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Cone invariance and expansion, and convergence of the slope graph transforms to unique invariant stable and unstable line fields

Statement

Let $a \in R$ , $k \in (0, 1)$ and $L \geq 2$ . Let $s = (s_{n})_{n \in Z}$ be a real bilateral sequence with $∣ s_{n} ∣ \geq L$ for every $n$ , put $z_{n} : = (s_{n - 1}, s_{n})$ , and let $C_{k}^{u}$ , $C_{k}^{s}$ , $Φ_{y}$ , $Ψ_{x}$ , $D F_{a}$ and $F_{a}$ be as in Stable and unstable cones, invariant line slopes, stable and unstable sets, and transverse homoclinic points of the quadratic family. Write

$λ : = 2 L - k > 1, ρ : = λ^{- 2} < 1,$

Assume in addition the aperture condition $k (2 L - k) \geq 1$ , equivalently $1 / λ \leq k$ . This condition is needed for invariance of the stated cones and slope interval; $L \geq 2$ alone does not imply it for arbitrarily small $k$ . Then:

(cone invariance and expansion) for every $n$ : $D F_{a} (z_{n})$ maps $C_{k}^{u}$ into itself, and every $(u, v) \in C_{k}^{u}$ satisfies

$\max (∣ v ∣, ∣ - u + 2 s_{n} v ∣) \geq λ \max (∣ u ∣, ∣ v ∣);$

and $D (F_{a}^{- 1}) (z_{n})$ maps $C_{k}^{s}$ into itself, with $\max (∣ 2 s_{n - 1} u - v ∣, ∣ u ∣) \geq λ \max (∣ u ∣, ∣ v ∣)$ for every $(u, v) \in C_{k}^{s}$ ; 2. (slope contractions) for every real $y$ with $∣ y ∣ \geq L$ , the map $Φ_{y}$ carries $[- k, k]$ into $[- 1 / λ, 1 / λ] \subseteq [- k, k]$ and satisfies $∣ Φ_{y} (h) - Φ_{y} (h^{'}) ∣ \leq ρ ∣ h - h^{'} ∣$ there; the same holds for $Ψ_{x}$ for every $∣ x ∣ \geq L$ ; 3. (unstable line field) define $h_{n}^{(0)} : = 0$ and $h_{n}^{(m + 1)} : = Φ_{s_{n - 1}} (h_{n - 1}^{(m)})$ for $m \geq 0$ . Then $(h_{n}^{(m)})_{m}$ converges for every $n$ to a limit $h_{n} \in [- k, k]$ , uniformly in $n$ , and

$h_{n} = \frac{1}{2 s_{n - 1} - h_{n - 1}} (n \in Z);$

furthermore $D F_{a} (z_{n})$ maps the line ${(h_{n} v, v) : v \in R}$ onto the line ${(h_{n + 1} v, v)}$ for every $n$ , so the lines form the unique unstable line field along $s$ whose slopes all lie in $[- k, k]$ ; 4. (stable line field) similarly, $r_{n}^{(0)} : = 0$ and $r_{n}^{(m + 1)} : = Ψ_{s_{n}} (r_{n + 1}^{(m)})$ converge to limits $r_{n} \in [- k, k]$ with $r_{n} = 1 / (2 s_{n} - r_{n + 1})$ , and $D (F_{a}^{- 1}) (z_{n + 1})$ maps ${(u, r_{n + 1} u) : u \in R}$ onto ${(u, r_{n} u) : u \in R}$ ; this is the unique stable line field along $s$ with slopes in $[- k, k]$ .

Facts & Assumptions

Given: The parameter $a$ , the aperture $k \in (0, 1)$ , the bound $L \geq 2$ , the sequence $s$ with $∣ s_{n} ∣ \geq L$ , the cones, slope maps and matrix data of Stable and unstable cones, invariant line slopes, stable and unstable sets, and transverse homoclinic points of the quadratic family, and the recursively defined slope families $h^{(m)}$ , $r^{(m)}$ .

[F1]

Cone, slope and matrix data (Stable and unstable cones, invariant line slopes, stable and unstable sets, and transverse homoclinic points of the quadratic family): $D F_{a} (x, y) (u, v) = (v, - u + 2 y v)$ and $D (F_{a}^{- 1}) (x, y) (u, v) = (2 x u - v, u)$ ; $C_{k}^{u} = {∣ u ∣ \leq k ∣ v ∣}$ , $C_{k}^{s} = {∣ v ∣ \leq k ∣ u ∣}$ , $C_{k}^{u} \cap C_{k}^{s} = {0}$ . For $h \neq 2 y$ , the matrix gives $D F_{a} (x, y) (h v, v) = (v, (2 y - h) v)$ , so its image line has slope $Φ_{y} (h) = 1 / (2 y - h)$ . Likewise, for $r \neq 2 x$ , $D (F_{a}^{- 1}) (x, y) (u, r u) = ((2 x - r) u, u)$ , so its image line has slope $Ψ_{x} (r) = 1 / (2 x - r)$ .

