Self-Power Escort Dynamics on the Probability Simplex

1. Introduction

Let

$${\Delta }_{n-1}=\left\{p=(p_1,\dots ,p_n)\in {\mathbb{R}}_{\ge 0}^n:{\displaystyle \sum _{i=1}^n}{p}_i=1\right\}$$

be the standard probability simplex. This paper studies the nonlinear transformation

$$T_n{\left(p\right)}_i={\displaystyle \frac{p_i^{p_i}}{{\sum }_{j=1}^n{p}_j^{p_j}}},i=1,\dots ,n,$$

(1)

with the continuous boundary convention $0^0=1$. We call $T_n$ the self-power escort map. The terminology is meant to distinguish (1) from the ordinary power escort map

$$p_i⟼{\displaystyle \frac{p_i^q}{{\sum }_j{p}_j^q}},$$

where q is an external parameter; see, for example, the escort-distribution and generalized-statistical contexts in [1,2]. Here the exponent is endogenous: the mass $p_i$ supplies its own exponent.

The basic scalar identity is

$$p_i^{p_i}=exp(p_{i}log{p}_i).$$

Thus the scalar $p_{i}log{p}_i$ is exponentiated to produce the unnormalized weight $p_i^{p_i}$, and the resulting weights are then normalized back to the simplex. This elementary coordinate self-coupling produces a surprisingly rigid discrete dynamical system on the simplex. Our use of the Hilbert log-diameter is in the tradition of projective-metric contraction methods for positive maps [3,4,5].

The first main feature is boundary activation. Since $x^x\to 1$ as $x\to 0^{+}$, zero coordinates are not preserved. On the contrary, every boundary point is mapped into the interior. More strongly, every point enters in one step the compact core

$$K_n=\left\{p\in {\Delta }_{n-1}:{\displaystyle \frac{e^{-e^{-1}}}{n}}\le p_i\le {\displaystyle \frac{e^{e^{-1}}}{n}},i=1,\dots ,n\right\}.$$

This differs sharply from replicator-type dynamics, where the zero pattern is usually invariant; see, for instance, [6].

The second main feature is projective contraction. If

$$D\left(p\right)=\underset{i,j}{max}\left|log{\displaystyle \frac{p_i}{p_j}}\right|$$

denotes the Hilbert log-diameter on the interior of the simplex, then the identity

$$log{\displaystyle \frac{T_n{\left(p\right)}_i}{T_n{\left(p\right)}_j}}=p_{i}log{p}_i-p_{j}log{p}_j$$

turns the problem into a one-dimensional Lipschitz estimate for $xlogx$ in logarithmic coordinates. After the first step, the orbit lies in $K_n$, and on this compact core D contracts uniformly. Hence every orbit converges exponentially to the uniform state

$${\nu }_n=\left({\displaystyle \frac{1}{n}},\dots ,{\displaystyle \frac{1}{n}}\right).$$

The same contraction excludes all nontrivial periodic orbits.

The third feature is the interaction between local dynamics and dimension. On the tangent space ${\sum }_i{\delta }_i=0$, the linearization at ${\nu }_n$ is scalar multiplication by

$${\mu }_n={\displaystyle \frac{1-logn}{n}}.$$

Therefore ${\mu }_2>0$, while ${\mu }_n<0$ for $n\ge 3$. The binary system approaches the uniform point without first-order order reversal, while higher-dimensional systems have an alternating local component. The second-order normal form shows that the nonlinear correction is a centered quadratic feedback term.

The fourth feature is a finite uniformization fiber. We completely classify the solutions of

$$T_n\left(p\right)={\nu }_n.$$

The classification is controlled by the two monotone branches of $x^x$ on $(0,1)$. Apart from the vertices and the uniform point, there is no nonuniform interior solution when $n=2$. For every $n\ge 3$, all nonuniform interior one-step uniformizers are exactly the n permutations of a unique vector

$$({\alpha }_n,\dots ,{\alpha }_n,{\beta }_n),$$

where ${\alpha }_n$ occurs $n-1$ times and ${\beta }_n$ occurs once.

The fifth feature is arithmetic. In the binary case the map reduces to a real analytic interval map

$$F\left(x\right)={\displaystyle \frac{x^x}{x^x+{(1-x)}^{1-x}}},0<x<1.$$

Using the Gelfond–Schneider theorem, we prove that every algebraic irrational input is mapped to a transcendental value in one step. We also prove that every rational nonuniform input first becomes an algebraic irrational value and then becomes transcendental in the next step. Thus the binary self-power escort map exhibits the arithmetic chain

$$\mathbb{Q}\setminus \{1/2\}⟶\overline{\mathbb{Q}}\setminus \mathbb{Q}⟶\mathbb{R}\setminus \overline{\mathbb{Q}}.$$

The paper is organized as follows. Section 2 defines the map and establishes the log-ratio structure. Section 3 proves boundary activation, compact-core regularization, uniqueness of the uniform fixed point, global Hilbert contraction, exponential uniformization, and absence of nontrivial periodic orbits. Section 4 refines the contraction constant after entry into the core. Section 5 gives the local spectrum and second-order normal form. Section 6 classifies one-step uniformizers. Section 7 analyzes the binary map, including exact logit contraction and arithmetic escape. Section 8 concludes the paper.

2. The Self-Power Escort Map

This section fixes the notation used throughout the paper and records the scalar mechanism behind the transformation. All logarithms are natural logarithms. The letter T is used because the object studied here is a discrete-time transformation of the simplex, while the subscript n records the number of probability coordinates.

For $n\ge 2$, let

$${\Delta }_{n-1}=\left\{p\in {\mathbb{R}}_{\ge 0}^n:{\displaystyle \sum _{i=1}^n}{p}_i=1\right\}$$

be the standard probability simplex, and let

$${\Delta }_{n-1}^{\circ }=\left\{p\in {\Delta }_{n-1}:p_i>0\mathrm{for}\mathrm{all}i\right\}$$

be its relative interior. The uniform point is denoted by

$${\nu }_n=\left({\displaystyle \frac{1}{n}},\dots ,{\displaystyle \frac{1}{n}}\right).$$

Throughout the paper we use the continuous boundary convention

$$0^0=1,$$

which is consistent with

$$\underset{x\to 0^{+}}{lim}{x}^x=1.$$

The scalar function driving the dynamics is

$$h\left(x\right)=x^x=exp(xlogx),0\le x\le 1,$$

where the boundary convention is used at $x=0$. The non-monotonicity of h on $[0,1]$ is not a technical inconvenience. It is the source of the order reversal, the finite uniformization fiber, and the boundary activation phenomena proved below.

We define the self-power escort map by

$$T_n:{\Delta }_{n-1}⟶{\Delta }_{n-1},T_n{\left(p\right)}_i={\displaystyle \frac{h\left(p_i\right)}{H_0\left(p\right)}},$$

(2)

where

$$H_0\left(p\right)={\displaystyle \sum _{j=1}^n}h\left(p_j\right)={\displaystyle \sum _{j=1}^n}{p}_j^{p_j}.$$

The notation $H_0$ is only a normalization symbol. Thus $T_n\left(p\right)$ is obtained by applying h coordinatewise and then normalizing the resulting positive vector by $H_0\left(p\right)$.

The first structural observation is that all output coordinate comparisons are governed by the single scalar function $xlogx$. The normalization factor disappears when one passes to coordinate ratios.

Lemma 1

(Log-ratio identity). For every $p\in {\Delta }_{n-1}$ and every pair of indices $i,j$,

$$log{\displaystyle \frac{T_n{\left(p\right)}_i}{T_n{\left(p\right)}_j}}=p_{i}log{p}_i-p_{j}log{p}_j,$$

(3)

with the convention $0log0=0$.

