-m)(1-\rho)^q\right]
}{(q-1)\log(1/m)}.
\]
In this interval the only nonconstant one-coordinate maximizers are
\[
h=\frac{\1_{\{x\}}}{m},\qquad \mu(x)=m.
\]
Moreover, equality in the sharp product inequality on \(\Omega^n\) holds
exactly for functions
\[
f(x_1,\ldots,x_n)
=c\prod_{i\in I}\frac{\1_{\{x_i=a_i\}}}{m},
\]
where \(c>0\), \(I\subseteq[n]\), and each \(a_i\) is a least atom.
\end{theorem}

\begin{proof}
At \(q=2\),
\[
R_2(h)=
\frac{\log(1+\rho^2\Var(h))}{\log(1+\Var(h))}.
\]
For \(0<c<1\), the function \(z\mapsto\log(1+cz)/\log(1+z)\) is strictly
increasing on \([0,\infty)\): after differentiating, the numerator has sign
equal to
\[
c(1+z)\log(1+z)-(1+cz)\log(1+cz),
\]
whose derivative is \(c\log((1+z)/(1+cz))>0\).  Thus maximizing \(R_2\) is
the same as maximizing variance in \(\C\).

If \(M=\max_xh(x)\), then \(h^2\le Mh\), so \(\E h^2\le M\).  If
\(h(x_0)=M\), then \(1=\E h\ge\mu(x_0)M\), hence \(M\le1/m\).  Equality in
\(\E h^2\le1/m\) forces \(h\) to be zero off a set of total mass \(m\) and
constant on that set.  Since every atom has mass at least \(m\), this set is
a single least atom.  Therefore the only nonconstant \(q=2\) maximizers are
the normalized least-atom indicators.

For persistence, write
\[
 F_q(s)=\frac{\log M_{m,q}(\rho s)}{\log M_{m,q}(s)},\qquad
 F_q(0)=\rho^2.
\]
Taylor expansion at zero shows that \(F_q\) is jointly continuous in
\((q,s)\) on \([q_-,2]\times[0,1]\) for any \(q_->1\).  Its derivative
in \(s\) is jointly continuous on \([q_-,2]\times[\varepsilon,1]\)
for every \(\varepsilon>0\): the powers \((1-s)^q\) have continuous
first derivatives there, including at \(s=1\).
At \(q=2\),
\[
 M_{m,2}(s)=1+\frac{1-m}{m}s^2.
\]
The strict monotonicity proved above gives \(F'_2(s)>0\) for \(s>0\).
Choose \(\delta\in(0,1)\).  Its derivative has a positive minimum on
\([1-\delta,1]\), which stays positive for \(q<2\) sufficiently close
to two.  On the compact complement \([0,1-\delta]\), the value of
\(F_2\) is separated from \(F_2(1)\) by a positive gap.  Uniform
continuity preserves that gap for nearby \(q\).  Hence the unique
maximizing contrast remains \(s=1\).  Theorem~\ref{thm:finite} and its
strict equality statement give the claimed scalar formula and
classification.  Notice that this argument depends on \(\mu\) only
through \(m\).

It remains to classify product equality cases.  The induction assertion is:
for every \(k<n\), every nonzero equality case on \(\Omega^k\) is a positive
constant times a product of normalized least-atom indicators on a subset of
the coordinates.  The case \(k=1\) is the scalar classification just proved.
Assume the assertion through dimension \(n-1\).  Write
points of \(\Omega^n\) as \((y,z)\in\Omega^{n-1}\times\Omega\), put
\(U=T_\rho^{[n-1]}\), and set \(F(y,z)=Uf(\,\cdot\,,z)(y)\).  The induction
proof of Theorem~\ref{thm:sharp} factors as
\begin{align*}
\log\|T_\rho^{[n]}f\|_q
&=\log\|T_\rho^{(z)}F\|_q\\
&\le \lambda\log\|F\|_q
 +(1-\lambda)\log\|\E_zF\|_q\\
&\le \lambda\E_R\log\|\E(f\mid R\cup\{n\})\|_q
 +(1-\lambda)\E_R\log\|\E(f\mid R)\|_q,
\end{align*}
where \(R\subseteq[n-1]\) is sampled with inclusion probability \(\lambda\).
The last line is the desired right-hand side.  Every gap is nonnegative;
therefore equality for \(f\) forces equality in the first one-step estimate.