[F2]

Absolute value and order (Absolute value in an ordered field, Basic properties of the absolute value, The triangle inequality, Complete ordered field (least-upper-bound property)): $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ ; $∣ u - w ∣ \geq ∣ u ∣ - ∣ w ∣$ ; if $∣ u ∣ \leq k ∣ v ∣$ and $0 < k < 1$ then $\max (∣ u ∣, ∣ v ∣) = ∣ v ∣$ ; if $∣ v ∣ \leq k ∣ u ∣$ then $\max (∣ u ∣, ∣ v ∣) = ∣ u ∣$ ; and the stated aperture condition gives $1 / λ \leq k$ .

[F3]

Series and geometric sequences (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ): for $0 \leq ρ < 1$ the series $\sum_{j \geq 0} ρ^{j}$ converges, $ρ^{m} \to 0$ , and $\sum_{j \geq m} ρ^{j} = ρ^{m} / (1 - ρ)$ .

[F4]

Completeness and limits of real sequences (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Algebra of limits: sums, scalar multiples, products and quotients, Limits preserve non-strict inequalities, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ): every Cauchy real sequence converges; limits of convergent real sequences are compatible with addition, multiplication, division by nonzero limits, and with weak inequalities; a limit of numbers in a closed interval $[- k, k]$ lies in $[- k, k]$ .

Proof

technique · direct; verify the two cone computations, then solve the slope recurrences by a contraction iteration from the far past and the far future

1.1givenF2algebra

Constants: since $L \geq 2$ and $0 < k < 1$ , we have $2 L - k \geq 4 - k > 3$ , so $λ = 2 L - k > 1$ and $0 < ρ = λ^{- 2} < 1$ ; moreover the aperture hypothesis gives $k (2 L - k) \geq 1$ , hence $1 / λ \leq k$ .

2.1givenstep 1.1F1F2algebra

Unstable cone: let $(u, v) \in C_{k}^{u}$ , so $∣ u ∣ \leq k ∣ v ∣$ and $\max (∣ u ∣, ∣ v ∣) = ∣ v ∣$ by [F2]. Then $D F_{a} (z_{n}) (u, v) = (v, - u + 2 s_{n} v)$ and $∣ - u + 2 s_{n} v ∣ \geq 2 ∣ s_{n} ∣ ∣ v ∣ - ∣ u ∣ \geq (2 L - k) ∣ v ∣ = λ ∣ v ∣$ by [F2]. Also $∣ - u + 2 s_{n} v ∣ \geq λ ∣ v ∣ \geq ∣ v ∣$ since $λ > 1$ , so $\max (∣ v ∣, ∣ - u + 2 s_{n} v ∣) \geq λ \max (∣ u ∣, ∣ v ∣)$ . Finally $∣ v ∣ / ∣ - u + 2 s_{n} v ∣ \leq 1 / λ \leq k$ whenever $v \neq 0$ , so the image is in $C_{k}^{u}$ (the case $v = 0$ gives $u = 0$ and image $0$ ).

2.2givenstep 1.1F1F2algebra

Stable cone: let $(u, v) \in C_{k}^{s}$ , so $∣ v ∣ \leq k ∣ u ∣$ and $\max (∣ u ∣, ∣ v ∣) = ∣ u ∣$ by [F2]. Then $D (F_{a}^{- 1}) (z_{n}) (u, v) = (2 s_{n - 1} u - v, u)$ and $∣ 2 s_{n - 1} u - v ∣ \geq 2 L ∣ u ∣ - ∣ v ∣ \geq (2 L - k) ∣ u ∣ = λ ∣ u ∣$ ; hence $\max (∣ 2 s_{n - 1} u - v ∣, ∣ u ∣) \geq λ \max (∣ u ∣, ∣ v ∣)$ and $∣ u ∣ / ∣ 2 s_{n - 1} u - v ∣ \leq 1 / λ \leq k$ whenever $u \neq 0$ , with the zero case trivial.

2.3givenstep 1.1F1F2algebra

Slope contractions: let $∣ y ∣ \geq L$ and $h, h^{'} \in [- k, k]$ ; by step 1.1, $∣ 2 y - h ∣ \geq 2 L - k = λ > 0$ and $∣ 2 y - h^{'} ∣ \geq λ$ , so $Φ_{y}$ is defined on $[- k, k]$ , $∣ Φ_{y} (h) ∣ \leq 1 / λ \leq k$ , and $∣ Φ_{y} (h) - Φ_{y} (h^{'}) ∣ = \frac{∣ h - h^{'} ∣}{∣ 2 y - h ∣ ∣ 2 y - h^{'} ∣} \leq λ^{- 2} ∣ h - h^{'} ∣ = ρ ∣ h - h^{'} ∣ .$ The same computation with $Ψ_{x}$ proves the second assertion.

