LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The square-root recurrence $T_{s} (x)_{n} = s_{n} \sqrt{a + x_{n - 1} + x_{n + 1}}$ is a contraction of $B_{R}$ with a unique coded solution

Statement

Fix $a \geq 100$ and put

$R : = \sqrt{a} + 3, L : = \sqrt{a - 2 R} .$

Let $X_{\pm}$ and $B_{R}$ be the two-sign shift space and the closed ball of bounded bilateral sequences of The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences, and for $s \in X_{\pm}$ define

$T_{s} : B_{R} ⟶ R^{Z}, (T_{s} x)_{n} : = s_{n} \sqrt{a + x_{n - 1} + x_{n + 1}} (n \in Z) .$

For a nonzero real $r$ , define $sign r : = + 1$ when $r > 0$ and $sign r : = - 1$ when $r < 0$ . Once clause 2 establishes $T_{s} (B_{R}) \subseteq B_{R}$ , its iterates are defined by $T_{s}^{0} : = {id}_{B_{R}}$ and $T_{s}^{m + 1} : = T_{s} \circ T_{s}^{m}$ for $m \in N$ .

Then:

$a - 2 R = (\sqrt{a} - 1)^{2} - 7 \geq 74$ , so $L \geq \sqrt{74} > 1$ ; moreover $a + 2 R < R^{2}$ , hence $\sqrt{a + 2 R} < R$ ;
for every $s \in X_{\pm}$ and $x \in B_{R}$ the radicands $a + x_{n - 1} + x_{n + 1}$ lie in $[a - 2 R, a + 2 R] \subseteq (0, \infty)$ , so $T_{s} x$ is a well-defined bounded sequence with $∥ T_{s} x ∥_{\infty} \leq \sqrt{a + 2 R} < R$ ; in particular $T_{s} (B_{R}) \subseteq B_{R}$ ;
for all $x, y \in B_{R}$ and every $s \in X_{\pm}$ ,

$∥ T_{s} x - T_{s} y ∥_{\infty} \leq \frac{1}{L} ∥ x - y ∥_{\infty}, \frac{1}{L} < 1;$

hence $T_{s}$ has, for each $s \in X_{\pm}$ , exactly one fixed point $x (s) \in B_{R}$ , and for every $n \in Z$

$sign x_{n} (s) = s_{n}, ∣ x_{n} (s) ∣ \geq L, x_{n + 1} (s) + x_{n - 1} (s) = x_{n} (s)^{2} - a .$

Facts & Assumptions

Given: The real number $a \geq 100$ , the constants $R = \sqrt{a} + 3$ and $L = \sqrt{a - 2 R}$ , the shift space $X_{\pm}$ and the complete ball $B_{R}$ of The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences and The bounded functions on a set form a complete metric space in the supremum metric, and the coding map of the two-sign shift space is an isometry, and the operators $T_{s}$ .

[F1]

Square roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Complete ordered field (least-upper-bound property)): every $u \geq 0$ has a unique $u^{1 / 2} : = \sqrt{u} \geq 0$ with $(\sqrt{u})^{2} = u$ ; if $0 \leq u \leq v$ then $\sqrt{u} \leq \sqrt{v}$ , since $\sqrt{u}, \sqrt{v} \geq 0$ and $(\sqrt{u})^{2} = u \leq v = (\sqrt{v})^{2}$ ; in particular $\sqrt{u^{2}} = u$ for $u \geq 0$ and $\sqrt{x} > 0$ for $x > 0$ . Also $\sqrt{a} \geq 10$ because $a \geq 100 = 10^{2}$ and $t \mapsto t^{2}$ is increasing on $[0, \infty)$ .

[F2]

The ball $B_{R}$ is nonempty and, with the supremum distance $∥ x - y ∥_{\infty}$ , a complete metric space (The bounded functions on a set form a complete metric space in the supremum metric, and the coding map of the two-sign shift space is an isometry, clauses 1, 2 and 4; Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); the value $∥ x ∥_{\infty}$ satisfies $∥ x ∥_{\infty} \leq R$ exactly when $∣ x_{n} ∣ \leq R$ for all $n$ (The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences, Absolute value in an ordered field).

[F3]

Banach's fixed point theorem (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction): a contraction of a nonempty complete metric space into itself, with constant $q < 1$ , has exactly one fixed point.

[F4]

Absolute value and order: $∣ u ∣ \leq M$ exactly when $- M \leq u \leq M$ for $M \geq 0$ ; $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ ; and $s_{n} \in {- 1, + 1}$ for every $n$ (Basic properties of the absolute value, Absolute value in an ordered field, The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences).

[F5]

Order arithmetic on intervals (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Complete ordered field (least-upper-bound property)): if $∣ x_{j} ∣ \leq R$ for $j \in Z$ then $a - R - R \leq a + x_{n - 1} + x_{n + 1} \leq a + R + R$ ; and for reals $u \geq v > 0$ one has $\sqrt{u} \geq \sqrt{v} > 0$ by [F1].