Proof. 

The normalization factor $H_0\left(p\right)$ cancels in the quotient. Hence

$${\displaystyle \frac{T_n{\left(p\right)}_i}{T_n{\left(p\right)}_j}}={\displaystyle \frac{p_i^{p_i}}{p_j^{p_j}}}.$$

Taking logarithms gives (3). □

Let

$$f\left(x\right)=xlogx,0\le x\le 1,$$

with $f\left(0\right)=0$. Then

$$f^{\prime }\left(x\right)=1+logx.$$

Therefore f is strictly decreasing on $(0,e^{-1})$, strictly increasing on $(e^{-1},1)$, and attains its unique minimum $-e^{-1}$ at $x=e^{-1}$. This elementary V-shaped structure already explains the first qualitative feature of the dynamics: small coordinates and large coordinates are treated in opposite monotonic regimes.

Proposition 1

(One-step order rule). Let $q=T_n\left(p\right)$. For every pair $i,j$,

$$q_i>q_j⟺f\left(p_i\right)>f\left(p_j\right).$$

(4)

Consequently, $T_n$ reverses coordinate order among the indices satisfying $p_i<e^{-1}$, preserves coordinate order among the indices satisfying $p_i>e^{-1}$, and has at most two coordinates in the latter block.

Proof. 

Since all coordinates of $T_n\left(p\right)$ have the same normalization factor,

$$q_i>q_j$$

holds exactly when

$$p_i^{p_i}>p_j^{p_j}.$$

Because the logarithm is strictly increasing, the last inequality is equivalent to

$$p_{i}log{p}_i>p_{j}log{p}_j,$$

which is exactly $f\left(p_i\right)>f\left(p_j\right)$.

If $p_i,p_j<e^{-1}$, then f is strictly decreasing. Hence $p_i<p_j$ gives $f\left(p_i\right)>f\left(p_j\right)$, and therefore $q_i>q_j$. Thus the order is reversed inside the sublevel block. If $p_i,p_j>e^{-1}$, then f is strictly increasing. Hence $p_i<p_j$ gives $f\left(p_i\right)<f\left(p_j\right)$, and therefore $q_i<q_j$. Thus the order is preserved inside the superlevel block.

Finally, there cannot be three coordinates greater than $e^{-1}$. Indeed, three such coordinates would have total mass greater than

$$3e^{-1}>1,$$

which is impossible for a point of ${\Delta }_{n-1}$. Hence the block $\{i:p_i>e^{-1}\}$ has cardinality at most two. □

3. Boundary Activation and Global Uniformization

This section proves the global dynamical core of the paper. The argument is based on three elementary facts that interact strongly. The boundary convention $0^0=1$ makes every zero coordinate produce a positive self-power weight. The scalar bound on $xlogx$ then forces every first iterate into a compact subset of the interior of the simplex. Once the orbit has entered this compact core, the log-ratio identity from Lemma 1 gives a uniform contraction of the Hilbert log-diameter. These facts imply global exponential convergence to the uniform distribution.

The first observation is the boundary activation mechanism. Unlike multiplicative dynamics that preserve zero coordinates, the present map immediately turns every boundary state into a strictly positive probability vector.

Lemma 2

(Boundary activation). For every $p\in {\Delta }_{n-1}$,

$$T_n\left(p\right)\in {\Delta }_{n-1}^{\circ }.$$

(5)

Thus every boundary point enters the interior after one iteration.

Proof. 

For every $x\in [0,1]$, the value $x^x$ is strictly positive under the convention $0^0=1$. Hence every numerator in (2) is positive, and $H_0\left(p\right)$ is positive as a sum of positive terms. Therefore

$$T_n{\left(p\right)}_i>0$$

for every i, which proves the claim. □

The previous lemma gives immediate entrance into the relative interior, but the global argument needs a stronger statement. The first iterate does not merely become positive; it enters a fixed compact core whose distance from the boundary is controlled uniformly over all initial data.

Proposition 2

(Compact-core regularization). Let

$$c=e^{-e^{-1}}.$$

For every $p\in {\Delta }_{n-1}$ and every i,

$${\displaystyle \frac{c}{n}}\le T_n{\left(p\right)}_i\le {\displaystyle \frac{1}{nc}}.$$

(6)

Consequently,

$$T_n\left({\Delta }_{n-1}\right)\subset K_n,$$

(7)

where

$$K_n=\left\{p\in {\Delta }_{n-1}:{\displaystyle \frac{e^{-e^{-1}}}{n}}\le p_i\le {\displaystyle \frac{e^{e^{-1}}}{n}},i=1,\dots ,n\right\}.$$

(8)

Proof. 

The function $xlogx$ has minimum $-e^{-1}$ and maximum 0 on $[0,1]$. Therefore

$$e^{-e^{-1}}\le x^x\le 1,0\le x\le 1.$$

Since $H_0\left(p\right)={\sum }_{j=1}^n{p}_j^{p_j}$, this bound gives

$$nc\le H_0\left(p\right)\le n.$$

For each coordinate,

$$T_n{\left(p\right)}_i={\displaystyle \frac{p_i^{p_i}}{H_0\left(p\right)}}.$$

Using $p_i^{p_i}\ge c$ and $H_0\left(p\right)\le n$ gives

$$T_n{\left(p\right)}_i\ge {\displaystyle \frac{c}{n}}.$$

Using $p_i^{p_i}\le 1$ and $H_0\left(p\right)\ge nc$ gives

$$T_n{\left(p\right)}_i\le {\displaystyle \frac{1}{nc}}.$$

This proves (6), and the stated inclusion into $K_n$ follows immediately. □

The compact-core estimate already suggests that the transformation regularizes arbitrary initial data toward the center of the simplex. The next result identifies the only possible equilibrium of this process.

Proposition 3

(Uniqueness of the uniform fixed point). The unique fixed point of $T_n$ in ${\Delta }_{n-1}$ is the uniform distribution ${\nu }_n$; equivalently,

$$T_n\left(p\right)=p⟺p={\nu }_n.$$

(9)

Proof. 

The uniform point is fixed because all its coordinates are equal. Conversely, suppose that $T_n\left(p\right)=p$. If p had a zero coordinate, then Lemma 2 would imply that $T_n\left(p\right)$ has only positive coordinates, contradicting $T_n\left(p\right)=p$. Hence every fixed point lies in ${\Delta }_{n-1}^{\circ }$.

Put

$$Z=H_0\left(p\right).$$

The fixed point equation is

$$p_i={\displaystyle \frac{p_i^{p_i}}{Z}}.$$

Since $p_i>0$, division by $p_i$ gives

$$Z=p_i^{p_i-1}.$$

Taking logarithms yields

$$(1-p_i)log{p}_i=-logZ.$$

Thus every coordinate $p_i$ solves the same scalar equation

$$(1-x)logx=C.$$

Let

$$\psi \left(x\right)=(1-x)logx,0<x<1.$$

Then

$${\psi }^{\prime }\left(x\right)={\displaystyle \frac{1}{x}}-1-logx.$$

Writing $t=1/x>1$, this becomes

$${\psi }^{\prime }\left(x\right)=t-1+logt>0.$$

Therefore $\psi$ is strictly increasing on $(0,1)$. The preceding scalar equation then forces all coordinates $p_i$ to be equal. Since their sum is one, one has

$$p_i={\displaystyle \frac{1}{n}},i=1,\dots ,n.$$

Thus $p={\nu }_n$. □

We now pass from fixed points to global convergence. The appropriate quantity is the Hilbert projective diameter of a positive probability vector. For $p\in {\Delta }_{n-1}^{\circ }$, define

$$D\left(p\right)=\underset{i,j}{max}\left|log{\displaystyle \frac{p_i}{p_j}}\right|.$$

(10)

This quantity vanishes exactly at the uniform distribution and measures the logarithmic spread of the coordinates.

Before stating the contraction estimate, set

$$c_n^{-}={\displaystyle \frac{e^{-e^{-1}}}{n}},c_n^{+}={\displaystyle \frac{e^{e^{-1}}}{n}},J_n=[c_n^{-},c_n^{+}].$$

(11)

Thus $K_n$ is the part of the simplex whose coordinates all belong to $J_n$.

Theorem 1

(Core Hilbert contraction). There exists $0<{\lambda }_n<1$ such that

$$D\left(T_n\left(p\right)\right)\le {\lambda }_{n}D\left(p\right),p\in K_n.$$

(12)

One may take

$${\lambda }_n=\left\{\begin{array}{cc}{\displaystyle \frac{e^{e^{-1}}}{n}}\left(1+e^{-1}-logn\right),\hfill & n=2,3,\hfill \\ {\displaystyle \frac{e^{-e^{-1}}}{n}}\left(logn+e^{-1}-1\right),\hfill & n=4,5,\hfill \\ e^{-2},\hfill & 6\le n\le 10,\hfill \\ {\displaystyle \frac{e^{e^{-1}}}{n}}\left(logn-1-e^{-1}\right),\hfill & n\ge 11.\hfill \end{array}\right.$$

(13)

Proof. 

Let $q=T_n\left(p\right)$. By Lemma 1,

$$log{\displaystyle \frac{q_i}{q_j}}=f\left(p_i\right)-f\left(p_j\right).$$

We estimate the right-hand side on the logarithmic scale. Since

$$f^{\prime }\left(x\right)=1+logx,$$

one has

$${\displaystyle \frac{d}{dlogx}}f\left(x\right)=x(1+logx).$$

For $p\in K_n$, every coordinate $p_i$ belongs to $J_n$. Hence, by the mean value theorem applied to f as a function of $logx$,

$$|f\left(p_i\right)-f\left(p_j\right)|\le \left(\underset{x\in J_n}{sup}\left|x(1+logx)\right|\right)\left|log{\displaystyle \frac{p_i}{p_j}}\right|.$$

Therefore

$$\left|log{\displaystyle \frac{q_i}{q_j}}\right|\le \left(\underset{x\in J_n}{sup}\left|x(1+logx)\right|\right)\left|log{\displaystyle \frac{p_i}{p_j}}\right|.$$

Taking the maximum over $i,j$ proves (12) once the supremum on $J_n$ is evaluated.

Let

$$g_{+}\left(x\right)=x(1+logx),x>0.$$

Then

$$g_{+}^{\prime }\left(x\right)=2+logx.$$

Thus $g_{+}$ is strictly decreasing on $(0,e^{-2})$ and strictly increasing on $(e^{-2},\infty )$. Moreover,

$$g_{+}\left(e^{-2}\right)=-e^{-2},g_{+}\left(e^{-1}\right)=0.$$

Consequently, $|g_{+}|$ is increasing on $(0,e^{-2})$, decreasing on $(e^{-2},e^{-1})$, and increasing on $(e^{-1},\infty )$.

Recall that $J_n=[c_n^{-},c_n^{+}]$. The location of $J_n$ relative to the two critical points $e^{-2}$ and $e^{-1}$ is determined by the elementary equivalences

$$c_n^{+}>e^{-1}⟺n<e^{1+e^{-1}},$$

$$c_n^{-}>e^{-2}⟺n<e^{2-e^{-1}},$$

and

$$c_n^{+}\ge e^{-2}⟺n\le e^{2+e^{-1}}.$$

Since

$$3<e^{1+e^{-1}}<4,5<e^{2-e^{-1}}<6,10<e^{2+e^{-1}}<11,$$

the possible positions of $J_n$ split into four cases.

For $n=2,3$, the interval $J_n$ intersects the increasing branch to the right of $e^{-1}$ but does not contain $e^{-2}$. Hence the maximum of $|g_{+}|$ is attained at an endpoint, and a direct comparison of the two endpoint values gives

$$\underset{x\in J_n}{sup}|g_{+}\left(x\right)|=|g_{+}\left(c_n^{+}\right)|={\displaystyle \frac{e^{e^{-1}}}{n}}\left(1+e^{-1}-logn\right).$$

For $n=4,5$, one has $J_n\subset (e^{-2},e^{-1})$. Since $|g_{+}|$ is decreasing on this interval, the maximum is attained at $c_n^{-}$, and

$$\underset{x\in J_n}{sup}|g_{+}\left(x\right)|=|g_{+}\left(c_n^{-}\right)|={\displaystyle \frac{e^{-e^{-1}}}{n}}\left(logn+e^{-1}-1\right).$$

For $6\le n\le 10$, the interval $J_n$ contains $e^{-2}$ and lies below $e^{-1}$ at its upper endpoint. Therefore

$$\underset{x\in J_n}{sup}\left|g_{+}\left(x\right)\right|=e^{-2}.$$

For $n\ge 11$, one has $J_n\subset (0,e^{-2})$. Since $|g_{+}|$ is increasing on this interval, the maximum is attained at $c_n^{+}$, and

$$\underset{x\in J_n}{sup}|g_{+}\left(x\right)|=|g_{+}\left(c_n^{+}\right)|={\displaystyle \frac{e^{e^{-1}}}{n}}\left(logn-1-e^{-1}\right).$$

Combining the four cases gives (13). In every case the obtained value is strictly smaller than 1, so $0<{\lambda }_n<1$. □

The contraction estimate applies after the first step because of compact-core regularization. This gives the global convergence theorem.

Theorem 2

(Global exponential uniformization). For every $p\in {\Delta }_{n-1}$,

$$\underset{k\to \infty }{lim}{T}_n^k\left(p\right)={\nu }_n.$$

(14)

More precisely, with ${\lambda }_n$ given by (13),

$$D\left(T_n^k\left(p\right)\right)\le e^{-1}{\lambda }_n^{k-1},k\ge 1.$$

(15)

Proof. 

By Proposition 2, the first iterate satisfies

$$T_n\left(p\right)\in K_n.$$

In particular, $T_n\left(p\right)$ lies in ${\Delta }_{n-1}^{\circ }$, and the Hilbert log-diameter is well defined for all later iterates.

Applying Theorem 1 repeatedly gives

$$D\left(T_n^k\left(p\right)\right)\le {\lambda }_n^{k-1}D\left(T_n\left(p\right)\right),k\ge 1.$$

It remains to bound the first diameter uniformly. Let $q=T_n\left(p\right)$. By Lemma 1,

$$log{\displaystyle \frac{q_i}{q_j}}=p_{i}log{p}_i-p_{j}log{p}_j.$$

Since

$$-e^{-1}\le xlogx\le 0,0\le x\le 1,$$

one obtains

$$D\left(q\right)=\underset{i,j}{max}|p_{i}log{p}_i-p_{j}log{p}_j|\le e^{-1}.$$

Therefore

$$D\left(T_n^k\left(p\right)\right)\le e^{-1}{\lambda }_n^{k-1},k\ge 1,$$

which proves (15).

Since $0<{\lambda }_n<1$, the right-hand side tends to zero. Hence the logarithmic ratios between all coordinates of $T_n^k\left(p\right)$ tend to zero. Equivalently, all coordinate ratios tend to one. Because every $T_n^k\left(p\right)$ is a probability vector, the only possible limit is the uniform distribution ${\nu }_n$. Thus $T_n^k\left(p\right)$ converges to ${\nu }_n$. □

As a direct consequence, the global dynamics has no periodic behavior other than the equilibrium already identified.

