Banach Contractivity and Decay Profiles of Guarded Language Operators in Weighted First-Disagreement Ultrametric Spaces

1. Introduction

Fixed-point theory has long provided one of the basic mathematical tools for studying iterative and recursive processes. Since Banach’s classical contraction principle [1], contractive mappings on complete metric spaces have been used to establish existence and uniqueness of solutions together with convergence of successive approximations. The strength of this approach lies not only in the existence of a fixed point, but also in the possibility of obtaining quantitative information about the iteration leading to it.

Formal languages provide a natural setting in which recursive constructions and metric ideas can interact. The classical theory of formal languages, developed systematically in works such as [3,7], is primarily combinatorial and algebraic. At the same time, spaces of languages can also be studied topologically by measuring how early two languages can be distinguished. An important development in this direction was the study of the Bodnarchuk metric space of languages by Vianu [9], where the topology of language spaces was related to agreement and disagreement on words of bounded length. Related topological questions for classes of formal languages were later revisited, from a modern perspective, by Fulop and Kephart [4].

This point of view is especially well suited to ultrametric structures. Instead of measuring the total amount by which two languages differ, a first-disagreement ultrametric records the earliest word length at which a difference occurs. Such a metric reflects the hierarchical nature of formal languages: two languages are close when they agree on all sufficiently short words, regardless of what may happen at larger lengths. More generally, transformations associated with ultrametric structures form part of the broader theory of ultrametric-preserving functions; see, for example, [6].

Weighted approaches to formal languages have also appeared in other forms. For instance, Bruse, Herwig, and Lange [2] introduced a geometric weighting of words and used the resulting weighted symmetric difference to define distances between languages, with applications to automatic grading. Their construction and the one considered here address different aspects of language comparison. The former accumulates contributions from differing words, whereas the first-disagreement approach studied in the present paper retains only the earliest level at which two languages cease to agree.

A direct connection between first-disagreement ultrametrics and fixed-point theory was developed recently in [5]. There, guarded language operators were studied in the classical length-based ultrametric. The central observation was that fixed left and right contexts postpone possible disagreements between languages. For positively guarded recursive operators, this combinatorial displacement produces a Banach contraction in the classical metric. As a consequence, guarded recursive language equations admit unique fixed languages, Picard iteration converges globally, and finite-depth correctness can be controlled explicitly.

The classical metric used in that framework assigns geometrically decreasing weights to successive disagreement depths. This particular choice has an important consequence: postponing a disagreement by a fixed number of word positions automatically produces a uniform contraction factor. The combinatorial effect of the guards and the numerical decay of the metric are therefore tightly linked. This raises a natural question that is not visible in the classical geometric setting: what happens when the same first-disagreement structure is retained, but the numerical weights assigned to successive depths are allowed to decay according to a general profile?

The present paper addresses this question. We replace the classical geometric weighting by an arbitrary strictly decreasing positive sequence converging to zero and study the resulting family of weighted first-disagreement ultrametrics. One of the first conclusions is that the choice of weights does not alter the underlying finite-depth topology. All admissible profiles considered here lead to the same qualitative notion of convergence and preserve compactness and completeness of the language space. What changes, sometimes drastically, is the quantitative behavior of guarded operators.

To describe this behavior, we introduce a uniform decay profile that measures how much of the metric weight can remain after a disagreement is shifted by a prescribed number of positions. This quantity turns out to have a remarkably rigid structure. It is monotone and submultiplicative, and it leads to a dichotomy between two fundamentally different regimes. Either no fixed positive displacement produces a uniform decrease bounded away from one, or such a decrease appears from a certain finite depth onward. In the latter case, the uniform decay becomes asymptotically strong and necessarily admits an exponential upper envelope. This gives rise to a natural decay threshold associated solely with the chosen weight profile.

The same decay quantity also has a precise operator-theoretic meaning. We compute the exact Lipschitz modulus of a single fixed-context operator and show that the decay profile gives the best possible Lipschitz coefficient that can be guaranteed uniformly over all guarded wrapping operators whose representations satisfy a prescribed lower bound on guard depth. This leads to an exact criterion for universal Banach contractivity and identifies the decay threshold with the precise guard level at which Banach contractivity becomes guaranteed for the whole corresponding operator class.

An equally important conclusion is that the absence of a uniform Banach coefficient does not destroy the fixed-point mechanism. Positive guardedness still pushes every possible disagreement to progressively greater depth. Using this combinatorial fact directly, we prove that every positively guarded wrapping operator has a unique fixed language and that its Picard iteration converges globally for every admissible weight profile, including profiles for which the operator need not be a Banach contraction. The same argument yields explicit finite-depth stabilization bounds. Thus the weighted setting separates phenomena that coincide in the classical geometric metric: displacement of disagreement depth, strict decrease of individual nonzero distances, and uniform Banach contractivity.

The examples developed in the final part of the paper show that this distinction is genuine rather than merely formal. Geometric profiles recover the familiar uniformly contractive situation. Stretched-exponential weights exhibit a transition between uniform and nonuniform behavior. A block-structured construction shows that every prescribed finite decay threshold can occur, while polynomially decaying weights provide a natural example with no finite uniform threshold. In the latter case, a guarded operator may fail to be Banach contractive and nevertheless possess a unique fixed language together with a globally convergent Picard orbit having an explicit quantitative rate.

From a broader perspective, the results show that the fixed-point behavior of guarded language operators is governed by two distinct mechanisms: the combinatorial displacement produced by the guards and the numerical geometry chosen on the language space. Separating these mechanisms makes it possible to identify exactly which conclusions genuinely require uniform Banach contractivity and which follow already from the underlying guarded structure. This distinction also provides a natural starting point for the study of more general contractive conditions on weighted language spaces, although such extensions lie beyond the scope of the present paper.

The paper is organized as follows. Section 2 recalls the necessary background on formal languages, first-disagreement ultrametrics, Banach’s contraction principle, and the basic disagreement estimates for guarded language operators. Section 3 introduces weighted first-disagreement ultrametrics and develops the main theory, including the uniform decay profile, the decay dichotomy and threshold, the optimal universal Banach criterion, and the profile-independent fixed-point and convergence results. Finally, Section 4 studies several representative families of weight profiles and demonstrates the different decay and convergence regimes established by the general theory.

2. Preliminaries

In this section, we collect the standard notions and previously established results that will be used throughout the paper. We first recall the basic language-theoretic notation and the classical first-disagreement ultrametric on the space of formal languages. We then recall Banach’s contraction principle, establish the exact disagreement shift produced by a single fixed left-right context, and recall the finite-union and guarded wrapping estimates from the earlier guarded-language framework.

Throughout the paper, $\Sigma$ denotes a fixed finite nonempty alphabet, ${\Sigma }^{*}$ denotes the set of all finite words over $\Sigma$, including the empty word $\epsilon$, and $\left|w\right|$ denotes the length of a word $w\in {\Sigma }^{*}$. We write ${\mathbb{N}}_0=\{0,1,2,\dots \}.$

Definition 1 

([3,7]). A formal language over Σ is any subset $L\subseteq {\Sigma }^{*}.$ The family of all formal languages over Σ is denoted by $\mathcal{L}=\mathcal{P}\left({\Sigma }^{*}\right).$

Each language can equivalently be represented by its characteristic function.

Definition 2 

([3,7]). For $L\in \mathcal{L}$, the characteristic function of L is

$$11_L:{\Sigma }^{*}⟶\{0,1\},11_L\left(w\right)=\left\{\begin{array}{cc}1,\hfill & w\in L,\hfill \\ 0,\hfill & w\notin L.\hfill \end{array}\right.$$

For $u,v\in {\Sigma }^{*}$ and $L\in \mathcal{L}$, we use the standard notation $uLv=\{uwv:w\in L\}.$

For languages $A,B\subseteq {\Sigma }^{*}$, their concatenation is $AB=\{uv:u\in A,v\in B\}.$

The metric structure considered below is based on the shortest word on which two languages disagree. This viewpoint is related to the classical Bodnarchuk–Cantor topology of language spaces [4,9] and was used in the guarded-language framework of [5].

Definition 3. 

For $A,B\in \mathcal{L}$, define the first-disagreement depth by

$$\nu (A,B)=\left\{\begin{array}{cc}+\infty ,\hfill & A=B,\hfill \\ {\displaystyle min\left\{\left|w\right|:11_A\left(w\right)\ne 11_B\left(w\right)\right\},}\hfill & A\ne B.\hfill \end{array}\right.$$

Thus, if $\nu (A,B)=N<\infty ,$ then A and B agree on every word of length strictly less than N, while they disagree on at least one word of length N.

We use the natural convention $+\infty +m=+\infty$ for every $m\in {\mathbb{N}}_0$.

The first-disagreement depth satisfies the following elementary inequality.

Lemma 1. 

For all $A,B,C\in \mathcal{L}$, $\nu (A,C)\ge min\left\{\nu \right(A,B),\nu (B,C\left)\right\}.$

Proof. 

Let $N=min\left\{\nu \right(A,B),\nu (B,C\left)\right\}.$ For every $w\in {\Sigma }^{*}$ with $\left|w\right|<N$, the languages A and B agree at w, and the languages B and C also agree at w. Consequently, $11_A\left(w\right)=11_C\left(w\right)$ for every such word. Hence the first possible disagreement between A and C cannot occur before depth N. □

We use the convention $2^{-\infty }=0.$

Definition 4 

([5]). For $A,B\in \mathcal{L}$, define $d_0(A,B)=2^{-\nu (A,B)}.$ Equivalently,

$$d_0(A,B)=\left\{\begin{array}{cc}0,\hfill & A=B,\hfill \\ 2^{-N},\hfill & \nu (A,B)=N<\infty .\hfill \end{array}\right.$$

Recall that a metric $\rho$ on a set X is called an ultrametric if $\rho (x,z)\le max\left\{\rho \right(x,y),\rho (y,z\left)\right\}$ for all $x,y,z\in X$; see, for example, [8].

Theorem 1 

([5,9]). The function $d_0$ is an ultrametric on $\mathcal{L}$, and $(\mathcal{L},d_0)$ is compact and complete.

The topology can be interpreted directly in terms of finite words.

Proposition 1 

([5]). Let $A,B\in \mathcal{L}$ and let $N\in {\mathbb{N}}_0$. Then $d_0(A,B)\le 2^{-N}$ if and only if $11_A\left(w\right)=11_B\left(w\right)$ for every $w\in {\Sigma }^{*}$ satisfying $\left|w\right|<N$.

Thus convergence in the first-disagreement ultrametric is equivalent to finite-depth stabilization: below every prescribed word-length bound, membership decisions eventually stabilize.

We recall the fixed-point principle that will be used in the contractive part of the main results.

Definition 5 

([1]). Let $(X,\rho )$ be a metric space. A mapping $T:X⟶X$ is called a Banach contraction if there exists a constant $q\in [0,1)$ such that $\rho (Tx,Ty)\le q\rho (x,y)$ for all $x,y\in X$.

Theorem 2 

([1]). Let $(X,\rho )$ be a complete metric space and let $T:X\to X$ be a Banach contraction. Then T has a unique fixed point $x^{*}\in X$. Moreover, for every initial point $x_0\in X$, the Picard sequence $x_{n+1}=T\left(x_n\right),$ $n\ge 0,$ converges to $x^{*}$.

If q is a contraction coefficient, then $\rho (T^{n}x,x^{*})\le q^n\rho (x,x^{*})$ for every $n\ge 1$.

We now collect the language-theoretic disagreement estimates that form the structural input for the weighted theory. We first establish the exact effect of a single fixed left-right context on first-disagreement depth. We then recall the finite-union and guarded wrapping estimates used in the classical length-based framework of [5].

The following elementary lemma gives the exact effect of a fixed left-right context on the first-disagreement depth. It is a direct consequence of cancellation in the free monoid.

Lemma 2. 

Let $u,v\in {\Sigma }^{*}$ and let $A,B\in \mathcal{L}$ be distinct. Then $\nu (uAv,uBv)=\nu (A,B)+\left|u\right|+\left|v\right|.$

Proof. 

Since $A\ne B$, the first-disagreement depth $n:=\nu (A,B)$ is finite. By definition of n, there exists a word $w_0\in {\Sigma }^{*},$$|w_0|=n,$ such that $11_A\left(w_0\right)\ne 11_B\left(w_0\right).$ Without loss of generality, suppose that $w_0\in A\setminus B.$ Then $u{w}_{0}v\in uAv.$

We claim that $u{w}_{0}v\notin uBv.$ Indeed, if $u{w}_{0}v\in uBv$, then there would exist some $z\in B$ such that $uzv=u{w}_{0}v.$ Since the free monoid ${\Sigma }^{*}$ is both left- and right-cancellative, this equality implies $z=w_0,$ contradicting $w_0\notin B$. Hence $u{w}_{0}v$ distinguishes $uAv$ and $uBv$. Its length is $|u{w}_{0}v|=|u|+n+|v|,$ so $\nu (uAv,uBv)\le n+\left|u\right|+\left|v\right|.$

For the reverse inequality, let $x\in uAv▵uBv,$ where ▵ denotes symmetric difference. Suppose, for instance, that $x\in uAv\setminus uBv.$ Then there exists $z\in A$ such that $x=uzv.$ We must have $z\notin B.$ Indeed, if $z\in B$, then $uzv\in uBv$, contradicting the choice of x. Thus $z\in A▵B.$ By the definition of $n=\nu (A,B)$, $\left|z\right|\ge n.$ Consequently, $\left|x\right|=\left|u\right|+\left|z\right|+\left|v\right|\ge \left|u\right|+n+\left|v\right|.$

The same argument applies when $x\in uBv\setminus uAv.$ Therefore every distinguishing word for $uAv$ and $uBv$ has length at least $n+\left|u\right|+\left|v\right|.$ Hence $\nu (uAv,uBv)\ge n+\left|u\right|+\left|v\right|.$

Combining the two inequalities yields $\nu (uAv,uBv)=\nu (A,B)+\left|u\right|+\left|v\right|.$ □

For the classical length-based ultrametric, the preceding exact depth identity immediately gives

$$\begin{array}{cc}\hfill d_0(uAv,uBv)& =2^{-\nu (uAv,uBv)}\hfill \\ & =2^{-\left(\nu (A,B)+\left|u\right|+\left|v\right|\right)}\hfill \\ & =2^{-\left(\right|u|+|v\left|\right)}{d}_0(A,B).\hfill \end{array}$$

In particular, this recovers the fixed-context contraction estimate used in the guarded-language framework of [5].

Finite unions preserve common finite-depth agreement.

