Least-Atom Extremizers for Resampling Noise

Introduction

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

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:

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 $q>1$ [2 Thm. 1.1], and obtained the sharp parameter for integer $q\geq2$ [9 Thm. 1.1]. Abawonse, Hązła, and O’Donnell extended the sharp result to finite product probability spaces and real $q\geq2$ [1 Thms. 1 and 19]. 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 [1] below refers to its full arXiv version.

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 [12 Thm. 3.1] gives, in our order of exponents,

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 [13 Thm. 1.6 and Cor. 1.7]. They pose the following conjecture immediately after Figure 3 on page 8.

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 3.1). This identifies the mass as well as the number of levels. For a least atom $a$, the resulting mean-one family is

It interpolates between a constant and the normalized indicator of $a$. The terms soft and 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.

• (i) For $1 < q < 2$, the entire output $q$-moment profile at a fixed input $q$-moment is attained by (2). Consequently the sharp parameter in (1) is a maximum over the single interval $0 \le s \le1$, depending on $\mu$ only through $m ({xref}Theorems 4.3 and 5.1`). Every nonconstant scalar extremizer belongs to this family.

• (ii) 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 6.1). This proves Conjecture 1.2, and the sharp noise parameter is strictly increasing in $m$ (Corollary 6.2).

• (iii) For every nonuniform $\mu$, all product equality cases in (1) factor into scalar equality cases and constants (Theorem 5.3). 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 7.4 and 7.5).

• (iv) For the uniform binary law, Theorem 8.1 determines all scalar phases, with thresholds $(\sqrt{13}-1)/2$ and $1+(4\log2-1)^{-1}$, and classifies product equality throughout $1<q<2$.

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.

Related work and scope

The sharp Samorodnitsky parameter is the fixed-reference Rényi contraction coefficient of the resampling channel. This interpretation already appears for $q \ge2$ in [1 Thm. 3 and Rem. 20], whose Theorem 12 gives the scalar-to-product passage. Vandenbroucque, Esposito, and Gastpar extend that passage to arbitrary product channels [10 Thm. 10(a)]. 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 [7 Thm. 4 and Props. 17–18]. Positivity of KL extremizers for strictly positive reversible kernels follows more generally from [11 Lem. 31], and continuity in the Rényi order at one is known for arbitrary finite channels [10 Cor. 1]. 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 [12 Thm. 3.1]. The extra comparison here selects its smallest mass and identifies the norm even outside the hypercontractive region. The operator is precisely $\mathsf{E}_{\mu} + \rho(I - \mathsf{E}_{\mu})$; 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 [12 Eq. (2.1)].

A fixed reference law must be distinguished from optimization over both laws in a divergence ratio. The binary-input reductions of Ordentlich and Polyanskiy [6 Thms. 1–2], and their Rényi counterparts [5 Thm. 1] and [10 Thm. 6], 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 [4]; related Rényi bounds appear in [3, 5, 10], and constrained-input divergence bounds motivated by privacy amplification in [8].

The threshold $(\sqrt{13} - 1)/2$ already occurs in the binary Rényi analysis of Song et al. [14 Lem. 2]. The Riccati representation and negative second derivative used below also appear in trureturing’s paper-level analysis of Cachin Rényi information combining [15]. 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.

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

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 $S \subseteq[n]$, $\mathbb{E}(f \mid S)$ 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, Hazła, and O’Donnell [1] treats $q \in[2,\infty]$. For $n = 1$ it proves

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 Rényi formulation is made explicit in Theorem 3 and Remark 20.

The first item of [1 Sec. 1.4] 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

Hypercontractive norms in § 6 use separate exponents $1 < p < q < \infty$, explicitly replacing the restriction $q < 2$.

A fixed-moment least-atom principle

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

Proof. Write $M = \mathbb{E}f(h)$. On the compact mean-one simplex the strictly convex functional $\mathbb{E}f(h)$ is minimized only by $h = 1$ and maximized at a vertex. The vertex supported on an atom of mass $a$ has value

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 (6) has these endpoint values, and its derivative for $s > 0$ is

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

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

For $N(a) = ag(x) + (1-a)g(y)$, substitution yields

where $U$ has density $f''(u)/D$ on $(y,x)$. The second identity uses $F-f'(y)\Delta=\int_y^x(x-u)f''(u)\,du$ and its analogue for $g$. For independent copies $U,V$, twice the covariance is

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 $\mathbb{E}g(h)$ 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

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.