Corollary 1

(No nontrivial periodic orbit). If L is a positive integer, then

$$T_n^L\left(p\right)=p⟹p={\nu }_n.$$

(16)

Proof. 

Suppose that $T_n^L\left(p\right)=p$ for some positive integer L. By Lemma 2, the point

$$q=T_n\left(p\right)$$

lies in ${\Delta }_{n-1}^{\circ }$, and by Proposition 2 it lies in $K_n$. Moreover,

$$T_n^L\left(q\right)=q.$$

Applying Theorem 1 along one full period gives

$$D\left(q\right)\le {\lambda }_n^{L}D\left(q\right).$$

Since $0<{\lambda }_n<1$, this forces

$$D\left(q\right)=0.$$

Hence $q={\nu }_n$. Therefore the forward orbit of p is equal to ${\nu }_n$ from the first iterate onward. Since $T_n^L\left(p\right)=p$, it follows that $p={\nu }_n$. □

4. Sharp Hilbert Contraction After Core Entrance

The constant ${\lambda }_n$ in Theorem 1 gives a global post-core contraction estimate, but it is obtained from the coordinate-wise interval $J_n=[c_n^{-},c_n^{+}]$ alone. It does not use the simplex constraint ${\sum }_i{p}_i=1$. This section incorporates that constraint and obtains the sharp post-core contraction constant in the monotone-core range $n\ge 4$.

The restriction $n\ge 4$ is natural. In this range,

$$c_n^{+}={\displaystyle \frac{e^{e^{-1}}}{n}}<e^{-1},$$

so every coordinate of every point in $K_n$ lies below the critical point $e^{-1}$ of $f\left(x\right)=xlogx$. Hence f is strictly decreasing on the whole coordinate range of $K_n$, and the Hilbert contraction ratio is determined by the smallest and largest coordinates.

For $p\in K_n$, write

$$u=p_{min},v=p_{max}.$$

A pair $(u,v)$ is feasible for some point of $K_n\cap {\Delta }_{n-1}$ exactly when

$$c_n^{-}\le u\le {\displaystyle \frac{1}{n}}\le v\le c_n^{+}$$

and

$$v+(n-1)u\le 1\le u+(n-1)v.$$

The first line records that the proposed lower and upper coordinates lie in the core interval $J_n$. The second line is the simplex feasibility condition for a vector whose minimum is u and whose maximum is v: one coordinate attains the lower endpoint u, one coordinate attains the upper endpoint v, and all remaining coordinates lie between them.

Define

$${\mathcal{R}}_n=\left\{(u,v):c_n^{-}\le u\le {\displaystyle \frac{1}{n}}\le v\le c_n^{+},v+(n-1)u\le 1\le u+(n-1)v\right\}.$$

(17)

Theorem 3

(Sharp core contraction for $n\ge 4$). Assume $n\ge 4$. Define

$${\Lambda }_n^{*}=\underset{\begin{array}{c}(u,v)\in {\mathcal{R}}_n\\ u<v\end{array}}{sup}{\displaystyle \frac{ulogu-vlogv}{log\left(v/u\right)}}.$$

(18)

Then

$$D\left(T_n\left(p\right)\right)\le {\Lambda }_n^{*}D\left(p\right),p\in K_n\cap {\Delta }_{n-1}^{\circ }.$$

(19)

The constant ${\Lambda }_n^{*}$ is the sharp contraction constant on $K_n\cap {\Delta }_{n-1}^{\circ }$. Moreover,

$$0<{\Lambda }_n^{*}\le e^{-2}<1.$$

(20)

Proof. 

For $n\ge 4$, one has $c_n^{+}<e^{-1}$. Hence every coordinate of every point in $K_n$ lies in $(0,e^{-1})$. On this interval the function $f\left(x\right)=xlogx$ is strictly decreasing. Therefore, if $u=p_{min}$ and $v=p_{max}$, then

$$D\left(T_n\left(p\right)\right)=f\left(u\right)-f\left(v\right)=ulogu-vlogv,$$

while

$$D\left(p\right)=log{\displaystyle \frac{v}{u}}.$$

Consequently,

$${\displaystyle \frac{D\left(T_n\left(p\right)\right)}{D\left(p\right)}}={\displaystyle \frac{ulogu-vlogv}{log\left(v/u\right)}}.$$

For every $p\in K_n\cap {\Delta }_{n-1}^{\circ }$ with $D\left(p\right)>0$, the pair $(u,v)=(p_{min},p_{max})$ belongs to ${\mathcal{R}}_n$ and satisfies $u<v$. Hence the displayed quotient is bounded above by ${\Lambda }_n^{*}$. If $D\left(p\right)=0$, then $p={\nu }_n$ and the inequality is trivial. This proves (19).

It remains to justify sharpness. Let $(u,v)\in {\mathcal{R}}_n$ with $u<v$. The two feasibility inequalities imply

$$(n-2)u\le 1-u-v\le (n-2)v.$$

Hence the average

$$\rho ={\displaystyle \frac{1-u-v}{n-2}}$$

lies in $[u,v]$. Taking one coordinate equal to u, one coordinate equal to v, and the remaining $n-2$ coordinates equal to $\rho$ gives, up to permutation, a vector

$$p=(u,v,\rho ,\dots ,\rho ).$$

This vector belongs to $K_n\cap {\Delta }_{n-1}^{\circ }$, has minimum coordinate u, and has maximum coordinate v. For this vector the ratio $D\left(T_n\left(p\right)\right)/D\left(p\right)$ is exactly the quotient in (18). Thus no smaller constant can work uniformly on $K_n\cap {\Delta }_{n-1}^{\circ }$, and ${\Lambda }_n^{*}$ is sharp.

We now prove (20). For $0<u<v<e^{-1}$, Cauchy’s mean value theorem applied to $f\left(x\right)=xlogx$ and $logx$ gives a point $\xi \in (u,v)$ such that

$${\displaystyle \frac{f\left(u\right)-f\left(v\right)}{logv-logu}}=\xi \left(-1-log\xi \right).$$

Let

$$g_{-}\left(x\right)=x\left(-1-logx\right),0<x<e^{-1}.$$

Then

$$g_{-}^{\prime }\left(x\right)=-2-logx.$$

Thus $g_{-}$ increases on $(0,e^{-2})$, decreases on $(e^{-2},e^{-1})$, and attains its maximum at $x=e^{-2}$. Since

$$g_{-}\left(e^{-2}\right)=e^{-2},$$

every quotient in (18) is at most $e^{-2}$. This proves (20). □

Remark 1

(Low-dimensional cores). The restriction $n\ge 4$ only marks the range in which the whole core lies below $e^{-1}$. For $n=2,3$, the compact core crosses the critical point of $f\left(x\right)=xlogx$, so the monotone-core formula (18) no longer applies. In dimension two, the binary logit representation gives a sharper global contraction result. In dimension three, one can still write a sharp compact variational formula, but it must involve all three values of f rather than only $p_{min}$ and $p_{max}$.

5. Local Dynamics Near the Uniform State

The global theorem gives convergence to ${\nu }_n$. The next question is how an orbit approaches this point once it is already close to uniformity. We write a nearby point as

$$p={\nu }_n+\delta .$$

Since points of the simplex have total mass one, the first-order perturbation must satisfy

$${\displaystyle \sum _{i=1}^n}{\delta }_i=0.$$

Thus the tangent space of the simplex at the uniform point is the mass-zero subspace

$$T_{{\nu }_n}{\Delta }_{n-1}=\left\{\delta \in {\mathbb{R}}^n:{\displaystyle \sum _{i=1}^n}{\delta }_i=0\right\}.$$

We first compute the derivative of $T_n$ on this subspace and then record the second-order normal form. In this section $\parallel \delta \parallel$ denotes the Euclidean norm in ${\mathbb{R}}^n$.