Lemma 3 

([5]). Let I be a finite nonempty index set, and let ${\left(X_i\right)}_{i\in I}$ and ${\left(Y_i\right)}_{i\in I}$ be families in $\mathcal{L}$. If $\nu (X_i,Y_i)\ge N$ for every $i\in I$, then $\nu \left({\bigcup }_{i\in I}{X}_i,{\bigcup }_{i\in I}{Y}_i\right)\ge N.$

In particular, $\nu (A\cup S,B\cup S)\ge \nu (A,B)$ for all $A,B,S\in \mathcal{L}$.

We next recall the guarded wrapping construction.

Definition 6 

([5]). A finite guard family is a set $G=\{(u_r,v_r):r=1,\dots ,p\}\subseteq {\Sigma }^{*}\times {\Sigma }^{*},$ $p\ge 1.$ Its minimal total guard length is $m\left(G\right)={min}_{1\le r\le p}\left(|u_r|+|v_r|\right).$

The seed-free guarded recursive part associated with G is $F_G\left(L\right)={\bigcup }_{r=1}^p{u}_{r}L{v}_r.$

For a fixed seed language $S\in \mathcal{L}$, the corresponding guarded wrapping operator is

$$T_{S,G}\left(L\right)=S\cup F_G\left(L\right)=S\cup {\displaystyle \bigcup _{r=1}^p}{u}_{r}L{v}_r.$$

The essential previously established property is the following disagreement-shift estimate.

Proposition 2 

([5]). Let G be a finite guard family and put $m=m\left(G\right).$ Then, for every pair of distinct languages $A,B\in \mathcal{L}$, $\nu (F_G\left(A\right),F_G\left(B\right))\ge \nu (A,B)+m.$

More generally, for every fixed seed $S\in \mathcal{L}$ and every pair of distinct languages $A,B\in \mathcal{L}$,

$$\nu (T_{S,G}\left(A\right),T_{S,G}\left(B\right))\ge \nu (A,B)+m.$$

When $A=B$, one has $F_G\left(A\right)=F_G\left(B\right)$ and $T_{S,G}\left(A\right)=T_{S,G}\left(B\right),$ so the corresponding metric estimates below hold trivially. Thus the restriction to distinct languages in Proposition 2 merely avoids unnecessary arithmetic with the value $+\infty$.

For the classical metric, Proposition 2 therefore yields $d_0(T_{S,G}\left(A\right),T_{S,G}\left(B\right))\le 2^{-m}{d}_0(A,B)$ for all $A,B\in \mathcal{L}$.

In particular, if $m\ge 1,$ then $2^{-m}<1,$ and hence every positively guarded wrapping operator is a Banach contraction in the classical dyadic metric.

3. Main Results

The purpose of this section is to separate two ingredients that coincide in the classical metric: the combinatorial displacement of a disagreement produced by a guard, and the numerical decay assigned to successive disagreement depths.

Definition 7. 

A sequence $\mathbf{w}={\left(w_n\right)}_{n\in {\mathbb{N}}_0}$ is called anadmissible weight profileif $1=w_0>w_1>w_2>\dots >0$ and ${lim}_{n\to \infty }{w}_n=0.$

The normalization $w_0=1$ causes no loss of generality. Indeed, multiplying every weight by the same positive constant rescales all distances by that constant and therefore does not change the induced topology, the first-disagreement structure, or any Lipschitz ratio.

Remark 1. 

The strict decrease required in Definition 7 is stronger than what is needed merely to obtain the finite-depth topology.

Indeed, suppose more generally that $1=w_0\ge w_1\ge w_2\ge \dots >0$ and $w_n⟶0.$ Then the formula

$$d_{\mathbf{w}}(A,B)=\left\{\begin{array}{cc}0,\hfill & A=B,\hfill \\ w_{\nu (A,B)},\hfill & A\ne B,\hfill \end{array}\right.$$

still defines an ultrametric. Moreover, the resulting metric still induces the usual finite-depth topology on $\mathcal{L}$.

To see the latter point, fix $L\in \mathcal{L}$. Given $\epsilon >0$, choose N such that $w_N<\epsilon .$ Then $\nu (L,K)\ge N$ implies $d_{\mathbf{w}}(L,K)\le w_N<\epsilon .$ Conversely, for every $N\ge 1$, if $d_{\mathbf{w}}(L,K)<w_{N-1},$ then a first disagreement cannot occur at any depth $j<N$, because monotonicity would give $w_j\ge w_{N-1}.$ Hence $\nu (L,K)\ge N.$ Thus the metric and finite-depth cylinder topologies still coincide even when plateaus are permitted.

The strict decrease imposed in Definition 7 is needed for the stronger numerical properties developed in the present paper.

First, it makes disagreement depth exactly recoverable from the threshold values of the metric. If, for example, $w_N=w_{N+1},$ choose $a\in \Sigma$ and set $A=⌀,$ $B=\left\{a^N\right\}.$ Then $\nu (A,B)=N,$ but $d_{\mathbf{w}}(A,B)=w_N=w_{N+1}.$ Consequently, $d_{\mathbf{w}}(A,B)\le w_{N+1},$ although $\nu (A,B)<N+1.$ Hence the exact equivalence $d_{\mathbf{w}}(A,B)\le w_M⟺\nu (A,B)\ge M$ may fail when the profile contains plateaus.

Second, strict decrease is precisely what converts every positive disagreement shift into strict metric decrease. Under a plateau $w_N=w_{N+1},$ a shift from depth N to depth $N+1$ need not change the numerical distance. By contrast, for an admissible profile and every $n\in {\mathbb{N}}_0,$ $m\ge 1,$ one has $w_{n+m}<w_n.$ Therefore every positive displacement of first-disagreement depth strictly decreases the corresponding nonzero distance.

Thus strict monotonicity is not imposed because it is necessary for compactness, completeness, or the finite-depth topology. It is imposed because the present theory requires both an exact numerical encoding of disagreement depth and the pointwise strict contraction mechanism generated by positive guards.

Definition 8. 

Let $\mathbf{w}$ be an admissible weight profile. For $A,B\in \mathcal{L}$, define

$$d_{\mathbf{w}}(A,B)=\left\{\begin{array}{cc}0,\hfill & A=B,\hfill \\ w_{\nu (A,B)},\hfill & A\ne B.\hfill \end{array}\right.$$

Remark 2. 

The general study of transformations preserving ultrametrics belongs to the theory of ultrametric-preserving functions; see [6]. Weighted metrics on formal languages have also been considered from a different viewpoint. In particular, Bruse, Herwig, and Lange [2] define a distance by assigning geometric weights to words and measuring the weighted symmetric difference of two languages. The distance in Definition 8 is of a different type: it depends only on the depth of the first disagreement and therefore preserves the hierarchical finite-depth structure used by guarded recursive operators.

The first result shows that every admissible weight profile preserves the underlying hierarchical geometry of the language space. In particular, the numerical values assigned to the disagreement levels may change, while the finite-depth cylinder structure remains unchanged.

Theorem 3. 

Let $\mathbf{w}={\left(w_n\right)}_{n\in {\mathbb{N}}_0}$ be an admissible weight profile. Then $d_{\mathbf{w}}$ is an ultrametric on $\mathcal{L}$.

Moreover, for every $L\in \mathcal{L}$ and every $N\in {\mathbb{N}}_0$, the set ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:\nu (L,K)\ge N\right\}$ admits the equivalent descriptions ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:11_K\left(w\right)=11_L\left(w\right)\mathrm{for}\mathrm{every}w\in {\Sigma }^{*},\left|w\right|<N\right\},$ ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:d_{\mathbf{w}}(L,K)\le w_N\right\},$ and ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:d_0(L,K)\le 2^{-N}\right\}.$

The family $\left\{{\mathcal{C}}_N\left(L\right):N\in {\mathbb{N}}_0\right\}$ is a neighborhood basis at L for both $d_{\mathbf{w}}$ and $d_0$. Consequently, the two metrics induce the same topology on $\mathcal{L}$. In particular, $(\mathcal{L},d_{\mathbf{w}})$ is compact and complete.

Proof. 

We first verify the ultrametric property. By Definition 8, one has $d_{\mathbf{w}}(A,B)\ge 0$ for all $A,B\in \mathcal{L}$, and $d_{\mathbf{w}}(A,B)=0⟺A=B.$ Symmetry follows from $\nu (A,B)=\nu (B,A).$

It remains to prove the strong triangle inequality. Let $A,B,C\in \mathcal{L}$. If two of the three languages coincide, the required inequality is immediate. We may therefore assume that the relevant first-disagreement depths are finite. By Lemma 1, $\nu (A,C)\ge min\left\{\nu \right(A,B),\nu (B,C\left)\right\}.$ Set $r=\nu (A,B),s=\nu (B,C).$ If $r\le s$, then $\nu (A,C)\ge r.$ Since the weight profile is strictly decreasing, $d_{\mathbf{w}}(A,C)=w_{\nu (A,C)}\le w_r=d_{\mathbf{w}}(A,B).$ Similarly, if $s\le r$, then $d_{\mathbf{w}}(A,C)\le d_{\mathbf{w}}(B,C).$ Hence in all cases $d_{\mathbf{w}}(A,C)\le max\{d_{\mathbf{w}}(A,B),d_{\mathbf{w}}(B,C)\},$ so $d_{\mathbf{w}}$ is an ultrametric.

We next identify its finite-depth balls. Fix $L\in \mathcal{L}$ and $N\in {\mathbb{N}}_0$. By the definition of the first-disagreement depth, $\nu (L,K)\ge N$ holds if and only if L and K agree on every word whose length is strictly smaller than N. Thus ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:11_K\left(w\right)=11_L\left(w\right)\mathrm{for}\mathrm{every}\left|w\right|<N\right\}.$

We now compare this condition with the weighted distance. If $K=L$, then $d_{\mathbf{w}}(L,K)=0\le w_N.$ Suppose that $K\ne L$. Since $\left(w_n\right)$ is strictly decreasing, $d_{\mathbf{w}}(L,K)\le w_N$ is equivalent to $w_{\nu (L,K)}\le w_N,$ which is in turn equivalent to $\nu (L,K)\ge N.$ Therefore ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:d_{\mathbf{w}}(L,K)\le w_N\right\}.$

The same argument for the classical profile $2^{-n}$ gives ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:d_0(L,K)\le 2^{-N}\right\}.$

It remains to verify that the cylinder sets form neighborhood bases for both metrics.

Fix $L\in \mathcal{L}$. For every $N\in {\mathbb{N}}_0$, ${\mathcal{C}}_N\left(L\right)=\left\{K\in \mathcal{L}:d_{\mathbf{w}}(L,K)\le w_N\right\}$ is a closed $d_{\mathbf{w}}$-ball. It is also open. Indeed, let $K\in {\mathcal{C}}_N\left(L\right).$ Then $\nu (L,K)\ge N,$ so L and K agree on every word of length strictly less than N. If $M\in {\mathcal{C}}_N\left(K\right),$ then K and M also agree on every word of length strictly less than N. Hence L and M agree on every such word, and therefore $M\in {\mathcal{C}}_N\left(L\right).$ Thus ${\mathcal{C}}_N\left(K\right)\subseteq {\mathcal{C}}_N\left(L\right).$ By the same argument with L and K interchanged, ${\mathcal{C}}_N\left(L\right)\subseteq {\mathcal{C}}_N\left(K\right),$ and consequently ${\mathcal{C}}_N\left(K\right)={\mathcal{C}}_N\left(L\right).$ Hence every point of ${\mathcal{C}}_N\left(L\right)$ has a $d_{\mathbf{w}}$-neighborhood contained in ${\mathcal{C}}_N\left(L\right)$, so ${\mathcal{C}}_N\left(L\right)$ is open.

We now show that these cylinders form a neighborhood basis at L. Let $\epsilon >0$. Since $w_N⟶0,$ there exists $N\in {\mathbb{N}}_0$ such that $w_N<\epsilon .$ Then ${\mathcal{C}}_N\left(L\right)=\left\{K:d_{\mathbf{w}}(L,K)\le w_N\right\}\subseteq \left\{K:d_{\mathbf{w}}(L,K)<\epsilon \right\}.$ Thus the family $\left\{{\mathcal{C}}_N\left(L\right):N\in {\mathbb{N}}_0\right\}$ is a neighborhood basis at L for $d_{\mathbf{w}}$.

Exactly the same argument, using $2^{-N}⟶0,$ shows that the same family of cylinders is a neighborhood basis at L for the classical metric $d_0$.

Therefore the identity mapping ${id}_{\mathcal{L}}:(\mathcal{L},d_0)⟶(\mathcal{L},d_{\mathbf{w}})$ and its inverse are continuous. Hence $(\mathcal{L},d_0)$ and $(\mathcal{L},d_{\mathbf{w}})$ are homeomorphic via the identity map and induce exactly the same topology on $\mathcal{L}$.

By Theorem 1, $(\mathcal{L},d_0)$ is compact. Since compactness is preserved under homeomorphisms, $(\mathcal{L},d_{\mathbf{w}})$ is compact.

Finally, $d_{\mathbf{w}}$ is a metric and every compact metric space is complete. Therefore $(\mathcal{L},d_{\mathbf{w}})$ is complete. □

The preceding theorem shows that the weight profile changes the numerical geometry of the language space without changing its finite-depth topology.

Corollary 1. 

Let $\mathbf{w}$ be an admissible weight profile, let $A,B\in \mathcal{L}$, and let $N\in {\mathbb{N}}_0$. The following conditions are equivalent:

label=()

$d_{\mathbf{w}}(A,B)\le w_N;$

lbbel=()

$\nu (A,B)\ge N;$

lcbel=()

$11_A\left(w\right)=11_B\left(w\right)$ for every $w\in {\Sigma }^{*}$ with $\left|w\right|<N$.

Proof. 

If $A=B$, all three assertions hold immediately. Assume $A\ne B$. Then $d_{\mathbf{w}}(A,B)=w_{\nu (A,B)}$. Because the profile is strictly decreasing, $w_{\nu (A,B)}\le w_N⟺\nu (A,B)\ge N$. This proves the equivalence of (i) and (ii). The equivalence of (ii) and (iii) follows directly from the definition of $\nu$. □

Corollary 2. 

Let $\left(L_n\right)\subseteq \mathcal{L}$ and let $L\in \mathcal{L}$. Then $L_n⟶L\mathrm{in}(\mathcal{L},d_{\mathbf{w}})$ if and only if, for every $N\in {\mathbb{N}}_0$, there exists $n_0\in \mathbb{N}$ such that $11_{L_n}\left(w\right)=11_L\left(w\right)$ for all $n\ge n_0$ and every $w\in {\Sigma }^{*}$ with $\left|w\right|<N$.

Proof. 

Assume first that $L_n⟶L$ with respect to $d_{\mathbf{w}}$. Fix $N\in {\mathbb{N}}_0$. Since $w_N>0$, there exists $n_0$ such that $d_{\mathbf{w}}(L_n,L)<w_N$ for every $n\ge n_0$. In particular, $d_{\mathbf{w}}(L_n,L)\le w_N,$ and Corollary 1 gives $11_{L_n}\left(w\right)=11_L\left(w\right)$ for every word w with $\left|w\right|<N$.

Conversely, assume finite-depth stabilization. Let $\epsilon >0$. Since $w_N⟶0,$ we may choose N such that $w_N<\epsilon .$ By hypothesis, there exists $n_0$ such that, for every $n\ge n_0$, the languages $L_n$ and L agree on all words of length strictly less than N. Corollary 1 therefore yields $d_{\mathbf{w}}(L_n,L)\le w_N<\epsilon .$ Thus $L_n⟶L.$ □

Corollary 3. 

A sequence $\left(L_n\right)\subseteq \mathcal{L}$ is Cauchy with respect to $d_{\mathbf{w}}$ if and only if, for every $N\in {\mathbb{N}}_0$, there exists $n_0\in \mathbb{N}$ such that $11_{L_n}\left(w\right)=11_{L_k}\left(w\right)$ for all $n,k\ge n_0$ and every $w\in {\Sigma }^{*}$ satisfying $\left|w\right|<N$.

Consequently, the metrics $d_{\mathbf{w}}$ and $d_0$ have the same convergent sequences and the same Cauchy sequences.

Proof. 

Suppose first that $\left(L_n\right)$ is $d_{\mathbf{w}}$-Cauchy. Fix $N\in {\mathbb{N}}_0$. There exists $n_0$ such that $d_{\mathbf{w}}(L_n,L_k)<w_N$ for all $n,k\ge n_0$. Hence $d_{\mathbf{w}}(L_n,L_k)\le w_N,$ and Corollary 1 shows that $L_n$ and $L_k$ agree on every word of length strictly less than N.

Conversely, suppose that the stated finite-depth stabilization property holds. Let $\epsilon >0$. Choose N such that $w_N<\epsilon .$ For sufficiently large $n,k$, the languages $L_n$ and $L_k$ agree on every word of length strictly less than N. Therefore $d_{\mathbf{w}}(L_n,L_k)\le w_N<\epsilon .$ Thus $\left(L_n\right)$ is $d_{\mathbf{w}}$-Cauchy.

The final assertion follows because the same finite-depth characterizations hold for the classical metric $d_0$. □

Thus every admissible weight profile produces the same qualitative finite-depth convergence structure. What changes from one profile to another is the numerical rate assigned to increasing agreement depth.

We first record a general principle that transfers a combinatorial disagreement shift into a quantitative estimate for every admissible weight profile. This observation separates the depth mechanism from the particular numerical decay of the weights.

Proposition 3. 

Let $\mathbf{w}$ be an admissible weight profile and let $F:\mathcal{L}\to \mathcal{L}$. Suppose that there exists $m\in {\mathbb{N}}_0$ such that $\nu \left(F\right(A),F(B\left)\right)\ge \nu (A,B)+m$ for every pair of distinct languages $A,B\in \mathcal{L}.$

Then, for every $A\ne B$, putting $n=\nu (A,B)<\infty ,$ one has $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le w_{n+m}.$ Equivalently, $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le {\displaystyle \frac{w_{n+m}}{w_n}}{d}_{\mathbf{w}}(A,B).$

If $m\ge 1$, then F strictly decreases every nonzero distance: $A\ne B⟹d_{\mathbf{w}}(F\left(A\right),F\left(B\right))<d_{\mathbf{w}}(A,B).$

Proof. 

Fix distinct $A,B\in \mathcal{L}$ and put $n=\nu (A,B)<\infty .$

There are two cases.

If $F\left(A\right)=F\left(B\right),$ then $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))=0,$ and therefore $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le w_{n+m}.$

Assume now that $F\left(A\right)\ne F\left(B\right).$ By hypothesis, $\nu \left(F\right(A),F(B\left)\right)\ge n+m.$ Since the weight profile is strictly decreasing, $w_{\nu \left(F\right(A),F(B\left)\right)}\le w_{n+m}.$ By Definition 8, $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))=w_{\nu \left(F\right(A),F(B\left)\right)},$ and hence $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le w_{n+m}.$

Since $d_{\mathbf{w}}(A,B)=w_n>0,$ we may write the same estimate in the form $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le {\displaystyle \frac{w_{n+m}}{w_n}}{d}_{\mathbf{w}}(A,B).$

Finally, suppose that $m\ge 1$. Because the profile is strictly decreasing, $w_{n+m}<w_n.$ If $F\left(A\right)=F\left(B\right)$, then $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))=0<d_{\mathbf{w}}(A,B).$ If $F\left(A\right)\ne F\left(B\right)$, then $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le w_{n+m}<w_n=d_{\mathbf{w}}(A,B).$ Thus every nonzero distance is strictly decreased. □

The preceding proposition applies exactly to fixed left-right contexts.

Proposition 4. 

Let $\mathbf{w}$ be an admissible weight profile, let $u,v\in {\Sigma }^{*}$, and put $m=\left|u\right|+\left|v\right|.$ Define $C_{u,v}\left(L\right)=uLv.$ Then, for every pair of distinct languages $A,B\in \mathcal{L}$, $d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))=w_{\nu (A,B)+m}.$ Consequently, ${\displaystyle \frac{d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))}{d_{\mathbf{w}}(A,B)}}={\displaystyle \frac{w_{\nu (A,B)+m}}{w_{\nu (A,B)}}}.$

In particular, if $m\ge 1$, then $d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))<d_{\mathbf{w}}(A,B)$ for every $A\ne B$.

Proof. 

Let $A\ne B$ and put $n=\nu (A,B).$ By Lemma 2, fixed left and right concatenation shifts the first disagreement exactly by the total context length: $\nu (uAv,uBv)=n+\left|u\right|+\left|v\right|=n+m.$ Therefore $d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))=w_{n+m},$ which proves the first assertion.

Since $d_{\mathbf{w}}(A,B)=w_n,$ division by the positive number $w_n$ gives ${\displaystyle \frac{d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))}{d_{\mathbf{w}}(A,B)}}={\displaystyle \frac{w_{n+m}}{w_n}}.$

If $m\ge 1$, strict decrease of the weight profile gives $w_{n+m}<w_n,$ and therefore

$$d_{\mathbf{w}}(C_{u,v}\left(A\right),C_{u,v}\left(B\right))<d_{\mathbf{w}}(A,B).$$

□

For finite guarded families, exact equality need not hold. Distinct guarded terms may overlap, and the common seed may mask disagreements that would otherwise be visible. Nevertheless, the minimal total guard length gives a uniform lower bound on the displacement of every possible first disagreement.

Proposition 5. 

Let $\mathbf{w}$ be an admissible weight profile, and let $T_{S,G}\left(L\right)=S\cup {\bigcup }_{r=1}^p{u}_{r}L{v}_r$ be a guarded wrapping operator. Put $m=m\left(G\right).$ Then, for every pair of distinct languages $A,B\in \mathcal{L}$, with $n=\nu (A,B),$ one has $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le w_{n+m}.$ Equivalently, $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le {\displaystyle \frac{w_{n+m}}{w_n}}{d}_{\mathbf{w}}(A,B).$

In particular, if $m\ge 1,$ then $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)<d_{\mathbf{w}}(A,B)$ for every $A\ne B$.

Proof. 

By Proposition 2, $\nu \left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\ge \nu (A,B)+m.$ Thus the operator $T_{S,G}$ satisfies the hypothesis of Proposition 3. Applying that proposition with $F=T_{S,G}$ gives $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le w_{n+m},$ where $n=\nu (A,B).$

Since $d_{\mathbf{w}}(A,B)=w_n,$ the estimate can equivalently be written as $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le {\displaystyle \frac{w_{n+m}}{w_n}}{d}_{\mathbf{w}}(A,B).$

If $m\ge 1$, Proposition 3 also gives the strict inequality $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)<d_{\mathbf{w}}(A,B)$ for every distinct pair $A,B$. □

Remark 3. 

Proposition 5 distinguishes pointwise strict distance decrease from uniform Banach contractivity.

For every positive guard length and every admissible weight profile, one has

$$d_{\mathbf{w}}(T_{S,G}\left(A\right),T_{S,G}\left(B\right))<d_{\mathbf{w}}(A,B)$$

whenever $A\ne B$.

However, the factor ${\displaystyle \frac{w_{\nu (A,B)+m}}{w_{\nu (A,B)}}}$ depends on the disagreement depth of the particular pair. These pair-dependent factors may approach 1 as $\nu (A,B)\to \infty$. Therefore strict decrease of every individual distance does not by itself imply the existence of a uniform Banach coefficient strictly smaller than 1.

The decay profile introduced below measures precisely whether such a uniform coefficient exists.

The preceding estimates show that the relevant numerical quantity is not the absolute size of the weights but their uniform decay under a prescribed increase of disagreement depth. We therefore introduce the following profile.

Definition 9. 

Let $\mathbf{w}$ be an admissible weight profile. For $m\in {\mathbb{N}}_0$, define $Q_{\mathbf{w}}\left(m\right)={sup}_{n\ge 0}{\displaystyle \frac{w_{n+m}}{w_n}}.$ We call $Q_{\mathbf{w}}:{\mathbb{N}}_0⟶(0,1]$ theuniform decay profileassociated with $\mathbf{w}$.

Thus $Q_{\mathbf{w}}\left(m\right)$ measures the largest relative weight that may remain after a disagreement is shifted by m additional word positions.

For $m=0$, ${\displaystyle \frac{w_{n+m}}{w_n}}=1$ for every n, and hence $Q_{\mathbf{w}}\left(0\right)=1.$

For every fixed $m\ge 1$ and every individual n, strict decrease of the weight profile gives ${\displaystyle \frac{w_{n+m}}{w_n}}<1.$ However, this pointwise strict inequality does not imply $Q_{\mathbf{w}}\left(m\right)<1,$ because the corresponding ratios may approach 1 as $n\to \infty$.

The basic structural properties of $Q_{\mathbf{w}}$ are collected in the following proposition.

Proposition 6. 

Let $\mathbf{w}$ be an admissible weight profile. Then:

label=()

for every $m\in {\mathbb{N}}_0$, $0<Q_{\mathbf{w}}\left(m\right)\le 1;$

lbbel=()

$Q_{\mathbf{w}}\left(0\right)=1;$

lcbel=()

$Q_{\mathbf{w}}$ is nonincreasing, that is, $m_1\le m_2⟹Q_{\mathbf{w}}\left(m_2\right)\le Q_{\mathbf{w}}\left(m_1\right);$

ldbel=()

$Q_{\mathbf{w}}$ is submultiplicative: $Q_{\mathbf{w}}(m+\ell )\le Q_{\mathbf{w}}\left(m\right)Q_{\mathbf{w}}(\ell )$ for all $m,\ell \in {\mathbb{N}}_0$;

lebel=()

for every $m\in {\mathbb{N}}_0$ and every $k\in \mathbb{N}$, $Q_{\mathbf{w}}\left(km\right)\le Q_{\mathbf{w}}{\left(m\right)}^k.$

Proof. 

We prove the assertions separately.

For (i), positivity of the weights gives ${\displaystyle \frac{w_{n+m}}{w_n}}>0$ for every $m,n\in {\mathbb{N}}_0$. Hence $Q_{\mathbf{w}}\left(m\right)>0.$ Since the sequence $\left(w_n\right)$ is decreasing, $w_{n+m}\le w_n,$ and therefore ${\displaystyle \frac{w_{n+m}}{w_n}}\le 1.$ Taking the supremum over n yields $Q_{\mathbf{w}}\left(m\right)\le 1.$

Assertion (ii) follows because ${\displaystyle \frac{w_{n+0}}{w_n}}=1$ for every n, so $Q_{\mathbf{w}}\left(0\right)=1.$

For (iii), let $0\le m_1\le m_2.$ For every $n\in {\mathbb{N}}_0$, monotonicity of the weight profile gives $w_{n+m_2}\le w_{n+m_1}.$ Since $w_n>0$, division by $w_n$ yields ${\displaystyle \frac{w_{n+m_2}}{w_n}}\le {\displaystyle \frac{w_{n+m_1}}{w_n}}.$ Taking the supremum over n on both sides gives $Q_{\mathbf{w}}\left(m_2\right)\le Q_{\mathbf{w}}\left(m_1\right).$

We next prove (iv). Let $m,\ell \in {\mathbb{N}}_0$. For every n, ${\displaystyle \frac{w_{n+m+\ell }}{w_n}}={\displaystyle \frac{w_{n+m+\ell }}{w_{n+m}}}{\displaystyle \frac{w_{n+m}}{w_n}}.$ By Definition 9, ${\displaystyle \frac{w_{n+m+\ell }}{w_{n+m}}}\le Q_{\mathbf{w}}(\ell ),$ because the index $n+m$ is an admissible index in the supremum defining $Q_{\mathbf{w}}(\ell )$. Similarly, ${\displaystyle \frac{w_{n+m}}{w_n}}\le Q_{\mathbf{w}}\left(m\right).$ Consequently, ${\displaystyle \frac{w_{n+m+\ell }}{w_n}}\le Q_{\mathbf{w}}(\ell )Q_{\mathbf{w}}\left(m\right).$ Taking the supremum over all $n\ge 0$ yields $Q_{\mathbf{w}}(m+\ell )\le Q_{\mathbf{w}}\left(m\right)Q_{\mathbf{w}}(\ell ).$

Finally, (v) follows by induction on k. For $k=1$ the statement is trivial. Suppose that $Q_{\mathbf{w}}\left(km\right)\le Q_{\mathbf{w}}{\left(m\right)}^k.$ Then submultiplicativity gives $Q_{\mathbf{w}}\left((k+1)m\right)=Q_{\mathbf{w}}(km+m)\le Q_{\mathbf{w}}\left(km\right)Q_{\mathbf{w}}\left(m\right),$ and hence $Q_{\mathbf{w}}\left((k+1)m\right)\le Q_{\mathbf{w}}{\left(m\right)}^{k+1}.$ The proof is complete. □

The previous proposition leads to a useful dichotomy. It shows that uniform decay cannot occur only at isolated guard lengths: once it occurs at one positive depth, it persists for every larger depth and becomes asymptotically arbitrarily strong.

Definition 10. 

Let $\mathbf{w}$ be an admissible weight profile. If $\left\{m\in \mathbb{N}:Q_{\mathbf{w}}\left(m\right)<1\right\}\ne ⌀,$ we define theuniform decay thresholdof $\mathbf{w}$ by $m_{\mathrm{dec}}\left(\mathbf{w}\right)=min\left\{m\in \mathbb{N}:Q_{\mathbf{w}}\left(m\right)<1\right\}.$ If no such integer exists, we set $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$

Theorem 4. 