The one-coordinate coefficient

Proof. Since $\mu$ is fully supported, $0 \le h(x) \le1/\mu(x)$ for every $h \in C$. Hence $C$ is compact. If $h = 1 + u$, $\mathbb{E}u = 0$, and $\lVert u\rVert_\infty\to0$, Taylor expansion gives

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

and strict convexity of $x^q$ gives

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

Then $\mathbb{E}g=0$, $\mathbb{E}g^2=a/(1-a)$, and $\mathbb{E}g^3=a(1-2a)/(1-a)^2>0$. For small $t>0$, $h_t=1+tg$ lies in $C$. Writing $L(t)=\log\mathbb{E}(1+tg)^q$, expansion gives

Therefore

Since $q-2<0$, $\rho-1<0$, and $\mathbb{E}g^3>0$, the coefficient of $t$ is positive. Thus $R_q(h_t)>\rho^2$ for all sufficiently small positive $t$.

Proof. Apply Theorem 3.1 with $f(t)=t^q$ and $g(t)=(1-\rho+\rho t)^q$. Its decisive hypothesis is

which is strictly increasing for $t>0$. The function $h_s^a$ is taken by $T_\rho$ to $h_{\rho s}^a$, giving (10) 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 (11), with the value at zero supplied by Lemma 4.2.

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 (10). Conversely each maximizing profile attains the ratio. The constant equality follows directly from $T_\rho1=1$.

Sharp product tensorization

The scalar-to-product passage itself is already contained in the general induction theorem of Abawonse, Hązła, and O’Donnell [1 Theorem 12]. 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 Hölder 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-$q$ Rényi divergence is

For a channel $K$ and a fixed reference input law $\pi$, define the fixed-reference coefficient

The channel-level coefficient

also optimizes the reference law. Throughout this paper the reference law remains $\mu$. General properties and bounds for these two coefficients are discussed in [3, 5, 10]; 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 $h=d\nu/d\mu$.

Proof. The one-coordinate inequality is

for all nonzero $h\ge0$. By homogeneity, with $u=h/\mathbb{E}h$, this is equivalent to

for all $u \in\mathbb{C}$. Hence it holds if and only if $\lambda\ge\lambda_*$. If $\mathrm{d}\nu/\mathrm{d}\mu=h$, then $\mu K=\mu$, $\mathrm{d}(\nu K)/\mathrm{d}\mu=1-\rho+\rho h$, and the same quotient is

If $\lambda=1$, the asserted product inequality is simply the $L_q$-contraction $\lVert T_\rho^{[n]}f\rVert_q\le\lVert f\rVert_q$. 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

For $m(y)>0$, put $h_y(x)=F(x,y)/m(y)$, and ignore zero fibers. Then $\mathbb{E}_xh_y=1$ and

If $A_y=\mathbb{E}_xh_y^q$, the scalar inequality gives

Hölder’s inequality, in the form

gives

But $\mathbb{E}_ym(y)^qA_y=\lVert F\rVert_q^q$ and $\mathbb{E}_ym(y)^q=\lVert\mathbb{E}_xF\rVert_q^q$. Taking logarithms and dividing by $q$,

The induction uses the following auxiliary-space statement. For every finite probability space $(Y,\nu)$ and every nonnegative $F:\Omega^n\times Y\to\mathbb{R}$,

where $\mathbb{E}(F\mid S,Y)$ 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

where $R\subseteq[n-1]$ includes each coordinate independently with probability $\lambda$. This is exactly the expectation over $S\sim\lambda$. Necessity follows from $n=1$.

(b) $A_y$ is constant on $\{y:m(y)>0\}$.

If one positive fiber is constant, then all positive fibers are constant.

Proof. Write $B_y=\mathbb{E}_x(1-\rho+\rho h_y)^q$. The two inequalities in the one-step proof are

All summands in the difference in (16) 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 (17), Hölder 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 $\mathbb{E}h_y=1$ force every such $h_y$ to be constant.

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 5.2 both for $F=f$ and for $F=Uf$. To see the first assertion, apply (14) 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 $b(y)=\mathbb{E}_x f(x,y)$. If one positive normalized raw fiber is constant, Lemma 5.2 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

where each $a(y)$ is a least atom. A nonuniform law has an atom $x_0$ with $\mu(x_0)>m$. At that atom,

The kernel of $U$ is strictly positive and $b\ne0$, 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

A mixture of the vectors $1-s+(s/m)\mathbf{1}_{\{a(y')\}}$ 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 $f(x,y)=h^a_s(x)b(y)$.

Both sides of the product inequality add under products. The factor $h^a_s$ 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.

Hypercontractive norms

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.

Proof. Assume first $0<\rho<1$ and $p\leq2$. Positivity gives $\lvert T_\rho f\rvert\leq T_\rho\lvert f\rvert$, so nonnegative functions suffice even for the complex norm. Normalize a nonzero such function to $\mathbb{E}f=1$. In Theorem 3.1, use the functions $F(t)=t^p$ and $G(t)=(1-\rho+\rho t)^q$. Their derivative ratio is positive, and

The theorem replaces $f$ by $h^a_s$ 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 (18). 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

Here $1<q'<p'<2$, so the case just proved applies, first on $\mu$ and then on $\nu_m$, proving (18). To transfer the extremizer statement, normalize $f\geq0$ by $\lVert f\rVert_p=1$ and take $g=(T_\rho f)^{q-1}/\lVert T_\rho f\rVert_q^{q-1}$. Then $\lVert g\rVert_{q'}=1$ and

Every inequality is an equality. Thus $g$ is a dual extremizer and Hölder 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 $\lVert f\rVert_p^p$ by $\mu_i\varepsilon^p=o(\varepsilon)$. The strictly positive kernel gives $T_\rho f>0$, and the corresponding change in $\lVert T_\rho f\rVert_q^q$ 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 (19). Finally, all norms in question are continuous in $\rho$, which includes $\rho=0,1$.

Corollary 6.2 (Sharp boundary and its dependence on the least atom). For every $1<p<q<\infty$,

The function $m\mapsto\sigma_{p,q}(m)$ is strictly increasing on $(0,1/2]$. Thus Conjecture 1.2 holds.

Proof. The identity follows immediately from (18). For strict monotonicity, duality reduces the proof to $p\leq2$. Fix $0<m<1/2$, put $Z=m^{-1}\mathbf{1}_{\{a\}}-1$ on $\nu_m$, and set $\rho_0=\sqrt{(p-1)/(q-1)}$. Then $\mathbb{E}Z=0$ and $\mathbb{E}Z^3>0$. Taylor expansion at the constant function gives

At $\rho=\rho_0$, the quadratic term vanishes, while the cubic bracket is

For $q\geq2$ its positivity follows from $p\leq2$; 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 (20) handles a neighborhood of the constant, uniformly in the two signs of $t$. On the compact complement, $\lVert T_0f\rVert_q=1<\lVert f\rVert_p$, 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 6.1 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 3.1, 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 $\rho_0$ [12 Eq. (2.1)], and the strict inequality for biased laws was proved above. Duality completes the proof.

Endpoint regimes

The $q\downarrow1$ soft-dictator regime

For a mean-one density $h$, define

and

with $0\log0 = 0$. Let $Th = 1-\rho+\rho h$ and set

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 [4] for general background on KL contraction coefficients. For uniform $\mu$, this kernel is the Potts semigroup, and Gu and Polyanskiy [7 Thm. 4 and Props. 17–18] already obtained an exact one-variable formula and the corresponding one-coordinate entropy extremizers. Proposition 7.3 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 [10 Cor. 1]. Positivity of KL maximizers for this strictly positive reversible kernel is also a special case of [11 Lem. 31]. The direct arguments below additionally control the ratios uniformly and use the two-level classification to obtain persistence of interior extremizers in Theorem 7.4.

Proof. Let $M=1/\mu^*$. Uniformly for $x\in[0,M]$,

so $D_{1+\varepsilon}(h)\to\operatorname{Ent}(h)$ uniformly on $C$; the same holds for $Th$.

Ratios near constants require a separate argument. Fix $q_+\in(1,2)$. If $h=1+u$, $\mathbb{E}u=0$, and $\lVert u\rVert_\infty\le s$, Taylor’s formula with a remainder bounded on the finite coordinate range gives, uniformly for $q\in(1,q_+]$,

The same expansion at $q=1$ follows from $(1+x)\log(1+x)=x+x^2/2+O(\lvert x\rvert^3)$. 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

It is compact and contains no constant density, so $d_s=\min_{h\in K_s}\operatorname{Ent}(h)>0$. 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 $K_s$. Division is therefore uniform there, and

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.

Proof. The perturbation from Lemma 4.2, with entropy in place of $\log L_q$, gives $\eta_1>\rho^2$. Indeed,

and the numerator has $t$ replaced by $\rho t$, so the quotient is

for small positive $t$.

Let $h$ maximize the KL ratio and put $\lambda=\eta_1$. Then

is at most 0 on $\mathcal{C}$ 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

Then $h_s\geq0$ and $\mathbb{E}h_s=\mathbb{E}h=1$. At the newly positive coordinate,

while the change at coordinate $j$ is $O(s)$. Consequently $\operatorname{Ent}(h_s)-\operatorname{Ent}(h)=s\log s+O(s)$. Since $Th\geq1-\rho>0$ coordinatewise and $T(h_s-h)=O(s)$, differentiability of $r\mapsto r\log r$ on this positive range gives $\operatorname{Ent}(Th_s)-\operatorname{Ent}(Th)=O(s)$. Hence

for small $s>0$, a contradiction. Thus $h>0$.

The Lagrange equation on the relative interior says that

for each value $r=h(x)$. The derivative has sign determined by

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.

Proposition 7.3

Proof. For a fixed density $h$, choose $s_q$ such that $M_{m,q}(s_q)=\mathop{\rm E}h^q$. Uniformly on $[0,1]$,

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 $K_m(s)=\operatorname{Ent}(h)$. Subtracting one from (10), dividing by $q-1$, and taking the limit yields $\operatorname{Ent}(T_\rho h)\le K_m(\rho s)$. Every $s$ is attained by $h_s$, proving (21).

For $m<1/2$, Lemma 7.2 makes every nonconstant KL extremizer positive and two-valued. Apply the fixed-moment comparison from the proof of Theorem 3.1 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,

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 $s<1$.

Proof. Let $\mathcal{K}_1$ be the compact set of KL maximizers. By Lemma 7.2, every element of $\mathcal{K}_1$ is strictly positive, so

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 7.1, 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 4.4 then rules out subset indicators and gives the two-level soft form.

The $q\uparrow2$ least-atom regime

Proof. At $q=2$,

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

whose derivative is $c\log((1+z)/(1+cz))>0$. Thus maximizing $R_2$ is the same as maximizing variance in $\mathcal{C}$.

If $M=\max_x h(x)$, then $h^2\le Mh$, so $\mathbb{E}h^2\le M$. If $h(x_0)=M$, then $1=\mathbb{E}h\ge\mu(x_0)M$, hence $M\le1/m$. Equality in $\mathbb{E}h^2\le1/m$ 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

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 $q=2$,

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 4.3 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 5.1 factors as

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 $f_z\ne0$, and is identically zero otherwise. Since $f\ne0$, $r(y)=\mathbb{E}_zF(y,z)>0$ for every $y$. Lemma 5.2, 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 $J=\{z:f_z\ne0\}$. 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 $\chi_{a_n}=m^{-1}\mathbf{1}_{\{a_n\}}$ and $g(y)=m f(y,a_n)$, one has

Independence of coordinates gives

whereas the conditional-expectation side equals

The scalar equality $\log\lVert T_\rho\chi_{a_n}\rVert_q=\lambda\log\lVert\chi_{a_n}\rVert_q$ 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.

The uniform binary phase diagram

The least-atom reduction leaves one scalar contrast to optimize. For $m=1/2$, its full behavior can be described. Put

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.

Proof. Write $a=q-1\in(0,1)$. We first establish the needed shape of $E$. The auxiliary function

is analytic at zero. Differentiating gives the Riccati equation

Its first three nonconstant Taylor coefficients are

For instance, substitution in (23) gives $(2j+1)c_j=(2j-3)c_{j-1}-\sum_{i=1}^{j-1}c_ic_{j-i}$ 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 (23) give

At a first zero of $C'$, equation (24) would give $C''<0$, contrary to its nonnegative derivative at that zero. Therefore $C'<0$. Equation (25) excludes a first zero of $C''$ in the same way, giving $C''<0$. This is the negative-curvature argument in [15]. The next identity, (26), excludes a first zero of $C'''$, since $(9-6C')C''<0$. Consequently

Let $D(z)=C(z)+az$. The substitution $s=\tanh t$ yields

Using $\cosh(qt)=\cosh(at)\cosh t+\sinh(at)\sinh t$, this gives

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

Direct differentiation gives

The strict sign uses $2D-zD'>0$ and (27).

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

Since $D''<0$, its derivative has a unique zero $z_0$, where $N(z_0)=z_0D(z_0)D''(z_0)<0$. Equation (29) shows that $N$ crosses zero exactly once.

By (28),

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 $(sF'(s))'=F'(s)V(s)$,

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 (30) proves that $E$ has one strict minimum $s_0$. Direct endpoint evaluation gives

For $s>0$, logarithmic differentiation gives

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 (22). 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 5.3, both the raw fibers in a coordinate and the fibers after fully noising all other coordinates satisfy Lemma 5.2. 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.

Concluding remarks

The least-atom principle removes the alphabet and all other atom masses from both sharp constants. For the Rényi 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 8.1. 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.