Proposition 4

(Linear spectrum at the uniform state). On $T_{{\nu }_n}{\Delta }_{n-1}$, the derivative of $T_n$ at ${\nu }_n$ is

$$D{T}_n\left({\nu }_n\right)\left[\delta \right]={\mu }_n\delta ,{\mu }_n={\displaystyle \frac{1-logn}{n}}.$$

(21)

Thus the only tangent multiplier is ${\mu }_n$, with multiplicity $n-1$.

Proof. 

Recall that

$$T_n{\left(p\right)}_i={\displaystyle \frac{h\left(p_i\right)}{{\sum }_{j=1}^{n}h\left(p_j\right)}},h\left(x\right)=x^x.$$

Let $\overline{p}=1/n$ and write

$$p_i=\overline{p}+{\delta }_i,{\displaystyle \sum _{i=1}^n}{\delta }_i=0.$$

Since

$$h^{\prime }\left(x\right)=x^x(1+logx),$$

Taylor expansion at $\overline{p}$ gives

$$h(\overline{p}+{\delta }_i)=h\left(\overline{p}\right)+h^{\prime }\left(\overline{p}\right){\delta }_i+{O(\parallel \delta \parallel }^2).$$

Summing this expansion over i gives

$${\displaystyle \sum _{j=1}^n}h(\overline{p}+{\delta }_j)=nh\left(\overline{p}\right)+h^{\prime }\left(\overline{p}\right){\displaystyle \sum _{j=1}^n}{\delta }_j+{O(\parallel \delta \parallel }^2).$$

The middle term vanishes because ${\sum }_j{\delta }_j=0$. Hence

$${\displaystyle \sum _{j=1}^n}h(\overline{p}+{\delta }_j)=nh\left(\overline{p}\right)+{O(\parallel \delta \parallel }^2).$$

Therefore

$$T_n{\left(p\right)}_i={\displaystyle \frac{h\left(\overline{p}\right)+h^{\prime }\left(\overline{p}\right){\delta }_i+{O(\parallel \delta \parallel }^2)}{nh\left(\overline{p}\right)+{O(\parallel \delta \parallel }^2)}}={\displaystyle \frac{1}{n}}+{\displaystyle \frac{h^{\prime }\left(\overline{p}\right)}{nh\left(\overline{p}\right)}}{\delta }_i+{O(\parallel \delta \parallel }^2).$$

It follows that

$$D{T}_n\left({\nu }_n\right){\left[\delta \right]}_i={\displaystyle \frac{h^{\prime }\left(\overline{p}\right)}{nh\left(\overline{p}\right)}}{\delta }_i.$$

Finally,

$${\displaystyle \frac{h^{\prime }\left(\overline{p}\right)}{h\left(\overline{p}\right)}}=1+log\overline{p}=1-logn.$$

Thus

$$D{T}_n\left({\nu }_n\right){\left[\delta \right]}_i={\displaystyle \frac{1-logn}{n}}{\delta }_i$$

for every i, which gives

$$D{T}_n\left({\nu }_n\right)\left[\delta \right]={\mu }_n\delta ,{\mu }_n={\displaystyle \frac{1-logn}{n}}.$$

Since $T_{{\nu }_n}{\Delta }_{n-1}$ has dimension $n-1$, the multiplier ${\mu }_n$ has multiplicity $n-1$ on the tangent space. □

Since

$${\mu }_n={\displaystyle \frac{1-logn}{n}},$$

one has ${\mu }_2>0$, while ${\mu }_n<0$ for every $n\ge 3$. Thus the linearized dynamics changes sign between dimension two and dimension three.

Proposition 5

(Second-order local normal form). Let

$$p_i={\displaystyle \frac{1}{n}}+{\delta }_i,{\displaystyle \sum _{i=1}^n}{\delta }_i=0,$$

(22)

and put

$$m_2={\displaystyle \frac{1}{n}}{\displaystyle \sum _{j=1}^n}{\delta }_j^2.$$

(23)

Then, near ${\nu }_n$,

$$T_n{\left(p\right)}_i={\displaystyle \frac{1}{n}}+{\displaystyle \frac{1-logn}{n}}{\delta }_i+{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{{(1-logn)}^2}{n}}\right)({\delta }_i^2-m_2)+{O(\parallel \delta \parallel }^3).$$

(24)

Proof. 

Let

$$\overline{p}={\displaystyle \frac{1}{n}},c_1=1+log\overline{p}=1-logn,c_2={\displaystyle \frac{1}{2\overline{p}}}={\displaystyle \frac{n}{2}}.$$

Since $f\left(x\right)=xlogx$, we have

$$f^{\prime }\left(\overline{p}\right)=c_1,{\displaystyle \frac{1}{2}}{f}^{\prime \prime }\left(\overline{p}\right)=c_2.$$

Hence Taylor expansion gives

$$f(\overline{p}+{\delta }_i)=f\left(\overline{p}\right)+c_1{\delta }_i+c_2{\delta }_i^2+O\left({\delta }_i^3\right).$$

Exponentiating and expanding once more gives

$$e^{f(\overline{p}+{\delta }_i)}=e^{f\left(\overline{p}\right)}\left(1+c_1{\delta }_i+\left(c_2+{\displaystyle \frac{c_1^2}{2}}\right){\delta }_i^2+{O(\parallel \delta \parallel }^3)\right).$$

Put

$$B=c_2+{\displaystyle \frac{c_1^2}{2}}.$$

Then

$$e^{f(\overline{p}+{\delta }_i)}=e^{f\left(\overline{p}\right)}\left(1+c_1{\delta }_i+B{\delta }_i^2+{O(\parallel \delta \parallel }^3)\right).$$

Summing over i and using ${\sum }_i{\delta }_i=0$ yields

$${\displaystyle \sum _{j=1}^n}{e}^{f(\overline{p}+{\delta }_j)}=e^{f\left(\overline{p}\right)}\left(n+B{\displaystyle \sum _{j=1}^n}{\delta }_j^2+{O(\parallel \delta \parallel }^3)\right).$$

Let $S_2={\sum }_{j=1}^n{\delta }_j^2$. Since $S_2={O(\parallel \delta \parallel }^2)$, we have

$${\displaystyle \frac{1}{n+B{S}_2+{O(\parallel \delta \parallel }^3)}}={\displaystyle \frac{1}{n}}-{\displaystyle \frac{B}{n^2}}{S}_2+{O(\parallel \delta \parallel }^3).$$

Therefore

$$\begin{array}{cc}\hfill T_n{\left(p\right)}_i& =\left(1+c_1{\delta }_i+B{\delta }_i^2+{O(\parallel \delta \parallel }^3)\right)\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{B}{n^2}}{S}_2+{O(\parallel \delta \parallel }^3)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{n}}+{\displaystyle \frac{c_1}{n}}{\delta }_i+{\displaystyle \frac{B}{n}}{\delta }_i^2-{\displaystyle \frac{B}{n^2}}{S}_2+{O(\parallel \delta \parallel }^3).\hfill \end{array}$$

Since $m_2=S_2/n$, this becomes

$$T_n{\left(p\right)}_i={\displaystyle \frac{1}{n}}+{\displaystyle \frac{c_1}{n}}{\delta }_i+{\displaystyle \frac{B}{n}}({\delta }_i^2-m_2)+{O(\parallel \delta \parallel }^3).$$

Finally,

$${\displaystyle \frac{B}{n}}={\displaystyle \frac{1}{n}}\left({\displaystyle \frac{n}{2}}+{\displaystyle \frac{{(1-logn)}^2}{2}}\right)={\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{{(1-logn)}^2}{n}}\right).$$

This proves (24). □

6. The Uniform Fiber and One-Step Uniformizers

The global convergence theorem shows that ${\nu }_n$ is the asymptotic endpoint of every orbit. A different finite-time question is whether a point can land on ${\nu }_n$ in a single step. This section classifies all solutions of $T_n\left(p\right)={\nu }_n$. The classification is finite and rigid because the scalar equation $x^x=y$ has at most two real solutions in $[0,1]$. The proof below uses only this two-branch geometry.