Let $\mathbf{w}$ be an admissible weight profile. Exactly one of the following two alternatives occurs.

label=()

$Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$. Equivalently, $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$

lbbel=()

There exists a finite integer $m_0=m_{\mathrm{dec}}\left(\mathbf{w}\right)$ such that $Q_{\mathbf{w}}\left(m\right)=1\mathrm{for}1\le m<m_0,$ while $Q_{\mathbf{w}}\left(m\right)<1\mathrm{for}\mathrm{every}m\ge m_0.$ Moreover, ${lim}_{m\to \infty }{Q}_{\mathbf{w}}\left(m\right)=0.$

More precisely, in case(B), put $q_0=Q_{\mathbf{w}}\left(m_0\right)<1.$ Then, for every $m\ge m_0$, $Q_{\mathbf{w}}\left(m\right)\le q_0^{⌊m/m_0⌋}.$ Consequently, if $\gamma =q_0^{1/m_0}\in (0,1),$ then $Q_{\mathbf{w}}\left(m\right)\le q_0^{-1}{\gamma }^m$ for every $m\ge m_0$. Thus the uniform decay profile has an exponentially decaying tail whenever its decay threshold is finite.

Proof. 

Suppose first that there is no positive integer m satisfying $Q_{\mathbf{w}}\left(m\right)<1.$ By Proposition 6, $Q_{\mathbf{w}}\left(m\right)\le 1$ for every $m\in {\mathbb{N}}_0$. Hence $Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$, and alternative (A) holds.

Assume now that there exists at least one positive integer m such that $Q_{\mathbf{w}}\left(m\right)<1.$ By the well-ordering principle, the nonempty set $\left\{m\in \mathbb{N}:Q_{\mathbf{w}}\left(m\right)<1\right\}$ has a smallest element. Denote it by $m_0=m_{\mathrm{dec}}\left(\mathbf{w}\right).$ By the minimality of $m_0$, $Q_{\mathbf{w}}\left(m\right)=1$ for every integer $1\le m<m_0.$

Since $Q_{\mathbf{w}}$ is nonincreasing, $Q_{\mathbf{w}}\left(m\right)\le Q_{\mathbf{w}}\left(m_0\right)<1$ for every $m\ge m_0$. Thus uniform decay, once it occurs, persists for every larger disagreement shift.

We now establish the quantitative tail estimate. Set $q_0=Q_{\mathbf{w}}\left(m_0\right)<1$ and fix $m\ge m_0.$ Put $k=⌊{\displaystyle \frac{m}{m_0}}⌋\ge 1.$ Then $k{m}_0\le m.$ Since $Q_{\mathbf{w}}$ is nonincreasing, $Q_{\mathbf{w}}\left(m\right)\le Q_{\mathbf{w}}\left(k{m}_0\right).$ By Proposition 6, $Q_{\mathbf{w}}\left(k{m}_0\right)\le Q_{\mathbf{w}}{\left(m_0\right)}^k=q_0^k.$ Therefore $Q_{\mathbf{w}}\left(m\right)\le q_0^{⌊m/m_0⌋}.$

Since $⌊{\displaystyle \frac{m}{m_0}}⌋\ge {\displaystyle \frac{m}{m_0}}-1$ and $0<q_0<1,$ the function $x\mapsto q_0^x$ is decreasing. Hence $q_0^{⌊m/m_0⌋}\le q_0^{m/m_0-1}.$ Define $\gamma =q_0^{1/m_0}.$ Then $0<\gamma <1$ and $q_0^{m/m_0-1}=q_0^{-1}{\gamma }^m.$ Consequently, $Q_{\mathbf{w}}\left(m\right)\le q_0^{-1}{\gamma }^m$ for every $m\ge m_0$.

Since ${\gamma }^m⟶0,$ we conclude that ${lim}_{m\to \infty }{Q}_{\mathbf{w}}\left(m\right)=0.$ This proves alternative (B), together with the stated quantitative estimates. □

Corollary 4. 

For an admissible weight profile $\mathbf{w}$, the following conditions are equivalent:

label=()

there exists $m\ge 1$ such that $Q_{\mathbf{w}}\left(m\right)<1;$

lbbel=()

$m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty ;$

lcbel=()

${lim}_{m\to \infty }{Q}_{\mathbf{w}}\left(m\right)=0;$

ldbel=()

there exists $m_0\ge 1$ such that $Q_{\mathbf{w}}\left(m\right)<1$ for every $m\ge m_0$;

lebel=()

there exist constants $C>0$ and $\gamma \in (0,1)$ such that $Q_{\mathbf{w}}\left(m\right)\le C{\gamma }^m$ for every sufficiently large m.

Proof. 

The implication $\left(i\right)⟹\left(ii\right)$ follows directly from Definition 10.

Assume (ii) and write $m_0=m_{\mathrm{dec}}\left(\mathbf{w}\right).$ By Theorem 4, if $q_0=Q_{\mathbf{w}}\left(m_0\right)<1$ and $\gamma =q_0^{1/m_0}\in (0,1),$ then $Q_{\mathbf{w}}\left(m\right)\le q_0^{-1}{\gamma }^m$ for every $m\ge m_0$. Thus (ii) implies (v).

Assume now (v). Since $0<\gamma <1,$ one has $C{\gamma }^m⟶0\mathrm{as}m\to \infty .$ Therefore $0<Q_{\mathbf{w}}\left(m\right)\le C{\gamma }^m⟶0,$ and hence ${lim}_{m\to \infty }{Q}_{\mathbf{w}}\left(m\right)=0.$ Thus $\left(v\right)⟹\left(iii\right).$

If (iii) holds, then there exists $m_0\ge 1$ such that $Q_{\mathbf{w}}\left(m\right)<1$ for every $m\ge m_0$. Hence $\left(iii\right)⟹\left(iv\right).$

Finally, $\left(iv\right)⟹\left(i\right)$ is immediate.

Thus all five conditions are equivalent. □

Remark 4. 

Corollary 4 shows that convergence $Q_{\mathbf{w}}\left(m\right)⟶0$ cannot occur arbitrarily slowly.

Indeed, once $Q_{\mathbf{w}}\left(m_0\right)<1$ at a single positive depth, submultiplicativity forces an exponentially decaying upper envelope for the entire tail of the uniform decay profile. More precisely, there exist constants $C>0\mathrm{and}\gamma \in (0,1)$ such that $Q_{\mathbf{w}}\left(m\right)\le C{\gamma }^m$ for all sufficiently large m.

Thus there is no intermediate regime in which $Q_{\mathbf{w}}\left(m\right)\to 0$ but the decay is slower than every exponential upper bound.

Remark 5. 

The quantity $m_{\mathrm{dec}}\left(\mathbf{w}\right)$ depends only on the numerical geometry of the weighted first-disagreement space and not on a particular language operator.

If $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty ,$ then no fixed positive shift in disagreement depth produces a uniform relative decay bounded away from 1.

If $m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty ,$ then every shift of at least $m_{\mathrm{dec}}\left(\mathbf{w}\right)$ positions produces a uniform relative decay strictly smaller than 1, and larger shifts make this decay arbitrarily strong.

In the next theorem, this purely metric threshold becomes exactly the uniform guard-length threshold that guarantees Banach contractivity for the entire class of guarded wrapping operators.

We now turn to Banach contractivity. The purpose is not only to obtain a sufficient contraction estimate for an individual guarded operator, but also to identify the optimal coefficient that can be guaranteed uniformly over all guarded wrapping operators with a prescribed minimal guard depth.

Definition 11. 

Let $\mathbf{w}$ be an admissible weight profile and let $F:\mathcal{L}\to \mathcal{L}$. TheLipschitz modulusof F with respect to $d_{\mathbf{w}}$ is defined by ${Lip}_{\mathbf{w}}\left(F\right)={sup}_{A\ne B}{\displaystyle \frac{d_{\mathbf{w}}(F\left(A\right),F\left(B\right))}{d_{\mathbf{w}}(A,B)}}\in [0,+\infty ],$ where the supremum is taken in the extended nonnegative real numbers.

Thus ${Lip}_{\mathbf{w}}\left(F\right)$ is the smallest extended nonnegative constant that bounds all pairwise distance ratios of F. In particular, F is Lipschitz with a finite Lipschitz constant if and only if ${Lip}_{\mathbf{w}}\left(F\right)<+\infty .$

Moreover, $F\mathrm{is}\mathrm{a}\mathrm{Banach}\mathrm{contraction}⟺{Lip}_{\mathbf{w}}\left(F\right)<1.$ Indeed, if ${Lip}_{\mathbf{w}}\left(F\right)=q<1,$ then, by the definition of the supremum, $d_{\mathbf{w}}(F\left(A\right),F\left(B\right))\le q{d}_{\mathbf{w}}(A,B)$ for all distinct $A,B\in \mathcal{L}$, while the same inequality is trivial when $A=B$. Hence q is an admissible Banach contraction coefficient.

Conversely, if F is a Banach contraction with some coefficient $q\in [0,1)$, then ${\displaystyle \frac{d_{\mathbf{w}}(F\left(A\right),F\left(B\right))}{d_{\mathbf{w}}(A,B)}}\le q$ for every $A\ne B$. Taking the supremum gives ${Lip}_{\mathbf{w}}\left(F\right)\le q<1.$

We first compute the modulus exactly for a single fixed context.

Theorem 5. 

Let $\mathbf{w}$ be an admissible weight profile, let $u,v\in {\Sigma }^{*}$, and put $m=\left|u\right|+\left|v\right|.$ Define $C_{u,v}:\mathcal{L}\to \mathcal{L},C_{u,v}\left(L\right)=uLv.$ Then ${Lip}_{\mathbf{w}}\left(C_{u,v}\right)=Q_{\mathbf{w}}\left(m\right).$

Equivalently,

$$\underset{A\ne B}{sup}{\displaystyle \frac{d_{\mathbf{w}}(uAv,uBv)}{d_{\mathbf{w}}(A,B)}}=\underset{n\ge 0}{sup}{\displaystyle \frac{w_{n+m}}{w_n}}.$$

In particular, if $m\ge 1$, then $C_{u,v}\mathrm{is}\mathrm{a}\mathrm{Banach}\mathrm{contraction}⟺Q_{\mathbf{w}}\left(m\right)<1.$

Proof. 

Fix distinct languages $A,B\in \mathcal{L}$ and write $n=\nu (A,B).$ By Proposition 4, $d_{\mathbf{w}}(uAv,uBv)=w_{n+m},$ whereas $d_{\mathbf{w}}(A,B)=w_n.$ Therefore ${\displaystyle \frac{d_{\mathbf{w}}(uAv,uBv)}{d_{\mathbf{w}}(A,B)}}={\displaystyle \frac{w_{n+m}}{w_n}}.$ By the definition of $Q_{\mathbf{w}}\left(m\right)$, ${\displaystyle \frac{w_{n+m}}{w_n}}\le Q_{\mathbf{w}}\left(m\right).$ Hence ${Lip}_{\mathbf{w}}\left(C_{u,v}\right)\le Q_{\mathbf{w}}\left(m\right).$

To obtain the reverse inequality, we show that every ratio occurring in the definition of $Q_{\mathbf{w}}\left(m\right)$ is realized by a pair of formal languages.

Fix an arbitrary $n\in {\mathbb{N}}_0$. Since $\Sigma$ is nonempty, choose a symbol $a\in \Sigma$ and set $z_n=a^n.$ For $n=0$, this means $z_0=\epsilon .$ Define $A_n=⌀,B_n=\left\{z_n\right\}.$ The two languages agree on every word of length strictly smaller than n, and they disagree on $z_n$, which has length n. Consequently, $\nu (A_n,B_n)=n.$ Applying the exact fixed-context formula gives $d_{\mathbf{w}}(u{A}_{n}v,u{B}_{n}v)=w_{n+m},$ while $d_{\mathbf{w}}(A_n,B_n)=w_n.$ Thus ${\displaystyle \frac{d_{\mathbf{w}}(u{A}_{n}v,u{B}_{n}v)}{d_{\mathbf{w}}(A_n,B_n)}}={\displaystyle \frac{w_{n+m}}{w_n}}.$

Since this holds for every $n\ge 0$, ${Lip}_{\mathbf{w}}\left(C_{u,v}\right)\ge {sup}_{n\ge 0}{\displaystyle \frac{w_{n+m}}{w_n}}=Q_{\mathbf{w}}\left(m\right).$ Combining the two inequalities yields ${Lip}_{\mathbf{w}}\left(C_{u,v}\right)=Q_{\mathbf{w}}\left(m\right).$

Finally, by Definition 5, $C_{u,v}$ is a Banach contraction exactly when its Lipschitz modulus is strictly smaller than one. Hence $C_{u,v}\mathrm{is}\mathrm{Banach}\mathrm{contractive}⟺Q_{\mathbf{w}}\left(m\right)<1.$ □

The preceding exact computation provides the extremal example needed to identify the best possible coefficient for the entire class of guarded wrapping operators.

For each $m\in \mathbb{N}$, define

$${\mathfrak{T}}_m=\left\{T:\mathcal{L}\to \mathcal{L}:\begin{array}{c}\mathrm{there}\mathrm{exist}S\in \mathcal{L}\mathrm{and}\mathrm{a}\mathrm{finite}\mathrm{guard}\mathrm{family}G\hfill \\ \mathrm{such}\mathrm{that}T=T_{S,G}\mathrm{and}m\left(G\right)\ge m\hfill \end{array}\right\}.$$

Thus membership $T\in {\mathfrak{T}}_m$ means that the mapping T admits at least one guarded representation

$$T\left(L\right)=T_{S,G}\left(L\right)=S\cup {\displaystyle \bigcup _{r=1}^p}{u}_{r}L{v}_r$$

whose guard family satisfies $m\left(G\right)\ge m.$

The quantity $m\left(G\right)$ belongs to the chosen guarded representation and is not assumed to be an intrinsic invariant of the underlying mapping $T:\mathcal{L}\to \mathcal{L}$. If a mapping admits more than one guarded representation, membership in ${\mathfrak{T}}_m$ requires only the existence of one representation with guard depth at least m.

Whenever an estimate below is applied to an operator $T\in {\mathfrak{T}}_m$, a representation $T=T_{S,G},$ $m\left(G\right)\ge m,$ witnessing this membership is chosen explicitly. The resulting estimate depends only on the lower bound m and therefore remains a statement about the mapping T itself.

If $m_1\le m_2$ and $T\in {\mathfrak{T}}_{m_2},$ then T admits a guarded representation satisfying $m\left(G\right)\ge m_2\ge m_1.$ Hence $T\in {\mathfrak{T}}_{m_1},$ and therefore ${\mathfrak{T}}_{m_2}\subseteq {\mathfrak{T}}_{m_1}.$ Thus ${\left({\mathfrak{T}}_m\right)}_{m\ge 1}$ is a decreasing hierarchy of guarded mappings classified by the existence of a representation with the prescribed minimal guard depth.