Proof

technique · direct; bound the radicands on the ball, estimate the square-root Lipschitz constant from below by $2L$, and apply Banach's fixed point theorem

1.1givenF1algebra

Constants: by [F1], $\sqrt{a} \geq 10$ , so $R = \sqrt{a} + 3 \geq 13$ and $a - 2 R = a - 2 \sqrt{a} - 6 = (\sqrt{a} - 1)^{2} - 7 \geq 81 - 7 = 74$ ; hence $L = \sqrt{a - 2 R} \geq \sqrt{74} > 1$ by [F1]. Further $R^{2} = a + 6 \sqrt{a} + 9$ and $R^{2} - (a + 2 R) = a + 6 \sqrt{a} + 9 - a - 2 \sqrt{a} - 6 = 4 \sqrt{a} + 3 > 0$ , so $a + 2 R < R^{2}$ and [F1] gives $\sqrt{a + 2 R} < R$ .

2.1givenstep 1.1F1F2F4F5

Well-definedness and self-mapping: let $s \in X_{\pm}$ , $x \in B_{R}$ and $n \in Z$ . Since $∣ x_{n \pm 1} ∣ \leq R$ by [F2], [F5] gives $a - 2 R \leq a + x_{n - 1} + x_{n + 1} \leq a + 2 R$ ; by step 1.1 this is a positive interval, so [F1] makes $\sqrt{a + x_{n - 1} + x_{n + 1}}$ a well-defined positive real, and $T_{s} x$ is a function $Z \to R$ . Moreover $∣ (T_{s} x)_{n} ∣ = ∣ s_{n} ∣ \sqrt{a + x_{n - 1} + x_{n + 1}} = \sqrt{a + x_{n - 1} + x_{n + 1}} \leq \sqrt{a + 2 R} < R$ by [F4], [F1] and step 1.1. Taking the supremum over $n$ gives $∥ T_{s} x ∥_{\infty} \leq \sqrt{a + 2 R} < R$ , so $T_{s} (B_{R}) \subseteq B_{R}$ .

2.2givenstep 1.1F1F2F4F5algebra

Contraction: let $x, y \in B_{R}$ and $n \in Z$ , and put $u : = a + x_{n - 1} + x_{n + 1}$ , $v : = a + y_{n - 1} + y_{n + 1}$ ; by step 1.1 both lie in $[a - 2 R, a + 2 R] \subseteq [L^{2}, \infty)$ , so $\sqrt{u}, \sqrt{v} \geq L > 0$ by [F1]. Then $∣ \sqrt{u} - \sqrt{v} ∣ = \frac{∣ u - v ∣}{\sqrt{u} + \sqrt{v}} \leq \frac{∣ u - v ∣}{2 L} \leq \frac{∣ x_{n - 1} - y_{n - 1} ∣ + ∣ x_{n + 1} - y_{n + 1} ∣}{2 L} \leq \frac{∥ x - y ∥_{\infty}}{L},$ where the first identity is the algebraic identity $u - v = (\sqrt{u} - \sqrt{v}) (\sqrt{u} + \sqrt{v})$ ([F1], [F4]), the second inequality uses [F4] and [F5], and the last uses [F2]. Since $∣ (T_{s} x)_{n} - (T_{s} y)_{n} ∣ = ∣ \sqrt{u} - \sqrt{v} ∣$ by [F4], taking the supremum over $n$ gives $∥ T_{s} x - T_{s} y ∥_{\infty} \leq ∥ x - y ∥_{\infty} / L$ , and $1 / L < 1$ by step 1.1.

3.1step 2.1step 2.2F2F3

Fixed point: $B_{R}$ is nonempty ( $0 \in B_{R}$ ) and complete in the sup distance by [F2]; $T_{s}$ maps $B_{R}$ into itself by step 2.1 and is a contraction with constant $1 / L < 1$ by step 2.2; [F3] therefore gives exactly one $x (s) \in B_{R}$ with $T_{s} x (s) = x (s)$ .

4.1step 3.1step 1.1F1F4F5∎

Properties of the fixed point: fix $n$ and write $r : = \sqrt{a + x_{n - 1} (s) + x_{n + 1} (s)} > 0$ , so that $x_{n} (s) = s_{n} r$ by the fixed point equation. If $s_{n} = + 1$ then $x_{n} (s) = r > 0$ , and if $s_{n} = - 1$ then $x_{n} (s) = - r < 0$ ; hence in both cases $sign x_{n} (s) = s_{n}$ and $∣ x_{n} (s) ∣ = r$ . Squaring the fixed point equation gives $x_{n} (s)^{2} = r^{2} = a + x_{n - 1} (s) + x_{n + 1} (s)$ , which is the recurrence, and $∣ x_{n} (s) ∣ = r \geq L$ by step 1.1, [F1] and [F5].

Remarks

The choice $a \geq 100$ is a convenient explicit margin: it gives $\sqrt{a} \geq 10$ , hence $L^{2} \geq 74$ and the self-mapping inequality $\sqrt{a + 2 R} < R$ with room to spare. The argument works for every $a$ with $a - 2 R > 1$ and $a + 2 R < R^{2}$ .
No choice principle is used: the contraction and the completeness of $B_{R}$ are choice-free, and the fixed point is unique.

The square-root recurrence $T_{s} (x)_{n} = s_{n} \sqrt{a + x_{n - 1} + x_{n + 1}}$ is a contraction of $B_{R}$ with a unique coded solution

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The square-root recurrence Ts(x)n=sna+xn−1+xn+1Ts​(x)n​=sn​a+xn−1​+xn+1​​ is a contraction of BRBR​ with a unique coded solution

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The square-root recurrence $T_{s} (x)_{n} = s_{n} \sqrt{a + x_{n - 1} + x_{n + 1}}$ is a contraction of $B_{R}$ with a unique coded solution