3.1givenstep 2.3F2F3F4algebra

Unstable slopes, uniform bounds and contraction: define $h_{n}^{(0)} : = 0$ , $h_{n}^{(m + 1)} : = Φ_{s_{n - 1}} (h_{n - 1}^{(m)})$ . By induction, step 2.3 gives $∣ h_{n}^{(m)} ∣ \leq 1 / λ$ for all $m \geq 1$ and $n$ , and $∣ h_{n}^{(m + 1)} - h_{n}^{(m)} ∣ \leq ρ^{m} / λ$ for all $m \geq 0$ . Hence for $m^{'} \geq m$ , $∣ h_{n}^{(m^{'})} - h_{n}^{(m)} ∣ \leq \sum_{j = m}^{m^{'} - 1} ρ^{j} / λ \leq \frac{ρ^{m}}{λ (1 - ρ)},$ uniformly in $n$ . Thus $(h_{n}^{(m)})_{m}$ is Cauchy; completeness in [F4] gives its limit $h_{n}$ , with $∣ h_{n} ∣ \leq 1 / λ \leq k$ . Passing $m^{'} \to \infty$ in the same estimate gives $∣ h_{n} - h_{n}^{(m)} ∣ \leq ρ^{m} / (λ (1 - ρ))$ for every $n$ , proving uniform convergence.

3.2givenstep 2.3F2F3F4

Stable slopes: the same argument with $r_{n}^{(0)} : = 0$ , $r_{n}^{(m + 1)} : = Ψ_{s_{n}} (r_{n + 1}^{(m)})$ gives $∣ r_{n}^{(m + 1)} - r_{n}^{(m)} ∣ \leq ρ^{m} / λ$ , limits $r_{n} \in [- k, k]$ satisfying $r_{n} = Ψ_{s_{n}} (r_{n + 1}) = 1 / (2 s_{n} - r_{n + 1})$ .

4.1step 3.1step 2.3F4

Unstable recurrence: fixing $n$ and passing to the limit in $h_{n}^{(m + 1)} = Φ_{s_{n - 1}} (h_{n - 1}^{(m)})$ , continuity of $Φ_{s_{n - 1}}$ on $[- k, k]$ (step 2.3) and the limit rules of [F4] give $h_{n} = Φ_{s_{n - 1}} (h_{n - 1}) = 1 / (2 s_{n - 1} - h_{n - 1})$ .

4.2step 3.1step 3.2F2F3F4

Uniqueness: let $({\tilde{h}}_{n})$ be another family with $∣ {\tilde{h}}_{n} ∣ \leq k$ and ${\tilde{h}}_{n} = Φ_{s_{n - 1}} ({\tilde{h}}_{n - 1})$ for all $n$ . Comparing with the construction gives, by induction on $m$ , $∣ {\tilde{h}}_{n} - h_{n}^{(m)} ∣ \leq ρ^{m} ∣ {\tilde{h}}_{n - m} - h_{n - m}^{(0)} ∣ \leq ρ^{m} \cdot 2 k$ (step 2.3, [F2]); letting $m \to \infty$ and using [F3] gives ${\tilde{h}}_{n} = h_{n}$ . The stable case is identical with the future-pointing iteration.

5.1step 4.1step 3.2F1

Invariance: by [F1] and step 4.1, $D F_{a} (z_{n})$ carries the line ${(h_{n} v, v)}$ to the line with slope $Φ_{s_{n}} (h_{n}) = 1 / (2 s_{n} - h_{n}) = h_{n + 1}$ ; since both lines pass through the origin and the first is mapped onto a line (the differential is invertible), the image is exactly ${(h_{n + 1} v, v) : v \in R}$ . Similarly, $D (F_{a}^{- 1}) (z_{n + 1})$ uses the first coordinate $s_{n}$ of $z_{n + 1}$ ; by [F1] and step 3.2 it carries ${(u, r_{n + 1} u) : u \in R}$ onto ${(u, r_{n} u) : u \in R}$ .

6.1step 2.1step 2.2step 2.3step 3.1step 3.2step 4.1step 4.2step 5.1∎

Conclusion: steps 2.1 and 2.2 prove clause 1, step 2.3 proves clause 2, steps 3.1, 4.1 and 5.1 prove clause 3, steps 3.2 and 5.1 prove clause 4, and step 4.2 proves the uniqueness assertions of clauses 3 and 4.

Remarks

The convergence of the slope iteration is uniform in $n$ , with the rate $ρ^{m} / (λ (1 - ρ))$ ; no compactness of the sequence space is used.
Both recurrences are one-sided compositions: $h_{n}$ is determined by the symbols in the infinite past, $r_{n}$ by the symbols in the infinite future. This is the sense in which the unstable direction has finite memory of the past and the stable direction finite foresight of the future.

Cone invariance and expansion, and convergence of the slope graph transforms to unique invariant stable and unstable line fields

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources