(1)Definition 1 A mapf isCr-structurally stable if there isε >0such that every map g with kf−gkCr &lt

Definition 1 A mapf isC^r-structurally stable if there isε >0such that every map g with kf−gk_C^r < εis conjugate tof. Herekhk_C^r is theC^r-norm, i.e., khk_C^r = max_i=0,...,rsup_x|h⁽ⁱ⁾|, where h⁽ⁱ⁾ denotes the i-th derivative.

Theorem 2 The map Q_a(x) = ax(1−x) isC²-structurally stable for a∈(0,1).

Proof. Compute Q⁰_a(x) = a(1−2x) andQ⁰⁰_a(x) =−2a. Fix a∈(0,1). If kQ_a−gk_C² < ε, then we can write g = Q_a+εh where h = h_ε and khk_C² < 1. We need to find a homeomorphism ψ :R→R such thatψ◦Q_a =g ◦ψ.

ClearlyQ_ahas a fixed point at 0. SetF(x, ε) =Q_a(x)+εh(x)−x. Then ^∂F_∂x(0,0) =a−16= 0, so the Implicit Function Theorem tells us that there is a neighbourhoodU 30 and a continuous function p:U →R with p(0) = 0, such that F(p(ε), ε) =g(p(ε))−p(ε) = 0. That is, p=p(ε) is a fixed point of g. Set ψ(0) =p(ε).

Also assume thatε <min{a,1−a}is so small that g⁰(p(ε)) = a(1−2p(ε)) +εh⁰(p(ε))<1.

Therefore, the fixed point p(ε) of g is attracting.

Next note that g⁰⁰(x) = −2a+ εh⁰⁰(x) < 0 (because sup_x|h⁰⁰(x)| < 1). Therefore f is strictly convex, and it has therefore a unique global maximum m =m(ε) with m(0) = ¹₂. Set ψ(¹₂) = m(ε).

Since g⁰(x) is decreasing on (p, m) and 0 < g⁰(p) < 1, we have p < g(m) < m. We call (g(m), m] a fundamental domain forg.

Note that for ε = 0, so g = Q_a, Q_a(m) = Q_a(¹₂) = ^a₄. Set ψ(^a₄) = g(m) and extend ψ : [^a₄,¹₂]→[g(m), m] in some homeomorphic (e.g. affine) way.

For every x ∈ (0,¹₂], there is a unique y ∈ (^a₄,¹₂] and n ≥ 0 such that Qⁿ_a(y) = x. Set ψ(x) = g^m(ψ(y)). This defines ψ continuously on [0,¹₂] so that ψ◦Q_a=g◦ψ.

Since m > p and g⁰(x) < 0, there is a unique q = q(ε) > m such that g(q) = p. Clearly q(0) = 1, andg : [m, q]→[g(m), p] is a strictly decreasing homeomorphism. For x ∈(^a₄,1], let ψ(x) the unique point in (m, q] such that ψ(Q_a(x)) =g(ψ(x)). We now have ψ : [0,1]→[p, q]

continuous (and in fact strictly increasing) such that ψ◦Q_a=g◦ψ.

Using similar arguments, we can extendψ to (−∞,0)∪(1,∞).

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