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Weighted half-tail contractions for $q_{n} = M_{s_{n}} (q_{n - 1}, q_{n + 1})$ give a unique $C^{1}$ curve with tangent slope bounded by $c / (1 - q)$

Statement

Let $θ \in (0, 1)$ , $c > 0$ and put

$q : = c (θ + θ^{- 1}), and assume q < 1.$

Let $J \subseteq R$ be a nonempty compact interval, let $s \in X_{\pm}$ , and let $O \subseteq R^{2}$ be open with $J \times J \subseteq O$ . For $u \in {- 1, + 1}$ let ${\tilde{M}}_{u} : O \to R$ be of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms); write $M_{u}$ for its restriction to $J \times J$ , assume $M_{u} (J \times J) \subseteq J$ , and suppose that

$∣ \partial_{1} M_{u} ∣ \leq c, ∣ \partial_{2} M_{u} ∣ \leq c on J \times J .$

Let $(y_{n})_{n \geq 0}$ be a sequence with $y_{n} \in J$ and let $b_{*} \in J$ satisfy

$y_{n} = M_{s_{n}} (y_{n - 1}, y_{n + 1}) (n \geq 1), y_{0} = M_{s_{0}} (b_{*}, y_{1}),$

and assume there is $\overset{ˉ}{ρ} > 0$ with $[y_{n} - \overset{ˉ}{ρ}, y_{n} + \overset{ˉ}{ρ}] \subseteq J$ for every $n \geq 0$ . Choose $δ_{*} > 0$ such that

$[b_{*} - δ_{*}, b_{*} + δ_{*}] \subseteq J, δ_{*} < \frac{(1 - q) \overset{ˉ}{ρ}}{c},$

and put $I : = (b_{*} - δ_{*}, b_{*} + δ_{*})$ . Write

$ℓ_{θ}^{\infty} (N) : = {u : N \to R : ∥ u ∥_{θ} : = \sup_{n \geq 0} θ^{- n} ∣ u_{n} ∣ < \infty}$

with metric $∥ u - v ∥_{θ}$ , and define its closed ball

$B_{\overset{ˉ}{ρ}} : = {u \in ℓ_{θ}^{\infty} (N) : ∥ u ∥_{θ} \leq \overset{ˉ}{ρ}} .$

For $u \in B_{\overset{ˉ}{ρ}}$ and $b \in I$ , put $z_{- 1} : = b$ and $z_{k} : = y_{k} + u_{k}$ for every $k \geq 0$ , and define $T_{b} : B_{\overset{ˉ}{ρ}} \to B_{\overset{ˉ}{ρ}}$ by

$(T_{b} u)_{n} : = M_{s_{n}} (z_{n - 1}, z_{n + 1}) - y_{n} (n \geq 0) .$

Then:

$ℓ_{θ}^{\infty} (N)$ is a complete metric space and $B_{\overset{ˉ}{ρ}}$ is complete in the distance $∥ u - v ∥_{θ}$ ; for every $b \in I$ the map $T_{b}$ is a contraction with constant $q < 1$ , so it has exactly one fixed point $u (b) \in B_{\overset{ˉ}{ρ}}$ ;
for every $b \in I$ one has the a priori bound $∥ u (b) ∥_{θ} \leq c ∣ b - b_{*} ∣ / (1 - q)$ , and the map $b \mapsto u (b)$ is Lipschitz on $I$ with

$∥ u (b^{'}) - u (b) ∥_{θ} \leq \frac{c}{1 - q} ∣ b^{'} - b ∣;$

$b \mapsto u (b)$ is continuously differentiable on $I$ into $ℓ_{θ}^{\infty} (N)$ : with $z_{k}^{*} : = y_{k} + u_{k} (b)$ for $k \geq 0$ and $z_{- 1}^{*} : = b$ , the derivative $d (b) : = d u (b) / d b \in ℓ_{θ}^{\infty} (N)$ is the unique element of $ℓ_{θ}^{\infty} (N)$ satisfying

$d_{n} (b) = \partial_{1} M_{s_{n}} (z_{n - 1}^{*}, z_{n + 1}^{*}) d_{n - 1} (b) + \partial_{2} M_{s_{n}} (z_{n - 1}^{*}, z_{n + 1}^{*}) d_{n + 1} (b) (n \geq 1),$

$d_{0} (b) = \partial_{1} M_{s_{0}} (b, z_{1}^{*}) + \partial_{2} M_{s_{0}} (b, z_{1}^{*}) d_{1} (b), d_{- 1} (b) : = 1,$

and $∥ d (b) ∥_{θ} \leq c / (1 - q)$ ; 4. consequently the local phase curve

$γ (b) : = (b, y_{0} + u_{0} (b))$

defined for $b \in I$ is $C^{1}$ with $γ^{'} (b) = (1, d_{0} (b))$ , and so $∣ d_{0} (b) ∣ \leq c / (1 - q)$ .