Proposition 6

(Structure of the uniform fiber). Let $p\in {\Delta }_{n-1}^{\circ }$. Then $T_n\left(p\right)={\nu }_n$ exactly when

$$p_1^{p_1}=p_2^{p_2}=\dots =p_n^{p_n}.$$

(25)

Moreover, if such a point is not uniform, then, after a permutation of coordinates, it has the form

$$p=(\underset{n-1}{\underbrace{\alpha ,\dots ,\alpha }},\beta ),$$

(26)

where

$$0<\alpha <e^{-1}<\beta <1,{\alpha }^{\alpha }={\beta }^{\beta },(n-1)\alpha +\beta =1.$$

(27)

All nontrivial equal self-power pairs are parametrized by

$$\alpha =t^{-t/(t-1)},\beta =t^{-1/(t-1)},t>1.$$

(28)

Equivalently, with $t=1+1/s$,

$$\alpha ={\left({\displaystyle \frac{s}{s+1}}\right)}^{s+1},\beta ={\left({\displaystyle \frac{s}{s+1}}\right)}^s,s>0.$$

(29)

Proof. 

The equality $T_n\left(p\right)={\nu }_n$ means that all normalized weights are equal. Since every coordinate of $T_n\left(p\right)$ has the same positive denominator, this is exactly

$$p_1^{p_1}=p_2^{p_2}=\dots =p_n^{p_n}.$$

Let

$$h\left(x\right)=x^x=e^{xlogx}.$$

The function $f\left(x\right)=xlogx$ is strictly decreasing on $(0,e^{-1})$ and strictly increasing on $(e^{-1},1)$. Hence each level set of h in $(0,1)$ contains at most two points. Therefore an interior solution of $T_n\left(p\right)={\nu }_n$ has at most two distinct coordinate values. If it is not uniform, then it has two values, one below $e^{-1}$ and one above $e^{-1}$. Thus, after a permutation of coordinates,

$$p=(\underset{k}{\underbrace{\alpha ,\dots ,\alpha }},\underset{\ell }{\underbrace{\beta ,\dots ,\beta }}),k+\ell =n,$$

with

$$0<\alpha <e^{-1}<\beta <1,{\alpha }^{\alpha }={\beta }^{\beta }.$$

We next parametrize the equal self-power pairs. Write

$$\beta =t\alpha ,t>1.$$

The equation ${\alpha }^{\alpha }={\beta }^{\beta }$ is equivalent to

$$\alpha log\alpha =\beta log\beta .$$

Substituting $\beta =t\alpha$ gives

$$\alpha log\alpha =t\alpha (logt+log\alpha ).$$

After division by $\alpha >0$, we obtain

$$(1-t)log\alpha =tlogt.$$

Therefore

$$\alpha =t^{-t/(t-1)},\beta =t\alpha =t^{-1/(t-1)}.$$

The parametrization by s follows from setting

$$t={\displaystyle \frac{s+1}{s}}=1+{\displaystyle \frac{1}{s}}.$$

It remains to prove that the larger coordinate $\beta$ has multiplicity one. We claim that every pair in (28) satisfies

$$\alpha +2\beta >1.$$

Since $\beta =t\alpha$ and $\alpha =t^{-t/(t-1)}$, this inequality is equivalent to

$$(1+2t)t^{-t/(t-1)}>1.$$

Taking logarithms gives the equivalent inequality

$$(t-1)log(1+2t)>tlogt.$$

Define

$$A\left(t\right)=(t-1)log(1+2t)-tlogt.$$

Then $A\left(1\right)=0$, and direct differentiation gives

$$A^{\prime }\left(t\right)=log\left(2+{\displaystyle \frac{1}{t}}\right)-{\displaystyle \frac{3}{1+2t}}.$$

Set

$$y=2+{\displaystyle \frac{1}{t}}.$$

Then $2<y<3$, and

$$A^{\prime }\left(t\right)=logy-3+{\displaystyle \frac{6}{y}}.$$

The function

$$\varphi \left(y\right)=logy-3+{\displaystyle \frac{6}{y}}$$

is strictly decreasing on $(2,3)$, and

$$\varphi \left(3\right)=log3-1>0.$$

Hence $\varphi \left(y\right)>0$ for $2<y<3$, so $A^{\prime }\left(t\right)>0$ for $t>1$. Since $A\left(1\right)=0$, we get $A\left(t\right)>0$ for $t>1$. This proves

$$\alpha +2\beta >1.$$

If $\ell \ge 2$, then the total mass of the vector is at least

$$\alpha +2\beta >1,$$

which contradicts $p\in {\Delta }_{n-1}$. Therefore $\ell =1$. Since $k+\ell =n$, we have $k=n-1$, and the simplex condition gives

$$(n-1)\alpha +\beta =1.$$

This proves the stated structure. □

Theorem 4

(Complete classification of one-step uniformizers). The solutions of $T_n\left(p\right)={\nu }_n$ are as follows.

1.

The boundary solutions are exactly the vertices $e_1,\dots ,e_n$.

2.

The uniform point ${\nu }_n$ is always a solution.

3.

For $n=2$, there is no nonuniform interior solution.

4.

For every $n\ge 3$, there is a unique number $t_n>1$ solving

$$(n-1+t)t^{-t/(t-1)}=1.$$

(30)

If

$${\alpha }_n=t_n^{-t_n/(t_n-1)},{\beta }_n=t_n^{-1/(t_n-1)},$$

(31)

then all nonuniform interior solutions are exactly the n permutations of

$$({\alpha }_n,\dots ,{\alpha }_n,{\beta }_n),$$

(32)

where ${\alpha }_n$ appears $n-1$ times.

Proof. 

A boundary point p satisfying $T_n\left(p\right)={\nu }_n$ must have all weights $p_i^{p_i}$ equal. If some coordinate is zero, then its weight is $0^0=1$ under the boundary convention. Hence all weights must be equal to 1. Since $x^x=1$ on $[0,1]$ only for $x=0$ and $x=1$, every coordinate of such a boundary point is either 0 or 1. The simplex condition then forces exactly one coordinate to be equal to 1 and all the others to be equal to 0. Thus the boundary solutions are exactly the vertices. The uniform solution is immediate.

Now consider a nonuniform interior solution. By Proposition 6, after a permutation of coordinates it has the form

$$p=(\alpha ,\dots ,\alpha ,\beta ),$$

where $\alpha$ is repeated $n-1$ times,

$$0<\alpha <e^{-1}<\beta <1,{\alpha }^{\alpha }={\beta }^{\beta },$$

and the larger coordinate has multiplicity one. The same proposition parametrizes every nontrivial equal self-power pair by

$$\alpha =t^{-t/(t-1)},\beta =t^{-1/(t-1)},t>1.$$

Therefore the simplex equation $(n-1)\alpha +\beta =1$ becomes precisely (30).