For a last-coordinate slice \(f_z(y)=f(y,z)\), the kernel of \(U\) is
strictly positive.  Thus \(Uf_z\) is positive everywhere if \(f_z\not\equiv0\),
and is identically zero otherwise.  Since \(f\ne0\),
\(r(y)=\E_zF(y,z)>0\) for every \(y\).  Lemma~\ref{lem:fiber-equality},
applied to the normalized fibers \(h_y(z)=F(y,z)/r(y)\), now says that each
\(h_y\) is either constant or a normalized least-atom indicator, and that all
their \(q\)-moments coincide.  The two possible moments are respectively
\(1\) and \(m^{1-q}>1\), so all fibers have the same type.

Suppose first that all fibers are constant.  Then \(F(y,z)=r(y)\) is
independent of \(z\), so \(Uf_z=r\) for every \(z\).  The operator \(U\) is
invertible: a one-coordinate \(T_\rho\) acts as the identity on constants and
as multiplication by \(\rho>0\) on the mean-zero subspace, and \(U\) is a
tensor product of these operators.  Hence all slices \(f_z\) are equal, and
\(f(y,z)=g(y)\).  For such a function, both sides of the product inequality
are exactly their \((n-1)\)-dimensional counterparts for \(g\).  The induction
hypothesis gives the asserted product form with coordinate \(n\) inactive.

Suppose instead that all fibers are least-atom indicators.  Let
\(J=\{z:f_z\not\equiv0\}\).  Strict positivity of \(U\) shows that the support
of \(z\mapsto F(y,z)\) is exactly \(J\), independently of \(y\).  Each
normalized fiber is supported on one point, so \(J=\{a_n\}\) for a least atom
\(a_n\).  With \(\chi_{a_n}=m^{-1}\1_{\{a_n\}}\) and
\(g(y)=m f(y,a_n)\), one has
\[
f(y,z)=g(y)\chi_{a_n}(z).
\]
Independence of coordinates gives
\[
\log\|T_\rho^{[n]}f\|_q
=\log\|T_\rho^{[n-1]}g\|_q+\log\|T_\rho\chi_{a_n}\|_q,
\]
whereas the conditional-expectation side equals
\[
\E_R\log\|\E(g\mid R)\|_q
+\lambda\log\|\chi_{a_n}\|_q.
\]
The scalar equality
\(\log\|T_\rho\chi_{a_n}\|_q=\lambda\log\|\chi_{a_n}\|_q\)
therefore shows that equality for \(f\) is equivalent to equality for \(g\).
The induction hypothesis adds the last active factor and proves necessity.

Conversely, for every function in the displayed family, the product noise,
the \(L_q\)-norm, and the conditional expectations factor coordinatewise.
Each active factor is a scalar equality case and every inactive factor is
constant, so equality follows.
\end{proof}

\begin{example}[Uniform three-point space]
Let \(\Omega=\{1,2,3\}\) with the uniform measure and let \(\rho=1/2\).  As
\(q\downarrow1\), a singleton indicator has limiting KL ratio
\[
\frac{\frac13\log2}{\log3}<\frac14,
\]
and the two-point indicator has limiting ratio
\[
\frac{5\log(5/4)-\log2}{6\log(3/2)}<\frac14.
\]
The local constant perturbation value is \(\rho^2=1/4\), and the asymmetric
cubic perturbation from Lemma~\ref{lem:exist} improves strictly above
\(1/4\).  Hence, for all \(q>1\) sufficiently close to \(1\), no subset
indicator is extremal on the uniform three-point space.  The optimizer is a
strictly positive two-level soft dictator.  By Theorem~\ref{thm:neartwo}, for
\(q<2\) sufficiently close to \(2\), the optimizer is instead a singleton
indicator.  Thus the optimizer type genuinely changes inside \(1<q<2\).

For a concrete numerical illustration, maximizing the single
one-variable objective in Theorem~\ref{thm:finite} at \(q=1.2\) gives
\[
 \lambda_*(1.2,\mu,1/2)\approx0.2610444999,
 \qquad
 h\approx(2.29520,0.35240,0.35240),
\]
up to permutation of the coordinates.  The corresponding KL endpoint from
Proposition~\ref{prop:kl-finite} is
\(\eta_1\approx0.2564934769>\rho^2=1/4\).  These decimal values are not used
in any proof; they display the soft, non-indicator optimizer well inside the
simplex.
\end{example}

\section{The uniform binary phase diagram}\label{sec:binary}

The least-atom reduction leaves one scalar contrast to optimize.  For
\(m=1/2\), its full behavior can be described.  Put
\[
 F(s)=\log\frac{(1+s)^q+(1-s)^q}{2},\qquad
 R_\rho(s)=\frac{F(\rho s)}{F(s)},\qquad R_\rho(0)=\rho^2,
\]
and define the elasticity \(E(s)=sF'(s)/F(s)\), with \(E(0)=2\).
The following theorem is about the ratio \(R_\rho\); the first threshold
and the Riccati ingredients have the precedents discussed in the
introduction.

\begin{theorem}[Uniform binary phases]\label{thm:binary-phase}
Let \(\mu=(1/2,1/2)\), \(0<\rho<1\), and set
\[
 q_0=\frac{\sqrt{13}-1}{2},\qquad
 q_1=1+\frac1{4\log2-1}.
\]
The scalar optimization in \eqref{eq:least-ratio} has these three regimes.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.3em]
\item If \(1<q\le q_0\), then \(R_\rho\) is strictly decreasing on
\((0,1]\).  Hence \(\lambda_*=\rho^2\), and the only nonzero scalar
equality cases are constants.
\item If \(q_0<q\le q_1\), then \(R_\rho\) has a unique maximizer
\(s_*\in(0,1)\) for every \(\rho\in(0,1)\).
\item If \(q_1<q<2\), there is a unique \(\rho_c(q)\in(0,1)\) such that
the unique maximizer is \(s_*=1\) for \(0<\rho\le\rho_c(q)\), and
\(s_*\in(0,1)\) for \(\rho_c(q)<\rho<1\).
\end{enumerate}
For \(q>q_0\), the elasticity \(E\) has exactly one minimum, at some
\(s_0\in(0,1)\).  In case \textup{(iii)}, the critical value is the
unique \(\rho_c\in(0,s_0)\) satisfying
\begin{equation}\label{eq:binary-critical}
 E(\rho_c)=E(1)=\frac{q}{2(q-1)\log2}.
\end{equation}
In all cases every nonzero product equality case factors into scalar
equality cases and constants.  In case \textup{(i)} it is constant.
\end{theorem}

\begin{proof}
Write \(a=q-1\in(0,1)\).  We first establish the needed shape of \(E\).
The auxiliary function
\[
 C(z)=a\sqrt z\,\coth\bigl(a\operatorname{artanh}\sqrt z\bigr),
 \qquad C(0)=1,
\]
is analytic at zero.  Differentiating gives the Riccati equation
\begin{equation}\label{eq:riccati}
 2z(1-z)C'=(1-z)C-C^2+a^2z.
\end{equation}
Its first three nonconstant Taylor coefficients are
\begin{align*}
 c_1&=-\frac{1-a^2}{3},&
 c_2&=-\frac{(1-a^2)(4-a^2)}{45},&
 c_3&=-\frac{(1-a^2)(4-a^2)(11-2a^2)}{945}.
\end{align*}
For instance, substitution in \eqref{eq:riccati} gives
\((2j+1)c_j=(2j-3)c_{j-1}-\sum_{i=1}^{j-1}c_i c_{j-i}\) for \(j\ge2\).
Thus \(C'(0),C''(0),C'''(0)<0\).