Theorem 6. 

Let $\mathbf{w}$ be an admissible weight profile.

If $T_{S,G}$ is a guarded wrapping operator with $M=m\left(G\right)\ge 1,$ then ${Lip}_{\mathbf{w}}\left(T_{S,G}\right)\le Q_{\mathbf{w}}\left(M\right).$ Equivalently, $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le Q_{\mathbf{w}}\left(M\right)d_{\mathbf{w}}(A,B)$ for all $A,B\in \mathcal{L}$.

Moreover, for every $m\ge 1$, ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)=Q_{\mathbf{w}}\left(m\right).$

Consequently, $Q_{\mathbf{w}}\left(m\right)$ is the optimal universal Lipschitz coefficient for the class ${\mathfrak{T}}_m$.

Proof. 

Let $T_{S,G}$ be a guarded wrapping operator and put $M=m\left(G\right)\ge 1.$

If $A=B$, then $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)=0,$ so the required estimate is immediate.

Assume that $A\ne B$ and put $n=\nu (A,B).$ By Proposition 5, $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le w_{n+M}.$ By Definition 9, ${\displaystyle \frac{w_{n+M}}{w_n}}\le Q_{\mathbf{w}}\left(M\right).$ Since $d_{\mathbf{w}}(A,B)=w_n,$ we obtain $d_{\mathbf{w}}\left(T_{S,G}\left(A\right),T_{S,G}\left(B\right)\right)\le Q_{\mathbf{w}}\left(M\right)d_{\mathbf{w}}(A,B).$ Taking the supremum over all distinct $A,B$ gives ${Lip}_{\mathbf{w}}\left(T_{S,G}\right)\le Q_{\mathbf{w}}\left(M\right).$

We now prove the optimality statement. Fix $m\ge 1$ and let $T\in {\mathfrak{T}}_m.$ By the definition of ${\mathfrak{T}}_m$, there exist a seed language $S\in \mathcal{L}$ and a finite guard family G such that $T=T_{S,G}$ and $M:=m\left(G\right)\ge m.$

Applying the first part of the theorem to this chosen representation gives ${Lip}_{\mathbf{w}}\left(T\right)={Lip}_{\mathbf{w}}\left(T_{S,G}\right)\le Q_{\mathbf{w}}\left(M\right).$ Since $Q_{\mathbf{w}}$ is nonincreasing and $M\ge m,$ we have $Q_{\mathbf{w}}\left(M\right)\le Q_{\mathbf{w}}\left(m\right).$ Consequently, ${Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(m\right).$

Since $T\in {\mathfrak{T}}_m$ was arbitrary, ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(m\right).$

It remains to prove the reverse inequality.

Since $\Sigma$ is nonempty, choose $a\in \Sigma .$ Consider the one-guard family $G_m=\left\{(a^m,\epsilon )\right\}$ and the empty seed $S=⌀.$ Then $m\left(G_m\right)=m,$ so the corresponding operator $T_m\left(L\right)=T_{⌀,G_m}\left(L\right)=a^{m}L$ belongs to ${\mathfrak{T}}_m$.

Moreover, $T_m=C_{a^m,\epsilon }.$ By Theorem 5, ${Lip}_{\mathbf{w}}\left(T_m\right)=Q_{\mathbf{w}}\left(m\right).$ Hence ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)\ge Q_{\mathbf{w}}\left(m\right).$

Combining the upper and lower bounds gives ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)=Q_{\mathbf{w}}\left(m\right).$ Thus $Q_{\mathbf{w}}\left(m\right)$ is the optimal universal Lipschitz coefficient for the class ${\mathfrak{T}}_m$. □

The optimal estimate yields an exact Banach classification of the operator class ${\mathfrak{T}}_m$.

Corollary 5. 

Let $\mathbf{w}$ be an admissible weight profile and let $m\ge 1.$ The following conditions are equivalent:

label=()

$Q_{\mathbf{w}}\left(m\right)<1;$

lbbel=()

every operator $T\in {\mathfrak{T}}_m$ is a Banach contraction on $(\mathcal{L},d_{\mathbf{w}})$;

lcbel=()

there exists a constant $q\in [0,1),$ depending only on $\mathbf{w}$ and m, such that $d_{\mathbf{w}}(T\left(A\right),T\left(B\right))\le q{d}_{\mathbf{w}}(A,B)$ for every $T\in {\mathfrak{T}}_m$ and all $A,B\in \mathcal{L}$.

When these equivalent conditions hold, $Q_{\mathbf{w}}\left(m\right)$ is the optimal universal Banach coefficient for ${\mathfrak{T}}_m$; that is, it is the smallest constant q for which the estimate in(iii)holds simultaneously for every $T\in {\mathfrak{T}}_m$.

Proof. 

Assume first that (i) holds. By Theorem 6, ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)=Q_{\mathbf{w}}\left(m\right)<1.$ Hence, for every $T\in {\mathfrak{T}}_m,$ one has ${Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(m\right)<1.$ Therefore every $T\in {\mathfrak{T}}_m$ is a Banach contraction. This proves (ii).

Moreover, choosing $q=Q_{\mathbf{w}}\left(m\right)$ gives $d_{\mathbf{w}}(T\left(A\right),T\left(B\right))\le q{d}_{\mathbf{w}}(A,B)$ for every $T\in {\mathfrak{T}}_m$ and all $A,B\in \mathcal{L}$. Thus (iii) also holds.

The implication $\left(iii\right)⟹\left(ii\right)$ is immediate.

It remains to prove $\left(ii\right)⟹\left(i\right).$ Suppose, to the contrary, that $Q_{\mathbf{w}}\left(m\right)=1.$ Since $\Sigma$ is nonempty, choose $a\in \Sigma$ and consider $T_m\left(L\right)=a^{m}L.$ This mapping admits the one-context guarded representation with empty seed and guard family $G_m=\left\{(a^m,\epsilon )\right\}.$ For this representation, $m\left(G_m\right)=m,$ and therefore $T_m\in {\mathfrak{T}}_m.$

By Theorem 5, ${Lip}_{\mathbf{w}}\left(T_m\right)=Q_{\mathbf{w}}\left(m\right)=1.$ Therefore $T_m$ is not a Banach contraction, contradicting (ii). Hence $Q_{\mathbf{w}}\left(m\right)<1.$

Finally, Theorem 6 gives ${sup}_{T\in {\mathfrak{T}}_m}{Lip}_{\mathbf{w}}\left(T\right)=Q_{\mathbf{w}}\left(m\right).$ Thus $Q_{\mathbf{w}}\left(m\right)$ is a valid universal coefficient, and no smaller constant can bound the Lipschitz moduli of all operators in ${\mathfrak{T}}_m$. Hence it is the optimal universal Banach coefficient. □

Combining the universal Banach criterion with the uniform-decay threshold identifies the exact level of the operator hierarchy at which Banach contractivity becomes universal.

Corollary 6. 

Let $\mathbf{w}$ be an admissible weight profile.

If $m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty ,$ then

$$m_{\mathrm{dec}}\left(\mathbf{w}\right)=min\left\{m\in \mathbb{N}:\mathrm{every}T\in {\mathfrak{T}}_m\mathrm{is}\mathrm{Banach}\mathrm{contractive}\mathrm{on}(\mathcal{L},d_{\mathbf{w}})\right\}.$$

Equivalently, if $m_0=m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty ,$ then ${\mathfrak{T}}_{m_0}$ consists entirely of Banach contractions, whereas for every $1\le m<m_0$ the class ${\mathfrak{T}}_m$ contains at least one operator that is not Banach contractive.

If $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty ,$ then no class ${\mathfrak{T}}_m$ consists entirely of Banach contractions. More precisely, for every $m\ge 1,$ the class ${\mathfrak{T}}_m$ contains a guarded wrapping operator with Lipschitz modulus equal to 1.

Proof. 

Assume first that $m_0=m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty .$ By Definition 10, $Q_{\mathbf{w}}\left(m_0\right)<1.$ Hence Corollary 5 implies that every $T\in {\mathfrak{T}}_{m_0}$ is a Banach contraction.

Now let $1\le m<m_0.$ By the minimality of $m_0$, $Q_{\mathbf{w}}\left(m\right)=1.$ Choose a symbol $a\in \Sigma$ and consider the canonical one-context operator $T_m\left(L\right)=a^{m}L.$ It admits the guarded representation with empty seed and guard family $G_m=\left\{(a^m,\epsilon )\right\},$ for which $m\left(G_m\right)=m.$ Therefore $T_m\in {\mathfrak{T}}_m.$ By Theorem 5, ${Lip}_{\mathbf{w}}\left(T_m\right)=Q_{\mathbf{w}}\left(m\right)=1.$ Thus $T_m$ is not a Banach contraction. Hence no integer smaller than $m_0$ has the required universal property.

Suppose now that $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$ Then, by Theorem 4, $Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$. For each such m, the same canonical operator $T_m\left(L\right)=a^{m}L$ belongs to ${\mathfrak{T}}_m$ and satisfies ${Lip}_{\mathbf{w}}\left(T_m\right)=1.$ Therefore no finite guard threshold yields universal Banach contractivity. □

Remark 6 

(Universal versus individual contractivity). The condition $Q_{\mathbf{w}}\left(m\right)<1$ is a statement about the entire operator class ${\mathfrak{T}}_m,$ not merely about a single guarded wrapping operator.

If $Q_{\mathbf{w}}\left(m\right)<1,$ then every $T\in {\mathfrak{T}}_m$ is Banach contractive, with the common estimate ${Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(m\right).$

By contrast, if $Q_{\mathbf{w}}\left(m\right)=1,$ then the class ${\mathfrak{T}}_m$ is not uniformly Banach contractive. Indeed, the canonical operator $T_m\left(L\right)=a^{m}L$ belongs to ${\mathfrak{T}}_m$ and satisfies ${Lip}_{\mathbf{w}}\left(T_m\right)=1.$

This does not imply that every operator in ${\mathfrak{T}}_m$ has Lipschitz modulus equal to one. For a particular guarded wrapping operator, overlaps between guarded terms or masking by the seed may increase the disagreement depth beyond the universal lower bound and thereby reduce its actual Lipschitz modulus below one.

Thus $Q_{\mathbf{w}}\left(m\right)=1$ expresses a sharp obstruction to auniversalBanach conclusion based solely on the prescribed guard depth, rather than an obstruction to Banach contractivity of every individual operator in the class.

We finally record the fixed-point and convergence consequences in the uniformly contractive regime. The optimal universal coefficient $Q_{\mathbf{w}}\left(m\right)$ yields a rate valid for the entire guard class, while the actual Lipschitz modulus of a particular operator may provide a strictly sharper estimate.

Corollary 7. 

Let $T=T_{S,G}:\mathcal{L}\to \mathcal{L}$ be a guarded wrapping operator with $m=m\left(G\right)\ge 1,$ and suppose that $q:=Q_{\mathbf{w}}\left(m\right)<1.$ Set ${\lambda }_T={Lip}_{\mathbf{w}}\left(T\right).$ Then $0\le {\lambda }_T\le q<1.$

Consequently, T has a unique fixed language $L^{*}\in \mathcal{L},$ and, for every initial language $L_0\in \mathcal{L}$, the Picard iteration $L_{n+1}=T\left(L_n\right),$$n\ge 0,$ converges to $L^{*}$.

Moreover, for every $n\ge 1$, $d_{\mathbf{w}}(L_n,L^{*})\le {\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*})\le q^n{d}_{\mathbf{w}}(L_0,L^{*}).$

Since $d_{\mathbf{w}}(L_0,L^{*})\le 1,$ one obtains, for every $n\ge 1$, the universal bound $d_{\mathbf{w}}(L_n,L^{*})\le q^n.$

More generally, let $N\in {\mathbb{N}}_0.$ For every $n\ge 1$, finite-depth correctness below depth N is guaranteed whenever ${\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*})\le w_N.$ In particular, the operator-independent sufficient condition $q^n\le w_N$ implies $11_{L_n}\left(w\right)=11_{L^{*}}\left(w\right)$ for every $w\in {\Sigma }^{*},\left|w\right|<N.$

For $n=0$, correctness below depth N is characterized directly by $d_{\mathbf{w}}(L_0,L^{*})\le w_N,$ or equivalently by $\nu (L_0,L^{*})\ge N.$

Proof. 

By Theorem 6, ${Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(m\right),$ and therefore $0\le {\lambda }_T\le q<1.$ Hence T is a Banach contraction on the complete metric space $(\mathcal{L},d_{\mathbf{w}})$, whose completeness follows from Theorem 3.

Banach’s contraction principle, Theorem 2, therefore gives a unique fixed language $L^{*}\in \mathcal{L}$ and convergence of the Picard iteration from every $L_0\in \mathcal{L}$.

Since $T\left(L^{*}\right)=L^{*},$ for every $n\ge 1$ repeated application of the Lipschitz estimate gives $d_{\mathbf{w}}(L_n,L^{*})=d_{\mathbf{w}}\left(T^n\left(L_0\right),T^n\left(L^{*}\right)\right)\le {\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*}).$

For $n\ge 1$, since $0\le {\lambda }_T\le q<1,$ we have ${\lambda }_T^n\le q^n,$ and therefore $d_{\mathbf{w}}(L_n,L^{*})\le {\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*})\le q^n{d}_{\mathbf{w}}(L_0,L^{*}).$

By the normalization $w_0=1,$ every distance in $(\mathcal{L},d_{\mathbf{w}})$ is at most one. Hence $d_{\mathbf{w}}(L_0,L^{*})\le 1,$ which gives $d_{\mathbf{w}}(L_n,L^{*})\le q^n$ for every $n\ge 1$.

Now let $N\in {\mathbb{N}}_0.$ If $n\ge 1$ and ${\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*})\le w_N,$ then $d_{\mathbf{w}}(L_n,L^{*})\le w_N.$ By Corollary 1, this is equivalent to agreement of $L_n$ and $L^{*}$ on every word of length strictly less than N.

Moreover, because $d_{\mathbf{w}}(L_0,L^{*})\le 1\mathrm{and}{\lambda }_T\le q,$ the simpler condition $q^n\le w_N$ is a universal sufficient criterion for every $n\ge 1$.

For $n=0$, Corollary 1 gives directly $d_{\mathbf{w}}(L_0,L^{*})\le w_N⟺\nu (L_0,L^{*})\ge N,$ which is exactly the finite-depth correctness condition for the initial language. □

The Banach property depends on the numerical decay of the weight profile. The underlying disagreement-shift mechanism, however, is independent of that numerical decay. We now show that positive guardedness alone already implies global Picard convergence and uniqueness of the fixed language for every admissible profile, even in the regime where no uniform Banach coefficient exists.

The proof also yields quantitative estimates for the entire Picard orbit, not only for its distance from the limiting fixed language.

Theorem 7. 

Let $\mathbf{w}$ be any admissible weight profile, and let $T=T_{S,G}:\mathcal{L}\to \mathcal{L}$ be a guarded wrapping operator with $m=m\left(G\right)\ge 1.$ For an arbitrary initial language $L_0\in \mathcal{L}$, define the Picard orbit by $L_{n+1}=T\left(L_n\right),n\ge 0.$

If $L_1=L_0,$ then $L_0$ is the unique fixed language of T and $L_n=L_0$ for every $n\ge 0$.

Assume now that $L_1\ne L_0$ and put $r_0=\nu (L_0,L_1)<\infty .$ Then the following assertions hold.

label=()

For every $n\ge 0$, $\nu (L_n,L_{n+1})\ge r_0+nm,$ where the left-hand side is understood as $+\infty$ if $L_n=L_{n+1}$.

In particular, if $L_s=L_{s+1}$ for some $s\ge 0$, then $L_s$ is the unique fixed language and $L_n=L_s$ for every $n\ge s$.

lbbel=()

For every pair of integers $k>n\ge 0$, $\nu (L_n,L_k)\ge r_0+nm,$ and consequently $d_{\mathbf{w}}(L_n,L_k)\le w_{r_0+nm}.$

lcbel=()

The Picard orbit converges in $(\mathcal{L},d_{\mathbf{w}})$ to a unique fixed language $L^{*}\in \mathcal{L}.$

ldbel=()

For every $n\ge 0$, $\nu (L_n,L^{*})\ge r_0+nm,$ and therefore $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}.$

Proof. 

Let $L_{n+1}=T\left(L_n\right),n\ge 0.$

We first consider the case $L_1=L_0.$ Then $T\left(L_0\right)=L_0,$ so $L_0$ is a fixed language and $L_n=L_0$ for every $n\ge 0$.

To prove uniqueness, suppose that $K^{*}\ne L_0$ is another fixed language. Since the two languages are distinct, $s:=\nu (L_0,K^{*})<\infty .$ Using the fixed-point identities and Proposition 2, we obtain $s=\nu (L_0,K^{*})=\nu (T\left(L_0\right),T\left(K^{*}\right))\ge \nu (L_0,K^{*})+m=s+m,$ which is impossible because $m\ge 1.$ Hence $L_0$ is the unique fixed language.

Assume from now on that $L_1\ne L_0,$ and define $r_0=\nu (L_0,L_1)<\infty .$

For every $n\ge 0$, let $r_n=\nu (L_n,L_{n+1})\in {\mathbb{N}}_0\cup \{+\infty \}.$

We first prove assertion (i).

Suppose that for some index s one has $r_s=+\infty .$ Then $L_s=L_{s+1}=T\left(L_s\right),$ so $L_s$ is a fixed language. Consequently, $L_n=L_s$ for every $n\ge s$, and hence $r_n=+\infty$ for every $n\ge s$.

Thus, once the orbit becomes stationary, assertion (i) is automatic for all subsequent indices.

It remains to consider indices for which $r_n<\infty .$ Then $L_n\ne L_{n+1},$ so Proposition 2 applies to the distinct input pair $L_n,L_{n+1}$. Since $L_{n+1}=T\left(L_n\right)$ and $L_{n+2}=T\left(L_{n+1}\right),$ we obtain $r_{n+1}=\nu (L_{n+1},L_{n+2})=\nu (T\left(L_n\right),T\left(L_{n+1}\right))\ge \nu (L_n,L_{n+1})+m=r_n+m.$ This inequality remains valid when $L_{n+1}=L_{n+2},$ because in that case $r_{n+1}=+\infty .$

We now prove by induction that $r_n\ge r_0+nm$ for every $n\ge 0$, with the inequality understood in the extended order when $r_n=+\infty$.

For $n=0$ there is equality. Assume that $r_n\ge r_0+nm.$ If $r_n=+\infty ,$ then the orbit is stationary from index n onward, and hence $r_{n+1}=+\infty \ge r_0+(n+1)m.$ If $r_n<\infty ,$ the estimate established above gives $r_{n+1}\ge r_n+m\ge r_0+(n+1)m.$ Therefore $r_n\ge r_0+nm$ for every $n\ge 0$.

Finally, if the orbit becomes stationary at some index s, then $L_s$ is a fixed language. The uniqueness argument given above applies verbatim and shows that this fixed language is unique. This proves all additional statements in (i).

We next prove (ii). Let $k>n.$ Apply Lemma 1 repeatedly to the finite chain $L_n,L_{n+1},\dots ,L_k.$ This gives $\nu (L_n,L_k)\ge min\left\{\nu (L_j,L_{j+1}):n\le j<k\right\}.$ By assertion (i), $\nu (L_j,L_{j+1})\ge r_0+jm.$ Since $j\ge n$, $r_0+jm\ge r_0+nm.$ Hence $\nu (L_n,L_k)\ge r_0+nm.$

By Definition 8, either $L_n=L_k$, in which case the corresponding metric estimate is trivial, or $d_{\mathbf{w}}(L_n,L_k)=w_{\nu (L_n,L_k)}.$ Since $\left(w_j\right)$ is decreasing, $d_{\mathbf{w}}(L_n,L_k)\le w_{r_0+nm}.$ This proves (ii).

We now establish the Cauchy property. Let $\epsilon >0.$ Because the weight profile is admissible, $w_j⟶0\mathrm{as}j\to \infty .$ Since $r_0+nm⟶\infty ,$ there exists $n_0$ such that $w_{r_0+n_{0}m}<\epsilon .$

Let $n,k\ge n_0$. If $n=k$, then $d_{\mathbf{w}}(L_n,L_k)=0<\epsilon .$

Assume now that $n\ne k$. By symmetry of the metric, we may suppose without loss of generality that $k>n.$ Then assertion (ii) yields $d_{\mathbf{w}}(L_n,L_k)\le w_{r_0+nm}.$ Since $n\ge n_0$ and the weight profile is decreasing, $w_{r_0+nm}\le w_{r_0+n_{0}m}.$ Hence $d_{\mathbf{w}}(L_n,L_k)\le w_{r_0+n_{0}m}<\epsilon .$

Therefore, for all $n,k\ge n_0$, $d_{\mathbf{w}}(L_n,L_k)<\epsilon ,$ and consequently $\left(L_n\right)$ is a Cauchy sequence in $(\mathcal{L},d_{\mathbf{w}})$.

By Theorem 3, $(\mathcal{L},d_{\mathbf{w}})$ is complete. Therefore there exists $L^{*}\in \mathcal{L}$ such that $L_n⟶L^{*}.$

We next prove directly that $L^{*}$ is a fixed language. The argument is carried out entirely in terms of first-disagreement depths.

Fix an arbitrary $N\in {\mathbb{N}}_0.$ Since $L_n⟶L^{*},$ Corollary 2 implies that there exists $n_1\in \mathbb{N}$ such that $\nu (L_n,L^{*})\ge N$ for every $n\ge n_1$.

The shifted sequence $\left(L_{n+1}\right)$ has the same limit $L^{*}$. Hence there exists $n_2\in \mathbb{N}$ such that $\nu (L_{n+1},L^{*})\ge N$ for every $n\ge n_2$.

Choose $n\ge max\{n_1,n_2\}.$ We claim that $\nu (T\left(L^{*}\right),L_{n+1})\ge N.$

There are two cases.

If $L_n=L^{*},$ then $T\left(L^{*}\right)=T\left(L_n\right)=L_{n+1}.$ Therefore $\nu (T\left(L^{*}\right),L_{n+1})=+\infty \ge N.$

Assume now that $L_n\ne L^{*}.$ Since the pair $L_n,L^{*}$ is distinct, Proposition 2 applies and gives $\nu (T\left(L^{*}\right),L_{n+1})=\nu (T\left(L^{*}\right),T\left(L_n\right))\ge \nu (L^{*},L_n)+m\ge N+m\ge N.$ Thus in either case, $\nu (T\left(L^{*}\right),L_{n+1})\ge N.$

On the other hand, by the choice of n, $\nu (L_{n+1},L^{*})\ge N.$ Applying Lemma 1 to $T\left(L^{*}\right),$$L_{n+1},$$L^{*},$ we obtain $\nu (T\left(L^{*}\right),L^{*})\ge min\left\{\nu (T\left(L^{*}\right),L_{n+1}),\nu (L_{n+1},L^{*})\right\}\ge N.$

Since $N\in {\mathbb{N}}_0$ was arbitrary, the first-disagreement depth between $T\left(L^{*}\right)$ and $L^{*}$ cannot be finite. Hence $\nu (T\left(L^{*}\right),L^{*})=+\infty .$ By Definition 3, $T\left(L^{*}\right)=L^{*}.$ Therefore $L^{*}$ is a fixed language.

We now prove uniqueness. Suppose that $K^{*}\in \mathcal{L}$ is another fixed language and $K^{*}\ne L^{*}.$ Put $s=\nu (L^{*},K^{*})<\infty .$ Since both languages are fixed, $T\left(L^{*}\right)=L^{*},$$T\left(K^{*}\right)=K^{*}.$ The disagreement-shift estimate therefore gives $s=\nu (L^{*},K^{*})=\nu (T\left(L^{*}\right),T\left(K^{*}\right))\ge \nu (L^{*},K^{*})+m=s+m,$ which is impossible because $m\ge 1$. Hence $L^{*}$ is the unique fixed language. This proves (iii).

It remains to establish the quantitative estimate in (iv).

Fix $n\ge 0$ and put $R_n=r_0+nm.$ By assertion (ii), for every $k>n$, $\nu (L_n,L_k)\ge R_n.$ Hence $L_k\in {\mathcal{C}}_{R_n}\left(L_n\right)$ for every $k>n$, where ${\mathcal{C}}_{R_n}\left(L_n\right)=\left\{K\in \mathcal{L}:\nu (L_n,K)\ge R_n\right\}.$

By Theorem 3, the cylinder ${\mathcal{C}}_{R_n}\left(L_n\right)$ is a closed subset of $(\mathcal{L},d_{\mathbf{w}})$. Since $L_k⟶L^{*}$ and every sufficiently late member of the sequence belongs to ${\mathcal{C}}_{R_n}\left(L_n\right)$, closedness implies $L^{*}\in {\mathcal{C}}_{R_n}\left(L_n\right).$ Therefore $\nu (L_n,L^{*})\ge R_n=r_0+nm.$

If $L_n=L^{*},$ then $d_{\mathbf{w}}(L_n,L^{*})=0\le w_{r_0+nm}.$ Otherwise, $d_{\mathbf{w}}(L_n,L^{*})=w_{\nu (L_n,L^{*})}\le w_{r_0+nm},$ because the weight profile is decreasing.

Thus $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm},$ which proves (iv). □

Remark 7. 

In the nonstationary initial case $L_1\ne L_0$, the estimate $\nu (L_n,L_k)\ge r_0+nm,k>n,$ is uniform in the later iterate k. Thus, once the nth approximation has been reached, every subsequent Picard iterate already agrees with it on all words of length strictly less than $r_0+nm.$ The same stabilized finite-depth region is inherited by the limiting fixed language.

The theorem gives a finite-depth certification bound that takes into account the agreement depth already gained after the first recursive update.

Corollary 8. 

Assume the hypotheses of Theorem 7 and suppose that $L_1\ne L_0.$ Set $r_0=\nu (L_0,L_1),$ and let $N\in {\mathbb{N}}_0.$

For every $n\ge 0$ satisfying $r_0+nm\ge N,$ the iterate $L_n$ agrees with the fixed language $L^{*}$ on every word of length strictly less than N; that is, $11_{L_n}\left(w\right)=11_{L^{*}}\left(w\right)$ for every $w\in {\Sigma }^{*},\left|w\right|<N.$

The smallest nonnegative integer guaranteed by this estimate is $n_N=max\left\{0,⌈{\displaystyle \frac{N-r_0}{m}}⌉\right\}.$ Thus every $n\ge n_N$ is sufficient for correctness below depth N.

In particular, if $N\le r_0,$ then $n_N=0,$ so the initial language $L_0$ already agrees with $L^{*}$ on every word of length strictly less than N.

Finally, since $r_0\ge 0,$ the simpler profile-independent estimate $n\ge ⌈{\displaystyle \frac{N}{m}}⌉$ is always sufficient.

Proof. 

By Theorem 7, $\nu (L_n,L^{*})\ge r_0+nm$ for every $n\ge 0$. Therefore, whenever $r_0+nm\ge N,$ one has $\nu (L_n,L^{*})\ge N.$ By Definition 3, this is equivalent to $11_{L_n}\left(w\right)=11_{L^{*}}\left(w\right)$ for every word w satisfying $\left|w\right|<N.$

The inequality $r_0+nm\ge N$ is equivalent to $n\ge {\displaystyle \frac{N-r_0}{m}}.$ Since n must be a nonnegative integer, the smallest value ensured by this sufficient condition is $n_N=max\left\{0,⌈{\displaystyle \frac{N-r_0}{m}}⌉\right\}.$

If $N\le r_0$, then $N-r_0\le 0,$ and hence $n_N=0.$

Finally, $r_0\ge 0$ implies $r_0+nm\ge nm.$ Thus the stronger requirement $nm\ge N$ is always sufficient, which gives $n\ge ⌈{\displaystyle \frac{N}{m}}⌉.$ □

Remark 8. 

The quantity $n_N$ is the smallest iteration index guaranteed by the general lower bound $\nu (L_n,L^{*})\ge r_0+nm.$ It need not be the actual first iteration at which correctness below depth N is attained for a particular guarded operator, since additional overlap or masking may shift disagreements farther than the minimal amount m.

Remark 9 

(Profile-dependent metric rate and sharpness). In the case $L_1\ne L_0,$ Part(iv)of Theorem 7 gives the general a priori estimate $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm},$$r_0=\nu (L_0,L_1)<\infty .$ Thus the underlying combinatorial agreement depth grows at least linearly with the iteration index, namely $\nu (L_n,L^{*})\ge r_0+nm,$ while the corresponding numerical convergence rate is determined by the decay of the chosen weight profile along the arithmetic progression $r_0,r_0+m,r_0+2m,\dots .$

In the nonstationary initial case $L_1\ne L_0$, the estimate is an upper bound valid for every guarded representation $T=T_{S,G}$ satisfying $m\left(G\right)=m.$ For a particular operator the actual convergence may be faster, since overlaps between guarded terms or masking by the seed can shift disagreements by more than the guaranteed amount m.

The general estimate is sharp for every prescribed positive guard length. Indeed, fix an integer $m\ge 1$. Since Σ is nonempty, choose $a\in \Sigma .$ Consider the guarded wrapping operator $T_m\left(L\right)=\left\{\epsilon \right\}\cup a^{m}L.$ It admits the guarded representation with $G_m=\left\{(a^m,\epsilon )\right\},$ for which $m\left(G_m\right)=m.$

Starting from $L_0=⌀,$ a direct induction gives $L_n=\{a^{jm}:0\le j<n\},n\ge 1.$ The unique fixed language is $L^{*}=\{a^{jm}:j\ge 0\}.$ Moreover, $L_1=\left\{\epsilon \right\},$ and hence $r_0=\nu (L_0,L_1)=0.$

For every $n\ge 0$, the first word belonging to $L^{*}$ but not to $L_n$ is $a^{nm},$ whose length is $nm$. Therefore $\nu (L_n,L^{*})=nm=r_0+nm.$ Consequently, for every admissible weight profile, $d_{\mathbf{w}}(L_n,L^{*})=w_{nm}=w_{r_0+nm}.$

Thus both the disagreement-depth estimate and the associated profile-dependent metric estimate in Theorem 7 are sharp for every prescribed positive guard length.

The Banach estimate and the profile-dependent depth estimate are linked by a simple but useful consequence of the definition of $Q_{\mathbf{w}}$.

Proposition 7. 

Let $\mathbf{w}$ be an admissible weight profile. Then, for every $r,m,n\in {\mathbb{N}}_0,$ one has $w_{r+nm}\le Q_{\mathbf{w}}{\left(m\right)}^n{w}_r.$

Equivalently, ${\displaystyle \frac{w_{r+nm}}{w_r}}\le Q_{\mathbf{w}}{\left(m\right)}^n.$

Proof. 

Fix $r,m,n\in {\mathbb{N}}_0.$

If $n=0,$ then $r+nm=r$ and $Q_{\mathbf{w}}{\left(m\right)}^0=1.$ Hence $w_{r+nm}=w_r=Q_{\mathbf{w}}{\left(m\right)}^0{w}_r,$ so the assertion holds with equality.

Assume now that $n\ge 1.$ Using the identity ${\displaystyle \frac{w_{r+nm}}{w_r}}={\displaystyle \frac{w_{r+m}}{w_r}}{\displaystyle \frac{w_{r+2m}}{w_{r+m}}}\dots {\displaystyle \frac{w_{r+nm}}{w_{r+(n-1)m}}},$ we obtain the telescoping representation ${\displaystyle \frac{w_{r+nm}}{w_r}}={\prod }_{j=0}^{n-1}{\displaystyle \frac{w_{r+(j+1)m}}{w_{r+jm}}}.$

For every $j=0,\dots ,n-1,$ the index $r+jm$ belongs to ${\mathbb{N}}_0$. Therefore, by Definition 9, ${\displaystyle \frac{w_{r+(j+1)m}}{w_{r+jm}}}={\displaystyle \frac{w_{(r+jm)+m}}{w_{r+jm}}}\le Q_{\mathbf{w}}\left(m\right).$ Multiplying these n inequalities gives ${\displaystyle \frac{w_{r+nm}}{w_r}}\le Q_{\mathbf{w}}{\left(m\right)}^n.$ Since $w_r>0,$ we conclude that $w_{r+nm}\le Q_{\mathbf{w}}{\left(m\right)}^n{w}_r.$ Thus the estimate holds for every $n\in {\mathbb{N}}_0.$ □

Corollary 9 

(Compatibility of the depth and Banach rates). Let $T=T_{S,G}$ be a guarded wrapping operator with $m=m\left(G\right)\ge 1,$ and suppose that $q:=Q_{\mathbf{w}}\left(m\right)<1.$ Let $L_{n+1}=T\left(L_n\right),n\ge 0,$ be the Picard orbit generated by an arbitrary initial language $L_0\in \mathcal{L}$, and let $L^{*}$ denote the unique fixed language of T.

If $L_1=L_0,$ then $L_n=L^{*}=L_0$ for every $n\ge 0$, and hence $d_{\mathbf{w}}(L_n,L^{*})=0.$

Assume now that $L_1\ne L_0$ and put $r_0=\nu (L_0,L_1)<\infty .$ Then, for every $n\ge 0$, $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}\le q^n{w}_{r_0}.$ Since $w_{r_0}=d_{\mathbf{w}}(L_0,L_1)\le 1,$ one obtains the chain $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}\le q^n{d}_{\mathbf{w}}(L_0,L_1)\le q^n.$

Proof. 

If $L_1=L_0,$ then $L_0$ is a fixed language. By uniqueness of the fixed point, $L^{*}=L_0,$ and the Picard orbit is constant, so the first assertion is immediate.

Assume now that $L_1\ne L_0.$ Then $r_0=\nu (L_0,L_1)<\infty .$ By Theorem 7, $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}$ for every $n\ge 0$.

Proposition 7, applied with $r=r_0,$ gives $w_{r_0+nm}\le Q_{\mathbf{w}}{\left(m\right)}^n{w}_{r_0}=q^n{w}_{r_0}.$ Since $L_0\ne L_1,$ Definition 8 yields $w_{r_0}=d_{\mathbf{w}}(L_0,L_1).$ Finally, the normalization $w_0=1$ implies $d_{\mathbf{w}}(L_0,L_1)\le 1.$ Therefore $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}\le q^n{d}_{\mathbf{w}}(L_0,L_1)\le q^n.$ □

Remark 10. 

The preceding corollary clarifies the relation between the two convergence mechanisms. In the nonstationary initial case $L_1\ne L_0,$ put $r_0=\nu (L_0,L_1)<\infty .$ Then the purely depth-based estimate $d_{\mathbf{w}}(L_n,L^{*})\le w_{r_0+nm}$ is available for every admissible profile, independently of whether $Q_{\mathbf{w}}\left(m\right)$ is strictly smaller than one.

When $Q_{\mathbf{w}}\left(m\right)=q<1,$ the same estimate is automatically compatible with the geometric Banach scale: $w_{r_0+nm}\le q^n{w}_{r_0}.$ Hence the depth estimate is always at least as strong as the coarse normalized universal bound $d_{\mathbf{w}}(L_n,L^{*})\le q^n.$

This should not be confused with the sharper operator-specific Banach estimate $d_{\mathbf{w}}(L_n,L^{*})\le {\lambda }_T^n{d}_{\mathbf{w}}(L_0,L^{*}),n\ge 1,$ where ${\lambda }_T={Lip}_{\mathbf{w}}\left(T\right).$ The quantity ${\lambda }_T$ may be substantially smaller than $Q_{\mathbf{w}}\left(m\right)$ for an individual operator.

Thus, for $n\ge 1$, the three quantities $w_{r_0+nm},Q_{\mathbf{w}}{\left(m\right)}^n,{Lip}_{\mathbf{w}}{\left(T\right)}^n$ encode, respectively, the profile-sensitive depth rate, the optimal universal Banach scale determined by guard length alone, and the actual Banach behavior of the particular operator.

Remark 11 

(Three levels of guarded contraction). The preceding results separate three levels of behavior that coincide in the classical geometric setting but need not coincide for a general admissible weight profile.

First, positive guardedness is a purely combinatorial property. If $m=m\left(G\right)\ge 1,$ then $\nu (T_{S,G}\left(A\right),T_{S,G}\left(B\right))\ge \nu (A,B)+m$ for every pair of distinct languages $A,B$. Thus every application of the operator postpones a possible first disagreement by at least m word positions.

Second, because the weight profile is strictly decreasing, this positive depth shift implies pointwise strict decrease of every nonzero distance: $A\ne B⟹d_{\mathbf{w}}(T_{S,G}\left(A\right),T_{S,G}\left(B\right))<d_{\mathbf{w}}(A,B).$ This conclusion holds for every admissible weight profile and does not require $Q_{\mathbf{w}}\left(m\right)<1.$ The corresponding contraction factor may, however, depend on the particular disagreement depth $\nu (A,B)$ and may approach 1 along pairs whose first disagreement occurs at increasingly large depths.

Third, the stronger condition $Q_{\mathbf{w}}\left(m\right)<1$ provides a uniform numerical factor valid independently of the pair $A,B$. In that case, $d_{\mathbf{w}}(T_{S,G}\left(A\right),T_{S,G}\left(B\right))\le Q_{\mathbf{w}}\left(m\right)d_{\mathbf{w}}(A,B),$ so every mapping in ${\mathfrak{T}}_m$ lies in the classical Banach class.

When $Q_{\mathbf{w}}\left(m\right)=1,$ such a uniform Banach conclusion cannot be inferred from the guard length alone. This does not exclude Banach contractivity for a particular operator, whose actual Lipschitz modulus may be smaller because of overlaps or masking. It means only that no contraction coefficient strictly below one can be guaranteed uniformly over the entire class from the prescribed guard depth.

Finally, even in this nonuniform regime, Theorem 7 shows that the combinatorial depth shift itself is sufficient to guarantee a unique fixed language, global Picard convergence, and explicit finite-depth stabilization.

Thus the weighted theory distinguishes

$$\begin{array}{c}\mathrm{depth}\mathrm{displacement},\mathrm{pointwise}\mathrm{strict}\mathrm{distance}\mathrm{decrease},\\ \mathrm{uniform}\mathrm{Banach}\mathrm{contractivity},\end{array}$$

and identifies precisely the additional decay condition required to pass from one level to the next.

4. Examples and Model Decay Regimes

We now illustrate the preceding theory on several representative classes of admissible weight profiles. The examples show that the uniform decay threshold can equal one, can be any prescribed finite positive integer, or can be infinite. They also demonstrate that the finite-depth convergence mechanism remains effective in all three regimes.

Example 1. 

Let $w_n=r^n,0<r<1.$ Then, for every $m,n\in {\mathbb{N}}_0$, ${\displaystyle \frac{w_{n+m}}{w_n}}={\displaystyle \frac{r^{n+m}}{r^n}}=r^m.$ Hence $Q_{\mathbf{w}}\left(m\right)=r^m.$

In particular, $Q_{\mathbf{w}}\left(m\right)<1$ for every $m\ge 1$, and therefore $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1.$

Since $Q_{\mathbf{w}}\left(m\right)=r^m<1$ for every $m\ge 1$, Theorem 6 implies that every positively guarded wrapping operator is a Banach contraction.

More precisely, if $m\left(G\right)=M\ge 1,$ then ${Lip}_{\mathbf{w}}\left(T_{S,G}\right)\le Q_{\mathbf{w}}\left(M\right)=r^M.$

Furthermore, for every prescribed $m\ge 1,$ the optimal uniform estimate over the entire class of guarded wrapping operators satisfying $m\left(G\right)\ge m$ is $sup\left\{{Lip}_{\mathbf{w}}\left(T_{S,G}\right):m\left(G\right)\ge m\right\}=r^m.$ Thus $r^m$ is not merely an admissible universal contraction coefficient: it is the smallest coefficient that can hold uniformly for all $T\in {\mathfrak{T}}_m.$

The classical length-based ultrametric is recovered for $r={\displaystyle \frac{1}{2}}.$ In this case, $w_n=2^{-n},Q_{\mathbf{w}}\left(m\right)=2^{-m},$ which recovers the contraction coefficient established in [5].

The geometric family represents uniform exponential decay. A broader one-parameter family exhibits a sharp transition between the uniform and nonuniform regimes.

Proposition 8. 

Let $c>0,\alpha >0,$ and define $w_n=exp(-c{n}^{\alpha }),n\in {\mathbb{N}}_0.$ Then $\mathbf{w}$ is an admissible weight profile.

If $\alpha \ge 1,$ then, for every $m\in {\mathbb{N}}_0$, $Q_{\mathbf{w}}\left(m\right)=exp(-c{m}^{\alpha }).$ Consequently, $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1.$

If $0<\alpha <1,$ then $Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$, and hence $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$

Proof. 

Clearly, $w_0=1,$ the sequence $\left(w_n\right)$ is strictly decreasing and positive, and $w_n⟶0.$ Thus $\mathbf{w}$ is admissible.

For fixed $m\ge 0$, ${\displaystyle \frac{w_{n+m}}{w_n}}=exp\left(-c\left({(n+m)}^{\alpha }-n^{\alpha }\right)\right).$

We first consider the case $\alpha =1.$ Then ${(n+m)}^{\alpha }-n^{\alpha }=(n+m)-n=m,$ and therefore ${\displaystyle \frac{w_{n+m}}{w_n}}=exp(-cm)$ for every $n\in {\mathbb{N}}_0$. Hence $Q_{\mathbf{w}}\left(m\right)=exp(-cm)=exp(-c{m}^{\alpha }).$ In particular, $Q_{\mathbf{w}}\left(m\right)<1$ for every $m\ge 1$, and thus $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1.$

Assume now that $\alpha >1.$ For fixed $m\in {\mathbb{N}}_0$, define ${\varphi }_m\left(x\right)={(x+m)}^{\alpha }-x^{\alpha },x\ge 0.$ For $x>0$, ${\varphi }_m^{\prime }\left(x\right)=\alpha \left({(x+m)}^{\alpha -1}-x^{\alpha -1}\right).$ Since $\alpha -1>0$ and $x+m\ge x,$ we have ${(x+m)}^{\alpha -1}\ge x^{\alpha -1},$ so ${\varphi }_m^{\prime }\left(x\right)\ge 0$ for every $x>0$. Therefore ${\varphi }_m$ is nondecreasing on $(0,\infty )$ and, by continuity at 0, on $[0,\infty )$.

Hence ${\varphi }_m\left(n\right)\ge {\varphi }_m\left(0\right)=m^{\alpha }$ for every $n\in {\mathbb{N}}_0$, that is, ${(n+m)}^{\alpha }-n^{\alpha }\ge m^{\alpha }.$ Consequently, ${\displaystyle \frac{w_{n+m}}{w_n}}=exp\left(-c\left({(n+m)}^{\alpha }-n^{\alpha }\right)\right)\le exp(-c{m}^{\alpha }).$

At $n=0$ equality holds: ${\displaystyle \frac{w_m}{w_0}}=exp(-c{m}^{\alpha }).$ Therefore $Q_{\mathbf{w}}\left(m\right)=exp(-c{m}^{\alpha }).$ Again, $Q_{\mathbf{w}}\left(m\right)<1$ for every $m\ge 1$, and hence $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1.$

Suppose now that $0<\alpha <1.$ Fix $m\ge 1.$ We claim that ${(n+m)}^{\alpha }-n^{\alpha }⟶0\mathrm{as}n\to \infty .$

For every integer $n\ge 1,$ the function $x⟼x^{\alpha }$ is continuous on $[n,n+m]$ and differentiable on $(n,n+m)$. Hence, by the mean value theorem, there exists ${\xi }_n\in (n,n+m)$ such that ${(n+m)}^{\alpha }-n^{\alpha }=\alpha m{\xi }_n^{\alpha -1}.$

Since ${\xi }_n>n$ and $\alpha -1<0,$ we have $0<{\xi }_n^{\alpha -1}<n^{\alpha -1}.$ Because $n^{\alpha -1}={\displaystyle \frac{1}{n^{1-\alpha }}}⟶0,$ it follows that $0<{(n+m)}^{\alpha }-n^{\alpha }<\alpha m{n}^{\alpha -1}⟶0.$ Therefore ${(n+m)}^{\alpha }-n^{\alpha }⟶0.$

Therefore ${\displaystyle \frac{w_{n+m}}{w_n}}=exp\left(-c\left({(n+m)}^{\alpha }-n^{\alpha }\right)\right)⟶1.$ Every individual ratio is strictly smaller than one, while their supremum is one. Hence $Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$, and therefore $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$ □

Remark 12. 

Proposition 8 exhibits a sharp change in the uniform contraction behavior at $\alpha =1.$

If $\alpha =1,$ then $Q_{\mathbf{w}}\left(m\right)=e^{-cm},$ so the dependence on the guard length is geometric. This is the same qualitative regime as the profile $w_n=r^n,$ with $r=e^{-c}.$

If $\alpha >1,$ then $Q_{\mathbf{w}}\left(m\right)=e^{-c{m}^{\alpha }}.$ Hence the optimal universal Lipschitz coefficient tends to zero faster than any geometric sequence in the guard length. More precisely, for every $r\in (0,1),$ there exists $m_0$ such that $e^{-c{m}^{\alpha }}\le r^m$ for every $m\ge m_0$.

Indeed, writing $r=e^{-\lambda },\lambda >0,$ the required inequality is equivalent to $c{m}^{\alpha }\ge \lambda m,$ or $c{m}^{\alpha -1}\ge \lambda ,$ which holds for all sufficiently large m because $\alpha -1>0.$

By contrast, when $0<\alpha <1,$ one has $Q_{\mathbf{w}}\left(m\right)=1$ for every fixed $m\ge 1$. Thus no finite guard depth yields a universal Banach coefficient strictly smaller than one.

Nevertheless, consider a Picard orbit satisfying $L_1\ne L_0$ and put $r_0=\nu (L_0,L_1)<\infty .$ Then Theorem 7 gives $d_{\mathbf{w}}(L_n,L^{*})\le exp\left(-c{(r_0+nm)}^{\alpha }\right).$ If instead $L_1=L_0,$ then the orbit is already stationary at the unique fixed language.

Therefore the absence of uniform Banach contractivity does not prevent a stretched-exponential metric convergence rate along every nonstationary Picard orbit, while the stationary case is immediate.

This example illustrates an important distinction: the behavior of $Q_{\mathbf{w}}\left(m\right)$ describes uniform contraction as the guard depth varies, whereas the quantity $w_{r_0+nm}$ describes the metric convergence rate along a fixed Picard orbit.

The next example shows that the uniform decay threshold need not be one or infinity. In fact, any prescribed finite threshold can occur.

Example 2. 

Fix an integer $h\ge 2$ and choose a constant $c\in (0,1).$ Let $b_k={\displaystyle \frac{k+1}{k+2}},k\in {\mathbb{N}}_0.$ Then $0<b_k<1$ and $b_k⟶1.$

Define a sequence ${\left({\rho }_n\right)}_{n\ge 0}$ blockwise as follows: ${\rho }_{kh+j}=b_k$ for $j=0,\dots ,h-2,$ and ${\rho }_{kh+h-1}=c$ for every $k\in {\mathbb{N}}_0$.

Starting from $w_0=1,$ define recursively $w_{n+1}={\rho }_n{w}_n.$

Then $\mathbf{w}=\left(w_n\right)$ is an admissible weight profile and $m_{\mathrm{dec}}\left(\mathbf{w}\right)=h.$ More precisely, $Q_{\mathbf{w}}\left(m\right)=1\mathrm{for}1\le m<h,$ whereas $Q_{\mathbf{w}}\left(h\right)=c<1.$

Proof. 

Since $0<{\rho }_n<1$ for every n, the sequence $\left(w_n\right)$ is positive and strictly decreasing.

At the end of the kth block one has ${\displaystyle \frac{w_{(k+1)h}}{w_{kh}}}=c{b}_k^{h-1}.$ Since $0<b_k<1,$ it follows that ${\displaystyle \frac{w_{(k+1)h}}{w_{kh}}}\le c.$ Starting from $w_0=1,$ an induction on k therefore gives $w_{kh}\le c^k$ for every $k\in {\mathbb{N}}_0$.

It remains to verify convergence of the full sequence, not only of the subsequence $\left(w_{kh}\right)$. Let $n\in {\mathbb{N}}_0$, and write $n=kh+j,$$0\le j<h.$ Since $\left(w_n\right)$ is decreasing, $w_n=w_{kh+j}\le w_{kh}.$ Hence $0<w_n\le w_{kh}\le c^k.$ As $n⟶\infty ,$ one necessarily has $k=⌊{\displaystyle \frac{n}{h}}⌋⟶\infty .$ Because $0<c<1,$ we obtain $c^k⟶0.$ Therefore, by the squeeze theorem, $w_n⟶0.$

Together with positivity and strict decrease of $\left(w_n\right)$, this proves that $\mathbf{w}$ is an admissible weight profile.

Let $1\le m<h.$ Fix a block index $k\in {\mathbb{N}}_0$ and choose the starting position $n=kh.$ By the definition of the one-step ratios, ${\rho }_{kh+j}=b_k$ for every $j=0,1,\dots ,h-2.$ Since $m<h,$ the indices $kh,kh+1,\dots ,kh+m-1$ all lie among the first $h-1$ positions of the kth block. Therefore ${\rho }_{kh+j}=b_k$ for every $j=0,\dots ,m-1.$

Using $w_{n+1}={\rho }_n{w}_n,$ we obtain ${\displaystyle \frac{w_{kh+m}}{w_{kh}}}={\prod }_{j=0}^{m-1}{\rho }_{kh+j}=b_k^m.$

Since $b_k={\displaystyle \frac{k+1}{k+2}}⟶1,$ it follows that $b_k^m⟶1.$ Hence, for every $\epsilon >0$, there exists k such that ${\displaystyle \frac{w_{kh+m}}{w_{kh}}}>1-\epsilon .$ Therefore $Q_{\mathbf{w}}\left(m\right)={sup}_{n\ge 0}{\displaystyle \frac{w_{n+m}}{w_n}}\ge 1.$

On the other hand, monotonicity of the weight profile gives $w_{n+m}\le w_n$ for every n, so ${\displaystyle \frac{w_{n+m}}{w_n}}\le 1.$ Thus $Q_{\mathbf{w}}\left(m\right)\le 1.$

Combining the two inequalities yields $Q_{\mathbf{w}}\left(m\right)=1$ for every $1\le m<h.$

We now consider a shift of exactly h positions. By the definition of the one-step ratios, the factor c occurs precisely at those indices j satisfying $j\equiv h-1(modh).$ Consequently, among any h consecutive integers $n,n+1,\dots ,n+h-1,$ there is exactly one index congruent to $h-1$ modulo h. Hence the product ${\displaystyle \frac{w_{n+h}}{w_n}}={\prod }_{j=n}^{n+h-1}{\rho }_j$ contains exactly one factor equal to c.

Every remaining factor is of the form $b_k$ for some k, and therefore lies strictly between 0 and 1. It follows that ${\displaystyle \frac{w_{n+h}}{w_n}}\le c$ for every $n\in {\mathbb{N}}_0$. Taking the supremum over n yields $Q_{\mathbf{w}}\left(h\right)\le c.$

To prove the reverse inequality, choose the block starting points $n=kh,k\in {\mathbb{N}}_0.$ For these indices, the first $h-1$ one-step ratios in the block are all equal to $b_k$, while the last one equals c. Therefore ${\displaystyle \frac{w_{kh+h}}{w_{kh}}}={\prod }_{j=kh}^{kh+h-1}{\rho }_j=b_k^{h-1}c.$ Since $b_k={\displaystyle \frac{k+1}{k+2}}⟶1,$ we obtain $b_k^{h-1}c⟶c.$ Thus, for every $\epsilon >0$, there exists k such that ${\displaystyle \frac{w_{kh+h}}{w_{kh}}}>c-\epsilon .$ Hence $Q_{\mathbf{w}}\left(h\right)\ge c.$

Combining the upper and lower bounds gives $Q_{\mathbf{w}}\left(h\right)=c<1.$

Hence $Q_{\mathbf{w}}\left(m\right)=1$ for every $1\le m<h$, while $Q_{\mathbf{w}}\left(h\right)<1.$ By Definition 10, $m_{\mathrm{dec}}\left(\mathbf{w}\right)=h.$ □

Remark 13. 

For the profile in Example 2, positive guards of length less than h strictly decrease every individual nonzero distance, but no uniform Banach coefficient can be guaranteed over the entire class from guard length alone.

By contrast, every guarded wrapping operator satisfying $m\left(G\right)\ge h$ is a Banach contraction, and h is the smallest guard depth with this universal property.

We finally return to a natural slowly decaying profile for which no finite uniform decay threshold exists.

Example 3. 

Fix $p>0$ and define $w_n={\displaystyle \frac{1}{{(n+1)}^p}}.$ Then $\mathbf{w}$ is admissible, and ${\displaystyle \frac{w_{n+m}}{w_n}}={\left({\displaystyle \frac{n+1}{n+m+1}}\right)}^p.$ For every fixed $m\ge 1$, ${lim}_{n\to \infty }{\displaystyle \frac{w_{n+m}}{w_n}}=1.$ Therefore $Q_{\mathbf{w}}\left(m\right)=1$ for every $m\ge 1$, and hence $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$

Thus no finite positive guard depth is sufficient for a universal Banach conclusion.

Consider, more concretely, a symbol $a\in \Sigma$ and the guarded wrapping operator $T\left(L\right)=\left\{\epsilon \right\}\cup aL.$ It has the displayed guarded representation with $G=\left\{\right(a,\epsilon \left)\right\},$ for which $m\left(G\right)=1.$

For $A_n=⌀,B_n=\left\{a^n\right\},$ one has $\nu (A_n,B_n)=n$ and hence $d_{\mathbf{w}}(A_n,B_n)=w_n.$ Moreover, $T\left(A_n\right)=\left\{\epsilon \right\},$ whereas $T\left(B_n\right)=\{\epsilon ,a^{n+1}\}.$ Their first disagreement occurs at $a^{n+1}$, so $d_{\mathbf{w}}(T\left(A_n\right),T\left(B_n\right))=w_{n+1}.$ Consequently, ${\displaystyle \frac{d_{\mathbf{w}}(T\left(A_n\right),T\left(B_n\right))}{d_{\mathbf{w}}(A_n,B_n)}}={\displaystyle \frac{w_{n+1}}{w_n}}={\left({\displaystyle \frac{n+1}{n+2}}\right)}^p⟶1.$ Hence, by the definition of the Lipschitz modulus, ${Lip}_{\mathbf{w}}\left(T\right)\ge 1.$

On the other hand, applying Theorem 6 to the displayed representation, for which $m\left(G\right)=1,$ gives ${Lip}_{\mathbf{w}}\left(T\right)\le Q_{\mathbf{w}}\left(1\right).$ For the present polynomial profile, $Q_{\mathbf{w}}\left(1\right)=1.$ Thus ${Lip}_{\mathbf{w}}\left(T\right)\le 1.$

Combining the lower and upper bounds yields ${Lip}_{\mathbf{w}}\left(T\right)=1.$ Consequently, T is not a Banach contraction.

Nevertheless, Theorem 7 applies. Starting from $L_0=⌀,$ the Picard orbit satisfies $L_1=\left\{\epsilon \right\}$ and therefore $r_0=\nu (L_0,L_1)=0.$ A direct induction gives $L_n=\{a^k:0\le k<n\}.$ The unique fixed language is $L^{*}=\{a^n:n\ge 0\}=a^{*}.$

The first disagreement between $L_n$ and $L^{*}$ occurs exactly at the word $a^n$. Thus $\nu (L_n,L^{*})=n$ and $d_{\mathbf{w}}(L_n,L^{*})=w_n={\displaystyle \frac{1}{{(n+1)}^p}}.$

Thus the Picard orbit converges to the fixed language at the exact polynomial rate $d_{\mathbf{w}}(L_n,L^{*})={\displaystyle \frac{1}{{(n+1)}^p}}.$ At the same time, ${Lip}_{\mathbf{w}}\left(T\right)=1,$ so the operator is not Banach contractive.

This example therefore shows that failure of uniform Banach contractivity does not prevent global convergence with an explicit quantitative rate. In the present polynomial profile, the rate is determined directly by the growth of the first-disagreement depth: $\nu (L_n,L^{*})=n$ and $d_{\mathbf{w}}(L_n,L^{*})=w_n.$

The preceding examples show that all possible values of the uniform decay threshold genuinely occur.

Corollary 10. 

For every $h\in \mathbb{N}\cup \{\infty \},$ there exists an admissible weight profile $\mathbf{w}$ such that $m_{\mathrm{dec}}\left(\mathbf{w}\right)=h.$

Proof. 

For $h=1,$ any geometric profile $w_n=r^n,$$0<r<1,$ has $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1$ by Example 1.

For every finite $h\ge 2,$ Example 2 constructs an admissible profile satisfying $m_{\mathrm{dec}}\left(\mathbf{w}\right)=h.$

Finally, the polynomial profile in Example 3 satisfies $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty .$ □

Remark 14. 

The examples reveal three genuinely different geometric regimes.

When $m_{\mathrm{dec}}\left(\mathbf{w}\right)=1,$ every positively guarded wrapping operator is Banach contractive, and the guard length alone yields a uniform contraction coefficient.

When $1<m_{\mathrm{dec}}\left(\mathbf{w}\right)<\infty ,$ guard lengths below the threshold guarantee, at the level of the whole class, only pointwise strict decrease of nonzero distances. They do not provide a universal Banach coefficient strictly smaller than one. Individual guarded operators with such shorter guards may nevertheless be Banach contractive because overlaps between guarded terms or masking by the seed can reduce their actual Lipschitz modulus.

By contrast, once $m\left(G\right)\ge m_{\mathrm{dec}}\left(\mathbf{w}\right),$ every guarded wrapping operator is Banach contractive, and the decay profile provides a uniform contraction coefficient valid for the entire class.

When $m_{\mathrm{dec}}\left(\mathbf{w}\right)=\infty ,$ no finite guard depth yields a universal Banach coefficient strictly smaller than one. For every prescribed finite guard length, canonical one-context maps show that Banach contractivity can fail. Nevertheless, positive guardedness still shifts every disagreement to greater depth and therefore guarantees uniqueness of the fixed language, global Picard convergence, and explicit finite-depth stabilization for every admissible weight profile.

Thus the decisive distinction is not merely between guarded and unguarded recursion. It is the interaction between the combinatorial depth shift imposed by the guards and the uniform numerical decay of the chosen first-disagreement weight profile.