It remains to prove the existence and uniqueness of $t_n$. Define

$$L_n\left(t\right)=log(n-1+t)-{\displaystyle \frac{t}{t-1}}logt,t>1.$$

Then (30) is equivalent to $L_n\left(t\right)=0$. One has

$$\underset{t\to 1^{+}}{lim}{L}_n\left(t\right)=logn-1.$$

Thus the limit is negative for $n=2$ and positive for $n\ge 3$. Also,

$$L_n\left(t\right)={\displaystyle \frac{n-1-logt}{t}}+o\left({\displaystyle \frac{1}{t}}\right)(t\to \infty ),$$

so $L_n\left(t\right)<0$ for sufficiently large t, and $L_n\left(t\right)\to 0$ from below as $t\to \infty$.

Put $u=t-1$. A direct calculation gives

$$L_n^{\prime }\left(t\right)=-{\displaystyle \frac{{\phi }_n\left(u\right)}{u^2}},$$

where

$${\phi }_n\left(u\right)=u-log(1+u)-{\displaystyle \frac{u^2}{u+n}}.$$

Moreover,

$${\phi }_n^{\prime }\left(u\right)={\displaystyle \frac{u\left(n\right(n-2)-u)}{(1+u){(u+n)}^2}}.$$

For $n\ge 3$, the derivative ${\phi }_n^{\prime }\left(u\right)$ is positive on $(0,n(n-2\left)\right)$ and negative on $\left(n\right(n-2),\infty )$. Hence ${\phi }_n$ first increases and then decreases. Since ${\phi }_n\left(0\right)=0$ and ${\phi }_n\left(u\right)\to -\infty$ as $u\to \infty$, the function ${\phi }_n$ has exactly one zero in $(0,\infty )$. Consequently $L_n^{\prime }$ is first negative and then positive. Thus $L_n$ decreases once and then increases toward 0 from below. Since $L_n\left(1^{+}\right)>0$ and $L_n\left(t\right)<0$ for large t, there is exactly one zero for $n\ge 3$. For $n=2$, one has

$${\phi }_2^{\prime }\left(u\right)=-{\displaystyle \frac{u^2}{(1+u){(u+2)}^2}}<0,u>0.$$

Since ${\phi }_2\left(0\right)=0$, it follows that ${\phi }_2\left(u\right)<0$ for every $u>0$. Hence $L_2^{\prime }\left(t\right)>0$ on $(1,\infty )$. Together with

$$\underset{t\to 1^{+}}{lim}{L}_2\left(t\right)=log2-1<0$$

and the fact that $L_2\left(t\right)\to 0$ from below as $t\to \infty$, this excludes any zero for $n=2$. This proves the classification. □

Example 5.

For $n=3$, equation (30) has the solution $t_3=2$. Hence

$${\alpha }_3={\displaystyle \frac{1}{4}},{\beta }_3={\displaystyle \frac{1}{2}},$$

and the nonuniform interior one-step uniformizers are the permutations of

$$\left({\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}}\right).$$

7. Binary Dynamics and Arithmetic Escape

The binary case is special because the simplex is an interval. It gives a complete one-dimensional model of the self-power mechanism and also supports an arithmetic analysis that is not visible from the purely metric convergence theorem.

For $n=2$, write $p=(x,1-x)$ with $0<x<1$. Then

$$T_2\left(p\right)=\left(F\left(x\right),1-F\left(x\right)\right),$$

where

$$F\left(x\right)={\displaystyle \frac{x^x}{x^x+{(1-x)}^{1-x}}}.$$

The natural coordinate on the open interval $(0,1)$ is the logit coordinate

$$y=log{\displaystyle \frac{x}{1-x}},$$

because it turns probability ratios into additive real variables. Its inverse is the logistic function

$$s\left(y\right)={\displaystyle \frac{e^y}{1+e^y}}.$$

In this coordinate, one step of the binary dynamics is given by

$$y^{+}=log{\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}=xlogx-(1-x)log(1-x).$$

Consequently the binary system is conjugate on $\mathbb{R}$ to

$$y_{k+1}=\mathcal{T}\left(y_k\right),$$

where

$$\mathcal{T}\left(y\right)=s\left(y\right)logs\left(y\right)-\left(1-s\left(y\right)\right)log\left(1-s\left(y\right)\right).$$

Proposition 7

(Binary logit contraction). For all $y,z\in \mathbb{R}$,

$$|\mathcal{T}\left(y\right)-\mathcal{T}\left(z\right)|\le {\displaystyle \frac{1-log2}{2}}|y-z|.$$

(33)

Consequently, for every $0<x<1$ and every $k\ge 0$,

$$\left|log{\displaystyle \frac{F^k\left(x\right)}{1-F^k\left(x\right)}}\right|\le {\left({\displaystyle \frac{1-log2}{2}}\right)}^k\left|log{\displaystyle \frac{x}{1-x}}\right|.$$

(34)

Proof. 

Let

$$s\left(y\right)={\displaystyle \frac{e^y}{1+e^y}}.$$

Then

$$s^{\prime }\left(y\right)=s\left(y\right)\left(1-s\left(y\right)\right),$$

and

$$\mathcal{T}\left(y\right)=s\left(y\right)logs\left(y\right)-\left(1-s\left(y\right)\right)log\left(1-s\left(y\right)\right).$$

Differentiating this expression gives

$${\mathcal{T}}^{\prime }\left(y\right)=s\left(y\right)\left(1-s\left(y\right)\right)\left(2+log\left(s\left(y\right)\left(1-s\left(y\right)\right)\right)\right).$$

Put

$$u=s\left(y\right)\left(1-s\left(y\right)\right).$$

Then $0<u\le 1/4$, and

$${\mathcal{T}}^{\prime }\left(y\right)=\eta \left(u\right),\eta \left(u\right)=u(2+logu).$$

It remains to bound $\left|\eta \right|$ on $(0,1/4]$. Since

$${\eta }^{\prime }\left(u\right)=3+logu,$$

the only critical point in $(0,1/4]$ is $u=e^{-3}$. At this point,

$$\eta \left(e^{-3}\right)=-e^{-3}.$$

At the endpoint,

$$\eta (1/4)={\displaystyle \frac{1}{4}}(2-log4)={\displaystyle \frac{1-log2}{2}}.$$

Moreover,

$$\underset{u\to 0^{+}}{lim}\eta \left(u\right)=0.$$

Since

$$e^{-3}<{\displaystyle \frac{1-log2}{2}},$$

we obtain

$$\left|\eta \left(u\right)\right|\le {\displaystyle \frac{1-log2}{2}},0<u\le {\displaystyle \frac{1}{4}}.$$

Hence

$$|{\mathcal{T}}^{\prime }\left(y\right)|\le {\displaystyle \frac{1-log2}{2}},y\in \mathbb{R}.$$

The mean value theorem gives (33).

Finally, $\mathcal{T}\left(0\right)=0$. If

$$y_0=log{\displaystyle \frac{x}{1-x}},y_{k+1}=\mathcal{T}\left(y_k\right),$$

then $y_k$ is the logit coordinate of $F^k\left(x\right)$. Applying (33) with $z=0$ and iterating gives

$$|y_k|\le {\left({\displaystyle \frac{1-log2}{2}}\right)}^k\left|y_0\right|,$$

which is exactly (34). □

We next consider the arithmetic nature of binary iterates when the initial coordinate is algebraic. The following identity will be used repeatedly:

$${\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}={\displaystyle \frac{x^x}{{(1-x)}^{1-x}}}={\displaystyle \frac{{\left(x(1-x)\right)}^x}{1-x}},0<x<1.$$

(35)

The transcendence input is the Gelfond–Schneider theorem in its standard form; see the original sources [7,8] and standard accounts such as [9,10]: if $\alpha$ is algebraic with $\alpha \ne 0,1$, and $\beta$ is algebraic irrational, then every value of ${\alpha }^{\beta }$ is transcendental.

Proposition 8

(Algebraic irrational one-step escape). Let $x\in \overline{\mathbb{Q}}\cap (0,1)$ be algebraic irrational. Then

$$F\left(x\right)\notin \overline{\mathbb{Q}}.$$

(36)

Proof. 

By (35),

$${\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}={\displaystyle \frac{{\left(x(1-x)\right)}^x}{1-x}}.$$

The base $x(1-x)$ is algebraic, nonzero, and not equal to one. The exponent x is algebraic irrational. Hence

$${\left(x(1-x)\right)}^x$$

is transcendental by the Gelfond–Schneider theorem. Multiplication by the nonzero algebraic number ${(1-x)}^{-1}$ preserves transcendence, so

$${\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}$$

is transcendental. If $F\left(x\right)$ were algebraic, then the quotient $F\left(x\right)/(1-F(x\left)\right)$ would be algebraic, a contradiction. Therefore $F\left(x\right)$ is transcendental. □

Theorem 6

(Rational two-step arithmetic escape). Let $x\in \mathbb{Q}\cap (0,1)$, $x\ne 1/2$. Then

$$F\left(x\right)\in \overline{\mathbb{Q}}\setminus \mathbb{Q},F^2\left(x\right)\notin \overline{\mathbb{Q}}.$$

(37)

Proof. 

Write

$$x={\displaystyle \frac{a}{b}},$$

where $a,b$ are coprime positive integers and $0<a<b$. Set

$$W={\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}.$$

By (35),

$$W={\displaystyle \frac{x^x}{{(1-x)}^{1-x}}}.$$

Substituting $x=a/b$ and raising to the b-th power gives

$$W^b={\displaystyle \frac{a^a{b}^{b-2a}}{{(b-a)}^{b-a}}}.$$

(38)

Thus W satisfies a polynomial equation over $\mathbb{Q}$, and hence $W\in \overline{\mathbb{Q}}$.

We first prove that W is not rational. Assume, for contradiction, that $W\in \mathbb{Q}$. Then the rational number on the right-hand side of (38) is equal to $W^b$ with $W\in \mathbb{Q}$, so it is a b-th power in $\mathbb{Q}$. By the elementary prime-valuation criterion for rational perfect powers [11], a rational number is a b-th power in $\mathbb{Q}$ only if every prime valuation is divisible by b.

The integers a, b, and $b-a$ are pairwise coprime. Indeed, $gcd(a,b)=1$ by assumption, and then

$$gcd(a,b-a)=1,gcd(b,b-a)=1.$$

Let ℓ be a prime divisor of a. Since ℓ divides neither b nor $b-a$, its valuation in the rational number (38) comes only from the factor $a^a$. Therefore that valuation is

$$a{v}_{\ell }\left(a\right).$$

Since (38) is assumed to be a rational b-th power, this exponent must be divisible by b. As $gcd(a,b)=1$, it follows that

$$b\mid v_{\ell }\left(a\right).$$

This holds for every prime divisor ℓ of a, so every prime valuation of a is divisible by b. Hence

$$a=A^b$$

for some integer $A\ge 1$.

Similarly, let ℓ be a prime divisor of $b-a$. Since ℓ divides neither a nor b, its valuation in (38) comes only from the denominator ${(b-a)}^{b-a}$. Hence this valuation is

$$-(b-a)v_{\ell }(b-a).$$

Again it must be divisible by b. Since $gcd(b-a,b)=1$, this forces

$$b\mid v_{\ell }(b-a).$$

Thus every prime valuation of $b-a$ is divisible by b, and therefore

$$b-a=B^b$$

for some integer $B\ge 1$.

Adding the two identities gives

$$b=a+(b-a)=A^b+B^b.$$

If $b\ge 3$ and at least one of $A,B$ is greater than one, then

$$A^b+B^b\ge 2^b+1>b,$$

which contradicts $b=A^b+B^b$. If $A=B=1$, then the same identity gives

$$b=1^b+1^b=2.$$

Hence the only possible rational case is

$$b=2,a=1,$$

namely $x=1/2$. This contradicts the assumption $x\ne 1/2$. Therefore

$$W\notin \mathbb{Q}.$$

Since $W\in \overline{\mathbb{Q}}$ and

$$F\left(x\right)={\displaystyle \frac{W}{1+W}},$$

we have

$$F\left(x\right)\in \overline{\mathbb{Q}}.$$

If $F\left(x\right)$ were rational, then

$$W={\displaystyle \frac{F\left(x\right)}{1-F\left(x\right)}}$$

would also be rational, contradicting $W\notin \mathbb{Q}$. Hence

$$F\left(x\right)\in \overline{\mathbb{Q}}\setminus \mathbb{Q}.$$

Finally, $F\left(x\right)\in (0,1)$ and is algebraic irrational. Applying Proposition 8 to the algebraic irrational input $F\left(x\right)$ gives

$$F^2\left(x\right)\notin \overline{\mathbb{Q}}.$$

This proves both assertions. □

8. Conclusions

We have introduced and analyzed a self-power escort transformation on the probability simplex. The map is generated by the scalar rule $x\mapsto x^x$, equivalently by exponentiating $xlogx$ coordinatewise and normalizing. Although the definition is elementary, the resulting dynamics has a rigid structure that combines boundary regularization, projective contraction, local spectral simplicity, exact uniform-fiber geometry, and arithmetic behavior in the binary subsystem.

The first main conclusion is global uniformization. Every orbit enters the interior after one step, then enters a compact core that is independent of the initial condition, and from that point on contracts exponentially in Hilbert log-diameter toward the uniform distribution ${\nu }_n$. This yields global convergence of all iterates to ${\nu }_n$ and excludes all nontrivial periodic orbits. The contraction mechanism is quantitative: it is controlled first by an explicit core constant ${\lambda }_n$, and, for $n\ge 4$, by a sharper post-core constant ${\Lambda }_n^{*}$ obtained from the simplex constraint.

The second main conclusion is that the local dynamics at the uniform point is completely explicit. The derivative of $T_n$ on the tangent space is scalar, with multiplier

$${\mu }_n={\displaystyle \frac{1-logn}{n}}.$$

This explains the dimension-dependent sign change between the binary case and the higher-dimensional cases. The second-order normal form shows that the first nonlinear correction is governed by a centered quadratic term. Thus the local approach to the uniform state is controlled by a scalar first-order multiplier together with an explicit centered quadratic feedback.

A third conclusion is the rigidity of the uniform fiber. The equation $T_n\left(p\right)={\nu }_n$ reduces to equality of the values $p_i^{p_i}$. Since the function $x^x$ has only two monotone branches on $(0,1)$, every nonuniform interior one-step uniformizer must be two-level. The full classification shows that, for $n\ge 3$, the only nonuniform interior one-step uniformizers are the n permutations of a vector of the form

$$({\alpha }_n,\dots ,{\alpha }_n,{\beta }_n),$$

where ${\alpha }_n$ appears $n-1$ times and ${\beta }_n$ appears once. Thus the fiber of the uniform state is finite and explicitly parametrized.

In the binary case, the system admits a sharper one-dimensional description through the logit coordinate. The logit coordinate is contracted globally by a dimension-specific scalar bound, giving a precise one-dimensional mechanism for convergence to the uniform point. The same binary subsystem also reveals an arithmetic escape phenomenon: algebraic irrational inputs leave the algebraic numbers after one step, while nonuniform rational inputs leave the rational numbers after one step and the algebraic numbers after two steps. This gives the binary subsystem an additional arithmetic structure beyond its dynamical behavior.

Taken together, these results provide a complete first analysis of the minimal self-power escort map. The system is simple to define, but it already exhibits a nontrivial combination of global contraction, sharp local structure, exact fiber classification, and arithmetic escape.