Facts & Assumptions

Given: The parameters $θ, c, q$ , the compact interval $J$ , the open neighborhood $O$ , the $C^{2}$ extensions ${\tilde{M}}_{- 1}, {\tilde{M}}_{+ 1}$ and their restrictions $M_{- 1}, M_{+ 1}$ , the two-sign sequence $s$ , the reference data $y, b_{*}$ , the radius $\overset{ˉ}{ρ}$ and parameter interval $I$ , the weighted space $ℓ_{θ}^{\infty} (N)$ , its closed ball $B_{\overset{ˉ}{ρ}}$ , and the operators $T_{b}$ .

[F1]

Each ${\tilde{M}}_{u}$ is $C^{2}$ on the open set $O$ , its restriction maps $J \times J$ into $J$ , and $∣ \partial_{1} M_{u} ∣, ∣ \partial_{2} M_{u} ∣ \leq c$ on $J \times J$ (given). The box $J \times J$ is compact by Heine--Borel; hence the first and second partial derivatives are bounded there (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, $C^{k}$ Euclidean maps and diffeomorphisms, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case). If $z, z^{'} \in J \times J$ and $Δ : = z^{'} - z$ , the segment $z + t Δ$ lies in $J \times J$ . The function $g (t) : = {\tilde{M}}_{u} (z + t Δ)$ is $C^{2}$ on an open interval containing $[0, 1]$ , and the chain rule gives $∣ g^{'} (t) ∣ \leq c ∥ Δ ∥_{1}$ and $∣ g^{''} (t) ∣ \leq K ∥ Δ ∥_{1}^{2}$ with a common finite $K$ for both signs (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The mean-value inequality (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ) applied to $g$ gives

$∣ M_{u} (z^{'}) - M_{u} (z) ∣ \leq c ∥ z^{'} - z ∥_{1},$

For the Taylor remainder, the mean-value inequality applied to $g^{'}$ on $[0, t]$ gives $∣ g^{'} (t) - g^{'} (0) ∣ \leq K t ∥ Δ ∥_{1}^{2}$ ; thus $ϕ (t) : = g (t) - g (0) - t g^{'} (0)$ has $∣ ϕ^{'} (t) ∣ \leq K ∥ Δ ∥_{1}^{2}$ . Applying the same inequality to $ϕ$ on $[0, 1]$ gives

$∣ M_{u} (z^{'}) - M_{u} (z) - D M_{u} (z) (z^{'} - z) ∣ \leq K ∥ z^{'} - z ∥_{1}^{2} .$

Here $∥ \cdot ∥_{1}$ is the sum of the absolute values of the two coordinates; the derivative bounds follow from the chain rule and the uniform second-derivative bound above (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[F2]

The map $W : ℓ_{θ}^{\infty} (N) \to ℓ^{\infty} (N)$ , $(W u)_{n} : = θ^{- n} u_{n}$ , is a bijection with $∥ W u ∥_{\infty} = ∥ u ∥_{θ}$ ; hence $(ℓ_{θ}^{\infty} (N), ∥ \cdot - \cdot ∥_{θ})$ is a complete metric space, as an isometric copy of the complete space $ℓ^{\infty} (N) = B (N)$ . It maps $B_{\overset{ˉ}{ρ}}$ isometrically onto the closed sup-norm ball $B_{\overset{ˉ}{ρ}} (N)$ , so that ball and therefore $B_{\overset{ˉ}{ρ}}$ are complete as well (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; Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[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 map of a nonempty complete metric space into itself with constant $q < 1$ has exactly one fixed point, and it is the limit of the iterates from every starting point.

[F4]

Absolute value and order arithmetic (Absolute value in an ordered field, Basic properties of the absolute value, Complete ordered field (least-upper-bound property)): $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u + w ∣ \leq ∣ u ∣ + ∣ w ∣$ , and $θ^{- n} > 0$ for all $n$ .

[F5]

Metric convergence and continuity (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

Proof

technique · direct; contract in the weighted supremum norm, estimate the boundary forcing, and linearise by a Taylor remainder estimate in the same norm

1.1givenF1F4algebra

Contraction: let $u, v \in B_{\overset{ˉ}{ρ}}$ and $b \in I$ . Every argument pair of $M_{s_{n}}$ lies in $J \times J$ : $b \in J$ and, for $k \geq 0$ , $∣ u_{k} ∣, ∣ v_{k} ∣ \leq \overset{ˉ}{ρ}$ with $[y_{k} - \overset{ˉ}{ρ}, y_{k} + \overset{ˉ}{ρ}] \subseteq J$ . The line segment between the two argument pairs also lies in $J \times J$ because $J$ is an interval, so the derivative bound in [F1] gives for $n \geq 1$ $∣ (T_{b} u)_{n} - (T_{b} v)_{n} ∣ \leq c (∣ u_{n - 1} - v_{n - 1} ∣ + ∣ u_{n + 1} - v_{n + 1} ∣) \leq θ^{n} c (θ^{- 1} + θ) ∥ u - v ∥_{θ} = θ^{n} q ∥ u - v ∥_{θ} .$ For $n = 0$ only the argument $z_{1}$ varies, so $∣ (T_{b} u)_{0} - (T_{b} v)_{0} ∣ \leq c ∣ u_{1} - v_{1} ∣ \leq q ∥ u - v ∥_{θ}$ because $c θ \leq q$ . Multiplying by $θ^{- n}$ and taking the supremum gives $∥ T_{b} u - T_{b} v ∥_{θ} \leq q ∥ u - v ∥_{θ}$ .

2.1givenstep 1.1F1F4algebra

Ball invariance: let $b \in I$ and $u \in B_{\overset{ˉ}{ρ}}$ . For every $k \geq 0$ , $z_{k} = y_{k} + u_{k} \in J$ by $∣ u_{k} ∣ \leq \overset{ˉ}{ρ}$ and the margin hypothesis; also $z_{- 1} = b \in J$ . For $n \geq 1$ , using the reference recurrence of the statement and [F1], $∣ (T_{b} u)_{n} ∣ = ∣ M_{s_{n}} (z_{n - 1}, z_{n + 1}) - M_{s_{n}} (y_{n - 1}, y_{n + 1}) ∣ \leq c (∣ u_{n - 1} ∣ + ∣ u_{n + 1} ∣) \leq θ^{n} q ∥ u ∥_{θ} \leq θ^{n} q \overset{ˉ}{ρ} .$ For $n = 0$ , $(T_{b} u)_{0} = M_{s_{0}} (b, z_{1}) - M_{s_{0}} (b_{*}, y_{1})$ , so $∣ (T_{b} u)_{0} ∣ \leq c ∣ b - b_{*} ∣ + c ∣ u_{1} ∣ \leq (1 - q) \overset{ˉ}{ρ} + q \overset{ˉ}{ρ} = \overset{ˉ}{ρ}$ , since $∣ b - b_{*} ∣ < δ_{*} < (1 - q) \overset{ˉ}{ρ} / c$ . Hence $∥ T_{b} u ∥_{θ} \leq \overset{ˉ}{ρ}$ .

2.2givenstep 1.1F1F3F4

Linearised equation: fix $b$ and put $z_{k}^{*} : = y_{k} + u_{k} (b)$ for $k \geq 0$ and $z_{- 1}^{*} : = b$ . Define the affine map $L$ on $ℓ_{θ}^{\infty} (N)$ by $(L d)_{n} : = \partial_{1} M_{s_{n}} (z_{n - 1}^{*}, z_{n + 1}^{*}) d_{n - 1} + \partial_{2} M_{s_{n}} (z_{n - 1}^{*}, z_{n + 1}^{*}) d_{n + 1} (n \geq 1), (L d)_{0} : = \partial_{2} M_{s_{0}} (b, z_{1}^{*}) d_{1} + \partial_{1} M_{s_{0}} (b, z_{1}^{*}),$ where $d_{- 1} : = 1$ is held fixed. The homogeneous part $d \mapsto L d - d_{- 1} -part$ is a contraction of constant $q$ by the computation of step 1.1, so [F3] gives a unique $d (b) \in ℓ_{θ}^{\infty} (N)$ with $L d (b) = d (b)$ , i.e. the displayed system; taking the fixed point of the homogeneous contraction started at $0$ gives $∥ d (b) ∥_{θ} \leq c / (1 - q)$ because the forcing $\partial_{1} M_{s_{0}} (b, z_{1}^{*}) e_{0}$ has norm at most $c$ .