For \(0<z<1\), one has \(C(z)>a\): setting
\(u=\sqrt z=\tanh t\), the inequality follows from
\(\coth(at)>\coth(t)=1/u\).  Successive derivatives of
\eqref{eq:riccati} give
\begin{align}
 2z(1-z)C''&=a^2-C+(-1+3z-2C)C',\label{eq:riccati-one}\\
 2z(1-z)C'''&=(-3+7z-2C)C''+2C'(1-C'),\label{eq:riccati-two}\\
 2z(1-z)C''''&=(-5+11z-2C)C'''+(9-6C')C''.
 \label{eq:riccati-three}
\end{align}
At a first zero of \(C'\), equation~\eqref{eq:riccati-one} would give
\(C''<0\), contrary to its nonnegative derivative at that zero.
Therefore \(C'<0\).  Equation~\eqref{eq:riccati-two} excludes a first
zero of \(C''\) in the same way, giving \(C''<0\).  This is the
negative-curvature argument in \cite{TrureTuring}.  The next identity,
\eqref{eq:riccati-three}, excludes a first zero of \(C'''\), since
\((9-6C')C''<0\).  Consequently
\begin{equation}\label{eq:c-signs}
 C'<0,\qquad C''<0,\qquad C'''<0\qquad(0<z<1).
\end{equation}

Let \(D(z)=C(z)+az\).  The substitution \(s=\tanh t\) yields
\[
 F(s)=\log\cosh(qt)-q\log\cosh t,\qquad
 F'(s)=\frac{q\cosh t\sinh(at)}{\cosh(qt)}.
\]
Using \(\cosh(qt)=\cosh(at)\cosh t+\sinh(at)\sinh t\), this gives
\begin{equation}\label{eq:binary-D}
 \frac{F'(s)}s=\frac{qa}{D(s^2)}.
\end{equation}
In particular \(D>0\), \(D''<0\), \(D'''<0\), and
\(D'(0)=(a^2+3a-1)/3\).  Concavity implies
\(D(z)-zD'(z)\ge D(0)=1\).  Define
\[
 r(z)=\frac{zD'(z)}{D(z)},\qquad
 N(z)=D(z)\bigl(D'(z)+zD''(z)\bigr)-zD'(z)^2.
\]
Direct differentiation gives
\begin{equation}\label{eq:binary-N}
 r'(z)=\frac{N(z)}{D(z)^2},\qquad
 N'(z)=D''(z)\bigl(2D(z)-zD'(z)\bigr)+zD(z)D'''(z)<0.
\end{equation}
The strict sign uses \(2D-zD'>0\) and \eqref{eq:c-signs}.

If \(q\le q_0\), then \(N(0)=D'(0)\le0\), so \(r'<0\) on
\((0,1)\).  If \(q>q_0\), then \(N(0)>0\).  In this case \(D'\) crosses
zero: indeed \(C'(z)\to-\infty\) as \(z\uparrow1\).  To verify the limit
directly, put \(u=\sqrt z\) and \(v=(1-u)/(1+u)\).  Then
\[
 C'(z)=\frac{a(1+v^a)}{2u(1-v^a)}
       -\frac{2a^2v^{a-1}}{(1+u)^2(1-v^a)^2}\longrightarrow-\infty.
\]
Since \(D''<0\), its derivative has a unique zero \(z_0\), where
\(N(z_0)=z_0D(z_0)D''(z_0)<0\).  Equation~\eqref{eq:binary-N} shows that
\(N\) crosses zero exactly once.

By \eqref{eq:binary-D},
\[
 V(s):=1+\frac{sF''(s)}{F'(s)}=2-2r(s^2).
\]
It follows that \(V\) is strictly increasing if \(q\le q_0\), and strictly
decreasing then strictly increasing if \(q>q_0\).  Also \(V(0)=2\) and
\(V(s)\to+\infty\) at one.  The latter follows from \(D'\to-\infty\)
and \(D(1)=2a>0\).

The same shape transfers to \(E\).  Since \((sF'(s))'=F'(s)V(s)\),
\begin{equation}\label{eq:elasticity-average}
 E(s)=\frac{\int_0^s F'(u)V(u)\,\dd u}
            {\int_0^s F'(u)\,\dd u},\qquad
 E'(s)=\frac{F'(s)}{F(s)}\bigl(V(s)-E(s)\bigr).
\end{equation}
If \(V\) is increasing, its average is smaller than its endpoint value,
so \(E\) is strictly increasing.  In the one-valley case, \(E'<0\) on
the decreasing branch of \(V\).  On its increasing branch, the function
\(Q(s)=V(s)F(s)-sF'(s)\) satisfies \(Q'(s)=V'(s)F(s)>0\).
It starts negative and tends to \(+\infty\) at one, because \(F(1)>0\)
and \(F'(1)<\infty\).  Hence it crosses zero exactly once.  Formula
\eqref{eq:elasticity-average} proves that \(E\) has one strict minimum
\(s_0\).  Direct endpoint evaluation gives
\[
 E(1)=\frac{q}{2(q-1)\log2}.
\]

For \(s>0\), logarithmic differentiation gives
\begin{equation}\label{eq:binary-ratio-sign}
 \frac{sR_\rho'(s)}{R_\rho(s)}=E(\rho s)-E(s).
\end{equation}
If \(E\) is increasing this is negative, proving case~(i).
Otherwise the difference is positive when \(s\le s_0\), negative when
\(\rho s\ge s_0\), and strictly decreasing on the intermediate region
\(\rho s<s_0<s\).  It has at most one zero, changing only from positive
to negative.  Therefore the unique maximum is at \(s=1\) exactly when
\(E(\rho)\ge E(1)\), and is interior otherwise.  If \(q\le q_1\),
then \(E(1)\ge E(0)=2\), and the one-valley shape gives
\(E(\rho)<E(1)\) for every \(\rho<1\).  If \(q>q_1\), then
\(E(1)<2\), so the decreasing branch crosses \(E(1)\) exactly once,
at the value \(\rho_c\) specified in \eqref{eq:binary-critical}.
The increasing branch is below \(E(1)\) until its endpoint.  Thus
\(E(\rho)\ge E(1)\) holds exactly for \(\rho\le\rho_c\), proving
(ii)--(iii), including uniqueness at the transition.

Finally consider equality on products.  As in the two tensorization
orders used in Theorem~\ref{thm:nonuniform-equality}, both the raw fibers
in a coordinate and the fibers after fully noising all other coordinates
satisfy Lemma~\ref{lem:fiber-equality}.  In case~(i), all raw fibers are
constant in every coordinate, so \(f\) is constant.  For \(q>q_0\),
if a coordinate has nonconstant raw fibers, all its positive normalized
raw fibers have the unique contrast \(s_*\).  The fully mixed fibers
are either all constant or all nonconstant by their common moment.
They cannot all be constant: invertibility of the other-coordinate
noise would then make the raw function independent of the selected
coordinate.  Thus the mixed fibers also have contrast \(s_*\).
Every mixed normalized fiber is a strictly positive convex combination
of all positive raw normalized fibers.  Its \(q\)-moment equals their
common value \(M_{1/2,q}(s_*)\).  Strict convexity of the \(q\)-moment
forces all of those raw fibers to coincide.  One scalar factor therefore
splits off.  Induction, with coordinatewise additivity of both sides,
proves necessity and sufficiency of the product form.
\end{proof}

\section{Concluding remarks}

The least-atom principle removes the alphabet and all other atom masses
from both sharp constants.  For the R\'enyi ratio, the remaining question
is scalar: for general \(m<1/2\), which contrast maximizes
\(\log M_{m,q}(\rho s)/\log M_{m,q}(s)\), and is it unique?
Explicit transition curves would quantify the soft and hard regimes.
The nonuniform product equality proof uses an atom larger than the
minimum.  The uniform binary case is settled by its unique contrast in
Theorem~\ref{thm:binary-phase}.  For uniform alphabets of size at least
three, an additional scalar rigidity argument is needed outside the
hard regime; the fixed-moment theorem alone does not supply it.

\small
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\end{thebibliography}

\end{document}