3.1step 1.1step 2.1F2F3

Fixed point: by [F2], $B_{\overset{ˉ}{ρ}}$ is a nonempty complete metric space. For every $b \in I$ , step 2.1 shows $T_{b}$ maps this ball into itself and step 1.1 shows that it is a contraction with constant $q < 1$ ; [F3] therefore gives exactly one fixed point $u (b) \in B_{\overset{ˉ}{ρ}}$ .

3.2givenstep 2.2F1F4algebra

Taylor remainder: fix $b \in I$ and take $h$ small enough that $b + h \in I$ . Put $u^{'} : = u (b + h)$ and $Δ : = u^{'} - u (b)$ , and for $k \geq 0$ put $z_{k} : = y_{k} + u_{k} (b)$ and $z_{k}^{'} : = y_{k} + u_{k}^{'}$ , with $z_{- 1} : = b$ and $z_{- 1}^{'} : = b + h$ . For $n \geq 1$ , [F1] applied at $z = (z_{n - 1}^{*}, z_{n + 1}^{*})$ and $z^{'} = (z_{n - 1}^{'}, z_{n + 1}^{'})$ gives $∣ (T_{b + h} u^{'})_{n} - (T_{b} u)_{n} - \partial_{1} M_{s_{n}} Δ_{n - 1} - \partial_{2} M_{s_{n}} Δ_{n + 1} ∣ \leq K (∣ Δ_{n - 1} ∣ + ∣ Δ_{n + 1} ∣)^{2} \leq 2 K (∣ Δ_{n - 1} ∣^{2} + ∣ Δ_{n + 1} ∣^{2}),$ and multiplying by $θ^{- n}$ bounds this by $2 K θ^{n} (θ^{- 2} + θ^{2}) ∥ Δ ∥_{θ}^{2}$ . For $n = 0$ the two input differences are $h$ and $Δ_{1}$ ; [F1] gives a remainder bounded by $K^{'} (∣ h ∣ + ∣ Δ_{1} ∣)^{2} \leq 2 K^{'} (h^{2} + ∣ Δ_{1} ∣^{2})$ with a uniform $K^{'}$ , so altogether $∥ u^{'} - (T_{b} u) - [D_{u} T_{b} (u)] Δ - D_{b} T_{b} (u) h ∥_{θ} \leq C_{0} (∥ Δ ∥_{θ}^{2} + h^{2})$ with $C_{0} < \infty$ .

4.1step 1.1step 3.1F1F3F4algebra

Lipschitz dependence and a priori bound: for $u \in B_{\overset{ˉ}{ρ}}$ and $b, b^{'} \in I$ , $(T_{b^{'}} u)_{n} = (T_{b} u)_{n}$ for $n \geq 1$ , while $∣ (T_{b^{'}} u)_{0} - (T_{b} u)_{0} ∣ \leq c ∣ b^{'} - b ∣$ by [F1]. Applying this with $u = u (b^{'})$ and using step 1.1 gives $∥ u (b^{'}) - u (b) ∥_{θ} \leq q ∥ u (b^{'}) - u (b) ∥_{θ} + c ∣ b^{'} - b ∣$ , hence $∥ u (b^{'}) - u (b) ∥_{θ} \leq c ∣ b^{'} - b ∣ / (1 - q)$ . At $b = b_{*}$ , $u (b_{*}) = 0$ because the zero sequence solves the fixed-point equation; taking one parameter to be $b_{*}$ gives $∥ u (b) ∥_{θ} \leq c ∣ b - b_{*} ∣ / (1 - q)$ .

5.1step 4.1step 2.2step 3.2F1F4algebra

Differentiability: write $D_{u} T : = D_{u} T_{b} (u (b))$ and $D_{b} T : = D_{b} T_{b} (u (b))$ for the two partial derivatives of the map $(b, u) \mapsto T_{b} u$ at $(b, u (b))$ . By step 3.2 applied to the fixed point equations $u (b + h) = T_{b + h} u (b + h)$ and $u (b) = T_{b} u (b)$ , $u (b + h) = u (b) + D_{u} T Δ + D_{b} T h + R_{1}, ∥ R_{1} ∥_{θ} \leq C_{0} (∥ Δ ∥_{θ}^{2} + h^{2}),$ with $Δ = u (b + h) - u (b)$ . By step 2.2, $h d (b) = D_{u} T (h d (b)) + D_{b} T h$ ; subtracting $h d (b)$ from the displayed equation and putting $R : = Δ - h d (b)$ therefore yields $R = D_{u} T R + R_{1}$ , whence $∥ R ∥_{θ} \leq q ∥ R ∥_{θ} + ∥ R_{1} ∥_{θ}$ . By step 4.1, $∥ Δ ∥_{θ} \leq c ∣ h ∣ / (1 - q)$ , so $∥ R_{1} ∥_{θ} \leq C_{1} h^{2}$ for $∣ h ∣$ small and hence $∥ R ∥_{θ} \leq C_{1} h^{2} / (1 - q)$ : that is, $u (b + h) = u (b) + h d (b) + O (h^{2})$ in $ℓ_{θ}^{\infty} (N)$ . This proves differentiability with derivative $d (b)$ and identifies it with the solution of clause 3.

5.2step 4.1step 2.2F1F4

Continuity of the derivative: by step 2.2 the difference of the linear operators at $b^{'}$ and $b$ has coefficients $\partial_{i} M_{s_{n}} (z_{n \pm 1}^{'}) - \partial_{i} M_{s_{n}} (z_{n \pm 1})$ , which tend to zero uniformly in $n$ as $b^{'} \to b$ by uniform continuity of the $\partial_{i} M_{u}$ on the compact $J \times J$ and by the uniform estimate $∣ z_{k}^{'} - z_{k} ∣ \leq c ∣ b^{'} - b ∣ / (1 - q)$ of step 4.1; the same holds for the forcing. Applying the difference estimate of step 4.1 to the fixed points $d (b^{'}), d (b)$ of the affine maps gives $∥ d (b^{'}) - d (b) ∥_{θ} \to 0$ .

6.1step 2.2step 5.1step 5.2F4F5∎

The curve: the map $b \mapsto γ (b) = (b, y_{0} + u_{0} (b))$ has components the identity and the affine function of $u_{0}$ , and $b \mapsto u_{0} (b)$ is continuously differentiable by steps 5.1 and 5.2 with derivative $d_{0} (b)$ ; hence $γ$ is $C^{1}$ with the displayed derivative ([F5]). The bound $∣ d_{0} (b) ∣ \leq ∥ d (b) ∥_{θ} \leq c / (1 - q)$ is step 2.2.

Remarks

The lemma isolates the exact contractive mechanism used for both the unperturbed square-root recurrence and its perturbations: only the two partial-derivative bounds of the update and the existence of one reference half-orbit are used, not the explicit shape of $M_{u}$ .
The weighted growth $∣ u_{n} ∣ \leq θ^{n} ∥ u ∥_{θ}$ is what makes the solution track the reference half-orbit exponentially; for the phase curve $γ$ this means that the forward orbits of its points converge to the same limit as the reference orbit, which is the stable-set membership used later.
No choice principle is used: completeness of the weighted space comes from the complete space of bounded sequences, and the fixed points are unique.

Weighted half-tail contractions for $q_{n} = M_{s_{n}} (q_{n - 1}, q_{n + 1})$ give a unique $C^{1}$ curve with tangent slope bounded by $c / (1 - q)$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Weighted half-tail contractions for qn=Msn(qn−1,qn+1)qn​=Msn​​(qn−1​,qn+1​) give a unique C1C1 curve with tangent slope bounded by c/(1−q)c/(1−q)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Weighted half-tail contractions for $q_{n} = M_{s_{n}} (q_{n - 1}, q_{n + 1})$ give a unique $C^{1}$ curve with tangent slope bounded by $c / (1 - q)$